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Differential Geometry

of Curves and Surfaces

Differential Geometry

of Curves and Surfaces

Second Edition

Thomas Banchoff

Stephen Lovett

CRC PressTaylor & Francis Group6000 Broken Sound Parkway NW, Suite 300Boca Raton, FL 33487-2742

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Contents

Preface vii

Acknowledgements xv

1 Plane Curves: Local Properties 1

1.1 Parametrizations . . . . . . . . . . . . . . . . . . . 1

1.2 Position, Velocity, and Acceleration . . . . . . . . . 13

1.3 Curvature . . . . . . . . . . . . . . . . . . . . . . . 25

1.4 Osculating Circles, Evolutes, and Involutes . . . . 33

1.5 Natural Equations . . . . . . . . . . . . . . . . . . 40

2 Plane Curves: Global Properties 47

2.1 Basic Properties . . . . . . . . . . . . . . . . . . . 47

2.2 Rotation Index . . . . . . . . . . . . . . . . . . . . 52

2.3 Isoperimetric Inequality . . . . . . . . . . . . . . . 61

2.4 Curvature, Convexity, and the Four-Vertex Theorem 64

3 Curves in Space: Local Properties 71

3.1 Definitions, Examples, and Differentiation . . . . . 71

3.2 Curvature, Torsion, and the Frenet Frame . . . . . 80

3.3 Osculating Plane and Osculating Sphere . . . . . . 90

3.4 Natural Equations . . . . . . . . . . . . . . . . . . 97

4 Curves in Space: Global Properties 103

4.1 Basic Properties . . . . . . . . . . . . . . . . . . . 103

4.2 Indicatrices and Total Curvature . . . . . . . . . . 106

4.3 Knots and Links . . . . . . . . . . . . . . . . . . . 115

v

vi Contents

5 Regular Surfaces 125

5.1 Parametrized Surfaces . . . . . . . . . . . . . . . . 1255.2 Tangent Planes and Regular Surfaces . . . . . . . . 1335.3 Change of Coordinates . . . . . . . . . . . . . . . . 1505.4 The Tangent Space and the Normal Vector . . . . 1555.5 Orientable Surfaces . . . . . . . . . . . . . . . . . . 159

6 The First and Second Fundamental Forms 165

6.1 The First Fundamental Form . . . . . . . . . . . . 1656.2 Map Projections (Optional) . . . . . . . . . . . . . 1806.3 The Gauss Map . . . . . . . . . . . . . . . . . . . . 1926.4 The Second Fundamental Form . . . . . . . . . . . 1986.5 Normal and Principal Curvatures . . . . . . . . . . 2096.6 Gaussian and Mean Curvatures . . . . . . . . . . . 2216.7 Developable Surfaces and Minimal Surfaces . . . . 231

7 The Fundamental Equations of Surfaces 247

7.1 Gauss’s Equations and the Christoffel Symbols . . 2487.2 Codazzi Equations and the Theorema Egregium . 2587.3 The Fundamental Theorem of Surface Theory . . . 268

8 The Gauss-Bonnet Theorem and Geometry of Geodesics 273

8.1 Curvatures and Torsion . . . . . . . . . . . . . . . 2748.2 Gauss-Bonnet Theorem, Local Form . . . . . . . . 2848.3 Gauss-Bonnet Theorem, Global Form . . . . . . . 2958.4 Geodesics . . . . . . . . . . . . . . . . . . . . . . . 3058.5 Geodesic Coordinates . . . . . . . . . . . . . . . . 3238.6 Applications to Plane, Spherical, and Elliptic Ge-

ometry . . . . . . . . . . . . . . . . . . . . . . . . . 3358.7 Hyperbolic Geometry . . . . . . . . . . . . . . . . 342

9 Curves and Surfaces in n-dimensional Euclidean Space 355

9.1 Curves in n-dimensional Euclidean Space . . . . . 3559.2 Surfaces in n-dimensional Euclidean Space . . . . . 364

A Tensor Notation 377

A.1 Tensor Notation . . . . . . . . . . . . . . . . . . . 377

Bibliography 405

Preface

What Is Differential Geometry?

Differential geometry studies properties of curves, surfaces, andhigher-dimensional curved spaces using tools from calculus and linearalgebra. Just as the introduction of calculus expands the descriptiveand predictive abilities of nearly every field of scientific study, sothe use of calculus in geometry brings about avenues of inquiry thatextend far beyond classical geometry.

Before the advent of calculus, much of geometry consisted ofproving consequences of Euclid’s postulates. Even conics, whichcame into vogue in the physical sciences after Kepler observed thatplanets travel around the sun in ellipses, arise as the intersection ofa double cone and a plane, two shapes which fit comfortably withinthe paradigm of Euclidean geometry. One cannot underestimatethe impact of geometry on science, philosophy, and civilization as awhole. The geometric proofs in Euclid’s Elements served as modelsof mathematical proof for over two thousand years in the Westerntradition of a liberal arts education. Geometry also produced anunending flow of applications in surveying, architecture, ballistics,astronomy, and natural philosophy more generally.

The objects of study in Euclidean geometry (points, lines, planes,circles, spheres, cones, and conics) are limited in what they can de-scribe. A boundless variety of curves and surfaces and manifoldsarise naturally in areas of inquiry that employ geometry. To ad-dress these new classes of objects, various branches of mathematicsbrought their tools to bear on the expanding horizons of geometry,each with a different bent and set of fruitful results. Techniques fromcalculus and analysis led to differential geometry, pure set theoreticmethods led to topology, and modern algebra contributed the fieldof algebraic geometry.

vii

viii Preface

The types of questions one typically asks in differential geometryextend far beyond what one can ask in classical geometry and yetthe former do not entirely subsume the latter. Differential geometryquestions often fall into two categories: local properties, by whichone means properties of a curve or surface defined in the neighbor-hood of a point, or global properties, which refer to properties ofthe curve or surface taken as a whole. As a comparison to func-tions of one variable, the derivative of a function f at a point a isa local property since one only needs information f near a whereasthe integral of f between a and b is a global property. Some of themost interesting theorems in differential geometry relate local prop-erties to global ones. The culminating theorem in this book, theGauss-Bonnet Theorem, relates global properties of curves and sur-faces to the topology of a surface and leads to fundamental resultsin non-Euclidean spherical and hyperbolic geometry.

Using This Textbook

This new edition is intended as a textbook for a single semesterundergraduate course in the differential geometry of curves and sur-faces, with only multivariable calculus and linear algebra as prereq-uisites. The interactive computer graphics applets that are providedfor this book can be used for computer labs, in-class illustrations,exploratory exercises, or simply as intuitive aids for the reader.Each section concludes with a collection of exercises which rangefrom perfunctory to challenging, suitable for daily or weekly problemsets.

However, the self-contained text, the careful introduction of con-cepts, the many exercises, and the interactive computer graphicsalso make this text well-suited for self-study. Such a reader shouldfeel free to primarily follow the textbook and use the software assupporting material; primarily follow the presentation in the soft-ware package and consult the textbook for definitions, theorems,and proofs; or try to follow both with equal weight. Either way, theauthors hope that the dual nature of software applets and classictextbook structure will offer the reader both a rigorous and intuitiveintroduction to the field of differential geometry.

Preface ix

This book is the first in a pair of books which together are in-tended to bring the reader through classical differential geometryinto the modern formulation of the differential geometry of mani-folds. The second book in the pair, by Lovett, is entitled DifferentialGeometry of Manifolds [24]. Neither book directly relies on the otherbut knowledge of the content of this book is quite beneficial for [24].

Computer Applets

An integral part of this book is the access to on-line computer graph-ics applets that illustrate many concepts and theorems introducedin the text. Though one can explore the computer demos indepen-dently of the text, the two are intended as complementary modesof studying the same material: a visual/intuitive approach and ananalytical/theoretical approach. Though the text does its best toexplain the reason for various definitions and why one might be in-terested in such and such a topic, the graphical applets can oftenprovide motivation for certain definitions, allow the reader to exploreexamples further, and give a visual explanation for complicated the-orems. The ability to change the choice of the parametric curve orthe parametrized surface in an applet or to change other propertiesallows the reader to explore the concepts far beyond what a staticbook permits.

Any element in the text (Example, Problem, Definition, Theo-rem, etc.) that has an associated applet is indicated by the symbolshown in this margin. Each demo comes with some explanation text.The authors intended the applets to be intuitive enough so that afterusing just one or two (and reading the supporting text) any readercan quickly understand their functionality. At present the appletsrun on a Java platform but we may add other formats possibly basedon standard computer algebra systems. As an interest to an instruc-tor of this course, the Java applets are extensible in that they aredesigned with considerable flexibility so that the reader can oftenchange whether certain elements are displayed or not. Often, thereare additional elements that one can display either by accessing theControls menu on the Demo window or the Plot/Add Plot menu onany display window. Applets given in a computer algebra system arenaturally extensible through the capabilities of the given program.

x Preface

The authors encourage the reader to consult the general usageinstruction pages for the applets. All of the applet materials areavailable on-line at http://diffgeo.akpeters.com/.

Organization of Topics

Chapters 1 through 4 cover alternately the local and global theory ofplane and space curves. In the local theory, we introduce the funda-mental notions of curvature and torsion, construct various associatedobjects (e.g., the evolute, osculating circle, osculating sphere), andpresent the fundamental theorem of plane or space curves, whichis an analogue of the fundamental theorem of calculus. The globaltheory studies how local properties (esp. curvature) relate to globalproperties such as closedness, concavity, winding numbers, and knot-tedness. The topics in these chapters are particularly well suited forcomputer investigation. The authors know from experience in teach-ing how often students make discoveries on their own by being ableto quickly manipulate curves and their associated objects and prop-erties.

Chapter 5 rigorously introduces the notion of a regular surface,the type of surface on which the techniques of differential geome-try are well-defined. Here one first sees the tangent plane and theconcept of orientability.

Chapter 6 introduces the local theory of surfaces in R3, focusingon the metric tensor and the Gauss map from which one defines theessential notions of principal, Gaussian, and mean curvatures. Inaddition, we introduce the study of surfaces that have Gaussian cur-vature or mean curvature identically 0. One cannot underestimatethe importance of this chapter. Even a reader primarily interestedin the advanced topic of differentiable manifolds should be comfort-able with the local theory of surfaces in R3 because it provides manyvisual and tractable examples of what one generalizes in the theoryof manifolds. Here again, as in Chapter 8, the use of the softwareapplets is an invaluable aid for developing a good geometric intuition.

Chapter 7 first introduces the reader to the component notationfor tensors. It then establishes the famous Theorema Egregium, thecelebrated classical result that the Gaussian curvature depends only

Preface xi

on the metric tensor. Finally, it outlines a proof for the fundamentaltheorem of surface theory.

Another title commonly used for Chapter 8 is Intrinsic Geometry.Just as Chapter 1 considers the local theory of plane curves, Chapter8 starts with the local theory of curves on surfaces. Of particularimportance in this chapter are geodesics and geodesic coordinates.The highlight of this chapter is the famous Gauss-Bonnet Theorem,both in its local and global forms, without which no elementarycourse in differential geometry is complete. The chapter finisheswith applications to spherical and hyperbolic geometry.

The book concludes in Chapter 9 with a brief discussion on curvesand surfaces in Euclidean n-space. This chapter emphasizes in whatways definitions and formulas given for objects in R3 extend andpossibly change when adapted to Rn.

A Comment on Prerequisites

The mathematics or physics student often first encounters differentialgeometry at the graduate level. Typically, at that point, one isimmediately exposed to the formalism of manifolds, thereby skippingthe intuitive and visual foundation that informs the deeper theory.The advent of computer graphics has added a new dimension toand renewed the interest in classical differential geometry but thispedagogical habit remains. The authors wish to provide a book thatintroduces the undergraduate student to an interesting and visuallystimulating mathematical subject that is accessible with only the fullcalculus sequence and linear algebra as prerequisites.

In calculus courses, students usually do not study all the analysisthat underlies the theorems one uses. Similarly, in keeping withthe stated requirements, this textbook does not always provide allthe topological and analytical background for some theorems. Thereader who is interested in all the supporting material is encouragedto consult [24].

A few key results presented in this textbook rely on theoremsfrom the theory of differential equations but either the calculationsare all spelled out or a reference to the appropriate theorem hasbeen provided. Therefore, experience with differential equations is

xii Preface

occasionally helpful though not necessary. In a few cases, the authorschose not to supply the full proofs of certain results but instead referthe reader to the more complete text [24]. A few exercises requiresome skills with differential equations but these are clearly markedwith the prefix (ODE).

Problems marked with (*) indicate difficulty, which may be re-lated to technical ability, insight, or length.

Notation

As a comment on vector notation, this book and [24] consistently usethe following conventions. A vector or vector function in a Euclideanvector space is denoted by ~v, ~X(t), or ~X(u, v). Often γ indicatesa curve parametrized by ~X(t) while writing ~X(t) = ~X(u(t), v(t))indicates a curve on a surface. The unit tangent and the binormalvectors of a curve in space are written in the standard notation~T (t) and ~B(t) but the principal normal is written ~P (t), reserving~N(t) to refer to the unit normal vector to a curve on a surface.For a plane curve, ~U(t) is the vector obtained by rotating ~T (t) by apositive quarter turn. Furthermore, we denote by κg(t) the curvatureof a plane curve since one identifies this curvature as the geodesiccurvature in the theory of curves on surfaces.

In this book, we often work with matrices of functions. The func-tions themselves are denoted, for example, by aij , and we denote thematrix by (aij). Furthermore, it is essential to distinguish betweena linear transformation between vector spaces T : V → W and itsmatrix with respect to given bases in V and W . Following notationthat is common in current linear algebra texts, if B is a basis in V

and B′ is a basis in W , then we denote by[T]B′B the matrix of T

with respect to these bases. If the bases are understood by context,we simply write [T ] for the matrix associated to T .

Occasionally, there arise irreconcilable discrepancies in defini-tions or notations (e.g., the definition of a critical point for a functionRn → Rm, or how one defines θ and φ in spherical coordinates). Inthese instances the authors made a choice that best suits their pur-poses and indicated commonly used alternatives.

Preface xiii

Section Dependency

This book is primarily organized on the theme of a pendulum backand forth between local theory and global theory, first of planecurves, then space curves, and then surfaces in R3. The authorswrote the book from the perspective that the Gauss-Bonnet The-orem (presented in Sections 8.2 and 8.3) serves as the culminatingtheorem of the textbook. Sections depend on each other accordingto the following chart.

1.1–1.5

2.1–2.4 3.1–3.4

4.1–4.3 5.1–5.5

6.1, 6.3–6.6

6.2 6.7

A.1

7.1–7.2

7.3 8.1–8.5

8.6–8.7 9.1–9.2

Note that Appendix A.1 on tensor notation is valuable for back-ground on the indices of tensor notation but it is not essential forChapter 7 and subsequent chapters.

Changes in the Second Edition

Many faculty who adopted the first edition of the textbook gavehelpful feedback on their experiences. Following this feedback, inthis second edition, the authors preserved the intent and attemptedto improve on the execution of the first edition. This includes thefollowing changes:

xiv Preface

• Reworking the presentation in many places to make it moreapproachable;

• Adding more exercises, both introductory and advanced;

• Including a section on the application of differential geometryto cartography;

• Adding a few investigative projects ideas;

• Significantly reorganizing the last third of the book in order toreach more quickly the Gauss-Bonnet Theorem;

• Adding two sections dedicated to hyperbolic and spherical ge-ometry as applications of intrinsic geometry;

• Adding a brief chapter on curves and surfaces in n-dimensionEuclidean space.

Acknowledgements

Thomas Banchoff

Our work at Brown University on computer visualizations in dif-ferential geometry goes back more than forty-five years, and I ac-knowledge my collaborator computer scientist Charles Strauss forthe first fifteen years of our projects. Since 1982, an impressivecollection of students have been involved in the creation and devel-opment of the software used to produce the applets for this book.The site for my 65th birthday conference http://www.math.brown.edu/TFBCON2003 lists dozens of them, together with descriptionsof their contributions. Particular thanks for the applets connectedwith this book belong to David Eigen, Mark Howison, Greg Baltazar,Michael Schwarz, and Michael Morris. For his work as a student, anassistant, and now as a co-author, I am extremely grateful to SteveLovett. Special thanks for help in the Brown University mathematicsdepartment go to Doreen Pappas, Natalie Johnson, Audrey Aguiar,and Carol Oliveira, and to Larry Larrivee for his invaluable computerassistance. Finally, I thank my wife Kathleen for all her support andencouragement.

Stephen Lovett

I would first like to thank Thomas Banchoff my teacher, mentor, andfriend. After one class, he invited me to join his team of studentson developing electronic books for differential geometry and mul-tivariable calculus. Despite ultimately specializing in algebra, theexciting projects he led and his inspiring course in differential geom-etry instilled in me a passion for differential geometry. His ability tointroduce differential geometry as a visually stimulating and mathe-

xv

xvi Acknowledgements

matically interesting topic served as one of my personal motivationsfor writing this book.

I am grateful to the students and former colleagues at EasternNazarene College. In particular I would like to acknowledge theundergraduate students who served as a sounding board for the firstfew drafts of this manuscript: Luke Cochran, David Constantine,Joseph Cox, Stephen Mapes, and Christopher Young. Special thanksare due to my colleagues Karl Giberson, Lee Hammerstrom, andJohn Free. In addition, I am indebted to Ellie Waal who helpedwith editing and index creation.

The continued support from my colleagues at Wheaton Collegemade writing this book a gratifying project. In particular, I mustthank Terry Perciante, Chair of the Department of Mathematics andComputer Science, for his enthusiasm and his interest. I am indebtedto Dorothy Chapell, Dean of the Natural & Social Sciences, andto Stanton Jones, Provost of the College, for their encouragementand for a grant which freed up my time to finish writing. I amalso grateful to Thomas VanDrunen and Darren Craig for helpfulcomments.

Finally, I cannot adequately express in just a few words how muchI am grateful to my wife Carla Favreau Lovett and my daughterAnne. While I was absorbed in this project, they provided a lovinghome, they braved the significant time commitment and encouragedme at every step. They also kindly put up with my occasional ge-ometry comments such as how to see the Gaussian curvature in thereflection of “the Bean” in Chicago.

Second Edition Acknowledgments

In preparation for this second edition, we wish to thank the manypeople who contributed corrections, improvements, and suggestions.These include but are not limited to Teddy Parker for careful edit-ing, Daniel Flath for many suggestions for improvements, JudithArms for errata and suggestions, and Robert Ferreol for bringing toour attention an excellent on-line encyclopedia of curves at http://www.mathcurve.com/ (in French). We would also like to thank NateVeldt, Gary Babbatz, Nathan Bliss, Matthew McMillan, and ColeAdams.

CHAPTER 1

Plane Curves: Local Properties

Just as calculus courses introduce real functions of one variable be-fore tackling multivariable calculus, so it is natural to study curvesbefore addressing surfaces and higher-dimensional objects. This firstchapter presents local properties of plane curves, where by local prop-erty we mean properties that are defined in a neighborhood of a pointon the curve. For the sake of comparison with calculus, the deriva-tive f ′(a) of a function f at a point a is a local property of thefunction since we only need knowledge of f(x) for x in (a− ε, a+ ε),where ε is any positive real number, to define f ′(a). In contrast, thedefinite integral of a function over an interval is a global propertysince we need knowledge of the function over the whole interval tocalculate the integral. In contrast to this present chapter, Chapter2 introduces global properties of plane curves.

1.1 Parametrizations

Borrowing from a physical understanding of motion in the plane, wecan think about plane curves by specifying the coordinates x and yas functions of a time variable t, which give the position of a pointtraveling along the curve. Thus we need two functions x(t) and y(t).Using vector notation to locate a point on the curve, we often write~X(t) = (x(t), y(t)) for this pair of coordinate functions and call ~X(t)a vector function into R2. From a mathematical standpoint, t doesnot have to refer to time and is simply called the parameter of thevector function.

Example 1.1.1 (Lines). Euclid’s first postulate of geometry is thatthrough two distinct points there passes exactly one line. The pointslope formula gives a Cartesian equation of a line through two points.

1

2 1. Plane Curves: Local Properties

x

y

~p

~v~p+ 2~v

Figure 1.1. A line in the plane.

In analytic geometry, the approach to describing a line through twopoints is slightly different. Given two distinct points ~p1 = (x1, y1)and ~p2 = (x2, y2), the vector

~v = ~p2 − ~p1 = (x2 − x1, y2 − y1)

is called a direction vector of the line because all vectors along thisline are multiples of ~v. Then every point on the line can be writtenwith a position vector ~p1 + t~v for some t ∈ R. Figure 1.1 showsan example with t = 2. Therefore, we find that a line can also bedescribed by providing a point and a direction vector.

Using the coordinates of vectors, given a point ~p = (x0, y0) anda direction vector ~v = (v1, v2), a line through ~p in the direction of ~vis the image of the following vector function:

~X(t) = ~p+ t~v = (x0 + v1t, y0 + v2t) for t ∈ R.

Note that just as the same line can be determined by two differentpairs of points, so the same line may be specified by different setsof these equations. For example, using any point ~p on the line willultimately trace out the same line as t varies through all of R. Sim-ilarly, if we replace ~v with any nonzero multiple of itself, the set ofpoints traced out as t varies through R is the same line in R2 plane.

Example 1.1.2 (Circles). The pair of functions

~X(t) = (R cos t+ a,R sin t+ b)

1.1. Parametrizations 3

traces out a circle of radius R > 0 centered at the point (a, b). Tosee this, note that by the definition of the sin t and cos t functions,(cos t, sin t) are the coordinates of the point on the unit circle thatis also on the ray out of the origin that makes an angle t with thepositive x-axis. Thus,

~X1(t) = (cos t, sin t), with t ∈ [0, 2π],

traces out the unit circle in a counterclockwise manner. Multiplyingboth coordinate functions by R stretches the circle out by a factorof R away from the origin. Thus, the vector function

~X2(t) = (R cos t, R sin t), with t ∈ [0, 2π],

has as its image the circle of radius R centered at the origin. Noticealso that by writing ~X2(t) = (x(t), y(t)), we deduce that x(t)2 +y(t)2 = R2 for all t, which is the algebraic equation of the circle. Inorder to obtain a vector function that traces out a circle centeredat the point (a, b), we must simply translate ~X2 by the vector (a, b).This is vector addition, and so we get

~X(t) = (R cos t+ a,R sin t+ b), with t ∈ [0, 2π].

Two different vector functions can have the same image in R2.For example, if ω > 0,

~X(t) = (cosωt, sinωt), with t ∈ [0, 2π/ω],

also has the unit circle as its image. Referring to vocabulary inphysics, this latter vector function corresponds to a point movingaround the unit circle at an angular velocity of ω.

When trying to establish a suitable mathematical definition ofwhat one usually thinks of as a curve, one does not wish to consideras a curve points that jump around or pieces of segments. We wouldlike to think of a curve as unbroken in some sense. In calculus, oneintroduces the notion of continuity to describe functions without“jumps” or holes, but one must exercise a little care in carrying overthe notion of continuity to vector functions. More generally, we needto define the notion of a limit of a vector function as the parametert approaches a fixed value. First, however, we remind the reader ofthe Euclidean distance formula.


Definition 1.1.3. Let ~v be a vector in R2 with coordinates ~v = (v1, v2)in the standard basis. The (Euclidean) length of ~v is given by

‖~v‖ =√~v · ~v =

√v2

1 + v22.

If p and q are two points in Rn with coordinates given by vectors ~vand ~w, then the Euclidean distance between p and q is ‖~w − ~v‖.

Definition 1.1.4. Let ~X be a vector function from a subset of R intoRn. We say that the limit of ~X(t) as t approaches a is a vector ~w,and we write

limt→a

~X(t) = ~w,

if for all ε > 0 there exists a δ > 0 such that 0 < |t− a| < δ implies‖ ~X(t)− ~w‖ < ε.

Definition 1.1.5. Let I be an open interval of R, let a ∈ I, and let~X : I → R2 be a vector function. We say that ~X(t) is continuous ata if the limit as t approaches a of ~X(t) exists and

limt→a

~X(t) = ~X(a).

The above definitions mirror the usual definition of a limit ofa real function but must use the length of a vector difference todiscuss the proximity between ~X(t) and a fixed vector ~w. Though atthe outset this definition appears more complicated than the usualdefinition of a limit of a real function, the following proposition showsthat it is not.

Proposition 1.1.6. Let ~X be a vector function from a subset of R intoR2 that is defined over an interval containing a, though perhaps notat a itself. Suppose in coordinates we have ~X(t) = (x(t), y(t)) wher-ever ~X is defined. If ~w = (w1, w2), then

limt→a

~X(t) = ~w if and only if limt→a

x(t) = w1 and limt→a

y(t) = w2.

Proof: Suppose first that limt→a ~X(t) = ~w. Let ε > 0 be arbitraryand let δ > 0 satisfy the definition of the limit of the vector function.


Note that |x(t)−w1| ≤ ‖ ~X(t)− ~w‖ and that |y(t)−w2| ≤ ‖ ~X(t)− ~w‖.Hence, 0 < |t − a| < δ implies |x(t) − w1| < ε and |y(t) − w2| < ε.Thus, limt→a x(t) = w1 and limt→a y(t) = w2.

Conversely, suppose that limt→a x(t) = w1 and limt→a y(t) = w2.Let ε > 0 be an arbitrary positive real number. By definition, thereexist δ1 and δ2 such that 0 < |t − a| < δ1 implies |x(t) − w1| <ε/√

2 and 0 < |t − a| < δ2 implies |y(t) − w2| < ε/√

2. Takingδ = min(δ1, δ2) we see that 0 < |t− a| < δ implies that

‖ ~X(t)− ~w‖ =√|x(t)− w1|2 + |y(t)− w2|2 <

√ε2

2+ε2

2= ε.

This finishes the proof of the proposition.

Corollary 1.1.7. Let I be an open interval of R, let a ∈ I, and considera vector function ~X : I → R2 with ~X(t) = (x(t), y(t)). Then ~X(t)is continuous at t = a if and only if x(t) and y(t) are continuous att = a.

Definition 1.1.5 and Corollary 1.1.7 provide the mathematicalframework for what one usually thinks of as a curve in physicalintuition. This motivates the following definition.

Definition 1.1.8. Let I be an interval of R. A parametrized curve (orparametric curve) in the plane is a continuous function ~X : I → R2.If we write ~X(t) = (x(t), y(t)), then the functions x : I → R andy : I → R are called the coordinate functions or parametric equationsof the parametrized curve. We call the locus of ~X(t) the image of~X(t) as a subset of R2.

It is important to note the distinction in this definition betweena parametrized curve and its locus. For example, in Example 1.1.1we show that the parametrization ~X(t) = t~v + ~p traces out a line Lthat goes through the point ~p with direction vector ~v. However, theline is the locus of ~X : R→ R2 and not the vector-valued function ~Xitself. The vector-valued function is the parametrized curve. Sincethe function t 7→ t3−t is a bijection from R to itself, the parametrizedcurve ~Y (t) = (t3− t)~v+~p has the same locus, the line L, but is quitedifferent as a function.


The following examples begin to provide a library of parametriccurves and illustrate how to construct parametric curves to describea particular shape or trajectory.

Example 1.1.9 (Graphs of Functions). The graph of a continuous func-tion f : [a, b]→ R over an interval [a, b] can be viewed as a parametriccurve. In order to view the graph of a continuous function as a pa-rametrized curve, we use the coordinate functions ~X(t) = (t, f(t)),with t ∈ [a, b].

Example 1.1.10 (Circles Revisited). Another parametrization for theunit circle (sometimes used in number theory) is

~X = (x(t), y(t)) =

(1− t2

1 + t2,

2t

1 + t2

)for t ∈ R. (1.1)

It is easy to see that for all t ∈ R, x(t)2+y(t)2 = 1, which means thatthe locus of ~X is on the unit circle. However, this parametrizationdoes not trace out the entire circle as it misses the point (−1, 0). Weleave it as an exercise to determine a geometric interpretation of theparameter t and to show that

limt→∞

~X = limt→−∞

~X = (−1, 0).

This example emphasizes that it is not possible to tell what theparametrization is simply by looking at the locus. A counterclock-wise circle is indistinguishable from the circle covered twice, or fromthe circle traced out in a clockwise fashion, or, as in this example,from a circle covered in a completely different way.

Example 1.1.11 (Ellipses). Without repeating all the reasoning of theprevious exercise, it is not hard to see that

~X(t) = (a cos t, b sin t)

provides a parametrization for the ellipse centered at the origin withaxes along the x- and y-axes, with respective half-axes of length |a|and |b|. Note that these coordinate functions do indeed satisfy

x(t)2

a2+y(t)2

b2= 1 for all t.


Example 1.1.12 (Lissajous Figures). It is sometimes amusing to see howcos t and sin t relate to each other if we change their respective peri-ods. Lissajous figures, which arise in the context of electronics, arecurves parametrized by

~X(t) = (cosmt, sinnt),

where m and n are positive integers. See Figure 1.2 for an exampleof a Lissajous figure with m = 5 and n = 3.

Figure 1.2. A Lissajous figure.

Example 1.1.13 (Cycloids). We can think of a usual cycloid as the locustraced out by a point of light affixed to a bicycle tire as the bicyclerolls forward. We can establish a parametrization of such a curve asfollows.

Assume the wheel of radius a begins with its center at (0, a) sothat the part of the wheel touching the x-axis is at the origin. Weview the wheel as rolling forward on the positive x-axis. As the wheelrolls, the position of the center of the wheel is ~f(t) = (at, a), where tis the angle measuring how much (many times) the wheel has turnedsince it started. At the same time, the light – at a distance a fromthe center of the wheel and first positioned straight down from thecenter of the wheel – rotates in a clockwise motion around the centerof the wheel. See Figure 1.3. The motion of the light with respectto the center of the wheel is

~g(t) =(a cos

(−t− π

2

), a sin

(−t− π

2

))= (−a sin t,−a cos t).


x

y

Figure 1.3. Cycloid.

The locus of the cycloid is the vector function that is the sum of ~f(t)and ~g(t). Thus, a parametrization for the cycloid is

~X(t) = (at− a sin t, a− a cos t).

One can point out that reflectors on bicycle wheels are usuallynot attached directly on the tire but on a spoke of the wheel. We caneasily modify the above discussion to obtain the relevant parametricequations for when the point of light is located at a distance b fromthe center of the rolling wheel. One obtains

~X(t) = (at− b sin t, a− b cos t).

If 0 < b < a, one obtains the curve of a realistic bicycle tire reflector,and this locus is called a curtate cycloid. In contrast, the locusobtained by letting b > a is called a prolate cycloid.

Example 1.1.14 (Heart Curve). Arguably the most popular curve aroundValentine’s Day is the heart. Here are some parametric equationsthat trace out such a curve:

~X(t) = ((1− cos2 t) sin t, (1− cos3 t) cos t).

We encourage the reader to visit this example in the accompanyingsoftware and to explore ways of modifying these equations to cre-ate other interesting curves. On the website http://www.mathcurve.com/ entitled an Online Encyclopedia of Curves, the host creditsthis curve to Raphael Laporte who designed this curve for his “pe-tite amie” (translated “girl friend”).

Example 1.1.15 (Polar Functions). Functions in polar coordinates areusually given in terms of the radius r as a function of the angle


θ by r = f(θ). The graphs of such functions can be written asparametrized curves. Recall the coordinate transformation

x = r cos θ,

y = r sin θ.

Then take θ as the parameter t, and the parametric equations forthe graph of r = f(θ) are

~X(t) = (f(t) cos t, f(t) sin t).

As an example, the polar function r = sin 3θ traces out a curvethat resembles a three-leaf flower. (See Figure 1.4.) As a parametriccurve, it is given by

~X(t) = (sin 3t cos t, sin 3t sin t).

x

y

Figure 1.4. Three-leaf flower.

Example 1.1.16 (Cardioid). Another common polar function is the car-dioid, which is the locus of r = 1 − cos θ. In parametric equations,we have

~X(t) = ((1− cos t) cos t, (1− cos t) sin t).

As mentioned in Example 1.1.2, for a parametric curve ~X : I →R2, the set of points C = ~X(t) | t ∈ I as a subset of R2 doesnot depend uniquely on the functions x(t) and y(t). In fact, it isimportant to make a careful distinction between the notion of a


parametrized curve as defined above, the image of the parametrizedcurve as a subset of R2 (also called the locus of the curve), and thenotion of a curve, eventually defined as a one-dimensional manifold.(See [24, Chapter 3].)

Definition 1.1.17. Given a parametrized curve ~X : I → R2 and anycontinuous functions g from an interval J onto the interval I, wecan produce a new vector function ~ξ : J → R2 defined by ~ξ =~X g. The image of ~ξ is again the set C, and ~ξ = ~X g is called areparametrization of ~X.

If g is not onto I, then the image of ~ξ may be a proper subsetof C. In this case, we usually do not call ~ξ a reparametrization as itdoes not trace out the same locus of ~X.

Problems

1.1.1. Using linear algebra, it is possible to prove that the shortest distancebetween a point (x0, y0) and a line with equation ax+ by + c = 0 is

d =|ax0 + by0 + c|√

a2 + b2.

However, a calculus proof of this result is also possible using param-etrized curves.

(a) Show that ~X(t) = (−bt− a/c, at) with t ∈ R is a parametriza-tion of the line with equation ax+ by + c = 0.

(b) Find the value t0 of t that minimizes the function

f(t) = ‖ ~X(t)− (x0, y0)‖ =√

(−bt− a/c− x0)2 + (at− y0)2,

which gives the distance between a point ~X(t) on the line andthe point (x0, y0).

(c) Show that f(t0) simplifies to the distance formula given above.

1.1.2. Let ~p be a fixed point, and let ~l(t) = ~at + ~b be the parametricequations of a line. Prove that the distance between ~p and the lineis √

‖~b− ~p‖2 − (~a · (~b− ~p))2‖~a‖2

=‖~a× (~b− ~p)‖

‖~a‖,

where for vectors ~v = (v1, v2, 0) and ~w = (w1, w2, 0) in the plane, wecall ~v × ~w = (0, 0, v1w2 − v2w1), which is the cross product between~v and ~w when viewed as vectors in R3. [See the previous problem.]


1.1.3. Equation (1.1) gave the following parametric equations for a circle:

~X = (x(t), y(t)) =

(1− t2

1 + t2,

2t

1 + t2

), for t ∈ R.

Prove that the parameter t is equal to tan θ, where θ is the angleshown in the following picture of the unit circle.

θ

1.1.4. Let C be the circle of center (0, 2) and radius 1. Let S be the setof points in R2 that are the same distance from the outside of thecircle C as they are from the x-axis.

(a) Prove that S has a parametrization

~X(t) =

(3 sin t

1 + cos t, 2− 3 cos t

1 + cos t

).

[Hint: Use the parameter t as the angle as shown below.]

(b) Prove that S is a parabola and find the Cartesian equation forit.

(0, 2)

t

~X(t)

1.1.5. Find the closest point to (16, 0.5) on the curve ~X(t) = (t, t2).

1.1.6. An epicycloid is defined as the locus of a point on the edge of acircle of radius b as this circle rolls on the outside of a fixed circleof radius a. Supposing that at t = 0, the moving point is located


at (a, 0). Prove that the following are parametric equations for anepicycloid:

~X(t)=

((a+ b) cos t− b cos

(a+ b

bt

), (a+ b) sin t− b sin

(a+ b

bt

)).

1.1.7. A hypocycloid is defined as the locus of a point on the edge of acircle of radius b as this circle rolls on the inside of a fixed circle ofradius a. Assuming that a > b and that at t = 0, the moving pointis located at (a, 0), find parametric equations for a hypocycloid.

1.1.8. Consider the parametrized curve ~X : R→ R2 with equations ~X(t) =(12 (et + e−t), 12 (et − e−t)

). Prove that the locus is one branch of a

unit hyperbola x2 − y2 = 1. Give a parametrization for the otherbranch of the hyperbola.

1.1.9. Consider the vectors ~a = (3, 3) and ~b = (−1, 1). Explain why the

parametric curve ~X(t) = (cos t)~a+(sin t)~b is an ellipse. Furthermore,give a Cartesian equation of the locus of this parametric curve.

1.1.10. Consider the parametric curve ~X(t) = (t3 − 5t, 3t2). Graphing itshows that it intersects itself. By solving for t1 and t2 the equation~X(t1) = ~X(t2), find the parameters where the curve intersects itselfand give the coordinates of the point on the locus where this occurs.

1.1.11. Consider the parametrized curve ~X : [0, 2π]→ R2 that gives a defor-

mation of a cardioid ~X(t) = ((a+ cos t) cos t, (a+ cos t) sin t), wherea is a real number. Show that the curve intersects itself for |a| < 1.Describe what happens for a = 0.

1.1.12. (*) Consider a curve in R2 that is the solution of a Cartesian equationof the form ax2 + 2bxy + cy2 = d, where a, b, c, d are real numbers.Note that the left-hand side can be expressed in matrices as

ax2 + 2bxy + cy2 =(x y

)(a bb c

)(xy

).

Recall the Spectral Theorem from linear algebra, namely that a sym-metric matrix can be orthogonally diagonalized. Prove that if theeigenvalues of the above symmetric matrix are both nonzero andhave the same sign, then the curve is an ellipse. Find a parametricequation for this ellipse.

1.2. Position, Velocity, and Acceleration 13

1.2 Position, Velocity, and Acceleration

In physics, one interprets the vector function ~X(t) = (x(t), y(t)) asproviding the location along a curve at time t in reference to somefixed frame. A frame is a (usually orthonormal) basis attached to afixed origin. The point O = (0, 0) along with the basis ~ı,~, where

~ı = (1, 0) and ~ = (0, 1),

form the standard reference frame. We call the vector function ~X(t)the position vector. When one uses the standard reference frame, itis not uncommon to write

~X(t) = x(t)~ı+ y(t)~.

Directly imitating Newton’s approach to finding the slope of acurve at a certain point, it is natural to ask the question, “What isthe direction and rate of change of a curve ~X at point t0?” If welook at two points on the curve, say ~X(t0) and ~X(t1), the change inposition is given by the vector ~X(t1) − ~X(t0), while to specify therate of change in position, we would need to scale this vector by afactor of t1 − t0. Hence, the vectorial rate of change between ~X(t0)and ~X(t1) is

1

t1 − t0

(~X(t1)− ~X(t0)

).

To define an instantaneous rate of change along the curve at thepoint t0, we need to calculate (if it exists and if it has meaning) thelimit

~X ′(t0) = limh→0

1

h

(~X(t0 + h)− ~X(t0)

). (1.2)

According to Proposition 1.1.6, if ~X(t) = (x(t), y(t)), then the limitin Equation (1.2) exists if and only if x(t) and y(t) are both dif-ferentiable at t0. Furthermore, if this limit exists, then ~X ′(t0) =(x′(t0), y′(t0)). This leads us to the following definition.

Definition 1.2.1. Let ~X : I → R2 be a vector function with coordinates~X(t) = (x(t), y(t)). We say that ~X is differentiable at t = t0 if x(t)and y(t) are both differentiable at t0. If J is the common domain tox′(t) and y′(t), then we define the derivative of the vector function~X as the new vector function ~X ′ : J → R2 defined by ~X ′(t) =(x′(t), y′(t)).


If x(t) and y(t) are both differentiable functions on their domain,the derivative vector function ~X ′(t) = (x′(t), y′(t)) is called the ve-locity vector. Mathematically, this is just another vector functionand traces out another curve when placed in the standard referenceframe. However, because it illustrates the direction of motion alongthe curve, one often visualizes the velocity vector corresponding tot = t0 as based at the point ~X(t0) on the curve.

Following physics language, we call the second derivative of thevector function ~X ′′(t) = (x′′(t), y′′(t)) the acceleration vector relatedto ~X(t).

In general, our calculations often require that our vector func-tions can be differentiated at least once and sometimes more. Con-sequently, when establishing theorems, we like to succinctly describethe largest class of functions for which a particular result holds.

Definition 1.2.2. Let I be an interval of R, and let f : I → R be afunction or ~X : I → R2 a parametrized curve. We say that f or ~X isof class Cr on I if the rth derivative of ~X exists and is continuous onI. We denote by C∞ the class of functions or parametrized curvesthat have derivatives of all orders on I. It is also common to writef ∈ Cr(I,R) to say that f is a real-valued function of class Cr andto write ~X ∈ Cr(I,R2) to say that ~X is a parametrized curve in theplane of class Cr.

To say that a function is of class C0 over the interval I meansthat it is continuous. By basic theorems in calculus, the conditionthat a function be of a certain class is an increasingly restrictivecondition. In other words, as sets of functions defined over the sameinterval I, the classes are nested according to

C0 ⊃ C1 ⊃ C2 ⊃ · · · ⊃ C∞.

Whenever we impose a condition that a function is of class Cr forsome r, such a supposition is generically called a “smoothness con-dition.”

Proposition 1.2.3. Let ~v(t) and ~w(t) be vector functions defined anddifferentiable over an interval I ⊂ R. Then the following hold:

1. If ~X(t) = c~v(t), where c ∈ R, then ~X ′(t) = c~v′(t).


2. If ~X(t) = c(t)~v(t), where c : I → R is a real function, then

~X ′(t) = c′(t)~v(t) + c(t)~v′(t).

3. If ~X(t) = ~v(t) + ~w(t), then

~X ′(t) = ~v′(t) + ~w′(t).

4. If ~X(t) = ~v (f(t)) is a vector function and f : J → I is a realfunction into I, then

~X ′(t) = f ′(t)~v′ (f(t)) .

5. If f(t) = ~v(t) · ~w(t) is the dot product between ~v(t) and ~w(t),then

f ′(t) = ~v′(t) · ~w(t) + ~v(t) · ~w′(t).

Example 1.2.4. Consider the spiral defined by ~X(t) = (t cos t, t sin t)for t ≥ 0. The velocity and acceleration vectors are

~X ′(t) = (cos t− t sin t, sin t+ t cos t),

~X ′′(t) = (−2 sin t− t cos t, 2 cos t− t sin t).

If we wish to calculate the angle between ~X and ~X ′ as a function oftime, we use the dot product method. In this example, we calculate

‖ ~X(t)‖ =√t2 sin2 t+ t2 cos2 t = |t|,

‖ ~X ′(t)‖ =√

(cos t− t sin t)2 + (sin t+ t cos t)2 =√

1 + t2,

~X(t) · ~X ′(t) = t cos2 t− t2 cos t sin t+ t sin2 t+ t2 cos t sin t = t.

Thus, the angle θ(t) between ~X(t) and ~X ′(t) is defined for all t 6= 0and is equal to

θ(t) = cos−1

(~X(t) · ~X ′(t)‖ ~X(t)‖ ‖ ~X ′(t)‖

)= cos−1

(t

|t|√

1 + t2

).


O

~X(t0)

~X(t1)∆ ~X

~X ′(t0)

Figure 1.5. Arc length segment.

Let C be the locus of a vector function ~X(t) for t ∈ [a, b]. Thenwe approximate the length of the arc l(C) with the Riemann sum

l(C) ≈n∑i=1

‖ ~X(ti)− ~X(ti−1)‖ ≈n∑i=1

‖ ~X ′(ti)‖∆t.

(See Figure 1.5 as an illustration for the approximation ‖ ~X(ti) −~X(ti−1)‖ ≈ ‖ ~X ′(ti)‖∆t.) Taking the limit of this Riemann sum, weobtain the following formula for the arc length of C:

l =

∫ b

a

√(x′(t))2 + (y′(t))2 dt. (1.3)

A rigorous justification for the arc length formula, using the MeanValue Theorem, can be found in most calculus textbooks.

In light of Equation (1.3), we define the arc length function s :[a, b]→ R related to ~X(t) as the arc length along C over the interval[a, t]. Thus,

s(t) =

∫ t

a‖ ~X ′(u)‖ du =

∫ t

a

√(x′(u))2 + (y′(u))2 du. (1.4)

By the Fundamental Theorem of Calculus, we then have

s′(t) = ‖ ~X ′(t)‖ =√

(x′(t))2 + (y′(t))2,

which, still following the vocabulary from the trajectory of a movingparticle, we call the speed function of ~X(t).


Example 1.2.5. Consider the parametrized curve defined by ~X(t) =(t2, t3). The velocity vector is ~X ′(t) = (2t, 3t2), and thus the arclength function from t = 0 is

s(t) =

∫ t

0‖ ~X ′(u)‖ du =

∫ t

0

√4u2 + 9u4 du =

∫ t

03|u|

√4

9+ u2 du

= sign(t)

(4

9+ t2

)3/2

,

where sign(t) = 1 if t > 0 and −1 if t < 0 and sign(0) = 0. Thelength of ~X(t) between t = 0 and t = 1 is

s(1)− s(0) =1

27

(133/2 − 8

).

The speed function allows us to understand that a parametrizedcurve ~X : I → R2 does not only contain the information that de-scribes the locus of the curve but it also contains information aboutthe speed of travel s(t) along the locus. By looking at the locus, itis impossible to discern the speed function.

Consider a reparametrization of the parametrized curve ~X f ,where f : J → I is a differentiable surjective (i.e., onto) function. Ift0 = f(u0) where f is a differentiable function at u0, then ~X(f(u0)) =~X(t0) and∥∥∥∥ ddt ~X(f(u0))

∥∥∥∥ = ‖f ′(u0) ~X ′(f(u0))‖

= |f ′(u0)| ‖ ~X ′(t0)‖ = |f ′(u0)|s′(t0).

Up to a change in sign, the direction of ~X ′(t), by which we meanthe unit vector associated with ~X ′(t), does not change under repa-rametrization. More precisely, if ~ξ = ~X f , at any parameter valuet where f ′(t) 6= 0 and ~X ′(f(t)) 6= ~0, we have

~ξ′(t)

‖~ξ′(t)‖=

f ′(t)

|f ′(t)|~X ′(f(t))

‖ ~X ′(f(t))‖, (1.5)

where f ′(t)/|f ′(t)| is +1 or −1. It bears repeating that in order forthe right-hand side to be well defined, one needs f ′(t) 6= 0.


Definition 1.2.6. Let ~X : I → R2 be a parametrized curve. For acontinuously differentiable function f from an interval J onto I, wecall ~ξ = ~X f a regular reparametrization if for all t ∈ J , f ′(t) iswell defined and never 0. In addition, a regular reparametrizationis called positively oriented (resp. negatively oriented) if f ′(t) > 0(resp. f ′(t) < 0) for all t ∈ J .

By the Mean Value Theorem, if f ′(t) 6= 0 for all t ∈ J , thenf : J → I is an injective function. Thus since f is by definitionsurjective, regular reparametrizations involve a bijective function f :J → I.

Equation (1.5) shows that the unit vector ~X ′(t)/‖ ~X ′(t)‖ is invari-ant under a positively oriented reparametrization and simply changesunder a negatively oriented reparametrization. Consequently, thisunit vector is an important geometric object associated to a curveat a point.

Definition 1.2.7. Let ~X : I → R2 be a plane parametric curve. A pointt0 ∈ I is called a critical point of ~X(t) if ~X(t) is not differentiableat t0 or if ~X ′(t0) = ~0. If t0 is a critical point, then ~X(t0) is called acritical value. A point t = t0 that is not critical is called a regularpoint. A parametrized curve ~X : I → R2 is called regular if it is ofclass C1 (i.e., continuously differentiable) and ~X ′(t) 6= ~0 for all t ∈ I.Finally, if t is a regular point of ~X, we define the unit tangent vector~T (t) as

~T (t) =~X ′(t)

‖ ~X ′(t)‖.

It is not uncommon to call a property of a curve C or a propertyof a point P on a curve C a geometric property if it does not dependon a parametrization of the locus near the point. In particular, ageometric property does not depend on the speed of travel along thecurve through the point.

At any point of a curve where ~T (t) is defined, that is to say atany regular point, we can write the velocity vector as

~X ′(t) = s′(t)~T (t). (1.6)


This expresses the velocity vector as the product of its magnitude(speed) and direction (unit tangent vector). In most differentialgeometry texts, authors simplify their formulas by reparametrizing byarc length. From the perspective of coordinates, this means pickingan “origin” O on the curve C occurring at some fixed t = t0 andusing the arc length along C between O and any other point P asthe parameter with which to locate P on C. In fact, this can alwaysbe done.

Proposition 1.2.8 (Reparametrization by Arc Length). If ~X(t) is a regu-lar parametrized curve, then there is a regular reparametrization of~X by arc length. Furthermore, if ~X is of class Ck, then the arclength reparametrization is also of class Ck.

Proof: Let s = f(t) be the arc length function with s = 0 corre-sponding to some point on the curve. Since ~X is regular over itsdomain, then f ′(t) = ‖ ~X ′(t)‖ > 0 for all t. Since f(t) is strictly in-creasing, it has an inverse function t = h(s). By the Inverse FunctionTheorem, since f ′(t) 6= 0, then h(s) is differentiable with

h′(s) =1

f ′(h(s))=

1

f ′(t). (1.7)

Note that the composite function ~Y (s) = ~X(h(s)) satisfies

~Y ′(s) = ~X ′(h(s))h′(s) = ~X ′(t)1

f ′(t)=

~X ′(t)

‖ ~X ′(t)‖.

If ~X is of class Ck, then s′(t) = f ′(t) = ‖ ~X ′(t)‖ is of class Ck−1.Hence s(t) itself is of class Ck. The Inverse Function Theorem statesthat the inverse function is also of class Ck. Consequently, the arclength parametrization ~y(s) is of class Ck since it is the compositionof ~X(t) of class Ck and t = h(s) of class Ck.

The Inverse Function Theorem as used in the above proof (and itsmore general multivariable counterpart) appear in most introductorytexts on analysis. (See [7] for example.) However, the key derivativein Equation (1.7) follows from the chain rule for invertible functionsand is a standard differentiation formula.


The habit of reparametrizing by arc length has benefits and draw-backs. The great benefit of this approach is that along a curve pa-rametrized by arc length, the speed function s′ is identically 1 andthe velocity vector is exactly the tangent vector

~X ′(s) = ~T (s).

This formulation simplifies proofs and difficult calculations. Themain drawback is that, in practice, most curves do not admit asimple formula for their arc length function, let alone a formula thatcan be written using elementary functions. For example, given anexplicit parametrized curve ~X(t), it is often very challenging to finds(t) as defined in Equation (1.4). Furthermore, to reparametrize byarc length, it would be necessary to find the inverse function t(s),representing the original parameter t as a function of s. Even if it ispossible to find s(t), determining this inverse function usually cannotbe written with elementary functions. Then the parametrization byarc length is ~X(s) = ~X(t(s)). Even using computer algebra systems,reparametrizing by arc length remains an intractable problem.

The notion of “regular” in many ways mirrors geometric proper-ties of continuously differentiable single-variable functions. In par-ticular, if a parametrized curve ~X(t) is regular at t0, then locallythe curve looks linear. As we will see again in Chapter 3, with theformalism of vector functions it becomes particularly easy to expressthe equation of the tangent line to a curve at a point in any numberof dimensions. If ~X(t) is a curve and t0 is not a critical point for thecurve, then the equation ~Lt0(t) of the tangent line at t0 is

~Lt0(t) = ~X(t0) + (t− t0) ~X ′(t0), with t ∈ R,

or alternatively

~Lt0(u) = ~X(t0) + u~T (t0), with u ∈ R, (1.8)

if we do not wish to confuse the parameter of the parametrized curve~X(t) and the parameter of the tangent line.

On the other hand, if t0 ∈ I is a critical point for the curve~X(t), the curve may or may not have a tangent line at t = t0. Ifthe following one-sided limits exist, we can define two unit tangent


vectors at ~X(t0):

~T (t+0 ) = limt→t+0

~T (t) and ~T (t−0 ) = limt→t−0

~T (t).

Consequently, we can determine the angle the curve makes with itselfat the corner t0 as

α0 = cos−1(~T (t+0 ) · ~T (t−0 )

).

To be more precise, the angle from ~T (t−0 ) to ~T (t+0 ) is the exteriorangle of the curve at the corner t = t0. One can only define a tangentline to ~X(t) at t = t0 if ~T (t−0 ) = ~T (t+0 ).

Example 1.2.9. Let ~X(t)=(t2, t3). We calculate that ~X ′(t)=(2t, 3t2),and therefore t = 0 is a critical point because ~X ′(0) = ~0. However,if t 6= 0, the unit tangent vector is

~T (t) =1√

4t2 + 9t4(2t, 3t2) =

t

|t|√

4 + 9t2(2, 3t).

Then the right-hand and left-hand side unit tangent vectors are

~T (0−) = (−1, 0) and ~T (0+) = (1, 0).

These calculations indicate that as t approaches 0, ~X(t) lies in thefourth quadrant but approaches (0, 0) in the horizontal direction(−1, 0). From an intuitive standpoint, we could say that ~X stops att = 0, spins around by 180, and moves away from the origin in thedirection (1, 0), remaining in the first quadrant.

Example 1.2.10. Let ~X(t) = (t, | tan t|). We calculate that ~X ′(t) =(1, sign(t) sec2 t) and deduce that t = 0 is a critical point becausesign(t) is not defined at 0, and hence ~X is not differentiable there.However, we also find that

~T (0−) =1√2

(1,−1) and ~T (0+) =1√2

(1, 1).

Thus,~T (0−) · ~T (0+) = 0,

which shows that ~X makes a right angle with itself at t = 0. However,one can tell from the explicit values for ~T (t−0 ) and ~T (t+0 ) that theexterior angle of the curve at t = 0 is π

2 .


If t0 is a critical point, it may still happen that

limt→t0

~T (t)

exists, namely when ~T (t+0 ) = ~T (t−0 ), which also means α0 = 0. Thenthrough an abuse of language, we can still talk about the unit tangentvector at that point. As an example of this possibility, consider thecurve ~X(t) = (t3, t4). We can quickly calculate that

limt→0

~T (t) = (1, 0) =~ı.

When this limit exists, even though t0 is a critical point, one canuse Equation (1.8) as parametric equations for the tangent line att0, replacing ~T (t0) in Equation (1.8) with limt→t0

~T (t).

In geometry, physics, and other applications, we must sometimesintegrate a function along a curve. To this end, we use path andline integrals of scalar functions or vector functions, depending onthe particular problem. Since we will make use of them, we remindthe reader of the notation for such integrals. Let C be a curve pa-rametrized by ~X(t) over the interval [a, b]. Let f : R2 → R be a real(scalar) function and ~F : R2 → R2 a vector field in the plane. Thenthe path integral of f and the line integral of ~F over the curve C arerespectively

∫Cf ds means

∫ b

af(x(t), y(t))

√(x′(t))2 + (y′(t))2dt,∫

C

~F · d~s means

∫ b

a

~F (x(t), y(t)) · ~T (t) dt.

Problems

1.2.1. Calculate the velocity, the acceleration, the speed, and, where de-fined, the unit tangent vector function of the following parametriccurves:

(a) The circle ~X(t) = (R cosωt,R sinωt).


(b) The circle as parametrized in Equation (1.1) in Example 1.1.10.

(c) The epicycloids defined in Problem 1.1.6, Section 1.1.

1.2.2. A quadratic Bezier curve with control points ~p0, ~p1, and ~p2 is acurve ~X : [0, 1] → R2 with end points ~X(0) = ~p0 and ~X(1) = ~p2and such that the control point ~p1 satisfies ~X ′(0) = ~p1 − ~p0 and~X ′(1) = ~p2 − ~p1. Find the parametrization for this quadratic Beziercurve.

1.2.3. Find the tangent line to ~X(t) = (cos(2t), sin t) at t = π/6. Also givethe components of the unit tangent vector there.

1.2.4. The parametrized curve ~X(t) = ((1 + 2 cos t) cos t, (1 + 2 cos t) sin t)intersects itself at one point. Find this point of intersection andfind the angle of self-intersection (i.e., the acute angle between thetangent lines corresponding to the two different parameters t1 andt2 of self-intersection).

1.2.5. For how many points on the Lissajous curve ~X(t) = (cos(3t), sin(2t))does the tangent line go through the point (3, 0)?

1.2.6. What can be said about a parametrized curve ~X(t) that has the

property that ~X ′′(t) is identically 0?

1.2.7. Find the arc length function along the parabola y = x2, using as theorigin s = 0.

1.2.8. Consider the parametrized curve ~X(t) = (t2, t3) with t ∈ R. Param-etrize this curve by arc length with s = 0 corresponding to t = 0.

[Ans: ~X(s) =(

19 (27s+ 8)2/3 − 4

9 , sign(s)(19 (27s+ 8)2/3 − 4

9

)3/2).

This is one of the few curves for which this is a tractable problem.]

1.2.9. (*) Consider the cycloid introduced in Example 1.1.13 given by~X(t) = (t− sin t, 1− cos t). Prove that the path taken by a point onthe edge of a rolling wheel of radius 1 during one rotation has length8.

1.2.10. Calculate the arc length function of the curve ~X(t) = (t2, ln t), de-fined for t > 0.

1.2.11. Let ~X : I → R2 be a regular parametrized curve, and let ~p be a fixedpoint. Suppose that the closest point on the curve ~X to ~p occurs att = t0, which is neither of the ends of I. Prove that the line between~p and the point closest to ~p, namely ~X(t0), is perpendicular to the

curve ~X at t = t0.


1.2.12. We say that a plane curve C parametrized by ~X : I → R2 intersectsitself at a point p if there exist u 6= t in I such that ~X(t) = ~X(u) = p.

Consider the parametric curve ~X(t) = (t2, t3− t) for t ∈ R. Find thepoint(s) of self-intersection. Also determine the angle at which thecurve intersects itself at those point(s) by finding the angle betweenthe unit tangent vectors corresponding to the distinct parameters.

1.2.13. Prove the differentiation formulas in Proposition 1.2.3.

1.2.14. Prove that if ~X(t) is a curve that satisfies ~X · ~X ′ = 0 for all values of

t, then ~X is a circle. [Hint: Use ‖ ~X‖2 = ~X · ~X and apply Proposition

1.2.3 to calculate the derivative of || ~X||2.]

1.2.15. Consider the ellipse given by ~X(t) = (a cos t, b sin t). Find the ex-tremum values of the speed function.

1.2.16. Consider the linear spiral of Example 1.2.4. Let n ≥ 0 be a non-negative integer. Prove that the length of the nth derivative vectorfunction is given by

‖ ~X(n)(t)‖ =√n2 + t2.

1.2.17. Consider the exponential spiral ~x(t) = (aebt cos t, aebt sin t) where aand b are constants. Calculate the arc length s(t) function of ~x(t).Reparametrize the spiral by arc length.

1.2.18. Consider again the exponential spiral ~x(t) = (aebt cos t, aebt sin t),with a > 0 and b < 0.

(a) Prove that as t→ +∞, lim ~X(t) = (0, 0).

(b) Show that ~X ′(t)→ (0, 0) as t→ +∞ and that for any t0,

limt→∞

∫ t

t0

‖ ~X ′(u)‖ du <∞,

i.e., any part of the exponential spiral that “spirals” in towardthe origin has finite arc length.

1.2.19. Recall that polar and Cartesian coordinate systems are related asfollows:

x = r cos θ,

y = r sin θ,and

r =

√x2 + y2,

tan θ = yx

Suppose C is a curve in the plane parametrized using polar coordi-nate functions r = r(t) and θ = θ(t) so that one has a parametriza-tion in Cartesian coordinates as

~x(t) = (x(t), y(t)) = (r(t) cos(θ(t)), r(t) sin(θ(t))).

1.3. Curvature 25

(a) Express x′ and y′ in terms of r, θ, r′, and θ′.

(b) Express r′ and θ′ in terms of x, y, x′, and y′.

(c) Express ||~x|| and ||~x′|| in terms of polar coordinate functions.

(All coordinate functions are viewed as functions of t and so r′, forexample, is a shorthand for r′(t), the derivative of r with respectto t.)

1.2.20. (ODE) Suppose a curve C is parametrized by ~x(t) such that ~x(t)and ~x′(t) always make a constant angle with each other. Find theshape of this curve. [Hint: Use polar coordinates and the results ofProblem 1.2.19.]

1.3 Curvature

Let ~X : I → R2 be a twice-differentiable parametrization of a curveC. As we saw in the previous section, the decomposition of the ve-locity vector ~X ′ = s′ ~T into magnitude and unit tangent directionseparates the geometric invariant (the unit tangent ~T ) from the dy-namical aspect (the speed s′(t)) of the parametrization. Taking onemore derivative, we obtain the decomposition

~X ′′ = s′′ ~T + s′ ~T ′. (1.9)

The first component describes a tangential acceleration, while thesecond component describes a rate of change of the tangent direction,or in other words, how much the curve is “curving.” Since ~T is aunit vector, we always have ~T · ~T = 1. Therefore,

(~T · ~T )′ = 0 =⇒ 2~T · ~T ′ = 0,

=⇒ ~T · ~T ′ = 0.

Thus, ~T ′ is perpendicular to ~T .

Just as there are two unit tangent vectors at a regular point of thecurve, there are two unit normal vectors as well. Given a particularparametrization, there is no naturally preferred way to define “the”unit normal vector, so we make a choice.


Definition 1.3.1. Let ~X : I → R2 be a regular parametrized curve and~T = (T1, T2) the tangent vector at a regular value ~X(t). The unitnormal vector ~U is

~U(t) = (−T2(t), T1(t)).

Equivalently, ~U is the vector function obtained by rotating ~T byπ2 . In still other words, if we view the xy-plane in three-dimensionalspace with x-, y-, and z-axes oriented in the usual way, and if we call~k = (0, 0, 1) the unit vector along the positive z-axis, then ~U = ~k× ~T .

Since ~T ′ ⊥ ~T , the vector function ~T ′ is a multiple of ~U at all t.This leads to the definition of curvature.

Definition 1.3.2. Let ~X : I → R2 be a regular twice-differentiableparametric curve. The curvature function κg(t) is the unique real-valued function defined by

~T ′(t) = s′(t)κg(t)~U(t). (1.10)

This definition only gives κg(t) implicitly but we can obtain aformula for it as follows. Equation (1.9) becomes

~X ′′ = s′′ ~T + (s′)2κg ~U. (1.11)

Viewing the plane as the xy-plane in three-space, we have ~T× ~U = ~k.Thus,

~X ′ × ~X ′′ = (s′ ~T )× (s′′ ~T + (s′)2κg ~U)

= s′s′′ ~T × ~T + (s′)3κg ~T × ~U

= (s′)3κg~k.

But s′(t) = ‖ ~X ′(t)‖, which leads to

κg(t) =( ~X ′(t)× ~X ′′(t)) · ~k

‖ ~X ′(t)‖3=x′(t)y′′(t)− x′′(t)y′(t)

(x′(t)2 + y′(t)2)3/2. (1.12)

The reader might well wonder why in the definition of κg(t) weinclude the factor s′(t). It is not hard to confirm (see Problem 1.3.15)

1.3. Curvature 27

that including the s′(t) factor renders κg(t) independent of any reg-ular reparametrization (except perhaps up to a change of sign).

One should note at this point that if a curve ~X(s) is parametrizedby arc length, then by Equations (1.6) and (1.9), the velocity andacceleration have the simple expressions

~X ′(s) = ~T (s),

~X ′′(s) = κg(s)~U(s).

Example 1.3.3. Consider the vector function that describes a particlemoving on a circle of radius R with a nonzero and constant angularvelocity ω > 0 given by ~X(t) = (R cosωt,R sinωt). In order tocalculate the curvature, we need

~X ′(t) = (−Rω sinωt,Rω cosωt),

~X ′′(t) = (−Rω2 cosωt,−Rω2 sinωt).

Thus, using the coordinate form (right-most expression) in Equation(1.12), we get

κg(t) =R2ω3 sin2 t+R2ω3 cos2 t

(R2ω2 sin2 t+R2ω2 cos2 t)3/2=R2ω3

R3ω3=

1

R.

The curvature of the circle is a constant function that is equal to thereciprocal of the radius for all t, regardless of the nonzero angularvelocity ω.

The study of trajectories in physics gives particular names forthe components of the first and second derivatives of a vector func-tion in the basis ~T , ~U. We already saw that the function s′(t) inEquation (1.6) is called the speed. In Equation (1.11), however, thefunction s′′(t) is called the tangential acceleration, while the quan-tity s′(t)2κg(t) is called the centripetal acceleration. Example 1.3.3connects the curvature function to the reciprocal of a radius, so ifκg(t) 6= 0, then we define the radius of curvature to ~X(t) at t as thefunction R(t) = 1

κg(t) . Using the common notation v(t) = s′(t) for

the speed function, one recovers the common formula for centripetalacceleration of

(s′)2κg =v2

R.


A

B

C

D

E

F

G

H

A

B

C

D

E

F

G

H

Figure 1.6. Curvature function example.

In introductory physics courses, this formula is presented only inthe context of circular motion. With differential geometry at ourdisposal, we see that centripetal acceleration is always equal to v2/Rat all points on a curve where κg 6= 0, where by R one means theradius of curvature.

In contrast to physics where one tends to refer to the radius ofcurvature, the curvature function is a more geometrically naturalquantity to study. Indeed, the radius of curvature of a line segmentis undefined even though line segments are such useful geometricobjects. On the other hand, a radius of curvature is 0 (and hencethe curvature is undefined) at degenerate curves that are points orat critical points.

Example 1.3.4. To help build an intuition for the curvature function ofa plane curve, consider the parametric curve ~X(t) = (2 cos t, sin(2t)+sin t) for t ∈ [0, 2π].

Figure 1.6 shows side by side the locus of the curve and the graphof its curvature function, along with a few labeled points to serveas references. One can see that the curvature function is positivewhen the curve turns to the left away from the unit tangent vector~T and negative when the curve turns to the right. The curvaturefunction is 0 when the curve changes from curving to the left of~T to curving to the right of ~T , or vice versa. The curve is locallycurving the “most” when κg(t) has a maximum and κg(t) > 0 or

1.3. Curvature 29

when κg(t) has a minimum and κg(t) < 0. These cases correspondrespectively to curving the most to the left and to the right. Besidesa situation when the curvature is 0, the curve is locally curving the“least” when κg(t) has a minimum and κg(t) > 0 or when κg(t)has a maximum and κg(t) < 0. Figure 1.6 illustrates all five of thedifferent possibilities just described.

Proposition 1.3.5. A regular parametrized curve ~X : I → R2 has cur-vature κg(t) = 0 for all t ∈ I if and only if the locus of ~X is a linesegment.

Proof: If the locus of ~X traces out a line segment, then (perhapsafter a regular reparametrization) we can write ~X(t) = ~a + ϕ(t)~b,where ϕ(t) is a differentiable real function with ϕ′(t) 6= 0. Then

~X ′(t) = ϕ′(t)~b,

~X ′′(t) = ϕ′′(t)~b.

Thus, by Equation (1.12),

κg(t) =(ϕ′(t)ϕ′′(t)~b ×~b) · ~k

(|ϕ′(t)| ‖~b‖)3/2= 0

because ~b×~b = ~0.Conversely, if ~X(t) = (x(t), y(t)) is a curve such that κg(t) = 0,

thenx′(t)y′′(t)− x′′(t)y′(t) = 0.

We need to find solutions to this differential equation or determinehow solutions are related. Since ~X is regular, ~X ′ 6= ~0 for all t ∈ I.Let I1 be an interval where y′(t) 6= 0. Over I1, we have

x′y′′ − x′′y′

(y′)2=

d

dt

(x′

y′

)= 0 =⇒ x′

y′= C,

where C is a constant. Thus, x′ = Cy′ for all t. Integrating withrespect to t we deduce x(t) = Cy(t)+D. Similarly, over an interval I2

where x′(t) 6= 0, we deduce that y(t) = Ax(t)+B for some constantsA and B. Since I can be covered with intervals where x′(t) 6= 0 ory′(t) 6= 0, we deduce that the locus of ~X is a piecewise linear curve.However, since ~X is regular, it has no corners, and hence its locus isa line segment.


Note that the curvature function arose from calculating ~X ′′ asthe derivative of the expression ~X ′ = s′ ~T , which involved finding anexpression for d

dt~T (t). Taking the third derivative of ~X(t) and using

the decomposition in Equation (1.11), we get

~X ′′′ = s′′′ ~T + (3s′′s′κg + (s′)2κ′g)~U + (s′)2κg ~U′.

Consequently, we need an expression for the derivative ~U ′(t). Bydefinition, for all t, the set ~T , ~U forms an orthonormal basis andhence

~U ′ = (~U ′ · ~T )~T + (~U ′ · ~U)~U.

However, since ~U(t) is a unit vector function, the identity ~U · ~U ′ = 0holds for all t. Furthermore, since ~U · ~T = 0 for all t, we also deducethat

~U ′ · ~T = −~U · ~T ′ = −s′κg.

Consequently,~U ′(t) = −s′(t)κg(t)~T (t).

It is not hard to see that if a curve ~X : [−a, a] → R2 is repara-metrized by ~X(−t), then the modified curvature function would be−κg(t). Thus, the sign of the curvature depends on what one mightcall the “orientation” of the curve, a notion ultimately arbitrary inthis case. Excluding this technicality, curvature has a physical in-terpretation that one can eyeball on particular curves. If a curve isalmost a straight line, then the curvature is close to 0, but if a regu-lar curve bends tightly along a certain section, then the curvature ishigh (in absolute value). Of particular interest are points where thecurvature reaches a local extremum.

Definition 1.3.6. Let C be a regular curve parametrized by ~X : I → R2

with curvature function κg(t). A vertex of the curve C is a point

P = ~X(t0) where the corresponding curvature function κg(t) attainsan extremum.

By the First Derivative Test, extrema of κg(t) occur at t = t0,if κ′g(t) changes sign through t0. This obviously must occur whereκ′g(t0) = 0 but the converse is not true.

From the above discussion, one should notice that since ~T , ~U |form an orthonormal basis for all t, every higher derivative of ~X(t),

1.3. Curvature 31

if it exists, can be expressed as a linear combination of ~T and ~U . Thecomponents of ~X(n)(t) in terms of ~T and ~U involve sums of productsof derivatives of s(t) and κg(t). Furthermore, if ~X is parametrized

by arc length, then the coefficients of ~X(n)(s) only involve sums ofpowers of derivatives of κg(s).

Problems

1.3.1. Find the curvature function of the curve ~X(t) = (tm, tn) for t ≥ 0.

1.3.2. Find the curvature function of the ellipse ~X(t) = (a cos t, b sin t).

1.3.3. Find the curvature function of the cycloid ~X(t) = (t−sin t, 1−cos t).

1.3.4. Find the curvature function of the curve ~X(t) = (ecos t, esin t) witht ∈ [0, 2π].

1.3.5. Calculate the curvature function of the flower curve parametrized by

~X(t) = (sin(nt) cos t, sin(nt) sin t).

1.3.6. Consider the curve ~X(t) = (t2, t3 − at), where a is a real number.Calculate the curvature function κg(t) and determine where κg(t)has extrema.

1.3.7. Consider the graph of a function y = f(x) parametrized as a curve

by ~X(t) = (t, f(t)). Find a formula for the curvature. Show that fmust be of class C2 for the curvature to exist and to be continuous.Prove that κg(t) = 0 if t is an inflection point. [Hint: Recall that x0is the inflection point of a function y = f(x) if f ′′(x) changes signat x0.]

1.3.8. Find an equation in f(t) that describes where the vertices occur for

a function graph ~X(t) = (t, f(t)). Find an example of a functiongraph where κ′g(t) = 0 but f ′(t) 6= 0 and vice versa.

1.3.9. Prove that the graph of a polar function r = f(θ) at angle θ hascurvature

κg(θ) =2f ′(θ)2 + f(θ)2 − f(θ)f ′′(θ)

(f ′(θ)2 + f(θ)2)3/2.

1.3.10. Find the vertices of an ellipse with half-axes a 6= b and calculate thecurvature at those points.

1.3.11. Calculate the curvature function for all the Lissajous figures: ~x(t) =(cos(mt), sin(nt)), with t ∈ [0, 2π].


1.3.12. Prove by direct calculation that the following formula holds for theellipse ~X(t) = (a cos t, b sin t):

∫ 2π

0

κg ds = 2π.

[Hint: Use a substitution involving tan θ = ab tan t.]

1.3.13. Calculate the curvature function for the cardioid:

~x(t) = ((1− cos t) cos t, (1− cos t) sin t) .

1.3.14. Find the vertices of the exponential curve ~X(t) = (t, et) for t ∈ R.Interpret this result in terms of radius of curvature on the curve.

1.3.15. Let ~X : I → R2 be a parametrized curve and f : J → I a surjectivefunction so that f makes ~ξ = ~X f a regular reparametrization of~X. Call t0 = f(u0) so that ~ξ(u0) = ~X(t0). Prove that

κg, ~X(t0) =

κg,~ξ(u0) if f ′(u0) > 0,

−κg,~ξ(u0) if f ′(u0) < 0.

1.3.16. Let I be a closed and bounded interval. We define a parallel curveto a parametrized curve ~X : I → R2 as a curve that can be param-etrized by ~γε(t) = ~X(t) + ε~U(t), where ε is a real number. Suppose

that ~X is a regular curve of class C2 and assume ε 6= 0. Prove that~γε is regular if and only if 1

ε /∈ [κm, κM ], where

κm = mint∈Iκg(t) and κM = max

t∈Iκg(t).

1.3.17. Let C be a curve in R2 defined as the solution to an equationF (x, y) = 0. Use implicit differentiation to prove that at any pointon this curve, the curvature of C is given by

κg =FxxF

2y − 2FxyFxFy + FyyF

2x

(F 2x + F 2

y )3/2.

1.3.18. Find the vertices of the curve given by the equation x4 + y4 = 1.

1.4. Osculating Circles, Evolutes, and Involutes 33

x

y

sinx

tanx

π2

Figure 1.7. Functions sinx and tanx have contact order 1 at the origin.

1.3.19. Pedal Curves. Let C be a regular curve and A a fixed point inthe plane. The pedal curve of C with respect to A is the locus ofpoints of intersection of the tangent lines to C and lines through Aperpendicular to these tangents. Given the parametrization ~X(t) ofa curve C, provide a parametric formula for the pedal curve to Cwith respect to A. Find an explicit parametric formula for the pedalcurve of the unit circle with respect to (1, 0).

1.4 Osculating Circles, Evolutes, and Involutes

Classical differential geometry introduces the notion of order of con-tact to measure the degree of intersection between two curves orsurfaces at a particular intersection point. See Struik in [32, p. 23]for this classical definition. In this text, we provide an alternate butequivalent definition that is more relevant to our approach to alwaysdescribing curves via a parametrization.

As a motivating example, consider the real functions f(x) = sinxand g(x) = tanx near x0 = 0. The graphs of f and g intersect at(0, 0). However, we notice in Figure 1.7 that not only do they inter-sect, but they share the same tangent line. Even further, considertheir sixth-order Taylor polynomial at 0:

sinx ≈ x− 1

6x3 +

1

120x5

tanx ≈ x+1

3x3 +

2

15x5.


In particular, both sinx and tanx have the same second-order Taylorpolynomial at x0 = 0, but their third-order Taylor polynomials differ.We say that f(x) and g(x) have contact of order 2 at x0 = 0.

Definition 1.4.1. Two functions f(x) and g(x) defined on a neighbor-hood of x0 have contact of order k if for i = 0, 1, . . . , k, the deriva-tives f (i)(x0) and g(i)(x0) exist and f (i)(x0) = g(i)(x0). We say thatf and g have contact of strict order k at x0 if they have contact oforder k but do not have contact of order k + 1.

Note that a simple monomial function f(x) = xn with the x-axisat (0, 0) has contact of strict order n−1. This is because f (i)(0) = 0for i = 0, 1, . . . , n− 1 but f (n)(0) = n!, which is not equal to 0.

As we attempt to generalize this notion to order of contact be-tween two parametrized curves in the plane, we are immediatelymet with two obstacles. First, in general the parameter for one pa-rametrized curve has no connection, geometric or otherwise, to theparameter for the second curve. In the definition of order of contactfor real-valued functions over the reals, the parameter x in an openinterval of R is used for both functions f(x) and g(x). Secondly, pa-rametrized curves are usually not described as functions with respectto some frame. Consequently, we must make a choice of parameteron both curves simultaneously that has geometric meaning. The typ-ical approach is to use one of the curves parametrized by arc lengthas a reference.

Definition 1.4.2. Suppose C1 : ~α(t) and C2 : ~β(u) are two parame-trized curves that intersect at a point P , which corresponds to wheret = t0 and u = u0. Reparametrize C1 by arc length and let s0 besuch that P = ~α(s0). Let u(s) be the function such that the projec-tion of ~α(s) onto C2 is located at ~β(u(s)). Then we say that C1 andC2 have contact of order n at P if they are of class Cn over an openinterval around P and

~α(i)(s0) =di

dsi~β(u(s))

∣∣∣s0

for all 0 ≤ i ≤ n. Furthermore, C1 and C2 have contact of strictorder n if they do not also have contact of order n+ 1.


The intuitive picture for order of contact indicates that two in-tersecting curves with contact of order 1 have the same tangent lineat the intersection point. In particular, we leave it as an exercise toprove the fact that a curve and its tangent line have order of contact1. In contrast, an intersection point between two curves with contactof strict order 0 is said to be a transversal intersection.

Note that since the concept of order of contact refers to ordersof differentiation, if two curves are not both of class Cn near a pointP it does not make sense to discuss contact of order n.

At first glance, this definition of order of contact seems asymmet-rical. However, it is possible to prove that using ~β as the referencecurve and the arc length of C2 as the reference parameter is equiva-lent to Definition 1.4.2. (In fact, for a given nonnegative integer n,on the set of parametrized curves in the plane, the relation of con-tact of order n is an equivalence relation, i.e., is reflexive, symmetric,and transitive. In contrast, the relation of contact of strict order nis reflexive and symmetric but not necessarily transitive.)

Definition 1.4.3. Let C be a curve parametrized by ~X : I → R2, andlet t0 be a regular value of the curve. The osculating circle to C atthe point t0 is a circle that has contact of order 2 with C at ~X(t0) .

Proposition 1.4.4. Let C : ~X : I → R2 be a parametrized curve and t0a regular value. Suppose that ~X(t) is twice differentiable at t0 andthat κg(t0) 6= 0. Then,

1. There exists a unique osculating circle to C at ~X(t0);

2. It is given by the following vector function:

~γ(t) = ~X(t0)+1

κg(t0)~U(t0)+

1

κg(t0)

((sin t)~T (t0)− (cos t)~U(t0)

).

Proof: Without loss of generality, we can assume that ~X is parame-trized by arc length s and that we are looking for the osculating circleat s = 0. Call ~γ(u) the parametrization of the proposed osculatingcircle, which must have the form

~γ(u) = (a+R cos(u), b+R sin(u)).


For ease of the proof, we will allow R to be any nonzero real number.We assume that u = u0 at the point of contact and that near u0,the function u(s) gives the projection of ~X(s) onto the curve ~γ. ByDefinition 1.4.2, in order for there to exist an osculating circle, theparameters a, b, and R and the function u(s) must satisfy

~X(0) = ~γ(u(0)), ~X ′(0) =d

ds~γ(u(s))

∣∣∣0, ~X ′′(0) =

d2

ds2~γ(u(s))

∣∣∣0,

which in coordinates is equivalent to

a+R cos(u(0)) = x(0), (1.13)

b+R sin(u(0)) = y(0), (1.14)

−Ru′(0) sin(u(0)) = x′(0), (1.15)

Ru′(0) cos(u(0)) = y′(0), (1.16)

−Ru′′(0) sin(u(0))−Ru′(0)2 cos(u(0)) = x′′(0), (1.17)

Ru′′(0) cos(u(0))−Ru′(0)2 sin(u(0)) = y′′(0). (1.18)

Since ~X(s) is parametrized by arc length, x′(0)2+y′(0)2 = 1, fromwhich we conclude that |u′(0)| = 1

R . Without loss of generality, wecan assume that u′(0) and R have the same sign, so that u′(0) = 1

R .Since the curve is parametrized by arc length, x′(s)2 + y′(s)2 = 1for all s. Taking a derivative of this equation with respect to s andevaluating at s = 0, we deduce that x′(0)x′′(0) + y′(0)y′′(0) = 0.This relation, along with Equations (1.15) through (1.18) leads tou′′(0) = 0.

We obtain the value of R by noting that after calculation

κg(0) = x′(0)y′′(0)− x′′(0)y′(0) = R2(u′(0))3 =1

R.

Using Equations (1.15) and (1.16), we find that

(cos(u(0)), sin(u(0))) = (y′(0),−x′(0)) = −~U,

since ~X is parametrized by arc length. Thus the center of the circle~γ is

(a, b) = (x(0), y(0))−R (cos(u(0)), sin(u(0))) = ~X(0) +1

κg(0)~U(0).


Finally, a quick check that we leave for the reader is to show thatgiven the above values for a, b, and R, Equations (1.13) through(1.18) are redundant given the fact that ~X(s) is parametrized by arclength. These results prove part 1 of the proposition.

Part 2 follows easily from part 1. Since ~X is parametrized by arclength, the unit tangent vector is just ~T (0) = (x′(0), y′(0)), givingalso ~U(0) = (−y′(0), x′(0)). In the solutions for Equations (1.13)and (1.14), we see that the center of the osculating circle is at

(a, b) = (x(0)−R cos(τ(0)), y(0)−R sin(τ(0))) = ~X(0) +R~U(0).

Furthermore, for any curve parametrized by arc length, κg(s) =x′(s)y′′(s)− x′′(s)y′(s). Thus, 1

R = κg(0).

In light of Proposition 1.4.4 and Example 1.3.3, the curvaturefunction κg(t) of a plane curve ~X has a nice physical interpreta-tion, namely, the reciprocal of the radius of the osculating circle. Ahigher order of contact indicates a better geometric approximation,and hence, since there is a unique osculating circle, it is, in a ge-ometric sense, the best approximating circle to a curve at a point.Furthermore, in Problem 1.4.6 of this section, we prove that to acurve ~X at t = t0, there does not necessarily exist a touching circlewith order of contact greater than 2. Thus, the curvature is the in-verse of the radius of the best approximating circle to a curve at apoint.

(From a physics point of view, we obtain an additional confir-mation of the above interpretation by considering the units of thecurvature function. If we view the coordinate functions x(t) and y(t)with the unit of meters and t in any unit, Equation (1.12) gives theunit of 1/meter for κg(t).)

Definition 1.4.5. Let ~X be a parametrized curve, and let t = t0 be aregular point that satisfies the conditions in Proposition 1.4.4. Thecenter of the osculating circle at t = t0 is called the center of curva-ture.

Definition 1.4.6. The evolute of a curve C is the locus of the centersof curvature.


Proposition 1.4.7. Let ~X : I → R2 be a regular parametric plane curvethat is of class C2, i.e., has a continuous second derivative. Let I ′

be a subinterval of I over which κg(t) 6= 0. Then over the interval

I ′, the evolute of ~X has the following parametrization:

~E(t) = ~X(t) +1

κg(t)~U(t).

Example 1.4.8. Consider the parabola y = x2 with parametric equa-tions ~X(t) = (t, t2). We calculate the curvature with the followingsteps:

~X ′(t) = (1, 2t),

~X ′′(t) = (0, 2),

s′(t) =√

1 + 4t2,

κg(t) =2

(1 + 4t2)3/2.

Thus, the parametric equations of the evolute are

~E(t) = (t, t2) +1

2(1 + 4t2)3/2 1√

1 + 4t2(−2t, 1)

=

(−4t3,

1

2+ 3t2

).

The Cartesian equation for the evolute of the parabola is then

y =1

2+ 3∣∣∣x4

∣∣∣2/3.A closely related curve to the evolute of ~X(t) is the involute,

though we leave the exact nature of this relationship to the problems.The involute is defined as follows.

Definition 1.4.9. Let ~X : I → R2 be a regular parametrized curve.We call an involute to ~X any parametrized curve ~ι such that for allt ∈ I, ~ι(t) meets the tangent line to ~X at t at a right angle.


x

y

~E

Figure 1.8. Evolute of a parabola.

For all t ∈ I, the point on the involute ~ι(t) is on the tangentline to ~X at t, so it is natural to write the parametric equationsas ~ι(t) = ~X(t) + λ(t)~T (t). Since we will wish to calculate ~ι′(t),which involves the derivative ~T ′(t), we must assume that the curve~X(t) is of class C2, i.e., with a second derivative that exists and iscontinuous. Definition 1.4.9 requires that the vector ~ι′(t) be in aperpendicular direction to ~T (t), so we have

~T · ~ι′ = 0 =⇒ ~T ·(s′ ~T + λ′ ~T + λs′κg ~U

)=⇒ s′ + λ′ = 0

=⇒ λ(t) = C − s(t),

where C is some constant of integration. Therefore, if ~X is a regularparametrized curve of class C2, then the formula for the involute is

~ι(t) = ~X(t) + (C − s(t))~T (t).

Problems

1.4.1. Prove the claim that the tangent line to a parametrized curve at aregular point has contact of order 1.

1.4.2. Consider the cubic curve ~X(t) = (t, t3) defined over R.

(a) Find the osculating circle of ~X at t = 1.

(b) Determine the parametric equations for the evolute of ~X.


1.4.3. Let ~X be a regular parametrized curve and let t0 be an inflectionpoint, i.e. a point where κg(t) = 0. Prove that the tangent line to~X(t) at t = t0 has contact of order 2.

1.4.4. Prove that the evolute of the ellipse ~X(t) = (a cos t, b sin t) has para-metric equations

~γ(t) =

((a2 − b2

a

)cos3 t,

(b2 − a2

b

)sin3 t

).

1.4.5. Consider the catenary given by the parametric equation ~X(t) =(t, cosh t) for t ∈ R.

(a) Prove that the curvature of the catenary is κg(t) = 1cosh2 t

.

