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Given a Riemannian manifold $(M,g)$ with Ricci tensor $ R_{mn} = k g_{mn} $. Suppose the Ricci scalar you get is

$$ R > 0 $$

What can you tell about the manifold $globally$ ? In particular, can you say anything about the topology of this manifold (e.g is this compact?) ?

This question arise in a Physics situation: in 11-dimensional supergravity, one can find solutions to equations with a factorised metric describing $M_4 \times M_7$, where the Riemannian manifold $M_7$ has the geometry described above (Einstein manifold with positive Ricci curvature). These solutions are said to furnish a spontaneous conpactification because $M_7$ is "automatically" compact. But I don't really understand why this is the case.

PS: Useful references where to study these topics in differential geometry? I just know basics (in order to understand General Relativity and String Theory)

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    $\begingroup$ Regarding references, Einstein Manifolds by A. Besse is worth a look (despite its age) if you have access to a university library. $\endgroup$ – Andrew D. Hwang Oct 31 '15 at 1:50
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Myers's Theorem says that if a manifold has positive lower bound for Ricci curvature, then it must be compact. In particular, if $M$ is Einstein with positive scalar curvature, we have $$Ric=\frac{R}{n}g.$$ Note that Einstein manifold must have constant scalar curvature (which follows from Bianchi identity). Combining all these, we can conclude that $M$ is compact.

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    $\begingroup$ Isn't that the constant scalar curvature follows from the equation $\text{Ric} = kg$? $\endgroup$ – user99914 Oct 27 '15 at 8:34
  • $\begingroup$ Yes. It really depends on your definition of Einstein manifold. You can define Einstein manifold to satisfy $Ric=kg$ for some constant $k$. Or you can define Einstein manifold to satisfy $Ric=\frac{R}{n}g$, or equivalently, the traceless Einstein tensor is zero. But they are equivalent by the Bianchi identity, at least for $n\geq 3$. $\endgroup$ – Paul Oct 27 '15 at 8:40
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    $\begingroup$ In Myers's theorem, completeness is an important assumption too (otherwise, of course, one could say take a round sphere minus a point, this is Einstein with positive scalar curvature and it is non-compact). $\endgroup$ – Malkoun Oct 29 '15 at 14:23

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