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Let $(M, \nabla)$ be a Riemannian manifold and $\gamma: [0,1] \to M$ a smooth curve and let $X$ be a vector field. I would like to prove

$$|X(\gamma(1))-X_{\parallel} | \le L(\gamma) \| \nabla X \|_{\infty}$$

where $X_{\parallel}$ is the parallel transport of $X(\gamma(0))$ onto $T_{\gamma(1)}M$ along $\gamma$.

I have no idea of how to do that.

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  • $\begingroup$ What is your background? What do you know about parallel transport? $\endgroup$ – Amitai Yuval Mar 1 at 11:42
  • $\begingroup$ I had a course in riemannian geometry. I know parallel transport is unique,, it is an isomorphism also isometric if the connection is the Levi Civita one. $\endgroup$ – Bremen000 Mar 1 at 11:45
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For $t\in[0,1],$ let $X_\|(t)$ denote the parallel transport of $X(\gamma(t))$ into $T_{\gamma(1)}M$ along $\gamma$. By the relation between the connection and parallel transport, we have $$\frac{d}{dt}X_\|(t)=P_{t\to1}\nabla_{\dot{\gamma}(t)}X,$$where $P_{t\to1}$ denotes parallel transport from $\gamma(t)$ to $\gamma(1)$ along $\gamma$. Hence, $$\left|\frac{d}{dt}X_\|(t)\right|\le\left|\dot{\gamma}(t)\right|\cdot\|\nabla X\|_\infty,$$and the desired inequality follows by integration.

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  • $\begingroup$ Thank you for your answer, I'm little bit confused about the definition of $\| \nabla X\|_{\infty} $. Could you please clarify it to me? $\endgroup$ – Bremen000 Mar 1 at 13:32
  • $\begingroup$ @Bremen000 Do you know what $\nabla X$ is? $\endgroup$ – Amitai Yuval Mar 1 at 17:03
  • $\begingroup$ The map $TM \to TM$ s.t. $ v \mapsto \nabla_v X $ $\endgroup$ – Bremen000 Mar 1 at 17:05
  • $\begingroup$ @Bremen000 So it has an operator norm at every point, and you can take the supremum of that. $\endgroup$ – Amitai Yuval Mar 1 at 17:07
  • $\begingroup$ Ok, now it is clear to me. Thank you! $\endgroup$ – Bremen000 Mar 1 at 17:08

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