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 positivedimensional manifolds does not seem to satisfy the second requirement because of Künneth).

8$\begingroup$ 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 nontrivial and I claim that $\pi_2$ is also nontrivial. Indeed: Consider the $2$sphere $S$, where we glued $P$ and $P$. As $P$ without a disk (or equivalently without a point) has nontrivial $\pi_1$, it is noncontractible. By Prop 3.10 of pi.math.cornell.edu/~hatcher/3M/3Mfds.pdf the sphere $S$ represents a nontrivial element in $\pi_2$. I do not know about higher homotopy groups. $\endgroup$ – Lennart Meier Apr 16 '19 at 8:44

3$\begingroup$ Narrowing the search: since $H_1(M) = \pi_{1}^{\mathrm{ab}}(M)$, $\pi_1(M)$ must be a non0 perfect group (equal to its commutator). It's worth pointing out that $\pi_1(M)$ must be non0. Otherwise the Hurewicz theorem implies that if $\pi_d(M)$ is the first non0 homotopy group, then $H_i(M) \simeq \pi_i(M)$ for $i \leq d$. $\endgroup$ – cgodfrey Apr 16 '19 at 8:51

1$\begingroup$ @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 $\pi_2$ nontrivial without altering its homology? $\endgroup$ – user137767 Apr 16 '19 at 9:27

2$\begingroup$ @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 nontrivial. Cf. mathoverflow.net/questions/64131/homotopygroupsofs2 for the analogous question about $S^2$. $\endgroup$ – Danny Ruberman Apr 16 '19 at 13:49

3$\begingroup$ @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. $\endgroup$ – 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 treelike 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 welldefined (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, 337344. 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 4stem 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 nonsimplyconnected homology 3spheres with a little more work to cover the case that the summands of $M$ have infinite fundamental groups.

$\begingroup$ Yes, the universal covers of these manifolds will all be homeomorphic. So you've described all 3dimensional examples. $\endgroup$ – Ian Agol Apr 24 '19 at 2:56