Theorem. In terms of the coordinate frame, the Levi-Civita connection is given by:

Similar documents
7 The cigar soliton, the Rosenau solution, and moving frame calculations

UNIVERSITY OF DUBLIN

Practice Final Solutions

1. Geometry of the unit tangent bundle

Part IB GEOMETRY (Lent 2016): Example Sheet 1

Differential Geometry Exercises

Part IB. Geometry. Year

Intrinsic Geometry. Andrejs Treibergs. Friday, March 7, 2008

Exercise 1 (Formula for connection 1-forms) Using the first structure equation, show that

Mathematical Relativity, Spring 2017/18 Instituto Superior Técnico

Hyperbolic Geometry on Geometric Surfaces

Equidistant curve coordinate system. Morio Kikuchi

Part IB Geometry. Theorems. Based on lectures by A. G. Kovalev Notes taken by Dexter Chua. Lent 2016

Rule ST1 (Symmetry). α β = β α for 1-forms α and β. Like the exterior product, the symmetric tensor product is also linear in each slot :

worked out from first principles by parameterizing the path, etc. If however C is a A path C is a simple closed path if and only if the starting point

Tutorial Exercises: Geometric Connections

SELECTED SAMPLE FINAL EXAM SOLUTIONS - MATH 5378, SPRING 2013

William P. Thurston. The Geometry and Topology of Three-Manifolds

d F = (df E 3 ) E 3. (4.1)

Problem 1, Lorentz transformations of electric and magnetic

Math 23b Practice Final Summer 2011

Chapter 7 Curved Spacetime and General Covariance

Homework for Math , Spring 2012

General Relativity ASTR 2110 Sarazin. Einstein s Equation

Hadamard s Theorem. Rich Schwartz. September 10, The purpose of these notes is to prove the following theorem.

Hyperbolic Discrete Ricci Curvature Flows

Plane hyperbolic geometry

The Geometry of Relativity

8.821 String Theory Fall 2008

MTH4101 CALCULUS II REVISION NOTES. 1. COMPLEX NUMBERS (Thomas Appendix 7 + lecture notes) ax 2 + bx + c = 0. x = b ± b 2 4ac 2a. i = 1.

Chapter 14. Basics of The Differential Geometry of Surfaces. Introduction. Parameterized Surfaces. The First... Home Page. Title Page.

Practice Problems for Exam 3 (Solutions) 1. Let F(x, y) = xyi+(y 3x)j, and let C be the curve r(t) = ti+(3t t 2 )j for 0 t 2. Compute F dr.

The Geometry of Relativity

Classical differential geometry of two-dimensional surfaces

INTRODUCTION TO GEOMETRY

(b) Find the range of h(x, y) (5) Use the definition of continuity to explain whether or not the function f(x, y) is continuous at (0, 0)

Math 5378, Differential Geometry Solutions to practice questions for Test 2

HYPERBOLIC GEOMETRY AND PARALLEL TRANSPORT IN R 2 +

DIFFERENTIAL GEOMETRY HW 5

A local characterization for constant curvature metrics in 2-dimensional Lorentz manifolds

MAT 211 Final Exam. Spring Jennings. Show your work!

Review problems for the final exam Calculus III Fall 2003

Math 20C Homework 2 Partial Solutions

Metrics and Curvature

Physics 236a assignment, Week 2:

Arc Length and Riemannian Metric Geometry

Physics 325: General Relativity Spring Final Review Problem Set

9 Gauss Differential Geometry and the Pseudo-Sphere

Tensors - Lecture 4. cos(β) sin(β) sin(β) cos(β) 0

EQUATIONS OF MOTION: NORMAL AND TANGENTIAL COORDINATES (Section 13.5)

Differential Geometry II Lecture 1: Introduction and Motivation

f dr. (6.1) f(x i, y i, z i ) r i. (6.2) N i=1

1 + f 2 x + f 2 y dy dx, where f(x, y) = 2 + 3x + 4y, is

DO NOT BEGIN THIS TEST UNTIL INSTRUCTED TO START

CURVILINEAR MOTION: NORMAL AND TANGENTIAL COMPONENTS

Syllabus. May 3, Special relativity 1. 2 Differential geometry 3

Introduction to Differential Geometry

MLC Practice Final Exam

Christoffel Symbols. 1 In General Topologies. Joshua Albert. September 28, W. First we say W : λ n = x µ (λ) so that the world

ENGI 4430 Parametric Vector Functions Page dt dt dt

Derivatives in General Relativity

j=1 ωj k E j. (3.1) j=1 θj E j, (3.2)

8.821 F2008 Lecture 12: Boundary of AdS; Poincaré patch; wave equation in AdS

Euler Characteristic of Two-Dimensional Manifolds

Geometric Modelling Summer 2016

Spacetime and 4 vectors

MATHS 267 Answers to Stokes Practice Dr. Jones

Instructions: No books. No notes. Non-graphing calculators only. You are encouraged, although not required, to show your work.

