Objective Parallel transport along a hyperbolic triangle Compare - PowerPoint PPT Presentation

Hyperbolic Geometry and Parallel Transport in R 2 + Tia Burden 1 Vincent Glorioso 2 Brittany Landry 3 Phillip White 2 1 Department of Mathematics Southern University Baton Rouge, LA, USA 2 Department of Mathematics Southeastern Louisiana University Hammond, LA, USA 3 Department of Mathematics University of Alabama Tuscaloosa, AL, USA SMILE VIGRE Program, July, 2013 Burden, Glorioso, Landry, White (Southern University ,Southeastern Louisiana University, and University of Alabama) Reyes Project 1 Smile 2013 1 / 33

Objective Parallel transport along a hyperbolic triangle Compare angle of initial and final vector Compute area of hyperbolic triangle Compare area and angles of parallel transports of hyperbolic triangles Burden, Glorioso, Landry, White (Southern University ,Southeastern Louisiana University, and University of Alabama) Reyes Project 1 Smile 2013 2 / 33

Albert Einstein and Hermann Minkowski Burden, Glorioso, Landry, White (Southern University ,Southeastern Louisiana University, and University of Alabama) Reyes Project 1 Smile 2013 3 / 33

Applications Complex variables Topology of two and three dimensional manifolds Finitely presented infinite groups Physics Computer science Burden, Glorioso, Landry, White (Southern University ,Southeastern Louisiana University, and University of Alabama) Reyes Project 1 Smile 2013 4 / 33

Postulate 1 A straight line segment can be drawn joining any two points. B A Burden, Glorioso, Landry, White (Southern University ,Southeastern Louisiana University, and University of Alabama) Reyes Project 1 Smile 2013 5 / 33

Postulate 2 Any straight line segment can be extended indefinitely in a straight line A B C Burden, Glorioso, Landry, White (Southern University ,Southeastern Louisiana University, and University of Alabama) Reyes Project 1 Smile 2013 6 / 33

Postulate 3 Given any straight line segment, a circle can be drawn having the segment as a radius and one endpoint as center. B A Burden, Glorioso, Landry, White (Southern University ,Southeastern Louisiana University, and University of Alabama) Reyes Project 1 Smile 2013 7 / 33

Postulate 4 All right angles are congruent. C A D B Burden, Glorioso, Landry, White (Southern University ,Southeastern Louisiana University, and University of Alabama) Reyes Project 1 Smile 2013 8 / 33

Parallel Postulate Through any given point not on a line there passes exactly one line that is parallel to that given line in the same plane. B C A Burden, Glorioso, Landry, White (Southern University ,Southeastern Louisiana University, and University of Alabama) Reyes Project 1 Smile 2013 9 / 33

Parallel Transport w v Burden, Glorioso, Landry, White (Southern University ,Southeastern Louisiana University, and University of Alabama) Reyes Project 1 Smile 2013 10 / 33

Parallel Transport About a Euclidean Triangle Burden, Glorioso, Landry, White (Southern University ,Southeastern Louisiana University, and University of Alabama) Reyes Project 1 Smile 2013 11 / 33

Upper Half-Plane Definition + = { ( x , y ) ∈ R 2 : y > 0 } . Upper half-plane : R 2 Complex plane: H 2 = { x + iy : x , y ∈ R , y > 0 } . Burden, Glorioso, Landry, White (Southern University ,Southeastern Louisiana University, and University of Alabama) Reyes Project 1 Smile 2013 12 / 33

Geodesic Definition Geodesic is a line, curved or straight, between two points such that the acceleration of the line is 0. So given a curve γ defined on an open d γ interval I it must satisfy D dt = 0 ∈ T γ ( t ) for all t ∈ I . dt Euclidean Geometry Hyperbolic Geometry Burden, Glorioso, Landry, White (Southern University ,Southeastern Louisiana University, and University of Alabama) Reyes Project 1 Smile 2013 13 / 33

Covariant Derivative Definition There is a unique correspondence that associates the vector field DV dt along the differentiable curve γ to the vector field V . The vector field V is referred to as the covariant derivative . Burden, Glorioso, Landry, White (Southern University ,Southeastern Louisiana University, and University of Alabama) Reyes Project 1 Smile 2013 14 / 33

Results from the Covariant Derivative From the Covariant Derivative we obtain a set of two differential equations, using the curve γ = ( γ 1 ( t ) , γ 2 ( t )) and the vector field V = ( f ( t ) , g ( t )) , that a parallel vector field must satisfy. f ′ ( t ) = γ ′ γ 2 ( t ) f ( t ) + γ ′ 2 ( t ) 1 ( t ) γ 2 ( t ) g ( t ) g ′ ( t ) = γ ′ γ 2 ( t ) f ( t ) + γ ′ 1 ( t ) 2 ( t ) γ 2 ( t ) g ( t ) Burden, Glorioso, Landry, White (Southern University ,Southeastern Louisiana University, and University of Alabama) Reyes Project 1 Smile 2013 15 / 33

Original Vector: V 0 γ 2 ( t ) = ( t , 2 ) 2 V 0 η 0 ( t ) = ( π 2 , t ) 2 ( t ) = ( π 2 , t ) η π π 8 1 γ 1 ( t ) = ( t , 1 ) π 2 Burden, Glorioso, Landry, White (Southern University ,Southeastern Louisiana University, and University of Alabama) Reyes Project 1 Smile 2013 16 / 33

