topics in di erential geometry, 2015 - ucsd mathematicsbdriver/257a-winter2015/lecture...

Bruce K. Driver

Topics in Differential Geometry, 2015

March 11, 2015 File:geotopics.tex

Contents

Part I Integral Curves and Surfaces

1 Calculus on Banach Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.1 The Differential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2 Product and Chain Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.3 Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.4 Higher Order Differentials* . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2 Linear Ordinary Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.1 Operator Logarithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

3 Ordinary Differential Equations in a Banach Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233.2 Uniqueness Theorem and Continuous Dependence on Initial Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253.3 Local Existence (Non-Linear ODE) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253.4 Global Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273.5 Semi-Group Properties of time independent flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 313.7 Smooth Dependence of ODE’s on Initial Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323.8 Derivative Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343.9 Commutators of Vector Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 353.10 Inverse Function Theorem by O.D.E.s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

4 Classical Frobenius Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

4 Contents

5 Applications of Frobenius . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 435.1 Frobenius for special F . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 435.2 Representations of Lie Groups and Lie Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 445.3 Parallel translation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 455.4 Riemann’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

6 Matrix Lie Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 496.1 Logarithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 496.2 Closed matrix sub-groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 516.3 BCHD Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

7 Imbedded Submanifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 557.1 Tangent Planes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 597.2 Lagrange Multipliers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 607.3 Tangent Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

8 Vector Bundles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 678.1 Fiber bundles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 678.2 Vector Bundles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 678.3 Sub-bundles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

9 Covariant derivatives on Vector Bundles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 739.1 Basics of covariant derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 739.2 Curvature of ∇ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 769.3 On the structure of covariant derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 789.4 Exercises on Transferring Covariant Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 799.5 Covariant derivatives on TM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 809.6 Cartan’s Rolling Map and a geometric interpretation of torsion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 809.7 Riemann’s Theorem revisited . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

10 Riemannian Manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8510.0.1 The Levi-Civita Connection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

10.1 Left Invariant Metrics on Lie Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9010.2 A General Left Invariant Ricci Curvature Computations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92

11 Riemannian curvatures on Embedded Submanifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9511.1 Induced Riemannian metrics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9511.2 Hyperbolic Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10011.3 Inversions and the upper half space model of Hyperbolic space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

12 Vector Field Commutators and Logarithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10712.1 Differentials of Flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10712.2 π – related vector fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109

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Contents 5

13 Approximate Logarithms of flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11313.1 Specialization to the Lie group case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

14 Principal Bundles and Connections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11914.1 Principal bundle basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11914.2 Curvature and Torsion considerations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12014.3 Horizontal Lifts and Curvature in Principal Bundles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12114.4 Structure Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124

15 Geometric Frobenius’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127

Part II Appendices

16 Spaces of Bounded Linear Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13116.1 Inverting Elements in L(X) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13216.2 Integral Operators as Examples of Bounded Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133

17 Contraction Mapping Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135

18 Inverse and Implicit Function Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13918.1 The Inverse Function Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13918.2 Application to Local Existence of ODE’s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14218.3 Implicit Function Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143

19 Holomorphic Functional Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147


Course Description: This will not be a topics course on only one subject.Rather we will concentrate on perhaps 4 or 5 different topics throughout thequarter. The main goal is to introduce you to some hands on computationsinvolving differential geometry while at the same time covering some topicsthat are often not covered in detail in Math 250. Some topics that we mightcover are;

1. ODE review.2. Frobenius’ integrability theorem and its relationship to curvature.3. Hypoellipiticity results at least for nilpotent lie groups.4. Densities and integration theory on manifolds and Lie groups.5. An introduction to some nilpotent Lie groups which arise in the theory of

rough path analysis and regularity structures.6. Riemannian geometry in the context of embedded submanifolds.7. The basics of Hamiltonian mechanics.8. Some calculus of Pseudo-differential operators on manifolds.

Part I

Integral Curves and Surfaces

1

Calculus on Banach Spaces

In this chapter, X and Y will be Banach spaces, U will be an open subsetof X, and L (X,Y ) will denote the continuous linear operators Λ : X → Y.

Notation 1.1 (ε, O, and o notation) Let 0 ∈ U ⊂o X, and f : U → Y be afunction. We will write:

1. f (x) = ε (x) if limx→0 ‖f (x) ‖ = 0.2. f (x) = O (x) if there are constants C < ∞ and r > 0 such that‖f (x) ‖ ≤ C‖x‖ for all x ∈ B(0, r). This is equivalent to the conditionthat lim supx→0

(‖x‖−1‖f (x) ‖

)<∞, where

lim supx→0

‖f (x) ‖‖x‖

:= limr↓0

sup‖f (x) ‖ : 0 < ‖x‖ ≤ r.

3. f (x) = o (x) if f (x) = ε (x)O (x) , i.e. limx→0 ‖f (x) ‖/‖x‖ = 0.

Example 1.2. Here are some examples of properties of these symbols.

1. A function f : U ⊂o X → Y is continuous at x0 ∈ U if f(x0 + h) =f(x0) + ε(h).

2. If f (x) = ε (x) and g (x) = ε (x) then f (x) + g (x) = ε (x) .Now let g : Y → Z be another function where Z is another Banach space.

3. If f (x) = O (x) and g (y) = o (y) then g f (x) = o (x) .4. If f (x) = ε (x) and g (y) = ε (y) then g f (x) = ε (x) .

1.1 The Differential

Definition 1.3. If f : J → X is a function we say limt→t0 f (t) = x ∈ X ifffor all ε > 0 there exists δ > 0 such that

‖f (t)− x‖ ≤ ε whenever 0 < |t− t0| ≤ δ.

A function f : J → X is said to be differentiable at t0 ∈ J iff limt→t0f(t)−f(t0)

t−t0exists in X. We denote the limit by f (t0) or f ′ (t0) or by df

dt (t0) , i.e.

f (t0) = limt→t0

f (t)− f (t0)

t− t0

Definition 1.4. A function f : U ⊂o X → Y is differentiable at x0 ∈ U ifthere exists a linear transformation Λ ∈ L (X,Y ) such that

f(x0 + h)− f(x0)− Λh = o(h). (1.1)

We denote Λ by f ′(x0) or Df(x0) if it exists. As with continuity, f is differ-entiable on U if f is differentiable at all points in U.

Remark 1.5. The linear transformation Λ in Definition 1.4 is necessarily unique.Indeed if Λ1 is another linear transformation such that Eq. (1.1) holds with Λreplaced by Λ1, then

(Λ− Λ1)h = o(h),

i.e.

lim suph→0

‖(Λ− Λ1)h‖‖h‖

= 0.

On the other hand, by definition of the operator norm,

lim suph→0

‖(Λ− Λ1)h‖‖h‖

= ‖Λ− Λ1‖.

The last two equations show that Λ = Λ1.

Exercise 1.1. Show that a function f : (a, b) → X is a differentiable at t ∈(a, b) in the sense of Definition 1.3 iff it is differentiable in the sense of Definition1.4. Also show Df (t) v = vf (t) for all v ∈ R.

Example 1.6. If T ∈ L (X,Y ) and x, h ∈ X, then

T (x+ h)− T (x)− Th = 0

which shows T ′ (x) = T for all x ∈ X.

Example 1.7. Assume that GL (X,Y ) is non-empty. Then by Corollary 16.9,GL (X,Y ) is an open subset of L (X,Y ) and the inverse map f : GL (X,Y )→GL (Y,X) , defined by f(A) := A−1, is continuous. We will now show that f isdifferentiable and

f ′(A)B = −A−1BA−1 for all B ∈ L (X,Y ) .

6 1 Calculus on Banach Spaces

This is a consequence of the identity,

f(A+H)− f(A) = (A+H)−1 (A− (A+H))A−1 = −(A+H)−1HA−1

which may be used to find the estimate,∥∥f(A+H)− f(A) +A−1HA−1∥∥ =

∥∥[A−1 − (A+H)−1]HA−1

∥∥≤∥∥A−1 − (A+H)−1

∥∥ ‖H‖ ∥∥A−1∥∥

≤ ‖A−1‖3 ‖H‖2

1− ‖A−1‖ ‖H‖= O

(‖H‖2

)wherein we have used the bound in Eq. (16.5) of Corollary 16.9 for the lastinequality.

1.2 Product and Chain Rules

The following theorem summarizes some basic properties of the differential.

Theorem 1.8. The differential D has the following properties:

1. Linearity: D is linear, i.e. D(f + λg) = Df + λDg.2. Product Rule: If f : U ⊂o X → Y and A : U ⊂o X → L(Y,Z) are

differentiable at x0 then so is x→ (Af) (x) := A (x) f (x) and

D(Af)(x0)h = (DA(x0)h)f(x0) +A(x0)Df(x0)h.

[In terms of partial derivatives introduced below the above formula may bewritten more intelligibly as;

∂h(Af)(x0) = ∂hA(x0)f(x0) +A(x0)∂hf(x0).]

3. Chain Rule: If f : U ⊂o X → V ⊂o Y is differentiable at x0 ∈ U, andg : V ⊂o Y → Z is differentiable at y0 := f(x0), then g f is differentiableat x0 and (g f)′(x0) = g′(y0)f ′(x0).

4. Converse Chain Rule: Suppose that f : U ⊂o X → V ⊂o Y is continu-ous at x0 ∈ U, g : V ⊂o Y → Z is differentiable at y0 := f(x0) and g′(y0)is invertible, and h := g f is differentiable at x0, then f is differentiableat x0 and

f ′(x0) := [g′(y0)]−1(g f)′(x0). (1.2)

5. Converse to implicit differentiation: Suppose that U ⊂o X andV ⊂o Y and f : U × V → Y is a function such that fy (x0, y0) ∈ L (Y ) isinvertible. Further suppose that Y : U → V is continuous, Y (x0) = y0, andf (x, Y (x)) = h (x) is C1 in x, then Y is C1 near x0 as well.

Proof. Linearity. Let f, g : U ⊂o X → Y be two functions which aredifferentiable at x0 ∈ U and λ ∈ R, then for ∆x ∈ X,

(f + λg)(x0 +∆x)

= f(x0) +Df(x0)∆x+ o(∆x) + λ(g(x0) +Dg(x0)∆x+ o(∆x)

= (f + λg)(x0) + (Df(x0) + λDg(x0))∆x+ o(∆x),

which implies that (f + λg) is differentiable at x0 and that

D(f + λg)(x0) = Df(x0) + λDg(x0).

Product Rule. The computation,

A(x0 +∆x)f(x0 +∆x)

= (A(x0) +DA(x0)∆x+ o(∆x))(f(x0) + f ′(x0)∆x+ o(∆x))

= A(x0)f(x0) +A(x0)f ′(x0)∆x+ [DA(x0)∆x]f(x0) + o(∆x),

verifies the product rule holds. This may also be considered as a special case ofProposition 1.10.

Chain Rule. Using f(x0 + ∆x) − f(x0) = O(∆x) (see Eq. (1.1)) ando(O(∆x)) = o(∆x),

(gf)(x0 +∆x)

= g(f(x0)) + g′(f(x0))(f(x0 +∆x)− f(x0)) + o(f(x0 +∆x)− f(x0))

= g(f(x0)) + g′(f(x0))(Df(x0)x0 + o(∆x)) + o(f(x0 +∆x)− f(x0)

= g(f(x0)) + g′(f(x0))Df(x0)∆x+ o(∆x).

Converse Chain Rule. Suppose that x0 ∈ U and g′(f (x0)) is invertible. Let∆f = f (x0 +∆x)−f (x0) and notice that ∆f = ε (∆x) because f is continuousat x0. Then on one hand,

∆h := g (f (x0) +∆f)− g (f (x0)) = g′ (f (x0))∆f + o (∆f)

while on the other hand,

∆h = h (x0 +∆x)− h (x0) = h′ (x0)∆x+ o (∆x) .

Comparing these equations implies,

h′ (x0)∆x+ o (∆x) = g′ (f (x0))∆f + o (∆f)

or equivalently,

∆f = [g′ (f (x0))]−1

[h′ (x0)∆x+ o (∆x)− o (∆f)]

= [g′ (f (x0))]−1h′ (x0)∆x+ o (∆x) + o (∆f) . (1.3)


1.2 Product and Chain Rules 7

Since f is continuous, ∆f → 0 as ∆x → 0. Therefore there exists δ > 0 suchthat ‖o (∆f)‖ ≤ 1

2 ‖∆f‖ if ‖∆x‖ ≤ δ. Using this in Eq. (1.3) implies, for someC <∞ and ‖∆x‖ ≤ δ that

‖∆f‖ =∥∥∥g′ (f (x0))

−1h′ (x0)∆x+ o (∆x) + o (∆f)

∥∥∥≤∥∥∥g′ (f (x0))

−1h′ (x0)∆x

∥∥∥+ o (∆x) +1

2‖∆f‖

≤ C ‖∆x‖+1

2‖∆f‖ ,

i.e. ‖∆f‖ ≤ 2C ‖∆x‖ for ‖∆x‖ ≤ δ. From this we may now conclude thato (∆f) = o (∆x) and therefore Eq. (1.3) asserts,

f (x0 +∆x)− f (x0) = [g′ (f (x0))]−1h′ (x0)∆x+ o (∆x) .

i.e. f ′ (x0) exists and is give by Eq. (1.2).Converse to implicit differentiation: Let

∆Y := Y (x0 +∆x)− Y (x0) = Y (x0 +∆x)− y0.

On one hand,

∆h := h (x0 +∆x)− h (x0) = h′ (x0)∆x+ o (∆x)

while on the other

∆h = f (x0 +∆x, Y (x0 +∆x))− f (x0, Y (x0))

= f (x0 +∆x, y0 +∆Y )− f (x0, Y (x0))

= fx (x0, y0)∆x+ fy (x0, y0)∆Y + o ((∆x,∆Y )) .

Solving these identities of ∆Y implies,

∆Y = fy (x0, y0)−1

[h′ (x0)∆x− fx (x0, y0)∆x+ o (∆x) + o ((∆x,∆Y ))]

= fy (x0, y0)−1

[h′ (x0)− fx (x0, y0)]∆x+ o ((∆x,∆Y )) (1.4)

Since ∆Y → 0 as ∆x→ 0, for ∆x small enough we may conclude that

‖∆Y ‖ ≤∥∥∥fy (x0, y0)

−1[h′ (x0)− fx (x0, y0)]

∥∥∥op‖∆x‖+

1

2(‖∆x‖+ ‖∆Y ‖)

from which it follows that ‖∆Y ‖ ≤ C ‖∆x‖ . Hence the term o ((∆x,∆Y )) inEq. (1.4) may be replaced by o (∆x) and we have shown,

∆Y = fy (x0, y0)−1

[h′ (x0)− fx (x0, y0)]∆x+ o (∆x) .

This shows that Y ′ (x0) exists and

Y ′ (x0) = fy (x0, y0)−1

[h′ (x0)− fx (x0, y0)] .

Corollary 1.9 (Chain Rule). Suppose that σ : (a, b) → U ⊂o X is differen-tiable at t ∈ (a, b) and f : U ⊂o X → Y is differentiable at σ (t) ∈ U. Thenf σ is differentiable at t and

d

dt(f σ) (t) = f ′(σ (t))σ (t) . (1.5)

Exercise 1.2 (First Derivative Test Again). Suppose that F : Rn → R hasa minimum at x ∈ Rn and F is differentiable at x. Show F ′ (x) = 0.

Proposition 1.10 (Product Rule II). Suppose that X := X1×· · ·×Xn witheach Xi being a Banach space and T : X1×· · ·×Xn → Y is a multilinear map,i.e.

xi ∈ Xi → T (x1, . . . , xi−1, xi, xi+1, . . . , xn) ∈ Y

is linear when x1, . . . , xi−1, xi+1, . . . , xn are held fixed. Then the following areequivalent:

1. T is continuous.2. T is continuous at 0 ∈ X.3. There exists a constant C <∞ such that

‖T (x)‖Y ≤ Cn∏i=1

‖xi‖Xi (1.6)

for all x = (x1, . . . , xn) ∈ X.4. T is differentiable at all x ∈ X1 × · · · ×Xn.

Moreover if T the differential of T is given by

T ′ (x)h =

n∑i=1

T (x1, . . . , xi−1, hi, xi+1, . . . , xn) (1.7)

where h = (h1, . . . , hn) ∈ X.

Proof. Let us equip X with the norm

‖x‖X := max‖xi‖Xi

.

If T is continuous then T is continuous at 0. If T is continuous at 0, usingT (0) = 0, there exists a δ > 0 such that ‖T (x)‖Y ≤ 1 whenever ‖x‖X ≤ δ. Now

if x ∈ X is arbitrary, let x′ := δ(‖x1‖−1

X1x1, . . . , ‖xn‖−1

Xnxn

). Then ‖x′‖X ≤ δ

and hence ∥∥∥∥∥(δn

n∏i=1

‖xi‖−1Xi

)T (x1, . . . , xn)

∥∥∥∥∥Y

= ‖T (x′)‖ ≤ 1



from which Eq. (1.6) follows with C = δ−n.Now suppose that Eq. (1.6) holds. For x, h ∈ X and ε ∈ 0, 1n let |ε| =∑ni=1 εi and

xε (h) := ((1− ε1)x1 + ε1h1, . . . , (1− εn)xn + εnhn) ∈ X.

By the multi-linearity of T,

T (x+ h) = T (x1 + h1, . . . , xn + hn) =∑

ε∈0,1nT (xε (h))

= T (x) +

n∑i=1

T (x1, . . . , xi−1, hi, xi+1, . . . , xn)

+∑

ε∈0,1n:|ε|≥2

T (xε (h)) . (1.8)

From Eq. (1.6), ∥∥∥∥∥∥∑

ε∈0,1n:|ε|≥2

T (xε (h))

∥∥∥∥∥∥ = O(‖h‖2

),

and so it follows from Eq. (1.8) that T ′ (x) exists and is given by Eq. (1.7).This completes the proof since it is trivial to check that T being differentiableat x ∈ X implies continuity of T at x ∈ X.

Definition 1.11. Let X and Y be Banach spaces and let L1(X,Y ) :=L (X,Y ) and for k ≥ 2 let Lk(X,Y ) be defined inductively by Lk+1(X,Y ) =L(X,Lk(X,Y )). For example L2(X,Y ) = L(X,L (X,Y )) and L3(X,Y ) =L (X,L(X,L (X,Y ))) .

Suppose f : U ⊂o X → Y is a function. If f is differentiable on U, then itmakes sense to ask if f ′ = Df : U → L (X,Y ) = L1(X,Y ) is differentiable. IfDf is differentiable on U then f ′′ = D2f := DDf : U → L2(X,Y ). Similarlywe define f (n) = Dnf : U → Ln(X,Y ) inductively.

Definition 1.12. Given k ∈ N, let Ck (U, Y ) denote those functions f : U →Y such that f (j) := Djf : U → Lj (X,Y ) exists and is continuous for j =1, 2, . . . , k.

Example 1.13. Let us continue on with Example 1.7 but now let X = Y tosimplify the notation. So f : GL(X)→ GL(X) is the map f(A) = A−1 and

f ′(A) = −LA−1RA−1 , i.e. f ′ = −LfRf .

where LAB = AB and RAB = BA for all A,B ∈ L(X). As the reader mayeasily check, the maps

A ∈ L(X)→ LA, RA ∈ L(L(X))

are linear and bounded. So by the chain and the product rule we find f ′′(A)exists for all A ∈ L(X) and

f ′′(A)B = −Lf ′(A)BRf − LfRf ′(A)B .

More explicitly

[f ′′(A)B]C = A−1BA−1CA−1 +A−1CA−1BA−1. (1.9)

Working inductively one shows f : GL(X) → GL(X) defined by f(A) := A−1

is C∞.

1.3 Partial Derivatives

Definition 1.14 (Partial or Directional Derivative). Let f : U ⊂o X → Ybe a function, x0 ∈ U, and v ∈ X. We say that f is differentiable at x0 inthe direction v iff d

dt |0(f(x0 + tv)) =: (∂vf)(x0) exists. We call (∂vf)(x0) thedirectional or partial derivative of f at x0 in the direction v.

Remark 1.15 (Another form of the chain rule). Notice that if f is differentiableat x0, then

∂vf(x0) =d

dt|0f (x0 + tv) = f ′ (x0) v

exists by the chain rule, see Corollary 1.9. More generally if c (t) is a curve inX such that c (0) = x0 and c (0) exists, then

d

dt|0f (c (t)) = f ′ (x0) c (0) =

(∂c(0)f

)(x0) .

Proposition 1.16. Let f : U ⊂o X → Y be a continuous function. We nowassume that ∂vf (x) := d

dt |0f (x+ tv) exists for x ∈ U and v ∈ X and thereexists a continuous function A : U → L(X,Y ) such that ∂vf (x) = A (x) v forall v ∈ X and x ∈ U. Then f ∈ C1(U, Y ) and Df = f ′ = A.

Proof. First observe that for x ∈ U and v ∈ X sufficiently small that

d

dtf (x+ tv) = ∂vf (x+ tv) = A (x+ tv) v

exists and is continuous in t. Therefore by the fundamental theorem of calculus,


1.3 Partial Derivatives 9

f(x+ v)− f (x) =

∫ 1

0

df(x+ tv)

dtdt

=

∫ 1

0

(∂vf)(x+ tv) dt =

∫ 1

0

A(x+ tv) v dt.

Therefore,

f(x+ v)− f (x)−A (x) v =

∫ 1

0

[A(x+ tv)−A (x)] v dt

and we have the estimate,

‖f(x+ v)− f (x)−A (x) v‖ ≤∫ 1

0

‖A(x+ tv)−A (x) v ‖ dt ≤ ‖v‖ ε (v) .

where

ε (v) :=

∫ 1

0

‖A(x+ tv)−A (x)‖ dt ≤ sup‖h‖≤‖v‖

‖A(x+ h)−A (x)‖ .

It now now follows from the continuity of A that limv→0 ε (v) = 0 and therefore,

‖f(x+ v)− f (x)−A (x) v‖ = o (v)

and so f ′ (x) exists and f ′ (x) = A (x) .

Corollary 1.17. Suppose now that X = Rd, f : U ⊂o X → Y be a continuousfunction such that ∂if (x) := ∂eif (x) exists and is continuous on U for i =

1, 2, . . . , d, where eidi=1 is the standard basis for Rd. Then f ∈ C1(U, Y ) andDf (x) ei = ∂if (x) for all i.

Proof. For x ∈ U, let A (x) : Rd → Y be the unique linear map suchthat A (x) ei = ∂if (x) for i = 1, 2, . . . , d. Then A : U → L(Rd, Y ) is a con-tinuous map (why?). Now let v ∈ Rd and v(i) := (v1, v2, . . . , vi, 0, . . . , 0) fori = 1, 2, . . . , d and v(0) := 0. Then for t ∈ R near 0, using the fundamentaltheorem of calculus and the definition of ∂if (x) ,

f (x+ tv)− f (x) =

d∑i=1

[f(x+ tv(i)

)− f

(x+ tv(i−1)

)]=

d∑i=1

∫ 1

0

d

dsf(x+ tv(i−1) + stviei

)ds

=

d∑i=1

tvi

∫ 1

0

∂if(x+ tv(i−1) + stviei

)ds

=

d∑i=1

tvi

∫ 1

0

A(x+ tv(i−1) + stviei

)eids.

Using the continuity of A, it now follows that

limt→0

f (x+ tv)− f (x)

t=

d∑i=1

vi limt→0

∫ 1

0

A(x+ tv(i−1) + stviei

)eids

=

d∑i=1

vi

∫ 1

0

A (x) eids = A (x) v

which shows ∂vf (x) exists and ∂vf (x) = A (x) v. The result now follows froman application of Proposition 1.16.

Here is a variant of the above proof which bypasses the use ofProposition 1.16. For v ∈ Rd, let v(i) := (v1, v2, . . . , vi, 0, . . . , 0) for i =1, 2, . . . , d and v(0) := 0 and let σv (s) be a path from x to x+ v such that

σv (s) = x+ v(i−1) + (d · s− (i− 1)) viei fori− 1

d≤ s ≤ i

d,

see Figure 1.1.Then for v sufficiently small,

x1

xx+ v(1)

x+ v(2)

x+ v

x2

x3

Fig. 1.1. Traversing from x to x+ v along a path with segments always parallel to acoordinate axis.



f (x+ v) = f (x) +

∫ 1

0

d

dsf (σv (s)) ds

= f (x) +

d∑i=1

∫ id

i−1d

d

dsf (σv (s)) ds

= f (x) +

d∑i=1

∫ id

i−1d

A (σv (s))σ′v (s) ds

= f (x) +

∫ 1

0

A (σv (s))σ′v (s) ds

wherein we have used σ′v (s) = dviei and hence

d

dsf (σv (s)) =

(∂σ′v(s)f

)(σv (s)) = A (σ (s))σ′v (s)

for i−1d ≤ s ≤

id . Hence we conclude

‖f (x+ v)− f (x)−A (x) v‖ =

∥∥∥∥∫ 1

0

A (σv (s))σ′v (s) ds−A (x) v

∥∥∥∥=

∥∥∥∥∫ 1

0

[A (σv (s))−A (x)]σ′v (s) ds

∥∥∥∥≤∫ 1

0

‖A (σv (s))−A (x)‖ ‖σ′v (s)‖ ds

≤ ε (v) · d ·maxi|vi| = o (v)

where

ε (v) :=

∫ 1

0

‖A (σv (s))−A (x)‖ ds→ 0 as v → 0.

Exercise 1.3 (Rudin #6, page 239). Let

f (x, y) =

xyx2+y2 if (x, y) 6= (0, 0)

0 if (x, y) = (0, 0) .

Show ∂f∂x (x, y) and ∂f

∂y (x, y) exists at all (x, y) ∈ R2 even though f is discon-

tinuous at (0, 0) .

Notation 1.18 (Jacobian matrix) Suppose that f : Rm → Rn is differen-tiable at x ∈ Rm then f ′ (x) is a linear map from Rm to Rn which may berepresented by a m×n matrix, Jf (x) often referred to as the Jacobian matrix.That is

Jf (x) = [f ′ (x) e1| . . . |f ′ (x) em]

= [(∂e1f) (x) | . . . | (∂emf) (x)]

=

[∂f

∂x1(x) | . . . | ∂f

∂xm(x)

]where eimi=1 is the standard basis for Rm. [I will typically just identify f ′ (x)with Jf (x) in this case.]

Example 1.19. If f : R3 → R2 is defined by

f (x) =

[x1x2 + ex3

sin (x2x3)

]where x =

x1

x2

x3

,then

∂f

∂x1(x) =

∂

∂x1

[x1x2 + ex3

sin (x2x3)

]=

[x2

0

]∂f

∂x2(x) =

∂

∂x2

[x1x2 + ex3

sin (x2x3)

]=

[x1

x3 cos (x2x3)

]∂f

∂x3(x) =

∂

∂x3

[x1x2 + ex3

sin (x2x3)

]=

[ex3

x2 cos (x2x3)

]so that

Jf (x) =

[x2 x1 ex3

0 x3 cos (x2x3) x2 cos (x2x3)

].

Example 1.20. Let T (r, θ) := (r cos θ, r sin θ)tr

for (r, θ) ∈ R2. In this case,

∂T

∂r(r, θ) =

∂

∂r

[r cos θr sin θ

]=

[cos θsin θ

]and

∂T

∂θ(r, θ) =

∂

∂θ

[r cos θr sin θ

]=

[−r sin θr cos θ

]so that

JT (r, θ) =[∂T∂r (r, θ) ∂T

∂θ (r, θ)]

=

[cos θ −r sin θsin θ r cos θ

].

We have computed the differential of T as this matrix clearly depends contin-uously on (r, θ) . For later purposes let us note that

det JT (r, θ) = r(cos2 θ + sin2 θ

)= r.

Exercise 1.4. Compute Jf (x) where f : R2 → R3 is the define by

f (x) =

ex1

sin(x1 + x2

2

)ln(1 + x2

2

) where x =

[x1

x2

].


1.3 Partial Derivatives 11

Corollary 1.21 (Chain Rule). Suppose σ : (a, b)→ U ⊂o Rd is differentiableat t ∈ (a, b) and f : U → Y is differentiable at σ (t) ∈ U. Then

d

dtf (σ1 (t) , . . . , σd (t)) =

d∑i=1

(∂eif) (σ (t))σi (t) . (1.10)

where eidi=1 is the standard basis for Rd and σi (t) = σ (t) · ei.

Proof. It only remains to prove Eq. (1.10) which follows by using σ (t) =∑di=1 σi (t) ei in Eq. (1.5);

d

dt(fσ) (t) = f ′(σ (t))

d∑i=1

σi (t) ei =

d∑i=1

σi (t) f ′(σ (t))ei =

d∑i=1

(∂eif) (σ (t))σi (t) .

Remark 1.22. When doing explicit computation it is often useful to mentallyrewrite Eq. (1.10) as,

d

dtf (σ1 (t) , . . . , σd (t)) =

d∑i=1

d

ds|0f (σ1 (t) , . . . σi (t+ s) , . . . , σd (t)) .

In words when differentiating a function where t appears in the function inmultiple places, we hold but one of these t’s fixed and differentiate with respectto the t not fixed and then sum these results over all possible ways of doingthis.

Example 1.23. Suppose that f (t) = exp (cos t+ t sin (et)) , then

f (t) =d

ds|0[

exp (cos (t+ s) + t sin (et)) + exp (cos t+ (t+ s) sin (et))+ exp (cos t+ t sin (et+s))

]= f (t) ·

[− sin (t) + sin

(et)

+ t cos(et)· et].

Exercise 1.5 (Not to be collected). Suppose that f : R2 → R2 and g :R2 → R3 are differentiable functions which we write as

f (x) = (f1 (x1, x2) , f2 (x1, x2))tr

and g (y) = (g1 (y1, y2) , g2 (y1, y2) , g3 (y1, y2))tr.

Let h = g f so that

h (x) = g (f (x))

= (g1 (f1 (x1, x2) , f2 (x1, x2)) , g2 (f1 (x1, x2) , f2 (x1, x2)) , g3 (f1 (x1, x2) , f2 (x1, x2)))tr.

Then on one hand

Jh (x) =

∂h1

∂x1(x) ∂h1

∂x2(x)

∂h2

∂x1(x) ∂h2

∂x2(x)

∂h3

∂x1(x) ∂h3

∂x2(x)

.Write out Jf (x) and Jg (y) and then use the chain rule to show

Jh (x) =

∂g1∂y1

(y) ∂f1∂x1(x) + ∂g1

∂y2(y) ∂f2∂x1

(x) ∂g1∂y1

(y) ∂f1∂x2(x) + ∂g1

∂y2(y) ∂f2∂x2

(x)∂g2∂y1

(y) ∂f1∂x1(x) + ∂g2

∂y2(y) ∂f2∂x1

(x) ∂g2∂y1

(y) ∂f1∂x2(x) + ∂g2

∂y2(y) ∂f2∂x2

(x)∂g3∂y1

(y) ∂f1∂x1(x) + ∂g3

∂y2(y) ∂f2∂x1

(x) ∂g3∂y1

(y) ∂f1∂x2(x) + ∂g3

∂y2(y) ∂f2∂x2

(x)

with the understanding that x = (x1, x2) and y = f (x) in this formula. So forexample

∂h2

∂x1(x) =

∂g2

∂y1(y)

∂f1

∂x1(x) +

∂g2

∂y2(y)

∂f2

∂x1(x) .

Remark 1.24. In general if h (x) = g (f (x)) where f (x) = (f1 (x) , . . . , fn (x))with x = (x1, . . . , xm) , then the chain rule may be written as

∂h

∂xi(x) =

∑k

∂g

∂yk(f (x))

∂fk (x)

∂xi=∑k

∂g

∂yk(y)

∂fk (x)

∂xi

provided f is differentiable at x and g is differentiable at y := f (x) .

Exercise 1.6. Let det : L (Rn) → R be the determinant function on n × nmatrices and for A ∈ L (Rn) we will let Ai denote the ith – column of A andwrite A = (A1|A2| . . . |An) . [Recall that det (A) is a polynomial in the matrixentries and hence is a smooth function so in order to compute the differentialof det it suffices to compute directional derivatives.]

1. Show det′ (A) exists for all A ∈ L (Rn) and

det ′ (A)H =

n∑i=1

det (A1| . . . |Ai−1|Hi|Ai+1| . . . |An) (1.11)

for all H ∈ L (Rn) . Hint: recall that det (A) is a multilinear function of itscolumns.

2. Use Eq. (1.11) along with basic properties of the determinant to showdet′ (I)H = tr(H).

3. Suppose now that A ∈ GL (Rn) , show

det ′ (A)H = det (A) tr(A−1H).

Hint: Notice that det (A+H) = det (A) det(I +A−1H

).



4. If A ∈ L (Rn) , show det(eA)

= etr(A). Hint: use the previous item andCorollary 1.9 to show

d

dtdet(etA)

= det(etA)

tr(A).

Exercise 1.7. Let f be a function on an open neighborhood of (0, 0) ∈ R2 withvalued in a Banach space X. Suppose that f is continuous, fs and ft exists andare continuous and fs,t and ft,s exist and are continuous. For (s, t) near (0, 0) ,let

ε (s, t) := max|σ|≤|s|,|τ |≤t

‖fs,t (σ, τ)− fs,t (0, 0)‖X+ max|σ|≤|s|,|τ |≤t

‖ft,s (σ, τ)− ft,s (0, 0)‖X

and then show;

1.∫ s

0dσ∫ t

0dτfs,t (σ, τ) = f (s, t)− f (s, 0)− f (0, t) + f (0, 0) and∥∥∥∥∫ t

0

dτ

∫ s

0

dσfs,t (σ, τ)− fs,t (0, 0) st

∥∥∥∥X

≤ |st| ε (s, t) .

2.∫ t

0dτ∫ s

0dσft,s (σ, τ) = f (s, t)− f (s, 0)− f (0, t) + f (0, 0) , and∥∥∥∥∫ t

0

dτ

∫ s

0

dσft,s (σ, τ)− ft,s (0, 0) st

∥∥∥∥X

≤ |st| ε (s, t) .

3. Use these results to conclude;

fs,t (0, 0) = ft,s (0, 0) .

Exercise 1.8. Let X and Y be Banach spaces, U be an open subset of X, andv, w ∈ X. Suppose that F : U → Y is a continuous function such that ∂vF and∂wF exist and are continuous on U. Further suppose that ∂w∂vF and ∂v∂wFexist and are continuous on U. Show ∂w∂vF = ∂v∂wF on U. Hint: given x ∈ Ulet f (s, t) := F (x+ sv + tw) for (s, t) near (0, 0) ∈ R2. Now show that Exercise1.7 may be applied to f.

1.4 Higher Order Differentials*

It is somewhat inconvenient to work with the Banach spaces Lk(X,Y ) in Def-inition 1.11. For this reason we will introduce an isomorphic Banach space,Mk(X,Y ).

Definition 1.25. For k ∈ 1, 2, 3, . . ., let Mk(X,Y ) denote the set of functionsf : Xk → Y such that

1. For i ∈ 1, 2, . . . , k, v ∈ X → f〈v1, v2, . . . , vi−1, v, vi+1, . . . , vk〉 ∈ Y islinear 1 for all vini=1 ⊂ X.

2. The norm ‖f‖Mk(X,Y ) should be finite, where

‖f‖Mk(X,Y ) := sup‖f〈v1, v2, . . . , vk〉‖Y‖v1‖‖v2‖ · · · ‖vk‖

: viki=1 ⊂ X \ 0.

Lemma 1.26. There are linear operators jk : Lk(X,Y ) → Mk(X,Y ) definedinductively as follows: j1 = IdL(X,Y ) (notice that M1(X,Y ) = L1(X,Y ) =L (X,Y )) and

(jk+1A)〈v0, v1, . . . , vk〉 = (jk(Av0))〈v1, v2, . . . , vk〉 ∀vi ∈ X.

(Notice that Av0 ∈ Lk(X,Y ).) Moreover, the maps jk are isometric isomor-phisms.

Proof. To get a feeling for what jk is let us write out j2 and j3 explicitly.If A ∈ L2(X,Y ) = L(X,L (X,Y )), then (j2A)〈v1, v2〉 = (Av1)v2 and if A ∈L3(X,Y ) = L(X,L(X,L (X,Y ))), (j3A)〈v1, v2, v3〉 = ((Av1)v2)v3 for all vi ∈X. It is easily checked that jk is linear for all k. We will now show by inductionthat jk is an isometry and in particular that jk is injective. Clearly this is trueif k = 1 since j1 is the identity map. For A ∈ Lk+1(X,Y ),

‖jk+1A‖Mk+1(X,Y )

:= sup‖(jk(Av0))〈v1, v2, . . . , vk〉‖Y‖v0‖‖v1‖‖v2‖ · · · ‖vk‖

: viki=0 ⊂ X \ 0

= sup‖(jk(Av0))‖Mk(X,Y )

‖v0‖: v0 ∈ X \ 0

= sup‖Av0‖Lk(X,Y )

‖v0‖: v0 ∈ X \ 0

= ‖A‖L(X,Lk(X,Y )) := ‖A‖Lk+1(X,Y ),

wherein the second to last inequality we have used the induction hypothesis.This shows that jk+1 is an isometry provided jk is an isometry. To finish theproof it suffices to show that jk is surjective for all k. Again this is true fork = 1. Suppose that jk is invertible for some k ≥ 1. Given f ∈Mk+1(X,Y ) wemust produce A ∈ Lk+1(X,Y ) = L(X,Lk(X,Y )) such that jk+1A = f. If suchan equation is to hold, then for v0 ∈ X, we would have jk(Av0) = f〈v0, · · · 〉.That is Av0 = j−1

k (f〈v0, · · · 〉). It is easily checked that A so defined is linear,bounded, and jk+1A = f.

From now on we will identify Lk with Mk without further mention. Inparticular, we will view Dkf as function on U with values in Mk(X,Y ).

1 I will routinely write f〈v1, v2, . . . , vk〉 rather than f(v1, v2, . . . , vk) when the func-tion f depends on each of variables linearly, i.e. f is a multi-linear function.


1.4 Higher Order Differentials* 13

Theorem 1.27 (Differentiability). Suppose k ∈ 1, 2, . . . and D is a densesubspace of X, f : U ⊂o X → Y is a function such that (∂v1∂v2 · · · ∂vlf) (x)exists for all x ∈ D ∩ U, vili=1 ⊂ D, and l = 1, 2, . . . k. Further assumethere exists continuous functions Al : U ⊂o X →Ml(X,Y ) such that such that(∂v1∂v2 · · · ∂vlf) (x) = Al (x) 〈v1, v2, . . . , vl〉 for all x ∈ D ∩ U, vili=1 ⊂ D,and l = 1, 2, . . . k. Then Dlf (x) exists and is equal to Al (x) for all x ∈ U andl = 1, 2, . . . , k.

Proof. We will prove the theorem by induction on k. We have alreadyproved the theorem when k = 1, see Proposition 1.16. Now suppose that k > 1and that the statement of the theorem holds when k is replaced by k−1. Hencewe know that Dlf (x) = Al (x) for all x ∈ U and l = 1, 2, . . . , k− 1. We are alsogiven that

(∂v1∂v2 · · · ∂vkf) (x) = Ak (x) 〈v1, v2, . . . , vk〉 ∀x ∈ U ∩D, vi ⊂ D. (1.12)

Now we may write (∂v2 · · · ∂vkf) (x) as (Dk−1f) (x) 〈v2, v3, . . . , vk〉 so that Eq.(1.12) may be written as

∂v1(Dk−1f) (x) 〈v2, v3, . . . , vk〉)= Ak (x) 〈v1, v2, . . . , vk〉 ∀x ∈ U ∩D, vi ⊂ D. (1.13)

So by the fundamental theorem of calculus, we have that

((Dk−1f)(x+ v1)− (Dk−1f) (x))〈v2, v3, . . . , vk〉

=

∫ 1

0

Ak(x+ tv1)〈v1, v2, . . . , vk〉 dt (1.14)

for all x ∈ U∩D and vi ⊂ D with v1 sufficiently small. By the same argumentgiven in the proof of Proposition 1.16, Eq. (1.14) remains valid for all x ∈U and vi ⊂ X with v1 sufficiently small. We may write this last equationalternatively as,

(Dk−1f)(x+ v1)− (Dk−1f) (x) =

∫ 1

0

Ak(x+ tv1)〈v1, · · · 〉 dt. (1.15)

Hence

(Dk−1f)(x+ v1)− (Dk−1f) (x)−Ak (x) 〈v1, · · · 〉

=

∫ 1

0

[Ak(x+ tv1)−Ak (x)]〈v1, · · · 〉 dt

from which we get the estimate,

‖(Dk−1f)(x+ v1)− (Dk−1f) (x)−Ak (x) 〈v1, · · · 〉‖ ≤ ε(v1)‖v1‖ (1.16)

where ε(v1) :=∫ 1

0‖Ak(x+tv1)−Ak (x) ‖ dt. Notice by the continuity of Ak that

ε(v1)→ 0 as v1 → 0. Thus it follow from Eq. (1.16) that Dk−1f is differentiableand that (Dkf) (x) = Ak (x) .

Example 1.28. Let f : GL (X,Y )→ GL (Y,X) be defined by f(A) := A−1. Weassume that GL (X,Y ) is not empty. Then f is infinitely differentiable and

(Dkf)(A)〈V1, V2, . . . , Vk〉

= (−1)k∑σ

B−1Vσ(1)B−1Vσ(2)B

−1 · · ·B−1Vσ(k)B−1, (1.17)

where sum is over all permutations of σ of 1, 2, . . . , k.

Let me check Eq. (1.17) in the case that k = 2. Notice that we have alreadyshown that (∂V1

f)(B) = Df(B)V1 = −B−1V1B−1. Using the product rule we

find that

(∂V2∂V1

f)(B) = B−1V2B−1V1B

−1 +B−1V1B−1V2B

−1 =: A2(B)〈V1, V2〉.

Notice that ‖A2(B)〈V1, V2〉‖ ≤ 2‖B−1‖3‖V1‖ · ‖V2‖, so that ‖A2(B)‖ ≤2‖B−1‖3 <∞. Hence A2 : GL (X,Y )→M2(L (X,Y ) , L (Y,X)). Also

‖(A2(B)−A2(C))〈V1, V2〉‖ ≤ 2‖B−1V2B−1V1B

−1 − C−1V2C−1V1C

−1‖≤ 2‖B−1V2B

−1V1B−1 −B−1V2B

−1V1C−1‖

+ 2‖B−1V2B−1V1C

−1 −B−1V2C−1V1C

−1‖+ 2‖B−1V2C

−1V1C−1 − C−1V2C

−1V1C−1‖

≤ 2‖B−1‖2‖V2‖‖V1‖‖B−1 − C−1‖+ 2‖B−1‖‖C−1‖‖V2‖‖V1‖‖B−1 − C−1‖

+ 2‖C−1‖2‖V2‖‖V1‖‖B−1 − C−1‖.

This shows that

‖A2(B)−A2(C)‖ ≤ 2‖B−1 − C−1‖‖B−1‖2 + ‖B−1‖‖C−1‖+ ‖C−1‖2.

Since B → B−1 is differentiable and hence continuous, it follows that A2(B) isalso continuous in B. Hence by Theorem 1.27 D2f(A) exists and is given as inEq. (1.17)

Example 1.29. Suppose that f : R → R is a C∞– function and F (x) :=∫ 1

0f(x (t)) dt for x ∈ X := C([0, 1],R) equipped with the norm ‖x‖ :=

maxt∈[0,1] |x (t) |. Then F : X → R is also infinitely differentiable and

(DkF ) (x) 〈v1, v2, . . . , vk〉 =

∫ 1

0

f (k)(x (t))v1 (t) · · · vk (t) dt, (1.18)

for all x ∈ X and vi ⊂ X.



To verify this example, notice that

(∂vF ) (x) :=d

ds|0F (x+ sv) =

d

ds|0∫ 1

0

f(x (t) + sv (t)) dt

=

∫ 1

0

d

ds|0f(x (t) + sv (t)) dt =

∫ 1

0

f ′(x (t))v (t) dt.

Similar computations show that

(∂v1∂v2 · · · ∂vkf) (x) =

∫ 1

0

f (k)(x (t))v1 (t) · · · vk (t) dt =: Ak (x) 〈v1, v2, . . . , vk〉.

Now for x, y ∈ X,

|Ak (x)〈v1, v2, . . . , vk〉 −Ak (y) 〈v1, v2, . . . , vk〉|

≤∫ 1

0

|f (k)(x (t))− f (k)(y (t))| · |v1 (t) · · · vk (t) |dt

≤k∏i=1

‖vi‖∫ 1

0

|f (k)(x (t))− f (k)(y (t))|dt,

which shows that

‖Ak (x)−Ak (y) ‖ ≤∫ 1

0

|f (k)(x (t))− f (k)(y (t))|dt.

This last expression is easily seen to go to zero as y → x in X. Hence Ak iscontinuous. Thus we may apply Theorem 1.27 to conclude that Eq. (1.18) isvalid.

1.5 Exercises

Definition 1.30 (Vector Fields). Let Ω be an open subset of Rd. A function,Z : Ω → Rd, is called a vector field on Ω. We will typically assume that Z isa smooth function, i.e. partial derivatives exist and are continuous to all orders.

Definition 1.31 (Divergence). Let Z be a smooth vector field on Ω. The di-vergence of Z is the smooth function, div (Z) : Ω → R, define by div (Z) (x) :=tr (Z ′ (x)) for all x ∈ Ω.

Remark 1.32. If eidi=1 is the standard basis for Rd, then

div (Z) (x) = tr (Z ′ (x)) =

d∑i=1

Z ′ (x) ei · ei

=

d∑i=1

∂eiZ (x) · ei =

d∑i=1

∂

∂xiZi (x) .

This last expression is often written as, (∇ · Z) (x) where one lets ∇ denote the

list of differential operators, (∂1, . . . , ∂d) =(

∂∂x1

, . . . , ∂∂xd

).

Exercise 1.9. Let Z : Ω → Rd be a smooth vector field and for x ∈ Ω, letϕt (x) denote the solution to the ordinary differential equation,

d

dtϕt (x) = Z (ϕt (x)) with ϕ0 (x) = x ∈ Ω.

Fact: There exists is a unique solution to this equation defined (t, x) in aneighborhood of 0 × Ω ⊂ R × Rd. Moreover the function (t, x) → ϕt (x) isjointly smooth on this neighborhood.

Show for t near zero that

d

dtdet [ϕ′t (x)] = det [ϕ′t (x)] (∇ · Z) (ϕt (x)) .

Hints: use Exercise 1.6 and the fact that mixed partial derivatives commutealong with the chain rule to verify,

d

dtϕ′t (x) v = Z ′ (ϕt (x))ϕ′t (x) v for v ∈ Rd.

Exercise 1.10. Let ψ : Ω → Rd be a smooth map such that ψ′ (x)−1

exists forall x ∈ Ω and define

Z (x) = det (ψ′ (x))ψ′ (x)−1Z (ψ (x))

where Z is a vector field on Rd. Show, using the outline below, that

∇ · Z = det (ψ′) (∇ · Z) (ψ) . (1.19)

Let A (x) := ψ′ (x) .

1. Using the product rule write 1det(A)∇ · Z as a sum of three terms.

2. One of the terms you get will be the right side of Eq. (1.19).3. You now need to show that the other two terms cancel and this should boil

down to showing for every v ∈ Rd that

d∑i=1

tr(A−1∂iA

)v · ei =

d∑i=1

[A−1 (∂iA) v

]· ei (1.20)


To prove this last identity show each side of the equation is equal totr(A−1∂vA

). In working with the right you might first prove that

(∂iA) (x) v = (∂vA) (x) ei. (1.21)

Exercise 1.11 (Curl of Z). Suppose that d = 3 and Z : Ω → R3 is a vectorfield and recall that the curl of Z is defined by;

∇× Z =

∣∣∣∣∣∣e1 e2 e3

∂1 ∂2 ∂3

Z1 Z2 Z3

∣∣∣∣∣∣= (∂2Z3 − ∂3Z2) e1 − (∂1Z3 − ∂3Z1) e2 + (∂1Z2 − ∂2Z1) e3

=

∂2Z3 − ∂3Z2

∂3Z1 − ∂1Z3

∂1Z2 − ∂2Z1

.Show that for all v ∈ R3 we have

(∇× Z)× v =(Z ′ − [Z ′]

tr)v. (1.22)

Hence ∇× Z contains the information about the skew symmetric part of Z ′.

2

Linear Ordinary Differential Equations

Let X be a Banach space, J = (a, b) ⊂ R be an open interval with 0 ∈ J,h ∈ C(J,X) and A ∈ C(J, L(X)). In this section we are going to consider theordinary differential equation,

y (t) = A (t) y (t) + h (t) and y(0) = x ∈ X, (2.1)

where y is an unknown function in C1(J,X). This equation may be written inits equivalent (as the reader should verify) integral form, namely we are lookingfor y ∈ C(J,X) such that

y (t) = x+

∫ t

0

h (τ) dτ +

∫ t

0

A(τ)y(τ)dτ. (2.2)

In what follows, we will abuse notation and use ‖·‖ to denote the operator normon L (X) associated to the norm, ‖·‖ , on X and let ‖ϕ‖∞ := maxt∈J ‖ϕ (t)‖for ϕ ∈ BC(J,X) or BC(J, L (X)).

Notation 2.1 For t ∈ R and n ∈ N, let

∆n (t) =

(τ1, . . . , τn) ∈ Rn : 0 ≤ τ1 ≤ · · · ≤ τn ≤ t if t ≥ 0(τ1, . . . , τn) ∈ Rn : t ≤ τn ≤ · · · ≤ τ1 ≤ 0 if t ≤ 0

and also write dτ = dτ1 . . . dτn and∫∆n(t)

f(τ1, . . . , τn)dτ : = (−1)n·1t<0

∫ t

0

dτn

∫ τn

0

dτn−1 . . .

∫ τ2

0

dτ1f(τ1, . . . , τn)

where

1t<0 =

1 if t < 00 if t ≥ 0

.

Lemma 2.2. Suppose that ψ ∈ C (R,R) , then

(−1)n·1t<0

∫∆n(t)

ψ(τ1) . . . ψ(τn)dτ =1

n!

(∫ t

0

ψ(τ)dτ

)n. (2.3)

Proof. Let Ψ (t) :=∫ t

0ψ(τ)dτ. The proof will go by induction on n. The

case n = 1 is easily verified since

(−1)1·1t<0

∫∆1(t)

ψ(τ1)dτ1 =

∫ t

0

ψ(τ)dτ = Ψ (t) .

Now assume the truth of Eq. (2.3) for n− 1 for some n ≥ 2, then

(−1)n·1t<0

∫∆n(t)

ψ(τ1) . . . ψ(τn)dτ

=

∫ t

0

dτn

∫ τn

0

dτn−1 . . .

∫ τ2

0

dτ1ψ(τ1) . . . ψ(τn)

=

∫ t

0

dτnΨn−1(τn)

(n− 1)!ψ(τn) =

∫ t

0

dτnΨn−1(τn)

(n− 1)!Ψ(τn)

=

∫ Ψ(t)

0

un−1

(n− 1)!du =

Ψn (t)

n!,

wherein we made the change of variables, u = Ψ(τn), in the second to lastequality.

Remark 2.3. Eq. (2.3) is equivalent to∫∆n(t)

ψ(τ1) . . . ψ(τn)dτ =1

n!

(∫∆1(t)

ψ(τ)dτ

)n.

and another way to understand this equality is to view∫∆n(t)

ψ(τ1) . . . ψ(τn)dτ

as a multiple integral rather than an iterated integral. Indeed, taking t > 0 forsimplicity and letting Sn be the permutation group on 1, 2, . . . , n we have

[0, t]n = ∪σ∈Sn(τ1, . . . , τn) ∈ Rn : 0 ≤ τσ1 ≤ · · · ≤ τσn ≤ t

with the union being “essentially” disjoint. Therefore, making a change of vari-ables and using the fact that ψ(τ1) . . . ψ(τn) is invariant under permutations,we find

18 2 Linear Ordinary Differential Equations(∫ t

0

ψ(τ)dτ

)n=

∫[0,t]n

ψ(τ1) . . . ψ(τn)dτ

=∑σ∈Sn

∫(τ1,...,τn)∈Rn:0≤τσ1≤···≤τσn≤t

ψ(τ1) . . . ψ(τn)dτ

=∑σ∈Sn

∫(s1,...,sn)∈Rn:0≤s1≤···≤sn≤t

ψ(sσ−11) . . . ψ(sσ−1n)ds

=∑σ∈Sn

∫(s1,...,sn)∈Rn:0≤s1≤···≤sn≤t

ψ(s1) . . . ψ(sn)ds

= n!

∫∆n(t)

ψ(τ1) . . . ψ(τn)dτ.

Theorem 2.4. Let ϕ ∈ BC(J,X), then the integral equation

y (t) = ϕ (t) +

∫ t

0

A(τ)y(τ)dτ (2.4)

has a unique solution given by

y (t) = ϕ (t) +

∞∑n=1

(−1)n·1t<0

∫∆n(t)

A(τn) . . . A(τ1)ϕ(τ1)dτ (2.5)

and this solution satisfies the bound

‖y‖∞ ≤ ‖ϕ‖∞ e

∫J‖A(τ)‖dτ

.

Proof. Define Λ : BC(J,X)→ BC(J,X) by

(Λy) (t) =

∫ t

0

A(τ)y(τ)dτ.

Then y solves Eq. (2.2) iff y = ϕ + Λy or equivalently iff (I − Λ)y = ϕ. Aninduction argument shows

(Λnϕ) (t) =

∫ t

0

dτnA(τn)(Λn−1ϕ)(τn)

=

∫ t

0

dτn

∫ τn

0

dτn−1A(τn)A(τn−1)(Λn−2ϕ)(τn−1)

...

=

∫ t

0

dτn

∫ τn

0

dτn−1 . . .

∫ τ2

0

dτ1A(τn) . . . A(τ1)ϕ(τ1)

= (−1)n·1t<0

∫∆n(t)

A(τn) . . . A(τ1)ϕ(τ1)dτ.

Taking norms of this equation and using the triangle inequality along withLemma 2.2 gives,

‖(Λnϕ) (t)‖ ≤ ‖ϕ‖∞ ·∫∆n(t)

‖A(τn)‖ . . . ‖A(τ1)‖dτ

≤‖ϕ‖∞ ·1

n!

(∫∆1(t)

‖A(τ)‖dτ

)n≤‖ϕ‖∞ ·

1

n!

(∫J

‖A(τ)‖dτ)n

.

Therefore,

‖Λn‖op ≤1

n!

(∫J

‖A(τ)‖dτ)n

(2.6)

and∞∑n=0

‖Λn‖op ≤ e∫J‖A(τ)‖dτ

<∞

where ‖·‖op denotes the operator norm on L (BC(J,X)) . An application of

Proposition 16.8 now shows (I − Λ)−1 =∞∑n=0

Λn exists and

∥∥(I − Λ)−1∥∥op≤ e∫J‖A(τ)‖dτ

.

It is now only a matter of working through the notation to see that theseassertions prove the theorem.

Corollary 2.5. Suppose h ∈ C(J,X) and x ∈ X, then there exists a uniquesolution, y ∈ C1 (J,X) , to the linear ordinary differential Eq. (2.1).

Proof. Let

ϕ (t) = x+

∫ t

0

h (τ) dτ.

By applying Theorem 2.4 with J replaced by any open interval J0 such that0 ∈ J0 and J0 is a compact subinterval1 of J, there exists a unique solutionyJ0 to Eq. (2.1) which is valid for t ∈ J0. By uniqueness of solutions, if J1 isa subinterval of J such that J0 ⊂ J1 and J1 is a compact subinterval of J, wehave yJ1 = yJ0 on J0. Because of this observation, we may construct a solutiony to Eq. (2.1) which is defined on the full interval J by setting y (t) = yJ0 (t)for any J0 as above which also contains t ∈ J.1 We do this so that φ|J0 will be bounded.


2.1 Operator Logarithms 19

Corollary 2.6. Suppose that A ∈ L(X) is independent of time, then the solu-tion to

y (t) = Ay (t) with y(0) = x

is given by y (t) = etAx where

etA =

∞∑n=0

tn

n!An. (2.7)

Moreover,e(t+s)A = etAesA for all s, t ∈ R. (2.8)

Proof. The first assertion is a simple consequence of Eq. 2.5 and Lemma 2.2with ψ = 1. The assertion in Eq. (2.8) may be proved by explicit computationbut the following proof is more instructive. Given x ∈ X, let y (t) := e(t+s)Ax.By the chain rule,

d

dty (t) =

d

dτ|τ=t+se

τAx = AeτAx|τ=t+s

= Ae(t+s)Ax = Ay (t) with y (0) = esAx.

The unique solution to this equation is given by

y (t) = etAx (0) = etAesAx.

This completes the proof since, by definition, y (t) = e(t+s)Ax.We also have the following converse to this corollary whose proof is outlined

in Exercise 2.10 below.

Theorem 2.7. Suppose that Tt ∈ L(X) for t ≥ 0 satisfies

1. (Semi-group property.) T0 = IdX and TtTs = Tt+s for all s, t ≥ 0.2. (Norm Continuity) t → Tt is continuous at 0, i.e. ‖Tt − I‖L(X) → 0 ast ↓ 0.

Then there exists A ∈ L(X) such that Tt = etA where etA is defined in Eq.(2.7).

2.1 Operator Logarithms

Our goal in this section is to find an explicit local inverse to the exponentialfunction, A → eA for A near zero. The existence of such an inverse can bededuced from the inverse function theorem although we will not need this facthere. We begin with the real variable fact that

ln (1 + x) =

∫ 1

0

d

dsln (1 + sx) ds =

∫ 1

0

x (1 + sx)−1ds.

Definition 2.8. When A ∈ L (X) satisfies 1 + sA is invertible for 0 ≤ s ≤ 1we define

ln (1 +A) =

∫ 1

0

A (1 + sA)−1ds. (2.9)

The invertibility of 1 + sA for 0 ≤ s ≤ 1 is satisfied if;

1. A is nilpotent, i.e. AN = 0 for some N ∈ N or more generally if2.∑∞n=0 ‖An‖ <∞ (for example assume that ‖A‖ < 1), of

3. if X is a Hilbert space and A∗ = A with A ≥ 0.

In the first two cases

(1 + sA)−1

=

∞∑n=0

(−s)nAn.

Proposition 2.9. If 1 + sA is invertible for 0 ≤ s ≤ 1, then

∂B ln (1 +A) =

∫ 1

0

(1 + sA)−1B (1 + sA)

−1ds. (2.10)

If 0 = [A,B] := AB −BA, Eq. (2.10) reduces to

∂B ln (1 +A) = B (1 +A)−1. (2.11)

Proof. Differentiating Eq. (2.9) shows

∂B ln (1 +A) =

∫ 1

0

[B (1 + sA)

−1 −A (1 + sA)−1sB (1 + sA)

−1]ds

=

∫ 1

0

[B − sA (1 + sA)

−1B]

(1 + sA)−1ds.

Combining this last equality with

sA (1 + sA)−1

= (1 + sA− 1) (1 + sA)−1

= 1− (1 + sA)−1

gives Eq. (2.10). In case [A,B] = 0,

(1 + sA)−1B (1 + sA)

−1= B (1 + sA)

−2

= Bd

ds

[−A−1 (1 + sA)

−1]

and so by the fundamental theorem of calculus

∂B ln (1 +A) = B

∫ 1

0

(1 + sA)−2ds = B

[−A−1 (1 + sA)

−1]s=1

s=0

= B[A−1 −A−1 (1 +A)

−1]

= BA−1[1− (1 +A)

−1]

= B[A−1 (1 +A)−A−1

](1 +A)

−1= B (1 +A)

−1.


20 2 Linear Ordinary Differential Equations

Corollary 2.10. Suppose that t→ A (t) ∈ L (X) is a C1 – function 1 + sA (t)is invertible for 0 ≤ s ≤ 1 for all t ∈ J = (a, b) ⊂ R. If g (t) := 1 + A (t) andt ∈ J, then

d

dtln (g (t)) =

∫ 1

0

(1− s+ sg (t))−1g (t) (1− s+ sg (t))

−1ds. (2.12)

Moreover if [A (t) , A (τ)] = 0 for all t, τ ∈ J then,

d

dtln (g (t)) = A (t) (1 +A (t))

−1. (2.13)

Proof. Differentiating past the integral and then using Eq. (2.10) gives

d

dtln (g (t)) =

∫ 1

0

(1 + sA (t))−1A (t) (1 + sA (t))

−1ds

=

∫ 1

0

(1 + s (g (t)− 1))−1g (t) (1 + s (g (t)− 1))

−1ds

=

∫ 1

0

(1− s+ sg (t))−1g (t) (1− s+ sg (t))

−1ds.

For the second assertion we may use Eq. (2.11) instead Eq. (2.10) in orderto immediately arrive at Eq. (2.13).

Theorem 2.11. If A ∈ L (X) satisfies, 1 + sA is invertible for 0 ≤ s ≤ 1, then

eln(I+A) = I +A. (2.14)

If C ∈ L (X) satisfies∑∞n=1

1n! ‖C

n‖n < 1 (for example assume ‖C‖ < ln 2, i.e.

e‖C‖ < 2), thenln eC = C. (2.15)

This equation also holds of C is nilpotent or if X is a Hilbert space and C = C∗

with C ≥ 0.

Proof. For 0 ≤ t ≤ 1 let

C (t) = ln (I + tA) = t

∫ 1

0

A (1 + stA)−1ds.

Since [C (t) , C (τ)] = 0 for all τ, t ∈ [0, 1] , if we let g (t) := eC(t), then

g (t) =d

dteC(t) = C (t) eC(t) = A (1 + tA)

−1g (t) with g (0) = I.

Noting that g (t) = 1 + tA solves this ordinary differential equation, it followsby uniqueness of solutions to ODE’s that eC(t) = g (t) = 1 + tA. Evaluatingthis equation at t = 1 implies Eq. (2.14).

Now let C ∈ L (X) as in the statement of the theorem and for t ∈ R set

A (t) := etC − 1 =

∞∑n=1

tn

n!Cn.

Therefore,

1 + sA (t) = 1 + s

∞∑n=1

tn

n!Cn

with ∥∥∥∥∥s∞∑n=1

tn

n!Cn

∥∥∥∥∥ ≤ s∞∑n=1

tn

n!‖Cn‖n < 1 for 0 ≤ s, t ≤ 1.

Because of this observation, ln(etC)

:= ln (1 +A (t)) is well defined and because[A (t) , A (τ)] = 0 for all τ and t we may use Eq. (2.13) to learn,

d

dtln(etC)

:= A (t) (1 +A (t))−1

= CetCe−tC = C with ln(e0C)

= 0.

The unique solution to this simple ODE is ln(etC)

= tC and evaluating this att = 1 gives Eq. (2.15).

2.2 Exercises

Exercise 2.1. To each A ∈ L (X) , we may define LA, RA : L (X)→ L (X) by

LAB = AB and RAB = BA for all B ∈ L (X) .

Show LA, RA ∈ L (L (X)) and that

‖LA‖L(L(X)) = ‖A‖L(X) = ‖RA‖L(L(X)) .