(b) Prove that the evolute of the catenary is

~E(t) = (t− sinh t cosh t, 2 cosh t).

1.4.6. Prove that if the osculating circle to a regular parametrized curveC at a point P has contact of order 3, then P is a vertex of C.Give an example where this does not occur, thereby proving that ata regular point on a parametrized curve, there does not necessarilyexist a circle with contact of order 3.

1.4.7. Prove that an osculating circle to a curve C at a point P that is nota vertex lies on either side of the curve. [See Problem 1.4.6.]

1.4.8. Let ~X : I → R2 be a parametrized curve that does not have anyinflection points (i.e., κg(t) 6= 0 for all t ∈ I). Prove that the evolute

of the curve has a critical point at t0 if and only if ~X(t0) is a vertexof the curve.

1.4.9. Let ~x(t) = (t, t2) be the parabola. Define a new curve ~ι as theinvolute of ~x such that ~ι(0) = ~x(0) = (0, 0). Calculate parametricequations for ~ι(t). [Hint: This will involve an integral.]

1.4.10. (*) Continuation of the last problem: Calculate parametric equa-tions for the evolute to ~ι.

1.5 Natural Equations

An isometry of the plane is any function F : R2 → R2 such that forany two points ~p, ~q ∈ R2, the distance between them is preserved,i.e.,

‖~q − ~p‖ = ‖F (~q)− F (~p)‖.

1.5. Natural Equations 41

A well-known result about isometries is that F is an isometry if andonly if

F (~v) = A~v + ~C,

where A is any 2×2 orthogonal matrix and ~C is any constant vector.Isometries of the plane include rotations, translations, reflections,and glide reflections. It is also not hard to show that a compositionof isometries is again an isometry.

The condition of orthogonality A>A = I implies that det(A) =±1. Consequently, isometries come in two flavors depending on thesign of the determinant of A. If det(A) = 1, we call the isometrya positive isometry. These include rotations, translations, and com-positions thereof. If det(A) = −1, we call the isometry a negativeisometry. These include reflections and glide reflections. Positiveisometries are also called rigid motions because any shape (imagin-ing it to be a solid physical object) can be moved from its original toits image under a positive isometry without bringing the shape outof the plane.

Recall now that for a parametrized curve ~X : I → R2 with~X(t) = (x(t), y(t)), the curvature function is given by

κg(t) =x′(t)y′′(t)− x′′(t)y′(t)

(x′(t)2 + y′(t)2)3/2.

Since any rigid motion does not stretch distances, such a trans-formation should not distort a plane curve. Therefore, as one mightexpect, the curvature is preserved under rigid motions, a fact thatwe now prove.

Theorem 1.5.1. Let ~X : I → R2 be a regular plane curve that is ofclass C2. Let F : R2 → R2 be a rigid motion of the plane given byF (~v) = A~v+ ~C, where A is a rotation matrix and ~C is any vector inR2. The vector function ~ξ = F ~X is a regular parametrized curvethat is of class C2, and the curvature function κg(t) of ~ξ is equal to

the curvature function κg(t) of ~X.

Proof: A rotation matrix in R2 is of the form

A =

(a −bb a

),


where a2 + b2 = 1. Let ~C = (e, f). We can write the parametricequation for ~ξ as

~ξ(t) = (ax(t)− by(t) + e, bx(t) + ay(t) + f).

Then

~ξ′(t) = (ax′(t)− by′(t), bx′(t) + ay′(t)),

~ξ′(t) = (ax′′(t)− by′′(t), bx′′(t) + ay′′(t)),

and therefore the curvature function of ~ξ is

κg(t) =(ax′(t)− by′(t))(bx′′(t) + ay′′(t))− (bx′(t) + ay′(t))(ax′′(t)− by′′(t))

((ax′(t)− by′(t))2 + (bx′(t) + ay′(t))2)3/2

=abx′x′′ + a2x′y′′ − b2y′x′′ − aby′y′′ − abx′x′′ + b2x′y′′ − a2y′x′′ + aby′y′′

(a2(x′)2 − 2abx′y′ + b2(y′)2 + b2(x′)2 + 2abx′y′ + a2(y′)2)3/2

=x′y′′ − y′x′′

((x′)2 + (y′)2)3/2= κg(t).

The curvature function is invariant under any positive isometry,i.e., a composition of rotations and translations. Furthermore, inProblem 1.3.15, we saw that the curvature function is invariant underany regular reparametrization, except up to a sign that depends on“the direction of travel” along the curve. Consequently, |κg| is ageometric invariant that only depends on the shape of the curve ata particular point and not how the curve is parametrized or wherethe curve sits in R2.

It is natural to ask whether a converse relation holds, namely,whether the curvature function is sufficient to determine the param-etrized curve up to a positive plane isometry. As posed, the questionis not well defined since a curve can have different parametrizations.However, we can prove the following fundamental theorem.

Theorem 1.5.2 (Fundamental Theorem of Plane Curves). For a given piece-wise continuous function κg(s), there exists a regular curve of class

C2 parametrized by arc length by ~X : I → R2 with the curvaturefunction κg(s). Furthermore, the curve is uniquely determined up toa rigid motion in the plane.


Proof: If a regular curve ~X(s) is parametrized by arc length, thenthe curvature formula is κg(s) = x′(s)y′′(s) − y′(s)x′′(s). The proofof this theorem consists of exhibiting a parametrization that satisfiesthis differential equation.

For a regular curve of class C2 that is parametrized by arc length,we have ~X ′ = ~T , and we can view ~T as a vector function of class C1

from the interval of definition of ~X into the unit circle. Therefore,

~T = (cos(θ(s)), sin(θ(s))) (1.19)

for some continuous function θ(s). However, ~T ′(s) = κg(s)~U , andEquation (1.19) show that

κg(s)~U = θ′(s) (− sin(θ(s)), cos(θ(s))) .

Thus, we deduce that κg(s) = θ′(s).The above remarks lead to the following result. Suppose we are

given the curvature function κg(s). Performing two integrations, we

see that the only curves ~X(s) with curvature function κg(s) must be

~X(s) =

(∫cos(θ(s)) ds+ e,

∫sin(θ(s)) ds+ f

), (1.20)

where

θ(s) =

∫κg(s) ds+ θ0, (1.21)

and where θ0, e, and f are constants of integration. Note that thetheorem requires κg(s) to be piecewise continuous in order to beintegrable. Furthermore, the trigonometric addition formulas showthat a nonzero constant θ0 changes ~X by a rotation of θ0 and thenonzero constants e and f correspond to a translation along thevector ~C = (e, f).

Because of the geometric nature of the curvature function κg(s)(with respect to arc length) and because there exists a unique planecurve (up to a rigid motion) for a given curvature function, we callκg(s) the natural equation of its corresponding curve. The Funda-mental Theorem of Plane Curves is surprising because, a priori, weexpect a curve to require two functions (coordinate functions) to de-fine it. However, only one function, κg(s), is required to uniquelydefine the shape of a curve.


If we assume that ~X(s) is of class C∞ and is equal to its powerseries in an open interval around s = 0, then we can see why Theorem1.5.2 holds using Taylor series. (This condition is called real analyticat s = 0.)

Let ~X : I → R2 be a real analytic curve parametrized by arclength and assume without loss of generality that 0 ∈ I. We canexpand the Taylor series of ~X about 0 to get

~X(s) = ~X(0) + s ~X ′(0) +s2

2!~X ′′(0) +

s3

3!~X ′′′(0) + · · · .

However, since the curve is parametrized by arc length, ~X ′(s) =~T (s), ~X ′′(s) = κg(s)~U(s), ~X ′′′(s) = κ′g(s)

~U(s) − (κg(s))2 ~T (s), and

so on. The first few terms look like

~X(s) = ~X(0) +

(s− 1

6κg(0)2s3 + · · ·

)~T (0) +

(1

2s2 +

1

6κ′g(0)s3 + · · ·

)~U(0).

Thus, since the normal vector ~U(s) is just a rotation of ~T (s) byπ2 , given a function κg(s), once one chooses for initial conditions

the point ~X(0) and the direction ~T (0), the Taylor series is uniquelydetermined. The intersection of the intervals of convergence of thetwo Taylor series in the ~T (0) and the ~U(0) components is a newinterval J that contains s = 0. It is possible that J could be trivial,J = 0, but if it is not, then the Taylor series uniquely defines~X(s) over J . Choosing a different ~T (0) amounts to a rotation ofthe curve in the plane, and choosing a different ~X(0) amounts to atranslation. Therefore, we see again that making different choices forthe initial conditions corresponds to a rigid motion of the curve inthe plane.

For even simple functions for κg(s), it is often difficult to usethe approach in the proof of Theorem 1.5.2 to explicitly solve for~X(s). However, using a computer algebra system (CAS) with toolsfor solving differential equations, it is possible to produce a pictureof curves that possess a given curvature function κg(s). One cancreate a numerical solution using the solution in Equation (1.20)with (1.21). As an equivalent technique, using a CAS, one can solve


Figure 1.9. Curve with κg(s) = 1 + 2/(1 + s2). Figure 1.10. Curve with κg(s) = 2 + sin s.

the system of differential equationsx′(s) = cos(θ(s)),

y′(s) = sin(θ(s)),

θ′(s) = κg(s),

and only plot the solution for the pair (x(s), y(s)). A choice of initialconditions determines the position and orientation of the curve inthe plane. Figures 1.9 and 1.10 give a few interesting examples ofparametric curves with a given κg(s).

The code in Maple (a common computer algebra system) forFigure 1.9 is the following.

> with(DEtools):

> with(plots):

> kappa:=s->1+2/(1+s^2);

> sys := D(x)(s)=cos(th(s)), D(y)(s)=sin(th(s)), D(th)(s)=kappa(s);

> DEplot([sys],[x(s),y(s),th(s)], s=-10..10, [[x(0)=0,y(0)=0,th(0)=0]],

scene=[x(s),y(s)], stepsize=0.01, iterations=10, linecolor=black,

scaling=constrained, axes=normal);

Problems

1.5.1. Suppose a curve has κg(s) = A. Prove by directly expanding theTaylor series that such a curve is a circle.


1.5.2. Suppose a curve has κg(s) = 2As. Prove by directly expanding the

Taylor series that if such a curve has ~X(0) = ~0 and ~X ′(0) = (1, 0),then such a curve is

~X(s) =

(∫ s

0

cos(As2) ds,

∫ s

0

sin(As2) ds

).

1.5.3. Find parametric equations for a curve satisfying κg(s) = 11+s2 by

direct integration, following the method in the proof of Theorem1.5.2. Show that this curve corresponds to the catenary y = coshx.

Investigative Projects

Project I. What can you discover about the properties of thepedal curve based on a starting curve and a given point?

Project II. Since κg(t) is the inverse of the radius of curvature, wefeel the curvature function when we drive. More precisely, fromthe centripetal acceleration formula v2

R where v is the speed,if we travel in a car along a curve at constant speed, thenthe centripetal acceleration is proportional to the curvaturefunction. In the proof of Theorem 1.5.2, we pointed out thatκg(s) = θ′(s) for some angle of direction θ(s). Use this torelate the curvature to the position of the steering wheel of acar. Then consider the problem of turning a corner. Turninga corner means starting on a straight line, turning the steeringwheel for a bit and then straightening out so that the car is nowgoing on a straight line that is perpendicular to the original lineof travel. Discuss curvature functions that make a reasonablygood turn on normal streets.

CHAPTER 2

Plane Curves: Global Properties

Most of the properties of curves we have studied so far are called localproperties. By definition, a local property of a curve (or surface) is aproperty that is related to a point on the curve based on informationcontained just in a neighborhood of that point. By contrast, globalproperties concern attributes about the curve taken as a whole. InChapter 1, the arc length of a curve was the only notion introducedthat we might consider a global property.

Analytically speaking, local properties of a curve at a point in-volve derivatives of the parametric equations while global propertiesdeal with integration along the curve and topological properties andgeometric properties of how the curve lies in R2. Some of the proofsof global properties rely on theorems from topology. These are sup-plied concisely in the appendix on topology in [24]. (The interestedreader is encouraged to consult some references on basic topology,such as Gemignani [16] or Armstrong [1].)

2.1 Basic Properties

Definition 2.1.1. A parametrized curve C is called closed if there ex-ists a parametrization ~X : [a, b] → R2 of C such that ~X(a) = ~X(b).A closed curve is of class Ck if, in addition, all the (one-sided) deriva-tives of ~X at a and at b are equal of order i = 0, 1, . . . , k; in otherwords, if as one-sided derivatives ~X ′(a) = ~X ′(b), ~X ′′(a) = ~X ′′(b),and so on up to ~X(k)(a) = ~X(k)(b).

We recall that the left-derivative (resp. right-derivative) of a func-tion f at a is the limit

limh→0−

f(a+ h)− f(a)

h

(resp. lim

h→0+

f(a+ h)− f(a)

h

).

47

48 2. Plane Curves: Global Properties

The notion of left- and right-sided derivatives of real-valued functionsnaturally extends to parametrized curves.

The conditions on the derivatives of ~X in the above definitionseem awkward but they attempt to establish the fact that the vectorfunction ~X behaves identically at a and at b. A more topologicalapproach involves using a circle S1 rather than an interval as thedomain for the map ~X. We define a topological circle S1 as theset [0, 1] with the points 0 and 1 identified. The topology of S1 isthe identification topology, which in this case means that an openneighborhood U of p contains, for some ε, the subsets

(p− ε, p+ ε) with ε < minp, 1− p, if p 6= 0,

[0, ε) ∪ (1− ε, 1], if p = 0 (= 1).

Then we say that a curve C is closed if there exists a continuoussurjective (onto) function ϕ : S1 → C.

Definition 2.1.2. A curve C is called simple if there exists a parametr-ization ~X : I → R2 of C such that ~X is an injective (i.e., one-to-one)function. A closed curve C is called simple if there exists a parame-trization ~X : [a, b]→ R2 of C such that ~X(t1) = ~X(t2) with t1 < t2only when t1 = a and t2 = b.

Using the language associated to a topological circle, we say thata closed curve C is simple if there exists a bijection ϕ : S1 → C thatis continuous and such that ϕ−1 is also continuous.

If a curve is not simple, we intuitively think of the curve asintersecting itself. Using parametrizations, we can give the conceptof a self-intersection a precise definition.

Definition 2.1.3. A curve C is said to have a self-intersection at a pointP ∈ C if for every parametrization ~X : I → R2 of C, there existt1 6= t2 that are not endpoints of I such that ~X(t1) = ~X(t2) = P .

Proposition 2.1.4. Let C be a closed curve with parametrization ~X :[a, b]→ R2. Then C is bounded as a subset of R2.

Proof: Let ~X : [a, b] → R2 be a parametrization of C with coordi-nate functions x(t) and y(t). By the Extreme Value Theorem, since

2.1. Basic Properties 49

x(t) and y(t) are continuous functions over an interval [a, b], thenthey respectively attain extrema xmin minimum and xmax maximumof x(t) and ymin minimum and ymax maximum of y(t). The param-etrized curve ~X lies entirely in the rectangle xmin ≤ x ≤ xmax andymin ≤ y ≤ ymax. Therefore, the curve lies inside a disc centeredaround the origin. More precisely, let

Mx = max|xmin|, |xmax| and My = max|ymin|, |ymax|.

Then for all t ∈ [a, b],

x(t)2 ≤M2x and y(t)2 ≤M2

y .

Thus,

‖ ~X(t)‖ ≤√M2x +M2

y ,

and so the curve C is contained in a disk of finite radius. Hence, Cis bounded.

One of the most fundamental properties of global geometry ofplane curves is that a simple, closed plane curve C separates theplane into two open connected components, each with the commonboundary of C.

Theorem 2.1.5 (Jordan Curve Theorem). Let C be a simple closed curvein R2. Then C separates the plane into precisely two componentsW1 and W2 such that R2 − C = W1 ∪W2 and W1 ∩W2 = ∅. Onecomponent is bounded and the other is unbounded.

This intuitive fact turns out to be rather difficult to prove. Fur-thermore, a rigorous proof requires a precise definition of compo-nent. We refer the reader to Munkres [26] for a detailed discussion.In [24, Section A.4], there is a more readable proof of the weakerstatement that only assumes that the curve is regular. The compo-nent of R2 − C that is bounded is called the interior of C and thecomponent of C that is unbounded is called the exterior.

We remind the reader of the following theorem from multivariablecalculus, which one may view as a global property of plane curves.


Theorem 2.1.6 (Green’s Theorem). Let C be a simple, closed, piecewiseregular, positively oriented plane curve with interior region R, andlet ~F = (P (x, y), Q(x, y)) be a differentiable vector field. Then,∫

CP dx+Qdy =

∫∫R

(∂Q

∂x− ∂P

∂y

)dx dy.

We remind the reader that a curve is positively oriented if as anobject travels along the curve, the interior is to the left. With unittangent and unit normal vectors, we can say that C is positivelyoriented if it is parametrized by ~X : [a, b] → R2 such that U(t)points toward the interior of the curve. We also remind the readerthat if ~X : [a, b]→ R2 is a regular parametrization for C with ~X(t) =(x(t), y(t)), then∫

CP dx+Qdy =

∫C

~F · d ~X

=

∫ b

a

~F (x(t), y(t)) · ~X ′(t) dt

=

∫ b

aP (x(t), y(t))x′(t) +Q(x(t), y(t))y′(t) dt.

Corollary 2.1.7. Let C be a simple closed regular plane curve with in-terior region R. Then the area A of R is

A =

∫Cx dy = −

∫Cy dx =

1

2

∫C−y dx+ x dy. (2.1)

Proof: In each of these integrals, one simply chooses a vector field~F = (P,Q) such that

∂Q

∂x− ∂P

∂y= 1.

For the three integrals, these are, respectively, ~F = (0, x), ~F =(−y, 0), and ~F = 1

2(−y, x). Then apply Green’s Theorem.

Example 2.1.8. Consider the ellipse given by ~X(t) = (a cos t, b sin t)for 0 ≤ t ≤ 2π. The techniques from introductory calculus used tocalculate area would lead us to evaluate the integral

A = 4

∫ a

0b

√1− x2

a2dx.


Though one can calculate this by hand, it is not simple. Much easierwould be to use Green’s Theorem, which gives

A =

∫Cx dy =

∫ 2π

0a cos t b cos t dt

= ab

∫ 2π

0cos2 t dt = ab

∫ 2π

0

1

2+

1

2cos(2t) dt

= ab[1

2t+

1

4sin(2t)

]2π

0

= πab.

Problems

2.1.1. Calculate the area of the region enclosed by the cardioid ~X(t) =((1− cos t) cos t, (1− cos t) sin t).

2.1.2. Use Green’s Theorem to calculate the area of one loop of ~X(t) =(cos t, sin(2t)).

2.1.3. Consider the function graph of a curve in polar coordinates r = f(θ)

parametrized by ~X(t) = (f(t) cos t, f(t) sin t). Suppose that for 0 ≤t ≤ 2π, the curve is closed and encloses a region R. Use Equation(2.1) and Green’s Theorem to prove the following area formula inpolar coordinates:

Area(R) =

∫∫Rr dr dθ.

2.1.4. The diameter of a curve C is defined as the maximum distancebetween two points, i.e.,

diam(C) = maxd(p1, p2) | p1, p2 ∈ C .

We call a diameter any chord of C whose length is diam(C). Let~X(t) be the parametrization of a closed differentiable curve C. Let

f(t, u) = ‖ ~X(t)− ~X(u)‖ be the distance function between two pointson the curve.

(a) Prove that if a chord [p1, p2] of C is a diameter of C, then theline (p1, p2) is simultaneously perpendicular to the tangent lineto C at p1 and the tangent line to C at p2.

(b) Under what other situations is the line (p1, p2) simultaneouslyperpendicular to the tangent line to C at p1 and the tangentline to C at p2?


2.1.5. Prove that the diameter of the cardioid ~X(t) = ((1− cos t) cos t, (1−cos t) sin t) is 3

√3/2. [Hint: First show that the maximum distance

between ~X(t) and ~X(u) occurs when u = 2π − t.]

2.1.6. (*) This problem studies the relation between the curvature κg(s)of a curve C and whether it is closed.

(a) Prove that if C is closed, then its curvature κg(s), as a functionof arc length, is periodic.

(b) Assume that κg(s) is not constant and is periodic with small-est period p. Use the Fundamental Theorem of Plane Curves(Theorem 1.5.2) to prove that if

1

2π

∫ p

0

κ(s) ds

is an element of Q − Z, i.e., a rational that is not an integer,then C is a closed curve. [This problem is discussed in [3],though the authors’ use of complex numbers is not necessaryfor this problem.]

2.2 Rotation Index

As a leading example of what we shall term the rotation index ofa curve, consider the circle C : ~X(t) = (R cos t, R sin t) with thedefining interval of I = [0, 2π]. A simple calculation shows that forall t ∈ [0, 2π], we have a curvature of κg(t) = 1

R . Then it is easy tosee that ∫

Cκg ds =

∫ 2π

0κg(t)s

′(t) dt =

∫ 2π

0

1

R·Rdt = 2π.

On the other hand, suppose that we use the defining interval[0, 2πn] with the same parametrization. In that case, were we toevaluate the same integral, we would obtain

1

2π

∫Cκg ds = n.

Now consider any regular, closed, plane curve ~X : I → R2. Tothis curve, we associate the unit tangent vector ~T (t). Placing thebase of this vector at the origin, we see that ~T itself draws out a

2.2. Rotation Index 53

curve ~T : I → R2. It is called the tangential indicatrix of the planecurve ~X.

The tangential indicatrix of a curve lies entirely on the unit cir-cle in the plane but with possibly a complicated parametrization.Depending on the shape of ~X, the tangential indicatrix may changespeed, stop, and double back along a portion of the circle. Nonethe-less, we can use the theory of usual plane curves to study the tan-gential indicatrix. We are in a position to state the main propositionfor this section.

Proposition 2.2.1. Let C be a closed, regular, plane curve parametrizedby ~X : I → R2 of class C2. Then the quantity

1

2π

∫Cκg ds

is an integer. This integer is called the rotation index of the curve.

Proof: Rewrite the above integral as follows:∫Cκg ds =

∫Iκg(t)s

′(t) dt =

∫Iκg(t)s

′(t)‖~U(t)‖ dt.

The curvature function κg(t) of plane curves is not always posi-

tive, but since ~X(t) is of class C2, then κg(t) is at least continuous,and the intermediate value theorem applies. Therefore, define twoclosed subsets (unions of closed subintervals) of the interval I asfollows:

I+ = t ∈ I : κg(t) ≥ 0 and I− = t ∈ I : κg(t) ≤ 0.

Using the definition of curvature from Equation (1.10), we splitthe above integral into the following two parts:∫

Cκg ds =

∫I+‖~T ′(t)‖ dt−

∫I−‖~T ′(t)‖ dt.

However, the two integrals on the right-hand side of the above equa-tion are integrals of the arc length of the tangential indicatrix trav-eling in a counterclockwise (resp. clockwise) direction. Since ~X isa closed curve, ~T is as well, and these two integrals represent 2πtimes the number of times ~T travels around the circle, with a signindicating the direction.


Figure 2.1. The curve ~X(t) = (cos t, sin(2t)).

Example 2.2.2. Consider ~X(t) = (cos t, sin(2t)) as the paramtrizationfor the curve C depicted in Figure 2.1. The curvature function isgiven by

κg(t)s′(t) =

4 sin t sin(2t) + 2 cos t cos(2t)

sin2 t+ 4 cos2(2t)=

6 cos t+ 2 cos(3t)

5− cos(2t) + 4 cos(4t).

Thus, the rotation index n of ~X is

n =1

2π

∫Cκg(t)s

′(t)dt =1

2π

∫ 2π

0

6 cos t+ 2 cos(3t)

5− cos(2t) + 4 cos(4t)dt.

Since the integrand is periodic of period 2π, using the substitutionu = t+ π

2 and recognizing that we can integrate over any interval oflength 2π, we have

n =1

2π

∫ π

−π

6 sinu− 2 sin(3u)

5 + cos(2u) + 4 cos(4u)du.

However, the integrand is now an odd function, so the rotation indexof ~X is n = 0.

The proof of Proposition 2.2.1 analyzed the integral∫C κg ds as

the signed arc length of the tangential indicatrix ~T (t), which one canview as a map from an interval I to the unit circle S1. However, the


result of Proposition 2.2.1 follows also from a more general result thatwe wish to explain in detail here. Some of the concepts below comefrom topology and illustrate the difficulty of analyzing functions froman interval to a circle. The reader should feel free to either skip thetechnical details of the proofs in the rest of this section or referto [24, Appendix A] for background material.

Any path f : I → S1 on the unit circle may be described by theangle function ϕ(t), so that

f(t) = (cos(ϕ(t)), sin(ϕ(t))) .

However, the angle ϕ(t) is defined only up to a multiple of 2π, andhence it need not be a well-defined function, let alone a continuousfunction. However, since f is continuous, for all t ∈ I there existsan interval (t − ε, t + ε) and a continuous function ϕ(t) such thatfor all u ∈ (t− ε, t+ ε), ϕ(u) differs from ϕ(u) by a multiple of 2π.If ϕ′(t) exists, it is a well-defined function, regardless of any choicemade in selecting ϕ(t). Finally, if I = [a, b], we define the total anglefunction related to f as the function

ϕ(t) =

∫ t

aϕ′(u) du+ ϕ(a).

By construction, ϕ(t) is continuous, satisfies

f(t) = (cos(ϕ(t)), sin(ϕ(t))) ,

and keeps track of how many times and in what direction the path ftravels around S1. The quantity ϕ(b)− ϕ(a) is called the total angleswept out by f .

Definition 2.2.3. Let I be an interval of R. Given a continuous func-tion f : I → S1 on the unit circle, the lifting of f is the functionϕ(t), with ϕ(a) chosen so that 0 ≤ ϕ(a) < 2π. The lifting of f isdenoted by f to indicate its unique dependence on f .

For any continuous function between circles f : S1 → S1, viewingS1 as an interval [0, 2π] with the endpoints identified, we also viewf as a continuous function f : [0, 2π] → S1. Since f(0) = f(2π) arepoints on S1, then f(2π)− f(0) is a multiple of 2π. This leads to thefollowing definition.


Definition 2.2.4. Let f : S1 → S1 be a continuous map between circles.Let f : [0, 2π] → R be the lifting of f . Then the degree of f is theinteger

deg f =1

2π

(f(2π)− f(0)

).

Intuitively speaking, the degree of a function f : S1 → S1 betweenunit circles is how many times around f winds its domain S1 ontoits target S1.

Returning to the example of the tangential indicatrix ~T of aregular closed curve C, since C is closed, ~T can be viewed as afunction S1 → S1.

Proposition 2.2.5. Let C be a regular closed curve parametrized by ~X :I → R2. The rotation index of C is equal to the degree of ~T .

Proof: Since ~T is a map ~T : I → S1, using Equation (2.2), we canwrite

~T = (cos(ϕ(t)), sin(ϕ(t))) ,

so~T ′ =

(−ϕ′(t) sin(ϕ(t)), ϕ′(t) cos(ϕ(t))

)= ϕ′(t)~U(t).

But ~T ′(t) = κg(t)s′(t)~U(t), so

κg(t)s′(t) = ϕ′(t). (2.2)

Thus, if we call I = [a, b], we have∫Cκg ds =

∫ b

aϕ′(t) dt = ϕ(b)− ϕ(a),

which concludes the proof.

The notion of degree of a continuous function S1 → S1 can beapplied in a wider context than can the rotation index of a curve sincethe latter requires a curve to be regular, while the former concept, aspresented here, only requires ϕ′(t) to be integrable. Another closelyrelated formula to calculate the degree of a curve is the following.


Proposition 2.2.6. Suppose that a function f : S1 → S1 is parametrizedas ~γ : [a, b]→ R2. If ~γ(t) = (γ1(t), γ2(t)), then

deg f =1

2π

∫ b

a

γ′2(t)

γ1(t)dt = − 1

2π

∫ b

a

γ′1(t)

γ2(t)dt.

Proof: (Left as an exercise for the reader. See Problem 2.2.7.)

In practice, since many functions f : S1 → S1 as described in theabove proposition rely on a parametrization that involves cosine andsine functions, it is very common for the interval [a, b] to be [0, 2πn]where n is some positive integer.

Among the many uses of the notion of degree, we can make pre-cise the notion of how often a curve turns around a point.

Definition 2.2.7. Let C be a closed, regular, plane curve parametrizedby ~X : I → R2, and let ~p be a point in the plane. Since C is a closedcurve, we may view ~X as a function on S1. We define the windingnumber w(p) of C around ~p as the degree of the function f : S1 → S1

that has the parametrization ~γ : IR2 with

~γ(t) =~X(t)− ~p|| ~X(t)− ~p ||

.

Proposition 2.2.6 gives us a direct method to calculate the wind-ing number of a curve around a point.

Proposition 2.2.8. Let C be a closed, regular, plane curve parametrizedby ~X : [a, b] → R2 with ~X(t) = (x(t), y(t)), and let ~p be a point inthe plane. The winding number of C around ~p is∫ b

a

(x(t)− px)y′(t)− (y(t)− py)x′(t)‖ ~X(t)− ~p‖2

dt.

Proof: Applying Proposition 2.2.6 to the calculation of the windingnumber of C around a point ~p, we use the parametrization

~γ(t) = (γ1(t), γ2(t)) =~X(t)− ~p‖ ~X(t)− ~p‖

=

(x(t)− px‖ ~X(t)− ~p‖

,y(t)− py‖ ~X(t)− ~p‖

)


as a parametrization γ : [a, b]→ R2. We calculate

γ′2(t) =y′(t)‖ ~X(t)− ~p‖ − (y(t)− py)1

2‖ ~X(t)− ~p‖−1(2 ~X ′(t) · ( ~X(t)− ~p))‖ ~X(t)− ~p‖2

.

After simplification this becomes

γ′2(t) =y′(t)(x(t)− px)2 − (x(t)− px)(y(t)− py)x′(t)

‖ ~X(t)− ~p‖3.

Thenγ′2(t)

γ1(t)=

(x(t)− px)y′(t)− (y(t)− py)x′(t)‖ ~X(t)− ~p‖2

and the proposition follows from the result of Proposition 2.2.6.

Example 2.2.9. Consider the ellipse ~X(t) = (2 cos t, sin t) with 0 ≤ t ≤2π and consider the winding number of this ellipse around the point~p = (1, 0). This is a simple configuration and we expect that thewinding number should be 1 (or possibly −1 based on the orientationof the parametrization of the ellipse). By Proposition 2.2.8, thewinding number is

1

2π

∫ 2π

0

(2 cos t− 1) cos t− (sin t)(−2 sin t)

(2 cos t− 1)2 + sin2 tdt

=1

2π

∫ 2π

0

2− cos t

(2 cos t− 1)2 + sin2 tdt.

This integral is challenging to evaluate by hand. A computer alge-bra system evaluates it as exactly 1. This confirms our geometricintuition.

In practice, it is often hard to explicitly calculate the windingnumber of a curve around a point either directly from Definition 2.2.7or by hand from Proposition 2.2.8. With Proposition 2.2.8 the calcu-lation becomes tractable with a computer algebra system. Interest-ingly enough, it is usually easy to see what the degree is by plottingout the image ϕ(t). (See Figure 2.2, where in Figure 2.2(b) we haveallowed ϕ(t) to come off the circle in order to see its graph moreclearly.)


~p

~X(t)− ~p~X(0)− ~p

(a) ~X(t)

O

ϕ(t) ϕ(0)

(b) ϕ(t)

Figure 2.2. Winding number of 2.

One might expect that the winding number of a curve around apoint p depends only on what connected component of the curve’scomplement p lies in.

Proposition 2.2.10. Let C be a closed, regular, plane curve parame-trized by ~X : I → R2, and let p0 and p1 be two points in the sameconnected component of R2 − C. The winding number of C aroundp0 is equal to the winding number of C around p1.

Proof: Let β : [0, 1]→ R2 be a path such that β(0) = p0, β(1) = p1,and β(u) 6= ~X(t) for any u ∈ [0, 1] and t ∈ I. Define the two-variablefunction H : I × [0, 1]→ S1 by

H(t, u) =~X(t)− β(u)

‖ ~X(t)− β(u)‖.

The function H is continuous since ~X(t) − β(u) is continuous and‖ ~X(t) − β(u)‖ is also continuous and never 0. For all u ∈ [0, 1], weconsider H(t, u) to be a function of t from I to S1 and define H(t, u)as its lifting, which is a continuous function H : I × [0, 1]→ R. Butthen the function of u given by

1

2π

(H(2π, u)− H(0, u)

)


is continuous and discrete. Thus, it is constant, and hence

1

2π

(H(2π, 1)− H(0, 1)

)=

1

2π

(H(2π, 0)− H(0, 0)

),

which means that the winding numbers of C around p0 and aroundp1 are equal.

The winding number of a regular curve around a point offers astrategy to prove the Jordan Curve Theorem (Theorem 2.1.5) in thecase when the curve is regular. The strategy is to show that near apoint p on the curve C, points on one side of C have a winding num-ber of 0, while on the other side of C points have a winding numberof 1 or −1, depending on the orientation of the parametrization givenfor C. This establishes that a regular, simple, closed curve has atleast two connected components. The proof of the Regular JordanCurve Theorem given in [24, Section A.4] also establishes a fact thatis useful in its own right.

Proposition 2.2.11. Let ~X : [a, b] → R2 be a simple, closed, regular,plane curve. The rotation index of ~X is ±1.

Problems

2.2.1. Consider the figure-eight curve ~X(t) = (cos t, sin(2t)) with 0 ≤ t ≤2π. Show that the rotation index is 0. [Hint: When performing anappropriate integration, reparametrize by t = u+ π

2 and change thebounds of integration to [−π, π].]

2.2.2. Show that the lemniscate given by

~X(t) =

(a cos t

1 + sin2 t,a cos t sin t

1 + sin2 t

)has a rotation index of 0.

2.2.3. Calculate the rotation index of the limacon de Pascal given by

~X(t) = ((1− 2 cos t) cos t, (1− 2 cos t) sin t) .

2.2.4. Let ~x(t) = (cos 3t cos t, cos 3t sin t) be the trefoil curve. Show thatI = [0, π] is enough of a domain to make ~x into a closed curve. Provethat the rotation index of ~x is 2.

2.3. Isoperimetric Inequality 61

2.2.5. Let C be a regular plane curve. Suppose that p is a point in theplane such that there exists a ray from p that does not intersect C.Show that the winding number of C around p is 0.

2.2.6. Give an example of a curve C (by either a sketch or parametrization)and a point ~p ∈ R2 such that the winding number of C around ~p is 2but that every ray based at ~p intersects C in more than two points.

2.2.7. Prove Proposition 2.2.6.

2.2.8. The curve C pictured below has the parametrization

~X(t) = ((3 + cos(5t)) cos(3t), (3 + cos(5t)) sin(3t))

for t ∈ [0, 2π].

(a) Calculate explicitly the integral formula for the winding num-ber of C around a point ~p as given in Proposition 2.2.8.

(b) Use a computer algebra system to calculate this winding num-ber for various points in the plane in all the different connectedcomponents of R2 − C.

(c) Give an intuitive justification for your results.

2.3 Isoperimetric Inequality

With the Jordan Curve Theorem at our disposal, we can use Green’sTheorem to prove an inequality between the length of a simple closedcurve and the area contained in the interior. This inequality, calledthe isoperimetric inequality, is another example of a global theoremsince it relates quantities that take into account the entire curve atonce.


Theorem 2.3.1. Let C be a simple, closed, plane curve with length l,and let A be the area bounded by C. Then

l2 ≥ 4πA, (2.3)

and equality holds if and only if C is a circle.

Proof: Let L1 and L2 be parallel lines that are both tangents to Cand such that C is contained in the strip between them. Let S1 be acircle that is also tangent to both L1 and L2 such that S1 does notintersect C. We call r the radius of this circle, and we set up thecoordinate axes in the plane so that the origin is at the center of S1,the x-axis is perpendicular to L1 (and L2), and the y-axis is parallelto L1 and L2. See Figure 2.3.

s = s0s = 0

~α

~γ

Figure 2.3. Isoperimetric inequality.

Assume that C is parametrized by arc length ~α(s) = (x(s), y(s))and that the parametrization is positively oriented at the points oftangency with L1 and L2. Also, let’s call s = 0 the parameterlocation for the tangency point C ∩L1 and s = s0 the parameter for

2.3. Isoperimetric Inequality 63

C ∩ L2. We can also assume that S1 is parametrized by ~γ(s), where~γ(s) is the intersection of

• the upper half of S1 with the vertical line through ~α(s) if 0 ≤s ≤ s0;

• the lower half of S1 with the vertical line through ~α(s) if s0 ≤s ≤ l.

Notice that in this parametrization for the circle, writing ~γ(s) =(x(s), y(s)), we have x(s) = x(s).

Let’s call A the area enclosed in the curve C, and let A be thearea of S1. Using Green’s Theorem, we have the following formulasfor the area of the circle of the interior of the curve:

A =

∫ l

0xy′ ds.

We would like to apply the same formula with the parametrization~γ(s) = (x(s), y(s)) of the circle. However, the parametrization isin general not simple. On the other hand, using a strategy similarto that in the proof of Proposition 2.2.1, we can also use Green’sTheorem and justify the claim that the area of the circle is

A = πr2 = −∫ l

0x′y ds.

Adding the two areas, we get

A+ πr2 =

∫ l

0xy′ − x′y ds ≤

∫ l

0

√(xy′ − x′y)2 ds

≤∫ l

0

√(x2 + y2)((x′)2 + (y′)2) ds =

∫ l

0

√x2 + y2 ds = lr.

(2.4)

We now use the fact that the geometric mean of two positive realnumbers is less than the arithmetic mean. Thus,

√A · πr2 ≤ 1

2(A+ πr2) =

1

2lr. (2.5)

Squaring both sides and dividing by r2, we get 4πA ≤ l2. This provesthe first part of Theorem 2.3.1.


In order for equality to hold in Equation (2.3), we need equalitiesto hold in both the inequalities in Equation (2.4). From Equation(2.5), we conclude that A = πr2 and l = 2πr. Furthermore, sinceA does not change, regardless of the direction of L1 and L2, neitherdoes the distance r. We also must have

(xy′ − x′y)2 = (x2 + y2)((x′)2 + (y′)2) = r2,

which, after expanding both sides, leads to (xx′+yy′)2 = 0. However,differentiating the relationship x2 + y2 = r2 for all parameter valuess, we find that xx′ + yy′ = 0 for all s. Thus, for all s, we havey′(s) = y′(s), and thus y(s) = y(s) +D for some constant D. Sinceby construction x(s) = x(s), then C is a circle – a translate of S1 inthe direction (0, D).

Since nothing in the above proof uses second derivatives, we onlyneed the curve ~α to be C1, i.e., that it have a continuous first deriva-tive. Furthermore, one can generalize the proof to apply to curvesthat are only piecewise C1.

2.4 Curvature, Convexity, and the Four-VertexTheorem

In this section, all curves are assumed to be of class C2.

Definition 2.4.1. A subset of S of Rn is called convex if for all p and qin S, the line segment [p, q] is a subset of S, i.e., lies entirely insideS. If C is not convex, it is called concave. (See Figure 2.4.)

Figure 2.4. A concave curve.

2.4. Curvature, Convexity, and the Four-Vertex Theorem 65

With the Jordan Curve Theorem, we can affirm that a simple,closed curve has an interior. Hence, we can discuss convexity prop-erties of a plane curve if we use the following definition.

Definition 2.4.2. A closed, regular, simple, parametrized curve C iscalled convex if the union of C and the interior of C form a convexsubset of R2.

In the context of regular curves, it is possible to provide vari-ous characterizations for when a curve is convex based on a curve’sposition with respect to its tangent lines.

Proposition 2.4.3. A regular, closed, plane curve C is convex if andonly if it is simple and its curvature κg does not change sign.

Proof: The proof involves showing that if the curvature functionκg(s) with respect to arc length changes sign at s0, then there exista value s0 − ε near and before s0 and a value s0 − ε near and afters0 such that the segment connecting the points corresponding to s0

and s0 − ε and the segment connecting the points corresponding tos0 and s0 +ε cannot be in the same connected component of R2−C.

Let ~α : [0, l] → R2 be a parametrization by arc length withpositive orientation for the curve C. Let ~T : [0, l] → S1 be thetangential indicatrix, and let θ : [0, l]→ R be the total angle functionassociated to the rotation index. By Equation (2.2), κg(s) = θ′(s),so the condition that κg(s) does not change sign is equivalent to θ(s)being monotonic.

We first prove that convexity implies that C is simple. This fol-lows from the definition of convexity, which assumes that C possessesan interior and, therefore, that R2−C has only two components. Bythe Jordan Curve Theorem, since we already assume C is regular andclosed, for it to possess an interior, it must be simple.

Assume from now on that C is simple. Suppose that κg(s) changessign on [0, l]. At a point s = a, define the height function for alls ∈ [0, l] by

ha(s) = (~α(s)− ~α(a)) · ~U(a).

This measures the height of the point ~α(s) from the tangent line Lato C at ~α(a), with ~U(a) being considered the positive direction. The


derivatives of this height function are

h′(s) = ~α′(s) · ~U(a) = ~T (s) · ~U(a),

h′′(s) = κg(s)~U(s) · ~U(a).

Consequently, at s = a, the height function satisfies h′(a) = 0 andh′′(a) = κg(a).

From the definition of differentiation, one deduces that in a neigh-boorhood of a, say over the interval (a − δ, a + δ), the projectionpa : C → La of C onto La is a bicontinuous bijection (i.e., both thefunction and its inverse are continuous). Let

pa(a− δ) = ~α(a)− ε1~T (a) and pa(a+ δ) = ~α(a) + ε2

~T (a),

where ε1, ε2 > 0. Define the function g : (−ε1, ε2)→ [0, l] as follows.Let u(s) = (~α(a+s)−~α(a))· ~T (a). This corresponds to the projectionof ~α(a + s) − ~α(a) onto the tangent line, or in other words the ~Tcomponent of the curve near ~α(a) in the frame with center ~α(a)with ordered basis (~T (a), ~U(a)). As discussed above, this projectionis a bijection over (−δ, δ) with its image. Let g : (ε1, ε2) → (−δ, δ)be the inverse function of u(s).

Finally, define f : (−ε1, ε2)→ R by f = ha g. The graph of thefunction f placed in the frame (~T (a), ~U(a)) with origin ~α(a) tracesout the curve C in a neighborhood of s = a (Figure 2.5).

C

La

~T

~U

(a) The curve with tangent line La.

u

f(u)

~T

~U

(b) The height function.

Figure 2.5. Height function f(u).


The derivatives of f are

f ′(u) = h′(g(u))g′(u),

f ′′(u) = h′′(g(u))(g′(u))2 + h′(g(u))g′′(u).

Since g(0) = a and h′(a) = 0, without knowing g′(0), we deduce thatf(0) = f ′(0) = 0 and that f ′′(0) has the same sign as κg(a).

If κg(a) < 0, then from basic calculus we deduce that f is concavedown over (−ε1, ε2), which implies that every line segment betweenpoints on the graph of f forms a chord that is below the graph.As an example, use the segment between (u1, f(u1)) and (u2, f(u2)).Furthermore, since the curve C has a positive orientation, the interiorof C is in the direction of ~U(a), which corresponds to being above thegraph of f . Thus, the line segment between the points ~α(g(u1)) and~α(g(u2)) is in the exterior of the curve. Consequently, this provesthat κg changes sign if and only if C is not convex and the propositionfollows.

This proposition and a portion of the proof leads to another moregeometric characterization of convex curves.

Proposition 2.4.4. A regular, simple, closed curve C is convex if andonly if it lies on one side of every tangent line to C.

Proof: Suppose we parametrize C by arc length with a positive ori-entation. From the proof of Proposition 2.4.3, one observes that atany point p of the curve C where the curvature is negative (given thepositive orientation), in a neighborhood of p, the tangent line L toC at p is in the interior of C. Consequently, since C is bounded andL is not, L must intersect the curve C at another point. By Propo-sition 2.4.3, this proves that C is concave if and only if there existsa tangent line that has points in the interior of C and the exteriorof C.

In Section 1.3, we gave the term vertex for a point on the curvewhere the curvature reaches an extremum. More precisely, for anyregular parametrized curve ~X : I → R2, we defined a vertex to bea point ~X(t0) where κ′g(t0) = 0. Note that since the curvature of acurve is independent of the parametrization up to a possible changeof sign, vertices are independent of any parametrization of C.


(a) ~X(t) = (a cos t, b sin t). (b) ~X(t) = (ecos t, esin t).

Figure 2.6. Two closed curves and their vertices.

The concept of a vertex is obviously a local property of the curve,but if one were to experiment with a variety of closed curves, onewould soon guess that there must be a restriction on the number ofvertices. In Problem 1.3.10, the reader calculated that the noncir-cular ellipse has four vertices. Figure 2.6(b) shows another simpleclosed curve with six vertices. (In each figure, the dots indicate thevertices of the curve.)

Theorem 2.4.5 (Four-Vertex Theorem). Every simple, closed, regular,convex, plane curve C has at least four vertices.

Proof: Let ~α : [0, l]→ R2 be a parametrization by arc length for thecurve C. Since [0, l] is compact and κg : [0, l] → R is continuous,then by the Extreme Value Theorem, κg attains both a maximumand a minimum over [0, l]. These values already assure that C hastwo vertices. Call these points p and q and call L the line betweenthem. Let β be the arc along C from p to q, and let γ be the arcalong C from q to p.

We claim that β and γ are contained in opposite half-planesdefined by the line L. Assume one of the arcs is not contained in oneof the half-planes. Then C meets L at another point r. By convexity,in order for the line segments [p, q], [p, r], and [q, r] to all lie inside C,


all three points would need to have L as their tangent line to C. ByProblem 2.4.1, this is a contradiction. Assume now that the arcs βand γ are contained in half-planes but in the same half-plane. Thenagain by convexity, the only possibility is that one of the arcs is aline segment along L. Then along that line segment, the curvatureκg(s) is identically 0. However, this implies that the curvature at pand q is 0, but since p and q are extrema of κg, this would force C tobe a line. This is a contradiction since C, being closed, is bounded.

If there are no other vertices on the curve, then κ′g(s) does notchange sign on β or on γ. If L has equation Ax+By +C = 0, then∫ l

0(Ax+By + C)

dκgds

ds (2.6)

is positive. However, this leads to a contradiction because for allreal constants A, B, and C, the integral in Equation (2.6) is always0 (Problem 2.4.2).

This proves that there is at least one other vertex. But then,since κ′g(s) changes sign at least three times, it must change signs afourth time as well. Hence, κ′g(s) has at least four 0s.

Problems

2.4.1. A bitangent line L to a regular curve C is a line that is tangent to Cat a minimum of two points p and q such that between p and q onC, there are points that do not have L as a tangent line. Prove thata simple, regular curve is not convex if and only if it has a bitangentline.

2.4.2. Let ~X : [0, l] → R2 be a regular, closed, plane curve parametrized

by arc length and write ~X(s) = (x(s), y(s)).

(a) Show that x′′ = −κy′ and y′′ = κx′.

(b) Prove that ∫ l

0

x(s)κ′g(s) ds = −∫ l

0

κg(s)x′(s) ds,

and do the same for y(s) instead of x(s).

(c) Use the above to show that for any constants A, B, and C, wehave ∫ l

0

(Ax+By + C)dκgds

ds.


2.4.3. If a closed curve C is contained inside a disk of radius r, prove thatthere exists a point P ∈ C such that the curvature κg(P ) of C at Psatisfies

|κg(P )| ≥ 1

r.

2.4.4. Let ~α : I → R2 be a regular, closed, simple curve with positiveorientation. Define the parallel curve at a distance r as

~β(t) = ~α(t)− r~U(t).

Call l(~α) the length of the curve ~α and A(~α) the area of the interiorof the curve. Prove the following:

(a) ~β is regular if and only if it is simple and if and only if κg(t) >− 1r for all t ∈ I.

(b) If ~β is regular, then l(~β) = l(~α) + 2πr.

(c) If ~β is regular, then A(~β) = A(~α) + rl(~α) + πr2.

(d) At any regular point of ~β, we have

κ~β(t) =κ~α(t)

1 + rκ~α(t).

2.4.5. Show that the limacon ~X(t) = ((1− 2 cos t) cos t, (1− 2 cos t) sin t)for t ∈ [0, 2π] has exactly two vertices. Explain why it does notcontradict the Four-Vertex Theorem.

2.4.6. (*) Consider the curve in Figure 2.6(b). Determine the coordinates

of the vertices of ~X(t).

2.4.7. Let [A,B] be a line segment in the plane, and let l > AB. Show thatthe curve C that joins A and B and has length l such that, togetherwith the segment [A,B], bounds the largest possible area is an arcof a circle passing through A and B.

BA

C

BA

C

CHAPTER 3

Curves in Space: Local Properties

The local theory of space curves is similar to the theory for planecurves, but differences arise in the richer variety of configurationsavailable for loci of curves in R3. In the study of plane curves, weintroduced the curvature function, a function of fundamental impor-tance that measures how much a curve deviates from being a straightline. As we shall see, in the study of space curves, we again intro-duce a curvature function that measures how much a space curvedeviates from being linear, but we also introduce a torsion functionthat measures how much the curve twists away from being planar.

3.1 Definitions, Examples, and Differentiation

As in the study of plane curves, one must take some care in definingwhat one means by a space curve. If I is an interval of R, to call aspace curve any function ~X : I → R3 (or the image thereof) wouldallow for separate pieces or even a set of scattered points. By acurve, one typically thinks of a connected set of points, and, just aswith plane curves, the desired property is continuity.

Instead of repeating the definitions provided in Section 1.1, wepoint out that the definition for the limit of a vector function (Defi-nition 1.1.4), the definition for continuity of a vector function (Defi-nition 1.1.5), Proposition 1.1.6, and Corollary 1.1.7 continue to holdfor vector functions into Rn.

Definition 3.1.1. Let I be an interval of R. A parametrized curve inRn is a continuous function ~X : I → Rn. If for all t ∈ I we have

~X(t) = (x1(t), x2(t), . . . , xn(t)) ,

then the functions xi : I → R for 1 ≤ i ≤ n are called the coordinatefunctions or parametric equations of the parametrized curve. The

71

72 3. Curves in Space: Local Properties

locus is the image ~X(I) of the parametrized curve. If n = 3, we call~X a space curve.

In order to develop an intuition for space curves, we present anumber of examples. These illustrate only some of the great vari-ability in the shape of parametric curves but provide a short libraryfor examples we revisit later.

Example 3.1.2 (Lines). In space or in Rn, a line is uniquely defined bya point ~p on the line and a nonzero direction vector ~v. Parametricequations for a line are then ~X : R→ Rn with ~X(t) = ~vt+ ~p.

Example 3.1.3 (Planar Curves). We define a planar curve as any spacecurve whose image lies in a plane. A parametrized plane curve ~X :I → R2 with ~X(t) = (x(t), y(t)) can be considered as a parametrizedspace curve by setting ~X(t) = (x(t), y(t), 0). This becomes a planarcurve.

More generally, recall that if ~a and ~b are linearly independentvectors of R3 and ~p is any point in R3, then the plane spanned by~a,~b through ~p can be described by ~p + u~a + v~b, where u, v ∈ R.Note that a normal vector to this plane is ~N = ~a×~b. Consequently,a planar curve in this plane will be given by a parametrized curve ofthe form

~X(t) = ~p+ u(t)~a+ v(t)~b,

where u(t) and v(t) are continuous functions I → R, where I is someinterval of R.

In order to properly devise parametrizations for specific curveson planes in R3, a judicious choice of the vectors ~a and ~b may benecessary. The usual basis vectors of (1, 0) and (0, 1) of R2 are notonly linearly independent but they form an orthonormal set, bothmutually perpendicular and of unit length. So for example, a spacecurve of the form

~X(t) = ~p+ r(cos t)~a+ r(sin t)~b

will not in general be a circle but an ellipse.As a specific example, we can obtain the equations for a circle

of radius 7 in the plane 3x − 2y + 2z = 6 with center ~p = (2, 1, 1)as follows. A normal vector to the plane is ~N = (3,−2, 2). Two

3.1. Definitions, Examples, and Differentiation 73

vectors that are not collinear and in this plane are ~a = (2, 3, 0)and ~b = (2, 0,−3). We use the Gram-Schmidt orthonormalizationprocess to find an orthonormal basis of Span(~a,~b). The first vectorin the orthonormal basis is

~u1 =~a

‖~a‖=

(2√13,

3√13, 0

).

Then calculate

~v2 = ~b− proj~u1~b = ~b− (~u1 ·~b)~u1 =

(18

13,−12

13,−3

)and get a second basis vector of

~u2 =~v2

‖~v2‖=

(18√1729

,− 12√1729

,− 39√1729

).

Finally, a parametrization for the circle we are looking for is givenby

~X(t) = ~p+ (7 cos t) ~u1 + (7 sin t)~u2.

Example 3.1.4 (Twisted Cubic). One of the simplest nonplanar curvesis called the twisted cubic and has the parametrization ~X(t) =(t, t2, t3) (see Figure 3.1(a)).

We might wonder why this should be considered the simplestnonplanar curve. After all, could we not create a nonplanar curvewith just quadratic polynomials? In fact, the answer is no. A curve~X(t) with quadratic polynomials for coordinate functions can bewritten as

~X(t) = ~at2 +~bt+ ~c,

where ~a, ~b, and ~c are linearly independent vectors in R3. If ~a and ~bare different and nonzero, then the image of ~X(t) lies in the planethrough ~c with direction vectors ~a and ~b, in other words, the planethrough ~c with normal vector ~a×~b.

To see clearly that the twisted cubic is not planar, consider~X(−1) = (−1, 1,−1), ~X(0) = (0, 0, 0), ~X(1) = (1, 1, 1), and ~X(2) =(2, 4, 8). The first three points lie in the plane x = z but the fourthdoes not.


x y

z

(a) The twisted cubic.

x y

z

(b) A helix.

Figure 3.1. Two important space curves.

Example 3.1.5 (Cylindrical Helix). A cylindrical helix is a space curvethat wraps around a circular cylinder, climbing in altitude at a con-stant rate. We have for equations

~X(t) = (a cos t, a sin t, bt).

See Figure 3.1(b) for an example with ~X(t) = (2 cos t, 2 sin t, 0.2t).

Example 3.1.6. The coordinate functions of the parametrized curve~X(t) = (at cos t, at sin t, bt) satisfy the algebraic equation

x2

a2+y2

a2− z2

b2= 0

for all t. Thus, the image of ~X lies on the circular cone described bythis equation.

Example 3.1.7. Consider the parametric curve ~X(t) = (a cosh t cos t,a cosh t sin t, b sinh t). It is not hard to see that the coordinate func-tions of ~X satisfy

x2

a2+y2

a2− z2

b2= 1


x y

z

(a) Curve on a hyperboloid.

x

y

z

(b) Space cardioid.

Figure 3.2. Examples of parametric curves.

so that the image of ~X lies on the hyperboloid of one sheet. SeeFigure 3.2(a) for an example with

~X(t) = (cosh t cos(10t), cosh t sin(10t), sinh t).

(Recall that hyperbolic trigonometric functions are defined as

cosh t =et + e−t

2and sinh t =

et − e−t

2

and that they satisfy the relation cosh2 t− sinh2 t = 1 for all t ∈ R.)

Example 3.1.8 (Space Cardioid). The parametric curve with equation

~X(t) = ((1− cos t) cos t, (1− cos t) sin t, sin t)

is called the space cardioid (see Figure 3.2(b)). One interesting prop-erty of this curve is that it is a closed curve in R3. Projected ontothe xy-plane it gives the cardioid and (as we shall see) has no criticalpoints. In an intuitive sense, we have stretched the cardioid out ofthe plane and removed its critical point (or cusp).

Using the same vocabulary as in Section 1.1, we call a reparame-trization of a parametrized curve ~X : I → Rn any other continuousfunction ~ξ : J → Rn defined by

~ξ = ~X g


for some surjective function g : J → I. Then the image C ⊆ Rnof ~ξ is the same as the image of ~X. When g is not surjective, theimage of ~ξ could be a proper subset of C, and we do not call ~ξ areparametrization. In addition, the reparametrization ~ξ of ~X is

• regular if g is continuously differentiable and g′(t) 6= 0;

• positively oriented if it is regular and g′(t) > 0 for all t ∈ J ;

• negatively oriented if it is regular and g′(t) < 0 for all t ∈ J .

The definition in Equation (1.2) of the derivative for a vectorfunction was purposefully presented irrespective of the dimension.We restate the definition in an alternate form that is sometimesbetter suited for proofs.

Definition 3.1.9. Let I be an interval in R, and let t0 ∈ I. If ~f : I →Rn is a continuous vector function, we say that ~f is differentiable att0 with derivative ~f ′(t0) if there exists a vector function ~ε such that

~f(t0 + h) = ~f(t0) + ~f ′(t0)h+ h~ε(h) and limh→0‖~ε(h)‖ = 0.

It follows as a consequence of Proposition 1.1.6 (modified to vec-tor functions in Rn) that a vector function ~X : I → Rn is differen-tiable at a point if and only if all its coordinate functions are differ-entiable at that point. Furthermore, if ~X : I → Rn is a continuousvector function that is differentiable at t = t0 and if the coordinatefunctions of ~X are

~X(t) = (x1(t), x2(t), . . . , xn(t)) ,

then the derivative ~X ′(t0) is

~X ′(t0) =(x′1(t0), x′2(t0), . . . , x′n(t0)

).

As in R2, for any vector function ~X : I → Rn, borrowing from thelanguage of trajectories in mechanics, we often call ~X the positionfunction, ~X ′ the velocity function, and ~X ′′ the acceleration function.Furthermore, we define the speed function associated to ~X as thefunction s′ : I → R defined by

s′(t) = ‖| ~X ′(t)‖.


It is also a simple matter to define equations for the tangent lineto a parametrized curve ~X(t) at t = t0 as long as ~X ′(t0) 6= ~0. Theparametric equations for the tangent line are

~L(u) = ~X(t0) + u ~X ′(t0).

Proposition 3.1.10. Proposition 1.2.3 holds for differentiable vector func-tions ~v and ~w from I ⊂ R into Rn. Furthermore, suppose that ~v(t)and ~w(t) are vector functions into R3 that are defined on and differ-entiable over an interval I ⊂ R. If ~X(t) = ~v(t) × ~w(t), then ~X isdifferentiable over I and

~X ′(t) = ~v′(t)× ~w(t) + ~v(t)× ~w′(t).


As in Section 1.1, in this introductory section for space curves,the problems focus on properties of vectors and vector functions inR3.

Problems

3.1.1. Calculate the velocity, acceleration, and speed of the space cardioid~X(t) = ((1− cos t) cos t, (1− cos t) sin t, sin t).

3.1.2. Calculate the velocity, acceleration, and speed of the twisted cubic~X(t) = (t, t2, t3).

3.1.3. Calculate the velocity, acceleration, and speed of the parametrizedcurve ~X(t) = (tan−1(t), sin t, cos 2t).

3.1.4. Find a parametrization of the intersection of the cylinder x2+z2 = 1with the plane x+ 2y + z = 2.

3.1.5. Let ~a, ~b, and ~c be three vectors in R3. Prove that

(~a×~b) · ~c = (~b× ~c) · ~a = (~c× ~a) ·~b.

3.1.6. Let ~a, ~b, ~c, and ~d be four vectors in R3. Prove that

(~a×~b) · (~c× ~d) =

∣∣∣∣∣(~a · ~c) (~b · ~c)(~a · ~d) (~b · ~d)

∣∣∣∣∣ .


3.1.7. Let ~a,~b,~c be constant vectors in R3. Calculate the derivatives

(a) ~X(t) = (2t~a+ t2~b)× (~b+ (4− t3)~c);

(b) f(t) = (2t~a+ t2~b) · (~b+ (4− t3)~c).

3.1.8. Modify the parametrization of a helix to give the parametrization~X : R → R3 of a curve that lies on the cylinder x2 + y2 = 1 suchthat the z-component has (0,∞) as its image.

3.1.9. Give parametric equations of the tangent lines to ~X(t) = (t2−1, 3t−t3, t4 − 4t2) at (−1, 0, 0) and also where t = 2.

3.1.10. Consider the parametrized space curve ~X(t) = (t cos t, t sin, t) witht ∈ R.

(a) Find the equation of the tangent line where t = π/3.

(b) Do any tangent lines of ~X(t) intersect the x-axis and if so, thenfor what values of t and at what points on the x-axis?

3.1.11. Consider the parametrization space curve ~X(t) = (cos t, sin, sin 2t)with t ∈ [0, 2π]. Find the equations of the tangent lines at the pointswhere the curve intersects the xy-plane.

3.1.12. Let ~α(t) be a regular, plane, parametrized curve. View the xy-planeas a subset of R3. Let ~p be a fixed point in the plane and ~u a fixedvector. Let θ(t) be the angle ~α(t) − ~p makes with the direction ~u.Prove that (up to a change in sign)

θ′(t) =‖~α′(t)× (~α(t)− ~p)‖‖~α(t)− ~p‖2

.

Conclude that the angle function θ(t) of ~α(t)− ~p with respect to thedirection ~u has a local extremum at a point t0 if and only if ~α′(t0)is parallel to ~α(t0)− ~p.

3.1.13. Consider the space curve ~X(t) = (t cos t, t sin t, t). Find where the

extremal values of the angle between ~X ′ and ~X ′′ occur and determinewhether they are maxima or minima.

3.1.14. Consider the plane in R3 given by the equation ax+ by+ cz+ f = 0and a point P = (x0, y0, z0). Prove that the distance between theplane and the point P is

d =|ax0 + by0 + cz0 + f |√

a2 + b2 + c2.


3.1.15. Determine the angle of intersection between the lines ~u(t) = (2t −1, 3t+ 2,−2t+ 3) and ~v(t) = (3t− 1, 5t+ 2, 3t+ 3).

3.1.16. Consider the two lines given by ~u(t) = (2t − 1, 3t + 2,−2t + 3) and~v(t) = (3t + 1,−5t − 3, 3t − 1). Find the shortest distance betweenthese two lines. [Hint: Consider the function f(s, t) = ‖~u(s)−~v(t)‖.]

3.1.17. Consider two nonparallel lines given by the equations

~l1(s) = ~a+ s~u and ~l2(t) = ~b+ t~v,

where ~u× ~v 6= ~0. Prove that the distance d between these two linesis

d =|(~u× ~v) · (~b− ~a)|

‖~u× ~v‖.

3.1.18. Let ~α(t) and ~β(t) be two differentiable parametrized curves R→ R3.

Prove that if ~γ(t) = ~α(t)×~β(t), then ~γ′(t) = ~α′(t)×~β(t)+~α(t)×~β′(t).

3.1.19. Let ~X(t) be any parametrized curve that is of class C3. Prove that

d

dt

(~X(t) · ( ~X ′(t)× ~X ′′(t))

)= ~X(t) · ( ~X ′(t)× ~X ′′′(t)).

[Hint: Use Problem 3.1.18.]

3.1.20. Consider a sphere of radius R and center ~p. Prove by a geometricargument that the closest point on this sphere to a point ~q 6= ~p is

~p+R~q − ~p‖~q − ~p‖

.

We call this closest point the projection of ~q to the sphere. Let~X : I → R3 be a parametrized space curve that does not go throughthe origin. Deduce that ~X(t)/‖ ~X(t)‖ parametrizes the projection ofthe space curve to the unit sphere.

3.1.21. Prove that no four distinct points of the twisted cubic lie in a com-mon plane. [Hint: Prove that for all a, b, c, d ∈ R,

∣∣∣∣∣∣b− a c− a d− ab2 − a2 c2 − a2 d2 − a2b3 − a3 c3 − a3 d3 − a3

∣∣∣∣∣∣ =

∣∣∣∣∣∣∣∣1 1 1 1a b c da2 b2 c2 d2

a3 b3 c3 d3

∣∣∣∣∣∣∣∣and use the Vandermonde determinant identity.]


3.2 Curvature, Torsion, and the Frenet Frame

Let I be an interval of R, and let ~X : I → R3 be a differentiable spacecurve. Following the setup with plane curves, we can talk about theunit tangent vector ~T (t) defined by

~T (t) =~X ′(t)

‖ ~X ′(t)‖. (3.1)

Obviously, the unit tangent vector is not defined at a value t = t0if ~X is not differentiable at t0 or if ~X ′(t0) = 0. This leads to thefollowing definition (which can be generalized to curves in Rn).

Definition 3.2.1. Let ~X : I → Rn be a parametrized curve. We callany point t = t0 a critical point if ~X ′(t0) is not defined or if ~X ′(t0) =~0. A point t = t0 that is not critical is called a regular point. Aparametrized curve ~X : I → Rn is called regular if it is of class C1

(i.e., continuously differentiable) and if ~X ′(t) 6= ~0 for all t ∈ I.

In practice, if ~X ′(t0) = ~0 but ~X ′(t) 6= ~0 for all t in some intervalJ around t0, it is possible for

limt→t0

~X ′(t)

‖ ~X ′(t)‖(3.2)

to exist. In such cases, it is common to think of ~T as completed bycontinuity by calling ~T (t0) the limit in Equation (3.2).

Since ~T (t) is a unit vector for all t ∈ I, we have ~T · ~T = 1 for allt ∈ I. Therefore, if ~T is itself differentiable,

d

dt

(~T (t) · ~T (t)

)= 2~T ′(t) · ~T (t) = 0.

Thus, the derivative of the unit tangent vector ~T ′ is perpendicularto ~T . At any point t on the curve, where ~T ′(t) 6= ~0, we define theprincipal normal vector ~P (t) to the curve at t to be the unit vector

~P (t) =~T ′(t)

‖~T ′(t)‖.

3.2. Curvature, Torsion, and the Frenet Frame 81

~T

~P

~B

~T

~P

~B ~T~P

~B

~T~P

~B

~T

~P~B

Figure 3.3. Moving Frenet frame on the space cardioid.

Again, it is still possible sometimes to complete ~P by continuityeven at points t = t0, where ~T ′(t0) = ~0. In such cases, it is commonto assume that ~P (t) is completed by continuity wherever possible.Nonetheless, by Equation (3.1), the requirement that ~T be differen-tiable is tantamount to ~X being twice differentiable.

Finally, for a space curve defined over any interval J , where ~T (t)and ~P (t) exist (perhaps when completed by continuity), we completethe set ~T , ~P to an orthonormal frame by adjoining the binormalvector ~B(t) given as

~B = ~T × ~P .

Definition 3.2.2. Let ~X : I → R3 be a continuous space curve of classC2 (i.e., has a continuous second derivative). To each point ~X(t)on the curve, we associate the Frenet frame as the ordered triple ofvectors (~T , ~P , ~B).

Figure 3.3 illustrates the Frenet frame as it moves through t = 0on the space cardioid

~X(t) = ((1− cos t) cos t, (1− cos t) sin t, sin t). (3.3)


(In this figure, the basis vectors of the Frenet frame were scaled downby a factor of 1

2 to make the picture clearer.) In this example, it isinteresting to see that near t = 0, though the unit tangent vectordoesn’t change too quickly, the vectors ~P and ~B rotate quickly aboutthe tangent line. The functions that measure how fast ~T (resp. ~B)changes are called the curvature (resp. torsion) functions of the spacecurve.

Similar to the basis ~T (t), ~U(t) for plane curves, the Frenetframe provides a geometrically natural basis in which to study localproperties of space curves. We now analyze the derivatives of a spacecurve in reference to the Frenet frame.

Recall that the speed s′(t) is ‖ ~X ′(t)‖. By definition of the unittangent vector, we have

~X ′(t) = s′(t)~T (t).

If ~x is twice differentiable at t, then ~T ′(t) exists. If, in addition,~P is defined at t, then by definition of the principal normal vector,~T ′ is parallel to ~P . This allows us to define the curvature of a spacecurve as follows.

Definition 3.2.3. Let ~X be a regular parametrized curve of class C2.The curvature κ : I → R+ of a space curve is

κ(t) =‖~T ′(t)‖s′(t)

.

Note that at any point where P (t) is defined, κ(t) is the nonnegativenumber defined by

~T ′(t) = s′(t)κ(t)~P (t).

We would like to determine how the other unit vectors of theFrenet frame behave under differentiation. Remember that ~B =~T × ~P . Taking a derivative of this cross product, we obtain

~B′ = ~T ′ × ~P + ~T × ~P ′ = ~T × ~P ′,

since ~T ′ is parallel to ~P . However, just as ~T ′ ⊥ ~T , we have the samefor ~P ′ ⊥ ~P . Thus, since we are in three dimensions, we can write


~P ′(t) = f(t)~T (t) + g(t) ~B(t) for some continuous functions f, g : I →R. Consequently, we deduce that

~B′ = ~T × (f(t)~T + g(t) ~B) = f(t)~T × ~T + g(t)~T × ~B = −g(t)~P .

Thus, the derivative of the unit binormal vector is parallel to theprincipal normal vector.

Definition 3.2.4. Let ~x : I → R3 be a regular space curve of class C2

for which the Frenet frame is defined everywhere. We define thetorsion function τ : I → R as the unique function such that

~B′(t) = −s′(t)τ(t)~P (t).

We now need to determine ~P ′(t), and we already know that ithas the form f(t)~T (t) + g(t) ~B(t). However, we can say much morewithout performing any specific calculations. Since (~T , ~P , ~B) forman orthonormal frame for all t, we have the following equations:

~T · ~P = 0 and ~P · ~B = 0.

Taking derivatives with respect to t, we have

~T ′ · ~P + ~T · ~P ′ = 0 and ~P ′ · ~B + ~P · ~B′ = 0,

that is,

~P ′ · ~T = −~T ′ · ~P and ~P ′ · ~B = −~P · ~B′.

Thus, we deduce that

~P ′(t) = −s′(t)κ(t)~T (t) + s′(t)τ(t) ~B(t). (3.4)

If we assume, as one does in linear algebra, that ~T , ~P , and ~B are

column vectors and write(~T ~P ~B

)as the matrix that has these

vectors as columns, we can summarize Definition 3.2.3, Definition3.2.4, and Equation (3.4) in matrix form by

d

dt

(~T ~P ~B

)=(~T ~P ~B

) 0 −s′κ 0s′κ 0 −s′τ0 s′τ 0

. (3.5)


As provided, the definitions for curvature and torsion of a spacecurve do not particularly lend themselves to direct computationswhen given a particular curve. We now obtain formulas for κ(t) andτ(t) in terms of ~x(t). In order for our formulas to make sense, weassume for the remainder of this section that the parametrized curve~x : I → R3 is regular and of class C3.

First, to find the curvature of a curve, we take the followingderivatives:

~X ′(t) = s′(t)~T (t), (3.6)

~X ′′(t) = s′′(t)~T (t) + (s′(t))2κ(t)~P (t). (3.7)

Taking the cross product of these two vectors, we now obtain

~X ′(t)× ~X ′′(t) = (s′(t))3κ(t) ~B(t). (3.8)

However, by definition of curvature for a space curve, κ(t) is a non-negative function. Furthermore, s′(t) = ‖ ~X ′(t)‖, so we get

κ(t) =‖ ~X ′(t)× ~X ′′(t)‖

s′(t)3. (3.9)

Secondly, to obtain the torsion function from ~X(t) directly, wewill need to take the third derivative

~X ′′′(t) = s′′′(t)~T + s′′(t)s′(t)κ(t)~P + 2s′(t)s′′(t)κ(t)~P

+ (s′(t))2κ′(t)~P + (s′(t))3κ(t)(−κ(t)~T + τ(t) ~B),

which, without writing the dependence on t explicitly, reads

~X ′′′ =(s′′′ − (s′)3κ2

)~T +

(3s′′s′κ+ (s′)2κ′

)~P + (s′)3κτ ~B. (3.10)

Taking the dot product of ~X ′× ~X ′′ with ~X ′′′ eliminates all the termsof ~X ′′′ associated to ~T and ~P . We get

( ~X ′ × ~X ′′) · ~X ′′′ = (s′(t))6(κ(t))2τ(t),

from which we deduce a formula for τ(t) only in terms of the curve~X(t) as follows:

τ(t) =( ~X ′(t)× ~X ′′(t)) · ~X ′′′(t)‖ ~X ′(t)× ~X ′′(t)‖2

.


Example 3.2.5 (Helices). We wish to calculate the curvature and tor-sion of the circular helix ~X(t) = (a cos t, a sin t, bt). We need thefollowing:

~X ′(t) = (−a sin t, a cos t, b),

s′(t) =√a2 sin2 t+ a2 cos2 t+ b2 =

√a2 + b2,

~X ′′(t) = (−a cos t,−a sin t, 0),

~X ′(t)× ~X ′′(t) = (ab sin t,−ab cos t, a2),

~X ′′′(t) = (a sin t,−a cos t, 0).

Note that one can easily parametrize the helix by arc length sinces′(t) =

√a2 + b2 is a constant function, so s(t) = t

√a2 + b2. Thus,

the parametrization by arc length for the circular helix is

~X(s) =

(a cos

(sc

), a sin

(sc

),b

cs

), where c =

√a2 + b2.

We now calculate the curvature and torsion as

κ(t) =‖ ~X ′ × ~X ′′‖

(s′)3=|a|√a2 + b2

(a2 + b2)3/2=

|a|a2 + b2

,

τ(t) =( ~X ′ × ~X ′′) · ~X ′′′

‖ ~X ′ × ~X ′′‖2=

a2b

a2(a2 + b2)=

b

a2 + b2.

Consequently, we find that this circular helix has constant curvatureand constant torsion.

The circular helix, however, is a particular case of a larger classof curves simply called helices. These are defined by requiring thatthe unit tangent make a constant angle with a fixed line in space.Thus, ~X(t) is a helix if and only if for some unit vector ~u,

~T · ~u = cosα = const. (3.11)

Taking a derivative of Equation (3.11), we obtain

~P · ~u = 0.

Hence for all t, ~u is in the plane determined by ~T and ~B, so we canwrite

~u = cosα~T + sinα~B or ~u = cosα~T − sinα~B


t

κ

2π

(a) κ(t).

t

κ

2π

(b) τ(t).

Figure 3.4. Curvature and torsion of the space cardioid.

for all t. Taking the derivative, we obtain

0 = s′κ cosα~P − s′τ sinα~P ,

which implies thatκ

τ= tanα,

and thus, for any helix, the ratio of curvature to torsion is a constant.This ratio κ

τ = ab is called the pitch of the helix.

One can follow the above discussion in reverse and also concludethat the converse is true. Therefore, a curve is a helix if and only ifthe ratio of curvature to torsion is a constant.

Example 3.2.6 (Space Cardioid). We consider the space cardioid againas a follow-up to Figure 3.3. Figure 3.3 shows that in the vicinity oft = 0, the Frenet frame twists quickly about the tangent line, evenwhile the tangent line does not move much. This indicates that near0, κ(t) is not large, while τ(t) is relatively large.

We leave the precise calculation of the curvature and torsionfunctions to the space cardioid as an exercise for the reader but plottheir graphs in Figure 3.4. The graphs of κ(t) and τ(t) justify theintuition provided by Figure 3.3 concerning how the Frenet framemoves through t = 0. In particular, the torsion function has a high


peak at t = 0, which indicates that the Frenet frame rotates quicklyabout the tangent line.

In some proofs that we will encounter later, it is often useful toassume that a curve is parametrized by arc length. In this case, inall of the above formulas, one has s′ = 1 and s′′ = 0 as functions.The transformation properties of the Frenet frame then read

d

ds

(~T ~P ~B

)=(~T ~P ~B

)0 −κ 0κ 0 −τ0 τ 0

. (3.12)

If ~X(s) is parametrized by arc length and is of class C3, then Equa-tion (3.7) gives us ~X ′′(s) = κ(s)~P (s). Hence, the curvature is givenby

κ(s) = ‖ ~X ′′(s)‖.

Furthermore, at any point where κ(s) 6= 0, the torsion function is

τ(s) =( ~X ′(s)× ~X ′′(s)) · ~X ′′′(s)

‖ ~X ′′(s)‖2.

A key property of the curvature and torsion functions of a spacecurve is summarized in the following proposition.

Proposition 3.2.7. Let ~X : I → R3 be a regular parametric curve.

1. Suppose that ~X is of class C2. If the curvature κ(t) is identi-cally 0, then the locus of ~X is a line segment.

2. Suppose that ~X is of class C3. If the torsion τ(t) is identically0, then the locus of ~X lies in a plane.


Problems

3.2.1. Calculate the curvature and torsion of the twisted cubic ~X(t) =(t, t2, t3).

3.2.2. Calculate the curvature and torsion of the space cardioid parame-trized in Equation (3.3).


3.2.3. Calculate the curvature and torsion of ~X(t) = (t, f(t), g(t)), wheref and g are functions of class C3.

3.2.4. Calculate the curvature and torsion of ~X = (a(t − sin t), a(1 −cos t), bt).

3.2.5. Calculate the curvature and torsion of ~X(t) = (t, 1+tt ,1−t2t ).

3.2.6. Consider the parametrized curve ~X(t) = (cos(t), sin(t), sin(2t)).

(a) Calculate the curvature and torsion of ~X(t).

(b) Prove that the curvature is never 0.

(c) Find the exact locations of the vertices, i.e., where κ′(t) = 0.

3.2.7. In Chapter 5, we will encounter a surface called a torus. Curvesthat lie on torus are often knotted. Define a torus knot as a curveparametrized by

~X(t) = ((a+ b cos(qt)) cos(pt), (a+ b cos(qt)) sin(pt), b sin(qt))

where a and b are real numbers, with a > b > 0, and p and q arerelatively prime positive integers. Let a = 2 and b = 1. Calculatethe curvature functions of the corresponding torus knot.

3.2.8. Let ~x be a parametrized curve, and let ~ξ = ~x g be a regular repa-rametrization of ~X.

(a) Prove that the curvature function κ is unchanged under a reg-ular reparametrization, i.e., that κ~x(g(u)) = κ~ξ(u).

(b) Prove that the torsion function τ is unchanged under a posi-tively oriented reparametrization and becomes −τ under a neg-atively oriented reparametrization.

3.2.9. Consider the curve ~X(t) = (a sin t cos t, a sin2 t, a cos t).

(a) Prove that the locus of ~X lies on a sphere.

(b) Calculate the curvature and torsion functions of ~X(t).

3.2.10. If ~X(t) is a parametrization for a planar curve, then for some fixed

vectors ~a, ~b, and ~c, with ~b and ~c not collinear, and for some realfunctions f(t) and g(t), we can write

~X(t) = ~a+ f(t)~b+ g(t)~c.

Prove that for a planar curve, its torsion function is identically 0.


3.2.11. (ODE) Prove Proposition 3.2.7.

3.2.12. Similar to the case of plane curves, we define the evolute to a spacecurve ~X(t) as the curve

~γ(t) = ~X(t) +1

κ(t)~P (t).

Prove that the evolute to the circular helix ~X(t) = (a cos t, a sin t, bt)is another circular helix. Find the pitch of this new helix.

3.2.13. Consider the circular cone with equation

x2

a2+y2

a2=z2

b2.

(a) Find parametric equations for a curve that lies on this conesuch that the tangent makes a constant angle with the z-axis.

(b) Prove that they project to logarithmic spirals on the xy-plane.

(c) Find the curvature and torsion functions.

3.2.14. Let ~α : I → R3 be a parametrized regular curve with κ(t) 6= 0 andτ(t) 6= 0 for t ∈ I. The curve ~α is called a Bertrand curve if there

exists another curve ~β : I → R3 such that the principal normal linesto ~α and ~β are equal at all t ∈ I. The curve ~β is called the Bertrandmate of ~α.

(a) Prove that we can write

~β(t) = ~α(t) + r ~P (t)

for some constant r.

(b) Prove that ~α is a Bertrand curve if and only if there exists alinear relation

aκ(t) + bτ(t) = 1 for all t ∈ I,

where a and b are nonzero constants and κ(t) and τ(t) are thecurvature and the torsion of ~α respectively.

(c) Prove that a curve ~α has more than one Bertrand mate if andonly if ~α is a circular helix.


3.3 Osculating Plane and Osculating Sphere

As with plane curves, if ~X(t) is a regular space curve of class C2, onecan talk about osculating circles to ~X(t) at a point t = t0. Recallthat an osculating circle to a curve at a point is a circle with contactof order 2 at that point. Following the proof of Proposition 1.4.4, onedetermines that at any point t = t0, where κ(t0) 6= 0, the osculatingcircle exists and a parametric formula for it is

~γ(t) = ~X(t0) +1

κ(t0)~P (t0) +

1

κ(t0)

((sin t)~T (t0)− (cos t)~P (t0)

).

(3.13)Even without reference to osculating circles, given any parametric

curve ~X : I → R3, the second-order Taylor approximation to ~X att = t0 is a planar curve with contact of order 2. Furthermore, if~T ′(t0) 6= ~0, this second-order approximation ~f of ~X is

~f(t) = ~X(t0) + (t− t0) ~X ′(t0) +1

2(t− t0)2 ~X ′′(t0)

= ~X(t0) +

(s′(t0)(t− t0) +

1

2s′′(t0)(t− t0)2

)~T (t0) +

1

2s′(t0)2κ(t0)(t− t0)2 ~P (t0).

The vector function ~f is a planar curve that lies in the plane thatgoes through the point ~X(t0) and has ~T (t0) and ~P (t0) as directionvectors. (If κ(t0) = 0, then ~P (t0) is not strictly defined and mightnot even be defined by completing by continuity. In this case, thesecond-order approximation ~f to ~X at t0 lies on a line.) This leadsto the following definition.

Definition 3.3.1. Let ~X : I → R3 be a parametrized curve, and lett0 ∈ I. Suppose that ~X is of class C2 over an open interval containingt0 and that κ(t0) 6= 0. The osculating plane to ~X at t = t0 is theplane through ~X(t0) spanned by ~T (t0) and ~P (t0). In other words,the osculating plane is the set of points ~u ∈ R3 such that

~B(t0) · (~u− ~X(t0)) = 0.

Note that from Equation (3.8), ~B(t0) is parallel to ~X ′(t0) ×~X ′′(t0), so the osculating plane also has the equation ( ~X ′(t0) ×~X ′′(t0)) · (~u− ~X(t0)) = 0 in points ~u ∈ R3.

3.3. Osculating Plane and Osculating Sphere 91

We introduced the notion of order of contact between two curvesin Section 1.4, but we can also talk about the order of contact be-tween a curve C and a surface Σ by defining this latter notion asthe order of contact between C and C2, where C2 is the orthogonalprojection of C onto Σ.

Proposition 3.3.2. Let ~X : I → R3 be a regular parametrized curve ofclass C2. The osculating plane to ~X at t = t0 is the unique planein R3 of contact order 2 or greater. Furthermore, assuming ~X is ofclass C3, the osculating plane has contact order 3 or greater if andonly if τ(t0) = 0 or κ(t0) = 0.

Proof: Set A = ~X(t0) and reparametrize ~X by arc length so thats = 0 corresponds to the point A. The orthogonal distance f(s)between ~X(s) and the osculation plane P is

f(s) = | ~B(0) · ( ~X(s)− ~X(0))|.

Using the Taylor approximation of ~X(s) near s = 0; Equations (3.6),(3.7), and (3.10); and the fact that ~B · ~T = ~B · ~P = 0, we deducethat

f(s) =∣∣∣16κ(0)τ(0)s3 + higher order terms

∣∣∣.Thus,

lims→0

f(s)

s3=|κ(0)τ(0)|

6

and the proposition follows.

One can now interpret the sign of the torsion function τ(t) of aparametrized curve in terms of the curve’s position with respect toits osculating plane at a point. In fact, τ(t0) > 0 at ~x(t0) when thecurve comes up through the osculating plane (where the binormal ~Bdefines the up direction) and τ(t0) < 0 when the curve goes downthrough the osculating plane (see Figure 3.5).

The osculating plane along with two other planes form what iscalled the moving trihedron, which consists of the coordinate planesin the Frenet frame. The plane through ~X(t0) and spanned by theprincipal normal and binormal is called the normal plane and is theset of points ~u that satisfy

~T (t0) · (~u− ~X(t0)) = 0.


~T

~P

~B

(a) Positive torsion.

~T

~P

~B

(b) Negative torsion.

Figure 3.5. Torsion and the osculating plane.

The plane through the tangent and binormal is called the rectifying

plane and is the set of points ~u that satisfy

~P (t0) · (~u− ~X(t0)) = 0.

Figure 3.6 shows the osculating plane, the normal plane, and the

rectifying plane together at a point on the space cardioid.

We now apply the theory of order of contact from Section 1.4 to

find the osculating sphere – a sphere that has order of contact 3 or

higher to a curve at a point.Suppose that a sphere has center ~c and radius r so that its points

~Z satisfy the equation

‖~Z − ~c‖2 = r2.

Consider a curve C parametrized by arc length by ~X : I → R3. The

distance f(s) between the point ~X(s) on the curve and the sphere is

f(s) =∣∣ ‖ ~X(s)− ~c‖ − r

∣∣.Since derivatives of G(t) =

√g(t) are equal to 0 if and only if

g′(t) = 0, then the derivatives of f(s) are equal to 0 if and only

if the derivatives of

h(s) = ( ~X(s)− ~c) · ( ~X(s)− ~c)


osculating

rectifying

normal

Figure 3.6. Moving trihedron.

are equal to 0. The first three derivatives of h(s) lead to

h′(s) = 0⇐⇒ ( ~X − ~c) · ~T = 0,

h′′(s) = 0⇐⇒ ( ~X − ~c) · κ~P + 1 = 0,

h′′′(s) = 0⇐⇒ ( ~X − ~c) · (κ′ ~P − κ2 ~T + κτ ~B) = 0.

