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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.

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1 Answer 1

$x$ 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.

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