Exercise 2.2. Suppose that A : R→ L(X) is a continuous function and U, V :R→ L(X) are the unique solution to the linear differential equations

V (t) = A (t)V (t) with V (0) = I (2.16)

andU (t) = −U (t)A (t) with U(0) = I. (2.17)

Prove that V (t) is invertible and that V −1 (t) = U (t), where by abuse of no-

tation I am writing V −1 (t) for [V (t)]−1. Hints: 1) show d

dt [U (t)V (t)] = 0(which is sufficient if dim(X) < ∞) and 2) show y (t) := V (t)U (t) solves alinear differential ordinary differential equation that has y ≡ I as an obvioussolution. (The results of Exercise 2.1 may be useful here.) Then use the unique-ness of solutions to linear O.D.E.s


2.2 Exercises 21

Exercise 2.3. Suppose that (X, ‖·‖) is a Banach space, J = (a, b) with −∞ ≤a < b ≤ ∞ and fn : R→ X are continuously differentiable functions such thatthere exists a summable sequence an∞n=1 satisfying

‖fn (t)‖+∥∥∥fn (t)

∥∥∥ ≤ an for all t ∈ J and n ∈ N.

Show:

1. sup∥∥∥ fn(t+h)−fn(t)

h

∥∥∥ : (t, h) ∈ J × R 3 t+ h ∈ J and h 6= 0≤ an.

2. The function F : R→ X defined by

F (t) :=

∞∑n=1

fn (t) for all t ∈ J

is differentiable and for t ∈ J,

F (t) =

∞∑n=1

fn (t) .

Exercise 2.4. Suppose that A ∈ L(X). Show directly that:

1. etA define in Eq. (2.7) is convergent in L(X) when equipped with the op-erator norm.

2. etA is differentiable in t and that ddte

tA = AetA.

Exercise 2.5. Suppose that A ∈ L(X) and v ∈ X is an eigenvector of A witheigenvalue λ, i.e. that Av = λv. Show etAv = etλv. Also show that if X = Rnand A is a diagonalizable n× n matrix with

A = SDS−1 with D = diag(λ1, . . . , λn)

then etA = SetDS−1 where etD = diag(etλ1 , . . . , etλn). Here diag(λ1, . . . , λn)denotes the diagonal matrix Λ such that Λii = λi for i = 1, 2, . . . , n.

Exercise 2.6. Suppose that A,B ∈ L(X) and [A,B] := AB − BA = 0. Showthat e(A+B) = eAeB .

Exercise 2.7. Suppose A ∈ C(R, L(X)) satisfies [A (t) , A(s)] = 0 for all s, t ∈R. Show

y (t) := e

(∫ t0A(τ)dτ

)x

is the unique solution to y (t) = A (t) y (t) with y(0) = x.

Exercise 2.8. Compute etA when

A =

(0 1−1 0

)and use the result to prove the formula

cos(s+ t) = cos s cos t− sin s sin t.

Hint: Sum the series and use etAesA = e(t+s)A.

Exercise 2.9. Compute etA when

A =

0 a b0 0 c0 0 0

with a, b, c ∈ R. Use your result to compute et(λI+A) where λ ∈ R and I is the3× 3 identity matrix. Hint: Sum the series.

Exercise 2.10. Prove Theorem 2.7 using the following outline.

1. Using the right continuity at 0 and the semi-group property for Tt, showthere are constants M and C such that ‖Tt‖L(X) ≤MCt for all t > 0.

2. Show t ∈ [0,∞)→ Tt ∈ L(X) is continuous.3. For ε > 0, let Sε := 1

ε

∫ ε0Tτdτ ∈ L(X). Show Sε → I as ε ↓ 0 and conclude

from this that Sε is invertible when ε > 0 is sufficiently small. For theremainder of the proof fix such a small ε > 0.

4. Show

TtSε =1

ε

∫ t+ε

t

Tτdτ

and conclude from this that

limt↓0

(Tt − It

)Sε =

1

ε(Tε − IdX) .

5. Using the fact that Sε is invertible, conclude A = limt↓0 t−1 (Tt − I) exists

in L(X) and that

A =1

ε(Tε − I)S−1

ε .

6. Now show, using the semigroup property and step 5., that ddtTt = ATt for

all t > 0.7. Using step 6., show d

dte−tATt = 0 for all t > 0 and therefore e−tATt =

e−0AT0 = I.


Exercise 2.11 (Duhamel’ s Principle I). Suppose that A : R → L(X) isa continuous function and V : R → L(X) is the unique solution to the lineardifferential equation in Eq. (2.16). Let x ∈ X and h ∈ C(R, X) be given. Showthat the unique solution to the differential equation:

y (t) = A (t) y (t) + h (t) with y(0) = x (2.18)

is given by

y (t) = V (t)x+ V (t)

∫ t

0

V (τ)−1h(τ) dτ. (2.19)

Hint: compute ddt [V

−1 (t) y (t)] (see Exercise 2.2) when y solves Eq. (2.18).

Exercise 2.12 (Duhamel’ s Principle II). Suppose that A : R → L(X) isa continuous function and V : R → L(X) is the unique solution to the lineardifferential equation in Eq. (2.16). Let W0 ∈ L(X) and H ∈ C(R, L(X)) begiven. Show that the unique solution to the differential equation:

W (t) = A (t)W (t) +H (t) with W (0) = W0 (2.20)

is given by

W (t) = V (t)W0 + V (t)

∫ t

0

V (τ)−1H(τ) dτ. (2.21)

3

Ordinary Differential Equations in a Banach Space

Let X be a Banach space, U ⊂o X, J = (a, b) 3 0 and Z ∈ C (J × U,X) .The function Z is to be interpreted as a time dependent vector-field on U ⊂ X.In this section we will consider the ordinary differential equation (ODE forshort)

y (t) = Z(t, y (t)) with y(0) = x ∈ U. (3.1)

The reader should check that any solution y ∈ C1(J, U) to Eq. (3.1) gives asolution y ∈ C(J, U) to the integral equation:

y (t) = x+

∫ t

0

Z(τ, y(τ))dτ (3.2)

and conversely if y ∈ C(J, U) solves Eq. (3.2) then y ∈ C1(J, U) and y solvesEq. (3.1).

Remark 3.1. For notational simplicity we have assumed that the initial condi-tion for the ODE in Eq. (3.1) is taken at t = 0. There is no loss in generality indoing this since if y solves

dy

dt(t) = Z(t, y (t)) with y(t0) = x ∈ U

iff y (t) := y(t+ t0) solves Eq. (3.1) with Z(t, x) = Z(t+ t0, x).

3.1 Examples

Let X = R, Z (x) = xn with n ∈ N and consider the ordinary differentialequation

y (t) = Z(y (t)) = yn (t) with y(0) = x ∈ R. (3.3)

If y solves Eq. (3.3) with x 6= 0, then y (t) is not zero for t near 0. Therefore upto the first time y possibly hits 0, we must have

t =

∫ t

0

y(τ)

y(τ)ndτ =

∫ y(t)

y(0)

u−ndu =

[y(t)]1−n−x1−n

1−n if n > 1

ln∣∣∣y(t)x

∣∣∣ if n = 1

and solving these equations for y (t) implies

y (t) = y(t, x) =

x

n−1√

1−(n−1)txn−1if n > 1

etx if n = 1.(3.4)

The reader should verify by direct calculation that y(t, x) defined above does in-deed solve Eq. (3.3). The above argument shows that these are the only possiblesolutions to the Equations in (3.3).

Notice that when n = 1, the solution exists for all time while for n > 1, wemust require

1− (n− 1)txn−1 > 0

or equivalently that

t <1

(1− n)xn−1if xn−1 > 0 and

t > − 1

(1− n) |x|n−1 if xn−1 < 0.

Moreover for n > 1, y(t, x) blows up as t approaches (n− 1)−1x1−n. The reader

should also observe that, at least for s and t close to 0,

y(t, y(s, x)) = y(t+ s, x) (3.5)

for each of the solutions above. Indeed, if n = 1 Eq. (3.5) is equivalent to thewell know identity, etes = et+s and for n > 1,

24 3 Ordinary Differential Equations in a Banach Space

y(t, y(s, x)) =y(s, x)

n−1√

1− (n− 1)ty(s, x)n−1

=

xn−1√

1−(n−1)sxn−1

n−1

√1− (n− 1)t

[x

n−1√

1−(n−1)sxn−1

]n−1

=

xn−1√

1−(n−1)sxn−1

n−1

√1− (n− 1)t xn−1

1−(n−1)sxn−1

=x

n−1√

1− (n− 1)sxn−1 − (n− 1)txn−1

=x

n−1√

1− (n− 1)(s+ t)xn−1= y(t+ s, x).

Example 3.2. Let β > 1 and let y (t) = sgn (t) |t|β , so that

y (t) = β |t|β−1sgn (t) · sgn (t) = β |t|β−1

= β |y (t)|β−1β .

Thus we have shown y (t) solves

y (t) = β |y (t)|β−1β (3.6)

with y (0) = 0. This is not the only solution to this equation as y (t) ≡ 0 alsosolves the equation as well as

y+ (t) = |t|β 1t≥0 and y− (t) = − |t|β 1t≤0.

Moreover, if y (t) is any solution to Eq. (3.6) then so is y (t+ t0) for any t0 ∈ R.In particular it follows for any t0 ≥ 0 that

y (t) := y+ (t− t0) = |t− t0|β 1t≥t0

solves Eq. (3.6) with y (0) = 0. So there is a high degree of non-uniqueness here.

Example 3.3. Now suppose Z (x) = |x|α with 0 < α < 1 and we now considerthe ordinary differential equation

y (t) = Z(y (t)) = |y (t)|α with y(0) = x ∈ R. (3.7)

Working as above we find, if x 6= 0 that

t =

∫ t

0

y(τ)

|y (t)|αdτ =

∫ y(t)

y(0)

|u|−α du =[y (t)]

1−α − x1−α

1− α,

where u1−α := |u|1−α sgn(u). Since sgn(y (t)) = sgn (x) the previous equationimplies

sgn (x) (1− α)t = sgn (x)[sgn(y (t)) |y (t)|1−α − sgn (x) |x|1−α

]= |y (t)|1−α − |x|1−α

and therefore,

y(t, x) = sgn (x)(|x|1−α + sgn (x) (1− α)t

) 11−α

(3.8)

is uniquely determined by this formula until the first time t where |x|1−α +sgn (x) (1 − α)t = 0.As before y (t) = 0 is a solution to Eq. (3.7) when x = 0,however it is far from being the unique solution. For example letting x ↓ 0 inEq. (3.8) gives a function

y(t, 0+) = ((1− α)t)1

1−α

which solves Eq. (3.7) for t > 0. Moreover if we define

y (t) :=

((1− α)t)

11−α if t > 0

0 if t ≤ 0,

(for example if α = 1/2 then y (t) = 14 t

21t≥0) then the reader may easily checky also solve Eq. (3.7). Furthermore, ya (t) := y(t − a) also solves Eq. (3.7) forall a ≥ 0, see Figure 3.1 below.

Fig. 3.1. Three different solutions to the ODE y(t) = |y(t)|1/2 with y(0) = 0.

With these examples in mind, let us now go to the general theory. The caseof linear ODE’s has already been studied in Chapter 2 above.


3.3 Local Existence (Non-Linear ODE) 25

3.2 Uniqueness Theorem and Continuous Dependence onInitial Data

Lemma 3.4 (Gronwall’s Lemma). Suppose that f, ε, and k are non-negativelocally integrable functions of t ∈ [0,∞) such that

f (t) ≤ ε (t) +

∣∣∣∣∫ t

0

k(τ)f(τ)dτ

∣∣∣∣ . (3.9)

Then

f (t) ≤ ε (t) +

∣∣∣∣∫ t

0

k(τ)ε(τ)e

∣∣∫ tτk(s)ds

∣∣dτ

∣∣∣∣ , (3.10)

and in particular if ε and k are constants we find that

f (t) ≤ εek|t|. (3.11)

Proof. I will only prove the case t ≥ 0. The case t ≤ 0 can be derived byapplying the t ≥ 0 to f (t) = f(−t), k (t) = k(−t) and ε (t) = ε(−t).

Set F (t) =∫ t

0k(τ)f(τ)dτ . Then by (3.9) and the Lebesgue version of the

fundamental theorem of calculus,

F = kf ≤ kε+ kF a.e.

Hence,

d

dt

(e−∫ t0k(s)ds

F (t)

)a.e.= e

−∫ t0k(s)ds

(F (t)−k (t)F (t))a.e.≤ k (t) ε (t) e

−∫ t0k(s)ds

.

Integrating this last inequality from 0 to t and then solving for F yields:

F (t) ≤ e∫ t0k(s)ds ·

∫ t

0

dτk(τ)ε(τ)e−∫ τ0k(s)ds

=

∫ t

0

dτk(τ)ε(τ)e

∫ tτk(s)ds

.

But by the definition of F and Eq. (3.9) we have,

f (t) ≤ ε (t) + F (t) ≤ ε (t) +

∫ t

0

dτk(τ)ε(τ)e

∫ tτk(s)ds

which is Eq. (3.10). Equation (3.11) follows from (3.10) by a simple integration.

Corollary 3.5 (Continuous Dependence on Initial Data). Let U ⊂o X,0 ∈ (a, b) and Z : (a, b)×U → X be a continuous function which is K–Lipschitzfunction on U, i.e. ‖Z(t, x)−Z(t, x′)‖ ≤ K‖x−x′‖ for all x and x′ in U. Supposey1, y2 : (a, b)→ U solve

dyi (t)

dt= Z(t, yi (t)) with yi(0) = xi for i = 1, 2. (3.12)

Then‖y2 (t)− y1 (t) ‖ ≤ ‖x2 − x1‖eK|t| for t ∈ (a, b) (3.13)

and in particular, there is at most one solution to Eq. (3.1) under the aboveLipschitz assumption on Z.

Proof. Let f (t) := ‖y2 (t) − y1 (t) ‖. Then by the fundamental theorem ofcalculus,

f (t) = ‖y2(0)− y1(0) +

∫ t

0

(y2(τ)− y1(τ)) dτ‖

≤ f(0) +

∣∣∣∣∫ t

0

‖Z(τ, y2(τ))− Z(τ, y1(τ))‖ dτ∣∣∣∣

= ‖x2 − x1‖+K

∣∣∣∣∫ t

0

f(τ) dτ

∣∣∣∣ .Therefore by Gronwall’s inequality we have,

‖y2 (t)− y1 (t) ‖ = f (t) ≤ ‖x2 − x1‖eK|t|.

3.3 Local Existence (Non-Linear ODE)

Lemma 3.6. Suppose that K (t) is a locally integrable function of t ∈ [0,∞)and εn (t)∞n=0 is a sequence of non-negative continuous functions such that

εn+1 (t) ≤∫ t

0

K (τ) εn (τ) dτ for all n ≥ 0 (3.14)

and ε0 (t) ≤ δ <∞ for all t ∈ [0,∞). Then

εn (t) ≤ δ

n!

[∫ t

0

K (s) dτ

]n. (3.15)

Proof. The proof is by induction. Notice that

ε1 (t) ≤∫ t

0

K (τ) ε0 (τ) dτ ≤ δ∫ t

0

K (τ) dτ

as desired. If Eq. (3.15) holds for level n, then



εn+1 (t) ≤∫ t

0

K (τ) εn (τ) dτ ≤ δ

n!

∫ t

0

K (τ)

[∫ τ

0

K (s) dτ

]ndτ

=δ

n!

∫ t

0

1

n+ 1

d

dτ

[∫ τ

0

K (s) dτ

]n+1

dτ

=δ

(n+ 1)!

[∫ t

0

K (s) dτ

]n+1

which is Eq. (3.15) at level n+ 1.We now show that Eq. (3.1) has a unique solution when Z satisfies the

Lipschitz condition in Eq. (3.17). See Exercise ?? below for another existencetheorem.

Theorem 3.7 (Local Existence). Let T > 0, J = (−T, T ), x0 ∈ X, r > 0and

C(x0, r) := x ∈ X : ‖x− x0‖ ≤ r

be the closed r – ball centered at x0 ∈ X. Assume

M = sup ‖Z(t, x)‖ : (t, x) ∈ J × C(x0, r) <∞ (3.16)

and there exists K <∞ such that

‖Z(t, x)− Z(t, y)‖ ≤ K ‖x− y‖ for all x, y ∈ C(x0, r) and t ∈ J. (3.17)

Let T0 < min r/M, T and J0 := (−T0, T0), then for each x ∈ B(x0, r −MT0)there exists a unique solution y (t) = y(t, x) to Eq. (3.2) in C (J0, C(x0, r)) .Moreover y(t, x) is jointly continuous in (t, x), y(t, x) is differentiable in t,y(t, x) is jointly continuous for all (t, x) ∈ J0 × B(x0, r −MT0) and satisfiesEq. (3.1).

Proof. The uniqueness assertion has already been proved in Corollary 3.5.To prove existence, let Cr := C(x0, r), Y := C (J0, C(x0, r)) and

Sx (y) (t) := x+

∫ t

0

Z(τ, y(τ))dτ. (3.18)

With this notation, Eq. (3.2) becomes y = Sx (y) , i.e. we are looking for a fixedpoint of Sx. If y ∈ Y, then

‖Sx (y) (t)− x0‖ ≤ ‖x− x0‖+

∣∣∣∣∫ t

0

‖Z(τ, y (τ))‖ dτ∣∣∣∣ ≤ ‖x− x0‖+M |t|

≤ ‖x− x0‖+MT0 ≤ r −MT0 +MT0 = r,

showing Sx (Y ) ⊂ Y for all x ∈ B(x0, r −MT0). Moreover if y, z ∈ Y,

‖Sx (y) (t)− Sx(z) (t)‖ =

∥∥∥∥∫ t

0

[Z(τ, y (τ))− Z(τ, z (τ))] dτ

∥∥∥∥≤∣∣∣∣∫ t

0

‖Z(τ, y (τ))− Z(τ, z (τ))‖ dτ∣∣∣∣

≤ K∣∣∣∣∫ t

0

‖y(τ)− z(τ)‖ dτ∣∣∣∣ . (3.19)

Let y0(t, x) = x and yn(·, x) ∈ Y defined inductively by

yn(·, x) := Sx(yn−1(·, x)) = x+

∫ t

0

Z(τ, yn−1(τ, x))dτ. (3.20)

From Eq. (3.19) we have,

‖ yn+1 (t, x)− yn (t, x) ‖ ≤ K∣∣∣∣∫ t

0

‖yn (τ, x)− yn−1 (τ, x)‖ dτ∣∣∣∣ .

Thus applying Lemma 3.6 with εn (t) := ‖yn+1 (t, x)− yn (t, x)‖ and K (s) = Kwe find,

‖ yn+1 (t, x)− yn (t, x) ‖ ≤ Kn |t|n

n!δ (t, x)

where

δ (t, x) = max|τ |≤t

‖y1 (τ, x)− y0 (τ, x)‖

= max|τ |≤t

∥∥∥∥∫ τ

0

Z(σ, x)dσ

∥∥∥∥ ≤ ∫ |t|−|t|‖Z(σ, x)‖ dσ ≤MT0.

In summary we have shown,

∞∑n=0

‖yn+1(·, x)− yn(·, x)‖∞,J0 ≤∞∑n=0

MT0KnTn0n!

= MT0eKT0 <∞

where

‖yn+1(·, x)− yn(·, x)‖∞,J0 := sup ‖yn+1(t, x)− yn(t, x)‖ : t ∈ J0 .

So y(t, x) := limn→∞ yn(t, x) exists uniformly for t ∈ J and using Eq. (3.17) wealso have

sup ‖Z(t, y (t, x))− Z(t, yn−1 (t, x))‖ : t ∈ J0≤ K ‖y(·, x)− yn−1(·, x)‖∞,J0 → 0 as n→∞.

Now passing to the limit in Eq. (3.20) shows y (·, x) solves Eq. (3.2). From thisequation it follows that y(t, x) is differentiable in t and y satisfies Eq. (3.1). The


3.4 Global Properties 27

continuity of y(t, x) is a consequence that yn (·, ·) is continuous and convergesuniformly to y (·, ·) . Alternatively and more quantitatively we also find, usingCorollary 3.5 and the fundamental theorem of calculus, that

‖y(t, x)− y(t′, x′)‖ ≤ ‖y(t, x)− y(t, x′)‖+ ‖y(t, x′)− y(t′, x′)‖

= ‖y(t, x)− y(t, x′)‖+

∥∥∥∥∫ t

t′Z(τ, y(τ, x′))dτ

∥∥∥∥≤ ‖y(t, x)− y(t, x′)‖+

∣∣∣∣∫ t

t′‖Z(τ, y(τ, x′))‖ dτ

∣∣∣∣≤ ‖x− x′‖eKT +

∣∣∣∣∫ t

t′‖Z(τ, y(τ, x′))‖ dτ

∣∣∣∣ (3.21)

≤ ‖x− x′‖eKT +M |t− t′| .

The continuity of y(t, x) is now a consequence Eq. (3.1) and the continuity ofy and Z.

Corollary 3.8. Let J = (a, b) 3 0 and suppose Z ∈ C(J ×X,X) satisfies

‖Z(t, x)− Z(t, y)‖ ≤ K ‖x− y‖ for all x, y ∈ X and t ∈ J. (3.22)

Then for all x ∈ X, there is a unique solution y(t, x) (for t ∈ J) to Eq. (3.1).Moreover y(t, x) and y(t, x) are jointly continuous in (t, x).

Proof. Let J0 = (a0, b0) 3 0 be a precompact subinterval of J and Y :=BC (J0, X) . By compactness, M := supt∈J0 ‖Z(t, 0)‖ < ∞ which combinedwith Eq. (3.22) implies

supt∈J0‖Z(t, x)‖ ≤M +K ‖x‖ for all x ∈ X.

Using this estimate and Lemma ?? one easily shows Sx(Y ) ⊂ Y for all x ∈ X.The proof of Theorem 3.7 now goes through without any further change.

3.4 Global Properties

Definition 3.9 (Local Lipschitz Functions). Let U ⊂o X, J be an openinterval and Z ∈ C(J ×U,X). The function Z is said to be locally Lipschitz inx if for all x ∈ U and all compact intervals I ⊂ J there exists K = K(x, I) <∞and ε = ε(x, I) > 0 such that B(x, ε(x, I)) ⊂ U and

‖Z(t, x1)− Z(t, x0)‖ ≤ K(x, I)‖x1 − x0‖ ∀ x0, x1 ∈ B(x, ε(x, I)) & t ∈ I.(3.23)

For the rest of this section, we will assume J is an open interval containing0, U is an open subset of X and Z ∈ C(J×U,X) is a locally Lipschitz function.

Lemma 3.10. Let Z ∈ C(J × U,X) be a locally Lipschitz function in X andE be a compact subset of U and I be a compact subset of J. Then there existsε > 0 such that Z(t, x) is bounded for (t, x) ∈ I × Eε and and Z(t, x) is K –Lipschitz on Eε for all t ∈ I, where

Eε := x ∈ U : dist(x,E) < ε .

Proof. Let ε(x, I) and K(x, I) be as in Definition 3.9 above. Since E is com-pact, there exists a finite subset Λ ⊂ E such that E ⊂ V := ∪x∈ΛB(x, ε(x, I)/2).If y ∈ V, there exists x ∈ Λ such that ‖y − x‖ < ε(x, I)/2 and therefore

‖Z(t, y)‖ ≤ ‖Z(t, x)‖+K(x, I) ‖y − x‖ ≤ ‖Z(t, x)‖+K(x, I)ε(x, I)/2

≤ supx∈Λ,t∈I

‖Z(t, x)‖+K(x, I)ε(x, I)/2 =: M <∞.

This shows Z is bounded on I × V. Let

ε := d(E, V c) ≤ 1

2minx∈Λ

ε(x, I)

and notice that ε > 0 since E is compact, V c is closed and E ∩ V c = ∅.If y, z ∈ Eε and ‖y − z‖ < ε, then as before there exists x ∈ Λ such that‖y − x‖ < ε(x, I)/2. Therefore

‖z − x‖ ≤ ‖z − y‖+ ‖y − x‖ < ε+ ε(x, I)/2 ≤ ε(x, I)

and since y, z ∈ B(x, ε(x, I)), it follows that

‖Z(t, y)− Z(t, z)‖ ≤ K(x, I)‖y − z‖ ≤ K0‖y − z‖

where K0 := maxx∈ΛK(x, I) < ∞. On the other hand if y, z ∈ Eε and‖y − z‖ ≥ ε, then

‖Z(t, y)− Z(t, z)‖ ≤ 2M ≤ 2M

ε‖y − z‖ .

Thus if we let K := max 2M/ε,K0 , we have shown

‖Z(t, y)− Z(t, z)‖ ≤ K‖y − z‖ for all y, z ∈ Eε and t ∈ I.

Proposition 3.11 (Maximal Solutions). Let Z ∈ C(J × U,X) be a locallyLipschitz function in x and let x ∈ U be fixed. Then there is an interval Jx =(a (x) , b (x)) with a ∈ [−∞, 0) and b ∈ (0,∞] and a C1–function y : J → Uwith the following properties:



1. y solves ODE in Eq. (3.1).2. If y : J = (a, b) → U is another solution of Eq. (3.1) (we assume that

0 ∈ J) then J ⊂ J and y = y| J .

The function y : J → U is called the maximal solution to Eq. (3.1).

Proof. Suppose that yi : Ji = (ai, bi) → U, i = 1, 2, are two solutions toEq. (3.1). We will start by showing that y1 = y2 on J1 ∩ J2. To do this1 letJ0 = (a0, b0) be chosen so that 0 ∈ J0 ⊂ J1∩J2, and let E := y1(J0)∪y2(J0) – acompact subset of X. Choose ε > 0 as in Lemma 3.10 so that Z is Lipschitz onEε. Then y1|J0 , y2|J0 : J0 → Eε both solve Eq. (3.1) and therefore are equal byCorollary 3.5. Since J0 = (a0, b0) was chosen arbitrarily so that [a0, b0] ⊂ J1∩J2,we may conclude that y1 = y2 on J1∩J2. Let (yα, Jα = (aα, bα))α∈A denote thepossible solutions to (3.1) such that 0 ∈ Jα. Define Jx = ∪Jα and set y = yαon Jα. We have just checked that y is well defined and the reader may easilycheck that this function y : Jx → U satisfies all the conclusions of the theorem.

Notation 3.12 For each x ∈ U, let Jx = (a (x) , b (x)) be the maximal intervalon which Eq. (3.1) may be solved, see Proposition 3.11. Set D(Z) := ∪x∈U (Jx×x) ⊂ J × U and let ϕ : D(Z) → U be defined by ϕ(t, x) = y (t) where y isthe maximal solution to Eq. (3.1). (So for each x ∈ U, ϕ(·, x) is the maximalsolution to Eq. (3.1).)

Proposition 3.13. Let Z ∈ C(J×U,X) be a locally Lipschitz function in x andy : Jx = (a (x) , b (x)) → U be the maximal solution to Eq. (3.1). If b (x) < b,then either lim supt↑b(x) ‖Z(t, y (t))‖ = ∞ or y(b (x)−) := limt↑b(x) y (t) existsand y(b (x)−) /∈ U. Similarly, if a > a (x) , then either lim supt↓a(x) ‖y (t) ‖ =∞or y(a (x) +) := limt↓a(x) y (t) exists and y(a (x) +) /∈ U.

Proof. Suppose that b < b (x) and M := lim supt↑b(x) ‖Z(t, y (t))‖ < ∞.Then there is a b0 ∈ (0, b (x)) such that ‖Z(t, y (t))‖ ≤ 2M for all t ∈ (b0, b (x)).Thus, by the usual fundamental theorem of calculus argument,

1 Here is an alternate proof of the uniqueness. Let

T ≡ supt ∈ [0,minb1, b2) : y1 = y2 on [0, t].

(T is the first positive time after which y1 and y2 disagree.)Suppose, for sake of contradiction, that T < minb1, b2. Notice that y1(T ) =

y2(T ) =: x′. Applying the local uniqueness theorem to y1(· − T ) and y2(· − T )thought as function from (−δ, δ) → B(x′, ε(x′)) for some δ sufficiently small, welearn that y1(·−T ) = y2(·−T ) on (−δ, δ). But this shows that y1 = y2 on [0, T +δ)which contradicts the definition of T. Hence we must have the T = minb1, b2, i.e.y1 = y2 on J1∩J2∩[0,∞).A similar argument shows that y1 = y2 on J1∩J2∩(−∞, 0]as well.

‖y (t)− y(t′)‖ ≤

∣∣∣∣∣∫ t′

t

‖Z(t, y(τ))‖ dτ

∣∣∣∣∣ ≤ 2M |t− t′|

for all t, t′ ∈ (b0, b (x)). From this it is easy to conclude that y(b (x)−) =limt↑b(x) y (t) exists. If y(b (x)−) ∈ U, by the local existence Theorem 3.7, thereexists δ > 0 and w ∈ C1 ((b (x)− δ, b (x) + δ), U) such that

w (t) = Z(t, w (t)) and w(b (x)) = y(b (x)−).

Now define y : (a, b (x) + δ)→ U by

y (t) =

y (t) if t ∈ Jxw (t) if t ∈ [b (x) , b (x) + δ)

.

The reader may now easily show y solves the integral Eq. (3.2) and hence alsosolves Eq. 3.1 for t ∈ (a (x) , b (x) + δ).2 But this violates the maximality of yand hence we must have that y(b (x)−) /∈ U. The assertions for t near a (x) areproved similarly.

Example 3.14. Let X = R2, J = R, U =

(x, y) ∈ R2 : 0 < r < 1

where r2 =x2 + y2 and

Z(x, y) =1

r(x, y) +

1

1− r2(−y, x).

Then the unique solution (x (t) , y (t)) to

d

dt(x (t) , y (t)) = Z(x (t) , y (t)) with (x(0), y(0)) = (

1

2, 0)

is given by

(x (t) , y (t)) =

(t+

1

2

)(cos

(1

1/2− t

), sin

(1

1/2− t

))for t ∈ J(1/2,0) = (−1/2, 1/2) . Notice that ‖Z(x (t) , y (t))‖ → ∞ as t ↑ 1/2 anddist((x (t) , y (t)), U c)→ 0 as t ↑ 1/2.

Example 3.15. (Not worked out completely.) Let X = U = `2, ψ ∈ C∞(R2)be a smooth function such that ψ = 1 in a neighborhood of the line segmentjoining (1, 0) to (0, 1) and being supported within the 1/10 – neighborhood ofthis segment. Choose an ↑ ∞ and bn ↑ ∞ and define

Z (x) =

∞∑n=1

anψ(bn(xn, xn+1))(en+1 − en). (3.24)

2 See the argument in Proposition 3.16 for a slightly different method of extendingy which avoids the use of the integral equation (3.2).


3.4 Global Properties 29

For any x ∈ `2, only a finite number of terms are non-zero in the above sumin a neighborhood of x. Therefor Z : `2 → `2 is a smooth and hence locallyLipschitz vector field. Let (y (t) , J = (a, b)) denote the maximal solution to

y (t) = Z(y (t)) with y(0) = e1.

Then if the an and bn are chosen appropriately, then b < ∞ and there willexist tn ↑ b such that y(tn) is approximately en for all n. So again y(tn) doesnot have a limit yet supt∈[0,b) ‖y (t)‖ <∞. The idea is that Z is constructed to“blow” the particle from e1 to e2 to e3 to e4 etc. etc. with the time it takes totravel from en to en+1 being on order 1/2n. The vector field in Eq. (3.24) is afirst approximation at such a vector field, it may have to be adjusted a littlemore to provide an honest example. In this example, we are having problemsbecause y (t) is “going off in dimensions.”

Here is another version of Proposition 3.13 which is more useful whendim(X) <∞.

Proposition 3.16. Let Z ∈ C(J × U,X) be a locally Lipschitz function in xand y : Jx = (a (x) , b (x))→ U be the maximal solution to Eq. (3.1).

1. If b (x) < b, then for every compact subset K ⊂ U there exists TK < b (x)such that y (t) /∈ K for all t ∈ [TK , b (x)).

2. When dim(X) <∞, we may write this condition as: if b (x) < b, then either

lim supt↑b(x)

‖y (t)‖ =∞ or lim inft↑b(x)

dist(y (t) , U c) = 0.

Proof. 1) Suppose that b (x) < b and, for sake of contradiction, there existsa compact set K ⊂ U and tn ↑ b (x) such that y(tn) ∈ K for all n. Since Kis compact, by passing to a subsequence if necessary, we may assume y∞ :=limn→∞ y(tn) exists in K ⊂ U. By the local existence Theorem 3.7, there existsT0 > 0 and δ > 0 such that for each x′ ∈ B (y∞, δ) there exists a unique solutionw(·, x′) ∈ C1((−T0, T0), U) solving

w(t, x′) = Z(t, w(t, x′)) and w(0, x′) = x′.

Now choose n sufficiently large so that tn ∈ (b (x)− T0/2, b (x)) and y(tn) ∈B (y∞, δ) . Define y : (a (x) , b (x) + T0/2)→ U by

y (t) =

y (t) if t ∈ Jxw(t− tn, y(tn)) if t ∈ (tn − T0, b (x) + T0/2).

wherein we have used (tn−T0, b (x) +T0/2) ⊂ (tn−T0, tn+T0). By uniquenessof solutions to ODE’s y is well defined, y ∈ C1((a (x) , b (x) + T0/2) , X) and ysolves the ODE in Eq. 3.1. But this violates the maximality of y.

2) For each n ∈ N let

Kn := x ∈ U : ‖x‖ ≤ n and dist(x, U c) ≥ 1/n .

Then Kn ↑ U and each Kn is a closed bounded set and hence compact ifdim(X) < ∞. Therefore if b (x) < b, by item 1., there exists Tn ∈ [0, b (x))such that y (t) /∈ Kn for all t ∈ [Tn, b (x)) or equivalently ‖y (t)‖ > n ordist(y (t) , U c) < 1/n for all t ∈ [Tn, b (x)).

Remark 3.17 (This remark is still rather rough.). In general it is not true thatthe functions a and b are continuous. For example, let U be the region in R2

described in polar coordinates by r > 0 and 0 < θ < 3π/2 and Z(x, y) = (0,−1)as in Figure 3.2 below. Then b(x, y) = y for all x ≥ 0 and y > 0 while b(x, y) =∞ for all x < 0 and y ∈ R which shows b is discontinuous. On the other handnotice that

b > t = x < 0 ∪ (x, y) : x ≥ 0, y > t

is an open set for all t > 0. An example of a vector field for which b (x)is discontinuous is given in the top left hand corner of Figure 3.2. The mapψ (r (cos θ, sin θ)) :=

(ln r, tan

(23θ −

π2

)), would allow the reader to find an ex-

ample on R2 if so desired. Some calculations shows that Z transferred to R2 bythe map ψ is given by the new vector

Z(x, y) = −e−x(

sin

(3π

8+

3

4tan−1 (y)

), cos

(3π

8+

3

4tan−1 (y)

)).

(Bruce: Check this!)

Theorem 3.18 (Global Continuity). Let Z ∈ C(J × U,X) be a locally Lip-schitz function in x. Then D(Z) is an open subset of J × U and the func-tions ϕ : D(Z) → U and ϕ : D(Z) → U are continuous. More precisely, forall x0 ∈ U and all open intervals J0 such that 0 ∈ J0 @@ Jx0 there existsδ = δ(x0, J0, Z) > 0 and C = C(x0, J0, Z) < ∞ such that for all x ∈ B(x0, δ),J0 ⊂ Jx and

‖ϕ(·, x)− ϕ(·, x0)‖BC(J0,U) ≤ C ‖x− x0‖ . (3.25)

Proof. Let |J0| = b0 − a0, I = J0 and E := y(J0) – a compact subset of Uand let ε > 0 and K < ∞ be given as in Lemma 3.10, i.e. K is the Lipschitzconstant for Z on Eε. Also recall the notation: ∆1 (t) = [0, t] if t > 0 and∆1 (t) = [t, 0] if t < 0. Suppose that x ∈ Eε, then by Corollary 3.5,

‖ϕ(t, x)− ϕ(t, x0)‖ ≤ ‖x− x0‖eK|t| ≤ ‖x− x0‖eK|J0| (3.26)

for all t ∈ J0 ∩ Jx such that such that ϕ (∆1 (t) , x) ⊂ Eε. Letting δ :=εe−K|J0|/2, and assuming x ∈ B(x0, δ), the previous equation implies



Fig. 3.2. Manufacturing vector fields where b(x) is discontinuous.

‖ϕ(t, x)− ϕ(t, x0)‖ ≤ ε/2 < ε ∀ t ∈ J0 ∩ Jx 3 ϕ (∆1 (t) , x) ⊂ Eε.

This estimate further shows that ϕ(t, x) remains bounded and strictly awayfrom the boundary of U for all such t. Therefore, it follows from Proposition3.11 and “continuous induction” that J0 ⊂ Jx and Eq. (3.26) is valid for allt ∈ J0. This proves Eq. (3.25) with C := eK|J0|. Suppose that (t0, x0) ∈ D(Z)and let 0 ∈ J0 @@ Jx0

such that t0 ∈ J0 and δ be as above. Then we have justshown J0×B(x0, δ) ⊂ D(Z) which proves D(Z) is open. Furthermore, since theevaluation map

(t0, y) ∈ J0 ×BC(J0, U)e→ y(t0) ∈ X

is continuous (as the reader should check) it follows that ϕ = e (x→ ϕ(·, x)) :J0 × B(x0, δ) → U is also continuous; being the composition of continuousmaps. The continuity of ϕ(t0, x) is a consequence of the continuity of ϕ and thedifferential equation 3.1 Alternatively using Eq. (3.2),

‖ϕ(t0, x)− ϕ(t, x0)‖ ≤ ‖ϕ(t0, x)− ϕ(t0, x0)‖+ ‖ϕ(t0, x0)− ϕ(t, x0)‖

≤ C ‖x− x0‖+

∣∣∣∣∫ t0

t

‖Z(τ, ϕ(τ, x0))‖ dτ∣∣∣∣

≤ C ‖x− x0‖+M |t0 − t|

where C is the constant in Eq. (3.25) and M = supτ∈J0 ‖Z(τ, ϕ(τ, x0))‖ < ∞.This clearly shows ϕ is continuous.

3.5 Semi-Group Properties of time independent flows

To end this chapter we investigate the semi-group property of the flow associatedto the vector-field Z. It will be convenient to introduce the following suggestivenotation. For (t, x) ∈ D(Z), set etZ (x) = ϕ(t, x). So the path t→ etZ (x) is themaximal solution to

d

dtetZ (x) = Z(etZ (x)) with e0Z (x) = x.

This exponential notation will be justified shortly. It is convenient to have thefollowing conventions.

Notation 3.19 We write f : X → X to mean a function defined on some opensubset D(f) ⊂ X. The open set D(f) will be called the domain of f. Given twofunctions f : X → X and g : X → X with domains D(f) and D(g) respectively,we define the composite function f g : X → X to be the function with domain

D(f g) = x ∈ X : x ∈ D(g) and g (x) ∈ D(f) = g−1(D(f))

given by the rule f g (x) = f(g (x)) for all x ∈ D(f g). We now write f = giff D(f) = D(g) and f (x) = g (x) for all x ∈ D(f) = D(g). We will also writef ⊂ g iff D(f) ⊂ D(g) and g|D(f) = f.

Theorem 3.20. For fixed t ∈ R we consider etZ as a function from X to Xwith domain D(etZ) = x ∈ U : (t, x) ∈ D(Z), where D(ϕ) = D(Z) ⊂ R × U,D(Z) and ϕ are defined in Notation 3.12. Conclusions:

1. If t, s ∈ R and t · s ≥ 0, then etZ esZ = e(t+s)Z .2. If t ∈ R, then etZ e−tZ = IdD(e−tZ).

3. For arbitrary t, s ∈ R, etZ esZ ⊂ e(t+s)Z .

Proof. Item 1. For simplicity assume that t, s ≥ 0. The case t, s ≤ 0 is leftto the reader. Suppose that x ∈ D(etZ esZ). Then by assumption x ∈ D(esZ)and esZ (x) ∈ D(etZ). Define the path y(τ) via:

y(τ) =

eτZ (x) if 0 ≤ τ ≤ se(τ−s)Z (x) if s ≤ τ ≤ t+ s

.

It is easy to check that y solves y(τ) = Z(y(τ)) with y(0) = x. But since, eτZ (x)is the maximal solution we must have that x ∈ D(e(t+s)Z) and y(t + s) =e(t+s)Z (x) . That is e(t+s)Z (x) = etZ esZ (x) . Hence we have shown thatetZ esZ ⊂ e(t+s)Z . To finish the proof of item 1. it suffices to show thatD(e(t+s)Z) ⊂ D(etZ esZ). Take x ∈ D(e(t+s)Z), then clearly x ∈ D(esZ). Sety(τ) = e(τ+s)Z (x) defined for 0 ≤ τ ≤ t. Then y solves


3.6 Exercises 31

y(τ) = Z(y(τ)) with y(0) = esZ (x) .

But since τ → eτZ(esZ (x)) is the maximal solution to the above initial valuedproblem we must have that y(τ) = eτZ(esZ (x)), and in particular at τ =t, e(t+s)Z (x) = etZ(esZ (x)). This shows that x ∈ D(etZ esZ) and in facte(t+s)Z ⊂ etZ esZ .

Item 2. Let x ∈ D(e−tZ) – again assume for simplicity that t ≥ 0. Sety(τ) = e(τ−t)Z (x) defined for 0 ≤ τ ≤ t. Notice that y(0) = e−tZ (x) andy(τ) = Z(y(τ)). This shows that y(τ) = eτZ(e−tZ (x)) and in particular thatx ∈ D(etZ e−tZ) and etZ e−tZ (x) = x. This proves item 2.

Item 3. I will only consider the case that s < 0 and t + s ≥ 0, the othercases are handled similarly. Write u for t+ s, so that t = −s+u. We know thatetZ = euZ e−sZ by item 1. Therefore

etZ esZ = (euZ e−sZ) esZ .

Notice in general, one has (f g) h = f (g h) (you prove). Hence, the abovedisplayed equation and item 2. imply that

etZ esZ = euZ (e−sZ esZ) = e(t+s)Z ID(esZ) ⊂ e(t+s)Z .

The following result is trivial but conceptually illuminating partial converseto Theorem 3.20.

Proposition 3.21 (Flows and Complete Vector Fields). Suppose U ⊂o X,ϕ ∈ C(R× U,U) and ϕt (x) = ϕ(t, x). Suppose ϕ satisfies:

1. ϕ0 = IU ,2. ϕt ϕs = ϕt+s for all t, s ∈ R, and3. Z (x) := ϕ(0, x) exists for all x ∈ U and Z ∈ C(U,X) is locally Lipschitz.

Then ϕt = etZ .

Proof. Let x ∈ U and y (t) := ϕt (x) . Then using Item 2.,

y (t) =d

ds|0y(t+ s) =

d

ds|0ϕ(t+s) (x) =

d

ds|0ϕs ϕt (x) = Z(y (t)).

Since y(0) = x by Item 1. and Z is locally Lipschitz by Item 3., we know byuniqueness of solutions to ODE’s (Corollary 3.5) that ϕt (x) = y (t) = etZ (x) .

3.6 Exercises

Exercise 3.1. Find a vector field Z such that e(t+s)Z is not contained in etZ esZ .

Definition 3.22. A locally Lipschitz function Z : U ⊂o X → X is said to be acomplete vector field if D(Z) = R × U. That is for any x ∈ U, t → etZ (x) isdefined for all t ∈ R.

Exercise 3.2. Suppose that Z : X → X is a locally Lipschitz function. Assumethere is a constant C > 0 such that

‖Z (x) ‖ ≤ C(1 + ‖x‖) for all x ∈ X.

Then Z is complete. Hint: use Gronwall’s Lemma 3.4 and Proposition 3.13.

Exercise 3.3. Suppose y is a solution to y (t) = |y (t)|1/2 with y(0) = 0. Showthere exists a, b ∈ [0,∞] such that

y (t) =

14 (t− b)2 if t ≥ b

0 if −a < t < b− 1

4 (t+ a)2 if t ≤ −a.

Exercise 3.4. Using the fact that the solutions to Eq. (3.3) are never 0 if x 6= 0,show that y (t) = 0 is the only solution to Eq. (3.3) with y(0) = 0.

Exercise 3.5 (Higher Order ODE). Let X be a Banach space, , U ⊂o Xn

and f ∈ C (J × U , X) be a Locally Lipschitz function in x = (x1, . . . , xn). Showthe nth ordinary differential equation,

y(n) (t) = f(t, y (t) , y (t) , . . . , y(n−1) (t)) with y(k)(0) = yk0 for k < n (3.27)

where (y00 , . . . , y

n−10 ) is given in U , has a unique solution for small t ∈ J. Hint:

let y (t) =(y (t) , y (t) , . . . , y(n−1) (t)

)and rewrite Eq. (3.27) as a first order

ODE of the form

y (t) = Z(t,y (t)) with y(0) = (y00 , . . . , y

n−10 ).

Exercise 3.6. Use the results of Exercises 2.9 and 3.5 to solve

y (t)− 2y (t) + y (t) = 0 with y(0) = a and y(0) = b.

Hint: The 2 × 2 matrix associated to this system, A, has only one eigenvalue1 and may be written as A = I +B where B2 = 0.



Exercise 3.7 (Non-Homogeneous ODE). Suppose that U ⊂o X is openand Z : R × U → X is a continuous function. Let J = (a, b) be an intervaland t0 ∈ J. Suppose that y ∈ C1(J, U) is a solution to the “non-homogeneous”differential equation:

y (t) = Z(t, y (t)) with y(to) = x ∈ U. (3.28)

Define Y ∈ C1(J − t0,R×U) by Y (t) := (t+ t0, y(t+ t0)). Show that Y solvesthe “homogeneous” differential equation

Y (t) = Z(Y (t)) with Y (0) = (t0, y0), (3.29)

where Z(t, x) := (1, Z (x)). Conversely, suppose that Y ∈ C1(J − t0,R×U) is asolution to Eq. (3.29). Show that Y (t) = (t+t0, y(t+t0)) for some y ∈ C1(J, U)satisfying Eq. (3.28). (In this way the theory of non-homogeneous O.D.E.’s maybe reduced to the theory of homogeneous O.D.E.’s.)

Exercise 3.8 (Differential Equations with Parameters). Let W be an-other Banach space, U × V ⊂o X × W and Z ∈ C(U × V,X) be a locallyLipschitz function on U × V. For each (x,w) ∈ U × V, let t ∈ Jx,w → ϕ(t, x, w)denote the maximal solution to the ODE

y (t) = Z(y (t) , w) with y(0) = x. (3.30)

ProveD := (t, x, w) ∈ R× U × V : t ∈ Jx,w (3.31)

is open in R× U × V and ϕ and ϕ are continuous functions on D.Hint: If y (t) solves the differential equation in (3.30), then v (t) := (y (t) , w)

solves the differential equation,

v (t) = Z(v (t)) with v(0) = (x,w), (3.32)

where Z(x,w) := (Z(x,w), 0) ∈ X×W and let ψ(t, (x,w)) := v (t) . Now applythe Theorem 3.18 to the differential equation (3.32).

Exercise 3.9 (Abstract Wave Equation). For A ∈ L(X) and t ∈ R, let

cos(tA) :=∞∑n=0

(−1)n

(2n)!t2nA2n and

sin(tA)

A:=

∞∑n=0

(−1)n

(2n+ 1)!t2n+1A2n.

Show that the unique solution y ∈ C2 (R, X) to

y (t) +A2y (t) = 0 with y(0) = y0 and y(0) = y0 ∈ X (3.33)

is given by

y (t) = cos(tA)y0 +sin(tA)

Ay0.

Remark 3.23. Exercise 3.9 can be done by direct verification. Alternatively andmore instructively, rewrite Eq. (3.33) as a first order ODE using Exercise 3.5.In doing so you will be lead to compute etB where B ∈ L(X ×X) is given by

B =

(0 I−A2 0

),

where we are writing elements of X×X as column vectors,

(x1

x2

). You should

then show

etB =

(cos(tA) sin(tA)

A−A sin(tA) cos(tA)

)where

A sin(tA) :=

∞∑n=0

(−1)n

(2n+ 1)!t2n+1A2(n+1).

Exercise 3.10 (Duhamel’s Principle for the Abstract Wave Equation).Continue the notation in Exercise 3.9, but now consider the ODE,

y (t) +A2y (t) = f (t) with y(0) = y0 and y(0) = y0 ∈ X (3.34)

where f ∈ C(R, X). Show the unique solution to Eq. (3.34) is given by

y (t) = cos(tA)y0 +sin(tA)

Ay0 +

∫ t

0

sin((t− τ)A)

Af(τ)dτ (3.35)

Hint: Again this could be proved by direct calculation. However it is moreinstructive to deduce Eq. (3.35) from Exercise 2.11 and the comments in Remark3.23.

3.7 Smooth Dependence of ODE’s on Initial Conditions

In this subsection, let X be a Banach space, U ⊂o X and J be an open intervalwith 0 ∈ J.

Lemma 3.24. If Z ∈ C(J×U,X) such that DxZ(t, x) exists for all (t, x) ∈ J×U and DxZ(t, x) ∈ C(J ×U,X) then Z is locally Lipschitz in x, see Definition3.9.


3.7 Smooth Dependence of ODE’s on Initial Conditions 33

Proof. Suppose I @@ J and x ∈ U. By the continuity of DZ, for every t ∈ Ithere an open neighborhood Nt of t ∈ I and εt > 0 such that B(x, εt) ⊂ U and

sup ‖DxZ(t′, x′)‖ : (t′, x′) ∈ Nt ×B(x, εt) <∞.

By the compactness of I, there exists a finite subset Λ ⊂ I such that I ⊂ ∪t∈INt.Let ε(x, I) := min εt : t ∈ Λ and

K(x, I) := sup ‖DZ(t, x′)‖(t, x′) ∈ I ×B(x, ε(x, I)) <∞.

Then by the fundamental theorem of calculus and the triangle inequality,

‖Z(t, x1)− Z(t, x0)‖ ≤(∫ 1

0

‖DxZ(t, x0 + s(x1 − x0)‖ ds)‖x1 − x0‖

≤ K(x, I)‖x1 − x0‖

for all x0, x1 ∈ B(x, ε(x, I)) and t ∈ I.

Theorem 3.25 (Smooth Dependence of ODE’s on Initial Conditions).Let X be a Banach space, U ⊂o X, Z ∈ C(R× U,X) such that DxZ ∈ C(R×U,X) and ϕ : D(Z) ⊂ R×X → X denote the maximal solution operator to theordinary differential equation

y (t) = Z(t, y (t)) with y(0) = x ∈ U, (3.36)

see Notation 3.12 and Theorem 3.18. Then ϕ ∈ C1(D(Z), U), ∂tDxϕ(t, x) existsand is continuous for (t, x) ∈ D(Z) and Dxϕ(t, x) satisfies the linear differentialequation,

d

dtDxϕ(t, x) = [(DxZ) (t, ϕ(t, x))]Dxϕ(t, x) with Dxϕ(0, x) = IX (3.37)

for t ∈ Jx.

Proof. Let x0 ∈ U and J be an open interval such that 0 ∈ J ⊂ J @@ Jx0 ,y0 := y(·, x0)|J and

Oε := y ∈ BC(J, U) : ‖y − y0‖∞ < ε ⊂o BC(J,X).

By Lemma 3.24, Z is locally Lipschitz and therefore Theorem 3.18 is applicable.By Eq. (3.25) of Theorem 3.18, there exists ε > 0 and δ > 0 such that G :B(x0, δ) → Oε defined by G (x) := ϕ(·, x)|J is continuous. By Lemma 3.26below, for ε > 0 sufficiently small the function F : Oε → BC(J,X) defined by

F (y) := y −∫ ·

0

Z(t, y (t))dt (3.38)

is C1 and

DF (y) v = v −∫ ·

0

DyZ(t, y (t))v (t) dt. (3.39)

By the existence and uniqueness for linear ordinary differential equations, The-orem 2.4, DF (y) is invertible for any y ∈ BC(J, U). By the definition of ϕ,F (G (x)) = h (x) for all x ∈ B(x0, δ) where h : X → BC(J,X) is defined byh (x) (t) = x for all t ∈ J, i.e. h (x) is the constant path at x. Since h is a boundedlinear map, h is smooth and Dh (x) = h for all x ∈ X. We may now apply theconverse to the chain rule in Theorem 1.8 to conclude G ∈ C1 (B(x0, δ),O) andDG (x) = [DF (G (x))]−1Dh (x) or equivalently, DF (G (x))DG (x) = h whichin turn is equivalent to

Dxϕ(t, x)−∫ t

0

[DZ(ϕ(τ, x)]Dxϕ(τ, x) dτ = IX .

As usual this equation implies Dxϕ(t, x) is differentiable in t, Dxϕ(t, x) is con-tinuous in (t, x) and Dxϕ(t, x) satisfies Eq. (3.37).

Lemma 3.26. Continuing the notation used in the proof of Theorem 3.25 andfurther let

F (y) :=

∫ ·0

Z(τ, y(τ)) dτ for y ∈ Oε.

Then F ∈ C1(Oε, Y ) and for all y ∈ Oε,

F ′ (y)h =

∫ ·0

DxZ(τ, y(τ))h(τ) dτ =: Λyh.

Proof. Let h ∈ Y be sufficiently small and τ ∈ J, then by fundamentaltheorem of calculus,

Z(τ,y(τ) + h(τ))− Z(τ, y(τ))

=

∫ 1

0

[DxZ(τ, y(τ) + rh(τ))−DxZ(τ, y(τ))]dr

and therefore,

F (y + h)− F (y)− Λyh (t)

=

∫ t

0

[Z(τ, y(τ) + h(τ))− Z(τ, y(τ))−DxZ(τ, y(τ))h(τ) ] dτ

=

∫ t

0

dτ

∫ 1

0

dr[DxZ(τ, y(τ) + rh(τ))−DxZ(τ, y(τ))]h(τ).

Therefore,



‖(F (y + h)− F (y)− Λyh)‖∞ ≤ ‖h‖∞δ(h) (3.40)

where

δ(h) :=

∫J

dτ

∫ 1

0

dr ‖DxZ(τ, y(τ) + rh(τ))−DxZ(τ, y(τ))‖ .

With the aid of Lemmas 3.24 and Lemma 3.10,

(r, τ, h) ∈ [0, 1]× J × Y → ‖DxZ(τ, y(τ) + rh(τ))‖

is bounded for small h provided ε > 0 is sufficiently small. Thus it follows fromthe dominated convergence theorem that δ(h) → 0 as h → 0 and hence Eq.(3.40) implies F ′ (y) exists and is given by Λy. Similarly,

‖F ′(y + h)− F ′ (y) ‖op

≤∫J

‖DxZ(τ, y(τ) + h(τ))−DxZ(τ, y(τ))‖ dτ → 0 as h→ 0

showing F ′ is continuous.

Remark 3.27. If Z ∈ Ck(U,X), then an inductive argument shows thatϕ ∈ Ck(D(Z), X). For example if Z ∈ C2(U,X) then (y (t) , u (t)) :=(ϕ(t, x), Dxϕ(t, x)) solves the ODE,

d

dt(y (t) , u (t)) = Z ((y (t) , u (t))) with (y(0), u(0)) = (x, IdX)

where Z is the C1 – vector field defined by

Z (x, u) = (Z (x) , DxZ (x)u) .

Therefore Theorem 3.25 may be applied to this equation to deduce: D2xϕ(t, x)

and D2xϕ(t, x) exist and are continuous. We may now differentiate Eq. (3.37) to

find D2xϕ(t, x) satisfies the ODE,

d

dtD2xϕ(t, x) = [

(∂Dxϕ(t,x)DxZ

)(t, ϕ(t, x))]Dxϕ(t, x)

+ [(DxZ) (t, ϕ(t, x))]D2xϕ(t, x)

with D2xϕ(0, x) = 0.

3.8 Derivative Formulas

Corollary 3.28. Continuing the notation in Exercise 3.8 and further assumethat Z (t, y, w) is C1 in all of its variables. Then ϕ (t, x, w) solving

ϕ (t, x, w) = Z (t, ϕ (t, x, w) , w) with ϕ (0, x, w) = x

is C1 in all of its variables.

Proof. Following the hint in Exercise 3.8, let

ψ (t, x, w) = (ϕ (t, x, w) , w)

and observe that

ψ (t, x, w) = (Z (t, ϕ (t, x, w) , w) , 0)

= (Z (t, π1 ψ (t, x, w) , w) , 0) with ψ (0, x, w) = (x,w) .

Thus it follows from Theorem 3.25 that ψ is C1 in all of its variables and henceso is ϕ = π1 ψ.

Theorem 3.29. Suppose that Z (t, s, x) is a vector field on a Banach space Xdepending on parameter t, s ∈ R such that Z is jointly C1 in all of its variables.If ϕ (t, s, x) denotes the solution to the ODE,

ϕ (t, s, x) = Z (t, s, ϕ (t, s, x)) with ϕ (0, s, x) = x, (3.41)

then

ϕ′ (t, s, x) = (Dϕ) (t, s, x)

∫ t

0

(Dϕ (τ, s, x))−1Z ′ (τ, s, ϕ (τ, s, x)) dτ (3.42)

where Dϕ denote the differential of ϕ with respect to x and ϕ′ denote the deriva-tive of ϕ with respect to s.

Proof. Differentiating Eq. (3.41) relative to s and to x we learn, respectively,that

ϕ′ = Z ′ ϕ+ [(DZ) ϕ]ϕ′ with ϕ′ (0, s, x) = 0

andDϕ = [(DZ) ϕ]Dϕ with Dϕ (0, s, x) = I. (3.43)

From the second equation we conclude that

d

dt(Dϕ)

−1= − (Dϕ)

−1[(DZ) ϕ] . (3.44)

Using the first and third equations along with the product rule allows us toshow,

d

dt

[(Dϕ)

−1ϕ′]

= − (Dϕ)−1

[(DZ) ϕ]ϕ′ + (Dϕ)−1

[Z ′ ϕ+ [(DZ) ϕ]ϕ′]

= (Dϕ)−1Z ′ ϕ with

[(Dϕ)

−1ϕ′]

(0, s, x) = 0.

Integrating this equation shows,[(Dϕ)

−1ϕ′]

(t, s, x) =

∫ t

0

(Dϕ (τ, s, x))−1Z ′ (τ, s, ϕ (τ, s, x)) dτ

from which Eq. (3.42) easily follows.


3.9 Commutators of Vector Fields 35

Corollary 3.30. Suppose that Zs is a smooth vector field on X dependingsmoothly on a parameter s, then

d

dseZs =

(DeZs

)(x)

∫ 1

0

(DeτZs

)(x)−1Z ′s eτZs (x) dτ.

Proof. In this case ϕZ (t, s, x) = etZs (x) and the result follows immediatelyfrom Theorem 3.29.

3.9 Commutators of Vector Fields

Notation 3.31 If v ∈ X and f ∈ C1 (X,Y ) , let

(∂vf) (x) := f ′ (x) v =d

dt|0f (x+ tv) .

Also if v, w ∈ X and f ∈ C2 (X,Y ) , let(∂2v⊗wf

)(x) := (∂v∂wf) (x) = (∂w∂vf) (x) =

d

dt|0d

ds|0f (x+ sv + tw) .

Notation 3.32 If V is a vector field on X and f ∈ C1 (X,Y ) let

(V f) (x) :=(∂V (x)f

)(x) = f ′ (x)V (x) .

Remark 3.33. If V and W are vector-fields on X and f ∈ C2 (X,Y ) , then

(VWf) (x) =(∂2V (x)⊗W (x)f

)(x) + (∂∂VW f) (x) .

In particular if we let [V,W ] := VW −WV we find,

[V,W ] f = ∂(∂VW−∂WV )f

and hence[V,W ] = ∂VW − ∂WV.

Proposition 3.34. Suppose that Z (t, x) is a smoothly varying time dependentvector field on X and ϕ (t, x) = ϕZ (t, x) is the corresponding flow. Then forany other smooth vector field, Y, on X we have

d

dt

[(Dϕ (t, x))

−1Y (ϕ (t, x))

]= (Dϕ (t, x))

−1[Z (t, ·) , Y ] (ϕ (t, x)) . (3.45)

Consequently, if [Z (t, ·) , Y ] = 0 for all t, then

Dϕ (t, x)Y (x) = Y (ϕ (t, x)) for all (t, x) . (3.46)

Proof. Using the differential equation of ϕ and Eq. (3.44) we find,

d

dt

[(Dϕ)

−1Y ϕ

]= − (Dϕ)

−1[(DZ) ϕ]Y ϕ+ (Dϕ)

−1∂ϕY ϕ

= (Dϕ)−1

[(∂ZY ) ϕ− (∂Y Z) ϕ]

= (Dϕ)−1

[Z, Y ] ϕ.

If [Z (t, ·) , Y ] = 0, the above equation shows (Dϕ (t, x))−1Y (ϕ (t, x)) is con-

stant in t, i.e.

(Dϕ (t, x))−1Y (ϕ (t, x)) = (Dϕ (0, x))

−1Y (ϕ (0, x)) = Y (x) .

Corollary 3.35. Suppose that Y (t, x) and Z (s, x) are smoothly varying timedependent vector field on X and ϕY (t, x) and ϕZ (s, x) . If [Y (t, ·) , Z (s, ·)] = 0for all t and s, then

ϕY (t, ·) ϕZ (s, ·) = ϕZ (s, ·) ϕY (t, ·) .

Proof. Let

U (t, s) := ϕY (t, ·) ϕZ (s, ·)− ϕZ (s, ·) ϕY (t, ·) .

Then using Proposition 3.34 to conclude that DϕYt Zs = Zs ϕYt , we learn

∂sU = DϕYt Zs ϕZs − Zs ϕZs ϕYt= Zs ϕYt ϕZs − Zs ϕZs ϕYt= Zs

(ϕZs ϕYt + U (t, s)

)− Zs ϕZs ϕYt with U (t, 0) = 0.

But the unique solution to this ODE is U (t, s) ≡ 0.

Corollary 3.36. Let Z (t, s, x) be a vector field on X depending on (t, s) andassume that

[Z ′ (t, s, ·) , Z (τ, s, ·)] = 0 for all τ and t.

Further let ϕ (t, s, x) = ϕZ (t, s, x) and

Z (t, s, x) =

∫ t

0

Z (τ, s, x) dτ.

Thenϕ′ (t, s, x) = (Dϕ) (t, s, x) Z ′ (t, s, x) = Z ′ (t, s, ϕ (t, s, x)) .



Proof. From Theorem 3.29 and Proposition 3.34 we find,

ϕ′ (t, s, x) = (Dϕ) (t, s, x)

∫ t

0

(Dϕ (τ, s, x))−1Z ′ (τ, s, ϕ (τ, s, x)) dτ

= (Dϕ) (t, s, x)

∫ t

0

Z ′ (τ, s, x) dτ = (Dϕ) (t, s, x) Z ′ (t, s, x) .

Moreover we have[Z ′ (t, s, ·) , Z (τ, s, ·)

]=

∫ t

0

[Z ′(t, s, ·

), Z (τ, s, ·)

]dt = 0

and hence by Proposition 3.34,

(Dϕ) (t, s, x) Z ′ (t, s, x) = Z ′ (t, s, ϕ (t, s, x)) .

Alternatively,

(Dϕ) (t, s, x)

∫ t

0

Z ′ (τ, s, x) dτ =

∫ t

0

(Dϕ) (t, s, x)Z ′ (τ, s, x) dτ

=

∫ t

0

Z ′ (τ, s, ϕ (t, s, x)) dτ = Z ′ (t, s, ϕ (t, s, x)) .

Corollary 3.37. If [Z (t, ·) , Z (τ, ·)] = 0 for all t and τ then

ϕZ (t, x) = eZt (x) . (3.47)

Proof. According to Corollary 3.36,

d

dteZt (x) =

(d

dtZt

) eZt (x) = Zt

(eZt (x)

)with eZ0 (x) = e0 (x) = x. Thus by uniqueness of solutions to ODEs we concludethat Eq. (3.47) holds.

