# Does the possibility of linear coordinate changes imply that the manifold is Euclidean?

Question. Let $M$ be a smooth manifold admitting an atlas $\mathcal{A}$ (i.e. a collection of coordinate charts that covers the whole manifold) with the following property:

for every pair $(U, \mathbf{x}),\ (U', \mathbf{x}')\in \mathcal{A}$, the transition function $$\mathbf{x'}\circ\mathbf{x}^{-1}\colon \mathbf{x}(U\cap U')\subset \mathbb{R}^n\to \mathbf{x}'(U\cap U')\subset\mathbb{R}^n$$ is (the restriction of) an isometry of $\mathbb{R}^n$ onto itself.

Is $M$ diffeomorphic to $\mathbb{R}^n$?

Motivation for this conjecture comes from the fact that in such a manifold one can define a Riemannian inner product in exactly the same way as one does in Euclidean space: $$\text{if }\mathbf{v}=v^\alpha\partial_\alpha,\ \mathbf{w}=w^\beta\partial_\beta,\quad \text{then } g(\mathbf{v}, \mathbf{w})=\sum_\alpha v^\alpha w^\alpha.$$ This definition is well posed because it is independent of the chosen coordinate chart in the atlas $\mathcal{A}$. The Riemannian manifold $(M, g)$ is flat and locally isometric to $\mathbb{R}^n$. I don't know if this extends to a global diffeomorphism, though.

Secondary question. What if the transition functions $\mathbf{x}'\circ\mathbf{x}^{-1}$ are merely affine transformations of $\mathbb{R}^n$ onto itself?

In this case, the previous assignment for a Riemannian metric $g$ is ill-posed. However, one can still define a flat affine connection as follows: $$\nabla_\mathbf{w} \mathbf{v}= w^\alpha\frac{\partial v^\beta}{\partial x^\alpha} \frac{\partial}{\partial x^\beta}.$$ It would be interesting to know if this suffices to have a global diffeomorphism onto $\mathbb{R}^n$.

• Don't bother with explicit transition functions, it is OK. Even the $\mathbb{S}^1$ example is fine for me. The only question I have is: is there any relationship between the Riemannian metric $g(v, w)=\sum_\alpha v^\alpha w^\alpha$ defined above and the actual Riemannian metric of the manifold (which is the one coming from the embedding into Euclidean space)? Take the case of $\mathbb{S}^1$. – Giuseppe Negro Jul 14 '14 at 16:08