Consequently, at any point ~X(s0) on the curve such that κ(s0) 6= 0and τ(s0) 6= 0, the first three derivatives can be equal to 0 if we have

( ~X − ~c) · ~T = 0, ( ~X − ~c) · ~P = −1

κ, ( ~X − ~c) · ~B =

κ′

κ2τ. (3.14)

The equations in (3.14) give a decomposition of the center of theosculating sphere to a curve ~X(s) at the point s = s0. Isolating ~c,the center is

~c = ~X(s0) +R(s0)~P (s0) +R′(s0)

τ(s0)~B(s0) (3.15)


where R(s) = 1κ(s) . The radius of the osculating sphere is

r =

√R(s0)2 +

(R′(s0)

τ(s0)

)2

.

That the first three derivatives of the distance function f(s) are 0implies that the curve C and the osculating sphere have contact oforder 3.

If τ(s0) = 0, then h′′′(s) may still be 0 as long as κ′(s0) = 0 at thesame time. In that case, the curve admits a one-parameter family ofosculating spheres at ~x(s0), where the ~B component of the center ~ccan be anything. This discussion leads to the following proposition.

Proposition 3.3.3. Let ~X : I → R3 be a regular parametrized curve ofclass C3. Let t0 ∈ I be a point on the curve where κ(t0) 6= 0. Thecurve ~X admits an osculating sphere at t = t0 if either (1) τ(t0) 6= 0or (2) τ(t0) = 0 and κ′(t0) = 0. Define R(t) = 1

κ(t) . If τ(t0) 6= 0,

then at t0 the curve admits a unique osculating sphere with center ~cand radius r, where

~c = ~X(t0) +R(t0)~P (t0) +R′(t0)

s′(t0)τ(t0)~B(t0) and r =

√R(t0)2 +

(R′(t0)

s′(t0)τ(t0)

)2

.

If τ(t0) = 0 and κ′(t0) = 0, then at t0 the curve admits as an oscu-lating sphere any sphere with center ~c and radius r where

~c = ~X(t0) +R(t0)~P (t0) + cB ~B(t0) and r =√R(t0)2 + c2

B,

where cB is any real number.

Proof: The only matter to address beyond the previous discussion isto see how the various quantities in Equation (3.15) change under areparametrization.

Let J ⊂ R be an interval, and let f : J → I be of class C3 suchthat f ′ doesn’t change sign. Then ~ξ = ~x f is a regular reparametr-


Figure 3.7. Osculating sphere.

ization of ~X. It is not hard to check that

~Tξ = sign(f ′)~T ,

~Pξ = ~P ,

~Bξ = sign(f ′) ~B.

Also, in Problem 3.2.8, the reader showed that

κξ = κ,

τξ = sign(f ′)τ.

Consquently, we have Rξ = R and also that 1τ~B is invariant under

any regular reparametrization.On the other hand, if t = f(u) so that ~ξ(u) = ~X(t), then

Rξ(u) = R(f(u)) and therefore, R′ξ(u) = R′(f(u))f ′(u) = R′(t)f ′(u).

In particular, dRds = 1

s′(t)dRdt .

The proposition then follows from the prior discussion.

Figure 3.7 illustrates an example of a curve and its osculatingsphere at a point. The figure also shows the Frenet frame at thepoint in question, along with the orthogonal projection of the curveonto the sphere. This projection onto the sphere has the parametricequations

~γ(t) =~X(t)− ~c‖ ~X(t)− ~c‖

r.


Problems

3.3.1. Find the osculating plane to the space cardioid ~X(t) = ((1−cos t) cos t,(1− cos t) sin t, sin t) at t = π.

3.3.2. Calculate the osculating circle to twisted cubic ~X(t) = (t, t2, t3) att = 1.

3.3.3. Consider the curve ~X(t) = (cos t, sin t, sin 2t) for t ∈ [0, 2π]. Notethat this curve lies on the cylinder x2 + y2 = 1.

(a) Find equations for the osculating plane at t0.

(b) For what values of t is the osculating plane tangent to thecylinder (i.e., the osculating plane has a normal vector of theform (λ cos θ, λ sin θ, 0)?

3.3.4. Let ~X(t) : I → R3 be the parametrization by arc length of a curve

C. Consider a circle that passes through the three points ~X(s),~X(s + h1), and ~X(s + h2). Prove that as (h1, h2) → (0, 0), thelimiting position of this circle is precisely the osculating circle to Cat the point ~X(s).

3.3.5. Let ~X(t) : I → R3 be a parametrized curve, and let t ∈ I be afixed point where κ(t) 6= 0. Define π : R3 → R2 as the orthogonal

projection of R3 onto the osculating plane to ~X at t. Define γ = π ~Xas the orthogonal projection of the space curve ~X into the osculatingplane. Prove that the curvature κ(t) of ~X is equal to the curvatureκg(t) of the plane curve ~γ.

3.3.6. Calculate the osculating sphere of the twisted cubic ~X(t) = (t, t2, t3)at the point (0, 0, 0).

3.3.7. Prove that if all the normal planes to a curve C pass through a fixedpoint ~p, then C lies on a sphere of center ~p.

3.3.8. Let C be a curve whose curvature and torsion in terms of arc lengthare κ(s) and τ(s). Set R(s) = 1

κ(s) . Suppose that τ(s) and R′(S)

are never 0. Prove that C lies on a sphere if and only if

R(s)2 +

(R′(s)

τ(s)

)2

= const.

3.3.9. Determine a condition where the osculating circle has contact oforder 3 or higher.

3.3.10. Let C be a regular curve with a parametrization ~X : I → R3 of classC3. Define the osculating helix to C at a point P = ~X(t0) as theunique helix that goes through P and has curvature κ(t0) and τ(t0).


(a) Find the parametric equations of the osculating helix.

(b) Prove that the osculating helix has contact of order 3 with Cat P .

3.3.11. Let C be a curve that lies on a sphere of radius r and suppose thatits curvature and torsion functions with respect to arc length areκ(s) and τ(s). Since the curve lies on a sphere, this sphere is itsosculating sphere at all points on C. Thus, by Proposition 3.3.3, if

R(s) = 1/κ(s), then R(s)2+(R′(s)τ(s)

)2is a constant function equation

to r2. Prove the converse: that if C is a curve with κ(s) and τ(s)

such that R(s)2+(R′(s)τ(s)

)2is constant, then C lies on a sphere. [Hint:

Prove that the center of the osculating sphere does not move.]

3.4 Natural Equations

In Section 1.5, we showed that the curvature (in terms of arc length)function uniquely specifies a regular curve up to its location andorientation in R2. For space curves, one must introduce the torsionfunction for a measurement of how much the curve twists away frombeing planar, and Equation (3.10) shows how the torsion functionappears as a component of ~X ′′′ in the Frenet frame. Since we knowhow ~T , ~P , and ~B change with respect to t, we can express all higherderivatives ~X(n)(t) of ~X(t) in the Frenet frame in terms of s′(t), κ(t),and τ(t) and their derivatives. This leads one to posit that a curveis to some degree determined uniquely by its curvature and torsionfunctions. The following theorem shows that this is indeed the case.

Theorem 3.4.1 (Fundamental Theorem of Space Curves). Given functionsκ(s) ≥ 0 and τ(s) continuously differentiable over some intervalJ ⊆ R containing 0, there exists an open interval I containing 0and a regular vector function ~X : I → R3 that parametrizes its locusby arc length, with κ(s) and τ(s) as its curvature and torsion func-tions, respectively. Furthermore, any two curves C1 and C2 withcurvature function κ(s) and torsion function τ(s) can be mappedonto one another by a rigid motion of R3.

Proof: Let κ(s) and τ(s) be functions defined over an interval J ⊂ Rwith κ(s) ≥ 0. Consider the following system of 12 linear first-order


differential equations:

x′1(s) = t1(s),

x′2(s) = t2(s),

x′3(s) = t3(s),

t′1(s) = κ(s)p1(s),

t′2(s) = κ(s)p2(s),

t′3(s) = κ(s)p3(s),

p′1(s) = −κ(s)t1(s) + τ(s)b1(s),

p′2(s) = −κ(s)t2(s) + τ(s)b2(s),

p′3(s) = −κ(s)t3(s) + τ(s)b3(s),

b′1(s) = −τ(s)p1(s),

b′2(s) = −τ(s)p2(s),

b′3(s) = −τ(s)p3(s)

(3.16)

where xi(s), ti(s), pi(s), and bi(s), with i = 1, 2, 3, are unknownfunctions. With the stated conditions on κ(s) and τ(s), accordingto the existence and uniqueness theorem for first-order systems ofdifferential equations (see [2, Section 31.8]), there exists a solution tothe above system defined for s in a neighborhood of 0. Furthermore,there exists a unique solution with specified initial conditions

x1(0) = x10, x2(0) = x20, x3(0) = x30,

t1(0) = t10, t2(0) = t20, t3(0) = t30,

p1(0) = p10, p2(0) = p20, p3(0) = p30,

b1(0) = b10, b2(0) = b20, b3(0) = b30.

Define the two matrices of functions

A(s) =

0 −κ(s) 0κ(s) 0 −τ(s)

0 τ(s) 0

and M(s) =

t1(s) p1(s) b1(s)t2(s) p2(s) b2(s)t3(s) p3(s) b3(s)

.

Recall from Equation (3.12) that M ′(s) = M(s)A(s). It is possi-ble to show that since A(s) is antisymmetric, the function f(s) =M(s)TM(s) is constant as a matrix of functions. Therefore, any so-lution to Equation (3.16) is such that M(s)TM(s) remains constant.In particular, if we choose initial conditions such thatt10

t20

t30

,

p10

p20

p30

,

b10

b20

b30

(3.17)

form an orthonormal basis of R3, then solutions to Equation (3.16)


are such that, for all s in the domain of the solution, the vectorst1(s)t2(s)t3(s)

,

p1(s)p2(s)p3(s)

,

b1(s)b2(s)b3(s)

(3.18)

form an orthonormal basis.

Therefore, we have shown that with a choice of initial conditionssuch that the vectors in Equation (3.17) form an orthonormal basis,the corresponding solution to Equation (3.16) is such that ~X(s) =(x1(s), x2(s), x3(s)) is a regular curve of class C3 with curvature κ(s)and torsion τ(s). In addition, the vector functions in Equation (3.18)are the ~T , ~P , and ~B vectors of the Frenet frame associated to ~X(s).This proves existence.

The existence and uniqueness theorem of systems of differentialequations states that a solution is unique once (the right numberof) initial conditions are specified. However, we have imposed theadditional condition that the vectors in Equation (3.17) form anorthonormal set. Any different choice for the initial conditions inEquation (3.17) corresponds to a rotation in R3. Also, two differentchoices of initial conditions x10, x20, x30 correspond to a translationin R3. Therefore, different allowed initial conditions correspond to arigid motion of the locus of ~X(s) in R3. This proves the theorem.

Because of the Fundamental Theorem of Space Curves, the pairof functions κ(s) and τ(s), where κ(s) is a positive function, arecalled the natural equations of a curve. Taken as a pair, they definethe curve uniquely up to a rigid motion in the plane.

Just as in Section 1.5, the proof of Theorem 3.4.1 gives an al-gorithm to reconstruct a space curve from its curvature and torsion(with respect to arc length) functions. The code in Maple for Fig-ure 3.8 is the following.

> with(DEtools):

> with(plots):

> kappa:=s->(1-s^2)/(1+s^2):

> tau:=s->sin(s)/5:

> sys := D(x1)(s)=t1(s), D(x2)(s)=t2(s), D(x3)(s)=t3(s),

D(t1)(s)=kappa(s)*p1(s), D(t2)(s)=kappa(s)*p2(s), D(t3)(s)=kappa(s)*p3(s),

D(p1)(s)=-kappa(s)*t1(s)-tau(s)*b1(s), D(p2)(s)=-kappa(s)*t2(s)-tau(s)*b2(s),


Figure 3.8. A ball of yarn by natural equations.

D(p3)(s)=-kappa(s)*t3(s)-tau(s)*b3(s), D(b1)(s)=tau(s)*p1(s),

D(b2)(s)=tau(s)*p2(s), D(b3)(s)=tau(s)*p3(s):

> DEplot3d(sys,x1(s),x2(s),x3(s),t1(s),t2(s),t3(s),

p1(s),p2(s),p3(s),b1(s),b2(s),b3(s),s=0..85,[[x1(0)=0,x2(0)=0,

x3(0)=0,t1(0)=1,t2(0)=0,t3(0)=0,p1(0)=0,p2(0)=1,p3(0)=0,

b1(0)=0,b2(0)=0,b3(0)=1]],scene=[x1(s),x2(s),x3(s)],

stepsize=0.02,linecolor=black,scaling=constrained);

This code constructs a curve with κ(s) = (1 − s2)/(1 + s2) andτ(s) = 1

5 sin(s), but only for s ≥ 0. Note that

lims→∞

κ(s) = 1,

so as s gets large, the space curve has approximately constant cur-vature κ = 1 but with a torsion that oscillates between −1

5 and 15 .

From Proposition 3.3.3, we expect that as s grows, the curve wouldlie more and more on a sphere of radius 1.

Problems

3.4.1. Assume that κ(s) and τ(s) are of class C∞ over an interval I and

assume that we only consider space curves ~X : I → R3 that are alsoof class C∞. Use the Taylor expansion of ~X(s) to prove Theorem3.4.1.

3.4.2. Prove that if κ(s) and τ(s) are nonzero constant functions, then theresulting curve must be a helix.


Investigative Projects

Project I. Problem 3.2.7 introduced the torus knot. Explore thetorsion function for various values of p and q. In particular, canyou create a torus knot whose torsion function is everywherenegative?

Project II. Using a CAS to draw the curves from the naturalequations, attempt to find some nontrivial examples of curva-ture and torsion functions κ(s) and τ(s) that generated closed,nonplanar space curves. Describe the reasons behind your at-tempts.

CHAPTER 4

Curves in Space: Global Properties

Paralleling our presentation of curves in the plane, we now turnfrom local properties of space curves to global properties. As before,global properties of curves are properties that involve the curve as awhole as opposed to properties that are defined in the neighborhoodof a point on the curve.

The Jordan Curve Theorem does not apply to curves in R3, soGreen’s Theorem, the isoperimetric inequality, and theorems con-necting curvature and convexity do not have an equivalent for spacecurves. On the other hand, curves in space exhibit new types ofglobal properties, in particular, knottedness and linking.

4.1 Basic Properties

Definition 4.1.1. A parametrized space curve C is called closed if thereexists a parametrization ~X : [a, b] → R3 of C such that ~X(a) =~X(b). A closed curve is of class Ck if, in addition, all the (one-sided)derivatives of ~X at a and at b are equal of order i = 0, 1, . . . , k;in other words, if as one-sided derivatives ~X ′(a) = ~X ′(b), ~X ′′(a) =~X ′′(b), and so on up to ~X(k)(a) = ~X(k)(b). A space curve C is calledsimple if it has no self-intersections, and a closed curve is calledsimple if it has no self-intersections except at the endpoints.

As discussed in Section 2.1, a closed space curve can be under-stood as a function f : S1 → R3 that is continuous as a functionbetween topological spaces (see [24, Appendix A]). Again, to saythat a closed curve is simple is tantamount to saying that as a con-tinuous function f : S1 → R3, it is injective.

Our first result has an equivalent in the theory of plane curves.

Proposition 4.1.2. If a regular curve is closed, then it is bounded.

103

104 4. Curves in Space: Global Properties

Proof: (Left as an exercise for the reader.)

Stokes’ Theorem presents a global property of curves in space inthat it relates a quantity calculated along the curve with a quan-tity that depends on any surface that has that curve as a boundary.However, the theorem involves a vector field in R3, and it does notaddress “geometric” properties of the curve, by which we mean prop-erties that are independent of the curve’s location and orientation inspace. Therefore, we simply state Stokes’ Theorem as an exampleof a global theorem and trust the reader has seen it in a previouscalculus course.

Theorem 4.1.3 (Stokes’ Theorem). Let S be an oriented, piecewise reg-ular surface bounded by a closed, piecewise regular curve C. Let~F : R3 → R3 be a vector field over R3 that is of class C1 on S.Supposing that C is oriented according to the right-hand rule,∫

C~F · d~s =

∫∫S

(~∇× ~F ) · d~S. (4.1)

If S has no boundary curve, then the line integral on the left is takenas 0.

Note as a reminder that the line element d~s stands for ~γ′(t)dt,where ~γ(t) is a parametrization of C, and d~S stands for ~ndA where~n is the unit normal and dA is the surface element at a point on thesurface. If a surface is parametrized by a vector function ~X(u, v)into R3, then d~S = ~Xu × ~Xv du dv.

The following is an interesting corollary to Stokes’ Theorem.

Corollary 4.1.4. Let C be a simple, regular, closed space curve param-etrized by ~γ : I → R3. There exists no function f : R3 → R of classC2 such that the gradient satisfies ~∇f = ~T (t) at all points of thecurve.

Proof: Recall that for all functions f : R3 → R of class C2, the curlof the gradient is identically 0, namely, ~∇ × ~∇f = ~0. If S is anyorientable piecewise regular surface that has C as a boundary, then∫∫

S(~∇× ~∇f) · d~S =

∫∫S~0 · d~S = ~0.


If f did satisfy ~∇f = ~T (t) at all points of the curve C, then∫C~∇f · d~s =

∫I

~T (t) · ~T (t)s′(t)dt =

∫Cds,

which is the length of the curve. By Stokes’ Theorem, this leads toa contradiction, assuming the curve has length greater than 0.

Example 4.1.5. Stokes’ Theorem is a very powerful theorem but, asthis example illustrates, it is important to check that the conditionsare satisfied. Consider the vector field ~F (x, y, z) defined by

~F (x, y, z) =

(0,− y

y2 + z2,

z

y2 + z2

)

and the curve C parametrized by ~X(t) = (cos(2t), cos t, sin t). Thecurl of the vector field ~F is ∇× ~F = (0, 0, 0) so the flux through anysurface S over which ~F is C1 is∫∫

S∇× ~F · d~S = 0.

On the other hand, the vector field is defined along C and the circu-lation of ~F along C is ∫

C~F · d~s = 2π.

This may seem like a contradiction because there certainly exists asurface S whose boundary is C. However, we cannot forget to observethat ~F is well-defined on all of R3 except for the x-axis. The reasonthis example does not give a contradiction to Stokes’ Theorem isbecause there is no surface S whose boundary is C and which doesnot intersect the x-axis.

Intuitively speaking, the curve C and the x-axis are “linked,”in the sense that there is no way to move and deform one curvecontinuously in such a way as to move C and the x-axis into twoseparate half-spaces without making the curves intersect. We discusslinked curves more in Section 4.3.


Problems

4.1.1. Verify the flux and circulation calculations in Example 4.1.5.

4.1.2. Let ~α : I → R3 be a regular, closed space curve, and let ~p be a point.Prove that a point t0 of maximum distance on the curve away from~p is such that ~α(t0)− ~p is in the normal plane to the curve at ~α(t0).

4.1.3. Recall that the diameter of a curve ~α is the maximum of the function

f(t, u) = ‖~α(t)− ~α(u)‖.

Use the second derivative test in multivariable calculus to show thatif (t0, u0) gives a diameter of a space curve, then the following hold:

(a) The vector ~α(u0) − ~α(t0) is in the intersection of the normalplanes of α(t0) and α(u0).

(b) The vector ~α(u0)− ~α(t0) is on the side of the rectifying plane

of ~α(t0) that makes an acute angle with ~P (t0) (and similarlyfor α(u0)).

(c) The diameter ‖~α(t0) − ~α(u0)‖ is greater than or equal tomax1/κ(t0), 1/κ(u0).

[Hint: One can assume that ~α is parametrized by arc length. Also,the extrema of f(t, u) occur at and have the same properties as theextrema of g(t, u) = f(t, u)2.]


4.1.5. By using the vector field ~F (x, y, z) = (0, x, 0), Stokes’ Theorem es-tablishes the familiar Green’s Theorem formula for area of the inte-rior of a curve C in the xy-plane:

A =

∫C(0, x, 0) · d~s =

∫Cx dy.

Suppose that we consider a curve C on the sphere of radius R andcentered at the origin. Prove that there does not exist a vectorfield ~F (x, y, z) in R3 that could be used to calculate the area of the“interior” of the curve as a line integral. [Hint: Recall relationsamong vector differential operators.]

4.2 Indicatrices and Total Curvature

Definition 4.2.1. Given a space curve ~X : I → R3, define the tangent,principal, and binormal indicatrices respectively, as the loci of the

4.2. Indicatrices and Total Curvature 107

space curves on the unit sphere given by ~T (t), ~P (t), and ~B(t), asdefined in Section 3.2.

Since the vectors of the Frenet frame have a length of 1, theindicatrices are curves on the unit sphere S2 in R3. In contrast tothe tangent indicatrix for plane curves, there are more possibilitiesfor curves on the sphere than for curves on a circle. Therefore,results such as the theorem on the rotation index or results aboutthe winding number do not have an immediate equivalent for curvesin space.

Example 4.2.2 (Lines). The tangent indicatrix of a line is a single point~v/‖~v‖ on the sphere, where the line is parametrized by

~X(t) = ~vt+ ~p.

Example 4.2.3 (Helices). Consider a helix around an axis L throughthe origin. The tangent indicatrix of this helix is the circle on the unitsphere S2, given as the intersection of S2 with the plane perpendicularto L at a distance of 1/

√1 + (κ/τ)2 from the origin, where κ

τ is thepitch of the helix. Notice that this result is true for general helices(see Example 3.2.5) and not just circular helices.

Example 4.2.4 (Space Cardioid). The space cardioid, given by the pa-rametrization

~X(t) = ((1− cos t) cos t, (1− cos t) sin t, sin t) , t ∈ [0, 2π],

is a closed curve. The tangent indicatrix is again a closed curve andis shown in Figure 4.1. The figure shows that the tangent indicatrixhas a double point. In Problem 4.2.3, the reader is invited to showthis and a few other interesting properties of the tangent indicatrixof the space cardioid.

Definition 4.2.5. The total curvature of a closed curve C parametrizedby ~X : [a, b]→ R3 is ∫

Cκ ds =

∫ b

aκ(t)s′(t) dt.

This is a nonnegative real number since κ(t) ≥ 0 for space curves.


~X(t2)

~X(t1)

x

y

z

~T (t1)

~T (t2)

Figure 4.1. The space cardioid and its tangent indicatrix.

Though we cannot define the concept of a rotation index, Fen-chel’s Theorem gives a lower bound for the total curvature of a spacecurve. Since ~T ′ = s′κ~P , the speed of the tangent indicatrix iss′(t)κ(t). Therefore, the total curvature of ~X is the length of thetangent indicatrix. One can also note that the tangent indicatrixhas a critical point where κ(t) = 0.

Theorem 4.2.6 (Fenchel’s Theorem). The total curvature of a regularclosed space curve C is greater than or equal to 2π. It is equal to 2πif and only if C is a convex plane curve.

Before we prove Fenchel’s Theorem, we need to discuss the con-cept of distance between points on the unit sphere S2. We can definethe distance between two points p and q on S2 as

d(p, q) = inflength(Γ) |Γ is a curve connecting p and q.

Let O be the center of the unit sphere. It is not difficult to show thatthe path connecting p and q of shortest distance is an arc between pand q on the circle defined by the intersection of the sphere and theplane containing O, p, and q. (We will obtain this result in Example8.4.8 when studying geodesics, but it is possible to prove this claim


O

A

B

Figure 4.2. Spherical distance.

without the techniques of geodesics.) Then the spherical distancebetween two points on the unit sphere is

d(p, q) = cos−1(p · q),

where we view p and q as vectors in R3.We will denote by AB the Euclidean distance between two points

A and B as elements in R3. If A,B ∈ S2, we use the notation ABto denote the distance between A and B on the sphere. Since OABforms an isosceles triangle (see Figure 4.2), we see that the sphericaland Euclidean distance are related via

AB = 2 sin

(AB

2

)and AB = 2 sin−1

(AB

2

).

Lemma 4.2.7 (Horn’s Lemma). Let Γ be a regular closed curve on theunit sphere S2. If Γ has length less than 2π, then there exists a greatcircle C on the sphere such that Γ does not intersect C.

Proof: The proof we give to this lemma is due to R. A. Horn [20].Let A and B be two points of Γ that divide the curve into two arcs

of equal length L/2. The points A and B divide Γ into two curvesΓ1 and Γ2 of equal length. Since L < 2π, the spherical distancebetween A and B is less than the length of Γ1, which is strictly lessthan π. Consequently, P and Q are not antipodal (i.e., the segment[P,Q] is not a diameter of the sphere) and therefore there exists a


unique point M on the minor arc from P to Q midway between Pand Q. We claim that Γ does not meet the equator C that has M asthe north pole, so the curve lies in the hemisphere centered at M .

To show that Γ1 does not meet the equator consider a copy Γ′1of Γ1 rotated one half turn about M . The curve Γ′1 is a curve fromB to A of length L/2. Define the closed curve Γ′′ as the curve thatfollows Γ1 from A and B and then follows Γ′1 from B back to A.The curve Γ′′ has length L. Furthermore, if Γ1 intersected C, then sowould Γ′1. Hence, Γ′′ would contain antipodal points R and R′. Butthe spherical distance RR′ = π. Therefore, the distance from R toR′ along Γ′′ would be greater than or equal to π and, by symmetryof Γ′′, similarly for the path from R′ to R. This contradicts the factthat the length of Γ′′ is less than 2π.

Thus any curve with length less than 2π lies in an open hemi-sphere.

We are now in a position to prove Fenchel’s Theorem.

Proof (of Theorem 4.2.6): Let ~γ : [a, b] → R3 be a parametrizationfor the regular closed curve C, and let p be any point on the unitsphere. Consider the function g(t) = p · ~γ(t). Since [a, b] is a closedand bounded interval and g(t) is continuous, then it attains a maxi-mum and minimum value in [a, b]. This value occurs where

g′(t) = p · ~γ′(t) = 0.

We remark that the set of points ~x ∈ S2 such that p · ~x = 0 isthe great circle on S2 that is on the plane perpendicular to the line(Op), where O is the center of the sphere. Since ~γ′(t) and ~T (t) arecollinear and since p was chosen arbitrarily, we conclude that thetangent indicatrix intersects every great circle on S2. Consequently,by Lemma 4.2.7, the length of the tangent indicatrix is greater thanor equal to 2π. Thus, since the length of the tangent indicatrix isthe total curvature,∫ b

aκ(t)s′(t) dt =

∫Cκ ds ≥ 2π.

To prove the second part of the theorem, first note that if ~γ(t)traces out a convex plane curve, then κ(t) = |κg(t)| for ~γ as a plane


curve. Furthermore, by Propositions 2.4.3 and 2.2.11, the total cur-vature is 2π. Therefore, to finish proving the theorem, we only needto prove the converse.

Suppose that the total curvature of ~γ is 2π. The length of thetangent indicatrix is therefore 2π. Furthermore, since by the abovereasoning the tangent indicatrix must intersect every great circle,the tangent indicatrix must itself be a great circle. Thus ~T (t), ~T ′(t),and ~T ′′(t) are coplanar and so τ(t) = 0. Thus, by Proposition 3.2.7,~γ(t) is planar. In this case, it is not hard to check that κ(t) = |κg(t)|,where κg(t) is the curvature of ~γ(t) as a plane curve. Thus,

2π =

∫C|κg| ds

is the total distance that ~T (t) travels on the unit circle, and thisis the length of the unit circle. Consequently, if t1, t2 ∈ [a, b), with~T (t1) = ~T (t2), then ~T (t) is constant over [t1, t2] because, otherwise,the total distance ~T (t) travels on the unit circle would exceed 2π byat least ∫ t2

t1

|κg(t)| ds.

We conclude then that ~γ has no bitangent lines, and by Problem2.4.1, we conclude that C is a convex plane curve.

As an immediate corollary, we obtain the following result.

Corollary 4.2.8. If a regular, closed space curve has a curvature func-tion κ that satisfies

κ ≤ 1

R,

then C has length greater than or equal to 2πR.

Proof: Suppose that a regular, closed space curve has a curvaturefunction κ(s), given in terms of arc length, that satisfies the givenhypothesis. Then if L is the length of the curve, we have

L =

∫ L

0ds ≥

∫ L

0Rκ(s) ds = R

∫ L

0κ(s) ds ≥ 2πR,

where the last inequality follows from Fenchel’s Theorem.


Though it is outside the scope of our current techniques, we wishto state here Jacobi’s Theorem since it is a global theorem on spacecurves. The proof follows as an application of the Gauss-BonnetTheorem (Problem 8.3.6).

Theorem 4.2.9 (Jacobi’s Theorem). Let ~α : I → R3 be a closed, regularparametrized curve whose curvature is never 0. Suppose that theprincipal normal indicatrix ~P : I → S2 is simple. Then the locus~P (I) of the principal normal indicatrix separates the sphere into tworegions of equal area.

The reader should note that Jacobi’s Theorem is quite profoundin the following sense. Given a parametrized curve ~γ : I → R3 suchthat ‖~γ(t)‖ = 1 for all t ∈ I, the problem of calculating the surfacearea of the sphere lying on one side or the other of the locus of thiscurve is not a tractable problem.

We now present Crofton’s Theorem, which we will use in the nextsection to prove a theorem by Fary and Milnor on the total curvatureof a knot.

Let O be the center of the unit sphere S2. Each great circle C isuniquely defined by a line L through the origin of the sphere as theintersection between S2 and the plane through O perpendicular to L.Furthermore, L is defined by two “poles,” the points of intersection ofL with S2. However, there exists a bijective correspondence betweenoriented great circles and points on S2 by associating to C the poleof L that is in the direction on L that is positive in the sense of theright-hand rule of motion on C.

Consider a set Z of oriented great circles on S2. We call themeasure m(Z) of the set Z the area of the region traced out on S2

by the positive poles of the oriented circles in Z.

Theorem 4.2.10 (Crofton’s Theorem). Let Γ be a curve of class C1 onS2. The measure of the great circles of S2 that meet Γ is equal tofour times the length of Γ.

Proof: Suppose that ~e1 : [0, L] → S2 parametrizes Γ by arc length.Complete ~e1(s) to form an orthonormal basis ~e1(s), ~e2(s), ~e3(s)so that ~ei(s) for i = 2, 3 are of class C1. Without loss of generality,we can construct ~e2 and ~e3 so that

det(~e1, ~e2, ~e3) = 1


for all s ∈ [0, L]. Using the same reasoning that established Equation(3.5) or the result of Problem 9.1.10, we have

d

ds

(~e1 ~e2 ~e3

)=(~e1 ~e2 ~e3

) 0 a2 a3

−a2 0 a1

−a3 −a1 0

(4.2)

for some continuous functions ai : [0, L] → R. Furthermore, since~e1 is parametrized by arc length we know that it has unit speed,so a2(s)2 + a3(s)2 = 1. Consequently, there exists a function α :[0, L]→ R of class C1 such that

~e ′1(s) = cos(α(s))~e2 + sin(α(s))~e3.

The set of oriented great circles that meet Γ at ~e1(s) is parame-trized by its positive poles

(cos θ)~e2(s) + (sin θ)~e3(s)

for θ ∈ [0, 2π]. Therefore, the region traced out by the positive polesof oriented great circles that meet Γ is parametrized by

~Y (s, θ) = (cos θ)~e2(s) + (sin θ)~e3(s) for (s, θ) ∈ [0, L]× [0, 2π].

We wish to determine the area element |dA| for the set of poles tracedout by the set of oriented great circles meeting Γ. However,

|dA| = ‖~Ys × ~Yθ‖ dθ ds,

so after some calculation and using Equation (4.2), one obtains

|dA| = |a2(s) cos θ + a3(s) sin θ| dθ ds = | cos(α(s)− θ)| dθ ds.

Call Cθ,s the oriented great circle with positive pole ~Y (θ, s) anddenote by n(Cθ,s) the number of points in Cθ,s∩Γ. Then the measurem of oriented great circles in S2 that meet Γ is

m =

∫∫n(Cθ,s)|dA| =

∫ L

0

∫ 2π

0| cos(α(s)− θ)| dθ ds. (4.3)

However, for all fixed α0, we have∫ 2π

0| cos(α0 − θ)| dθ = 4,

so we conclude that m = 4L, which establishes the theorem.


Problems

4.2.1. Consider the twisted cubic ~X(t) = (t, t2, t3) with t ∈ R.

(a) Prove that the closure of the tangential indicatrix of the twistedcubic is a closed curve, in particular, that

limt→∞

~T (t) = limt→−∞

~T (t) = (0, 0, 1).

(b) Prove that the tangential indicatrix does not have a corner at(0, 0, 1), namely, that

limt→∞

~T ′(t) = limt→−∞

~T ′(t).

(c) Prove directly that the total curvature of the twisted cubic isbounded.

4.2.2. Consider the helix ~X(t) = (a cos t, a sin t, bt) for t ∈ R. Determine

the locus of the tangent indicatrix of ~X(t).

4.2.3. Figure 4.1 appears to show that the tangent indicatrix of the spacecardioid has a double point.

(a) Find the value t0 of t at which the double point occurs.

(b) Show that the antipodal point to ~X(t0), i.e., − ~X(t0) is also onthe tangent indicatrix of the space cardioid.

(c) Show that the curve obtained by projecting the tangent indica-

trix onto the plane perpendicular to ~X(t0) through the originhas three cusps.

4.2.4. Calculate directly the total curvature of the space cardioid

~X(t) = ((1− cos t) cos t, (1− cos t) sin t, sin t) for t ∈ [0, 2π].

4.2.5. Prove Fenchel’s Theorem as a corollary to Crofton’s Theorem.

4.2.6. (*) Let ~α : [0, L] → R3 be a regular closed curve (parametrized byarc length) whose image lies on a sphere. Suppose also that κ(t) 6= 0.Prove that ∫ L

0

τ(s) ds = 0.

4.3. Knots and Links 115

4.3 Knots and Links

The reader should be forewarned that the study of knots and linksis a vast and fruitful area that one usually considers as a subbranchof topology. In this section, we only have space to give a cursoryintroduction to the concept of a knotted curve in R3 or two linkedcurves in R3. However, the property of being knotted or linked is aglobal property of a curve or curves (since it depends on the curveas a whole), which motivates us to briefly discuss these topics in thischapter. This section presents two main theorems: the Fary-MilnorTheorem, which shows how the property of knottedness imposes acondition on the total curvature of a curve, and Gauss’s formula forthe linking number of two curves.

4.3.1 Knots

Intuitively speaking, a knot is a simple closed curve in R3 that cannotbe deformed into a circle without breaking the curve and reconnect-ing it. In other words, a knot cannot be deformed into a circle by acontinuous process without passing through a stage where it is nota simple curve.

The following gives a precise definition to the above intuition.

Definition 4.3.1. A simple closed curve Γ in R3 is called unknotted ifthere exists a continuous function H : S1 × [0, 1] → R3 such thatH(S1 × 0) = Γ and H(S1 × 1) = S1 and such that Γt = H(S1 ×t) is a curve that is homeomorphic to a circle. If there does notexist such a function H, the curve Γ is called knotted .

The function H described in the above definition is called a ho-motopy in R3 between Γ and a circle S1. Figure 4.3 illustrates fourintermediate stages of a homotopy between a space curve and a cir-cle.

Figure 4.4 shows the trefoil knot realized as a curve in space,along with a two-dimensional diagram. As it is somewhat tedious toplot general knotted curves, even with the assistance of a computeralgebra system, one often uses a diagram that shows the “crossings,”i.e., which part of the curve passes above the other whenever thediagram would intersect in the given perspective. (The interested


Figure 4.3. An unknotting homotopy in R3.

Figure 4.4. A trefoil knot and its diagram.

reader is encouraged to read [23] for an advanced introduction toknot theory.)

The following theorem gives a necessary relationship between aknotted curve and the curvature of a space curve.

Theorem 4.3.2 (Fary-Milnor Theorem). The total curvature of a knot isgreater than or equal to 4π.


~Y (s, θ)

~T (s)

~α(s)

Figure 4.5. Proof of the Fary-Milnor Theorem.

Proof: Let ~x : [0, L] → R3 be a closed, regular space curve parame-trized by arc length, and let ~T : [0, L]→ S2 be the tangent indicatrix.Call C the image of ~x in R3 and Γ the image of ~T on the sphere.

Recall the notations in the proof of Crofton’s Theorem (Theorem4.2.10) and, in particular, that n(Cθ0,s0) is the number of times thatthe great circle Cθ0,s0 intersects the curve Γ (see Figure 4.5). By partof the proof of Crofton’s Theorem, n(Cθ0,s0) > 0 over the domain(θ, s) ∈ [0, 2π]× [0, L], and n(Cθ0,s0) is even.

Define a height function h(s) = ~Y (θ0, s0) ·~x(s). Then the deriva-tive is h′(s) = ~Y (θ0, s0)·~T (s) so that the relative maxima and minimaof h(s) occur where Γ intersects Cθ0,s0 , so n(Cθ0,s0) is the number ofcritical points of h(s).

Suppose that the total curvature of the curve ~x, which is alsothe length of Γ, is less than 4π. By Equation (4.3) in the proof ofCrofton’s Theorem, ∫∫

n(Cθ,s)|dA| < 16π.

Since the area of the sphere is 4π, there exists (θ0, s0) such thatn(Cθ0,s0) = 2, which means that the height function corresponding

to ~Y (θ0, s0) has two critical points, namely, one maximum and one


minimum occuring at s1 and s2. The points ~x(s1) and ~x(s2) partitionC into two curves over which the height function h(s) is monotonic,one increasing and the other decreasing. Consequently, any planeperpendicular to ~Y (θ0, s0) (that intersects ~x between the points ~x(s1)and ~x(s2) of extremal height function) intersects ~x in exactly twopoints. This fact allows us to construct a homotopy between C anda circle, which we describe below.

Call v the height parameter for a point on the curve ~x(s). Callvmin and vmax the minimum and maximum values of h(s) over thedomain [0, L] of ~x. Every plane perpendicular to ~Y (θ0, s0) is de-termined uniquely by the height parameter v as the unique planeperpendicular to ~Y (θ0, s0) and going through the point v~Y (θ0, s0).According to the discussion in the previous paragraph, the plane Pvperpendicular to ~Y (θ0, s0) at height v intersects C in two points.Therefore, C can be expressed as the union of the locus of two con-tinuous curves

~γ1 : [vmin, vmax] −→ R3,

~γ2 : [vmin, vmax] −→ R3,

such that Pv ∩ C = ~γ1(v), ~γ2(v). Note that

~γ1(vmax) = ~γ2(vmax) = ~x(s1), and

~γ1(vmin) = ~γ2(vmin) = ~x(s2).

Call ~γ(v) = 12(~γ1(v) + ~γ2(v)) the midpoint of the pair Pv ∩ C =

~γ1(v), ~γ2(v) (which degenerates to a singleton set for v = vmin orvmax).

Complete ~Y (θ0, s0) to an orthonormal basis ~Y (θ0, s0), ~e2, ~e3.In this basis, we can write ~γ(v) as

~γ(v) = (v, a(v) cos(α(v)), a(v) sin(α(v)))

for some functions a(v) ≥ 0 and α(v) defined on [vmin, vmax]. Fur-thermore, since ~γ(v), ~γ1(v), and ~γ2(v) as position vectors all lie inPv, there exist functions b(v) and θ(v) such that we can write

~γ1(v) = ~γ(v) + b(v) ((cos θ(v))~e2 + (sin θ(v))~e3) ,

~γ2(v) = ~γ(v)− b(v) ((cos θ(v))~e2 + (sin θ(v))~e3) .


Call v0 = 12(vmin + vmax) and R = 1

2(vmax − vmin) and definethe function v(u) = v0 + R cos(u). Finally, define the function H :[0, 2π]× [0, 1] by

H(u, t) = v(u)~Y (θ0, s0)

+ (1− t)a(v(u)) cos(α(v(u))) + sinu ((1− t)r(u) + tR) cos ((1− t)θ(v(u)))~e2

+ (1− t)a(v(u)) sin(α(v(u))) + sinu ((1− t)r(u) + tR) sin ((1− t)θ(v(u)))~e3,

(4.4)

where r : [0, 2π]→ R≥0 is a function such that r(u)| sinu| = b(v(u)).We now leave it as an exercise to the reader to prove that H(u, t)is a homotopy between C and a circle of radius R such that forall t0 ∈ (0, 1), the image of H(u, t0) is homeomorphic to a circle(Problem 4.3.1).

By Definition 4.3.1, the existence of H as defined above allowsus to conclude that C is unknotted. Therefore, if C is a knot, weconclude that the total curvature of C is greater than or equal to4π.

4.3.2 Links

A simple, closed space curve in itself resembles a circle. Knottednessis a property that concerns how a simple closed curve “sits” in theambient space. The technical language for this scenario is that anysimple closed curve is homeomorphic to a circle, but knottednessconcerns how the curve is embedded in R3. In a similar way, thenotion of linking between two simple closed curves is a notion thatconsiders how the curves are embedded in space in relation to eachother.

Intuitively speaking, we would like the linking number betweentwo simple closed curves C1 and C2 in R3 to be the minimum numberof crossings required in order to continuously move the curves inspace (including deformations that do not break the curves) untilthey are separated by a plane in R3. We cannot use this intuitionas a definition since counting the number of times it takes to breakone curve in order to pass through another is too vague.

Figure 4.6 illustrates a basic scenario of linked curves. The figureon the left shows a rendering of two thick circles in space that are


Figure 4.6. Linked curves; linking number +1.

linked. Our intuitive picture leads us to think of this curve as havinga single link. Since it is difficult to accurately sketch curves in R3

and to see which curve is in front of which, it is common to use adiagram to depict the curves. The diagram on the right is obtainedby projecting the curves into a plane (perpendicular to some vector~n), careful to show the white space along the curve diagram to com-municate which curve is above (in reference to ~n) the other at theintersections that occur when projecting. The diagram is called thelink diagram.

In order to give a definition of the linking number, we need togive orientations to the curves. The standard definition of the linkingnumber of two curves uses the link diagram along with the signassociated to each crossing, as shown below:

−1 +1

The arrows indicate the orientation on the respective curves ata crossing point. In Figure 4.6, the picture on the left makes noreference to orientation, but the diagram on the right imposes anorientation by the depicted arrows.

Definition 4.3.3. The linking number link(C1, C2) of two simple closedoriented curves C1 and C2 is half the sum of the signs of all thecrossings in the diagram of the pair (C1, C2).

For example, in the oriented diagram on the right in Figure 4.6,both crossings have sign +1. Thus half the sum of the signs of thecrossings is +1.


One of the surprising results about linking is Gauss’s formulafor the linking number between two curves. Though linking and thelinking number are obviously global properties of the curves, Gauss’sformula calculates the linking number in terms of the parametriza-tions of the curves, thereby connecting the global property to localproperties.

Theorem 4.3.4 (Gauss’s Linking Formula). Let C1 and C2 be two closed,regular space curves parametrized by ~α : I → R3 and ~β : J → R3,respectively. The linking number between C1 and C2 is

link(C1, C2) =1

4π

∫I

∫J

det(~α(u)− ~β(v), ~α′(u), ~β′(v))

‖~α(u)− ~β(v)‖3du dv. (4.5)

Though we cannot give a proof for this formula at this time, webriefly sketch a few of the concepts and techniques that go into it.

In Section 2.2, we introduced the notion of degree for a continuousmap between circles, f : S1 → S1. Intuitively speaking, the degreeof f counts how many times f covers its codomain and with whatorientation. The degree of f takes into account when f might doubleback. For example, if q ∈ S1 has six preimages from f and, for fourof the preimages, f passes through q with the same orientation ason the domain and, for the other two preimages, f passes through qin an opposite orientation, then the degree of f is 4− 2 = 2.

Similarly (and leaving many of the technical details for later),one can define the degree of a continuous map between spheresf : S2 → S2 as how often f covers S2. Again, one should note the dif-ficulty inherent in making this definition precise since one must takeinto account a form of orientation and doubling back, i.e., whether ffolds back over itself over some region of the codomain. In the sameway, one can define the degree of a continuous map F : S → S2,where S is a regular surface without boundary. (We give the defi-nition for a regular surface in Chapter 5.) The conditions that S isregular and has no boundary guarantee that F generically covers S2

by the same amount at all its points, so long as we take into accountorientation and assign a negative covering value to F when F coversS2 negatively.

As discussed in Definition 2.2.7 and in the following, it is moreappropriate to view the parametrization of a simple closed curve C


as a continuous function ~x : S1 → R3. So consider two simple closedcurves with parametrizations ~α : S1 → R3 and ~β : S1 → R3. Thefunction

Ψ : S1 × S1 → S2 given by Ψ(t, u) =~β(u)− ~α(t)

‖~β(u)− ~α(t)‖

defines a continuous function from the torus S1 × S1 to the unitsphere S2. The technical definition of the linking number of twocurves is the degree of Ψ(t, u). The proof of Theorem 4.3.4 consistsof calculating the degree of Ψ(t, u).

Equation (4.5) is difficult to use in general, and in the exercises,we often content ourselves with using a computer algebra systemto calculuate the integrals. Some basic facts about linking are notdifficult to show with the appropriate topological background butare difficult using Equation (4.5). For example, if two simple closedcurves can be separated by a plane, then link(C1, C2) = 0. Thisfact is simple once one has a few facts about the degrees of maps toS2, but using Gauss’s formula to show the same result is a difficultproblem.

Problems

4.3.1. This exercise finishes the proof of the Fary-Milnor Theorem. Con-sider the function H(u, t) defined in Equation (4.4).

(a) Prove that, for all t0 ∈ [0, 1], H(u, t0) is an injective functionfor u ∈ (0, 2π) and that H(0, t0) = H(2π, t0).

(b) Show that the locus of H(u, 1) is a circle.

(c) Show that the locus of H(u, 0) is the curve C.

(d) Use the previous parts of the exercise to conclude that H(u, t)is a homotopy between C and a circle.

4.3.2. Consider the trefoil knot parametrized by

~α(t) = ((3 + cos 3t) cos 2t, (3 + cos 3t) sin 2t, sin 3t) .

Using a CAS, calculate the total curvature of this trefoil knot. Showhow to create another simple closed curve that is homotopic to thistrefoil knot, with a total curvature of 4π + ε for any ε > 0.


Figure 4.7. An interesting link.

4.3.3. Let C1 and C2 be two simple, closed, oriented space curves. Supposethat C−2 is the same locus as C2 but with the opposite orientation.Prove that link(C1, C

−2 ) = −link(C1, C2).

4.3.4. Let C1 and C2 be two simple, closed, oriented space curves. Provethat link(C1, C2) = link(C2, C1).

4.3.5. Consider the two linked curves depicted in Figure 4.7. Impose anorientation on the curves, sketch the link diagram, and then calculatethe linking number.

4.3.6. Let C1 and C2 be the two circles in R3 parametrized by

~α(t) = (cos t, sin t, 0), for t ∈ [0, 2π],

~β(t) = (1− cos t, 0, sin t), for t ∈ [0, 2π].

(a) Using Equation (4.5), write an integral that gives the linkingnumber of these two curves.

(b) Use a computer algebra system to calculate this linking num-ber.

(c) (*) Calculate the integral explicitly.

4.3.7. A rigid motion in R3 is a transformation F : R3 → R3 definedby F (~x) = A~x + ~b, where A is an orthogonal matrix. Prove fromEquation (4.5) that the linking number is a geometric invariant, i.e.,


that if a rigid motion is applied to two simple closed curves C1 andC2, then their linking number does not change.

4.3.8. Consider the two simple closed curves

~α(t) = (3 cos t, 3 sin t, 0), for t ∈ [0, 2π],

~β(t) = ((3 + cos(nt)) cos t, (3 + cos(nt)) sin t, sin(nt)) , for t ∈ [0, 2π].

(a) Give the link diagram of these two space curves, including theorientation, and show that the linking number of these twocurves is n.

(b) Gauss’s formula in Equation (4.5) is quite difficult to use, but,using a computer algebra system, give support for the aboveanswer.

CHAPTER 5

Regular Surfaces

5.1 Parametrized Surfaces

There are many approaches that one can take to introduce surfaces.Some texts immediately build the formalism of differentiable mani-folds, some texts first encounter surfaces in R3 as the solution set toan algebraic equation F (x, y, z) = 0 with three variables, and othertexts present surfaces as the images of vector functions of two vari-ables. In this section, we introduce the last of these three optionsand occasionally refer to the connection with the surfaces as solu-tion sets to algebraic equations. Only later will we show why onerequires more technical definitions to arrive at a workable definitionthat matches what one typically means by a “surface.”

We imitate the definition of parametrized curves in Rn and beginour study of surfaces (yet to be defined) by considering continuousfunctions ~X : U → R3, where U is a subset of R2. Below are someexamples of such functions whose images in R3 are likely to appearin a multivariable calculus course.

Example 5.1.1 (Planes). If ~a and ~b are linearly independent vectors inR3, then the plane through the point ~p and parallel to the vectors ~aand ~b can be expressed as the image of the following function:

~X(u, v) = ~p+ u~a+ v~b

for (u, v) ∈ R2. In other words, we can write

~X(u, v) = (p1 + a1u+ b1v, p2 + a2u+ b2v, p3 + a3u+ b3v)

for constants pi, ai, bi, with 1 ≤ i ≤ 3 that make (a1, a2, a3) and(b1, b2, b3) noncollinear vectors.

125

126 5. Regular Surfaces

Example 5.1.2 (Graphs of Functions). If z = f(x, y) is a real functionof two variables defined over some set U ⊂ R2, one can obtain thegraph of this function as the image of a function ~X : U → R3 bysetting

~X(u, v) = (u, v, f(u, v)).

Example 5.1.3 (Spheres). One can obtain a sphere of radius R centeredat the origin as the image of

~X(u, v) = (R cosu sin v,R sinu sin v,R cos v),

with (u, v) ∈ [0, 2π] × [0, π]. This expression should not appear toomysterious since it merely puts the equation of spherical coordinatesr = R into Cartesian coordinates, with u = θ and v = φ. Thismethod of parametrizing the sphere is the astronomical one, wherev measures the colatitude, i.e., the angle down from the north pole.(Unfortunately, there is no uniformity in the parametrization usedfor spheres in calculus books.)

Example 5.1.4 (Conics). Besides getting the sphere, one could also ob-tain an ellipsoid as the image of some vector function ~X : U → R2

by modifying the coefficients in front of the x, y, and z coordinatefunctions of the parametrization for the sphere. In particular, theparametrization

~X(u, v) = (a cosu sin v, b sinu sin v, c cos v),

with (u, v) ∈ [0, 2π]× [0, π] has for its image the ellipsoid of equation

x2

a2+y2

b2+z2

c2= 1.

One can obtain all of the other conic surfaces as follows. Thecircular cone

x2

a2+y2

a2=z2

b2

is the image of

~X(u, v) = (au cos v, au sin v, bu), with (u, v) ∈ R× [0, 2π].

5.1. Parametrized Surfaces 127

Figure 5.1. Hyperboloid of one sheet.

The function

~X(u, v) = (coshu cos v, coshu sin v, sinhu), with (u, v) ∈ R×[0, 2π],

traces out the hyperboloid of one sheet. (See Figure 5.1 for twoperspectives of a hyperboloid of one sheet.) For the hyperboloid oftwo sheets, we need two separate functions

~X(u, v) = (sinhu cos v, sinhu sin v, coshu)

and~X(u, v) = (sinhu cos v, sinhu sin v,− coshu).

Example 5.1.5 (Surfaces of Revolution). A surface of revolution is a setS in R3 obtained by rotating, in R3, a regular plane curve C in aplane P about a line L in P that does not meet C. The curve C iscalled the generating curve, and the line l is called the rotation axis.The circles described by the rotation locus of the points of C arecalled the parallels of S, and the various positions of C are calledthe meridians of S. A surface of revolution can be parametrized


Figure 5.2. Surface of revolution.

naturally by the parameter on the curve C and the angle of rotationu about the axis.

Let ~x : I → R2 be a parametric curve in the xy-plane, with ~x(t) =(x(t), y(t)), and suppose we want to find the surface of revolution ofthis curve about either the x- or y-axis. About the y-axis, we take~X(t, u) = (x(t) cosu, y(t), x(t) sinu), and about the x-axis, we wouldtake ~X(t, u) = (x(t), y(t) cosu, y(t) sinu). Both of these options havedomains of I × [0, 2π]. (See Figure 5.2 for an example of a surface ofrevolution about the x-axis based on a curtate cycloid.)

Looking more closely at Example 5.1.3, one should note thatthough the function ~X(u, v) maps onto the sphere of radius R cen-tered at the origin, this function exposes two potential problems withthe intuitive approach. The first problem is that ~X is not injective.In particular, for any point (x, y, z) on this sphere with y = 0 andx > 0, we have

v = cos−1( zR

)and u = 0 or 2π,

and so there exist two preimages for such points. Worse still, ifwe consider the north and south poles (0, 0, 1) and (0, 0,−1), thepreimages are the points (u, 0) and (u, π), respectively, for all u ∈[0, 2π]. With the given parametrization, the arc on the sphere definedby u = 0 is the set of points with more than one preimage. Thislatter remark leads to the second problem because the points on thisarc have no particular geometric significance, and one would wish toavoid any formulation of a surface that confers special properties onsome points that are geometrically ordinary.


~XU

x

y

~X(U)

S

Figure 5.3. Coordinate patch.

Before addressing these and other issues necessary to be able to“do calculus” on a surface, we need to define the class of subsetsof R3 that one can even hope to study with differential geometry.The primary criteria for such a subset is that one must be able todescribe its points using continuous functions.

Definition 5.1.6. A subset S ⊆ R3 is called a parametrized surface iffor each p ∈ S, there exists an open set U ⊂ R2, an open neigh-borhood V of p in R3, and a continuous function ~X : U → R3 suchthat ~X(U) = V ∩ S. Each such ~X is called a parametrization of aneighborhood of S. We call a parametrized surface of class Cr if itcan be covered by parametrizations ~X of class Cr.

Note that in contrast to space curves, where we use intervalsas the domain for parametrizations, this definition uses open sets.Furthermore, this definition does not insist that S be the image ofa single parametrization but, rather that it be covered by images ofparametrizations.

For any such function ~X : U → V ∩ S, where V ∩ S is an openneighborhood of p in S, we call ~X a parametrization of the coordinatepatch V ∩S. (See Figure 5.3 for an illustration of the relation of thedomain U ⊂ R2 and the coordinate patch on the surface V ∩ S.)

Definition 5.1.6 states that, for every point p ∈ S, there is an openneighborhood of p on S that is the image of some vector function.

With the tools of curves in the plane or space, one can studyproperties of a surface S by considering various families of curves on


Figure 5.4. Coordinate lines on a torus.

the surface. If one wishes to study S near a point p, one could lookfor common properties of all the curves on S passing through p.

There are two natural classes of curves on any given parame-trized surface S that arise naturally. First, we can consider what aregenerically called the coordinate lines of the surface S. A coordinateline on S is the image of a space curve defined by fixing one of thevariables in a particular parametrization of a coordinate patch of S.Namely, if ~X : U → R3 parametrizes a patch of S, then a curve ofthe form

~γ1(u) = ~X(u, v0), respectively ~γ2(v) = ~X(u0, v),

is called a coordinate line of the variable u, and respectively forv. (See Figure 5.4 for an example of coordinate lines of a torusparametrized by ~X(u, v) = ((3 + cos v) cosu, (3 + cos v) sinu, sin v)for (u, v) ∈ [0, 2π]× [0, 2π].)

Example 5.1.7. With the surfaces of revolution described in Example5.1.5, the coordinate lines of t and u, respectively, are precisely theparallels and the meridians.

Another family of curves we might study are the slices of S, bywhich we mean the intersection of S with a family of parallel planes.As an example, Figure 5.5 shows the parametric surface

~X(u, v) = (cos v, sin 2v cosu, sin 2v sinu),


Figure 5.5. Slices of a surface of revolution.

with domain [0, 2π] × [0, π] along with two sets of slices. Thoughany set of parallel planes can produce a set of slices on S, we oftenconsider the intersection of S with planes of the form x = a, y = b,or z = c. In the figure on the left, we sliced the surface with planesx = a, which are perpendicular to the axis of revolution. These slicesresult in a set of circles. If we fix x = a, with −1 ≤ a ≤ 1, we getcos v = a, and there exists a value v0 ∈ [0, π] satisfying this equation.Then the slice of ~X(u, v) intersecting the x = a plane gives a circleparallel to the yz-plane, given by

(x, y, z) = (cos(v0), sin(2v0) cosu, sin(2v0) sinu).

On the other hand, if we consider planes perpendicular to the z-axis, that is planes of the form z = c, the z = c slice correspondsto the set of points (cos v, sin(2v) cosu) where (u, v) ∈ [0, 2π]× [0, π]satisfy sin(2v) sin(u) = c. This defines a curve implicitly. In thiscase, the best we can do is to solve for sinu = c/ sin(2v) and thenget a parametrization for the slices by

(cos v,±√

sin2(2v)− c2, c).

These are shown on the right in Figure 5.5.


Problems

5.1.1. Find a parametrization ~X for the cylinder (x, y, z) ∈ R3 |x2 + y2 =

1. Can the domain of this parametrization be chosen so that ~X isbijective onto the cylinder? Explain why or why not.

5.1.2. Prove that the following functions provide parametrizations of thehyperbolic paraboloid x2 − y2 = z. Provide appropriate domainsof definitions. Which ones are injective? Which ones are surjectiveonto the surface?

(a) ~X(u, v) = (v coshu, v sinhu, v2).

(b) ~X(u, v) = ((u+ v), (u− v), 4uv).

(c) ~X(u, v) = (uv, u(1− v), u2(2v − 1)).

5.1.3. Let ~α : I → R2 be a regular, plane curve, and let L be a line that doesnot intersect the image of ~α. Suppose that L goes through the point~p and has direction vector ~u so that ~p+ t~u is a parametrization of L.Find a formula for the parametrization of the surface of revolutionobtained by rotating ~α(I) about L.

5.1.4. Let ~α, ~β : I → R3 be two parametrized space curves with the samedomain. Define the secant surface by the parametrization ~X : I ×R→ R3 with

~X(t, u) = (1− u)~α(t) + u~β(t).

(a) Let ~α(t) = (2 cos t, sin t, 1) and ~β(t) = (cos t, 2 sin t,−1). Provethat every slice by a constant u coordinate is an ellipse andgive the eccentricity of the ellipse as a function of u. (Recallthat the eccentricity of an ellipse with half-axes a > b > 0 is

e =√

1− b2

a2 .)

(b) Now let ~α(t) = (cos t, sin t, 1) and ~β(t) = (− sin t, cos t,−1).Prove that every slice by a constant u coordinate is a circleand give the radius as a function of u.

5.1.5. Prove that on the unit sphere parametrized by ~X(θ, φ) = (cos θ sinφ,sin θ sinφ, cosφ), the curves given by

(θ, φ) = (ln t, 2 tan−1 t)

intersect every meridian with an angle of π4 . (Note that a meridian

is a curve such that θ = const. A curve on the sphere that intersectsthe meridians at a constant angle is called a loxodrome.)

5.2. Tangent Planes and Regular Surfaces 133

5.1.6. Consider a curve in the plane ~α(t) = (x(t), y(t)) and the surfaceof revolution obtained by rotating the image of ~α about the y-axis.This surface of revolution is parametrized by ~X(t, u) = (x(t) cos(u),y(t), x(t) sin(u)).

(a) Consider u-coordinate lines, i.e., the space curves

γ1(t) = (x(t) cos(u0), y(t), x(t) sin(u0)) ,

where u0 is fixed. Calculate the space curvature and torsion ofγ1.

(b) Repeat the above question with the t-coordinate lines, i.e., thespace curves γ2(u) = (x(t0) cos(u), y(t0), x(t0) sin(u)), where t0is fixed.

5.1.7. Consider the set of points S = (x, y, z) ∈ R3 |x4 + y4 + z4 = 1.Modify the usual parametrization for a sphere to find parametriza-tions that cover S.

5.2 Tangent Planes and Regular Surfaces

In the local theory of curves, we called a curve C regular at a point pif there is a parametrization ~x(t) of C near p = ~x(t0) such that ~x′(t0)exists and ~x′(t0) 6= ~0. In Section 1.2, we saw that this definition istantamount to requiring that

limt→t0

~x′(t)

‖~x′(t)‖

exists and, hence, that there exists a tangent line to C at p. Imitatingthis latter geometric property, we will eventually call a point p ∈ Sregular if one can define a tangent plane to p at S, but we must delaya precise definition until after we define the tangent plane to S at p.

Definition 5.2.1. Let S be a parametrized surface and p a point in S.Consider the set of space curves ~γ : (−ε, ε)→ R3 such that ~γ(0) = p,and the image of ~γ lies entirely in S. A tangent vector to S at p isany vector in R3 that can be expressed as ~γ′(0), where ~γ is such aspace curve.

From the view point of intuition, one often considers tangentvectors to S at p to be based at p. The set of tangent vectors,


~γ(t)

~γ′(0)

Figure 5.6. Double cone.

as a subset of R3, gives a sort of approximation of the behavior ofS at p. By rescaling the parameter t, one notices that the set oftangent vectors to S at a point p contains ~0 and is closed underscalar multiplication. Hence, the set of tangent vectors will be a(usually infinite) union of lines, and, as we shall see, it is often, butnot always, a plane.

Example 5.2.2. As an example of a set of tangent vectors that is nota plane, consider the cone with equation

x2 + y2 − z2 = 0,

with parametrization

~X(u, v) = (v cosu, v sinu, v),

for (u, v) ∈ [0, 2π] × R (see Figure 5.6). We note that ~X(u, 0) =(0, 0, 0) for all u ∈ [0, 2π] and consider the point p = ~X(u, 0) =(0, 0, 0). A curve on the cone through p has a parametric equationof the form ~γ(t) = ~X(u(t), v(t)), where v(0) = 0 and u is any real in[0, 2π]. For such a curve, using the multivariable chain rule, we have

~γ′(t) =∂ ~X

∂u

du

dt+∂ ~X

∂v

dv

dt= u′(t)

∂ ~X

∂u+ v′(t)

∂ ~X

∂v,

and since ~Xu(u, 0) = (0, 0, 0),

~γ′(0) = v′(0) (cosu, sinu, 1).


Since v′(0) and u can be any real values, we determine that the unionof the endpoints the set of tangent vectors to the cone at (0, 0, 0) isprecisely the cone itself.

Definition 5.2.3. Let S be a parametrized surface and p a point inS. If the set of tangent vectors to S at p forms a two-dimensionalsubspace of R3, we call this subspace the tangent space to S at p anddenote it by TpS. If TpS exists, we call the tangent plane the set ofpoints

p+ ~v |~v ∈ TpS.

We wish to find a condition that determines when the set oftangent vectors is a two-dimensional vector subspace.

Let p be a point on a surface S, and let ~X : U → R3 be aparametrization of a neighborhood V ∩ S of p for some U ⊆ R2.Suppose that p = ~X(u0, v0) and consider the coordinate lines ~γ1(t) =~X(u0 + t, v0) and ~γ2(t) = ~X(u0, v0 + t) through p. Both ~γ1 and ~γ2

lie on the surface and, by the chain rule,

~γ′1(0) =∂ ~X

∂u(u0, v0) and ~γ′2(0) =

∂ ~X

∂v(u0, v0).

Furthermore, for any curve ~γ(t) on the surface with ~γ(0) = p, wecan write

~γ(t) = ~X(u(t), v(t))

for some functions u(t) and v(t), with u(0) = u0 and v(0) = v0. Bythe chain rule,

~γ′(t) = u′(t)∂ ~X

∂u(u(t), v(t)) + v′(t)

∂ ~X

∂v(u(t), v(t)),

so at p,

~γ′(0) = u′(0)∂ ~X

∂u(u0, v0)+v′(0)

∂ ~X

∂v(u0, v0) = u′(0)~γ′1(0)+v′(0)~γ′2(0).

Thus, ~γ′(0) is a linear combination of ∂ ~X∂u (u0, v0) and ∂ ~X

∂v (u0, v0).

Definition 5.1.6 does not assume that the parametrization ~X isinjective. Therefore, the set of tangent vectors is a plane if and only


if the union of all linear subspaces

Span

(∂ ~X

∂u(u0, v0),

∂ ~X

∂v(u0, v0)

),

where ~X(u0, v0) = p is a plane. In particular, if there exists only one(u0, v0) such that ~X(u0, v0) = p, then a tangent plane exists at p if∂ ~X∂u (u0, v0) and ∂ ~X

∂v (u0, v0) are not collinear.

To simplify notations, one often uses the following abbreviatednotation for partial derivatives:

~Xu(u, v) =∂ ~X

∂u(u, v) and ~Xv(u, v) =

∂ ~X

∂v(u, v).

One then writes succinctly that if (u0, v0) = ~X−1(p), then S hasa tangent plane at p if and only if ~Xu(u0, v0) × ~Xv(u0, v0) 6= ~0.Furthermore, the tangent plane to S at p is the unique plane throughp with normal vector ~Xu(u0, v0) × ~Xv(u0, v0) 6= ~0. This leads us toa formula for the tangent plane.

Proposition 5.2.4. Let S be a surface parametrized near a point p by~X : U → R3. Suppose also that ~X is injective at p (i.e., p hasonly one preimage) and that p = ~X(u0, v0). Then the set of tan-gent vectors forms a two-dimensional subspace of R3 if and only if~Xu(u0, v0)× ~Xv(u0, v0) 6= ~0. In this case, the tangent plane to S atp satisfies the following equation for position vectors ~x = (x, y, z):

(~x− ~X(u0, v0)) · ( ~Xu(u0, v0)× ~Xv(u0, v0)) = 0. (5.1)

Through an abuse of language, even if ~X is not injective at pand ~Xu(u0, v0) × ~Xv(u0, v0) 6= 0 for some (u0, v0) ∈ U , we will callEquation (5.1) the equation of the tangent plane to ~X at q = (u0, v0).This is an abuse of language since if p = ~X(u0, v0) and S is the imageof ~X, then p might have more than one preimage and, therefore, theset of tangent vectors to S of p might not be a plane but be a unionof planes. Problem 5.2.9 gives an example of a parametrized surfacethat has points where the set of tangent vectors is the union of twoplanes.


Figure 5.7. A tangent plane.

Example 5.2.5. Consider the function ~X(u, v) = (u, v, 4 + 3u2 − u4 −2v2). We calculate

~Xu × ~Xv = (1, 0, 6u− 4u3)× (0, 1,−4v)

= (−6u+ 4u3, 4v, 1).

Figure 5.7 shows the surface along with the tangent plane at thepoint ~X(−1.4,−1). At this specific point, we have p = ~X(−1.4,−1) =(−1.4,−1, 4.0384) and ~Xu × ~Xv(−1.4,−1) = (−2.576,−4, 1). Since~X is in fact the graph of an injective two-variable function, by Propo-sition 5.2.4, the tangent plane to ~X at p is

− 2.576(x+ 1.4)− 4(y + 1) + (z − 4.0384) = 0

⇐⇒ − 2.576x− 4y + z = 11.6448.

Example 5.2.6. As a second example, consider the hyperboloid of onesheet given by the parametrization

~X(u, v) = (coshu cos v, coshu sin v, sinhu).

A calculation produces

~Xu × ~Xv = (− cosh2 u cos v,− cosh2 u sin v, coshu sinhu).


Figure 5.8. A tangent plane on a hyperboloid of one sheet.

Then at the point p = ~X(1, 0) = (cosh 1, 0, sinh 1), the tangent planeexists, and its equation is

− cosh2 1(x− cosh 1) + cosh 1 sinh 1(z − sinh 1) = 0 ⇐⇒ (cosh 1)x− (sinh 1)z = 1.

Figure 5.8 shows two perspectives of this hyperboloid along with itstangent plane at p. It is interesting to contrast this picture withFigure 5.7, in which the tangent plane does not intersect the surfaceexcept at p. The shape of the hyperboloid forces it to intersect itstangent plane at all points.

Using the notion of the differential of the parametrization ~X, wecan summarize Proposition 5.2.4 in another way that is more conve-nient when we discuss parametrized surfaces in higher dimensions.(See Section 5.2 in [24] for a longer explanation of the differential ofa function from Rm to Rn.)

If ~F is a function from an open set U ⊂ Rn to Rm, we write~F = (F1, F2, . . . , Fm)T and think of ~F as a column vector of functionsFi, each in n variables. For all ~a ∈ U , we say that F is differentiableat ~a if each Fi is differentiable at ~a. Furthermore, we define thedifferential of ~F at ~a, written d~F~a, as the linear transformation Rn →Rm that sends the jth standard vector ~ej to ∂ ~F

∂xj(~a). We denote by

[d~F~a] the matrix of d~F~a with respect to the standard bases of Rn and


Rm, which we can write explicitly as

[d~F~a

]=

(∂ ~F

∂x1

∂ ~F

∂x2· · · ∂ ~F

∂xn

)=

∂F1∂x1

∂F1∂x2

· · · ∂F1∂xn

∂F2∂x1

∂F2∂x2

· · · ∂F2∂xn

......

. . ....

∂Fm∂x1

∂Fm∂x2

· · · ∂Fm∂xn

. (5.2)

By a slight abuse of notation, we will sometimes write [d~F ] oreven d~F , without the subscript, to mean the matrix of functions with∂ ~F/∂xj as the jth column.

We can now restate Proposition 5.2.4 as follows.

Corollary 5.2.7. Let S be a surface, where p ∈ S, and suppose thatin an open neighborhood of p, the surface S is parametrized by ~X :U → R3. The tangent space (and tangent plane) to S at p existsif ~X−1(p) = q (a singleton set), the differential d ~Xq exists, and

d ~Xq has maximal rank. Furthermore, the tangent space is given by

TpS = Im(d ~Xq).

Proof: Note that with respect to the standard bases in R2 and R3,if q = (u0, v0), the matrix of d ~Xq is the 3× 2 matrix

[d ~Xq

]=

∂X1∂u (q) ∂X1

∂v (q)∂X2∂u (q) ∂X2

∂v (q)∂X3∂u (q) ∂X3

∂v (q)

, (5.3)

where we have written ~X(u, v) = (X1(u, v), X2(u, v), X3(u, v)).First note that for the tangent plane to exist, the partial deriva-

tives of ~X must exist, so the differential exists. From Proposition5.2.4, we know that the surface is a plane if ~X−1(p) is a singletonset, which we’ll denote by q ⊂ U , and ~Xu(q) × ~Xv(q) exists andis nonzero. This is equivalent to saying that ~Xu(q) and ~Xv(q) arelinearly independent, which means that d ~Xq has maximal rank.

The condition in Corollary 5.2.7 that requires d ~Xq to have max-imal rank occurs often enough in various contexts that we give it aname, as follows.


Definition 5.2.8. Let F be a function from an open set U ⊂ Rn to Rm.A point q ∈ U is called a critical point if dFq does not exist or doesnot have maximal rank, i.e., rank dFq < min(m,n). If q is a criticalpoint of F , we call F (q) ∈ Rm a critical value of F . If p ∈ Rm is nota critical value of F (even if p is not in the image of F ), then we callp a regular value of F .

Definition 5.2.9. Let S be a parametrized surface in R3. A point p ∈ Sis called a singular point of S if no tangent plane exists at p.

In light of Corollary 5.2.7 and Definition 5.2.8 a point p of asurface S is a singular point if every parametrization ~X : U → R3 ofan open neighborhood of p on S is a critical value of ~X or ~X−1(p)is not a singleton.

We have belabored the issue of whether or not one can definea tangent plane to S at a point p because, in an intuitive sense,if a tangent plane does not exist locally near p, then the surfaceS does not “resemble” a two-dimensional plane and thus does notlook like what we would expect in a surface. Just as with curves weintroduced the notion of a regular curve (i.e., a curve ~x : I → Rnsuch that ~x′(t) 6= ~0 for all t ∈ I), so with surfaces, we would liketo define a class of surfaces that is smooth enough so that we can“do calculus” on it. The points in the above discussion lead to thefollowing definition.

Definition 5.2.10. A subset S ⊆ R3 is a regular surface if for eachp ∈ S, there exists an open set U ⊆ R2, an open neighborhood V ofp in R3, and a surjective continuous function ~X : U → V ∩ S suchthat

1. ~X is continuously differentiable: if we write ~X(u, v) = (x(u, v),y(u, v), z(u, v)), then the functions x(u, v), y(u, v), and z(u, v)have continuous partial derivatives with respect to u and v;

2. ~X is a homeomorphism: ~X is continuous and has an inverse~X−1 : V ∩ S → U such that ~X−1 is continuous;

3. ~X satisfies the regularity condition: for each (u, v) ∈ U , thedifferential d ~X(u,v) : R2 → R3 is a one-to-one linear transfor-mation.


The parametrization ~X is called a system of coordinates (in aneighborhood) of p, and the neighborhood V ∩S of p in S is called acoordinate neighborhood . One should observe that a regular surfaceis defined as a subset of R3 and not as a function from a subset ofR2 to R3. However, one may view a regular surface as the union ofthe images of an appropriate set of systems of coordinates.

Definition 5.2.11. A regular surface is called of class Cr if all the sys-tems of coordinates ~X are of class Cr, i.e., all partial derivatives oforder up to r exist and are continuous. A regular surface is called ofclass C∞ if all its systems of coordinates ~X are such that all partialderivatives of all orders exist and are continuous.

Example 5.2.12. We can prove directly that the unit sphere, denotedS2, is a regular surface in a few different ways. We’ll use rectangularcoordinates first. Consider a point p = (x, y, z) ∈ S2, and let U =(u, v) |u2 + v2 < 1. If z > 0, then the mapping ~X(1) : U → R3

defined by (u, v,√

1− u2 − v2) is clearly a bijection between U andS2 ∩ (x, y, z)|z > 0. ~X(1) is also a homeomorphism because it

is continuous, and its inverse ~X −1(1) is simply the vertical projection

of the upper unit sphere onto R2 and since projection is a lineartransformation, it is continuous. Furthermore, ~X(1) satisfies all theconditions in Definition 5.2.10. We leave it to the reader to checkCondition 1, while for Condition 3, we calculate the differential

d ~X(1) =

1 00 1

− u√1− u2 − v2

− v√1− u2 − v2

.

For all q = (u, v) ∈ U , the entries in this matrix are well defined andproduce a matrix that defines a one-to-one linear transformation.Consequently, [d ~X(1)q] is of maximal rank.

The mapping ~X(1) : U → R3 so far only tells us that the openupper half of the sphere is a regular surface, but we wish to showthat the whole sphere is a regular surface.

For other points p = (x, y, z) ∈ S2, we use the following parame-trizations ~X(i) : U → R3 of coordinate patches around p:


~X(1)

~X(2)

~X(3)

~X(4)

~X(5)

~X(6)

Figure 5.9. Six coordinate patches on the sphere.

if z > 0, ~X(1)(u, v) = (u, v,√

1− u2 − v2), if z < 0, ~X(2)(u, v) = (u, v,−√

1− u2 − v2),

if y > 0, ~X(3)(u, v) = (u,√

1− u2 − v2, v), if y < 0, ~X(4)(u, v) = (u,−√

1− u2 − v2, v),

if x > 0, ~X(5)(u, v) = (√

1− u2 − v2, u, v), if x < 0, ~X(6)(u, v) = (−√

1− u2 − v2, u, v).

These six parametrizations give six coordinate patches such thatfor all p ∈ S2, p is in at least one of these coordinate patches (seeFigure 5.9 for an illustration of the six patches covering the sphere).Since the domain for each parametrization is open, all six patchesare necessary.

Example 5.2.13. Carefully using two colatitude–longitude parametr-izations, we get another way to show that the unit sphere S2 is aregular surface. Let H1 be the closed half-plane

H1 = (x, y, z) ∈ R3| x ≥ 0 and y = 0.

Let p ∈ S2 −H1, and let U = (0, 2π) × (0, π). Then ~X(1) : U → R3

defined by

~X(1)(u, v) = (cosu sin v, sinu sin v, cos v)


is a homeomorphism between U and S2 − H1 that satisfies the re-maining conditions in the definition of a regular surface. (The readershould check this claim.)

On the other hand, if p ∈ S2 −H2, where H2 is the closed half-plane

H2 = (x, y, z) ∈ R3| x ≤ 0 and z = 0,

then a homeomorphism between U → S2 −H2 is

~X(2)(u, v) = (− cosu sin v, cos v,− sinu sin v).

This also satisfies the differentiability and regularity conditions. Fur-thermore, (S2 −H1) ∪ (S2 −H2) = S2, so the two parametrizationsof coordinate patches cover the whole sphere.

The three conditions in Definition 5.2.10 each eliminate varioussituations we do not wish to let into the class of surfaces of inter-est in differential geometry. The condition that ~X be differentiableeliminates the possibility of corners or folds. A cube, for example, isnot a regular surface because for whatever parametrization is usedin the neighborhood of an edge where two faces meet, at least oneof the partial derivatives will not exist.

The second condition, that ~X be a homeomorphism, might ini-tially appear the least intuitive. Since we assume that in the neigh-borhood of each point p ∈ S there is a parametrization ~X : U →V ∩ S that is a bijection, we already eliminate a surface that inter-sects itself or degenerates to a curve. Figure 5.10 (see Problem 5.2.9for the parametrization) shows the image of a parametrized surface~X that intersects itself along a ray. The tangent surface to this sur-face at any point along this ray is the union of two planes and not asingle plane.

Requiring not only that each patch ~X : U → V ∩ S of a parame-trized surface be a bijection but also bicontinuous (i.e., be a home-omorphism) means that if ~X(q1) and ~X(q2) are arbitrarily close inR3, then q1 and q2 are arbitrarily close. This eliminates situationssimilar to that depicted in Figure 5.11, where an open strip is twistedback onto itself so that it does not intersect itself but that the dis-tance between one end and another part in the middle of the stripis 0.


Figure 5.10. Not a bijection.

We now explain why this is not a homeomorphism. Consider anopen disk U inside the open strip, as shown in the figure. Then ~X(U)is not open because a set U ′ is open in S if and only if U ′ = S ∩ Vfor some open set V in R3. But any open set V ⊂ R3 that contains~X(U) must contain other points of S, as shown. Thus, U open in thedomain does not imply that ~X(U) is open, which shows that ~X−1 isnot continuous and, hence, that ~X is not a homeomorphism.

~X

Figure 5.11. Not a homeomorphism.


The third condition of Definition 5.2.10 is necessary because,even with the first two conditions satisfied, it is still possible tocreate a parametrized surface ~X that has a point p such that thetangent space to the surface at p is not a plane but a line. (SeeProblem 5.2.12 for an example of a surface that satisfies the firsttwo conditions for a regular surface but not the third.)

From the above explanations of the criteria for a regular surface,it would seem hard to determine at the outset whether or not a givensurface is regular. However, the next two propositions show underwhat conditions function graphs and surfaces given as solutions toone equation are regular.

Proposition 5.2.14. Let U ⊂ R2 be open. Then if a function f :U → R is differentiable, the subset S = (x, y, x) ∈ R3 | (x, y, z) =(u, v, f(u, v)) is a regular surface.

Proof: The inverse function F−1 : S → U is the projection onto thexy-plane. F−1 is continuous, and since f is differentiable, F is alsocontinuous. Thus, F is a homeomorphism between U and S. Theregularity condition is clearly satisfied because

dF(u,v) =

1 00 1fu fv

is always one-to-one regardless of the values of fu and fv.

Another class of surfaces in R3 one studies arises as level surfacesof functions F : R3 → R, that is, as points that satisfy the equa-tion f(x, y, z) = a. The following proposition involves a little moreanalysis than that presented in most multivariable calculus courses.However, it shows where the terminology “regular surface” comesfrom in light of the notion of a regular value of multivariable func-tions presented in Definition 5.2.8. (Most problems in this textbookinvolve explicit parametrizations, but the proof of this propositionillustrates the value of the Implicit Function Theorem, usually intro-duced in a course on mathematical analysis.)

Proposition 5.2.15. Let f : U ⊂ R3 → R be a differentiable function,and let a ∈ R be a regular value of f , i.e., a real number such that


for all p ∈ f−1(a), dfp is not ~0. Then the surface defined by S =(x, y, z) ∈ R3 | f(x, y, z) = a is a regular surface. Furthermore, thegradient dfp = ~∇f(p) is normal to S at p.

Proof: Let p ∈ S. Since dfp 6= ~0, after perhaps relabeling the axes,we have fz(p) 6= 0. We consider the function G : R3 → R3 definedby G(x, y, z) = (x, y, f(x, y, z)) and notice that

dGp =

1 0 00 1 0fx fy fz

.

Consequently, det(dGp) = fz(p) 6= 0, so dGp is invertible. TheImplicit Function Theorem allows us to conclude that there ex-ists a neighborhood U ′ of p in U such that G is one-to-one on U ′,V = G(U ′) is open, and the inverse G−1 : V → U ′ is differen-tiable. However, since (u, v, w) = G(x, y, z) = (x, y, f(x, y, z)), wewill be able to write the inverse function as (x, y, z) = G−1(u, v, w) =(u, v, g(u, v, w)) for some differentiable function g : V → R.

Furthermore, if V ′ is the projection of V onto the uv-plane, thenh : V ′ → R defined by ha(u, v) = g(u, v, a) is differentiable and

G(f−1(a) ∩ U ′) = V ∩ (u, v, w) ∈ R3 |w = a ⇐⇒f−1(a) ∩ U ′ = (u, v, ha(u, v)) | (u, v) ∈ V ′.

Thus, f−1(a) ∩ U ′ is the graph of ha, and by Proposition 5.2.14,f−1(a) ∩ U ′ is a coordinate neighborhood of p. Thus, every pointp ∈ S has a coordinate neighborhood, and so f−1(a) is a regularsurface in R3.

To prove that dfp is normal to S at p, suppose that ~X(u, v) =(x(u, v), y(u, v), z(u, v)) parametrizes regularly a neighborhood of Saround p. Then, differentiating the defining equation f(x, y, z) = awith respect to u and v one obtains

∂f

∂x

∂x

∂u+∂f

∂y

∂y

∂u+∂f

∂z

∂z

∂u= 0,

∂f

∂x

∂x

∂v+∂f

∂y

∂y

∂v+∂f

∂z

∂v

∂u= 0.

Thus, dfp = (fx(p), fy(p), fz(p)) is perpendicular to ~Xu and ~Xv

at p.


(Note: See Section 6.4 in [24] for a statement of the ImplicitFunction Theorem and some examples.)

We point out that Definition 5.2.10 can be easily modified toprovide a definition of regularity for a surface S in Rn. The onlyrequired changes are that each parametrization of a neighborhood ofS be a continuous function ~X : R2 → Rn and that at each point p =~X(u, v), the differential d ~X(u,v) be an injective linear transformationR2 to Rn.

Problems

5.2.1. Consider the parametrized surface

~X(u, v) = ((v2+1) cosu, (v2+1) sinu, v), where (u, v) ∈ [0, 2π]× R.

Find the equation of the tangent plane to this surface at p = (2, 0, 1) =~X(0, 1).

5.2.2. Calculate the equation of the tangent plane to the torus

~X(u, v) = ((2 + cos v) cosu, (2 + cos v) sinu, sin v)

at the point (u, v) =(π6 ,

π3

).

5.2.3. Show that the equation of the tangent plane at (x0, y0, z0) at a reg-ular surface given by f(x, y, z) = 0, where 0 is a regular value of f ,is

fx(x0, y0, z0)(x−x0)+fy(x0, y0, z0)(y−y0)+fz(x0, y0, z0)(z−z0) = 0.

5.2.4. Determine the tangent planes to x2 + y2 − z2 = 1 at the points(x, y, 0) and show that they are all parallel to the z-axis.

5.2.5. Show that the cylinder (x, y, z) ∈ R3 |x2 + y2 = 1 is a regularsurface and find parametrizations whose coordinate neighborhoodscover it.

5.2.6. Show that the two-sheeted cone with equation x2 + y2 − z2 = 0 isnot a regular surface.

5.2.7. Surfaces of Revolution. Consider the surface of revolution S withparametrization ~X(u, v) = (f(v) cosu, f(v) sinu, g(v)), where (f(v),g(v)) is a regular plane curve (in the xz-plane). Prove that S isregular if and only if for all v in the domain of f , f(v) 6= 0 andg′(v) 6= 0 or f(v) 6= 0 and f ′(v) 6= 0.


Figure 5.12. Parametrization of Problem 5.2.12.

5.2.8. Let f(x, y, z) = (x+ 2y + 3z − 4)2.

(a) Locate the critical points and the critical values of f .

(b) For what values of c is the set f(x, y, z) = c a regular surface?

(c) Repeat (a) and (b) for the function f(x, y, z) = xy2z3.

5.2.9. Consider the surface parametrized by ~X(u, v) = (uv, u2−v2, u3−v3).

Prove that ~X intersects itself along the ray (x, 0, 0), with x ≥ 0. Findthe points (u, v) that map to this ray. Use these to prove that the

tangent space to ~X at any point on the open ray (x, 0, 0), with x > 0,is the union of two planes. (See Figure 5.10 for the plot of a portionof this surface near (0, 0, 0).)

5.2.10. Consider the parametrized ~X(u, v) = (u, v3,−v2). Prove that for

any point on the surface p = ~X(q) such that q = (u, 0), d ~Xq doesnot have maximal rank.

5.2.11. Consider ~X(u, v) = (v cosu, v sinu, e−v) for (u, v) ∈ [0, 2π]× [0,∞).

Find all points where ~X fails the regularity condition.

5.2.12. Show that the parametrized surface ~X(u, v) = (u3+v+u, u2+uv, v3)satisfies the first and second conditions of Definition 5.2.10 but doesnot satisfy the third at p = (0, 0, 0). (See Figure 5.12.)

