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

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).

• The connected sum $M$ of two Poincare spheres $P$ might be a candidate of such a closed $3$-manifold. This is a homology sphere, $\pi_1$ is non-trivial and I claim that $\pi_2$ is also non-trivial. Indeed: Consider the $2$-sphere $S$, where we glued $P$ and $P$. As $P$ without a disk (or equivalently without a point) has non-trivial $\pi_1$, it is non-contractible. By Prop 3.10 of pi.math.cornell.edu/~hatcher/3M/3Mfds.pdf the sphere $S$ represents a non-trivial element in $\pi_2$. I do not know about higher homotopy groups. – Lennart Meier Apr 16 '19 at 8:44
• Narrowing the search: since $H_1(M) = \pi_{1}^{\mathrm{ab}}(M)$, $\pi_1(M)$ must be a non-0 perfect group (equal to its commutator). It's worth pointing out that $\pi_1(M)$ must be non-0. Otherwise the Hurewicz theorem implies that if $\pi_d(M)$ is the first non-0 homotopy group, then $H_i(M) \simeq \pi_i(M)$ for $i \leq d$. – cgodfrey Apr 16 '19 at 8:51
• @ThiKu could you clarify your expression "does the job"? The answer to the question can not have $\pi_2=0$. Or is there some way to modify Poincare sphere to make $\pi_2$ non-trivial without altering its homology? – user137767 Apr 16 '19 at 9:27
• @StepanBanach Maybe you could clarify your question: do you mean $\pi_i(M) \neq 0$ for all $i>0$ or just some $i>0$? I'm guessing the former from the responses to the comments. If so then it's not clear if all of the homotopy groups in the connected sum of Poincaré spheres are non-trivial. Cf. mathoverflow.net/questions/64131/homotopy-groups-of-s2 for the analogous question about $S^2$. – Danny Ruberman Apr 16 '19 at 13:49
• @ThiKu yes, I think you are right. But you do not really need to know them up to isomorphism, just that they do not vanish. – user137767 Apr 16 '19 at 15:12

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.