hyperbolic geometry and parallel transport in r+2
TRANSCRIPT
Hyperbolic Geometry and Parallel Transport in R2+
Tia Burden1 Vincent Glorioso2 Brittany Landry3
Phillip White2
1Department of MathematicsSouthern University
Baton Rouge, LA, USA
2Department of MathematicsSoutheastern Louisiana University
Hammond, LA, USA
3Department of MathematicsUniversity of AlabamaTuscaloosa, AL, USA
SMILE VIGRE Program, July, 2013
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Objective
Parallel transport along a hyperbolic triangleCompare angle of initial and final vectorCompute area of hyperbolic triangleCompare area and angles of parallel transports of hyperbolictriangles
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Albert Einstein and Hermann Minkowski
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Applications
Complex variablesTopology of two and three dimensional manifoldsFinitely presented infinite groupsPhysicsComputer science
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Postulate 1
A straight line segment can be drawn joining any two points.
A
B
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Postulate 2
Any straight line segment can be extended indefinitely in a straight line
A B C
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Postulate 3
Given any straight line segment, a circle can be drawn having thesegment as a radius and one endpoint as center.
AB
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Postulate 4
All right angles are congruent.
A B
C
D
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Parallel Postulate
Through any given point not on a line there passes exactly one linethat is parallel to that given line in the same plane.
A
B
C
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Parallel Transport
w
v
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Parallel Transport About a Euclidean Triangle
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Upper Half-Plane
DefinitionUpper half-plane: R2
+ = {(x , y) ∈ R2 : y > 0}.Complex plane: H2 = {x + iy : x , y ∈ R, y > 0}.
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Geodesic
DefinitionGeodesic is a line, curved or straight, between two points such that theacceleration of the line is 0. So given a curve γ defined on an openinterval I it must satisfy D
dtdγdt = 0 ∈ Tγ(t) for all t ∈ I.
Euclidean Geometry Hyperbolic Geometry
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Covariant Derivative
DefinitionThere is a unique correspondence that associates the vector field DV
dtalong the differentiable curve γ to the vector field V . The vector field Vis referred to as the covariant derivative.
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Results from the Covariant Derivative
From the Covariant Derivative we obtain a set of two differentialequations, using the curve γ = (γ1(t), γ2(t)) and the vector fieldV = (f (t),g(t)), that a parallel vector field must satisfy.
f ′(t) =γ′2(t)γ2(t)
f (t) +γ′1(t)γ2(t)
g(t)
g′(t) =γ′1(t)γ2(t)
f (t) +γ′2(t)γ2(t)
g(t)
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Original Vector: V0
π2
1
2
γ1(t) = (t , 1)
ηπ2(t) = (π2 , t)
γ2(t) = (t , 2)
η0(t) = (π2 , t)
V0
π8
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Transport Along γ1(t) to V1
π2
1
2
γ1(t) = (t , 1)
ηπ2(t) = (π2 , t)
γ2(t) = (t , 2)
η0(t) = (π2 , t)
V0
π8
V1
3π8
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Transport Along the Geodesic ηπ2
to V2
π2
1
2
γ1(t) = (t , 1)
ηπ2(t) = (π2 , t)
γ2(t) = (t , 2)
η0(t) = (π2 , t)
V0
π8
V1
3π8
V2
7π8
3π8
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Transport Back Along γ2(t) to V3
π2
1
2
γ1(t) = (t , 1)
ηπ2(t) = (π2 , t)
γ2(t) = (t , 2)
η0(t) = (π2 , t)
V0
π8
V1
3π8
V2
7π8
3π8
V33π8
7π8
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Transport Down the Geodesic η0 to Vf
π2
1
2
γ1(t) = (t , 1)
ηπ2(t) = (π2 , t)
γ2(t) = (t , 2)
η0(t) = (π2 , t)
V0
π8
V1
3π8
V2
7π8
3π8
V33π8
7π8
Vf3π8
π4
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Area Calculation
To find the area of the rectangle, we use the following integral:
A =
∫ π2
0
∫ 2
1
dydxy2
=
∫ π2
0(−1y
∣∣∣∣21) dx
=
∫ π2
0
12
dx
=12
x∣∣∣∣π20
=π
4
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General Equation
Given the curve c(t) = (t ,mt + b), we used Mathematica to find theequation for the parallel vector field:V1(t) = e−
tb+mt
(e
tb+mt − 1
),
V2(t) = e−t
b+mt
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Line: y = mx + b
Parallel Transport about the line 2t + 1.