3.10 Inverse Function Theorem by O.D.E.s

Theorem 3.38 (Inverse Function Theorem by ODE methods). SupposeU ⊂o X and f : U → Y is a C2 – function.

1. If f ′ (x) is invertible for all x ∈ U then f : U → Y is an open mapping.

2. If x0 ∈ U and R > 0 such that BX(x0, R) ⊂ U,∥∥f ′(x0)−1f ′ (x)− I∥∥ ≤ 1/2 for all x ∈ BX(x0, R), (3.48)

then f |BX(x0,R) : BX(x0, R) → f(BX(x0, R)

)⊂o Y is a C1 – diffeomor-

phism.

Proof. Let x0 ∈ U and R > 0 be chosen so that Eq. (3.48) is valid. By thefundamental theorem of calculus, for any x, y ∈ BX(x0, R),

f (y)− f (x) =

[∫ 1

0

f ′(x+ t(y − x))dt

](y − x)

= f ′(x0) [R(x, y) + I] (y − x) (3.49)

where

R(x, y) :=

∫ 1

0

[f ′(x0)−1f ′(x+ t(y − x))− I

]dt. (3.50)

Because of Eq. (3.48), ‖R(x, y)‖ ≤ 1/2 for all x, y ∈ BX(x0, R) and so by

Proposition 16.8, [R(x, y) + I]−1

exists and∥∥∥[R(x, y) + I]

−1∥∥∥ ≤ 2 for all x, y ∈

BX(x0, R). Combining this result with Eq. (3.49) then shows

‖y − x‖ =∥∥∥[R(x, y) + I]

−1f ′(x0)−1 [f (y)− f (x)]

∥∥∥≤ 2

∥∥f ′(x0)−1∥∥ ‖f (y)− f (x)‖ . (3.51)

In particular this shows f is injective on BX(x0, R) and f |−1BX(x0,R)

:

f(BX(x0, R)

)→ BX(x0, R) is continuous. By Proposition 16.8 and Eq. (3.48)

f ′(x0)−1f ′ (x) is invertible and∥∥∥[f ′(x0)−1f ′ (x)

]−1∥∥∥ ≤ 2 for all x ∈ BX(x0, R).

Since f ′ (x) = f ′(x0)[f ′(x0)−1f ′ (x)

]this implies f ′ (x) is invertible and, for all

x ∈ BX(x0, R),∥∥∥f ′ (x)−1∥∥∥ =

∥∥∥[f ′(x0)−1f ′ (x)]−1

f ′(x0)−1∥∥∥ ≤ 2

∥∥f ′(x0)−1∥∥ =: M.

Let y0 := f(x0). We will now show there exists ε > 0 and a C1 – functiong : BY (y0, ε)→ U such that g(y0) = x0 and f g (y) = y for all y ∈ BY (y0, ε).To motivate our construction of g notice that if such a g exists and xy(t) :=g(y0 + ty) for y ∈ BY (0, ε), then

y0 + ty = f(xy(t)).

Differentiating this equation in t then implies y = f ′(xy(t))xy(t), i.e. xy (t) isdetermined as the unique solution to the ODE,


xy(t) = Z(xy(t), y) := f ′(xy(t))−1y with xy(0) = x0. (3.52)

Notice that Z(x, y) is a C1 – function of (x, y). We now reverse the logic anduse the ODE to construct g. Let ε := R

2M and define xy(t) as the solution toEq. (3.52) Then ∥∥f ′(xy(t))−1y

∥∥ ≤M ‖y‖ ≤ R/2and hence

‖xy(t)‖ ≤M ‖y‖ |t| ≤ R |t| /2 (3.53)

from which we may conclude xy exists for t ∈ (−2, 2). By Exercise ??, g (y) :=xy(1) is a C1 – function of y and f(g(y0 + y)) = y0 + y for all y ∈ BY (0, ε)which follows by integrating the identity,

d

dtf(xy(t)) = f ′(xy(t))xy(t) = f ′(xy(t))f ′(xy(t))−1y = y.

(If dim(X) = dim(Y ) < ∞, we would not need the differentiability of f ′ toconstruct g and we could conclude the full force of the inverse function theo-rem.) Moreover Eq. (3.53) shows ‖g (y)‖ ≤ M ‖y‖ from which it follows thatBY (y0, δ) ⊂ f(BX(x0,Mδ)) for all δ < ε, i.e. f(x0) is in the interior of f(W )for any open set W ⊂ U with x0 ∈ W. Since x0 ∈ U was arbitrary, it noweasily follows that f is an open mapping. The above argument also shows thatf |−1BX(x0,R)

: f(BX(x0, R)

)→ BX(x0, R) is C1 – although we could also use

the converse to the chain rule for this point as well.

Remark 3.39. From the above proof we have constructed g (y) for y ∈ BY (y0, ε)where ε := R

2M . We can easily show that g(BY (y0, ε)

)is an open neighborhood

of x0 ∈ X as follows. Let y ∈ BY (y0, ε) and x := g (y) ∈ BX(x0, R). For anyx′ ∈ BX(x0, R) sufficiently close to x we will have y′ = f (x′) near y = f (x)and hence back in BY (y0, ε) . Thus it follows that x′, g (y′) ∈ BX(x0, R) andf (x′) = y′ = f (g (y′)) . As f is injective on BX(x0, R) we may conclude thatg (y′) = x′, i.e. x′ ∈ g

(BY (y0, ε)

).

4

Classical Frobenius Theorem

Here is the problem that we now want to consider. Suppose that X and Yare two Banach spaces and F : X × Y → L (X,Y ) is a C1 – map (at least).Given (x0, y0) ∈ X × Y we wish to find a function y ∈ C2 (X → Y ) such that

y′ (x) = F (x, y (x)) with y(x0) := y0. (4.1)

The equation, y′ (x) = F (x, y (x)) , is equivalent to

(∂wy) (x) = F (x, y (x))w for all w ∈W. (4.2)

Notation 4.1 Let Fw (x, y) := F (x, y)w for all x,w ∈ X and y ∈ Y.

If Eq. (4.2) holds, then we must also have,

(∂v∂wy) (x) = ∂v [x→ Fw (x, y (x))]

= (∂vFw) (x, y (x)) +(∂y′(x)vFw

)(x, y (x))

= (∂vFw) (x, y (x)) +(∂F (x,y(x))vFw

)(x, y (x)) . (4.3)

Definition 4.2 (Curvature of F ). Let RF (x, y) be the bilinear form on Xdefined at (x, y) ∈ X × Y by

RF (x, y) 〈v, w〉 :=[(∂vFw) (x, y) +

(∂F (x,y)vFw

)(x, y)

]− [v ←→ w]

= (∂vFw) (x, y)− (∂wFv) (x, y)

+(∂F (x,y)vFw

)(x, y)−

(∂F (x,y)wFv

)(x, y) . (4.4)

Proposition 4.3 (Necessary Condition). If Eq. (4.1) has a (local) solutionfor any choice of (x0, y0) ∈ X × Y, then RF ≡ 0.

Proof. Since (∂v∂wy) (x) = (∂w∂vy) (x) , it follows from Eq. (4.3) that

0 = (∂v∂wy) (x0)− (∂w∂vy) (x0) = RF (x0, y (x0)) 〈v, w〉= RF (x0, y0) 〈v, w〉 .

Since (x0, y0) ∈ X × Y was arbitrary the result follows.

Remark 4.4 (A manifold ready way to find RF ). It is useful to give the argumentin Proposition 4.3 in more general terms of vector fields, V and W, on X. Inthis setting if y (x) satisfies,

y′ (x) = F (x, y (x)) ,

then formally we have

RF (x, y (x)) (V (x) ,W (x)) = (VWy −WV y) (x)− ([V,W ] y) (x) .

To see this is the case observe that

(VWy) (x) = V (x) [x→ F (x, y (x))W (x)]

=(∂V (x)F

)(x, y (x))W (x) +

(∂y′(x)V (x)F

)(x, y (x))W (x)

+ F (x, y (x))(∂V (x)W

)(x)

=(∂V (x)F

)(x, y (x))W (x) +

(∂F (x,y(x))V (x)F

)(x, y (x))W (x)

+ F (x, y (x))(∂V (x)W

)(x)

and hence

(VWy −WV y) (x) =RF (x, y (x)) (V (x) ,W (x))

+ F (x, y (x))[(∂V (x)W

)(x)−

(∂W (x)V

)(x)]

=RF (x, y (x)) (V (x) ,W (x)) + F (x, y (x)) [V,W ] (x)

=RF (x, y (x)) (V (x) ,W (x)) + ([V,W ] y) (x) .

Our next goal is to prove the converse of Proposition 4.3. To see how wemight go about constructing y : X → Y, suppose first that y satisfying Eq. (4.1).If σ (t) is a curve in X such that σ (0) = x0 then the path yσ (t) := y (σ (t)) inY solves

yσ (t) = y′ (σ (t)) σ (t) = Fσ(t) (σ (t) , y (σ (t))) with yσ (0) = y0. (4.5)

Thus we see that we should try to define y (x) := yσ (1) where σ (t) ∈ X suchthat σ (0) = x0 and σ (1) = x. In order for this to work we need to show thatyσ (1) only depends on σ (1) and not on the whole path σ. In fact, we shouldhave for h (t) ∈ X with h (0) = 0 (so that (σ + sh) (0) = x0 for all s) is that

∂hyσ (t) = y′ (σ (t))h (t) = Fh(t) (σ (t) , yσ (t)) .

That is, if RF ≡ 0, we hope to show

(Dhyσ) (t) := (∂hyσ) (t)− Fh(t) (σ (t) , yσ (t)) ≡ 0.

40 4 Classical Frobenius Theorem

Theorem 4.5. Suppose σ (t) is a curve in X such that σ (0) = x0 and yσ (t)denotes the solution to Eq. (4.5). Moreover, if h (t) ∈ X such that h (0) = 0,then

d

dt(Dhyσ) (t) =

(∂(Dhyσ)(t)Fσ(t)

)(σ (t) , yσ (t)) +RF (σ (t) , yσ (t)) 〈h (t) , σ (t)〉

(4.6)with (Dhyσ) (0) = 0. Thus is g (t) ∈ End (Y ) solves,1

g (t) =(∂g(t)[·]Fσ(t)

)(σ (t) , yσ (t)) with g (0) = I, (4.7)

then

(Dhyσ) (t) = g (t)

∫ t

0

g (τ)−1RF (σ (τ) , yσ (τ)) 〈h (τ) , σ (τ)〉 dτ. (4.8)

and in particular if RF = 0 we have (Dhyσ) (t) = 0, i.e.

(∂hyσ) (t) = Fh(t) (σ (t) , yσ (t)) . (4.9)

Proof. Let

u (t) := (Dhyσ) (t) := (∂hyσ) (t)− Fh(t) (σ (t) , yσ (t)) .

Then

d

dt∂hyσ (t) =

d

dt

d

ds|0yσ+sh (t) =

d

ds|0yσ+sh (t)

=d

ds|0Fσ(t)+sh(t) (σ (t) + sh (t) , yσ+sh (t))

=(∂∂hyσ(t)Fσ(t)

)(σ (t) , yσ (t))

+(∂h(t)Fσ(t)

)(σ (t) , yσ (t)) + Fh(t) (σ (t) , yσ (t))

=(∂u(t)Fσ(t)

)(σ (t) , yσ (t)) +

(∂Fh(t)(σ(t),yσ(t))Fσ(t)

)(σ (t) , yσ (t))

+(∂h(t)Fσ(t)

)(σ (t) , yσ (t)) + Fh(t) (σ (t) , yσ (t))

while

d

dtFh(t) (σ (t) , yσ (t))

=(∂σ(t)Fh(t)

)(σ (t) , yσ (t)) +

(∂yσ(t)Fh(t)

)(σ (t) , yσ (t))

+ Fh(t) (σ (t) , yσ (t))

=(∂σ(t)Fh(t)

)(σ (t) , yσ (t)) +

(∂F (σ(t),yσ(t))σ(t)Fh(t)

)(σ (t) , yσ (t))

+ Fh(t) (σ (t) , yσ (t)) .

1 In fact by Theorem 3.25, g (t) := (Dϕt) (y0) where ϕt is the differential of the flowassociated to the ODE in Eq. (4.5).

Subtracting the last two equations gives Eq. (4.6).Alternate (better) proof. Let ϕt = ϕσt denote the flow associated to the

ODE in Eq. (4.5) and let gt := (Dϕt) (y0) be the differential of this flow whichaccording to Theorem 3.25 solves Eq. (4.7). Then using Theorem 3.29 we learnthat

(∂hyσ) (t) = g (t)

∫ t

0

g (τ)−1 (

∂h[σ → Fσ(τ) (σ (τ) , ·)

])(yσ (τ)) dτ

= g (t)

∫ t

0

g (τ)−1((∂h(τ)F

)σ(τ)

(σ (τ) , yσ (τ)) + Fh(τ) (σ (τ) , yσ (τ)))dτ.

To finish the proof we will use integration by parts to get the derivative of theh (τ) term;

g (t)

∫ t

0

g (τ)−1F (σ (τ) , yσ (τ)) h (τ) dτ

=g (t) g (τ)−1F (σ (τ) , yσ (τ))h (τ) |t0

− g (t)

∫ t

0

(d

dτ

[g (τ)

−1F (σ (τ) , yσ (τ))

])h (τ) dτ

=F (σ (t) , yσ (t))h (t)− g (t)

∫ t

0

(d

dτ

[g (τ)

−1F (σ (τ) , yσ (τ))

])h (τ) dτ.

From Eq. (3.45) of Proposition 3.34 and the chain rule we find

d

dτ

[g (τ)

−1F (σ (τ) , yσ (τ))

]h (τ)

= g (τ)−1 [

Fσ(τ) (σ (τ) , ·) , Fh(τ) (σ (τ) , ·)]

(yσ (τ))

+ g (τ)−1 (

∂σ(τ)Fh(τ)

)(σ (τ) , yσ (τ)) .

Putting this all together shows,

(∂hyσ) (t) = F (σ (t) , yσ (t))h (t)

+ g (t)

∫ t

0

g (τ)−1

(∂h(τ)F

)σ(τ)

(σ (τ) , yσ (τ))

−(∂σ(τ)Fh(τ)

)(σ (τ) , yσ (τ))

−[Fσ(τ) (σ (τ) , ·) , Fh(τ) (σ (τ) , ·)

] dτ

= Fh(t) (σ (t) , yσ (t))

+ g (t)

∫ t

0

g (τ)−1RF (σ (τ) , yσ (τ)) 〈h (τ) , σ (τ)〉 dτ.

Theorem 4.6 (Classical Frobeneous Theorem). Suppose that RF ≡ 0 and(x0, y0) ∈ X × Y is given. Then there exits U ⊂o X with x0 ∈ U and y ∈C1 (U, Y ) such that y (xo) = y0 and y′ (x) = F (x, y (x)) for all x ∈ U.


4 Classical Frobenius Theorem 41

Proof. Using the continuity of F we may choose ε > 0 and δ > 0 so that

M := sup‖F (x, y)‖L(X,Y ) : (x, y) ∈ BX (x0, ε)× BY (y0, δ)

<∞.

By shrinking ε > 0 if necessary we may further assume that Mε < δ. Then forx ∈ BX (x0, ε) , let

σx (t) := x0 + t (x− x0)

and yx (t) denote the solution to

yx (t) = Fσx(t) (σx (t) , yx (t)) = F(x−x0) (σx (t) , yx (t)) with yx (0) = x.

Observe that

‖yx (t)‖ ≤ ‖σx (t)‖X ‖F (σx (t) , yx (t))‖L(X,Y ) ≤M · ‖x− x0‖ = Mε < δ.

Hence it follows that

‖yx (t)− x0‖ =

∥∥∥∥∫ t

0

yx (τ) dτ

∥∥∥∥ ≤ ∫ t

0

‖yx (τ)‖ dτ ≤Mεt

and we learn that yx (t) exists for 0 ≤ t ≤ 1. Hence we may define y (x) := yx (1)and we know y ∈ C1

(BX (x0, ε) , B

Y (y0, δ)). So it only remains to show that

y′ (x) = F (x, y (x)) .Let v ∈ X and h (t) := tv so that

σx+sv (t) = x0 + t (x+ sv − x0) = σx (t) + sh (t) .

Hence from Theorem 4.5 (with RF = 0), we find,

(∂vy) (x) =d

ds|0yσx+sh (1) = (∂hy)σx+sh (1) = Fh(1) (x, y (x)) = Fv (x, y (x))

as required.Before applying this result to a few different situations, let us globalize the

above construction.

Proposition 4.7. Suppose that U ⊂o X, (xo, y0) ∈ U × Y, and F : U × Y →L (X,Y ) is a C1 – map such that yσ (t) solving Eq. (4.5) has global solutionsfor all C1 – paths (σ) into U such that σ (0) = x0. If RF = 0 and σs is smoothlyvarying one parameter family of paths into U with σs (0) = x0, then

d

dsyσs (t) = F (σs (t) , yσs (t))σ′s (t) . (4.10)

In particular, if σs (1) = x for all s, then yσs (1) is constants in s.

Proof. The result in Eq. (4.10) follows directly from the proof of Theorem4.5 (with RF = 0). The second assertion follows directly from this result asσ′s (1) = 0 if σs (1) = x for all s.

Lemma 4.8 (Smoothing Lemma). Suppose that Π =0 = t0 < t1 < · · · < tn = 1 is a partition of [0, 1] and γ : [0, 1] → [0, 1]is a continuous path such that γ (0) = 0, γ (1) = 1, and γ|[ti−1,ti] (t) = ait + bifor some ai ≥ 0 and bi ∈ R. Then for ε > 0 sufficiently small, there exists asmooth function γε (t) such that γε (0) = 0, γε (1) = 1, γε (t) ≥ 0 for all t, andγε|[ti−1+ε,ti−ε] (t) = ait+ bi for some ai ≥ 0 and bi ∈ R for 1 ≤ i ≤ n.

Proof. Let ϕ ∈ C∞c ((−1, 1) , [0,∞)) be an even function such that∫ 1

−1ϕ (t) dt = 1. Then for ε > 0 let ϕε (t) := 1

εϕ(tε

). Now extend γ to all

of R by setting γ (t) = a1t for t ≤ 0 and γ (t) = ant+ bn for t ≥ 1. Then defineγε (t) := γ ∗ ϕε (t) . It then follows for t ∈ [ti−1 + ε, ti − ε] ,

γε (t) =

∫Rγ (t− τ)ϕε (τ) dτ =

∫ 1

−1

γ (t− ετ)ϕ (τ) dτ

=

∫ 1

−1

(ai (t− ετ) + bi)ϕ (τ) dτ = ait+ bi

wherein we have used∫ 1

−1

ϕ (τ) dτ = 1 and

∫ 1

−1

τϕ (τ) dτ = 0.

So the only thing left to remark is that

γε (t) =

∫Rγ (t− τ)ϕε (τ) dτ ≥ 0.

Corollary 4.9 (Global Frobenius’ Theorem I). Continue the assumptionsin Proposition 4.7 and further assume that U is connected and simply connected.Then, there exists y ∈ C1 (U, Y ) such that y′ (x) = F (x, y (x)) with y (x0) = y0.

Proof. If σ0 and σ1 are two C1 – curves in U such σ0 (0) = σ1 (0) = x0

and σ0 (1) = σ1 (1) = x, then there exists a homotopy σs (t) ∈ U interpolatingbetween these two curves with σs (0) = x0 and σs (1) = x for 0 ≤ s ≤ 1. Itthen follows by Proposition 4.7 that yσs (1) is constant in s and in particularyσ0 (1) = yσ1 (1) . Thus we may define y (x) := yσ (1) where σ ∈ C1 ([0, 1] , U)is any curve such that σ (0) = x0 and σ (1) = x.

We now verify that the function y (·) satisfies y′ (x) = F (x, y (x)) . To thisend suppose that x1 ∈ U is fixed and ε > 0 so that B (x1, ε) ⊂ U. We then


choose a curve σ ∈ C1([

0, 12

], U)

such that σ (0) = x0 and σ(

12

)= x1. Then

for x ∈ B (x1, ε) , let

σx (t) :=

σ (t) if 0 ≤ t ≤ 1

2x1 + 2

(t− 1

2

)(x− x1) if 1

2 ≤ t ≤ 1.

In order to smooth σ out use Lemma 4.8 to construct a smooth path γ : [0, 1]→[0, 1] such that γ (0) = 0, γ (1) = 1, γ (t) ≥ 0, γ (t) = 0 for t near 1

2 , andγ (1) = 1. Then set σx (t) := σx γ (t) .

Since y (x) = yσx (1) and for v ∈ Rd, we have

d

dsσx+sv (1) = v

we may conclude from Eq. (4.10) that

(∂vy) (x) =d

ds|0yσx+sv (1) = F (σx (1) , yσx (1)) v = F (x, y (x)) v.

5

Applications of Frobenius

5.1 Frobenius for special F

Corollary 5.1. If F : X → L (X,Y ) is a C1 – function, then y′ (x) = F (x)has local solutions iff

∂vFw = ∂wFv for all v, w ∈W.

If X = Rk, we can state the result as: the system of equations,

∂iy (x) = Fi (x) for 1 ≤ i ≤ k

has local solutions iff ∂iFj = ∂jFi for all 1 ≤ i, j ≤ k.

Proof. In this case,RF (x, y) 〈v, w〉 = (∂vFw − ∂wFv) (x, y) and so the resultnow follows from Theorem 4.6.

Corollary 5.2. If F : Y → L (X,Y ) is a C1 – function, then

y′ (x) = F (y (x)) with y (x0) = y0 (5.1)

has (local) solutions for arbitrary starting points iff [Fv, Fw] = 0 for all v, w ∈ X.Moreover, the solution y is given by

y (x) = eF(x−x0) (y0) (5.2)

for those x in which the right side is well defined.

Proof. In this case RF (x, y) 〈v, w〉 = (∂FvFw − ∂FwFv) (y) = [Fv, Fw] (y)and again the result follows by Frobenius’ Theorem 4.6. Moreover, from theproof of Theorem 4.6, we have y (x) = yσ (1) where σ ∈ C1 ([0, 1] , X) such thatσ (0) = x0 and s (1) = x. Since

[Fσ(t), Fσ(τ)

]= 0 for all t and τ , the solution

yσ (t) to the ODE,

yσ (t) = Fσ(t) (yσ (t)) with yσ (0) = y0

is give (according to Corollary 3.37) by

yσ (t) = e

∫ t0Fσ(τ)dτ (y0) = eFσ(t)−x0 (y0) .

Hence y (x) = yσ (1) is given by Eq. (5.2) as claimed. Moreover we may useCorollary 3.36 to verify y (·) as defined in Eq. (5.2) does indeed solve Eq. (5.1);

(∂vy) (x) = ∂v[eFx−x0 (y0)

]= (∂vFx−x0

)eFx−x0 (y0) = FveFx−x0 (y0) = Fv (y (x)) .

Remark 5.3. Before continuing one with the examples let us show that we mayuse Corollary 5.2 to give another proof of Theorem 4.6. So suppose that U ⊂o Xwith x0 ∈ U and F : U × Y → L (X,Y ) is a C1 – map. We then define vector-fields, Zvv∈X , on U × Y by

Zv (x, y) := (v, Fv (x, y)) .

There are now two important observations:

1. Firstly we have [Zv, Zw] (x, y) =(0, RF (x, y) 〈v, w〉

)for all (x, y) ∈ U × Y

and v, w ∈ X. Indeed,

(∂ZvZw) (x, y) =d

dt|0Zw ((x, y) + tZv (x, y))

=d

dt|0 (v, Fw (x+ tv, y + tFv (x, y)))

= (0, (∂vFw) (x, y) + (∂FvFw) (x, y))

from which the claim easily follows.2. If σ (t) is a curve in U starting at x0 and zσ (t) = (x (t) , y (t)) ∈ U × Y

solveszσ (t) = Zσ(t) (zσ (t)) with zσ (0) = (x0, y0) , (5.3)

then denote then zσ (t) = (σ (t) , yσ (t)) where

yσ (t) = Fσ(t) (σ (t) , yσ (t)) with yσ (0) = y0.

Indeed, it suffices to check that zσ (t) = (σ (t) , yσ (t)) solves Eq. (5.3);

d

dt(σ (t) , yσ (t)) = (σ (t) , yσ (t))

=(σ (t) , Fσ(t) (σ (t) , yσ (t))

)= Zσ(t) (σ (t) , yσ (t)) with (σ (0) , yσ (0)) = (x0, y0) .

44 5 Applications of Frobenius

From observation 1. it follows that if RF = 0 then [Zv, Zw] = 0. Then fromobservation 2. and Corollary 5.2 that

(x, yσ (1)) = eZx−x0 (x0, y0) .

Thus we should takey (x) := πY eZx−x0 (x0, y0)

provided the right hand side exists. Again from Corollary 3.36 we have directlythat

(∂vy) (x) = πY ∂veZx−x0 (x0, y0) = πY Zv eZx−x0 (x0, y0)

= Fv eZx−x0 (x0, y0) = Fv (x, y (x)) .

x0

xU

D(y)

x0

x

U

π1 etZx−x0 (x0, y0)

Fig. 5.1. Notice that π1 etZx−x0 (x0, y0) /∈ U and hence the advantage to our firstproof of Frobenius’s theorem.

5.2 Representations of Lie Groups and Lie Algebras

Definition 5.4. A Lie algebra, g, is a real or complex vector space equippedwith a multiplication law, [·, ·] : g× g→ g satisfying the following properties:

1. [A,B] = − [B,A] for all A,B ∈ g.2. [A, ·] : g→ g and [·, A] : g→ g is linear for all A ∈ g.3. [·, ·] satisfied the Jacobi identity,

[A, [B,C]] + (cyclic permutations) = 0.

Example 5.5. 1. Suppose that V is a vector space and End (V ) denotes thelinear transformations on V. Then defining [A,B] := AB−BA for all A,B ∈End (V ) make End (V ) into a Lie algebra.

2. If g is a Lie algebra and h ⊂ g is a subspace such that [h, h] ⊂ h, then h isa Lie algebra as well.

The reader unfamiliar with Lie groups may wish to visit the two appendicesof this section first and then read what follows.

Exercise 5.1 (Jacobi Identity). Suppose that X,Y, Z ∈ Γ∞(TM), prove theJacobi identity,

[X, [Y,Z]] + cyclic ≡ [X, [Y,Z]] + [Y, [Z,X]] + [Z, [X,Y ]] = 0 (5.4)

Also show that the Jacobi identity may be written as

L[X,Y ] = [LX , LY ] := LXLY − LY LX . (5.5)

Proposition 5.6. Let H be a simply connected Lie group, h := TeH be its Liealgebra, and α : h → End (Z) be a linear map where dimZ < ∞. Then thereexists a Lie group homomorphism,

ρ : H → Aut (Z) =: G

such that dρe = α iff α : h→ End (Z) is a Lie algebra homomorphism, i.e.

α ([A,B]) = [α (A) , α (B)] for all A,B ∈ h. (5.6)

Proof. For A ∈ h we let

A (h) :=d

dt|0hetA for all h ∈ H.

Assuming ρ exists, we must have,(Aρ)

(h) =d

dt|0ρ(hetA

)=

d

dt|0ρ (h) ρ

(etA)

= ρ (h)α (A) .

This is a first order system of equation for ρ which will have a solution iff theassociated curvature is zero. To find the curvature condition we note(

ABρ)

= A [ρα (B)] = ρα (A)α (B)

and therefore,

ρ · α ([A,B]) =(

[A,B]ρ)

=([A, B

]ρ)

= ρ · [α (A) , α (B)] .

Thus the curvature condition is precisely as stated in Eq. (5.6).


5.3 Parallel translation 45

Remark 5.7. Frobenius’ theorem tells us how to construct ρ above. Namely,given h ∈ H, choose any path σ (t) ∈ H such that σ (0) = e and σ (1) = h, andthen ρ (h) = g (1) where

g (t) =d

dtρ (σ (t)) =

d

ds|0ρ(σ (t)σ (t)

−1σ (t+ s)

)= ρ (σ (t))

d

ds|0ρ(σ (t)

−1σ (t+ s)

)= ρ (σ (t))α

(d

ds|0[σ (t)

−1σ (t+ s)

])= g (t)α

(A (t)

)with g (0) = I ∈ End (Z)

whereA (t) := σ (t)

−1σ (t) .

Corollary 5.8. If H = SU (2) – the space of 2 × 2 complex unitary matriceswhose determinant is 1. As SU (2) is diffeomorphic to S3 which is simply con-nected, it follows that the finite dimensional Lie group representations of SU (2)are in one to one correspondence with the finite dimensional Lie algebra rep-resentations of h := LieSU (2) – the space of traceless 2 × 2 complex skew –adjoint matrices.

Example 5.9. Let ρ : H := SO (3)→ Aut(L2(R3, dm

))be defined by

(ρ (g) f) (x) := f(g−1x

).

Then ρ is a unitary representation albeit infinite dimensional. The Lie algebraof H is so (3) – the space of real 3×3 -skew symmetric matrices. For A ∈ so (3) ,we have

(dρe (A) f) (x) =d

dt|0f(e−tAx

)= −Ax · ∇f (x)

so that α (A) = −Ax · ∇ in this case. Observe that that

[α (A) , α (B)] = [−Ax · ∇,−Bx · ∇] = BAx · ∇ −ABx · ∇=− [A,B]x · ∇ = α ([A,B])

as one would expect. In this infinite dimensional setting α = dρe is an un-bounded first order differential operator and our previous theory is not directlyapplicable.

5.3 Parallel translation

Proposition 5.10. Let X and Z be Banach spaces, Y = End (Z) , A be a Y –valued one form on X and

RA := dA+A ∧A (5.7)

wheredA 〈vx, wx〉 := ∂v [x→ A (wx)]− ∂w [x→ A (vx)] (5.8)

andA ∧A (vx, wx) = [A (vx) , A (wx)] . (5.9)

Then the equation,

∂vy (x) +A (vx) y (x) = 0 ∀ v ∈ X with y (x0) = y0 (5.10)

has a unique global solution iff RA ≡ 0.

Proof. In this example, F (x, y) v = −A (vx) y and therefore

RF (x, y) (v, w) =[−∂vA (wx) y + ∂−A(vx)y (−A (wx) y)

]− [v ⇐⇒ w]

= [−∂vA (wx) y +A (wx)A (vx) y]− [v ⇐⇒ w]

= [−dA (vx, wx)−A ∧A (vx, wx)] y = −RA 〈vx, wx〉 y.

Remark 5.11. If X = Rd, we may write

A =

d∑i=1

Aidxi

and then

A ∧A =

d∑i,j=1

AiAjdxi ∧ dxj .

Usingα ∧ β (v, w) := α (v)β (w)− α (w)β (v)

we learn

A∧A (v, w) =

d∑i,j=1

AiAj [viwj − wivj ] = A 〈v〉A 〈w〉−A 〈w〉A 〈v〉 = [A 〈v〉 , A 〈w〉] .

The reader should also verify that the definition of dA in Eq. (5.8) may also bewritten as

dA =

d∑i=1

dAi ∧ dxi.



Remark 5.12. Some times you may see RA written as

RA = dA+1

2[A ∧A]

where

[A ∧A] :=

d∑i,j=1

[Ai, Aj ] dxi ∧ dxj .

This agrees with our previous formula since,

d∑i,j=1

[Ai, Aj ] dxi ∧ dxj =

d∑i,j=1

(AiAj −AjAi) dxi ∧ dxj

=

d∑i,j=1

AiAj[dxi ∧ dxj − dxj ∧ dxi

]= 2

d∑i,j=1

AiAjdxi ∧ dxj = 2A ∧A.

Proposition 5.13. Let X and Z be Banach spaces, Y = End (Z) , and A be aY – valued one form on X and RA be as in Eq. (5.7). For any curve σ (t) ∈ Xsuch that σ (0) = x0, let yσ (t) solve the ODE,

yσ (t) +A(σ (t)σ(t)

)yσ (t) = 0 with yσ (0) = I ∈ End (Z) . (5.11)

To each C1− curve h (t) ∈ X with h (0) = 0, let

u (t) := (∂hyσ) (t) +A(h (t)σ(t)

)yσ (t) .

Then

u (t) +Aσ(t) (σ (t))u (t) = RA(σ (t)σ(t) , h (t)σ(t)

)yσ (t) with u (0) = 0 (5.12)

or equivalently,

u (t) = yσ (t)

∫ t

0

yσ (τ)−1RA

(σ (τ)σ(τ) , h (τ)σ(τ)

)yσ (τ) dτ. (5.13)

Proof. Differentiating Eq. (5.11) with respect to σ shows,

d

dt(∂hyσ) (t) =∂hyσ (t) = −∂h

[Aσ(t) (σ (t)) yσ (t)

]=−Ah(t) (σ (t)) yσ (t)−

(∂h(t)Aσ(t)

)(σ (t)) yσ (t)

−Aσ(t) (σ (t)) ∂hyσ (t)

=−Ah(t) (σ (t)) yσ (t)−(∂h(t)Aσ(t)

)(σ (t)) yσ (t)

−Aσ(t) (σ (t))[u (t)−A

(h (t)σ(t)

)yσ (t)

]

while

d

dt

[Ah(t) (σ (t)) yσ (t)

]=Ah(t) (σ (t)) yσ (t) +

(∂σ(t)Ah(t)

)(σ (t)) yσ (t)

+Ah(t) (σ (t)) yσ (t)

=Ah(t) (σ (t)) yσ (t) +(∂σ(t)Ah(t)

)(σ (t)) yσ (t)

−Ah(t) (σ (t))A(σ (t)σ(t)

)yσ (t)

Therefore adding these two equations together shows,

u (t) =[(∂σ(t)Ah(t)

)(σ (t))−

(∂h(t)Aσ(t)

)(σ (t))

]yσ (t)

+[Aσ(t) (σ (t)) , A

(h (t)σ(t)

)]yσ (t)−Aσ(t) (σ (t))u (t)

from which Eq. (5.12) easily follows.Next we compute,

d

dt

[yσ (t)

−1u (t)

]= yσ (t)

−1A (σ (t))u (t) + yσ (t)

−1u (t)

= RA(σ (t)σ(t) , h (t)σ(t)

)yσ (t) .

Integrating this equation on t using u (0) = 0 and then multiplying the resulton the left by yσ (t) gives Eq. (5.13).

5.4 Riemann’s Theorem

Let Pd be the space of positive definite d× d real matrices and g : Rd → Pd besmooth function. Associated to g is a Riemannian metric on Rd defined by

〈vx, wx〉g := g (x) v · w for all v, w, x ∈ Rd.

Riemann’s Question. When is it true that there exists an isometry from

ϕ :(Rd, 〈·, ·〉g

)→(Rd, 〈·, ·〉I

). In more detail we wish to require that

〈ϕ′ (x) v, ϕ′ (x)w〉I = 〈vx, wx〉g = 〈g (x) v, w〉I for all x, v, w ∈ Rd. (5.14)

The condition in Eq. (5.14) is equivalent to finding a diffeomorphism, ϕ : Rd →Rd, such that

g (x) = ϕ′ (x)trϕ′ (x) . (5.15)


5.4 Riemann’s Theorem 47

Theorem 5.14 (Riemann’s Theorem). Let Γ be the d × d matrix–valuedone form on Rd determined uniquely by

[g (x)Γ (vx)] a · b =1

2

[(∂vg) (x) a · b+ (∂ag) (x) v · b

− (∂bg) (x) v · a

]∀ a, b ∈ Rd. (5.16)

Then g (x) is locally of the form ϕ′ (x)trϕ′ (x) with ϕ′ (x0) a given isometry, iff

the curvature RΓ = dΓ + Γ ∧ Γ is zero. [In this context we say that g is flat ifRΓ = 0.]

Proof. LetA (vx) := g (x)Γ (vx)

so that

A (vx) a · b =1

2[(∂vg) (x) a · b+ (∂ag) (x) v · b− (∂bg) (x) v · a] ∀ a, b ∈ Rd.

If ϕ : Rd → Rd satisfies Eq. (5.15), then

∂vg (x) = (∂vϕ)′(x)

trϕ′ (x) + ϕ′ (x)

tr∂vϕ

′ (x) for all v ∈ Rd.

This last equation is equivalent to

∂vg (x) a · b =[(∂vϕ)

′(x)

trϕ′ (x) a+ ϕ′ (x)

tr∂vϕ

′ (x) a]· b

= ϕ′ (x) a · (∂vϕ)′(x) b+ ∂vϕ

′ (x) a · ϕ′ (x) b

holding for all a, b, v ∈ Rd, i.e. .

(∂vg) a · b = ∂aϕ · ∂v∂bϕ+ ∂v∂aϕ · ∂bϕ.

Taking this equation and cyclically permuting the indices (v, a, b) gives twomore identities;

(∂ag) b · v = ∂bϕ · ∂a∂vϕ+ ∂a∂bϕ · ∂vϕ(∂bg) v · a = ∂vϕ · ∂b∂aϕ+ ∂b∂vϕ · ∂aϕ

and from these three equations we may conclude,

(∂vg) a · b+ (∂ag) b · v − (∂bg) v · a = 2∂v∂aϕ · ∂bϕ.

In other words we have shown

A (v) a · b = ∂v∂aϕ · ∂bϕ (5.17)

where A (vx) := g (x)Γ (vx) is defined in Eq. (5.16). Since

∂v∂aϕ · ∂bϕ = ∂vϕ′a · ϕ′b = [ϕ′]

tr∂vϕ

′a · b,

we may rewrite Eq. (5.17) as

ϕ′ tr (x) (∂vϕ′) (x) = A (vx) .

Multiplying the last equation on the left by g−1 = ϕ′−1 [ϕ′ tr]−1

then shows

ϕ′−1∂vϕ′ = g−1A (v) =⇒ ∂vϕ

′ = ϕ′g−1A (v) = ϕ′Γ (v) .

Choose ψ0 ∈ Aut(Rd)

so that g (0) = ψtr0 ψ0, for example we might take

ψ0 =√g (0).1 We can solve the equation

∂vψ = ψΓ (v) with ψ (0) = ψ0 (5.18)

for ψ : Rd → End(Rd)

by Frobenius” theorem iff Γ has zero curvature. Notethat solving Eq. (5.18) is equivalent to solving

∂vψ−1 + Γ (v)ψ−1 = 0 with ψ−1 (0) = ψ−1

0

from which can be done iff RΓ = dΓ + Γ ∧ Γ = 0 by Proposition 5.10.2

We now must verify ψ satisfies ψtr (x)ψ (x) = g (x) away from x = 0. Tothis end we compute,

∂v[ψtrψ

]= (∂vψ)

trψ + ψtr∂vψ = Γ (v)

trψtrψ + ψtrψΓ (v) . (5.19)

On the other hand

Γ (v)trg + gΓ (v) = [gΓ (v)]

tr+ gΓ (v) = ∂vg, (5.20)

where the last equality is deduced by showing Eq. (5.16) implies

gΓ (v) a · b+ gΓ (v) b · a = (∂vg) a · b.1 Notice that g (0) = ψtr

0 ψ0 is equivalent to[ψ−1

0

]trg (0)ψ−1

0 = I which is equivalentto

⟨ψ−1

0 ei, ψ−10 ej

⟩g(0)

= δij and so a chose of ψ0 is equivalent to choosing an

orthonormal baseui = ψ−1

0 eidi=1

for the inner product 〈·, ·〉g(0) .

2 Alternatively,

∂v∂wψ = ∂v [ψΓw] = (∂vψ)Γw + ψ∂vΓw = ψΓvΓw + ψ∂vΓw

and so0 = [∂v, ∂w]ψ = ψ ([Γv, Γw] + ∂vΓw − ∂wΓv) = ψRΓ (v, w) .



From the uniqueness of solutions to ordinary differential equations we mayconclude from Eqs. (5.19) and (5.20) that ψtr (x)ψ (x) = g (x) .

Up to now we have shown that, if ϕ exists, then ϕ′ = ψ where ψ solves Eq.(5.18). We still have to show we can solve the equation ϕ′ = ψ for ϕ. This isnow a second application of the Frobenius theorem. We need only check thenecessary zero curvature condition;3 i.e. the equation ∂vϕ = ψv for all v ∈ Rdis locally solvable iff

∂wψv = ∂w∂vϕ = ∂v∂wϕ = ∂vψw.

Now from Eq. (5.18) we have

∂vψw = ψΓ (v)w = ψΓ (w) v = ∂wψv

wherein we have used Eq. (5.16) to see that Γ (v)w = Γ (w) v.Let us now see how this result relates to elliptic partial differential equations.

To this end suppose that

L = gij∂i∂j + lower order terms. (5.21)

Corollary 5.15. There is a local diffeomorphism, ϕ : Rd → Rd, such thatL [f ϕ] = (∆f + Sf) ϕ where ∆ is the standard Laplacian on Rd and S is a

linear first order differential operator, iff (gij) :=(gij)−1

is flat.

Proof. If ϕ : Rd → Rd is a diffeomorphism, then

L [f ϕ] = gij∂i [(f ′ ϕ) ∂jϕ] = gij [(f ′′ ϕ) ∂iϕ⊗ ∂jϕ+ (f ′ ϕ) ∂i∂jϕ]

= gij (f ′′ ϕ) ∂iϕ⊗ ∂jϕ+ lower order terms.

We would like to choose ϕ if possible so that

gij∂iϕ⊗ ∂jϕ =∑`

e` ⊗ e`

so thatL [f ϕ] = (∆f) ϕ+ lower order terms.

We havegij∂iϕ⊗ ∂jϕ = gij 〈∂iϕ, em〉〈∂jϕ, en〉 em ⊗ en

and so we wish to require

gij 〈∂iϕ, em〉〈∂jϕ, en〉 = δmn

3 One might have tried to construct ψ as the positive square-root of g, i.e. ψ =√g.

It would then certainly be the case that ψtrψ = g. However, the equation ϕ′ = ψwould in general not be solvable with this choice of ψ.

and so if we define Anj := 〈∂jϕ, en〉 the above equation may be written as

AgAtr = I =⇒ g = A−1[Atr]−1

=⇒ g−1 = AtrA.

But an an operator,

Av = enAnjvj = 〈∂vϕ, en〉 en = ∂vϕ = ϕ′v

and hence we need to solve the equation,

g−1 = [ϕ′]trϕ′

for ϕ which is only possible by Riemann’s Theorem 5.14 if g−1 is flat.


6

Matrix Lie Groups

In this section we pause to introduce the reader to an important class of Liegroups and some of their basic properties. We first need a short digression onmatrix logarithms.

6.1 Logarithms

Suppose that B is a Banach algebra with identity and G := B× is the group ofunits in B. For A ∈ B let eA :=

∑∞n=0

1n!A

n or alternatively define etA as thesolution to the ODE,

d

dtetA = AetA with e0A = 1.

Our goal here is to find inverse of the function A → eA for A near zero.One way to do this is through the use of power series arguments. I would liketo avoid those here. We begin with the real variable fact that

log (1 + x) =

∫ 1

0

d

dslog (1 + sx) ds =

∫ 1

0

x (1 + sx)−1ds.

Hence, provided 1 + sA is invertible for 0 ≤ s ≤ 1 we define

log (1 +A) =

∫ 1

0

A (1 + sA)−1ds. (6.1)

One way to satisfy this invertibility requirement is to assume1 thatA is nilpotentor to assume that

∑∞n=0 ‖An‖ < ∞ (for example assume that ‖A‖ < 1). In

either of these cases,

(1 + sA)−1

=

∞∑n=0

(−s)nAn

and then

log (1 +A) =

∞∑n=0

∫ 1

0

(−s)nAnds =

∞∑n=0

(−1)n

n+ 1An.

1 If we are working with the bounded operators on a Hilbert space we could requireA to be a non-negative operator, i.e. A = A∗ and A ≥ 0.

Remark 6.1. From the last equation it easily follows that

log (1 +A) = log (1) + Id (A) +O(∥∥A2

∥∥)and therefore log′ (I) = Id.

Lemma 6.2. Suppose A ∈ B such that (1 + sA)−1

exists for 0 ≤ s ≤ 1, then∫ 1

0

(1 + sA)−2ds = (1 +A)

−1.

Proof. For a ∈ R\ 0 , we will use the following anti-derivative, f, of

(1 + ta)−2

;

f (t) :=

∫ t

0

1

(1 + sa)2 ds = −1

a

1

1 + sa|s=ts=0 =

1

a

[1− 1

1 + ta

]=

t

1 + ta.

Notice this formula is valid for a = 0 as well. It is now easy to check that

d

dtt (1 + tA)

−1= (1 + tA)

−2

and hence by the fundamental theorem of calculus,∫ 1

0

(1 + sA)−2ds = t (1 + tA)

−1 |t=1t=0 = (1 +A)

−1.

Theorem 6.3. With the definitions above,

log(etA)

= A for all A ∈ B (6.2)

and

elog(1+A) = 1 +A ∀ A ∈ B 3 (1 + sA)−1

exists for 0 ≤ s ≤ 1. (6.3)


50 6 Matrix Lie Groups

∂B log (1 +A) =

∫ 1

0

[B (1 + sA)

−1 −A (1 + sA)−1sB (1 + sA)

−1]ds

=

∫ 1

0

[B − sA (1 + sA)

−1B]

(1 + sA)−1ds.

Combining this last equality with the identity,

I − sA (1 + sA)−1

= [(1 + sA)− sA] (1 + sA)−1

= (1 + sA)−1,

shows,

∂B log (1 +A) =

∫ 1

0

(1 + sA)−1B (1 + sA)

−1ds. (6.4)

If we now further assume that [A,B] = 0, then Eq. (6.4) reduces to

∂B log (1 +A) = B

∫ 1

0

(1 + sA)−2ds = B (I +A)

−1,

wherein we have used Lemma 6.2 for the last equality.So for 1 +A (t) = etC (so [A (t) , A (s)] = 0 for all s, t) we have

d

dtlog(etC)

=d

dtlog (1 +A (t)) = A (t) (1 +A (t))

−1= CetCe−tC = C

from which we conclude directly that log(etC)

= log (I) + tC = tC showing

log(eC)

= C.If

C (t) = log (I + tA) =

∫ 1

0

tA (1 + stA)−1ds =

∫ t

0

A (1 + τA)−1dτ,

then C (t) commutes with A and C (s) for all s and t. Therefore,

d

dteC(t) = C (t) eC(t) = A (1 + tA)

−1eC(t) with eC(0) = I.

On the other hand g (t) := 1 + tA solves,

g (t) = A = A (1 + tA)−1g (t) with g (0) = I.

Therefore, by the uniqueness theorem for ODE, we conclude that eC(t) = g (t) =1 + tA and we have shown elog(I+A) = I +A.

Remark 6.4. If g (t) = 1 +A (t) , then using Eq. (6.4) we learn

d

dtlog (g (t)) =

d

dtlog (1 +A (t)) =

(∂A(t) log

)(1 +A (t))

=

∫ 1

0

(1 + sA (t))−1A (t) (1 + sA (t))

−1ds

=

∫ 1

0

(1 + s (g (t)− 1))−1g (t) (1 + s (g (t)− 1))

−1ds

=

∫ 1

0

(1− s+ sg (t))−1g (t) (1− s+ sg (t))

−1ds. (6.5)

Theorem 6.5 (Differentiating the exponential). For any z, w ∈ B, wehave,

∂wez = ez

[I − e−adz

adzw

]=

[eadz − Iadz

w

]ez. (6.6)

In this formula, the reader should interpret

I − e−adzadz

= f (−adz) andeadz − Iadz

= f (adz) (6.7)

where

f (w) :=ew − 1

w=

∞∑k=1

1

k!wk−1. (6.8)

Proof. By the usual techniques we find,

d

dt∂we

tz =∂w[zetz

]= wetz + z∂we

tz

with ∂wetz|t=0 = 0.

It then follows (deriving Du Hammel’s principle) that

d

dt

[e−tz∂we

tz]

= −e−tzz∂wetz + e−tz[wetz + z∂we

tz]

= e−tzwetz

= e−tadz (w) =d

dt

[e−tadz − I−adz

](w) .

Integrating this equation for 0 ≤ t ≤ 1 shows,

e−z∂wez =

e−adz − I−adz

w =I − e−adz

adzw

from which the first equality in Eq. (6.6) follows. For the second we have

ez[I − e−adz

adzw

]= ez

[I − e−adz

adzw

]e−zez

= eadz[I − e−adz

adzw

]ez =

[eadz − Iadz

w

]ez.


6.2 Closed matrix sub-groups 51

6.2 Closed matrix sub-groups

Suppose that V is a finite dimensional vector space over R or C. Let End (V )denote the linear transformations on V and Aut (V ) denote those g ∈ End (V )which are invertible. For A,B ∈ End (V ) , let [A,B] := AB − BA. We also letadA ∈ End (End (V )) (i.e. adA is a linear transformation on End (V )) be definedby

adAB = [A,B] = AB −BA. (6.9)

Definition 6.6. If G is a subgroup of Aut (V ) we let

TIG :=

g (0) ∈ End (V ) :

[g ∈ C1 ((−ε, ε) ,Aut (V )) such that

g (0) = I & g ((−ε, ε)) ⊂ G

].

Theorem 6.7. If G is a (topologically) closed subgroup of Aut (V ) , then

TIG =A ∈ End (V ) : etA ∈ G for all t ∈ R

.

Proof. If g (t) := etA ∈ G for all t, then clearly A = g (0) ∈ TIG. Nowsuppose that A = g (0) for some g ∈ C1 ((−ε, ε) ,Aut (V )) with g (0) = Iand g ((−ε, ε)) ⊂ G. Let A (t) := log g (t) so that A (0) = log (I) = 0 andA (0) = d

dt |0 log g (t) = log′ (I) g (0) = A wherein we have used Remark 6.1.Hence by the fundamental theorem of calculus we learn that

A (t) =

∫ t

0

A (τ) dτ = tA+

∫ t

0

(A (τ)−A

)dτ = t [A+ ε (t)]

where

ε (t) :=1

t

∫ t

0

(A (τ)−A

)dτ → 0 as t→ 0.

From all of this we may now conclude that

g (t) = eA(t) = et[A+ε(t)].

from this is now easily follows that[g

(t

n

)]n=(etn [A+ε( tn )]

)n= et[A+ε( tn )] → etA as n→∞.

Since[g(tn

)]n ∈ G for all n and G is closed we conclude that

etA = limn→∞

[g

(t

n

)]n∈ G.

Proposition 6.8. If G is a closed subgroup of Aut (V ) , then TIG is a realsubspace of End (V ) .

Proof. From Theorem 6.7, it follows that A ∈ TIG implies cA ∈ TIG forall c ∈ R. The real point is to show that if A,B ∈ TIG, then A+B ∈ TIG. Butthis is easy as g (t) = etAetB ∈ G for all t and A+B = g (0) ∈ TIG.

Proposition 6.9. If G is a closed subgroup of Aut (V ) , then TIG is a realsubspace of End (V ) which is closed under the bracket operation, [A,B] = AB−BA, i.e. if A,B ∈ TIG then [A,B] ∈ TIG. In short, TIG is a Lie sub-algebraof End (V ) .

Proof. If A,B ∈ TIG, then etAesBe−tA ∈ G for all s, t ∈ R. As g (s) =etAesBe−tA is a path in G such that g (0) = I we know that

etABe−tA = g′ (0) ∈ TIG for all t ∈ R.

Differentiating this equation at t = 0 then shows that AB − BA ∈ TIG andTIG is a finite dimensional subspace of End (V ) and is therefore closed.

Fact: If G is a Lie group and H ⊂ G is a closed subgroup. Then H is anembedded submanifold and in particular H is a Lie subgroup of G, see Theorem?? below.

Proposition 6.10. Suppose that G is a closed sub-group of Aut (V ) . To A ∈TIG, let A be the vector-field on G defined by A (g) = d

dt |0getA. [This the so

called left invariant vector field on G such that A (I) = A.] Then for A,B ∈ TIG,[A, B

]= [A,B].

Proof. By the above fact, G is an embedded submanifold of Aut (V ) andtherefore if f ∈ C∞ (G) there exists F ∈ C∞ (Aut (V )) such that f = F |G. Wethen have(ABf

)(g) =

(ABF

)(g) = A [g → F ′ (g) 〈gB〉] = F ′′ (g) 〈gA, gB〉+F ′ (g) 〈gAB〉

and hence ([A, B

]f)

(g) = F ′ (g) 〈g (AB −BA)〉 =(

[A,B]f)

(g) .

Lemma 6.11. Let V be a finite dimensional vector space and let A,B ∈End (V ) and define adA (B) = AB −BA = [A,B] . Then;

1. eABe−A = eadA (B) for all A,B ∈ End (V ) and



2. eAeBe−A = eeadA (B) for all A,B ∈ End (V ) .

Proof. 1. Let B (t) := etABe−tA and observe that

B (t) = AB (t)−B (t)A = adAB (t) with B (0) = B.

The solution to this equation is B (t) = etadA (B) and taking t = 1 proves thefirst assertion.

2. Now let g (t) := eAetBe−A and observe that

g (t) = eABetBe−A = eABe−AeAetBe−A = eadA (B) g (t) with g (0) = I.

The solution to this equation is g (t) = eteadA (B) and then taking t = 1 gives

the claim in item 2.

Notation 6.12 For g ∈ Aut (V ) , let Adg ∈ Aut (End (V )) (so Adg is a linearisomorphism on End (V )) be defined by

AdgA = gAg−1.

Remark 6.13. Notice that item 1. of Eq. (6.11) states that

AdeA (B) = eadA (B) for all A,B ∈ End (V ) .

Also notice that for g, h ∈ Aut (V ) we have

Adgh (B) = ghB (gh)−1

= ghBh−1g−1 = AdgAdh (B) ,

i.e.Adgh = AdgAdh for all g, h ∈ Aut (V ) .

6.3 BCHD Formula

To really justify everything that follows it is useful to have developed the holo-morphic functional calculus, see for example Appendix A of Michael Taylor’snotes on Banach Algebras which may be found at

http://www.unc.edu/math/Faculty/met/banalg.pdf.

It is also possible to prove all the necessary results used below by hand usingpower series arguments alone.

Theorem 6.14 (Baker-Campbell-Dynkin-Hausdorff Formula). Supposenow that G is a closed subgroup of Aut (V ) where V = Rn or Cn for some n.Let g = Lie (G) = TIG and for Given A,B ∈ g, with e‖adA‖+‖adB‖ < 2, then

1.

Γ (A,B) := log(eAeB) = A+

∫ 1

0

Ψ(eadAetadB )Bdt, (6.10)

where Ψ(z) ≡ z log zz−1 . Alternatively we can write

Γ (A,B) := log(eAeB) = B +

∫ 1

0

f(etadAeadB )Bdt (6.11)

where f(z) = log(z)z−1 .

2. Let Ik :=

(m,n) ∈ Z2k+ : m+ n > 0

, where m+n > 0 means mi+ni > 0

for all i. As usual set m! = m1! . . .mk!, n! = n1! . . . nk−1! and |n| = n1 +· · ·+ nk. Then Eq. (6.10) implies

Γ (A,B) = A+B

+

∞∑k=1

(−1)k+1

k(k + 1)

∑(m,n)∈Ik

1

m!n! (|n|+ 1)admkA adnkB . . . adm2

A adn2

B adm1

A adn1

B B

(6.12)

= A+B

+

∞∑k=1

(−1)k+1

k(k + 1)

∑(m,n)∈Ik

1n1=01

m!n! (|n|+ 1)admkA adnkB . . . adm2

A adn2

B adm1

A B

(6.13)

where the series in Eqs. (6.12) and (6.13) are absolutely convergent.3. We may use Eq. (6.11) to find the alternative formula,

log(eAeB) = A+B+

∞∑k=1

(−1)k

k + 1

∑(m,n)∈Ik

1

m!n! (|m|+ 1)admkA adnkB . . . adm1

A adn1

B A.

(6.14)Note: the formula in Eq. (6.13) has better convergence properties than Eq.(6.14) owing to the two different prefactors, 1/k (k + 1) versus 1/ (k + 1) .

Proof. Let A,B ∈ g and g(t) := eAetB so that

g (t) = g (t)B with g (0) = A (6.15)

Assume that A,B are sufficiently close to 0 ∈ g so that we can define C (t) for0 ≤ t ≤ 1 as

C(t) := log(eAetB) = log(g(t)).

Since g (t) = eC(t), it follows from Theorem 6.5 that


http://www.unc.edu/math/Faculty/met/banalg.pdf

6.3 BCHD Formula 53

g (t) = eC(t)

[I − e−adC(t)

adC(t)C(t)

]= g (t)

[I − e−adC(t)

adC(t)C(t)

]This equation along with Eq. (6.15) then shows C (t) satisfies,[

I − e−adC(t)

adC(t)C(t)

]= B with C (0) = A.

This equation can be put in more standard form as

C(t) =adC(t)

I − e−adC(t)B =

adC(t)eadC(t)

eadC(t) − IB with C (0) = A. (6.16)

Moreover using item 1. of Lemma 6.11 in the form of Remark 6.13, we have,

eadC(t) = AdeC(t) = Adg(t) = AdeAAdetB = eadAetadB .

Using this identity, it follows that

adC(t)eadC(t)

eadC(t) − I=

log(eadAetadB

)eadAetadB

eadAetadB − I= Ψ(eadAetadB )

where Ψ(z) = z log zz−1 is analytic in the unit disc D ⊂ C. Using this result back

in Eq. (6.16) then gives,

C(t) = Ψ(eadAetadB ) (B) with C (0) = A

and then integrating this result in t gives Eq. (6.10).In order to prove Eq. (6.12) we will start by computing the power series for

ψ(1 + z) :

ψ(1 + z) =(1 + z) log (1 + z)

z=

1 + z

z

∞∑k=1

(−1)k−1

kzk

= (1 + z)

∞∑k=1

(−1)k−1

kzk−1 =

∞∑k=0

(−1)k

k + 1zk +

∞∑k=1

(−1)k−1

kzk

= 1 +

∞∑k=1

(−1)k

(1

k + 1− 1

k

)zk = 1 +

∞∑k=1

(−1)k+1

k(k + 1)zk (6.17)

= 1 +1

2z − 1

6z2 +

1

12z2 + · · ·+ (−1)

k+1

k(k + 1)zk + . . . .

From this expansion we see that ψ (1 + z) is analytic for |z| < 1. Since

eadAetadB =

∞∑m,n=0

tn

m!n!admAad

nB = I +

∑m+n>0

tn

m!n!admAad

nB , (6.18)

if e‖adA‖+‖adB‖ < 2, then

∥∥eadAetadB − I∥∥ =

∥∥∥∥∥ ∑m+n>0

tn

m!n!admAad

nB

∥∥∥∥∥ ≤∞∑

m+n>0

tn

m!n!‖adA‖m ‖adB‖n

≤ et‖adA‖+‖adB‖ − 1 < 2− 1 = 1

and therefore

ψ(eadAetadB

)= ψ

(I +

[eadAetadB − I

])= I +

∞∑k=1

(−1)k+1

k(k + 1)

( ∑m+n>0

tn

m!n!admAad

nB

)kis well defined. Expanding all of this out in full detail then shows,

ψ(eadAetadB

)B

= B +

∞∑k=1

(−1)k+1

k(k + 1)

( ∑m+n>0

tn

m!n!admAad

nB

)kB

= B +

∞∑k=1

(−1)k+1

k(k + 1)

∑mi+ni>0

tn1+···+nk

m1! . . .mk!n1! . . . nk!adm1

A adn1

B . . . admkA adnkB B

= B +

∞∑k=1

(−1)k+1

k(k + 1)

∑mi+ni>0

1nk=0tn1+···+nk

m1! . . .mk!n1! . . . nk−1!adm1

A adn1

B . . . admkA B.

Integrating this equation on t ∈ [0, 1] and using the result in Eq. (6.10) provesEq. (6.12) provided A,B are sufficiently close to 0 ∈ g.

Aside: Let us now remove the extra condition that A,B are sufficientlyclose to 0 ∈ g and replace it with the condition that e‖adA‖+‖adB‖ < 2. To dothis we notice by what we have already just proved, we may make G into anreal analytic manifold in such a way that exp : g→ G is an analytic mapping.Moreover for A,B ∈ g close to 0 we have shown eAeB = eΓ (A,B) where

Γ (A,B) := A+B+

∞∑k=1

(−1)k+1

k(k + 1)

∑m,n∈Zk

+

ck(m,n)adm1

A adn1

B . . . admk−1

A adnk−1

B admkA B,

and ck (m,n) are chosen so that the above expression is equal to the rightmember in Eq. (6.12). Now Γ is a power series in A and B hence is analytic fore‖adA‖+‖adB‖ < 2. So we have (A,B) → eAeB and (A,B) → eΓ (A,B) are realanalytic maps on the connected open set,

(A,B) ∈ g2 : e‖adA‖+‖adB‖ < 2,



which are equal for A,B in a neighborhood of 0. The technique of analyticcontinuation then implies eAeB = eΓ (A,B) whenever e‖adA‖+‖adB‖ < 2.

Proof of the alternate formulas (optional!). Now let C(t) =log(etAeB), i.e. etAeB = eC(t) and now use

AeC(t) = AetAeB =d

dteC(t) =

[eadC(t) − IadC(t)

C(t)

]eC(t) with C (0) = B.

This then implies, (working as above) that

C(t) =adC(t)

eadC(t) − IA =

log(etadAeadB

)etadAeadB − I

A = f(etadAeadB

)A

=

∞∑k=0

(−1)k

k + 1

∑(m,n)∈Ik

t|m|

m!n!admkA adnkB . . . adm1

A adn1

B A.

where f(z) = log(z)z−1 . Using

f(1 + z) =log(1 + z)

z=

∞∑k=0

(−1)k

k + 1zk

we may conclude that

C(t) = f(I +

[etadAeadB − I

])A

=∞∑k=0

(−1)k

k + 1

( ∑m+n>0

tm

m!n!admAad

nB

)kA

=

∞∑k=0

(−1)k

k + 1

∑(m,n)∈Ik

t|m|

m!n!admkA adnkB . . . adm1

A adn1

B A,

where Ik :=

(m,n) ∈ Z2k+ : m+ n > 0

as above. Integrating this equation

then gives,

log(eAeB) = C (1)

= B +

∞∑k=0

(−1)k

k + 1

∑(m,n)∈Ik

1


A adn1

B A

= A+B +

∞∑k=1

(−1)k

k + 1

∑(m,n)∈Ik

1


A adn1

B A.

(6.19)

Remark 6.15. Let us work out a few terms in Eq. (6.12) in terms of the degreep counted in the sum of factors of A and B that appear in the expression. Hereare the terms to order p = 3,

log(eAeB) = A+B +1

2

∞∑m=1

1

m!admAB

− 1

6

∑m+n>0,q≥1

1

m!n!q!(n+ 1)admAad

nBad

qAB + . . .

= A+B +1

2

[A,B] +

1

2[A, [A,B]] + . . .

− 1

6

∑m+n>0

1

m!(n+ 1)!admAad

nB [A,B] + . . .

= A+B +1

2

[A,B] +

1

2[A, [A,B]] + . . .

− 1

6

1

2[B, [A,B]] + [A, [A,B]] + . . .

+ . . .

= A+B +1

2[A,B] +

1

4[A, [A,B]]− 1

12[B, [A,B]]− 1

6[A, [A,B]] + . . .

= A+B +1

2[A,B] +

1

12[A, [A,B]]− 1

12[B, [A,B]] + . . . .

