university of washington seattle, wa 98195-4350 usa [email protected] branko grünbaum...

University of WashingtonSeattle, WA [email protected]

Branko Grünbaum

SMALL POLYHEDRAL MODELS

OF THE TORUS, THE PROJECTIVE

PLANE, AND THE KLEIN BOTTLE

For which integers n are there polyhedra that are topologically

equivalent to a torus, with all n faces triangles ? Or all faces

quadrangles, or pentagons, ... ?




We need to stop at hexagons, since no torus can have all faces

with seven or more sides.




We need to stop at hexagons, since no torus can have all faces

with seven or more sides.

Joint work with Lajos Szilassi

The question is: For which n can the torus be geometrically

(i) triangulated

(ii) quadrangulated

(iii) quintangulated

(iv) hexangulated

with n faces ?

The torus should be acoptic, that is, with simple faces and

without selfintersections.

TRIANGULATIONS

It is well known that combinatorial (topological) triangulations of the torus with n faces exist for all even n with n ≥ 2, and only for such n.Geometric triangulations exist if and only if n is even and n ≥ 14.

n = 14: Császár torus

For triangulations with n ≥ 16:

Start with a triangulation with n–2 faces and attach a tetrahedron

on one of the faces for a net increase of two faces.

For triangulations with n ≥ 16:

Start with a triangulation with n–2 faces and attach a tetrahedron

on one of the faces for a net increase of two faces.

There exist isogonal triangulations of the torus with n faces

if and only if n = 4m, m ≥ 5.

OVERARCHING FACES:

The intersection of two faces has more than one connectedcomponent.

A far-reaching generalization of these results is the

following theorem of Archdeacon et al.:

Every topological triangulation of the torus with no

overarching faces is isomorphic to a geometric

triangulation.

QUADRANGULATIONS

Quadrangulations with n non-overarching faces exist for

all n ≥ 9, except possibly for n = 10, 11.

QUADRANGULATIONS

Quadrangulations with n non-overarching faces exist for

all n ≥ 9, except possibly for n = 10, 11.

Two basic constructions

First construction:

For all integers p ≥ 3 and q ≥ 3 we can construct "picture

frames" for p-sided "pictures", with q-sided cross-sections.

These give quadrangulations with n = p q, thus yielding the

values n = 9, 12, 15, 16, 18, 20, ... .

A picture frame for a triangular picture with pentagonal cross section, and a frame for a pentagonal picture with a triangular cross section.

Second construction:To a face of a given quadrangulation attach a suitable image of a cube. This increases the number of faces by 4.

From n = 9 we get n = 13 and then n = 17.Then the consecutive values n = 15, 16, 17, 18 are available, hence adding multiples of 4 yields all n ≥ 15.

The only still missing value is n = 14.

An example is shown here:

If overarching faces are admitted, then quadrangulations are

possible for all n > 9.

Example with n = 10.

An example with n = 11 faces.

Conjecture. There exist no geometric acoptic quadrangulations with n ≤ 8 quadrangles.

A far-reaching conjecture is:

Every topological quadrangulation with no overarching

faces can be realized geometrically.

Only a few examples of such topological

quadrangulations are known for which it has been

proved that they cannot be realized by polyhedra with

convex faces. They all seem to be realizable with simple

faces that are not necessarily convex.

QUINTANGULATIONS

For every even n ≥ 12 there are convex-faced

quintangulations of the torus, except possibly for n = 14.

Basic construction:

Simple variants of this construction yield

quintangulations with p-fold rotational symmetry

and with n = 2 p q faces, for all p ≥ 3, q ≥ 2. In

particular, n = 12, 16, 18, 20, 24, 28, ... are obtained.

Attaching a copy of a dodecahedron yields an

increase of 10 faces, thus establishing the claim.

In the previous slide p = 7, q = 2.

CONJECTURE:

There are no quintangulations with n convex faces

where n = 14 or n ≤ 10.

It is not clear what happens if one does not insist that

the faces are convex.

HEXANGULATIONS

Acoptic hexangulations with n faces, with no overarching faces, exist for n = 7, and for all n = p q with p ≥ 3, q ≥ 3.

