# Isometry $f:M\to M$ has a fixed point $p$ with $df_p=\text{id} \Rightarrow f=\text{id}$

I'm trying to prove whether or not the following statement is true:

If $$(M,g)$$ is a connected Riemannian manifold and $$f:M\to M$$ is an isometry, then $$f=\text{id}_M\Leftrightarrow$$ there is some $$p\in M$$ with $$f(p)=p$$ and $$df_p=\text{id}$$.

$$(\Rightarrow)$$ is obvious.

For $$(\Leftarrow)$$, I know how to prove it when $$(M,g)$$ is complete: for any $$q\in M$$ there is a geodesic $$\gamma$$ joining $$p,q$$. Since $$f$$ is an isometry, $$f\circ\gamma$$ is a geodesic starting at $$f\circ\gamma(0)=f(p)=p$$ with velocity $$(f\circ\gamma)'(0)=df_p(\gamma'(0))=\gamma'(0)$$ so $$f\circ\gamma=\gamma$$ by uniqueness of geodesics. In particular, $$f(q)=q$$, which means $$f=\text{id}_M$$ since $$q$$ is arbitrary.

For the general case, I don't know how to prove it or find a counterexample.

• It is true. For the other implication, prove that $C=\{q\in M | f(q)=q, T_q f = Id_{T_q M}\}$ is both open, closed and nonempty. For the openness, use that if $q\in C$, then $f \circ \exp_q = \exp_{f(q)} \circ T_q f = \exp_q \circ T_q f=\exp_q$.
– Laz
Dec 7, 2019 at 20:46

If the manifold is connected, the goal is to show that the agreement set $$A = \{p \in M \mid f(p) = p \quad\mbox{and}\quad {\rm d}f_p = {\rm Id}_{T_pM}\}$$is non-empty, open, and closed. It is non-empty by assumption. Closed by smoothness of $$f$$ (which implies continuity of $$f$$ and of its differential). It remains to show that $$A$$ is open. So let $$p \in A$$, and let $$D_p\subseteq T_pM$$ be an open and starshaped domain for which $$\exp_p\colon D_p \to \exp_p[D_p]\subseteq M$$ is a diffeomorphism. Then $$p \in \exp_p[D_p]$$ is open, and we can show that $$\exp_p[D_p] \subseteq A$$.
If $$q \in \exp_p[D_p]$$, there is $$v \in D_p$$ such that the geodesic segment $$\gamma\colon [0,1] \to M$$ given by $$\gamma(t) = \exp_p(tv)$$ always lies inside $$\exp_p[D_p]$$, with $$\gamma(1) =q$$. By the uniqueness argument you mentioned yourself, we have that $$f\circ \gamma = \gamma$$. This in particular shows that $$f(q) = q$$. But since $$f$$ is an isometry, we have that
$$f\circ \exp_p = \exp_{f(p)} \circ \,{\rm d}f_p\implies f \circ \exp_p = \exp_q.$$Differentiating at $$0$$ and using that $${\rm d}(\exp_p)_0 = {\rm Id}_{T_pM}$$ and $${\rm d}(\exp_q)_0 = {\rm Id}_{T_qM}$$ gives that $${\rm d}f_q={\rm Id}_{T_qM}$$ as well.
The real consequence here is that if you have two isometries $$f,\widetilde{f}\colon M \to M'$$ between (connected) pseudo-Riemannian manifolds (positiveness of the metric was irrelevant here), then if there is $$p \in M$$ with $$f(p) = \widetilde{f}(p)$$ and $${\rm d}f_p = {\rm d}\widetilde{f}_p$$, we conclude that $$f=\widetilde{f}$$.
• Can you give the details on "Differentiating at $0$ and using that ${\rm d}(\exp_p)_0 = {\rm Id}_{T_pM}$ and ${\rm d}(\exp_q)_0 = {\rm Id}_{T_qM}$ gives that ${\rm d}f_q={\rm Id}_{T_qM}$ as well."? I'd really appreciate it!!! May 11, 2020 at 15:23