5.2.13. Consider the hyperbolic paraboloid defined by S = (x, y, z) ∈R3 | z = x2 − y2. Check that the following provide systems ofcoordinates for parts of S.

(a) ~X(u, v) = (u+ v, u− v, 4uv) for (u, v) ∈ R2.

(b) ~X(u, v) = (u cosh v, u sinh v, u2) for (u, v) ∈ R2.


x

y

z

N p

π(p)

Figure 5.13. Stereographic projection.

5.2.14. Find a parametrization for the hyperboloid of two sheets x2 + y2 −z2 = 1.

5.2.15. Stereographic Projection. One way to define coordinates on the sur-face of the sphere S2 given by x2 + y2 + z2 = 1 is to use the stere-ographic projection of π : S2 − N → R2, where N = (0, 0, 1),defined as follows. Given any point p ∈ S2, the line (pN) intersectsthe xy-plane at exactly one point, which is the image of the func-tion π(p). If (x, y, z) are the coordinates for p in S2, let us writeπ(x, y, z) = (u, v) (see Figure 5.13).

(a) Prove that π(x, y, z) =(

x1−z ,

y1−z

).

(b) Prove that

π−1(u, v) =

(2u

u2 + v2 + 1,

2v

u2 + v2 + 1,u2 + v2 − 1

u2 + v2 + 1

).

(c) Prove that by using stereographic projections, it is possible tocover the sphere with two coordinate neighborhoods.

5.2.16. Let ~α : I → R2 be a plane curve and let ~β : I → R3 be the pullback of~α onto the unit sphere via the stereographic projection as describedin Problem 5.2.15. In other words, ~β(t) = π−1(~α(t)). Prove that

points on the locus of ~β where the torsion τ is zero correspond tovertices in ~α, that is points where the plane curvature κg of ~α satisfiesκ′g(t) = 0.


5.2.17. Cones. Let C be a regular planar curve parametrized by ~α : (a, b)→R3 lying in a plane P that does not contain the origin O. The coneΣ over C is the union of all the lines passing through O and p forall points p on C.

(a) Find a parametrization ~X of the cone Σ over C.

(b) Find the points where Σ is not regular.

5.2.18. Central Projection. The central projection used in cartography in-volves a function f : S2 → (0, 2π) × R defined geometrically asfollows. View (0, 2π) × R as a cylinder wrapping around the unitsphere S2 in such a way that the slit in the cylinder falls in the half-plane P with y = 0 and x ≥ 0. The central projection f sends apoint p ∈ S2 − P to f(p) on the cylinder by defining f(p) as theunique point on the cylinder on ray [O, p), where O = (0, 0, 0) is theorigin.

(a) Let (x, y, z) be the coordinates of p ∈ S2−P and call f(x, y, z) =(u, v). Find a formula for f .

(b) Calculate a formula for f−1.

(c) If one uses central projections for coordinate neighborhoods onthe sphere, how many are necessary to cover the sphere?

5.3 Change of Coordinates

A particular patch of a surface can be parametrized in a variety ofways. In particular, we could use a different coordinate system inR2 to describe the domain U of ~X.

Example 5.3.1 (Two Parametrizations of the Sphere). As an example, con-sider the following two parametrizations of the upper half of the unitsphere in R3:

1. ~X(u, v) = (u, v,√

1− u2 − v2), with (u, v) ∈ U where the do-main U is the closed unit disk, i.e., U = (u, v) ∈ R2 |u2 +v2 ≤1. To be more specific (if we needed to provide intervals for(u, v), say to calculate integrals), we could say

U = (u, v) ∈ R2 | −√

1− u2 ≤ v ≤√

1− u2 and −1 ≤ u ≤ 1.

2. ~Y (θ, ϕ) = (cos θ sinϕ, sin θ sinϕ, cosϕ), with (θ, ϕ) ∈ U ′ wherewe have U ′ = [0, 2π]× [0, π2 ].

5.3. Change of Coordinates 151

In this case, it’s relatively easy to verify that ~X and ~Y parametrizethe same surface patch by setting the change of coordinates (u, v) =F (θ, ϕ), with

F : U ′ −→ U

(θ, ϕ) 7−→ (cos θ sinϕ, sin θ sinϕ).

We notice that F (U ′) = U (i.e., F is a surjective function) and that,as vector functions, ~Y = ~X F .

The function F : R2 → R2 is not a bijection between U and U ′,but it is a bijection between

U2 = (u, v) ∈ U |u2 + v2 < 1 and v = 0 implies u < 0

andU ′2 = (0, 2π)×

(0,π

2

),

and the inverse function (θ, ϕ) = F−1(u, v) is given by

θ =

tan−1(vu

), if u > 0 and v ≥ 0,

π2 , if u = 0 and v > 0,

tan−1(vu

)+ π, if u < 0,

3π2 , if u = 0 and v < 0,

tan−1(vu

)+ 2π, if u > 0 and v ≤ 0.

andϕ = sin−1(

√u2 + v2).

It is tedious but not hard to verify that both F and F−1 are differ-entiable on the open domains U2 and U ′2. Figure 5.14 illustrates F .

We now consider a general change of coordinates F in R2. Con-sider two parametrized surfaces defined by ~X : U → R3 and ~Y =~X F : U ′ → R3, where F : U ′ → U for subsets U ′ and U in R2. Theabove example illustrates three progressively stringent conditions onthe change-of-coordinates function F . First, a sufficient conditionfor ~Y = ~X F and ~X to have the same image in R3 is for F to besurjective. Second, if one wishes for the coordinates on a surface tobe unique or if we are concerned with ~Y being a homeomorphism,


θ

ϕ

U ′2

π 2π

π2

F

v

1 u

Figure 5.14. Coordinate change by F .

we might wish to require that F be a bijection, in which case, ifF is already surjective, one merely must restrict the domain of Fto make it a bijection. Finally, requiring that both F and F−1 bedifferentiable adds a “smoothness” condition to how the coordinatestransform. This latter condition has a name.

Definition 5.3.2. Let U and V be subsets of Rn. We call a functionF : U → V a diffeomorphism if it is a bijection, and both F andF−1 are differentiable.

Proposition 5.3.3. Let U and V be subsets of Rn, and let F : U →V be a diffeomorphism. Suppose that q = F (q′). Then the lineartransformation dFq′ is invertible, and

dF−1q′ = d(F−1)q.

Proof: Suppose that F : Rm → Rn and G : Rn → Rs are differ-entiable, multivariable, vector-valued functions. (For what follows,one can also assume that these functions have small domains, justas long as the range of F is a subset of the domain of G.) Usingthe notion of the differential defined in Equation (5.2), it is an easyexercise to show that the formula for the chain rule on G F in amultivariable context can be written as

d(G F )a = dGF (a)dFa,

where we mean matrix multiplication.

5.3. Change of Coordinates 153

Applying this to the situation of this proposition, where F−1 F = idU , we get

d(F−1)F (q′) dFq′ = In,

where In is the n × n identity matrix. Furthermore, the same rea-soning holds with F F−1, and we conclude that dFq′ is invertibleand that (dFq′)

−1 = d(F−1)q.

Suppose again that we have parametrized surfaces ~X : U → R3

and ~Y = ~X F : U ′ → R3, where F : U ′ → U is a diffeomorphismbetween open subsets U ′ and U in R2. Let q′ ∈ U ′ and defineq = F (q′). Then, just as in the above proof, the chain rule in multiplevariables gives

d~Yq′ = d( ~X F )q′ = d ~Xq dFq′ .

If for the change of coordinates F we write

(u, v) = F (s, t) = (f(s, t), g(s, t)),

then the matrix of the differential of this coordinate change is

[dF ] =

∂u

∂s

∂u

∂t∂v

∂s

∂v

∂t

=

∂f

∂s

∂f

∂t∂g

∂s

∂g

∂t

.

With matrices, the chain rule is written as

[d~Y(s,t)] = [d ~XF (s,t)] ·

∂u

∂s

∂u

∂t∂v

∂s

∂v

∂t

. (5.4)

Writing ~X = (X1, X2, X3) and ~Y = (Y1, Y2, Y3) if necessary, we findthat Equation (5.4) is equivalent to the relations

∂~Y

∂s=∂ ~X

∂u

∂u

∂s+∂ ~X

∂v

∂v

∂s=∂u

∂s~Xu +

∂v

∂s~Xv,

∂~Y

∂t=∂ ~X

∂u

∂u

∂t+∂ ~X

∂v

∂v

∂t=∂u

∂t~Xu +

∂v

∂t~Xv.

(5.5)


Problems

5.3.1. Prove the claim in Example 5.3.1 that both F and F−1 are differ-entiable on the open domains U2 and U ′2.

5.3.2. Consider the change of coordinates from polar to Cartesian describedby (x, y) = F (r, θ) = (r cos θ, r sin θ).

(a) Calculate the matrix of the differential [dF ].

(b) Determine F−1 and calculate [dF−1].

(c) Verify Proposition 5.3.3 that [d(F−1)(x,y)] = [dF(r,θ)]−1.

5.3.3. Consider the coordinate change in two variables

F (s, t)→ (s2 − t2, 2st).

Let U = (s, t) | s2 + t2 ≤ 1 and t ≥ 0. Show that F (U) is the

whole unit disk. Prove that if ~X(x, y) = (x, y,√

1− x2 − y2), then~X F : U → R3 is a parametrization for the upper half of the unitsphere.

5.3.4. Consider the coordinate change in the previous problem but withdomain U = (s, t) | t > 0.

(a) Determine the set F (U) and show that F : U → F (U) is abijection.

(b) Determine the function F−1 explicitly.

(c) Show in what sense [d(F−1)] = [dF ]−1.

5.3.5. Consider spherical and cylindrical coordinate systems. Consider thecoordinate change function from spherical to cylindrical defined by

(r, θ, z) = F (ρ, θ, ϕ) = (ρ sinϕ, θ, ρ cosϕ).

(Note that ϕ is the angle down from the positive z-axis.) Show thatwith

U ′ = (0,∞)× (0, 2π)× (0, π)

U = (0,∞)× (0, 2π)× R

the coordinate change function F : U ′ → U is a diffeomorphism.

5.3.6. Prove that F (x, y) = (x3, y3) is a bijection from R2 to R2 but nota diffeomorphism. Prove also that F (x, y) = (x3 + x, y3 + y) is adiffeomorphism of R2 onto itself.

5.4. The Tangent Space and the Normal Vector 155

5.3.7. Prove that the function F : R2 → R2 defined by

F (x, y) =(x cos(x2 + y2)− y sin(x2 + y2), x sin(x2 + y2) + y cos(x2 + y2)

)is a diffeomorphism of R2 onto itself.

5.3.8. Find an example of a diffeomorphism of R2 onto an open square andshow why it is a diffeomorphism.

5.4 The Tangent Space and the Normal Vector

Let S be a regular surface parametrized in the neighborhood of apoint p by ~X : U → R3. Suppose that p = ~X(q). Since S is regular,the set of tangent vectors to S at p is a subspace TpS, and we saw in

Corollary 5.2.7 that TpS = Im(d ~Xq). The condition that S be regular

at p ensures that the columns of d ~Xq are linearly independent. Thus,

given the parametrization ~X of the neighborhood of p, the set ofvectors ~Xu(q), ~Xv(q) forms a basis of TpS.

Comparing surfaces to curves, recall that at every regular pointon a curve, the unit tangent vector to a curve at a point is invari-ant under a positively oriented reparametrization of the curve, andthe tangent line to the curve at a point is an entirely geometric ob-ject, completely unchanged under reparametrizations. Implicit inDefinition 5.2.1, the set of tangent vectors to a surface at a point,and hence the tangent space, is a geometric object, invariant underreparametrization.

In contrast, the change of basis formulas Equations (5.5) and(5.4) show that when ~Y = ~X F and if q = F (q′) and p = ~X(q), thesets of vectors ~Xu(q), ~Xv(q) and ~Ys(q′), ~Yt(q′) are not necessarilyequal. Consequently, though TpS is invariant under a reparametri-zation, the basis induced from the parametrization is not invariant.

Proposition 5.4.1. Let ~X : U → R3 be a parametrization of a co-ordinate patch on a regular surface S. Consider a diffeomorphismF : U ′ ⊂ R2 → U such that ~Y = ~X F is a regular parametrizationof an open subset of S. Set q = F (q′) and p = ~X(q). If ~a ∈ TpS hascoordinates

~a = a1~Ys + a2

~Yt = a1~Xu + a2

~Xv,


then (a1

a2

)= [dFq′ ]

−1

(a1

a2

)= [d(F−1)q]

(a1

a2

).

Proof: By the chain rule in Equations (5.5) and (5.4), we have

a1~Ys + a2

~Yt = a1

(∂u

∂s~Xu +

∂v

∂s~Xv

)+ a2

(∂u

∂t~Xu +

∂v

∂t~Xv

)=

(a1∂u

∂s+ a2

∂u

∂t

)~Xu +

(a1∂v

∂s+ a2

∂v

∂t

)~Xv.

Since ~a = a1~Xu + a2

~Xv and ~Xu, ~Xv is a basis of TpS, we deducethat (

a1

a2

)=

∂u

∂s

∂u

∂t∂v

∂s

∂v

∂t

(a1

a2

).

The last claim of the proposition follows in light of Proposition5.3.3.

Using the language of linear algebra, Proposition 5.4.1 showsthat [d(F−1)q] is the change of coordinates matrix between the basis

corresponding to the parametrization ~X and the basis correspondingto the reparametrization ~Y = ~X F .

At this point, we wish to restate Proposition 5.4.1 using notationsthat we will adopt later, in particular in Chapter 7. We label thecoordinates (u, v) as (x1, x2), and we label the coordinates (s, t) as(x1, x2). We write the diffeomorphism that relates the coordinatesas

(x1, x2) = F (x1, x2) and (x1, x2) = F−1(x1, x2).

Then according to Proposition 5.4.1, the change of coordinates onTpS satisfies (

a1

a2

)=

∂x1

∂x1

∂x1

∂x2

∂x2

∂x1

∂x2

∂x2

(a1

a2

). (5.6)

5.4. The Tangent Space and the Normal Vector 157

The matrix product in Equation (5.6) is often written in summationnotation as

ai =

2∑k=1

∂xi∂xk

ak.

According to Proposition 5.2.4, if ~X is a regular parametrizationof a neighborhood of p = ~X(q) of a regular surface S, the tangentplane has ~Xu(q)× ~Xv(q) as a normal vector. Since the tangent planeis invariant under reparametrization, this cross product must onlybe rescaled under reparametrization.

Proposition 5.4.2. Let ~X : U → R3 be a parametrization of a coor-dinate patch on a regular surface S. Consider a function F : U ′ ⊂R2 → U such that ~Y = ~X F is a regular parametrization of an opensubset of S. Then if q = F (q′) and p = ~X(q),

~Ys(q′)× ~Yt(q

′) = det(dFq′)( ~Xu(q)× ~Xv(q)).


We remind the reader that the determinant det(dFq′) of the dif-ferential matrix of the change of coordinate function is called theJacobian of F at q′. (The reader may recall that the Jacobian ap-pears in the change of variables formula for multiple integrals.)

Proposition 5.4.2 motivates us to define the unit normal to aregular surface S at point p as follows. If ~X is a parametrization ofa neighborhood of p and p = ~X(q), then the unit normal vector is

~N(q) =~Xu × ~Xv

‖ ~Xu × ~Xv‖(q). (5.7)

This vector is invariant under reparametrization except up to achange in sign, namely, the sign of the Jacobian of the correspondingchange of coordinates.

Definition 5.4.3. With the setup as in Proposition 5.4.2, we say that aregular reparametrization ~Y = ~X F is a positively (resp. negatively)oriented reparametrization if det(dFq′) > 0 (resp. det(dFq′) < 0) forall q′ ∈ U ′.


This unit normal vector, among other things, allows one to easilyanswer questions that involve the angle at which two surfaces inter-sect at a point. In other words, let S1 and S2 be two regular surfaces,and let p ∈ S1 ∩ S2. Then S1 intersects with S2 at p, with an angleθ, where ~N1 · ~N2 = cos θ. Of course, the particular parametrizationmight change the sign of one of the unit normal vectors, but theacute angle between Tp(S1) and Tp(S2) is well defined, regardless ofsigns.

Problems

5.4.1. Calculate the unit normal vector for function graphs ~X(u, v) =(u, v, f(u, v)).

5.4.2. Suppose that a coordinate neighborhood of a regular surface can beparametrized by

~X(u, v) = ~α(u) + ~β(v),

where ~α and ~β are regular parametrized curves defined over intervalsI and J , respectively. Prove that along coordinate lines (either u =u0 or v = v0) all the tangent planes to the surface are parallel to afixed line.

5.4.3. Let ~X : U → R3 be a parametrization of an open set of a regularsurface, and let ~p be a point in R3 that is not in S. Consider thefunction F : U → R defined by

F (u, v) = ‖ ~X(u, v)− ~p‖.

Prove that if ~q = (u0, v0) is a critical point of F , then the vector~X(~q)− ~p is normal to the surface S at ~X(~q).

5.4.4. Tangential Surfaces. Let ~α : I → R3 be a regular parametrizedcurve with curvature κ(t) 6= 0. Call the tangential surface to ~α theparametrized surface

~X(t, u) = ~α(t) + u~α′(t), with t ∈ I and u 6= 0.

Prove that for any fixed t0 ∈ I, along any curve ~X(t0, u), the tangentplanes are all equal.

5.4.5. Tubes. Let ~α : I → R3 be a regular parametrized curve. Let r be apositive real constant. Define the tube of radius r around ~α as theparametrized surface

~X(t, u) = ~α(t)+r(cosu)~P (t)+r(sinu) ~B(t), with (t, u) ∈ I× [0, 2π].

5.5. Orientable Surfaces 159

(a) Prove that a necessary (though not sufficient) condition for ~Xto be a regular surface is that

r < 1/(maxt∈I

κ(t)).

(b) Show that when ~X is regular, the unit normal vector is

~N(t, u) = − cosu~P (t)− sinu~B(t).


5.4.7. Let f and g be real functions such that f(v) > 0 and g′(v) 6= 0.Consider a parametrized surface given by

~X(u, v) = (f(v) cosu, f(v) sinu, g(v)) .

Show that all the normal lines to this surface pass through the z-axis.[Hint: See Problem 5.2.7.]

5.4.8. Two regular surfaces S1 and S2 intersect transversely if for all p ∈S1 ∩ S2, Tp(S1) 6= Tp(S2). Prove that if S1 and S2 intersect trans-versely, then the set S1 ∩ S2 is a disjoint union of regular curves.

5.5 Orientable Surfaces

Let S be a regular surface. Each point p of S has a coordinate neigh-borhood Vα ⊂ S parametrized by some vector function ~X : Uα → R3

satisfying the differentiability, homeomorphism, and regularity con-ditions of Definition 5.2.10. Thus S is covered by a collection ofcoordinate patches, and if S is compact, then we can cover S by afinite such collection.

The concept of an orientable surface encapsulates the notion ofbeing able to define an inside and an outside to the surface. At anypoint p of S, there are two unit normal vectors to the surface. If itmakes sense to distinguish between “two sides” of the surface, thenspecifying the unit normal ~Np at p eliminates all the options for all

the other points on the surface and uniquely defines ~Np′ for all otherpoints of S.

Definition 5.5.1. Let S be a regular surface in R3. We say that S isorientable if it can be covered by a collection of coordinate patches


Ri, with i ∈ I, given as the images of parametrizations ~X(i) : Ui →R3, where Ui ⊂ R2, such that if

~N(i)(q) =

∂ ~X(i)

∂u×∂ ~X(i)

∂v∥∥∥∥∥∂ ~X(i)

∂u×∂ ~X(i)

∂v

∥∥∥∥∥(q),

then ~N(i)(q) = ~N(j)(q) for all q ∈ Ui ∩ Uj , for all i, j ∈ I. If for thesurface S it is impossible to find such a covering by parametrizations,then we call S nonorientable.

By Proposition 5.4.2, we can provide an alternative formulationof this definition.

Proposition 5.5.2. A regular surface S is orientable if and only if it ispossible to cover it with coordinate patches Ri, given as the imagesof parametrizations ~X(i) : Ui → R3, such that if ~X(j) = ~X(i) F fora change of coordinates F : Ui ∩ Uj → Ui ∩ Uj, then det(dFq) > 0for all q ∈ Ui ∩ Uj.

Example 5.5.3. The commonly given example of a nonorientable sur-face in R3 is the Mobius strip M . Intuitively, the Mobius strip is asurface obtained by taking a long and narrow strip of paper and glu-ing the short ends together as though to make a cylinder but puttingone twist in the strip before gluing (see Figure 5.15).

Figure 5.15. Mobius strip.


By looking at Figure 5.15, one can imagine a normal vector point-ing outward on the surface of the Mobius strip, but if one followsthe direction of this normal vector around on the surface, when itcomes back around, it will be pointing inward this time instead ofoutward.

We wish to be more specific in this case. One can parametrizemost (an open dense subset) of the Mobius strip by

~X(u, v) =((

2− v sinu

2

)sinu,

(2− v sin

u

2

)cosu, v cos

u

2

),

with (u, v) ∈ (0, 2π)× (−1, 1). This coordinate neighborhood omitsthe boundary of the closed Mobius strip as well as the coordinateline with u = 0.

Let p be the point on M given by limu→0~X(u, 1

2

)= (0, 2, 1

2).To obtain the same point p as a limit with u → 2π, we must havea different value for v, namely, p = limu→2π

~X(u,−1

2

). With some

calculations, one can find that

~Xu × ~Xv =(−v

2cosu−

(2− v sin

u

2

)sinu cos

u

2

)~i

+(v

2sinu−

(2− v sin

u

2

)cosu cos

u

2

)~j

−((

2− v sinu

2

)sin

u

2

)~k

and

‖ ~Xu × ~Xv‖2 =v2

4+(

2− v sinu

2

)2.

From this, it is not hard to show that

limu→0

~N

(u,

1

2

)=

(− 1√

65,− 8√

65, 0

)and lim

u→2π~N

(u,−1

2

)=

(1√65,

8√65, 0

).

Consequently, no collection of systems of coordinates of M that in-cludes ~X can satisfy the conditions of orientability. However, thespecific system of coordinates ~X is not the problem. Using any col-lection of coordinate patches to cover M , as soon as one tries to “goall the way around” M , there are two overlapping coordinate patchesfor which the conditions of Definition 5.5.1 fail.


The Mobius strip has a boundary at v = ±1 in the above pa-rametrization. Then there are many examples of nonorientable reg-ular surfaces that have a boundary. For example, one can attacha “handle” to the Mobius strip, a tube connecting one part of thestrip to another. However, for reasons that can only be explainedby theorems in topology, there is not “enough room” in R3 to fita closed nonorientable surface that does not have a boundary anddoes not intersect itself. However, because there is “more room” inhigher dimensions, there exist many other closed nonorientable sur-faces without boundary in R4 that do not intersect themselves. SeeChapter 9.2 for a few examples.

Any particular choice of unit normal vector at each point on aregular surface defines a function n : S → S2, where S2 is the two-dimensional sphere. (The notation Sk refers to the k-dimensionalsphere. From a topological perspective, the dimension of the ambi-ent space is irrelevant, but in basic differential geometry, one oftenpictures Sk as the unit k-dimensional sphere in Rk+1.) We can nowstate a new criterion for orientability.

Proposition 5.5.4. Let S be a regular surface in R3. S is orientable ifand only if there exists a continuous function n : S → S2 such thatn(p) is normal to S at p for all p ∈ S.

Proof: (⇒) Suppose that S is a regular orientable surface. Supposethat S is covered by a collection of coordinate patches Ri, with i ∈I, given as the images of parametrizations ~X(i) : Ui → R3 thatsatisfy the condition of Definition 5.5.1. Suppose that p is in thecoordinate patch Rk on S and p = ~X(k)(q). Define n : S → S2 as

n(p) = ~N(k)(q). The criteria for Definition 5.5.1 ensure that n iswell defined, regardless of the chosen coordinate patch Rk. Since nis continuous over each Ri and the collection Rii∈I cover, S, n iscontinuous.

(⇐) Let n : S → S2 be a continuous function such that n(p)is normal to S at p for all p ∈ S. Let Rii∈I be a collection ofcoordinate patches of S parametrized by ~X(i) : Ui → R3 that coversS and satisfies the definition for a regular surface. Over each open


set Ri, the function

~N(i)(u, v) =

∂ ~X(i)

∂u×∂ ~X(i)

∂v∥∥∥∂ ~X(i)

∂u×∂ ~X(i)

∂v

∥∥∥is defined and continuous. Now for all i ∈ I, if ~X(i)(q) = p for some

p ∈ Ri, the normal vector ~N(i)(q) may be equal to n(p) or −n(p).

Furthermore, since ~Ni and n are both continuous functions and

‖n(p)− (−n(p))‖ = 2,

we deduce that ~N(i)(q) = n ~X(i)(q) or ~N(i)(q) = −n ~X(i)(q) for allq ∈ Ui.

Define a new collection of coordinate patches as follows: For alli ∈ I, if ~N(i) = n ~X(i) over Ui, then set ~Yi = ~Xi. If ~N(i) = −n ~X(i)

over Ui, then define ~Y(i)(u, v) = ~X(i)(−u, v), with the domain of ~Y(i)

modified accordingly. Then

~N~Y(i)= n ~Y(i)

for all i ∈ I, and thus the collection of parametrizations ~Y(i)i∈Isatisfies the Definition 5.5.1 for orientability.

Definition 5.5.5. Let S be an orientable regular surface in R3. A choiceof continuous function n : S → S2 of unit normal vectors on S iscalled an orientation on S. An orientable regular surface equippedwith a choice of such a function n is called an oriented surface. If Sis an oriented regular surface, a pair of vectors (~v, ~w) of the tangentplane TpS is called a positively oriented basis if the vectors form abasis and

(~v × ~w) · n(p) > 0.

A parametrization ~X : U → R3 of an open set of S is also calledpositively oriented if the ordered pair ( ~Xu, ~Xv) forms a positivelyoriented basis of the tangent plane of S at ~X(u, v) for all (u, v) ∈ U .


If we consider surfaces in Rk, the cross product is not available,so Definition 5.5.1 no longer applies. However, Proposition 5.5.2 hasno explicit dependency on the dimension of the ambient space andtherefore is taken as the definition of orientability for regular surfacesin Rk.

In diagrams depicting orientable regular surfaces, if the regularsurface S is in R3, one can often sketch the surface and indicate aunit normal vector at a point on the surface. Because the surfaceis orientable, this choice of unit normal vector uniquely determinesthe unit normal vectors at all points on the surface. However, in adiagram that depicts a neighborhood of a regular surface S in Rk,where k is not necessarily 3, we indicate the orientation of S withan oriented loop. The orientation of the loop indicates that in thisdiagram, any parametrization of S is such that one must sweep thatplane in this direction so that the angle between ~Xu and ~Xv is in theinterval (0, π). See Figure 5.16.

~Xu

~Xv

~Xu

~Xv

Figure 5.16. Orientation diagram.

Problems

5.5.1. Supply the details for Example 5.5.3.

5.5.2. Let S be a regular surface in R3 given as the set of solutions to theequation F (x, y, z) = a, where F : U ⊂ R3 → R is differentiable anda is a regular value of F . Prove that S is orientable.

CHAPTER 6

The First and Second

Fundamental Forms

Recall that the local geometry of space curves is completely deter-mined by two geometric invariants: the curvature and the torsion.Similarly, as we shall see, the local geometry of a regular surface Sin R3 is determined by the first and second fundamental forms.

The value of restricting attention to regular surfaces is that atall points on a regular surface, there is an open neighborhood reg-ularly homeomorphic to R2 via a parametrization ~X. Thus, at apoint p ∈ S, with p = ~X(q), the differential d ~Xq provides a naturalisomorphism between R2 and TpS. Whenever we consider vectors onS based at the point p, we must consider them as elements of TpS,and we can “do geometry” locally on S by identifying TpS with R2.

The first fundamental form establishes a natural direct producton TpS that ultimately leads to formulas for length of vectors in TpS,arc length, angle between vectors in TpS, and area formulas on S.The second fundamental form provides a measurement of how thenormal vector changes as one moves over the surface S, thereby, inan intuitive sense, describing how S sits in R3.

6.1 The First Fundamental Form

Definition 6.1.1. Let S be a regular surface and p ∈ S. The firstfundamental form Ip(·, ·) is the restriction of the usual dot product

in R3 to the tangent plane TpS. Namely, for ~a,~b in TpS = d ~Xq(R2),

Ip(~a,~b) = ~a ·~b.

Note that for each point p ∈ S, the first fundamental form Ip(·, ·)is defined only on the tangent space. There exists a unique matrix

165

166 6. The First and Second Fundamental Forms

that represents Ip(·, ·) with respect to the standard basis on TpS, butit is important to remember that this matrix is a matrix of functionswith components depending on the point p ∈ S. For any other pointp2 ∈ S, the first fundamental form is still defined the same way, butthe corresponding matrix is most likely different.

This definition draws on the ambient space R3 for an inner prod-uct on each tangent plane. However, if around p the surface S canbe parametrized by ~X, there is a natural basis on TpS, namely ~Xu

and ~Xv. We wish to express the inner product Ip(·, ·) in terms ofthis basis on S. Let

~a = a1~Xu(q) + a2

~Xv(q) and ~b = b1 ~Xu(q) + b2 ~Xv(q).

Then we calculate that

~a ·~b = a1b1 ~Xu(q) · ~Xu(q) + a1b2 ~Xu(q) · ~Xv(q)

+ a2b1 ~Xu(q) · ~Xv(q) + a2b2 ~Xv(q) · ~Xv(q).

This proves the following proposition.

Proposition 6.1.2. Let Ip(·, ·) : TpS2 → R be the first fundamental

form at a point p on a regular surface S. Given a regular parametri-zation ~X : U → R3 of a neighborhood of p, the matrix associated withthe first fundamental form Ip(·, ·) with respect to the basis ~Xu, ~Xv is

g =

(g11 g12

g21 g22

),

where

g11 = ~Xu · ~Xu and g12 = ~Xu · ~Xv,

g21 = ~Xv · ~Xu and g22 = ~Xv · ~Xv,

evaluated at (u0, v0) when p = ~X(u0, v0).

The quantity g is a matrix of real functions from the open domainU ⊂ R2. With this notation, if p = ~X(q), then the first fundamentalform Ip(·, ·) at the point p can be expressed as the bilinear form

Ip(~a,~b) = ~aT g(q)~b =(a1 a2

)g(q)

(b1b2

)for all ~a,~b ∈ TpS.

6.1. The First Fundamental Form 167

One should remark immediately that for any differentiable func-tion ~X : U → R3, one has ~Xu · ~Xv = ~Xv · ~Xu. Thus, g12 = g21 andthe matrix g is symmetric.

Example 6.1.3 (The xy-Plane). As the simplest possible example, con-sider the xy-plane. It is a regular surface parametrized by a singlesystem of coordinates ~X(u, v) = (u, v, 0). Obviously, we obtain

g =

(1 00 1

).

This should have been obvious from the definition of the first fun-damental form. The xy-plane is its own tangent space for all p, andthe parametrization ~X induces the basis

~Xu, ~Xv = (1, 0, 0), (0, 1, 0).

Therefore, for any ~v =(v1v2

)and ~w =

(w1

w2

), with coordinates given in

terms of the standard basis, we have

~v · ~w = v1w1 + v2w2 =(v1 v2

)(1 00 1

)(w1

w2

).

Example 6.1.4 (Cylinder). Consider the right circular cylinder param-etrized by ~X(u, v) = (cosu, sinu, v), with (u, v) ∈ (0, 2π) × R. Wecalculate

~Xu = (− sinu, cosu, 0) and ~Xv = (0, 0, 1),

and thus,

g =

(1 00 1

).

Obviously a cylinder and the xy-plane are not the same surface.As we will see later, a plane and a cylinder share many properties.However, the second fundamental form, which carries informationabout how the normal vector behaves on the cylinder, will distinguishbetween the cylinder and the xy-plane.


x

y

z

~Xv

~Xu

Figure 6.1. Coordinate basis on the sphere.

Example 6.1.5 (Spheres). Consider the longitude–colatitude parame-trization on the sphere ~X(u, v) = (cosu sin v, sinu sin v, cos v), with(u, v) ∈ (0, 2π)× (0, π). The first derivatives of ~X are

~Xu = (− sinu sin v, cosu sin v, 0),

~Xv = (cosu cos v, sinu cos v,− sin v).

We deduce that for this parametrization

g =

(g11 g21

g12 g22

)=

(sin2 v 0

0 1

).

Figure 6.1 shows the sphere with this parametrization along withthe basis ~Xu, ~Xv at the point ~X(π/2, π/4).

Some classical differential geometry texts use the letters E = g11,F = g12, G = g22. The classical notation was replaced by the “tensornotation” gij as the notion of a tensor became more prevalent indifferential geometry. A discussion of tensor notation is provided inthe Appendix A.1 since it is not essential in this book.

As a first application of how the first fundamental form allowsone to do geometry on a regular surface, consider the problem ofcalculating the arc length of a curve on the surface. Let S be aregular surface with a coordinate neighborhood parametrized by ~X :


U → R3, where U ⊂ R2 is open. Consider a curve on S givenby ~γ(t) = ~X(u(t), v(t)), where (u(t), v(t)) = ~α(t) is a differentiableparametrized plane curve in the domain U . The arc length formulaover the interval [t0, t] is

s(t) =

∫ t

t0

‖~γ′(τ)‖ dτ.

After some reworking, we can rewrite the formula as

s(t) =

∫ t

t0

√g11(u′(τ))2 + 2g12u′(τ)v′(τ) + g22(v′(τ))2 dτ (6.1)

or, more explicitly,

s(t) =

∫ t

t0

√g11(u(τ), v(τ))(u′(τ))2 + 2g12(u(τ), v(τ))u′(t)v′(τ) + g22(u(τ), v(τ))(v′(τ))2 dτ .

Using the first fundamental form, we can rewrite the arc length for-mula as

s(t) =

∫ t

t0

√I~γ(t) (~α′(τ), ~α′(τ)) dτ . (6.2)

The reader should note that this is an abuse of notation since, forall p ∈ S, Ip(·, ·) is a bilinear form on the tangent space TpS, but~α′(t) is a vector function in R2. However, the justification behindthis notation is that for all t, the coordinates of ~α′(t) in the stan-dard basis of R2 are precisely the coordinates of ~γ′(t) in the basis ~Xu, ~Xv based at the point ~γ(t). (To be more precise, one couldwrite Iγ(t)([~α

′(t)], [~α′(t)]) but this notation can become rather bur-densome.)

In essence, Equation (6.2) relates the geometry in the particularcoordinate neighborhood of S to the plane geometry in the tangentplane TpS. More precisely, while doing geometry in U , by usingIp(·, ·) instead of the usual dot product in R2, one obtains informa-tion about what happens on the regular surface S as opposed tosimply what happens on the tangent plane TpS. This approach doesnot work only for calculating arc length.

Consider two plane curves ~α(t) and ~β(t) in the domain of ~X,with ~α(t0) = ~β(t0) = q. Also let ~γ = ~X ~α and ~δ = ~X ~β be the


corresponding curves on the regular surface S. Then at the pointp = ~X(q), the curves ~γ(t) and ~δ(t) form an angle of θ, with

cos θ =~γ′(t0) · ~δ′(t0)

‖~γ′(t0)‖ ‖~δ′(t0)‖=

Ip(~α′(t0), ~β′(t0))√

Ip(~α′(t0), ~α′(t0))√Ip(~β′(t0), ~β′(t0))

.

(6.3)(Again, the comment following Equation (6.2) applies here as well.)

In the basis ~Xu, ~Xv, the vectors ~Xu and ~Xv have the coordi-nates (1, 0) and (0, 1), respectively. Therefore, Equation (6.3) im-plies that the angle ϕ of the coordinate curves of a parametrization~X(u, v) is given by

cosϕ =~Xu · ~Xv

‖ ~Xu‖ ‖ ~Xv‖=

Ip((1, 0), (0, 1))√Ip((1, 0), (1, 0))

√Ip((0, 1), (0, 1))

=g12√g11g22

.

Thus, all the coordinate curves of a parametrization are orthogonalif and only if g12(u, v) = 0 for all (u, v) in the domain of ~X. If ~Xsatisfies this property, we call ~X an orthogonal parametrization.

Example 6.1.6 (Loxodromes). Consider the longitude–colatitude para-metrization of the sphere, given in Example 6.1.5. Recall that ameridian of the sphere is any curve on the sphere with u fixed andthat the coefficients for the first fundamental form are

g11 = sin2 v, g12 = g21 = 0, g22 = 1.

A loxodrome on the sphere is a curve that makes a constantangle β with every meridian. Let ~α(t) = (u(t), v(t)) be a curve inthe domain (0, 2π) × (0, π) and ~γ(t) = ~X(~α(t)). If ~γ(t) makes thesame angle β with all the meridians, then by Equation (6.3),

cosβ =

(u′ v′

)(sin2 v 00 1

)(01

)√

(u′)2 sin2 v + (v′)2√

1=

v′√(u′)2 sin2 v + (v′)2

.


Then

(u′)2 sin2 v cos2 β + (v′)2 cos2 β = (v′)2

⇐⇒(v′)2 sin2 β = (u′)2 sin2 v cos2 β

⇐⇒(u′)2 sin2 v = (v′)2 tan2 β

=⇒± u′ cotβ =v′

sin v,

where the ± makes sense in that a loxodrome can travel either to-ward the north pole or toward the south pole. Now integrating bothsides of the last equation with respect to t and then performing asubstitution, one obtains∫

±u′ cotβ dt =

∫v′

sin vdt⇐⇒

∫± cotβ du =

∫1

sin vdv

⇐⇒ (± cotβ)u+ C = ln tan(v

2

),

where we have chosen a form of antiderivative of 1sin v that suits our

calculations best. This establishes an equation between u and v thatany loxodrome must satisfy. From this, one can deduce the followingparametrization for the loxodrome:

(u(t), v(t)) =((± tanβ)(ln t− C), 2 tan−1 t

).

(Figure 6.2 was plotted using tanβ = 4 and C = 0.)

Figure 6.2. Loxodrome on the sphere.


Not only does the first fundamental form allow us to talk aboutlengths of curves and angles between curves on a regular surface, butit also provides a way to calculate the area of a region on the regularsurface.

Proposition 6.1.7. Let ~X : U ⊂ R2 → S be the parametrization for acoordinate neighborhood of a regular surface. Then if Q is a compactsubset of U and R = ~X(Q) is a region of S, then the area of R isgiven by

A(R) =

∫∫Q

√det(g) du dv. (6.4)

Proof: Recall the formula for surface area introduced in multivari-able calculus that gives the surface area of the region R on S as

A(R) =

∫∫R

dA =

∫∫Q

‖ ~Xu × ~Xv‖ du dv. (6.5)

(See [25, Section 7.4] or [30, Section 17.7] for an explanation of Equa-tion (6.5).)

By Problem 3.1.6, for any two vectors ~v and ~w in R3, the followingidentity holds:

(~v × ~w) · (~v × ~w) = (~v · ~v)(~w · ~w)− (~v · ~w)2.

Therefore,

‖ ~Xu× ~Xv‖2 = ( ~Xu · ~Xu)( ~Xv · ~Xv)− ( ~Xu · ~Xv)2 = g11g22− g2

12, (6.6)

and the proposition follows because g12 = g21.

Proposition 6.1.7 is interesting in itself as it leads to another prop-erty of the first fundamental form. It is obvious from the definition ofthe first fundamental form that g11(u, v) ≥ 0 and g22(u, v) ≥ 0 for all(u, v) ∈ U . However, since ‖ ~Xu× ~Xv‖2 = det(g) and since ~Xu× ~Xv isnever ~0 on a regular surface, we deduce that det(g) = g11g22−g12 > 0for all (u, v) ∈ U . In other words, the matrix of functions g is alwaysa positive definite matrix.


S

R

Figure 6.3. Surface area patch for Example 6.1.8.

Example 6.1.8. Consider the regular surface of revoluation S param-etrized by ~X(u, v) = (v cosu, v sinu, ln v), with (u, v) ∈ [0, 2π] ×(0,∞) = U . (From a practical standpoint, it is not so importantthat U is not an open subset of R2, but it should be understood thatS may require two coordinate neighborhoods to satisfy the criteriafor a regular surface. See Figure 6.3.) Let Q = [0, 2π] × [1, 2], andlet’s calculate the area of R = ~X(Q).

It’s not hard to show that

(gij) =

(v2 00 1 + 1

v2

).

Then Proposition 6.1.7 gives

A(R) =

∫ 2π

0

∫ 2

1

√v2 + 1 dv du = 2π

[1

2v√v2 + 1 +

1

2sinh−1 v

]2

1

= π

(2√

5−√

2 + ln

(2 +√

5

1 +√

2

)).

As we have done for many quantities on curves and surfaces, wewish to see how the functions in the first fundamental form changeunder a coordinate transformation. In order to avoid confusion with


various coordinate systems, we introduce some notations that willbecome more common in future topics. Suppose that an open subsetV of a regular surface S can be parametrized by two different sets ofcoordinates (x1, x2) and (x1, x2). If we write ~X(x1, x2) and ~Y (x1, x2)for the specific parametrizations, then the function F = ~Y −1 ~X isa diffeomorphism between two open subsets of R2 and

(x1, x2) = F (x1, x2).

For convenience, we will often simply writex1 = x1(x1, x2),

x2 = x2(x1, x2)

to indicate the dependency of one set of variables on the other. Viceversa, we can write

(x1, x2) = F−1(x1, x2) or simply

x1 = x1(x1, x2),

x2 = x2(x1, x2).

Now ~Y (x1, x2) = ~X(x1(x1, x2), x2(x1, x2)) for the respective parame-trizations. Let gij be the coefficient functions in the first fundamen-tal form for S parametrized by x1, x2, and let gij be the coefficientfunctions for S parametrized by x1, x2. Then by the chain rule,

gij = ~Yxi · ~Yxj =

(~Xx1

∂x1

∂xi+ ~Xx2

∂x2

∂xi

)·(~Xx1

∂x1

∂xj+ ~Xx2

∂x2

∂xj

).

After some reorganization, we find

gij =

2∑k=1

2∑l=1

∂xk∂xi

∂xl∂xj

gkl. (6.7)

This type of transformation of coordinates is our first (nontriv-ial) encounter with tensors. Because the gij functions satisfy thisparticular identity under a change of coordinates, we call the ma-trix of functions g = (gij) the components of a tensor of type (0, 2).The collection of the quantities gij defined at each point of p ∈ S isreferred to as the components of metric tensor of S.


The value of the first fundamental form is that, given the metrictensor, one can use the appropriate formulas for arc length, for anglesbetween curves, and for areas of regions without knowing the specificparametrization of the surface. An additional benefit of the firstfundamental form and the metric tensor is that one can still usethem to study the geometry of surfaces in higher dimensions, as wewill see in Section 9.2.

Because of Equation (6.2), a few mathematicians and most physi-cists use an alternative notation for the metric tensor. These authorsdescribe the metric tensor by saying that the “line element” in a co-ordinate patch is given by

ds2 = g11(u, v) du2 + 2g12(u, v) du dv + g22(u, v) dv2.

This notation has the advantage that one can write it concisely ona single line instead of in matrix format. However, one should notforget that it not only leads to Equation (6.2) for arc length of acurve on a surface but that it also leads to Equation (6.3) for anglesbetween curves on the surface and the formula in Proposition 9.2.2that gives the area of regions on the surface.

Example 6.1.9 (Intrinsic Geometry). If we are given the metric coeffi-cients gij(u, v) of a surface as functions of u and v, we do not knowthe actual equations of the surface in space in order to computequantities like the length of a curve or the area of a coordinate patchin the domain of the metric coefficients. Such computations are anexample of “intrinsic geometry of a surface,” namely the study ofgeometric computations that only depend on the metric coefficients.

One of the most important examples of intrinsic geometry is givenby the metric coefficient functions g11(u, v) = 1/v2, g12(u, v) = 0 =g21(u, v), and g22(u, v) = 1/v2 defined on the domain consisting ofall points in the (u, v) plane with v > 0, the “upper half-plane.”

For example, we can compute the length of the curve given by(u(t), v(t)) = (t, c) where a ≤ t ≤ b by calculating the integral∫ b

a

√1

c2u′(t)2 + 2(0)u′(t)v′(t) +

1

c2v′(t)2 dt

=

∫ b

a

√1

c2+ 0 + 0 dt =

∫ b

a(1/c) dt =

1

c(b− a).


We can also calculate the length of the curve (u(t), v(t)) = (a, t)where 0 < c ≤ t ≤ d by∫ d

c

√1

t20 + 0 +

1

t2(1) dt =

∫ d

c1/tdt = [ln(t)]dc = ln(d)− ln(c).

Note that the limit of this length as d goes to infinity is infinity andthe same is true as c goes to 0. It follows that the length of theboundary of the domain a ≤ u ≤ b, and c ≤ v ≤ d, is given by

(b− a)(1/c)− (b− a)(1/d) + 2(ln(d)− ln(c)).

Since the area of this coordinate patch depends only on the metriccoefficient gij(u, v), we can compute the area of the coordinate patchin this metric by integrating

√g11(u, v)g22(u, v) =

√(1/v2)(1/v2) =

1/v2 to get ∫ b

a

∫ d

c

1v2

dv du = (b− a)

(1

c− 1

d

).

Note that the limit of the area of this domain using these metriccoefficients goes to (b − a)(1/c) as d goes to infinity, and as c goesto 0, the area becomes infinitely large.

The upper half-plane with this alternative metric tensor is calledthe Poincare Upper Half-Plane.

Problems

6.1.1. Calculate the metric tensor gij for the plane parametrized by ~X(s, t)= ~u+ s~v + t~w with (s, t) ∈ R3.

6.1.2. Calculate the metric tensor gij for the surface with parametrization~X(u, v) = (u2 − v, uv, u).

6.1.3. Let a be a nonzero real constant. Calculate the metric tensor for thehelicoid parametrized by ~X(u, v) = (v cosu, v sinu, au).

6.1.4. Calculate the metric tensor for the following families of conic sur-faces:

(a) ~X(u, v) = (a cosu sin v, b sinu sin v, c cos v), ellipsoid.

(b) ~X(u, v) = (av cosu, bv sinu, v2), elliptic paraboloid.


(c) ~X(u, v) = (au cosh v, bu sinh v, u2), hyperbolic paraboloid.

(d) ~X(u, v) = (a cosh v cosu, b cosh v sinu, c sinh v), hyperboloid ofone sheet.

(e) ~X(u, v) = (a cosu sinh v, b sinu sinh v, c cosh v), hyperboloid oftwo sheets.

6.1.5. Let 0 < r < R be real numbers and consider the torus parametrizedby

~X(u, v) = ((R+ r cos v) cosu, (R+ r cos v) sinu, r sin v).

(a) Show that if the domain of ~X is U = (0, 2π)× (0, 2π), then theparametrization is regular and that if U = [0, 2π]× [0, 2π] theparametrization is surjective.

(b) Calculate the metric tensor of this torus.

(c) Use the metric tensor to calculate the area of the torus.

6.1.6. Let U be an open subset of R2, and let f : U → R be a two-variablefunction. Explicitly calculate the metric tensor for the graph off , which can be parametrized by ~X : U → R3, with ~X(u, v) =(u, v, f(u, v)). Prove that the coordinate lines are orthogonal if andonly if fufv = 0.

6.1.7. Let γ be a loxodrome on the sphere as described in Example 6.1.6.Prove that the arc length of the loxodrome between the north andsouth pole is π secβ (regardless of the constant of integration C).

6.1.8. Consider the surface of revolution parametrized by

~X(u, v) = (f(v) cosu, f(v) sinu, g(v)),

with f and g chosen so that the surface is regular (see Problem5.2.7). Calculate the metric tensor.

6.1.9. Let ~α(t) be a regular space curve and consider the tangential surface

S parametrized by ~X(t, u) = ~α(t) + u~T (t). Calculate the metric

tensor for ~X.

6.1.10. Tubes. Let ~α(t) be a regular space curve. We call the tube of

radius r around ~α the surface that is parametrized by ~X(t, u) =

~α(t) + (r cosu)~P (t) + (r sinu) ~B(t). Calculate the metric tensor for~X. Supposing that the tube is regular, prove that the area of thetube is 2πr times the length of ~α.

6.1.11. Consider the right circular cone parametrized by ~X(u, v) = (v cosu,v sinu, v).


(a) Let ~α(t) be a curve in the plane such that for all t, the vectors~α(t) and ~α′(t) make a constant angle of β. Prove that ~α(t) =(f(t) cos t, f(t) sin t), where f(t) = Re(cot β)t for any real R > 0.

(b) Consider the spiral curve ~γ(t) = ~X(t, f(t)) on the cone. Cal-culate the arc length of ~γ for 0 ≤ t ≤ 2π. (It will depend on Rand β.)

6.1.12. Consider the parametrization ~Y : R2 → R3 of the unit sphere bystereographic projection (see Problem 5.2.15 and use the parame-

trization ~Y (u, v) = π−1(u, v)).

(a) Calculate the corresponding metric tensor component func-tions gij .

(b) Consider the usual parametrization ~X of the sphere (as pre-sented in Example 6.1.5.) Give appropriate domains for andexplicitly determine the formula for the change of coordinatesystem given by F−1 = ~Y −1 ~X = π ~X.

(c) Explicitly verify Equation (6.7) in this context.

6.1.13. Consider the parametrization ~X : R2 − (0, 0) → R3 defined by~X(x, y) = (x, y, 12 ln(x2 + y2)).

(a) Show that ~X is another parametrization for the Log Trumpetdiscussed in Example 6.1.8.

(b) Calculate the components of the metric tensor associate to thisparametrization.

(c) Use this and the parametrization given in Example 6.1.8 toillustrate how your result exemplifies the coordinate changeproperties of the metric tensor components as described inEquation 6.7.

6.1.14. Suppose that the first fundamental form has a matrix of the form

g =

(1 00 f(u, v)

).

Prove that all the v-coordinate lines have equal arc length over anyinterval u ∈ [u1, u2]. In this case, the v-coordinate lines are calledparallel.

6.1.15. Let (gij) be the metric tensor of some surface S parametrized bythe coordinates (x1, x2) ∈ U , where U is some open subset in R2.


Suppose that we reparametrize the surface with coordinates (x1, x2)in such a way that

x1 = f(x1, x2),

x2 = x2,

where f : V → U is a function with V ⊂ R. Let (gkl) be the metrictensor to S under this parametrization.

(a) Prove that (gkl) is a diagonal matrix if and only if the functionf satisfies the differential equation

∂f

∂x2= −g12 (f(x1, x2), x2)

g11 (f(x1, x2), x2).

(b) (ODE) Use the above result and the existence theorem fordifferential equations to prove that every regular surface admitsan orthogonal parametrization.

6.1.16. Let C be a regular value of the function F (x, y, z) and consider theregular surface S defined by F (x, y, z) = C. Without loss of gener-ality, suppose that the variables x and y can be used to parametrizea neighborhood U of some point p ∈ S. Use implicit differentiationto calculate the metric tensor functions gij in terms of derivatives ofF .

6.1.17. Consider the parametrization ~X(u, v) = (cosu sin v, sinu sin v, cos v)of the unit sphere. It is easy to show that the unit normal vector~N to the sphere at (u, v) is again the vector ~N(u, v) = ~X(u, v). Letf(u, v) be a nonnegative real function in two variables such thatf(u, 0) is constant, f(u, π) is constant, and for all fixed v0, f(u, v0)is periodic 2π. A normal variation to a given surface S is a surfacecreated by going out a distance of f(u, v) along the normal vector~N of S, given by

~Y (u, v) = ~X(u, v) + f(u, v) ~N(u, v).

A normal variation of the unit sphere is therefore given by

~Y (u, v) = (1 + f(u, v)) ~X(u, v).

(a) Calculate the metric tensor for the parametrization ~Y .

(b) Use the explicit function f(u, v) = cos2(2u) cos2(2v − π/2).

Calculate the metric tensor for ~Y .

6.1.18. Consider the metric coefficients g11(u, v) = 1/v2, g12(u, v) = 0 =g21(u, v), and g22(u, v) = 1/v2 on the upper half-plane in Exam-ple 6.1.9.


(a) Compute the length of the curve (u, v) = (√

2 cos t,√

2 sin t)with π/4 ≤ t ≤ 2π/4 from (1, 1) to (−1, 1) using these metriccoefficients.

(b) Show that the length of the curve is less than the length ofthe segment (u(t), v(t)) = (t, 1) with −1 ≤ t ≤ 1 using thesemetric coefficients.

(c) Show that the length of the curve (u(t), v(t)) = (R cos t, R sin t)where R is a constant and pi/4 ≤ t ≤ 3π/4 is independent of R.

6.1.19. Consider the metric defined in the previous problem on the upperhalf-plane.

(a) Compute the area above the curve (u(t), v(t)) = (R cos t +m,R sin t) with π/4 ≤ t ≤ 3π/4 and below the line v = d,for some d > R. Observe that the area is independent of theconstant m. What happens to the area of this region as dapproaches infinity?

(b) Repeat the same question with (u, v) = (R cos t + m,R sin t)but with 0 ≤ t ≤ π. [Hint: This will involve an improperintegral.]

6.1.20. Consider the general metric coefficients g11(u, v) = g22(u, v) = f(v)and g12(u, v) = g21(u, v) = 0 defined on the upper half-plane. Findan expression for the length of the curve (u(t), v(t)) for a ≤ t ≤ b.

6.2 Map Projections (Optional)

6.2.1 Metric Properties of Maps of the Earth

In order to illustrate properties and uses of the first fundamentalform, we propose to briefly discuss maps of the Earth. Throughhistory, maps of portions of the Earth have helped political leadersunderstand the geography of regions over which they exert influence.Starting particularly in the age of exploration, traders and naviga-tors needed maps that describe large regions of the Earth. For nav-igation purposes, it would be ideal if a map accurately representeddistances between points, angles between directions, and areas ofregions. Distances, angles, and areas are the metric properties ofa map. However, because a map is typically drawn on a flat pieceof paper whereas the Earth is (very close to) spherical, no map canaccurately reflect all metric properties at the same time.

6.2. Map Projections (Optional) 181

Points on the Earth are usually located using coordinates of lat-itude then longitude. In other places in this text (Example 6.1.5),we introduced the mathematicians’ customary way of parametrizingthe unit sphere: the polar coordinates angle θ, which is essentiallythe longitude, and then the colatitude ϕ, which is the angle downfrom the positive z-axis. To cover the sphere, we use 0 ≤ θ < 2πand 0 ≤ ϕ ≤ π. There is no absolute consensus on which coordi-nates is listed first. If we denote geographer’s latitude by ϕg, thenϕ + ϕg = π/2. Hence, the North Pole has ϕg = π/2, the equator isat ϕg = 0, and the South Pole has ϕg = −π/2. (Of course, latitudeand longitude are typically described with degrees.)

For the purposes of our discussion, we will use the parametriza-tion of the sphere

~X(ϕ, θ) = (R sinϕ cos θ,R sinϕ sin θ,R cosϕ) (6.8)

with colatitude 0 ≤ ϕ ≤ π and longitude −π < θ ≤ π and where R isthe radius of the Earth. Recall that coordinate lines with θ constantare called meridians. Note that colatitude lines are precisely latitudelines.

Points on a map are given by a pair of coordinates, say x andy, which are standard Cartesian coordinates of the plane. A mapprojection corresponds to a function from (ϕ, θ) coordinates to (x, y)coordinates,

M(ϕ, θ) = (x(ϕ, θ), y(ϕ, θ)). (6.9)

Note that maps generally do not show the x and y grid but ratherthe curves in the xy-plane that correspond to constant θ and ϕ co-ordinates.

Using the standard parametrization of the Earth (6.8), the metricproperties of the Earth are encapsulated by the first fundamentalform given by its metric tensor

gEarth =

(~Xϕ · ~Xϕ

~Xϕ · ~Xθ

~Xθ · ~Xϕ~Xθ · ~Xθ

)=

(R2 00 R2 sin2 ϕ

).

From this, we deduce that the line element ds satisfies

ds2 = R2(dϕ2 + sin2 ϕdθ2),


which means in other words

dsEarth = R

√(ϕ′)2 + sin2 ϕ(θ′)2 dt.

The surface area element is

dSEarth =√

det g dϕ dθ = R2 sinϕdϕdθ.

For the purposes of calculations, some authors set R = 1 so thatthe radius of the Earth becomes the unit length in length and areacalculations. By keeping R explicit, we can give it any necessaryvalue in any given system of units.

Now the map function M in (6.9) is a function M : [0, π] ×[−π, π]→ R2. This can be understood as an alternate parametriza-tion of a portion of the plane using the coordinates ϕ and θ. It hasa first fundamental form defined by

gMap =

(Mϕ ·Mϕ Mϕ ·Mθ

Mθ ·Mϕ Mθ ·Mθ

),

where the dot products require us to view M(ϕ, θ) and its deriva-tives as a vector. Note that at each point p on a surface, the firstfundamental form is an inner product Ip( , ) on the tangent planeTpS to the surface at p. However, in the case of a map, the surfaceis a subset of R2 and the tangent plane is R2 itself.

Definition 6.2.1. A conformal map is a map in which the angles ofintersection of paths on the map are the same as the angles of inter-section of paths on the Earth.

Conformal maps are particularly useful for navigation since theycan tell a pilot that at least he or she is heading in a desired direction.

Proposition 6.2.2. A map is conformal if there is a positive functionw(ϕ, θ) such that

gMap = w(ϕ, θ)gEarth.

Proof: Let ~a and ~b be vectors in the tangent plane to the Earth at apoint p = (ϕ, θ), with components given with respect to the ( ~Xϕ, ~Xθ)


ordered basis of the tangent plane. The angle α between them hasa cosine of

cosα =Ip(~a,~b)√

Ip(~a,~a)√Ip(~b,~b)

.

If there is a positive function w(ϕ, θ) as described in the proposition,then the first fundamental form of the map is

IMapp ( , ) = w(p)Ip( , ).

Then cosine of the angle between directions in the map is

IMapp (~a,~b)√

IMapp (~a,~a)

√IMapp (~b,~b)

=w(p)Ip(~a,~b)√

w(p)Ip(~a,~a)√w(p)Ip(~b,~b)

=Ip(~a,~b)√


= cosα.

The proposition follows.

Maps that preserve area are maps such that there exists a con-stant D so that the area of any region measured in the map is afactor D times that surface area of the corresponding region on theEarth. Hence, for all regions R in the ϕθ-plane, we have∫∫

RdSMap = D

∫∫RdSEarth

⇐⇒∫∫R

√det gMap dϕ dθ = D

∫∫R

√det gEarth dϕ dθ.

The Mean Value Theorem for integrals gives the following proposi-tion.

Proposition 6.2.3. A map of the Earth is area-preserving if and onlyif there is a positive constant D such that

det(gMap) = D2 det(gEarth).

The arc length element in a map is ds =√

(x′(t))2 + (y′(t))2 dt,which, in particular, does not depend on the position in the plane


but only on the components of the velocity vector. The explicitdependence of dsEarth on ϕ shows that no map can calculate arclengths of all paths correctly. In other words, there is no map witha given proportionality factor such that the arc length of the pathin the map is that factor times the arc length of the correspondingpath on the Earth. However, it is still possible that certain sets ofpaths in a given map calculation have the same relative arc lengthsas on the Earth.

6.2.2 Azimuthal Projections

A first category of map projection is the azimuthal projections. Onepoint P on the Earth is chosen as the center of the map and serves asthe origin in the xy-coordinate system. If P is the North Pole, thenthe longitude lines through P appear on the map as equally spacedrays radiating from the origin, and the latitude circles appear on themap as circles centered on the map with radii determined by theparticular projection. (If P is the South Pole, longitude and latitudelines have a similar representation on the map. However, if P isnot one of the poles, then the radial lines in the map and circles atgiven radii around the origin have more complicated interpretationsin latitude and longitude on the Earth.)

Choosing P as the North Pole, any azimuthal projection mapsthe (ϕ, θ) coordinates to polar coordinates (r, θ) = (r(ϕ), θ) on themap. Hence, the map function has the form

(x, y) = M(ϕ, θ) = (r(ϕ) cos θ, r(ϕ) sin θ),

where r is a nonnegative increasing function with r(0) = 0. The firstfundamental form of any azimuthal projection is

gMap =

(r′(ϕ)2 0

0 r(ϕ)

).

Different choices of the function r(ϕ) correspond to different az-imuthal projections. Even if we always pick the center of the mapto be the North Pole, there are a variety of classical choices. Somegive a map for the whole Earth (except for the South Pole), whileothers only map the northern hemisphere.


Example 6.2.4 (Orthographic Projection). One of the geometrically sim-plest azimuthal projections is the orthographic projection. It mapsonly the northern hemisphere. We imagine the map as the tangentplane T to the North Pole. A point Q in the northern hemisphere ofthe Earth is mapped to the point Q′ in the tangent via orthogonalprojection onto the tangent plane T . The resulting map fills a circu-lar disc in the plane. This disc is then scaled uniformly by a factorc to any reasonable radius for a map. (If the disc were not scaled, itwould have a radius equal to the radius of the Earth, which wouldmake for a very large map!)

In this case, the function r(ϕ) is simply r(ϕ) = cR sinϕ for 0 ≤ϕ ≤ π/2.

Example 6.2.5 (Gnomic Projection). Another simple azimuthal projec-tion is the gnomic projection. It also maps only the northern hemi-sphere and this time excludes the equator. We imagine the map asthe tangent plane T to the North Pole. Let O be the center of theEarth. A point Q in the northern hemisphere of the Earth is mappedto the point Q′ in the tangent plane where Q′ is the intersection of

the ray−−→OQ with the tangent plane T . The resulting map fills the

whole plane. As usual, the tangent plane is then scaled uniformly bya reasonable factor c. In practice, one only depicts a certain boundedportion of the map.

In this case, the function r(ϕ) is simply r(ϕ) = cR tanϕ for0 ≤ ϕ < π/2.

One interesting property about the gnomic projection is thatevery arc of a great circle on the Earth is mapped to a straightline segment. Indeed, a great circle on a sphere corresponds to theintersection of the sphere and a plane P through the origin. Viathe gnomic projection, the arc of such a great circle maps to theintersection of P ∩ T , which is a line.

Example 6.2.6 (Stereographic Projection). Problem 5.2.15 introducedstereographic projection as an alternate parametrization of the sphere.However, as presented in that problem, it defines an azimuthal pro-jection with the South Pole as the center. In this example, we makea few adjustments to the presentation in Problem 5.2.15 to conformto the presentation here.


Like the gnomic projection, the stereographic projection assumesthe map is the tangent plane T to the North Pole. (This a variationfrom the presentation in Problem 5.2.15, where the map was assumedto be the plane through the equator.) A point Q on the Earth(except for the South Pole) is mapped to the point Q′, which is theintersection of T and the ray out of the South Pole through Q. Asusual, the tangent plane is then scaled uniformly by a reasonablefactor c. The stereographic projection maps the whole Earth exceptfor the South Pole and covers the whole plane.

We leave it as an exercise to show that in this case, the functionr(ϕ) is

r(ϕ) =2cR sinϕ

1 + cosϕfor 0 ≤ ϕ < π.

Example 6.2.7 (Area-Preserving Azimuthal). Suppose that we want todevise an area-preserving azimuthal projection. Then by Proposi-tion 6.2.3, the radial function r(ϕ) must be such that

det(gMap) = r(ϕ)2(r′(ϕ))2 = D2R4 sin2 ϕ

for some positive constant D. Taking a square root of both sides, weget r(ϕ)r′(ϕ) = DR2 sinϕ. Integrating both sides with respect to ϕand using substitution, we find that

−DR2 cosϕ =

∫r(ϕ)r′(ϕ) dϕ =

∫r dr =

1

2r2 + C,

for some constant of integration. Since we want r(0) with ϕ = 0, wehave C = −DR2 and thus

r(ϕ) = R√

2D(1− cosϕ) = 2√DR sin

ϕ

2

for 0 ≤ ϕ < π. Hence, there exists an area-preserving azimuthalprojection that covers the whole Earth except for the South Poleand its map is a disc of radius 2

√DR.

6.2.3 Cylindrical Projections

A cylindrical projection map is one based off of the intuition of takinga globe, wrapping a sheet of paper as a cylindrical tube around the


Figure 6.4. Cylindrical projection.

equator of the globe, projecting the surface of the globe onto thesheet of paper, and then unwrapping the cylinder of paper to lay itflat. See Figure 6.4 for a visual.

More precisely, one point P on the Earth is chosen as a pole.If P is the North Pole, then longitude lines through P appear onthe map as equally spaced vertical parallel lines and latitude circlesappear on the map as horizontal lines, with spacing determined bythe particular projection. The choice of point P corresponds to achoice of an axis for the cylinder because we assume the cylinder has

axis←→OP , where O is the center of the sphere.

We always scale the map horizontally so that a unit in the xdirection corresponds to a unit in θ. With P as the North Pole, then(ϕ, θ) on the Earth maps to (x, y) = M(ϕ, θ) = (θ, h(ϕ)), whereh is a decreasing function with h(π/2) = 0. That h is decreasingcorresponds to setting the North Pole ϕ = 0 as up on the map. Thefirst fundamental form of such maps is

gMap =

((h′(ϕ))2 0

0 1

).

Different choices of the function h(ϕ) correspond to different cylin-drical projections. Also h(ϕ) may be scaled linearly simply to adjustthe vertical size of the map. As with the azimuthal projections, evensetting P as the North Pole, there are a number of classical choices.


Example 6.2.8 (Radial Cylindrical). Let O represent the center of theEarth. The radial cylindrical projection is a geometrically simpleprojection in which each point Q on the Earth, except for the NorthPole and the South Pole, is mapped to the point Q′ on the cylinder

that is the intersection with the ray←−−OQ. The cylinder is then un-

rolled and, as usual, scaled by a reasonable factor c. The map thenconsists of the strip [0, 2π]× R.

In this projection, we have h(ϕ) = c cotϕ, where c is a constant.

Example 6.2.9 (Mercator). The well-known Mercator projection is acylindrical projection that is conformal. By Proposition 6.2.2, theMercator projection must be such that there is a function w(ϕ, θ)such that (

(h′(ϕ))2 00 1

)= w(ϕ, θ)

(R2 00 R2 sin2 ϕ

).

Obviously, we must have w(ϕ, θ) = 1/R2 sin2 ϕ and then we alsoneed

h′(ϕ) = − 1

R sinϕwith h

(π2

)= 0.

The choice of the negative sign for h′(ϕ) comes from the requirementthat h(ϕ) is decreasing. Integrating, we find that

h(ϕ) = −∫ ϕ

π/2

dπ

sinϕ= ln

(1 + cosϕ

sinϕ

)= ln

(cot

π

2

),

for 0 < ϕ < π/2. Hence, the map for the Mercator projection isM(ϕ, θ) =

(θ, ln

(cot π2

)).

Many other classes of map projections exist. For example, pseu-docylindrical projections attempt to fix the distortions that invari-ably occur near the poles in cylindrical projections. Indeed, on asphere, latitude circles decrease in radius the closer one gets to thepoles. However, under any cylindrical projection every latitude cir-cle becomes a horizontal line segment of the same length. A pseu-docylindrical projection has for its map

(x, y) = M(ϕ, θ) =

(w(ϕ)θ

2π, h(θ)

)where


• as for cylindrical projections, h(ϕ) is a decreasing function withh(π/2) = 0;

• w(ϕ) is a nonnegative function that gives the width of the mapat a given latitude ϕ.

For many maps, the function w(ϕ) satisfies w(0) = w(π) = 0 andw(π/2) = 1, so the width is 0 at the North and South Pole and 1 atthe equator.

6.2.4 Coordinate Changes on the Sphere

With various maps for the Earth (sphere) at our disposal, we takethe opportunity to illustrate the coordinate change transformationproperty of the metric tensor as given in Equation (6.7). To locatepoints on the sphere (Earth), we will consider the standard latitudeand longitude (ϕ, θ) coordinates and the (x, y) coordinates as givenby the stereographic projection described in Example 6.2.6.

For simplicity, let us use R = c = 1. Then the stereographic mapis

(x, y) = M(ϕ, θ) =

(2 sinϕ cos θ

1 + cosϕ,2 sinϕ sin θ

1 + cosϕ

).

Using similar geometry as required by Problem 5.2.15, we can showthat the (x, y) coordinates parametrize the unit sphere by

~Y (x, y) =

(4x

4 + x2 + y2,

4y

4 + x2 + y2,4− x2 − y2

4 + x2 + y2

).

In order to verify Equation (6.7), let us call the (x1, x2) coordi-nates the (x, y) coordinates and we call the (x1, x2) coordinates the(ϕ, θ) coordinates. In these labels, we already know that

(gij) =

(1 00 sin2 ϕ

).

Using the parametrization ~Y (x, y), we find that

(gij) =

(~Yx · ~Yx ~Yx · ~Yy~Yy · ~Yx ~Yy · ~Yy

)=

16

(4 + x2 + y2)20

016

(4 + x2 + y2)2

.

(6.10)


Furthermore, the differential [dM(ϕ,θ)], also known as the Jacobianmatrix of the coordinate change for (ϕ, θ) coordinates to (x, y) co-ordinates, is

[dM(ϕ,θ)] =

∂x

∂ϕ

∂x

∂θ

∂y

∂ϕ

∂y

∂θ

=

2 cos θ

1 + cosϕ−2 sinϕ sin θ

1 + cosϕ

2 sin θ

1 + cosϕ

2 sinϕ cos θ

1 + cosϕ

.

Recall that Equation (6.7) states that

gij =2∑

k=1

2∑l=1

∂xk∂xi

∂xl∂xj

gkl.

This can be rewritten more explicitly as

gij =∂x1

∂xi

∂x1

∂xjg11 +

∂x1

∂xi

∂x2

∂xjg12 +

∂x2

∂xi

∂x1

∂xjg21 +

∂x2

∂xi

∂x2

∂xjg22.

We consider our specific example. However, note that as a simpli-fication, we have g12 = g21 = 0. Also, we point out ahead of timethat

x2 + y2 + 4 =

(2 sinϕ cos θ

1 + cosϕ

)2

+

(2 sinϕ sin θ

1 + cosϕ

)2

+ 4 =8

1 + cosϕ.

Thus16

(x2 + y2 + 4)2=

(1 + cosϕ)2

4.

So we get the following confirmation:

g11 =∂x1

∂x1

∂x1

∂x1g11 +

∂x2

∂x1

∂x2

∂x1g22

=

(2 cos θ

1 + cosϕ

)2 16

(4 + x2 + y2)2+

(2 sin θ

1 + cosϕ

)2 16

(4 + x2 + y2)2

=4 cos2 θ

(1 + cosϕ)2

(1 + cosϕ)2

4+

4 sin2 θ

(1 + cosϕ)2

(1 + cosϕ)2

4

= 1;


g12 =∂x1

∂x1

∂x1

∂x2g11 +

∂x2

∂x1

∂x2

∂x2g22

=

(2 cos θ

1 + cosϕ

)(−2 sinϕ sin θ

1 + cosϕ

)16

(4 + x2 + y2)2+

(2 sin θ

1 + cosϕ

)(2 sinϕ cos θ

1 + cosϕ

)16

(4 + x2 + y2)2

= 0;

we can verify that g21 = g12 = 0; and finally

g22 =∂x1

∂x2

∂x1

∂x2g11 +

∂x2

∂x2

∂x2

∂x2g22

=

(−2 sinϕ sin θ

1 + cosϕ

)2 16

(4 + x2 + y2)2+

(2 sinϕ cos θ

1 + cosϕ

)2 16

(4 + x2 + y2)2

=4 sin2 ϕ

(1 + cosϕ)2

(1 + cosϕ)2

4

= sin2 ϕ.

These calculations illustrate the coordinate change transformationdescribed by Equation (6.7)

Problems

6.2.1. Find the possible maps for cylindrical projections that are area-preserving.

6.2.2. Verify the calculations for the metric tensor given in Equation (6.10).

6.2.3. Show that the metric tensor of the map for the general pseudocylin-drical projection is

gMap =1

4π2

(4π2h′(ϕ) + θ2w′(ϕ) θw(ϕ)w′(ϕ)

θw(ϕ)w′(ϕ) w(ϕ)2

).

6.2.4. The Sinusoidal Projection is a pseudocylindrical map projection thatuses w(ϕ) = sinϕ and h(ϕ) = c

(π2 − ϕ

), where c is a positive con-

stant.

(a) Prove that the lengths of the latitude lines on the map are afixed multiple of the lengths of the latitude lines on the Earth.

(b) Prove that this map also preserves areas.

(c) Decide if the map is conformal.

(d) Describe the shape of the image of the map.


6.3 The Gauss Map

In Section 2.2, given any closed, simple, regular curve γ : I → R2

in the plane, we defined the tangential indicatrix to be the curvegiven by ~T : I → R2, the unit tangent vector of γ(t). The imageof ~T lies on the unit circle, and though ~T is a closed curve, it neednot be regular as its locus may stop and double back. Regardless ofthe parametrization, the tangential indicatrix is well defined up to achange in sign.

On any regular surface S in R3, the tangent plane TpS at a pointp ∈ S is a two-dimensional subspace of R3, and hence the vectorsthat are normal to S at p form a one-dimensional subspace. Thus,there exist exactly two possible choices for a unit normal vector.Proposition 5.5.4 implies that if the surface is orientable, we canspecify its orientation by a continuous function n : S → S2, whereS2 means the unit sphere in R3.

Definition 6.3.1. Let S be an oriented regular surface in R3 with anorientation n : S → S2. In the classical theory of surfaces, thefunction n is also called the Gauss map.

It is important to remain aware of the distinction between thefunctions n and ~N . Let S be an oriented surface with orientationn : S → S2. Suppose that ~X : U → R3, where U is an opensubset of R2, parametrizes a neighborhood of S, then the function~N : U → S2 is defined in reference to ~X by Equation (5.7). Thefunctions n and ~N are related by the fact that if ~X is a positivelyoriented parametrization, then

~N = n ~X.

Example 6.3.2. Consider as an example a sphere S in R3 equippedwith the orientation n, with the unit vectors normal to S pointingaway from the center ~c of the sphere. The sphere can be given asthe solution to the equation

‖~x− ~c‖2 = R2. (6.11)

Recall that ‖~v‖2 = ~v · ~v. If ~x(t) is any curve on the sphere, then bydifferentiating the relationship in Equation (6.11), one obtains

~x′(t) · (~x(t)− ~c) = 0.

6.3. The Gauss Map 193

A

B

D

xy

A

B D

xy

z

Figure 6.5. Gauss map of the elliptic paraboloid.

The tangent plane to the sphere S at a point ~p consists of all possiblevectors ~x′(t0), where ~x(t) is a curve on S, with ~x(t0) = ~p. Therefore,at any point ~p on the sphere, there are two options for unit normalvectors:

~p− ~c‖~p− ~c‖

or − ~p− ~c‖~p− ~c‖

.

However, it is the former that provides the outward-pointing orien-tation n. Thus the Gauss map for the sphere of radius R and center~c is explicitly

n(~p) =~p− ~cR

.

Furthermore, if S is itself the unit sphere centered at the origin, thenthe Gauss map is the identity function.

Example 6.3.3. Consider the elliptic paraboloid S defined by the equa-tion z = x2 + y2. This is an orientable surface, and suppose itis oriented with the unit normal always pointing in a positive z-direction. (Figure 6.5 shows this is possible.) Using the functionF (x, y, z) = z− x2− y2, the elliptic paraboloid is given by the equa-tion F (x, y, z) = 0. By Proposition 5.2.15, for all (x, y, z) ∈ S, thegradient ~∇F (x, y, z) is a normal vector, so (since the z-direction is


positive) the Gauss map is

n(x, y, z) =(−2x,−2y, 1)√4x2 + 4y2 + 1

.

Figure 6.5 shows how the Gauss map acts on the patch

(x, y, z) ∈ S | − 1 ≤ x ≤ 1 and − 1 ≤ y ≤ 1.

It is not hard to show that the image of the Gauss map on theunit sphere is the upper hemisphere. Indeed, using cylindrical coor-dinates, the Gauss map is

n(r, θ, z) =(−2r cos θ,−2r sin θ, 1)√

4r2 + 1.

The function f(r) = 1/√

4r2 + 1 is a bijection from [0,+∞) to (0, 1]and it is decreasing. Thus, for any z0 ∈ (0, 1], there exists a uniquer0 ∈ [0,+∞) with f(r0) = z0. Then, with θ ∈ [0, 2π], the imageof n(r0, θ, z0) is a circle on the unit sphere at height z0. Thus, theimage of n is

n(S) = (x, y, z) ∈ S2 | z > 0.

Example 6.3.4. In contrast to Example 6.3.3, consider the hyperbolicparaboloid given by the parametrization ~X : R2 → R3, with

~X(u, v) = (u, v, u2 − v2).

We calculate

~Xu = (1, 0, 2u) and ~Xv = (0, 1,−2v),

and thus,

~N =~Xu × ~Xv

‖ ~Xu × ~Xv‖=

(−2u, 2v, 1)√4u2 + 4v2 + 1

.

Like Figure 6.5, Figure 6.6 shows the Gauss map for a “square” onthe hyperbolic paraboloid. These two examples manifest a centralbehavior of the Gauss map. On the elliptic paraboloid, the Gaussmap preserves the orientation of the square in the sense that if onetravels along the boundary in a clockwise sense on the surface, onealso travels along the boundary of the image of the square mappedby n in a clockwise sense. On the other hand, with the hyperbolicparaboloid, the Gauss map reverses the orientation of the square.


A

B

C

Dyz

A

BD

x

y

z

Figure 6.6. Gauss map on the hyperbolic paraboloid.

Proposition 6.3.5. Let S be a regular surface in R3, and let ~X : U →R3 be a parametrization of a coordinate patch V = ~X(U), where Uis an open set in R2. Define the vector function ~N : U → R3 by

~N =~Xu × ~Xv

‖ ~Xu × ~Xv‖.

If ~X is of class Cr, then ~N is of class Cr−1.

Proof: By Problem 3.1.18, we deduce that if ~F : U → R3, ~G :U → R3, and h : U → R are of class C1, then the three functions~F · ~G : U → R, ~F × ~G : U → R3, and h~F : U → R3 are also of classC1, and their partial derivatives follow appropriate product rules.By definition,

~N =~Xu × ~Xv√

( ~Xu × ~Xv) · ( ~Xu × ~Xv).

One can see that any partial derivative of any order of ~N will involvethe partial derivative of a combination of the above three types ofproducts and the derivative of an expression of the form

f(u, v) =(

( ~Xu × ~Xv) · ( ~Xu × ~Xv))−k/2

,


where k is an odd positive integer. Since S is regular, ‖ ~Xu × ~Xv‖ isnever 0, so a particular higher derivative of ~N will exist if that higherderivative exists for ~Xu and for ~Xv. The proposition follows.

One would be correct to think that the different behaviors inFigures 6.5 and 6.6 can be illustrated and quantified by a functionbetween tangent spaces to S and S2, respectively. This perspectiveis encapsulated in the notion of the differential of functions betweenregular surfaces. We develop the more general theory of differentialsof functions between manifolds in Chapter 11 of [24]. However, thedifferential of the Gauss map, though only a particular case of thedifferentials of functions between surfaces, serves a central role in therest of this chapter.

Let S be a regular oriented surface with orientation n, and let~X : U → R3 be a regular positively oriented parametrization of acoordinate patch ~X(U) of S. Suppose also that S is of class C2,which, according to Proposition 6.3.5, implies that ~N is of class C1.Since ~N · ~N = 1 for all (u, v) ∈ U , we have

~N · ~Nu = 0 and ~N · ~Nv = 0 (6.12)

for all (u, v) ∈ U . Therefore, both derivatives ~Nu and ~Nv are vectorfunctions such that ~Nu(q) and ~Nv(q) are in TpS when ~X(q) = p.

The vector function ~N : U → R3 is itself a parametrized surface,with its image lying in the unit sphere, though not necessarily givinga regular parametrization of S2. The simple fact in Equation (6.12)indicates that if ~N admits a tangent plane at q ∈ U , then

TpS = T ~N(q)(S2) (6.13)

as subspaces of R3. In other words, the sets of vectors ~Xu, ~Xv and ~Nu, ~Nv span the same subspace TpS.

The differential of the Gauss map at a point p ∈ S is a lineartransformation dnp : TpS → Tn(p)(S2), but by virtue of Equation(6.13), one identifies it as a linear transformation dnp : TpS → TpS.For all ~v ∈ TpS, one defines dnp(~v) = ~w if there exists a curve


~α : I → U such that~α(0) = q, with ~X(q) = p,d

dt

(~X ~α

)(t)∣∣t=0

= ~v, and

d

dt

(~N ~α

)(t)∣∣t=0

= ~w.

(6.14)

In other words, by writing ~X(t) = ~X ~α(t) and ~N(t) = ~N ~α(t), thedifferential of the Gauss map satisfies dnp( ~X

′(t)) = ~N ′(t).

Using coordinate lines ~X(t, v0) or ~X(u0, t) for the curve ~α(t) inthe above definition, it is easy to see that

dnp( ~Xu) = ~Nu,

dnp( ~Xv) = ~Nv.(6.15)

Though the Gauss map n : S → S2 is independent of any co-ordinate systems on S, the differential dnp requires reference to aregular positively oriented parametrization of a neighborhood of p.However, though different parametrizations of a neighborhood of pinduce different coordinate bases on TpS, the definition in Equation(6.14) remains unchanged and consequently, as a linear transforma-tion of TpS to itself, is independent of the parametrization.

Problems

6.3.1. Describe the region of the unit sphere covered by the image of theGauss map for the following surfaces:

(a) The hyperbolic paraboloid given by z = x2 − y2.

(b) The hyperboloid of one sheet given by x2 + y2 − z2 = 1.

6.3.2. Describe the region of the unit sphere covered by the image of theGauss map for the following surfaces:

(a) The cone with opening angle α given by z2 tan2 α = x2 + y2.

(b) The right circular cylinder x2 + y2 = R2, with R a constant.

6.3.3. What regular surface S has a single point on S2 as the image of theGauss map?


6.3.4. Show that the graph of a differentiable function z = f(x, y) has aGauss map that lies inside either the upper hemisphere or the lowerhemisphere.

6.3.5. Prove that the image of the Gauss map of a regular surface S is anarc of a great circle if and only if S is a generalized cylinder. (SeeExample 6.7.2.)

6.3.6. Consider the surface S obtained as a one-sheeted cone over a regu-lar plane curve C (see Problem 5.2.17). Prove that the image of theGauss map for S is a curve on the unit sphere. Find a parametriza-tion for the image of the Gauss map in terms of a parametrizationof C. [Hint: Without loss of generality, suppose that C lies in theplane z = a.]

6.4 The Second Fundamental Form

The examples in the previous section illustrate how the differentialof the Gauss map dnp qualifies how much the surface S is curving atthe point p. Despite the abstract definition for dnp, it is not difficultto calculate the matrix for dnp : TpS → TpS. However, in orderfor the differential dnp to exist at all points p ∈ S, we will need toassume that S is a surface of class C2.

Let S be a regular oriented surface of class C2 with orientationn, and let ~X : U → R3 be a positively oriented parametrization of aneighborhood V of a point p on S. Since ~N satisfies

~N · ~Xi = 0 for i = 1, 2,

then by differentiating with respect to another variable, the productrule gives

~Nj · ~Xi + ~N · ~Xij = 0 for i, j = 1, 2.

Note that since ~Xij = ~Xji, by interchanging i and j, one deducesthat

~Ni · ~Xj = − ~N · ~Xij = ~Nj · ~Xi. (6.16)

(Recall that the notation ~fi in Equation (6.16) and in what followsrefers to taking the derivative of the multivariable vector function ~fwith respect to the ith variable.) This leads to the following propo-sition.

6.4. The Second Fundamental Form 199

Proposition 6.4.1. Using the above setup, the linear map dnp is a self-adjoint operator with respect to the first fundamental form Ip(·, ·).

Proof: Let ~v, ~w ∈ TpS and write ~v = v1~Xu + v2

~Xv and ~w = w1~Xu +

w2~Xv. Recall that the first fundamental form Ip(~v, ~w) is the dot

product ~v · ~w when viewing TpS as a subset of R3. Also recall from

Equation (6.15) that dnp( ~Xu) = ~Nu and similarly for v. Hence,

Ip(dnp(~v), ~w) = Ip(v1~Nu + v2

~Nv, ~w)

= (v1~Nu + v2

~Nv) · (w1~Xu + w2

~Xv)

=

2∑i,j=1

viwj ~Ni · ~Xj

=2∑

i,j=1

viwj ~Nj · ~Xi (by Equation (6.16))

= (v1~Xu + v2

~Xv) · (w1~Nu + w2

~Nv)

= Ip(~v, dnp(~w)).

Definition 6.4.2. Let S be an oriented regular surface of class C2 withorientation n, and let p be a point of S. We define the second fun-damental form as the quadratic form on TpS defined by

IIp(~v) = −dnp(~v) · ~v = −Ip(dnp(~v), ~v).

The first fundamental form allows one to measure lengths, angles,and area of regions on a parametrized surface. The second funda-mental form provides a measure for how much the normal vectorchanges if one travels away from p in a particular direction ~v, with~v ∈ TpS.

Recall from linear algebra that every quadratic form Q on a vec-tor space V of dimension n is of the form

Q(~v) = ~vtMv

for some n× n matrix. Define the functions Lij : U → R by

IIp(~v) = ~vT(L11(q) L12(q)L21(q) L22(q)

)~v,


where p = ~X(q), for all ~v ∈ TpS. Since

IIp(~v) = −dnp(~v) · ~v = −~vTdnp(~v),

by writing ~v =(ab

)in the coordinate basis ~Xu, ~Xv, one obtains

IIp

(a

b

)= −(a ~N1 + b ~N2) · (a ~X1 + b ~X2)

= −(a2 ~N1 · ~X1 + ab ~N1 · ~X2 + ab ~N2 · ~X1 + b2 ~N2 · ~X2).