= A+B +1

2[A,B] +

1

12

(ad2AB + ad2

BA)

+ . . .

that is to say eAeB = eΓ (A,B) with

Γ (A,B) = log(eAeB) = A+B +1

2[A,B] +

1

12

(ad2AB + ad2

BA)

+ . . . . (6.20)

For more details, see Strichartz [?].


7

Imbedded Submanifolds

This chapter will introduce the reader to imbedded submanifolds1 of Eu-clidean space E = RN . For later purposes of the stating the divergence theoremwe will be most interested in hypersurfaces, i.e. submanifolds of codimensionone. We will equip E = RN with the standard inner product,

〈a, b〉 = 〈a, b〉RN :=

N∑i=1

aibi.

In general, we will denote inner products in these notes by 〈·, ·〉.

Definition 7.1. A subset M of E (see Figure 7.1) is a d – dimensionalimbedded submanifold (without boundary) of E iff for all m ∈M, there is afunction z : E → RN such that:

1. D(z) is an open neighborhood of E containing m,2. R(z) is an open subset of RN ,3. z : D(z)→ R(z) is a diffeomorphism (a smooth invertible map with smooth

inverse), and4. z(M ∩ D(z)) = R(z) ∩ (Rd × 0) ⊂ RN .

(We write Md if we wish to emphasize that M is a d – dimensional mani-fold.)

Notation 7.2 Given an imbedded submanifold and diffeomorphism z as in theabove definition, we will write z = (z<, z>) where z< is the first d componentsof z and z> consists of the last N − d components of z. Also let x : M → Rddenote the function defined by D (x) := M ∩D(z) and x := z<|D(x). Notice that

R (x) := x(D (x)) is an open subset of Rd and that x−1 : R (x)→ D (x) , thoughtof as a function taking values in E, is smooth. The bijection x : D (x)→ R (x)is called a chart on M. Let A = A(M) denote the collection of charts on M.The collection of charts A = A(M) is often referred to as an atlas for M.

1 There is an abstract theory of manifolds in which imbedded submanifolds forma subclass of objects. We are not going to discuss this abstract theory here. Thereader should take comfort in knowing that the abstract theory is, in the end, nomore general than the theory presented here thanks to the Whitney embeddingtheorem, see for example Theorem 6-3 in Auslander and MacKenzie [?].

Fig. 7.1. An imbedded one dimensional submanifold in R2.

Remark 7.3. The imbedded submanifold M is made into a topological spaceusing the induced topology from E. With this topology, each chart x ∈ A(M)is a homeomorphism from D (x) ⊂o M to R (x) ⊂o Rd.

Remark 7.4. Item 4. in Definition 7.1 may be restated as saying

M ∩ D(z) = z> = 0 := ξ ∈ D (z) : z> (ξ) = 0 . (7.1)

Indeed, we have

z> = 0 =ξ ∈ D (z) : z (ξ) ∈ Rd × 0

=ξ ∈ D (z) : z (ξ) ∈ R (z) ∩

(Rd × 0

)= z−1

(R (z) ∩

(Rd × 0

)).

Combining this observation with the fact that item 4. is equivalent to the state-ment that

M ∩ D(z) = z−1(R (z) ∩

(Rd × 0

))verifies Eq. (7.1).

Theorem 7.5 (A Basic Construction of Manifolds). Let F : E → RN−dbe a smooth function and M := F−1(0) ⊂ E which we assume to be non-empty. Suppose that F ′ (m) : E → RN−d is surjective for all m ∈ M. Then Mis a d – dimensional imbedded submanifold of E.

56 7 Imbedded Submanifolds

Proof. Let m ∈ M, we will begin by constructing a smooth function G :E → Rd such that (G,F )

′(m) : E → RN = Rd × RN−d is invertible. To do

this, let X = Nul(F ′ (m)) and Y be a complementary subspace (for exampleY = X⊥) so that E = X ⊕ Y and let P : E → X be the associated projectionmap, see Figure 7.2. Notice that F ′ (m) : Y → RN−d is a linear isomorphism ofvector spaces and hence

dim (X) = dim(E)− dim(Y ) = N − (N − d) = d.

In particular, X and Rd are isomorphic as vector spaces. Set G (m) = APmwhere A : X → Rd is an arbitrary but fixed linear isomorphism of vector spaces.Then for x ∈ X and y ∈ Y,

(G,F )′(m) (x+ y) = (G′ (m) (x+ y), F ′ (m) (x+ y))

= (AP (x+ y), F ′ (m) y) = (Ax, F ′ (m) y) ∈ Rd × RN−d

from which it follows that (G,F )′(m) is an isomorphism.

Fig. 7.2. Constructing charts for M using the inverse function theorem. For simplicityof the drawing, m ∈M is assumed to be the origin of E = X ⊕ Y.

By the inverse function theorem, there exists a neighborhood U ⊂o E of msuch that V := (G,F )(U) ⊂o RN and (G,F ) : U → V is a diffeomorphism.Let z = (G,F ) with D(z) = U and R(z) = V. Then z is a chart of E aboutm satisfying the conditions of Definition 7.1. Indeed, items 1) – 3) are clear byconstruction. If p ∈ M ∩ D(z) then z(p) = (G(p), F (p)) = (G(p), 0) ∈ R(z) ∩(Rd × 0). Conversely, if p ∈ D(z) is a point such that z(p) = (G(p), F (p)) ∈R(z) ∩ (Rd × 0), then F (p) = 0 and hence p ∈ M ∩ D(z); so item 4) ofDefinition 7.1 is verified.

Second proof. Let m ∈M be given. By the rank-nullity theorem,

N = dim Nul (F ′ (m)) + dim RanF ′ (m) = dim Nul (F ′ (m)) +N − d

from which it follows that dim NulF ′ (m) = d. Hence we may choose a lineartransformation G : E → Rd so that G|NulF ′(m) : NulF ′ (m)→ Rd is an isomor-

phism. The differential of (G,F ) : E → RN = Rd×RN−d at m is the linear map(G,F )

′(m) : E → RN = Rd × RN−d is given by (G,F )

′(m) v = (Gv, F ′ (m) v)

for all v ∈ RN . By our construction of G we have,

Nul((G,F )

′(m)

)= Nul (G) ∩Nul (F ′ (m)) = 0

and hence (G,F )′(m) is invertible.

By the inverse function theorem, there exists a neighborhood U ⊂o E of msuch that V := (G,F )(U) ⊂o RN and (G,F ) : U → V is a diffeomorphism.Let z = (G,F ) with D(z) = U and R(z) = V. Then z is a chart of E aboutm satisfying the conditions of Definition 7.1. Indeed, items 1) – 3) are clear byconstruction. If p ∈ M ∩ D(z) then z(p) = (G(p), F (p)) = (G(p), 0) ∈ R(z) ∩(Rd × 0). Conversely, if p ∈ D(z) is a point such that z(p) = (G(p), F (p)) ∈R(z) ∩ (Rd × 0), then F (p) = 0 and hence p ∈ M ∩ D(z); so item 4) ofDefinition 7.1 is verified.

Lemma 7.6. Suppose g ∈ GL(n,R) and A ∈ gl(n,R), then

det′(g)A = det(g)tr(g−1A). (7.2)

Proof. You were asked to prove this statement in Exercise 1.6. Here isalternative proof for those who did not do the exercise.

By definition we have

det ′(g)A =d

dt|0 det(g + tA) = det(g)

d

dt|0 det(I + tg−1A).

So it suffices to prove ddt |0 det(I + tB) = tr(B) for all matrices B. If B is upper

triangular, then det(I + tB) =∏ni=1(1 + tBii) and hence by the product rule,

d

dt|0 det(I + tB) =

n∑i=1

Bii = tr(B).

This completes the proof because; 1) every matrix can be put into upper trian-gular form by a similarity transformation, and 2) “det” and “tr” are invariantunder similarity transformations.

Example 7.7. Let gl(n,R) denote the set of all n×n real matrices. The followingare examples of imbedded submanifolds.

1. Any open subset M of E.


7 Imbedded Submanifolds 57

2. The graph,

Γ (f) :=

(x, f (x)) ∈ Rd × RN−d : x ∈ D (f)⊂ D (f)× RN−d ⊂ RN ,

of any smooth function f : Rd → RN−d as can be seen by applying Theorem7.5 with F (x, y) := y − f (x) . In this case, is a good idea for the reader tocheck that z (x, y) := (x, y − f (x)) is a chart on D (z) = R (z) = D (f) ×RN−d which could be used verifying Definition 7.1.

3. The unit sphere, SN−1 := x ∈ RN : 〈x, x〉RN = 1, as is seen by applyingTheorem 7.5 with E = RN and F (x) := 〈x, x〉RN − 1. Alternatively, expressSN−1 locally as the graph of smooth functions and then use item 2.

4. GL(n,R) := g ∈ gl(n,R)|det(g) 6= 0, see item 1.5. SL(n,R) := g ∈ gl(n,R)|det(g) = 1 as is seen by taking E = gl(n,R)

and F (g) := det(g) and then applying Theorem 7.5 with the aid of Lemma7.6 below which implies F ′ (g) [gA] = tr (A) for all A ∈ gl(n,R).

6. O(n) := g ∈ gl(n,R)|gtrg = I where gtr denotes the transpose of g. Inthis case take F (g) := gtrg − I thought of as a function from E = gl(n,R)to S(n), where

S(n) :=A ∈ gl(n,R) : Atr = A

is the subspace of symmetric matrices. To show F ′(g) is surjective, show

F ′(g)(gB) = B +Btr for all g ∈ O(n) and B ∈ gl(n,R).

7. SO(n) := g ∈ O(n)|det(g) = 1, an open subset of O(n).8. M × N ⊂ E × V, where M and N are imbedded submanifolds of E andV respectively. The reader should verify this by constructing appropriatecharts for E × V by taking “tensor” products of the charts for E and Vassociated to M and N respectively.

9. The n – dimensional torus,

Tn := z ∈ Cn : |zi| = 1 for i = 1, 2, . . . , n = (S1)n,

where z = (z1, . . . , zn) and |zi| =√zizi. This follows by induction

using items 3. and 8. Alternatively apply Theorem 7.5 with F (z) :=(|z1|2 − 1, . . . , |zn|2 − 1

).

Notation 7.8 Given a non-empty subset Λ of 1, . . . , N with k – elements,let πΛ : RN → Rk be the projection map defined by πΛ (x) = (xi1 , . . . , xik) whereΛ = i1, . . . , ik with 1 ≤ i1 < i2 < · · · < ik ≤ N.

Corollary 7.9 (Local Graph Description). If M is a d – dimensionalimbedded submanifold of E. Then M is locally the graph of a smooth functionof d of the coordinates of RN to the remaining N − d coordinates of RN .

[For example of N = 3 and d = 2, then we are saying that locally M mustbe the graph of a function of one of the forms,

x1 = f (x2, x3) or x2 = f (x1, x3) or x3 = f (x1, x2)

where the domain of f is some open subset R2 in this case. More precisely, Mis locally of the form,

Γ1 = (f (a) , a1, a2) : (a1, a2) ∈ D or Γ2 = (a1, f (a) , a2) : (a1, a2) ∈ D or

Γ1 = (a1, a2, f (a)) : (a1, a2) ∈ D

for some D ⊂o R2 and some smooth function, f : D → R.]

Proof. It is always possible to choose G in the second proof of Theorem 7.5above to be of the form πΛ for some subset Λ ⊂ 1, . . . , N with d – elements.

Indeed as F ′ (m) is surjective we have F ′ (m) eiNi=1 spans RN−d and thereforethere exists Λ ⊂ 1, . . . , N with d – elements such that F ′ (m) ei : i /∈ Λ is abasis for RN−d. Now suppose that ξ ∈ Nul (πΛ)∩Nul (F ′ (m)) . Since πΛ (ξ) = 0,ξ =

∑i/∈Λ aiei for some ai ∈ R and since F ′ (m) ξ = 0,

0 = F ′ (m) ξ =∑i/∈Λ

aiF′ (m) ei.

But F ′ (m) ei : i /∈ Λ is a basis for RN−d and therefore ai = 0 for all i ∈ Λcand therefore ξ = 0. This shows Nul (πΛ) ∩ Nul (F ′ (m)) = 0 and therefore(πΛ, F )

′(m) : RN → RN is invertible.

For notational simplicity, let us reorder the coordinates on RN so that Λ =1, 2, . . . , d . We now define D :=

a ∈ Rd : (a, 0) ∈ R (z)

(an open subset of

Rd) and define Γ : D → RN by Γ (a) := (πΛ, F )−1

(a, 0) . Since πΛ Γ (a) = a,if we define f := πΛc Γ : D → RN−d, we will have Γ (a) = (a, f (a)) for alla ∈ D. Hence it follows that Γ (D) = Γ (f) – the graph of f.

We now finish the proof by showing Γ (D) = M ∩D (z) . For a ∈ D, Γ (a) =

(πΛ, F )−1

(a, 0) ∈ D (z) ∩M since F Γ (a) = 0. Conversely, if ξ ∈M ∩ D (z) ,then (πΛ, F ) (ξ) = (a, 0) for some a ∈ Rd such that (a, 0) ∈ R (z) , i.e. a ∈ Dand ξ = (πΛ, F )

−1(a, 0) = Γ (a) ∈ Γ (D) .

Theorem 7.10 (Parametrized Surfaces I.). Let D be an open neighborhoodof a0 ∈ Rd with 1 ≤ d < N and suppose that Σ : D → RN is a Ck – map suchthat Σ′ (a0) : Rd → RN is injective. Then there exists an open ball, V ⊂ D,centered at a0 such that

1. Σ|V : V → RN is injective and2. M := Σ (V ) is a an imbedded surface inside of RN covered by a single chartz.



Proof. Let Y = RanΣ′ (a0) which is a d – dimensional subspace of RN .Suppose that ud+1, . . . , uN is a basis for Y ⊥ – the orthogonal compliment toY. Then D × RN−d is an open subset of RN and the map

F : D × RN−d 3 (a, b)→ F (a, b) := Σ (a) +

N∑l=d+1

blul. (7.3)

is a Ck – map. According to Exercise 7.1, F ′ (a0, 0) : RN → RN is invertible.By the inverse function, theorem there exists an open ball B ⊂ D×Rd ⊂ RN

centered at (a0, 0) of radius r > 0 such that F (B) is an open subset of E andF : B → F (B) is a Ck – diffeomorphism. We now take

V =x ∈ Rd : (x, 0) ∈ B

=x ∈ Rd : ‖x− a0‖Rd < r

(7.4)

and set D (z) := F (B) , R (z) := B, and z := F |−1B . One then shows

M = Σ (V ) = z> = 0 = ξ ∈ D (z) = F (B) : z> (ξ) = 0 . (7.5)

Exercise 7.1. Verify the two assertions made in the proof of Theorem 7.10.That is show;

1. F ′ (a0, 0) is invertible where F is given as in Eq. (7.3) and2. M = Σ (V ) = z> = 0 as in Eq. (7.5).

Corollary 7.11 (Parametrized Surfaces II.). Let D be an open subset ofRd and Σ : D → E be a Ck – mapping and set M := Σ (D) . Further assume;

1. Σ : D → M is a homeomorphism where the metric on M is defined byρ (m1,m2) := ‖m2 −m1‖E .

2. For each a ∈ D, Σ′ (a) : Rd → E is an injective linear map.

Then M is an imbedded submanifold of E.

Proof. Let m0 = Σ (a0) be a given point in M where a0 ∈ D. Then byTheorem 7.10, there exists an open ball V ⊂ D centered at a0 and a chart z onE with m0 ∈ D (z) such that Σ (V ) = z> = 0 . Since Σ is a homeomorphismwe know that Σ (V ) is open in M and therefore there exists δ > 0 such thatBE (m0, δ) ∩M (an open ρ – ball in M) is contained in Σ (V ) . Let us nownotice thatz>|D(z)∩BE(m0,δ) = 0

=ξ ∈ D (z) ∩BE (m0, δ) : z> (ξ) = 0

= D (z) ∩BE (m0, δ) ∩Σ (V ) = D (z) ∩BE (m0, δ) ∩M

since

D (z)∩BE (m0, δ)∩Σ (V ) ⊂ D (z)∩BE (m0, δ)∩M ⊂ D (z)∩BE (m0, δ)∩Σ (V ) .

Hence by replacing D (z) by D (z)∩BE (m0, δ) ,R (z) by z(D (z) ∩BE (m0, δ)

),

and z by z|D(z)∩BE(m0,δ), we will have found a chart, z, on E such that z0 ∈D (z) and M ∩ D (z) = z> = 0 .

Example 7.12. Let Σ (θ) =(eiθ, eirθ

)in which case Σ′ (θ) is injective for all θ.

Notice that Σ need not be a homeomorphism in general as Σ (−ε, ε) need notrelatively open in T2 for any ε > 0 if the orbits of Σ are dense.

Definition 7.13. Let E and V be two finite dimensional vector spaces andMd ⊂ E and Nk ⊂ V be two imbedded submanifolds. A function f : M → Nis said to be smooth if for all charts x ∈ A(M) and y ∈ A(N) the functiony f x−1 : Rd → Rk is smooth.

Exercise 7.2. Show the composition of smooth maps between imbedded sub-manifolds is smooth. That is if M, N, and P are imbedded submanifolds ofEuclidean spaces and g : M → N and f : N → P are smooth maps, thenf g : M → P is a smooth map.

Exercise 7.3. Let Md ⊂ E and Nk ⊂ V be two imbedded submanifolds as inDefinition 7.13.

1. Show that a function f : Rk → N is smooth iff f is smooth when thoughtof as a function from Rk to E.

2. If F : E → V is a smooth function such that F (M ∩D(F )) ⊂ N, show thatf := F |M : M → N is smooth.

Proposition 7.14. Assuming the notation in Definition 7.13, a function f :M → N is smooth iff there is a smooth function F : D (F ) ⊂o E → V suchthat M ⊂ D (F ) and f = F |M .

Proof. (Sketch.) Suppose that f : M → N is smooth, m ∈ M and n =f(m). Let z be as in Definition 7.1 and w be a chart on N such that n ∈ D(w).By shrinking the domain of z if necessary, we may assume that R(z) = U ×Wwhere U ⊂o Rd and W ⊂o RN−d in which case z(M ∩ D(z)) = U × 0 . Forξ ∈ D(z), let F (ξ) := f(z−1(z<(ξ), 0)) with z = (z<, z>) as in Notation 7.2.Then F : D(z)→ N is a smooth function such that F |M∩D(z) = f |M∩D(z). Thefunction F is smooth. Indeed, letting x = z<|D(z)∩M ,

w< F = w< f(z−1(z<(ξ), 0)) = w< f x−1 (z<(·), 0)

which, being the composition of the smooth maps w< f x−1 (smooth byassumption) and ξ → (z<(ξ), 0), is smooth as well. Hence by definition, F


7.1 Tangent Planes 59

is smooth as claimed. Using a standard partition of unity argument (which weomit), it is possible to piece this local argument together to construct a globallydefined smooth function F : E → V such that f = F |M .

Definition 7.15. A function f : M → N is a diffeomorphism if f is smoothand has a smooth inverse. The set of diffeomorphisms f : M → M is a groupunder composition which will be denoted by Diff(M).

7.1 Tangent Planes

Definition 7.16. Given an imbedded submanifold M ⊂ E and m ∈ M, letτmM ⊂ E denote the collection of all vectors v ∈ E such there exists a smoothpath σ : (−ε, ε) → M with σ(0) = m and v = d

ds |0σ(s). The subset τmM iscalled the tangent plane to M at m and v ∈ τmM is called a tangent vector,see Figure 7.3.

Fig. 7.3. Tangent plane, τmM, to M at m and a vector, v, in τmM.

Theorem 7.17. For each m ∈M, τmM is a d – dimensional subspace of E. Ifz : E → RN is as in Definition 7.1, then τmM = Nul(z′>(m)). If x is a charton M such that m ∈ D (x) , then

dds|0x−1(x(m) + sei)di=1

is a basis for τmM, where eidi=1 is the standard basis for Rd.

Proof. Let σ : (−ε, ε) → M be a smooth path with σ(0) = m and v =dds |0σ(s) and z be a chart (for E) around m as in Definition 7.1 such thatx = z<. Then z>(σ(s)) = 0 for all s and therefore,

0 =d

ds|0z>(σ(s)) = z′>(m)v

which shows that v ∈ Nul(z′>(m)), i.e. τmM ⊂ Nul(z′>(m)). In preparation forthe proof of the converse inclusion let us also notice that

d

ds|0z (σ (s)) =

d

ds|0 (z<(σ(s)), z>(σ(s))) = (z′< (p)σ′ (0) , 0) = (z′< (p) v, 0) .

Now suppose that v ∈ Nul(z′>(m)). We wish to find σ (s) ∈ M withσ (0) = m and σ′ (0) = v. According to the previous equation we must havethat d

ds |0z (σ (s)) = (z′< (m) v, 0) = z′ (m) v which suggests we define

σ (s) = z−1 (z (m) + sz′ (m) v)

= z−1 (z (m) + s (z′< (m) v, 0)) = x−1 (x (m) + sz′< (m) v) .

For this choice of σ (s) we have z (σ (s)) = z (m) + sz′ (m) v and thereforez′ (m)σ′ (0) = z′ (m) v from which it follows that σ′ (0) = v as desired. Wehave now shown Nul(z′>(m)) ⊂ τmM which completes the proof that τmM =Nul(z′>(m)).

Since z′<(m) : τmM = Nul(z′>(m)) → Rd is a linear isomorphism2, theabove argument also shows

d

ds|0x−1(x(m) + sw) = (z′<(m)|τmM )

−1w ∈ τmM ∀ w ∈ Rd.

In particular it follows that

dds|0x−1(x(m) + sei)di=1 = (z′<(m)|τmM )

−1eidi=1

is a basis for τmM, see Figure 7.5 below.The following proposition is an easy consequence of Theorem 7.17 and the

proof of Theorem 7.5.

Proposition 7.18. Suppose that M is an imbedded submanifold constructed asin Theorem 7.5. Then τmM = Nul(F ′(m)) .

Proof. Recall from the proof of Theorem 7.5 that we constructed the chartson E so that z> = F. Therefore τmM = Nul(z′>(m)) = Nul(F ′(m)) .

Exercise 7.4. Referring to Example 7.7. Show:

1. τmM = E, if M is an open subset of E.2. τgGL(n,R) = gl(n,R), for all g ∈ GL(n,R).

2 The reader should convice herself of this fact if it is not evident.



3. τmSN−1 = m⊥ for all m ∈ SN−1.

4. Let sl(n,R) be the traceless matrices,

sl(n,R) := A ∈ gl(n,R)| tr(A) = 0. (7.6)

ThenτgSL(n,R) = A ∈ gl(n,R)|g−1A ∈ sl(n,R)

and in particular τISL(n,R) = sl(n,R).5. Let so (n,R) be the skew symmetric matrices,

so (n,R) := A ∈ gl(n,R)|A = −Atr.

ThenτgO(n) = A ∈ gl(n,R)|g−1A ∈ so (n,R)

and in particular τIO (n) = so (n,R) .6. If M ⊂ E and N ⊂ V are imbedded submanifolds then

τ(m,n)(M ×N) = τmM × τnN ⊂ E × V.

It is quite possible that τmM = τm′M for some m 6= m′, with m and m′

in M (think of the sphere). Because of this, it is helpful to label each of thetangent planes with their base point.

7.2 Lagrange Multipliers

Let N ≥ 2, 1 ≤ d < N, V = RN−d, and suppose that F : RN → V andE : RN → R are smooth functions. Let M0 := F−1 (0) = F = 0 andsuppose that m ∈ M0 is a local minimizer (or maximizer) of E|M0

. Then forany smooth curve, σ (t) ∈ RN with σ (0) = m and σ (−ε, ε) ⊂ M0 for someε > 0, we know by the first derivative test that E′ (m) σ (0) = d

dt |0E (σ (t)) = 0.This is the beginning of the proof of the following theorem.

Theorem 7.19. Suppose that m ∈ M0 is a local minimizer (or maximizer) ofE|M0

and that F ′ (m) : TmRN → V is surjective. Then there exists a uniqueλ ∈ V ∗ (called the Lagrange multiplier) such that E′ (m) = λ F ′ (m) on RN .

Proof. Since the map F ′ (x) being surjective is an open condition, we mayfind an open neighborhood, O ⊂ RN of m such that F ′ (x) is surjective for allx ∈ O. It then follows that M0 ∩ O is an embedded submanifold of O ⊂ RNand τxM0 = Nul (F ′ (x)) for all x ∈M0∩O. Thus if E|M0 has a local minimum(maximum) at m ∈ M0, then by the first derivative test, E′ (m) v = 0 forall v ∈ τmM0 = Nul (F ′ (m)) . As F ′ (m) is surjective, we may factor E′ (m)through F ′ (m) , i.e. there exists a unique λ ∈ V ∗ such that E′ (m) = λF ′ (m) .

Alternative more direct proof. As in the second proof of Theorem7.5, choose a linear transformation G : RN → Rd such that (G,F )

′(m) =

(G,F ′ (m)) is invertible. Given v ∈ Nul (F ′ (m)) we wish to find a curveσ (t) ∈ M := F−1 (0) such that σ (0) = m and σ (0) = v. If this curveexists it satisfies (G,F ) (σ (t)) = (Gσ (t) , 0) where Gσ (t) = Gm+ tGv + o (t) .So let us define σ (t) by requiring (G,F ) (σ (t)) = (Gm+ tGv, 0) , i.e. σ (t) =

(G,F )−1

(Gm+ tGv, 0) which is well defined by the inverse function theorem.Since F (σ (t)) = 0, σ (t) ∈ M. Since (G,F ) (σ (0)) = (Gm, 0) = (G,F ) (m) ,σ (0) = m and since

(G,F )′(m) σ (0) =

d

dt|0 (G,F ) (σ (t)) =

d

dt|0 (Gm+ tGv, 0)

= (Gv, 0) = (G,F )′(m) v

(wherein we have used F ′ (m) v = 0 for the last equality), we may concludeσ (0) = v. Hence if m ∈ M0 is a local minimizer of E we will have by thefirst derivative test that 0 = d

dt |0E (σ (t)) = E′ (m) v. And so we have shownE′ (m) v = 0 whenever F ′ (m) v = 0.As F ′ (m) is surjective, we may factorE′ (m) through F ′ (m) , i.e. there exists a unique λ ∈ V ∗ such that E′ (m) =λ F ′ (m) .

Let us consider two cases where the condition F ′ (m) being surjective is notmet.

Example 7.20. Suppose that F : R2 → R is defined by F (x, y) = xy. In thiscase F ′ (x, y) (v, w)(x,y) = yv + xw and hence F ′ (x, y) is surjective everywhere

except at (0, 0) . Moreover M0 = R × 0 ∪ 0 × R is not a submanifold ofR2 because of the cross structure at (0, 0) . Now suppose that E : R2 → R is asmooth function such that E|M0

has a local minimum at (0, 0) . In this case itfollows, using the observation before Theorem 7.19, that

∂E

∂x(0, 0) = 0 =

∂E

∂y(0, 0)

and therefore E′ (0.0) ≡ 0. In particular, for any λ ∈(R2)∗

we will haveE′ (0.0) = λ F ′ (0, 0) . Hence there still is a Lagrange multiplier in this ex-ample but it is certainly not unique.

Example 7.21. Suppose that F : R2 → R is defined by F (x, y) =(y − x2

) (y + x2

)= y2 − x4 so that

M0 =y = x2

∪y = −x2

.

In this case F ′ (x, y) (v, w)(x,y) = 2yv − 4x3w and F ′ (x, y) is surjective every-

where except at (0, 0) . Also notice that M0 is not an embedded submanifold ofR2. Now let E (x, y) = x2 + 1

2y and observe that


7.3 Tangent Space 61

Fig. 7.4. The subset, M0 ⊂ R2.

E(x, x2

)=

(1 +

1

2

)x2 and E

(x,−x2

)=

(1− 1

2

)x2

from which it follows that (0, 0) is a minimum of E|M0 . However sinceE′ (0, 0) (v, w) = 1

2w 6= 0 and F ′ (0, 0) = 0, there is no λ ∈(R2)∗

such thatE′ (0.0) = λ F ′ (0, 0) . Thus this is an example where no Lagrange multiplier,λ, exists and we would miss the local minimum of E|M0

at (0, 0) .

Exercise 7.5 (Compare with Exercise ??). Let F : RN → R be the map

defined by F (x) = ‖x‖2 − 1 = x · x − 1 so that M0 =x ∈ RN : ‖x‖ = 1

.

Further let E (x) = Qx · x where Q is a symmetric real N × N matrix andlet x0 ∈ M0 be a point where E (x0) := max E (x) : x ∈M0 . Show that themethod of Lagrange multipliers implies there exists λ0 ∈ R such that Qx0 =λ0x0.

7.3 Tangent Space

Definition 7.22. The tangent space (TmM) to M at m is given by

TmM := m × τmM ⊂M × E.

LetTM := ∪m∈MTmM,

and call TM the tangent space (or tangent bundle) of M. A tangentvector is a point vm := (m, v) ∈ TM and we let π : TM → M denote thecanonical projection defined by π(vm) = m. Each tangent space is made intoa vector space with the vector space operations being defined by: c(vm) := (cv)mand vm + wm := (v + w)m.

Exercise 7.6. Prove that TM is an imbedded submanifold of E × E. Hint:suppose that z : E → RN is a function as in the Definition 7.1. Define D(Z) :=D(z) × E and Z : D(Z) → RN × RN by Z(x, a) := (z (x) , z′ (x) a). Use Z’s ofthis type to check TM satisfies Definition 7.1.

Notation 7.23 In the sequel, given a smooth path σ : (−ε, ε) → M, we willabuse notation and write σ′ (0) for either

d

ds|0σ(s) ∈ τσ(0)M

or for

(σ(0),d

ds|0σ(s)) ∈ Tσ(0)M = σ(0) × τσ(0)M.

Also given a chart x = (x1, x2, . . . , xd) on M and m ∈ D (x) , let ∂/∂xi|m denotethe element TmM determined by ∂/∂xi|m = σ′(0), where σ(s) := x−1(x(m) +sei), i.e.

∂

∂xi|m = (m,

d

ds|0x−1(x(m) + sei)), (7.7)

see Figure 7.5.

Fig. 7.5. Forming a basis of tangent vectors.

The reason for the strange notation in Eq. (7.7) will be explained afterNotation 7.25. By definition, every element of TmM is of the form σ′ (0) whereσ is a smooth path into M such that σ (0) = m. Moreover by Theorem 7.17,∂/∂xi|mdi=1 is a basis for TmM.

Definition 7.24. Suppose that f : M → V is a smooth function, m ∈ D(f)and vm ∈ TmM. Write



vmf = df(vm) :=d

ds|0f(σ(s)),

where σ is any smooth path in M such that σ′(0) = vm. The function df :TM → V will be called the differential of f.

Notation 7.25 If M and N are two manifolds f : M × N → V is a smoothfunction, we will write dMf (·, n) to indicate that we are computing the differ-ential of the function m ∈M → f(m,n) ∈ V for fixed n ∈ N.

To understand the notation in (7.7), suppose that f = F x =F (x1, x2, . . . , xd) where F : Rd → R is a smooth function and x is a charton M. Then

∂f(m)

∂xi:=

∂

∂xi|mf = (DiF )(x(m)),

where Di denotes the ith – partial derivative of F. Also notice that

dxj(∂∂xi |m

)= δij so that

dxi|TmM

di=1

is the dual basis of ∂/∂xi|mdi=1

and therefore if vm ∈ TmM then

vm =

d∑i=1

dxi (vm)∂

∂xi|m. (7.8)

This explicitly exhibits vm as a first order differential operator acting on“germs” of smooth functions defined near m ∈M.

Remark 7.26 (Product Rule). Suppose that f : M → V and g : M → End(V )are smooth functions, then

vm(gf) =d

ds|0 [g(σ(s))f(σ(s))] = vmg · f(m) + g(m)vmf

or equivalently

d(gf)(vm) = dg(vm)f(m) + g(m)df(vm).

This last equation will be abbreviated as d(gf) = dg · f + gdf.

Definition 7.27. Let f : M → N be a smooth map of imbedded submanifolds.Define the differential, f∗, of f by

f∗vm = (f σ)′(0) ∈ Tf(m)N,

where vm = σ′(0) ∈ TmM, and m ∈ D(f).

Lemma 7.28. The differentials defined in Definitions 7.24 and 7.27 are welldefined linear maps on TmM for each m ∈ D(f).

Fig. 7.6. The differential of f.

Proof. I will only prove that f∗ is well defined, since the case of df is similar.By Proposition 7.14, there is a smooth function F : E → V, such that f = F |M .Therefore by the chain rule

f∗vm = (f σ)′(0) :=

[d

ds|0f(σ(s))

]f(σ(0))

= [F ′(m)v]f(m) , (7.9)

where σ is a smooth path in M such that σ′(0) = vm. It follows from (7.9) thatf∗vm does not depend on the choice of the path σ. It is also clear from (7.9),that f∗ is linear on TmM.

Remark 7.29. Suppose that F : E → V is a smooth function and that f := F |M .Then as in the proof of Lemma 7.28,

df(vm) = F ′(m)v (7.10)

for all vm ∈ TmM , and m ∈ D(f). Incidentally, since the left hand sides of (7.9)and (7.10) are defined “intrinsically,” the right members of (7.9) and (7.10) areindependent of the possible choices of functions F which extend f.

Lemma 7.30 (Chain Rules). Suppose that M, N, and P are imbedded sub-manifolds and V is a finite dimensional vector space. Let f : M → N,g : N → P, and h : N → V be smooth functions. Then:

(g f)∗vm = g∗(f∗vm), ∀ vm ∈ TM (7.11)

andd(h f)(vm) = dh(f∗vm), ∀ vm ∈ TM. (7.12)

These equations will be written more concisely as (g f)∗ = g∗f∗ and d(hf) =dhf∗ respectively.



Proof. Let σ be a smooth path in M such that vm = σ′(0). Then, see Figure7.7,

(g f)∗vm := (g f σ)′(0) = g∗(f σ)′(0)

= g∗f∗σ′(0) = g∗f∗vm.

Similarly,

d(h f)(vm) :=d

ds|0(h f σ)(s) = dh((f σ)′(0))

= dh(f∗σ′(0)) = dh(f∗vm).

Fig. 7.7. The chain rule.

If f : M → V is a smooth function, x is a chart on M , and m ∈ D(f)∩D (x) ,we will write ∂f(m)/∂xi for df

(∂/∂xi|m

). Combining this notation with Eq.

(7.8) leads to the pleasing formula,

df =

d∑i=1

∂f

∂xidxi, (7.13)

by which we mean

df(vm) =

d∑i=1

∂f(m)

∂xidxi(vm).

Suppose that f : Md → Nk is a smooth map of imbedded submanifolds,m ∈ M, x is a chart on M such that m ∈ D (x) , and y is a chart on N suchthat f(m) ∈ D (y) . Then the matrix of

f∗m := f∗|TmM : TmM → Tf(m)N

relative to the bases ∂/∂xi|mdi=1 of TmM and ∂/∂yj |f(m)kj=1 of Tf(m)N is

(∂(yj f)(m)/∂xi). Indeed, if vm =∑di=1 v

i∂/∂xi|m, then

f∗vm =

k∑j=1

dyj(f∗vm)∂/∂yj |f(m)

=

k∑j=1

d(yj f)(vm)∂/∂yj |f(m) (by Eq. (7.12))

=

k∑j=1

d∑i=1

∂(yj f)(m)

∂xi· dxi(vm)∂/∂yj |f(m) (by Eq. (7.13))

=

k∑j=1

d∑i=1

∂(yj f)(m)

∂xivi∂/∂yj |f(m).

Example 7.31. Let M = O(n), k ∈ O(n), and f : O(n) → O(n) be defined byf(g) := kg. Then f is a smooth function on O(n) because it is the restrictionof a smooth function on gl(n,R). Given Ag ∈ TgO(n), by Eq. (7.9),

f∗Ag = (kg, kA) = (kA)kg

(In the future we denote f by Lk; Lk is left translation by k ∈ O(n).)

Definition 7.32. A Lie group is a manifold, G, which is also a group such thatthe group operations are smooth functions. The tangent space, g := Lie (G) :=TeG, to G at the identity e ∈ G is called the Lie algebra of G.

Exercise 7.7. Verify that GL(n,R), SL(n,R), O(n), SO(n) and Tn (see Ex-ample 7.7) are all Lie groups and

Lie (GL(n,R)) ∼= gl(n,R),

Lie (SL(n,R))) ∼= sl(n,R)

Lie (O(n))) = Lie (SO(n))) ∼= so(n,R) and

Lie (Tn)) ∼= (iR)n ⊂ Cn.

See Exercise 7.4 for the notation being used here.



Suppose that Q ∈ GLn (R) and q (a, b) := Qa · b for all a, b ∈ Rn. Let

GQ := g ∈ GLn (R) : q (ga, gb) = q (a, b) for all a, b ∈ Rn .

Let us observe that if g1, g2 ∈ GQ then g1g2 ∈ GQ so that GQ is a group. Since

q (ga, gb) = Qga · gb = gtrQga · b

we can rewrite GQ as,

GQ =g ∈ GLn (R) : gtrQg = Q

.

From this description of GQ we easily see that GQ is a closed subgroup ofGLn (R) and is therefore a Lie group. We then know that

TIGQ =A ∈ gln (R) : AtrQ+QA = 0

.

I have not determined the dimension of this space in general. The next theoremconsiders two special cases of the above construction.

Theorem 7.33. Let us continue the notation and assumptions as above withthe further assumption that Q is invertible. Also let

F (g) := gtrQg −Q.

1. If Qtr = Q, then F : GLn (R) → Sn – the n × n symmetric matrices hasa surjective differential at all points in GQ. Hence the above results followfrom the inverse function theorem and we know that

LieGQ = TIGQ =A ∈ gln (R) : [QA]

tr= −QA

.

2. If Qtr = −Q, then n ∈ 2N, F : GLn (R)→ An – the n× n skew-symmetricmatrices has a surjective differential at all points in GQ. Hence the aboveresults follow from the inverse function theorem and we know that

LieGQ = TIGQ =A ∈ gln (R) : [QA]

tr= QA

.

Proof. First suppose that A ∈ gln (R) and g ∈ GQ, then

F ′ (g) [gA] =d

dt|0F

(getA

)=

d

dt|0[etA

tr

gtrQgetA −Q]

= AtrgtrQg + gtrQgA = AtrQ+QA.

We now consider each of the cases.

1. If Qtr = Q we can set A = Q−1B where B is a symmetric matrix to learn

F ′ (g) [gA] = Btr[Q−1

]trQ+QQ−1B = Btr +B

which is the arbitrary symmetric matrix.2. If Qtr = −Q we set A = Q−1B where B is a skew-symmetric matrix to

learnF ′ (g) [gA] = Btr

[Q−1

]trQ+QQ−1B = −Btr +B

which is the arbitrary skew-symmetric matrix.

Exercise 7.8 (Continuation of Exercise 7.6). Show for each chart x on Mthat the function

ϕ(vm) := (x(m), dx(vm)) = x∗vm

is a chart on TM. Note that D(ϕ) := ∪m∈D(x)TmM.

The following lemma gives an important example of a smooth function onM which will be needed when we consider M as a “Riemannian manifold.”

Lemma 7.34. Suppose that (E, 〈·, ·〉) is an inner product space and the M ⊂ Eis an imbedded submanifold. For each m ∈ M, let P (m) denote the orthogonalprojection of E onto τmM and Q(m) := I−P (m) denote the orthogonal projec-tion onto τmM

⊥. Then P and Q are smooth functions from M to gl(E), wheregl(E) denotes the vector space of linear maps from E to E.

Proof. Let z : E → RN be as in Definition 7.1. To simplify notation, letF (p) := z>(p) for all p ∈ D(z), so that τmM = Nul (F ′(m)) for m ∈ D (x) =D(z) ∩M. Since F ′(m) : E → RN−d is surjective, an elementary exercise inlinear algebra shows

(F ′(m)F ′(m)∗) : RN−d → RN−d

is invertible for all m ∈ D (x) . The orthogonal projection Q (m) may be ex-pressed as;

Q(m) = F ′(m)∗(F ′(m)F ′(m)∗)−1F ′(m). (7.14)

Since being invertible is an open condition, (F ′(·)F ′(·)∗) is invertible in an openneighborhood N ⊂ E of D (x) . Hence Q has a smooth extension Q to N givenby

Q (x) := F ′ (x)∗

(F ′ (x)F ′ (x)∗)−1F ′ (x) .

Since Q|D(x) = Q|D(x) and Q is smooth on N , Q|D(x) is also smooth. Since zas in Definition 7.1 was arbitrary and smoothness is a local property, it followsthat Q is smooth on M. Clearly, P := I −Q is also a smooth function on M.



Definition 7.35. A local vector field Y on M is a smooth function Y : M →TM such that Y (m) ∈ TmM for all m ∈ D(Y ), where D(Y ) is assumed to bean open subset of M. Let Γ (TM) denote the collection of globally defined (i.e.D(Y ) = M) smooth vector-fields Y on M.

Note that ∂/∂xi are local vector-fields on M for each chart x ∈ A(M) andi = 1, 2, . . . , d. The next exercise asserts that these vector fields are smooth.

Exercise 7.9. Let Y be a vector field on M, x ∈ A(M) be a chart on M andY i := dxi(Y ). Then

Y (m) :=

d∑i=1

Y i (m) ∂/∂xi|m ∀ m ∈ D (x) ,

which we abbreviate as Y =∑di=1 Y

i∂/∂xi. Show the condition that Y issmooth translates into the statement that each of the functions Y i is smooth.

Exercise 7.10. Let Y : M → TM, be a vector field. Then

Y (m) = (m, y(m)) = y(m)m

for some function y : M → E such that y(m) ∈ τmM for all m ∈ D(Y ) = D (y) .Show that Y is smooth iff y : M → E is smooth.

Example 7.36. Let M = SL(n,R) and A ∈ sl(n,R) = τISL(n,R), i.e. A is an×n real matrix such that tr (A) = 0. Then A(g) := Lg∗Ae = (g, gA) for g ∈Mis a smooth vector field on M.

Example 7.37. Keep the notation of Lemma 7.34. Let y : M → E be any smoothfunction. Then Y (m) := (m,P (m)y(m)) for all m ∈M is a smooth vector-fieldon M.

Definition 7.38. Given Y ∈ Γ (TM) and f ∈ C∞(M), let Y f ∈ C∞(M) bedefined by (Y f)(m) := df(Y (m)), for all m ∈ D(f) ∩ D(Y ). In this way thevector-field Y may be viewed as a first order differential operator on C∞(M).

Notation 7.39 The Lie bracket of two smooth vector fields, Y and W, on Mis the vector field [Y,W ] which acts on C∞(M) by the formula

[Y,W ]f := Y (Wf)−W (Y f), ∀ f ∈ C∞(M). (7.15)

(In general one might suspect that [Y,W ] is a second order differential operator,however this is not the case, see Exercise 7.11.) Sometimes it will be convenientto write LYW for [Y,W ].

Exercise 7.11. Show that [Y,W ] is again a first order differential operatoron C∞(M) coming from a vector-field. In particular, if x is a chart on M,

Y =∑di=1 Y

i∂/∂xi and W =∑di=1W

i∂/∂xi, then on D (x) ,

[Y,W ] =

d∑i=1

(YW i −WY i)∂/∂xi. (7.16)

You have shown in the previous exercise that [Γ (TM) , Γ (TM)] ⊂Γ (TM) and hence Γ (TM) is a Lie algebra by item 2. of Example 5.5 withg =End(C∞(M)) and h := Γ (TM) .

Proposition 7.40. If Y (m) = (m, y(m)) and W (m) = (m,w(m)) and y, w :M → E are smooth functions such that y(m), w(m) ∈ τmM, then we mayexpress the Lie bracket, [Y,W ](m), as

[Y,W ](m) = (m, (Y w −Wy)(m)) = (m, dw(Y (m))− dy(W (m))). (7.17)

Proof. Let f be a smooth function M which we may take, by Proposition7.14, to be the restriction of a smooth function on E. Similarly we we mayassume that y and w are smooth functions on E such that y(m), w(m) ∈ τmMfor all m ∈M. Then

(YW −WY )f = Y [f ′w]−W [f ′y]

= f ′′(y, w)− f ′′(w, y) + f ′ (Y w)− f ′ (Wy)

= f ′ (Y w −Wy) (7.18)

wherein the last equality we have use the fact that mixed partial derivativescommute to conclude

f ′′(u, v)− f ′′(v, u) := (∂u∂v − ∂v∂u) f = 0 ∀ u, v ∈ E.

Taking f = z> in Eq. (7.18) with z = (z<, z>) being a chart on E as inDefinition 7.1, shows

0 = (YW −WY )z> (m) = z′> (dw(Y (m))− dy(W (m)))

and thus (m, dw(Y (m))− dy(W (m))) ∈ TmM. With this observation, we thenhave

f ′ (Y w −Wy) = df ((m, dw(Y (m))− dy(W (m))))

which combined with Eq. (7.18) verifies Eq. (7.17).

Exercise 7.12. Let M = SL(n,R) and A,B ∈ sl(n,R) and A and B be theassociated left invariant vector fields on M as introduced in Example 7.36. Show[A, B

]= [A,B] where [A,B] := AB −BA is the matrix commutator of A and

B.


8

Vector Bundles

8.1 Fiber bundles

Definition 8.1. Let F be a fixed smooth manifold. A fiber bundle with modelfiber F is a triple (E,M, π) with the following properties:

1. E and M are smooth manifolds and π : E → M is a smooth map of Eonto M. We refer to E as the total space and M as the base space.For m ∈ M , set Em ≡ π−1(m) which we refer to as the fiber over m.For each U ⊂M we write EU ≡ ∪m∈UEm = π−1(U).

2. For each point m ∈M , there is an open neighborhood U ⊂M of m and asmooth (local trivialization) map Ψ : EU → F, such that (π, Ψ) : EU →U × F is a diffeomorphism of manifolds.

We often summarize the data presented in the above definition as

F → Eπ→M or by

F → E↓ πM

.

Example 8.2. The trivial bundle associate to M and F has total space of theform E = M × F.

Definition 8.3. A fiber bundle F → Eπ→ M is said to be trivial if there is a

local trivialization of F which is defined on all of M.

Definition 8.4. Let F → Eπ→ M be a fiber bundle. A section S of E is a

function S : M → E such that π S = id|M . Let Γ (E) denote the collection ofall sections on E. If U ⊂o M and S : U → E is a map such that π S = id|U ,we call S a local section of E. Write Γ (EU ) for the collection of local sectionsdefined on U. In the above definitions we write S ∈ Γ∞(E) or Γ∞(EU ) ifS : M → E or S : U → E, respectively, is smooth.

We are not going to work with such general bundles here but only with twospecial cases of vector and principal bundles where F is either a vectorspace of a Lie group. In each of these cases we will assume more structure onthe local trivialization maps.

8.2 Vector Bundles

Definition 8.5. Suppose that W is a finite dimensional vector space. A vectorbundle E over M with model space W is a fiber bundle W → E

π→ M suchthat,

1. each fiber, Em, of π is endowed with a vector space structure,2. for each m ∈ M there exists a local trivialization, (U,ψ) of EU so thatψx := ψ|Ex : Ex →W is a linear isomorphism for all x ∈ U.

3. We call uψ (x) := ψ−1x : W → Ex the frame associated to ψ.

4. If (V, ϕ) is another local trivialization of E, we say that gψ,ϕ (x) :=

uψ (x)−1uϕ (x) defined for x ∈ U ∩ V the transition function from ϕ to

ψ.

Remark 8.6. Here are a few basic remarks.

1. The transition functions, gψ,ϕ : U ∩ V → GL (W )m are smooth since

gψ,ϕ (x)w = uψ (x)−1uϕ (x)w = ψ

((π, ϕ)

−1(x,w)

)is smooth.

2. If (η,O) is another local trivialization of E, then for x ∈ V ∩U ∩O we have

gψ,ϕ (x) gϕ,η (x) = uψ (x)−1uϕ (x)uϕ (x)

−1uη (x)

−1= gψ,η (x) .

This last equation is referred to as the cocycle condition which incidentallyimplies the obvious property that g−1

ψ,ϕ = gϕ,ψ.3. Rather than specifying local trivializations ψ, we often specify the local

frames uψ. Given a local frame, u, over U ⊂o M we can recover ψ using

ψ (u (x)w) := w for all w ∈W and x ∈ U.

Theorem 8.7. Suppose that M is a manifold, Ukk∈Λ is a finite or countableopen cover of M, and gkl ∈ C∞ (Uk ∩ Ul, GL (W )) are given maps. If gklk,l∈Λsatisfy the cocycle condition,

gklglm = gkm on Uk ∩ Ul ∩ Um for all k, l,m ∈ Λ,

then there exists a vector bundle E over M modeled on W having these transi-tion functions.

68 8 Vector Bundles

Proof. Here is a sketch of the construction. Let

E := ∪k∈Λ [k × Uk ×W ]

and define an equivalence relation on E by saying (k, x, w) ∼ (l, y, v) iff x = yand w = gkl (x) v. Then take E := E/ ∼ and let π : E → M be define byπ ([(k, x, w)]) = x. Notice that

EUk := π−1 (Uk) = [(k, x, w)] : (x,w) ∈ Uk ×W

and so we may define ψk : EUk →W by

ψk ([(k, x, w)]) = w for all w ∈W.

Notice that if x ∈ Ul, then

ψl ([(k, x, w)]) = ψl ([(l, x, glk (x)w)]) = glk (x)w = glk (x)ψk ([(k, x, w)]) .

This shows that glk are the desired transition functions. We may put a vectorspace structure on Ex by requiring the maps ψk|Ex to be a linear isomorphism.More details may be found in Steenrod [S].

Example 8.8. E = M ×W, Em = m×W , and u(x)ξ = (x, ξ) for all x ∈M .E = M ×W is called a trivial vector bundle.

Example 8.9. Tangent bundle TM of M with model space Rd where d is thedimension of M . If m ∈ M and x is a chart on M with m ∈ U := D (x) , wemay define ψx : TUM → Rd by

ψx (vm) =(dx1, . . . , dxd

)(vm) .

Example 8.10. If M = G is a Lie group that TG is the trivial bundle. Indeedlet ψ : TG→ TIG be defined by ψ (g) := Lg−1∗.

Remark 8.11. There are lots of constructions one can do with vector bundles tomake new one out of old ones. For example, if E and F are vector bundles overM then we can define E ⊕ F and E ⊗M so that [E ⊕ F ]m = Em ⊕ Fm and[E ⊗ F ]m = Em ⊗ Fm. Similarly we may can define E∗ so that E∗m = (Em)

∗

and for k ∈ N we can define Λk (E) so that[Λk (E)

]m

= Λk (Em) .

Definition 8.12. Let W → E →M be a vector bundle.

1. A section (S) of E is a smooth function S : M −→ E such that S(m) ∈Em ∀m ∈M . Let Γ (E) denote the sections of E.

2. A section along a parameterized curve (s −→ σ(s)) in M is a function(s −→ S(s)) such that S(s) ∈ Eσ(s) for all s.

3. A section along a parameterized surface ((s, t) −→ γ(s, t)) in M is a func-tion ((s, t) −→ S(s, t)) such that S(s, t) ∈ Eγ(s,t) for all (s, t).

Definition 8.13. A vector field X on M is a smooth section of the tangentbundle TM. A differential k - form is a section of the bundle Λk(T ∗M) to bedefined later.

Exercise 8.1. Show X ∈ Γ (TM) is smooth iff for each x ∈ A(M), there aresmooth functions ai : U → R (i = 1, . . . ,dim(M)) such that X =

∑i ai ∂∂xi .

(You should note that the functions ai are necessarily given by ai = Xxi =dxi〈X〉.)

Exercise 8.2. Show X ∈ Γ (TM) is smooth iff for each f ∈ C∞(M), Xf ∈C∞(M); where (Xf)(m) ≡ X(m)f = df〈X(m)〉.

1. Definition 8.14. Suppose that V → E → M and V → F → M are vectorbundles over M. A smooth function ϕ : E → F is a vector bundle iso-morphism if ϕ(Em) = Fm for all m ∈ M and ϕ|Em : Em → Fm is anisomorphism for all m ∈M.

Exercise 8.3. Suppose that ϕ : E → F is a vector bundle isomorphism. DefineΨ = ϕ−1, so that Ψ |Fm = ϕ|−1

Em. Show that Ψ : F → E is smooth and hence is

also a vector bundle isomorphism.

8.3 Sub-bundles

In this section let V → Eπ→ M be a vector bundle. All vector spaces and

manifolds are finite dimensional unless otherwise stated.

Definition 8.15. A sub-bundle F of E is an embedded submanifold (F ) suchthat Fm ≡ Em∩F is a subspace of Em for all m ∈M and dimFm is a constanton M. (This need only be required if M is not connected.)

Proposition 8.16. Suppose that F ⊂ E is a subset of E and V has a directsum decomposition V = X⊕Y such that for all o ∈M ∃U ∈ τo(M) and a localtrivialization ϕ : π−1(U) → V of E, such that ϕ(F ∩ Em) = X for all m ∈ U.Then F is a sub-bundle of E. Furthermore, F is a vector-bundle.

Proof. Write Fm ≡ F ∩ Em = π|−1F (m). For each local trivialization ϕ :

π−1(U) → V such that ϕ(Fm) = X for all m ∈ U, set ϕ ≡ ϕ|F∩π−1(U). LetΨ : π−1(W )→ V be another such local trivialization, for x ∈ X

ϕm Ψ−1m (x) = ϕm Ψ−1

m (x) = gϕΨ (m)(x) ∈ X.


8.3 Sub-bundles 69

Therefore gϕΨ (m) = gϕΨ (m)|X , and hence the ϕ’s form local trivializations ofF which make F into a vector bundle. It is easy to check that Fm is a vector-subspace of Em and the vector space structure on Fm is that induced from Em.So it only remains to show that F is an embedded submanifold of E.

For this let Q ∈End(V ) denote projection onto Y, i.e. Q(x + y) = y forall x ∈ X and y ∈ Y . Suppose that ϕ is a local trivialization of E as in thestatement of the proposition. Since F ∩ π−1(U) = Qϕ = 0 and Qϕ hassurjective differential (check), it follows that F is an embedded submanifold ofE.

Definition 8.15 of a sub-bundle is somewhat inconvenient. In fact, it is noteven clear from this definition that a sub-bundle F is a vector-bundle. Laterin this section we will prove that any sub-bundle F may be described as inthe above proposition, see Proposition 8.29 below. Because of Proposition 8.29below, the reader is well advised to consider Proposition 8.16 as a workingdefinition of a sub-bundle.

Example 8.17. Let Fm ≡ 0m ⊂ Em for all m ∈M, and F = ∪m∈MFm. ThenF is the trivial sub-bundle.

Example 8.18. Suppose that E = M × V is the a trivial bundle and X ⊂ V isa subspace, then F = M ×X is a sub-bundle of E.

Example 8.19. Let V → Fπ→ M be any fiber bundle. For each ξ ∈ F, let

Vξ = kerπ∗ξ. Then V ≡ ∪ξ∈FVξ is a sub-bundle of TF. The sub-bundle V iscalled the vertical sub-bundle of TF.

Proof. Let p : TF → F denote the canonical projection map, d = dimM,and k = dimV . Suppose that ξ ∈ F is given and m ≡ π(ξ). Choose x ∈ Am(M)and shrink the domain D(x) if necessary so that there is a local trivializationϕ : π−1(D(x)) → V of F. Let y ∈ Aϕ(ξ)(V ) and set x ≡ x π and y ≡y ϕ. The (x, y) is a chart on F with domain U ≡ ϕ−1(D(y)). Therefore,(dx, dy) : p−1(U) → Rd × Rk is a local trivialization TF. It is easy to checkthat (dx, dy)(Vξ) = 0 × Rk, which shows by Proposition 8.16 that V is asub-bundle of TF.

In order to give some more examples of sub-bundles, we need some prepara-tory lemmas.

Lemma 8.20. Let V be a Hilbert space and P ∈End(V ) (End(V ) denotesthe space of bounded linear operators on V ) be an orthogonal projection. IfP ′ ∈End(V ) is another orthogonal projection such that ‖P − P ′‖ < 1/2, thenP are P ′ are conjugate. Moreover, if J ≡ P ′P + Q′Q, where Q ≡ I − P andQ′ ≡ I − P ′, then J is invertible and JP = P ′J.

Proof. First notice that JP = P ′P and P ′J = P ′P. Hence, JP = P ′J. Soit suffices to show that J is invertible. But,

J ≡ P ′P +Q′Q = P ′P + (I − P ′)(I − P )

= I − P ′ − P + 2P ′P

= I + P ′(P − P ′) + (P ′ − P )P.

So J will be invertible provided δ ≡ P ′(P −P ′)+(P ′−P )P has norm less than1. But ‖δ‖ ≤ 2‖P − P ′‖ < 2 1

2 = 1, where we have used ‖P‖ = ‖P ′‖ = 1 (or 0if P or P ′ is projection onto 0).

In the above lemma, we do not have to use the fact that P and P ′ areorthogonal projections.

Lemma 8.21. Let V be a Banach space and P ∈End(V ) be a projection,i.e. P 2 = P. Set ε ≡ ε(P ) = (

√‖P‖2 + 1 − ‖P‖). Then for all projections

P ′ ∈End(V ) satisfying ‖P − P ′‖ < ε, P and P ′ are conjugate. Moreover, ifJ ≡ P ′P + Q′Q, where Q ≡ I − P and Q′ ≡ I − P ′, then J is invertible andJP = P ′J.

Proof. Follow the same proof as above to show that JP = P ′J, and J isinvertible provided δ ≡ P ′(P − P ′) + (P ′ − P )P has norm less than 1. Now

‖δ‖ ≤ (2‖P‖+ ε)‖P − P ′‖ < (2‖P‖+ ε)ε.

The proof is finished with the aid of the quadratic formula which shows that(2‖P‖+ ε)ε = 1 when ε ≡ ε(P ) = (

√‖P‖2 + 1− ‖P‖).

Corollary 8.22. Suppose that V → E →M is a vector bundle and P : E → Eis a smooth map such that: i) P P = P, ii) πP = π, and iii) P |Em ∈End(Em)for all m ∈ M. We further assume that M is connected. Then kerP ≡ ξ ∈E|Pξ = 0π(ξ) and ranP ≡ P (E) are sub-bundles of E.

Proof. First notice that ranP = kerQ, where Q ≡ I −P . Since Q satisfiesthe same assumptions as P it suffices to show that kerP is a sub-bundle.

With the aid of a local trivialization, it is easy to reduce the proof to thecase where E = M × V is a trivial vector bundle. By the assumptions on P, itis easily seen that P (m, v) = (m, p(m)v), where p : M →End(V ) is a smoothfunction and p(m) is a projection for all m ∈M. Furthermore,

kerP = (m, v)|m ∈M,v ∈ ker p(m).

Fix an o ∈M, and choose ε = ε(p(o)) as in the above lemma. Let

U ≡ m ∈M |‖p(m)− p(o)‖ < ε ∈ τo(M)

and for m ∈M, set

Jm = I − p(m)− p(o) + 2p(m)p(o).


70 8 Vector Bundles

By the above lemma Jm is invertible and

Jmp(o) = p(m)Jm (8.1)

for all m ∈ U. It is clear from the formula for Jm the m → Jm is smooth. Wenow define a local trivialization ϕ : π−1(U) = U × V → V of E via:

ϕ(m, v) ≡ J−1m v.

For m ∈ U, the following statements are equivalent: i) (m, v) ∈ kerP , ii)p(m)v = 0, iii) J−1

m p(m)v = 0, iv) (by (8.1)) p(o)J−1m v = 0, and v) J−1

m v ∈ker p(o). Therefore, it follows that ϕ(kerP |m) = ker p(o) for all m ∈ U. It nowfollows from Proposition 8.16 that kerP is a sub-bundle.

Example 8.23. Suppose that M is an embedded submanifold of RN and i :M → RN is the inclusion map. Let E ≡ M × RN be the trivial bundle. Foreach m ∈M, let

Fm ≡ m × di〈TmM〉 and F⊥m ≡ m × (di〈TmM〉)⊥.

Then F ≡ ∪m∈MFm and F⊥ ≡ ∪m∈MF⊥m are sub-bundles of E. Moreover F isisomorphic to TM. (F⊥ is called the normal bundle.)

Proof. Let p(m) ∈End(RN ) denote orthogonal projection onto di〈TmM〉and define P : E → E by P (m, v) = (m, p(m)v). It is easy to check P is smooth.Indeed, use Gram-Schmidt to find a smooth local frame Eidi=1 (defined onU ⊂o M) of TM such that di〈Ej(m)〉dj=1 is an orthonormal set in RN , thenwrite

p(m)v =

d∑j=1

(di〈Ej(m)〉, v)RN di〈Ej(m)〉.

P is easily seen to satisfy the assumptions of Corollary 8.22 above, so thatranP = F and kerP = F⊥ are sub-bundles of E. The proof is finished by

observing that the map (vm → (m, di〈vm〉)) : TM → F is a vector bundleisomorphism.

In example 8.19 we constructed a bundle by considering the kernel of avector bundle morphism. Our next goal is to abstract this construction.

Lemma 8.24. Suppose that V and W are inner product spaces and A : M →Hom(V,W ) is a smooth map such that rankAm is a constant on M. Let Rmdenote orthogonal projection onto ranAm ⊂ W and Km denote orthogonalprojection onto kerAm ⊂ V. Then K : M →End(V ) and R : M →End(W ) issmooth.

Proof. We start by proving that Km is smooth. Let o ∈M. By continuity,it follows that RoAm : V → ranAo is surjective for m near o. Since rankAm isa constant, it follows by the rank-nullity theorem that kerRoAm = kerAm form near o. Since the condition that Km is smooth is a local condition, we mayassume by the above remarks that Am : V → W is surjective for all m ∈ M.The remainder of the proof that Km is smooth follows easily from the followingclaim.

Claim: If Am is surjective then Km = I −A∗m(AmA∗m)−1Am.

Let Bm ≡ Am|(kerAm)⊥ , then Bm : (kerAm)⊥ →W is an isomorphism since

Am is surjective. Set Qm ≡ I −Km — orthogonal projection onto (kerAm)⊥.Then it is clear that Qm = B−1

m Am. We also know that A∗m : W → V is aninjective map with ranA∗m = (kerAm)⊥, and so it follows that Cm ≡ BmA∗m =AmA

∗m : W →W is an isomorphism. Therefore, I = BmA

∗mC−1m , which implies

thatB−1m = A∗mC

−1m = A∗m(AmA

∗m)−1.

Hence Km = I −Qm = I −A∗m(AmA∗m)−1Am as claimed.

It is now clear that Km is smooth since, Am, A∗m, and m → (AmA

∗m)−1

are all smooth. For the proof that Rm is smooth, it suffices to notice thatranAm = (kerA∗m)⊥. Replacing Am by A∗m in the above proof shows that Rm

is smooth.

Corollary 8.25. Suppose that V → E → M and W → F → M are vec-tor bundles and H : E → F is a vector-bundle morphism, i.e. H is smooth,H(Em) ⊂ Fm for all m ∈M, and Hm ≡ H|Em : Em → Fm is a linear map forall m ∈M. If dimH(Em) is a constant for all m ∈M, then

kerH ≡ ξ ∈ E : H(ξ) = 0 ∈ Fπ(ξ) ⊂ E

andranH ≡ H(E) ⊂ F

are sub-bundles of E and F respectively.

