Einstein manifolds and topology

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)

• Regarding references, Einstein Manifolds by A. Besse is worth a look (despite its age) if you have access to a university library. – Andrew D. Hwang Oct 31 '15 at 1:50

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.
• Isn't that the constant scalar curvature follows from the equation $\text{Ric} = kg$? – user99914 Oct 27 '15 at 8:34
• 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 $n\geq 3$. – Paul Oct 27 '15 at 8:40