Therefore,

Lij = − ~Nj · ~Xi for all 1 ≤ i, j ≤ 2, (6.17)

and by Equation (6.16),

Lij = ~N · ~Xij . (6.18)

This provides a convenient way to calculate the Lij functionsand, hence, the second fundamental form. By Proposition 6.4.1,the matrix (Lij) is a symmetric matrix, so L12 = L21. In classicaldifferential geometry texts, authors often refer to the coefficients ofthe second fundamental form using the letters e, f , and g, as follows:

e = L11, f = L12 = L21, g = L22.

Example 6.4.3 (Spheres). Let S be the sphere of radius R with out-ward orientation and consider the coordinate patch parametrized by~X(u, v) = (R cosu sin v,R sinu sin v,R cos v). The unit normal vec-tor is

~N = (cosu sin v, sinu sin v, cos v),

and the second derivatives are

~X11 = (−R cosu sin v,−R sinu sin v, 0),

~X12 = ~X21 = (−R sinu cos v,R cosu cos v, 0),

~X22 = (−R cosu sin v,−R sinu sin v,−R cos v).

Thus, the matrix for the second fundamental form is

(Lij) =

(−R sin2 v 0

0 −R

).


One should have expected the negative signs since along any curvethrough a point p with direction ~v, the unit normal vector ~N changeswith a differential in the direction of ~v and not opposite to it, so byDefinition 6.4.2, the second fundamental form is always negative.

At a point p ∈ S in a coordinate neighborhood parametrizedpositively by ~X(u, v), if ~X(q) = p, the function

f(s, t) =1

2IIp

(s

t

)=

1

2(L11(q)s2 + 2L12(q)st+ L22(q)t2)

is called the osculating paraboloid . This paraboloid provides thesecond-order approximation of the surface near p in reference to thenormal vector ~N(q) in the following sense. Setting q = (u0, v0), thesecond-order Taylor approximation of ~X is

~X(u, v) ≈ ~X(u0, v0) + ~Xu(u0, v0)(u− u0) + ~Xv(u0, v0)(v − v0)

+1

2~Xuu(u0, v0)(u− u0)2 + ~Xuv(u0, v0)(u− u0)(v − v0)

+1

2~Xvv(u0, v0)(v − v0)2. (6.19)

Since ~N is perpendicular to ~Xu and ~Xv, setting s = u − u0 andt = v − v0,

f(s, t) =(~X(s+ u0, t+ v0)− ~X(u0, v0)

)· ~N(u0, v0).

Furthermore, from the second derivative test in multivariable calcu-lus, one knows that the point (s, t) = (0, 0) is a local extremum ifL11L22 − L2

12 > 0 and is a saddle point if L11L22 − L212 < 0. This

leads to the following definition. (See also Figure 6.7.)

Definition 6.4.4. Let S be a regular orientable surface of class C2. Apoint p on S is called:

1. elliptic if det(Lij) > 0;

2. hyperbolic if det(Lij) < 0;

3. parabolic if det(Lij) = 0 but not all Lij = 0; and

4. planar if Lij = 0 for all i, j.


(a) Elliptic point. (b) Hyperbolic point. (c) Parabolic point.

Figure 6.7. Elliptic, hyperbolic, and parabolic points.

It is not hard to check (Problem 6.4.7) that if (x1, x2) and (x1, x2)are two coordinate systems for the same open set of an orientedsurface S, then

det(Lkl) =

(∂(x1, x2)

∂(x1, x2)

)2

det(Lij). (6.20)

Furthermore, the change of coordinates between (x1, x2) and (x1, x2)is a diffeomorphism between two open sets in R2. Therefore, theJacobian in Equation (6.20) is never 0 and Definition 6.4.4 is inde-pendent of any particular parametrization of S.

Example 6.4.5. Consider the torus parametrized by

~X(u, v) = ((2 + cos v) cosu, (2 + cos v) sinu, sin v) .

The unit normal vector and the second derivatives of the vectorfunction ~X are

~N = (− cosu cos v,− sinu cos v,− sin v),

~X11 = (−(2 + cos v) cosu,−(2 + cos v) sinu, 0),

~X12 = (sinu sin v,− cosu sin v, 0),

~X22 = (− cos v cosu,− cos v sinu,− sin v).

Thus,

L11 = (2 + cos v) cos v, L12 = 0, L22 = 1.


xy

z

Elliptic points

Hyperbolic points

Parabolic points

Figure 6.8. Torus.

Thus, in this example it is easy to see that det(Lij) = (2+cos v) cos v.Since −1 ≤ cos v ≤ 1, we know 2 + cos v ≥ 1, so the sign of det(Lij)is the sign of cos v. Hence, the parabolic points on the torus arewhere v = ±π

2 , the elliptic points are where −π2 < v < π

2 , and thehyperbolic points are where π

2 < v < 3π2 (see Figure 6.8).

As Figure 6.7 implies, the only quadratic surface possessing atleast one planar point is a plane. However, similar to the “unde-cided” case in the second derivative test from multivariable calculus,on a general surface S, the existence of a planar point does not implythat S is a plane; it merely implies that third-order behavior, as op-posed to second-order, governs the local geometry of the surface withrespect to the normal vector. In this case, a variety of possibilitiescan occur.

Example 6.4.6 (Monkey Saddle). The simplest surface that illustratesthird-order behavior is the monkey saddle parametrized by ~X(u, v) =(u, v, u3 − 3uv2). We calculate that

~Xuu = (0, 0, 6u),

~Xuv = (0, 0,−6v),

~Xvv = (0, 0,−6u),


Figure 6.9. Monkey saddle.

and hence, even without calculating the unit normal vector function,we deduce that Lij(0, 0) = 0 for all i, j. Consequently, (0, 0, 0) is aplanar point (see Figure 6.9 for a picture). Near (0, 0, 0), the bestapproximating quadratic to the surface is in fact the plane z = 0, butobviously such an approximation describes the surface poorly. (Theterminology “monkey saddle” comes from the shape of the surfacehaving room for two legs and a tail.)

Example 6.4.5 illustrates a general fact about the local shape ofa surface encapsulated in the following proposition.

Proposition 6.4.7. Let S be a regular surface of class C2, and let V bea coordinate neighborhood on S.

1. If p ∈ V is an elliptic point, then there exists a neighborhoodV ′ of p such that TpS does not intersect V ′ − p.

2. If p ∈ V is a hyperbolic point, then TpS intersects every deletedneighborhood V ′ of p.

Proof: Let ~X : U → R3 be a parametrization of the coordinateneighborhood V and suppose that ~X(0, 0) = p. Consider the real-valued function on U

h(u, v) = ( ~X(u, v)− ~X(0, 0)) · ~N(0, 0).


Since ~N(0, 0) is a unit vector perpendicular to TpS, then h(u, v)is the height function of signed distance between the surface S atp′ = ~X(u, v) and the tangent plane TpS.

The second-order Taylor series of ~X near (0, 0) is given by Equa-tion (6.19), with a remainder function ~R(u, v) such that

lim(u,v)→(0,0)

~R(u, v)

u2 + v2= ~0.

Thus

h(u, v) =1

2~Xuu(0, 0) · ~N(0, 0)u2 + ~Xuv(0, 0) · ~N(0, 0)uv

+1

2~Xvv(0, 0) · ~N(0, 0)v2 + ~R(u, v) · ~N(0, 0)

=1

2(L11(0, 0)u2 + 2L12(0, 0)uv + L22(0, 0)v2)

+ ~R(u, v) · ~N(0, 0)

=1

2IIp

((u

v

))+ ~R(u, v) · n(p).

Solving the quadratic equation

L11u2 + 2L12uv + L22v

2 = 0 (6.21)

for u in terms of v leads to

L11u =(−L12 ±

√(L12)2 − L11L22

)v.

If we assume that (u, v) 6= (0, 0), Equation (6.21) has no solutionsif det(Lij) > 0 and has solutions if det(Lij) < 0. Hence, if p ishyperbolic, IIp(

(uv

)) changes sign in a neighborhood of (0, 0), and if

p is elliptic, it does not.Define functions R1 and R2 by

Ri(u, v) =Ri(u, v)

u2 + v2for i = 1, 2.

Then solving h(u, v) = 0 amounts to solving

(L11 + R1(u, v))u2 + 2L12uv + (L22 + R2(u, v))v2 = 0.


Since both R1 and R2 have a limit of 0 as (u, v) approaches (0, 0),

lim(u,v)→(0,0)

(L11(0, 0) + R1(u, v))(L22(0, 0) + R2(u, v))− L12(0, 0)2

= L11(0, 0)L22(0, 0)− L12(0, 0)2

Thus, if p is an elliptic point, there is a neighborhood of (0, 0) inwhich h(u, v) = 0 does not have solutions except for (u, v) = (0, 0),and if p is hyperbolic, every neighborhood of (0, 0) has points throughwhich h(u, v) changes sign.

Let us return now to considering the differential of the Gaussmap. With Equation (6.18), it is now possible to explicitly calcu-late the matrix for dnp : TpS → TpS in terms of the oriented basis

( ~Xu, ~Xv), where ~X : U → R3 is a positively oriented parametrizationof a neighborhood around p. Since ~Nu and ~Nv lie in TpS, there existfunctions aij(u, v) defined on U such that

~Nu = a11~Xu + a2

1~Xv,

~Nv = a12~Xu + a2

2~Xv.

(6.22)

Using numerical indices to represent corresponding derivatives, Equa-tion (6.22) can be written as

~Nj =

2∑i=1

aij~Xi, (6.23)

where we denote ~X1 = ~Xu and ~X2 = ~Xv. (It is important to remem-ber that the superscripts in aij also correspond to indices and not topowers. This notation, though perhaps awkward at first, is the stan-dard notation for components of a tensor. It is useful to rememberthat the superscript represents a row index while the subscript is acolumn index.

For any vector ~w ∈ TpS with coordinates ~w =(w1

w2

)= w1

~Xu +

w2~Xv, the differential of the Gauss map satisfies

dnp(~w) = w1~N1 + w2

~N2

= (a11w1 + a1

2w2) ~X1 + (a21w1 + a2

2w2) ~X2

=

(a1

1 a12

a21 a2

2

)(w1

w2

).


However, from Equations (6.16), (6.18), and (6.23),

−Lij = ~Ni · ~Xj =

(2∑

k=1

aki ~Xk

)· ~Xj ,

and so

−Lij =

2∑k=1

aki gkj =

2∑k=1

gjkaki . (6.24)

In matrix notation, Equation (6.24) means that

−(L11 L12

L21 L22

)=

(g11 g12

g21 g22

)(a1

1 a12

a21 a2

2

). (6.25)

Since the metric tensor is a positive definite matrix at any point ona regular surface, one can multiply both sides of Equation (6.25) bythe inverse of (gij), and conclude that the matrix for dnp given in

terms of the basis ~Xu, ~Xv is(a1

1 a12

a21 a2

2

)= −

(g11 g12

g21 g22

)−1(L11 L12

L21 L22

). (6.26)

This matrix formula is more common in modern texts but clas-sical differential geometry texts refer to the individual componentequations implicit in the above formula as the Weingarten equations.

Note that the (gij) and (Lij) matrices are symmetric but thatthe (aij) matrix need not be symmetric.

Since all the above matrices are 2 × 2, det(Lij) = det(−Lij)and since det(gij) > 0, the determinant det(Lij) has the same signas det(aij). Therefore, one deduces the following reformulation ofDefinition 6.4.4.

Proposition 6.4.8. Let S be an oriented regular surface of class C2

with orientation n. Then, a point p ∈ S is called

1. elliptic if det(dnp) > 0;

2. hyperbolic if det(dnp) < 0;

3. parabolic if det(dnp) = 0 but dnp 6= 0;

4. planar if dnp = 0.


Problems

6.4.1. Calculate the second fundamental form (i.e., the matrix of functions

(Lij)) for the ellipsoid ~X(u, v) = (a cosu sin v, b sinu sin v, c cos v).


(Lij)) for the parabolic hyperboloid ~X(u, v) = (au, bv, uv).


(Lij)) for the catenoid ~X(u, v) = (cosh v cosu, cosh v sinu, v).

6.4.4. Calculate the second fundamental form (i.e., the matrix of functions(Lij)) using the following parametrizations of the right cylinder:

(a) ~X(u, v) = (cosu, sinu, v).

(b) ~Y (u, v) = (cos(u+ v), sin(u+ v), v).

Using the parametrization ~Y , find all vectors ~v ∈ TpS such thatIIp(~v) = 0. Show that these correspond to the straight lines on thecylinder.

6.4.5. Consider Enneper’s surface parametrized by

~X(u, v) =

(u− u3

3+ uv2, v − v3

3+ vu2, u2 − v2

).

Show that

(a) the coefficients of the first fundamental form are

g11 = g22 = (1 + u2 + v2)2, g12 = 0;

(b) the coefficients of the second fundamental form are

L11 = 2, L12 = 0, L22 = −2.

6.4.6. Suppose that an open set V of a regular oriented surface has twosystems of coordinates (x1, x2) and (x1, x2) and suppose that V is

parametrized by ~X(x1, x2) in terms of (x1, x2) ∈ U and by ~Y (x1, x2)in terms of (x1, x2). Call Lij the terms of the second fundamentalform in terms of the (x1, x2) coordinates and call Lij the terms ofthe second fundamental form in terms of the (x1, x2) coordinates.Prove that

Lkl =2∑

i,j=1

∂xi∂xk

∂xj∂xl

Lij .

6.5. Normal and Principal Curvatures 209

6.4.7. Use the previous exercise to show that under the same conditions

det(Lkl) =

(∂(x1, x2)

∂(x1, x2)

)2

det(Lij).

6.4.8. Suppose we are under the same conditions as in Problem 6.4.6. Call[dF ] the 2 × 2 matrix of the differential of the coordinate change,that is

[dF ] =

(∂xi∂xj

)i,j=1,2

.

Call aji the coefficients of dnp in terms of the coordinate system(x1, x2). Prove that, as matrices,

(aji ) = (dF ) (alk) (dF )−1.

6.4.9. Calculate the second fundamental form (Lij) and the matrix for

dnp for function graphs ~X(u, v) = (u, v, f(u, v)). Determine whichpoints are elliptic, hyperbolic, or parabolic. (This exercise coupledwith Proposition 6.4.8 is a proof of the second derivative test frommultivariable calculus.)

6.4.10. Prove that all points on the tangential surface of a regular spacecurve are parabolic.

6.4.11. Give an example of a surface with an isolated parabolic point.

6.4.12. Consider a regular surface S given by the equation F (x, y, z) = 0.Use implicit differentiation to provide a criterion for determiningwhether points are elliptic, hyperbolic, parabolic, or planar. [Hint:Assume that z can be expressed as a function of x and y. Then byimplicit differentiation,

∂z

∂x= −Fx

Fzand

∂z

∂y= −Fy

Fz.

Then using the parametrization of S by ~X(x, y) = (x, y, z(x, y)) and

these implicit derivatives, it is possible to calculate ~N and then alsothe Lij matrix.]

6.5 Normal and Principal Curvatures

One way to analyze the shape of a regular surface S near a point p isto consider curves on S through p and analyze the normal component


of their principal curvature vector. In order to use the techniques ofdifferential geometry, in this section we will always assume that thesurface S is of class C2 and that the curves on S are regular curvesalso of class C2.

Definition 6.5.1. Let S be a regular surface of class C2, and let ~X :U → R3 be a parametrization of a coordinate neighborhood V of S.Let ~γ : I → R3 be a parametrization of class C2 for a curve C thatlies on S in V . The normal curvature of S along C is the function

κn(t) =1

s′~T ′ · ~N = κ(~P · ~N) = κ cos θ,

where θ is the angle between the principal normal vector ~P of thecurve and the normal vector ~N of the surface.

Interestingly enough, though the curvature of a space curve ingeneral depends on the second derivative of the curve, once one hasthe second fundamental form for a surface, the normal curvature ata point depends only on the direction.

Proposition 6.5.2. Let p be a point on S in the neighborhood V andsuppose that ~γ is the parametrization of a curve C on S such that~γ(0) = p and ~T (0) = ~w ∈ TpS. Then at p, we write

κn(0) = IIp(~w).

Proof: Suppose that ~γ = ~X ~α, where ~α(t) = (u(t), v(t)) is a curvein the domain U . The normal vector of S along the curve is givenby the function ~N(t) = ~N(~α(t)). Since ~T (t) · ~N(t) = 0 for all t ∈ I,then

~T ′ · ~N = −~T · ~N ′,and, hence,

κn(t) = − 1

s′(t)~T · ~N ′.

Since ~N(t) = ~N(u(t), v(t)), then ~N ′(t) = ~Nuu′(t) + ~Nvv

′(t), and

therefore, ~N ′(0) = dnp(u′(0)v′(0)

)= dnp(s

′(0)~w). Thus, at the point p,

the normal curvature of S along C is

κn = − 1

s′(0)~T (0) · dnp(s′(0)~w) = −~w · dnp(~w) = IIp(~w).


Corollary 6.5.3. All curves C on S passing through the point p withdirection ~w ∈ TpS have the same normal curvature at p.

Because of Corollary 6.5.3, given a point p on a regular surfaceS, one knows everything about the local change in the Gauss mapby knowing the values of the second fundamental form on the unitcircle in TpS around p. In particular, most interesting are the optimalvalues of IIp(~w) for ‖~w‖ = 1 in TpS. To find these optimal values,

set ~w =(ab

)in the coordinate basis ~Xu, ~Xv and optimize

IIp(~w) = L11a2 + 2L12ab+ L22b

2,

with variables a, b subject to the constraint

Ip(~w, ~w) = g11a2 + 2g12ab+ g22b

2 = 1.

Using Lagrange multipliers, one finds that there exists λ such thatthe optimization problem is solved when

L11a+ L12b = λ(g11a+ g12b),

L21a+ L22b = λ(g21a+ g22b),

or, in matrix form,(L11 L12

L21 L22

)(ab

)= λ

(g11 g12

g21 g22

)(ab

). (6.27)

This remark leads to the following fundamental proposition.

Proposition 6.5.4. The maximum and minimum values κ1 and κ2 ofIIp(~w) restricted to the unit circle are the negatives of the eigenvaluesof dnp. Furthermore, there exists an orthonormal basis ~e1, ~e2 ofTpS such that dnp(~e1) = −κ1~e1 and dnp(~e2) = −κ2~e2.

Proof: By multiplying on the left by the inverse g−1 of the metrictensor, Equation (6.27) becomes

g−1L

(a

b

)= λ

(a

b

).

Since the matrix of the differential of the Gauss map is[dnp]

=−g−1L, the first part of the proposition follows.


Since the first fundamental form Ip(·, ·) is positive definite andsince, by Proposition 6.4.1, dnp is self-adjoint with respect to thisform, then by the Spectral Theorem, dnp is diagonalizable. (Manylinear algebra textbooks discuss the Spectral Theorem exclusivelyusing the dot product as the positive definite form. See [21, Sec-tion 7, Chapter XV] for a more general presentation of the SpectralTheorem, which we use here.)

Now, since dnp is self-adjoint with respect to the first fundamen-tal form,

Ip(dnp(~e1), ~e2) = Ip(−κ1~e1, ~e2) = −κ1Ip(~e1, ~e2),

‖Ip(~e1, dnp(~e2)) = Ip(~e1,−κ2~e2) = −κ2Ip(~e1, ~e2),

and thus,(κ1 − κ2)Ip(~e1, ~e2) = 0.

Hence, if κ1 6= κ2, then Ip(~e1, ~e2) = ~e1 · ~e2 = 0, where in the latterdot product we view ~e1 and ~e2 as vectors in R3, and if κ1 = κ2, thenby the Spectral Theorem, any orthonormal basis satisfies the claimof the proposition.

Definition 6.5.5. Let S be a regular surface, and let p be a point onS. The maximum and minimum normal curvatures κ1 and κ2 atp are called the principal curvatures of S at p. The correspondingdirections, i.e., unit eigenvectors ~e1 and ~e2 with dnp(~ei) = −κi~ei, arecalled principal directions at p.

In a plane, the second fundamental form is identically 0, all nor-mal curvatures including the principal curvatures are 0, and hence,all directions at all points are principal directions. Similarly, it is nothard to show that at every point of the sphere the normal curvaturein every direction is the same, and hence, all directions are principal.

Example 6.5.6 (Ellipsoids). As a contrast to the plane or sphere, con-sider the ellipsoid parametrized by

~X(u, v) = (a cosu sin v, b sinu sin v, c cos v).

It turns out that the formulas for κ1 and κ2 as functions of (u, v) arequite long so we shall calculate the principal curvatures at the point


corresponding to (u, v) =(π3 ,

π6

). It is not too hard to calculate the

coefficients of the first and second fundamental forms

g11 = a2 sin2 u sin2 v + b2 cos2 u sin2 v, L11 =abc sin2 v√

det g,

g12 = (b2 − a2) sinu cosu sin v cos v, L12 = 0,

g22 = a2 cos2 u cos2 v + b2 sin2 u cos2 v + c2 sin2 v, L22 =abc√det g

.

At (u, v) =(π3 ,

π6

), this leads to

(aji )=8abc

(12a2b2 + 3a2c2 + b2c2)3/2

(−3a2 − 9b2 − 4c2 −12a2 + 12b2

−3a2 + 3b2 −12a2 − 4b2

),

and, after some algebra simplifications, the eigenvalues of this matrixare

λ=4abc

(−15a2 − 13b2 − 4c2±

√(9a2 − 5b2 − 4c2)2 + 36(a2 − b2)2

)(12a2b2 + 3a2c2 + b2c2)3/2

.

From this we see that at this particular point for (u, v), the eigen-values are equal if and only if a2 − b2 = 0 and 9a2 − 5b2 − 4c2 = 0,which is equivalent to a2 = b2 = c2.

If a curve on the surface given by γ(t) = ~X(u(t), v(t)) is suchthat at γ(t0) its direction γ′(t0) is a principal direction with principalcurvature κi, then by Equation (6.27), with λ = −κi, we have

L11u′ + L12v

′ = −κi(g11u′ + g12v

′),

L21u′ + L22v

′ = −κi(g21u′ + g22v

′).

Eliminating κi from these two equations leads to the relationship

(L11g21 − L21g11)(u′)2 + (L11g22 − L22g11)u′v′ + (L12g22 − L22g12)(v′)2 = 0, (6.28)

or equivalently, ∣∣∣∣∣∣(v′)2 −u′v′ (u′)2

g11 g12 g22

L11 L12 L22

∣∣∣∣∣∣ = 0.


Also, in light of the Weingarten equations in Equation (6.26), we cansummarize the above two formulas by(

v′

−u′

)· dnp

(u′

v′

)= 0.

Definition 6.5.7. Any regular curve on a regular oriented surface givenby γ(t) = ~X(u(t), v(t)) in a coordinate neighborhood parametrizedby ~X : U → R3 that satisfies Equation (6.28) for all t is called a lineof curvature.

The lines of curvature on a surface form an orthogonal family ofcurves on the surface, that is, two sets of curves intersecting at rightangles. With an appropriate change of variables and in an interval

of t where u′(t) 6= 0, using the chain rule dvdu = v′(t)

u′(t) , (6.28) can berewritten as

(L12g22−L22g12)

(dv

du

)2

+(L11g22−L22g11)dv

du+(L11g21−L21g11) = 0.

(6.29)If in the neighborhood we are studying u′(t) = 0 for some t, thensince we assume the curve is regular, v′(t) cannot also be 0, andhence, we can rewrite Equation (6.28) with u as a function of v.In general, solving the above differential equation is an intractableproblem. Of course, as long as the coefficients of the first and secondfundamental form are not proportional to each other, one can easilysolve algebraically for dv

du in Equation (6.29) and obtain two distinct

solutions of dvdu as a function of u and v. Then according to the

theory of differential equations (see [2, Section 9.2] for a reference),given any point (u0, v0) and for each of the two solutions of dv

du as afunction of u and v, there exists a unique function v = f(u) solvingEquation (6.29).

Consequently, a regular oriented surface of class C2 can be cov-ered by lines of curvature wherever the coefficients of the first andsecond fundamental form are continuous and are not proportional toeach other. Points where this cannot be done have a special name.

Definition 6.5.8. Let S be a regular oriented surface of class C2. Anumbilical point is a point p on S such that, given a parametriza-tion ~X of a neighborhood of p, the corresponding first and second


fundamental forms have coefficients that are proportional, namely,

L11g12 − L12g11 = 0, L11g22 − L22g11 = 0, and L22g12 − L12g22 = 0.

Proposition 6.5.9. Let S be a regular oriented surface. A point p on Sis an umbilical point if and only if the eigenvalues of dnp are equal.

Proof: Let ~X be a parametrization of a neighborhood of p. Withrespect to this parametrization and the associated basis ~Xu, ~Xvon TpS, the matrix of dnp is

[dnp]

= (aji ). Proposition 6.5.4 implies

that (aji ) is diagonalizable.

The matrix (aji ) has equal eigenvalues if and only if there existsan invertible matrix B such that

(aji ) = B

(λ 00 λ

)B−1,

and then(aji ) = B(λI)B−1 = λBB−1 = λI.

Consequently (aji ) has equal eigenvalues if and only if it is alreadydiagonal, with elements on the diagonal being equal. Then λI =(aji ) = −g−1L, and therefore, L = −λg, which is tantamount tosaying that the coefficients of the first and second fundamental formsare proportional. The proposition follows.

The proof of Proposition 6.5.4 shows that if κ1 6= κ2, then theorthonormal basis of principal directions is unique up to signs, whileif κ1 = κ2, any orthogonal basis is principal. Therefore, in light ofProposition 6.5.9, at umbilical points, there is no preferred basis ofprincipal directions, and thus it makes sense that the only solutionsto Equation (6.28) at umbilical points have u′(t) = v′(t) = 0, thatis, (u(t), v(t)) = (u0, v0). Note that at an umbilical point, Equation(6.28) degenerates to the trivial equation 0 = 0.

Solving Equation (6.28) gives the lines of curvature on a surface.At every point p on the surface that is not umbilical, there are twolines of curvature through p, and they intersect at right angles. Thissimple remark leads to the following nice characterization.

Proposition 6.5.10. The coordinate lines of a parametrization ~X of asurface are curvature lines if and only if g12 = L12 = 0.


Proof: Suppose that coordinate lines are curvature lines. Since anytwo lines of curvature intersect at a point at a right angle, we knowthat g12 = ~Xu · ~Xv = 0. Furthermore, coordinate lines are given by~γ(t) = ~X(t, v0) or ~γ(t) = ~X(u0, t). If ~γ(t) = ~X(t, v0), then u′(t) = 1and v′(t) = 0, and if ~γ(t) = ~X(u0, t), then u′(t) = 0 and v′(t) = 1.Then Equation (6.28) implies that both of the following hold:

L11g21 − L21g11 = 0 and L12g22 − L22g12 = 0.

Since g12 = 0, we have L21g11 = 0 and L12g22 = 0. However, sincedet(g) = g11g22 − g2

12 > 0, then at all points on S the functions g11

and g22 cannot both be 0 at the same time. Since L12 = L21, wededuce that L12 = 0.

Conversely, if g12 = L12 = 0, then (gij) and (Lij) are both di-

agonal matrices, making (aij) a diagonal matrix, and hence, ~Xu and~Xv are eigenvectors of dnp. Hence, coordinate lines are curvaturelines.

As a linear transformation from TpS to itself, dnp is independentof a parametrization of a neighborhood of p. Therefore, if p is notan umbilical point, the principal directions ~e1, ~e2 as eigenvectorsof dnp provide a basis of TpS that possesses more geometric mean-ing than the coordinate basis of any particular parametrization of aneighborhood of p.

In the orthonormal basis ~e1, ~e2, a unit vector ~w ∈ TpS is writtenas ~w = cos θ~e1 + sin θ~e2 for some angle θ. Using these coordinates,the normal curvature of S at p in the direction of ~w is

IIp(~w) = −~w · dnp(~w)

= −(cos θ~e1 + sin θ~e2) · dnp(cos θ~e1 + sin θ~e2)

= −(cos θ~e1 + sin θ~e2) · (cos θdnp(~e1) + sin θdnp(~e2))

= −(cos θ~e1 + sin θ~e2) · (− cos θκ1~e1 − sin θκ2~e2)

IIp(~w) = (cos2 θ)κ1 + (sin2 θ)κ2. (6.30)

Equation (6.30) is called Euler’s curvature formula.Another useful geometric characterization of the behavior of S

near p is called the Dupin indicatrix . The Dupin indicatrix consistsof all vectors ~w ∈ TpS such that IIp(~w) = ±1. If ~w = (w1, w2) =


(ρ cos θ, ρ sin θ) are expressions of ~w in Cartesian and polar coordi-nates referenced in terms of the orthonormal frame ~e1, ~e2, thenEuler’s curvature formula gives

±1 = IIp(~w) = ρ2IIp(~w) = κ1ρ2 cos2 θ + κ2ρ

2 sin2 θ.

Thus, the Dupin indicatrix satisfies the equation

κ1w21 + κ2w

22 = ±1. (6.31)

Since the principal curvatures at a point p are the negatives of theeigenvalues of dnp, Proposition 6.4.8 provides a characterization ofwhether a point is elliptic, hyperbolic, parabolic, or planar in termsof the principal curvatures. More precisely, a point p ∈ S is

1. elliptic if κ1 and κ2 have the same sign;

2. hyperbolic if κ1 and κ2 have opposite signs;

3. parabolic if exactly one of κ1 and κ2 is 0;

4. planar if κ1 = κ2 = 0.

The Dupin indicatrix justifies this terminology because p is el-liptic or hyperbolic if and only if Equation (6.31) is the equationfor a single ellipse or two hyperbolas, respectively. Furthermore, thehalf-axes for the corresponding ellipse or hyperbola are√

1

|κ1|and

√1

|κ2|.

See Figure 6.10 for an illustration of the Dupin indicatrix in theelliptic and hyperbolic cases. Also observe that this figure has |κ1| >|κ2|.

If a point is parabolic, then the Dupin indicatrix is simply a pairof parallel lines equidistant from one of the principal direction lines,and if a point is planar, the Dupin indicatrix is the empty set.

When a point is hyperbolic, it is possible to find the asymptotesof the Dupin indicatrix without referring directly to the principalcurvatures.


~e1

~e2

p

ρ

pθ

(a) Elliptic point.

~e1

~e2

pθ

(b) Hyperbolic point.

Figure 6.10. The Dupin indicatrix.

Definition 6.5.11. Let p be a point on a regular oriented surface S ofclass C2. An asymptotic direction of S at p is a unit vector ~w in TpSsuch that the normal curvature is 0. An asymptotic curve on S isa regular curve C such that at every point p ∈ C, the unit tangentvector at p is an asymptotic direction.

Since the principal curvatures at a point p are the maximumand minimum normal curvatures, at an elliptic point, there exist noasymptotic directions, which one can see from the Dupin indicatrix.If, on the other hand, p is hyperbolic, then the principal curvatureshave opposite signs. If a unit vector ~w is an asymptotic direction atp and makes an angle θ with ~e1, then by Euler’s formula we deducethat

cos2 θ = − κ2

κ1 − κ2.

The right-hand side is always positive and less than 1, so this equa-tion leads to four solutions for ~w, i.e., two mutually negative pairs,each representing one of the asymptotes of the Dupin indicatrix hy-perbola. As we see in Problem 6.5.11, the asymptotic curves in factprovide a more subtle description of the behavior of a surface neara point than the Dupin indicatrix does since it describes more thansecond-order phenomena.

As with the lines of curvature, it is not difficult to find a dif-ferential equation that characterizes asymptotic curves on a surface.Suppose that ~X : U → R3 parametrizes a neighborhood of p ∈ Sand that ~γ : I → S is a curve on S defined by ~γ = ~X ~α, with


~α : I → U and ~α(t) = (u(t), v(t)). By Proposition 6.5.2, if we callκn(t) the normal curvature of S at ~γ(t) in the direction of ~γ′(t), then

κn(t) = II~γ(t)

(~T (t)

)=

1

(s′(t))2II~γ(t)

((u′

v′

)).

Thus, since an asymptotic curve must satisfy κn(t) = 0 for all t, thenany asymptotic curve must satisfy the differential equation

L11(u′)2 + 2L12u′v′ + L22(v′)2 = 0.

Example 6.5.12 (Surfaces of Revolution). Consider a surface of revolu-tion parametrized by

~X(u, v) = (f(v) cosu, f(v) sinu, h(v)),

with f(v) > 0, u ∈ (0, 2π), and v ∈ (a, b). It is not hard to show (seeProblem 6.6.11) that the coefficients of the first fundamental formare

g11 = f(v)2, g12 = g21 = 0, g22 = (f ′(v))2 + (h′(v))2,

and the coefficients of the second fundamental form are

L11 = − fh′√(f ′)2 + (h′)2

, L12 = L21 = 0, L22 =f ′′h′ − f ′h′′√(f ′)2 + (h′)2

.

By Problem 6.5.6, we deduce that the meridians (where u = const.)and the parallels (where v = const.) are lines of curvature. We canalso conclude that the principal curvatures are

κ1(u, v) = − h′(v)

f(v)√

(f ′(v))2 + (h′(v))2,

κ2(u, v) =f ′′(v)h′(v)− f ′(v)h′′(v)

(f ′(v))2 + (h′(v))3/2,

(6.32)

though, as written, one can make no assumption that κ1 > κ2. Ob-viously, the first and second fundamental forms depend only on thecoordinate v, and hence, all properties of points, such as whetherthey are elliptic, hyperbolic, parabolic, planar, or umbilical, dependonly on v. Setting κ1 = κ2 in Equation (6.32) produces an equationthat determines for what v the points on the surface are umbilicalpoints.


Problems

6.5.1. Find the principal directions and the principal curvatures of thequadric surface z = ax2 + 2bxy + cy2 at (0, 0, 0) in terms of theconstants a, b, c.

6.5.2. Find the principal directions and the principal curvatures of thesurface z = 4x2 − x4 − y2 at its critical points (as a function in twovariables).

6.5.3. Provide the details for Example 6.5.6.

6.5.4. Consider the ellipsoid with half-axes a, b, and c.

(a) Prove the ellipsoid has four umbilical points when a, b, and care distinct.

(b) Calculate the coordinates of the umbilical points when all half-axes have different length.

(c) What happens when two of the half-axes are equal?

6.5.5. (ODE) Determine the asymptotic curves and the lines of curvatureof the helicoid

~X(u, v) = (v cosu, v sinu, cu).

6.5.6. Let ~X be the parametrization for a neighborhood of S. Prove that if(gij) and (Lij) are diagonal matrices, then the lines of curvature arethe coordinate curves (curves on S where u =const. or v =const.).

6.5.7. Let ~X be the parametrization for a neighborhood of S. Prove thatthe coordinate curves are asymptotic curves if and only if L11 =L22 = 0.

6.5.8. (ODE) Determine the asymptotic curves of the catenoid

~X(u, v) = (cosh v cosu, cosh v sinu, v).

6.5.9. Consider Enneper’s surface. Use the results of Problem 6.4.5 to showthe following:

(a) The principal curvatures are

κ1 =2

(1 + u2 + v2)2, κ2 = − 2

(1 + u2 + v2)2.

(b) The lines of curvature are the coordinate curves.

(c) The asymptotic curves are u+ v =const. and u− v =const.

6.6. Gaussian and Mean Curvatures 221

6.5.10. Find equations for the lines of curvature when the surface is givenby z = f(x, y).

6.5.11. Consider the monkey saddle given by the graph of the functionz = x3 − 3xy2. Prove that the set of asymptotic curves that pos-sesses (0, 0, 0) as a limit point consists of three straight lines through(0, 0, 0) with equal angles between them.

6.5.12. Let S1 and S2 be two regular surfaces that intersect along a regularcurve C. Let p be a point on C, and call λ1 and λ2 the normalcurvatures of S at p in the direction of C. Prove that the curvatureκ of C at p satisfies

κ2 sin2 θ = λ21 + λ22 − 2λ1λ2 cos2 θ,

where θ is the angle between S1 and S2 at p (calculated using thenormals to S1 and S2 at p).

6.5.13. Let S be a regular oriented surface and p a point on S. Two nonzerovectors ~u1, ~u2 ∈ TpS are called conjugate if

Ip(dnp(~u1), ~u2) = Ip(~u1, dnp(~u2)) = 0. (6.33)

Prove the following:

(a) A curve C on S parametrized by ~γ : I → S is a line of curvature

if and only if the unit tangent vector ~T and any normal vectorto ~T in T~γ(t)(S) are conjugate to each other.

(b) Let ~γ1(t) and ~γ2(t) be regular space curves, and define a sur-

face S by the parametrization ~X(u, v) = ~γ1(u) + ~γ2(v). Sur-faces constructed in this manner are called translation surfaces.Show that the coordinate lines of S are conjugate lines.

6.6 Gaussian and Mean Curvatures

We now arrive at two fundamental geometric invariants that encap-sulate a considerable amount of useful information about the localshape of a surface. Again, in this section, we must assume that S isa regular surface of class C2.

Definition 6.6.1. Let κ1 and κ2 be the principal curvatures of a regularoriented surface S at a point p. Define

1. the Gaussian curvature of S at p as the product K = κ1κ2;


2. the mean curvature of S at p as the average H = κ1+κ22 of the

principal curvatures.

By Problem 6.4.8, the matrix for the Gauss map in terms of anyparticular coordinate system on a neighborhood of p on S is conju-gate to the corresponding matrix for a different coordinate system via

the change of basis matrix(∂xi∂xj

). Therefore, since det(BAB−1) =

det(A) and Tr(BAB−1) = Tr(A), which is a standard result in linearalgebra, the eigenvalues of the Gauss map, the principal curvatures,the Gaussian curvature, and the mean curvature are invariant undercoordinate changes. Furthermore, since det(−A) = det(A) when Ais a square matrix with an even number of rows, then

K = κ1κ2 = det(aji ) = det(dnp),

H =κ1 + κ2

2= −1

2Tr(aji ) = −1

2Tr(dnp).

Consequently, as claimed above, the Gaussian curvature and themean curvature (up to a sign) of S at p are geometric invariants inthat they do not depend on the orientation or position of S in space,and they do not depend on any particular coordinate system on Sin a neighborhood of the point p.

In order to calculate the mean curvature H(u, v), one has nochoice but to calculate the matrix of the differential of the Gauss map(aij) = [dnp]. However, from a computational perspective, Equation(6.26) leads to a much simpler formula for the Gaussian curvaturefunctionK(u, v). Recall that det(AB) = det(A) det(B) for all squarematrices of the same size and also that det(A−1) = 1/ det(A). ThenEquation (6.26) implies that

K =det(Lij)

det(gij)=L11L22 − L2

12

g11g22 − g212

. (6.34)

This equation lends itself readily to calculations since one does notneed to fully compute the matrix of the Gauss map, let alone findits eigenvalues. However, we can also obtain an alternate character-ization of the Gaussian curvature.


Proposition 6.6.2. Let S be a regular oriented surface of class C2, andlet V be a neighborhood parametrized by ~X : U ⊂ R2 → R3. Definethe unit normal vector ~N as

~N =~Xu × ~Xv

‖ ~Xu × ~Xv‖.

Then over the domain U , the Gaussian curvature is the unique func-tion K(u, v) satisfying

~Nu × ~Nv = K(u, v) ~Xu × ~Xv.

Proof: From the definition of the (aij) functions in Equation (6.22),we have

~Nu × ~Nv = (a11~Xu + a2

1~Xv)× (a1

2~Xu + a2

2~Xv).

It is then easy to see that

~Nu × ~Nv = det(aij) ~Xu × ~Xv.

The result follows since K = det(aij).

Corollary 6.6.3. If V is a region of a regular oriented surface S ofclass C2, then ∫∫

VK dS =

∫∫n(V )

dS,

where the latter integral is the signed area on the unit sphere of theimage of V under the Gauss map.

Example 6.6.4 (Spheres). Consider the sphere parametrized by the vec-tor function


with (u, v) ∈ (0, 2π)× (0, π). Example 6.1.5 and Example 6.4.3 gaveus

(gij) =

(R2 sin2 v 0

0 R2

)and (Lij) =

(−R sin2 v 0

0 −R

).


Using Equation (6.34), it is then easy to see that the Gaussian cur-vature function on the sphere is

K(u, v) =R2 sin2 v

R4 sin2 v=

1

R2,

which is a constant function. (Note that we assumed that v 6= 0, π,which correspond to the north and south poles of the parametri-zation. However, another parametrization that includes the northand south poles would show that at these points as well we haveK = 1/R2.)

The matrix of the Gauss map is then

[dnp]

= −g−1L =

(− 1R 0

0 − 1R

),

from which one immediately deduces that the principal curvaturesare κ1(u, v) = κ2(u, v) = 1

R . This shows that all points on thesphere are umbilical points. In addition, the mean curvature is alsoa constant function H(u, v) = 1

R .

Example 6.6.5 (Function Graphs). Consider the graph of a function f :U ⊂ R2 → R. The graph can be parametrized by ~X : U → R3, with

~X(u, v) = (u, v, f(u, v)).

Problem 6.4.9 asks the reader to calculate the matrix (Lij). It is nothard to show that

(gij) =

(1 + (fu)2 fufvfvfu 1 + (fv)

2

)and

(Lij) =1√

1 + f2u + f2

v

(fuu fuvfvu fvv

),

where fu is the typical shorthand to mean ∂f∂u . But then det(g) =

1 + f2u + f2

v , and we find that the Gaussian curvature function on afunction graph is

K(u, v) =1

(1 + f2u + f2

v )2

∣∣∣∣fuu fuvfvu fvv

∣∣∣∣ =fuufvv − f2

uv

(1 + f2u + f2

v )2.


This result allows us to rephrase the second derivative test in thecalculus of a function f from R2 to R as follows: If f has contin-uous second partial derivatives and (u0, v0) is a critical point (i.e.,fu(u0, v0) = fv(u0, v0) = 0), then

1. (u0, v0) is a local maximum if K(u0, v0) > 0 and fuu(u0, v0) <0;

2. (u0, v0) is a local minimum ifK(u0, v0) > 0 and fuu(u0, v0) > 0;

3. (u0, v0) is a saddle point if K(u0, v0) < 0;

4. the test is inconclusive if K(u0, v0) = 0.

In the language we have introduced in this chapter, local minima andlocal maxima of the function z = f(u, v) are elliptic points, saddlepoints are hyperbolic points, and points where the second derivativetest is inconclusive are either parabolic or planar points.

Example 6.6.6 (Pseudosphere). A tractrix is a curve in the plane withparametric equations

~α(t) = (sech t, t− tanh t),

and it has the y-axis as an asymptote. The pseudosphere is definedas half of the surface of revolution of a tractrix about its asymptote.More precisely, we can parametrize the pseudosphere by

~X(u, v) = (sech v cosu, sech v sinu, v − tanh v),

with u ∈ [0, 2π) and v ∈ [0,∞). (See Figure 6.11 for a picture of thetractrix and the pseudosphere.) We leave it as an exercise for thereader to prove that

(gij) =

(sech2 v 0

0 tanh2 v

)and

(Lij) =

(− sech v tanh v 0

0 sech v tanh v

).

The Gaussian curvature for the pseudosphere is K = −1, whichmotivates the name “pseudosphere” since it is analogous to the unitsphere but with a constant Gaussian curvature of −1 instead of 1.


Figure 6.11. Tractrix and pseudosphere.

Proposition 6.6.7. Let S be a regular oriented surface of class C2 withGauss map n : S → S2, and let p be a point on S. Call K(p) theGaussian curvature of S at p, and suppose that K(p) 6= 0. Let Bε bethe ball of radius ε around p and define Vε = Bε ∩ S. Then

|K(p)| = limε→0

Area(n(Vε))

Area(Vε).

In other words, the absolute value of the Gaussian curvature at apoint p is the limit around p of the ratio of the surface area on S2

mapped under the Gauss map to the corresponding surface area on S.

Proof: Let ~X : U → R3 be a regular parametrization of a neighbor-hood of p = ~X(u0, v0), where U is an open subset of R2. Since ~X(U)is an open set, Vε ⊂ ~X(U) for all ε small enough. Furthermore,since we assumed that K(p) 6= 0, by the continuity of the Gaussiancurvature function K : U → R, we deduce that K does not changesign for ε chosen small enough. Therefore, from now on, we assumethat K does not change sign.

Define Uε = ~X−1(Vε), the preimage of the neighborhood Vε under~X. Since ~X is bijective as a regular parametrization, (u0, v0) isthe unique point in all Uε for all ε > 0. Then by the formula forsurface area,

Area(Vε) =

∫∫Vε

dA =

∫∫Uε

‖ ~Xu × ~Xv‖ du dv.


Similarly, on the unit sphere,

Area(n(Vε)) =

∫∫Uε

‖ ~Nu × ~Nv‖ du dv.

However, by Proposition 6.6.2, we also have

Area(n(Vε)) =

∫∫Uε

|K(u, v)|‖ ~Xu × ~Xv‖ du dv.

By the Mean Value Theorem for double integrals, for every ε > 0,there exist points (uε, vε) and (u′ε, v

′ε) in the open set Uε such that∫∫

Uε

‖ ~Xu × ~Xv‖ du dv = ‖ ~Xu(uε, vε)× ~Xv(uε, vε)‖

and∫∫Uε

‖ ~Nu × ~Nv‖ du dv = |K(u′ε, v′ε)| ‖ ~Xu(u′ε, v

′ε)× ~Xv(u

′ε, v′ε)‖.

Thus, since limε→0(uε, vε) = limε→0(u′ε, v′ε) = (u0, v0), we have

limε→0

Area(n(Vε))

Area(Vε)= lim

ε→0

|K(u′ε, v′ε)| ‖ ~Xu(u′ε, v

′ε)× ~Xv(u

′ε, v′ε)‖

‖ ~Xu(uε, vε)× ~Xv(uε, vε)‖

=|K(u0, v0)|‖ ~Xu(u0, v0)× ~Xv(u0, v0)‖

‖ ~Xu(u0, v0)× ~Xv(u0, v0)‖

= |K(u0, v0)| = |K(p)|.

Problems

6.6.1. In Example 6.6.5 we calculated the Gaussian curvature of functiongraphs. Calculate the mean curvature.

6.6.2. Using an appropriate parametrization, find the Gaussian curvatureof the hyperboloid of one sheet, x2 + y2 − z2 = 1.

6.6.3. Calculate the Gaussian and mean curvature functions of the generalellipsoid

~X(u, v) = (a cosu sin v, b sinu sin v, c cos v).


6.6.4. Calculate the Gaussian curvature of the torus parametrized by

~X(u, v) = ((a+ b cos v) cosu, (a+ b cos v) sinu, b sin v)

where a > b are constants and (u, v) ∈ (0, 2π)× (0, 2π).

6.6.5. Consider a regular space curve ~γ : I → R3 and the tangential surfacedefined by

~X(t, u) = ~γ(t) + u~γ′(t)

for (t, u) ∈ I × R. Calculate the mean and Gaussian curvaturefunctions.

6.6.6. Let ~α(t) and ~β(t) be differentiable vector functions with commondomain I. Define the secant surface between the two resulting curvesby

~X(t, u) = (1− u)~α(t) + u~β(t)

for (t, u) ∈ I ×R. Assume that the corresponding surface is regular.

(a) Prove that K(u, v) = 0 for any point with u = 12 .

(b) Prove that K(u, v) = 0 if and only if u = 12 or (~β(t)− ~α(t)) is

in the plane spanned by ~α′(t) and ~β′(t).

6.6.7. If M is a nonorientable surface, one can define the Gaussian curva-ture on a coordinate patch of M by using ~N = ~Xu × ~Xv‖ ~Xu × ~Xv‖and Equation (6.34) without reference to dnp, which is not well de-fined since no Gauss map n : M → S2 exists. Consider the Mobiusstrip M depicted in Figure 5.15. Show that, using the parametriza-tion of the Mobius strip in Example 5.5.3, the Gaussian curvatureis

K = − 1(14v

2 + (2− v sin u2 )2)2 .

6.6.8. Tubes. Let ~γ : I → R3 be a regular space curve and let r be apositive real number. Consider the tube of radius r around ~γ(t)parametrized by

~X(u, v) = ~γ(u) + (r cos v)~P (u) + (r sin v) ~B(u).

(a) Calculate the second fundamental form (Lij) and the matrixfor dnp.

(b) Calculate the Gaussian curvature function K(u, v) on the tube.

(c) Prove that all the points with K = 0 are either points on

curves ~γ(t) ± r ~B(t) for all t ∈ I or points on circles ~γ(u0) +

(r cos v)~P (u0) + (r sin v) ~B(u0), where u0 satisfies κ(u0) = 0.


6.6.9. Consider the pseudosphere and the parametrization provided in Ex-ample 6.6.6.

(a) Prove the statements about the pseudosphere in Example 6.6.6.

(b) Find the mean curvature function on the pseudosphere.

(c) Modify the given parametric equations to find a parametriza-tion of a surface with constant Gaussian curvature K = − 1

R2 .

(d) Determine the lines of curvature on the pseudosphere.

6.6.10. Consider the monkey saddle ~X(u, v) = (u, v, u3 − 3uv2). Calculate

the Gaussian curvature function of ~X and the points where K > 0,K < 0, or K = 0.

6.6.11. Surfaces of revolution. Consider the surface of revolution defined byrevolving the parametrized curve (f(t), h(t)) in the xy-plane aboutthe y-axis. Its paramtrization is

~X(u, v) = (f(v) cosu, f(v) sinu, h(v))

for u ∈ [0, 2π) and v ∈ I, where I is some interval. Assume that fand h are such that the surface of revolution is a regular surface.

(a) Calculate the second fundamental form (Lij).

(b) Calculate the coefficients of the matrix for dnp.

(c) Calculate the Gaussian curvature function.

(d) Determine which points are elliptic, hyperbolic, parabolic, orplanar.

(e) Prove that the lines of curvature are the coordinate lines.

6.6.12. Normal Variations. Let ~X(u, v) be the parametrization for a coordi-nate patch V of a regular surface S. Consider the normal variationof S over V that is parametrized by

~Yr(u, v) = ~X(u, v) + r ~N(u, v)

for some constant r ∈ R and where ~N is the unit normal vectorassociated to ~X.

(a) Prove that the unit normal ~NY to ~Y is everywhere equal to ~N(except perhaps up to a sign).

(b) Call KY the Gaussian curvature for ~Yr and K the Gaussian

curvature of ~X. Prove that

K( ~Xu × ~Xv) = KY∂~Yr∂u× ∂~Yr

∂v.


(c) Call VY the corresponding coordinate patches on the normalvariation. Conclude that∫∫

V

K dA =

∫∫VY

KY dA.

6.6.13. Consider plane curves ~α(t) = (α1(s), α2(s)) and ~β(t) = (β1(t), β2(t))both parametrized by arc length. Assume we are in R3 with standardbasis ~i,~j,~k. Consider the parametrized surface S given by

~X(s, t) = ~α(s) + β1(t)~U(s) + β2(t)~k,

where ~U(s) is the usual unit normal vector for plane curves.

(a) Prove that if S is regular, then 1−β1(t)κα(s) 6= 0 for all (s, t),where κα(s) is the curvature of the plane curve ~α.

(b) Calculate the Gaussian curvature of S.

(c) Prove that κα(s) = 0 or κβ(t) = 0 imply that K(s, t) = 0, butexplain why the converse is not true.

6.6.14. Let S be a regular oriented surface and p a point of S. Let ~u be anyfixed unit vector in TpS. Show that the mean curvature H at p isgiven by

H =1

π

∫ π

0

κn(θ) dθ,

where κn(θ) is the normal curvature of S along a direction makingan angle θ with ~u.

6.6.15. (*) Theorem of Beltrami-Enneper . Prove that the absolute valueof the torsion at any point on an asymptotic curve with nonzerocurvature is given by

|τ | =√−K,

where K is the Gaussian curvature of the surface at that point.

6.6.16. Let S be a regular surface in R3 parametrized by ~X, and consider thelinear transformation T : R3 → R3 given by T (~x) = A~x with respectto the standard basis, where A is an invertible matrix. It is usuallyan intractable problem to determine the Gaussian or mean curvatureof the image surface S′ = T (S) from those of S. Nonetheless, it ispossible to answer the following question: Prove that T preserves thesign of the Gaussian curvature of any surface in R3, more precisely,if p ∈ S and q = T (p) is the corresponding point on S′, then K(p) =0 ⇔ K(q) = 0 and signK(p) = signK(q). [Hint: Use Equation(6.18).]

6.7. Developable Surfaces and Minimal Surfaces 231

6.6.17. (*) Consider a regular surface S ∈ R3 defined by F (x, y, z) = 0 and apoint p on S. Use implicit differentiation to find a formula for Gaus-sian curvature K at the point p. (The formula is not particularlypretty but can be written concisely using determinants.)

6.7 Developable Surfaces and Minimal Surfaces

When studying plane curves, we showed in Proposition 1.3.5 thatif the curvature κg(t) of a regular plane curve is always 0, then thecurve is a line segment. In this section, we wish to study propertiesof surfaces with either Gaussian curvature everywhere 0 or meancurvature everywhere 0. Since there exist formulas for the meancurvature and Gaussian curvature in terms of a particular parame-trization, the equations K = 0 and H = 0 are partial differentialequations. However, since they are nonlinear differential equationsthat a priori involve three unknown functions in two variables, theyare intractable in general. Nonetheless, we shall study various classesof surfaces that satisfy K = 0 or H = 0.

Planes satisfy K = 0, but other surfaces do as well, for example,cylinders and cones. In a geometric sense, cylinders and cones in factresemble a plane because, as any elementary school student knows,one can create a cylinder or a cone out of a flat paper without folding,stretching, or crumpling. In the first half of this section, we introduceruled surfaces and determine the conditions for these ruled surfacesto have Gaussian curvature everywhere 0.

As of yet, we have neither seen a particularly intuitive interpre-tation of the mean curvature nor have presented surfaces that satisfyH = 0. However, as we shall see, surfaces that satisfy H = 0 haveminimal surface area in the following sense. Given a simple closedcurve C in R3, among surfaces S that have C = ∂S as a boundary,a surface with minimal surface area will have mean curvature H ev-erywhere 0. Consequently, a surface that satisfies H = 0 is called aminimal surface. Many articles and books are devoted to the studyof minimal surfaces (see [22], [28], or [27] to name a few; an Inter-net search will reveal many more), so in the interest of space, thesecond half of this section gives only a brief introduction to minimalsurfaces.


6.7.1 Developable Surfaces

Geometrically, we define a ruled surface as the union of a (differen-tiable) one-parameter family of straight lines in R3. One can specifyeach line in the family by a point ~α(t) and by a direction given byanother vector ~w(t). The adjective “differentiable” means that both~α and ~w are differentiable vector functions over some interval I ⊂ R.We parametrize a ruled surface by

~X(t, u) = ~α(t) + u ~w(t), with (t, u) ∈ I × R. (6.35)

We call the lines Lt passing through ~α(t) with direction ~w(t) therulings, and the curve ~α(t) is called the directrix of the surface. Weshould note that this definition does not insist that ruled surfaces beregular; in particular, we allow singular points, that is, points where~Xt × ~Xu = ~0.

Definition 6.7.1. A developable surface is a surface S such that at eachpoint P on S, there is a line (called generator) through P that lieson S and such that S has the same tangent plane at all points onthis generator.

A developable surface is a surface that can be formed by bendinga portion of the plane into space. Intuitively speaking, each bendline corresponds to a generator. The set of generator lines on thesurface define a one-parameter family of lines in R3 that sweep outthe surface. Hence, every developable surface is a ruled surface. Nowsuppose we can choose a directrix ~α(t) of the developable with thegenerators as the rulings. Then the normal vector is constant alongeach ruling and hence the partial derivative in that direction is 0.Then by Proposition 6.6.2, a developable surface has a Gaussiancurvature that is identically 0.

Consequently, an equivalent definition of a developable surface isthat it is a ruled surface that has Gaussian curvature that is con-stantly 0.

Example 6.7.2 (Cylinders). The simplest example of a ruled surface isa cylinder. The general definition of a cylinder is a ruled surface thatcan be given as a one-parameter family of lines ~α(t), ~w(t), where~α(t) is planar and ~w(t) is a constant vector in R3. It is easy to show


(a) Cone. (b) Hyperboloid.

Figure 6.12. Ruled surfaces.

that this has Gaussian curvature that is identically 0, so the cylinderis a developable surface.

Example 6.7.3 (Cones). The general definition of a cone is a surfacethat can be given as a one-parameter family of lines ~α(t), ~w(t),where ~α(t) lies in a plane P and the rulings Lt all pass through somecommon point p /∈ P . Therefore, the cone over ~α(t) through p canbe parametrized by

~X(t, u) = ~α(t) + u(p− ~α(t)).

Again, it is possible to show that the Gaussian curvature is identi-cally 0.

Example 6.7.4. As a perhaps initially surprising example, the hyper-boloid of one sheet is also a ruled surface. The standard parametri-zation for the hyperboloid of one sheet is

~X(u, v) = (cosh v cosu, cosh v sinu, sinh v).

Consider now the ruled surface with directrix ~α(t) = (cos t, sin t, 0)and with ruling directions ~w(t) = ~α′(t) + ~k = (− sin t, cos t, 1) (seeFigure 6.12(b)). We obtain the parametrized surface

~Y (t, u) = (cos t− u sin t, sin t+ u cos t, u).


Though it is not particularly easy to express ~Y as a reparametriza-tion of ~X, it is not hard to see that both of these parametrizationssatisfy the following usual equation that gives the hyperboloid as aconic surface:

x2 + y2 − z2 = 1.

This ruled surface is not developable since the Gaussian curvature isnot identically 0.

In order to identify developable surfaces, we determine the Gaus-sian curvature for ruled surfaces generally. However, in order to sim-plify subsequent calculations, we make two additional assumptionsthat do not lose generality. First, note that one can impose the con-dition that ‖~w(t)‖ = 1 without changing the definition or the imageof the parametrization in Equation (6.35). In defining particularruled surfaces, it is usually easier not to make this assumption, butit does simplify calculations since ‖~w(t)‖ = 1 for all t ∈ I impliesthat ~w(t) · ~w(t)′ = 0. Second, note that different curves ~α(t) canserve as the directrix for the same ruled surface, so we wish to em-ploy a curve ~β(t) as a directrix, which will simplify the algebra inour calculations. We choose a curve ~β(t) satisfying ~β′(t) · ~w′(t) = 0.

Since ~β(t) lies on ~X, we can write

~β(t) = ~α(t) + u(t)~w(t). (6.36)

Then~β′(t) = ~α′(t) + u′(t)~w(t) + u(t)~w′(t),

and since ~β′ · ~w′ = 0, we have

~α′(t) · ~w′(t) + u(t)~w′(t) · ~w′(t) = 0.

Thus, we determine that

u(t) = −~α′(t) · ~w′(t)‖~w′(t)‖2

. (6.37)

Furthermore, it is easy to prove (see Problem 6.7.4) that, with ourpresent assumptions, ~β(t) is unique. We call the curve ~β(t) the lineof stricture of the ruled surface.


We now use the parametrization for the ruled surface ~X = ~β(t)+u~w(t) and proceed to calculate the first and second fundamentalforms. Understanding that ~β and ~w are functions of t, we find that

~Xt = ~β′ + u ~w′,

~Xu = ~w,

~Xt × ~Xu = ~β′ × ~w + u ~w′ × ~w,

and

~Xtt = ~β′′ + u ~w′′,

~Xtu = ~w′,

~Xuu = ~0.

One can already notice that if ~w(t) is constant, then L12 = L21 =L22 = 0, which leads to K = 0 and proves that all cylinders haveGaussian curvature K = 0. We will assume now that ~w′(t) is notidentically 0.

Because of the conditions ~w · ~w′ = 0 and ~β′ · ~w′ = 0, we canconclude that ~β′× ~w is parallel to ~w′. Thus, ~β′× ~w is perpendicularto ~w × ~w′, so

‖ ~Xt × ~Xu‖2 = ‖~β′ × ~w‖2 + u2‖~w′‖2.

Furthermore, since ~β′ × ~w is parallel to ~w′ it is its own projectiononto ~w′. Hence, ‖~β′ × ~w‖ = |(~β′ × ~w) · ~w′|/‖~w′‖ and hence,

‖ ~Xt × ~Xu‖2 = (~β′ ~w~w′)2/‖~w′‖2 + u2‖~w′‖2 (6.38)

=1

‖~w′‖2(

(~β′ ~w~w′)2 + u2‖~w′‖4),

where we use the notation (~β′ ~w~w′) for (~β′× ~w) · ~w′, the triple-vectorproduct in R3. Consequently, we can write the unit normal vector~N as

~N(t, u) =‖~w′‖√

(~β′ ~w~w′)2 + u2‖~w′‖4

(~β′ × ~w + u ~w′ × ~w

),

and therefore, again using the conditions that ~w · ~w′ = 0 and ~β′ · ~w′ =0, we get

(gij) =

(‖~β′‖2 + u2‖~w′‖2 ~β′ · ~w

~β′ · ~w 1

)and


(Lij) =‖~w′‖√

(~β′ ~w~w′)2 + u2‖~w′‖4

((~β′′ + u~w′′) · (~β′ × ~w + u~w′ × ~w) (~β′ ~w~w′)

(~β′ ~w~w′) 0

).

We cannot say much for the entry L11, but thanks to Equation(6.34) for the Gaussian curvature and the fact that ‖ ~Xu × ~Xv‖2 =det(gij) (see Equation (6.6)), we can calculate the Gaussian curva-ture for a ruled surface as

K =det(Lij)

det(gij)= − ‖~w′‖4(~β′ ~w~w′)2(

(~β′ ~w~w′)2 + u2‖~w′‖4)2 . (6.39)

This formula for the Gaussian curvature of a ruled surface makesa few facts readily apparent. First, K ≤ 0 for all points of a ruledsurface. Second, by Equation (6.38), all singular points of a ruledsurface, i.e., points where ~Xt× ~Xu = ~0, must have u = 0 and thereforeoccur on the line of stricture. Finally, by Equation (6.39), a ruledsurface satisfies K(t, u) = 0 if and only if (~β′ ~w~w′) = 0 for all t ∈ I.

Let us return now to the general definition of a ruled surface fromEquation (6.35) with regular curves ~α(t) and ~w(t) with no conditions.Since both ~w and ~w′ are perpendicular to ~w × ~w′,

(~α′ ~w~w′) = ~α′ · ~w × ~w′ = ~β′ · ~w × ~w′ = (~β′ ~w~w′).

Furthermore, if we call w = ~w/‖~w‖, we remark that

(~α′ ~w~w′) = ‖~w‖2(~α′ww′) = ‖~w‖2(~β′ww′).

Consequently, (~β′ww′) = 0 if and only if (~α′ ~w~w′) = 0. We haveproven the following proposition.

Proposition 6.7.5. A ruled surface in R3 with directrix ~α(t) and suchthat each ruling has direction ~w(t) is a developable surface if andonly if (~α′ ~w~w′) = 0.

Essentially by definition, every developable surface has Gaussiancurvature that is identically 0. Surprisingly, the converse is true.

Theorem 6.7.6. A regular surface S of class C2 in R3 has K = 0identically if and only if the surface is a developable surface.


Proof: Suppose that S is a regular surface of class C2 such thatK = 0. Let ~X : U → R3 be a parametrization ~X(u, v) of a co-ordinate patch of S. The following reasoning will occur on thiscoordinate patch but the result with extend to the whole surface.By Proposition 6.6.2, K(u, v) = 0 if and only if the set ~Nu, ~Nvare everywhere linearly dependent. Also notice that K = 0 if andonly if L11L22 − L2

12 = 0. Obviously, in order for this equality tohold L11 and L22 must have equal signs everywhere. Without lossof generality in what follows, we assume that they are nonnegative.

Consider the asymptotic curves on the surface whose parametri-zation (u(t), v(t)) satisfy the differential equation

L11(u′)2 + 2L12u′v′ + L22(v′)2 = 0

⇐⇒L11(u′)2 + 2√L11L22u

′v′ + L22(v′)2 = 0

⇐⇒(√L11u

′ +√L22v

′)2 = 0.

Not both L11 and L22 can be zero for L11L22−L212 = 0 to hold. If we

suppose that L22 6= 0, then by substitution, this gives the differentialequation

dv

du= −

√L11

L22.

By the Theorem of Existence and Uniqueness for differential equa-tions, for each point (u0, v0) ∈ U , there is an asymptotic curvethrough (u0, v0). Consequently, at each point p ∈ S, there is a neigh-borhood of p for which it is possible to change parametrizations touse coordinates (u, v) such that the asymptotic curves correspondingto v are constant.

Using the (u, v) coordinates, the equation for asymptotic curvesis simply v′ = 0 or more precisely L22(v′)2 = 0. From this, wededuce that the components of the second fundamental form haveL11 = L12 = 0. Thus

~Xu · ~Nu = ~Xv · ~Nu = 0.

Thus ~Nu = 0 at all points on the surface and hence the normal vectordepends on only one parameter, v.

Now suppose that the tangent planes of S depend on only oneparameter v. Then there exist a differentiable vector function ~a(v)


Figure 6.13. A developable surface.

and a differential real-valued function f(v) such that the tangentspaces of S (possibly on some open coordinate neighborhood of S)satisfy the equation

~x · ~a(v) = f(v), where ~x = (x, y, z).

Given a fixed v0 we have ~x · ~a(v0) = f(v0) as well as ~x · ~a(v0 + h) =f(v0 + h) as h → 0. In particular, this leads to the fact the normalvectors to the surface S that are also in the tangent plane to S at apoint that has v = v0 must also satisfy

~x · ~a′(v0) = f ′(v0),

which is another plane. Hence, such normal vectors are perpendicu-lar to two planes. Consequently, the surface is a ruled surface, suchthat along the rulings, the normal vectors are constant. Thus thesurface S is developable.

It is interesting to remark that a ruled surface is a cone if ~α′(t) =0 and is a cylinder if and only if ~w′(t) = 0, showing again that bothcones and cylinders are developable and have K = 0. The exercisespresent examples of developable surfaces that are neither cones norcylinders. Figure 6.13 also illustrates a developable surface that has~α = (0, t, cos t).

Developable surfaces are particularly interesting for design andmanufacturing. Because developable surfaces are ruled and haveGaussian curvature 0, they can be created by bending a region ofthe plane, without stretching it. Consequently, developable surfacescan be made out of sheet metal or any inelastic material that starts


out flat. Some postmodern architecture exemplifies the use of de-velopable surfaces (such as Frank Gehry’s Guggenheim Museum inBilbao or City of Wine Hotel Marques de Riscal).

6.7.2 Minimal Surfaces

Definition 6.7.7. A minimal surface is a parametrized surface of classC2 that satisfies the regularity condition and for which the meancurvature is identically 0.

We first wish to justify the name “minimal.”Let ~X : U → R3 be a coordinate neighborhood of a regular

parametrized surface of class C2. Let D′ be a connected compactset in U , and let D = ~X(D′). Let h : U → R be a differentiablefunction. A normal variation of ~X over D′ determined by h is thefamily of surfaces with t ∈ (−ε, ε) defined by

~Xt : D′ −→ R3

(u, v) 7−→ ~X(u, v) + th(u, v) ~N(u, v).

Proposition 6.7.8. Let ~Xt be a normal variation of ~X over a compactregion D′ and determined by some function h. For ε small enough,~Xt satisfies the regularity condition for all t ∈ (−ε, ε). In this case,the area of ~Xt is

A(t) =

∫∫D′

√1− 4thH + t2R

√det(gij) du dv (6.40)

for some function R(u, v, t) that is polynomial in t.

Proof: Denote gtij as the coefficients of the metric tensor. The sur-faces of the normal variation have

~Xtu = ~Xu + thu ~N + th ~Nu,

~Xtv = ~Xv + thv ~N + th ~Nv.

Thus, we calculate,

gt11 = g11 + 2th ~Xu · ~Nu + t2h2 ~Nu · ~Nu + t2h2u,

gt12 = g12 + th( ~Xu · ~Nv + ~Xv · ~Nu) + t2h2 ~Nu · ~Nv + t2huhv,

gt22 = g22 + 2th ~Xv · ~Nv + t2h2 ~Nv · ~Nv + t2h2v.


However, by Equation (6.17), one can summarize the above equa-tions as

gtij = gij − 2thLij + t2h2 ~Ni · ~Nj + t2hihj ,

where we use the notation h1 (resp. h2) to indicate the partial deriva-tive hu (resp. hv). Therefore, we calculate that

det(gtij) = det(gij)− 2th(g11L22 − 2g12L12 + g22L11) + t2R,

where R(u, v, t) is a function of the form A0(u, v) + tA1(u, v) +t2A2(u, v) for continuous functions Ai defined over D′. However,the mean curvature is

H =

(1

2

)g11L22 − 2g12L12 + g22L11

g11g22 − g212

,

which leads to

det(gtij) = det(gij)(1− 4thH) + t2R = det(gij)(1− 4thH + t2R),(6.41)

where R = R/ det(gij). Since D′ is compact, the functions h, H,and R are bounded over D′, which shows that

limt→0

det(gtij) = det(gij) for all (u, v) ∈ D′.

Hence, if ε > 0 is small enough, then det(gtij) 6= 0 for all t ∈ (−ε, ε),and thus, all normal variations satisfy the regularity condition. Therest of the proposition follows from Equation (6.41) since

A(t) =

∫∫D′

√det(gtij) du dv.

Proposition 6.7.9. Let ~X : U → R3 be a parametrized surface of classC2, and let D′ ⊆ U be a compact set. Let A(t) be the area functiondefined in Equation (6.40). Then ~X parametrizes a minimal surfaceif and only if A′(0) = 0 for all D′ and all normal variations of ~Xover D′.


Proof: We calculate

A′(t) =

∫∫D′

−4hH + 2tR+ t2Rt

2√

1− 4thH + t2R

√det(gij) du dv,

which implies that

A′(0) = −2

∫∫D′

hH√

det(gij) du dv.

Obviously, if ~X parametrizes a minimal surface, which means thatH = 0 for all (u, v) ∈ U , then A′(0) = 0 regardless of the functionh(u, v) or the compact set D′.

To prove the converse, suppose that A′(0) = 0 for all continu-ous functions h(u, v) and all compact D′ ⊂ U . Choosing h(u, v) =H(u, v), since det(gij) > 0, we have

A′(0) = −2

∫∫D′

H2√

det(gij) du dv,

so A′(0) ≤ 0 for all choices of D′. Since H(u, v) is continuous,if H(u0, v0) 6= 0 for any point (u0, v0), there is a compact set D′

containing (u0, v0) such that H(u, v)2 > 0 over D′. Thus, if His anywhere nonzero, there exists a compact subset D′ such thatA′(0) < 0. This is a contradiction, so we conclude that A′(0) = 0 forall h and all D′ implies that H(u, v) = 0 for all (u, v) ∈ U .

Proposition 6.7.9 shows that a parametrized surface ~X that sat-isfies the regularity condition ‖ ~Xu× ~Xv‖ 6= 0 is minimal (has H = 0everywhere) if it is a surface such that over every patch of surfacethere is no way to deform it along normal vectors to obtain a surfaceof lesser surface area.

Minimal surfaces have also enjoyed considerable popular atten-tion, especially in museums of science, as soap films on a wire frame.When a wire frame is dipped into a soapy liquid and pulled out, thesurface tension on the soap film pulls the film into the state of leastpotential energy, which turns out to be the surface such that no nor-mal variation can lead to a surface with less surface area. Of course,


when experimenting with soap films, one can experiment with wireframes that are nonregular curves or not even curves at all (the skele-ton of a cube, for example) and investigate what kinds of soap filmsurfaces result.

The problem of determining a minimal surface with a given closedregular space curve C as a boundary was raised by Lagrange in 1760.However, the mathematical problem became known as Plateau’sproblem, after Joseph Plateau who specifically studied soap filmsurfaces. The 19th century saw a few specialized solutions to theproblem, but it was not solved until 1930. (See [27] for a more com-plete history of Plateau’s problem.)

Though one can easily construct a minimal surface with a soapfilm on a wire frame, either checking that a surface is minimal orfinding a parametrization for a minimal surface is quite difficult. Infact, the study of minimal surfaces continues to provide new areas ofresearch and connections with other branches of analysis. Perhapsamong the most interesting results is a connection between complexanalytic functions and minimal surfaces [11, p. 206] or the use ofelliptic integrals to parametrize special minimal surfaces [27, Chapter1], topics that go beyond the scope of this book. However, it isquite likely that having both the simple experimentative aspect andconnections to advanced mathematics has fueled interest in minimalsurfaces.

Problems

6.7.1. Prove that the surface given by z = kxy, where k is a constant, is aruled surface and give it as a one-parameter family of lines.

6.7.2. Figure 6.13 shows a developable surface that has the line of stricture~α(t) = (0, t, cos t). Suppose the rulings are ~w(t) = (1, w2(t), w3(t)).Find the equations or differential equation required by w2(t) andw3(t) to produce a developable surface. [In Figure 6.13, the rulingshave ~w(t) = (1,−0.3t,−0.3 cos t).]

6.7.3. The tangential surface to a space curve ~α(t) was presented in Prob-lem 6.6.5. Show that the curve ~α is the line of stricture for thetangential surface. Show that the tangential surface to a regularspace curve is a developable surface.

6.7.4. Suppose that ~α1(t), ~w(t) and ~α2(t), ~w(t), where ‖~w(t)‖ = 1, aretwo one-parameter families of lines that trace out the same ruled


surface. Prove that Equation (6.36) with (6.37), using either ~α1 or~α2, produce the same line of stricture.

6.7.5. Suppose that a ruled surface has ~α(t) = (cos t, sin t, 0) as the line ofstricture. Suppose also that w(t) = (w1(t), w2(t), 1). Find algebraicor differential equations that w1(t) and w2(t) must satisfy so thatthe ruled surface is a developable surface.

6.7.6. Let S be an orientable surface, and let ~α(s) be a curve on S pa-rametrized by arc length. Assume that ~α is nowhere tangent to theasymptotic direction on S, and define ~N(s) as the unit normal vectorto S along ~α(s). Consider the ruled surface

~X(s, u) = ~α(s) + u~N(s)× ~N ′(s)

‖ ~N ′(s)‖.

The assumption that ~α′(s) is not an asymptotic direction ensures

that ~N ′(s) 6= ~0. Prove that ~X(s, v) is a developable surface. (Thiskind of surface is called the envelope of a family of tangent planesalong a curve of a surface.)

6.7.7. Let ~X(t, v) = ~α(t) + v ~w(t) be a developable surface. Prove that ata regular point

~Nv · ~Xt = 0 and ~Nv · ~Xv = 0.

Use this result to prove that the tangent plane of a developablesurface is constant along a line of ruling.

6.7.8. Show that a surface F (x, y, z) = 0 is developable if∣∣∣∣∣∣∣∣Fxx Fxy Fxz FxFyx Fyy Fyz FyFzx Fzy Fzz FzFx Fy Fz 0

∣∣∣∣∣∣∣∣ = 0.

6.7.9. Consider the helicoid parametrized by ~X(u, v) = (v cosu, v sinu, cu),where c is a fixed number.

(a) Determine the asymptotic curves.

(b) Determine the lines of curvature.

(c) Show that the mean curvature of the helicoid is 0.

6.7.10. Prove that the catenoid parametrized by

~X(u, v) = (a cosh v cosu, a cosh v sinu, av)

is a minimal surface.


6.7.11. Prove that Enneper’s minimal surface ~X(u, v) = (u− u3

3 + uv2, v −v3

3 + vu2, u2 − v2) deserves its name.

6.7.12. Show that for all constants D1 and D2, the surface parametrized bythe function graph

~X(u, v) = (u, v,− ln(cos(u+D1)) + ln(cos(−v +D2)))

where it is defined is a minimal surface.

6.7.13. (ODE) Prove that the only nonplanar minimal surfaces of the form~X(u, v) = (u, v, h(u) + k(v)) have

h(u) = − 1

Cln(cos(Cu+D1)) and k(v) =

1

Cln(cos(−Cv +D2))

for some nonzero constant C and any constants D1 and D2.

6.7.14. Let U ⊂ R2, and suppose that ~X : U → R3 and ~Y : U → R3

parametrize two minimal surfaces defined over the same domain.Prove that for all t ∈ [0, 1], the surfaces parametrized by ~Zt(u, v) =

(1− t) ~X(u, v) + t~Y (u, v) are also minimal surfaces.

6.7.15. (ODE) Prove that the only surface of revolution that is a minimalsurface is a catenoid (see Problem 6.7.10).

6.7.16. Suppose that S is a minimal regular surface with no planar points.Prove that at all points p ∈ S, the Gauss map n : S → S2 satisfies

dnp(~w1) · dnp(~w2) = −K(p)~w1 · ~w2

for all ~w1, ~w2 ∈ TpS, where K(p) is the Gaussian curvature of S at p.Use this result to show that on a minimal surface the angle betweentwo intersecting curves on S is the angle between their images on S2under the Gauss map n.

6.7.17. Let ~X(u, v) be a parametrization of a regular, orientable surface,

and let ~N(u, v) = ~Xu× ~Xv/‖ ~Xu× ~Xv‖ be the orientation. A parallel

surface to ~X is a surface parametrized by

~Y (u, v) = ~X(u, v) + a ~N(u, v).

(a) If K and H are the Gaussian and mean curvatures of ~X, provethat

~Yu × ~Yv = (1− 2Ha+Ka2) ~Xu × ~Xv.


(b) Prove that at regular points, the Gaussian curvature of ~Y is

K

1− 2Ha+Ka2,

and the mean curvature is

H −Ka1− 2Ha+Ka2

.

(c) Use the above to prove that if ~X is a surface with constantmean curvature H = c 6= 0, then there is a parallel surface to~X that has constant Gaussian curvature.

CHAPTER 7

The Fundamental Equations

of Surfaces

In the previous chapter, we began to study the local geometry of asurface and saw how many computations depend on the coefficientsof the first and second fundamental forms. In fact, in Section 6.1we saw that many metric calculations (angles between curves, arclength, area, etc.) depend only on the first fundamental form. Wefirst approached such concepts from the perspective of objects in R3,but with the first fundamental form, one can perform all the calcu-lations by using only the coordinates that parametrize the surface,without referring to the ambient space. In fact, it is not at all dif-ficult to imagine a regular surface as a subset of Rn, with n > 3,and the formulas that depend on the first fundamental form wouldremain unchanged. See Chapter 9. Concepts that rely only on thefirst fundamental form are called intrinsic properties of a surface.

On the other hand, concepts such as the second fundamentalform, the Gauss map, Gaussian curvature, and principal curvatureswere defined in reference to the unit normal vector, and these are notnecessarily intrinsic properties. To illustrate this point, consider aparametrized surface S that is a subset of R4, and let p be a point ofS. One can still define the tangent plane as the span of two linearlyindependent tangent vectors, but there no longer exists a unique (upto sign) unit normal vector to S at p. In this situation, one cannotdefine the Gauss map. Properties that we presented as dependingon the Gauss map either cannot be defined in this situation or needto be defined in an alternate way.

This chapter studies what kind of information about a surfaceone can know from just the first fundamental form versus that whichcan be determined from knowledge of both the first and the secondfundamental forms.

247

248 7. The Fundamental Equations of Surfaces

Section 7.1 introduces the Christoffel symbols and studies rela-tions between the coefficients of the first and second fundamentalforms. In Section 7.2, we present the famous Theorema Egregium,which proves that the Gaussian curvature of a surface is in fact an in-trinsic property. More precisely, the theorem expresses the Gaussiancurvature in terms of the gij functions. In Section 7.3, we present an-other landmark result for surfaces in R3, the Fundamental Theoremof Surface Theory, which proves that under appropriate conditions,the coefficients of the first and second fundamental forms determinethe surface up to position and orientation in space.

Some of the quantities we encounter in this chapter involve multi-ple indices. Some of these quantities represent a new mathematicalobject called a tensor. Tensors simultaneously generalize vectors,matrices, inner products, and many other objects that arise in linearalgebra. Furthermore, just as it is possible to define a vector fieldthat to each point in a region of the plane (or space) one associatesa vector, so it is possible to define a tensor field.

An introduction to tensor notation and the classical description ofa tensor is given in the appendix in Section A.1. The only notationalconvention we underscore here is the Einstein summation convention:In a tensor notation expression, whenever a superscript index alsoappears as a subscript index, it is understood that we sum overthat index. For example, suppose that Aijkl and Bα

βγ are differentcollections of quantities with each index showing running from 1 ton, where n is the dimension of the ambient space Rn. Then by theexpression AijklB

kim we mean the set of quantities

Cjlm = AijklBkim

def=

n∑i=1

n∑k=1

AijklBkim.

The collection Aijkl of quantities consists of n4 quantities, the collec-

tion of Bαβγ consists of n3 quantities, and the collection Cjlm consists

again of n3 quantities.

7.1 Gauss’s Equations and the Christoffel Symbols

We now return to the study of surfaces. As we shall see, the resultsof this section assume that one can take the third derivative of a

7.1. Gauss’s Equations and the Christoffel Symbols 249

coordinate parametrization. Therefore, in this section and in theremainder of the chapter, unless otherwise stated we consider onlyregular surfaces of class C3.

If one compares the theory of surfaces developed so far to thetheory of curves, one will point out one major gap in the presentationof the former. In the theory of space curves, we discussed naturalequations, namely the curvature κ(s) and torsion τ(s) with respectto arc length, and we proved that these two functions locally define aunique curve up to its position in space. Implicit in the proof of thisresult for natural equations of curves was the fact that in generalthere do not exist algebraic relations between the functions κ(s) andτ(s).

In the theory of surfaces, the problem of finding natural equationscannot be quite so simple. For example, even when restricting ourattention to the first fundamental form, we know that given any threefunctions E(u, v), F (u, v), and G(u, v), there does not necessarilyexist a surface with (

g11 g12

g21 g22

)=

(E FF G

)because we need the nontrivial requirement that EG − F 2 > 0.Furthermore, one might suspect that, because of the smoothnessconditions that imply that ~Xuuv = ~Xuvu = ~Xvuu, the (gij) and (Lij)coefficients may satisfy some inherent relations.

Also, when discussing natural equations for space curves, one of-ten uses the Frenet frame as a basis for R3, with origin at a pointp of the curve. In particular, one must use the Frenet frame toobtain a parametrization of a curve in the neighborhood of p fromthe knowledge of the curvature κ(s) and torsion τ(s). When per-forming calculations in the neighborhood of a point p on a surfaceS parametrized by ~X : U → R3, the basis ~X1, ~X2, ~N is the mostnatural, but, unlike the Frenet frame, this basis is not orthonormal.In addition, the basis ~X1, ~X2, ~N depends significantly on the pa-rametrization ~X, while a reparametrization of a curve can at mostchange the sign of ~T and ~B.

One might propose ~e1, ~e2, ~N, where ~e1 and ~e2 are the principaldirections, as a basis more related to the geometry of a surface.The orthonormality has advantages, but calculations using this basis


quickly become intractable. Furthermore, the eigenvectors ~e1 and ~e2

are not well-defined at umbilic points since the eigenspaces of dnpat umbilic points consist of the whole tangent space TpS. For these

reasons, it remains more natural to use the basis ~X1, ~X2, ~N for R3

in a neighborhood of p and study the relations that arise betweenthe (gij) and (Lij) coefficients.

Recapping earlier definitions, we have ~X1 · ~X1~X1 · ~X2

~X1 · ~N~X2 · ~X1

~X2 · ~X2~X2 · ~N

~N · ~X1~N · ~X2

~N · ~N

=

g11 g12 0g21 g22 00 0 1

.

Every vector in R3 can be expressed as a linear combinationin this basis. In particular, one would like to express the secondderivatives ~X11, ~X12, ~X21, and ~X22 as linear combinations of thisbasis. From Equation (6.18), we know that Lij = ~Xij · ~N , and since~N · ~Xi = 0, we deduce that Lij is the coordinate of ~Xij along basis

vector ~N . However, we do not know the coordinates of ~Xij along

the other two basis vectors ~X1 and ~X2.

Definition 7.1.1. The collection of eight functions Γijk : U → R withindices 1 ≤ i, j, k ≤ 2 are defined as the unique functions that satisfy

~Xjk = Γ1jk~X1 + Γ2

jk~X2 + Ljk ~N. (7.1)

Definition 7.1.1 names the functions Γijk implicitly, but we nowproceed to find formulas for them in terms of the metric coefficients.

Note that though we have eight combinations for three indices,each ranging between 1 and 2, we do not in fact have eight distinctfunctions because ~Xjk = ~Xkj . Thus, we already know that

Γijk = Γikj .

Before determining formulas for the Γijk functions, we will first

establish expressions for ~Xij · ~Xk that are easier to find. Since thesequantities occur frequently, there is a common shorthand symbol forthem, namely,

[ij, k] = ~Xij · ~Xk.


Begin by fixing j = k = 1. Using the formulas

∂g11

∂x1=

∂

∂x1( ~X1 · ~X1) = 2 ~X11 · ~X1,

∂g12

∂x1=

∂

∂x1( ~X1 · ~X2) = ~X11 · ~X2 + ~X21 · ~X1,

∂g11

∂x2=

∂

∂x2( ~X1 · ~X1) = 2 ~X12 · ~X1,

we deduce that

[11, 1] =1

2

∂g11

∂x1and [11, 2] =

∂g12

∂x1− 1

2

∂g11

∂x2.