Proof. By choosing local trivializations of E and F and using the abovelemma, it is easily seen for each o ∈ M there exists vector-bundle morphismsK : EU ≡ ∪m∈UEm → EU and R : FU ≡ ∪m∈UFm → FU such that KK = K,RR = R, kerK = EU ∩kerH, and ranH∩FU = kerR. Corollary 8.22 impliesthat EU ∩kerH is a sub-bundle of EU and H ∩FU is a sub-bundle of FU . Sinceo ∈M was arbitrary, it follows that kerH and ranH are sub-bundles of E andF respectively.

Definition 8.26. Suppose that π : E →M is a vector bundle and that F,K ⊂E are sub-bundles. We say that E = F ⊕K if Em = Fm ⊕Km for all m ∈M.


8.3 Sub-bundles 71

Theorem 8.27. Let Z = X ⊕ Y be the fiber model space for E with X and Ythe fiber model space for F and K respectively. Suppose that E = F⊕K, then forall m ∈M, there exists a frame (U,O) of E such that U |X is a frame of F andU |Y is a frame of K. Furthermore, if P (m) : Em → Fm is projection relativeto the decomposition Em = Fm ⊕ Km, then P : E → F is a vector-bundlemorphism.

Proof. Choose frames (V,O) and (W,O) of E such that V |X and W |Y areframes of F and K respectively over an open set O ∈ τ(M). Set U(m)(x+y) =V (m)x + W (m)y for all x ∈ X and y ∈ Y. Clearly U(m) : Z → Em is anisomorphism for all m ∈ O. Furthermore,

V (m)−1U(m)(x+ y) = x+ V (m)−1W (m)y,

which shows that the frame U(m) is smooth. Let p : Z → X be projectionrelative to the decomposition Z = X ⊕ Y. Then clearly

P (m) = U(m)pU(m)−1

which shows that P is smooth.

Theorem 8.28. Let π : E →M be a vector bundle with model fiber V = X⊕Y.We now assume that V is an inner product space, and V = X ⊕ Y is anorthogonal direct sum decomposition. Also suppose that h is a fiber-wise innerproduct on E, such that h〈S, T 〉 ∈ C∞(M) for all smooth sections S and T ofE. . If F ⊂ E is a sub-bundle with model fiber X then

K ≡ F⊥ ≡ ∪m∈MF⊥mis a sub-bundle of E and E = F ⊕ K = F ⊕ F⊥. Furthermore, it is alwayspossible to choose a frames (U,O) such that U(m) is an isometry for all m ∈ O,U(m)(X) = Fm and U(m)(Y ) = Km = F⊥m for all m ∈ O.

Proof. Suppose that (W,O) is a frame on E such that W |X is a frame onF. Choose an orthonormal basis e1, . . . , ek for X and ek+1, . . . , en for Y.Set E′i(m) = W (m)ei for m ∈ O. Notice that E′i(m) ∈ Fm for all m ∈ O. Nowperform Gram-Schmidt on the basis E′i(m)ni=1 to get an orthonormal basisEi(m)ni=1 of Em. Notice that Ei(m) =

∑j≤i cij(m)E′j(m), where functions

cij : O → R are smooth functions, such that cii(m) 6= 0 for all m ∈ O andi = 1, . . . , n. In particular, it is now clear that U(m) : Z → Em defined byU(m)

∑aiei ≡

∑aiEi(m) is a frame on E such that U(m) : V → Em is an

isometry for all m ∈ O, U(m)(X) = Fm, and U(m)(Y ) = F⊥m .

Proposition 8.29. Suppose that F is a sub-bundle (as in Definition 8.15) of avector bundle V → E → M. Decompose V into a direct sum V = X ⊕ Y suchthat dimX = dimFm for all m ∈M. Then for all o ∈M there is a neighborhoodU ∈ τo(M) and local trivialization ϕ : π−1(U)→ V of E, such that ϕ(Fm) = Xfor m ∈ U.

Proof. Let us start by showing that π ≡ π|F : F →M is a submersion. LetS : M → E be the zero section in E, i.e. for each m ∈ M, S(m) = 0 ∈ Em.(In the future I will write 0m for the 0–vector in Em.) It is easily seen that Sis smooth and clearly S(M) ⊂ F. Therefore S thought of a function from M toF is also smooth. Since π S = id|M , TmM = π∗S∗TmM ⊂ π∗T0mF ⊂ TmM.Therefore π∗0m is surjective for all m ∈M. By continuity, it follows that π∗ξ issurjective for all ξ sufficiently close to 0m. Given ξ ∈ Fm, choose ε > 0 such thatπ∗εξ is surjective. Since µε : E → E, given by µε(ξ) = εξ, is a diffeomorphism ofE with inverse µε−1 , and µε(F ) = F, it follows that µε is also a diffeomorphismof F. Because π µε = π and µε∗TξF = TεξF, it follows that

π∗TξF = π∗µε∗TξF = π∗TεξF = TmM.

Since π is a submersion, it follows that Fm ≡ π−1(m) = F ∩ Em is anembedded submanifold of F and dimFm = dimF − dimM. Notice that Fm issubmanifold of E since F is a submanifold of E. Since Em is also a submanifoldof E and Fm ⊂ Em, it is easy to check that Fm is a submanifold of Em.Because Fm is a subspace of Em, it now follows that the manifold structure onFm which makes Fm into a submanifold of F coincides with manifold structureon Fm induced from the vector space structure of Fm. Therefore the manifolddimension of Fm and the vector space dimension coincide. Thus we have shownthat

dimF = dimM + dimFm = dimM + dimX.

For the rest of the proof, we assume with out loss of generality, that E =M × V, and π(m, v) = m. We then have that Fm = m × V ′m for some vectorsubspace V ′m of V with dimV ′m = dimX and S(m) = (m, 0) is the zero sectionin this notation.

Claim: For all m ∈M,

T(m,0)F = (vm, w0)|vm ∈ TmM,w ∈ V ′m = TmM × (V ′m)0.

Since S : M → F is smooth,

S∗TmM = TmM × 00 ⊂ T(m,0)F.

Similarly the map k : V ′m → E given by k(w) = (m,w) is smooth with image inF, so that k : V ′m → F is also smooth. Therefore, k∗T0V

′m = (0m, w0) : w ∈ V ′0

is also in T(m,0)F. Hence it follows that TmM × (V ′m)0 ⊂ T(m,0)F. A dimensioncount now shows that in fact TmM × (V ′m)0 = T(m,0)F, which proves the claim.

Fix a point o ∈ M. Using the fact that F is an embedded submanifold ofE, choose neighborhoods U ∈ τo(M), W ∈ τ0(V ) and a smooth submersionρ : U ×W → Y such that ρ = 0) = F ∩ (U ×W ). Since T(m,0)F = TmM ×(V ′m)0 = ker ρ∗(m,0), it follows that V ′m = kerD2ρ(m, 0), where D2ρ(m, 0)w ≡


ddt |0ρ(m, tw). Therefore, if we define H : U × V → U × Y by H(m, v) =(m,D2ρ(m, 0)v), then H is a vector bundle morphism such that rankHm =dimY (Hm ≡ H|m×V ) is constant and F ∩ (U × V ) = kerH. The existenceof a “good” (as in the statement of the proposition) local trivialization ϕ of Eis now guaranteed by an application of Corollary 8.25.

9

Covariant derivatives on Vector Bundles

The basic goal of this chapter is to develop the notion of a derivative ofsections of a vector bundle. We could of course consider S ∈ Γ (E) as a mapfrom M → E in which case S∗ : TM → TE. This differential however cariesa lot of redundant information owing to the fact that we know π S = id andtherefore π∗S∗ = ITM . Hence we do not really need to store the “horizontal”information encoded in S∗. It however turns out that making a choice of thespitting of TξE into horizontal and vertical subspaces require a choice which weare going to encode into a covariant derivative.

9.1 Basics of covariant derivatives

Definition 9.1. A covariant derivative ∇ on E is a map ∇ : TM×Γ (E) −→ E(we will write ∇vS for ∇(v, S)) such that

1. For v ∈ TmM and S ∈ Γ (E), ∇vS ∈ Em.2. v −→ ∇vS is linear on TmM for all m ∈M .3. For sections S1, S2 ∈ Γ (E), f ∈ C∞(M) and v ∈ TmM ; ∇v satisfies

∇v(S1 + fS2) = ∇vS1 + (vf)S2(m) + f(m)∇vS2.

4. If X ∈ Γ (TM) and S ∈ Γ (E) , then ∇XS ∈ Γ (E) , i.e. (∇XS) (m) :=∇X(m)S defines a smooth section.

Remark 9.2. If S ∈ Γ (E) and S = 0 near some point m ∈ M, then ∇vmS = 0for all vm ∈ TmM. To see this is the case, choose ϕ ∈ C∞c (M) such that ϕ = 1near m and supp (ϕ) is contained in the neighborhood of m where S = 0. ThenϕS ≡ 0 and hence

0 = ∇v0 = ∇v (ϕS) = (vϕ)S (m) + ϕ (m)∇vS = (vϕ) 0 + 1∇vS = ∇vS.

This shows that in order to compute ∇vmS we need only know S in a smallneighborhood of m.

Definition 9.3. Let π : E →M be a vector bundle with fiber W. A local frameu on E is a local section of the bundle Aut(V,E)→M, i.e. for m ∈ U := D(u)(the domain of u) u(m) : W → Em = π−1(m) is a linear isomorphism of vectorspaces.

Notice that any local section S of E may be written as S(m) = u(m)s(m),where s ∈ C∞(M,W ).

Notation 9.4 If ∇ is a covariant derivative on E, define for v ∈ TmM , alinear transformation ∇vu : W → Em by

(∇vu)w := ∇v(u(·)w) for all w ∈W.

Notice that W 3 w → (∇vu)w ∈ W is linear. We now define the connection1–form associated to the local frame u by

A(v) = Au(v) := u(m)−1 (∇vu) for all vm ∈ TmM. (9.1)

Lemma 9.5. If S ∈ Γ (EU ) and s : U →W is defined by s := u−1S, then

∇vS = ∇v [us] = u (m) [ds (vm) +Au (vm) s (m)] .

In particular this shows that

u(m)−1∇v(us) = ∂vs+A(v)s(m),

where the local representation of ∇ is ∇ = d + A, where A is a one form withvalues in End(W ).

Proof. Let s =∑Ni=1 siwi where wiNi=1 is a basis for W. Then S =∑N

i=1 siuwi and we have,

∇vS =

N∑i=1

∇v [siuwi] =

N∑i=1

((vmsi)u (m)wi + si (m)∇v [uwi])

= u (m) vms+

N∑i=1

si (m)u (m)Au (vm)wi

= u (m) [ds (vm) +Au (vm) s (m)] .

Lemma 9.6. Suppose that u and u are two local frames for E → M. Thenthere exists g : M → Aut (M) such that u = ug. With this notation we have

Au = Adg−1Au + g−1dg. (9.2)

74 9 Covariant derivatives on Vector Bundles

Proof. By definition,

Au := u−1∇u = (ug)−1∇ (ug) = g−1u−1 [(∇u) g + udg] = Adg−1Au + g−1dg.

Proposition 9.7. Given a path S(t) ∈ E, let σ(t) = π(S(t)) and s(t) =u(σ(t))−1S(t) where u is a local frame with σ (t) ∈ D (u) . Then defining∇S(t)/dt by the formula,

∇S(t)/dt := u(σ(t)) (s(t) +A(σ(t))s(t)) , (9.3)

is well defined. In this way we define an operator (which in a local frame isgiven by ∇dt = d

dt +A(σ(t))),

∇dt

: C∞ ((a, b)→ E)→ C∞ ((a, b)→ E) (9.4)

satisfying:

1. ∇S(t)/dt ∈ Eπ(S(t)) for all t.2. If S is a local section of E defined near σ (t) , then

∇dtS (σ (t)) = ∇σ(t)S. (9.5)

3. If we let

Γσ (E) :=S ∈ C∞ ((a, b)→ E) : S (t) ∈ Eσ(t) for all t ∈ (a, b)

,

then ∇dt : Γσ (E)→ Γσ (E) is a linear map which satisfies the product rule,

∇dt

[f (t)S (t)] = f (t)S (t) + f (t)∇dtS (t) (9.6)

for all f ∈ C∞ ((a, b)→ F) and C∞ ((a, b)→ E) .

In fact, the properties in items 1. – 3. uniquely determines the operator ∇dt .

Proof. It is easily seen that ∇dt defined in Eq. (9.3) has the properties initems 1. – 3. above. For example if S is a local section of E and let s := u−1S :D (u)→W. Then by definition,

∇S(σ(t)))/dt = u(σ(t))

(d

dts(σ(t)) +A(σ(t))s(σ(t))

)= u(σ(t))

(∂σ(t)s+A(σ(t))s(σ(t))

)= ∇σ(t)S.

To see that ∇dt is well defined independent of the choice of local frame, u. Write

S (t) = u (σ (t)) s (t) =∑i

si (t)u (σ (t)) ei

where eidim(W )i=1 is a basis for W and s (t) =

∑i si (t) ei. Then using only the

properties in items 1. –3. above we learn that

∇S(t)/dt =∑i

si (t)u (σ (t)) ei +∑i

si (t)∇σ(t) [u (·) ei]

= u (σ (t))

[∑i

si (t) ei

]+∑i

si (t)u (σ (t))Au (σ (t)) ei

= u (σ (t)) s (t) + u (σ (t))Au (σ (t))∑i

si (t) ei

= u (σ (t)) [s (t) +Au (σ (t)) s (t)]

which shows that the definition in Eq. (9.3) is well defined.

Exercise 9.1. Suppose g : D (u) → Aut (V ) is a smooth function, u = ug and

a (t) := u (σ (t))−1S (t) . Show by direct computation that

u(σ(t))

(d

dts(t) +Au(σ(t))s(t)

)= u(σ(t))

(d

dta(t) +Au(σ(t))a(t)

)in order to give a direct proof that ∇dt is well defined.

Lemma 9.8. If S(t) is a smooth path in E and t = τ(s), then

∇S(τ(s))/ds = τ ′(s)∇S(t)/dt|t=τ(s). (9.7)

Proof. This property is easily verified from Eq. (9.3). Indeed, locally wehave

∇S(τ(s))/ds =

(d

ds+A

(d

ds(σ τ) (s)

))S (τ (s))

= S (τ (s)) τ ′ (s) +A (τ ′ (s) σ (τ (s)))S (τ (s))

= τ ′ (s)[S (τ (s)) +A (σ (τ (s)))S (τ (s))

]= τ ′ (s)

[d

dt+Aσ (t) S (t)

]t=τ(s)

= τ ′(s)∇S(t)/dt|t=τ(s).

Definition 9.9. We say that a path S(t) ∈ E is parallel provided that∇S(t)/dt = 0 for all t.


9.1 Basics of covariant derivatives 75

Lemma 9.10. Given a curve σ(t) in M and a point S0 ∈ Eσ(0), there is aunique path S(t) ∈ E such that π(S(t)) = σ(t) and ∇S(t)/dt = 0.

Proof. This path is constructed by solving (locally) the linear equation

s(t) +A(σ(t))s(t) = 0 with s(0) = u(σ(0))−1S0 (9.8)

and then setting S(t) = u(σ(t))s(t). One then easily pieces these local solutionstogether to get a global solution.

It is easy to check that the map S0 ∈ Eσ(0) → S(t) ∈ Eσ(t) is linear.

Definition 9.11 (Parallel translation). Let //t(σ) : Eσ(0) → S(t) ∈ Eσ(t) bethe unique linear map such that S(t) = //t(σ)S0 is parallel for all S0 ∈ Eσ(0).We refer to //t(σ) as parallel translation along σ up to times t.

Parallel translation is uniquely characterized as the solution to the differen-tial equation

∇//t(σ)/dt = 0 with //0(σ) = id ∈ End(Eσ(0)), (9.9)

In fact//t(σ) = u(σ(t))g(t)u(σ(0))−1 (9.10)

where g(t) ∈ End(W ) is the unique solution to the linear differential equation,

g(t) +A(σ(t))g(t) = 0 with g(0) = id ∈ End(W ). (9.11)

(If U(t) ∈ Hom(Eσ(0), Eσ(t)) for each t, then ∇U(t)/dt is by definition thelinear transformation from Eσ(0) to Eσ(t) determined by (∇U(t)/dt) ξ :=∇ (U(t)ξ) /dt for all ξ ∈ E0.) We have the following properties of paralleltranslation which follow from the uniqueness theorem for ordinary differentialequations and the chain rule for covariant derivatives.

Theorem 9.12. Parallel translation depends smoothly on the curve σ : [0, 1]→M.

Proof. Choose a partition 0 = t0 < t1 < · · · < tn = 1 of [0, 1] such thatfor each 1 ≤ i ≤ n, σ ([ti−1, ti]) is contained in an open set, Ui ⊂o M suchthat EUi is trivializable. Let ui be a frame for EUi , i.e. ui (x) : V → Ex is alinear isomorphism for all x ∈ Ui. Further let Ai := Aui and let gi : [ti−1, ti]→GL (W ) solve the ODE,

gi (t) +Ai 〈σ (t)〉 gi (t) = 0 with gi (ti−1) = I.

We then have,

//1 (σ) =[un (σ (tn)) gtn (σ)un (σ (tn−1))

−1]·

·[un−1 (σ (tn−1)) gtn−1

(σ)un−1 (σ (tn−2))−1]. . .

· · · ·[u1 (σ (t1)) gt1 (σ)u1 (σ (t0))

−1].

This may be rewritten as

un (σ (tn))−1//1 (σ)u1 (σ (t0))

= gtn (σ) ·(u−1n un−1

)(σ (tn−1)) · gtn−1

(σ) (un−1un−2) (σ (tn−2)) . . .

. . .(u−1

2 u1

)(σ (t1)) gt1 (σ)

all of which are products of functions of σ into GL (W ) depending smoothly onσ.

Proposition 9.13. Let σ(t) ∈ M be a smooth curve for t ∈ [0, T ] and letτ : [0, S]→ [0, T ] be a smooth function such that τ(0) = 0. Then

//s(σ τ) = //τ(s)(σ),

i.e. parallel translation does not depend on how the underlying curve is param-eterized. Secondly, let σ(t) := σ(T − t), then

//t(σ) = //T−t(σ)//T (σ)−1.

In particular //T (σ)−1 = //T (σ).

Proof. We have that

∇//s(σ τ)/ds = 0 with //0(σ τ) = id ∈ End(Eσ(0))

and

∇//τ(s)(σ)/ds = ∇//t(σ)/dt|t=τ(s)τ′(s) = 0 with

//τ(s)(σ)|s=0 = id ∈ End(Eσ(0))

and hence by uniqueness of solutions to O.D.E.’s we must have that //s(στ) =//τ(s)(σ). Similarly,

∇//t(σ)/dt = 0 with //0(σ) = id ∈ Eσ(T )

and

∇//T−t(σ)//T (σ)−1/dt = −∇//s(σ)//T (σ)−1/ds|s=T−t = 0 with

//T−t(σ)//T (σ)−1|t=0 = id ∈ Eσ(T ).

Hence again by uniqueness of solutions to O.D.E.’s we must have that //t(σ) =//T−t(σ)//T (σ)−1.



9.2 Curvature of ∇

Definition 9.14. The curvature tensor R = R∇ of the covariant derivative ∇is defined by,

R∇〈X,Y 〉S := [∇X ,∇Y ]S −∇[X,Y ]S, (9.12)

where X,Y and S are vector fields on M.

The reason for the second term in (9.12) is to make R “tensorial.” Thisshould become clear from the next lemma.

Lemma 9.15. Let u be a local frame of E over an open set O and A := Au =u−1∇u be the “local connection 1-form” associated to u. Then

R∇〈X,Y 〉S = uRA〈X,Y 〉u−1S, (9.13)

whereRA = dA+A ∧A

and A ∧ A (v, w) := [A (v) , A (w)] . In particular this shows that(R∇〈X,Y 〉S)(m) depends only on the values X,Y, and S at m and noton X,Y, and S away from m.

Proof. To simplify notation, set s = u−1S. We know that

∇Y S = u(Y s+A〈Y 〉s),

and hence

∇X∇Y S = uX(Y s+A〈Y 〉s) +A〈X〉(Y s+A〈Y 〉s)= uXY s+ (XA〈Y 〉)s+A〈Y 〉Xs+A〈X〉Y s+A〈X〉A〈Y 〉s.

Thus

[∇X ,∇Y ]S = u[X,Y ]s+ (XA〈Y 〉 − Y A〈X〉+ [A〈X〉, A〈Y 〉])s.

We also have∇[X,Y ]S = u([X,Y ]s+A〈[X,Y ]〉s),

so that

R∇〈X,Y 〉S = uXA〈Y 〉 − Y A〈X〉 −A〈[X,Y ]〉+ [A〈X〉, A〈Y 〉]s= udA〈X,Y 〉+ [A〈X〉, A〈Y 〉]s = uRA〈X,Y 〉u−1S

as claimed.

Proposition 9.16. Let S(t, s) be a section of E along γ(t, s). Write γ for ddt

and γ′ for dds . Then [

∇dt,∇ds

]S = R∇〈γ, γ′〉S (9.14)

Proof. Choose a local frame u and let u(t, s) ≡ u(γ(t, s)) and s(t, s) ≡u(t, s)−1S(t, s). Then by definition,

∇dtS = u

[(d

dt+A〈γ〉)s

]and

∇dsS = u

[(d

ds+A〈γ′〉)s

].

Therefore, [∇dt,∇ds

]S = u[

d

dt+A〈γ〉, d

ds+A〈γ′〉]s

= u

[d

dtA〈γ′〉 − d

dsA〈γ〉+ [A〈γ〉, A〈γ′〉]

]s

= u [dA〈γ′, γ〉+ [A〈γ〉, A〈γ′〉]] s= uRA〈γ, γ′〉s = R∇〈γ, γ′〉us = R∇〈γ, γ′〉S.

where we used Lemma ?? in the second equality.The following theorem shows explicitly how curvature controls the path

dependence the parallel translation operator.

Theorem 9.17 (Differentiating Parallel translation). Let γss∈R be aone parameter family of paths in M such that γs (0) = m ∈ M and γs (t) =γ (s, t) where γ is smooth in both s and t. Then

∇ds//∇1 (γs) = //∇1 (γ(s))B∇〈γs〉, (9.15)

where B∇ is the GL(Eo) -valued 1-form on the path space on M defined by

B∇〈X〉 ≡∫ 1

0

//∇t (σ)−1R∇〈σ′(t), X(t)〉//∇t (σ)dt (9.16)

for each X ∈ Γσ(M) and path σ ∈M.

Proof. By Theorem 9.12 we know that parallel translation dependssmoothly on σ and so it makes sense to compute its derivative. By Proposi-tion 9.16 and the definition of parallel transport, ∇//At (γs)/dt = 0,

∇dt

∇ds//∇t (γs) =

[∇dt,∇ds

]//∇t (γs) = R∇〈γ(s, t), γ′(s, t)〉//∇t (γs). (9.17)


9.2 Curvature of ∇ 77

Therefore,

d

dt

[//∇t (γs)

−1∇ds//∇t (γs)

]=

[//∇t (γs)

−1∇dt

∇ds//∇t (γs)

]= //∇t (γs)

−1R∇〈γ′(s, t), γ(s, t)〉//∇t (γs) (9.18)

The theorem is now easily proved by integrating (9.18) relative to t using∇//∇0 (γs)/ds = 0.

We get as an immediate corollary, the Frobenius theorem in this setting.

Corollary 9.18 (A Frobenius Theorem). If R∇ ≡ 0, then //1(σ) = //1(τ)provided σ can be deformed into τ while keeping the end points fixed.

Proof. Let γs denote a smooth homotopy of σ to τ with end points fixed.By Theorem 9.17,

∇ds//∇1 (γs) = 0.

Since γs(1) is assumed to be a constant independent of s, it follows that //∇1 (γs)is a linear map from Eσ(0) → Eσ(1) for each s. Furthermore, in this case∇ds//

∇1 (γs) = d

ds//∇1 (γs). (Why? Answer: because d

dsγs (1) = 0.) So we con-

clude that dds//

∇1 (γs) = 0, that is //∇1 (γs) is a constant.

Theorem 9.19 (Parallel Translation around small loops). Suppose γs (t)is a homotopy of loops such that γ0 (t) = m for all t and let X (t) := d

ds |0γs (t) ∈TmM for all t. Then

d

dε|0//∇1 (γ√ε) =

1

2

∫ 1

0

R∇〈X (t) , X(t)〉dt. (9.19)

Proof. This is a local theorem and so in the proof we may assume thatM = Rd and then write

//∇1 (γs) = I + sg1 + s2g2 + . . . and

γs (t) = m+ sγ1 (t) + s2γ2 (t) + . . . , (9.20)

From Eq. (9.15) we have

//∇1 (γs)−1 d

ds//∇1 (γs) = B∇〈γs〉 =

∫ 1

0

//∇t (γs)−1R∇γs(t)〈γs (t) , γ′s(t)〉//∇t (γs)dt.

Plugging the expansions indicated in Eq. (9.20) into this identity along expan-sions for Rγs we find,

//∇1 (γs)−1 d

ds//∇1 (γs) =

∫ 1

0

//∇t (γs)−1R∇γs(t)〈γs (t) , γ′s(t)〉//∇t (γs)dt

= s

∫ 1

0

//∇t (γs)−1R∇γs(t)〈γ

1 (t) , γ1 (t)〉//∇t (γs)dt+O(s2)

= s

∫ 1

0

Rm⟨γ1 (t) , γ1 (t)

⟩dt+O

(s2).

On the other hand we have

//∇1 (γs)−1 d

ds//∇1 (γs) = (I − sg1) (g1 + 2sg2) +O

(s2)

= g1 + s[2g2 − g2

1

]+O

(s2)

and comparing these last twos expansions we learn, g1 = 0 and

2g2 =

∫ 1

0

R∇〈γ1 (t) , γ1(t)〉dt.

Thus we conclude that

//∇1 (γs) = I +s2

2

∫ 1

0

R∇〈γ1 (t) , γ1(t)〉dt+O(s3).

Next observe that γ1(t) = dds |0γs (t) = X (t) so that the previous formula may

be expressed as

//∇1 (γs) = I +s2

2

∫ 1

0

R∇〈X (t) , X(t)〉dt+O(s3)

= es2

2

∫ 1

0R∇〈γ′0(t),γ′0(t)〉dt+O(s3) for s small.

Taking s =√ε in this expression shows,

//∇1 (γ√ε) = I +ε

2

∫ 1

0

R∇〈X (t) , X(t)〉dt+O(ε3/2

)which upon differentiating at ε = 0 gives Eq. (9.19).

Proposition 9.20 (Another Curvature Formula). Suppose σ is an M -valued function of two real variables; s and t and let p(s, t) := //∇ (σs,t) denoteparallel translation around the closed curve σs,t (see Figure 9.1)determined byordered sequence of paths:

1. σ (0, τ) with τ going from 0 to t.2. σ (τ, t) with τ going from 0 to s.



Fig. 9.1. A path determined by a parameterized surface, σ, in M.

3. σ (s, τ) with τ going from t to 0.4. σ (τ, 0) with τ going from s to 0.

Then

R∇ (σ′ (0, 0) , σ (0, 0)) =d

dt|0d

ds|0//∇(σs,t). (9.21)

Proof. Using the basic concatenation properties of parallel translation weknow that

p (s, t) = //s (σ (·, 0))−1//t (σ (s, ·))−1

//s (σ (·, t)) //t (σ (0, ·))

or equivalently that

//s (σ (·, t)) //t (σ (0, ·)) = //t (σ (s, ·)) //s (σ (·, 0)) p (s, t) . (9.22)

Let v := σ (0, 0) and w = σ′ (0, 0) . Then applying

∇dt|0∇ds|0 =

∇ds|0∇dt|0 +R 〈v, w〉

to Eq. (9.22) implies

0 =∇ds|0∇dt|0 [//t (σ (s, ·)) //s (σ (·, 0)) p (s, t)] +R 〈v, w〉 p (0, 0)

=∇ds|0 [//s (σ (·, 0)) p (s, 0)] +R 〈v, w〉 = p′ (0, 0) +R 〈v, w〉

which is equivalent to Eq. (9.21).

Remark 9.21. One often sees a version of the above theorem stated as,

//∇(σ√s,

√s

)= I + s ·R∇ (σ′ (0, 0) , σ (0, 0)) +O

(s3/2

),

i.e.

R∇ (σ′ (0, 0) , σ (0, 0)) =1

s

[//∇

(σ√s,

√s

)− I]

+O(√s).

Example 9.22. Suppose that X and Y are commuting vector fields on M andm ∈M is given. Let σs,t denote the path following

1. s′ → es′Y (m) where s′ goes from 0 to s,

2. t′ → et′X esY (m) where t′ goes from 0 to t,

3. s′ → es′Y etX esY (m) where s′ goes from s to 0, and

4. t′ → et′X esY etX esY (m) where t′ goes from t to 0.

(Notice that the above path is closed because [X,Y ] = 0.) Then

R∇〈Xm, Ym〉 =d

dt|0d

ds|0//∇(σs,t) (9.23)

This should be compared with interpretation of the Lie bracket given in Propo-sition ??.

9.3 On the structure of covariant derivatives

Lemma 9.23. If ∇ and D are two covariant derivatives on a vector bundleE

π→M, then there exists an End (E) – valued one form on M so that

∇vS = DvS +A 〈v〉S (m) for all vm ∈ TM.

Proof. This is easily seen to be true by passing to a local trivialization.Alternative, if f ∈ C∞ (M) we have

(∇v −Dv) [fS] = f (m) (∇v −Dv)S + vmf · S (m)− vmf · S (m)

= f (m) (∇v −Dv)S

from which it follows that ∇v −Dv is tensorial.

Lemma 9.24. Every vector bundle has a covariant derivative.

Proof. We may easily construct a covariant derivative ∇i over EUi whereUii∈Λ is an open cover of M so that EUi

∼= Ui × F for all i. Now let ϕii∈Λbe a partition of unity subordinate to Uii∈Λ and set ∇ :=

∑i∈Λ ϕi∇i.


9.4 Exercises on Transferring Covariant Derivatives 79

Definition 9.25. Suppose that E is equipped with a smoothly inner product,gm := 〈·, ·〉m , on each fiber Em. We say that ∇ is compatible with 〈·, ·〉 if∇g = 0 where vm ∈ TmM and S, T ∈ Γ (E) ,

(∇vg) (S, T ) = v [g (S, T )]− g (∇vS, T (m))− g (S (m) ,∇vT ) .

Definition 9.26. Suppose that g is a fiber metric on E and W is an innerproduct space. We say a frame, u (m) : W → Em for m ∈ D (u) is orthonormalif u (m) is an isometry for all m ∈M. Orthonormal frames always exist by theGram–Schmidt orthogonalization procedure.

Theorem 9.27. Suppose that g is a fiber metric on E and ∇ is a covariantderivative on E. Then the following are equivalent:

1. ∇ is g – compatible.2. If u is an orthonormal frame, then Au := u−1∇u is an so (W ) – valued one

form.3. For all smooth curves σ (t) ∈M and all S, T ∈ Γσ(t) (M) ,

d

dtg (S (t) , T (t)) = g

(∇dtS (t) , T (t)

)+ g

(S (t) ,

∇dtT (t)

). (9.24)

4. Parallel translation, //t (σ) : Eσ(0) → Eσ(t) is always an isometry.

Proof. 1. =⇒ 2. Let u be an orthonormal frame, i.e. u (m) : W → Em isan isometry for all m ∈ D (u) . Then for a, b ∈W we have

(∇vmg) (ua, ub) = vv [g (ua, ub)]− g (∇vmua, ub)− g (ua,∇vmub)= 0−

(u−1∇vmua, b

)−(a, u−1∇vmub

)= − (Au 〈vm〉 a, b)− (a,Au 〈vm〉 b)

from which it follows that ∇vmg = 0 iff Au 〈vm〉 ∈ so (W ) .2. =⇒ 3. Writing S (t) = u (σ (t)) a (t) and T (t) = u (σ (t)) b (t) , we have

d

dtg (S (t) , T (t)) =

d

dt(a (t) , b (t)) = (a (t) , b (t)) +

(a (t) , b (t)

).

As Au 〈σ (t)〉 is skew symmetric we also know

0 = (Au 〈σ (t)〉 a (t) , b (t)) + (a (t) , Au 〈σ (t)〉 b (t)) .

Adding these last two displayed equations together givens Eq. (9.24) whereinwe have used

g

(∇dtS (t) , T (t)

)= g (u (σ (t)) [a (t) +Au 〈σ (t)〉 a (t)] , u (σ (t)) b (t))

= ([a (t) +Au 〈σ (t)〉 a (t)] , b (t)) .

3. =⇒ 4. Equation (9.24) implies that ddtg (S (t) , T (t)) = 0 whenever

S and T are parallel. From this we conclude by taking S (t) := //t (σ) v andT (t) := //t (σ)w, then

g (//t (σ) v, //t (σ)w) = g (S (t) , T (t)) = g (v, w) .

4. =⇒ 1. Suppose that S, T ∈ Γ (E) and v ∈ TmM. Choose σ (t) ∈ M suchthat v = σ (0) and let

a (t) := //t (σ)−1S (σ (t)) and b (t) := //t (σ)

−1T (σ (t)) .

Then we have

v [g (S, T )] =d

dt|0gσ(t) (S (σ (t)) , T (σ (t)))

=d

dt|0gσ(0) (a (t) , b (t)) = gσ(0) (a (0) , b (0)) + gσ(0)

(a (0) , b (0)

)= g ((∇vS) , T (m)) + g (S (m) ,∇vT )

from which it follows that ∇vg = 0.

9.4 Exercises on Transferring Covariant Derivatives

Let E → M and F → N be vector bundles over manifolds M and N respec-tively. Let f : M → N be a diffeomorphism and Ψ : E → F be a vector-bundleisomorphism covering f. More explicitly: Ψ : E → F is a smooth map such thatΨm ≡ Ψ |Em : Em → Ff(m) is a linear isomorphism for each m ∈M.

Exercise 9.2. Let ∇ be a covariant derivative on E. Given a section S ∈ Γ (F )and v ∈ TN set DvS ≡ Ψ(∇f−1

∗ v[Ψ−1S f ]). Show that D is a covariant

derivative on F .

Exercise 9.3. Let σ : [0, 1]→ N be a smooth curve in N. Show that

//Dt (σ) = Ψf−1σ(t)//∇t (f−1 σ)Ψ−1

f−1σ(0).

which we will abbreviate by

//Dt (σ)Ψ = Ψ//∇t (f−1 σ).

Alternatively, writing σ ≡ f−1 σ — a curve in M, we have

//Dt (σ)Ψσ(t) = Ψσ(t)//∇t (σ).



Exercise 9.4. Let E → M and F → N be vector bundles over manifolds Mand N respectively. Let f : M → N be a diffeomorphism and Ψ : E → F be avector-bundle isomorphism covering f. More explicitly: Ψ : E → F is a smoothmap such that Ψm ≡ Ψ |Em : Em → Ff(m) is a linear isomorphism for eachm ∈ M. Show that R∇〈v, w〉 = Ψ−1

m RD〈f∗v, f∗w〉Ψm for all v, w ∈ TmM andm ∈M.

9.5 Covariant derivatives on TM

Definition 9.28. Given a covariant derivative, ∇ on TM, the torsion tensorof ∇ is defined by

T∇ (X,Y ) := ∇XY −∇YX − [X,Y ] for all X,Y ∈ Γ (TM) .

Proposition 9.29. Suppose that x is a chart on M and u = ux is the framedefined by

u (m) a := a · ∂ :=

d∑i=1

ai∂i (9.25)

where ∂i := ∂/∂xi. (Note: 1. this is in general not an orthogonal frame and 2.

u (m)−1vm = dx (vv) .) Then, with Γ := Au = dx ∇u being the Christoffel

symbols for ∇,

T∇ (X,Y ) = (Γ (X) dx (Y )− Γ (Y ) dx (X)) · ∂. (9.26)

In particular this shows that T∇ is indeed a tensor. Moreover, if Σ (s, t) ∈ M,then (

∇dt

d

ds− ∇ds

d

dt

)Σ = T∇

(Σ′, Σ

)(9.27)

Proof. In this frame we have

∇Y Z = u ([Y + Γ (Y )] dx (Z)) .

Therefore

∇XY −∇YX = u ([X + Γ (X)] dx (Y )− [Y + Γ (Y )] dx (X))

= u (Xdx (Y )− Y dx (X)) + u (Γ (X) dx (Y )− Γ (Y ) dx (X))

= [X,Y ] + u (Γ (X) dx (Y )− Γ (Y ) dx (X))

from which it follows that

T∇ (X,Y ) = u (Γ (X) dx (Y )− Γ (Y ) dx (X)) = (Γ (X) dx (Y )− Γ (Y ) dx (X))·∂.

For the second assertion, observe that

∇dt

d

dsΣ = u (Σ)

[d

dt+ Γ

(Σ)]dx (Σ′) = u (Σ)

[d

dtdx (Σ′) + Γ

(Σ)dx (Σ′)

]= u (Σ)

[d

dt

d

dsx Σ + Γ

(Σ)dx (Σ′)

].

Hence we learn that(∇dt

d

ds− ∇ds

d

dt

)Σ = u (Σ)

([d

dt,d

ds

]x Σ + Γ

(Σ)dx (Σ′)− Γ (Σ′) dx

(Σ))

= u (Σ)(Γ(Σ)dx (Σ′)− Γ (Σ′) dx

(Σ))

= T∇(Σ′, Σ

),

(9.28)

wherein we have used Eq. (9.26) in the last equality.

Lemma 9.30. If M is a manifold, there exists a Torsion free covariant deriva-tive ∇ on TM.

Proof. Let D be any covariant derivative on TM and let A be a End (TM)– valued one form on M and set ∇ := D +A. Then

T∇ (X,Y ) = ∇XY −∇YX − [X,Y ]

= DXY +A 〈X〉Y − (DYX +A 〈Y 〉X)− [X,Y ]

= TD (X,Y ) +A 〈X〉Y −A 〈Y 〉X.

Now let A 〈vm〉wm := − 12T

D 〈vm, wm〉 which used in the above equation shows,

T∇ (X,Y ) = TD (X,Y )− 1

2TD (X,Y ) +

1

2TD (Y,X) = 0.

9.6 Cartan’s Rolling Map and a geometric interpretationof torsion

In order to find a geometric interpretation of the torsion tensor, we must in-troduce a new global construction which can only be done on TM and not anarbitrary vector bundle. The construction is Cartan’s rolling map. In this sec-tion ∇ will be a fixed covariant derivative on TM, o ∈M a given point, and welet C∞0 ([0, 1] , ToM) and C∞o ([0, 1] ,M) denote those b ∈ C∞([0, 1] , ToM) andσ ∈ C∞o ([0, 1] ,M) respectively so that b (0) = 0 and σ (0) = o.


9.6 Cartan’s Rolling Map and a geometric interpretation of torsion 81

Definition 9.31 (Cartan’s rolling map). Given the above notation we defineCartan’s rolling map,

ϕ = ϕ∇ : C∞0 ([0, 1] , ToM)→ C∞o ([0, 1] ,M)

to be the solution to the functional differential equation,

ϕt (b) = //t (ϕ· (b)) bt with ϕ0 (b) = o.

Alternatively, we may put this into a more standard ODE format by lettingut (b) := //t (ϕ· (b)) , in which case we must solve the equations,

ϕt (b) = ut (b) bt with ϕ0 (b) = o and

∇ut (b)

dt= 0 with u0 (b) = IToM .

Our next goal is to compute the differential of the Cartan Rolling map withrespect to b. First it will be convenient to introduce a little more notation.

Notation 9.32 For u ∈ O (ToM,TmM) , let T∇u : ToM × ToM → ToM andR∇u : ToM × ToM → End (ToM) be defined by

T∇u (v, w) := u−1T∇m (uv, uw) and R∇u (v, w) := u−1R∇m (uv, uw)u. (9.29)

Theorem 9.33 (Differentiating Cartan’s Rolling Map). For h, b ∈C∞0 ([0, 1] , ToM), bt (s) := bt + sht, and ut (b) := //t (ϕ· (b)) . If we define

at := ut (b)−1

(∂hϕt) (b) = ut (b)−1 d

ds|0ϕt (b (s)) , (9.30)

then at solves the functional linear differential equation,

at = T∇ut

(bt, at

)+

[∫ t

0

R∇uτ

(bτ , aτ

)dτ

]bt + ht with a0 = 0. (9.31)

Proof. We may rewrite Eq. (9.30) as

utat =d

ds|0ϕt (b (s)) where ut := ut (b) .

Taking ∇dt of this equation implies

utat =∇dt

[utat] =∇dt

d

ds|0ϕt (b (s))

= T∇(ϕt (b) ,

d

ds|0ϕt (b (s))

)+∇ds|0d

dtϕt (b (s))

= T∇(utbt, utat

)+∇ds|0[//t (ϕ· (b (s))) bt (s)

]= T∇

(utbt, utat

)+

[∇ds|0//t (ϕ· (b (s)))

]bt + utht

= T∇(utbt, utat

)+ ut

[∫ t

0

u−1τ R∇

(uτ bτ , uτaτ

)uτdτ

]bt + utht.

This equation is then equivalent to Eq. (9.31).

Theorem 9.34 (Cartan’s rolling map around small loops). If b ∈C∞0 ([0, 1] , ToM), s ∈ R, then

ToM 3 ut (sb)−1 d

dsϕt (sb) = bt + s

∫ t

0

T∇(bτ , bτ

)dτ +O

(s2). (9.32)

Proof. Let bt (s) := sbt so that ht = ddsbt (s) = bt and we set

at (s) := ut (sb)−1 d

dsϕt (sb) and

ut (s) := ut (b (s)) = //t (ϕ· (b (s))) .

Then from Theorem 9.33, it follows that

at (s) = sT∇ut(s)

(bt, at (s)

)+ s2

[∫ t

0

R∇uτ (s)

(bτ , aτ (s)

)dτ

]bt + bt. (9.33)

Now let us expand at (s) in its Taylor’s series approximation in s, i.e.

at (s) = a0t + sa1

t + s2a2t + . . . .

Plugging this expansion into Eq. (9.33) using T∇ut(s) = T∇o +O (s) allows us toconclude that

a0t = bt with a0

0 = 0 =⇒ a0t = bt

and

a1t = T∇

(bt, a

0t

)=⇒ a1

t =

∫ t

0

T∇(bτ , bτ

)dτ.

This then leads directly to Eq. (9.32).



Corollary 9.35. Let b ∈ C∞0 ([0, 1] , ToM) be a loop, i.e. b0 = b1 = 0. Then[0, 1) 3 ε→ ϕ1 (

√εb) is right differentiable at ε = 0 and

d

dε|0+ϕ1

(√εb)

=1

2

∫ 1

0

T∇(bτ , bτ

)dτ.

Proof. Evaluating Eq. (9.32) at t = 1 using b1 = 0 shows

//1 (ϕ· (sb))−1 d

dsϕ1 (sb) = s

∫ 1

0

T∇(bτ , bτ

)dτ +O

(s2).

Hence if f is a smooth function on M,

d2

ds2|0f (ϕ1 (sb)) =

d

ds|0df

⟨d

dsϕ1 (sb)

⟩=

d

ds|0

[df //1 (ϕ· (sb))] //1 (ϕ· (sb))−1 d

dsϕ1 (sb)

= df

⟨∫ 1

0

T∇(bτ , bτ

)dτ

⟩.

Therefore we have shown,

f (ϕ1 (sb)) = f (o) +s2

2df

⟨∫ 1

0

T∇(bτ , bτ

)dτ

⟩+O

(s3)

and hence

f(ϕ1

(√εb))

= f (o) +ε

2df

⟨∫ 1

0

T∇(bτ , bτ

)dτ

⟩+O

(ε3/2

).

Therefore it follows that

d

dε|0+f

(ϕ1

(√εb))

=1

2df

⟨∫ 1

0

T∇(bτ , bτ

)dτ

⟩.

Corollary 9.36 (Riemann’s Theorem). Suppose (M, g) is a Riemannianmanifold and ∇g is the Levi-Civita covariant derivative, see Definition 9.37below. If R∇ = 0, then the rolling map is a homotopy invariant and we maydefine ψ (x) := ϕ1 ((·)x) for x ∈ ToM. This map is an isometry.

Proof. Let bx (t) := tx for all x ∈ ToM. From Theorem 9.33 it follows that

ψ∗vx = (∂vψ) (x) = u1 (bx) (1v) = //1 (ϕ· (bx)) v.

This shows that ψ : ToM →M is an isometry where it is defined.

9.7 Riemann’s Theorem revisited

Definition 9.37 (Levi–Civita covariant derivative). Let (M, g) be a Rie-mannian manifold. The Levi–Civita covariant derivative (∇ = ∇g) is theunique metric compatible and Torsion free covariant derivative on (TM, g).

Proposition 9.38. Let (M, g) be a Riemannian manifold. Then there is aunique covariant derivative ∇ on TM such that ∇g = 0 and T∇ = 0. Moreover,∇ is determined by the equation

2g(∇XY,Z) = Xg(Y, Z) + Y g(Z,X)− Zg(X,Y )

+ g(Z, [X,Y ])− g(Y, [X,Z])− g(X, [Y,Z]). (9.34)

Proof. For the moment assume that ∇ exists and let X,Y, Z be vector fieldson M. Since ∇g = 0, we have

Xg(Y,Z) = g(∇XY,Z) + g(Y,∇XZ)

Y g(Z,X) = g(∇Y Z,X) + g(Z,∇YX)

Zg(X,Y ) = g(∇ZX,Y ) + g(X,∇ZY ).

Subtracting the bottom equation from the sum of the top two equations impliesthat

Xg(Y,Z) + Y g(Z,X)− Zg(X,Y )

= g(∇XY,Z) + g(Y,∇XZ) + g(∇Y Z,X) + g(Z,∇YX)

− g(∇ZX,Y )− g(X,∇ZY )

= g(∇XY +∇YX,Z) + g(∇XZ −∇ZX,Y )

+ g(X,∇Y Z −∇ZY )

= g(∇XY +∇XY + [Y,X], Z)

+ g([X,Z], Y ) + g(X, [Y,Z]).

From this equation we learn that

2g(∇XY,Z) = Xg(Y, Z) + Y g(Z,X)− Zg(X,Y )

+ g([X,Y ], Z)− g([X,Z], Y )− g(X, [Y,Z]).

This shows that ∇ is unique if it exists. To prove existence, define ∇XY byequation (9.34) above. We must only check that ∇ satisfies the condition of acovariant derivative.

To this end let α(X,Y, Z) denote the right side of Eq. (9.34). Recall that[X, fY ] = Xf · Y + f [X,Y ]. Therefore for f ∈ C∞(M),we have


9.7 Riemann’s Theorem revisited 83

α(X,Y, fZ) = Xg(Y, fZ) + Y g(fZ,X)− fZg(X,Y )

+ g(fZ, [X,Y ])− g(Y, [X, fZ])− g(X, [Y, fZ])

= fα(X,Y, Z) +Xf · g(Y,Z) + Y f · g(Z,X)

−Xf · g(Y, [X,Z])− Y f · g(X, [Y,Z])

= fα(X,Y, Z),

α(fX, Y, Z) = fα(X,Y, Z) + Y f · g(Z,X)− Zf · g(X,Y )

− Y f · g(X,Z) + Zf · g(X,Y )

= fα(X,Y, Z),

and

α(X, fY, Z) = fα(X,Y, Z) +Xf · g(Y, Z)− Zf · g(X,Y )

+Xf · g(Y,Z) + Zf · g(X,Y )

= fα(X,Y, Z) + 2Xf · g(Y,Z).

The first equation shows that there exists ∇XY ∈ Γ (TM) such that Eq. (9.34)holds, the second equation shows that ∇fXY = f∇XY, and the third showsthat ∇X(fY ) = f∇XY +Xf · Y. That is ∇ is a covariant derivative.

To finish the proof we must show that ∇g = 0 and T∇ = 0.We may now suppose that X,Y and Z commute, in which case Eq. (9.34)

reduces to the equation

2g(∇XY, Z) = Xg(Y,Z) + Y g(Z,X)− Zg(X,Y ).

Interchanging X and Y gives

2g(∇YX,Z) = Y g(X,Z) +Xg(Z, Y )− Zg(X,Y ).

Subtracting the previous two equations shows that

2g(T∇(X,Y ), Z) = 2g(∇XY −∇YX,Z) = 0,

which shows that T∇ = 0. To verify ∇g = 0, we must shows that

Xg(Y,Z) = g(∇XY,Z) + g(Y,∇XZ).

Now as above

2g(∇XZ, Y ) = Xg(Y,Z) + Zg(X,Y )− Y g(Z,X),

so that

2g(∇XY, Z) + 2g(∇XZ, Y ) = Xg(Y, Z) + Y g(Z,X)− Zg(X,Y )

+Xg(Y, Z) + Zg(X,Y )− Y g(Z,X)

= 2X [g(Y, Z)]

as desired.

Definition 9.39. A smooth map, f : M → N between two Riemannian mani-folds, (M, g) and (N,h) , is isometric if f∗m : TmM → Tf(m)N is isometric forall m ∈M.

Theorem 9.40. Suppose f : M → N is a diffeomorphism of manifolds, ∇ is acovariant derivative on TN, and let f∗∇ be defined by

(f∗∇)vX := f−1∗ ∇f∗vXf for all X ∈ Γ (TM) and v ∈ TM. (9.35)

Here X is a vector field on M and Xf ∈ Γ (TN) is defined by Xf := f∗X f−1.Then:

1. f∗∇ is a covariant derivative on TM.2. If X (t) ∈ TM is a smooth curve then, f∗∇

dt = f−1∗∇dtf∗, i.e.

f∗∇dt

X (t) = f−1∗∇dt

[f∗X (t)] . (9.36)

3. The torsion tensors, T∇ and T f∗∇ satisfy the identity,

T f∗∇ (v, w) = f−1

∗mT∇ (f∗v, f∗w) for all v, w ∈ TmM and m ∈M. (9.37)

In particular T f∗∇ = 0 iff T∇ = 0.

4. The curvature tensors, R∇ and Rf∗∇ satisfy the identity,

Rf∗∇ (v, w) = f−1

∗mR∇ (f∗v, f∗w) f∗m for all v, w ∈ TmM and m ∈M.

(9.38)5. The relation between parallel translations on TM and TN are given by

//f∗∇t (σ) = f−1

∗σ(t)//∇t (f σ) f∗σ(0). (9.39)

6. Further suppose that (M, g) and (N,h) are Riemannian manifolds, f :M → N is an isometric diffeomorphism, and ∇ is the Levi-Civita covariantderivative on N. Then f∗∇ is the Levi-Civita covariant derivative on N.

Proof. 1. & 2. As usual the main point is to show that f∗∇ defined in Eq.(9.35) satisfies the product rule. For this let ϕ ∈ C∞(M), then



(f∗∇)v [ϕX] := f−1∗ ∇f∗v

[ϕ f−1 ·Xf

]= ϕ f−1 (f (m)) · f−1

∗ ∇f∗vXf + f−1∗((f∗v)

[ϕ f−1

]·Xf (f (m))

)= ϕ (m) · (f∗∇)vX + v

[ϕ f−1 f

]·X (m)

= ϕ (m) · (f∗∇)vX + vϕ ·X (m) .

The remaining properties of a covariant derivative are all easily verified. It isalso verified that f−1

∗∇dtf∗X (t) satisfies the basic properties in Proposition 9.7

and hence Eq. (9.36) is verified as well.3. Let Σ (s, t) ∈M be a smooth function, then

f∗Tf∗∇

(Σ,Σ′

)=

(f∗∇dt

d

ds− f∗∇

ds

d

dt

)Σ =

∇dt

[f∗Σ′]− ∇

ds

[f∗Σ

]=∇dt

d

ds(f Σ)− ∇

ds

d

dt(f Σ)

= T∇(d

dt(f Σ) ,

d

ds(f Σ)

)= T∇

(f∗Σ, f∗Σ

′)

which finishes the proof of Eq. (9.37).4. and 5. Similarly if we let S (s, t) ∈ TΣ(s,t)M, we find

Rf∗∇(Σ,Σ′

)S =

[f∗∇dt

,f∗∇ds

]S = f∗

[f−1∗∇dtf∗, f

−1∗∇dsf∗

]S

= f−1∗

[∇dt,∇ds

]f∗S = f−1

∗ R(f∗Σ, f∗Σ

′)f∗S.

As for parallel translation, simply observe that

f∗∇dt

[f−1∗σ(t)//

∇t (f σ) f∗σ(0)

]= f−1∗σ(t)

∇dtf∗

[f−1∗σ(t)//

∇t (f σ) f∗σ(0)

]= 0

and[f−1∗σ(t)//

∇t (f σ) f∗σ(0)

]t=0

= ITσ(0)M .

6. Given the results already proved, we need only show that f∗∇ is g –compatible. To see this let X (t) , Y (t) ∈ Tσ(t)M, then

d

dtg (X (t) , Y (t)) =

d

dth (f∗X (t) , f∗Y (t))

= h

(∇dtf∗X (t) , f∗Y (t)

)+ h

(f∗X (t) ,

∇dtf∗Y (t)

)= g

(f−1∗∇dtf∗X (t) , Y (t)

)+ g

(X (t) , f−1

∗∇dtf∗Y (t)

)= g

(f∗∇dt

X (t) , Y (t)

)+ g

(X (t) ,

f∗∇dt

Y (t)

)and the proof is complete.

Theorem 9.41. Suppose that(Md, g

)is a Riemannian manifold, ∇ is the Levi-

Civita covariant derivative on TM and R∇ = 0. Then(Md, g

)is locally iso-

metric to Rd with its standard flat metric.

Proof. Let o ∈ M be a fixed point eidi=1 be an orthonormal frame at o.Let U be a simply connected open neighborhood of o ∈M and for any x ∈ U, letEi (x) := // (σ) ei where σ is any path in U such that σ (0) = o and σ (1) = x.Since R∇ = 0 we know that Ei (x) is well defined. Moreover if vx ∈ TxM wemay choose σ such that σ (1) = vx and use this to conclude,

∇vxEi =∇dt|0Ei (σ (t)) =

∇dt|0//t (σ) ei = 0.

Thus we know that ∇Ei = 0 for all i. As ∇ is metric compatible we know thatEidi=1 is an orthonormal frame for TU and since T∇ = 0 we may concludethat

0 = ∇EiEj −∇EjEi − [Ei, Ej ] = − [Ei, Ej ] .

This shows that Eidi=1 are commuting vector fields. Thus the function,ϕ (t1, . . . , td) := et1E1+···+tdEd (o) satisfies

ϕ′ (t) ei :=∂

∂tiϕ (t1, . . . , td)

=∂

∂tietiEi et1E1+···+tiEi+···+tdEd (o)

= Ei ϕ (t1, . . . , td) .

From this we learn that ϕ′ (t) : Rd → Tϕ(t)M is isometric and the proof iscomplete.


10

Riemannian Manifolds

This chapter and the next are devoted to some basic Riemannian geometryalong with some some examples. The reader not interested in this material mayskip directly to Chapter 12.

Definition 10.1. A Riemannian metric g on a manifold M is a section ofΓ (T ∗M ⊗ T ∗M) such that for all m ∈ M, gm : TmM ⊗ TmM is an innerproduct. A manifold M is called a Riemannian manifold if M is equipped witha Riemannian metric g.

Definition 10.2. A Riemannian metric, 〈·, ·〉 (also denoted by g), on M isa smoothly varying choice of inner product, gm = 〈·, ·〉m, on each of the tangentspaces TmM, m ∈ M. The smoothness condition is the requirement that thefunction m ∈ M → 〈X(m), Y (m)〉m ∈ R is smooth for all smooth vector fieldsX and Y on M.

It is customary to write ds2 for the function on TM defined by

ds2(vm) := 〈vm, vm〉m = gm (vm, vm) . (10.1)

By polarization, the Riemannian metric 〈·, ·〉 is uniquely determined by thefunction ds2. Given a chart x on M and v ∈ TmM, by Eqs. (10.1) and theidentity,

vm =

d∑i=1

dxi (vm)∂

∂xi|m, (10.2)

implies

ds2(vm) =

d∑i,j=1

〈∂/∂xi|m, ∂/∂xj |m〉mdxi(vm)dxj(vm). (10.3)

We will abbreviate this equation in the future by writing

ds2 =

d∑i,j=1

gxijdxidxj (10.4)

wheregxi,j(m) := 〈∂/∂xi|m, ∂/∂xj |m〉m = g

(∂/∂xi|m, ∂/∂xj |m

).

Typically gxi,j will be abbreviated by gij if no confusion is likely to arise.

Example 10.3. Let M = RN and let x = (x1, x2, . . . , xN ) denote the standardchart on M, i.e. x(m) = m for all m ∈M. The standard Riemannian metric onRN is determined by

ds2 =

N∑i=1

(dxi)2 =

N∑i=1

dxi · dxi,

and so gx is the identity matrix here. The general Riemannian metric on RN isdetermined by ds2 =

∑Ni,j=1 gijdx

idxj , where g = (gij) is a smooth gl(N,R) –

valued function on RN such that g(m) is positive definite matrix for all m ∈ RN .

Let M be an imbedded submanifold of a finite dimensional inner productspace (E, 〈·, ·〉). The manifold M inherits a metric from E determined by

ds2(vm) = 〈v, v〉 ∀ vm ∈ TM.

It is a well known deep fact that all finite dimensional Riemannian manifoldsmay be constructed in this way, see Nash [?] and Moser [?, ?, ?]. To simplifythe exposition, in the sequel we will usually assume that (E, 〈·, ·〉) is an innerproduct space, Md ⊂ E is an imbedded submanifold, and the Riemannianmetric on M is determined in this way, i.e.

〈vm, wm〉 = 〈v, w〉RN , ∀ vm, wm ∈ TmM and m ∈M.

In this setting the components gxi,j of the metric ds2 relative to a chart x may

be computed as gxi,j(m) = 〈φ;i(x(m)), φ;j(x(m))〉, where eidi=1 is the standard

basis for Rd,φ := x−1 and φ;i(a) :=

d

dt|0φ(a+ tei).

Example 10.4. Let M = G := SL(n,R) and Ag ∈ TgM.

1. Thends2(Ag) := tr(A∗A) (10.5)

defines a Riemannian metric on G. This metric is the inherited metric fromthe inner product space E = gl(n,R) with inner product 〈A,B〉 := tr(A∗B).

86 10 Riemannian Manifolds

2. A more “natural” choice of a metric on G is

ds2(Ag) := tr((g−1A)∗g−1A). (10.6)

This metric is invariant under left translations, i.e. ds2(Lk∗Ag) = ds2(Ag),for all k ∈ G and Ag ∈ TG. According to the imbedding theorem of Nashand Moser, it would be possible to find another imbedding of G into aEuclidean space, E, so that the metric in Eq. (10.6) is inherited from aninner product on E.

Example 10.5. Let M = R3 be equipped with the standard Riemannian metricand (r, ϕ, θ) be spherical coordinates on M , see Figure 10.1. Here r, ϕ, and

Fig. 10.1. Defining the spherical coordinates, (r, θ, φ) on R3.

θ are taken to be functions on R3 \ p ∈ R3 : p2 = 0 and p1 > 0 definedby r(p) = |p|, ϕ(p) = cos−1(p3/|p|) ∈ (0, π), and θ(p) ∈ (0, 2π) is given byθ(p) = tan−1(p2/p1) if p1 > 0 and p

2> 0 with similar formulas for (p1, p2) in

the other three quadrants of R2. Since

x1 = r sinϕ cos θ, x2 = r sinϕ sin θ, and dx3 = r cosϕ,

it follows that,

dx1 =∂x1

∂rdr +

∂x1

∂ϕdϕ+

∂x1

∂θdθ

= sinϕ cos θdr + r cosϕ cos θdϕ− r sinϕ sin θdθ,

dx2 = sinϕ sin θdr + r cosϕ sin θdϕ+ r sinϕ cos θdθ,

anddx3 = cosϕdr − r sinϕdϕ.

An elementary calculation now shows that

ds2 =

3∑i=1

(dxi)2 = dr2 + r2dϕ2 + r2 sin2 ϕdθ2. (10.7)

From this last equation, we see that

g(r,ϕ,θ) =

1 0 00 r2 00 0 r2 sin2 ϕ

. (10.8)

Exercise 10.1. Let M := m ∈ R3 : |m|2 = ρ2, so that M is a sphere ofradius ρ in R3. Since r = ρ on M and dr (v) = 0 for all v ∈ TmM, it followsfrom Eq. (10.7) that the induced metric ds2 on M is given by

ds2 = ρ2dϕ2 + ρ2 sin2 ϕdθ2, (10.9)

and hence

g(ϕ,θ) =

[ρ2 00 ρ2 sin2 ϕ

]. (10.10)

10.0.1 The Levi-Civita Connection

Theorem 10.6. Suppose that(Md, g

)is a Riemannian manifold and ∇ is a

covariant derivative on TM. Then the following are equivalent:

1. ∇ is metric compatible.2. If u is an orthonormal frame, then Au := u−1∇u is an so (d) – valued one

form.3. For all smooth curves σ (t) ∈M and all X,Y ∈ Γσ(t) (M) ,

d

dtg (X (t) , Y (t)) = g

(∇dtX (t) , Y (t)

)+ g

(X (t) ,

∇dtY (t)

). (10.11)

4. Parallel translation, //t (σ) : Tσ(0)M → Tσ(t)M is always an isometry.

Proof. 1. ⇐⇒ 2. Let u be an orthonormal frame, i.e. u (m) : Rd → TmMis an isometry for all m ∈ D (u) . Then for a, b ∈ Rd we have

(∇vmg) (ua, ub) = vv [g (ua, ub)]− g (∇vmua, ub)− g (ua,∇vmub)= 0−

(u−1∇vmua, b

)−(a, u−1∇vmub

)= − (Au 〈vm〉 a, b)− (a,Au 〈vm〉 b)

from which it follows that ∇vmg = 0 iff Au 〈vm〉 ∈ so (d) .


10 Riemannian Manifolds 87

2. ⇐⇒ 3. Writing X (t) = u (σ (t)) a (t) and Y (t) = u (σ (t)) b (t) , we have

d

dtg (X (t) , Y (t)) =

d

dt(a (t) , b (t)) = (a (t) , b (t)) +

(a (t) , b (t)

)while on the other hand

g

(∇dtX (t) , Y (t)

)= g (u (σ (t)) [a (t) +Au (σ (t)) a (t)] , u (σ (t)) b (t))

= ([a (t) +Au (σ (t)) a (t)] , b (t))

and using a similar result for g(X (t) , ∇dtY (t)

), we find

g

(∇dtX (t) , Y (t)

)+ g

(X (t) ,

∇dtY (t)

)= (a (t) +Au (σ (t)) a (t) , b (t)) +

(a (t) , b (t) +Au (σ (t)) b (t)

).

Hence we see that Eq. (10.11) holds iff

(Au (σ (t)) a (t) , b (t)) + (a (t) , Au (σ (t)) b (t)) = 0

for all a and b and from this it follows that ∇ is metric compatible iff Au 〈vm〉 ∈so (d) iff Eq. (10.11) holds.

3. ⇐⇒ 4. Equation (10.11) implies that ddtg (X (t) , Y (t)) = 0 whenever

X and Y are parallel. From this we conclude by taking X (t) := //t (σ) v andY (t) := //t (σ)w, then

g (//t (σ) v, //t (σ)w) = g (X (t) , Y (t)) = g (v, w) .