Transport Along γ 1 ( t ) to V 1 γ 2 ( t ) = ( t , 2 ) 2 V 0 η 0 ( t ) = ( π 2 , t ) 2 ( t ) = ( π 2 , t ) η π π 8 1 3 π γ 1 ( t ) = ( t , 1 ) 8 V 1 π 2 Burden, Glorioso, Landry, White (Southern University ,Southeastern Louisiana University, and University of Alabama) Reyes Project 1 Smile 2013 17 / 33

Transport Along the Geodesic η π 2 to V 2 γ 2 ( t ) = ( t , 2 ) 7 π 8 2 3 π V 0 8 η 0 ( t ) = ( π 2 , t ) 2 ( t ) = ( π 2 , t ) η π π 8 V 2 1 3 π γ 1 ( t ) = ( t , 1 ) 8 V 1 π 2 Burden, Glorioso, Landry, White (Southern University ,Southeastern Louisiana University, and University of Alabama) Reyes Project 1 Smile 2013 18 / 33

Transport Back Along γ 2 ( t ) to V 3 V 3 3 π 7 π 8 8 γ 2 ( t ) = ( t , 2 ) 7 π 8 2 3 π V 0 8 η 0 ( t ) = ( π 2 , t ) 2 ( t ) = ( π 2 , t ) η π π 8 V 2 1 3 π γ 1 ( t ) = ( t , 1 ) 8 V 1 π 2 Burden, Glorioso, Landry, White (Southern University ,Southeastern Louisiana University, and University of Alabama) Reyes Project 1 Smile 2013 19 / 33

Transport Down the Geodesic η 0 to V f V 3 3 π 7 π 8 8 7 π γ 2 ( t ) = ( t , 2 ) 8 2 3 π V 0 8 η 0 ( t ) = ( π 2 , t ) 2 ( t ) = ( π 2 , t ) η π V f π π 8 V 2 4 3 π 8 1 3 π γ 1 ( t ) = ( t , 1 ) 8 V 1 π 2 Burden, Glorioso, Landry, White (Southern University ,Southeastern Louisiana University, and University of Alabama) Reyes Project 1 Smile 2013 20 / 33

Area Calculation To find the area of the rectangle, we use the following integral: � 2 π � dydx 2 = A y 2 0 1 2 π ( − 1 � � 2 � = ) dx � y � 0 1 π � 1 2 = 2 dx 0 π 1 � 2 � = 2 x � � 0 π = 4 Burden, Glorioso, Landry, White (Southern University ,Southeastern Louisiana University, and University of Alabama) Reyes Project 1 Smile 2013 21 / 33

General Equation Given the curve c ( t ) = ( t , mt + b ) , we used Mathematica to find the equation for the parallel vector field: t t � � b + mt − 1 V 1 ( t ) = e − e , b + mt t V 2 ( t ) = e − b + mt Burden, Glorioso, Landry, White (Southern University ,Southeastern Louisiana University, and University of Alabama) Reyes Project 1 Smile 2013 22 / 33

Line: y = mx + b Parallel Transport about the line 2 t + 1. Burden, Glorioso, Landry, White (Southern University ,Southeastern Louisiana University, and University of Alabama) Reyes Project 1 Smile 2013 23 / 33

General Parallel Transport When a vector is being transported around a hyperbolic triangle it maintains the angle with the tangent vectors of the curve on which it is moving. Burden, Glorioso, Landry, White (Southern University ,Southeastern Louisiana University, and University of Alabama) Reyes Project 1 Smile 2013 24 / 33

Area Area Equation Step 1 For a hyperbolic triangle with angles α and β and γ = 0, which is an ideal triangle, we get the equation for the area: � cos β � ∞ dydx A = = π − ( α + β ) √ y 2 1 − x 2 − cos α C D 2 . 5 2 1 . 5 AB β 1 α B . 5 A . 5 1 1 . 5 2 Burden, Glorioso, Landry, White (Southern University ,Southeastern Louisiana University, and University of Alabama) Reyes Project 1 Smile 2013 25 / 33

Area Calculation For the triangle AB ∞ � cos β � ∞ dydx A = √ y 2 − cos α 1 − x 2 � cos β ∞ − 1 � � = 1 − x 2 dx √ � y � − cos α � cos β 1 = √ 1 − x 2 dx − cos α � cos β � = arcsin ( x ) − cos α arcsin ( sin ( π 2 − β )) − arcsin ( − sin ( π = 2 − α )) (( π 2 − β ) + ( π = 2 − α )) = π − ( α + β ) Burden, Glorioso, Landry, White (Southern University ,Southeastern Louisiana University, and University of Alabama) Reyes Project 1 Smile 2013 26 / 33

Area Area equation We 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 ) Y X D 2 . 5 180 − γ C 2 γ β 2 1 . 5 β 1 1 α B . 5 A . 5 1 1 . 5 2 Burden, Glorioso, Landry, White (Southern University ,Southeastern Louisiana University, and University of Alabama) Reyes Project 1 Smile 2013 27 / 33