Pursuing the calculations for the remaining cases, using the factthat (gij) is symmetric, and rewriting the results in a convenientmanner, one can prove the following lemma.

Lemma 7.1.2. For all indices 1 ≤ i, j, k ≤ 2, we have

[ij, k] = ~Xij · ~Xk =1

2

(∂gjk∂xi

+∂gki∂xj

− ∂gij∂xk

). (7.2)

Using this lemma, one can easily establish a formula for the func-tions Γijk. We remind the reader that we shall often use the Einsteinsummation convention when an index is repeated in one superscriptand one subscript position. (See Section A.1 for a more accurateexplanation of the convention.) Before we give the proof, we alsomention the useful artifice in tensorial notation called the Kroneckerdelta δji with 1 ≤ i ≤ n and 1 ≤ j ≤ n defined as

δji =

1 if i = j

0 if i 6= j.

The Kronecker delta is a way to represent the identity matrix.

Proposition 7.1.3. Let ~X : U → R3 be the parametrization of a regularsurface of class C2 in the neighborhood of some point p = ~X(q).Then the coefficients Γijk satisfy

Γijk =

2∑l=1

gil[jk, l] =

2∑l=1

gil1

2

(∂gkl∂xj

+∂glj∂xk−∂gjk∂xl

,

)(7.3)

where (gij), with the indices in superscript, is the inverse matrix(gij)

−1.


Proof: We have already determined formulas for [ij, k] = ~Xij · ~Xk.However, from Equation (7.1), taking a dot product with respect tothe derivative vectors ~Xi, we obtain

[ij, k] = Γlijglk.

Multiplying the matrix of the first fundamental form (glk) by itsinverse, we can write

glkgkα = δαl

where δαl is the Kronecker delta (defined in Equation (A.3)). Thenwe get

2∑k=1

gkα[ij, k] = Γlijglkgkα = Γlijδ

αl = Γαij .

The proposition follows from the symmetry of the (gij) matrix.

Definition 7.1.4. The symbols [ij, k] are called the Christoffel symbolsof the first kind, while the functions Γijk are called the Christof-fel symbols of the second kind or, more simply, just the Christoffelsymbols.

The formula in Proposition 7.1.3 along with Definition 7.1.1 iscalled Gauss’s formula for surfaces.

One of the first uses of Gauss’s formula is that knowing the func-tions Ljk and Γijk allows us to write not just the second partial

derivatives of the parametrization ~X with coordinates with respectto the basis ordered ( ~X1, ~X2, ~N), but also all higher derivatives of~X(x1, x2). We point out that knowing gij in the neighborhood of a

point allows us to determine the lengths of and angle between ~X1

and ~X2. Furthermore, as we shall see later, knowing the three dis-tinct functions Ljk and the six distinct functions Γijk for a particularsurface S in the neighborhood of p allows one to use a Taylor seriesexpansion in the two variables x1 and x2 to write an infinite sumthat provides a parametrization of S in a neighborhood of p oncep, ~X1, and ~N are given. However, we still have not identified anyrelations that must exist between Lij and gij , so we cannot yet statean equivalent to the natural equations theorem for space curves.


We point out that though we used a superscript and subscriptsfor the Christoffel symbol Γijk, these functions do not form the com-ponents of a tensor but rather transform according to the followingproposition.

Proposition 7.1.5. Let (x1, x2) and (x1, x2) be two coordinate systemsfor a neighborhood of a point p on a surface S. If we denote by Γmijand Γµαβ the Christoffel symbols in the respective coordinate systems,then they are related by

Γµαβ =∂xi

∂xα∂xj

∂xβ∂xµ

∂xmΓmij +

∂2xm

∂xα∂xβ∂xµ

∂xm.

Proof: We leave some of the details of this proof for the reader butpresent an outline here.

Consider two systems of coordinates (x1, x2) and (x1, x2) for aneighborhood of a point p on a surface S. We know that the metriccoefficients change according to

gαβ =∂xi

∂xα∂xj

∂xβgij and gαβ =

∂xα

∂xi∂xβ

∂xjgij . (7.4)

In the (x1, x2) coordinate system, we denote the Christoffel symbolsof the first kind by

[αβ, ν] =1

2

(∂gβν∂xα

+∂gνα∂xβ

−∂gαβ∂xν

), (7.5)

and we must first relate this to the Christoffel symbols [ij, k] in the(x1, x2) coordinate system. Note that in this proof, we make indices(i, j, k,m) in the (x1, x2) coordinate system correspond to indices(α, β, ν, µ). Using Equation (7.4), the first term in Equation (7.5)transforms according to

∂gβν∂xα

=∂2xj

∂xα∂xβ∂xk

∂xνgjk +

∂xj

∂xβ∂2xk

∂xα∂xνgjk +

∂xj

∂xβ∂xk

∂xν∂xi

∂xα∂gjk∂xi

.

(7.6)

Equation (7.6) and the corresponding results of the other twoterms in the sum in Equation (7.5) produce an expression for thetransformation property of the Christoffel symbol of the first kind.


The expression is not pleasing, especially since it involves the coeffi-cients gij explicitly. One must now calculate

Γµαβ =

2∑ν=1

gµν [αβ, ν] =

2∑ν=1

∂xµ

∂xm∂xν

∂xlgml[αβ, ν],

where we replace [αβ, ν] with the terms found in Equation (7.6) andsimilar equalities. After appropriate simplifications, we find that

Γµαβ =∂xµ

∂xm∂xi

∂xα∂xj

∂xβΓmij +

1

2

(∂xµ

∂xj∂2xj

∂xα∂xβ+∂xµ

∂xi∂2xi

∂xα∂xβ

)+

1

2

∂xν

∂xl∂xµ

∂xm

(∂xj

∂xα∂2xk

∂xα∂xνgmlgjk −

∂xj

∂xα∂2xi

∂xα∂xνgmlgij

)+

1

2

∂xν

∂xl∂xµ

∂xm

(∂xi

∂xα∂2xk

∂xβ∂xνgmlgki −

∂xi

∂xα∂2xj

∂xν∂xβgmlgji

).

(7.7)

However, it is important to remember that when one sums over anindex, the actual name of the index does not change the result ofthe summation. Applying this observation to Equation (7.7) andremembering that the components gij are symmetric in their indicesfinishes the proof of the proposition.

One should not view the fact that the Christoffel symbols Γijk donot form the components of a tensor as just a small annoyance; itis of fundamental importance in the theory of manifolds. From anintuitive perspective, one could understand tensors as objects thatare related to the tangent space to a surface at a point. However,since the Γijk functions explicitly involve the second derivatives ofa parametrization of a neighborhood of a point on a surface, onemight have been able to predict that the Γijk functions would notnecessarily form the components of a tensor.

Example 7.1.6 (Sphere). As a simple example, consider the usual pa-rametrization of the sphere,

~X(x1, x2) = (R cosx1 sinx2, R sinx1 sinx2, R cosx2).


In spherical coordinates, the variable names we used correspond tox1 = θ as the meridian and x2 = ϕ the latitude, measured as theangle down from the positive z-axis. A simple calculation leads to

gij =

(R2 sin2(x2) 0

0 R2

)and gij =

1

R2

(1

sin2(x2)0

0 1

).

By Equation (7.2), we find that

[11, 1] = 0,

[12, 1] = R2 sin(x2) cos(x2),

[21, 1] = R2 sin(x2) cos(x2),

[22, 1] = 0,

[11, 2] = −R2 sin(x2) cos(x2),

[12, 2] = 0,

[21, 2] = 0,

[22, 2] = 0.

Then, using Equation (7.3), we get

Γ111 = 0,

Γ112 = Γ1

21 = cot(x2),

Γ122 = 0,

Γ211 = − sin(x2) cos(x2),

Γ212 = Γ2

21 = 0,

Γ222 = 0.

As an example of an application of Gauss’s equation, recall fromEquation (6.33) that two vectors ~u1 and ~u2 in the tangent space TpSto a surface S at p are called conjugate if

Ip(dnp(~u1), ~u2) = Ip(~u1, dnp(~u2)) = 0. (7.8)

Intuitively speaking, Equation (7.8) states that conjugate directionsare such that the direction of change of the unit normal vector ~Nalong ~u1 is perpendicular to ~u2 and vice versa. Given a parametri-zation ~X of a regular surface in the neighborhood of a point p, wesay that ~X produces a conjugate set of coordinate lines if

dnp( ~X1) · ~X2 = 0 = dnp( ~X2) · ~X1

for all (x1, x2) in the domain of ~X. One can state this alternativelyas

~N1 · ~X2 = 0 = ~N2 · ~X1,


so ~X produces a set of conjugate coordinate lines if and only if L12 =0. In this situation, Gauss’s equation for ~X12 reduces to

~X12 = Γ112~X1 + Γ2

12~X2, (7.9)

so the parametrization ~X has a conjugate set of coordinate lines ifand only if Equation (7.9) holds.

In addition, Equation (7.9) encompasses three independent lineardifferential equations in the rectangular coordinate functions of theparametrization. Therefore, given functions P (x1, x2) and Q(x1, x2),any three linearly independent solutions to the equation

∂2f

∂x1∂x2− P (x1, x2)

∂f

∂x1−Q(x1, x2)

∂f

∂x2= 0, (7.10)

when taken as the coordinate functions, give a parametrization ofa surface with conjugate coordinate lines in which Γ1

12 = P (x1, x2)and Γ2

12 = Q(x1, x2). Equation (7.10) is, in general, not simple tosolve but since ~N does not appear in Equation (7.10), the coordinatefunction solutions are independent of each other.

Example 7.1.7 (Intrinsic Geometry). Since the Christoffel symbols aredefined using the metric coefficients and their partial derivatives,these functions are intrinsic. We can use the metric coefficientsg11(u, v) = 1/v2, g12(u, v) = 0 = g21(u, v), and g22(u, v) = 1/v2 fromExample 6.1.9 to compute the Christoffel symbols for the points inthe Poincare upper half-plane. Note that

0 =∂g11

∂u=∂g12

∂u=∂g22

∂u

since the coefficients only depend on v. Also

∂g11

∂v=∂g22

∂v= − 2

v3and

∂g12

∂v= 0.

From this information we can calculate the Christoffel symbols forthis metric.


Using Equation (7.2) we find

[11, 1] = 0, [11, 2] =1

v3,

[12, 1] = − 1

v3, [12, 2] = 0,

[21, 1] = − 1

v3, [21, 2] = 0,

[22, 1] = 0, [22, 2] = − 1

v3.

Then using Equation (7.2), we get

Γ111 = 0, Γ2

11 = 1/v,

Γ112 = −1/v, Γ2

12 = 0,

Γ121 = −1/v, Γ2

21 = 0,

Γ122 = 0, Γ2

22 = −1/v.

Problems

7.1.1. Fill in the details of the proof for Proposition 7.1.5.

7.1.2. Calculate the Christoffel symbols for the torus parametrized by

~X(u, v) = ((a+ b cos v) cosu, (a+ b cos v) sinu, b sin v) ,

where we assume that b < a.

7.1.3. Calculate the Christoffel symbols for functions graphs, i.e., surfacesparametrized by ~X(u, v) = (u, v, f(u, v)), where f is a function fromU ⊂ R2 to R.

7.1.4. Calculate the Christoffel symbols for surfaces of revolution (see Prob-lem 5.2.7).

7.1.5. Calculate the Christoffel symbols for the pseudosphere (see Example6.6.6).

7.1.6. Tubes. Let ~α(t) be a regular space curve and let r be small enoughthat the tube

~X(t, u) = ~α(t) + (r cosu)~P (t) + (r sinu) ~B(t)

is a regular surface. Calculate the Christoffel symbols for the tube ~X.


7.1.7. Prove that a parametrization ~X of a neighborhood on a surface S issuch that all the coordinate lines x2 = c are asymptotic lines if andonly if L11 = 0.

7.1.8. Defining the function D(x1, x2) as D2 = det(gij), show that

∂

∂x1(lnD) = Γ1

11 + Γ212 and

∂

∂x2(lnD) = Γ1

12 + Γ222.

7.1.9. Suppose that S is a surface with a parametrization that satisfiesΓ112 = Γ2

12 = 0. Prove that S is a translation surface.

7.1.10. Using D as defined in Problem 7.1.8, calling θ(x1, x2) the angle be-tween the coordinate lines, prove that

∂θ

∂x1= − D

g11Γ211 −

D

g22Γ211 and

∂θ

∂x2= − D

g11Γ212 −

D

g22Γ122.

(7.11)Show that if the parametrization is orthogonal, i.e., g12 = 0, Equa-tion (7.11) becomes

g22Γ211 + g11Γ1

12 = 0 and g22Γ212 + g11Γ1

22 = 0.

7.1.11. As in Example 7.1.7, calculate the Christoffel symbols of the firstand second kind for a general metric of the form g11(u, v) = f(v),g22(u, v) = f(v), and g12(u, v) = 0.

7.2 Codazzi Equations and the Theorema Egregium

Gauss formula in Equation (7.1) define expressions for any secondderivative ~Xij in terms of the coefficients of the first and second fun-

damental forms and the basis ~X1, ~X2, ~N. We remind the readerthat the symmetry of the dot product imposes gij = gji. Further-

more, from ~Xij = ~Xji, we also deduced that Lij = Lji.Definition 5.2.10 of a regular surface imposes the condition that

all the higher derivatives of any parametrization ~X of a coordinatepatch be continuous. Therefore, for a regular parametrization, theorder in which one takes derivatives with respect to given variablesis irrelevant. In particular, for the mixed third derivatives, we have~X112 = ~X121 and ~X221 = ~X122, which can be listed in the followingslightly more suggestive manner:

∂ ~X11

∂x2=∂ ~X12

∂x1and

∂ ~X22

∂x1=∂ ~X12

∂x2.

7.2. Codazzi Equations and the Theorema Egregium 259

Consequently, because of the nontrivial expressions in Gauss’s equa-tions, we obtain

∂

∂x2

(Γ1

11~X1 + Γ2

11~X2 + L11

~N)

=∂

∂x1

(Γ1

12~X1 + Γ2

12~X2 + L12

~N),

∂

∂x2

(Γ1

12~X1 + Γ2

12~X2 + L12

~N)

=∂

∂x1

(Γ1

22~X1 + Γ2

22~X2 + L22

~N).

(7.12)

It is natural to equate the basis components of each equation.However, when we take the partial derivatives, we will obtain ex-pressions involving second derivatives of ~X and derivatives of ~Nthat we need to put back into the usual ~X1, ~X2, ~N basis usingthe Weingarten equations given in Equation (6.26) or Gauss’s for-mula in Equation (7.1). Doing this transforms Equation (7.12) intosix distinct equations from which we deduce two significant theoremsin the theory of surfaces.

Theorem 7.2.1 (Codazzi Equations). For any parametrized surface ofclass C3, the following hold:

∂L11

∂x2− ∂L12

∂x1= L11Γ1

12 + L12Γ212 − L12Γ1

11 − L22Γ211,

∂L12

∂x2− ∂L22

∂x1= L11Γ1

22 + L12Γ222 − L12Γ1

12 − L22Γ212.

(7.13)

We can summarize these two equations in one by

∂Lij∂xl

− ΓkilLkj =∂Lil∂xj

− ΓkijLkl

for all 1 ≤ i, j, l ≤ 2, where Einstein summation is implied.

Proof: Equate the ~N coefficient functions in both equations fromEquation (7.12).

The classical formulation of the above relationships in Equation(7.13) are collectively called the Codazzi equations or sometimes theMainardi-Codazzi equations. These equations present a relationshipthat must hold between the coefficients of the first and second fun-damental forms that must hold for all regular surfaces. They are a


crucial part of the Fundamental Theorem of Surface Theory that wewill present in Section 7.3.

The following theorem is another result that stems from equatingcomponents along ~X1 and ~X2 in Equation (7.12).

Theorem 7.2.2 (Theorema Egregium). The Gaussian curvature of a sur-face is an intrinsic property of the surface, that is, it depends only onthe coefficients of the metric tensor and higher derivatives thereof.

Proof: Equating the ~X1 and ~X2 coefficient functions in the first equa-tion of Equation (7.12), we obtain

∂Γ111

∂x2+ Γ1

11Γ112 + Γ2

11Γ122 + L11a

12 =

∂Γ112

∂x1+ Γ1

12Γ111 + Γ2

12Γ112 + L12a

11,

∂Γ211

∂x2+ Γ1

11Γ212 + Γ2

11Γ222 + L11a

22 =

∂Γ212

∂x1+ Γ1

12Γ211 + Γ2

12Γ212 + L12a

21.

Since aij = −∑2

l=1 Ljlgli, where (gij) = (gij)

−1, after some simplifi-cation, we can write these two equations as

g21L11L22 − (L12)2

det g=∂Γ1

12

∂x1− ∂Γ1

11

∂x2+ Γ2

12Γ112 − Γ2

11Γ122, (7.14)

−g11L11L22 − (L12)2

det g=∂Γ2

12

∂x1− ∂Γ2

11

∂x2+ Γ1

12Γ211 + Γ2

12Γ212 − Γ1

11Γ212 − Γ2

11Γ222. (7.15)

SinceK = det(Lij)/ det(gij), we can use either one of these equationsto obtain a formula for K in terms of the function gij and Γijk. How-

ever, since the Christoffel symbols Γijk are themselves determined bythe coefficients of the metric tensor, then these equations provideformulas for the Gaussian curvature exclusively in terms of the firstfundamental form.

Gauss coined the name “Theorema Egregium,” which means “anexcellent theorem,” and indeed this result should seem rather sur-prising. A priori, the Gaussian curvature depends on the compo-nents of the first fundamental form and the second fundamentalform. After all, the Gaussian curvature at a point p on the sur-face is K = det(dnp), where dnp is the differential of the Gauss


map. Since the coefficients of the matrix dnp with respect to the

basis ~X1, ~X2 involve the Lij coefficients, one would naturally ex-pect that we would need the second fundamental form to calculateK. However, the Theorema Egregium shows that only the coefficientfunctions of the first fundamental form are necessary.

In the study of surfaces, we have described a geometric quantityas a “local” or “global property” of a curve or surface if it does notchange under the orientation and position of the curve or surface inspace. On the other hand, topology studies properties of point setsthat are invariant under any homeomorphism – a bijective functionthat is continuous in both directions. The concept of an intrinsicproperty lies somewhere between these two extremes. Any bijectivefunction between two regular surfaces preserving the metric tensordefines an isometry between the two surfaces. Then intrinsic geom-etry will treat these two surfaces as the same.

In the above proof, Equations (7.14) and (7.15) motivate thefollowing definition of the Riemann symbols:

Rlijk =∂Γlik∂xj

−∂Γljk∂xi

+ ΓmikΓlmj − ΓmjkΓ

lmi. (7.16)

As it turns out (see Problem 7.2.3), the Riemann symbols form thecomponents of a (1, 3)-tensor. Then, a closely associated tensor de-noted by

Rijkl = Rmijkgml (7.17)

has the interesting property that R1212 = det(Lij), and hence, theGaussian curvature of a surface at a point is given by the componentfunction

K =R1212

det(gij). (7.18)

The tensor associated to the components Rlijk is called the Riemanncurvature tensor and plays an important role in the analysis on man-ifolds.

Example 7.2.3 (Cone). As a simple example, consider the right circu-lar cone defined by the parametrization

~X(u, v) = (v cosu, v sinu, v),


where u ∈ [0, 2π] and v > 0, which we know, as a developable surface,has Gaussian curvature K = 0 everywhere. Simple calculations give,for the metric tensor,

g11 = v2, g12 = 0, g22 = 2.

For the Christoffel symbols, one obtains

Γ211 = −1

2v, Γ1

12 = Γ121 =

1

v

with all the other symbols of the second kind Γijk = 0. With thesedata, an application of Equation (7.16) shows that all but two ofthe Riemann symbols vanish simply because all the terms vanish.However, for the two nontrivial terms, one calculates

R1122 =

∂Γ112

∂v− ∂Γ1

22

∂u+ Γ1

12Γ112 + Γ2

12Γ122 − Γ1

22Γ111 − Γ2

22Γ121 = − 1

v2+

1

v2= 0,

R2121 =

∂Γ211

∂v− ∂Γ2

12

∂u+ Γ1

11Γ212 + Γ2

11Γ222 − Γ1

21Γ211 − Γ2

21Γ221 = −1

2−(

1

v

)(−1

2v

)= 0.

Consequently, though the Christoffel symbols are not identically 0,the Riemann curvature tensor is identically 0. One finds then thatR1212 = 0 is a function of u and v, and hence, using Equation (7.18),one recovers the fact that the cone has Gaussian curvature identically0 everywhere.

The Theorema Egregium (Theorem 7.2.2) excited the mathemat-ics community in the middle of the 19th century and sparked a searchfor a variety of alternative formulations for the Gaussian curvatureof a surface as a function of the gij metric coefficients. We leave afew such formulas for K for the exercises, but we present a few here.

Recall that Lij = ~Xij · ~N , det(gij) = ‖ ~Xu × ~Xv‖2, and ~N =~Xu × ~Xv/‖ ~Xu × ~Xv‖. From these facts, one can easily see that

K =L11L22 − L2

12

det(gij)=

( ~Xuu~Xu

~Xv)( ~Xvv~Xu

~Xv)− ( ~Xuv~Xu

~Xv)2

det(gij)2,

(7.19)where ( ~A~B ~C) is the triple-vector product in R3. However, the triple-vector product is a determinant and we know that for all matrices M1


and M2, one has det(M1M2) = det(M1) det(M2), and so Equation(7.19) becomes

K =1

det(gij)2

∣∣∣∣∣∣∣~Xuu · ~Xvv

~Xuu · ~Xu~Xuu · ~Xv

~Xu · ~Xvv~Xu · ~Xu

~Xu · ~Xv

~Xv · ~Xvv~Xv · ~Xu

~Xv · ~Xv

∣∣∣∣∣∣∣−∣∣∣∣∣∣∣~Xuv · ~Xuv

~Xuv · ~Xu~Xuv · ~Xv

~Xu · ~Xuv~Xu · ~Xu

~Xu · ~Xv

~Xv · ~Xuv~Xv · ~Xu

~Xv · ~Xv

∣∣∣∣∣∣∣ .

(7.20)

This equation involves only the metric tensor coefficients and Γijksymbols, except in the inner products ~Xuu· ~Xvv and ~Xuv · ~Xuv. At firstglance, this problem seems insurmountable, but if one performs theLaplace expansion on the determinants along the first row, factorsterms involving det(gij), and collects into determinants again, onecan show that Equation (7.20) is equivalent to

K =

1

det(gij)2

∣∣∣∣∣∣∣~Xuu · ~Xvv − ~Xuv · ~Xuv

~Xuu · ~Xu~Xuu · ~Xv

~Xu · ~Xvv~Xu · ~Xu

~Xu · ~Xv

~Xv · ~Xvv~Xv · ~Xu

~Xv · ~Xv

∣∣∣∣∣∣∣−∣∣∣∣∣∣∣

0 ~Xuv · ~Xu~Xuv · ~Xv

~Xu · ~Xuv~Xu · ~Xu

~Xu · ~Xv

~Xv · ~Xuv~Xv · ~Xu

~Xv · ~Xv

∣∣∣∣∣∣∣ .

The value in this is that though one cannot express ~Xuu · ~Xvv or~Xuv · ~Xuv using only the metric coefficients, it can be shown (Prob-lem 7.2.1) that

~Xuu · ~Xvv − ~Xuv · ~Xuv = −1

2g11,22 + g12,12 −

1

2g22,11, (7.21)

where by gij,kl we mean the function

∂2gij∂xk∂xl

.

This produces the following formula for K that is appealing on thegrounds that it illustrates symmetry of the metric tensor coefficientsin determining K:

K =1

det(gij)2

∣∣∣∣∣∣−1

2g11,22 + g12,12 − 12g11,22

12g11,1 g12,1 − 1

2g11,2

g12,2 − 12g22,1 g11 g12

12g22,2 g21 g22

∣∣∣∣∣∣−∣∣∣∣∣∣

0 12g11,2

12g22,1

12g11,2 g11 g1212g22,1 g21 g22

∣∣∣∣∣∣ .

(7.22)


In the situation where one considers an orthogonal parametriza-tion of a neighborhood of a surface, namely, when g12 = 0, Equation(7.22) takes on the particularly interesting form of

K = − 1√g11g22

(∂

∂x1

(1√g11

∂√g22

∂x1

)+

∂

∂x2

(1√g22

∂√g11

∂x2

)).

(7.23)

By historical habit, Equations (7.22) and (7.23) are referred toas Gauss’s equations.

We now prove a theorem that we could have introduced muchearlier, but it arises in part as an application of the Codazzi equa-tions.

Definition 7.2.4. A set S ⊂ Rn is called path-connected if for any pairof points p, q ∈ S, there exists a continuous path (curve) α : [0, 1]→Rn, with α(0) = p, α(1) = q, and α([0, 1]) ⊂ S.

Proposition 7.2.5. If S is a path-connected regular surface in which allits points are umbilical, then S is either contained in a plane or ina sphere.

Proof: Let U ⊂ R2 be an open set, and let ~X : U → R3 be theparametrization of a neighborhood V = ~X(U) of S. Since all pointsof S are umbilical, then the eigenvalues of dnp are equal to a value λthat a priori is a function of the coordinates (u, v) of the patch V .We first show that λ(u, v) is a constant over U .

By Proposition 6.5.4, there always exist two linearly independenteigenvectors to dnp, so dnp is always diagonalizable. Thus, at anypoint p of S, the whole tangent plane TpS is the eigenspace of theeigenvalue λ. Thus, in particular we have

~Nu = dnp( ~Xu) = λ ~Xu,

~Nv = dnp( ~Xv) = λ ~Xv.

Since ~Nuv = ~Nvu, a fact that is equivalent to the Codazzi equations(see Problem 7.2.2), we have

λv ~Xu + λ ~Xuv = λu ~Xv + λ ~Xvu.


Since ~Xu and ~Xv are linearly independent, we deduce that λu =λv = 0, and therefore λ is a constant function.

If λ = 0, then ~Nu = ~Nv = ~0, so the unit normal vector ~N is aconstant vector ~N0 over the domain U . Then

∂

∂u( ~X · ~N0) = ~Xu · ~N0 +~0 = ~0

and similarly for ∂∂v ( ~X · ~N0) = ~0, which shows that ~X · ~N0 = ~0,

which further proves that V = ~X(U) lies in the plane through pperpendicular to ~N0.

On the other hand, if λ 6= 0, then consider the vector function

~Y (u, v) = ~X(u, v)− 1

λ~N(u, v).

By a similar calculation as above, we see that the function ~Y (u, v)is a constant vector. Then we deduce that

‖ ~X(u, v)− ~Y ‖ =∥∥∥ 1

λ~N∥∥∥ =

1

|λ|,

which shows that V = ~X(U) lies on a sphere of center ~Y and ofradius 1

|λ| .So far, the above proof has established that any parametrization

of a neighborhood V of S either lies on a sphere or lies in a planebut not that all of S necessarily lies on a sphere or in a plane. Theassumption that S is path-connected extends the result to the wholesurface as follows.

By Example A.1.10, if V1 and V2 are two coordinate patches ofS with respective coordinate systems (x1, x2) and (x1, x2) such thatV1 ∩ V2 6= ∅, then on V1 ∩ V2 the coefficients of the Gauss map withrespect to the coordinate systems are related by

aij =∂xi

∂xk∂xl

∂xjakl .

This is tantamount to saying that the matrices for the Gauss maprelative to different coordinate systems are similar matrices. Hence,if V1 and V2 are two overlapping coordinate patches, then the eigen-value λ is constant over V1 ∪ V2, and then all of V1 ∪ V2 either lies ina plane or on a sphere.


Since S is path-connected, given any two points p and q on S,there exists a path α : [0, 1]→ S, with α(0) = p and α(1) = q. Frombasic facts in topology, we know that α([0, 1]) is a compact set andthat any open cover of a compact set has a finite subcover. Thus,given a collection of coordinate patches that cover α([0, 1]), thenonly a finite subcollection of these coordinate patches is necessary tocover α([0, 1]). Call this collection V1, V2, . . . , Vr. By the reasoningin the above paragraph, we see that V1 ∪ V2 ∪ · · · ∪ Vr lies either ina plane or on a sphere. Thus all points of S either lie in the sameplane or lie on the same sphere.

Problems

7.2.1. Prove Equation (7.21).

7.2.2. Show that one can obtain the Codazzi equations from the equation~N12 = ~N21 and by using the Gauss equations in Equation (7.1).

7.2.3. Prove that the Riemann symbols defined in Equation (7.16) formthe components of a (1, 3)-tensor. [Requires Appendix A.]

7.2.4. Prove Equation (7.18) and the claim that supports it.

7.2.5. Suppose that g12 = L12 = 0 (i.e., the coordinate lines are lines ofcurvature), so that the principal curvatures satisfy

κ1 =L11

g11and κ2 =

L22

g22.

Prove that the Codazzi equations are equivalent to

∂κ1∂x2

=g11,22g11

(κ2 − κ1) and∂κ2∂x1

=g22,12g22

(κ1 − κ2),

where by g11,2 we mean∂g11∂x2

and similarly for g22,1.

7.2.6. Prove that the Riemann symbols Rlijk defined in Equation (7.16) are

antisymmetric in the indices i, j. Conclude that for surfaces, Rlijkrepresent at most four distinct functions.

7.2.7. Prove Equation (7.23).


7.2.8. Prove the following formula by Blaschke for the Gaussian curvature:

K =− 1

4 det(gij)2

∣∣∣∣∣∣g11 g12 g22g11,1 g12,1 g22,1g11,2 g12,2 g22,2

∣∣∣∣∣∣− 1

2√

det(gij)

(∂

∂x1

(g22,1 − g12,2√

det(gij)

)− ∂

∂x2

(g12,1 − g11,2√

det(gij)

)).

7.2.9. Consider the two surfaces parametrized by

~X(u, v) = (v cosu, v sinu, ln v),

~Y (u, v) = (v cosu, v sinu, u).

Prove that these two surfaces have equal Gaussian curvature func-tions over the same domain but that they do not possess the samemetric tensor. (This gives an example of two surfaces with givencoordinate systems over which different metric tensor componentslead to the same Gaussian curvature function.)

7.2.10. Since the Gaussian curvature is intrinsic, depending only on the met-ric coefficients and their derivatives, compute that Gaussian curva-ture for the metric with g11(u, v) = g22(u, v) = 1/v2 and g12(u, v) =0 on the Poincare upper half-plane, as in Example 6.1.9 and Exam-ple 7.1.7.

7.2.11. Carry out the same computation as in the previous problem for thegeneral metric g11(u, v) = g22(u, v) = f(v) and g12(u, v) = 0 on theupper half-plane. [See Problems 6.1.20 and 7.1.11.]

7.2.12. Consider a surface with metric coefficients g11(u, v) = 1, g12(u, v) =0, and g22(u, v) = f(u, v), for some positive function f . Find theChristoffel symbols of both kinds for this metric and compute theGaussian curvature K.

7.2.13. Liouville surface. Consider a surface with metric coefficients definedby g11(u, v) = g22(u, v) = U(u)+V (v) and g12(u, v) = 0, where U(u)is a function of u and V (v) is a function of v. Find the Christoffelsymbols of both kinds for this metric and compute the Gaussiancurvature K. Such a surface is called a Liouville surface.

7.2.14. The coordinate curves of a parametrization ~X(u, v) form a Tchebysh-eff net if the lengths of the opposite sides of any quadrilateral formedby them are equal.


(a) Prove that a necessary and sufficient condition for a parame-trization to be a Tchebysheff net is

∂g11∂v

=∂g22∂u

= 0.

(b) (ODE) Prove that when a parametrization constitutes a Tche-bysheff net, there exists a reparametrization of the coordinateneighborhood so that the new components of the metric tensorare

g11 = 1, g12 = cos θ, g22 = 1,

where θ is the angle between the coordinate lines at the givenpoint on the surface.

(c) Show that in this case,

K = − θuvsin θ

.

7.3 The Fundamental Theorem of Surface Theory

The original proof of this theorem was provided by Bonnet in 1855in [5]. In more recent texts, one can find the proof in the Appendixto Chapter 4 in [11] or in Chapter VI of [31]. We will not provide acomplete proof of the Fundamental Theorem of Surface Theory heresince it involves solving a system of partial differential equations, butwe will sketch the main points behind it.

Suppose we consider a regular oriented surface S and coordinatepatch V parametrized by ~X : U → R3. We have seen that the co-efficients (gij) and (Lij) of the first and second fundamental formssatisfy det(gij) > 0 and the Gauss-Codazzi equations. That giventhese conditions, there exists an essentially unique surface with spec-ified first and second fundamental forms is a profound result, calledthe Fundamental Theorem of Surface Theory.

Theorem 7.3.1. If E, F , G and e, f , g are sufficiently differentiablefunctions of (u, v) that satisfy the Gauss-Codazzi equations (7.22)and (7.13) and EG−F 2 > 0, then there exists a parametrization ~Xof a regular orientable surface that admits

g11 = E, g12 = F, g22 = G,

L11 = e, L12 = f, L22 = g.

7.3. The Fundamental Theorem of Surface Theory 269

Furthermore, this surface is uniquely determined up to its positionin space.

The setup for the proof is to consider nine functions ξi(u, v),ϕi(u, v), and ψi(u, v) with 1 ≤ i ≤ 3, and think of these functionsas the components of the vector functions ~Xu, ~Xv, and ~N so that~Xu = (ξ1, ξ2, ξ3), ~Xv = (ϕ1, ϕ2, ϕ3), and ~N = (ψ1, ψ2, ψ3). With thissetup, the equations that define Gauss’s and Weingarten equations,namely, Equations (7.1) and (6.26), become the following system of18 partial differential equations: for i = 1, 2, 3,

∂ξi∂u

= Γ111ξi + Γ2

11ϕi + L11ψi,∂ξi∂v

= Γ112ξi + Γ2

12ϕi + L12ψi,

∂ϕi∂u

= Γ121ξi + Γ2

21ϕi + L21ψi,∂ϕi∂v

= Γ122ξi + Γ2

22ϕi + L22ψi,

(7.24)

∂ψi∂u

= a11ξi + a2

1ϕi,∂ψi∂v

= a12ξi + a2

2ϕi.

In general, when a system of partial differential equations involv-ing n functions ui(x1, . . . , xm) has n < m, the solutions may involvenot only constants of integration but also unknown functions thatcan be any continuous function from R to R (or some appropriateinterval). However, when n > m, i.e., when there are more func-tions in the system than there are independent variables, the systemmay be “overdetermined” and may either have less freedom in itssolution set or have no solutions at all. In fact, one cannot expectthe above system to have solutions if the mixed partial derivativesof ~ξ = (ξ1, ξ2, ξ3), ~ϕ = (ϕ1, ϕ2, ϕ3), and ~ψ = (ψ1, ψ2, ψ3) are notequal. This is usually called the compatibility condition for systemsof partial differential equations, and, as we see in the above system,this condition imposes relations between the functions Γijk(u, v) andLjk(u, v).

The key ingredient behind the Fundamental Theorem of SurfaceTheory is Theorem V in Appendix B of [31] that, applied to our con-text, states that if all second derivatives of the Γijk and Ljk functionsare continuous and if the compatibility condition holds in Equation(7.24), solutions to the system exist and are unique once values for~ξ(u0, v0), ~ϕ(u0, v0), and ~ψ(u0, v0) are given, where (u0, v0) is a point


in the common domain of Γijk(u, v) and Ljk(u, v). The compatibil-ity condition required in this theorem is satisfied if and only if thefunctions g11 = E, g12 = F , g22 = G, L11 = e, L12 = f , and L22 = gsatisfy the Gauss-Codazzi equations.

Solutions to Equation (7.24) can be chosen in such a way that

~ξ(u0, v0) · ~ξ(u0, v0) = E(u0, v0), ~ϕ(u0, v0) · ~ϕ(u0, v0) = G(u0, v0),

~ξ(u0, v0) · ~ϕ(u0, v0) = F (u0, v0), ~ψ(u0, v0) · ~ψ(u0, v0) = 1,

~ψ(u0, v0) · ~ξ(u0, v0) = 0, ~ψ(u0, v0) · ~ϕ(u0, v0) = 0,

~ξ(u0, v0)× ~ϕ(u0, v0)

‖~ξ(u0, v0)× ~ϕ(u0, v0)‖= ~ψ(u0, v0). (7.25)

The next step of the proof is to show that, given the above initialconditions, the following equations hold for all (u, v) where the so-lutions are defined:

~ξ(u, v) · ~ξ(u, v) = E(u, v), ~ϕ(u, v) · ~ϕ(u, v) = G(u, v),

~ξ(u, v) · ~ϕ(u, v) = F (u, v), ~ψ(u, v) · ~ψ(u, v) = 1,

~ψ(u, v) · ~ξ(u, v) = 0, ~ψ(u, v) · ~ϕ(u, v) = 0,

~ξ(u, v)× ~ϕ(u, v)

‖~ξ(u, v)× ~ϕ(u, v)‖= ~ψ(u, v).

From the solutions for ~ξ, ~ϕ, and ~ψ, we form the new system ofdifferential equations

~Xu = ~ξ,~Xv = ~ϕ.

One easily obtains a solution for the function ~X over appropriate(u, v) by

~X(u, v) =

∫ u

u0

~ξ(u, v) du+

∫ v

v0

~ϕ(u0, v) dv. (7.26)

The resulting vector function ~X is defined over an open set U ⊂ R2

containing (u0, v0), and ~X parametrizes a regular surface S. Byconstruction, the coefficients of the first fundamental form for thissurface are

g11 = E, g12 = F, g22 = G.

7.3. The Fundamental Theorem of Surface Theory 271

One then proves that it is also true that the coefficients of the secondfundamental form satisfy

L11 = e, L12 = f, L22 = g.

It remains to be shown that this surface is unique up to a rigidmotion in R3. It is not hard to see that the equalities in Equation(7.25) imposed on the initial conditions still allow one the freedomto choose the unit vector ξ(u0, v0) = ~ξ(u0, v0)/‖~ξ(u0, v0)‖ and thevector ~ψ(u0, v0), which must be perpendicular to ξ(u0, v0). The vec-tors

ξ(u0, v0), ~ψ(u0, v0), ξ(u0, v0)× ~ψ(u0, v0)

form a positive orthonormal frame, so any two choices allowed byEquation (7.25) differ from each other by a rotation in R3. Finally,the integration in Equation (7.26) introduces a constant vector ofintegration. Thus, two solutions to Gauss’s and Weingarten’s equa-tions differ from each other by a rotation and a translation, namely,any rigid motion in R3.

Problems

7.3.1. (ODE) Consider solutions to Gauss’s and Weingarten’s equationsfor which the coefficients of the first and second fundamental formsare constant.

(a) Let E, F , and G be constants such that EG−F > 0 and viewthem as constant functions. Prove that the Gauss-Codazziequations impose Ljk = 0.

(b) Prove that all solutions to the Gauss-Weingarten equations inthis situation are planes.

7.3.2. (ODE) Find all regular parametrized surfaces that have

g11 = 1, g12 = 0, g22 = cos2 u,

L11 = 1, L12 = 0, L22 = cos2 u.

7.3.3. Does there exist a surface ~X(u, v) with

g11 = 1, g12 = 0, g22 = cos2 u,

L11 = cos2 u, L12 = 0, L22 = 1?

CHAPTER 8

The Gauss-Bonnet Theorem and

Geometry of Geodesics

Historians of mathematics often point to Euclid as the inventor, orat least the father in some metaphorical sense, of the synthetic meth-ods of mathematical proofs. Euclid’s Elements, which comprises 13books, treats a wide variety of topics but focuses heavily on geome-try. In his Elements, Euclid presents 23 geometric definitions alongwith five postulates (or axioms), and in Books I-VI and XI-XIII,he proves around 250 propositions about lines, circles, angles, andratios of quantities in the plane and in space.

Most popular texts about the nature of mathematics agree thatEuclid’s geometric methods held a foundational importance in thedevelopment of mathematics (see, e.g., [12, pp. 76–77] or [18, p. 7]).Many such texts also retell the story of the discovery in the 19th

century of geometries that remain consistent yet do not satisfy thefifth and most debatable of Euclid’s postulates ([6], [9, pp. 214–227], [10], [18, pp. 217–223]). One can readily list elliptical geometryand hyperbolic geometry as examples of such geometries.

At this point in this book, with the methods from the theory ofcurves and surfaces now at our disposal, we stand in a position tostudy geometry on any regular surface. As we shall see, on a gen-eral regular surface the notion of “straightness” is not intuitive, andone might debate whether such a notion should exist at all. Hence,instead of only trying to consider lines and circles, we first studyregular curves on a regular surface in general and only later definenotions of shortest distance, straightness, and parallelism. Arguably,the most important theorem of this book is the Gauss-Bonnet Theo-rem, which relates the Gaussian curvature on a region R of the sur-face to a specific function along the boundary ∂R. One of the centralthemes of this chapter is to present the Gauss-Bonnet Theorem and

273

274 8. The Gauss-Bonnet Theorem and Geometry of Geodesics

introduce geodesics (curves that generalize the concept of shortestdistance and straightness). In the last two sections, we discuss appli-cations of the Gauss-Bonnet Theorem to non-Euclidean geometry.

8.1 Curvatures and Torsion

8.1.1 Natural Frames

Throughout this chapter, we let S be a regular surface of class C2

and let V be an open set of S parametrized by a vector function~X : U → R3, where U is an open set in R2. We consider a curve Cof class C2 of the form ~γ = ~X ~α, where ~α : I → U is a curve in thedomain of ~X, with ~α(t) = (u(t), v(t)).

Let p = ~γ(t0) be a point on the curve and on the surface. Usingthe theory of curves in R3, one typically performs calculations onquantities related to ~γ in the Frenet frame (~T , ~P , ~B). (Since ordermatters when discussing coordinates, we use the triple notation asopposed to set notation to describe an ordered basis.) This frameis orthonormal and hence has many nice properties. On the otherhand, using the local theory of surfaces, one would typically performcalculations in the ( ~Xu, ~Xv, ~N) reference frame. This frame is notorthonormal but is often the most practical. If a point p on a surfaceis not an umbilical point, the frame (~e1, ~e2, ~N), where the vectors ~eiare principal directions (well defined up to a change in sign), is anatural orthonormal frame associated to S at p. We could constructyet another orthonormal frame by first taking ~w1 = ~Xu/‖ ~Xu‖ andthen using the Gram-Schmidt algorithm on ~w1, ~Xv to obtain a unitvector ~w2 so that (~w1, ~w2, ~N) is an orthonormal frame.

Though more geometric in nature than ( ~Xu, ~Xv, ~N), the frame(~e1, ~e2, ~N) does not lend itself well to calculations with specific para-metrizations. Studying curves on surfaces, one could choose betweenthese three reference frames, but it turns out that a combination willbe most helpful.

Borrowing first from surface theory, we use the unit normal vec-tor ~N , which is invariant up to a sign under parametrization of S.We will often consider ~N(t) to be a single-variable vector functionover the interval I, by which we mean explicitly ~N(t) = ~N ~α(t).Borrowing from the theory of space curves, we use ~T , the unit tan-gent vector to the curve C at p, which is also invariant up to a sign

8.1. Curvatures and Torsion 275

under reparametrization of the curve. By construction, ~N and ~T areperpendicular to each other, and we just need to choose a third vec-tor to complete the frame. We cannot use ~P as a third vector in theframe because there is no guarantee that the principal normal vector~P (t) is perpendicular to ~N . In fact, Section 6.5 discusses how thesecond fundamental form measures the relationship between ~N and~P . Consequently, to complete a natural orthonormal frame relatedto a curve on a surface, we define the new vector function

~U(t) = ~N(t)× ~T (t).

We remind the reader that in the theory of plane curves, the unitnormal ~Uplane to a curve at a point is defined as the counterclockwise

rotation of the unit tangent ~T . One can rephrase this definition as

~Uplane = ~k × ~T .

Therefore, our definition of a normal vector in the theory of planecurves matches the above definition if one considers the xy-plane asurface in R3, with ~k as the unit surface normal vector and, in thiscontext, what we called ~Uplane we now denote as ~U .

From now on, for calculations related to a curve on a surface,we will use the frame (~T , ~U, ~N), which is often called the Darbouxframe.

8.1.2 Normal Curvature

As in the theory of space curves, we begin the study of curves onsurfaces with the derivative ~T ′ = s′κ~P . Since the principal normalvector is perpendicular to ~T , this vector ~T ′, often called the curva-ture vector , is perpendicular to ~T , and thus, in the (~T , ~U, ~N) frame,decomposes into a ~U component and an ~N component. (Alternately,the “curvature vector” sometimes refers to just κ~P .) In Definition6.5.1, we already introduced the notion of the normal curvature κn(t)as the component of ~T ′ along ~N . We now define an additional func-tion κg : I → R such that

~T ′ = s′(t)κ(t)~P = s′(t)κg(t)~U + s′(t)κn(t) ~N (8.1)

for all t ∈ I. We call κg the geodesic curvature of C at p on S.


Since ( ~N, ~T , ~U) is an orthonormal frame, we can already provideone way to calculate these curvatures, namely, using

κg = κ~P · ~U and κn = κ~P · ~N. (8.2)

If one is given parametric equations for the curve ~γ and the surface~X, then Equation (8.2) is often the most direct way of calculatingκn. If we are given the second fundamental form to the surface S,there is an alternative way to calculate κn. We remind the reader ofProposition 6.5.2, which relates the second fundamental form of Sto the normal curvature by

κn(t0) = IIp(~T (t0)).

Therefore, the second fundamental form of a unit vector ~T is thecomponent of a curve’s curvature vector ~T ′ in the normal direction. If~T (t0) makes an angle θ with ~e1, the principal direction associated tothe maximum principal curvature, then by Euler’s curvature formulain Equation (6.30),

κn(t0) = κ1 cos2 θ + κ2 sin2 θ.

We remind the reader that ~T at a point p on a regular curve Cmay change sign under a reparametrization and thus the principalnormal ~P may as well. For surfaces, the unit normal ~N , which iscalculated by

~N =~Xu × ~Xv

‖ ~Xu × ~Xv‖,

may change sign under a reparametrization. Thus, in Equation (8.2)one notices that κn and κg are unique only up to a possible signchange under a reparametrization of the curve or of the surface.On the other hand, none of these vectors changes direction underpositively oriented reparametrizations.

Figure 8.1 illustrates the principal normal ~P in the (~T , ~U, ~N)reference frame of a circle on the surface of a sphere. In this figure,however, since the surface is a sphere, all the points are umbilical,and there does not exist a unique basis ~e1, ~e2 of principal directions(even up to sign).


~T

~U

~N

~P

Figure 8.1. ~N , ~T , ~U , and ~P of a curve on a surface.

As an application of Euler’s formula in the context of curves onsurfaces, let us consider any two curves ~γ1 and ~γ2 on the surface S,each parametrized by arc length in such a way that they intersectorthogonally at the point p at s = s0. If we write

~γ′1(s0) = (cos θ)~e1 + (sin θ)~e2,

then

~γ′2(s0) = ±((sin θ)~e1 − (cos θ)~e2)

for either − or + signs as a choice on ±. If we denote κ(1)n and κ

(2)n

as the normal curvatures of ~γ1 and ~γ2, respectively, at p, then oneobtains the interesting fact that

κ(1)n + κ(2)

n = cos2 θκ1 + sin2 θκ2 + sin2 θκ1 + cos2 θκ2

= κ1 + κ2 = 2H.

In other words, for any two orthogonal unit tangent directions ~v1 and~v2 at a point p, the average of the associated normal curvatures isequal to the mean curvature and does not depend on the particulardirections of ~v1 and ~v2.

Note that since ~T · ~N = 0, it follows that ~N ′ · ~T = −~T ′ · ~N , andhence, we deduce that

~N ′ · ~T = −s′(t)κn(t).


We remind the reader that an asymptotic curve on a surface is acurve that satisfies κn(t) = 0 (see Definition 6.5.11) at all points onthe curve. Geometrically, this means that an asymptotic curve is acurve on which ~N ′ has no ~T component or, vice versa, a curve onwhich ~T ′ has no ~N component. Also, from an intuitive perspective,though we introduced κn as a measure of how much ~T changes in thenormal direction ~N , −κn gives a measure of how much ~N changesin the tangent direction ~T .

8.1.3 Geodesic Curvature

The geodesic curvature κg of a curve on a surface corresponds to thecomponent of the curvature vector occurring in the tangent plane.Equation (8.2) states this relationship as

κg = κ~P · ~U,

which is equivalent to

s′(t)κg = ~T ′ · ~U = ~T ′ · ( ~N × ~T ) = (~T ′ ~N ~T ) = (~T ~T ′ ~N), (8.3)

where (~u~v ~w) is the triple-vector product in R3. This already pro-vides a formula to calculate κg(t) in specific situations. It turns out,however, that the geodesic curvature is an intrinsic quantity, a factthat we show now.

Let us write ~α(t) = (u(t), v(t)) and ~γ = ~X ~α for the curve on thesurface. (To simplify the expression of certain formulas as we did inChapter 7, we will refer to the coordinates u and v as x1 and x2, andwe will use the expressions x1, x2, x1, x2 to refer to the derivatives ofu′, v′, u′′, v′′.) For the tangent vector, we have

s′(t)~T = ~Xuu′(t) + ~Xvv

′(t) =

2∑i=1

xi ~Xi. (8.4)

Taking another derivative of this expression, we get

s′′ ~T + s′ ~T ′

= ~Xuu(u′)2 + 2 ~Xuvu′v′ + ~Xvv(v

′)2 + ~Xuu′′ + ~Xvv

′′ (8.5)

=

2∑i=1

2∑j=1

xixj ~Xij +

2∑k=1

xk ~Xk.


The triple-vector product (~T ~T ′ ~N) can also be written as (~T× ~T ′) · ~N .Therefore, since ~T × ~T = ~0, taking the cross product of the twoexpressions in Equations (8.4) and (8.5) and then taking the dotproduct with ~N leads to

(s′)2(~T ~T ′ ~N) =[( ~Xu × ~Xuu)(u′)3 + (2 ~Xu × ~Xuv + ~Xv × ~Xuu)(u′)2v′

+ ( ~Xu × ~Xvv + 2 ~Xv × ~Xuv)u′(v′)2 + ( ~Xv × ~Xvv)(v

′)3]· ~N

+ ( ~Xu × ~Xv) · ~N(u′v′′ − u′′v′). (8.6)

It is possible to express all the coefficients of the terms (u′)3, (u′)2v′,and so forth using the metric tensor coefficients gij or their deriva-tives. For example, using the result of Problem 3.1.6,

( ~Xu × ~Xuu) · ~N = ( ~Xu × ~Xuu) ·~Xu × ~Xv√

det(g)

=( ~Xu · ~Xu)( ~Xuu · ~Xv)− ( ~Xu · ~Xv)( ~Xuu · ~Xu)√

det(g)

=g11[11, 2]− g12[11, 1]√

det(g)

=det(g)g22[11, 2] + det(g)g21[11, 1]√

det(g)

= Γ211

√det(g).

Repeating the calculations for all the relevant terms, we get

( ~Xu × ~Xuu) · ~N = Γ211

√det(g), ( ~Xv × ~Xuu) · ~N = −Γ1

11

√det(g),

( ~Xu × ~Xuv) · ~N = Γ212

√det(g), ( ~Xv × ~Xuv) · ~N = −Γ1

12

√det(g),

( ~Xu × ~Xvv) · ~N = Γ222

√det(g), ( ~Xv × ~Xvv) · ~N = −Γ1

22

√det(g).

Putting these expressions, along with ( ~Xu× ~Xv) · ~N =√

det(g), intoEquation (8.6) and using Equation (8.3) gives

(s′)3κg =(Γ2

11(u′)3 + (2Γ212 − Γ1

11)(u′)2v′ + (Γ222 − 2Γ1

12)u′(v′)2

−Γ122(v′)3 + u′v′′ − u′′v′

)√g11g22 − (g12)2. (8.7)


This shows that, as opposed to the normal curvature, the geodesiccurvature is an intrinsic quantity, depending only on the metric ten-sor and the parametric equations ~α(t) = (u(t), v(t)).

Though complete, Equation (8.7) is not written as concisely as itcould be using tensor notation. We define the permutation symbols

εh1h2···hn =

+1, if h1h2 · · ·hn is an even permutation of 1, 2, . . . , n,

−1, if h1h2 · · ·hn is an odd permutation of 1, 2, . . . , n,

0, if h1h2 · · ·hn is not a permutation of 1, 2, . . . , n.

(See also Section A.1 for more details on this tensor notation.) Weshall use this symbol in the simple case with n = 2, so with only twoindices. Then our indices satisfy 1 ≤ i, j ≤ 2 and εii = 0, ε12 = 1,and ε21 = −1. We can now summarize (8.7) as

κg =

√det(g)

(s′)3

(2∑

i,j,k,l=1

εilΓljkx

ixj xk +2∑

i,j=1

εij xixj

)or, in other words,

κg =

√det(g)

(s′)3

(εilΓ

ljkx

ixj xk + εij xixj),

where we use the Einstein summation notation convention.

Example 8.1.1 (Plane Curves). We modeled the (~T , ~U, ~N) frame off theframe (~T , ~U,~k), viewing the plane as a surface in R3 with normal vec-tor ~k. Consider then curves in the plane as curves on a surface. Usingan orthonormal basis in the plane, i.e., the trivial parametrization~X(u, v) = (u, v), we get g11 = g22 = 1 and g12 = 0. In this case,all the Christoffel symbols are 0 and of course

√det(g) = 1. Then

Equation (8.7) gives

κg =u′v′′ − u′′v′

(s′)3,

which is precisely the Equation (1.12) we obtained for the curvatureof a plane curve in Section 1.3.

It is an interesting fact, the proof of which we leave as an exerciseto the reader (Problem 8.1.6), that the geodesic curvature at a pointp on a curve C on a surface S is equal to the geodesic curvature of the


plane curve obtained by projecting C orthogonally onto TpS. Froman intuitive perspective, this property indicates why the geodesiccurvature should be an intrinsic quantity. Intrinsic properties dependon the metric tensor, which is an inner product on the tangent spaceTpS, so any quantity that measures something within the tangentspace is usually an intrinsic property.

8.1.4 Geodesic Torsion

Similar to how we viewed the Frenet frame in Chapter 3, we considerthe triple (~T , ~U, ~N) as a moving frame based at a point ~γ(t). Since~T , ~U , and ~N , are unit vectors, using the same reasoning as we didto establish Equation (3.5), we can determine that

d

dt

(~T ~U ~N

)=(~T ~U ~N

)A(t),

where A(t) is an antisymmetric matrix. We already know most ofthe coefficient functions in A(t). In Equation (8.1), our definitionsof the normal and geodesic curvature gave

~T ′ = s′(t)κg(t)~U + s′(t)κn(t) ~N (8.8)

and using the usual dot product relations among the vectors in anorthonormal basis, namely

~T · ~T = 1, ~U · ~U = 1, ~N · ~N = 1,

~N · ~T = 0, ~N · ~U = 0, ~T · ~U = 0,

we deduce the following equalities:

~T ′ · ~T = 0, ~U ′ · ~U = 0, ~N ′ · ~N = 0,

~N ′ · ~T = −~T ′ · ~N, ~N ′ · ~U = −~U ′ · ~N, ~T ′ · ~U = −~U ′ · ~T .

Consequently, from Equation (8.8),

~N ′ · ~T = −s′κn and ~U ′ · ~T = −s′κg.

Therefore, in order to describe the coefficients ofA(t), and therebyexpress the derivatives ~T ′, ~U ′, ~N ′ in the (~T , ~U, ~N) frame, we need


to label one more coefficient function. Define the geodesic torsionτg : I → R to be the unique function such that

~U ′ · ~N = s′(t)τg(t) for all t ∈ I.

With this function defined, we can imitate Equation (3.5) for thechange of the Frenet frame, and write

d

dt

(~T ~U ~N

)=(~T ~U ~N

)s′(t)

0 −κg(t) −κn(t)κg(t) 0 −τg(t)κn(t) τg(t) 0

.

The normal and geodesic curvatures both possess fairly intuitivegeometric explanations as to what they measure. However, it isharder to say precisely what the geodesic torsion measures. Usingthe formula

~N ′ = s′(−κn ~T + τg ~U),

we could say that the geodesic torsion measures the rate of changeof ~N in the ~U direction, which would mean the rate of change of ~Ntwisting around the direction of motion along the curve (the direc-tion in the tangent plane perpendicular to ~T ). This intuition is notparticularly instructive. In Section 8.4, we study geodesic curves ofa surface, which in some sense generalize the notion of straight lineson the surface. Problem 8.4.1 shows that the geodesic torsion is thetorsion function of the geodesic curves.

Problems

8.1.1. Let ~X = (R cosu sin v,R sinu sin v,R cos v), with (u, v) ∈ [0, 2π] ×[0, π], be a parametrization for a sphere. Consider the circle on the

sphere given by ~γ(t) = ~X(t, ϕ0), where ϕ0 is a fixed constant. Calcu-late the normal curvature, the geodesic curvature, and the geodesictorsion of ~γ on ~X (see Figure 8.1 for an illustration with ϕ0 = 2π/3).

8.1.2. Consider the cylinder x2 + y2 = 1, and consider the curve C on thecylinder obtained by intersecting the cylinder with the plane throughthe x-axis that makes an angle of θ with the xy-plane.

(a) Show that C is an ellipse.

(b) Compute the normal curvature, geodesic curvature, and geo-desic torsion of C on the cylinder.

(c) Is the geodesic torsion an intrinsic quantity?


8.1.3. Let ~γ(s) be a curve parametrized by arc length on a surface param-

etrized by ~X(u, v). Prove that

κg√

det(g) =∂

∂u(~T · ~Xv)−

∂

∂v(~T · ~Xu).

8.1.4. Consider the torus parametrized by ~X(u, v) = ((a+b sin v) cosu, (a+b sin v) sinu, b cos v), where a > b. The curve ~γ(mt, nt), where m andn are relatively prime, is called the (m,n)-torus knot. Calculate thegeodesic curvature of the (m,n)-torus knot on the torus.

8.1.5. Consider the surface that is a function graph z = f(x, y). Calculatethe normal curvature, geodesic curvature, and geodesic torsion of alevel curve, i.e., a curve of the form f(x, y) = c.

8.1.6. Let ~X be the parametrization for a coordinate patch of a regularsurface S, and let ~γ = ~X ~α be the parametrization for a curve C onS. Consider a point p on S, and let C ′ be the orthogonal projectionof C onto the tangent plane TpS. Prove that the geodesic curvatureκg at p of C on S is equal to the curvature κg of C ′ at p as a planecurve in TpS.

8.1.7. Bonnet’s Formula. Suppose that a curve on a surface is given byϕ(u, v) = C, where C is a constant. Prove that the geodesic curva-ture is given by

κg =1√

det(g)

[∂

∂u

(g12ϕv − g22ϕu

g11ϕ2v − 2g12ϕuϕv + g22ϕ2

u

)+∂

∂v

(g12ϕu − g11ϕv

g11ϕ2v − 2g12ϕuϕv + g22ϕ2

u

)].

8.1.8. Liouville’s Formula. (Comment: This result is of vital importancein the proof of the Gauss-Bonnet Theorem in the next section.) Let~X(u, v) be an orthogonal parametrization of a patch on a regularsurface S. Let C be a curve on this patch parametrized by arclength by ~γ(s) = ~X(u(s), v(s)). Let θ(s) be a function defined along

C that gives the angle between ~T and ~Xu. Prove that the geodesiccurvature of C is given by

κg =dθ

ds+ κ(u) cos θ + κ(v) sin θ,

where κ(u) is the geodesic curvature along the u-parameter curve(i.e., v = v0) and similarly for κ(v).

8.1.9. Let S be a regular surface parametrized by ~X(u, v), and let C be

a regular curve on S parametrized by ~X(t) = ~X(u(t), v(t)) for t ∈


[a, b]. Consider the normal tube around C parametrized by

~Yr(t, v) = ~X(t) + r cos v ~U(t) + r sin v ~N(t)

for some r > 0.

(a) Find the metric tensor associated to ~Y (t, v), and conclude that

det g = (s′)2r2(1− rκg cos v − rκn sin v)2.

(b) Show that if r is small enough but still positive, then ~Yr is aregular parametrization.

(c) Assuming that ~Yr is regular, calculate the coefficients Lij ofthe second fundamental form of the normal tube around C.Prove that the Gaussian curvature K satisfies

K(t, v) = − s′(t)√det(g)

(κn(t) sin v + κg(t) cos v).

8.2 Gauss-Bonnet Theorem, Local Form

No course on classical differential geometry is complete without theGauss-Bonnet Theorem, arguably the most profound theorem in thedifferential geometry study of surfaces. The Gauss-Bonnet Theoremsimultaneously encompasses a total curvature theorem for surfaces,the total geodesic curvature formula for plane curves, and other fa-mous results, such as the sum of angles formula for a triangle inplane, spherical, or hyperbolic geometry. Other applications of thetheorem extend much further and lead to deep connections betweentopological invariants and differential geometric quantities, such asthe Gaussian curvature.

In his landmark paper [15], Gauss proved an initial version ofwhat is now called the Gauss-Bonnet Theorem. The form in whichwe present this theorem was first published by Bonnet in 1848 [4].Various alternative proofs exist (e.g., [33]) and a search of the lit-erature turns up a large variety of generalizations (e.g., to poly-hedral surfaces and to higher-dimensional manifolds). Despite thefar-reaching consequences of the theorem, the difficulty of the proofresides in one essentially topological property of curves on surfaces.When we present this theorem, we will simply provide a reference.

8.2. Gauss-Bonnet Theorem, Local Form 285

The theorem then follows easily from Green’s Theorem for simpleclosed curves in the plane.

First, however, we motivate the Gauss-Bonnet Theorem with thefollowing intuitive example.

Example 8.2.1 (The Moldy Potato Chip). We particularly encourage thereader to consult the demo applet for this example.

Consider a region R on a regular surface S such that the bound-ary curve ∂R is a regular curve that is simple and simply connected(can be shrunk continuously to a point) on the surface S. We nowcreate the “moldy potato chip” as the surface that consists of takingthe region R and spreading it out over every possible normal direc-tion by a distance of r, where r is a fixed real number. As the demoshows, the surface of the moldy potato chip (MPC) consists of threepieces: two pieces that are the parallel surfaces of S “above” and“below” R, and one region that consists of a half-tube around ∂R.(In the applet, these portions are colored by magenta and green,respectively.)

The total Gaussian curvature∫∫

MPCK dS of the MPC is thesame as the area of the portion of the unit sphere covered by theGauss map of the MPC. Consider this area as the boundary ∂Ris shrunk continuously to a point; it must vary continuously withthe shrinking of ∂R. However, this area must always be an integralmultiple of 4π because the Gauss map of a surface without boundarycovers the unit sphere a fixed number of times. The only functionthat is continuous and discrete is a constant function. In the limit,the moldy surface around a point is simply a sphere whose Gauss mapimage is the unit sphere with area 4π. Hence, the total curvature ofthe moldy potato chip is always 4π.

However, Problem 6.6.12 showed that if r > 0 is small enough,the total Gaussian curvature for each of the two parallel surfacesabove and below R is

∫∫RK dS. Furthermore (and we leave the full

details of this until Example 8.2.6), the total Gaussian curvature ofthe half-tube around ∂R is 2

∫∂R κg ds. Hence, we conclude that∫∫

MPCK dS = 4π ⇐⇒ 2

∫∫RK dS + 2

∫∂R

κg ds = 4π

⇐⇒∫∫RK dS +

∫∂R

κg ds = 2π. (8.9)


The applet for this example colors the magenta and green regions ina lighter shade when the Gaussian curvature of the MPC is positiveand in a darker shade when the Gaussian curvature is negative. Inthis way, the user can see how, even if certain regions of the Gaussmap are covered more than once, the signed area cancels portionsout so that the total signed area covered is 4π.)

In what follows, we work to establish Equation (8.9) rigorouslyand expand the hypotheses under which it holds, the end resultbeing the celebrated global Gauss-Bonnet Theorem. Furthermore,we provide an intrinsic proof of the Gauss-Bonnet Theorem, whichestablishes it as long as we have a metric tensor and without theassumption that the surface is in an ambient Euclidean three-space.

Certain types of regions on surfaces play an important role inwhat follows. We call a region R on a regular surface S simple if itsboundary can be parametrized by a simple, closed, piecewise regularcurve.

Using theorems of existence and uniqueness to certain differentialequations, there are a variety of ways to see that near every pointp on S there exists a neighborhood of p that is parametrized by anorthogonal parametrization. In Section 8.5 we will see such an or-thogonal parametrization using what are called geodesic coordinatesystems on S.

On a regular surface of class C3 in R3, we can utilize extrinsicproperties to realize an orthogonal parametrization. As discussed inSection 6.5, if p = ~X(u0, v0) is not an umbilical point of S, thenthe lines of curvature provide orthogonal coordinate lines for a para-metrization of a neighborhood of p. More precisely, by the theoremon existence and uniqueness of solutions to differential equations,it is possible to find a solution to the first-order differential equa-tion that results from Equation (6.29) with the initial condition of(u0, v0) and for each point, say (u0 + h, v0) with −ε < h < ε, alongthe resulting line of curvature, we can calculate the perpendicularline of curvature as the solution to the other branch resulting fromEquation (6.29). This is not a tractable problem for specific sur-faces, but what matters for what follows is that at every point p, wecan parametrize an open neighborhood of p on S with an orthogonalparametrization.


Let S be a regular surface of class C3, and let ~X : U → R3

be an orthogonal parametrization of a neighborhood V = ~X(U) inS. Consider a regular curve ~γ(s) parametrized by arc length whoseimage lies in V . Liouville’s Formula (see Problem 8.1.8) states thatthe geodesic curvature of ~γ(s) satisfies

κg(s) =dϕ

ds+ κ(u) cosϕ+ κ(v) sinϕ,

where κ(u) is the geodesic curvature along the u-parameter curve(i.e., v = v0) and similarly for κ(v) and ϕ(s) is the angle ~γ′(s) makes

with ~Xu. Using Equation (8.7) to calculate κ(u) and κ(v) and writ-ing cosϕ and sinϕ in terms of the metric tensor, one can rewriteLiouville’s Formula as

κg(s) =dϕ

ds+

1

2√g11g22

∂g22

∂u

dv

ds− 1

2√g11g22

∂g11

∂v

du

ds. (8.10)

Now consider a simple region R in V with a boundary ∂R thatis parametrized by arc length by a regular curve ~α(s) defined overthe interval [0, `]. Additionally, suppose that ~α(s) = ~X(u(s), v(s))for coordinate functions u(s) and v(s). Finally, suppose that U ′ isthe subset of U such that ~X(U ′) = R.

Integrating the geodesic curvature around the curve ∂R, we get∫∂R

κg ds =

∫ `

0

(1

2√g11g22

∂g22

∂u

dv

ds− 1

2√g11g22

∂g11

∂v

du

ds

)ds+

∫ `

0

dϕ

dsds,

where ` is the length of C. Applying Green’s Theorem (Theorem2.1.6) to the first term, we get∫

∂Rκg ds =

∫∫U ′

1

2

(∂

∂u

(1

√g11g22

∂g22

∂u

)+

∂

∂v

(1

√g11g22

∂g11

∂v

))du dv +

∫ `

0

dϕ

dsds.

By Equation (7.23), this becomes∫∂R

κg ds = −∫∫

U ′K√g11g22 du dv +

∫ `

0

dϕ

dsds,

which leads to∫∂R

κg ds = −∫∫

U ′K√g11g22 du dv + 2π. (8.11)


The last term in Equation (8.11) is the content of the Theorem ofTurning Tangents, proved by H. Hopf in [19], which we state withoutproving.

Lemma 8.2.2 (Theorem of Turning Tangents, Regular). Let C be a sim-ple, closed, regular curve on a regular surface S of class C3 param-etrized by ~α. Let ϕ(s) be defined as above. Then∫

C

dϕ

dsds = ±2π,

where the sign of 2π is positive if the orientation of ~α is such thatthe normal vector ~U(s) to ~α(s) points into the region enclosed by thecurve and negative otherwise.

Putting together Equation (8.11) and Lemma 8.2.2 establishes afirst local version of the Gauss-Bonnet Theorem.

Theorem 8.2.3 (Local Gauss-Bonnet Theorem, Regular). Let ~X : U →R3 be an orthogonal parametrization of a neighborhood V = ~X(U) ofan oriented surface S of class C3. Let R ⊂ V be a simple region ofS, and suppose that the boundary is ∂R = ~α([0, `]) for some simple,closed, regular, positively oriented curve ~α : [0, `] → S of class C3

parametrized by arc length. Then∫∂R

κg ds+

∫∫RK dS = 2π.

We have called the above formulation of the Gauss-Bonnet The-orem a local version since, as stated, it requires that the region R ofS be inside a coordinate neighborhood of S that admits an orthogo-nal parametrization. Furthermore, the above theorem assumes that∂R is a regular curve. By Problem 6.1.15, we know that if S is aregular surface of class C2, then at every point p ∈ S there exists aneighborhood of p that can be parametrized by a regular orthogo-nal parametrization. We will first generalize this theorem to includeregions whose boundaries satisfy a looser condition than being reg-ular. Then, to obtain a global version of the theorem, we will “piecetogether” local instances of the above theorem.

The above proof of the local Gauss-Bonnet Theorem is almostan intrinsic proof but not quite. In order to use Liouville’s Formula,


we needed to refer to an orthogonal parametrization of a patch ofthe surface. At that point, we chose to refer to a parametrizationof a patch that has curvature lines as coordinate lines. However,curvature lines are along the direction of principal curvatures, whichare extrinsic properties. Proposition 8.5.5 proves the existence ofgeodesic coordinate systems on patches of a regular surface. Suchcoordinate systems do depend entirely on the components of themetric tensor and its derivatives, and they induce orthogonal para-metrizations. Citing this proposition that we will encounter laterestablishes the Gauss-Bonnet Theorem as a purely intrinsic result.

We now extend the local Gauss-Bonnet Theorem to the broaderclass of piecewise regular curves in order to set the scene for theglobal theorem.

In order to present the Gauss-Bonnet Theorem for piecewise reg-ular curves, we need to first establish a few definitions about thegeometry of such curves on surfaces.

We call a set of points tii∈K in R discrete if

inf|ti − tj |∣∣ i, j ∈ K and i 6= j > 0,

i.e., if any two distinct points are separated by at least some fixed,positive real number. The indexing set K may be finite, say K =1, . . . , k, or, if it is infinite, it may be taken as either the set ofnonnegative integers N or the set of integers Z.

Suppose that a curve ~X : I → Rm is regular near t0 ∈ I, i.e., onthe interval (t0 − ε, t0 + ε), where ε > 0, and also suppose that itis regular on (t0 − ε, t0) and (t0, t0 + ε). Then we can consider thelimits of the unit tangent ~T (t) as t approaches t0 from the right andfrom the left. Call

~T (t−0 ) = limt→t−0

~T (t) and ~T (t+0 ) = limt→t+0

~T (t).

To be precise, we say that a curve ~α : I → Rn is piecewiseregular if

• ~α is continuous over I;

• there exists a discrete set of points tii∈K such that ~α is reg-ular over each open interval in the set difference I − tii∈K;


• ~T (t−i ) and ~T (t+i ) exist for all ti ∈ tii∈K.

A regular curve is also piecewise regular, by virtue of considering thesituation in which the set tii∈K is empty. If ~α is regular over aninterval I ′, then the trace ~α(I ′) is called a regular arc of the curve.If ~T (t−i ) = −~T (t+i ), then a point ~α(ti) is called a cusp. Otherwise, if~T (t−i ) and ~T (t+i ) are not collinear, then ~α(ti) is called a corner .

We point out that our definition for piecewise regular curves ap-plies to any curve in Rm. However we now also assume that ~α hasthe properties that will interest us for the Gauss-Bonnet Theorem,namely that ~α : I → R3 is a simple, closed, piecewise regular curveon a regular oriented surface S with orientation n. In this case, I is aclosed and bounded interval, and there can be at most a finite num-ber of vertices. Furthermore, we impose the criterion that the curve~α traces out the image ~α(I) in the same direction as the orientationof S. See the surface and curve orientations in Figure 8.2.

~α(t1)

~α(t2)

θ1 < 0θ2 > 0

Figure 8.2. A regular piecewise curve.

For all corners ~α(ti), we define the external angle θi of the vertexas the angle −π < θi < π swept out from ~T (t−i ) to ~T ′(t+i ) in the

plane through the point ~α(ti) and spanned by ~T ′(t−i ) and ~T ′(t+i )(see Figure 8.2). Note that this external angle may be positive ornegative.


We are now in a position to approach the local Gauss-BonnetTheorem for piecewise regular curves.

Let S be a regular surface, and let V be a neighborhood of Sparametrized orthogonally by ~X : U → R3 with U ⊆ R2. We assumethat ~X : U → V is a homeomorphism. Consider a simple region R inthe neighborhood V on S such that its boundary ∂R is parametrizedby a simple, closed, piecewise regular curve ~α : I → R3. Supposethat ~α is parametrized by arc length so that I = [0, `] and has verticesat s1 < s2 < · · · < sk. Set s0 = 0 and sk+1 = `. Also call Ci theimage of ~α([si−1, si]) for 1 ≤ i ≤ k + 1; these are the regular arcs of∂R . Suppose, additionally, that ~α(s) = ~X(u(s), v(s)). Let U ′ bethe subset of U such that R = ~X(U ′).

As in the previous case, integrating the geodesic curvature aroundthe curve C, we get

∫∂R

κg ds =

k∑i=0

∫Ci

κg ds =

k∑i=0

∫ si+1

si

κg(s) ds

=k∑i=0

∫ si+1

si

(1

2√g11g22

∂g22

∂u

dv

ds− 1

2√g11g22

∂g11

∂v

du

ds

)ds+

k∑i=0

∫ si+1

si

dϕ

dsds.

Applying Green’s Theorem (generalized to simple closed piece-wise regular curves) to the first term, we get

∫∂R

κg ds =

∫∫U ′

1

2

(∂

∂u

(1

√g11g22

∂g22

∂u

)+

∂

∂v

(1

√g11g22

∂g11

∂v

))du dv +

k∑i=0

∫ si+1

si

dϕ

dsds.

By Equation (7.23), this becomes

∫∂R

κg ds = −∫∫

U ′K√g11g22 du dv +

k∑i=0

∫ si+1

si

dϕ

dsds,

which leads to∫∂R

κg ds = −∫∫

~X(U ′)K dS +

k∑i=0

(ϕ(si+1)− ϕ(si)) . (8.12)


It remains for us to interpret the last term in Equation (8.12).As before, this is precisely the more general version of the Theoremof Turning Tangents [19].

Lemma 8.2.4 (Theorem of Turning Tangents). Let ~α be a simple, closed,piecewise regular curve on a regular surface S of class C3. Let ~α(si)be the vertices with external angles θi, and let ϕ(s) be as definedabove. Then

k∑i=0

(ϕ(si+1)− ϕ(si)) = ±2π −k∑i=1

θi,

where the sign of 2π is positive if the orientation of ~α is such thatthe normal vector ~U(s) to ~α(s) points into the region enclosed by thecurve and negative otherwise.

Putting together Equation (8.12) and Lemma 8.2.4 establishes alocal version of the Gauss-Bonnet Theorem.

Theorem 8.2.5 (Local Gauss-Bonnet Theorem). Let ~X : U → R3 be anorthogonal parametrization of a region V = ~X(U) of an orientedsurface S of class C3. Let R ⊂ V be a simple region of S, andsuppose that the boundary is ∂R = ~α([0, `]) for some simple, closed,piecewise regular, positively oriented curve ~α : [0, `]→ S of class C2,parametrized by arc length. Let ~α(si), with 1 ≤ i ≤ k, be the verticesof ∂R, and let θi be their external angles. Call Ci the regular arcsof ∂R. Then

∫∂R

κg ds+

∫∫RK dS +

k∑i=1

θi = 2π.

Note that in this statement of the theorem, there are k verticesand k + 1 regular arcs because the formulation assumes that ~α(0)is not a vertex. Consequently, we must remember to interpret theintegral on the left as an integral over k + 1 regular arcs as

∫∂R

κg ds =k∑i=0

∫Ci

κg ds.


We typically call the more general Theorem 8.2.5 the Local Gauss-Bonnet Theorem and understand that Theorem 8.2.3 is a subcase ofthis theorem.

We illustrate this theorem in a similar vein as the Moldy PotatoChip example but now with a patch of surface bounded by a piece-wise regular curve.

Example 8.2.6 (The Moldy Patch). The motivating example, Example8.2.1 with the moldy potato chip, falls just shy of giving an extrinsic(assumes we have/know a normal vector to the surface) proof of thelocal Gauss-Bonnet Theorem. We provide the details here.

Consider a simply connected region R on a regular surface Sas described in Theorem 8.2.5. Suppose that R is parametrized by~X : U → R3 for some U ⊂ R2, and call ~N the associated normalvector.

Define the surface Tr as the tubular neighborhood of R withradius r. This means that Tr consists of

1. two pieces for the normal variation to R parametrized respec-tively by ~X+r ~N and ~X−r ~N over U (which we call respectivelyU(+r) and U(−r));

2. k half-tubes of radius r around the smooth pieces of the bound-ary ∂R pointing “away” from the region R;

3. k lunes of spheres (of radius r) at the k vertices of ∂R.

Figure 8.3 shows the pieces of Tr for a patch on a torus. We assumefrom now on that r is small enough so that each of the pieces of Tris a regular surface.

Let K be the Gaussian curvature of S over R, and call KT theGaussian curvature of the moldy patch Tr. By Proposition 6.6.2,the quantity KT dST is a signed area element of the image on theunit sphere of Tr under its Gauss map. So in calculating the totalcurvature of the moldy patch, one is adding or subtracting area ofthe sphere depending on the sign of KT . We now reason why∫∫

TrKT dST = 4π. (8.13)


Figure 8.3. A “moldy patch.”

Since Tr has no boundary, the Gauss map of Tr covers the unitsphere an integer number of times, where we add area for positivecurvature and subtract for negative curvature. Thus, the integralin Equation (8.13) must be equal to 4πh, where h is an integer.Suppose that the region R is contractible, which means that it canbe shrunk continuously to a point while remaining on the surface.Now if R is a point, then Tr is a sphere of radius r. In this case,the Gauss map for Tr is a bijective map onto the unit sphere, andhence, the integral in Equation (8.13) gives precisely the surface ofthe unit sphere and so is equal to 4π. However, as one “uncon-tracts” from a point to R, the integral in Equation (8.13) mustvary continuously. However, a continuous function to the set of in-teger multiples of 4π is a constant function. Though we have notspelled out the whole topological background, this reasoning justifiesEquation (8.13).

We now break down Equation (8.13) according to various piecesof Tr: the normal variation patches, the half-tubes, and the lunes ofspheres.

By Problem 6.6.12, if we call K(+r) and K(−r) the respective

Gaussian curvatures of ~X + r ~N and ~X − r ~N over U , then

2

∫∫UK dS =

∫∫U(+r)

K(+r) dS +

∫∫U(−r)

K(−r) dS.

In Problem 8.1.9, one calculates the Gaussian curvature of thenormal tube around a curve on the surface. Consider a regular arc

8.3. Gauss-Bonnet Theorem, Global Form 295

C of ∂R, and assume that it is parametrized by t ∈ [a, b]. As a con-sequence of Problem 8.1.9, over the normal half-tube HT pointingaway from ~U (i.e., away from the inside of R), the total Gaussiancurvature is∫∫

HTK dS =

∫ b

a

∫ 3π/2

π/2−s′(t) ((sin v)κn(t) + (cos v)κg(t)) dt

= 2

∫ b

as′(t)κg(t) dt = 2

∫Cκg ds.

Finally, we consider the lunes of Tr that are around the verticesof ∂R. Under the image of the Gauss map, the lunes map to thesame corresponding lune on the unit sphere. Thus, in Equation(8.13), a lune around a vertex with exterior angle θi contributes(θi/2π)4π = 2θi to the surface of the unit sphere.

Combining each of these results, we find that∫∫TrKT dST = 2

∫∫UK dS + 2

k∑i=0

∫Ci

κg ds+ 2k∑i=1

θi,

and the local Gauss-Bonnet Theorem follows immediately from Equa-tion (8.13).

In our presentation of the local Gauss-Bonnet Theorem, we didnot allow corners to be cusps. The main difficulty lies in decidingwhether to assign a value of π or −π to the angle θi of any givencusp in order to retain the validity of the local Gauss-Bonnet Theo-rem. We can allow for cusps on the boundary ∂R if we employ thefollowing sign convention for the angles of cusps: If the cusp ~α(ti)points into the interior of the closed curve (i.e., ~α(t−i ) points into theinterior) then θi = −π; and if the cusp ~α(ti) points away from theinterior of the closed curve (i.e., ~α(t−i ) points away from the interior)then θi = π. (See Figure 8.4.) The orientation of the surface mat-ters since it determines the direction of travel around the boundarycurve ∂R.)

8.3 Gauss-Bonnet Theorem, Global Form

In order to extend the Gauss-Bonnet Theorem to a global presen-tation (i.e., outside of a single coordinate patch on the surface), we


θ2 = πθ1 = −π

Figure 8.4. External angles at cusps.

need to briefly discuss triangulations of surfaces, the classification oforientable surfaces, and the Euler characteristic of regions of surfacesin R3. These topics are typically considered in the area of topology,but we summarize the results that we need in order to give a fulltreatment to the global Gauss-Bonnet Theorem without insistingthat the reader have mastery of the supporting topology. (The au-thors include technical details behind these concepts in AppendicesA.5 and A.6 in [24]. Otherwise, the interested reader could consultChapters 6 and 7 of [1].)

In intuitive terms, a triangulation of a surface consists of a net-work of a finite number of regular curve segments on the surface suchthat any point on the surface either lies on one of the curves or liesin a region that is bounded by precisely three curve segments. Thefirst picture in Figure 8.5 depicts a triangulation on a torus. As anadditional technical requirement, one should be able to continuouslydeform the surface with its triangulation so that each “triangle” be-comes a true triangle without changing the topological nature of thesurface. (Compare the two pictures in Figure 8.5.)

In a triangulation, a vertex is an endpoint of one of the curvesegments on the surface. We call the curve segments edges, andthe regions enclosed by edges (the “triangles”) we call faces. Aninteresting and useful result, first proved by Rado in 1925, is thatevery regular compact surface admits a triangulation.

A result of basic topology is that given a compact regular surfaceS, the quantity

#(vertices)−#(edges) + #(faces)


Figure 8.5. Torus triangulation.

is the same regardless of any triangulation of S. This number iscalled the Euler characteristic of S and is denoted by χ(S). Fur-thermore, from the definition of triangulation, one can deduce thatthe Euler characteristic does not change if the surface is deformedcontinuously (no cutting or pinching). One often restates this lastproperty by saying that the Euler characteristic is a topological in-variant .

For example, the torus triangulation in Figure 8.5 has 16 ver-tices, 48 edges, and 32 faces. Thus, the Euler characteristic of thetorus is 0. As another example consider the tetrahedron, which ishomeomorphic to the sphere. A tetrahedron has four vertices, sixedges, and four faces, so its Euler characteristic, and therefore theEuler characteristic of the sphere, is χ = 4− 6 + 4 = 2.

A profound theorem in topology, the Classification Theorem ofSurfaces states that every orientable surface without boundary ishomeomorphic to a sphere or to a sphere with a finite number of“handles” added to it. Figure 8.6(a) shows a torus while Figure8.6(b) shows a sphere with one handle added to it. These two sur-faces are in fact the same under a continuous deformation, i.e., theyare homeomorphic. Figure 8.6(c) depicts a two-holed torus that, inthe language of the Classification Theorem of Surfaces, is called asphere with two handles.

It is not hard to show that the Euler characteristic of a spherewith g handles added is

χ(S) = 2− 2g. (8.14)


(a) Torus. (b) Sphere with one handle. (c) Two-holed torus.

Figure 8.6. Tori.

The notion of the Euler characteristic applies equally well to asurface with boundary as long as the boundary is completely coveredby edges and vertices of the triangulation. For example, we encour-age the reader to verify that a sphere with a small disk removed hasEuler characteristic of 1.

We must now discuss orientations on a triangulation. When con-sidering adjacent triangles, we can think of the orientation of a trian-gle as a direction of travel around the edges. Two adjacent triangleshave a compatible orientation if the orientation of the first leads oneto travel along the common edge in the opposite direction of the ori-entation on the second triangle (see Figure 8.7). It turns out that ifa surface is orientable, then it is possible to choose an orientation ofeach triangle in the triangulation such that adjacent triangles havecompatible orientations.

(a) compatible (b) incompatible

Figure 8.7. Adjacent oriented triangles.


A compact regular surface S is covered by a finite number of coor-dinate neighborhoods given by regular parametrizations. In general,if R is a regular region of S, it may not lie entirely in one coordinatepatch. However, it is possible to show that not only does a regularregion R admit a triangulation, but every regular region R admitsa triangulation such that each triangle is contained in a coordinateneighborhood. This comment and two lemmas show that it makessense to talk about the surface integral over the whole region R ⊂ S.

Lemma 8.3.1. Suppose that ~X1 : U1 → R3 and ~X2 : U2 → R3 are twosystems of coordinates of a regular surface S. Call (ui, vi) the coor-dinates of ~Xi, and call g(i) the corresponding metric tensor. Supposethat T ⊂ ~X1(U1) ∩ ~X2(U2). Then for any function f : T → R, wehave ∫∫

~X−11 (T )

f(u1, v1)√

det(g(1)) du1 dv1 =

∫∫~X−12 (T )

f(u2, v2)√

det(g(2)) du2 dv2.