The case n = 7 is the well-known Szilassi polyhedron

QuickTimeª and aTIFF (Uncompressed) decompressor

are needed to see this picture.

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Hexangulations with n = p q for p ≥ 3, q ≥ 3 can be constructedby starting with trapezohedra (Catalan polyhedra, that are polar tothe antiprisms). For each p ≥ 3 such a polyhedron P has 2p quadrangular faces, p of which meet at each of two apices of P. By intersecting such a polyhedron P with a p-sided prism, having its axis coinciding with the axis of P and rotated appropriately, the resulting tunnel has p hexagonal sides, and all 2p sides of P become hexagons as well.

Example: p = 3, q = 3 so n = 9

An example with p = 5, q = 3, so n = 15.

An example with p = 3, q = 5, so again n = 15.



Conjecture. The only acoptic hexangulations with n faces and no overarching that have rotational symmetry of order3 or more are those with n = p q, with p, q ≥ 3.

There are hexangulations lacking such symmetry for some other values of n. These values have not been characterized.

If overarching faces are allowed,some other symmetric possibilitiesarise. Here is a hexangulation with8 hexagons. It was found by J. Schwörbel.



In contrast to the situation concerning quintangulations, it is well known that no hexangulation of the torus can have only convex faces.

Conjecture. Every hexangulation of the torus has at least six non-convex faces.

Conjecture. Polyhedra of Kepler-Poinsot type admit hexangulations with n faces for all n ≥ 7.







Example. A hexangulation with n = 11 faces. It has two pairs of intersecting faces, and one selfintersecting face.

One may wonder how far all these constructions can be generalized. Long ago I proposed the following

Conjecture. Every cell-complex decomposition, without overarching elements, of an orientable 2-manifold is realizable by an acoptic polyhedron.

The theorem of Archdeacon et al. confirms this in a very special case, while a result of Bokowski et al. shows that it fails for triangulations of a 6-manifold.

As far as I know, there is no information in the case overarching faces are admitted.

No characterization of spherical acoptic polyhedra is known.

In the second part we shall be concerned with

polyhedral realizations of non-orientable manifolds.

By a basic topological result such polyhedra cannot

be acoptic.

In this part we shall look mainly for polyhedra that

realize or represent the projective plane, the Klein

bottle, and the Möbius band. These are the simplest

non-orientable 2-manifolds.

This will lead to considerations of non-convex

polyhedra, polyhedra with selfintersections, and most

significantly, polyhedra with selfintersecting faces.

During most of the Twentieth Century polyhedra with

selfintersections have been neglected, except for some

very special classes. In particular, their topological

properties have not been investigated. This is part of

our goal here.

Another part is to simply show some of these

polyhedra, in order to illustrate their mathematical and

esthetic aspects.

Selfintersecting polygons are of primary importance as faces of the polyhedra we consider. Here are some examples of suchpolygons, and the names I will use.

Bow-tie (quadrangle) Hexagram

Pentagram

In all cases we need to distinguish between the polygonal lineand the polygon as a 2-dimensional part of the plane.

USUAL WAYS OF PRESENTING THE REAL PROJECTIVE

PLANE P2

USUAL WAYS OF PRESENTING THE REAL PROJECTIVE

PLANE P2

Euclidean plane E2 together with points and line at infinity

Family of lines and planes through the origin of E3

Points and great circles of a sphere, with antipodal points identified

Circular disk in E2, with diametral points identified

Convex polyhedra with center, in which opposite elements (vertices, edges, faces) are identified. These are often calledhemi-polyhedra.

POLYHEDRAL MODELS OF THE PROJECTIVE PLANE

Regular dodecahedron withopposite elements identified

Convex polyhedra with center, in which opposite elements (vertices, edges, faces) are identified. These are often calledhemi-polyhedra.

POLYHEDRAL MODELS OF THE PROJECTIVE PLANE

Planar polyhedral maps, with boundary identification, are analogues of the circular disk model in E2, with diametral points identified.

