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Suppose that we have a Riemannian Manifold $(M,g)$ whose curvature vanishes in an open neighborhood U of a point p.

When does this imply that the metric is Flat ?

In particular, does it happen under some special or exceptional holonomy like $G_2$ ?

Are there any topological conditions on $M$ to make it Flat ?

Note that a manifold is Flat if and only if the restricted holonomy $Hol^0(M,g)=0$. This condition means that contractible loops have zero holonomy. Our case is somehow 'close' to this condition because if the loop is sufficiently small then its holonomy is zero. It actually includes any small loop contractible or not. So it is a strong condition. I guess Ricci-Flatness or something like that implies the global result. References also appreciated.

(Flat means the curvature is zero, in case of ambiguity.)

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There must be a glitch in the formulation because the answer to the first question is an obvious "no": take a round sphere and flatten it around the North Pole. Also note that every sufficiently small loop is contractible.

As to topological restrictions, by Bieberbach's theorem a compact manifold admits a flat Riemannian metric if and only if it can be finitely covered by a torus.

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    $\begingroup$ But what if the assumption is that the holonomy is, say, at most $G_2$? Or the metric is Kahler? Is the metric necessarily real analytic and the result follows? $\endgroup$
    – Deane Yang
    Aug 16, 2018 at 20:41
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    $\begingroup$ @DeaneYang, the question would make a lot more sense if it were : can a compact Riemannian manifold with U(n), SU(n), ..., Spin(7) holonomy be flat in some open set without being globally flat? Let's wait for Robert for that one ;-) $\endgroup$
    – alvarezpaiva
    Aug 16, 2018 at 21:09
  • $\begingroup$ Exactly. That is what I meant before the second question mark. Thanks for detailing though. Also real analyticity is a good point. Einstein metrics are real analytic as far as I know. So Ricci-flatness (e.g. as in $SU_n, G_2, Spin_7$ cases) should imply Flatness. It is the idea, however I am not sure this is a rigorous/careful proof. $\endgroup$
    – kalafat
    Aug 17, 2018 at 13:21
  • $\begingroup$ @kalafat, could you please edit the question to make it clearer. You'll probably get a bit more attention that way. $\endgroup$
    – alvarezpaiva
    Aug 17, 2018 at 18:27

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