Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I think I am a little confused about the notion of a local isometry of Riemannian manifolds.

Let's say I have a manifold $(M,g)$ where $g$ is the Riemannian metric. Take a chart $x:U \rightarrow x(U)$ where $U$ is an open subset of $M$ and $x(U)$ is an open subset of $\mathbb{R}^n$.

We can use the map $x^{-1}$ to pull back the metric on $U$ to $x(U)$. What I mean is, defining a new metric $h$ on $x(U)$ by $$ h_q(u,v) = g_{x^{-1}(q)}(Dx^{-1}(q)u,Dx^{-1}(q)v). $$ Now, $x$ is an isometry of $U$ and of $x(U)$ equipped with metrics $g$ and $h$. I think this should not be possible, because isometries can't exist between manifolds of different curvature, and $x(U)$ is flat (as a subspace of $\mathbb{R}^n$) and $U$ doesn't have to be flat.

share|cite|improve this question

$x$ is indeed an isometry between $(U,g)$ and $(x(U),h)$. Subspaces of $\mathbb R^n$ are only necessarily flat if you consider these with the metric induced by the usual metric on $\mathbb R^n$, but here you are using a different metric $(h)$ which need not be flat.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.