Proof: By Equation (6.7), one deduces that

det(g(1)) =

(∂(u2, v2)

∂(u1, v1)

)2

det(g(2)).

However, ∂(u2,v2)∂(u1,v1) is the Jacobian of the coordinate transformation

F : ~X−11 (T ) → ~X−1

2 (T ) defined by ~X−12 ~X1 restricted to ~X−1

1 (T ).The result follows as an application of the change of variables formulain double integrals.

Lemma 8.3.2. Let S be a regular oriented surface, and let R be a regu-lar compact region of S, possibly with a boundary. Given a collection ~Xii∈I of coordinate neighborhoods that cover S, Tjj∈J trianglesof a triangulation of R, and i : J → I such that Tj is in the image

of ~Xi(j), define the sum∑j∈J

∫∫~X−1i(j)

(Tj)f(ui(j), vi(j))

√det g(i(j)) dui(j) dvi(j).

This sum is independent of the choice of triangulation of R, collec-tion of coordinate patches, and function i.


Proof: That the sum is independent of the collection of coordinateneighborhoods follows from Lemma 8.3.1. That the sum does notdepend on the choice of triangulation is a little tedious to prove andis left as an exercise for the reader.

This leads to a definition of a surface integral over any region ona regular surface.

Definition 8.3.3. Let S be a regular oriented surface, and let R be aregular compact region of S, possibly with a boundary. We call thecommon sum described in Lemma 8.3.2 the surface integral of f overR and denote it by ∫∫

Rf dS.

We point out that if ~X : U → R3 is a regular parametrization of aregion of S such that ~X(U) is dense in R, then∫∫

Rf dS =

∫∫Uf( ~X(u, v))‖ ~Xu × ~Xv‖ dA,

where the right-hand side is the usual double integral.We can now state the main theorem of this section.

Theorem 8.3.4 (Global Gauss-Bonnet Theorem). Let S be a regular ori-ented surface of class C3, and let R be a compact region of S withboundary ∂R. Suppose that ∂R is a simple, closed, piecewise regu-lar, positively oriented curve. Suppose that ∂R has k regular arcs Ciof class C2, and let θi be the external angles of the vertices of ∂R.Then

k∑i=1

∫Ci

κg ds+

∫∫RK dS +

k∑i=1

θi = 2πχ(R),

where χ(R) is the Euler characteristic of R.

Proof: Let R be covered by a collection of coordinate patches. LetTjj∈J be the triangles of a triangulation of R in which all thetriangles Tj on S are subsets of some coordinate patch. Suppose alsothat every triangle in the set Tj is equipped with an orientationthat is compatible with the orientation of S.


R

Figure 8.8. A triangulation of a region R.

For each triangle Tj , for 1 ≤ l ≤ 3, call Ejl the edges of Tj (ascurves on S), call Vjl the vertices of Tj , and let βjl be the interiorangle of Tj at Vjl. For this triangulation, call a0 the number ofvertices, a1 the number of edges, and a2 the number of triangles.By construction, the local Gauss-Bonnet Theorem applies to eachtriangle Tj on S, so on each Tj , we have

3∑l=1

∫Ejl

κg ds+

∫∫Tj

K dS = 2π−3∑l=1

(π−βjl) = −π+

3∑l=1

βjl. (8.15)

Now consider the sum of Equation (8.15) over all the trianglesTj . Since each triangle has an orientation compatible with the ori-entation of S, then whenever two triangles share an edge, the edge istraversed in opposite orientations on the adjacent triangles (see Fig-ure 8.8). Consequently, in the sum of Equation (8.15), each integral∫Ejl

κg ds cancels out another similar integral along any edge that is

not a part of the boundary of R. Therefore, applying Lemma 8.3.2,the left-hand side of the sum of Equation (8.15) is precisely

k∑i=1

∫Ci

κg ds+

∫∫RK dS. (8.16)


The right-hand side of the sum of Equation (8.15) is

−πa2 +∑j

3∑i=1

βjl.

In the double sum∑∑

βjl, the sum of interior angles associatedto a vertex on the interior of R contributes 2π, and the sum ofangles associated to a vertex V on the boundary ∂R contributesπ − (exterior angle of V ). Thus,

∑j

(−π +

3∑l=1

βjl

)= −

k∑i=1

θi − πa2 + 2π(#interior vertices) + π(#exterior vertices).

Since there are as many vertices on the boundary as there are edges,we rewrite this as

∑j

(−π +

3∑l=1

βjl

)= −

k∑i=1

θi − πa2 + 2πa0 − π(#exterior edges).

(8.17)However, since each triangle has three edges,

−a2 = 2a2 − 3a2 = 2a2 − 2(#interior edges)− (#exterior edges).(8.18)

Consequently, taking the sum of Equation (8.15) and combiningEquations (8.16), (8.17), and (8.18) one obtains

∑i

∫Ci

κg ds+

∫∫RK dS = −

k∑i=1

θi + 2π(a2 − a1 + a0).

By definition of the Euler characteristic, χ(R) = a2 − a1 + a0. Thetheorem follows.

The global version of the Gauss-Bonnet Theorem directly gen-eralizes the local version so when one refers to the Gauss-BonnetTheorem, one means the global version. Even at a first glance, theGauss-Bonnet Theorem is profound because it connects local prop-erties of curves on a surface (the geodesic curvature κg, the Gaussian


curvature K, and angles associated to vertices) with global proper-ties (the Euler characteristic of a region of a surface). In fact, sincethe Euler characteristic is a topological invariant, the Gauss-BonnetTheorem connects local geometric properties to a topological prop-erty.

Let S be a compact regular surface (without boundary). An in-teresting particular case of the global Gauss-Bonnet Theorem occurswhen we consider R = S, which implies that ∂R is empty. This situ-ation leads to the following strikingly simple and profound corollary.

Corollary 8.3.5. Let S be an orientable, compact, regular surface ofclass C3 without boundary. Then∫∫

SK dS = 2πχ(S).

Example 8.3.6. Consider the torus T parametrized by ~X : [0, 2π]2 →R3, with

~X(u, v) = ((a+ b cos v) cosu, (a+ b cos v) sinu, b sin v),

where a > b. We note that ~X((0, 2π)2) covers all of T except for twocurves on the torus, so ~X((0, 2π)2) is dense in T . By the commentafter Definition 8.3.3, we can use the usual surface integral over thisone coordinate neighborhood to calculate

∫∫SK dS directly.

It is not hard to calculate that over the coordinate neighborhooddescribed above, we have

K(u, v) =cos v

b(a+ b cos v)and ‖ ~Xu × ~Xv‖ = b(a+ b cos v).

Thus,∫∫SK dS =

∫ 2π

0

∫ 2π

0

cos v

b(a+ b cos v)· b(a+ b cos v) du dv =

∫ 2π

0

∫ 2π

0cos v du dv = 0,

which proves that χ(T ) = 0. This agrees with the calculation pro-vided by the triangulation in Figure 8.5.

We shall now present a number of applications of the Gauss-Bonnet Theorem, as well as leave a few as exercises for the reader.


Proposition 8.3.7. Let S be a compact, regular, orientable surface ofclass C3 without boundary and with positive curvature everywhere.Then S is homeomorphic to a sphere.

Proof: Since K > 0 over the whole surface S, then for the totalGaussian curvature we have∫∫

SK dS > 0.

By Corollary 8.3.5, this integral is 2πχ(S). However, by the Classi-fication Theorem of Surfaces and Equation (8.14), since χ(S) > 0,we have χ(S) = 2, and therefore S is homeomorphic to a sphere.

Problems

8.3.1. Provide all the details that establish Equation (8.10).

8.3.2. Provide the details for the proof of Lemma 8.3.2.

8.3.3. Verify directly the Gauss-Bonnet Theorem for the rectangular regionR on a torus specified by u1 ≤ u ≤ u2 and v1 ≤ v ≤ v2, where theu-coordinate lines are the parallels and the v-coordinate lines are themeridians of the torus, as a surface of revolution.

8.3.4. Consider the surface of revolution parametrized by

~X(u, v) = ((2 + sin v) cosu, (2 + sin v) sinu, v)

with (u, v) ∈ [0, 2π]× R.

(a) Show that the geodesic curvature is constant along the coordi-nate line v = v0.

(b) Determine the geodesic curvature along a coordinate line u =u0.

(c) Use the above results and the Gauss-Bonnet Theorem to de-termine

∫∫RK dS over a region R defined by 0 ≤ u ≤ 2π and

v1 ≤ v ≤ v2, for constants v1 and v2.

8.3.5. Let S be a regular, orientable, compact surface without boundarythat has positive Gaussian curvature. Prove that the surface area ofS is less than 4π/Kmin, where Kmin > 0 is the minimum Gaussiancurvature.

8.4. Geodesics 305

8.3.6. Jacobi’s Theorem. Let ~α : I → R3 be a closed, regular, parametrizedcurve. Suppose also that ~α(t) has a curvature function κ(t) that isnever 0. Suppose also that the principal normal indicatrix, i.e., thecurve ~P : I → S2, is simple, that is, that it cuts the sphere into onlytwo regions. By viewing ~P (I) as a curve on the sphere S2, use the

Gauss-Bonnet Theorem to prove that ~P (I) separates the sphere intotwo regions of equal area.

8.3.7. Consider the surface parametrized ~X(u, v) = (u, v, uv) with (u, v) ∈R2.

(a) Show that the Gaussian curvature is K = −1/(1 + u2 + v2)2.

(b) Find the formula for the geodesic curvature κg(t)s′(t)3 in terms

of the coordinate functions (u(t), v(t)) for any curve on thesurface.

(c) Show that κg(t) = 0 along the coordinate lines.

(d) Use the Gauss-Bonnet Theorem for this surface and a regionR defined as u1 ≤ u ≤ u2 and v1 ≤ v ≤ v2 to prove the doubleintegral formula

∫ u2

u1

∫ v2

v1

du dv

(1 + u2 + v2)3/2= cos−1

(−u1v1√

(1 + u21)(1 + v21)

)

+ cos−1

(u2v1√

(1 + u22)(1 + v21)

)+ cos−1

(−u2v2√

(1 + u22)(1 + v22)

)

+ cos−1

(u1v2√

(1 + u21)(1 + v22)

)− 2π.

8.4 Geodesics

Classical geometry in the plane studies in great detail relationshipsbetween points, straight lines, and circles. One could characterizeour theory of surfaces until now as a local theory in that we haveconcentrated our attention on the behavior of curves on surfaces ata point. Consequently, the notion of straightness (a global notion)and the concept of a circle (also a curve defined by a global property)do not yet make sense in our theory of curves on surfaces.

Euclid defines a line as a “breadthless width” and a straightline as “a line which lies evenly with the points on itself.” These


definitions do not particularly help us generalize the concept of astraight line to a general surface. However, it is commonly knownthat given any two points P and Q in Rn, a line segment connectingP and Q provides the path of shortest distance between these twopoints.

On a regular surface S ⊂ R3 that is not planar, even the notionof distance in S between two points P and Q poses some difficultysince we cannot assume a straight line in R3 connecting P and Qlies in S. However, since it is possible to talk about the arc lengthof curves, we can define the distance on S between P and Q as

infarc length of C |C is a curve on S connecting P and Q.

Therefore, one might wish to take as a first intuitive formulation ofstraightness on a regular surface S the following definition: A curveC on S is “straight” if for all pairs of points on the curve, the arclength between those two points P and Q is equal to the distancePQ between them. For general regular surfaces, this proposed def-inition turns out to be unsatisfactory, but it does lead to a moresophisticated way of generalizing straightness to surfaces.

Let S be a regular surface, and let P and Q be points of S.Suppose that P and Q are in a coordinate patch that is parametrizedby ~X : U → R3, where U ⊂ R2. Consider curves on S parametrizedby ~γ(t) = ~X(u(t), v(t)) such that ~γ(0) = P and ~γ(1) = Q. Accordingto Equation (6.2), the arc length of such a curve is

s =

∫ 1

0

√g11(u′(t))2 + 2g12u′(t)v′(t) + g22(v′(t))2 dt, (8.19)

where we understand that the gij coefficients are functions of u andv, which are in turn functions of t. To find a curve that connects Pand Q with the shortest arc length, one must find parametric equa-tions (u(t), v(t)) that minimize the integral in Equation (8.19). Suchproblems are studied in calculus of variations, a brief introductionto which is presented in Appendix B of [24].

According to the Euler-Lagrange Theorem in calculus of varia-tions, if we set

f =√g11(u′(t))2 + 2g12u′(t)v′(t) + g22(v′(t))2,

8.4. Geodesics 307

then the parametric equations (u(t), v(t)) that optimize the arc lengths in Equation (8.19) must satisfy

Lu(f)def=∂f

∂u− d

dt

(∂f

∂u′

)= 0 and

Lv(f)def=∂f

∂v− d

dt

(∂f

∂v′

)= 0.

We call Lu the Euler-Lagrange operator with respect to u and simi-larly for v. For a function with two intermediate variables u and v,as we have in this instance, let us also define the operator

L(f)def= (Lu(f),Lv(f)) .

Proposition 8.4.1. The Euler-Langrange operator of

f =√g11(u′(t))2 + 2g12u′(t)v′(t) + g22(v′(t))2

satisfiesL(f) =

√det(g)κg(t)

(v′(t),−u′(t)

).

Proof: (The proof is left as an exercise for the persistent reader andrelies on the careful application of Euler-Lagrange equations andEquation (8.7). )

Corollary 8.4.2. The parametric equations (u(t), v(t)) optimize the in-tegral in Equation (8.19) if and only if κg(t) = 0 or (v′(t), u′(t)) = ~0.

Obviously, (v′(t),−u′(t)) = ~0 integrates to (u, v) = (c1, c2), wherec1 and c2 are constants, so the curve degenerates to a point. Theother part of Corollary 8.4.2 motivates the following definition, whichgeneralizes to a regular surface the notion of a straight line.

Definition 8.4.3. A geodesic is a curve on a surface with geodesic cur-vature κg(t) identically 0.

As a first remark, one notices that on a geodesic C, the curvaturevector is ~T ′ = s′κ~P = s′κn ~N , so at each point on the surface, κ =±κn and the curve’s principal normal vector is equal to the surface


unit normal vector up to a possible change of sign. The transitionmatrix between the Frenet frame and the ~N, ~T , ~U frame becomes

(~T ~U ~N

)=(~T ~P ~B

)1 0 00 0 ε0 ε 0

,

where ε = ±1. Consequently, along a geodesic, the osculating planeof the curve and the tangent plane to the surface are normal to eachother.

As a second remark, the defining property that the geodesic cur-vature κg is identically 0 along a geodesic simplifies applicationsinvolving the Gauss-Bonnet Theorem. For example, suppose that Sis a surface as in the conditions of the Gauss-Bonnet Theorem (The-orem 8.3.4) and that R is a region on S whose boundary consists ofarcs of geodesics. Then the integration of κg along the regular arcsvanishes and the Gauss-Bonnet Theorem becomes∫∫

RK dS +

k∑i=1

θi = 2πχ(R),

where θi are the exterior angles where the geodesic arcs meet. Thefollowing proposition gives another example of how the Gauss-BonnetTheorem can provide profound geometric properties about geodesicsjust from this simplification.

Proposition 8.4.4. Let S be a compact, connected, orientable, regularsurface without boundary and of positive Gaussian curvature. If thereexist two simple, closed geodesics γ1 and γ2 on S then they intersect.

Proof: By Proposition 8.3.7, S is homeomorphic to a sphere. Sup-pose that γ1 and γ2 do not intersect. Then they form the boundary ofregionR that is homeomorphic to a cylinder with boundary. It is nothard to verify by supplying R with a triangulation that χ(R) = 0.However, applying the Gauss-Bonnet Theorem to this situation, weobtain ∫∫

RK dS = 0,

which is a contradiction since K > 0.

8.4. Geodesics 309

We now address the problem of finding geodesics. From Equation(8.7), a curve ~γ = ~X ~α, with ~α(t) = (u(t), v(t)), is a geodesic if andonly if

Γ211(u′)3 + (2Γ2

12 − Γ111)(u′)2v′ + (Γ2

22 − 2Γ112)u′(v′)2 − Γ1

22(v′)3 + u′v′′ − u′′v′ = 0. (8.20)

This formula holds for any parameter t and not just when ~γ is pa-rametrized by arc length.

We can approach the task of finding equations for geodesics in analternative way. Suppose now that we consider curves on the surfaceparametrized by arc length so that s′ = 1 and s′′ = 0. Since ~T ′ isparallel to ~N , we conclude that

~T ′ · ~Xu = 0 and ~T ′ · ~Xv = 0.

Then using Equation (8.5) we deduce that

~Xuu · ~Xu(u′)2 + 2 ~Xuv · ~Xuu′v′ + ~Xvv · ~Xu(v′)2 + ~Xu · ~Xuu

′′ + ~Xv · ~Xuv′′ = 0,

~Xuu · ~Xv(u′)2 + 2 ~Xuv · ~Xvu

′v′ + ~Xvv · ~Xv(v′)2 + ~Xu · ~Xvu

′′ + ~Xv · ~Xvv′′ = 0.

Solving algebraically for u′′ and v′′ in the above two equations, weobtain the following classical equations for a geodesic curve whichwe will express using tensor notation for simplicity with variablesx1 = u and x2 = v:

d2xi

ds2+

2∑j,k=1

Γijkdxj

ds

dxk

ds= 0 for i = 1, 2. (8.21)

Since i can be 1 or 2, this equation represents a system of two dif-ferential equations.

At first sight, one might see a discrepancy between Equation(8.21), which involves two equations and Equation (8.20), which in-volves only one. However, it is essential to point out that the systemin Equation (8.21) holds for geodesics parametrized by arc length.Furthermore, Equation (8.21) is equivalent to the system of equa-tions

κg(t) = 0 and (s′)2 = g11(u′)2 + 2g12u′v′ + g22(v′)2.


Finding explicit parametric equations of geodesic curves on a sur-face is often difficult since it involves solving a system of nonlinearsecond-order differential equations. Interestingly enough, there is acommon strategy in differential equations that allows us to trans-form this system of second-order equations in two functions x1(s)and x2(s) into a system of first-order differential equations in fourfunctions. (See the proof of Theorem 8.4.10.) There are commoncomputational techniques to solve systems of first-order differentialequations numerically, even if they are nonlinear. Consequently,there are algorithms to solve Equation (8.21) numerically.

In specific situations, there sometimes exist simplifications forEquation (8.21) that allow one to explicitly compute the geodesicson a particular surface.

Example 8.4.5 (The Plane: Cartesian Coordinates). Consider the xy-planeparametrized with the usual Cartesian coordinates. Then the Christ-offel symbols are all Γijk = 0. Using Equation (8.21), the equationsfor a geodesic in the plane are simply

d2u

ds2= 0 and

d2v

ds2= 0.

Integrating both equations twice, one obtains u(s) = as + c andv(s) = bs + d. Furthermore, since (u′(s))2 + (v′(s))2 = 1, theseconstants must satisfy a2+b2 = 1. Therefore, geodesics parametrizedby arc length in the plane are given as ~γ(s) = ~p + s~u, where ~p is apoint and ~u is a unit vector.

Example 8.4.6 (The Plane: Polar Coordinates). In contrast to the pre-vious example, consider the xy-plane parametrized with polar coor-dinates so that as a surface in R3, the xy-plane is given by

~X(r, θ) = (r cos θ, r sin θ, 0).

A short calculation gives

Γ212 = Γ2

21 =1

rand Γ1

22 = −r

8.4. Geodesics 311

and the remaining five other symbols are 0. Equations (8.21) become

d2r

ds2− r

(dθ

ds

)2

= 0, (8.22)

d2θ

ds2+

2

r

dr

ds

dθ

ds= 0. (8.23)

We transform this system to obtain a differential equation relatingr and θ as follows. Note that by repeatedly using the chain rule, weget

dr

ds=dr

dθ

dθ

dsand

d2r

ds2=d2r

dθ2

(dθ

ds

)2

+dr

dθ

d2θ

ds2.

Putting these two into Equations (8.22) and (8.23) leads to(dθ

ds

)2(d2r

dθ2− 2

r

(dr

dθ

)2

− r

)= 0,

which breaks into the pair of equations

dθ

ds= 0 or

d2r

dθ2− 2

r

(dr

dθ

)2

− r = 0. (8.24)

One notices that the first equation in Equation (8.24) is satisfiedwhen θ is a constant, which corresponds to a line through the origin.The other equation in Equation (8.24) does not appear particularlytractable but a substitution simplifies it greatly. Using the new vari-able u = 1

r , one can check that

d2r

dθ2=

2

u3

(du

dθ

)2

− 1

u2

d2u

dθ2,

and hence the second equation in Equation (8.24) reduces to

d2u

dθ2+ u = 0. (8.25)

Using standard techniques in ordinary differential equations, the gen-eral solution to Equation (8.25) can be written as u(θ) = C cos(θ−θ0)where C and θ0 are constants. Therefore, in polar coordinates, equa-tions for geodesics in the plane (i.e., lines) are given by

θ = C or r(θ) = C sec(θ − θ0).


p

q

Figure 8.9. Two geodesics on a cylinder.

Example 8.4.7 (Cylinder). Consider a right circular cylinder. A pa-rametrization for this cylinder is ~X(u, v) = (cos(u), sin(u), v), with(u, v) in [0, 2π] × R. An easy calculation shows that Γijk = 0 for alli, j, k, and hence, that geodesics on a cylinder are curves of the formγ = ~X ~α where

~α(t) = (at+ b, ct+ d)

for a, b, c, d that are constant. As a result, the geodesics on a cylinderare either straight lines parallel to the axis of the cylinder, circles inplanes perpendicular to the axis of the cylinder, or helices aroundthat axis.

Figure 8.9 illustrates how on a surface, two different geodesicsmay connect two distinct points, a situation that does not occurin the plane. In fact, on a cylinder, there is an infinite number ofgeodesics that connect any two points. The difference between eachgeodesic connecting p and q is how many times the geodesic wrapsaround the cylinder and in which direction it wraps.

Example 8.4.8 (Sphere). Consider the parametrization of the spheregiven by

~X(x1, x2) = (R cosx1 sinx2, R sinx1 sinx2, R cosx2),

8.4. Geodesics 313

where x1 = u is the longitude θ in spherical coordinates, and x2 = vis the colatitude angle ϕ down from the positive z-axis. In Example7.1.6, we determined the Christoffel symbols for this parametriza-tion. Equations (8.21) for geodesics on the sphere become

d2x1

ds2+ 2 cot(x2)

dx1

ds

dx2

ds= 0,

d2x2

ds2− sin(x2) cos(x2)

(dx1

ds

)2

= 0.

(8.26)

A geodesic on the sphere is now just a curve of the form ~γ(s) =~X(x1(s), x2(s)), where x1(s) and x2(s) satisfy the system of differ-ential equations in Equation (8.26). Taking a first derivative of ~γ(s)gives

~γ′(s) = R

(− sinx1 sinx2dx

1

ds+ cosx1 cosx2dx

2

ds,

cosx1 sinx2dx1

ds+ sinx1 cosx2dx

2

ds,− sinx2dx

2

ds

),

and the second derivative, after simplification using Equation (8.26),is

d2~γ

ds2= −

[sin2(x2)

(dx1

ds

)2

+

(dx2

ds

)2 ]~γ(s).

However, the term R2[

sin2(x2)(dx1

ds

)2+(dx2

ds

)2 ]is the first funda-

mental form applied to

((x1)′(s), (x2)′(s)),

which is precisely the square of the speed of ~γ(s). However, sincethe geodesic is parametrized by arc length, its speed is identically 1.Thus, Equations (8.26) lead to the differential equation

~γ′′(s) +1

R2~γ(s) = 0.

Standard techniques with differential equations allow one to showthat all solutions to this differential equation are of the form

~γ(s) = ~a cos( sR

)+~b sin

( sR

),


where ~a and ~b are constant vectors. Note that ~γ(0) = ~a and that~γ′(0) = 1

R~b. Furthermore, to satisfy the conditions that ~γ(s) lie on

the sphere of radius R and be parametrized by arc length, we deducethat ~a and ~b satisfy

‖~a‖ = R, ‖~b‖ = R, and ~a ·~b = 0.

Therefore, we find that ~γ(s) traces out a great arc on the sphere thatis the intersection of the sphere and the plane through the center ofthe sphere spanned by ~γ(0) and ~γ′(0).

Example 8.4.9 (Surfaces of Revolution). Consider a surface of revolu-tion about the z-axis given by the parametric equations

~X(u, v) = (f(v) cosu, f(v) sinu, h(v)) ,

where f and h are functions defined over a common interval I suchthat over [0, 2π]×I. We assume that over the open interval (0, 2π)×I the function ~X is a regular parametrization, which implies thatf(v) > 0. A simple calculation reveals that the Christoffel symbolsof the second kind are

Γ111 = 0, Γ1

12 =f ′

f, Γ1

22 = 0,

Γ211 = − ff ′

(f ′)2 + (h′)2, Γ2

12 = 0, Γ222 =

f ′f ′′ + h′h′′

(f ′)2 + (h′)2.

Equations (8.21) for geodesics parametrized by arc length on a sur-face of revolution become

d2u

ds2+

2f ′

f

du

ds

dv

ds= 0

d2v

ds2− ff ′

(f ′)2 + (h′)2

(du

ds

)2

+f ′f ′′ + h′h′′

(f ′)2 + (h′)2

(dv

ds

)2

= 0.

(8.27)

As complicated as Equations (8.27) appear, it is possible to find a“solution” to this system of differential equations for u in terms of v.However, before establishing a general solution, we will determinewhich meridians (u =const.) and which parallels or latitude lines(v =const.) are geodesics.

8.4. Geodesics 315

Consider first the meridian lines that are defined by u = C,where C is a constant. Notice that the first equation in the systemin Equation (8.27) is trivially satisfied for meridians and that thesecond equation becomes

d2v

ds2+f ′f ′′ + h′h′′

(f ′)2 + (h′)2

(dv

ds

)2

= 0. (8.28)

It is easy to check that the first fundamental form on (u′(t), v′(t))on the surface of revolution is

f(v)2u′(t)2 + (f ′(v)2 + h′(v)2)v′(t)2. (8.29)

However, assuming that we have a meridian that is parametrized byarc length, then

(f ′(v)2 + h′(v)2)v′(s)2 = 1

since the speed function of a curve parametrized by arc length is 1.Consequently,

v′(s)2 =1

f ′(v)2 + h′(v)2,

and taking a derivative of this equation with respect to s, one obtains

2v′v′′ = −2f ′(v)f ′′(v)v′ + 2h′(v)h′′(v)v′

(f ′(v)2 + h′(v)2)2

= −2f ′(v)f ′′(v) + h′(v)h′′(v)

f ′(v)2 + h′(v)2(v′)3.

Since v′(s) 6= 0 on the meridian parametrized by arc length, then

d2v

ds2= −f

′(v)f ′′(v) + h′(v)h′′(v)

f ′(v)2 + h′(v)2

(dv

ds

)2

,

which shows that Equation (8.28) is identically satisfied on all merid-ians.

Now consider the parallel curves on a surface of revolution, whichare defined by v = v0, a real constant. (See Figure 8.10.) In Equation(8.27), the first equation leads to u′(s) = C, a constant, and thesecond equation becomes

ff ′

(f ′)2 + (h′)2

(du

ds

)2

= 0.


geodesic

geodesic

geodesic

not a geodesic

Figure 8.10. Geodesics on a surface of revolution.

Since the parallel curves, parametrized by arc length, are regularcurves, then C 6= 0, and the condition that the surface of revolutionbe a regular surface implies that f(v) 6= 0. Thus, the second equationis satisfied for parallels v = v0 if and only if f ′(v0) = 0.

Even if a geodesic is neither a meridian nor a parallel, the firstequation in Equation (8.27) is simple enough that one can nonethe-less deduce some interesting consequences. Note that by taking aderivative with respect to the arc length parameter s, we have

d

ds(f2u′) = 2ff ′v′u′ + f2u′′.

Multiplying the first equation of Equation (8.27) by f2, we see thatit can be written as

f(v)2u′(s) = C, (8.30)

where C is a constant. Note that when a curve on a surface isparametrized by arc length with coordinate functions (u′(s), v′(s)),the angle θ it makes with any given parallel curve satisfies

cos θ =Ip((u

′, v′), (1, 0))√Ip((u′, v′), (u′, v′))

√Ip((1, 0), (1, 0))

= u′f.

8.4. Geodesics 317

As a geometric interpretation, since f is the radius r of surface ofrevolution at a given point, Equation (8.30) leads to the relation

r cos θ = C (8.31)

for all nonparallel geodesics on a surface of revolution, where θ isthe angle between the geodesic and the parallels. Equation (8.31),which is equivalent to Equation (8.30) for nonparallel curves, is oftencalled the Clairaut relation. Note that a curve satisfying Clairaut’srelation is a meridian if and only if C = 0.

With the relation u′ = C/f2, since the speed is equal to 1, Equa-tion (8.29) leads to(

dv

ds

)2

((f ′)2 + (h′)2) = 1− C2

f2. (8.32)

Taking the derivative of this equation with respect to s, one obtains

2dv

ds

d2v

ds2((f ′)2 + (h′)2) +

(dv

ds

)3

2(f ′f ′′ + h′h′′) = 2C2f ′

f3

dv

ds

which is equivalent to

((f ′)2 + (h′)2)dv

ds

[d2v

ds2− ff ′

(f ′)2 + (h′)2

(du

ds

)2

+f ′f ′′ + h′h′′

(f ′)2 + (h′)2

(dv

ds

)2 ]= 0.

Therefore, if a geodesic is not a parallel, the first equation of (8.27),which is equivalent to Clairaut’s relation, implies the second equa-tion.

Assuming that a geodesic is not a meridian, then from Clairaut’srelation, we know that u′(s) is never 0 so one can define an inversefunction to u(s), namely s(u), and then v can be given as a functionof u by v = v(s(u)). Then in Equation (8.32), replacing dv

ds withdvdu

duds = dv

duCf2

, we obtain(dv

du

)2 C2

f4((f ′)2 + (h′)2) = 1− C2

f2,

and hence,

dv

du=f

C

√f2 − C2

(f ′)2 + (h′)2.


Over an interval where this derivative is not 0, it is possible to takean inverse function of u with respect to v, and, by integration, thisfunction satisfies

u = C

∫1

f(v)

√f ′(v)2 + h′(v)2

f(v)2 − C2dv +D

for some constant of integration D. The constants C and D leadto a two-parameter family of solutions that parametrize segments ofgeodesics.

Theorem 8.4.10. Let S be a regular surface of class C3. For everypoint p on S and every unit vector ~w ∈ TpS, there exists a uniquegeodesic on S through p in the direction of ~w.

Proof: Let ~X : U → R3 be a regular parametrization of a neighbor-hood of p on S, and let gij and Γijk be the components of the metrictensor and Christoffel symbols of the second kind, respectively, inrelation to ~X. Note that since S is of class C3, all of the functionsΓijk are of class C1 over their domain.

The proof of this proposition is an application of the existenceand uniqueness theorem for first-order systems of differential equa-tions (see [2, Section 31.8]), which states that if F : Rn → Rn is ofclass C1, then there exists a unique function ~x : I → Rn that satisfies

~x′ = F (~x) and ~x(t0) = ~C,

where I is an open interval containing t0, and ~C is a constant vector.

Consider now the system of differential equations given in Equa-tion (8.21), where we do not assume that a solution is parame-trized by arc length. Setting the dependent variables v1 = (x1)′

and v2 = (x2)′, then Equation (8.21) is equivalent to the system offirst-order differential equations

(x1)′ = v1,

(x2)′ = v2,

(v1)′ = −Γ111(v1)2 − 2Γ1

12v1v2 − Γ1

22(v2)2,

(v2)′ = −Γ211(v1)2 − 2Γ2

12v1v2 − Γ2

22(v2)2,

(8.33)

8.4. Geodesics 319

where the functions Γijk are functions of x1 and x2. Therefore,according to the existence and uniqueness theorem, there exists aunique solution to Equation (8.21) with specific values given forx1(s0), x2(s0), (x1)′(s0), and (x2)′(s0). Set ~α(s) = ~X(x1(s), x2(s)).Now Equation (8.21) is the formula for arc length parametrizationsof geodesics. However, it is not hard to prove that if

f(s) = g11(x1(s), x2(s))(x1)′(s)2 + 2g12(x1(s), x2(s))(x1)′(s)(x2)′(s)

+ g22(x1(s), x2(s))(x2)′(s)2,

where x1 and x2 satisfy Equation (8.21), then f ′(s) = 0, so f(s) isa constant function over its domain. Consequently, if (x1)′(s0) and(x2)′(s0) are given so that ~α′(s0) = ~w, then ‖~α(s)‖ = 1 for all s. If,in addition, x1(s0) and x2(s0) are chosen so that ~α(s0) = p, then~α(s) is an arc length parametrization of a geodesic passing throughp with direction ~w and is unique.

The proof of the above theorem establishes an additional factconcerning the equation for a geodesic curve.

Proposition 8.4.11. Let ~X be the parametrization of a neighborhood ofa regular surface of class C3. Any solution (x1(s), x2(s)) to Equation(8.21), where one makes no prior assumptions on the parameter s, issuch that the parametric curve ~γ(s) = ~X(x1(s), x2(s)), whose locusis a geodesic, has constant speed.

Example 8.4.12. Since the geodesic curvature is intrinsic, only de-pending on the metric coefficients gij(u, v), we can find the expressionfor the geodesic curvature of a curve in the Poincare upper half-planewith the metric g11(u, v) = g22(u, v) = 1/v2 and g12(u, v) = 0, usingthe computations from Example 7.1.7. Using Formula (8.7), we findthat

s′(t)3κg(t) =1

v(t)2

(1

v(t)u′(t)3 +

1

v(t)2u′(t)v′(t)2 + u′(t)v′′(t)− u′′(t)v′(t)

). (8.34)

Problems

8.4.1. Let S be a regular surface, and let ~γ be a geodesic on S. Provethat the geodesic torsion of γ is equal to ±τ , where τ is the torsion


function of ~γ(t) as a space curve. [Hint: The possible difference insign stems from the possible change in sign of κn which may comefrom a reparametrization of the surface. This property of τg justifiesits name as geodesic torsion.]

8.4.2. (ODE) Consider a right circular cone with opening angle α, where0 < α < π/2. Consider the coordinate patch parametrized by~X(u, v) = (v sinα cosu, v sinα sinu, v cosα), where we assume v >0. Determine equations for the geodesics on this cone. [Hint: Find adifferential equation that expresses dv

du in terms of u and v and thensolve this to find an equation for u in terms of v.]

8.4.3. Consider the torus parametrized by ~X(u, v) = ((a+b cos v) cosu, (a+b cos v) sinu, b sin v) where a > b. Show that the geodesics on a torussatisfy the differential equation

dr

du=

1

Cbr√r2 − C2

√b2 − (r − a)2,

where C is a constant and r = a+ b cos v.

8.4.4. Find the differential equations that determine geodesics on a func-tion graph z = f(x, y).

8.4.5. If ~X : U → R3 is a parametrization of a coordinate patch on aregular surface S such that g11 = E(u), g12 = 0 and g22 = G(u)show that

(a) the u-parameter curves (i.e., over which v is a constant) aregeodesics;

(b) the v-parameter curve u = u0 is a geodesic if and only ifGu(u0) = 0;

(c) the curve ~x(u, v(u)) is a geodesic if and only if

v = ±∫

C√E(u)√

G(u)√G(u)− C2

du,

where C is a constant.

8.4.6. Fill in the details in the proof of Proposition 8.4.10, namely, provethat if x1(s) and x2(s) satisfy Equation (8.21), then the function

f(s), which is the square of the speed of ~X(x1(s), x2(s)), is constant.

8.4.7. Liouville Surface. A regular surface is called a Liouville surface if itcan be covered by coordinate patches in such a way that each patchcan be parametrized by ~X(u, v) such that

g11 = g22 = U(u) + V (v) and g12 = 0,

8.4. Geodesics 321

where U is a function of u alone and V is a function of v alone. (Notethat Liouville surfaces generalize surfaces of revolution.) Prove thefollowing facts:

(a) Show that the geodesics on a Liouville surface can be given assolutions to an equation of the form∫

du√U(u)− c1

= ±∫

dv√V (v) + c1

+ c2,

where c1 and c2 are constants.

(b) Show that if ω is the angle a geodesic makes with the curvev =const., then

U sin2 ω − V cos2 ω = C

for some constant C.

8.4.8. Let ~α : I → R3 be a regular space curve parametrized by arc lengthwith nowhere 0 curvature. Consider the ruled surface parametrizedby

~X(s, t) = ~α(s) + t ~B(s),

where ~B is the binormal vector of ~α. Suppose that ~X is defined overI × (−ε, ε), with ε > 0.

(a) Prove that if ε is small enough, the image S of ~X is a regularsurface.

(b) Prove that if S is a regular surface, ~α(I) is a geodesic on S.

(This shows that every regular space curve that has nonzero curva-ture is the geodesic of some surface.)

8.4.9. Consider the elliptic paraboloid given by z = x2 + y2. Note thatthis is a surface of revolution. Consider the geodesics on this surfacethat are not meridians.

(a) Suppose that the geodesic intersects the parallel at z = z0with an angle of θ0. Find the lowest parallel that the geodesicreaches.

(b) Prove that any geodesic that is not a meridian intersects itselfan infinite number of times.

8.4.10. Consider the hyperboloid of one sheet given by the equation x2 +y2 − z2 = 1 and let p be a point in the upper half-space defined byz > 0. Consider now geodesic curves that go through p and makean acute angle of θ0 with the parallel of the hyperboloid passingthrough p. Call r0 the distance from p to the z-axis.


p

r0

θ0

p

r0

θ0

Figure 8.11. Geodesics on the hyperboloid of one sheet.

(a) Prove that if cos θ0 > 1/r0, then the geodesic remains in theupper half-space z > 0.

(b) Prove that if cos θ0 < 1/r0, then the geodesic crosses the z = 0plane, and descends indefinitely in the negative z-direction.

(c) Prove that if cos θ0 = 1/r0, then the geodesic, as it descendsfrom p, asymptotically approaches the parallel given by x2 +y2 = 1 at z = 0.

(d) Using the parametrization

~X(u, v) = (cosh v cosu, cosh v sinu, sinh v) ,

suppose that the initial conditions for the geodesic are u(0) = 0

and v(0) = v0, so that p = ~X(0, v0). Find the initial conditionsu′(0) and v′(0) so that the geodesic is parametrized by arclength and satisfies the condition in (c).

(See Figure 8.11 for examples of nonasymptotic behavior.)

8.4.11. Using the calculations of the formula for geodesic curvature fromExample 8.4.12, show that the curve u(t) = (R cos t + m,R sin t),where R > 0 and m are constants, is a geodesic in the Poincareupper half-plane. Also show that any curve (a, t), where a is aconstant, is a geodesic for this metric.

8.5. Geodesic Coordinates 323

8.4.12. Let S be a regular, orientable surface of class C3 in R3 that is home-omorphic to the sphere. Let γ be a simple closed geodesic in S. Thecurve γ separates S into two regions A and B that share γ as theirboundary. Let n : S → S2 be the Gauss map induced from a givenorientation of S. Prove that n(A) and n(B) have the same area.[Hint: Use the Gauss-Bonnet Theorem.]

8.4.13. Let S be an orientable surface with Gaussian curvature K ≤ 0, andlet p ∈ S.

(a) Let γ1 and γ2 be two geodesics that intersect at p. Prove thatγ1 and γ2 do not intersect at another point q in such a waythat γ1 and γ2 form the boundary of a simple region R.

(b) Prove also that a geodesic on S cannot intersect itself in sucha way as to enclose a simple region.

8.5 Geodesic Coordinates

8.5.1 General Geodesic Coordinates

Definition 8.5.1. Let S be a regular surface of class C3. A system ofgeodesic coordinates is an orthogonal regular parametrization ~X ofS such that, for one of the coordinates, all the coordinate lines aregeodesics.

There are many ways to define geodesic coordinates on a regularsurface of class C3. We introduce one general way first.

Suppose that ~α(t) for t ∈ [a, b] is a regular curve with image C onS of class C2. According to Theorem 8.4.10, for each t0 ∈ [a, b], thereexists a unique geodesic on S through ~α(t0) perpendicular to C. (SeeFigure 8.12.) Furthermore, one can parametrize the geodesics ~γt0(s)by arc length such that ~γt0(0) = ~α(t0) and t 7→ ~γ′t(0) is continuousin t. Define the function

~X(s, t) = ~γt(s). (8.35)

We wish to show that over an open set containing C, the function~X(s, t) is a regular parametrization of class C2. (One desires classC2 since this is required for the first and second fundamental formsto exist.) However, we must assume that S is of class C5.


~α(v)

u-c

oor

din

ate

lines

Figure 8.12. Geodesic coordinate system generated by ~α(v).

Proposition 8.5.2. If S is of class C5, there exists an ε > 0 such that~X(s, t), as defined in Equation (8.35), is a regular parametrizationof class C2 over (−ε, ε)× (a, b).

Proof: The proof relies on standard theorems of existence and unique-ness for differential equations as well as on some basic topology.

Let p be a point on C, and let ~X(x1, x2) be a regular parametri-zation of class C5 of a neighborhood Vp = ~X(Up) of S containing p.

Let α1(t) and α2(t) be functions such that ~α(t) = ~X(α1(t), α2(t)),the given parametrization of C on S in the neighborhood Vp. Call Ipthe domain of ~α such that ~α(Ip) ⊂ Vp, and suppose that p = ~α(t0).

Let x1(s, t) and x2(s, t) be functions that map onto the geodesics~γt(s) via

~X(x1(s, t), x2(s, t)) = ~γt(s).

By Theorem 8.4.10, for all parameters t, over an interval s ∈ (−εt, εt),there exists a unique solution for x1(s, t) and for x2(s, t) to Equation(8.21) given the initial conditions

x1(0, t) = α1(t), x2(0, t) = α2(t),

∂x1

∂s(0, t) = u(t),

∂x2

∂s(0, t) = v(t),


where u and v are any functions of class C1. Furthermore, thesesolutions are of class C2 in the variable s. Now since for each twe want ~γt(s) to be an arc length parametrization of a geodesicperpendicular to C, we impose the following two conditions on uand v for all t:

(i) g11u2 + 2g12uv + g22v

2 = 1,

(ii) g11udα1

dt+ g12u

dα2

dt+ g12v

dα1

dt+ g22v

dα2

dt= 0,

where (i), for example, means that

g11(α1(t), α2(t))u(t)2 + 2g12(α1(t), α2(t))u(t)v(t) + g22(α1(t), α2(t))v(t)2 = 1

for all t. The requirement to parametrize each ~γt(s) so that ~γ′t(0)is continuous in t is equivalent to requiring that u and v be contin-uous. These conditions completely specify u(t) and v(t) for all t.Notice that for (i) and (ii) to both be satisfied, ((α1)′(t), (α2)′(t))and (u(t), v(t)) cannot be linear multiples of each other, and hence,∣∣∣∣(α1)′(t) u(t)

(α2)′(t) v(t)

∣∣∣∣ 6= 0 for all t.

Now write the equations for geodesics as a first-order system, asin Equation (8.33). Since ~X is of class C5, then all the functions Γijkare of class C3. Then according to theorems of dependency of solu-tions of differential equations on initial conditions (see [2, Theorem32.4]), solutions to the system from Equation (8.33) are of class C2

in terms of initial conditions, which implies that

∂2x1

∂s∂t,

∂2x2

∂s∂t,

∂2x1

∂t2, and

∂2x2

∂t2

are continuous over a possibly smaller open neighborhood U ′p of(0, t0). Thus, we conclude that x1(s, t) and x2(s, t) are of class C2

over U ′p.

To prove regularity of ~X(s, t), note that∣∣∣∣(α1)′(t0) u(t0)(α2)′(t0) v(t0)

∣∣∣∣ =∂(x1, x2)

∂(s, t)

∣∣∣∣(0,t0)

,


the Jacobian of the change of variables x1(s, t) and x2(s, t) at thepoint (0, t0). Since this Jacobian is a continuous function and isnonzero at (0, t0), there is an open neighborhood U ′′p of (0, t0) such

that ∂(x1,x2)∂(s,t) 6= 0. By Proposition 5.4.2,

∂ ~X

∂s× ∂ ~X

∂t=∂(x1, x2)

∂(s, t)

(∂ ~X

∂x1× ∂ ~X

∂x2

),

so over U ′′p the parametrization ~X(s, t) is regular. Consequently,there exists an εp > 0 such that

Updef= (−εp, εp)× (t0 − εp, t0 + εp) ⊂ U ′p ∩ U ′′p ,

and over Up, the parametrization ~X(s, t) is regular and of class C2.

Call Vp = ~X(Up).Finally, consider the whole curve C on S. The curve C can be

covered by open sets of the form Vp for various p ∈ C. Let p and q

be two points on C, and let us write ~Xp : Up → Vp and ~Xq : Uq → Vqfor associated parametrizations of Vp and Vq. If

Up = (−εp, εp)× Ip and Uq = (−εq, εq)× Iq

overlap, since for all t ∈ Ip ∩ Iq we have ~Xp(s, t) = ~Xq(s, t) by the

uniqueness of geodesics, then we can extend ~Xp and ~Xq to a function~X over

(−min(εp, εq),min(εp, εq))× (Ip ∪ Iq).

Since [a, b] is compact, [a, b] can be covered by a finite number of setsUp. Therefore, there exists ε > 0 and a single function ~X defined

over (−ε, ε) × (a, b) such that ~X(0, t) = ~α(t), and for t ∈ (a, b),~X(s, t) parametrizes a geodesic by arc length.

Proposition 8.5.2 leads to the following general theorem.

Theorem 8.5.3. Let S be a regular surface of class C5, and let ~α :[a, b]→ S be a regular parametrization of a simple curve of class C2.Then there exists a system of geodesic coordinates ~X(u, v) of classC2 defined over −ε < u < ε and a < v < b such that ~X(0, v) = ~α(v),and the u-coordinate lines parametrize geodesics by arc length.


Proof: Proposition 8.5.2 constructs the function ~X(u, v) as desired.However, it remains to be proven that ~X(u, v) is an orthogonal pa-rametrization.

Consider the derivative of g12 with respect to u

g12,1 =∂g12

∂u=

∂

∂u( ~X1 · ~X2) = ~X11 · ~X2 + ~X1 · ~X12.

Since each u-coordinate line is parametrized by arc length, ~X1 · ~X1 =1, so by differentiating with respect to v, we find that g11,2 = 2 ~X1 ·~X12 = 0. By the same reasoning, over the u-coordinate lines, in the ~N, ~T , ~U frame one has

∂ ~X

∂u= ~T and

∂2 ~X

∂u2= κ~P = κg ~U + κn ~N.

But since ~X(u, v) with v fixed is a geodesic, κg = 0. Thus, ~X11 =

κn ~N and, in particular, ~X11 · ~X2 = 0.

Consequently, g12,1 = 0, and therefore, g12 is a function of vonly. We can write g12(u, v) = g12(0, v). However, by constructionof ~X(u, v) in Proposition 8.5.2, g12(0, v) = 0 for all v. Thus, g12 isidentically 0, and hence, ~X(u, v) is an orthogonal parametrization.

The class of geodesic coordinate systems described in Theorem8.5.3 is of a particular type. Not every geodesic coordinate sys-tem needs to be defined in reference to a curve C on S as doneabove. Let ~X(u, v) be any system of geodesic coordinates where theu-coordinate lines are geodesics. A priori, we know only that g12 = 0.However, much more can be said.

Proposition 8.5.4. Let S be a surface of class C3, and let ~X : U → Vbe a regular parametrization of class C3 of a neighborhood V of S.The parametrization ~X(u, v) is a system of geodesic coordinates inwhich the u-coordinate lines are geodesics if and only if the metrictensor is of the form

g =

(E(u) 0

0 G(u, v)

).


Proof: First, the parametrization ~X is an orthogonal if and only ifg12 = 0.

By Equation (8.7), along the u-coordinate lines, the geodesiccurvature is

κg =

(du

ds

)3

Γ211

√g11g22.

Since the metric tensor must be positive definite everywhere, g11g22

is never 0. Since du/ds 6= 0, along a u-coordinate line, κg = 0 if and

only if Γ211 = −1

2g22∂g11

∂v= 0. The result follows.

Along u-coordinate lines, since v′ = 0, the speed functionds

du=√

E(u) is independent of v. Regardless of v, the arc length formulabetween u = u0 and u is

s(u) =

∫ u

u0

√E(u) du.

Therefore, it is possible to reparametrize u along the u-coordinatelines by arc length, with u(s) = s−1(u). This leads to the followingproposition.

Proposition 8.5.5. Let S be a regular surface of class C3, and let ~X(u, v)be a system of geodesic coordinates in a neighborhood V of S. Overthe same neighborhood V , there exists a system of geodesic coor-dinates ~X(u, v) such that the u-coordinate lines are geodesics pa-rametrized by arc length. Furthermore, if this is the case, then thecoefficients of the metric tensor are of the form

g11(u, v) = 1, g12(u, v) = 0, g22(u, v) = G(u, v),

and the Gaussian curvature of S is given by

K = − 1√g22

∂2√g22

∂u2. (8.36)

Proof: The fact that g11(u, v) = 1 follows from the u-coordinate linesbeing parametrized by arc length and Equation (8.36) is a directapplication of Equation (7.23).


8.5.2 Geodesic Polar Coordinates

Let p be a point on a regular surface of class C3. By Theorem 8.4.10,for every ~v ∈ TpS, there exists a unique geodesic ~γp,~v : (−ε, ε)→ S,with ~γp,~v(0) = p. Furthermore, by Proposition 8.4.11, we know that‖~γ ′p,~v(t)‖ = ‖~v‖. We remark then that for any constant scalar λ,

~γp,~v(λt) = ~γp,λ~v(t)

for all t ∈ (−ε/λ, ε/λ).

Definition 8.5.6. Let S be a regular surface of class C3, and let p ∈ Sbe a point. For any tangent ~v ∈ TpS, we define the exponential mapat p as

expp(~v) =

p, if ~v = ~0,

~γp,~v(1), if ~v 6= ~0,

whenever ~γp,~v(1) is well defined.

The map expp : U → S, where U is a neighborhood of ~0 in TpScorresponds to traveling along the geodesic through p with direction~v over the distance ‖~v‖. We will show that the exponential map canlead to some nice parametrizations of a neighborhood of p on S, butwe first need to prove the following two propositions.

Proposition 8.5.7. For all p ∈ S, the exponential map is defined overan open neighborhood U of ~0. Furthermore, if S is of class C4, thenexpp is differentiable over U as a function from TpS into R3.

Proof: We first show that expp is defined over some open disk cen-

tered at ~0.Let C1 ⊂ TpS be the set of all vectors of unit length. Set R to

be a large positive real number. For each ~v ∈ C1, let ε(~v) be thelargest positive real ε ≤ R such that ~γp,~v : (−ε, ε)→ S parametrizesa geodesic, or in other words, solves the system of equations in Equa-tion (8.21). The theorems of existence and uniqueness of solutionsto differential equations tell us that the solutions to Equation (8.21)depend continuously on the initial conditions, so ε(~v) is continuousover C1. (Setting R as an upper bound ensures that ε(~v) is definedfor all ~v ∈ C1.)


Since C1 is a compact set, as a function to R, ε(~v) attains aminimum ε0 on C1. However, since ε(~v) > 0 for all ~v ∈ C1, thenε0 > 0. Consequently, expp is defined over the open ball Bε0(~0) andis continuous.

Let ~X : U ′ → S be a regular parametrization of a neighborhoodof p on S. According to the theorem of dependence of solutions todifferential equations on initial conditions, the function

~γp(~v, t) : U × (−ε, ε) −→ R3

is of class C1 in the initial conditions ~v if the Christoffel symbolfunctions are of class C2, which means that ~X needs to be of classC4. The result follows.

Proposition 8.5.8. Let S be a regular surface of class C4. There is aneighborhood U of ~0 in TpS such that expp is a homeomorphism ontoV = expp(U), which is an open neighborhood of p on S.

Proof: The tangent space TpS is isomorphic as a vector space to R2,and one can therefore view expp as a function from U ′ ⊂ R2 into R3.Let ~v ∈ TpS be a nonzero vector, and let ~α(t) = t~v be defined fort ∈ (−ε, ε) for some ε > 0. Consider the curve on S defined by

expp(~α(t)) = expp(t~v) = ~γp,t~v(1) = ~γp,~v(t).

According to the chain rule,

d

dt

(expp(t~v)

) ∣∣∣0

= d(expp)0d~α

dt

∣∣∣0

= d(expp)0(~v),

where ~v in this expression is viewed as an element in TpS, and hence,by isomorphism, in R2. However, by construction of the geodesics,~γ′p,~v(0) = ~v, where we view ~v as an element of R3. This result proves,in particular, that d(expp)0 is nonsingular.

Using the Implicit Function Theorem from analysis (see Theorem8.27 in [7]), the fact that d(expp)0 is invertible implies that there

exists an open neighborhood U of ~0 such that expp : U → expp(U)is a bijection. Furthermore, setting V = expp(U), by the ImplicitFunction Theorem, the inverse function exp−1

p : V → U is at least ofclass C1, and hence, it is continuous.

Hence, expp is a homeomorphism between U and expp(U).


The identification of TpS with R2 allows one to define parametr-izations of neighborhoods of p with some nice properties.

Definition 8.5.9. Let S be a surface of class C4. A system of Riemannnormal coordinates of a neighborhood of p is a parametrization de-fined by

~X(u, v) = expp(u~w1 + v ~w2),

where ~w1, ~w2 is an orthonormal basis of TpS.

Propositions 8.5.7 and 8.5.8 imply that Riemann normal coor-dinates provide a regular parametrization of a neighborhood V ofp. Furthermore, the theorem of the dependence of solutions to dif-ferential equations on initial conditions shows that Riemann normalcoordinates form a parametrization of class Cr if S is a surface ofclass Cr+3.

Definition 8.5.10. Let S be a surface of class C4. The geodesic polarcoordinates of a neighborhood of p give a parametrization defined by

~X(r, θ) = expp ((r cos θ)~w1 + (r sin θ)~w2) ,

where ~w1, ~w2 is an orthonormal basis of TpS. A curve on S thatcan be parametrized by

~γ(t) = expp ((R cos t)~w1 + (R sin t)~w2) , for t ∈ [0, 2π]

is called a geodesic circle of center p and radius R.

Figure 8.13 gives an example of coordinate lines of a geodesicpolar coordinate system on a torus. It is important to note that ageodesic circle is neither a geodesic curve on the surface nor a circlein R3.

We present the following propositions about the above coordinatesystems but leave the proofs as exercises for the reader.

Theorem 8.5.11. Let S be a regular surface of class C5, and let p bea point on S. There exists an open neighborhood U of (0, 0) in R2

such that the Riemann normal coordinates defined in Definition 8.5.9form a system of geodesic coordinates.


Figure 8.13. Geodesic polar coordinate lines on a torus.

Proof: This theorem follows immediately from Theorem 8.5.3, whereone uses the curve

~α(t) = expp(t~w1),

where ~w1 is a unit vector in TpS.

Proposition 8.5.12. Let S be a regular surface of class C5, and let pbe a point on S. Let ~X be a parametrization of a Riemann normalcoordinate system in a neighborhood of p so that ~X(0, 0) = p, andlet gij be the coefficients of the metric tensor associated to ~X. The

coefficients satisfy gij(0, 0) = δji , and all the first partial derivativesof all the gij functions vanish at (0, 0).


The next theorem discusses the existence of geodesic polar coor-dinate systems in a neighborhood of a point p ∈ S.

Theorem 8.5.13. Let S be a regular surface of class C5, and let p bea point on S. There exists an ε > 0 such that the parametrization~X(r, θ) in Definition 8.5.10, with 0 < r < ε and 0 < θ < 2π, is aregular parametrization and defines a geodesic coordinate system ina neighborhood V whose closure contains p in its interior.

The parametrization ~X(r, θ) is defined as a function for −r0 <r < r0 and θ ∈ R for some r0 > 0. However, similar to usual polar


coordinates in R2, one must restrict one’s attention to r > 0 and0 < θ < 2π to ensure that ~X is a homeomorphism. The image~X ((0, ε)× (0, 2π)) is then a “geodesic disk” on S centered at p witha “geodesic radius” removed. By definition, the parametrization~X(r, θ) is such that all the coordinate lines for one of the variablesare geodesics, but Theorem 8.5.13 asserts that, within a small enoughradius, the parametrization ~X(r, θ) is regular and orthogonal.

The existence of geodesic polar coordinates at any point p on asurface S of high enough class leads to interesting characterizationsof the Gaussian curvature K of S at p. First, we remind the readerthat if ~X(r, θ) is a system of geodesic polar coordinates, then thecoefficients of the associated metric tensor are of the form

g11(r, θ) = 1, g12(r, θ) = 0, g22(r, θ) = G(r, θ) (8.37)

for some function G(r, θ) defined over the domain of ~X.

Proposition 8.5.14. Let S be a surface of class C5, and let ~X(r, θ) bea system of geodesic polar coordinates at p on S. Then the functionG(r, θ) in Equation (8.37) satisfies√

G(r, θ) = r − 1

6K(p)r3 +R(r, θ),

where K(p) is the Gaussian curvature of S at pm and R(r, θ) is afunction that satisfies

limr→0

R(r, θ)

r3= 0.


Proposition 8.5.14 shows that the perimeter of a geodesic circleat p of radius R is

C =

∫ 2π

0

√G(R, θ) dθ = 2πR− 1

3K(p)πR3 + F (R)

where F (r) is a function of the radial variable r and satisfieslimr→0 F (r)/r3 = 0. This, and a similar consideration of the area ofgeodesic disks around p, leads to the following geometric character-ization of the Gaussian curvature at p.


Theorem 8.5.15. Let S be a surface of class C5, and let p ∈ S be apoint. Define C(r) (resp. A(r)) as the perimeter (resp. the area) ofthe geodesic circle (resp. disk) centered at p and of radius r. TheGaussian curvature K(p) of S at p satisfies

K(p) = limr→0

3

π

(2πr − C(r)

r3

)and K(p) = lim

r→0

12

π

(πr2 −A(r)

r4

).


Theorem 8.5.15 is particularly interesting because, like the The-orema Egregium, it establishes the Gaussian curvature to a surfaceas a point as an intrinsic property of the surface, but in an originalway. Geodesics and geodesic coordinate systems are intrinsic prop-erties of the surface, and hence geodesic circles around a point areintrinsically defined. Some authors go so far as to use the secondformula in Theorem 8.5.15 as a definition of the Gaussian curvature.

Problems

8.5.1. Let S be the unit sphere and suppose that p is a point on S.

(a) Find a formula (using vectors) for the exponential map expp :TpS → S.

(b) Use this to give a formula for the geodesic polar coordinates(Definition 8.5.10) of a patch of S around p in terms of some~w1 and ~w2.

(c) Show that if p is the north pole, then with a proper choice of~w1 and ~w2 we recover our usual colatitude–longitude parame-trization of the sphere.

8.5.2. Prove that if ~X : U → V is a parametrization in which both fami-lies of coordinate lines are families of geodesics, then the Gaussiancurvature satisfies K(u, v) = 0.

8.5.3. Prove Proposition 8.5.12. [Hint: First use Equation (8.21) to provethat all Γijk vanish at (0, 0), i.e., at p.]

8.5.4. (*) Prove Theorem 8.5.13.

8.5.5. Prove Proposition 8.5.14. [Hint: Use a Taylor series expansion ofG(r, θ) and Equation (8.36).]

8.5.6. Prove Theorem 8.5.15.

8.6. Applications to Plane, Spherical, and Elliptic Geometry 335

8.5.7. Consider the usual parametrization of the sphere S of radius R

~X(u, v) = (R cosu sin v,R sinu sin v,R cos v).

Define the new parametrization Y (r, θ) = ~X(θ, r/R).

(a) Prove that Y (r, θ) is a geodesic polar coordinate system of Sat p = (R, 0, 0).

(b) Prove the result of Proposition 8.5.14 directly and determinethe corresponding remainder function R(r, θ).

8.6 Applications to Plane, Spherical, and EllipticGeometry

8.6.1 Plane Geometry

In plane geometry, we consider the plane to be a surface with Gaus-sian curvature identically 0 and curves in the plane we consider tobe curves on a surface. Consider now a region R in the plane suchthat the boundary ∂R is a piecewise regular, simple, closed curve.It is easy to calculate that the Euler characteristic of a region thatis homeomorphic to a disk is χ(R) = 1.

As a first case, let ∂R be a polygon. Since the regular arcs arestraight lines, the geodesic curvature of each regular arc of ∂R isidentically 0. The Gauss-Bonnet formula then reduces to the well-known fact that the sum of the exterior angles θ1, θ2, . . . , θk arounda polygon is 2π. Also, since the exterior angle θi at any corner isπ−αi, where αi is the interior angle, we deduce that the sum of theinterior angles of an n-sided polygon is

n∑i=1

αi = (n− 2)π.

This statement is in fact equivalent to a theorem that occurs inevery high school geometry curriculum, namely, that the sum of theinterior angles of a triangle is π radians.

As a second case of the Gauss-Bonnet Theorem in plane geome-try, suppose that ∂R is a simple, closed, regular curve of class C2. Inthis case, ∂R has no corners or cusps and hence no exterior angles.


Figure 8.14. Triangle on a sphere.

Assuming the boundary is oriented so that the interior R is in thepositive ~U direction, then the Gauss-Bonnet Theorem reduces to theformula ∮

∂Rκg ds = 2π.

Therefore, the Gauss-Bonnet Theorem subsumes the propositionthat the rotation index of a simple, closed, regular curve is 1 (seePropositions 2.2.1 and 2.2.11).

8.6.2 Spherical Geometry

Ever since navigators confirmed that the Earth is (approximately)spherical, geometry of the sphere became an important area of study.Improper calculations could send explorers and navigators in drasti-cally wrong directions. A key result particularly valuable to naviga-tors is that the sum of the interior angles of a triangle is not equalto two right angles.

Figure 8.14 gives an example of a triangle on a sphere in whicheach vertex has a right angle, and hence, the sum of the interiorangles is 3π/2.

Example 8.4.8 showed that a geodesic on a sphere is a great arcof the sphere, i.e., an arc on the circle of intersection between thesphere and a plane through the center of the sphere. Therefore, any


two geodesic lines intersect in at least two points: the common in-tersection of the sphere and the two planes supporting the geodesics.This remark, along with the fact that geometry on the sphere satis-fies the first four postulates of Euclid’s Elements, justifies the claimthat geometry of the sphere is an elliptic geometry.

Consider now a triangle R on a sphere of radius R. By definitionof a geometric triangle, ∂R has three vertices, and its regular arcsare geodesic curves. The Gaussian curvature of a sphere of radius Ris 1

R2 . In this situation, the Gauss-Bonnet Theorem gives

(θ1 + θ2 + θ3) +A

R2= 2π, (8.38)

where A is the surface area of R, and θi are the exterior angles ofthe vertices. With the interior angles αi = π − θi, then Equation(8.38) reduces to

α1 + α2 + α3 = π +A

R2= π + 4π

(A

4πR2

).

This shows that in a spherical triangle, the excess sum of the angles,i.e., α1 +α2 +α3− π, is 4π times the ratio of the surface area of thetriangle to the surface area of the whole sphere. As an example, thespherical triangle in Figure 8.14 has an excess of π/2 and covers 1/8of the sphere. The quantity α1 + α2 + α3 − π is often referred to asthe angle excess of a triangle on sphere.

For navigators, the Gauss-Bonnet Theorem matters significantlybecause if they traveled in a circuit, generally following straight linesand only making sharp turns at discrete specific locations, the sumof the exterior angles of the circuit does not add up to 2π but to aquantity that is strictly less than 2π.

We propose to study (geodesic) circles on a sphere of radius R.Recall that a circle on a sphere is the set of points that are at afixed distance r from a point, where distance means geodesic dis-tance. Call D the closed disk inside this circle. An example of ageodesic circle is any latitude line, since points on a latitude line areequidistant from the north pole. Using the parametric equations forthe sphere,



a circle of radius r around the north pole (0, 0, R) is given by ~γ(t) =~X(t, v0). Note that the geodesic radius is the length of an arc ofradius R and angle equal to the colatitude, so r = Rv0. It is nothard to tell by symmetry that the geodesic curvature κg of a circleon a sphere is a constant. The Gauss-Bonnet Theorem then gives

κg(perimeter of ∂D) +1

R2(surface area of D) = 2π.

Using the surface area formula for a surface of revolution from single-variable calculus, it is possible to prove (Problem 8.6.1) that the areaof D is

surface area of D = 2πR2(1− cos v0) = 2πR2(

1− cos( rR

)).

(8.39)As a circle in R3, the radius is R sin v0, so the perimeter of D is

perimeter of D = 2πR sin v0 = 2πR sin( rR

).

Consequently, the Gauss-Bonnet Theorem gives the geodesic curva-ture of a circle on a sphere as

κg =1

Rcot( rR

).

Using a Maclaurin series for the even function x cotx, we can showthat this geodesic curvature is approximately

κg ≈1

R

(( rR

)−1− 1

3

( rR

)− 1

45

( rR

)3− · · ·

)≈ 1

r− 1

3R2r − 1

45R4r3 − · · ·

In this expression, as r → 0, then κg behaves asymptotically like1r , where 1

r is the usual curvature of a circle of radius r. Also, asR →∞, which corresponds to a sphere with a radius that grows solarge that the sphere approaches planar, we have

limR→∞

κg = limR→∞

1

Rcot( rR

)=

1

r,

which again recovers the familiar curvature of a circle of radius r.


8.6.3 Elliptic Geometry

As mentioned in the introduction to this chapter, in his Elements,Euclid proves all his propositions from 23 definitions and five postu-lates. The fifth postulate reads as follows:

If a straight line crossing two straight lines makes the interiorangles on the same side less than two right angles, the twostraight lines, if extended indefinitely, meet on that side onwhich are the angles less than the two right angles.

This postulate is much wordier than the other four, and manymathematicians over the centuries attempted to prove the fifth pos-tulate from the others, but never succeeded. Studying Euclid’s Ele-ments, it appears that Euclid himself uses the fifth postulate spar-ingly. Proposition I.29 is the first proposition to cite Postulate 5.Furthermore, some proofs could be simplified if Euclid cited Pos-tulate 5. One could speculate that Euclid avoided a liberal use ofPostulate 5 so that, were someone to prove it as a consequence ofthe other four, then the proofs of theorems would still be given ina minimal form. Indeed, many commonly known properties aboutplane geometry hold without reference to the fifth postulate. How-ever, examples of a few commonly known theorems that rely on itare:

1. Playfair Axiom: Given a line L and a point P not on L, thereexists a unique line through P that does not intersect L.

2. The sum of the interior angles of a triangle is π.

3. Pythagorean Theorem: the squared length of the hypotenuseof a right triangle is the sum of the squares of the lengths ofthe sides adjacent to the right angle.

In fact, in the above list, (1) is equivalent to the fifth postulateunder the assumption that the first four postulates hold. That is whythis theorem became known as the Playfair Axiom. In the 19th cen-tury, mathematicians Lobachevsky and Bolyai independently con-sidered geometries that retained the first four postulates of Euclid’sElements but assuming alternatives to the Playfair Axiom. The twological alternatives to the Playfair Axiom are:


Elliptic Given a line L and a point P not on L, there does not exista line through P that does not intersect L.

Hyperbolic Given a line L and a point P not on L, there existsmore than one line through P that does not intersect L.

Depending on whether we use one or the other of the above two al-ternatives we obtain alternate geometries that are respectively calledelliptic geometry and hyperbolic geometry. Collectively, elliptic andhyperbolic geometries were first called non-Euclidean geometry butthen this label soon encompassed all types of geometry where themetric differs from the flat Euclidean metric.

It is possible to prove many theorems in these different geometriesusing synthetic techniques, that is involving proofs that avoid the useof coordinates. (See [10], [29], or [8] for a comprehensive synthetictreatment of elliptic or hyperbolic geometry.) One well-known resultin these geometries is that the sum of the interior angles of a triangleis greater than π in the case of elliptic geometry and less than π inthe case of hyperbolic geometry.

When Lobachevsky first investigated alternatives to Euclidean heconsidered hyperbolic geometry but rejected the elliptic hypothesis.Spherical geometry satisfies the elliptic hypothesis. However, in ge-ometry on the sphere, any two great circles intersect at two points,and this does not satisfy the first axiom of Euclid. Consequently,it is possible that he believed that since spherical geometry has theelliptic hypothesis but fails Euclid’s first axiom, the elliptic geom-etry was inconsistent. However, in the late nineteenth century, itwas realized that on a real projective plane obtained by identifyingopposite points on the sphere, all of the first four Euclidean axiomsremain valid and that the elliptic hypothesis holds as well. A numberof the theorems that were known for spheres were valid in this ellipticgeometry on a non-orientable surface. (Because a proper topologicalintroduction to the real projective plane would take us too far afield,we do not describe it explicitly in this textbook but encourage thereader to discover it in subsequent studies in geometry.)

We would like to now consider applications of the Gauss-BonnetTheorem to elliptic and hyperbolic geometry. We first need to clarifythe use of the terms “line,” “right angle,” and “parallel.” Let usassume that we have a metric. In the geometry on a surface of class


C3, the word “line” (or straight line) means a geodesic. Two linesmeet at a right angle at a point p if the direction vectors of the linesat p have a first fundamental form that vanishes. The word “parallel”is more problematic. In Problem 1.3.16, we defined a parallel curveto a given curve ~γ as a curve whose locus is always a fixed orthogonaldistance r from the locus of ~γ. This is true of parallel lines in theplane. However, in this sense of parallelism, a parallel curve to aline (geodesic) need not be another line. Consequently, we do notdirectly use the word parallel but prefer to discuss whether two linesintersect or not.

Elliptic and hyperbolic geometry involve the notion of congru-ence. Recall that in any geometry that has a notion of congruence,the concept of area must satisfy the following three axioms:

Axiom 1 The area of any set must be nonnegative.

Axiom 2 The area of congruent sets must be the same.

Axiom 3 The area of the union of disjoint sets must equal the sumof the areas of these sets.

From Axiom 2 and Theorem 8.5.15, we can deduce that the Gaussiancurvature is constant in elliptic and hyperbolic geometries.

Consider a triangle R in a space of constant Gaussian curvatureK. Because the edges of the triangle are geodesics, by the Gauss-Bonnet Theorem,∫∫

RK dS + θ1 + θ2 + θ2 = 2πχ(R),

where θi is the exterior angle at the ith vertex. Note that θi = π−αi,where αi are the interior angles of the triangle. Furthermore, theEuler characteristic of any triangle is χ(R) = 3− 3 + 1 = 1. Thus

K ·Area(R) + (π − α1) + (π − α2) + (π − α3) = 2π

=⇒ α1 + α2 + α3 = π +K ·Area(R).

We mentioned earlier that in elliptic (resp. hyperbolic) geometry,the interior sum of angles of a triangle is greater (resp. less) thanpi. We deduce that in elliptic geometry the Gaussian curvature is


positive and constant, whereas in hyperbolic geometry, the Gaussiancurvature is negative and constant. It is possible to prove fromsynthetic geometry in either elliptic or hyperbolic geometry) thatthe sum of the interior angles of a triangle satisfies

α1 + α2 + α3 = π + c ·Area(R),

where c is a positive constant for elliptic geometry and a negativeconstant for hyperbolic geometry. However, the Gauss-Bonnet The-orem shows that this constant c is precisely the Gaussian curvature.

Problems

8.6.1. Calculate the surface area of a disk (Equation (8.39)) on a sphereusing the surface area formula for a surface of revolution. Calculatedirectly the geodesic curvature function of a circle on a sphere.

8.6.2. Using Equation (8.39) for the surface area of a disk on a sphereand Theorem 8.5.15 recover the familiar result that the Gaussiancurvature of a sphere of radius R is 1/R2.

8.6.3. Suppose that a sphere of radius R has the longitude–colatitude (u, v)system of coordinates. Suppose that a simply connected region Rof this sphere does not include the north or south pole and supposethat the boundary ∂R is a single regular, closed curve of class C2

parametrized by (u(t), v(t)) for t ∈ [a, b] in its coordinate plane.Prove that the area of the region is

2πR2 +R2

∫ b

a

(cos v)u′ +(cos v)u′ + u′′v′ − u′v′′

(sin2 v)(u′)2 + (v′)2dt.

8.6.4. Using Euclidean style constructions, describe a procedure for findingthe midpoint of a given great circle arc on the unit sphere withendpoints A and B.

8.7 Hyperbolic Geometry

8.7.1 Synthetic Hyperbolic Geometry

Recall from the previous section that hyperbolic geometry is a ge-ometry that involves the undefined terms of points, lines, and circlesand that satisfies the first four postulates of Euclid’s Elements along

8.7. Hyperbolic Geometry 343

`

Q

P

`2 `1α

β

Figure 8.15. Left and right sensed-parallels.

with the fifth hyperbolic axiom: (H) for any line ` and any point Pnot on `, there pass more than one line that does intersect `.

Synthetic hyperbolic geometry is like Euclidean geometry in itsstyle of proof, in that it does not use coordinate systems. Varioustexts (see Chapter 2 in [8] for example) explore this geometry indetail. Some theorems are identical to Euclidean geometry if theyonly involve the first four postulates and some theorems differ con-siderably from Euclidean geometry. We mention a few results with-out proof in order to motivate the applications of the Gauss-BonnetTheorem to hyperbolic geometry.

Let ` be a line and let P be a point not on `. We can invokeProposition I.12 in the Elements because its proof does not rely onthe fifth postulate. Hence, we can construct a line through P that is

perpendicular to `. In Figure 8.15, this is the line←→PQ. (Figure 8.15

only shows the segment PQ.)

Obviously, the diagram in Figure 8.15 is sketched in the Eu-clidean plane as a “local approximation” and we must understandthat as `1 and `2 are extended indefinitely, they do not intersect `.

Consider the lines through P as they sweep out angles away from←→PQ. There must exist a first line on the right side of

←→PQ that does

not intersect ` and similarly on the left side. These lines are calledthe right (resp. left) sensed-parallel of ` through P ; they are indicatedin Figure 8.15 by `1 (resp. `2). The angle α (resp. β) is called theangle of right (resp. left) parallelism. It is not hard to prove that


α = β and that this common angle is acute. All the lines throughP that lie in the shaded region in Figure 8.15 are called ultraparallelto ` through P .

In addition to usual triangles, in hyperbolic geometry, one alsoconsiders asymptotic triangles. If `1 and `2 are right (resp. left)sensed-parallels, then we say that they meet at an ideal point Ω. If

A is on `1 and B is on `2, then the figure consisting of the ray−→AΩ,

the segment AB, and the ray−→BΩ is called an asymptotic triangle

and denoted 4ABΩ. It is also possible to have asymptotic triangleswith two or three ideal points as we shall see in models below.

It is natural to try to prove theorems in hyperbolic geometry thatmirror the development of Euclidean geometry as presented in TheElements. A number of surprising results arise. For example, in anytriangle, the sum of the interior angles is less than two right angles.For a triangle 4ABC, the defect of 4ABC is the difference betweenπ and the sum of the interior angles. Furthermore, with a few basicaxioms of how area works, it is possible to prove that there exists apositive constant c such that for all triangles 4ABC, the area-defectformula holds

Area(4ABC) = c2 defect(4ABC).

In the previous section, we offered a reason why the Gaussiancurvature in elliptic and in hyperbolic geometry should be constant.Under this assumption, we saw that because the sum of the interiorangles of the triangle is less than π, then the Gaussian curvature inhyperbolic geometry must be negative. Let us suppose that K = − 1

c2

in imitation of what happens for a sphere. Then using the Gauss-Bonnet Theorem applied to triangle R in hyperbolic geometry, weget

− 1

c2Area(R) + (π − α1) + (π − α2) + (π − α3) = 2π,

where αi are the interior angles of the triangle. Thus

Area(R) = c2(π − (α1 + α2 + α3)),

which recovers the area-defect formula. The Gauss-Bonnet approachis not a synthetic approach but it leads to the same relationship.


As mentioned earlier, the diagram offered in Figure 8.15 is onlyof limited value because if we extend `1 and ` for example, they inter-sect. The integrity of a synthetic proof cannot rely on a diagram butit is unfortunate that such a diagram fails to capture key properties,such as the angle defect of a triangle.

This made it difficult for some mathematicians to initially ac-cept hyperbolic geometry since it deviated from sense perception.However, it was soon discovered that by using different metrics, it ispossible to depict the hyperbolic plane more effectively.

8.7.2 The Poincare Upper Half-Plane

As early as Section 6.1, we introduced the alternate metric

gij =

1

v20

01

v2

on the upper half-plane H = (u, v) ∈ R2 | v > 0. We called this thePoincare upper half-plane. In Example 6.1.9, we calculated certainlengths of curves and also showed the area of the region a ≤ u ≤ band c ≤ v ≤ d is

(b− a)

(1

c− 1

d

).

The area of any region R in the plane is

Area(R) =

∫∫R

1

v2du dv.

It is possible to prove that, given this metric, the Gaussian curvatureis constant with K = −1. Building on other examples in the text,in Exercise 8.4.11 we proved that geodesics in this space are eithervertical lines or semicircles of radius R and with origin on the u-axis. Consequently, the Poincare upper half-plane offers a model ofhyperbolic geometry in which “lines” consist of these geodesics.

Interestingly enough, unlike for the cylinder in which there existpoints such that there are many geodesics connecting them, given


A2

P2

Q2

A B

P

Q

Figure 8.16. Geodesics in the Poincare half-plane.

any two points in the Poincare, there exists only one geodesic con-necting them. In Problem 8.7.3, the reader is asked to prove a dis-tance formula between points by calculating the length between thepoints along the unique geodesic that connects them.

Finally, consider two vectors ~a and ~b based at a point p withcoordinates (u0, v0). We can understand ~a and ~b as vectors in thetangent plane to p. The angle θ between ~a and ~b satisfies

cos θ =Ip(~a,~b)√


=

1v20a1b1 + 1

v20a2b2√

1v20a2

1 + 1v20a2

2

√1v20a2

1 + 1v20a2

2

=a1b1 + a2b2√a2

1 + a22

√b21 + b22

.

This is the usual formula for the cosine of an angle between twovectors in the plane. Consequently, the Poincare upper half-planemodel of hyperbolic geometry accurately reflects angles as the anglesmeasured in the usual sense.

The following diagram depicts three lines (geodesics) in the Poin-care upper half-plane model of hyperbolic geometry that meet toform a triangle. This particular triangle is a usual and not an asymp-totic one.


v = 0

AB

C

The reader might have guessed that there is a right angle at A andthat would be correct. The circles in the above diagram were chosenspecifically so that the two smaller circles meet at a right angle.Because of the above comment, since they meet at right angles ascircles in the model shown in the Euclidean plane, the correspondinglines in the hyperbolic geometry meet at right angles as well. (It iseasy to prove in Euclidean geometry that two circles of radius r andR respectively intersect orthogonally if the distance between them is√r2 +R2.)

On the other hand, in the diagrams below, the first figure depictsan asymptotic triangle with one ideal point, the second figure depictsan asymptotic triangle with two ideal points for vertices, and thethird figure depicts an asymptotic triangle with three ideal points.

A

Ω

B

A

Ω1 Ω2

Ω1 Ω2 Ω3


8.7.3 The Poincare Disk

Another model for hyperbolic geometry is to model the points in thehyperbolic plane as the points in the open disk of radius R, namelyDR = (u, v) ∈ R2 |u2 + v2 < R2 but to use the metric tensor

gij =

4R4

(R2 − x2 − y2)20

0 4R4

(R2−x2−y2)2

.

The disk DR equipped with this metric is called the Poincare disk. If(u(t), v(t)) parametrizes a curve in DR, then the length of the curvewith t1 ≤ t ≤ t2 is∫ t2

t1

2R2

(R2 − u(t)2 − v(t)2)

√(u′(t))2 + (v′(t))2 dt.

Consequently, as the curve approaches the boundary of the disk DR,the length becomes arbitrarily large. Hence, we can intuitively thinkof the boundary ∂DR as heading far away, becoming unbounded. Weleave it as an exercise for the reader to compute the Christoffel sym-bols and to prove that the Gaussian curvature of the Poincare diskis −1/R2. Consequently, the Poincare disk has constant negativeGaussian curvature. (If R is unspecified, we assume that R = 1 andthat D is the unit disk.)

Using the so called Mobius transformations from complex anal-ysis (a technique that is just beyond the scope of this book) fromthe Poincare upper half-plane to the Poincare disk, it is possible toprove that the geodesics on the Poincare are either diameters of thedisk or arcs of circles that meet the boundary ∂D at right angles.


Figure 8.17. Tiling of the Poincare disk with regular pentagons.

The diagram above depicts a (non-asymptotic) triangle in thePoincare disk. An asymptotic triangle in the Poincare disk involvesgeodesic edges that meet at the boundary of D.

Because of the relationship between the defect of triangles andby extension the defect of other polygons, it is possible to tessellate(tile) the hyperbolic plane with regular polygons in ways that areimpossible in the Euclidean plane. For example, it is possible totessellate the hyperbolic plane with pentagons in such a way thatat each vertex of the tessellation, four regular pentagons meet. (SeeFigure 8.17.) Since four pentagons meet at a vertex, each interiorangle of the pentagon is π

2 . Each pentagon is the union of five isosce-les triangles, each congruent to each other, meeting at the center ofthe pentagon. Each of these triangles has for interior angles π

4 , π4 ,

and 2π5 . Hence the defect of each triangle is π

10 and its area is c2 π10 ,

when the Gaussian curvature is a constant K = − 1c2

. Thus, in or-der to tessellate the hyperbolic plane when K = −1 with pentagonssuch that four meet at each vertex, we simply must choose regularpentagons of area π

2 .

Artist M. C. Escher (1891–1972) was known for artwork thatexplores mathematical symmetries in creative ways. Among a few ofhis more esoteric works are symmetry patterns based on tessellationsof the hyperbolic plane.


Figure 8.18. A geodesic triangle on a pseudosphere.

8.7.4 The Pseudosphere Revisited

The Poincare upper half-plane and Poincare disk offer effective mod-els of the hyperbolic plane. One may feel still somewhat dissatisfiedbecause these models do not correspond to regular surfaces in R3.The pseudosphere introduced in Example 6.6.6 is a regular surface inR3 that has constant Gaussian curvature equal to K = −1. Conse-quently, the pseudosphere offers another model of hyperbolic space.

Figure 8.18 depicts a triangle (the edges are geodesics) on thepseudosphere. Problem 8.7.8 guides the reader to explore some ofthe properties of the pseudosphere starting from its metric.

Problems

8.7.1. In the Poincare upper half-plane, consider the isosceles triangle whosevertices are where the following three lines (geodesics) meet: x2 +y2 = 36, (x − 3)2 + y2 = 16, and (x + 3)2 + y2 = 16. Calculate thearea of this triangle.

8.7.2. Consider the Poincare upper half-plane.

(a) Find the geodesic curvature of the line (u(t), v(t)) = (t,mt),with t > 0, where m is some constant.

(b) Explicitly evaluate the line integral∫∂R κg ds, where R is the

region bounded by three Euclidean straight lines from (1, 1) to(2, 2) to (1, 2) and back to (1, 1).


(c) Without additional calculations, deduce∫∫RK dS.

(d) Directly evaluate the double integral in the previous part toconfirm your result. [Recall that K = −1.]

8.7.3. Consider the Poincare half-plane. Let P and Q be two points on H.We determine the distance d(P,Q) between these two points in thePoincare metric on H (see Figure 8.16).

(a) If P and Q lie on a geodesic that is a half-circle, prove that thedistance between them is

d(P,Q) =∣∣∣ ln |PA|/|PB||QA|/|QB|

∣∣∣,where A and B are points as shown in Figure 8.16 and |PA| isthe usual Euclidean distance between P and A and similarlyfor all the others.

(b) If P and Q lie on a geodesic that is a vertical line, prove thatthe distance between them is

d(P,Q) =∣∣∣ ln |P2A2||Q2A2|

∣∣∣,where A2 is again as shown in Figure 8.16.

8.7.4. Describe a procedure for finding the midpoint of a given hyperbolicsegment on the Poincare upper half-plane. In particular, find the

midpoint of the hyperbolic segment from (0, 1) to(

1√2, 1√

2

). [Hint:

See Problem 8.7.3.]

8.7.5. Show that the set of points at a fixed distance d from the vertical lineu = 0 in the hyperbolic plane consists of the points on two Euclideanlines of the form y = mx.

8.7.6. Using the Poincare upper half-plane, Figure 8.19 depicts two lines `1and `2 that are parallel (right-sensed). The line

←→AB is perpendicular

to `1 so the acute angle between←→AB and `2 is the angle of parallelism.

Suppose that in this diagram `1 is a circle of radius 1 and let h bethe distance between A and B. Calculate the angle of parallelism interms of the distance h. Show that as h→ 0, the angle of parallelismgoes to π/2, and as h→∞, the angle of parallelism goes to 0.

8.7.7. Prove that in the Poincare disk, the angle at which curves meet asobserved in the model in R2 is precisely the angle at which the curvesmeet in the Poincare disk metric.


A

B

Ω

`1

`2h

Figure 8.19. Angle of parallelism.

8.7.8. Consider the upper half-plane H = (x, y) ∈ R2 | y > 0 with themetric tensor

(gij) =

(e−2y 0

0 1

).

(a) Prove that the Gaussian curvature of such a surface has K =−1 everywhere.

(b) Show that the parametrization ~X : H → R3 defined by

~X(x, y) =(e−y cosx, e−y sinx, ln(1 +

√1− e−2y) + y −

√1− e−2y

)has the above metric coefficients.

