Composition of exponential with an isometry

Question

I have trouble understanding the following equation.

We are given an isometry $$F: M \rightarrow M$$ on a Riemannian manifold $M$.

Why does the following hold true? $$ F \circ exp_p = exp_{F(p)} \circ dF_p $$ Does someone have any ideas? Thank you.

Answer 1

Let $M$ be a Riemannian manifold and for $p \in M$, $D(p)$ be the open subset of the tangent space $T_{p}M$ such that :

$$ D(p) = \lbrace v \in T_{p}M, \; \gamma_{v}(1) \; \text{exists} \rbrace $$

where $\gamma_{v}$ is the unique maximal geodesic of $M$ with initial conditions: $\gamma_{v}(0) = p$ and $\dot{\gamma_{v}}(0) = v$. $D(p)$ is the definition domain of $\mathrm{exp}_{p}$.

Let $p \in M$ and $v \in D(p)$. Since $F$ is an isometry, it is distance-preserving and it sends geodesics of $M$ onto geodesics of $M$ (this can be seen using the length minimizing property of geodesics). Hence, $\eta \, : t \, \mapsto \, F\big( \mathrm{exp}_{p}(tv) \big)$ is a geodesic of $M$. It satisfies : $\eta(0) = F(p)$ and $\dot{\eta}(0) = dF(p)(v)$. The curve $t \mapsto \mathrm{exp}_{F(p)}\big( tdF(p)(v) \big)$ is another geodesic of $M$ which satisfies the same set of initial conditions. By unicity, these two curves are equal.