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Regular dodecahedron withopposite elements identified

The correspondingpolyhedral map

CAN A POLYHEDRAL MAP OF THE PROJECTIVE PLANEP2 BE REALIZED BY AN ACTUAL POLYHEDRON ?

This cannot be an acoptic (selfintersection-free) polyhedron, since

acoptic polyhedra are orientable, while the projective plane is

non-orientable.

Allowing intersections of faces, can all faces be simple polygons?


This cannot be an acoptic (selfintersection-free) polyhedron, since

acoptic polyhedra are orientable, while the projective plane is

non-orientable.

Allowing intersections of faces, can all faces be simple polygons?


YES !112233

(hemi-cuboctahedron)

The projective plane P2 is a 2-manifold, as is the polyhedral map

with boundary identification. However, the heptahedron (or

hemi-cuboctahedron) is not a manifold, since each neighborhood

of each of its six vertices has an essential selfintersection.

Vertex figure is a bow-tie quadrangle

Such a singularpoint is known as aWhitney umbrella

…

If the heptahedron were a 2-manifold, which manifold would it be ?

Every closed 2-manifold M is characterized by its orientability (orientable or non-orientable) and by its Euler characteristic (M) = V – E + F.

(M) = 2 ==> M is the orientable 2-sphere.

(M) = 1 ==> M is the non-orientable projective plane.

(M) = 0 ==> M is either the orientable torus, or else the non-orientable Klein bottle.

If the heptahedron were a 2-manifold, which manifold would it be ?

Every closed 2-manifold M is characterized by its orientability (orientable or non-orientable) and by its Euler characteristic (M) = V – E + F.

Since (H) = 1, if H denotes the heptahedron, the only manifoldit could be is the projective plane – but the Whitney umbrella isan impediment.

(M) = 2 ==> M is the orientable 2-sphere.

(M) = 1 ==> M is the non-orientable projective plane.

(M) = 0 ==> M is either the orientable torus, or else the non-orientable Klein bottle.

There are many possibilities for realizations of hemi-polyhedra(that is, the projective plane) and not all have Whitney umbrellas.

Hemi-dodecahedron

Reinhardt 1885

AMHFJDGKEBCFaces:ADGKBABEHCACFJDJEBKFGKFCHEHGDJ

M is the point at whichthree faces haveselfintersections; it isnot a vertex of the hexahedron

ADGKBEHCFJEHCFJ

Hemi-dodeca-hedron(L. Szilassi, 2007)

Conjecture (Szilassi): Every realization of the hemi-dodecahedron has at least one selfintersecting face.

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Face [1,2,3,4,5] isselfintersecting,faces [1,8,9,10,2],[3,2,10,6,7],[4,5,6,10,9],[1,5,6,7,8] and[3,7,8,9,4] aresimple polygons;the last two intersect.

Problem: The classification of 2-manifolds implies that every

manifold is homeomorphic to one of the standard representatives.

For (M) ≥ 0 this means that M is homeomorphic to one of the

four we have listed three slides back. But just as a Whitney

umbrella cannot be homeomorphic to a disk, neither can a

selfintersecting polygon.

?

The problem has nothing to do with non-orientability, or the

projective plane –– it is inherent in selfintersection of faces.

The prism with bow-tie basis is isomorphic to the cube,

but is it homeomorphic ?

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The regular dodecahedron and the great stellated dodecahedron are

isomorphic, as shown by the labels. But are they homeomorphic?

That is, is there is a 1-to-1 map that is continuous and has a continuous

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Dilemma: Do we stop thinking of polyhedra with selfintersecting

faces as manifolds (as topology does, for at least the last

century), or do we find a way to interpret selfintersecting

polygons as homeomorphic to simple polygons ?

The regular dodecahedron and the great stellated dodecahedron are

isomorphic, as shown by the labels. But are they homeomorphic?

That is, is there is a 1-to-1 map that is continuous and has a continuous

inverse.13

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A similar problem arose, and was solved, long ago in connectionwith curves and polygonal lines. It is obvious that as a set of points, the bow-tie quadrangular line is not homeomorphicto the boundary of the square (or to a circle).

