# Inequality involving parallel transport and vector fields

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.

• What is your background? What do you know about parallel transport? – Amitai Yuval Mar 1 at 11:42
• 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. – Bremen000 Mar 1 at 11:45

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.
• Thank you for your answer, I'm little bit confused about the definition of $\| \nabla X\|_{\infty}$. Could you please clarify it to me? – Bremen000 Mar 1 at 13:32
• @Bremen000 Do you know what $\nabla X$ is? – Amitai Yuval Mar 1 at 17:03
• The map $TM \to TM$ s.t. $v \mapsto \nabla_v X$ – Bremen000 Mar 1 at 17:05