Conversely if //t (σ) is always isometric, then

d

dtg (X (t) , Y (t)) =

d

dtg(//t (σ)

−1X (t) , //t (σ)

−1Y (t)

)=g

(d

dt

[//t (σ)

−1X (t)

], //t (σ)

−1Y (t)

)+ g

(//t (σ)

−1X (t) ,

d

dt

[//t (σ)

−1Y (t)

])=g

(//t (σ)

d

dt

[//t (σ)

−1X (t)

], Y (t)

)+ g

(X (t) , //t (σ)

d

dt

[//t (σ)

−1Y (t)

])=g

(∇dtX (t) , Y (t)

)+ g

(X (t) ,

∇dtY (t)

).

Example 10.7. Suppose that M ⊂ RN is an embedded submanifold, and P :M → End

(RN)

is a smooth map such that P (m)2

= P (m) and Ran(P (m)) =τmM for all m ∈ M. Then ∇ := Pd is a torsion free covariant derivative onTM. To prove this observe that

∂vX = ∂v [PX] = ∂vPX + P∂vX = (∂vP )X +∇vX

which shows that

∇vX = ∂vX − (∂vP )X = ∂vX + dQ (v) X

where Q = I−P. From this last expression one easily sees that ∇ is a covariantderivative. Moreover, we know that

R∇ = d [dQ] + dQ ∧ dQ = dQ ∧ dQ.

To see that T∇ = 0, observe(∇dt

d

ds− ∇ds

d

dt

)Σ =

∇dtΣ′ − ∇

dsΣ = P (Σ)

([d

dt,d

ds

]Σ

)= 0.

Definition 10.8 (Geodesics). A path σ in M satisfying

∇dtσ (t) = 0. (10.12)

will be called a ∇–geodesic. When (M, g) is a Riemannian manifold, and ∇ isthe Levi-Civita covariant derivative and σ satisfies Eq. (10.12), we will simplyrefer to σ as a geodesic. Given v ∈ TpM, let σv (t) denote the unique maximalsolution to Eq. (10.12) such that σv (0) = v.

Exercise 10.2. Suppose that f : M → N is an isometric diffeormorphism oftwo Riemannian manifolds, (M, g) and (N,h) . Show σ : [a, b]→M is a geodesicin M iff f σ is a geodesic in N. Show also that f(exp(tv)) = exp(tf∗v) for allv ∈ TmM.

Definition 10.9 (Exponential Map). For each m ∈M, let

D (expm) := vm ∈ TmM : σv (1) exists

and for vm ∈ D (expm) we let expm (vm) := σvm (1) . Finally, let

D (exp) := ∪m∈MD (expm) ⊂ TM.

To better understand σv (t) , let us notice that Equation (10.12) is equivalentto the statement that σv (t) = //t (σ) v from which it follows that, σv (t) ∈Tσv(t)TM satisfies



σv (t) =d

ds|0σv (t+ s) =

d

ds|0//t+s (σ) v

=d

ds|0//s (σ (t+ ·)) //t (σ) v

=d

ds|0//s (σ (t+ ·)) σv (t) . (10.13)

Recall that if X is a vector field on a manifold Q, we let D (X) := ∪q∈QJq×qwhere Jq is the maximal open interval containing 0 ∈ R such that etX is defined.The previous simple computation suggests the following theorem.

Theorem 10.10. Let X be the vector field on TM defined by

X (v) :=d

dt|0 [//t (σ) v] ∈ TvTM for all v ∈ TM (10.14)

where σ (t) ∈M is any C1– curve such that σ (0) = v. Then

etX (v) = σv (t) =d

dtexp (tv)

and in particular

D (exp) = vm ∈ TM : (vm, 1) ∈ D (X) ⊂0 TM

andexp (tv) = π

(etX (v)

)where π : TM → M is the natural projection map. Since X (0m) = 00m itfollows that etX (0m) = 0m for all m ∈ M and from all of this we learn thatD (exp) ⊂0 TM which contains 0m : m ∈M .

Proof. The only thing that needs to be proved is that X in Eq. (10.14) iswell defined and defines a smooth vector field on TM. One method to do thisis to work in a local coordinates. To this end, let x be a chart on M and define

Γ (vm) a = dxm

(∇dt|t=0

(dxσ(t)

)−1(a)

)= dxm

(d∑i=1

ai∇σ(0)∂

∂xi

)

where σ (0) = v. To be more precise, we have

(dxσ(t)

)−1(a) =

d∑i=1

ai∂

∂xi|σ(t)

so that

∇dt|t=0

(dxσ(t)

)−1(a) =

d∑i=1

ai∇dt|t=0

∂

∂xi|σ(t) =

d∑i=1

ai∇v∂

∂xi

so that

Γ (vm) a =

d∑i=1

aidx

(∇v

∂

∂xi

)= dx

(∇v[a · ∂

∂x

])= dx

(a · ∇v

∂

∂x

)where

a · ∂∂x

:=

d∑i=1

ai∂

∂xi.

With this notation we have, for V ∈ Γσ (TM) , that

V (t) = dx (V (t)) · ∂∂x

and

dx

(∇dtV (t)

)= dx

(d

dt[dx (V (t))] · ∂

∂x+ [dx (V (t))] · ∇v

∂

∂x

)=

d

dt[dx (V (t))] + Γ (σ (t)) dx (V (t)) .

Writing c (t) := x (σ (t)) so that c (t) = dx (σ (t)) , then

dx

(∇dtσ (t)

)= c (t) + Γ (σ (t)) c (t)

and therefore the equations for a geodesic becomes,

c (t) + Γ(c (t)c(t)

)c (t) = 0 with c (0) = x (m) and c (0) = dx (vm) .

Hence if we let w (t) := (c (t) , c (t)) , then

w (t) = (c (t) , c (t))(c(t),c(t)) =(c (t) ,−Γ

(c (t)c(t)

)c (t)

)w(t)

which is a first order differential equation for w driven by the vector field

Y (a, b) = (b,−Γ (ba) b)(a,b) .

Since Y is the local representation of the vector field X and Y is well definedand smooth, it follows that X is well defined and smooth as well.

Our next goal is to show that the exponential map, expm, is a local dif-feomorphism around 0m ∈ TmM. This is in fact a simple consequence of theinverse function theorem and the observation that

(expm)∗ (wm)0m=

d

dt|0 expm (twm) = wm ∈ TmM

is invertible. The next theorem is a generalization of this result which providessome uniformity in m. To state this theorem we will use the following notation.


10 Riemannian Manifolds 89

Notation 10.11 For each p ∈M and δ > 0, let

(TpM)<δ := v ∈ TpM : |v| < δ

and similarly let(TpM)≤δ := v ∈ TpM : |v| ≤ δ

and(TpM)δ := v ∈ TpM : |v| = δ .

Moreover, if W ⊂0 M, let

W (δ) := ∪p∈W (TpM)δ ⊂o TM.

Theorem 10.12. To each m ∈ M, there exists a δ = δ (m) > 0 and an openneighborhood W = Wm of m such that

expp : (TpM)<δ → expp((TpM)<δ

)⊂o M

is a diffeomorphism for all p ∈ W. In particular, for every compact subset,K ⊂ M, there exists a δ > 0 such that expp is a diffeomorphism on (TpM)<δfor all p ∈ K.

Proof. Define

F (vm) := (m, exp (vm)) ∈M ×M for all vm ∈ D.

We now compute the differential of F at 0m as follows. Let V (t) ∈ Tσ(t)M suchthat V (0) = 0m and σ (0) = wm ∈ TmM, then

d

dt|0F (V (t)) =

(wm,

d

dt|0 exp (V (t))

).

To compute the latter derivative, write V (t) := //t (σ) v (t) with v (t) ∈ TmMin which case

d

dt|0 exp (V (t)) =

d

dt|0 exp (//t (σ) v (t))

=d

dt|0 exp (//t (σ) v (0)) +

d

dt|0 exp (v (t))

=d

dt|0 exp (//t (σ) 0m) +

d

dt|0 exp (v (t))

=d

dt|0σ (t) + exp∗ [v (t)]0m = wm + v (0) .

Thus we have shown that

F∗0m V (0) = (wm, wm + v (0)) =

(wm, wm +

∇dt|0V (t)

).

We further observe from this formula that F∗0m is surjective (for example letV (t) = //t (σ) tvm) and hence invertible. Alternatively, notice that F∗0m V (0) =0 iff wm = 0 and ∇dt |0V (t) = 0.

Since F is a local diffeomorphism about 0m for all m ∈ M, there exists aneighborhood W of m and δ > 0 such that F |W (δ) : W (δ)→ F (W (δ)) ⊂o M×M is a diffeomorphism. Then for p ∈ W, expp

((TpM)δ

)= i−1

p (F (W (δ))) ⊂oM and exp−1

p (q) = F−1 (p, q) is smooth in p and q and hence in q.

Exercise 10.3. Suppose that f : M → M is an isometric diffeomorphism ofa connected manifold, M. If there is a point m ∈ M such that f(m) = mand f∗m = idTmM , then f ≡ idM Use this to conclude if f, g : M → M areisometries of M such that f(m) = g(m) and f∗m = g∗m for some point m ∈M,then f ≡ g.

Definition 10.13. A vector-field X ∈ Γ (TM) is a Killing vector field if(etX)∗m is an isometry for all m ∈ D(etX) and t ∈ R. (We will see later, thatif (M, g) is “complete”, then every Killing vector field on M is complete.)

Exercise 10.4. Let X ∈ Γ (TM). Show the following are equivalent:

1. X is a Killing vector field.2. LXg = 0, where g denotes the metric on M.3. X〈Y, Y 〉 = 2〈[X,Y ], Y 〉 for all Y ∈ Γ (TM), where

〈Y, Y 〉 ≡ g〈Y, Y 〉.

4. 〈∇vX, v〉 = 0 for all v ∈ TmM.

Exercise 10.5. Let ΓK ⊂ Γ (TM) denote the set of Killing vector-fields on M.Show that ΓK is a Lie-subalgebra of Γ (TM).

Hint: First show for X,Y ∈ Γ∞(TM) that

[LX , LY ] = L[X,Y ], (10.15)

where LX denotes the Lie derivative acting on arbitrary tensor fields. In par-ticular, (10.15) may be applied to the metric g.

Exercise 10.6. Let(Md, g

)be a connected Riemannian manifold. Fix a point

m ∈M, and let

som := A ∈ End(TmM) : 〈Av, v〉 = 0 for all v ∈ TmM.

For each X ∈ ΓK let AX ∈ som be defined by AXv ≡ ∇vX for all v ∈ TmM.Prove the map



(X → (X(m), AX)) : ΓK → TmM × som

is injective. Conclude from this that

dim(ΓK) ≤ d(d+ 1)

2.

Hint: Let v ∈ T σ(t) = expm(tv), and f(s, t) = esX(σ(t)) — f(s, t) is welldefined for s sufficiently small. Then make use of Exercise 10.2 to conclude thatt→ f(s, t) is a geodesic.

10.1 Left Invariant Metrics on Lie Groups

Let G be a Lie group and g = (·, ·) be any Riemannian metric on G which isleft invariant, i.e. L∗kg = g for all k ∈ G. We may then define a global frame,u : G→ End (g, TG) by

u (g)A = Lg∗A = A (g) for all g ∈ G and A ∈ g. (10.16)

Example 10.14. Let M = G := SL(n,R) and Ag ∈ TgM.

1. Thends2(Ag) := tr(A∗A) (10.17)

defines a Riemannian metric on G. This metric is the inherited metric fromthe inner product space E = gl(n,R) with inner product 〈A,B〉 := tr(A∗B).

2. A more “natural” choice of a metric on G is

ds2(Ag) := tr((g−1A)∗g−1A). (10.18)

This metric is invariant under left translations, i.e. ds2(Lk∗Ag) = ds2(Ag),for all k ∈ G and Ag ∈ TG. According to the imbedding theorem of Nashand Moser, it would be possible to find another imbedding of G into aEuclidean space, E, so that the metric in Eq. (10.18) is inherited from aninner product on E.

Example 10.15. Relative to the frame in Eq. (10.16), the connection one formassociated to the Levi-Civita derivative on TG is given by

ω(A)B =

1

2([A,B]− ad∗AB − ad∗BA) , (10.19)

where ad∗A denotes the adjoint of adA on (g, (·, ·)) . So if we let θ (vg) := Lg−1∗vg

ω (vg)B =1

2

([θ (vg) , B]− ad∗θ(vg)B − ad

∗Bθ (vg)

)(10.20)

Proof. Let A,B,C ∈ g, then, according to Eq. (9.34),

2g(∇AB, C) = Ag(B, C) + Bg(C, A)− Cg(A, B)

+ g(C, [A, B])− g(B, [A, C])− g(A, [B, C])

= (C, [A,B])− (B, [A,C])− (A, [B,C])

= (adAB − ad∗AB − ad∗BA,C)

= g(

(adAB − ad∗AB − ad∗BA)˜, C)

from which we conclude that

∇A [uB] = ∇AB =1

2(adAB − ad∗AB − ad∗BA)

˜

=1

2u (adAB − ad∗AB − ad∗BA) .

Thus we have shown

ωu(A)B =

1

2u (adAB − ad∗AB − ad∗BA)

as claimed in Eq. (10.19).

Proposition 10.16. Now suppose that σ and v are C1 – curves in G and TGrespectively and

a(t) ≡ Lσ(t)−1∗v(t).

If ∇v(t)/dt denotes the covariant differential of v along σ, then

∇v(t)

dt= Lσ(t)∗

(da(t)

dt+ ω (σ (t)) a(t)

). (10.21)

Proof. Let S is an orthonormal basis for g, then

v (t) =∑A∈S〈a(t), A〉A(σ(t))

and therefore,

∇v(t)

dt=∑A∈S〈da(t)

dt, A〉A(σ(t)) + 〈a(t), A〉∇σ(t)A

=∑A∈S〈da(t)

dt, A〉A(σ(t)) + 〈a(t), A〉 [ω (σ (t))A]

˜

= Lσ(t)∗

(da(t)

dt+ ω (σ (t)) a(t)

).


10.1 Left Invariant Metrics on Lie Groups 91

Corollary 10.17. Let σ (t) ∈ G be a C2 curve and a (t) := θ (σ (t)) . Then σ isa geodesic iff

a (t) = ad∗a(t)a (t) . (10.22)

Proof. From the equation for a geodesic, we have σ is a geodesic iff

0 =∇σ(t)

dt= Lσ(t)∗

(da(t)

dt+ ω (σ (t)) a(t)

)which happens iff

0 =da(t)

dt+ ω (σ (t)) a(t)

= a (t) +1

2

([a (t) , a (t)]− ad∗a(t)a (t)− ad∗a(t)a (t)

)= a (t)− ad∗a(t)a (t) .

Example 10.18. Suppose that (·, ·) is an AdG – invariant inner product. In thiscase ad∗A = −adA and Eqs. (10.20) and (10.22) become

ω (vg)B =1

2[θ (vg) , B] =

1

2adθ(vg)B and a (t) = 0

respectively. Hence in this case, σ (t) ∈ G is a geodesic iff σ (t) = A (σ (t)) , i.e.σ (t) = getA for some g ∈ G and A ∈ g. Moreover we have

dω(A, B

)C = Aω

(B)− Bω

(A)− ω

([A, B

])= AadB − BadA −

1

2ad[A,B] = −1

2ad[A,B]

while

ω ∧ ω(A, B

)=[ω(A), ω(B)]

=

[1

2adA,

1

2adB

]=

1

4ad[A,B].

Therefore we have

R∇(A, B

)= u

([dω + ω ∧ ω]

(A, B

))u−1 = −1

4uad[A,B]u

−1.

and in particular we have

R∇(A, B

)C = −1

4

(ad[A,B]C

)˜.

From this we compute the “sectional curvature” as

(R∇

(A, B

)B, A

)= −1

4

(ad[A,B]B,A

)=

1

4(adB [A,B] , A)

=1

4([A,B] ,−adBA) =

1

4([A,B] , [A,B])

=1

4|[A,B]|2 ≥ 0. (10.23)

Alternatively we may compute the curvature using,

R∇(A, B

)C = [∇A,∇B ] C −∇[A,B]C

=

([1

2adA,

1

2adB

]C − 1

2ad[A,B]C

)˜

=

(1

4ad[A,B]C −

1

2ad[A,B]C

)˜

= −1

4

(ad[A,B]C

)˜.

Let us finally observe that ∇R∇ = 0 in this case, i.e. that R∇ is covariantlyflat. Indeed, we have(

∇DR∇) (A, B) C=∇D

[R∇

(A, B

)C]−R∇

(∇DA, B

)C

−R∇(A,∇DB

)C −R∇

(A, B

)∇DC

=1

8

−adDad[A,B]C + ad[adDA,B]C+ad[A,adDB]C + ad[A,B]adDC

˜

=1

8

−adD [[A,B] , C] + [[adDA,B] , C]+ [[A, adDB] , C] + [[A,B] , adDC]

˜

.

Since

adD [[A,B] , C]

=d

dt|0AdetD [[A,B] , C] =

d

dt|0 [[AdetDA,AdetDB] , AdetDC]

= [[adDA,B] , C] + [[A, adDB] , C] + [[A,B] , adDC] ,

it now follows that(∇DR∇

) (A, B

)C ≡ 0.

This may also be understood by looking at the equations for parallel trans-lation in this case, namely //t (σ) = Lσ(t)∗U (t) where

U (t) +1

2adc(t)U (t) = 0 with U (0) = I



and c (t) := θ (σ (t)) . Suppose that we let g (t) ∈ G be defined by

g (t) =1

2Lg(t)∗c (t) with g (0) = e,

then we find thatd

dtAdg(t) =

1

2Adg(t)adc(t)

and hence that

d

dtAdg(t)−1 = −1

2adc(t)Adg(t)−1 with Adg−1(0) = Id.

From this observation, it follows that U (t) = Adg(t)−1 and we learn that//t (σ) = Lσ(t)∗Adg(t)−1 . Therefore

R//t(σ) (A,B)C

= //t (σ)−1R (//t (σ)A, //t (σ)B) //t (σ)C

= Adg(t)L−1σ(t)∗R

(Lσ(t)∗Adg(t)−1A,Lσ(t)∗Adg(t)−1B

)Lσ(t)∗Adg(t)−1C

= −1

4Adg(t)

[ad[Adg(t)−1A,Adg(t)−1B]Adg(t)−1C

]= −1

4

[ad[A,B]C

]which is independent of t.

10.2 A General Left Invariant Ricci CurvatureComputations

Proposition 10.19. Suppose that G is a Lie group, 〈·, ·〉 = (·, ·) is a left in-variant Riemannian metric on G, then for A ∈ g = TeG = Lie(G),

(RicA,A) =∑B∈Γ

tr(adB) (adAB,A)− 1

2tr(ad2A

)+

1

4

∑B,C∈Γ

|(A, [B,C])|2 − 1

2

∑B∈Γ|[A,B]|2 (10.24)

where Γ ⊂ g be an orthonormal basis.

Remark 10.20. If G is unimodular, tr (adB) = 0 for all B ∈ g and Eq. (10.24)reduces to

(RicA,A) = −1

2tr(ad2A

)+

1

4

∑B,C∈Γ

|(A, [B,C])|2 − 1

2

∑B∈Γ|[A,B]|2 . (10.25)

Moreover, if g is nilpotent, then adA is nilpotent and hence has only zero eigen-values. Hence both tr (adA) = 0 and tr

(ad2A

)= 0 and therefore Eq. (10.24)

reduces to

(RicA,A) =1

4

∑B,C∈Γ

|(A, [B,C])|2 − 1

2

∑B∈Γ|[A,B]|2 . (10.26)

Proof. Since the Levi-Civita covariant derivative is metric compatible andsince (A,B) is constant for A,B ∈ g,

(R(A,B)B,A) =(∇A∇BB −∇B∇AB −∇[A,B]B,A

)= −(∇BB,∇AA) + (∇AB,∇BA)− (∇[A,B]B,A). (10.27)

In order to compute the Ricci tensor we need to sum this last equation onB ∈ Γ. From Eq. (10.19)∑B∈Γ∇BB = −

∑B∈Γ

ad∗BB = −∑B∈Γ

∑C∈Γ

(ad∗BB,C)C

= −∑B∈Γ

∑C∈Γ

(B, adBC)C =∑C∈Γ

∑B∈Γ

(B, adCB)C =∑C∈Γ

tr(adC)C.

(10.28)

Hence we have

−∑B∈Γ

(∇BB,∇AA) = −∑C∈Γ

tr(adC) (C,−ad∗AA) =∑C∈Γ

tr(adC) (adAC,A) .

(10.29)Similarly, using Eq. (10.19), the second term in Eq. (10.27) becomes

(∇AB,∇BA) =1

4(adAB − ad∗AB − ad∗BA, adBA− ad∗BA− ad∗AB)

= −1

4(adAB − ad∗AB − ad∗BA, adAB + ad∗AB + ad∗BA)

= −1

4

(|adAB|2 − |ad∗AB + ad∗BA|

2)

and the third becomes

(∇[A,B]B,A) =1

2

(ad[A,B]B − ad∗[A,B]B − ad

∗B [A,B], A

)=

1

2

(ad[A,B]B,A)− (B, ad[A,B]A)− ([A,B], adBA)

=

1

2

−(adB [A,B], A) + (B, adA[A,B]) + |[A,B]|2

=

1

2

(ad2

BA,A) + (B, ad2AB) + |[A,B]|2

.


10.2 A General Left Invariant Ricci Curvature Computations 93

Combining all of these results implies

(RicA,A) =∑B∈Γ


4

∑B∈Γ

(|adAB|2 − |ad∗AB + ad∗BA|

2)

− 1

2

∑B∈Γ

(ad2

BA,A) + (B, ad2AB) + |[A,B]|2

=∑B∈Γ


2tr(ad2A

)− 3

4

∑B∈Γ|[A,B]|2

+1

4

∑B∈Γ|ad∗AB + ad∗BA|

2 − 1

2

∑B∈Γ

(ad2BA,A). (10.30)

Since∑B∈Γ|ad∗AB + ad∗BA|

2=

∑B,C∈Γ

|(ad∗AB + ad∗BA,C)|2

=∑

B,C∈Γ|(B, adAC) + (A, adBC)|2

=∑C∈Γ|adAC|2 + 2

∑B,C∈Γ

(B, adAC)(A, adBC) +∑

B,C∈Γ|(A, [B,C])|2

=∑B∈Γ|[A,B]|2 + 2

∑C∈Γ

(A, adadACC) +∑

B,C∈Γ|(A, [B,C])|2

=∑B∈Γ|[A,B]|2 + 2

∑C∈Γ

(A, ad2CA) +

∑B,C∈Γ

|(A, [B,C])|2

=∑B∈Γ|[A,B]|2 + 2

∑B∈Γ

(A, ad2BA) +

∑B,C∈Γ

|(A, [B,C])|2

Eq. (10.30) may be written as

(RicA,A) =∑B∈Γ


2tr(ad2A

)+

1

4

∑B,C∈Γ

|(A, [B,C])|2 − 1

2

∑B∈Γ|[A,B]|2 .

Example 10.21. Suppose that G is compact and 〈·, ·〉 is a bi-invariant Rieman-nian metric on G, then 〈·, ·〉g is an AdG invariant inner product on g. Therefore∑

B,C∈Γ|(A, [B,C])|2 =

∑B,C∈Γ

|(A, adBC)|2 =∑

B,C∈Γ|(adBA,C)|2

=∑B∈Γ|adBA|2 =

∑B∈Γ|[A,B]|2

and ∑B∈Γ|[A,B]|2 =

∑B∈Γ

(adAB, adAB) = −∑B∈Γ

(ad2AB,B) = − tr

(ad2A

)so that Eq. (10.24) becomes

(RicA,A) = −1

4tr(ad2A

)+

1

2tr(ad2A

)− 1

2tr(ad2A

)= −1

4tr(ad2A

)which agrees with a direct computation using Eq. (10.23).


11

Riemannian curvatures on Embedded Submanifolds

11.1 Induced Riemannian metrics

Notation 11.1 In this subsection, we assume that Md is an imbedded subman-ifold of an inner product space (E = RN , 〈·, ·〉), and that M is equipped with theinherited Riemannian metric. Also let P (m) denote orthogonal projection of Eonto τmM for all m ∈M and Q(m) := I −P (m) be orthogonal projection onto(τmM)⊥.

The following elementary lemma will be used throughout the sequel.

Lemma 11.2. The differentials of the orthogonal projection operators, P andQ, satisfy

0 = dP + dQ,

PdQ = −dPQ = dQQ and

QdP = −dQP = dPP.

In particular,QdPQ = QdQQ = PdPP = PdQP = 0.

Proof. The first equality comes from differentiating the identity, I = P+Q,the second from differentiating 0 = PQ and the third from differentiating 0 =QP.

Definition 11.3 (Levi-Civita Covariant Derivative). Let V (s) =(σ(s), v(s)) = v(s)σ(s) be a smooth path in TM (see Figure 11.1), then thecovariant derivative, ∇V (s)/ds, is the vector field along σ defined by

∇V (s)

ds:= (σ(s), P (σ(s))

d

dsv(s)). (11.1)

Proposition 11.4 (Properties of ∇/ds). Let W (s) = (σ(s), w(s)) andV (s) = (σ(s), v(s)) be two smooth vector fields along a path σ in M. Then:

1. ∇W (s)/ds may be computed as:

∇W (s)

ds:= (σ(s),

d

dsw(s) + (dQ(σ′(s)))w(s)). (11.2)

Fig. 11.1. The Levi-Civita covariant derivative.

2. ∇ is metric compatible, i.e.

d

ds〈W (s), V (s)〉 = 〈∇W (s)

ds, V (s)〉+ 〈W (s),

∇V (s)

ds〉. (11.3)

Now suppose that (s, t) → σ(s, t) is a smooth function into M, W (s, t) =(σ(s, t), w(s, t)) is a smooth function into TM, σ′(s, t) := (σ(s, t), ddsσ(s, t))

and σ(s, t) = (σ(s, t), ddtσ(s, t)). (Notice by assumption that w(s, t) ∈Tσ(s,t)M for all (s, t).)

3. ∇ has zero torsion, i.e.∇σ′

dt=∇σds

. (11.4)

4. If R is the curvature tensor of ∇ defined by

R(um, vm)wm = (m, [dQ(um), dQ(vm)]w), (11.5)

then [∇dt,∇ds

]W := (

∇dt

∇ds− ∇ds

∇dt

)W = R(σ, σ′)W. (11.6)

Proof. Differentiate the identity, P (σ(s))w(s) = w(s), relative to s implies

96 11 Riemannian curvatures on Embedded Submanifolds

(dP (σ′(s)))w(s) + P (σ(s))d

dsw(s) =

d

dsw(s)

from which Eq. (11.2) follows.For Eq. (11.3) just compute:

d

ds〈W (s), V (s)〉 =

d

ds〈w(s), v(s)〉

=

⟨d

dsw(s), v(s)

⟩+

⟨w(s),

d

dsv(s)

⟩=

⟨d

dsw(s), P (σ(s))v(s)

⟩+

⟨P (σ(s))w(s),

d

dsv(s)

⟩=

⟨P (σ(s))

d

dsw(s), v(s)

⟩+

⟨w(s), P (σ(s))

d

dsv(s)

⟩=

⟨∇W (s)

ds, V (s)

⟩+

⟨W (s),

∇V (s)

ds

⟩,

where the third equality relies on v(s) and w(s) being in τσ(s)M and the fourthequality relies on P (σ(s)) being an orthogonal projection.

From the definitions of σ′, σ, ∇/dt, ∇/ds and the fact that mixed partialderivatives commute,

∇σ′(s, t)dt

=∇dt

(σ(t, s), σ′(s, t)) = (σ(t, s), P (σ(s, t))d

dt

d

dsσ(t, s))

= (σ(t, s), P (σ(s, t))d

ds

d

dtσ(t, s)) = ∇σ(s, t)/ds,

which proves Eq. (11.4).Using

∇dt

=∂

∂t+ dQ (σ) and

∇ds

=∂

∂s+ dQ (σ′)

we find,

R∇ (σ, σ′) =

[∇dt,∇ds

]=

[∂

∂t+ dQ (σ) ,

∂

∂s+ dQ (σ′)

]=

∂

∂t[dQ (σ′)]− ∂

∂s[dQ (σ)] + [dQ (σ) dQ (σ′)]

=(d2Q

)(σ, σ′) + [dQ (σ) dQ (σ′)] = [dQ (σ) dQ (σ′)] .

Lemma 11.5. A curve σ (t) ∈M is a geodesic in this setting iff

0 = P (σ (t)) σ (t) = σ (t) + dQ (σ (t)) σ (t) =∇σ (t)

dt. (11.7)

Example 11.6. Let M = m ∈ RN : |m| = ρ be the sphere of radius ρ. In thiscase Q(m) = 1

ρ2mmtr for all m ∈M. Therefore

dQ(vm) =1

ρ2vmtr +mvtr ∀ vm ∈ TmM

and hence

dQ(um)dQ(vm) =1

ρ4umtr +mutrvmtr +mvtr

=1

ρ4ρ2uvtr + 〈u, v〉Q(m).

So the curvature tensor is given by

R(um, vm)wm = (m,1

ρ2uvtr − vutrw) = (m,

1

ρ2〈v, w〉u− 〈u,w〉v).

Moreover, according to Eq. (11.7) σ (t) ∈M is a geodesic iff

0 = σ (t) + dQ (σ (t)) σ (t) = σ (t) +1

ρ2σ (t)σ (t)

tr+ σ (t) σ (t)

trσ (t)

= σ (t) +1

ρ2|σ (t)|2 σ (t) = σ (t) +

1

ρ2|vm|2 σ (t)

where σ (0) = vm ∈ TmM. The solutions to this equation is given by

σv (t) = cos

(|vm|ρt

)m+

sin(|vm|ρ t)

|vm|ρ

v (11.8)

= cos

(|vm|ρt

)m+ ρ sin

(|vm|ρt

)v (11.9)

where v := v/ |v| and in particular we have

exp (vm) = cos

(|vm|ρ

)m+ sin

(|vm|ρ

)ρv. (11.10)

Proposition 11.7. Let S be the unit sphere in Rd, ρ > 0, and let m ∈ ρS befixed point. For all v ∈ TmM and w ∈ B (0, πρ) , define

ds2 (vw) := (w · v)2

+ sinc2

(|w|ρ

)|Qwv|2 (11.11)

where Qw is orthogonal projection onto w⊥ and

sinc (x) :=sin (x)

x.

Then ds2 is a Riemannian metric on B (0, πρ) ⊂ TmM such that expm :B (0, πρ)→ ρS \ −m is an isometric diffeomorphism.


11.1 Induced Riemannian metrics 97

Proof. Let h (w) := expm (w) for all v, w ∈ TmM,

h∗vw =d

ds|0h (w + sv)

=d

ds|0[cos

(|w + sv|

ρ

)m+ sin

(|w + sv|

ρ

)ρw + sv

|w + sv|

]=

[− sin

(|w|ρ

)m

ρ+ cos

(|w|ρ

)w

]w · v + sin

(|w|ρ

)ρ

[v |w| − (w · v)w

|w|2

]

=

[− sin

(|w|ρ

)m

ρ+ cos

(|w|ρ

)w

]w · v + sin

(|w|ρ

)ρ

|w|[v − (w · v) w]

= − (w · v) sin

(|w|ρ

)m

ρ+ cos

(|w|ρ

)(w · v) w + sin

(|w|ρ

)ρ

|w|Qwv.

From this we learn that

(h∗g) (vw, vw) =h∗vw · h∗vw

=

∣∣∣∣(w · v) sin

(|w|ρ

)m

ρ

∣∣∣∣2 +

∣∣∣∣cos

(|w|ρ

)(w · v) w

∣∣∣∣2+

∣∣∣∣sin( |w|ρ)

ρ

|w|Qwv

∣∣∣∣2= (w · v)

2sin2

(|w|ρ

)+ (w · v)

2cos2

(|w|ρ

)+ sin2

(|w|ρ

)ρ2

|w|2|Qwv|2

= (w · v)2

+ sin2

(|w|ρ

)ρ2

|w|2|Qwv|2

= (w · v)2

+ sinc2

(|w|ρ

)|Qwv|2 .

Exercise 11.1. Show the curvature tensor of the cylinder

M = (x, y, z) ∈ R3 : x2 + y2 = 1

is zero.

If Y (m) = (m, y(m)), then

∇vmY = (m,P (m)dy(vm)) = (m, dy(vm) + dQ(vm)y(m)),

from which it follows ∇vmY is well defined, i.e. ∇vmY is independent of thechoice of σ such that σ′ (0) = vm. The following proposition relates curvatureand torsion to the covariant derivative ∇ on vector fields.

Proposition 11.8. Let m ∈ M, v ∈ TmM, X, Y, Z ∈ Γ (TM), and f ∈C∞(M), then the following relations hold.

1. Product Rule ∇v(f ·X) = df(v) ·X(m) + f(m) · ∇vX.2. Zero Torsion ∇XY −∇YX − [X,Y ] = 0. Equivalently put, if Σ (s, t) ∈M is

a smooth parametrized surface, that

∇dt

d

dsΣ (s, t) =

∇ds

d

dtΣ (s, t) . (11.12)

3. Zero Torsion For all vm, wm ∈ TmM, dQ(vm)wm = dQ(wm)vm.4. Curvature Tensor R(X,Y )Z = [∇X ,∇Y ]Z −∇[X,Y ]Z, where

[∇X ,∇Y ]Z := ∇X(∇Y Z)−∇Y (∇XZ).

Moreover if u, v, w, z ∈ TmM, then R has the following symmetries

a R(um, vm) = −R(vm, um)b [R(um, vm)]

tr= −R(um, vm) and

c if zm ∈ τmM, then

〈R(um, vm)wm, zm〉 = 〈R(wm, zm)um, vm〉. (11.13)

5. Ricci Curvature Tensor For each m ∈ M, let Ricm : TmM → TmM be de-fined by

Ricm vm :=∑a∈S

R(vm, a)a, (11.14)

where S ⊂ TmM is an orthonormal basis. Then Rictrm = Ricm and Ricm

may be computed as

〈Ricmu, v〉 = tr(dQ(dQ(u)v)− dQ(v)dQ(u)) for all u, v ∈ TmM. (11.15)

Proof. The product rule is easily checked and may be left to the reader.For the second and third items, write X(m) = (m,x(m)), Y (m) = (m, y(m)),and Z(m) = (m, z(m)) where x, y, z : M → RN are smooth functions such thatx(m), y(m), and z(m) are in τmM for all m ∈M. Then using Eq. (??), we have

(∇XY −∇YX)(m) = (m,P (m)(dy(X(m))− dx(Y (m))))

= (m, (dy(X(m))− dx(Y (m)))) = [X,Y ](m), (11.16)

which proves the second item. The proof of the equivalent zero torsion condition,Eq. (11.12) is similar;

∇dt

d

dsΣ (s, t)− ∇

ds

d

dtΣ (s, t) = P (Σ (s, t))

[d

dt

d

dsΣ (s, t)− d

ds

d

dtΣ (s, t)

]= 0.



Since (∇XY )(m) may also be written as

(∇XY )(m) = (m, dy(X(m)) + dQ(X(m))y(m)),

Eq. (11.16) may be expressed as dQ(X(m))y(m) = dQ(Y (m))x(m) which im-plies the third item.

Similarly for fourth item:

∇X∇Y Z = ∇X(·, Y z + (Y Q)z)

= (·, XY z + (XYQ)z + (Y Q)Xz + (XQ)(Y z + (Y Q)z)),

where Y Q := dQ(Y ) and Y z := dz(Y ). Interchanging X and Y in this lastexpression and then subtracting gives:

[∇X ,∇Y ]Z = (·, [X,Y ]z + ([X,Y ]Q)z + [XQ,Y Q]z)

= ∇[X,Y ]Z +R(X,Y )Z.

The anti-symmetry properties in items 4a) and 4b) follow easily from Eq. (11.5).For example for 4b), dQ (um) and dQ(vm) are symmetric operators and hence

[R(um, vm)]tr

= [dQ(um), dQ(vm)]tr = [dQ(vm)tr, dQ(um)tr]

= [dQ(vm), dQ(um)] = −[dQ(um), dQ(vm)] = −R(um, vm).

To prove Eq. (11.13) we make use of the zero - torsion condition dQ(vm)wm =dQ(wm)vm and the fact that dQ (um) is symmetric to learn

〈R(um, vm)w, z〉 = 〈[dQ(um), dQ(vm)]w, z〉= 〈[dQ(um)dQ(vm)− dQ(vm)dQ(um)]w, z〉= 〈dQ(vm)w, dQ(um)z〉 − 〈dQ(um)w, dQ(vm)z〉= 〈dQ(w)v, dQ(z)u〉 − 〈dQ(w)u, dQ(z)v〉 (11.17)

= 〈[dQ(z), dQ(w)] v, u〉 = 〈R (z, w) v, u〉 = 〈R (w, z)u, v〉

where we have used the anti-symmetry properties in 4a. and 4b. By Eq. (11.17)with v = w = a,

〈Ricu, z〉 =∑a∈S〈R(u, a)a, z〉

=∑a∈S

[〈dQ(a)a, dQ(u)z〉 − 〈dQ(u)a, dQ(a)z〉]

=∑a∈S

[〈a, dQ(a)dQ(u)z〉 − 〈dQ(u)a, dQ(z)a〉]

=∑a∈S

[〈a, dQ(dQ(u)z)a〉 − 〈dQ(z)dQ(u)a, a〉]

= tr(dQ(dQ(u)z)− dQ(z)dQ(u))

which proves Eq. (11.15). The assertion that Ricm : TmM → TmM is a sym-metric operator follows easily from this formula and item 3.

Notation 11.9 To each v ∈ RN , let ∂v denote the vector field on RN definedby

∂v(at x) = vx =d

dt|0(x+ tv).

So if F ∈ C∞(RN ), then

(∂vF )(x) :=d

dt|0F (x+ tv) = F ′ (x) v

and(∂v∂wF ) (x) = F ′′ (x) (v, w) ,

see Notation ??.

Notice that if w : RN → RN is a function and v ∈ RN , then

(∂v∂wF ) (x) = ∂v [F ′ (·)w (·)] (x) = F ′ (x) ∂vw (x) + F ′′ (x) (v, w (x)) .

The following variant of item 4. of Proposition 11.8 will be useful in provingthe key Bochner-Weitenbock identity.

Proposition 11.10. Suppose that Z ∈ Γ (TM) , v, w ∈ TmM and let X,Y ∈Γ (TM) such that X (m) = v and Y (m) = w. Then

1. ∇2v⊗wZ defined by

∇2v⊗wZ := (∇X∇Y Z −∇∇XY Z) (m) (11.18)

is well defined, independent of the possible choices for X and Y.2. If Z(m) = (m, z(m)) with z : RN → RN a smooth function such z (m) ∈τmM for all m ∈M, then

∇2v⊗wZ = dQ (v) dQ (w) z (m)+P (m) z′′ (m) (v, w)−P (m) z′ (m) [dQ (v)w] .

(11.19)3. The curvature tensor R (v, w) may be computed as

∇2v⊗wZ −∇2

w⊗vZ = R (v, w)Z (m) . (11.20)

4. If V is a smooth vector field along a path σ (s) in M, then the followingproduct rule holds,

∇ds

(∇V (s)Z

)=(∇ ∇

dsV (s)Z)

+∇2σ′(s)⊗V (s)Z. (11.21)


11.1 Induced Riemannian metrics 99

Proof. We will prove items 1. and 2. by showing the right sides of Eq.(11.18) and Eq. (11.19) are equal. To do this write X(m) = (m,x(m)),Y (m) = (m, y(m)), and Z(m) = (m, z(m)) where x, y, z : RN → RN aresmooth functions such that x(m), y(m), and z(m) are in τmM for all m ∈ M.Then, suppressing m from the notation,

∇X∇Y Z −∇∇XY Z = P∂x [P∂yz]− P∂P∂xyz= P (∂xP ) ∂yz + P∂x∂yz − P∂P∂xyz= P (∂xP ) ∂yz + Pz′′ (x, y) + Pz′ [∂xy − P∂xy]

= (∂xP )Q∂yz + Pz′′ (x, y) + Pz′ [Q∂xy] .

Differentiating the identity, Qy = 0 on M shows Q∂xy = − (∂xQ) y whichcombined with the previous equation gives

∇X∇Y Z −∇∇XY Z = (∂xP )Q∂yz + Pz′′ (x, y)− Pz′ [(∂xQ)Y ] (11.22)

= − (∂xP ) (∂yQ) z + Pz′′ (X,Y )− Pz′ [(∂xQ)Y ] .

Evaluating this expression at m proves the right side of Eq. (11.19).Equation (11.20) now follows from Eqs. (11.19) and (11.5), item 3. of Propo-

sition 11.8 and the fact the z′′ (v, w) = z′′ (w, v) because mixed partial deriva-tives commute.

We give two proofs of Eq. (11.21). For the first proof, choose local vector

fields Eidi=1 defined in a neighborhood of σ (s) such that Ei (σ (s))di=1 is a

basis for Tσ(s)M for each s. We may then write V (s) =∑di=1 Vi (s)Ei (σ (s))

and therefore,

∇dsV (s) =

d∑i=1

V ′i (s)Ei (σ (s)) + Vi (s)∇σ′(s)Ei

(11.23)

and

∇ds

(∇V (s)Z

)=∇ds

(d∑i=1

Vi (s) (∇EiZ) (σ (s))

)

=

d∑i=1

V ′i (s) (∇EiZ) (σ (s)) +

d∑i=1

Vi (s)∇σ′(s) (∇EiZ) .

Using Eq. (11.18),

∇σ′(s) (∇EiZ) = ∇2σ′(s)⊗Ei(σ(s))Z +

(∇∇σ′(s)EiZ

)and using this in the previous equation along with Eq. (11.23) shows

∇ds

(∇V (s)Z

)= ∇∑d

i=1V ′i (s)Ei(σ(s))+Vi(s)∇σ′(s)EiZ +

d∑i=1

Vi (s)∇2σ′(s)⊗Ei(σ(s))Z

=(∇ ∇

dsV (s)Z)

+∇2σ′(s)⊗V (s)Z.

For the second proof, write V (s) = (σ (s) , v (s)) = v (s)σ(s) and p (s) :=

P (σ (s)) , then

∇ds

(∇V Z)−(∇ ∇

dsVZ)

= pd

ds(pz′ (v))− pz′ (pv′)

= p [p′z′ (v) + pz′′ (σ′, v) + pz′ (v′)]− pz′ (pv′)= pp′z′ (v) + pz′′ (σ′, v) + pz′ (qv′)

= p′qz′ (v) + pz′′ (σ′, v)− pz′ (q′v)

= ∇2σ′(s)⊗V (s)Z

wherein the last equation we have made use of Eq. (11.22).

Exercise 11.2 (Stereographic Projection). Let X = Rn, X∗ := X ∪ ∞be the one point compactification of X, Sn := y ∈ Rn+1 : |y| = 1 be theunit sphere in Rn+1 and N = (0, . . . , 0, 1) ∈ Rn+1. Define f : Sn → X∗ byf(N) = ∞, and for y ∈ Sn \ N let f(y) = b ∈ Rn be the unique point suchthat (b, 0) is on the line containing N and y, see Figure 11.2 below. Find aformula for f and show f : Sn → X∗ is a homeomorphism. (So the one pointcompactification of Rn is homeomorphic to the n sphere.)

Fig. 11.2. Sterographic projection and the one point compactification of Rn.



11.2 Hyperbolic Space

let M = R1+d, with coordinates (t, x), where for a = (a0, a1, · · · ad); t(a) = a0

and xi(a) = ai for i = 1, . . . , d and for a, b ∈ R1+d let,

q〈a, b〉 =

d∑n=1

aibi − a0b0 = a · b− a0b0.

Fix a number ρ > 0, and define

H = Hρ := a ∈M |t2(a) = |x|2(a) + ρ2 and t(a) = a0 > 0= a ∈M |q〈a, a〉 = −ρ2 and a0 > 0.

Notice that H is an embedded hypersurface in M and that

TmH = vm : q (m, vm) = 0 = vm : v ∈ Nul(q (m, ·) .

Fig. 11.3. A plot of H2 when d = 2.

Define a Riemannian metric g on TH such that

ds2 = |dx|2 − dt2.

Notice that on TH, dt = xt · dx so that

ds2 = |dx|2 − (x

t· dx)2

≥ |dx2| − |xt|2|dx|2 = |dx2| − (1− ρ2

t2)|dx|2 =

ρ2

t2|dx|2 > 0.

Hence g is in fact a Riemannian metric on H. Let X ∈ Γ (TM) which weidentify with a function X : H → Rd+1 with the property that q〈a,X(a)〉 = 0for all a ∈ H.

Lemma 11.11. The Levi-Civita covariant derivative on (H, g) is given by

∇vX =

[dX〈v〉 − 1

ρ2q〈v,X(a)〉a

]a

for all v ∈ TaH. (11.24)

Alternatively if X (t) ∈ Tσ(t)H, then

∇dtX (t) = X (t)− 1

ρ2q〈σ (t) , X (t)〉σ (t) . (11.25)

Proof. For a ∈ H, let Q (a)w := − 1ρ2 q〈a,w〉a, and observe that

Q (a)2w = − 1

ρ2q〈a,w〉Q (a) a =

1

ρ2q〈a,w〉 1

ρ2q〈a, a〉a

= − 1

ρ2q〈a,w〉a = Q (a)w

and

q (Q (a)w, v) = − 1

ρ2q (a,w) q (a, v) = q (w,Q (a) v) .

Hence Q (a) and P (a) : M →M are projection operators with

P (a)w = w +1

ρ2q〈a,w〉a.

Since,

q (P (a)w, a) = q

(w +

1

ρ2q〈a,w〉a, a

)= q (w, a) +

q (a, a)

ρ2q〈a,w〉

= q (w, a)− q〈a,w〉 = 0,

it follows that P (a) is a projection operator onto τaH for all a ∈ H. Workingas in Proposition 11.4, one shows ∇vX = P (a) dX (v) and hence that ∇ haszero Torsion by Example 10.7. Moreover, if X (t) ∈ Tσ(t)H,

d

dtq (X (t) , X (t)) = 2q

(X (t) , X (t)

)= 2q

(X (t) , P (σ (t))X (t)

)= 2q

(P (σ (t)) X (t) , X (t)

)= 2q

(∇dtX (t) , X (t)

),

it follows that g = q|TH×TH is covariantly constant, i.e. that ∇ is metric com-patible. Hence, ∇ defined in Eq. (11.24) is the Levi-Civita covariant derivative.


11.2 Hyperbolic Space 101

Proposition 11.12. Continuing the notation above, for v = ba ∈ TaH we have

exp (v) = a cosh

(|v|ρ

)+ ρv sinh

(|v|ρ

). (11.26)

Moreover exp : TaH→ H is surjective with

exp−1a (c) = b =

c− a cosh(|v|ρ

)sinh

(|v|ρ

) with |v| := ρ cosh−1

(−q (a, c)

ρ2

)

and

d (a, c) = ρ cosh−1

(−q (a, c)

ρ2

).

Proof. Suppose that v = ba ∈ TaH and σ (t) := exp (tv) , then σ mustsatisfy

σ (t)− 1

ρ2q〈σ (t) , σ (t)〉σ (t) = 0 (11.27)

and as we know,

d

dtq〈σ (t) , σ (t)〉 = 2q〈∇

dtσ (t) , σ (t)〉 = 0

so that q〈σ (t) , σ (t)〉 = q (b, b) . Using this result in Eq. (11.27) we learn thatwe must solve

σ (t)− q (b, b)

ρ2σ (t) = 0 with σ (0) = a and σ (0) = b.

The solution to this equation is

exp (tv) = σ (t) = a cosh

(|v|ρt

)+

ρ

|v|b sinh

(|v|ρt

)= a cosh

(|v|ρt

)+ ρb sinh

(|v|ρt

)where

|v| = |ba| :=√q (b, b) =

√g (v, v).

Suppose that a, c ∈ H, and consider solving the equation,

c = exp (ba) = a cosh

(|v|ρ

)+ ρb sinh

(|v|ρ

), (11.28)

for b. By applying q (a, ·) to both sides of this equation, we learn that

q (a, c) = cosh

(|v|ρ

)from which it follows that

|b| = |v| = ρ cosh−1

(−q (a, c)

ρ2

). (11.29)

It will be shown below that − q(a,c)ρ2 ≥ 1 for all a, c ∈ H with equality iff a = c.

It now follows from Eq. (11.29) that

b =c− a cosh

(|v|ρ

)sinh

(|v|ρ

) with |v| := ρ cosh−1

(−q (a, c)

ρ2

).

It now follows that (H, g) is complete and that

d (a, c) = ρ cosh−1

(−q (a, c)

ρ2

).

It should be observed here that − q(a,c)ρ2 ≥ 1 for all a, c ∈ H with equality iffa = c. To see this notice that for any a, c ∈ H that the slope of the ac is alwaysstrictly between −1 and 1, i.e. |a0 − c0| < |a− c| for all a, c ∈ H, see Figure11.4. From this it follows that

Fig. 11.4. Here is a plot of H along with the “light cone.” Hopefully it is clear to thereader from this figure that any line joining two points of H must have slope strictlybetween −1 and +1.

0 < q (a− c, a− c) = q (a, a) + q (c, c)− 2q (a, c) = −2ρ2 − 2q (a, c) .



This certainly implies that − q(a,c)ρ2 ≥ 1, so that Eq. (11.29) always has a uniquepositive solution.

An algebraic proof is as follows. Let u = |a| and v = |c| , then ρ2 ≤ −q (a, c)is true iff

−q (a, c) = a0c0 − a · c ≥ a0c0 − |a| |c| =√

(ρ2 + u2) (ρ2 + v2)− uv

so that

−q (a, c)

ρ2≥√(

1 + (u/ρ)2)(

1 + (v/ρ)2)− u

ρ

v

ρ

so to finish the argument it suffices to show√(1 + (u/ρ)

2)(

1 + (v/ρ)2)− u

ρ

v

ρ≥ 1

or equivalently that(1 + (u/ρ)

2)(

1 + (v/ρ)2)≥(

1 +u

ρ

v

ρ

)2

= 1 + 2u

ρ

v

ρ+

(u

ρ

v

ρ

)2

which is equivalent to showing

(u/ρ)2

+ (v/ρ)2 ≥ 2

u

ρ

v

ρ

which is clearly true.

Proposition 11.13. The curvature tensor for (H, gρ) is given by

R (v, w) ξ = − 1

ρ2q〈w, ξ〉v +

1

ρ2q〈v, ξ〉w = − 1

ρ2(v ∧ w) ξ

for all v, w ∈ TaH. If v, w ⊂ TaH is an orthonormal set, we have

(R (v, w)w, v) = − 1

ρ2

so that (H, gρ) has constant sectional curvature of −1/ρ2.

Proof. We haveR (v, w) = [dQ (v) , dQ (w)]

where

Q (a)w := − 1

ρ2q〈a,w〉a.

Now

dQ (v)w = − 1

ρ2q〈v, w〉a− 1

ρ2q〈a,w〉v

so that

dQ (v) = − 1

ρ2q〈v, ·〉a− 1

ρ2q〈a, ·〉v

and hence for v, w, ξ ∈ TaH we have

dQ (v) dQ (w) ξ =

(− 1

ρ2q〈v, ·〉a− 1

ρ2q〈a, ·〉v

)(− 1

ρ2q〈w, ξ〉a− 1

ρ2q〈a, ξ〉w

)=

(− 1

ρ2q〈v, ·〉a− 1

ρ2q〈a, ·〉v

)(− 1

ρ2q〈w, ξ〉a

)=

1

ρ2

1

ρ2q〈w, ξ〉q〈a, a〉v = − 1

ρ2q〈w, ξ〉v

and hence it follows that

R (v, w) ξ = − 1

ρ2q〈w, ξ〉v +

1

ρ2q〈v, ξ〉w = − 1

ρ2(v ∧ w) ξ.

Thus the sectional curvature associated to the orthonormal set v, w ⊂ TaHis given by

(R (v, w)w, v) = − 1

ρ2

so that (H, g) has constant sectional curvature of −ρ−2.

Proposition 11.14. Let (H, gρ) be Hyperbolic space as above and m ∈ H be afixed point. For all v, w ∈ TmM define

ds2 (vw) := (w · v)2

+ρ2

|w|2sinh2

(|w|ρ

)|Qwv|2

where for all w ∈ TmM, |w| =√q (w,w), w = w/ |w| , and Qw is orthogonal

projection onto w⊥. Then ds2 is a Riemannian metric on TmM such that expm :TmM → H is an isometric diffeomorphism.

Proof. Let h (w) := expm (w) and Then for all v, w ∈ TmM,


11.2 Hyperbolic Space 103

h∗vw =d

ds|0h (w + sv)

=d

ds|0[cosh

(|w + sv|

ρ

)m+ sinh

(|w + sv|

ρ

)ρw + sv

|w + sv|

]=

[sinh

(|w|ρ

)m

ρ+ cosh

(|w|ρ

)w

]w · v + sinh

(|w|ρ

)ρ

[v |w| − (w · v)w

|w|2

]

=

[sinh

(|w|ρ

)m

ρ+ cosh

(|w|ρ

)w

]w · v + sinh

(|w|ρ

)ρ

|w|[v − (w · v) w]

= (w · v)

[sinh

(|w|ρ

)m

ρ+ cosh

(|w|ρ

)w

]+ sinh

(|w|ρ

)ρ

|w|Qwv.

From this we learn that

(h∗g) (vw, vw) =h∗vw · h∗vw = q (h∗vw, h∗vw)

= (w · v)2

[sinh2

(|w|ρ

)q

(m

ρ,m

ρ

)+

∣∣∣∣cosh

(|w|ρ

)(w · v) w

∣∣∣∣2]

+

∣∣∣∣sinh

(|w|ρ

)ρ

|w|Qwv

∣∣∣∣2= (w · v)

2

[− sinh2

(|w|ρ

)+ cosh2

(|w|ρ

)]+ sinh2

(|w|ρ

)ρ2

|w|2|Qwv|2

= (w · v)2

+ sinh2

(|w|ρ

)ρ2

|w|2|Qwv|2 .

We are now going to show that Iso (M) – the set of isometries of H isthe L+, the connected component of the Lohrenz group which preserves the Hconsisting of those Λ ∈ L such that Λ0,0 > 0. First off it is easily seen, sinceL+ preserves q, then L+ acts isometrically on H. Now suppose f : M → M isan isometric diffeomorphism and choose g ∈ L+ such that f (e0) = ge0, whichis possible to do since L+ acts transitively on H with stab (e0) = 1 × O (d) .Then by replacing f by g−1f, we may assume that f (e0) = e0. In this case weknow f∗e0 acts isometrically on Rd × 0 = Te0H. Let

h =

(1 00 f∗e0

)∈ L+,

then F := h−1g−1f ∈ Iso (M) such that F (e0) = F (e0) and F∗e0 = Id. Hence,as we have already seen, this implies that F = id, so that f = hg ∈ L+.

Fig. 11.5. Stereographic projection for the pseudosphere. Let a ∈ H be the pointwhere the two curves intersect and x be the point in Rn where the sloped line intersectsthe x – plane, i.e. where xn+1 = 0.

The Killing vector fields on H is now isomorphic to Lie (L+) , via A ∈Lie (L+)→ A∗ ∈ ΓK (TM) where

A∗ (p) :=d

dt|0e−tAp.

Let us now consider the ball model for hyperbolic space. For this, we needto define the analogue of stereographic projection in this context, see Figure11.5. We then have, for some t and x ∈ Rd,

a = −e0 + t (x+ e0) .

Now t must satisfy a0 = −1 + t and hence t = 1 + a0. From this it follows that

a = −e0 + (1 + a0) (x+ e0)

and hence that

x = (1 + a0)−1

(a+ e0)− e0

=a+ e0 − (1 + a0) e0

1 + a0=a− a0e0

1 + a0

=a

1 + a0=: f (a) . (11.30)



Theorem 11.15. Define a Riemannian metric on B (0, 1) ⊂ Rd by

ds2 :=4(

1− |x|2)2 |dx|

2.

Then f : (H, gρ=1)→(B (0, 1) , ds2

)is an isometric diffeomorphism.

Proof. We now compute f∗g using

f∗ (va) = (1 + a0)−2

[(1 + a0) v − v0a]

and hence

f∗ (va) · f∗ (va) = (1 + a0)−4 |(1 + a0) v − v0a|2

= (1 + a0)−4[(1 + a0)

2 |v|2 − 2 (1 + a0) v0a · v + v20 |a|

2].

We now make use of the facts that

0 = q (a, v) = a · v − a0v0 and

−1 = q (a, a) = |a|2 − a20

in the equation above to learn

(1 + a0)2 |v|2 − 2 (1 + a0) v0a · v + v2

0 |a|2

= (1 + a0)2 |v|2 − 2 (1 + a0) a0v

20 + v2

0

(a2

0 − 1)

= (1 + a0)2 |v|2 + v2

0 [(a0 − 1) (a0 + 1)− 2 (1 + a0) a0]

= (1 + a0)2 |v|2 + v2

0 (1 + a0) [(a0 − 1)− 2a0]

= (1 + a0)2(|v|2 − v2

0

)= (1 + a0)

2q (v, v)

so thatf∗ (va) · f∗ (va) = (1 + a0)

−2q (v, v)

or equivalently we have

(1 + a0)2f∗ (va) · f∗ (va) = q (v, v) . (11.31)

We now wish to find a0 in terms of x. From Eq. (11.30) we have

(1 + a0)2 |x|2 = |a|2 = a2

0 − 1 = (a0 − 1) (a0 + 1)

and hence we have (1 + a0) |x|2 = (a0 − 1) or equivalently that a0

(1− |x|2

)=

1 + |x|2 . Thus we have

(1 + a0)2

=

(1 +

1 + |x|2

1− |x|2

)2

=

(2

1− |x|2

)2

=4(

1− |x|2)2 .

Combining this equation with Eq. (11.31) we find,

ds2 (f∗ (va)) =4(

1− |x|2)2 f∗ (va) · f∗ (va) = q (v, v) = g (v, v)

which completes the proof of the theorem.

11.3 Inversions and the upper half space model ofHyperbolic space

Let M := Rd \ 0 , r > 0, and let ir : M → M be the diffeomorphims of Mdefined by

ir (x) =x

|x|2r2 = r2 x

|x|.

Notice that ir (x) is the unique vector on the ray determined by 0 and x suchthat |ir (x)| |x| = r2.

Theorem 11.16. The ir maps spheres in M to spheres in M with the conven-tion that a sphere with infinite radius is a hyperplane in M. More precisely letS = bd(B (0, 1)) be the unit sphere in Rd, a ∈M, and t > 0. Then

ir (a+ tS) =r2

|a|2 − t2a+

r2t∣∣∣|a|2 − t2∣∣∣S if t 6= |a| (11.32)

and

ir (a+ |a|S) =r2

2 |a|2a+ a⊥. (11.33)

Proof. Let S = bd(B (0, 1)) be the unit sphere in Rd. We wish to showthat T := ir (a+ tu) : u ∈ S is a sphere for all a ∈M and t > 0. Figure 11.6suggests that the center of T should be λa for some λ ∈ R. So we should tryto find λ such that |ir (a+ tu)− λa|2 = ρ2 is constant. To this end let α := |a|and s := u · a. Then


11.3 Inversions and the upper half space model of Hyperbolic space 105

Fig. 11.6. Some inversions about the unit circle. Notice that circles go to circleswhere a line is considered to be a circle through infinitity.

|ir (a+ tu)− λa|2 =

∣∣∣∣∣ a+ tu

|a+ tu|2r2 − λa

∣∣∣∣∣2

=r4

|a+ tu|2+ λ2 − 2λr2

|a+ tu|2(α+ st)

= λ2 +r4 − 2λr2 (α+ st)

|a+ tu|2.

In order for this expression to be independent of u, we need to require that

r4 − 2λr2 (α+ st) = C |a+ tu|2 = C(α2 + 2sαt+ t2

)for all s ∈ [−, 1, 1] .

Comparing coefficients of s and s0 we must require

−2λr2t = C2αt and r4 − 2λr2α = C(α2 + t2

).

The first of these equations implies that C = −λr2/α and putting this into thesecond equation implies

r4 − 2λr2α = −λr2

α

(α2 + t2

)= −λr2α− λr2t2

α

or equivalently that λr2(α− t2

α

)= r4 and hence we have

λ =r2α

α2 − t2so that λa =

r2

α2 − t2a =

r2

|a|2 − t2a.

With this choice of λ we have∣∣∣∣ir (a+ tu)− r2α

α2 − t2a

∣∣∣∣2 = λ2 + C

=

(r2α

α2 − t2

)2

− r2

α

r2α

α2 − t2

=r4

α2 − t2

[α2

α2 − t2− 1

]=

r4

α2 − t2

[α2

α2 − t2− 1

]=

r4t2

(α2 − t2)2 .

Hence we have proved Eq. (11.32).We now want to consider what happens to ir (a+ tS) as t ↑ |a| . Notice that

r2

|a|2 − t2a− r2t

|a|2 − t2a ∈ ir (a+ tS)

is the closest point to the origin in ir (a+ tS) and that

r2

|a|2 − t2a− r2t

|a|2 − t2a = r2 (|a| − t)

|a|2 − t2a =

r2

|a|+ ta ∈ ir (a+ tS) .

Letting t ↑ |a| in this expression shows that

r2

2 |a|a =

r2

2 |a|2a ∈ ir (a+ |a|S)

and therefore, this being the closes point to the origin in ir (a+ |a|S) we have

ir (a+ |a|S) =r2

2 |a|2a+ a⊥.

See http://www.xahlee.org/SpecialPlaneCurves dir/Inversion dir/inversion.htmlfor more properties and facts about inversions.

Let i = i1 and consider the function,

F (x) = −ed + 2x+ ed

|x+ ed|2= T−ed M2 i Ted ,

which is inversion about bd(B(−ed,

√2)) followed by translation by −ed. We

have from above that F (bd(B (0, 1)) is a hyperplane in Rd which contains thepoint



F (ed) = −ed + 2ed + ed

|ed + ed|2= 0

as its closest point to −ed. Hence we know that F (bd(B (0, 1)) = e⊥d = Rd−1×0 . Moreover, since F (0) = ed, it follows that F : B (0, 1) → Rd+ := Rd−1 ×(0,∞) is a diffeomorphism. Since

F = T−ed M2 i Ted

where Tx is translation by x, Mλ is multiplication by λ, and i (x) = x/ |x|2 isthe standard inversion, we see that

G = F−1 = T−ed i M1/2 Ted ,

i.e.

G (z) = −ed +12 (z + ed)∣∣ 12 (z + ed)

∣∣2 = −ed + 2z + ed

|z + ed|2= F (z) .

We now pull back the hyperbolic metric

ds2 :=4(

1− |x|2)2 |dx|

2on B (0, 1) ⊂ Rd

by the diffeomorphism F = G : Rd+ → B (0, 1) . To this end we observe that

F∗ (va) = 2|a+ ed|2 v − 2v · (a+ ed) (a+ ed)

|a+ ed|4

=2

|a+ ed|2[v − 2

(v · a+ ed

)a+ ed

].

