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I heard that there is a theorem of Liouville (Something like "Liouville's rigidity theorem") which states the following:

Every Riemannian isometry between open subset of $\mathbb{R}^n$ is affine.

Specifically I mean to $\mathbb{R}^n$ with the standard euclidean metric. (And my open subsets are Riemannian submanifolds of $\mathbb{R}^n$).

I would like to see a proof (or merely get a reference for a proof).

Asaf Shachar
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  • This can't be true without additional assumptions. For example, consider $(0,1)\cup (2,3)\subseteq \mathbb{R}$. As a Riemannian manifold, there is an isometry which fixes $(0,1)$ and flips $(2,3)$, but there is no affine isometry of $\mathbb{R}$ which accomplishes this. Perhaps the theorem is true if your open set is connected? – Jason DeVito - on hiatus Jul 20 '15 at 17:09

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As Jason Devito commented, there is an additional neede assumption of connectedness.

Here is a version of the theorem: Let $\Omega$ be an open connected subset of $\mathbb{R}^n$, and let $\phi:\Omega \to \mathbb{R}^n$ be a differentiable map, such that $d\phi(x) \in O(n) \, \,\forall x \in \Omega $. Then there exists a matrix $Q \in O(n)$ and a vector $v \in \mathbb{R}^n$ such that: $$ \phi(x)=Qx+v $$

A proof can be found in the book "Mathematical Elasticity, Volume 1L Three-dimensional elasticity" by Phillippe G.Cairlet.

The proof is quite elementary, the main idea is to shows $d\phi$ is locally constant, hence (since the domain is connected) globally constant.

Asaf Shachar
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