The relation of being locally isometric for Riemannian manifolds is reflexive and transitive. Is it symmetric? Can you give me an example?

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    $\begingroup$ Take $\mathbb{R}^n$ and a manifold with a flat piece that is however not globally flat. $\endgroup$ Oct 3, 2013 at 0:39

2 Answers 2


This depends on the precise definition of the relation "$M$ and $N$ are locally isometric". For the first two definitions that sprang to my mind, the answer is "no."

If one defines "$M$ and $N$ are locally isometric if there exists a local isometry $f:M\to N$", then the relation is not symmetric. Consider, say, a closed complete hyperbolic manifold $M$. In particular, the projection map $\mathbb{H}^n\to M$ is a local isometry, but there is no local isometry $M\to \mathbb{H}^n$, for there is not even a local diffeomorphism $M\to\mathbb{H}^n$ (choose a reference point $r\in\mathbb{H}^n$ and consider $df$ at a point $p\in M$ where $d(f(p),r)$ is maximized).

(This argument works for any closed complete Riemannian manifold with infinite-diameter universal cover.)

If one defines "$M$ and $N$ are locally isometric if for any $m\in M$ there exists an $n\in N$ and neighborhoods $U_p\ni p$, $U_q\ni q$ such that $U_p$ is isometric to $U_q$", then the relation is also not symmetric. Daniel Fischer provides a counterexample in the comments.

  • $\begingroup$ Neal: Please replace "locally symmetric" with "locally isometric". $\endgroup$ Oct 3, 2013 at 4:03
  • $\begingroup$ @studiosus Hah, good catch. Thanks. In my head, "isometric" became "symmetric". $\endgroup$
    – Neal
    Oct 3, 2013 at 11:14

The Answer to this question is according to the following definition of "locally isometric Riemannian manifolds":

DEFFINITION: Let $M$ and $N$ be Riemannian manifolds. A differentiable mapping $$f:M\rightarrow N $$ is a "local isometry" at $p\in M$ if there is a neighborhood $U\subset M$ of $p$ such that $$f:U\rightarrow f(U) $$ is a diffeomorphism satisfying $$\langle u,v\rangle_{p}=\langle d{f}_{p}(u),d{f}_{p}(v)\rangle_{f(p)}$$

Now a Riemannian manifold $M$ is "locally isometric" to a Riemannian manifold $N$ if for every $p$ in $M$ there exists a neighborhood $U$ of $p$ in $M$ and a local isometry $$f:U \rightarrow f(U)\subset N$$

It is possible to see that the natural projection $$\pi :S^{2}\rightarrow P^{2}(\mathbb{R})$$ is a local isometry. In fact with the aid of this map along with different "suitable neighborhoods" one can say that $S^{2}$ is locally isometric to $P^{2}(\mathbb{R})$.

Now assume that the "locally isometric" is a symmetric relation. Then one would have $P^{2}(\mathbb{R})$ locally isometric to $S^{2}$. Consequently, one gets $P^2\mathbb{R}$ locally diffeomorphic to $S^2$.

But this makes a contradiction:

$S^{2}$ is an orientable manifold and hence due to the existence of the local diffeomorphisms, $P^{2}(\mathbb{R})$ should be an orientable manifold which is a contradiction (Since $P^{2}(\mathbb{R})$ is not orientable).

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    $\begingroup$ I believe what could not exist is a global diffeomorphism. A local diffeomorphism indeed exists, and is given by the projection. Just take $f = \pi^{-1}$ restricted to suitable neighborhoods. $\endgroup$ Jun 10, 2020 at 18:51

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