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I am really stuck on the following question.

Let $ \ \gamma : I \longrightarrow M \ $ be a non-constant (i.e $ \ \gamma'\ $ is not identically zero) geodesic. Show that a reparametrization $\ \gamma \circ h : J \longrightarrow M \ $ is a geodesic if and only if $ \ h: J \longrightarrow I \ $ is of the form $h(t) = at+b \ $ with $ \ a, b \in \mathbb{R}$.

Does anyone have any idea how to prove this?

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What are $I$, $J$ and $M$? –  LostInMath Apr 6 '12 at 22:02
    
I am sorry I forgot to add this! I and J are just intervals contained in $\mathbb{R}$. While $M$ is a Riemannian manifold equipped with the Levi-Civita connection. –  Alex Kite Apr 6 '12 at 22:13
1  
If the reparametrization is a geodesic, then use the fact that $\nabla_{(\gamma\circ h)'}(\gamma\circ h)' = 0$ to derive an equation that $h$ must satisfy. –  treble Apr 6 '12 at 22:31

1 Answer 1

up vote 1 down vote accepted

For $h: J\rightarrow I$ such that $s=h(t)$, we use $'$ to denote $\frac{d}{dt}$ and $\cdot$ to denote $\frac{d}{ds}$. Then we have $$(\gamma\circ h)'(t)=h'(t)\dot{\gamma}(s)$$ by chain rule. Therefore, we have $$\nabla_{(\gamma\circ h)'(t)}(\gamma\circ h)'(t)= \nabla_{h'(t)\dot{\gamma}(s)}\Big(h'(t)\dot{\gamma}(s)\Big)$$ $$= h'(t)\nabla_{\dot{\gamma}(s)}\Big(h'(t)\dot{\gamma}(s)\Big)= h'(t)^2\nabla_{\dot{\gamma}(s)}\dot{\gamma}(s)+\frac{d}{ds}\big(h'(t)\big)\dot{\gamma}(s).$$ Since $\gamma(s)$ is geodesic, $\nabla_{\dot{\gamma}(s)}\dot{\gamma}(s)=0$, which implies that $$\tag{1}\nabla_{(\gamma\circ h)'(t)}(\gamma\circ h)'(t)=\frac{d}{ds}\big(h'(t)\big)\dot{\gamma}(s).$$

Therefore, if $(\gamma\circ h)(t)$ is geodesic, i.e. $\nabla_{(\gamma\circ h)'(t)}(\gamma\circ h)'(t)=0$ if and only if $$\tag{2}\frac{d}{ds}\big(h'(t)\big)=0.$$ Integrating it, we have $h'(t)=a$, which implies that $h(t)=at+b$ for some constant $a, b\in\mathbb{R}$. Conversely, if $h(t)=at+b$ for some constant $a, b\in\mathbb{R}$, then $(2)$ is satisfied, which implies that the expression in $(1)$ is zero, i.e. $(\gamma\circ h)(t)$ is geodesic.

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Thank you so much for a such a clear and concise answer! This makes perfect sense. –  Alex Kite Apr 7 '12 at 16:05
    
Actually there is one thing. For $h'(t)\nabla_{\dot{\gamma}(s)} \big(h'(t)\dot{\gamma}(s) \big )$ you have to use the product rule and does not this give you $h'(t) \big (h'(t) \nabla_{\dot{\gamma}(s)} \dot{\gamma}(s)) + \frac{d}{ds}(h'(t)) \dot{\gamma}(s) \big )$? So there should be a $h'(t)$ in the second term of your answer? –  Alex Kite Apr 7 '12 at 17:12

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