The map v → v − 2(v · a+ ed

)a+ ed is a reflection and in particular is norm

preserving, and thus we find

|F∗ (va)|2 =4

|a+ ed|4|v|2

and hence, with w := |a+ ed|2 we have

ds2 (F∗ (va)) =4(

1− |F (a)|2)2

4

|a+ ed|4|v|2

=4(

1− |F (a)|2)2

4

w2|v|2 (11.34)

Now, letting , we find

|F (a)|2 =

∣∣∣∣∣−ed + 2a+ ed

|a+ ed|2

∣∣∣∣∣2

=

∣∣∣∣−ed + 2a+ edw

∣∣∣∣2= 1− 4

ad + 1

w+

4

w= 1− 4

adw

from which we learn that(1− |F (a)|2

)2

=(

4adw

)2

= 42a2d/w

2

and feeding this back into Eq. (11.34) shows

ds2 (F∗ (va)) =|v|2

a2d

.

Putting this all together yields the following theorem.

Theorem 11.17 (Hyperbolic Space Models). Letting (H, gρ=1) be thePseudo-Sphere model for hyperbolic space than the following maps are isometricdiffeomorphisms of Riemannian spaces

(H, gρ=1)f→

B (0, 1) ,4(

1− |x|2)2 |dx|

2

F→

(Rd+,|dx|2

x2d

)

and hence gives three models of hyperbolic space. Here F is the inversion map,

F (x) = −ed + 2x+ ed

|x+ ed|2

and f is the stereographic projection map

f (a) =a

1 + a0.

See page 56 of Gallot, Hulin, and Lafontaine, where the same results are statedwith f being replaced by the “pseudo-inversion” on Minikowski space.


12

Vector Field Commutators and Logarithms

12.1 Differentials of Flows

In this section, suppose that M is a manifold, Xt ∈ Γ (TM) is a time dependentvector-field, ϕt = ϕXt is the flow associated to Xtt∈R , and ∇ is any torsionfree covariant derivative on TM. Recall that ϕt (m) is the solution to the ODE,

d

dtϕt (m) = Xt (ϕt (m)) with ϕ0 (m) = m (12.1)

which is defined on its maximal time interval.

Lemma 12.1 (Differentials of flows). Let ϕt = ϕXt be the flow associated toa time dependent vector field, Xt. Then

∇dt

[ϕt∗vm] = ∇ϕt∗vmXt for all vm ∈ TmM. (12.2)

Proof. If σ (s) be a curve in M so that σ′ (0) = vm, then

∇dt

[ϕt∗vm] =∇dt

d

ds|0ϕt (σ (s)) =

∇ds|0d

dtϕt (σ (s))

=∇ds|0Xt (ϕt (σ (s))) = ∇(ϕt)∗vm

Xt.

Theorem 12.2 (Commutator Theorem). Let ϕXt be the flow associated toX (t, ·) and Y (·) be another vector field on M. Then

d

dt

(ϕ−1t∗ Y ϕt

)= [ϕt]

−1∗ [Xt, Y ] ϕt. (12.3)

Proof. Let Yt := [ϕt]−1∗ Y ϕt so that Y ϕt = (ϕt)∗ Yt and hence,

∇dt

[(ϕt)∗ Yt] =∇dt

[Y ϕt] .

On one hand using Lemma 12.1,

∇dt

[(ϕt)∗ Yt] =

[∇dt

(ϕt)∗

]Yt + (ϕt)∗ Yt

= ∇(ϕt)∗YtXt + (ϕt)∗ Yt

= ∇Y ϕtXt + (ϕt)∗ Yt

= (∇YXt) ϕt + (ϕt)∗ Yt

while on the other

∇dt

[Y ϕt] = ∇ϕtY = ∇XtϕtY = (∇XtY ) ϕt.

Equating the last two equations and then solving the result for Yt gives,

Yt = (ϕt)−1∗ [(∇XtY ) ϕt − (∇YXt) ϕt]

= [ϕt]−1∗ [Xt, Y ] ϕt.

which is precisely Eq. (12.3).

Proposition 12.3 (More characterizations of ϕt∗). Let ϕt = ϕXt and α bea one form on M. Then

d

dtϕ∗tα = ϕ∗tLXtα = ϕ∗t [diXtα+ iXtdα] (12.4)

and in particular if α = df, then

d

dtϕ∗t df = ϕ∗t d [Xtf ] . (12.5)

Proof. We give a number of proofs.Proof 1. Let Y ∈ Γ (TM) and set Yt := ϕ−1

t∗ Y ϕt, Then one one hand,

(Xt [α 〈Y 〉]) ϕt =d

dtα 〈Y ϕt〉 =

d

dt[(ϕ∗tα) 〈Yt〉]

=

(d

dtϕ∗tα

)〈Yt〉+ (ϕ∗tα)

(Yt

)=

(d

dtϕ∗tα

)〈Yt〉+ α ([Xt, Y ]) ϕt.

108 12 Vector Field Commutators and Logarithms

While on the other hand,

(Xt [α 〈Y 〉]) ϕt = (LXtα) 〈Y 〉 ϕt + α ([Xt, Y ]) ϕt

and hence we learn, (d

dtϕ∗tα

)〈Yt〉 = (ϕ∗tLXtα) 〈Yt〉

Proof 2. First let us note that for X,Y ∈ Γ (TM) that,

X [α (Y )]− Y [α (X)] = (∇Xα) (Y )− (∇Y α) (X) + α (∇XY −∇YX)

= (∇Xα) (Y )− (∇Y α) (X) + α ([X,Y ])

so that(dα) (X,Y ) = (∇Xα) (Y )− (∇Y α) (X) .

Therefore,

d

dtα 〈ϕt∗vm〉 =

(∇ d

dtϕt(m)α)〈ϕt∗vm〉+ α

⟨∇dt

[ϕt∗vm]

⟩=(∇Xt(ϕt(m))α

)〈ϕt∗vm〉+ α

⟨∇dt

[ϕt∗vm]

⟩= (∇ϕt∗vmα) 〈Xt (ϕt (m))〉+ dα 〈Xt (ϕt (m)) , ϕt∗vm〉+ α 〈∇ϕt∗vmXt〉= d [α 〈Xt〉] 〈ϕt∗vm〉+ dα 〈Xt (ϕt (m)) , ϕt∗vm〉

and this shows,

d

dt(ϕ∗tα) = ϕ∗t d (iXtα) + ϕ∗t (iXtdα) = ϕ∗t (LXtα) .

Alternate proof of Eq. (12.5). If f ∈ C∞ (M) , then

d

dt[ϕ∗t df ] =

d

dtd [f ϕt] = d

[d

dtf ϕt

]= d [(Xtf) ϕt] = ϕ∗t d [Xtf ] .

Proof 3 of Eq. (12.4). Suppose that α = hdf, then

d

dtϕ∗tα =

d

dt([h ϕt] · ϕ∗t df) = Xth ϕt · ϕ∗t df + [h ϕt] · ϕ∗t d [Xtf ] (12.6)

whileLXtα = Xth · df + hLXtdf = Xth · df + hdXtf.

This equation shows the right member of Eq. (12.6) is precisely, ϕ∗t (LXtα) .

Definition 12.4. Let X and Y be two smooth vector fields on M, then the Liederivative of Y by X (LXY ) is the vector field

LXY =d

dt|0e−tX∗ Y etX =

d

dt|0d

ds|0e−tX esY etX . (12.7)

Remark 12.5. Since

d

ds|0e−sY e−tXesY etX =

d

ds|0e−tXesY etX+

d

ds|0e−sY =

[e−tX∗ Y etX − Y

],

we also have

LXY =d

dt|0d

ds|0e−sY e−tX esY etX . (12.8)

Corollary 12.6. If X,Y ∈ Γ (TM) , then LXY = [X,Y ] .

Proof. This follows directly from Theorem 12.2 with ϕXt = etX in this case.

The following elementary remark will be used repeatedly in the proofs inthis section.

Remark 12.7 (Chain Rule). Suppose that h(t, s) is a function of (t, s) ∈ R2.Then

d

dt|0h(t, t) =

d

dt|0h(t, 0) +

d

dt|0h(0, t).

For example if f(u, v, w) is a function on R3, then

d

dt|0d

ds|0f(t, s, t) =

d

ds|0d

dt|0f(0, s, t) +

d

ds|0d

dt|0f(t, s, 0)

=d

dt|0d

ds|0f(0, s, t) +

d

dt|0d

ds|0f(t, s, 0).

Lemma 12.8. Suppose that h: R×R→ V is a smooth function– V is a vectorspace. Assume that h(s, 0) = h(0, t) = h (0, 0) for all s, and t in R. Then

h(s, t) = h (0, 0) + st(h′(0, 0) +O(s, t)), (12.9)

where O(s, t) denotes a smooth function such that for some constant C > 0,|O(s, t)| ≤ C|(s, t)| for all (s, t) near (0, 0).

Proof. Using the fundamental theorem of calculus twice (and the assump-tions on h) we learn:

h(s, t) = h (0, t) + s

∫ 1

0

duh′(us, t)

= h (0, 0) + s

∫ 1

0

du

[h′(us, 0) + t

∫ 1

0

dvh′(us, vt)

]= h (0, 0) + st

∫ 1

0

du

∫ 1

0

dvh′(su, tv)

= h (0, 0) + st(h′(0, 0) +O(s, t))


12.2 π – related vector fields 109

where

O(s, t) =

∫ 1

0

du

∫ 1

0

dvh′(su, tv)− h′(0, 0)

=

∫ 1

0

du

∫ 1

0

dvh′(su, tv)− h′(0, 0).

Corollary 12.9. If X,Y ∈ Γ (TM) , m ∈M, and f ∈ C∞ (M) , then

f e−sY e−tX esY etX (m) = f (m) + st[X,Y ]mf +O(s, t), (12.10)

where O is a smooth function such that |O(s, t)| ≤ C |(s, t)| . In particular,

[X,Y ]mf = lim(s,t)→(0,0)

s6=0 6=t

f e−sY e−tX esY etX (m)− f (m)/st.

In particular,

[X,Y ]mf =d

dt|0+e

−√tY e−

√tX e

√tY e

√tX (m) . (12.11)

Proof. Let

h(s, t) = f e−sY e−tX esY etX (m) .

Thend

ds|0h (s, t) =

d

ds|0f e−tX esY etX (m) +

d

ds|0f e−sY

and so

d

dt|0d

ds|0h (s, t) =

d

dt|0d

ds|0f e−tX esY etX (m)

=d

dt|0d

ds|0f esY etX (m) +

d

dt|0d

ds|0f e−tX esY (m)

= (XY f) (m)− (Y Xf) (m) .

Now applying Lemma 12.8 to the h proves Eq. (12.10). The remaining equa-tions easily follow from Eq. (12.10).

Theorem 12.10 (Derivative formula). Let ϕXt be the flow associated toX (t, ·) and Y (t, ·) be another time dependent vector field. Then

∂Y[ϕXt]

=[ϕXt]∗

∫ t

0

[ϕXτ]−1

∗ Yτ ϕXτ dτ, (12.12)

where ∂Y[ϕXt]

:= dds |0ϕ

X+sYt .

Proof. For s ∈ R, let ϕst be the flow associated to X (t, ·)+sY (t, ·) . Furtherlet ϕt := ϕ0

t and

Vt := ϕ−1t∗ ∂Y

[ϕXt]

:= ϕ−1t∗

d

ds|0ϕst

so thatd

ds|0ϕst = ϕt∗Vt.

On one hand,

∇dt

d

ds|0ϕst =

∇ds|0d

dtϕst =

∇ds|0 [X (t, ·) + sY (t, ·)] ϕst

= Yt ϕt +∇ dds |0ϕ

stXt = Yt ϕt +∇ϕt∗VtXt

while on the other hand using Lemma 12.1,

∇dt

[ϕt∗Vt] =

(∇dtϕt∗

)Vt + ϕt∗Vt

= ∇ϕt∗VtXt + ϕt∗Vt.

Equating these last two equations shows,

ϕt∗Vt = Yt ϕt =⇒ Vt = ϕ−1t∗ Yt ϕt.

Integrating this equation on t then shows,

ϕ−1t∗ ∂Y

[ϕXt]

=

∫ t

0

[ϕXτ]−1

∗ Yτ ϕXτ dτ


12.2 π – related vector fields

Definition 12.11. Let π : M → N be a smooth map between two smooth man-ifolds and X be a vector field on M and that X be a vector field on N. The twovector fields X and X are said to be π−related if π∗X = X π.

Proposition 12.12. Let π : M → N be a smooth map between two smoothmanifolds.

1. Suppose that X and X are π – related, then

π etX(m) = etX π(m) (12.13)

provided both sides of this equation is defined. The converse is also true.


110 12 Vector Field Commutators and Logarithms

2. Suppose now that Y and Y are π – related then so are [X,Y ] and [X, Y ].

Proof. We take each item in turn.

1. Define σ(t) = π etX(m), so that σ(0) = π(m) and

σ(t) = π∗X etX(m) = X π etX(m) = X σ(t).

So by uniqueness of the flow of X we conclude that σ(t) = etX(π(m)),which proves the first item in the theorem.

2. Proof 1. Let f ∈ C∞(N), then

Y (f π) = (π∗Y ) f =(Y f) π

and similarly,

XY (f π) = X[(Y f) π]

=(XY f

) π.

Thus

(π∗[X,Y ]) f = [X,Y ] (f π) = (XY − Y X) (f π) =([XY − Y X

]f) π

showing again that π∗[X,Y ] =[XY − Y X

] π.

Proof 2. Let us compute π∗[X,Y ] using item 1. of this theorem repeatedly:

π∗[X,Y ] = π∗d

dt|0e−tX∗ Y etX =

d

dt|0π∗e−tX∗ Y etX

=d

dt|0(π e−tX)∗Y etX =

d

dt|0(etX π)∗Y etX

=d

dt|0etX∗ π∗Y etX =

d

dt|0etX∗ Y π etX

=d

dt|0etX∗ Y etX π = [X, Y ] π.

Alternatively, the computation can be carried out as follows:

π∗[X,Y ] = π∗LXY = π∗d

dt|0e−tX∗ Y etX

=d

dt|0π∗e−tX∗ Y etX

=d

dt|0d

ds|0π e−tX esY etX

=d

dt|0d

ds|0e−tX esY etX π

= [X, Y ] π,

where in the second to last line, item 1. of the proposition was used repeat-edly.

Corollary 12.13. Suppose that X is a smooth vector field on M and that α :M →M is a diffeomorphism. Then

α−1 etX α = etXα

, (12.14)

where Xα := α−1∗ X α. (Later we will see that we should write Adα−1X for

Xα.)

Proof. By construction Xα and X are α – related (i.e. α∗Xα = X α) and

therefore by Proposition 12.12 we have

α etXα

= etX α


Lemma 12.14. Suppose X,Y ∈ Γ (TM) and for the simplicity of the statement

assume that X is complete. Further let Yt := e−tX∗ Y etX = Y etX

, then

d

dtYt = e−tX∗ (LXY ) etX = LXYt

or equivalentlyd

dtY e

tX

= (LXY )etX

= LXYetX . (12.15)

Proof. These follow from the following two simple computations:

d

dtYt =

d

ds|0Yt+s =

d

ds|0e−tX∗ e−sX∗ Y esX etX

= e−tX∗d

ds|0[e−sX∗ Y esX

] etX = e−tX∗ (LXY ) etX

andd

dtYt =

d

ds|0Yt+s =

d

ds|0e−sX∗ e−tX∗ Y etX esX = LXY

etX .

Theorem 12.15. Suppose (for simplicity of the statement) that X and Y arecomplete vector fields. Then etX esY = esY etX for all s and t iff [X,Y ] = 0.More generally, we have

e−tX esY etX = esYt (12.16)

where Yt = e−tX∗ Y etX and Yt has the following asymptotic expansion;

Yt ≈ Y + tLXY +t2

2!L2XY +

t3

3!L3XY + . . . . (12.17)


Proof. Suppose that etX esY = esY etX holds for all s, t ∈ R. Then

esY = e−tX esY etX

for all s and t. Differentiating this equation with respect to s and then settings = 0 implies that Y = e−tX∗ Y etX for all t. Differentiating this last expressionat t = 0 implies that 0 = LXY = [X,Y ].

Now suppose that [X,Y ] ≡ 0. By Corollary 12.13 above,

e−tX esY etX = esYt (12.18)

where Yt = e−tX∗ Y etX = Y etX

. But because of Eq. (12.15) and the assumptionthat [X,Y ] = 0, it follows that Yt = Y0 = Y for all t. Feeding this result backinto Eq. (12.18) completes the proof of the first assertion.

Equation (12.16) is a consequence of Corollary 12.13 while Eq. (12.17) isTaylor’s theorem along with the observation that(

d

dt

)nYt = LnXYt

which follows from repeated use of Lemma 12.14.

13

Approximate Logarithms of flows

This section will be based on the following restatement of Theorem 12.10.

Theorem 13.1. If Yt ∈ Γ (TM) is a smoothly varying one parameter family ofvector fields on M such that Y0 ≡ 0 and ϕt := eYt : M →M, then

ϕt = Zt ϕt with ϕ0 = Id, (13.1)

where

Zt =

∫ 1

0

esYt∗ Yt e−sYtds. (13.2)

Lemma 13.2. Let f : [0, 1] → V (V is a vector space) be a smooth function.Then for all n ∈ N0,∫ 1

0

f (s) ds =

n∑k=0

1

(k + 1)!f (k) (0) +Rn (13.3)

where,

Rn :=1

(n+ 1)!

∫ 1

0

f (n+1) (s) (1− s)n+1ds.

Proof. First off notice by integration by parts,∫ 1

0

f (s) ds = −∫ 1

0

f (s) d (1− s) = f (0) +

∫ 1

0

f ′ (s) (1− s) ds

which gives Eq. (13.3) for n = 0. The proof is not finished by inductions uponnoting that

Rn := − 1

(n+ 2)!

∫ 1

0

f (n+1) (s) d (1− s)n+2

=1

(n+ 2)!f (n+1) (0) +

1

(n+ 2)!

∫ 1

0

f (n+2) (s) (1− s)n+2ds

=1

(n+ 2)!f (n+1) (0) +Rn+1.

Corollary 13.3. Let us continue the notation and assumptions in Theorem13.1 and so in particular,

ZYt :=

∫ 1

0


If n ∈ N0, then

Zt =

n∑k=0

1

(k + 1)!(−1)

k(adkYt Yt

)+Rn t,

= Yt +

n∑k=1

1

(k + 1)!(−1)

k(adkYt Yt

)+Rn t, where

Rn t =1

(n+ 1)!

∫ 1

0

esYt∗

(ad

(n+1)Yt

Yt

) e−sYt (s− 1)

n+1ds

=1

(n+ 1)!

∫ 1

0

esYt∗

(ad

(n+1)Yt

Yt

) e−sYt (s− 1)

n+1ds

Proof. We may write Zt as Zt =∫ 1

0f (s) ds, where f (s) = esYt∗ Yt e−sYt .

Then from repeaded use of the commutator Theorem 12.2 we find,

f (k) (s) = (−1)kesYt∗

(adkYt Yt

) e−sYt .

The results stated in the corollary now follow directly from these observationsand Lemma 13.2.

Given a time dependent vector - field Xt ∈ Γ (TM) , our goal is compute a“logarithm” of ϕXt . That is we wish to find Yt ∈ Γ (TM) so that ϕXt = eYt .

Proposition 13.4. Let Xt ∈ Γ (TM) be a time dependent vector field. If thereeixsts Yt ∈ Γ (TM) with Y0 = 0 such that ϕXt = eYt , then Yt solves,∫ 1

0

esYt∗ Yt e−sYtds = ZYt = Xt. (13.5)

Proof. Since

114 13 Approximate Logarithms of flows

d

dtϕXt = Xt ϕXt with ϕX0 = Id and

d

dteYt = ZYt eYt with eY0 = Id

where Zt is given in Eq. (13.4) we will get ϕXt = eYt iff Xt = ZYt .The difficulty here is that it may not be possible to solve Eq. (13.5) for Y.

Nevertheless we can hope to find an “approximate solution” to this equation.Before explaining how to do this, let us record a variant Proposition 13.4.

Proposition 13.5. Let Xt, Yt ∈ Γ (TM) be a time dependent vector fields withY0 = 0. Then

ϕXt = eYt ϕWt with Wt = e−Yt∗ Wt eYt (13.6)

where

Wt := Xt −∫ 1

0


Proof. Let ψt := e−Yt ϕXt , then

d

dtψt = Z−Yt ψt + e−Yt∗ Xt ϕXt

= Z−Yt ψt + e−Yt∗ Xt eYt ψt = Wt ψt

where

Wt = Z−Yt + e−Yt∗ Xt eYt = e−Yt∗ Xt eYt −∫ 1

0

e−sYt∗ Yt esYtds

= e−Yt∗

[Xt −

∫ 1

0

e(1−s)Yt∗ Yt e(s−1)Ytds

] eYt

= e−Yt∗

[Xt −

∫ 1

0

esYt∗ Yt e−sYtds] eYt .

In order to find an approximation to the desired Yt, let us replace Xt byεXt and Yt by Yt (ε) and try to find

Y ktnk=1⊂ Γ (TM) such that Yt (ε) :=∑n

k=1 εkY kt satisfies,

εXt −∫ 1

0

esYt(ε)∗ Yt (ε) e−sYt(ε)ds = O

(εn+1

).

Now as we have seen,∫ 1

0

esYt(ε)∗ Yt (ε) e−sYt(ε)ds =

n∑l=0

1

(l + 1)!

(−adYt(ε)

)lYt (ε) +Rn,t (ε)

where

Rn t (ε) =1

(n+ 1)!

∫ 1

0

esYt(ε)∗

(ad

(n+1)Yt(ε)

Yt (ε))e−sYt(ε) (s− 1)

n+1ds = O

(εn+1

).

Thus it suffices to consider the equation

εXt −n∑l=0

1

(l + 1)!

(−adYt(ε)

)lYt (ε) = O

(εn+1

)or equivalently stated,

n∑l=0

(−1)l

(l + 1)!

n∑k0,k1,...,kl=1

εk0+···+kladYk1tadYk2t. . . ad

Yklt

Y k0t = εXt +O(εn+1

)Looking at the coefficient of ε1, ε2, . . . εm in the above expression shows we mustrequire,

ε1 : Y 1t = Xt

ε2 : Y 2t −

1

2!adY 1

tY 1t = 0

...

εm : Y mt +

m∑l=1

(−1)l

(l + 1)!

∑k0+k1+···+kl=m

adYk1tadYk2t. . . ad

Yklt

Y k0t = 0 (13.8)

wherein the above sum, ki ≥ 1 for all i. In all cases we have Y m0 = 0. It is noweasy to see that the above system of equations may be solved recursively tofind,

Y 1t =

∫ t

0

Xτdτ

Y 2t =

1

2!

∫ t

0

adY 1τY 1τ dτ =

1

2

∫0≤τ1≤τ2≤t

[Xτ1 , Xτ2 ] dτ1dτ2.

...

Y mt =

m∑l=1

(−1)l+1

(l + 1)!

∑k0+k1+···+kl=m

∫ t

0

adYk1τadYk2τ. . . ad

YklτY k0τ dτ (13.9)

...

Making use of the formula we have already found for Y 1t and Y 2

t along with theequation for Y 3

t ;


13 Approximate Logarithms of flows 115

Y 3t −

1

2

[adY 2

tY 1t + adY 1

tY 2t

]+

1

3!ad2Y 1tY 1t = 0,

allows us to conclude,

Y 3t =

1

2

[adY 2

tY 1t +

1

2ad2Y 1tY 1t

]− 1

3!ad2Y 1tY 1t

=1

2adY 2

tY 1t +

(1

4− 1

6

)ad2Y 1tXt

=1

4

∫0≤τ1≤τ2≤t

[[Xτ1 , Xτ2 ] , Xt] dτ1dτ2 +1

12

∫[0,t]2

adXτ1adXτ2Xtdτ1dτ2

=1

4

∫0≤τ1≤τ2≤t

[adXτ1 , adXτ2

]Xtdτ1dτ2 +

1

12

∫[0,t]2

adXτ1adXτ2Xtdτ1dτ2

=1

3

∫0≤τ1≤τ2≤t

adXτ1adXτ2Xtdτ1dτ2 −1

6

∫0≤τ1≤τ2≤t

adXτ2adXτ1Xtdτ1dτ2.

Let us now work out Wt defined in Eq. (13.7) fairly explicitly in the twospecial cases where

Yt = Y 1t =

∫ t

0

Xτdτ and (13.10)

Yt = Y 1t + Y 2

t =

∫ t

0

Xτdτ +1

2

∫0≤τ1≤τ2≤t

[Xτ1 , Xτ2 ] dτ1dτ2. (13.11)

Theorem 13.6. Suppose that Xt ∈ Γ (TM) is a time dependent vector field

and Yt :=∫ t

0Xτdτ as in Eq. (13.10). Then Wt defined in Eq. (13.7) is given by

Wt =

∫ 1

0

esYt∗

[∫ t

0

[Xτ , Xt] dτ

] e−sYt (s− 1) ds = O (t) (13.12)

or with ε > 0 in the picture we have,

Wt (ε) = ε2

∫ 1

0

esYt(ε)∗

[∫ t

0

[Xτ , Xt] dτ

] e−sYt(ε) (s− 1) ds = O

(ε2). (13.13)

Proof. By definition and integration by parts,

Wt = Xt −∫ 1

0

esYt∗ Yt e−sYtds = Xt +

∫ 1

0

esYt∗ Xt e−sYtd (1− s)

= −∫ 1

0

esYt∗ [Xt, Yt] e−sYt (1− s) ds =

∫ 1

0

esYt∗ [Yt, Xt] e−sYt (1− s) ds

=

∫ 1

0

esYt∗

[∫ t

0

[Xτ , Xt] dτ

] e−sYt (s− 1) ds

which proves Eq. (13.12). Equation (13.13) follows from this equation by re-placing Xt by εX.

Theorem 13.7. Suppose that Xt ∈ Γ (TM) is a time dependent vector fieldand

Yt = Y 1t + Y 2

t =

∫ t

0

Xτdτ +1

2

∫0≤τ1≤τ2≤t

[Xτ1 , Xτ2 ] dτ1dτ2

as in Eq. (13.11). Then Wt defined in Eq. (13.7) is given by

Wt =1

4ad2Y 1tXt +

1

2

[Y 2t , Yt

]− 1

2

∫ 1

0

esYt∗

[ad2Y 1tYt

] e−sYt (1− s)2

ds. (13.14)

Moreover, by replacing X by εX we learn,

Wt (ε) = ε3

[ 14ad

2Y 1tXt + 1

2

[Y 2t , Xt + εY 2

t

]− 1

2

∫ 1

0esYt(ε)∗

[ad2Y 1tXt + εY 2

t

] e−sYt(ε) (1− s)2

ds

]. (13.15)


Yt := Y 1t + Y 2

t = Xt +1

2

∫ t

0

[Xτ , Xt] dτ = Xt +1

2

[Y 1t , Xt

].

From Corollary 13.3 with n = 1 we know that

ZYt :=

∫ 1

0

esYt∗ Yt e−sYtds = Yt −1

2!

[Yt, Yt

]+Rt

where ,

Rt =1

2

∫ 1

0

esYt∗

[ad2Yt Yt

] e−sYt (1− s)2

ds.

Moreover we have,

Yt −1

2!

[Yt, Yt

]= Xt +

1

2

[Y 1t , Xt

]− 1

2!

[Y 1t , Xt +

1

2

[Y 1t , Xt

]]− 1

2!

[Y 2t , Yt

]= Xt +

1

2

[Y 1t , Xt

]− 1

2!

[Y 1t , Xt +

1

2

[Y 1t , Xt

]]− 1

2!

[Y 2t , Yt

]= Xt −

1

4

[Y 1t ,[Y 1t , Xt

]]− 1

2!

[Y 2t , Yt

].

Combining these results then shows Wt := Xt − ZYt is given as in Eq. (13.14).

The above results suggest the following general theorems.



Notation 13.8 For n ∈ N and t > 0, let

∆n (t) := (τ1, . . . , τn) ∈ Rn : 0 ≤ τ1 < τ2 < · · · < τn ≤ t

and Sn denote the permutation group on 1, 2, . . . , n , i.e. σ ∈ Sn iff σ :1, 2, . . . , n → 1, 2, . . . , n is a bijective map.

Theorem 13.9. For all 1 ≤ m <∞ let there exists cm : Sm−1 → Q, such thatfor m

Y mt =

∑σ∈Sm−1

cm (σ)

∫∆m−1(t)

adXτσ1 ...adXτσ(m−1)dτ1 . . . dτm−1

Xt. (13.16)

By convention, Y 1t = Xt in the above formula.1

Proof. We have see that c1 = 1, c2 (id) = 12 and c3 (id) = 1

2 and c3 (flip) =− 1

6 are acceptable choices. So clearly we wish to finish the proof by induction.This is a bit of notational mess so let me just show c4 must exists. To this endwe have,

Y 4t =

4∑l=1

(−1)l+1

(l + 1)!

∑k0+...kl=4

adYk1tadYk2t. . . ad

Yklt

Y k0t

For example if l = 2 one the terms we have to consider is

adY 1tadY 2

tY 1t =

[∫ t

0

adXτ dτ

]1

2

∫0≤τ1≤τ2≤t

ad[Xτ1 ,Xτ2 ]dτ1dτ2Xt

which may be put into the desired format using the Jacobi identity to see

ad[Xτ1 ,Xτ2 ] = adXτ1adXτ2 − adXτ2adXτ1 .

Another l = 2 term is;

adY 1tadY 1

tY 2t =

[∫ t

0

adXτ dτ

]21

2

∫0≤τ≤t

adXτXtdτ

which may again be put into the desired form. The general induction prooffollows along these lines. The key point is that

adadXτ1 ...adXτkXτ0=[adXτ1 ,

[adXτ2 , [. . . adXτk , adXτ0 ] . . .

]]and so all of the terms appearing in ad

Yk1tadYk2t. . . ad

Yklt

Y k0t may be expressed

as a linear combination with coefficients in Q of terms appearing in Eq. (13.16).

This leads to the following theorem.

1 In light of Jacobi’s identity, the choice of the cl∞l=1 are not unique!

Theorem 13.10. Let Xt ∈ Γ (TM) be a time dependent vector field and letY lt ∈ Γ (TM) be as in Theorem 13.9 so that Eqs. (13.8) hold. Given n ∈ N, letYt (ε) =

∑nl=1 ε

lY lt and then define

Wt (ε) = εXt −∫ 1

0

esYt(ε)∗ Yt (ε) e−sYt(ε)ds

and Wt := Wt (1) . Then there exists Wt (ε) ∈ Γ (TM) such that Wt (ε) =εn+1Wt (ε) and there exists a locally bounded function such that

∣∣Wt (m)∣∣ ≤

C (m) tn.

Proof. Given a polynomial,∑Nl=1 ε

lal, in ε, let(N∑l=1

εlal

)>n

=

N∑l=n+1

εlal.

As we have seen

Wt (ε) = εXt −n∑l=0

1

(l + 1)!

(−adYt(ε)

)lYt (ε) +Rn,t (ε)

= −n∑l=0

1

(l + 1)!

[(−adYt(ε)

)lYt (ε)

]>n

+Rn,t (ε)

where

Rn t (ε) =1

(n+ 1)!

∫ 1

0

esYt(ε)∗

(ad

(n+1)Yt(ε)

Yt (ε)) e−sYt(ε) (s− 1)

n+1ds.

From these expression we see that we can divide Wt (ε) by εn+1 in order to finda smoothly varying vector-field, Wt (ε) , so that Wt (ε) = εn+1Wt (ε) .

Moreover taking ε = 1and setting Wt = Wt (1) , we learn that∣∣Wt (m)

∣∣ =O (tn) . The point is that the power of ε keeps track of the number of inte-grals involved in computing a term in our expansions. Thus the εk – coefficient

of(−adYt(ε)

)lYt (ε) involves expression with k − 1 – integrals and hence is

O(tk−1

).

13.1 Specialization to the Lie group case

Suppose now that M = G is a Lie group and we require

Xt = At ∈ Γ (TG) (13.17)


13.1 Specialization to the Lie group case 117

where R 3 t → At ∈ g := Lie (G) = TeG is a smooth function and At (g) :=Lg∗At is the left invariant vector field associated to At. Recall that for A,B ∈ gwe define

adAB := [A,B] :=[A, B

](e) .

As Lg∗A = A Lg and similarly for B we know that[A, B

]is also Lg – related

for all g ∈ G, i.e.[A, B

]is still left invariant. Hence it follows that[A, B

]= [A,B] for all A,B ∈ g.

Notation 13.11 For g ∈ G let Lg : G → G and Rg : G → G be the smoothmaps defined by

Lgx = gx and Rgx = xg for all x ∈ G.

Further let Adg ∈ End (g) be defined by

AdgA = Lg∗Rg−1∗A =d

dt|0getAg−1.

whereetA := etA (e) .

Lemma 13.12. If gXt = ϕXt (e) , then ϕXt (x) = xgXt for all x ∈ G, or alterna-tively put, ϕXt = RgXt .

Proof. Let gt := gXt and σt := xgt. By definition we have

gt = At (gt) = Lgt∗At with g0 = e

and similarly,

σt = Lx∗gt = Lx∗Lgt∗At = At (xgt) = At (σt) .

By uniqueness of solutions to ODEs it follows that ϕXt (x) = σt = xgt.

Corollary 13.13. If A ∈ g, then etAesA = e(t+s)A.

Proof. By definition and the group homomorphism properties of flows wehave,

e(t+s)A = e(t+s)A (e) = esA etA (e) = ResA(etA)

= etAesA.

Lemma 13.14. For all g ∈ G and A ∈ g we have geAg−1 = eAdgA.

Proof. Let σt := getAg−1, then

σt =d

ds|0σt+s =

d

ds|0ge(t+s)Ag−1 =

d

ds|0getAesAg−1

=d

ds|0getAg−1gesAg−1 = Lσt∗ [AdgA] with σ0 = e.

Thus we have shown σt = [AdgA]∼

(σt) with σ0 = e, i.e.

getAg−1 = σt = et[AdgA]∼ (e) = etAdgA.

Lemma 13.15. If A ∈ g, then AdeA = eadA .

Proof. Let B ∈ g, then

AdetAB = Re−tA∗LetA∗B = Re−tA∗B(etA)

= e−tA∗ B etA (e) .

Differentiating this equation then gives,

d

dtAdetAB = e−tA∗

[A, B

] etA (e) = e−tA∗ [A,B] etA (e)

= AdetA [A,B] = AdetAadAB with Ade0A = Ig.

From this we conclude that AdetA = etadA .

Lemma 13.16. If Yt = Bt where R 3 t → Bt ∈ g := Lie (G) = TeG is asmooth function, then

d

dteBt = Ct

(eBt)

where Ct :=

∫ 1

0

e−sadBt Btds.

Proof. Since eBt = eBt (e) , we may apply Theorem 12.10 in the form ofTheorem 13.1 to learn,

d

dteBt =

d

dteBt (e) = Zt

(eBt)

where

Zt (x) =

∫ 1

0

esBt∗˜Bt e−sBt (x) ds =

∫ 1

0

esBt∗˜Bt (xe−sBt) ds

=

∫ 1

0

ResBt∗L(xe−sBt )∗Btds = Lx∗

∫ 1

0

ResBt∗Le−sBt∗Btds

= Lx∗

∫ 1

0

Ade−sBt Btds = Ct (x)



where

Ct :=

∫ 1

0

Ade−sBt Btds =

∫ 1

0

e−sadBt Btds.

Proposition 13.17. There exists an open neighborhood U of 0 ∈ g such thatV :=

eA : A ∈ U

is an open neighborhood of e ∈ G and U 3 A → eA ∈ V is

a diffeomorphism.

Proof. Let ψ (A) := eA and observe that ψ∗B0 = dds |0ψ (0 + sB) =

dds |0e

sB = B ∈ g = TeG. From this it follows that ψ∗0 is an isomorphismand hence the result follows by the inverse function theorem.

Let log : V → U be the inverse to the map described in Proposition 13.17and let Xt = At be as in Eq. (13.17). Then there exists ε > 0 such that gXt ∈ Vfor all |t| ≤ ε and so we may define Bt := log

(gXt)∈ g so that gXt = eBt for

|t| ≤ ε. It then follows from Lemma 13.16 that

At = Ct =

∫ 1

0

e−sadBt Btds =

[e−sadBt

−adBt

]s=1

s=0

Bt =

[I − e−adBtadBt

]Bt.

Alternatively written, Bt ∈ g, must satisfy,

Bt =

[I − e−adBtadBt

]−1

At =

[adBt

I − e−adBt

]At with B0 = 0. (13.18)

As the coefficients on the right side of this equation are analytic functions ofBt one may argue that if Bt (ε) satisfies,

Bt (ε) =

[adBt(ε)

I − e−adBt(ε)

]εAt with B0 (ε) = 0

then Bt (ε) =∑∞k=1 ε

kBkt for someBkt∞k=1⊂ g. We may then conclude in this

setting that the formal expansion we had for Yt = Bt actually is convergent.Moreover, one can actually write down a series solution to Eq. (??). The pointis that

eadBt = AdeBt = AdgXt

and therefore,

d

dteadBt =

d

dtAdgt =

d

ds|0Adgt+s = Adgt

d

ds|0Adg−1

t gt+s= AdgtadAt

= eadBtadAt with eadB0 = I.

Thus letting Ut := eadBt and T (U)t :=∫ t

0UτadAτ dτ we have

I = Ut −∫ t

0

UτadAτ dτ = [(I − T ) (U)]t .

Thus the usual geometric series arguments shows,

eadBt =

∞∑n=0

T nI = I +

∫ t

0

adAτ dτ +

∫0≤τ1≤τ2≤t

adAτ1adAτ2dτ1dτ2 + . . .

=

∞∑n=0

∫0≤τ1≤τ2≤···≤τn≤t

adAτ1 . . . adAτndτ1 . . . dτn.

One may now use this result back in Eq. (13.18) in order to get a fairly explicitseries solution for Bt, see Strichartz [?].


14

Principal Bundles and Connections

14.1 Principal bundle basics

Definition 14.1. Suppose that G is a Lie group. A principal G – bundle isa fiber bundle fiber bundle G→ P

π→M such that:

1. G is a right action on P, i.e. there is a smooth map, P ×G 3 (p, g)→ p ·g ∈P, such that p · (g1g2) = (p · g1) · g2 for all g1, g2 ∈ G.

2. The right action G acts preserves the fibers, Pm of P, i.e. π (p · g) = π (p)for all g ∈ G and p ∈ P.

3. G acts faithfully and transitively on Pm for all m ∈ M. [This assumptionis redundant as it is implied by the next requirement.]

4. For each m ∈ M there exits U ⊂o M such that m ∈ U and there is a localtrivialization, ψ : PU → G such that ψ (pg) = ψ (p) g for all p ∈ PU andg ∈ G.

Example 14.2. As always the trivial bundle, P = M ×G is a principal bundle.

Example 14.3. If V → Eπ→M is a vector bundle, let Gl(V )→ P

π→M be theprincipal bundle of frames defined by P = ∪m∈MPm where Pm ≡ GL(V,Em) –the invertible linear maps from V to Em. If ψ : EU → V is a local trivializationof EU we define a local trivialization ψ of PU by ψ (p) := ψ p ∈ GL (V ) for allp ∈ PU .

Remark 14.4. Here are a few basic remarks.

1. If ψ : PU → G is a principal bundle local trivialization of P, then weconstruct the associated local section, uψ (x) := (π, ψ)

−1(x, e) , of P. Con-

versely if u ∈ Γ (PU ) we may define a principal bundle local trivialization,ψu, of PU by requiring ψu (u (x) · g) := g for all x ∈ U and g ∈ G. Thuslocal trivializations of P are in one to on correspondence with local sectionsof P.

2. If u ∈ Γ (PU ) and v ∈ Γ (PV ) then there exists a unique smooth transitionfunction gvu : U ∩ V → G such that u (x) = v (x) gvu (x) for all x ∈ U ∩ V.Moreover if w ∈ Γ (PW ) , we will have for x ∈ U ∩V ∩W on one hand that

u (x) = v (x) gvu (x) = w (x) gwv (x) gvu (x)

while on the other hand,

u (x) = w (x) gw,u (x) .

Comparing these two identities gives the following “cocylcle” condition;

gw,u (x) = gwv (x) gvu (x) for all x ∈ U ∩ V ∩W.

The next theorem is the principal bundle analogue of Theorem 8.7.

Theorem 14.5. Suppose that M is a manifold, G is a Lie group, Ukk∈Λ is afinite or countable open cover of M, and gkl ∈ C∞ (Uk ∩ Ul, G) are given maps.If gklk,l∈Λ satisfy the cocycle condition,

gklglm = gkm on Uk ∩ Ul ∩ Um for all k, l,m ∈ Λ,

then there exists a principal G - bundle P over M having these transition func-tions. In more detail, we may find P with local section uk ∈ Γ (PUk) for allk ∈ Λ such that ul (x) = uk (x) gkl (x) for all x ∈ Uk ∩ Ul.

Exercise 14.1. Suppose that V → Eπ→ M is a vector bundle and Gl(V ) →

Pπ→ M is the bundle of frames, i.e. Pm ≡ GL(V,Em). Suppose U ⊂o M, and

S ∈ Γ (PU ). Notice for each v ∈ V, the function Sv(m) ≡ S(m)v for m ∈ U is alocal section of E. Show that S is smooth iff Sv is smooth for all v ∈ V.

Exercise 14.2. Keeping the same notation as in the above problem. Show thatany smooth section S ∈ Γ (PU ) can be used to define a smooth local trivializa-tion ϕ of E via

ϕm ≡ ϕ|Em ≡ S(m)−1.

Conversely, show that if ϕ is a local trivialization of E over U, then S(m) ≡ ϕ−1m

is a smooth local section of P.

Exercise 14.3. Assume the same notation as problem 3. Let ρ : GL(V ) →GL(V ) be the identity map and let E′ be the associated vector bundle: E′ ≡P ×ρ V . Recall that E′m = p · v|p ∈ Pm, v ∈ V , where p · v ≡ (pg, ρ(g−1)v) =(pg, g−1v)|g ∈ G. Show that ϕ : E′ → E defined by ϕ(p · v) = pv is a vectorbundle isomorphism.

120 14 Principal Bundles and Connections

Exercise 14.4. Suppose that G → Pπ→ M is a principal bundle, ρ : G →

GL(V ) is a representation of G, and E is the associated bundle E ≡ P ×ρ V.Set

C∞ρ (P, V ) = f ∈ C∞(P, V )|f(pg) = ρ(g−1)f(p)∀p ∈ P, g ∈ G.

(A function f ∈ C∞ρ (P, V ) is said to be G – equivariant.) For f ∈ C∞ρ (P, V )define a section Ψ(f) of E via Ψ(f)(m) ≡ p · f(p), where p is any element ofPm. Show that Ψ(f) is well defined, Ψ(f) ∈ Γ∞(E) (i.e. is a smooth section),and Ψ : C∞ρ (P, V )→ Γ∞(E) is an isomorphism of vector spaces.

Exercise 14.5. Show that a principal bundle is trivial iff Γ∞(P ) is not empty.Is the same statement true for vector bundles?

14.2 Curvature and Torsion considerations

Since the results we are going to state are local in nature, we will mostly workin a local trivialized model. First though we set up the objects globally andthen use local models to give the proofs about the globally defined objects. [Seethe results surrounding 14.31 for a more global prospective of what is going onin this section.]

Let ∇ be a covariant derivative on a vector bundle W → E → M. Furtherlet P := GL (E) be the associated principal bundle whose fiber, Pm, over mconsists of the invertible linear transformations from W → Em.

Definition 14.6 (Horizontal Lifts). Given vm ∈ TmM and u ∈ Pm we letvu ∈ TuP be defined by

vu :=d

dt|0//t (σ)u where σ (t) ∈M 3 σ (0) = vm.

We refer to vu as the horizontal lift of vm to TuP.

Definition 14.7 (Vertical Vector Fields). For A ∈ End (W ) = LieGL (W ) ,let A ∈ Γ (TP ) be defined by

A (u) :=d

dt|0uetA = u ·A for all u ∈ P.

We refer to A as the vertical vector-field associated to A.

If E = M ×W, u = (m, g) ∈ P = M × Aut (W ) , and ∇ = d + Γ where Γis a connection one form on M with values in End (W ) , we have

vu =d

dt|0 (σ (t) , //t (σ)u)(m,g) = (v,−Γ 〈v〉 g)(m,g) ,

which shows that vu exists and is well defined independent of the choice ofσ (t) ∈M such that σ (0) = vm. Moreover if A ∈ End (W ) , then

A (u) =d

dt|0(m, getA

)= (m, gA)u .

Remark 14.8. Notice that π∗vu = vm. Thus if X ∈ Γ (TM) and X ∈ Γ (TP ) isit horizontal lift we have π∗X = X π, i.e. X and X are π – related.

Theorem 14.9. Let X,Y ∈ Γ (TM) and X, Y ∈ Γ (TP ) be their horizontallifts, then[X, Y

](u)−[X,Y ] (u) = −u·

[u−1R∇ (X (m) , Y (m))u

]∀ u ∈ Pm and m ∈M.

Proof. As this is a local statement we may assume that E = M ×W andP = M ×Aut (W ) and then taking u = (m, g) we have,

X (u) = (X (m) ,−Γ (X (m)) g)u .

Thus we learn(∂X Y

)= ((∂XY ) ,− [∂XΓ (Y )] g + Γ (Y )Γ (X) g)u

from which it follows that[X, Y

](u) = ([X,Y ] , [Y Γ (X)] g − [XΓ (Y )] g + [Γ (Y ) , Γ (X)] g)

= ([X,Y ] , (dΓ (Y,X) + Γ ([Y,X]) + [Γ (Y ) , Γ (X)]) g)

=(0, RΓ (Y,X) g

)+ ([X,Y ] ,−Γ ([X,Y ]) g)

= [X,Y ] (u) +(0, g g−1RΓ (Y,X) g

)= [X,Y ] (u)− u ·

[u−1R∇ (X,Y )u

]where all expression should be evaluated at m ∈M.

Remark 14.10 (Second proof of Proposition 9.20). Let X,Y ∈ Γ (TM) be com-muting vector fields so by Remark 14.8,

π e−sY e−tX esY etX(u) = e−sY e−tX esY etX(π (u)) = π (u) .

Thus it follows that e−sY e−tX esY etX(u) ∈ Pm for all s, t and hence thereexists g (s, t) ∈ Aut (W ) such that

e−sY e−tX esY etX(u) = ug (s, t) .

We then learn from Corollary 12.9 that


14.3 Horizontal Lifts and Curvature in Principal Bundles 121

d

ds|0d

dt|0e−sY e−tX esY etX(u) =

[X, Y

](u)

= −u ·[u−1R∇ (X (m) , Y (m))u

]= R∇ (Y (m) , X (m))u

which is another proof of Proposition 9.20 upon noting that

etX (u) = //t

(e(·)X

)u.

Now let suppose M is a d – dimensional manifold and ∇ is a covariantderivative on E = TM.

Definition 14.11 (Canonical horizontal vector fields). If a ∈ Rd letB∇a (u) := ua for all u ∈ P. Notice that ua ∈ Tπ(u)M in this case.

Theorem 14.12. If a, b ∈ Rd, then

[Ba, Bb] (u) = −BTu(a,b) (u)− u ·Ru (a, b)

where,

Tu (a, b) := u−1T (ua, ub) ∈ Rd and

Ru (a, b) = u−1R (ua, ub)u ∈ End(Rd).

Proof. Again this is a local statement so we may assume that M = Rd (oran open subset of U ⊂ Rd) and TRd = Rd × Rd and P = Rd × GL

(Rd). We

then have for u = (x, g) that

Ba (u) = (ua,−Γ ([ua]x)u)u.

Therefore

(∂BaBb) (u) =

(−Γ ([ua]x)ub,

Γ([Γ ([ua]x)ub]

x

)+Γ ([ub]x)Γ ([ua]x)− ∂ua [x→ Γ ([ub]x)]

u

)u

=(−Γ ([ua]x)ub, Γ

([Γ ([ua]x)ub]

x

)u)

+ (0, Γ ([ub]x)Γ ([ua]x)u− ∂ua [x→ Γ ([ub]x)u])

and it follows from this that

[Ba, Bb] (u) =(Γ ([ub]x)ua− Γ ([ua]x)ub, Γ

([Γ ([ua]x)ub− Γ ([ub]x)ua]

x

)u)u

+ (0, [Γ ([ub]x) , Γ ([ua]x)]u− dΓ ([ua]x , [ub]x)u)u

=(T∇ (ub, ua) ,−Γ

(T∇ (ub, ua)

))−(0, R∇ (ua, ub)u

)u

= −Bu−1T∇(ua,ub) − u ·Ru (a, b) .

Corollary 14.13 (Rolling interpetation of Torsion). For a, b ∈ Rd wehave,

d

ds|0d

dt|0π e−sBb e−tBa esBb etBa(u) = π∗ [Ba, Bb] (u) = −T∇ (ua, ub) .

To reinterpret this result suppose that vm = ua and wm = ub, s, t > 0, and Cs,tis the closed path in TmM which consists of the straight line segments joiningthe following points in the following prescribed order;

0→ wm → wm + vm → vm → 0.

If Σ (s, t) ∈ M is the result of ∇ – rolling (starting at m ∈ M) our manifoldalong the path Cst, then

T∇ (ua, ub) =d

ds|0d

dt|0Σ (s, t) .

To be more explicit, each of the paths t→ Σ (s, t) and s→ Σ (·, t) are geodesics.

14.3 Horizontal Lifts and Curvature in Principal Bundles

This section discusses parallel translation and curvature in the principal bundlesetting. However, first consider the trivial bundle P , P = U × RN π→ U, whereπ is the projection onto the first factor. A function Y : U → RN may beidentified with a section of P via Y → S(x) = (x, Y (x)). Using this notationwe see that differential equation Y ′(x) = F (x, Y (x)) of Section ?? is equivalentto the equation:

S∗vx = (vx, (F (x, Y (x))vx)Y (x)) = (vx, (F (S(x))v)Y (x)) ∀ vx ∈ TU. (14.1)

This suggests that we define

H(p)vx = (vx, (F (p)v)y), (14.2)

where p ≡ (x, y). Noting that π(p) = x, we may use (14.2) to rewrite (14.1) as:

S∗π(p) = H(p). (14.3)

Let us specialize this to the linear example in the last section. For this wetake P = M × G, where G is the Lie group of N × N invertible matrices andF (p) = −A(x)〈·〉y. In this case (14.2) becomes

H(p)vx = (vx, (−A(x)〈v〉g)g) = (vx,−Rg∗A(x)〈v〉), (14.4)



where p = (x, g) now. We let G act on P on the right via (x, g)h = (x, gh).Notice that

Rg∗H(p)vx = H(pg)vx. (14.5)

This is again the statement that F (x, y) is linear in y in this example. We willsay more about this all below.

Our next goal is to reformulate the integrability condition and curvaturetensor in the last section using this frame work. To be more suggestive we willnow write M for U. For this we start with a few definitions:

Definition 14.14. The horizontal subspace (Hp) of TpP is defined to be therange of the operator H(p) : Tπ(p)M → TpP. A vector ξ ∈ TpP is said to behorizontal if ξ ∈ Hp.

Notice that H = Hpp∈P is a distribution in TP, that is H is a sub-bundleof TP . This distribution is rather special in the fact that π∗ : Hp → Tπ(p)M is

an isomorphism for all p ∈ P. Notice that H(p) = π∗|−1Hp .

Definition 14.15. The horizontal lift of vector field V on M is the vector fieldV on P defined by V (p) ≡ H(p)V (π(p)).

Proposition 14.16. Let V and U be two tangent vector fields on M , then

([V , U ]− [V,U ])(p) = (0x, R(p)〈U(x), V (x)〉), (14.6)

where again p = (x, y).

The proof is a straight forward computations and so is left to the reader.We are now ready for the general definitions:

Definition 14.17. Let Pπ→ M be a fiber bundle. A connection on P is a

“horizontal” distribution H ⊂ TP, i.e. H is a sub-bundle of TP with the addi-tional property that π∗p : Hp → Tπ(p)M is an isomorphism for all p ∈ P. Asabove we define the horizontal lift map H(p) to be the inverse of π∗p|Hp

Definition 14.18. For each p ∈ P we define the vertical subspace (Vp) ofTpP to be the kernel of π∗p : TpP → Tπ(p)M.

Remark 14.19. It can be checked that for each p ∈ P, that TpP = Vp ⊕Hp, soas bundles TP = V ⊕H.

Definition 14.20. Given a connection H on P we define the curvature (Ω(p))at p ∈ P of this connection by

Ω(p)〈u, v〉 ≡ ([U , V ]− [U, V ])(p) (14.7)

where u, v ∈ Tπ(p)M are arbitrary, and U and V are any pair of tangent vectorfields on M such that u = U(π(p)) and v = V (π(p)). (It can be checked that Ω iswell defined independent of any choices made for U and V.) Another equivalentform of this tensor which is more common in the literature is the tensor (Ω)defined by

Ω〈ξ, η〉 ≡ Ω(p)〈π∗ξ, π∗η〉, (14.8)

where ξ, η ∈ TpP are now arbitrary. It is Ω that we will actually call the cur-vature tensor. Notice that

Ω〈U , V 〉 = [U , V ]− [U, V ]. (14.9)

The following theorem is trivial but a very useful observation.

Theorem 14.21. Let Pπ→M be a fiber bundle with connection H. Then H is

an involutive distribution iff Ω ≡ 0, where Ω is the curvature of H.

Remark 14.22. Let U and V be two vector fields on M and U and V be theirhorizontal lifts. By definition U and V are π−related to U and V respectively sothat [U , V ] is also π−related to [U, V ]. From this it easily follows that [U , V ]−[U, V ] is vertical. From this it follows that Ω〈ξ, η〉 is always a vertical vector.

We would now like to specialize to the case corresponding to Example ??.It turns out the nicest way to do this is to assume that P

π→ M is now aprincipal bundle. Let G denote the structure group of this bundle. With thisextra structure we now can deduce that the vertical bundle V is trivial:

Lemma 14.23. Let Pπ→M be a principal G−bundle, then V is isomorphic to

P × g where g = Lie(G) is the Lie algebra of G. This isomorphism is given bythe map:

((p,A)→ p ·A ≡ d

ds|0pesA) : P × g→ V. (14.10)

We may also write A(p) for p ·A.

The proof of this lemma is left to the reader.It turns out that following added condition on a connection H on P is the

analogue of assuming that F (x, y) at the beginning of this section is linear iny.

Definition 14.24. Let Pπ→M be a principal G−bundle. A connection H on

P is a connection as in Definition 14.17 with the additional assumption thatRg∗Hp = Hpg for all g ∈ G and p ∈ P.


14.3 Horizontal Lifts and Curvature in Principal Bundles 123

Because of the above isomorphism (Lemma 14.23) and Remark 14.19, wemay define a g−valued one form ω on P via:

ω〈p ·A+ ξ〉 ≡ A (14.11)

where ξ is an arbitrary element of Hp. It now trivially follows that

Hp = Ker ωp. (14.12)

One may also easily check that R∗gω = Ad−1g · ω for all g ∈ G. We make the

following definition:

Definition 14.25. A g−valued 1-form ω on P is a connection 1-form ifω〈p ·A〉 = A for all A ∈ g and R∗gω = Ad−1

g · ω for all g ∈ G.Lemma 14.26. The space of connection 1-forms on P are in one-to-one cor-respondence with the space of connections on P. The correspondence is given byEq. (14.12).

Remark 14.27. In the case of a principal bundle with connection H and corre-sponding connection 1-form ω, it is customary to replace the definition of thecurvature tensor Ω in Definition 14.20 by ω〈Ω〉. Because of Remark 14.22 andLemma 14.23, there is no information lost in making this change. We will writeΩ now for ω〈Ω〉 and when necessary write Ω = ω〈old−Ω〉.

The following theorem summarizes the relationship between the connection1-form ω and the curvature tensor Ω.

Theorem 14.28.Ω = dωh = dω + ω ∧ ω, (14.13)

where dωh〈ξ, η〉 ≡ dω〈ξh, ηh〉, ξh and ηh are the horizontal components of ξ andη respectively in TpP , and ω ∧ ω〈ξ, η〉 ≡ [ω〈ξ〉, ω〈η〉].

Finally, we will write (??) of the last section in this geometrical formalism.To set this up let σ : I → M be a curve in M based at o ∈ M. Let X be avector-field along σ such that X(0) = 0. We view X as a tangent vector basedat σ in the path space of M. Let Hs(σ) denote the horizontal lift (or paralleltranslation) of σ to P such that H0(σ) = uo, where uo is a given point in thefiber Po ≡ π−1(o).

The translation of (??) using this notation becomes:

Theorem 14.29.

H∗〈X〉(s) = Hs(σ) ·∫ s

0

Ω〈H〈σ′(s′)〉(Hs′(σ)),H〈X(s′)〉(Hs′(σ))〉ds′〉

+H〈X(s)〉

= Hs(σ) ·∫ s

0

ΩHs′ (σ)〈σ′(s′), X(s′)〉ds′ +H〈X(s)〉, (14.14)

where Ωp〈v, w〉 ≡ Ω〈H(p)v,H(p)w〉.

Proof. One can prove this by translating the results in last section to thiscase. Notice that the term −A〈X(s)〉Hs(σ) in Eq. (??) is the local form ofH(p)X(s). Compare with Eq. (14.4). We may also give a more geometric proofas follows. (Also see Driver [?] for a proof using the notation of covariant deriva-tives.)

Choose a one parameter family σ(t, ·) of curves in M such that σ(t, s = 0) =o for all t, σ(t = 0, s) = σ(s), and σ(t = 0, s) = X(s). Let u(t, s) ≡ Hs(σ(t, ·)),so that u(t, ·) is the horizontal lift of σ(t, ·). By definition of horizontal lift wehave: 0 = ω〈u′〉 for all (t, s). Differentiate this expression with respect to t tofind:

0 =d

dtω〈u′〉 = dω〈u, u′〉+

d

dsω〈u〉 = Ω〈u, u′〉+

d

dsω〈u〉,

where we have use the structure equation (and again the fact that 0 = ω〈u′〉)to get the second equality. Integrating this last equation relative to s yields:

ω〈u〉 =

∫Ω〈u, u′〉ds.

Now notice that π(u) = σ, and hence π∗u′ = σ′ and π∗u = σ. Hence it follows

thatΩ〈u(t, s), u′(t, s)〉 = ΩHs(σ(t,·))〈σ(t, s), σ′(t, s)〉,

and thatH〈σ〉 = uh

— the horizontal component of u. Now it is a general fact which may be easilyverified in our local model that

u = u · ω〈u〉+ uh,

no matter what the tangent vector u is in TP. Putting together the above fourequations yields that

u(t, s) = u(t, s) ·∫ s

0

Ωu(t,s′)〈σ(t, s′), σ′(t, s′)〉ds′ +H〈σ(t, s)〉,

which is exactly Eq. (14.14) after setting t = 0.I will end this section by specializing to the case where M is a Riemannian

manifold with metric g and P = O(M) is the orthogonal frame bundle of M.We assume that M is equipped with a metric compatible covariant derivative ∇— the corresponding connection 1-form on O(M) will be denoted by ω. Recallthat ω〈u〉 = u−1∇u/dt, where u(t) is any smooth curve in O(M). Because P isthe orthogonal frame bundle we now have some extra structure which we nowdefine. [Also see Section 14.2 for a more down to ear perspective.]

Definition 14.30. 1. The canonical 1-form θ on O(M) is the Rd−valued 1-form given by θ〈u〉 = u−1u, where u(·) is a smooth curve in O(M).



2. The standard horizontal vector-fields B〈a〉a∈Rd on O(M) are given by

B〈a〉(u) = H〈ua〉(u).

3. Fix a point o ∈ M and a frame uo ∈ Oo(M). The development map I :W (Rd)→W (O (M)) is defined by I(b) = u, where u solves:

u′(s) = B〈b′(s)〉(u(s)) with u(0) = uo.

We end this section with a computation of the differential of the developmentmap I.

Theorem 14.31. Let h ∈ H, where H is the Cameron-Martin space. For b ∈W ≡W (Rd), let hb ∈ TbW be defined by hb = d

dt |0(b+ th). Then

I∗(hb) = I(b) ·A+B〈a〉(I(b)),

where A(s) ∈ Lie(O(n)) = so(n) and a(s) ∈ Rd are solutions to the ordinarydifferential equations:

A′ = ΩI(b)〈b′, a〉 with A(0) = 0, (14.15)

anda′ = ΘI(b)〈b′, a〉+Ab′ + h′ with a(0) = 0. (14.16)

HereΩu〈a, b〉 ≡ Ω〈B〈a〉(u), B〈b〉(u)〉,

andΘu〈a, b〉 ≡ Θ〈B〈a〉(u), B〈b〉(u)〉,

where a, b ∈ Rd and u ∈ O(M). Here Θ = dθ+ω ∧ θ is the torsion tensor of ω.

Proof. Write u(t, s) for I(b+ th)(s) so that u solves:

u′(t, s) = B〈b′(s) + th′(s)〉(u(s)) with u(t, 0) = uo. (14.17)

We must compute u. To this end we first recall that u may be written as u =u ·ω〈u〉+B〈θ〈u〉〉. So we see that A(s) = ω〈u(0, s)〉, and that a(s) = θ〈u(0, s)〉.(Notice that A(0) = 0 and a(0) = 0, because u(t, 0) = 0.) We now find thedifferential equations for A and a.

We start with the computation of A′ (using Ω = dω + ω ∧ ω) :

d

dsω〈u〉 =

d

dtω〈u′〉+ dω〈u′, u〉

= 0 +Ω〈u′, u〉 − ω ∧ ω〈u′, u〉= Ω〈u′, u〉.

In the above computation we have repeatedly made use of the fact that ω〈u′〉 =0, since by its very definition u′ is horizontal. By restricting the above equationto t = 0 and using the fact that Ω only depends on the horizontal componentsof its arguments it follows that A′ = ΩI(b)〈b′, a〉, i.e. (14.15) holds.

Now to the computation of a′ (using Θ = dθ + ω ∧ θ) :

d

dsθ〈u〉 =

d

dtθ〈u′〉+ dθ〈u′, u〉

=d

dt(b′ + th′) + (Θ − ω ∧ θ)〈u′, u〉

= h′ +Θu〈b′ + th′, θ〈u〉〉+ ω〈u〉(b′ + th′).

Restricting this last equation to t = 0 proves (14.16).

14.4 Structure Equations

To match notation with Helgason’s book, let us write ω for A as a above. Thatis, we assume, u is a frame on E and ω (v) := u (m)

−1∇vu for v ∈ TmM. Wealso recall that if S = us is a general section, then

∇vS = u[u−1∇vu · s+ vs

]= u [vs+ ω (v) s]

so that the local form of ∇ is d+ ω. Let us now compute the curvature tensorin these terms. Since,

∇X∇Y S = u [(X + ω (X)) (Y + ω (Y )) s] ,

we have

u−1∇X∇Y S = (XY +Xω (Y ) + ω (X)ω (Y )) s+ ω (X)Y s+ ω (Y )Xs.

Since the last term is symmetric in X and Y it follows that

u−1 [∇X ,∇Y ]S − u−1∇[X,Y ]S = ([X,Y ] +Xω (Y )− Y ω (X) + [ω (X) , ω (Y )]) s

− ([X,Y ] + ω ([X,Y ])) s

= (Xω (Y )− Y ω (X)− ω ([X,Y ]) + [ω (X) , ω (Y )]) s

= (dω (X,Y ) + [ω (X) , ω (Y )]) s.

