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Given an arbitrary Riemannian manifold $\mathcal{M}$ with metric $g$, is it always possible to choose a diffeomorphism $\phi:\mathcal{M}\rightarrow\mathcal{M}$ such that the pullback $\phi^*g$ is flat?

If not, is it at least possible locally on some open set of $\mathcal{M}$? What if I only consider 2D manifolds?

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  • $\begingroup$ yes, locally it is always possible. in 2D the basic obstruction is the Euler number, but you can always decide how to distribute the curvature among the various open sets or charts. for example, for a sphere you can put all curvature at a point and keep the remaining manifold flat everywhere. or you can always choose your metric to be $dz d\bar{z}$ at a point without any assumptions about the curvature at that point, etc. $\endgroup$ Commented Aug 16, 2019 at 14:48
  • $\begingroup$ ( .. other than the assumption that the curvature is $C^\infty$ differentiable at that point) $\endgroup$ Commented Aug 16, 2019 at 15:01

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It is not always possible. There are some topological obstuctions. The Euler class of a closed manifold endowed with a flat metric vanishes identically. So for example the sphere $S^2$ cannot be endowed with a flat metric.

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  • $\begingroup$ If you puncture a hole in the sphere would you then be able to choose a diffeomorphism on the set to flatten the metric? That's my real question. I thought there might be topological obstructions which is why I added the second paragraph. Thank you for reminding me what they were. $\endgroup$
    – octonion
    Commented Aug 9, 2019 at 19:31
  • $\begingroup$ @octonion Locally any neighborhood is diffeomorphic to an open set in $\mathbb{R}^{n}$. That's the definition of a manifold. $\endgroup$
    – Valac
    Commented May 9, 2022 at 3:57

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