Assignment 2; Solutions

# Assignment 2; Solutions - PMATH 465/665 Riemannian Geometry...

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PMATH 465/665: Riemannian Geometry Assignment 2; Solutions [1] Let G be a Lie group, and let X be a left-invariant vector field on G . [a] Let Θ be the flow of X . Recall from Assignment #1, problem #1, we proved a lemma that said: if there exists an ε > 0 such that, for any point g G , the integral curve Θ ( g ) starting at g is defined for all t ( - ε, ε ), then X is complete. Consider the point p = e , the identity element of G . There exists some ε > 0 such that Θ ( e ) is defined on ( - ε, ε ). We claim that Θ ( g ) is defined on ( - ε, ε ). This would prove that X is complete. Define γ : ( - ε, ε ) G by γ ( t ) = g · Θ ( e ) ( t ) = ( L g Θ ( e ) )( t ), where · is the group multiplication in G . We have γ (0) = g · Θ ( e ) (0) = g · e = g . Furthermore, using the fact that Θ ( e ) is an integral curve of X , and that X is left-invariant, we find that γ 0 ( t 0 ) = d dt t = t 0 γ ( t ) = d dt t = t 0 ( L g Θ ( e ) )( t ) = ( ( L g ) * ) Θ ( e ) ( t 0 ) ( ( e ) ) 0 ( t 0 ) ) = ( ( L g ) * ) Θ ( e ) ( t 0 ) ( X Θ ( e ) ( t 0 ) ) = X g · Θ ( e ) ( t 0 ) = X γ ( t 0 ) . Thus γ is an integral curve of X and γ (0) = g , so γ = Θ ( g ) by uniqueness. Hence X is complete. [b] All integral curves are smooth maps from their domain into the manifold. So we just need to prove the group homomorphism property. Since X is left-invariant, its flow Θ is a global flow by part [a]. Now for any t, s R , we have that Θ ( e ) ( t + s ) = Θ t + s ( e ) = Θ t s ( e )) = Θ t ( e ) ( s )) = Θ ( e ) ( s )) ( t ) . In part [a] we showed that Θ ( g ) = g · Θ ( e ) for any g G , so we have Θ ( e ) ( t + s ) = Θ ( e ) ( s )) ( t ) = Θ ( e ) ( s ) · Θ ( e ) ( t ) . This says that Θ ( e ) is a group homomorphism. [c] Let γ : R G be a 1-parameter subgroup of G . That is, γ is smooth and a group homomor- phism. We want to show that γ is an integral curve of a left-invariant vector field. Since γ is a homomorphism, we have γ (0) = e G . Let X e = γ 0 (0) T e G , and define the left-invariant vector field X on G as usual by X g = ( ( L g ) * ) e ( X e ) T g G . We need to show that γ is an integral curve of X . Let t 0 R . By the chain rule, the fact that γ is a homomorphism, the fact that γ 0 (0) = X e , and the left-invariance of X , we have d dt t = t 0 γ ( t ) = d ds s =0 γ ( t 0 + s ) = d ds s =0 γ ( t 0 ) · γ ( s ) = d ds s =0 ( L γ ( t 0 ) γ )( s ) = ( ( L γ ( t 0 ) ) * ) γ (0) ( γ 0 (0)) = ( ( L γ ( t 0 ) ) * ) γ (0) ( X e ) = X γ ( t 0 ) · e = X γ ( t 0 ) , which is what we wanted to show. [2] Let X be a vector field on M , and let Θ be the flow of X . Then (Θ t ) * : T * Θ t ( p ) M T * p M is an isomorphism whenever Θ t ( p ) is defined. Given p M and a ( k, 0) tensor σ on M , we define the Lie derivative ( L X σ ) p to be the ( k, 0) tensor given by ( L X σ ) p = lim t 0 t ) * ( σ Θ t ( p ) ) - σ p t = d dt t =0 t ) * ( σ Θ t ( p ) ) . 1

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[a] Let k = 0, and let σ = f be a smooth function on M . We want to prove that L X f = Xf . Recall first that in the general case, if σ = 1 ⊗ · · · ⊗ α k , where each α j is a 1-form, then h * ( σ ) = ( f h ) h * ( α 1 ) ⊗ · · · ⊗ h * ( α k ). Thus, when k = 0, we get h * f = f h . [Another way to argue this is as follows. Given a linear map A : V W between vector spaces, it induces a linear map k A : k V → ⊗ k W , usually also simply denoted by A , by A ( v 1 ⊗· · ·⊗ v k ) = ( Av 1 ) ⊗ · · · ⊗ ( Av k ). When k = 0, we have 0 V = 0 W = R , and 0 A = Id : R R .] Thus we have shown that (Θ t ) * ( σ Θ t ( p ) ) = ( σ Θ t )( p ) = ( σ Θ ( p ) )( t ) when σ = f is a function.
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