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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).

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    $\begingroup$ 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 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. $\endgroup$ – Lennart Meier Apr 16 '19 at 8:44
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    $\begingroup$ 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 $i \leq d$. $\endgroup$ – cgodfrey Apr 16 '19 at 8:51
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    $\begingroup$ @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 $\pi_2$ non-trivial without altering its homology? $\endgroup$ – user137767 Apr 16 '19 at 9:27
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    $\begingroup$ @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 non-trivial. Cf. mathoverflow.net/questions/64131/homotopy-groups-of-s2 for the analogous question about $S^2$. $\endgroup$ – Danny Ruberman Apr 16 '19 at 13:49
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    $\begingroup$ @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
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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.

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

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