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.

  • 2
    $\begingroup$ 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 $f \circ \exp_q = \exp_{f(q)} \circ T_q f = \exp_q \circ T_q f=\exp_q$. $\endgroup$
    – Laz
    Dec 7, 2019 at 20:46

1 Answer 1


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 $f=\widetilde{f}$.

The result remains true if we replace "isometries" by "pseudo-Riemannian submersions", provided that the fundamental tensors associated to both submersions are also equal (that's the price one pays for freeing themselves of the dimension constraint). See the paper The Fundamental Equations of a Submersion, by Barrett O'Neill.

  • $\begingroup$ 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!!! $\endgroup$ May 11, 2020 at 15:23

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