locally isometric is not a symmetric relation.

Question

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

Answer 1

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.