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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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    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.2012-04-06

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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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    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?2012-04-07