12341234x

To establish a homeomorphism, the concept of immersion was accepted as the solution: A crossing-point (such as x) is considered

as representing two distinct points, one on each branch of the

curve or polygonal line. This makes possible a parametrization of

the selfintersecting curve or polygonal line.





With this understanding, and obvious generalizations, selfintersection points of polygonal lines (or curves) do not

prevent homeomorphisms of the parametrized line with simple

polygonal lines, or with a circle.





With this understanding, and obvious generalizations, selfintersection points of polygonal lines (or curves) do not

prevent homeomorphisms of the parametrized line with simple

polygonal lines, or with a circle.

Note: we are dealing with a homeomorphism of the parametrized

lines or curves –– not with a homeomorphism of the plane.

The concept of immersion and the appropriate parametrization was long ago extended to cover segments of multiple points, aswell as whole regions of such points. An example was providedby Reinhardt, in comments to the 1886 edition of Möbius’ collected works. His drawing of the Möbius band (as proposedby Möbius) is shown at left, while a version of the band, and ofthe hemi-icosahedron obtained by constructing a pyramid overthe band, are shown in the other parts. These are taken from abook by Apéry (1987).

We need to validate the use of the Euler characteristic in deciding which manifold is represented by a polyhedron that may contain selfintersecting faces. By relying on the immersion andparametrization concepts, we know that a simple polygonal line ishomeomorphic with the polygonal line that determines a face, evenif the polygonal line is selfintersecting. Now we need to extend thishomeomorphism in a way that the interior of the simple polygonmaps onto a kind of “interior” of the selfintersecting polygon.Where is the quadrangular “interior” of the bow-tie that ishomeomorphic to the interior of the square?

12341234x

We need a different kind of immersion. To avoid misunder-

standings, we call it infolding. In the infolding of the bow-tie

quadrangle, or of the selfintersecting pentagon in the Szilassi

pentagon, we need each crossing point to represent a segment of

points. How such a parametrization can be done in a consistent and

bicontinuous way is illustrated in the next slide. It shows an isotopy

(continuous deformation) from a parametrized bow-tie to a square.

The point parameters throughout the isotopy are (x, c), the isotopy

parameter is t, and the points have coordinates (x, y). All

parameters go from –1 to +1. The coordinate y is given by

y = ((t – 1)x + t + 1)c/2

(1,–t)

(1,t)

(-1,1)

(-1,-1)

(-1,c)

(1,t c)

t = –1

(1,–t)

(1,t)

(-1,1)

(-1,-1)

(-1,c)

(1,t c)

t = –0.5

(1,t)

(-1,1)

(-1,-1)

(-1,c)

(1,t c)

t = 0

(1,t)

(1,–t)

(-1,1)

(-1,-1)

(-1,c)

(1,t c)

t = 0.5

(-1,1) (1,1)

(-1,-1) (1,–1)

(-1,c) (1,c)

t = 1

y = ((t – 1)x + t + 1)c/2

For a different variety of infolding, consider the regular

dodecahedron, and the great stellated dodecahedron.

They are isomorphic, so the great stellated dodecahedron

should be homeomorphic to a sphere –– even though

it is well known that the density at its center is 3.

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Here an isotopy connects a pentagram to a bow-tie quadrangle, which is isotopic to a square, which is isotopic to a pentagon. Thus the pentagram is homeomorphic to the pentagon, hence the great stellated dodecahedron is indeed homeomorphic to the Platonic dodecahedron.

It can be shown that the infolding approach generalizes to all

polyhedra, and that therefore the criteria of orientability and

Euler characteristic are sufficient to establish each polyhedron

(in suitably parametrized form) as homeomorphic to a (compact)

2-manifold. All it takes is to interpret each selfintersection point as

representing a whole segment in a suitable parametrization.

A simple way to formulate this conclusion is:

For suitably parametrized polyhedra

ISOMORPHIC => HOMEOMORPHICHere are a few examples of the application of this result.

Examples of connels (= conical tunnels) formed by bow-ties.Like tunnels, they can be used to replace pairs of faces that are related by symmetry in a point. If the pair was part of an orientable polyhedron, the resulting polyhedron is orientable as well. Each replacement reduces the Euler characteristic by 2.A simple example is shown in the next slide.

