TITLE: locally isometric is not a symmetric relation.
QUESTION [6 upvotes]: The relation of being locally isometric for Riemannian manifolds is reflexive and transitive. Is it symmetric? Can you give me an example?

REPLY [6 votes]: 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).