# locally isometric is not a symmetric relation.

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

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

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.

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

• 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. Jun 10, 2020 at 18:51