Closed manifold with non-vanishing homotopy groups and vanishing homology groups

Question

Is there a closed connected $n$-dimensional topological manifold $M$ ($n\geq 2$) such that $\pi_i(M)\neq 0$ for all $i>0$ and $H_i(M, \mathbb{Z})=0$ for $i\neq 0$, $n$? The manifold $S^1\times S^2$ satisfies the first requirement but not the second (generally, the direct product of two positive-dimensional manifolds does not seem to satisfy the second requirement because of Künneth).

Answer 1

As suggested by Lennart Meier, the connected sum $M=P\#P$ of two copies of the Poincaré homology sphere gives an example. The homotopy groups $\pi_n(M)$ are nonzero for all $n>1$ because the universal cover $\widetilde{M}$ has a retraction (not a deformation retraction) onto $S^2$ and it is known that all the higher homotopy groups of $S^2$ are nontrivial.

A retraction $r:\widetilde M \to S^2$ can be obtained in the following way. The connected sum $P\#P$ is constructed by removing an open ball from $P$ to obtain a manifold $P'$, then gluing two copies of $P'$ together by identifying their boundary spheres. The universal cover $\widetilde {P'}$ is $S^3$ with $|\pi_1(P)|=120$ disjoint balls removed. The universal cover $\widetilde M$ is obtained by gluing infinitely many copies of $\widetilde {P'}$ together in a tree-like pattern, identifying boundary spheres in pairs. Each copy of $\widetilde {P'}$ retracts onto any one of its boundary spheres since $\widetilde {P'}$ is $S^2\times I$ with 118 balls removed and $S^2\times I$ retracts onto one of its boundary spheres hence $\widetilde {P'}$ also retracts onto this boundary sphere by restriction. We can build $\widetilde M$ as an infinite sequence of attachments of one copy of $\widetilde {P'}$ at a time, starting with a single copy. Each stage of this construction retracts onto the previous stage. The infinite composition of these retractions is well-defined (and continuous) since any compact subset of $\widetilde M$ is contained in a finite stage. The infinite composition gives a retraction of $\widetilde M$ onto $\widetilde {P'}$ which in turn retracts onto $S^2$.

The fact that all the higher homotopy groups of $S^2$ are nontrivial was shown in a paper by S. O. Ivanov, R. Mikhailov, and Jie Wu in Homology Homotopy Appl. 18 (2016), no. 2, 337--344. The corresponding result holds also for $S^3$ since $\pi_n(S^3)=\pi_n(S^2)$ for $n\geq 3$, and the result is known for $S^4$ and $S^5$ as well. It fails for other spheres since $\pi_{n+4}(S^n)=0$ for $n\geq 6$ since this is in the stable range and the stable 4-stem is trivial. However, it's not clear how to apply these results to obtain homology $n$-spheres with all homotopy groups nontrivial when $n>3$.

The argument for $P\# P$ can probably be extended to connected sums of arbitrary nonsimply-connected homology 3-spheres with a little more work to cover the case that the summands of $M$ have infinite fundamental groups.