Hyperbolic Analytic Geometry

MATH 4030 Differential Geometry Lecture Notes Part 4 last revised on December 4, Elementary tensor calculus

Homework for Math , Spring 2008

Solutions to old Exam 3 problems

MTH4101 Calculus II. Carl Murray School of Mathematical Sciences Queen Mary University of London Spring Lecture Notes

274 Curves on Surfaces, Lecture 4

Is there a magnification paradox in gravitational lensing?

MAY THE FORCE BE WITH YOU, YOUNG JEDIS!!!

On the cosmic no-hair conjecture in the Einstein-Vlasov setting

Review Sheet for the Final

for 2 1/3 < t 3 1/3 parametrizes

Math 132 Information for Test 2

Sec. 1.1: Basics of Vectors

Math 211, Fall 2014, Carleton College

x + ye z2 + ze y2, y + xe z2 + ze x2, z and where T is the

7a3 2. (c) πa 3 (d) πa 3 (e) πa3

1. (30 points) In the x-y plane, find and classify all local maxima, local minima, and saddle points of the function. f(x, y) = 3y 2 2y 3 3x 2 + 6xy.

1. (a) (5 points) Find the unit tangent and unit normal vectors T and N to the curve. r (t) = 3 cos t, 0, 3 sin t, r ( 3π

1 First and second variational formulas for area

27. Folds (I) I Main Topics A What is a fold? B Curvature of a plane curve C Curvature of a surface 10/10/18 GG303 1

NIL. Let M be R 3 equipped with the metric in which the following vector. fields, U = and. are an orthonormal basis.

Quantum Black Hole and Information. Lecture (1): Acceleration, Horizon, Black Hole

Without fully opening the exam, check that you have pages 1 through 12.

Einstein Toolkit Workshop. Joshua Faber Apr

Math 350 Solutions for Final Exam Page 1. Problem 1. (10 points) (a) Compute the line integral. F ds C. z dx + y dy + x dz C

Special Relativity. Chapter The geometry of space-time

NEW LIFTS OF SASAKI TYPE OF THE RIEMANNIAN METRICS

Properties of the Pseudosphere

Math 234 Exam 3 Review Sheet

Without fully opening the exam, check that you have pages 1 through 10.

Tensors, and differential forms - Lecture 2

Solutions to Practice Test 3

Transcription:

THE LEVI-CIVITA CONNECTION FOR THE POINCARÉ METRIC We denote complex numbers z = x + yi C where x, y R. Let H 2 denote the upper half-plane with the Poincaré metric: {x + iy x, y R, y > 0} g = dz 2 y 2 We compute the Levi-Civita connection with respect to several different frames. 1. The usual coordinate system on H 2 Let x, y be the coordinate vector fields, so that: 1) 2) g x, x ) = g y, y ) = y 1 g x, y ) = g y, x ) = 0. The vector fields define an orthonormal frame field. ξ := y 1 x η := y 1 y Theorem. In terms of the coordinate frame, the Levi-Civita connection is given by: 3) 4) 5) 6) x x = y 1 y x y = y 1 x y x = y 1 x y y = y 1 y Date: wmg, July 17, 2014. 1

2 LEVI-CIVITA CONNECTION In terms of the orthonormal frame, the Levi-Civita connection is given by: ξ ξ = η ξ η = ξ η ξ = 0 η η = 0 Proof. Orthonormality implies gy x, y x ) = 1 is constant, whence and 0 = y gy x, y x ) = 2g y y x ), y x ) = 2yg y y x ), x ) = 2yg x, x ) + yg y x, x )) = 2yy 1 + yg y x ), x ) 7) g y x, x ) = y 3. Differentiating the constant gy y, y ) with respect to x: 8) 0 = 2 x g x y, x ) whence 9) g y x, x ) = 0. Since is symmetric, 10) x y = y x. Combining 9) with 10): g x y, y ) = y 3 and combining with 8), yields 4). Applying 10) again yields 5). Differentiating gy x, y x ) = 1 with respect to x yields: 11) g x x, x ) = 0 Similarly, differentiating with respect to y ; 12) g x x, x ) = y 3 which implies 3). Differentiating gy x, y y ) = 0 with respect to y yields: 13) g y y, x ) = 0 Similarly, 0 = y g y, y ) implies; 14) g y y, y ) = y 3

