# Is the universal cover of an integral homology sphere again an integral homology sphere?

Let $\Sigma$ be an integral homology sphere (from now on I will drop the word 'integral').

If $\Sigma$ is simply connected, then it is homotopy equivalent to a sphere by Whitehead's Theorem, and hence homeomorphic to a sphere by the solution of the topological Poincaré conjecture.

If $\pi_1(\Sigma)$ is infinite, then $\widetilde{\Sigma}$ is non-compact; in particular, it is not a homology sphere.

Suppose now that $\pi_1(\Sigma)$ is finite but non-trivial. If $\dim\Sigma$ is even, then $\chi(\widetilde{\Sigma}) = |\pi_1(\Sigma)|\chi(\Sigma) = 2|\pi_1(\Sigma)|$, so $\widetilde{\Sigma}$ is not a homology sphere. If $\dim\Sigma$ is odd, then there are examples where $\widetilde{\Sigma}$ is a homology sphere, e.g. the Poincaré homology sphere which has $S^3$ as its universal cover.

If $\Sigma$ is an odd-dimensional homology sphere and $\pi_1(\Sigma)$ is finite, is $\widetilde{\Sigma}$ necessarily a homology sphere?

If the answer were yes, then $\widetilde{\Sigma}$ would be a simply connected homology sphere and hence $\widetilde{\Sigma}$ is a sphere by the argument above. In dimension three, the only homology sphere with finite fundamental group is the Poincaré homology sphere, so if there is a counterexample (which I suspect there is), it must have dimension at least five.

• A simply connected homology sphere is $S^n$, so you're essentially asking about group actions on $S^n$. You can approach this via the classification which shows that the Poincare sphere is special. I was thinking about it the other day but I don't have a simple proof that it's the only integer homology sphere spherical space form. – user98602 May 12 '17 at 18:58
• (In fact, you can use Kervaire to build 5-manifolds which are homology spheres with specified fundamental group whose first and second homology vanishes. There are many of these that don't act freely on a sphere.) – user98602 May 12 '17 at 19:00

## 1 Answer

To be concrete, take any finite simple noncyclic group with trivial Schur multiplier, e.g. $G=A_2(9)$ (see here for more examples). Then, by Kervaire's theorem, for every $k\ge 5$ there exists a $k$-dimensional integer homology sphere $M$ with $\pi_1(M)=G$. Milnor proved (see here) that if $G$ is a finite group acting freely on a sphere then every involution in $G$ has to be central. All finite noncyclic simple groups contain involutions; by simplicity, such involutions cannot be central. It follows that the universal cover of $M$ cannot be a homology sphere.