(c) Prove that ~X(x, y) is a regular reparametrization of the pseu-dosphere as described in Example 6.6.6.

(d) (ODE,*) Find parametric equations (u(t), v(t)) for the geo-desic curves on H with this given metric.

8.7.9. Consider the Poincare half-plane. We can consider R2 as the setof complex numbers C. In this context, H = z ∈ C | Im(z) > 0.Consider a fractional linear transformation of C of the form

w =az + b

cz + d,

where a, b, c, d ∈ R and ad−bc = 1. Write z = u+iv and w = x+iy.

(a) Prove that the function w = f(z) sends H into H bijectively.


(b) Prove that the metric tensor for H is unchanged under thistransformation, more precisely, that the metric coefficients ofH in the w-coordinate are

(gkl) =

(y−2 00 y−2

).

[We say that the Poincare metric on the upper half-plane isinvariant under the action of SL(2,R).]

(c) Show explicitly that this fractional linear transformation sendsgeodesics to geodesics.

CHAPTER 9

Curves and Surfaces in n-dimensional

Euclidean Space

Up to this point, this text emphasized curves and surfaces in R3.However, our only reason to do so was based on a discriminationby dimensionality: As beings living in three-dimensional space, weare accustomed to visualizing in R3. In this chapter, we generalizea number of definitions and theorems about curves and surfaces toEuclidean n-dimensional space.

9.1 Curves in n-dimensional Euclidean Space

In Chapters 1 and 3, we studied local properties of curves with an em-phasis on notions of velocity, acceleration, speed, curvature, and tor-sion. Along the way, we introduced the natural orthonormal framesattached to curves at a point, namely the (~T , ~U) for a regular planecurve and the Frenet frame (~T , ~P , ~B) for a regular space curve ofclass C2 in R3.

Many local properties and definitions have immediate generaliza-tions to curves in n-dimensional Euclidean space, Rn. Let I be aninterval of R and let ~X : I → Rn be a parametrized curve. Then

• the velocity is the vector function ~X ′(t);

• the speed is real-valued function s′(t) = ‖ ~X ′(t)‖;

• the arc length function with origin t = t0, is the antiderivativeof s′(t) with s(t0) = 0;

• the acceleration is the vector function ~X ′′(t);

355

356 9. Curves and Surfaces in n-dimensional Euclidean Space

• the parametrization ~X is regular at t = t0 if ~X ′(t0) exists andis nonzero;

• the parametrization ~X is regular if it is regular at all t0 in itsdomain;

• a curve C in Rn is regular if it is the locus (image) of a regularparametrized curve;

• at any regular point t = t0, the unit tangent vector is

~T (t0) =~X ′(t0)

‖ ~X ′(t0)‖.

One hitch that occurs in generalizing our theory from curves inspace to curves in Rn comes from the fact that for n 6= 3 there doesnot exist a cross product in Rn. So any formula or reasoning mustavoid that nice property about R3.

9.1.1 Curvatures and the Frenet Frame

Recall that any unit vector function ~Y : I → Rn satisfies ~Y (t)·~Y (t) =1 so after differentiating this expression, for all t ∈ I,

2~Y ′(t) · ~Y (t) = 0 =⇒ ~Y ′(t) ⊥ ~Y (t). (9.1)

Thus, if the derivative exists, ~T ′(t) is perpendicular to ~T (t), though~T ′(t) is not necessarily of unit length.

Suppose that ~X(t) is a regular parametrized curve of class C2.We define the first curvature κ1(t) of ~X(t) implicitly as the uniquenonnegative function such that

~T ′(t) = s′(t)κ1(t)~P1(t),

and where ~P1(t) is a unit vector function. This implicit definitionmight not produce a well-defined vector at ~P1(t0) if κ1(t0) = 0,i.e., if ~T ′(t0) = ~0. This problem can be remedied if ~T ′(t) 6= ~0 for0 < |t − t0|ε, for some positive ε. In this case, ~P1(t) is a well-defined and continuous vector function on the unit sphere in Rnfor 0 < |t − t0| < ε. Then a value for ~P1(t0) can be assigned by

9.1. Curves in n-dimensional Euclidean Space 357

completing ~P1(t) by continuity. On the other hand, if κ1(t) = 0over an interval of t, then for the points in this interval, the vectorfunction ~P1(t) is undefined.

As with curves in R3, assuming the derivatives exist, we have

~X ′(t) = s′(t)~T (t)

~X ′′(t) = s′′(t)~T (t) + s′(t)2κ1(t)~P1(t).

The vector function ~P1(t) is called the first normal or first principalnormal vector function. As we consider higher derivatives of ~X(t)in reference to this type of decomposition, we need to make senseof the derivative of ~P1(t) and the derivatives of subsequent vectorfunctions that arise in a similar way.

Now if ~Y (t) and ~Z(t) are unit vector functions with the samedomain interval I and are everywhere perpendicular to each other,then for all t ∈ I,

~Y (t) · ~Z(t) = 0 =⇒ ~Y ′(t) · ~Z(t) + ~Y (t) · ~Z ′(t) = 0

=⇒ ~Y ′(t) · ~Z(t) = −~Y (t) · ~Z ′(t). (9.2)

Now consider the vector function ~P ′1(t). From Equation (9.1), weknow that ~P ′1(t) · ~P1(t) = 0 and from (9.2), we have ~P ′1(t) · ~T (t) =−s′(t)κ1(t). If n > 3, we now define the second curvature of ~X(t) asthe unique nonnegative function κ2 : I → R≥0 such that

~P ′1(t) = −s′(t)κ1(t)~T (t) + s′(t)κ2(t)~P2(t),

for some unit vector function ~P2(t) that is perpendicular to both ~Tand ~P1. As long as κ2(t) is not zero, ~P2(t) is well-defined. It is calledthe second normal or second principal normal vector function.

By repeating this process, we recursively define the ith curvaturefunction and the associated ith unit normal vector function, for 1 ≤i ≤ n−1. However, an exception is made for the (n−1)th “curvature”and normal vector. In defining ~T , ~P1, up to ~Pn−1 as above, weestablish an orthonormal set of vectors (unit length and mutuallyperpendicular). By construction, the n-tuple of vectors organizedinto a matrix as

A(t) =(~T (t) ~P1(t) · · · ~Pn−1(t)

)(9.3)


defines an orthogonal matrix. Orthogonal matrices can have a deter-minant that is 1 or −1. Knowing the vectors ~T through ~Pn−2 alongwith the condition that A(t) be orthogonal leaves exactly two pos-sibilities for ~Pn−1. For calculations and other reasons, it is desirablethat A(t) be a positive orthogonal matrix, i.e., satisfy detA(t) = 1.Thus, with the requirement that detA(t) = 1, the (n− 1)th normalvector ~Pn−1 is uniquely determined from ~T through ~Pn−2. However,because we use ~T through ~Pn−2 to determine ~Pn−1(t), then we cannotdefine the (n− 1)th curvature function κn−1(t) to be a nonnegativefunction.

Definition 9.1.1. Let ~X : I → Rn be a parametrized curve. The n-tuple of vector functions (~T (t), ~P1(t), · · · , ~Pn−1(t)) as constructedabove is called the Frenet frame of ~X.

As with Equation (3.5), we can summarize the derivatives of theFrenet frame vectors by

d

dtA(t) = A(t)

0 −s′κ1 0 · · · 0 0s′κ1 0 −s′κ2 · · · 0 0

0 s′κ2 0 · · · 0 0...

......

. . ....

...0 0 0 · · · 0 −s′κn−1

0 0 0 · · · s′κn−1 0

.

where A(t) is the positive orthogonal matrix defined in Equation(9.3). See [17] for the original introduction to the notion of highercurvatures of curves in Rn.

Equation (3.9) for a formula of the curvature of a space curverelied on taking the cross product of two vectors. However, this isnot a valid operation in Rn for n > 3. Consequently, we need anotherformula for κ1(t). Note that as before, we have

~X ′ = s′ ~T

~X ′′ = s′′ ~T + (s′)2κ1~P1.

The vector formula of Problem 3.1.6 shows

‖~a×~b‖2 =

∣∣∣∣∣~a · ~a ~a ·~b~b · ~a ~b ·~b

∣∣∣∣∣ .


Since Formula (3.9) involves ‖ ~X ′ × ~X ′′‖, we are inspired to considerthe following calculation:∣∣∣∣∣ ~X ′ · ~X ′ ~X ′ · ~X ′′

~X ′′ · ~X ′ ~X ′′ · ~X ′′

∣∣∣∣∣ =

∣∣∣∣(s′)2 s′s′′

s′s′′ (s′′)2 + (s′)4κ21

∣∣∣∣= (s′)2(s′′)2 + (s′)6κ2

1 − (s′)2(s′′)2

= (s′)6κ21.

This leads to the following formula for first curvature:

κ1(t) =1

‖ ~X ′‖3

√√√√∣∣∣∣∣ ~X ′ · ~X ′ ~X ′ · ~X ′′~X ′′ · ~X ′ ~X ′′ · ~X ′′

∣∣∣∣∣ =1

‖ ~X ′‖3

√det(B(t)>B(t)),

where B(t) is the n × 2 matrix(~X ′(t) ~X ′′(t)

). In order to prove

the more general formula, we must use the Cauchy-Binet formula.

Theorem 9.1.2 (Cauchy-Binet). Define [m] = 1, 2, . . . ,m. If M is ann ×m matrix and S ⊆ [n] and S′ ⊆ [m], use the notation MS,S′ tomean the submatrix consisting of the entries from the rows takenfrom the set S with columns taken from the set S′. If A is an m×nmatrix and B an n×m matrix with m ≤ n, then

det(AB) =∑S⊆[n]|S|=m

det(A[m],S) det(BS,[m]).

Proof: (A proof can be found in the appendices of [24].)

Proposition 9.1.3. Let ~X : I → Rn be a regular parametrization ofclass Cn of a curve in Rn for n ≥ 2. Define the matrix Bm(t) =(~X ′ ~X ′′ · · · ~X(m)

)for any integer m with 1 ≤ m ≤ n. Then if

1 ≤ m ≤ n− 2,

s′(t)(m+1)(m+2)/2κ1(t)mκ2(t)m−1 · · ·κm(t) =√

det(Bm+1(t)>Bm+1(t)).

and for the n− 1 curvature function κn−1(t), we have

s′(t)n(n+1)/2κ1(t)n−1κ2(t)n−2 · · ·κn−1(t) = det(Bn(t)).


Proof: Since the standard frame and the Frenet frame are orthonor-mal bases, there exists an orthogonal matrix M(t) that gives thetransition matrix from coordinates in the Frenet frame to stan-dard coordinates. By the definition of the Frenet frame, Bm(t) =M(t)Dm(t) where Dm(t) is an n×m upper triangular matrix (in thesense that entries are 0 below the main diagonal). For example, ifn = 4 then, using Equations (3.6), (3.7), and (3.10) as generalizedto R4, we have

D2 =

s′ s′′

0 (s′)2κ1

0 00 0

, D3 =

s′ s′′ s′′′ − (s′)3κ2

1

0 (s′)2κ1 3s′′(s′)2κ1 + (s′)2κ′10 0 (s′)3κ1κ2

0 0 0

.

By the manner in which we defined the curvatures, we can see thatthe (1, 1) entry of Dm(t) is s′ and that the jth diagonal entry ofDm(t) is ~X(j) · ~Pj−1. By repeatedly using the orthonormality prop-erty of the Frenet frame, we obtain

~X(j) · ~Pj−1 = s′ ~T (j−1) · ~Pj−1

= (s′)2κ1~P

(j−2)1 · ~Pj−1

= (s′)3κ1κ2~P

(j−3)2 · ~Pj−1

=...

= (s′)jκ1κ2 · · ·κj−1.

Since M(t) is an orthogonal matrix,

Bm(t)>Bm(t) = Dm(t)>M(t)>M(t)Dm(t) = Dm(t)>Dm(t).

Writing simply D for Dm(t), by the Cauchy-Binet Theorem,

det(D>D) =∑S⊆[n]|S|=m

det((DS,[m])>) det(DS,[m]) =

∑S⊆[n]|S|=m

det(DS,[m])2.

However, sinceD is upper triangular, if S 6= [m], thenDS,[m] contains

a row of 0s and hence its determinant vanishes. Thus det(D>D) =det(D[m],[m])

2, which is the square of the product of its diagonal


elements. Having determined the diagonal elements above, and sinceall the functions involved are nonnegative,√

det(Bm(t)>Bm(t)) = (s′)∑mj=1 jκm−1

1 κm−22 · · ·κm−1

= (s′)m(m+1)/2κm−11 κm−2

2 · · ·κm−1.

The first part of the proposition follows.We can use the same recursive formula for κn−1 simply because

κn−1(t) is not necessarily a nonnegative function. However, in thiscase Bn(t) is a square matrix. Also, Bn(t) = M(t)Dn(t) and sinceM(t) is orthogonal det(Bn(t)) = det(Dn(t)). However, Dn(t) is up-per triangular so its determinant is the product of its diagonal ele-ments. Thus

det(Bn(t)) = (s′)n(n+1)/2κn−11 κn−2

2 · · ·κn−1.

Proposition 9.1.3 gives a recursive formula for the higher curva-tures and consequently it is a straightforward exercise, though some-times tedious to calculate the curvature functions of a parametrizedcurve in Rn. The reader should also note that Proposition 9.1.3 di-rectly generalizes the formula for curvature of a plane curve (Equa-tion (1.12)), the formula for the curvature of a space curve (Equation(3.9)), and the formula for the torsion of a space curve (Equation(3.2)).

9.1.2 Osculating Planes, Circles, and Spheres

Following the local theory of plane curves and space curves as de-veloped earlier in the text, for curves in Rn we can consider oscu-lating k-planes or osculating k-spheres, where k is any integer with1 ≤ k ≤ n− 1.

Definition 9.1.4. Let ~X : I → Rn be a regular parametrized curveof class Cn. The osculating k-plane to ~X at t = t0 is the k-planethrough the point ~X(t0) and spanned by the vectors ~T (t0), ~P1(t0),. . . , ~Pk−1(t0).

Note that the osculating line to a curve at a point is simply thetangent line.


We call a k-sphere in Rn, any k-dimensional sphere in a (k + 1)dimensional plane in Rn. Note that in this indexing, a circle is a 1-sphere (in a 2-plane), a usual sphere is a 2-sphere (in a 3-plane), andso forth. We leave as exercises to the reader formulas for osculatingk-spheres.

9.1.3 The Fundamental Theorem of Curves in Rn

The Fundamental Theorem of Space Curves immediately generalizesto an n dimensional case. The proof of this generalization involvesno new strategy so we omit the proof.

Theorem 9.1.5 (Fundamental Theorem of Curves). Given curvature func-tions κi(s) ≥ 0, for 1 ≤ i ≤ n − 2 and κn−1(s) continuously differ-entiable over some interval J ⊆ R containing 0, there exists an openinterval I containing 0 and a regular vector function ~X : I → Rnthat parametrizes its locus by arc length, with κi(s) as the ith cur-vature functions. Furthermore, any two curves C1 and C2 with thesame (n − 1)-tuple of curvature functions can be mapped onto oneanother by a rigid motion of Rn.

For the same reason as in the case of space curves, we call the(n − 1)-tuple of curvature functions (κ1(s), κ2(s), . . . , κn−1(s)) thenatural equations of the curve in Rn.

Problems

9.1.1. In R4, consider the twisted quartic ~X(t) = (t, t2, t3, t4), for t ∈ R.Calculate the three curvature functions, namely κ1(t), κ2(t), andκ3(t). [This “twisted quartic” is usually called the rational normalcurve of dimension 4.]

9.1.2. Calculate all three curvature functions κ1(t), κ2(t), and κ3(t) asso-

ciated to the regular curve ~X(t) = (R cos t, R sin t, r cos t, r sin t) inR4, where R, r are positive real constants.

9.1.3. Calculate all three curvature functions κ1(t), κ2(t), and κ3(t) asso-

ciated to the regular curve ~X(t) = (cos t, sin t, cos(2t), sin(2t)).

9.1.4. Let m and n be positive integers. Calculate all three curvaturefunctions κ1(t), κ2(t), and κ3(t) associated to the regular curve~X(t) = (cos(mt), sin(mt), cos(nt), sin(nt)).


9.1.5. Consider regular curves in Rn. Use Proposition 9.1.3 to show thatthe curvature functions are invariant under a positively oriented re-parametrization. Prove that under a negatively oriented reparame-trization the curvatures κ1(t), . . . , κn−2(t) remain invariant and thatthe n− 1 curvature changes according to κn−1(t) = (−1)nκn−1(t).

9.1.6. Proposition 9.1.3 defines curvatures in a recursive way but does notshow any simplifications if such exist. Show an explicit formula forκ3(t) for curves in Rn, first when n = 4 and then for n ≥ 5.

9.1.7. Calculate the osculating 2-sphere to ~X(t) = (t, t2, t3, t4) in R4 att = 0. How is this similar or different from the osculating 2-sphereto ~X(t) = (t, t2, t3) in R3? [See Problem 3.3.6.]

9.1.8. Follow the ideas in Section 3.3 to find the center and the radius ofthe osculating 3-sphere to a curve in R4. [Note that the osculating3-sphere to a curve C at a point P will be a 3-sphere with contactof order 4 to C at P .]

9.1.9. Let A(t) and B(t) be an n × n matrix of differentiable real-valuedfunctions defined in an interval I ⊆ R. If A(t) = (aij(t)), thendenote by A′(t) the n× n matrix (a′ij(t)). Prove that:

(a) ddt (A(t)>) = (A′(t))>;

(b) ddt (A(t) +B(t)) = A′(t) +B′(t);

(c) ddt (A(t)B(t)) = A′(t)B(t) +A(t)B′(t);

(d) ddt (A(t)−1) = −A(t)−1A′(t)A(t)−1, if A(t) is invertible over I.

9.1.10. Suppose that A(t) is an n × n matrix of differentiable real-valuedfunctions defined in an interval I ⊆ R that is everywhere orthog-onal, i.e. A(t)>A(t) = I. Use the previous exercise to prove thatA′(t) = A(t)M(t), whereM(t) is an antisymmetric matrix (M(t)> =−M(t)).

9.1.11. Prove that a curve in Rn lies in a k-dimensional “plane” (a subspaceof Rn of the form ~p + W , where W is a k-dimensional subspace) ifand only if κk(s) = · · · = κn−1(s) = 0.

9.1.12. (*) Find a parametrized curve in R4 that has constant and nonzerocurvature functions κ1, κ2, and κ3. [Hint: This will be a four-dimensional equivalent of a helix.]


9.2 Surfaces in n-dimensional Euclidean Space

9.2.1 Regular Surfaces in Rn

In Chapter 5, we introduced the concept of a regular surface in R3.As an overarching intuition, a regular surface in R3 is a set of points Sobtained from a parametrization such that at each point p ∈ S the setof tangent vectors forms a plane. In order to make all the notions inthe previous sentence precise we needed to discuss parametrizationsof surfaces, which in turn made sense of tangent vectors to surfacesat a point, and address the conditions under which the set of tangentvectors forms a tangent plane. This culminated in Definition 5.2.10.

Subsequent to establishing a workable definition for a regularsurface, we proceeded to define orientability, the metric tensor, andthen a variety of intrinsic and extrinsic properties of surfaces. Ourinvestigations culminated in the Gauss-Bonnet Theorem, from whichwe gave a variety of applications.

Most of the presentation given for regular surfaces in R3 candirectly generalize to define regular surfaces in Rn with n > 3, withone main exception. The difference between the case of surfaces inR3 and surfaces in Rn comes from two reasons that are intimatelyrelated:

• there is no way to naturally define a vector cross product be-tween two vectors in Rn;

• the vector space of perpendicular (normal) vectors to a planeis not one-dimensional.

Given any two-dimensional plane of vectors in Rn, the set of vectorsperpendicular to the plane has dimension n − 2. So if n > 3, thenn− 2 > 1.

In Proposition 5.2.4, we proved that a parametrized surface ~X :U → R3, where U is an open set in R2, has a tangent plane atp if ~X−1(p) is a single point q = (u0, v0) ∈ U and ~Xu(u0, v0) ×~Xv(u0, v0) 6= ~0. Subsequently, we gave an alternate characterizationwhich led to the definition of a regular surface, Definition 5.2.10.Happily, that alternate characterization generalizes with hardly anychanges to a definition in Rn.

9.2. Surfaces in n-dimensional Euclidean Space 365

Definition 9.2.1. A subset S ⊆ Rn is a regular surface if for each p ∈ S,there exists an open set U ⊆ R2, an open neighborhood V of p inRn, and a surjective continuous function ~X : U → V ∩ S such that

1. ~X is continuously differentiable: if we write ~X(u, v) = (x1(u, v),x2(u, v), . . . , xn(u, v)), then for i = 1, 2, . . . , n, the functionsxi(u, v) have continuous partial derivatives with respect to uand v;

2. ~X is a homeomorphism: ~X is continuous and has an inverse~X−1 : V ∩ S → U such that ~X−1 is continuous;

3. ~X satisfies the regularity condition: for each (u, v) ∈ U , thedifferential d ~X(u,v) : R2 → Rn is a one-to-one linear transfor-mation.

Furthermore, a regular surface is said to be of class Cn (resp. C∞)if each function ~X : U → Rn has continuous n derivatives (resp.continuous derivatives of all orders).

The regularity condition can be restated to say that ~Xu(u, v) and~Xu(u, v) are linearly dependent over the domain.

The concepts of a regular, positively oriented, or negatively ori-ented reparametrization of a surface carry over in an identical fashionfrom surfaces in R3 to surfaces in Rn.

9.2.2 Intrinsic Geometry for Surfaces

Recall that intrinsic geometry is a property of surfaces that de-pend entirely on the metric coefficients (and the partial derivativesthereof) of the regular. Since intrinsic properties do not depend ona normal vector to a surface at a point, then such properties applyto any surface as long as we have a metric tensor.

For a regular surface in Rn, we define the first fundamental formin precisely the same way as we did for surfaces in R3. The firstfundamental form Ip( , ) to a regular surface S at p is the innerproduct on TpS obtained as the restriction of the dot product in Rnto the tangent plane TpS.


Suppose that a coordinate patch of S has a regular parame-trization ~X : U → Rn. Suppose also that p = ~X(u0, v0) with(u0, v0) ∈ U is a point on the surface. Consider the standard or-dered basis B = ( ~Xu(u0, v0), ~Xv(u0, v0)) of the tangent space TpS.

Suppose also that two vectors ~a,~b ∈ Tp(S) are expressed with thefollowing components with respect to B,

[~a]B =

(a1

a2

)and [~b]B =

(b1b2

).

Then the first fundamental form has

Ip(~a,~b) = ~a ·~b= (a1

~Xu(u0, v0) + a2~X(u0, v0)) · (b1 ~Xu(u0, v0) + b2 ~X(u0, v0))

=(a1 a2

)( ~Xu(u0, v0) · ~Xu(u0, v0) ~Xu(u0, v0) · ~Xv(u0, v0)~Xv(u0, v0) · ~Xu(u0, v0) ~Xv(u0, v0) · ~Xv(u0, v0)

)(b1b2

)=(a1 a2

)g(u0, v0)

(b1b2

),

where we define the metric tensor g = (gij) as the matrix of functionsdefined on U ⊆ R2

g(u, v) =

(~Xu(u, v) · ~Xu(u, v) ~Xu(u, v) · ~Xv(u, v)~Xv(u, v) · ~Xu(u, v) ~Xv(u, v) · ~Xv(u, v)

).

Since the metric tensor depends on dot products, which exist in anyRn, the usual definition of the metric tensor remains unchanged inany number of dimensions. Furthermore, the uses and interpretationof the metric remain unchanged.

Equations (6.1) and (6.2) for the arc length of a curve on a regularsurface remain identical for surfaces in Rn with exactly the sameproof. This is also true for Equation (6.3) to calculate angles betweentangent vectors. However, Proposition 6.1.7 concerning the area isstill true but its original proof referred to the cross product of ~Xu× ~Xv

in order to find the area element.


Proposition 9.2.2. Let S be a regular surface in Rn, with n ≥ 3, andlet ~X : U → Rn be a parametrization of a coordinate patch of S,where U is an open subset of R2. Let Q be a compact subset of Uand call R = ~X(Q) the compact subset of S. The area of R is givenby

A(R) =

∫∫Q

√det(g) du dv. (9.4)

Proof: The proof of Equation (6.4) relied on the fact that S is aregular surface in R3 and made reference to the cross product of ~Xu

and ~Xv, namely, using the formula∫∫Q

‖ ~Xu × ~Xv‖ du dv

from multivariable calculus. The Riemann sum behind this inte-gral involves approximating the surface area traced out by ~X over[ui, ui+1] × [vi, vi+1] by a parallelogram spanned on two sides by~Xu(u∗, v∗) and ~Xv(u

∗, v∗), where (u∗, v∗) is a selection point in thesubrectangle [ui, ui+1]× [vi, vi+1].

In order to find a formula for the surface area of a regular surfacein Rn, one cannot refer to any formula using the cross product sincethe cross product is only defined in R3. However, the approach usedby the Riemann sum is still the right one. Therefore, one must finda convenient formula for the area of the parallelogram spanned by~Xu and ~Xv in Rn.

Regardless of the dimension of the ambient Euclidean space, wecan calculate the area of these parallelograms as ‖ ~Xu‖ ‖ ~Xv‖ sin θ,where θ is the angle between the two vectors. One cannot get sin θdirectly but one can obtain cos θ from ~Xu · ~Xv = ‖ ~Xu‖ ‖ ~Xv‖ cos θ.Thus,

( ~Xu · ~Xv)2 = ( ~Xu · ~Xu)( ~Xv · ~Xv) cos2 θ

=⇒ ( ~Xu · ~Xv)2 = ( ~Xu · ~Xu)( ~Xv · ~Xv)(1− sin2 θ)

=⇒ ( ~Xu · ~Xu)( ~Xv · ~Xv)− ( ~Xu · ~Xv)2 = ( ~Xu · ~Xu)( ~Xv · ~Xv) sin2 θ

⇐⇒√

( ~Xu · ~Xu)( ~Xv · ~Xv)− ( ~Xu · ~Xv)2 = ‖ ~Xu‖ ‖ ~Xv‖ sin θ.


Thus, at each point (u, v) ∈ U , we have√

det(g) = ‖ ~Xu‖ ‖ ~Xv‖ sin θ.Taking the limit of the Riemann sum as the norm of a mesh thatgoes to 0 gives Equation (9.4).

Another proof for the above result relies on the geometric factthat the area of a parallelogram spanned by ~a,~b ∈ Rn is√

det(C>C) =

√√√√∣∣∣∣∣~a · ~a ~a ·~b~b · ~a ~b ·~b

∣∣∣∣∣,where C is the n × 2 matrix C =

(~a ~b

)with the vectors ~a and ~b

(expressed in standard coordinates) as columns.In Chapter 7, we introduced the Christoffel symbols and used

them to prove the Theorema Egregium, which affirms that the Gaus-sian curvature is an intrinsic property. In particular, Christoffel sym-bols and the Gaussian curvature can be computed for regular surfacesof class C2 in Rn in precisely the same way that we computed themin Chapter 7.

Example 9.2.3 (Flat Torus). The topological definition of a torus is anyset that is homeomorphic to S1 × S1, where S1 is a circle. In otherparts of this book, we discuss the torus in R3 given by the parame-trization

~X(u, v) = ((b+ a cos v) cosu, (b+ a cos v) sinu, sin v) for (u, v) ∈ [0, 2π]× [0, 2π],

and where 0 < a < b. This torus has a non-identity metric tensor andin Example 6.4.5, we saw that this torus has elliptic points whenever0 < v < π, hyperbolic points when π < v < 2π, and parabolic pointswhen v is a multiple of π. In particular, these are regions where theGaussian curvature is respectively positive, negative, and zero.

Consider the following parametrization of the torus in R4:

~X(u, v) = (cosu, sinu, cos v, sin v). (9.5)

The set traced out in R4 by the function ~X over [0, 2π] × [0, 2π]satisfies the topological definition of a torus. It is an easy exercise(Problem 9.2.3) to prove that the metric tensor is constant with

gij(u, v) =

(1 00 1

).


This is the same metric tensor as a plane equipped with an orthonor-mal basis. Furthermore, it is obvious that the Christoffel symbols ofthe first kind will be identically 0, since they involve partial deriva-tives of the gij , and hence the Christoffel symbols of the second kindand the Gaussian curvature will be 0.

This surface in Rn is called the flat torus. Speaking intuitively,we can say that there is “enough room” in R4 to embed a torus bybending but without stretching. A common way to imagine creatinga torus in R3, is to take a piece of paper and first roll it into a cylinderas shown below.

If the cylinder can be stretched, we can imagine bending it aroundso that the two circular ends meet up. This bending is required whenwe are in R3. In four dimensions, suppose that Π1 and Π2 are planesthat meet orthogonally at one point P . Suppose also that there is aline segment L1 in Π1 and another line segment L2 in Π2 that areperpendicular to each other that also meet at P . Let Π be a planeparallel and unequal to the plane spanned by L1 and L2. We canfirst roll Π around L1 along a circle that lies in Π2 and the resultingcylinder can be rolled without bending around L2 along a circle thatlies in Π2. The resulting torus never bends the plane Π.

We remind the reader that the formulas for the Gaussian curva-ture arise in Equations (7.16), (7.17), and (7.18). In particular,

K =R1212

det(gij)=

1

det(gij)

(R1

121g12 +R2121g22

), (9.6)

where the symbols Rlijk are defined in Equation (7.16). The Riemannsymbols satisfy certain symmetry relations which imply that R1212 =


R2121. Hence, we also have

K =R2121

det(gij)=

1

det(gij)

(R1

121g11 +R2121g21

).

9.2.3 Orientability

The property of orientability becomes more challenging to definefor surfaces in Rn precisely because in Rn with n > 3, there doesnot exist a one-dimensional normal line to a regular surface at apoint. Consequently, we cannot talk about a normal vector ~N(u, v).From an intuitive perspective, we would like to say that a surface isorientable if it is possible to cover the surface continuously with asense of clockwise rotation.

Without a normal vector, we cannot refer to a positive rotationaround a normal vector. Instead, we refer to the sign of a determi-nant. Recall that the definition of a regular surface implied that S iscovered by a collection of coordinate patches. Recall that for any twocoordinate patches U and U with coordinates (x1, x2) and (x1, x2),the change of coordinates corresponds to a regular reparametrizationof the surface. Consequently, the Jacobian

∂(x1, x2)

∂(x1, x2)=

∣∣∣∣∣∣∣∣∂x1

∂x1

∂x1

∂x2

∂x2

∂x1

∂x2

∂x2

∣∣∣∣∣∣∣∣is never zero. The sign of a 2× 2 determinant is the sign of the an-gle between the first column and the second column vector. Supposethat surface is parametrized by ~X : U → Rn and also by ~Y : U → Rn,then over subsets of the domains that correspond to the intersec-tion of ~X(U) and ~Y (U), the sign of the determinant corresponds towhether the angle from ~Xx1 to ~Xx2 has the same sign as the anglefrom ~Yx1 to ~Yx2 . It is this sign that is used to determine orientability.

Definition 9.2.4. A regular surface is called orientable if it is coveredby a collection of coordinate patches such that for any two over-lapping coordinate patches U and U with coordinates (x1, x2) and


(x1, x2), the Jacobian satisfies

∂(x1, x2)

∂(x1, x2)> 0.

Whenever this condition holds between two coordinate patches, wesay they have a compatible orientation.

This condition is challenging to check for surfaces in general.However, for some surfaces that arise as the image of a single pa-rametrized function, not necessarily a bijection, it may be easy tocheck if the surface is orientable. First, if a single parametrization~X : U → S provides the required homeomorphism between U andS, then the surface is automatically orientable.

Example 9.2.5 (Flat Torus). As a second example, consider the flattorus described in Example 9.2.3. The parametrization

~X(u, v) = (cosu, sinu, cos v, sin v)

with (u, v) ∈ R2 has the flat torus as its locus, though not bijectively.Now any point p on the flat torus has p = ~X(u1, v1) = ~X(u2, v2),where u2 − u1 and v2 − v1 are integer multiples of 2π. However, wehave

~Xu(u, v) = (− sinu, cosu, sin v, cos v) and

~Xv(u, v) = (cosu, sinu,− sin v, cos v)

and for any two pairs (u1, v1) with (u2, v2) with u2−u1, v2−v1 ∈ 2πZ,we have

~Xu(u1, v1) = ~Xu(u2, v2) and ~Xv(u1, v1) = ~Xv(u2, v2).

We can cover the flat torus with two coordinate patches U = (0, 2π)×(0, 2π) and U = (π, 3π) × (π, 3π), each using the same function ~X.Over regions where these coordinate patches overlap, the Jacobianof the change of coordinates is constant. Hence the flat torus isorientable.


Example 9.2.6 (Klein Bottle). An example of a non-orientable surfacein R4 is the Klein bottle. This is similar to the torus but with a twist.Consider the parametrization ~X : [0, 2π]× [0, 2π]→ R4 defined by

~X(u, v) =(

(b+ a cos v) cosu, (b+ a cos v) sinu, b sin v cos(u

2

), b sin v sin

(u2

)),

with 0 < a < b. The first two components of this should remindthe reader of the usual torus in R3. (See Problem 6.6.4.) However,the last two components are in the x3 and x4 direction and are bothperpendicular to the internal ring of the torus (b cosu, b sinu, 0, 0).In these components, we trace out a small ring as it goes around thelarger internal ring, but making a half-turn twist in the process. It isnot hard to see that the image of ~X over the domain [0, 2π]× [0, 2π]defines a regular surface without boundary. We leave it as an exerciseto check that over the domain [0, 2π) × [0, 2π), the parametrizedsurface is a bijection.

Now consider the coordinate patch with this parametrization butover the domain U = (0, 2π)× (0, 2π). Consider the point

~X(0, π/2) = (b, 0, b, 0) = ~X(2π, 3π/2),

which is not in ~X(U). We can calculate that

~Xu

(0,π

2

)=

(0, b, 0,

b

2

)and ~Xv

(0,π

2

)= (−a, 0, 0, 0)

and also that

~Xu

(2π,

3π

2

)=

(0, b, 0,

b

2

)and ~Xv

(2π,

π

2

)= (a, 0, 0, 0).

The pair of vectors ( ~Xu(u, v), ~Xv(u, v)) vary continuously on the sur-face. Assume there does exist a cover of the Klein bottle using coor-dinate patches that make it an orientable surface. Then the Jacobianof the coordinate change with respect to the coordinates defined by~X and that of any other patch must have the same sign. Considera coordinate patch ~Y : U → S such that p ∈ ~Y (U) and that ~Y (U)is connected. The image ~Y (U) ∩ ~X(U) consists of two connectedcomponents V1 and V2 with V1∩V2 = ∅. At p, the coordinate change


matrix on TpS over V1 and the coordinate change matrix on TpS

over V2 cannot have the same sign because ~Xu

(0, π2

)= ~Xu

(2π, 3π

2

),

whereas ~Xv

(0, π2

)= − ~Xv

(2π, 3π

2

).

9.2.4 The Gauss-Bonnet Theorem

In Chapter 8, we approached the Gauss-Bonnet Theorem startingfrom the perspective of curves on surfaces. After introducing the con-cept of geodesic curvature, we were able to prove the Gauss-BonnetTheorem. One key part of the proof relied on using an orthogonalcoordinate system. We originally cited the orthogonal coordinatesystem involving lines of curvature; however, lines of curvature areextrinsic properties and hence cannot be naturally generalized tosurfaces in Rn. On the other hand, geodesics are intrinsic propertiesof surfaces and in Section 8.5 we proved the existence of geodesic co-ordinate systems that are orthogonal. More precisely, around everypoint p on the surface S there exists a neighborhood U of p that isparametrized by a geodesic coordinate system.

Such a coordinate system, which is an intrinsic property, leadsto the proof of the global Gauss-Bonnet Theorem. Consequently,Theorem 8.3.4 is an intrinsic theorem and holds for regular surfacesin Rn. The Gauss-Bonnet Theorem requires the surface to be ori-entable in order to put a compatible orientation on the boundarycurve C. This orientation on C affects the sign of∫

Cκg ds,

which in turns affects the Gauss-Bonnet formula. If R is a region ofthe surface without boundary, then the orientation of the surface isirrelevant for the formula and we deduce the following corollary.

Corollary 9.2.7. If R is a regular surface without boundary of class C2

in Rn, then ∫∫RK dS = χ(R).

Problems

9.2.1. Consider the parametrized surface in R4 given by

~X(u, v) = (u cos v sin v, u sin2 v, u cos v, u)


with (u, v) ∈ R× [0, 2π].

(a) Prove that this surface is regular everywhere except at (0, 0, 0, 0).

(b) Find the tangent plane at (u0, v0) = (1, π/2).

(c) Find the normal plane to the surface at (u0, v0) = (1, π/2).

9.2.2. Consider the parametrized surface in R4 given by

~X(u, v) = (a cosu sin v, a sinu sin v, a cos v, bu)

for (u, v) ∈ R × [0, π], where a and b are positive constants. Wecan view this as a generalization of a helix in the sense that, insteadof parametrizing a circle that rises along a perpendicular axis, itparametrizes a sphere that rises along an axis perpendicular to theplane of the sphere.

(a) Prove that this surface is regular.

(b) Find the tangent plane at (u0, v0) = (π/2, π/2).

(c) Calculate the metric tensor components.

9.2.3. Prove the claim concerning the metric tensor of the flat torus inExample 9.2.3.

9.2.4. Consider the following generalization to the flat torus. Let a, b, c, dbe positive real constants and consider the surface in R4 parame-trized by

~X(u, v) = (a cosu, b sinu, c cos v, d sin v)

with (u, v) ∈ [0, 2π]× [0, 2π]. Show that in general the coefficients ofthe metric tensor are non-constant functions but that the Gaussiancurvature is identically 0.

9.2.5. A Veronese surface in R5 is a surface parametrized by ~X(u, v) =a(u, v, u2, uv, v2) for (u, v) ∈ R2. Calculate the metric tensor, theChristoffel symbols, and the Gaussian curvature.

9.2.6. Calculate the metric tensor and the Gaussian curvature for the sur-face in R4 parametrized by

~X(u, v) = (u3, u2v, uv2, v3) for (u, v) ∈ R2.

9.2.7. The graph of a change of coordinates in F : R2 → R2 is a parame-trized surface in R4. In particular, if F (u, v) = (f(u, v), g(u, v)) overa domain U , then the graph of f is parametrized by

~X(u, v) = (u, v, f(u, v), g(u, v)) for (u, v) ∈ U.


(a) Calculate the components of the metric tensor of this functiongraph.

(b) Using Equation (9.6), calculate the Gaussian curvature.

(c) As an application, give the metric tensor and the Gaussiancurvature for polar coordinate transformation, i.e., when u = r,v = θ, f(r, θ) = r cos θ, and g(r, θ) = r sin θ.

9.2.8. Let ~α : I → R2 and ~β : J → R2 be regular curves in the plane withcomponent functions ~α(t) = (α1(t), α2(t)) and ~β(t) = (β1(t), β2(t)).Calculate the metric tensor and the Gaussian curvature function ofthe surface parametrized by

~X(u, v) = (α1(u), α2(u), β1(v), β2(v))

for (u, v) ∈ I × J .

9.2.9. Consider the flat torus with the parametrization given in Equation(9.5). Use Equation (8.7) to show that curves of the form (u, v) =(at+ c, bt+ d), for a, b, c, d constants and t ∈ R, are geodesics.

9.2.10. Justify the claim in Example 9.2.6 that ~X : [0, 2π)× [0, 2π)→ R4 isbijective with its image.

9.2.11. Calculate the Gaussian curvature of the Klein bottle using the pa-rametrization given in Example 9.2.6.

9.2.12. There are many ways to define a two-dimensional tube around acurve ~γ : I → R4. Suppose that we define the tube as

~X(u, v) = ~γ(u) + (r cos v)~P1(u) + (r sin v)~P2(u),

with (u, v) ∈ I × [0, 2π], and for some radius r chosen small enoughso that the surface is regular.

(a) Determine the metric tensor of this tube.

(b) Calculate the Gaussian curvature function K(u, v).

(c) Find all the points where K(u, v) = 0.

[Compare to Problem 6.6.8.]

APPENDIX A

Tensor Notation

A.1 Tensor Notation

In Chapter 6, we gave to the first fundamental form the alternatename of metric tensor and delayed the explanation of what a tensoris. Mathematicians and physicists often present tensors and the ten-sor product in very different ways, sometimes making it difficult fora reader to see that authors in different fields are talking about thesame thing. In this section, we introduce tensor notation in whatone might call the “physics style,” which emphasizes how compo-nents of objects change under a coordinate transformation. Readersof mathematics who are well aquainted with tensor algebras on vec-tor spaces might find this approach unsatisfactory, but physicistsshould recognize it. (The reader who wishes to understand the fullmodern mathematical formulation of tensors and see how the physicsstyle meshes with the mathematical style should consult AppendixC in [24].)

The description of tensors we introduce below relies heavily ontransformations between coordinate systems. Though we discusscoordinates and transformations between them generally, one cankeep in mind as running examples Cartesian or polar coordinatesfor regions of the plane or Cartesian, cylindrical, and spherical co-ordinates for regions of R3. Ultimately, our discussion will applyto changes of coordinates between overlapping coordinate patcheson regular surfaces. (The reader should be aware that [24], whichfollows the present text, provides a rigorous introduction to theseconcepts, whereas our presentation here is a little more based onintuition.)

377

378 A. Tensor Notation

A.1.1 Curvilinear Coordinate Systems

Let S be an open set in Rn. A continuous surjective function f :U → S, where U is an open set in Rn, defines a coordinate sys-tem on S by associating to every point P ∈ S an n-tuple x(P ) =(x1(P ), x2(P ), . . . , xn(P )), such that f(x(P )) = P . In this notation,the superscripts do not indicate powers of a variable x but the ithcoordinate for that point in the given coordinate system. Though apossible source of confusion at the beginning, differential geometryliterature uses superscripts instead of the usual subscripts in order tomesh properly with subsequent tensor notation. One also sometimessays that f : U → S parametrizes S. As with polar coordinates,where (r0, θ0) and (r0, θ0 + 2π) correspond to the same point in theplane, the n-tuple need not be uniquely associated to the point P .

Let S be an open set in Rn, and consider two coordinate systemson S relative to which the coordinates of a point P are denoted by(x1, x2, . . . , xn) and (x1, x2, . . . , xn).

Suppose that the open set U ⊂ Rn parametrizes S using the n-tuple of coordinates (x1, x2, . . . , xn) and that the open set V ⊂ Rnparametrizes S using the coordinates (x1, x2, . . . , xn). We assumethat there exists a bijective change-of-coordinates function F : U →V so that we can write

(x1, x2, . . . , xn) = F (x1, x2, . . . , xn). (A.1)

Again using the superscript notation, we might write explicitlyx1 = F 1(x1, x2, . . . , xn),

...

xn = Fn(x1, x2, . . . , xn),

where F i are functions from U to R. We will assume from now onthat the change of variables function F is always of class C2, i.e., thatall the second partial derivatives are continuous. Unless it becomesnecessary for clarity, one often abbreviates the notation and writes

xi = xi(xj),

by which one understands that the coordinates (x1, x2, . . . , xn) func-

A.1. Tensor Notation 379

tionally depend on the coordinates (x1, x2, . . . , xn). Thus, we write

∂xi

∂xjfor

∂F i

∂xj,

and the matrix of the differential dFP (see Equation (5.2)) is givenby

[dFP ]ij =

(∂xi

∂xj

).

Just as the functions xi = xi(xj) represent the change of variablesF , we write xj = xj(xk) to indicate the component functions of theinverse F−1 : V → U . In this notation, Proposition 5.3.3 states that,as matrices, (

∂xj

∂xi

)=

(∂xi

∂xj

)−1

, (A.2)

where we assume that the functions in the first matrix are evaluatedat p in the (x1, . . . , xn)-coordinates, while the functions in the sec-ond matrix are evaluated at p in the (x1, . . . , xn)-coordinates. Onecan express the same relationship in an alternate way by writingxi = xi(xj(xk)) and applying the chain rule when differentiatingwith respect to xk as follows:

∂xi

∂xk=∂xi

∂x1

∂x1

∂xk+∂xi

∂x2

∂x2

∂xk+ · · ·+ ∂xi

∂xn∂xn

∂xk=

n∑j=1

∂xi

∂xj∂xj

∂xk.

However, by definition of a coordinate system in Rn, there must beno function dependence of one variable on another, so

∂xi

∂xk= δik,

where δik is the Kronecker delta symbol defined by

δij =

1, if i = j,

0, if i 6= j.(A.3)

Therefore, since δij are essentially the entries of the identity matrix,we conclude that

n∑j=1

∂xi

∂xj∂xj

∂xk= δik (A.4)

and hence recover Equation (A.2).


x

y

z

O

P

θ

ϕ

Figure A.1. Spherical coordinates.

The space Rn is a vector space, so to each point P , we can asso-

ciate the position vector ~r =−−→OP . Let (x1, x2, . . . , xn) be a coordinate

system of an open set containing P . The natural basis of Rn at apoint P associated to this coordinate system is the set of vectors ∂~r

∂x1,∂~r

∂x2, . . . ,

∂~r

∂xn

,

where all the derivatives are evaluated at P . Note that one oftenexpresses the vector ~r in terms of the Cartesian coordinate system,and this expression is precisely the transformation functions betweenthe given coordinate system and the Cartesian coordinate system.

Example A.1.1 (Spherical Coordinates). Spherical coordinates for pointsin R3 consist of a triple (ρ, θ, ϕ), where ρ is the distance of P to theorigin, θ is the angle between the xz-plane and the vertical planecontaining P , and ϕ is the angle between the positive z-axis and theray [OP ) (see Figure A.1).

In Cartesian coordinates, the position vector ~r for a point withspherical coordinates (ρ, θ, ϕ) is

~r = (ρ cos θ sinϕ, ρ sin θ sinϕ, ρ cosϕ). (A.5)


This corresponds to the coordinate transformation from spherical toCartesian coordinates. Then

∂~r

∂ρ= (cos θ sinϕ, sin θ sinϕ, cosϕ),

∂~r

∂θ= (−ρ sin θ sinϕ, ρ cos θ sinϕ, 0),

∂~r

∂ϕ= (ρ cos θ cosϕ, ρ sin θ cosϕ,−ρ sinϕ).

It is interesting to note that these three vectors are orthogonal forall triples (ρ, θ, ϕ). When this is the case, we say that the coordinatesystem is orthogonal . In this case, one obtains a natural orthonormalbasis at P associated to this coordinate system by simply dividing bythe length of each vector (or its negative). For spherical coordinates,the associated orthonormal basis at any point consists of the threevectors

~eρ = (cos θ sinϕ, sin θ sinϕ, cosϕ),

~eθ = (− sin θ, cos θ, 0),

~eϕ = (cos θ cosϕ, sin θ cosϕ,− sinϕ).

The factors 1, ρ sinϕ, and ρ between ∂~r/∂xi and its normalizedvector are sometimes called the scaling factors for each coordinate.Again, it is important to note that unlike the usual bases in linearalgebra, the basis ~eρ, ~eθ, ~eϕ depends on the coordinates (ρ, θ ϕ) ofa point in Rn.

We are now in a position to discuss how components of variousquantities defined locally, namely in a neighborhood U of a pointp ∈ R3, change under a coordinate transformation on U . As men-tioned before, our definitions do not possess the usual mathematicalflavor, and the supporting discussion might feel like a game of sym-bols. However, it is important to understand the transformationalproperties of tensor components even before becoming familiar withthe machinery of linear algebra of tensors. We begin with the sim-plest situation.


Definition A.1.2. Let p ∈ Rn, and let U be a neighborhood of p. Sup-pose that (x1, x2, . . . , xn) and (x1, x2, . . . , xn) are two systems of co-ordinates on U . A function f(x1, . . . , xn) given in the (x1, x2, . . . , xn)-coordinates is said to be a scalar if its expression f(x1, . . . , xn) inthe (x1, . . . , xn)-coordinates has the same numerical value. In otherwords, if

f(x1, . . . , xn) = f(x1, . . . , xn).

In Definition A.1.2, it is understood that the coordinates (x1, x2,. . . , xn) and (x1, x2, . . . , xn) refer to the same point p in Rn.

This definition might appear at first glance not to hold muchcontent in that every function defined in reference to some coor-dinate system should possess this property, but that is not true.Suppose that f is a scalar. The quantity that gives the derivative off in the first coordinate is not a scalar, for though f(x1, . . . , xn) =f(x1, . . . , xn), we have

∂f

∂x1=

∂f

∂x1

∂x1

∂x1+

∂f

∂x2

∂x2

∂x1+ · · ·+ ∂f

∂xn∂xn

∂x1.

As a second example of how quantities change under coordi-nate transformations, we consider the gradient ~∇f of a differentiablescalar function f . Recall that

~∇f =

(∂f

∂x1,∂f

∂x2, . . . ,

∂f

∂xn

)in usual Cartesian coordinates. The gradient is a vector field, or wemay simply consider the gradient of f at P , namely ~∇fP , which is avector. We highlight the transformational properties of the gradient.The chain rule gives

∂f

∂xj=

n∑i=1

∂xi

∂xj∂f

∂xi. (A.6)

However, it turns out that this is not the only way componentsof what we usually call a “vector” can change under a coordinatetransformation.


Again, consider two coordinate systems (x1, x2, . . . , xn) and (x1,x2, . . . , xn) on an open set of Rn. Let ~A be a vector in Rn, which weconsider based at P . The components of the vector ~A in the respec-tive coordinate systems is (A1, . . . , An) and (A1, . . . , An), where

~A =n∑i=1

Ai∂~r

∂xi=

n∑j=1

Aj∂~r

∂xj.

Since ∂~r∂xi

=∑

j∂~r∂xj

∂xj

∂xi, we find that the components of ~A in the two

systems of coordinates are related by

Aj =n∑i=1

∂xj

∂xiAi. (A.7)

Example A.1.3 (Velocity in Spherical Coordinates). Consider a spacecurve that is parametrized by ~r(t) for t ∈ I. The chain rule allowsus to write the velocity vector in spherical or Cartesian coordinatesas

~r′(t) = ρ′(t)∂~r

∂ρ+ θ′(t)

∂~r

∂θ+ ϕ′(t)

∂~r

∂ϕ

= x′(t)~i+ y′(t)~j + z′(t)~k.

However, differentiating Equation (A.5), assuming ρ, θ, and ϕ arefunctions of t, allows us to identify the Cartesian coordinates of thevelocity vector as:x′y′

z′

=

cos θ sinϕ −ρ sin θ sinϕ ρ cos θ cosϕsin θ sinϕ −ρ cos θ sinϕ ρ sin θ cosϕ

cosϕ 0 −ρ sinϕ

ρ′θ′ϕ′

.

If we label the spherical coordinates as (x1, x2, x3) and the Cartesiancoordinates as (x1, x2, x3), then the transition matrix between com-ponents of the velocity vector from spherical coordinates to Carte-

sian coordinates is precisely(∂xi

∂xj

). Thus, the velocity vector of any

curve ~γ(t) through a point P = ~γ(t0) does not change according toEquation (A.6) but according to Equation (A.7).


Though we have presented the notion of curvilinear coordinatesin general, one should keep in mind the linear coordinate changes

x1

x2

...xn

= M

x1

x2

...xn

where M is an n×n matrix. If either of these coordinate systems isgiven as coordinates in a basis of Rn, then the other system simplycorresponds to a change of basis in Rn. It is easy to show that thetransition matrix is then (

∂xi

∂xj

)= M,

the usual basis transition matrix. Furthermore,(∂xi

∂xj

)is constant

over all Rn.

A.1.2 Tensors: Definitions and Notation

The relations established in Equations (A.6) and (A.7) show thatthere are two different kinds of vectors, each following different trans-formational properties under a coordinate change. This distinctionis not emphasized in most linear algebra courses but essentially cor-responds to the difference between a column vector and a row vector,which in turn corresponds to vectors in Rn and its dual (Rn)∗. (The ∗

notation denotes the dual of a vector space. See Appendix C.3 in [24]for background.) We summarize this dichotomy in the following twodefinitions.

Definition A.1.4. Let (x1, . . . , xn) and (x1, . . . , xn) be two coordinatesystems in a neighborhood of a point p ∈ Rn. An n-tuple of realnumbers (A1, A2, . . . , An) is said to constitute the components ofa contravariant vector at a point p if these components transformaccording to the relation

Aj =n∑i=1

∂xj

∂xiAi,

where we assume the partial derivatives are evaluated at p.


Definition A.1.5. Under the same conditions as above, an n-tuple (B1,B2, . . . , Bn) is said to constitute the components of a covariant vectorat a point p if these components transform according to the relation

Bj =

n∑i=1

∂xi

∂xjBi,

where we assume the partial derivatives are evaluated at p.

A few comments are in order at this point. Though the abovetwo definitions are unsatisfactory from the modern perspective ofset theory, these are precisely what one is likely to find in a classicalmathematics text or a physics text presenting differential geometry.Nonetheless, we will content ourselves with these definitions and withthe more general Definition A.1.6. We defer until Chapter 4 in [24]what is considered the proper modern definition of a tensor on amanifold.

Next, we point out that the quantities (A1, A2, . . . , An) in Def-inition A.1.4 or (B1, B2, . . . , Bn) in Definition A.1.5 can either beconstant or be functions of the coordinates in a neighborhood of p.If the quantities are constant, one says they form the components ofan affine vector. If the quantities are functions in the coordinates,then the components define a different vector for every point in anopen set, and thus, one views these quantities as the components ofa vector field over a neighborhood of p.

Finally, in terms of notation, we distinguish between the twotypes of vectors by using subscripts for covariant vectors and su-perscripts for contravariant vectors. This convention of notation isconsistent throughout the literature and forms a central part of ten-sor calculus. This convention also explains the use of superscriptsfor the coordinates since (xi) represents the components of a con-travariant vector, namely, the position vector of a point.

As a further example to motivate the definition of a tensor, re-call the transformational properties of the components of the firstfundamental form given in Equation (6.7):

gij =2∑

k=1

2∑l=1

∂xk∂xi

∂xl∂xj

gkl,


where gkl (resp. gij) represents the coefficients of the first fundamen-tal form in the (x1, x2)- (resp. (x1, x2)-) coordinates. This formulamimics but generalizes the transformational properties in DefinitionA.1.4 and Definition A.1.5. Many objects of interest that arise indifferential geometry possess similar properties and lead to the fol-lowing definition of a tensor.

Definition A.1.6. Let (x1, . . . , xn) and (x1, . . . , xn) be two coordinatesystems in a neighborhood of a point p ∈ Rn. A set of nr+s quantitiesT i1i2···irj1j2···js is said to constitute the components of a tensor of type(r, s) if under a coordinate transformation these quantities transformaccording to

Tk1k2···krl1l2···ls =

n∑i1=1

n∑i2=1

· · ·n∑

ir=1

n∑j1=1

n∑j2=1

· · ·n∑

js=1

∂xk1

∂xi1∂xk2

∂xi2· · · ∂x

kr

∂xir∂xj1

∂xl1∂xj2

∂xl2· · · ∂x

js

∂xlsT i1i2···irj1j2···js ,

(A.8)

where we assume the partial derivatives are evaluated at p. The rankof the tensor is the integer r + s.

From the above definition, one could rightly surmise that basiccalculations with tensors involve numerous repeated summations.In order to alleviate this notational burden, mathematicians andphysicists who use the tensor notation as presented above utilize theEinstein summation convention. In this convention of notation, oneassumes that one takes a sum from 1 to n (the dimension of Rnor the number of coordinates) over any index that appears both ina superscript and a subscript of a product. Furthermore, for thisconvention, in a partial derivative ∂xi

∂xj, the index i is considered a

superscript, and the index j is considered a subscript.For example, if Aij form the components of a (0, 2)-tensor and

Bk constitutes the components of a contravariant vector, with theEinstein summation convention, the expression AijB

j means

n∑j=1

AijBj .

As another example, with the Einstein summation convention, thetransformational property in Equation (A.8) of a tensor is written


as

Tk1k2···krl1l2···ls =

∂xk1

∂xi1∂xk2

∂xi2· · · ∂x

ks

∂xis∂xj1

∂xl1∂xj2

∂xl2· · · ∂x

jr

∂xlrT i1i2···irj1j2···js ,

where the summations from 1 to n over the indices i1, i2, . . . ir, j1,j2, . . . js is understood. As a third example, if Cijkl is a tensor of type

(2, 2), then Cijkj meansn∑j=1

Cijkj .

On the other hand, with this convention, we do not sum over theindex i in the expression Ai + Bi or even in Ai + Bi. In fact, aswe shall see, though the former expression has an interpretation, thelatter does not.

In the rest of this book, we will use the Einstein summationconvention when working with components of tensors.

A.1.3 Operations on Tensors

It is possible to construct new tensors from old ones. (Again, thereader is encouraged to consult Appendix C.4 in [24] to see the un-derlying algebraic meaning of the following operations.)

First of all, if Si1i2···irj1j2···js and T i1i2···irj1j2···js are both components of ten-sors of type (r, s), then the quantities

W i1i2···irj1j2···js = Si1i2···irj1j2···js + T i1i2···irj1j2···js

form the components of another (r, s)-tensor. In other words, ten-sors of the same type can be added to obtain another tensor of thesame type. The proof is very easy and follows immediately from thetransformational properties and distributivity.

Secondly, if Si1i2···irj1j2···js and T k1k2···ktl1l2···lu are components of tensors oftype (r, s) and (t, u), respectively, then the quantities obtained bymultiplying these components as in,

W i1i2···irk1k2···ktj1j2···jsl1l2···lu = Si1i2···irj1j2···jsT

k1k2···ktl1l2···lu ,

form the components of another tensor but of type (r + t, s + u).Again, the proof is very easy, but one must be careful with the


plethora of indices. One should note that this operation of tensorproduct works also for multiplying a tensor by a scalar since a scalaris a tensor of rank 0.

Finally, another common operation on tensors is the contractionbetween two indices. We illustrate the contraction with an exam-ple. Let Aijkrs be the components of a (3, 2)-tensor, and define thequantities

Bijr = Aijkrk =

n∑k=1

Aijkrk by Einstein summation convention.

It is not hard to show (left as an exercise for the reader) that Bijr

constitute the components of a tensor of type (2, 1). More generally,starting with a tensor of type (r, s), if one sums over an index thatappears both in the superscript and in the subscript, one obtains thecomponents of a (r − 1, s − 1)-tensor. This is the contraction of atensor over the stated indices.

A.1.4 Examples

Example A.1.7. Following the terminology of Definition A.1.6, a co-variant vector is often called a (0, 1)-tensor, and similarly, a con-travariant vector is called a (1, 0)-tensor.

Example A.1.8. In Problem 6.4.6, one showed that the coefficientsLij of the second fundamental form constitute the components of a(0, 2)-tensor, just as the metric tensor does.

Example A.1.9 (Inverse of a (0, 2)-tensor). As a more involved exam-ple, consider the components Aij of a (0, 2)-tensor in Rn. Denote byAij the quantities given as the coefficients of the inverse matrix of(Aij). We prove that Aij form the components of a (2, 0)-tensor.

Suppose that the coefficients Aij given in a coordinate systemwith variables (x1, . . . , xn) and Ars are given in the (x1, . . . , xn)-coordinate system. That they are the inverse to the matrices (Aij)and (Ars) means that Aij and Ars are the unique quantities suchthat

AijAjk = δik, and

ArsAst = δrt , (A.9)


where the reader must remember that we are using the Einsteinsummation convention. Combining Equation (A.9) and the trans-formational properties of Ajk, we get

Ars∂xi

∂xs∂xj

∂xtAij = δrt .

Multiplying both sides by ∂xt

∂xα and summing over t, we obtain

Ars∂xi

∂xs∂xj

∂xtAij

∂xt

∂xα= δrt

∂xt

∂xα

⇐⇒Ars ∂xi

∂xsδjαAij =

∂xr

∂xα

⇐⇒Ars ∂xi

∂xsAiα =

∂xr

∂xα.

Multiplying both sides by Aαβ and then summing over α, we get

Ars∂xi

∂xsδβi = Ars

∂xβ

∂xs=∂xr

∂xαAαβ.

Finally, multiplying the rightmost equality by ∂xs

∂xβand summing over

β, one concludes that

Ars =∂xr

∂xα∂xs

∂xβAαβ.

This shows that the quantities Aij satisfy Definition A.1.6 and formthe components of a (2, 0)-tensor.

By a similar manipulation, one can show that if Bij are thecomponents of a (2, 0)-tensor, then the quantities Bij correspondingto the inverse of the matrix of Bij form the components of a (0, 2)-tensor.

Example A.1.10 (Gauss Map Coefficients). In differential geometry, onedenotes by gij the coefficients of the inverse of the matrix associatedto the first fundamental form. Example A.1.9 shows that gij is a(2, 0)-tensor. Furthermore, recall that the Weingarten equations inEquation (6.26) give the components (associated to the standard


basis on Tp(S) given by a particular parametrization) of the Gaussmap as

aij = −gikLkj .

By tensor product and contraction, we see that the functions aij formthe components of a (1, 1)-tensor.

Example A.1.11 (Metric Tensors). It is important to understand somestandard operations of vectors in the context of tensor notation.Consider two vectors in a vector space V of dimension n. Usingtensor notation, one refers to these vectors as affine contravariantvectors with components Ai and Bj , with i, j = 1, 2, . . . , n. We haveseen that addition of the vectors or scalar multiplication are the usualoperations from linear algebra. Another operation between vectorsin V is the dot product, which was originally defined as

n∑i=1

AiBi,

but this is not the correct way to understand the dot product inthe context of tensor algebra. The very fact that one cannot usethe Einstein summation convention is a hint that we must adjustour notation. The use of the usual dot product for its intendedgeometric purpose makes an assumption of the given basis of V ,namely, that the basis is orthonormal. When using tensor algebra,one makes no such assumption. Instead, one associates a (0, 2)-tensorgij , called the metric tensor, to the basis of V with respect to whichcoordinates are defined. Then the first fundamental form (or scalarproduct) between Ai and Bj is

gijAiBj (Einstein summation).

One immediately notices that because of tensor multiplication andcontraction, the result is a scalar quantity, and hence, will remainunchanged under a coordinate transformation. In this formulation,the assumption that a basis is orthonormal is equivalent to having

gij =

1, if i = j,

0, if i 6= j.


A.1.5 Symmetries

The usual operations of tensor addition and scalar multiplicationwere explained above. We should point out that, using distributivityand associativity, one notices that the set of affine tensors of type(r, s) in Rn form a vector space. The (r + s)-tuple of all the indicescan take on nr+s values, so this vector space has dimension nr+s.

However, it is not uncommon that there exist symmetries withinthe components of a tensor. For example, as we saw for the metrictensor, we always have gij = gji. In the context of matrices, we saidthat the matrix (gij) is a symmetric matrix, but in the context oftensor notation, we say that the components gij are symmetric in theindices i and j. More generally, if T i1i2···irj1j2···js are the components of atensor of type (r, s), we say that the components are symmetric in aset S of indices if the components remain equal when we interchangeany two indices from among the indices in S.

For example, let Aijkrs be the components of a (3, 2)-tensor. To saythat the components are symmetric in i, j, k affirms the equalities

Aijkrs = Aikjrs = Ajikrs = Ajkirs = Akijrs = Akjirs

for all i, j, k ∈ 1, 2, . . . , n. Note that, because of the additionalconditions, the dimension of the space of all (3, 2)-tensors that aresymmetric in their contravariant indices is smaller than n5 but notsimply n5/6 either. We can find the dimension of this vector spaceby determining the cardinality of

I =

(i, j, k) ∈ 1, 2, . . . , n3 | 1 ≤ i ≤ j ≤ k ≤ n.

We will see shortly that |I| =(n+2

3

), and therefore, the dimension

of the vector space of (3, 2)-tensors that are symmetric in their con-travariant indices is

(n+2

3

)n2. We provide the following proposition

for completeness.

Proposition A.1.12. Let Aj1···jrk1···ks be the components of a tensor over Rnthat is symmetric in a set S of its indices. Assuming that all the in-dices are fixed except for the indices of S, the number of independentcomponents of the tensor is equal to the cardinality of

I =

(i1, . . . , im) ∈ 1, 2, . . . , nm | 1 ≤ i1 ≤ i2 ≤ · · · ≤ im ≤ n.


This cardinality is (n− 1 +m

m

)=

(n− 1 +m)!

(n− 1)!m!.

Proof: Since the components are symmetric in the set S of indices,one gets a unique representative of equivalent components by im-posing that the indices in question be listed in nondecreasing order.This remark proves the first part of the proposition. To prove thesecond part, consider the set of integers 1, 2, . . . , n + m and pickm distinct integers l1, . . . , lm that are greater than 1 from amongthis set. We know from the definition of combinations that there are(n−1+m

m

)ways to do this. Assuming that l1 < l2 < · · · < lm, define

it = lt − t. It is easy to see that the resulting m-tuple (i1, . . . , im)is in the set I. Furthermore, since one can reverse the process bydefining lt = it + t for 1 ≤ t ≤ m, there exists a bijection between Iand the m-tuples (l1, . . . , lm) described above. This establishes that

|I| =(n− 1 +m

m

).

Another common situation with relationships between the com-ponents of a tensor is when components are antisymmetric in a setof indices. We say that the components are antisymmetric in a setS of indices if the components are negated when we interchange anytwo indices from among the indices in S. This condition imposesa number of immediate consequences. Consider, for example, thecomponents of a (0, 3)-tensor Aijk that are antisymmetric in all itsindices. If k is any value but i = j, then

Aijk = Aiik = Ajik = −Aijk,

and so Aiik = 0. Given any triple (i, j, k) in which at least twoof the indices are equal, the corresponding component is equal to0. As another consequence of the antisymmetric condition, considerthe component A231. One obtains the triple (2, 3, 1) from (1, 2, 3) byfirst interchanging 1 and 2 to get (2, 1, 3) and then interchanging thelast two to get (2, 3, 1). Therefore, we see that

A123 = −A213 = A231.


In modern algebra, a permutation (a bijection on a finite set) thatinterchanges two inputs and leaves the rest fixed is called a transpo-sition. We say that we used two transpositions to go from (1, 2, 3)to (2, 3, 1).

The above example illustrates that the value of the componentof a tensor indexed by a particular m-tuple (i1, . . . , im) of distinctindices determines the value of any component involving a permuta-tion (j1, . . . , jm) of (i1, . . . , im), as

Aj1...jm = ±Ai1...im ,

where the sign ± is + (resp. −) if it takes an even (resp. odd) numberof interchanges to get from (i1, . . . , im) to (j1, . . . , jm). A priori, ifone could get from (i1, . . . , im) to (j1, . . . , jm) with both an odd andan even number of transpositions, then Ai1...im and all componentsindexed by a permutation of (i1, . . . , im) would be 0. However, afundamental fact in modern algebra (see Theorem 5.5 in [14]) statesthat given a permutation σ on 1, 2, . . . ,m, if we have two ways towrite σ as a composition of transpositions,

σ = τ1 τ2 · · · τa = τ ′1 τ ′2 · · · τ ′b ,

then a and b have the same parity.

Definition A.1.13. We call a permutation even (resp. odd) if this com-mon parity is even (resp. odd) and the sign of σ is

sign(σ) =

1, if σ is even,

−1, if σ is odd.

The above discussion leads to the following proposition about thecomponents of an antisymmetric tensor.

Proposition A.1.14. Let Aj1···jrk1···ks be the components of a tensor over Rnthat is antisymmetric in a set S of its indices. If any of the indicesin S are equal, then

Aj1···jrk1···ks = 0.

If |S| = m, then fixing all but the indices in S, the number of inde-pendent components of the tensor is equal to(

n

m

)=

n!

m!(n−m)!.


Finally, if the indices of Ai1···irj1···js differ from Ak1···krl1···ls only by a permu-tation σ on the indices in S, then

Ak1···krl1···ls = sign(σ)Ai1···irj1···js .

A.1.6 Numerical Tensors

As a motivating example of what are called numerical tensors, notethat the quantities δij form the components of a (1, 1)-tensor. To

see this, suppose that the quantities δij are expressed in a system

of coordinates (x1, . . . , xn) and suppose that δkl are its transformedcoefficients in another system of coordinates (x1, . . . , xn). Obviously,for all fixed i and j, the values of δij are constant, and therefore

δkl =

1, if k = l,

0, if k 6= l.

But using the properties of the δij coefficients and the chain rule,

∂xk

∂xi∂xj

∂xlδij =

∂xk

∂xi∂xi

∂xl=∂xk

∂xl= δkl .

Therefore, δij is a (1, 1)-tensor in a tautological way.

A numerical tensor is a tensor of rank greater than 0 whosecomponents are constant in the variables (x1, . . . , xn) and hence also(x1, . . . , xn). The Kronecker delta is just one example of a numeri-cal tensor and we have already seen that it plays an important rolein many complicated calculations. The Kronecker delta is the sim-plest case of the most important numerical tensor, the generalizedKronecker delta. The generalized Kronecker delta of order r is atensor of type (r, r), with components denoted by δi1···irj1···jr defined asthe following determinant:

δi1···irj1···jr =

∣∣∣∣∣∣∣∣∣δi1j1 δi1j2 · · · δi1jrδi2j1 δi2j2 · · · δi2jr...

.... . .

...

δirj1 δirj2 · · · δirjr

∣∣∣∣∣∣∣∣∣ . (A.10)


It is not obvious from Equation (A.10) that the quantities δi1···irj1···jrform the components of a tensor. However, one can write the com-ponents of the generalized Kronecker delta of order 2 as

δijkl = δikδjl − δ

ilδjk,

which presents δijkl as the difference of two (2, 2)-tensors, which shows

that the coefficients δijkl indeed constitute a tensor. More generally,expanding out Equation (A.10) gives the generalized Kronecker deltaof order r as a sum of r! components of tensors of type (r, r), provingthat δi1···irj1···jr are the components of an (r, r)-tensor.

Properties of the determinant imply that δi1···irj1···jr is antisymmetricin both the superscript indices and the subscript indices. That is tosay, δi1···irj1···jr = 0 if any of the superscript indices are equal or if any ofthe subscript indices are equal. Hence, the value of a component isnegated if any two superscript indices are interchanged and similarlyfor subscript indices. We also note that if r > n where we assumeδi1···irj1···jr is a tensor in Rn, then δi1···irj1···jr = 0 for all choices of indices sinceat least two superscript (and at least two subscript) indices wouldbe equal.

We introduce one more symbol related to the generalized Kro-necker delta, namely the permutation symbol. Define

εi1···in = δi1···in1···n ,

εj1···jn = δ1···nj1···jn .

(A.11)

Note that the use of the maximal index n in Equation (A.11) asopposed to r is intentional. Because of the properties of the de-terminant, it is not hard to see that εi1···in = εi1···in is equal to1 (resp. −1) if (i1, . . . , in) is an even (resp. odd) permutation of(1, 2, . . . , n) and is equal to 0 if (i1, . . . , in) is not a permutation of(1, 2, . . . , n).

We are careful, despite the notation, not to call the permutationsymbols the components of a tensor, for they are not. Instead, wehave the following proposition.


Proposition A.1.15. Let (x1, . . . , xn) and (x1, . . . , xn) be two coordi-nate systems. The permutation symbols transform according to

εj1···jn = J∂xj1

∂xi1· · · ∂x

jn

∂xinεi1···in ,

εk1···kn = J−1∂xh1

∂xk1· · · ∂x

hn

∂xknεh1···hn ,

where J = det(∂xi

∂xj

)is the Jacobian of the transformation of coor-

dinates function.


Example A.1.16 (Cross Product). Consider two contravariant vectorsAi and Bj in R3. If we define Ck = εijkA

iBj , we easily find that

C1 = A2B3 −A3B2, C2 = A3B1 −A1B3, C3 = A1B2 −A2B1.

The values Ck are precisely the terms of the cross product of thevectors Ai and Bj . However, a quick check shows that the quantitiesCk do not form the components of a covariant tensor.

One explanation in relation to standard linear algebra for thefact that Ck does not give a contravariant vector is that if ~a and ~bare vectors in R3 given with coordinates in a certain basis and if Mis a coordinate-change matrix, then

(M~a)× (M~b) 6= M(~a×~b).

In many physics textbooks, when one assumes that we use theusual metric, (gij) being the identity matrix, one is not always care-ful with the superscript and subscript indices. This is because onecan obtain a contravariant vector Bj from a covariant vector Ai sim-ply by defining Bj = gijAi, and the components (B1, B2, B3) arenumerically equal to (A1, A2, A3). Therefore, in this context, onecan define the cross product as the vector with components

C l = gklεijkAiBj . (A.12)

However, one must remember that this is not a contravariant vectorsince it does not satisfy the transformational properties of a tensor.


The generalized Kronecker delta has a close connection to deter-minants, which we will elucidate here. Note that if the superscriptindices are exactly equal to the subscript indices, then δi1···irj1···jr is thedeterminant of the identity matrix. Thus, the contraction over allindices δj1···jrj1···jr counts the number of permutations of r indices takenfrom the set 1, 2, . . . , n. Thus,

δj1···jrj1···jr =n!

(n− r)!. (A.13)

Another property of the generalized Kronecker delta is that

εj1···jnεi1···in = δj1···jni1···in ,

the proof of which is left as an exercise for the reader (ProblemA.1.10). Now let aij be the components of a (1, 1)-tensor, which wecan view as the matrix of a linear transformation from Rn to Rn. Bydefinition of the determinant,

det(aij) = εj1···jna1j1 · · · a

njn .

Then, by properties of the determinant related to rearranging rowsor columns, we have

εi1···in det(aij) = εj1···jnai1j1 · · · ainjn.

Multiplying by εi1···in and summing over all the indices i1, . . . , in, wehave

εi1···inεi1···in det(aij) = δj1···jni1···in a

i1j1· · · ainjn .

Since εi1···inεi1···in counts the number of permutations of 1, . . . , n,

we have

n! det(aij) = δj1···jni1···in ai1j1· · · ainjn . (A.14)

Problems

A.1.1. Prove that

(a) δijδjkδkl = δil ,

(b) δijδjkδki = n.


A.1.2. Let Bi be the components of a covariant vector. Prove that thequantities

Cjk =∂Bj∂xk

− ∂Bk∂xj

form the components of a (0, 2)-tensor.

A.1.3. Let T i1i2···irj1j2···js be the components of a tensor of type (r, s). Prove

that the quantities T ii2···irij2···js , obtained by contracting over the firsttwo indices, form the components of a tensor of type (r − 1, s− 1).Explain why one still obtains a tensor when one contracts over anysuperscript and subscript index.

A.1.4. Let Sijk be the components of a tensor, and suppose they are an-tisymmetric in i, j. Find a tensor with components Tijk that isantisymmetric in j, k satisfying

−Tijk + Tjik = Sijk.

A.1.5. If Ajk is antisymmetric in its indices and Bjk is symmetric in itsindices, show that the scalar AjkBjk is 0.

A.1.6. Consider Ai, Bj , Ck the components of three contravariant vectors.Prove that εijkA

iBjCk is the value triple product ( ~A~B ~C) = ~A·( ~B×~C), which is the volume of the parallelopiped spanned by these threevectors.

A.1.7. Prove that εijkεrsk = δrsij . Assume that we use a metric gij that

is the identity matrix, and define the cross product ~C of two con-travariant vectors ~A = (Ai) and ~B = (Bj) as Ck = gklεijlA

iBj .Use what you just proved to show that

~A× ( ~B × ~C) = ( ~A · ~C) ~B − ( ~A · ~B)~C.

A.1.8. Let ~A, ~B, ~C, and ~D be vectors in R3. Use the εijk symbols to provethat

( ~A× ~B)× (~C × ~D) = ( ~A~B ~D)~C − ( ~A~B ~C) ~D.

A.1.9. Prove Proposition A.1.15.

A.1.10. Prove that εi1···inεj1···jn = δi1···inj1···jn .

A.1.11. Let Aij be the components of an antisymmetric tensor of type (0, 2),and define the quantities

Brst =∂Ast∂xr

+∂Atr∂xs

+∂Ars∂xt

.


(a) Prove that Brst are the components of a tensor of type (0, 3).

(b) Prove that the components Brst are antisymmetric in all theindices.

(c) Determine the number of independent components of antisym-metric tensors of type (0, 3) over Rn.

(d) Would the quantities Brst still be the components of a tensorif Aij were symmetric?

A.1.12. Let A be an n× n matrix with coefficients A = (Aji ), and considerthe coordinate transformation

xj =n∑i=1

Ajixi.

Recall that this transformation is called orthogonal if AAT = I,where AT is the transpose of A and I is the identity matrix. Theorthogonality condition implies that det(A) = ±1. An orthogonaltransformation is called special or proper if, in addition, det(A) = 1.A set of quantities T i1···irj1···js is called a proper tensor of type (r, s) ifit satisfies the tensor transformation property from Equation (A.8)for all proper orthogonal transformations.

(a) Prove that the orthogonality condition is equivalent to requir-ing that

ηij = ηhkAhi A

kj ,

where

ηij =

1, if i = j,

0, if i 6= j..

(b) Prove that the orthogonality condition is also equivalent tosaying that orthogonal transformations are the invertible lin-ear transformations that preserve the quantity (x1)2 +(x2)2 +· · ·+ (xn)2.

(c) Prove that (1) the space of proper tensors of type (r, s) form avector space over R, (2) the product of a proper tensor of type(r1, s1) and a proper tensor of type (r2, s2) is a proper tensorof type (r1 + r2, s1 + s2), and (3) contraction over two indicesof a proper tensor of type (r, s) produces a proper tensor oftype (r − 1, s− 1).

(d) Prove that the permutation symbols are proper tensors of type(n, 0) or (0, n), as appropriate.


(e) Use this to prove that the cross product of two contravariantvectors in R3 as defined in Equation (A.12) is a proper tensorof type (1, 0). [Hint: This explains that the cross product oftwo vectors transforms correctly only if we restrict ourselvesto proper orthogonal transformations on R3.]

(f) Suppose that we are in R3. Prove that the rotation with ma-trix

A =

cosα − sinα 0sinα cosα 0

0 0 1

is a proper orthogonal transformation.

(g) Again, suppose that we are in R3. Prove that the linear trans-formation with matrix given with respect to the standard basis

B =

cosβ sinβ 0sinβ − cosβ 0

0 0 1

is an orthogonal transformation that is not proper.

A.1.13. Consider the vector space Rn+1 with coordinates (x0, x1, . . . , xn),where (x1, . . . , xn) are called the space coordinates and x0 is thetime coordinate. The usual connection between x0 and time t isx0 = ct, where c is the speed of light. We equip this space withthe metric ηµν where η00 = −1, ηii = 1 for 1 ≤ i ≤ n, and ηij = 0if i 6= j. (The quantities ηµν do not give a metric in the sense wehave presented in this text so far because it is not positive definite.Though we do not provide the details here, this unusual metric givesa mathematical justification for why it is impossible to travel fasterthan the speed of light c.) This vector space equipped with themetric ηµν is called the n-dimensional Minkowski spacetime, andηµν is called the Minkowski metric.

Let L be an (n+1)×(n+1) matrix with coefficients Lαβ , and considerthe linear transformation

xj =n∑i=0

Ljixi.

A Lorentz transformation is an invertible linear transformation onMinkowski spacetime with matrix L such that

ηαβ = ηµνLµαL

νβ .


Finally, a set of quantities T i1···irj1···js , with indices ranging in 0, . . . , n,is called a Lorentz tensor of type (r, s) if it satisfies the tensor trans-formation property from Equation (A.8) for all Lorentz transforma-tions.

(a) Show that a transformation of Minkowski spacetime is a Lor-entz transformation if and only if it preserves the quantity

−(x0)2 + (x1)2 + (x2)2 + · · ·+ (xn)2.

(b) Suppose we are working in three-dimensional Minkowski space-time. Prove that the rotation matrix

A =

1 0 0 00 cosα − sinα 00 sinα cosα 00 0 0 1

represents a Lorentz transformation.

(c) Again, suppose we are working in three-dimensional Minkowskispacetime. Consider the matrix

L =

γ −βγ 0 0−βγ γ 0 0

0 0 1 00 0 0 1

,

where β is a positive real number satisfying −1 < β < 1and γ = 1/(

√1− β2). Prove that L represents a Lorentz

transformation.

(d) Prove that (1) the space of Lorentz tensors of type (r, s) forma vector space over R, (2) the product of a Lorentz tensor oftype (r1, s1) and a Lorentz tensor of type (r2, s2) is a Lorentztensor of type (r1 + r2, s1 + s2), and (3) contraction over twoindices of a Lorentz tensor of type (r, s) produces a Lorentztensor of type (r − 1, s− 1).

A.1.14. Let aij be the components of a (1, 1)-tensor, or in other words, thematrix of a linear transformation from Rn to Rn given with respectto some basis. Recall that the characteristic equation for the matrixis

det(aij − λδij) = 0. (A.15)

[Hint: The solutions to this equation are the eigenvalues of thematrix.] Prove that Equation (A.15) is equivalent to

λn +n∑r=1

(−1)ra(r)λn−r = 0


where

a(r) =1

r!δi1···irj1···jra

i1j1· · · airjr .

A.1.15. Moment of Inertia Tensor. Suppose that R3 is given a basis thatis not necessarily orthonormal. Let gij be the metric tensor corre-sponding to this basis, which means that the scalar product betweentwo (contravariant) vectors Ai and Bj is given by

〈 ~A, ~B〉 = gijAiBj .

In the rest of the problem, call (x1, x2, x3) the coordinates of theposition vector ~r.

Let S be a solid in space with a density function ρ(~r), and supposethat it rotates about an axis ` through the origin. The angularvelocity vector ~ω is defined as the vector along the axis `, pointingin the direction that makes the rotation a right-hand corkscrewmotion, and with magnitude ω that is equal to the radians persecond swept out by the motion of rotation. Let (ω1, ω2, ω3) be thecomponents of ~ω in the given basis. The moment of inertia of thesolid S about the direction ~ω is defined as the quantity

I` =

∫∫∫S

ρ(~r)r2⊥ dV,

where r⊥ is the distance from a point ~r with coordinate (x1, x2, x3)to the axis `.

The moment of inertia tensor of a solid is often presented usingcross products, but we define it here using a characterization thatis equivalent to the usual definition but avoids cross products. Wedefine the moment of inertia tensor as the unique (0, 2)-tensor Iijsuch that

(Iijωi)ωj

ω= I`ω, (A.16)

where ω = ‖~ω‖ =√〈~ω, ~ω〉.

(a) Prove that

r2⊥ = gijxixj − (gklω

kxl)2

grsωrωs.

(b) Prove that, using the metric gij , the moment of inertia tensoris given by

Iij =

∫∫∫S

ρ(x1, x2, x3)(gijgkl − gikgjl)xkxl dV.


(c) Prove that Iij is symmetric in its indices.

(d) Prove that if the basis of R3 is orthonormal (which means that(gij) is the identity matrix), one recovers the following usualformulas one finds in physics texts:

I11 =

∫∫∫S

ρ((x2)2 + (x3)2) dV, I12 = −∫∫∫S

ρx1x2 dV,

I22 =

∫∫∫S

ρ((x1)2 + (x3)2) dV, I13 = −∫∫∫S

ρx1x3 dV,

I33 =

∫∫∫S

ρ((x1)2 + (x2)2) dV, I23 = −∫∫∫S

ρx2x3 dV.

(We took the relation in Equation (A.16) as the defining property ofthe moment of inertia tensor because of the theorem that I`ω is thecomponent of the angular moment vector along the axis of rotation

that is given by (Iijωi)ω

j

ω . See [13, pp. 221–222], and in particular,Equation (9.7) for an explanation.

The interesting point about this approach is that it avoids the useof an orthonormal basis and provides a formula for the moment ofinertia tensor when one has an affine metric tensor that is not theidentity. Furthermore, since it avoids the cross product, the abovedefinitions for the moment of inertia tensor of a solid about an axisare generalizable to solids in Rn.)

A.1.16. This problem considers formulas for curvatures of curves or surfacesdefined implicitly by one equation.

(a) In Problem 1.3.17, the reader was asked to prove a formulafor the geodesic curvature κg at a point p on a curve givenimplicitly by the equation F (x, y) = 0. Show that the formulafound there can be written as

κg =1

(F 2x + F 2

y )3/2εi1i2εj1j2Fi1Fj1Fi2j2 , (A.17)

where, in the Einstein summation convention, F1 means ∂F∂x1 =

∂F∂x and F2 means ∂F

∂x2 = ∂F∂y , and where all the functions are

evaluated at the point p.

(b) Prove that Equation (A.17) can be written as

κg = − 1

(F 2x + F 2

y )3/2

∣∣∣∣∣∣Fxx Fxy FxFyx Fyy FyFx Fy 0

∣∣∣∣∣∣ .


(c) Problem 6.6.17 asked the reader to do the same exercise butto find a formula for the Gaussian curvature at a point on asurface given implicitly by the equation F (x, y, z) = 0. Showthat the Gaussian curvature K at a point p on a curve givenimplicity by the equation F (x, y, z) = 0 can be written as

K =1

(F 2x + F 2

y + F 2z )2

εi1i2i3εj1j2j3Fi1Fj1Fi2j2Fi3j3 (A.18)

with the same conventions as used for a curve defined implic-itly.

(d) Prove that Equation (A.18) can be written as

K = − 1

(F 2x + F 2

y + F 2z )2

∣∣∣∣∣∣∣∣Fxx Fxy Fxz FxFyx Fyy Fyz FyFzx Fzy Fzz FzFx Fy Fz 0

∣∣∣∣∣∣∣∣ . (A.19)

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