Fixed Points Set of an Isometry

Question

I'm reading Kobayashi's "Transformation Groups In Riemannian Geometry". I'm trying to understand the proof of the following theorem:

Theorem. Let $M$ be a Riemannian manifold and $K$ any set of isometries of $M$. Let $F$ be the set of points of $M$ which are left fixed by all elements of $K$. Then each connected component of $F$ is a closed totally geodesic submanifold of $M$.

In the proof first we consider $p\in F$ and we take $V$ to be the subspace of $T_pM$ of the vectors which are fixed by all the elements of $K$. The we take $U^{*}$ a neighborhood of the origin in $T_pM$ such that $\mathrm{exp}_{p}: U^{*} \rightarrow M$ is an injective diffeomorphism. The Kobayashi says we can further assume that $U=exp_{p}(U^{*})$ is convex (I'm assuming he means geodesically convex). Then my problem arises: He says it's easy to see that $U\cap F = \mathrm{exp}_p(U^{*}\cap V)$ but I have no clues as to how to prove this. Please help!

Answer 1

It is easy to see that inclusion $U\cap F \supset \exp_p(U^*\cap V)$: geodesics are uniquely specified by an initial point and an initial velocity. So if an isometry $\mathfrak{f}:M\to M$ fixes $\mathfrak{f}(p) = p$ and $d\mathfrak{f}(p)(v) = v$, the entire geodesic $\mathfrak{f}(\gamma_{p,v}) = \gamma_{p,v}$. (Because isometries send geodesics to geodesics, and the two geodesics have the same initial point and initial velocity.)

On the other hand, suppose there exists a point $q$ in $U\cap F$ such that unique minimizing geodesic $\gamma$ joining $p$ to $q$ is not in $F$ ($\gamma$ exists since $U$ is a normal neighbourhood). Then by definition $\mathfrak{f}(\gamma) \neq \gamma$ for some isometry, yet this isometry fixes the end-points $p,q$. This contradicts the assumption that we are in a normal neighbourhood (uniqueness of minimizing geodesics). Hence $U\cap F \subset \exp_p(U^*\cap V)$.