Metric of positive curvature and Homology group

Question

Is it possible to decide that whether the manifold $M$ admit a metric of positive curvature by knowing all Homology group of $M$?

Answer 1

No. For example, $S^9$ admits a metric of constant positive sectional curvature, but there is an exotic $9$-sphere $\Sigma$ (i.e. $\Sigma$ is a smooth manifold homeomorphic to $S^9$, but not diffeomorphic to it) which does not even admit a metric of positive scalar curvature. As $S^9$ and $\Sigma$ are homeomorphic, they have the same homology groups.

Answer 2

Definition. A manifold $M$ is called an $n$-dimensional homology sphere if its homology groups are isomorphic to that of $S^n$ (i.e. $M$ is connected and its homology is nonzero only in degrees 0 and n).

There are many examples of integer homology spheres in dimension 3 which are apherical (i.e. have contractible universal covering space). The simplest example I know is obtained as follows: Take the trefoil knot $K\subset S^3$, let $N$ be the complement in $S^3$ to a regular neighborhood of $K$. The boundary of $N$ is $T^2$. Now, glue two copies of $N$ by swapping the meridians and longitudes. MV sequence shows that the resulting manifold $M$ is a 3-dimensional homology sphere. With a bit more work, one verifies that $M$ is aspherical.

There are many other examples like this, for instance, among hyperbolic 3-manifolds.

According to Thm. 8.1 in

M. Gromov, H.B. Lawson, Positive scalar curvature and the Dirac operator on complete Riemannian manifolds, Inst. Hautes Etudes Sci. Publ. ´Math. No. 58 (1983), 83–196.

such a manifold cannot admit a metric of positive scalar curvature.