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General Parallel Transport
When a vector is being transported around a hyperbolic triangle itmaintains the angle with the tangent vectors of the curve on which it ismoving.
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Area
Area Equation Step 1For a hyperbolic triangle with angles α and β and γ = 0, which is anideal triangle, we get the equation for the area:
A =
∫ cosβ
− cosα
∫ ∞√
1−x2
dydxy2 = π − (α+ β)
.5 1 1.5 2
.51
1.52
2.5
AB
AB
CD
αβ
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Area Calculation
For the triangle AB∞
A =
∫ cosβ
− cosα
∫ ∞√
1−x2
dydxy2
=
∫ cosβ
− cosα
−1y
∣∣∣∣∞√1−x2
dx
=
∫ cosβ
− cosα
1√1− x2
dx
= arcsin(x)∣∣cosβ− cosα
= arcsin(sin(π
2− β))− arcsin(− sin(
π
2− α))
= ((π
2− β) + (
π
2− α))
= π − (α+ β)
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Area
Area equationWe can find the area of a general hyperbolic triangle with angles α, β,and γ by subtracting the areas of two ideal hyperbolic triangles:
A = π − (α+ β + γ)
.5 1 1.5 2
.5
1
1.5
2
2.5
AB
XY
C
D
αβ1
γβ2
180− γ
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Area Calculation
For the triangle ABC
A =
∫ cosβ
− cosα
∫ ∞√
1−x2
dydxy2 −
∫ cosβ2
− cos(π−γ)
∫ ∞√
1−x2
dydxy2
=
∫ cosβ
− cosα
−1y
∣∣∣∣∞√1−x2
dx −∫ cosβ2
− cos(π−γ)
−1y∣∣∞√
1−x2 dx
=
∫ cosβ
− cosα
1√1− x2
dx −∫ cosβ2
− cos(π−γ)
1√1− x2
dx
= arcsin(x)∣∣cosβ− cosα − arcsin(x)
∣∣cosβ2− cos(π−γ)
= ((π
2− β) + (
π
2− α))− ((
π
2− β2) + (
π
2− γ))
= π − (α+ β)− (π − (γ + β2))
= π − (α+ β1 + β2)− (π − (γ + β2))
= π − (α+ β + γ)
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Area vs. Angle of Initial to Transported Vector
TheoremThe area of any hyperbolic triangle in R2
+ is equal to the angle betweenthe initial and final transported vectors.
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Non-Normal Hyperbolic Triangle
1 2 3 4 5 6
1
2
3
4
0
A′
C′
B′
H
C
B
A
I
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What are Fractional Transformations?
A fractional transformation is a special representation of a matrix thatis used in Möbius Transformations.
Fractional Transformation
Given matrix M =
[a bc d
], where a,b, c,d ∈ R, and the point
p = (x , y). The fractional transformation of p in terms of M is writtenas:
fM(p) =ap + bcp + d
=a(x + yi) + bc(x + yi) + d
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Future Studies
Parallel Transport in Different Models of Hyperbolic GeometryPoincaré DiskHyperboloid
B
A
C
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Acknowledgments
We would like to give a big thanks to our graduate student AndrewHolmes for helping us with this project, in addition to Dr. Edgar Reyesfor providing the information needed for this project. Also, we wouldlike to thank the SMILE program for allowing us to have this greatopportunity.
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