$\mathbf {The \ Problem \ is}:$ Let $f:M\to M$ be a Riemannian isometry of a Riemannian manifold $(M,g).$
Let $F=\{p\in M\mid f(p)=p\}$ and $V_p=\{v\in T_pM\mid df_p(v)=v\}.$
Show if $p$ and $q$ lies in the same connected component of $F,$ then $\operatorname{dim}V_p=\operatorname{dim}V_q.$
Further, show each connected component of $F$ is a totally geodesic submanifold.
$\mathbf {My \ approach}:$ We know for each point $p\in F,$ there exists $\epsilon>0$ such that $exp_p(B_{\epsilon}(0_p)\cap V_p)=B_{\epsilon}(p)\cap F…..(1)$
Then when $p$ and $q$ are in same connected component , there exists an $\epsilon>0$ such that $(1)$ holds for both $p$ and $q$.
But, I can't get any lead after this ? Along what line shall I proceed ?
Thanks in advance for any hint .
Best Answer
First, note that $V_p = \varnothing$ unless $p\in F$, in which case $T_pF = V_p$. Now, if $p,q\in F$ are in the same connected component of $M$, fix a curve $\gamma\colon [0,1]\to M$ with $\gamma(0)=p$ and $\gamma(1)=q$, and consider the parallel transport operator $\Pi_\gamma\colon T_pM \to T_qM$. Since $$\Pi_\gamma\circ {\rm d}f_p = {\rm d}f_q\circ \Pi_\gamma \tag{1},$$we have that if $v\in V_p$, then ${\rm d}f_q(\Pi_\gamma(v)) = \Pi_\gamma({\rm d}f_p(v)) = \Pi_\gamma(v)$, so that $\Pi_\gamma(v) \in V_q$. Hence $\Pi_\gamma|_{V_p}\colon V_p \to V_q$ is an isomorphism (namely, this gives that $\dim V_p\leq \dim V_q$, and switching the roles of $p$ and $q$ and $\gamma$ with its reverse $\gamma^{\leftarrow}$ gives that $\dim V_q\leq \dim V_p$).
As for why $F$ is totally geodesic, it suffices to show that geodesics of $M$ which start at $F$ and tangent to $F$ remain in $F$ at least for small time. So, let $p\in F$ and $\alpha\colon (-\varepsilon,\varepsilon)\to M$ be a geodesic with $\alpha(0) = p$ and $v\doteq\alpha'(0)\in V_p$. Then the composition $f\circ \alpha$ is also a geodesic, with initial conditions $$(f\circ \alpha)(0) = f(\alpha(0)) = f(p) = p\quad\mbox{and}\quad (f\circ \alpha)'(0) = {\rm d}f_{\alpha(0)}(\alpha'(0)) = {\rm d}f_p(v) = v = \alpha'(0),$$so that $f\circ \alpha = \alpha$ and hence $\alpha(t) \in F$ for all $t\in (-\varepsilon,\varepsilon)$.