# Relation between exponential map and parallel transport [closed]

I'm starting to learn Riemannian geometry and have a question.

Let $\mathcal{M}$ be a Riemannian manifold, $p \in \mathcal{M}$; $\tau_{p}^{q}$ be a parallel transport from $p$ to $q$ and $\operatorname{exp}_p$ be an exponential map. The question is: given a smooth curve $\gamma : [a, b] \to \mathcal{M}$ such that $\gamma([a, b]) \subset \mathcal{U}$, where $\mathcal{U}$ is some neighborhood of $p$ where $\operatorname{exp}_p, \tau_{p}^{(\cdot)}$, are defined, is it true that \begin{equation} \left( \frac{\mathrm{d}}{\mathrm{d}t} \operatorname{exp}_p^{-1}\gamma \right)(t) = \tau_{\gamma(t)}^{p}\dot{\gamma}(t) \,? \end{equation}

## closed as off-topic by Alex M., HK Lee, user91500, TheSimpliFire, GibbsAug 31 '18 at 19:32

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• It certainly works in $\Bbb R^n$. What have you tried doing? – Olivier Bégassat Nov 28 '13 at 15:01
• You mean during what process arose this question or what have I tried to do to prove it? :) – Artem Malykh Nov 28 '13 at 15:11
• Your proof attempts. – Olivier Bégassat Nov 28 '13 at 15:24
• Let $g$ be a metric tensor on $\mathcal{M}$. I've tired to use countiniuty of $g$ together with the fact that for any $p \in \mathcal{M}$ and for $u, v \in T_p\mathcal{M}$, if $g(u, u)=g(v,v)$ and angle between $u$ and $v$ is zero, then $u=v$. – Artem Malykh Nov 28 '13 at 20:10
• Possible duplicate of Difference between parallel transport and derivative of the exponential map – Alex M. Aug 30 '18 at 4:37