# Nonexistence of local isometry between equidimensional Riemannian manifolds

Recall that all inner product spaces of the same dimension are isometric.

For example, if $(M,\mathrm{g})$ and $(N,\mathrm{h})$ are Riemannian manifolds of the same dimension, then $(T_pM,\mathrm{g}_p)$ and $(T_qN,\mathrm{h}_q)$ are isometric inner product spaces. Let $\beta:T_pM\to T_qN$ be a (linear) isometry.

Question: Why isn't $\exp_q\!\circ\,\beta\circ\exp_p^{-1}$ a local isometry wherever it is defined?

To clarify, I know that it isn't a local isometry, at least in general. If it was, then I'd have a local isometry between neighborhoods of every point on each equidimensional Riemannian manifold, which is clearly absurd. I'm looking for an explanation as to why this composition is not a local isometry. I'm not looking for a counterexample.

I think my mental impasse comes from Gauss' lemma, but I'm not sure. Using the chain rule, shouldn't $\exp_{q*}\!\circ\,\beta\circ\exp_{p*}^{-1}$, as a composition of linear isometries, itself be a linear isometry?

• You are talking about the composition of the linear map $\beta$ with two maps between a tangent space and the manifold itself. Only $\beta$ is a linear isometry, the exponential maps are not. (Or maybe, since you are mentioning the chain rule, you are only thinking of the differential at the base point which would indeed be an isometry. However, this argument does not work at any other point.) – Lukas Geyer Jun 28 '15 at 5:32