LEVI-CIVITA CONNECTION 3 which implies 6). The routine calculations for the orthonormal frame are omitted. 2. Connection 1-form for orthonormal frame Denote the coframe field dual to the orthonormal frame by: ξ := ydx η := ydy. The covariant differentials of the orthonormal frame are: so the connection form is: [ ] 0 ξ ξ 0 ξ = η ξ η = ξ ξ = y 1 dx [ ] 0 1 1 0 3. Geodesic curvature of a hypercycle We use these to calculate the geodesic curvature of a hypercycle. The positive imaginary axis ir + is a geodesic with endpoints 0,. The Euclidean rays from 0 subtending an angle θ with ir + are hypercycles re iθ = r cosθ) + i sinθ) ) = r tanhρ) + i sechρ) ) at distance ρ from ir +. The point x 0 + i has coordinates The curve x 0 = tanhρ) sechρ) = sinhρ). γt) := e t x 0 + i) has velocity vector and speed, respectively: 15) γ t) := e t x 0 x + y ) = x 0 ξ + η γ t) = 1 + x 2 0 We compute the accelaration of γt): D dt γt) = x 0 ξ+ηx 0 ξ + η) = x 2 0η x 0 ξ

4 LEVI-CIVITA CONNECTION Now reparametrize γt) by unit speed: γt) := 1 1 + x0 e t x 0 + i) and the geodesic curve is the tangential component of the acceleration: D dt γt) = 1 1 + x 0 x 2 0 η x 0 ξ ) which, since γt) has constant speed, equals: D dt γt) = x 0 1 + x 2 0 = sinθ) = tanhρ) Similarly the geodesic curvature of a metric circle of radius ρ equals cothρ). To see this, use the Poincare unit disc model: for z < 1, the metric tensor is: g := 4 dz 2 1 z 2 ) 2 and writing hyperbolic polar coordinates) z = e iθ tanhρ/2) the metric tensor is g = dρ 2 + sinh 2 ρ)dθ 2, with area form da = sinh 2 ρ)dρ dθ. Consider a disc D ρ with hyperbolic) radius ρ. Then its circumference equals 2π sinhρ) and its area 2πcoshρ) 1. Let k g be the geodesic curvature of the metric circle D ρ. Applying the Gauss-Bonnet theorem 2πχΣ) = KdA + k g ds to σ = D ρ obtaining: Σ Σ 2π = 2π coshρ) 1) + k g 2π sinhρ) ), that is, k g = cothρ) as desired.

LEVI-CIVITA CONNECTION 5 4. Fermi coordinates around a geodesic Another convenient coordinate system begins with a geodesic and considers the family of geodesics orthogonal to this one. Let u denote the parameter along the geodesic, and v the parameter along the perpendiculars: x = e u tanhv) y = e u sechv). Then sinhv) = x/y and z = x 2 + y 2 = e u and u = 1 2 logx2 + y 2 ). Writing with metric tensor The coordinate 1-forms are: and dual coodinate vector fields: z = e u tanhv) + i sechv)) dz = zdu + i sechv)dv) dz 2 = z 2 du 2 + sech 2 v)dv 2) g = dz 2 y 2 = cosh 2 v)du 2 + dv 2. du = x 2 + y 2 ) 1 x dx + y dy) dv = x 2 + y 2 ) 1/2 y dx x/y dy) u = x x + y y v = yx 2 + y 2 ) 1/2 y x x y ). 5. The Levi-Civita connection in Fermi coordinates Using the coordinates u, v) R 2 with metric tensor g = cosh 2 v)du 2 + dv 2 as above, we compute the Christoffel symbols of the Levi-Civita connection. Since u, v are coordinates, the corresponding vector fields u, v commute: [ u, v ] = 0 and since has zero torsion, u v = v u. Since all the inner products of this basis are constant except for g u, u ), 0 = u g u, v) = v g u, v) = u g v, v) = v g v, v) and g is parallel with respect to, implies v v = 0

6 LEVI-CIVITA CONNECTION and differentiating g u, u ) = cosh 2 v) yields u v = v u = tanhv) u u u = sinhv) coshv) v One can modify this coordinate basis to an orhonormal basis by replacing u by U := sechv) u, in which the covariant derivatives are: U U) = tanhv) v, v U) = 0, U v ) = tanhv)u, v v ) = 0. 6. Geodesic curvature in Fermi coordinates Let γt) = ut), vt) ) be a curve. Its velocity is: γ t) = u t) u + v t) v and the square of its speed is: ) 2 ds = u t) cosh 2 vt) ) + v t) 2 dt Its acceleration is: D dt γ t) = u t) + tanh ) vt))u t)v t) u + v t) + tanh vt))u t)v t) u t) 2 cosh vt) ) sinh vt) )) u