Torus: 4 squares, one connel with 4 bow-tie faces, 16 edges, 8 vertices, = 0, orientable

Faces: ABFE, BCGF, CDHG, DAEH, ADFG, DCEF, CBHE, BAGH

A B B C C D D A

E F F G G H H E

A B B C C D D A

E F F GG H H E

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C

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A selfdual torus, isomorphic to the preceding one: 4 rectangles, 4 bow-tie faces, 16 edges, 8 vertices (there are overarching faces).

Described by K. Merz (1935)

This torus, and the one in the preceding slide, have fewer edges than the minimum number established by U. Brehm (1990) fora class of “polyhedral maps” more restricted than allowed here.

Faces: ABHG, ADHE, CDFE, CBFG, AGFD, AEFB, CEHB, CGHDEAGCHDFBEAGCHDFB

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1 3 0 2 4 1

A selfdual torus consisting of five bow-tie quadrangles,with five vertices and ten edges. All faces are in one plane.Some pairs of faces are overarching (have intersection with more than one component).

A projective plane consisting of five coplanar convex quadrangles

(only one is shown filled in, for easier visualization). Some pairs

overlap on triangular regions. This is the hemi-5-sided-trapezohedron

(dual of antiprism). The center is a singular point (vertex figure has

rotation number 2).

Analogous projective planes exist for all odd n ≥ 5.

Projective plane

Modified from an example of F. Apéry (1987)

6 vertices, 14 edges, 9 faces (8 triangles, one bow-tie)

AAFFCCDDEBAAFFBBDDEECC

ABDEFC

ABDEFCABDEFC

Projective plane

Taking two copies with coinciding bow-ties, and eliminating

the bow-ties yields an ORIENTABLE polyhedron. This is a

hemi-polyhedron of a torus, which is itself a torus.

6 vertices, 14 edges, 9 faces (8 triangles, one bow-tie)

AAFFCCDDEBAAFFBBDDEECC ABDEFCABDEFC

6 quadrangles (3 convex, 3 bow-tie), 12 edges, 7 vertices, = 1

Projective plane, hemi-rhombic dodecahedron

This model has fewer edges than the minimum for a more restrictedkind of “polyhedral maps” considered by U. Brehm (1990)

Projective plane: Heptahedron, dual to the hexahedron of the previous slide.3 bow-tie quadrangles, 4 triangles, 12 edges, 6 verticesTwo copies joined at bases give a Klein bottle, with6 bow-tie quadrangles, 6 triangles, 21 edges, 9 vertices

Projective plane:6 quadrangles, 1 hexagon, 15 edges, 9 vertices

Put two copies together, and eliminate hexagons:

Klein bottle with 12 quadrangles, 24 edges, 12 vertices

Faces: ABCD, BAEF, FEHG, GHDC, AEDH, ADEH, BCFG, BFCD

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G

HE

FF

F

CGGF

C

C C B B

BB

A

A A

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Klein bottle: 8 faces, 16 edges, 8 vertices, = 0, non-orientable

ABCDEFGHBACDEFGH

Klein bottle: 12 quadrangles, 20 edges, 8 vertices, non-orientableno overarching faces

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Klein bottle: 12 quadrangles 8 quadrangles, 2 hexagons

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Six triangles of the mantle (only one shown at

right), a connel (conical tunnel) of three bow-ties

(two shown at left), 15 edges, six vertices.

Orientable. = 0. Torus with overarching faces.

Two triangles, six bow-ties, 15 edges, six vertices, = –1.

Overarching faces. Non-orientable, of genus 3

Various interesting polyhedra are possible with more

complicated selfintersecting faces. A few examples

are shown next. For simplicity, all are isogonal, that

is, have vertices that are equivalent under symmetries

of the polyhedron.

(3.6*.3#.6*)

An isogonal faceting of the truncated tetrahedron. This istopologically a torus, in which each vertex is incident with one small triangle, one large triangle and two hexagrams.

(3.6*.3#.6*)

An isogonal faceting of the truncated tetrahedron. This istopologically a torus, in which each vertex is incident with one small triangle, one large triangle and two hexagrams.

