Simply-connected $\mathbb{Z}_p$-homology spheres?

Question

Let $X$ be a $\mathbb{Z}$-homology $n$-sphere, i.e., a closed manifold with $\mathbb{Z}$-homology groups of the standard $n$-dimensional sphere. If $X$ is simply-connected, it is not difficult to see (and not difficult to look up on the internet) that such a space is in fact homotopy equivalent (hence homeomorphic, by the Poincare conjecture) to the $n$-sphere.

Now assume that $X$ is a $\mathbb{Z}_p$-homology sphere for some prime $p$. Is it true that if $X$ is simply-connected, then $X$ is homotopy equivalent to the $n$-sphere?

EDIT: OOPS, sorry! As pointed out in the comments, the first sentence didn't make any sense before. Added simply-connecteness as an assumption.

Answer 1

Edit #2: There is a $5$-dimensional counterexample, but I don't know how to produce it explicitly.

According to Smale's classification of closed, oriented, smooth, simply connected, spin $5$-manifolds, there is a simply connected closed $5$-manifold $X$ with $H_2(X) \cong \mathbb{Z}_p \oplus \mathbb{Z}_p$. By the same universal coefficient and Poincaré duality arguments as below, such a manifold $X$ is a $\mathbb{Z}_q$-homology sphere for any prime $q \neq p$, but not a $\mathbb{Z}$-homology sphere.

Universal coefficients also shows that a simply connected closed $n$-manifold which is a $\mathbb{Z}_p$-homology sphere for all primes $p$ is a $\mathbb{Z}$-homology sphere.

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Edit: This is not an answer. I just want to record the observation that the minimal dimension of a counterexample is $5$.

First let me record that a simply connected closed manifold cannot have a nontrivial orientation double cover, and hence is orientable.

The cases $n = 0, 1$ are degenerate. When $n = 2$ the $2$-sphere is already the only simply connected closed surface by the classification of surfaces.

When $n = 3$, let $X$ be a simply connected closed $3$-manifold. Since it's orientable, $H_3(X) \cong \mathbb{Z}$ and $H_2(X) \cong H^1(X) \cong 0$ by Poincaré duality, and hence $X$ is already a $\mathbb{Z}$-homology sphere.

When $n = 4$, let $X$ be a simply connected closed $4$-manifold. Since it's orientable, $H_4(X) \cong \mathbb{Z}$. By universal coefficients, $H^2(X)$ is torsion-free, so by Poincaré duality, $H_2(X) \cong H^2(X)$ is also torsion-free. And by Poincaré duality again, $H_3(X) \cong H^1(X) \cong 0$. It follows that

$$H_2(X, \mathbb{Z}_p) \cong H_2(X) \otimes \mathbb{Z}_p$$

vanishes iff $H_2(X)$ vanishes. Hence if $X$ is a simply connected closed $4$-manifold which is also a $\mathbb{Z}_p$-homology sphere for any $p$, then $X$ is a $\mathbb{Z}$-homology sphere.

Answer 2

You need the condition to hold at least hold for all primes $p$. Otherwise you could just take some space which has for example non trivial $\mathbb Z$ homology isomorphic to $\mathbb Z_q$ in some other degree, which would consequently vanish with $\mathbb Z_p$ coefficients (hence you will loose necessary information). Furthermore I don't think your claims are true, e.g. you would need $X$ to be an n-manifold. But even then you should now that Poincaré's original conjecture was the one you stated, he asked if every 3-manifold which is a $\mathbb Z$-homology sphere is a 3-sphere, but turned out to be wrong by the Poincaré sphere. One problem is that possibility to have a perfect fundamental group. I just wrote down some comments I hope it helps in some way.

Answer 3

As pointed out by Qiaochu Yuan, there are counterexamples to your claim in dimension five. The simplest such counterexample is the Wu manifold, constructed below, which is a simply connected $\mathbb{Z}_p$-homology sphere for every odd prime $p$ (in fact, every odd integer $p$) but is not homeomorphic to a sphere (in particular, it is not a $\mathbb{Z}_2$-homology sphere).

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As $SO(3)$ is a closed Lie subgroup of $SU(3)$, the set of left cosets of $SO(3)$, namely $M := SU(3)/SO(3)$, is a connected, closed manifold and the projection map $\pi : SU(3) \to M$ is a principal $SO(3)$-bundle. Note that $\dim M = \dim SU(3) - \dim SO(3) = 8 - 3 = 5$. From the long exact sequence in homotopy, we see that

$$\dots \to \pi_2(SU(3)) \to \pi_2(M) \to \pi_1(SO(3)) \to \pi_1(SU(3)) \to \pi_1(M) \to \pi_0(SO(3)) \to \dots$$

As $SU(3)$ is simply connected and $SO(3)$ is connected, it follows that $M$ is simply connected. Every Lie group has trivial second homotopy group, so $\pi_2(M) \cong \pi_1(SO(3)) \cong \mathbb{Z}_2$. By the Hurewicz Theorem, $H_1(M; \mathbb{Z}) = 0$ and $H_2(M; \mathbb{Z}) \cong \mathbb{Z}_2$.

Note that $M$ is orientable (because $M$ is simply connected), so by Poincaré duality $H_4(M; \mathbb{Z}) \cong H^1(M; \mathbb{Z}) = \operatorname{Hom}(\pi_1(M), \mathbb{Z}) = 0$. By the Universal Coefficient Theorem, $H^2(M; \mathbb{Z}) \cong \operatorname{Hom}(H_2(M; \mathbb{Z}_2), \mathbb{Z})\oplus\operatorname{Ext}(H_1(M; \mathbb{Z}_2), \mathbb{Z}) \cong \operatorname{Hom}(\mathbb{Z}_2, \mathbb{Z})\oplus\operatorname{Ext}(0, \mathbb{Z}) = 0$ so, again by Poincaré duality, $H_3(M; \mathbb{Z}) \cong H^2(M; \mathbb{Z}) = 0$.

Therefore $M$ is a simply connected, five-dimensional manifold with integral homology groups

$$\mathbb{Z}, 0, \mathbb{Z}_2, 0, 0, \mathbb{Z}.$$

Let $p$ be an odd integer (e.g. a prime other than $2$). As $\mathbb{Z}_2\otimes\mathbb{Z}_p = 0$ and $\operatorname{Tor}(\mathbb{Z}_2, \mathbb{Z}_p) = 0$, the Universal Coefficient Theorem for homology shows that $M$ has $\mathbb{Z}_p$ homology groups

$$\mathbb{Z}_p, 0, 0, 0, 0, \mathbb{Z}_p.$$

So $M$ is a simply connected $\mathbb{Z}_p$-homology sphere which is not homeomorphic to a sphere.