Letting Ω (v, w) := u−1F (v, w)u where F = F∇ is the curvature of ∇, we thenhave proven

Ω = dω + ω ∧ ω (14.18)

where


14.4 Structure Equations 125

ω ∧ ω (v, w) := [ω (v) , ω (w)] or ω = ωiωjdxi ∧ dxj

where ω = ωidxi. If we go ahead and compute d of Eq. (14.18) we learn that

dΩ = d (ω ∧ ω) .

Now, for commuting vector fields,

d (ω ∧ ω) (X,Y, Z) = X [(ω ∧ ω) (Y,Z)] + cyclic

= X [ω (Y ) , ω (Z)] + cyclic

= [Xω (Y ) , ω (Z)] + [ω (Y ) , Xω (Z)] + cyclic

= [Xω (Y ) , ω (Z)]− [Xω (Z) , ω (Y )] + cyclic

= [Xω (Y ) , ω (Z)]− [Y ω (X) , ω (Z)] + cyclic

= [Xω (Y )− Y ω (X) , ω (Z)] + cyclic

= [dω (X,Y ) , ω (Z)] + cyclic.

Using dω = Ω − ω ∧ ω in this last equation then implies, making use of theJacobi identity, that

d (ω ∧ ω) (X,Y, Z) = ([Ω (X,Y ) , ω (Z)]− [[ω (X) , ω (Y )] , ω (Z)]) + cyclic

= [Ω (X,Y ) , ω (Z)] + cyclic

= − [ω (Z) , Ω (X,Y )] + cyclic = − [ω (X) , Ω (Y,Z)] + cyclic

=: − [ω ∧Ω] (X,Y, Z)

Hence we have proved the second Bianchi identity following Bianchi iden-tity:

dΩ + [ω ∧Ω] = 0. (14.19)

The simpler proof of this fact is,

dΩ = dω ∧ ω − ω ∧ dω = − [ω ∧ dω] = − [ω ∧Ω − ω ∧ ω ∧ ω] = − [ω ∧Ω] .

We now want to restrict our attention to covariant derivatives on E = TM.In this case, Xi (m) = u (m) ei and let α (vm) := u (m)

−1vm so that α is

an Rd – valued one form on M. We begin by computing dα as follows. Lety := α (Y ) = u−1Y so that Y = uy, then

dα (X,Y ) = X [α (Y )]− Y [α (X)]− α ([X,Y ])

= Xy − Y x− u−1 [X,Y ] .

Now

T (X,Y ) = ∇XY −∇YX − [X,Y ]

= u ([X + ω (X)] y − [Y + ω (Y )]x)− [X,Y ]

= u ([X + ω (X)] y − [Y + ω (Y )]x)− [X,Y ]

= u (dα (X,Y ) + ω (X) y − ω (Y )x)

= u (dα (X,Y ) + ω (X)α (Y )− ω (Y )α (X))

= u (dα (X,Y ) + ω (X)α (Y )− ω (Y )α (X)) .

Thus we have proven the structure equation that

dα+ ω ∧ α = α T =: Θ

where T is the Torsion tensor. Letting Θ := α T, we have

Θ = dα+ ω ∧ α.

Again we should compute d of this equation to get the first Bianchi identity,

dΘ = d (ω ∧ α) .

Now (using the fact that d is an anti-derivation and ω has degree one) we learn

d (ω ∧ α) = dω ∧ α− ω ∧ dα.

Alternatively,

d (ω ∧ α) (X,Y, Z) = X [(ω ∧ α) (Y,Z)] + cyclic

= X [ω (Y )α (Z)− ω (Z)α (Y )] + cyclic

= X [ω (Y )α (Z)] + cyclic− Y [ω (X)α (Z)] + cyclic

= [Xω (Y )− Y ω (X)]α (Z) + cyclic

+ [ω (Y )Xα (Z)− ω (X)Y α (Z)] + cyclic

= dω (X,Y )α (Z) + cyclic

+ [ω (Y )Xα (Z)− ω (Y )Zα (X)] + cyclic

= [dω (X,Y )α (Z) + ω (Y ) dα (X,Z)] + cyclic

= [dω (X,Y )α (Z) + ω (X) dα (Z, Y )] + cyclic

= [dω (X,Y )α (Z)− ω (X) dα (Y,Z)] + cyclic

= (dω ∧ α− ω ∧ dα) (X,Y, Z) .

Thus we have shown that

dΘ = dω ∧ α− ω ∧ dα = (Ω − ω ∧ ω) ∧ α− ω ∧ (Θ − ω ∧ α)

= Ω ∧ α− ω ∧Θ

or equivalently put thatdΘ + ω ∧Θ = Ω ∧ α (14.20)

which is the first Bianchi identity.


15

Geometric Frobenius’ Theorem

In this chapter, suppose that M is a d - dimensional manifold. A distribution∆ = ∆mm∈M on M is a sub-bundle of TM, i.e. there exists 1 ≤ k ≤ d suchthat for each m ∈ M, ∆m is a k – dimensional subspace of TmM which issmoothly varying. This last statement may be quantified by requiring for eachm0 ∈M that there exist smooth vector fields Xlkl=1 defined in a neighborhoodof m0 so that ∆m = span Xl (m) : 1 ≤ l ≤ k for m near m0.

A vector field X defined (locally) on M is ∆−adapted if X(m) ∈ ∆m for allm ∈ M (m in the domain of X), i.e. X is a local section of ∆. The set of ∆ -adapted vector fields on M will be denoted by χ(∆) = Γ (∆) .

A submanifold (N, i) of M is said to be an integrable submanifoldof ∆ if forall m ∈ N, i∗TmN = ∆m.

Definition 15.1. The distribution ∆ is called involutive if [X,Y ] ∈ χ(∆) forall X,Y ∈ χ(∆). Alternatively stated, ∆ is integrable iff χ(∆) := Γ (∆) is a Liesub-algebra of χ (M) := Γ (TM) .

Remark 15.2. If Xlkl=1 is a local frame for ∆ near some point m0 ∈ M, then

∆ is involutive near m0 iff [Xl, Xk] =∑dj=1 c

jlkXj for some smooth functions

cjlk defined near m0. Indeed the direction ( =⇒ ) is trivial. For the conversedirection, suppose that Y,W ∈ Γ (∆) which may be written as,

Y =

k∑i=1

yiXi and W =

k∑j=1

wjXj

where yi and wj are smooth functions. We then have

[W,Y ] =

k∑i,j=1

[yiXi, w

jXj

]=

k∑i,j=1

yiXi

(wjXj

)− wjXj

(yiXi

)=

k∑i,j=1

yiwj [Xi, Xj ] +

k∑j=1

(Y wj)Xj −k∑i=1

(Wyi

)iXi ∈ ∆.

Theorem 15.3. If ∆ is a k - dimensional distribution on M. Then ∆ is inte-grable iff ∆ is involutive.

Proof. First suppose for all m ∈ M there exists an integral submanifoldi : N → M such m ∈ i (N) . Now let X,Y ∈ χ(∆) and let N be an integralsubmanifold of ∆. Let i : N →M denote the inclusion map, and set X = X iand Y = Y i. By assumption X and Y are vector fields on N. It should also beclear that X and X and Y and Y are i−related. Hence by Proposition 12.12,[X,Y ] and [X, Y ] are also i– related. In particular this implies that [X,Y ](n) ∈TnN = ∆n for all n ∈ N. Since this is true for all integral submanifolds N, wesee that [X,Y ] ∈ χ(∆) provided X,Y ∈ χ(∆).

Let us now assume that ∆ is involutive. By shrinking M if necessary, wemay assume that ∆m = span Xl (m) : 1 ≤ l ≤ k for m ∈ M. [Note that

Xl (m)kl=1 is a basis for ∆m under this assumption.] Given m0 ∈ M, chooseπ : M → Rk smooth so that π (m0) = 0 and π∗ : ∆m0 → T0Rk ∼= Rk is anisomorphism.1 It then follows that dπ : ∆m → Rk is an isomorphism for all mnear m0. By shrinking M more if necessary we may assume that dπ : ∆m → Rkis an isomorphism for all m ∈M. For a ∈ Rk let Xa :=

∑kl=1 alXl and set

u (m) a := dπ Xa (m) .

Then by the arranged properties of dπ and the fact that Xlkl=1 is a frame for∆, we conclude that u (m) ∈ GL

(Rk)

for all m, i.e. u : M → GL(Rk). We

then defineYa (m) := Xu(m)−1a (m)

so that dπYa (m) = u (m)u (m)−1a = a for all a ∈ Rk. [In short, we could

simply define Ya so that Ya (m) ∈ ∆m and dπYa (m) = a for all a ∈ Rk andm ∈M.]

By construction we have π∗Ya = ∂a π for all a ∈ Rk and thereforeπ∗ [Ya, Yb] = [∂a, ∂b] π = 0. On the other hand since ∆ is involutive, weknow that [Ya, Yb] (m) ∈ ∆m and since π∗ is injective on ∆m we may concludethat [Ya, Yb] = 0 for all a, b ∈ Rk. Given this we may now define

ψ (a) := eYa (m0)

will produce the desired embedded submanifold. Indeed,

1 For example, let ϕ : M → Rd be a chart, then dϕ (∆m0) is a k – dimensionalsubspace of Rd and by pre-composing ϕ with a rotation, R, we may assume thatdϕ (∆m0) = Rk × 0 . We may then take π :=

(ϕ1, . . . , ϕk

).

128 15 Geometric Frobenius’ Theorem

(∂bψ) (a) = eYa∗

∫ 1

0

dse−sYa∗ Yb esYa (m0) =

∫ 1

0

dse(1−s)Ya∗ Yb esYa (m0)

=

[∫ 1

0

dsesYa∗ Yb e−sYa] esYa (m0) = Yb (ψ (a)) .

Alternatively,

(∂bψ) (a) =d

ds|0eYa+sYb (m0) =

d

ds|0esYb eYa (m0) = Yb (ψ (a)) ∈ ∆ψ(a).

It is often the case that a distribution is described as follows. Associated toa linear map, Z : Rk → Γ (TM) such that

Rk 3 a→ Za (m) ∈ TmM

is injective for all m ∈M, is the distribution,

∆Zm :=

Za (m) : a ∈ Rk

for all m ∈M.

The assumptions guarantee that dim∆Zm = k for all m ∈M. In this case, ∆ is

involutive iff there exists a skew symmetric bilinear form α : Rk×Rk → C∞ (M)such that [Za, Zb] = Zα〈a,b〉.

Theorem 15.4. Suppose that Z : Rk → Γ (TM) be as above and assume that∆Z is involutive and α : Rk × Rk → C∞ (M) is chosen so that

[Za, Zb] = Zα〈a,b〉.

Given a path a (t) ∈ Rk and m0 ∈M, let γt (a) solve the differential equation,

γt (a) = Za(t) (γt (a)) with γ0 (a) = m0. (15.1)

If b (t) ∈ Rk is another curve in Rk, then

(∂bγt) (a) :=d

ds|0γt (a+ sb) = Zc(t) (γt (a)) (15.2)

where c (t) ∈ Rk is the unique solution to the linear differential equation,

c (t) = b (t) + α 〈c (t) , a (t)〉 (γt (a)) with c (0) = 0. (15.3)

Proof. Let us begin by showing how c (t) arises. We know that γt (a) shouldlie in the integral sub-manifold associated to ∆ which goes through m0 andtherefore we should have (∂bγt) (a) ∈ ∆γt(a), i.e. there should exists c (t) ∈ Rksuch that

(∂bγt) (a) = Zc(t) (γt (a)) . (15.4)

Since γ0 (a) = m0 for all a and hence (∂bγ0) (a) = 0m0 ∈ Tm0M we also see thatc (0) = 0. To determine the equations satisfied by c (t) ∈ Rk we will differentiateEq. (15.4) with respect to t.

Suppose that ∇ is a Torsion free covariant derivative. On one hand, we have

∇dt

(∂bγt) (a) =∇ds|0d

dtγt (a+ sb) =

∇ds|0Za(t)+sb(t) (γt (a+ sb))

= Zb(t) (γt (a)) +∇∂bγt(a)Za(t)

= Zb(t) (γt (a)) +(∇Zc(t)Za(t)

)(γt (a)) ,

while on the other,

∇dt

[Zc(t) (γt (a))

]= Zc(t) (γt (a)) +∇γt(a)Zc(t)

= Zc(t) (γt (a)) +(∇Za(t)Zc(t)

)(γt (a)) .

Comparing these two equations shows

Zc(t) (γt (a)) = Zb(t) (γt (a)) +(∇Zc(t)Za(t) −∇Za(t)Zc(t)

)(γt (a))

= Zb(t) (γt (a)) +[Zc(t), Za(t)

](γt (a))

= Zb(t) (γt (a)) + Zα〈c(t),a(t)〉(γt(a))

and hence we learn that c (t) has to solve Eq. (15.3).To finish the proof of the theorem we now work backwards. If c (t) solves

Eq. (15.4), then

∇dt

[Zc(t) (γt (a))

]= Zc(t) (γt (a)) +

(∇Za(t)Zc(t)

)(γt (a))

= Zb(t)+α〈c(t),a(t)〉(γt(a)) (γt (a)) +(∇Za(t)Zc(t)

)(γt (a))

= Zb(t) (γt (a)) +[Zc(t), Za(t)

](γt (a)) +

(∇Za(t)Zc(t)

)(γt (a))

= Zb(t) (γt (a)) +(∇Zc(t)Za(t)

)(γt (a)) =

∇dt

(∂bγt) (a) .

Equation (15.2) now follows from this identity and the fact that Zc(0) (γ0 (a)) =0 = (∂bγ0) (a) .


Part II

Appendices

16

Spaces of Bounded Linear Operators

In this chapter, X and Y will be normed spaces and L (X,Y ) will denotethe continuous linear operators Λ : X → Y. It turns out for later developmentsit is very useful to introduce a natural norm on L (X,Y ) .

Definition 16.1. Let X and Y be normed spaces and T : X → Y be a linearmap. Then T is said to be bounded provided there exists C < ∞ such that‖T (x)‖Y ≤ C‖x‖X for all x ∈ X. We denote the best constant by ‖T‖op =‖T‖L(X,Y ), i.e.

‖T‖L(X,Y ) = supx6=0

‖T (x) ‖Y‖x‖X

= supx 6=0‖T (x) ‖Y : ‖x‖X = 1 .

The number ‖T‖L(X,Y ) is called the operator norm of T.

The most useful thing retain from this definition is that

‖Tx‖Y ≤ ‖T‖op ‖x‖X for all x ∈ X. (16.1)

In the future, we will usually drop the garnishing on the norms and simply write‖x‖X as ‖x‖, ‖T‖L(X,Y ) as ‖T‖ , etc. The reader should be able to determine thenorm that is to be used by context. The following proposition is a restatementof Exercise ??.

Proposition 16.2. Suppose that X and Y are normed spaces and T : X → Yis a linear map. The the following are equivalent:

1. T is continuous.2. T is continuous at 0.3. T is bounded.

Proof. 1.⇒ 2. trivial. 2.⇒ 3. If T continuous at 0 then there exist δ > 0 suchthat ‖T (x) ‖ ≤ 1 if ‖x‖ ≤ δ. Therefore for any nonzero x ∈ X, ‖T (δx/‖x‖)‖ ≤ 1which implies that ‖T (x)‖ ≤ 1

δ ‖x‖ and hence ‖T‖ ≤ 1δ <∞. 3.⇒ 1. Let x ∈ X

and ε > 0 be given. Then

‖Ty − Tx‖ = ‖T (y − x)‖ ≤ ‖T‖ ‖y − x‖ < ε

provided ‖y − x‖ < ε/‖T‖ := δ.For the next three exercises, let X = Rn and Y = Rm and T : X → Y be a

linear transformation so that T is given by matrix multiplication by an m× nmatrix. Let us identify the linear transformation T with this matrix.

Exercise 16.1. Assume the norms on X and Y are the `1 – norms, i.e. forx ∈ Rn, ‖x‖ =

∑nj=1 |xj | . Then the operator norm of T is given by

‖T‖ = max1≤j≤n

m∑i=1

|Tij | .

Exercise 16.2. Assume the norms on X and Y are the `∞ – norms, i.e. forx ∈ Rn, ‖x‖ = max1≤j≤n |xj | . Then the operator norm of T is given by

‖T‖ = max1≤i≤m

n∑j=1

|Tij | .

Exercise 16.3. Assume the norms on X and Y are the `2 – norms, i.e. forx ∈ Rn, ‖x‖2 =

∑nj=1 x

2j . Show ‖T‖2 is the largest eigenvalue of the matrix

T trT : Rn → Rn. Hint: Use the spectral theorem for symmetric real matrices.

Lemma 16.3. Let X,Y be normed spaces, then the operator norm ‖·‖ onL (X,Y ) is a norm. Moreover if Z is another normed space and T : X → Yand S : Y → Z are linear maps, then ‖ST‖ ≤ ‖S‖‖T‖, where ST := S T.

Proof. As usual, the main point in checking the operator norm is a normis to verify the triangle inequality, the other axioms being easy to check. IfA,B ∈ L (X,Y ) then the triangle inequality is verified as follows:

‖A+B‖ = supx 6=0

‖Ax+Bx‖‖x‖

≤ supx 6=0

‖Ax‖+ ‖Bx‖‖x‖

≤ supx 6=0

‖Ax‖‖x‖

+ supx 6=0

‖Bx‖‖x‖

= ‖A‖+ ‖B‖ .

For the second assertion, we have for x ∈ X, that

‖STx‖ ≤ ‖S‖‖Tx‖ ≤ ‖S‖‖T‖‖x‖.

From this inequality and the definition of ‖ST‖, it follows that ‖ST‖ ≤ ‖S‖‖T‖.

Proposition 16.4. Suppose that X is a normed vector space and Y is a Banachspace. Then (L (X,Y ) , ‖ · ‖op) is a Banach space.

132 16 Spaces of Bounded Linear Operators

Proof. Let Tn∞n=1 be a Cauchy sequence in L (X,Y ) . Then for each x ∈ X,

‖Tnx− Tmx‖ ≤ ‖Tn − Tm‖ ‖x‖ → 0 as m,n→∞

showing Tnx∞n=1 is Cauchy in Y. Using the completeness of Y, there exists anelement Tx ∈ Y such that

limn→∞

‖Tnx− Tx‖ = 0.

The map T : X → Y is linear map, since for x, x′ ∈ X and λ ∈ F we have

T (x+ λx′) = limn→∞

Tn (x+ λx′) = limn→∞

[Tnx+ λTnx′] = Tx+ λTx′,

wherein we have used the continuity of the vector space operations in the lastequality. Moreover,

‖Tx− Tnx‖ ≤ ‖Tx− Tmx‖+ ‖Tmx− Tnx‖ ≤ ‖Tx− Tmx‖+ ‖Tm − Tn‖ ‖x‖

and therefore

‖Tx− Tnx‖ ≤ lim infm→∞

(‖Tx− Tmx‖+ ‖Tm − Tn‖ ‖x‖)

= ‖x‖ · lim infm→∞

‖Tm − Tn‖ .

Hence‖T − Tn‖ ≤ lim inf

m→∞‖Tm − Tn‖ → 0 as n→∞.

Thus we have shown that Tn → T in L (X,Y ) as desired.The reader is asked to prove the following continuity lemma in Exercise 16.4.

Lemma 16.5. Let X, Y and Z be normed spaces. Then the maps

(S, x) ∈ L (X,Y )×X −→ Sx ∈ Y

and(S, T ) ∈ L (X,Y )× L(Y,Z) −→ TS ∈ L(X,Z)

are continuous. More precisely if ∆x ∈ X, ∆S ∈ L (X,Y ) and ∆T ∈ L(Y, Z),then

‖(S +∆S) (x+∆x)− Sx‖Y ≤ ‖S‖op ‖∆x‖X +‖∆S‖op ‖x‖X +‖∆S‖op ‖∆x‖X

and

‖(T +∆T ) (S +∆S)− TS‖op ≤ ‖∆T‖op ‖S‖op+‖T‖op ‖∆S‖op+‖∆T‖op ‖∆S‖op .

Exercise 16.4. Prove Lemma 16.5.

16.1 Inverting Elements in L(X)

Definition 16.6. A linear map T : X → Y is an isometry if ‖Tx‖Y = ‖x‖Xfor all x ∈ X. T is said to be invertible if T is a bijection and T−1 is bounded.

Notation 16.7 We will write GL (X,Y ) for those T ∈ L (X,Y ) which areinvertible. If X = Y we simply write L(X) and GL(X) for L(X,X) andGL(X,X) respectively.

Proposition 16.8. Suppose X is a Banach space and Λ ∈ L(X) := L(X,X)

satisfies∞∑n=0‖Λn‖ <∞. Then I − Λ is invertible and

(I − Λ)−1 = “1

I − Λ” =

∞∑n=0

Λn and∥∥(I − Λ)−1

∥∥ ≤ ∞∑n=0

‖Λn‖.

In particular if ‖Λ‖ < 1 then the above formula holds and∥∥(I − Λ)−1∥∥ ≤ 1

1− ‖Λ‖.

Proof. Since L(X) is a Banach space and∞∑n=0‖Λn‖ < ∞, it follows from

Theorem ?? that

S := limN→∞

SN := limN→∞

N∑n=0

Λn

exists in L(X). Moreover, by Lemma 16.5,

(I − Λ)S = (I − Λ) limN→∞

SN = limN→∞

(I − Λ)SN

= limN→∞

(I − Λ)

N∑n=0

Λn = limN→∞

(I − ΛN+1) = I

and similarly S (I − Λ) = I. This shows that (I − Λ)−1 exists and is equal toS. Moreover, (I − Λ)−1 is bounded because∥∥(I − Λ)−1

∥∥ = ‖S‖ ≤∞∑n=0

‖Λn‖.

If we further assume ‖Λ‖ < 1, then ‖Λn‖ ≤ ‖Λ‖n and

∞∑n=0

‖Λn‖ ≤∞∑n=0

‖Λ‖n =1

1− ‖Λ‖<∞.

For a nice application of Proposition 16.8 to linear ordinary differentialequations, see the optional Chapter 2 below.


16.2 Integral Operators as Examples of Bounded Operators 133

Corollary 16.9. Let X and Y be Banach spaces. Then GL (X,Y ) is an open(possibly empty) subset of L (X,Y ) . More specifically, if A ∈ GL (X,Y ) andB ∈ L (X,Y ) satisfies

‖B −A‖ < ‖A−1‖−1 (16.2)

then B ∈ GL (X,Y )

B−1 =

∞∑n=0

[IX −A−1B

]nA−1 ∈ L (Y,X) , (16.3)

∥∥B−1∥∥ ≤ ‖A−1‖ 1

1− ‖A−1‖ ‖A−B‖(16.4)

and ∥∥B−1 −A−1∥∥ ≤ ‖A−1‖2 ‖A−B‖

1− ‖A−1‖ ‖A−B‖. (16.5)

In particular the map

A ∈ GL (X,Y )→ A−1 ∈ GL (Y,X) (16.6)

is continuous.

Proof. Let A and B be as above, then

B = A− (A−B) = A[IX −A−1(A−B))

]= A(IX − Λ)

where Λ : X → X is given by

Λ := A−1(A−B) = IX −A−1B.

Now

‖Λ‖ =∥∥A−1(A−B))

∥∥ ≤ ‖A−1‖ ‖A−B‖ < ‖A−1‖‖A−1‖−1 = 1.

Therefore I −Λ is invertible and hence so is B (being the product of invertibleelements) with

B−1 = (IX − Λ)−1A−1 =[IX −A−1(A−B))

]−1A−1.

Taking norms of the previous equation gives∥∥B−1∥∥ ≤ ∥∥(IX − Λ)−1

∥∥ ‖A−1‖ ≤ ‖A−1‖ 1

1− ‖Λ‖

≤ ‖A−1‖1− ‖A−1‖ ‖A−B‖

which is the bound in Eq. (16.4). The bound in Eq. (16.5) holds because∥∥B−1 −A−1∥∥ =

∥∥B−1 (A−B)A−1∥∥ ≤ ∥∥B−1

∥∥∥∥A−1∥∥ ‖A−B‖

≤ ‖A−1‖2 ‖A−B‖1− ‖A−1‖ ‖A−B‖

.

16.2 Integral Operators as Examples of BoundedOperators

In the examples to follow, all integrals are the standard Riemann integrals andwe will make use of the following notation.

Notation 16.10 Given an open set U ⊂ Rd, let Cc (U) denote the collectionof real valued continuous functions f on U such that

supp(f) := x ∈ U : f (x) 6= 0

is a compact subset of U.

Example 16.11. Suppose that K : [0, 1] × [0, 1] → C is a continuous function.For f ∈ C([0, 1]), let

Tf (x) =

∫ 1

0

K(x, y)f (y) dy.

Since

|Tf (x)− Tf(z)| ≤∫ 1

0

|K(x, y)−K(z, y)| |f (y)| dy

≤ ‖f‖∞maxy|K(x, y)−K(z, y)| (16.7)

and the latter expression tends to 0 as x → z by uniform continuity of K.Therefore Tf ∈ C([0, 1]) and by the linearity of the Riemann integral, T :C([0, 1])→ C([0, 1]) is a linear map. Moreover,

|Tf (x)| ≤∫ 1

0

|K(x, y)| |f (y)| dy ≤∫ 1

0

|K(x, y)| dy · ‖f‖∞ ≤ A ‖f‖∞

where

A := supx∈[0,1]

∫ 1

0

|K(x, y)| dy <∞. (16.8)

This shows ‖T‖ ≤ A <∞ and therefore T is bounded. We may in fact show‖T‖ = A. To do this let x0 ∈ [0, 1] be such that

supx∈[0,1]

∫ 1

0

|K(x, y)| dy =

∫ 1

0

|K(x0, y)| dy.

Such an x0 can be found since, using a similar argument to that in Eq. (16.7),

x→∫ 1

0|K(x, y)| dy is continuous. Given ε > 0, let


134 16 Spaces of Bounded Linear Operators

fε (y) :=K(x0, y)√

ε+ |K(x0, y)|2

and notice that limε↓0 ‖fε‖∞ = 1 and

‖Tfε‖∞ ≥ |Tfε(x0)| = Tfε(x0) =

∫ 1

0

|K(x0, y)|2√ε+ |K(x0, y)|2

dy.

Therefore,

‖T‖ ≥ limε↓0

1

‖fε‖∞

∫ 1

0

|K(x0, y)|2√ε+ |K(x0, y)|2

dy

= limε↓0

∫ 1

0

|K(x0, y)|2√ε+ |K(x0, y)|2

dy = A

since

0 ≤ |K(x0, y)| − |K(x0, y)|2√ε+ |K(x0, y)|2

=|K(x0, y)|√ε+ |K(x0, y)|2

[√ε+ |K(x0, y)|2 − |K(x0, y)|

]

≤√ε+ |K(x0, y)|2 − |K(x0, y)|

and the latter expression tends to zero uniformly in y as ε ↓ 0.We may also consider other norms on C([0, 1]). Let (for now) L1 ([0, 1])

denote C([0, 1]) with the norm

‖f‖1 =

∫ 1

0

|f (x)| dx,

then T : L1 ([0, 1], dm) → C([0, 1]) is bounded as well. Indeed, let M =sup |K(x, y)| : x, y ∈ [0, 1] , then

|(Tf) (x)| ≤∫ 1

0

|K(x, y)f (y)| dy ≤M ‖f‖1

which shows ‖Tf‖∞ ≤M ‖f‖1 and hence,

‖T‖L1→C ≤ max |K(x, y)| : x, y ∈ [0, 1] <∞.

We can in fact show that ‖T‖ = M as follows. Let (x0, y0) ∈ [0, 1]2 satisfying|K(x0, y0)| = M. Then given ε > 0, there exists a neighborhood U = I ×J of (x0, y0) such that |K(x, y)−K(x0, y0)| < ε for all (x, y) ∈ U. Let f ∈Cc(I, [0,∞)) such that

∫ 1

0f (x) dx = 1. Choose α ∈ C such that |α| = 1 and

αK(x0, y0) = M, then

|(Tαf)(x0)| =∣∣∣∣∫ 1

0

K(x0, y)αf (y) dy

∣∣∣∣ =

∣∣∣∣∫I

K(x0, y)αf (y) dy

∣∣∣∣≥ Re

∫I

αK(x0, y)f (y) dy

≥∫I

(M − ε) f (y) dy = (M − ε) ‖αf‖L1

and hence‖Tαf‖C ≥ (M − ε) ‖αf‖L1

showing that ‖T‖ ≥ M − ε. Since ε > 0 is arbitrary, we learn that ‖T‖ ≥ Mand hence ‖T‖ = M.

One may also view T as a map from T : C([0, 1])→ L1([0, 1]) in which caseone may show

‖T‖L1→C ≤∫ 1

0

maxy|K(x, y)| dx <∞.


17

Contraction Mapping Principle

Many problems in analysis may be formulated as a fixed point problem.That is given a function, S : X → X, we wish to find a solution to the equationS (x) = x, i.e. x is a fixed point of the map S.

Example 17.1 (ODEs as a fixed point problem). Suppose Z : R × Rd → Rdis a continuous function and x0 ∈ Rd is given and suppose we want to findx ∈ C1

((−T, T ) ,Rd

)which solves the ordinary differential equation (O.D.E.),

x (t) = Z(t, x (t)) with x(0) = x0. (17.1)

Using the fundamental theorem of calculus, it is easy to see that solving Eq.(17.1) is equivalent to finding x ∈ X := C

((−T, T ) ,Rd

)such that

x (t) = x0 +

∫ t

0

Z (τ, x (τ)) dτ for |t| < T.

This is then precisely the fixed point problem of finding x ∈ S such that x =S (x) where S : X → X is defined by

S (x) := x0 +

∫ ·0

Z (τ, x (τ)) dτ. (17.2)

Example 17.2 (Newton’s method as a fixed point problem). Suppose that wehave a function, f : [0,∞) → [0,∞) and b ∈ [0,∞) is given. We wish tofind a solution, a, to the equation f (a) = b. Newton proposes that given a“guess” x ∈ [0,∞) to the solution a, we should look for an “improved” guessby approximating f near x by f (x′) ∼= f (x) + f ′ (x) (x′ − x) . Thus we hopethat solving f (x′) = b is close to solving,

f (x) + f ′ (x) (x′ − x) = b for x′.

Solving this linear equation leads to the new guess for a to be,

x′ = x+b− f (x)

f ′ (x)=: Sb (x) .

Notice that if a is a fixed point of Sb then

a = Sb (a) = a+b− f (a)

f ′ (a)=⇒ f (a) = b.

In the special case where f (x) = x2 and b = 2, we are lead to looking for afixed point to the map,

S (x) = x+2− x2

2x=x2 + 2

2x=

1

2

(x+

2

x

). (17.3)

Fig. 17.1. Here is a graph of x and S (x) := 12

(x+ 2

x

). The fixed point,

√2, we are

looking for is graphically given as the x – coordinate of the point of intersection ofthese two curves.

Here is a basic general theorem which give sufficient conditions in order forthe map S : X → X to have a unique fixed point.

Theorem 17.3. Suppose that (X, ρ) is a complete metric space and S : X → Xis a contraction, i.e. there exists α ∈ (0, 1) such that ρ(S (x) , S (y)) ≤ αρ(x, y)for all x, y ∈ X. Then S has a unique fixed point in X, i.e. there exists a uniquepoint x ∈ X such that S (x) = x. Moreover, for every x0 ∈ X, define x1 = S (x0)and xn inductively using xn := S (xn−1) . Then

ρ (xn, x0) ≤ ρ(x1, x0)αn

1− α=ρ(S (x0) , x0)

1− ααn

so that xn converges to x0 geometrically fast.

136 17 Contraction Mapping Principle

Proof. For uniqueness suppose that x and x′ are two fixed points of S, then

ρ(x, x′) = ρ(S (x) , S(x′)) ≤ αρ(x, x′).

Therefore (1 − α)ρ(x, x′) ≤ 0 which implies that ρ(x, x′) = 0 since 1 − α > 0.Thus x = x′.

For existence, let x0 ∈ X be any point in X and define xn ∈ X inductivelyby xn+1 = S(xn) for n ≥ 0. We will show that x ≡ limn→∞ xn exists in X andbecause S is continuous this will imply,

x = limn→∞

xn+1 = limn→∞

S(xn) = S( limn→∞

xn) = S (x) ,

showing x is a fixed point of S.So to finish the proof, because X is complete, it suffices to show xn∞n=1 is

a Cauchy sequence in X. An easy inductive computation shows, for n ≥ 0, that

ρ(xn+1, xn) = ρ(S(xn), S(xn−1)) ≤ αρ(xn, xn−1) ≤ · · · ≤ αnρ(x1, x0).

Another inductive argument using the triangle inequality shows, for m > n,that,

ρ(xm, xn) ≤ ρ(xm, xm−1) + ρ(xm−1, xn) ≤ · · · ≤m−1∑k=n

ρ(xk+1, xk).

Combining the last two inequalities gives (using again that α ∈ (0, 1)),

ρ(xm, xn) ≤m−1∑k=n

αkρ(x1, x0) ≤ ρ(x1, x0)αn∞∑l=0

αl = ρ(x1, x0)αn

1− α.

This last equation shows that ρ(xm, xn) → 0 as m,n → ∞, i.e. xn∞n=0 is aCauchy sequence.

Exercise 17.1 (Finding the√

2 by Newton’s method. ). Show S :(0,∞) → R defined by S (x) = 1

2

(x+ 2

x

)(as in Eq. (17.3) of Example 17.2)

satisfies;

1. S ([1, 2]) ⊂ [1, 2] , and2. for all x, x′ ∈ [1, 2] ,

|S (x)− S (x′)| ≤ 1

2|x− x′| . (17.4)

Therefore the contraction mapping principle (Theorem 17.3) implies S|[1,2] :

[1, 2] → [1, 2] has a unique fixed point in [1, 2] which is in fact√

2. (Therate of convergence of the iterates to this

√2is order

(12

)n.)

Exercise 17.2. Continue the notation is Example 17.1, i.e. suppose that Z :R × Rd → Rd is a continuous function and x0 ∈ Rd. We now further assumethere exists, K <∞, such that

‖Z(t, x)− Z (t, x′)‖ ≤ K ‖x− x′‖ for all t ∈ R and x, x′ ∈ Rd.

For T > 0 let

XT := C([−T, T ]→ Rd

), ‖x‖T := max

|t|≤T‖x (t)‖ , MT := sup

|t|≤T‖Z (t, x0)‖

and for all x ∈ XT let

S (x) := x0 +

∫ ·0

Z (τ, x (τ)) dτ ∈ XT

be defined as in Eq. (17.2). Show

1. ‖S (x)− S (y)‖T ≤ KT · ‖x− y‖T for all x, y ∈ XT .2. Conclude if T < 1/K, then S has a unique fixed point x ∈ XT .3. Show x satisfies the ordinary differential equation (17.1) for |t| < T, i.e.

showx (t) = Z (t, x (t)) for |t| < T with x (0) = x0. (17.5)

[You may safely skip the remainder of this chapter on first reading.] Thefollowing theorem is a useful generalization of Theorem 17.3.

Corollary 17.4 (Contraction Mapping Principle II). Suppose that (X, ρ)is a complete metric space and S : X → X is a continuous map. If there existsn ∈ N such that

S(n) ≡n times︷︸︸︷

S S . . . S

is a contraction (i.e. there exists α ∈ (0, 1) such that ρ(S(n) (x) , S(n) (y)) ≤αρ(x, y) for all x, y ∈ X), then S has a unique fixed point in X.

Proof. Let T ≡ S(n), then T : X → X is a contraction and hence T has aunique fixed point x ∈ X. Since any fixed point of S is also a fixed point of T,we see if S has a fixed point then it must be x. Now

T (S (x)) = S(n)(S (x)) = S(S(n) (x)) = S(T (x)) = S (x) ,

which shows that S (x) is also a fixed point of T. Since T has only one fixedpoint, we must have that S (x) = x. So we have shown that x is a fixed pointof S and this fixed point is unique.


Lemma 17.5. Suppose that (X, ρ) is a complete metric space, n ∈ N, Z isa topological space, and α ∈ (0, 1). Suppose for each z ∈ Z there is a mapSz : X → X with the following properties:

Contraction property ρ(S(n)z (x) , S

(n)z (y)) ≤ αρ(x, y) for all x, y ∈ X and z ∈ Z.

Continuity in z For each x ∈ X the map z ∈ Z → Sz (x) ∈ X is continuous.

By Corollary 17.4 above, for each z ∈ Z there is a unique fixed point G(z) ∈X of Sz.

Conclusion: The map G : Z → X is continuous.

Proof. Let Tz ≡ S(n)z . If z, w ∈ Z, then

ρ(G(z), G(w)) = ρ(Tz(G(z)), Tw(G(w)))

≤ ρ(Tz(G(z)), Tw(G(z))) + ρ(Tw(G(z)), Tw(G(w)))

≤ ρ(Tz(G(z)), Tw(G(z))) + αρ(G(z), G(w)).

Solving this inequality for ρ(G(z), G(w)) gives

ρ(G(z), G(w)) ≤ 1

1− αρ(Tz(G(z)), Tw(G(z))).

Since w → Tw(G(z)) is continuous it follows from the above equation thatG(w)→ G(z) as w → z, i.e. G is continuous.

18

Inverse and Implicit Function Theorems

18.1 The Inverse Function Theorems

In this section, let X be a Banach space, R > 0, U = B = B(0, R) ⊂ X andε : U → X be a continuous function such that ε (0) = 0. Our immediate goal isto give a sufficient condition on ε so that F (x) := x+ε (x) is a homeomorphismfrom U to F (U) with F (U) being an open subset of X. Let’s start by lookingat the one dimensional case first. So for the moment assume that X = R,U = (−1, 1), and ε : U → R is C1. Then F will be injective iff F is eitherstrictly increasing or decreasing. Since we are thinking that F is a “small”perturbation of the identity function we will assume that F is strictly increasing,i.e. F ′ = 1 + ε′ > 0. This positivity condition is not so easily interpreted foroperators on a Banach space. However the condition that |ε′| ≤ α < 1 is easilyinterpreted in the Banach space setting and it implies 1 + ε′ > 0.

Lemma 18.1. Suppose that U = B = B(0, R) (R > 0) is a ball in X andε : B → X is a C1 function such that ‖Dε‖ ≤ α <∞ on U. Then

‖ε (x)− ε (y) ‖ ≤ α‖x− y‖ for all x, y ∈ U. (18.1)

Proof. By the fundamental theorem of calculus and the chain rule:

ε (y)− ε (x) =

∫ 1

0

d

dtε(x+ t(y − x))dt

=

∫ 1

0

[Dε(x+ t(y − x))](y − x)dt.

Therefore, by the triangle inequality and the assumption that ‖Dε (x) ‖ ≤ α onB,

‖ε (y)− ε (x) ‖ ≤∫ 1

0

‖Dε(x+ t(y − x))‖dt · ‖(y − x)‖ ≤ α‖(y − x)‖.

Remark 18.2. It is easily checked that if ε : U = B(0, R)→ X is C1 and satisfies(18.1) then ‖Dε‖ ≤ α on U.

Using the above remark and the analogy to the one dimensional example,one is lead to the following proposition.

Proposition 18.3. Suppose α ∈ (0, 1), R > 0, U = B(0, R) ⊂o X and ε : U →X is a continuous function such that ε (0) = 0 and

∗ ‖ε (x)− ε (y) ‖ ≤ α‖x− y‖ ∀ x, y ∈ U. (18.2)

Then F : U → X defined by F (x) := x+ ε (x) for x ∈ U satisfies:

1. F is an injective map and G = F−1 : V → U is continuous where V :=F (U) .

2. If x0 ∈ U, and r > 0 such that B(x0, r) ⊂ U (i.e. r ≤ R− ‖x0‖), then

B (F (x0) , (1− α)r) ⊂ F (B(x0, r)) ⊂ B(F (x0) , (1 + α)r). (18.3)

In particular, for all r ≤ R,

B(0, (1− α) r) ⊂ F (B(0, r)) ⊂ B(0, (1 + α) r), (18.4)

see Figure 18.1 below.3. V := F (U) is open subset of X and F : U → V is a homeomorphism.

Fig. 18.1. Nesting of F (B(x0, r)) between B(z0, (1−α)r) and B(z0, (1 +α)r) wherez0 := F (x0) .

Proof.

1. Using the definition of F and the estimate in Eq. (18.2),

140 18 Inverse and Implicit Function Theorems

‖x− y‖ = ‖(F (x)− F (y))− (ε (x)− ε (y))‖≤ ‖F (x)− F (y) ‖+ ‖ε (x)− ε (y) ‖≤ ‖F (x)− F (y) ‖+ α‖(x− y)‖

for all x, y ∈ U. This implies

‖x− y‖ ≤ (1− α)−1‖F (x)− F (y) ‖ (18.5)

which shows F is injective on U and hence shows the inverse function G =F−1 : V := F (U) → U is well defined. Moreover, replacing x, y in Eq.(18.5) by G (x) and G (y) respectively with x, y ∈ V shows

‖G (x)−G (y) ‖ ≤ (1− α)−1‖x− y‖ for all x, y ∈ V. (18.6)

Hence G is Lipschitz on V and hence continuous.2. Let x0 ∈ U, r > 0 and z0 = F (x0) = x0 + ε(x0) be as in item 2. The second

inclusion in Eq. (18.3) follows from the simple computation:

‖F (x0 + h)− z0‖ = ‖h+ ε (x0 + h)− ε (x0)‖≤ ‖h‖+ ‖ε (x0 + h)− ε (x0)‖≤ (1 + α) ‖h‖ < (1 + α) r

for all h ∈ B (0, r) .To prove the first inclusion in Eq. (18.3) we must find, for every z ∈B(z0, (1 − α)r), there exists an h ∈ B (0, r) such that z = F (x0 + h) .If we let k = z − z0 ∈ B(0, (1 − α)r) we are looking for an h ∈ B (0, r)solving the equation,

k = F (x0 + h)− F (x0) = h+ ε(x0 + h)− ε(x0), (18.7)

i.e. we are looking for h ∈ B (0, r) such that h = S (h) where

S (h) := k + ε (x0)− ε (x0 + h) .

If h ∈ B (0, r) , then

‖S (h)‖ ≤ ‖k‖+ ‖ε (x0)− ε (x0 + h)‖ ≤ ‖k‖+ α ‖h‖≤ ‖k‖+ αr =: ρ < (1− α) r + αr = r.

and so any fixed point h ∈ B (0, r) of S is in fact in C (0, ρ) = B (0, ρ) –the closed unit ball centered at 0 with radius ρ. The above computation hasjust shown S : C (0, ρ)→ C (0, ρ) . Moreover for h, h′ ∈ C (0, ρ) we have

‖S (h)− S (h′)‖ = ‖−ε (x0 + h) + ε (x0 + h′)‖ ≤ α ‖h− h′‖

which show S : C (0, ρ) → C (0, ρ) is a contraction. Therefore the con-traction mapping principle of Theorem 17.3, S has a unique fixed point inC (0, ρ) and hence Eq. (18.7) has a solution.

3. Given x0 ∈ U, the first inclusion in Eq. (18.3) shows that z0 = F (x0) isin the interior of F (U) . Since z0 ∈ F (U) was arbitrary, it follows thatV = F (U) is open. The continuity of the inverse function has already beenproved in item 1.

For the remainder of this section letX and Y be two Banach spaces, U ⊂o X,k ≥ 1, and f ∈ Ck(U, Y ).

Lemma 18.4. Suppose x0 ∈ U, R > 0 is such that BX(x0, R) ⊂ U and T :BX(x0, R)→ Y is a C1 – function such that T ′(x0) is invertible. Let

α (R) := supx∈BX(x0,R)

∥∥T ′(x0)−1T ′ (x)− I∥∥L(x)

(18.8)

and ε ∈ C1(BX(0, R), X

)be defined by

ε (h) = T ′(x0)−1 [T (x0 + h)− T (x0)]− h (18.9)

so thatT (x0 + h) = T (x0) + T ′(x0) (h+ ε(h)) . (18.10)

Then ε(h) = o(h) as h→ 0 and

‖ε(h′)− ε(h)‖ ≤ α (R) ‖h′ − h‖ for all h, h′ ∈ BX(0, R). (18.11)

If α (R) < 1 (which may be achieved by shrinking R if necessary), then T ′ (x)is invertible for all x ∈ BX(x0, R) and

supx∈BX(x0,R)

∥∥∥T ′ (x)−1∥∥∥L(Y,X)

≤ 1

1− α (R)

∥∥T ′(x0)−1∥∥L(Y,X)

. (18.12)

Proof. By definition of T ′ (x0) and using T ′ (x0)−1

exists,

T (x0 + h)− T (x0) = T ′(x0)h+ o(h)

from which it follows that ε(h) = o(h). In fact by the fundamental theorem ofcalculus,

ε(h) =

∫ 1

0

(T ′(x0)−1T ′(x0 + th)− I

)hdt

but we will not use this here. Let h, h′ ∈ BX(0, R) and apply the fundamentaltheorem of calculus to t→ T (x0 + t(h′ − h)) to conclude

ε(h′)− ε(h) = T ′(x0)−1 [T (x0 + h′)− T (x0 + h)]− (h′ − h)

=

[∫ 1

0

(T ′(x0)−1T ′(x0 + t(h′ − h))− I

)dt

](h′ − h).


18.1 The Inverse Function Theorems 141

Taking norms of this equation gives

‖ε(h′)− ε(h)‖ ≤[∫ 1

0

∥∥T ′(x0)−1T ′(x0 + t(h′ − h))− I∥∥ dt] ‖h′ − h‖

≤ α (R) ‖h′ − h‖ .

It only remains to prove Eq. (18.12), so suppose now that α (R) < 1. Then byProposition 16.8, T ′(x0)−1T ′ (x) = I −

(I − T ′(x0)−1T ′ (x)

)is invertible and∥∥∥[T ′(x0)−1T ′ (x)

]−1∥∥∥ ≤ 1

1− α (R)for all x ∈ BX(x0, R).

Since T ′ (x) = T ′(x0)[T ′(x0)−1T ′ (x)

]this implies T ′ (x) is invertible and∥∥∥T ′ (x)

−1∥∥∥ =

∥∥∥[T ′(x0)−1T ′ (x)]−1

T ′(x0)−1∥∥∥ ≤ 1

1− α (R)

∥∥T ′(x0)−1∥∥

for all x ∈ BX(x0, R).

Lemma 18.5. Suppose U ⊂o X, k ≥ 1, T ∈ Ck(U, Y ), x0 ∈ U, and R > 0 suchthat BX(x0, R) ⊂ U. Then (see Figure 18.2) for all r ≤ R,

T (BX(x0, r)) ⊂ T (x0) + T ′ (x0)BX (0, (1 + α (r))r) . (18.13)

Proof. Let ε ∈ C1(BX(0, R), X

)be as defined in Eq. (18.9). Using Eqs.

(18.10) and (18.4),

T(BX(x0, r)

)= T (x0) + T ′ (x0) (I + ε)

(BX (0, r)

)(18.14)

⊂ T (x0) + T ′ (x0)BX (0, (1 + α (r)) r)

which proves Eq. (18.13).

Theorem 18.6 (Inverse Function Theorem I). Suppose U ⊂o X, k ≥ 1and T ∈ Ck(U, Y ) such that T ′ (x) is invertible for all x ∈ U.1 Further assumex0 ∈ U and R > 0 such that BX(x0, R) ⊂ U. By shrinking R is necessary, wealso assume that α (R) < 1 where, as in Eq. (18.8),

α (r) := supx∈BX(x0,r)

∥∥T ′(x0)−1T ′ (x)− I∥∥L(X)

.

Then

1 Suppose we only assumed that T ′ (x0) is invertible at x0 ∈ U. Since T ′ (x) dependscontinuously on x, T ′ (x) is invertible in some neighborhood of x0. So by shrinkingU if necessary we can arrange for T ′ (x) to be invertible on U.

1. for all r ≤ R,

T (x0) + T ′ (x0)BX (0, (1− α (r))r) ⊂ T (BX(x0, r)), (18.15)

see Figure 18.2.2. T : U → Y is an open mapping.3. Also T |BX(x0,R) : BX(x0, R)→ T (BX(x0, R)) is invertible and T |−1

BX(x0,R):

T(BX(x0, R)

)→ BX(x0, R) is a Ck – map.

x0T

B(x0, (1− α)r)

B(x0, (1 + α)r)

B(x0, r)

T (x0)

T (B(x0, r))

T (x0) + T ′(x0)B(0, (1 + α)r)

T (x0) + T ′(x0)B(0, (1− α)r)

Fig. 18.2. The nesting of T (BX(x0, r)) between T (x0) + T ′ (x0)BX (0, (1− α (r))r)and T (x0) + T ′ (x0)BX (0, (1 + α (r))r) .

Proof. Let ε ∈ C1(BX(0, R), X

)be as defined in Eq. (18.9).

1. By Eqs. (18.10) and (18.4),

T (x0) + T ′ (x0)BX (0, (1− α (r)) r)

⊂ T (x0) + T ′ (x0) (I + ε)(BX (0, r)

)= T

(BX (x0, r)

)which proves Eq. (18.15).

2. Notice that h ∈ X → T (x0) + T ′ (x0)h ∈ Y is a homeomorphism. The factthat T is an open map follows easily from Eq. (18.15) which shows thatT (x0) is interior of T (W ) for any W ⊂o X with x0 ∈W.

3. The fact that T |BX(x0,R) : BX(x0, R) → T (BX(x0, R)) is invertible with acontinuous inverse follows from Eq. (18.10) and Proposition 18.3. It now fol-lows from the converse to the chain rule, Theorem 1.8, that g := T |−1

BX(x0,R):

T(BX(x0, R)

)→ BX(x0, R) is differentiable and

g′ (y) = [T ′ (g (y))]−1

for all y ∈ T(BX(x0, R)

).


142 18 Inverse and Implicit Function Theorems

This equation shows g is C1. Now suppose that k ≥ 2. Since T ′ ∈Ck−1(B,L(X)) and i(A) := A−1 is a smooth map by Example 1.28,g′ = i T ′ g is C1, i.e. g is C2. If k ≥ 2, we may use the same argu-ment to now show g is C3. Continuing this way inductively, we learn g isCk.

Corollary 18.7 (Inverse Function Theorem II). Suppose U ⊂o X, k ≥ 1and T ∈ Ck(U, Y ) satisfies, T is injective, and T ′ (x) is invertible for all x ∈ U.Then V := T (U) is an open set and T−1 : V → U is a Ck – map such that(

T−1)′

(y) =[T ′(T−1 (y))

]−1for all y ∈ V.

Proof. Since differentiability and smoothness are local conditions these re-sults follow directly from Theorem 18.6.

Example 18.8. Let F : R2 → R2 be the smooth map defined by F (x, y) =ex (cos y, sin y)

trwhich is secretly the complex function, z → ez. Then

F ′ (x, y) =

[∂

∂xF (x, y) | ∂

∂yF (x, y)

]= ex

[cos y − sin ysin y cos y

]which is always invertible since detF ′ (x, y) = e2x 6= 0. Thus the inverse func-tion theorem guarantees that F is locally invertible at all points (x, y) ∈ R2.On the other hand it is not the case that F is globally invertible. First offF(R2)

= R2 \ 0 6= R2 so F is not onto. Of course we should have just viewedF as a map from R2 to R2 \ 0 from the start to avoid this problem. What ismore serious is that F is not one to one. Indeed, F (x, y) = F (x, y + n2π) forall n ∈ Z and (x, y) ∈ R2.

Exercise 18.1. Let M2 be the collection of real 2 × 2 matrices, B ∈ M2 bean invertible matrix, and let T : M2 → M2 be the map defined by T (A) :=A2 +BA for all A ∈M2. Use the inverse function theorem to show there is anopen neighborhood, V, of 0 ∈M2 such that for ever C ∈ V there is a solution,A ∈M2, to the equation,

A2 +BA = C.

18.2 Application to Local Existence of ODE’s

We can use this theorem to prove existence of solutions to ODE as follows.Suppose that we wish to solve

y (t) = f(y (t)) with y(0) = y0.

Notice that if y solves this equation and ys (t) := y(st) solves the equation

ys (t) = sf(ys (t)) with y(0) = y0.

Rewriting this equation in integral form gives

ys (t) = y0 + s

∫ t

0

f (ys(τ)) dτ.

We might Jazz this up by considering the following equation instead

ys (t) = ϕ (t) + s

∫ t

0

f(ys(τ)dτ

where s is a parameter and ϕ is a given function. Notice that when s = 0, thisequation is easy to solve, namely we take y0 (t) = ϕ (t) .

Let us now show that this equation has a solution in a small neighborhoodof s = 0 and y = y0 by invoking the inverse function theorem. To do this, let

F (s, y) := (s, t→ y (t)− s∫ t

0

f(y(τ)dτ)

and notice the differential of F is given by

Fs(s, y)a := (a, t→ −a∫ t

0

f(y(τ)dτ) and

Fy(s, y)z := (0, t→ z (t)− s∫ t

0

f ′(y(τ)z(τ)dτ)

and in particular

F ′(0, ϕ)(a, z (·)) = (a, t→ z (·)− a∫ ·

0

f(ϕ(τ)dτ)

which is clearly invertible. Now F (0, ϕ) = (0, ϕ) and hence by the inverse func-tion theorem, there exists for (r, ξ) in a neighborhood of (0, ϕ), a unique solutionto the equation, F (s, y) = (r, ξ). Take r = s and ξ = ϕ, and letting (s, ys(·))denote the solution to

F (s, ys) := (s, t→ ys (t)− s∫ t

0

f(ys(τ)dτ) = (s, ϕ).

Then

ys (t)− s∫ t

0

f(ys(τ)dτ = ϕ (t)


18.3 Implicit Function Theorem 143

and hence if we apply this to the case that ϕ (t) = y0, we learn that there is asolution, ys, (for s sufficiently small) to

ys (t) = y0 + s

∫ t

0

f(ys(τ)dτ.

This may be written in differential form as

ys (t) = sf(ys (t)) with ys(0) = y0.

Let y (t) = ys(t/s), then

y (t) = s−1ys(t/s) = f(ys(t/s)) = f(y (t)) with y(0) = y0.

This gives another proof of local existence to ordinary differential equations.

18.3 Implicit Function Theorem

Theorem 18.9 (Implicit Function Theorem). Suppose that X, Y, and Ware three Banach spaces, k ≥ 1, A ⊂ X × Y is an open set, (x0, y0) is apoint in A, and f : A → W is a Ck – map such f(x0, y0) = 0. Assume thatD2f(x0, y0) := D(f(x0, ·))(y0) : Y → W is a bounded invertible linear trans-formation. Then there exists an open set O ⊂ X × Y and U ⊂ X with x0 ∈ Uand (x0, y0) ∈ O satisfying; to each x ∈ U there exists a unique y ∈ Y suchthat (x, y) ∈ O and f (x, y) = 0. For x ∈ U we define u (x) = y where y is theunique y ∈ Y such that (x, y) ∈ O and f (x, y) = 0. Then u : U → Y is a Ck

map such that u (x0) = y0, (x, u (x)) ∈ O and f (x, u (x)) = 0 for all x ∈ U, and

u′ (x) = −D2f(x, u (x))−1D1f(x, u (x)) for all x ∈ U. (18.16)

Proof. By replacing f by (x, y) → D2f(x0, y0)−1f(x, y) if necessary, wemay assume with out loss of generality that W = Y and D2f(x0, y0) = IY .Define F : A→ X × Y by F (x, y) := (x, f(x, y)) for all (x, y) ∈ A. Notice that

DF (x, y) =

[I 0

D1f(x, y) D2f(x, y)

]which is invertible iff D2f(x, y) is invertible and if D2f(x, y) is invertible then

DF (x, y)−1 =

[I 0

−D1f(x, y) [D2f(x, y)]−1

[D2f(x, y)]−1

].

Since D2f(x0, y0) = I is invertible, the inverse function theorem guaranteesthat there exists a neighborhood O of (x0, y0) ∈ X × Y such that F (O) is an

open neighborhood of F (x0, y0) = (x0, 0) ∈ X × Y and F : O → F (O) is a Ck

– diffeomorphismWe now let

U := x ∈ X : (x, 0) ∈ F (O) .

It is easily checked that U is open in X and that x0 ∈ U since (x0, 0) ∈ F (O) .For x ∈ U there exists a unique point (x′, y) ∈ O such that

(x, 0) = F (x′, y) = (x′, f (x′, y)) =⇒ x′ = x and f (x, y) = 0.

Conversely if (x, y′) ∈ O and f (x, y′) = 0 then F (x, y′) = (x, 0) = F (x, y) andtherefore y = y′ as F : O → F (O) is one to one. The function u may now bewritten as u (x) = πY F−1 (x, 0) where πY : X × Y → Y is the projectionmap, πY (x, y) = y. Since u is the composition of the smooth map πY with theCk – map F−1, it follows hat u is also a Ck – map. Differentiating the identity,f (x, u (x)) = 0 implies

D1f (x, u (x)) +D2f (x, u (x))u′ (x) = 0



19

Holomorphic Functional Calculus

A good reference for this section is Taylor’s notes, banalg.pdf – see thechapter with this title.

Let T be a bounded (closed) operator on a Banach space X. Suppose thatσ (T ) is a disjoint union of sets Σknk=1 which are surrounded by contoursCknk=1 and Ω is an open subset of C which contains the contours and theirinteriors, see Figure 19.1.

C1

C2

C3

Σ1Σ2

Σ3

Fig. 19.1. The spectrum of T is in red, the counter clockwise contours are in black,and Ω is the union of the grey sets.

Given a holomorphic function, f, on Ω we let

f (T ) :=1

2πi

∮C

f (z)

z − Tdz :=

n∑k=1

1

2πi

∮Ck

f (z)

z − Tdz,

where 1z−T := (z − T )

−1and C = ∪nk=1Ck.

Let us observe that f (T ) is independent of the possible choices of contoursC as described above. One way to prove this is to choose ` ∈ B (X)

∗and notice

that

` (f (T )) =1

2πi

∮C

f (z) `(

(z − T )−1)dz

where f (z) `(

(z − T )−1)

is a holomorphic function on Ω \ σ (T ) . Therefore

12πi

∮Cf (z) `

((z − T )

−1)dz remains constant over deformations of C which

remain in Ω \ σ (T ) . As ` is arbitrary it follows that 12πi

∮Cf(z)z−T dz remains

constant over such deformations as well.

Theorem 19.1. The map H (Ω) 3 f → f (T ) ∈ B (X) is an algebra homomor-

phism satisfying the consistency criteria; if f (z) =∑Nm=0 amz

m is a polynomialthen

f (T ) =

N∑m=0

amTm.

More generally, ρ > 0 is chosen so that ‖T‖ < ρ and f ∈ H (D (0, ρ)) , then

f (T ) =

∞∑m=0

fm (0)

m!Tm.

Proof. It is clear that H (Ω) 3 f → f (T ) ∈ B (X) is linear in f. Nowsuppose that f, g ∈ H (Ω) and for each k let Ck be another contour around Σkwhich is inside Ck for each k. Then

f (T ) g (T ) =

(1

2πi

)2 n∑k,l=1

∮Ck

f (z)

z − Tdz

∮Cl

g (ζ)

ζ − Tdζ

=

(1

2πi

)2 n∑k,l=1

∮Ck

dz

∮Cl

dζf (z)

z − Tg (ζ)

ζ − T.

=:

(1

2πi

)2 n∑k,l=1

Akl

Using the resolvent formula,

1

z − T− 1

ζ − T=

ζ − z(z − T ) (ζ − T )

,

we find

146 19 Holomorphic Functional Calculus

Akl :=

∮Ck

dz

∮Cl

dζf (z)

z − Tg (ζ)

ζ − T

=

∮Ck

dz

∮Cl

dζf (z) g (ζ)1

ζ − z

(1

z − T− 1

ζ − T

)If k 6= l then for z ∈ Ck, ζ → g (ζ) 1

ζ−z1

z−T is analytic for ζ inside Cl andtherefore ∮

Cl

dζg (ζ)1

ζ − z1

z − T= 0. (19.1)

Similarly if ζ ∈ Cl, then z → f (z) 1ζ−z

1ζ−T is analytic inside of Ck and therefore∮

Ck

dzf (z)1

ζ − z1

z − T= 0.

From these last two identities and Fubini’s theorem it follows that Ak,l = 0 ifk 6= l.

Now suppose that k = l so that

Ak,k =

∮Ck

dz

∮Ck

dζf (z) g (ζ)1

ζ − z

(1

z − T− 1

ζ − T

).

For z ∈ Ck, ζ → g (ζ) 1ζ−z

1z−T is still analytic for ζ inside Cl and therefore Eq.

(19.1) holds for l = k and

Ak,k =

∮Ck

dζg (ζ)

ζ − T

∮Ck

dzf (z)

z − ζ= 2πi

∮Ck

dζg (ζ) f (ζ)

ζ − T.

Thus we have shown

f (T ) g (T ) =

(1

2πi

)2 n∑k=1

Ak,k

=1

2πi

n∑k=1

∮Ck

dζg (ζ) f (ζ)

ζ − T= (f · g) (T )

which shows that T → f (T ) is an algebra homomorphismIf f (z) is an entire function of z we may replace the contour C by z = ρeiθ

for any ρ sufficiently large so that σ (T ) ⊂ D (0, ρ) . Since |z| = ρ with ρ > ‖T‖we have

1

z − T=

1

z

1

1− T/z=

1

z

∞∑n=0

1

znTn =

∞∑n=0

1

zn+1Tn,

it now follows that

f (T ) =1

2πi

∮C

f (z)

z − Tdz =

1

2πi

∮|z|=ρ

f (z)

z − Tdz

=1

2πi

∮|z|=ρ

∞∑m=0

f (z)

zm+1Tmdz

=

∞∑m=0

(1

2πi

∮|z|=ρ

f (z)

zm+1dz

)Tm =

∞∑m=0

f (m) (0)

m!Tm.

Exercise 19.1. Give a power series proof.

Theorem 19.2 (Spectral Mapping Theorem). Keeping the same notationas above, f (σ (T )) = σ (f (T )) .

Proof. Suppose that µ ∈ σ (T ) and define

g (z) :=

f(z)−f(µ)

z−µ if z 6= µ

f ′ (µ) if z = µ

so that g ∈ H (U) and f (z)− f (µ) = (z − µ) g (z) . Therefore f (T )− f (µ) =(T − µ) g (T ) and so if f (µ) /∈ σ (f (T )) then f (T ) − f (µ) is invertible andtherefore T − µ would be invertible contradicting µ ∈ σ (T ) . Thus we haveshown f (σ (T )) ⊂ σ (f (T )) . Convereley if α /∈ f (σ (T )) then g (z) := 1

f(z)−α is

holomorphich on a niegborhood of σ (T ) . Since (f (z)− α) g (z) = 1 it followsthat (f (T )− α) g (T ) = I and thefefore α /∈ σ (f (T )) and we have shown[f (σ (T ))]

c ⊂ [σ (f (T ))]c, i.e. σ (f (T )) ⊂ σ (f (T )) .

Theorem 19.3 (Composition formula). If h = f g, then h (T ) = f (g (T ))if this expression makes sense.

Proof. A good reference for this section is Taylor’s notes, banalg.pdf – seethe chapter with this title.

Exercise 19.2. Give a power series proof.


References

1. Robert S. Strichartz, The Campbell-Baker-Hausdorff-Dynkin formula and solutionsof differential equations, J. Funct. Anal. 72 (1987), no. 2, 320–345. MR 89b:22011

topics in di erential geometry, 2015 - ucsd mathematicsbdriver/257a-winter2015/lecture...

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