2134561,62,34,5123456123456123456123456{6}1,42,53,6{6/2}

The same type of polyhedron is possible for all hexagrams in theisotopy (metamorphosis) shown.

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A view of the metamorphosis

An isogonal faceting of the truncated dodecahedron involving decagrammatical faces. Each vertex meets one triangle, oneregular pentagon, and two decagrams, as shown at left. The resulting polyhedron is shown at right without the pentagons, to make it more intelligible.

(3.10*.5.10*)

(3.10*.5.10*)Pentagons not shown.

012345678901234567890123456789012345678901234567893,47,81,25,60,90123456789012345678901234567891,60,52,73,84,90123456789012345678901234567890,34,71,82,56,90123456789{10/2}=10/2-110/2-210/2-310/2-510/2-610/2-710/2-810/2-910/2-1010/2-4{10/3}=10/2-1110/2-210/2-210/2-410/2-6

As in the case with hexagons we have seen earlier, there is

a metamorphosis of decagons that connects two regular

decagons through a continuum of isogonal ones. All are

suitable to yield polyhedra of the same type.

Another isogonal faceting of the truncated dodecahedron

involving decagrammatical faces. At each vertex one regular

pentagon meets two decagrams. The resulting polyhedron

is shown without the pentagons, to make it more intelligible.

Orientable, 24 faces, 90 edges, 60 vertices, genus 4.

(5.10*.10*)

(3.10*.10*.10*.10*)

An isogonal faceting of the truncated dodecahedron in which

each vertex is incident with one triangle and four decagrams.

Orientable, 44 faces, 150 edges, 60 vertices, genus 24.

Isogonal, one triangle and four bow-ties at each vertex.

60 vertices, 150 edges, 80 faces, = –10.

Non-orientable, genus 12.

ö

The boundary of a Möbius band consist of a single

closed curve. A realization of the Möbius band in

which this curve is planar and bounds a topological

disk is often called a cross-cap. Any cross-cap must

have selfintersections. Here are two familiar versions.

Adapted fromApéry

ö

Each has two Whitney umbrellas. The homeomorphismto the Möbius band needs justification. This is possible,but is rarely given.

Each has two Whitney umbrellas. The homeomorphismto the Möbius band needs justification. This is possible,but is rarely given.

What is the importance of cross-caps?They are needed in the basic theorem on the classification of closed 2-manifolds,

Orientable 2-manifold M

Euler characteristic

(M) = 2 – 2g,

g = (orientable) genus of M

= (2 – )/2

Non-orientable 2-manifold M


(M) = 2 – g

g = (non-orientable) genus of M

= 2 -

Orientable 2-manifold M


(M) = 2 – 2g,

g = (orientable) genus of M

= (2 – )/2

Non-orientable 2-manifold M


(M) = 2 – g

g = (non-orientable) genus of M

= 2 -

Sphere: Orientable, g = 0, = 2.

Each orientable M of genus

g ≥ 0 is homeomorphic

to sphere with g handles.

Each non-orientable M of

genus g ≥ 0 is homeomorphic

to sphere with g cross-caps.

The classification of 2-manifolds:

A

B

C

D

E

F

ö Polyhedral Möbius band

that is a cross-cap

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FA

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Transition from the traditional Möbius band tothe polyhedral three-bow-ties cross-cap.

A variety of cross-caps

Each is homeomorphic to the Möbius band by infolding.

The three bow-tie quadrangles AA*BB*, BB*CC* and CC*AA*

overlap in the triangular region shown in white. The three bow-ties

form a cross-cap, just as in the case their crossover points coincide.

Another kind of cross-caps

A

B

C

B*

A*

C*

Despite the work of Brückner and others a

century or more ago (or possibly because of it !?),

the study of polyhedra with selfintersections and

with selfintersecting faces has been neglected

during much of the last 100 years. I hope the

examples of such polyhedra and their applications

to the topology of manifolds discussed here will

lead to an awakening of interest in their study.

university of washington seattle, wa 98195-4350 usa [email protected] branko grünbaum...

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