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Going off of this question: is there a general classification of homology 5-spheres?

If this is too ambitious, how about when we restrict to simply-connected homology 5-spheres?

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  • $\begingroup$ In all dimensions, a simply connected integral homology sphere is homeomorphic to a sphere, see here for example. $\endgroup$ May 6 at 19:59

2 Answers 2

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As it was already mentioned, non-simply connected case is hopeless.

Claim. If 5-dimensional simply connected manifold $W$ has integral homology of sphere, it's sphere (in any category — topological, PL or smooth).

Proof: there's a fundamental class $o: W \to K(\Bbb Z, 5)$. We can assume that 5-skeleton of $K(\Bbb Z, 5)$ is $S^5$ and that image of $o$ lies in it (by cellular approximation). $o': W \to S^5$ is a homology equivalence, both spaces are simply connected CWs and by Whitehead theorem it is homotopy equivalence. Generalized Poincare conjecture is settled in dimension $5$ in all categories, so $W$ is sphere.


However, there are rational homology spheres in dimension 5, i. e. 5-manifolds $W$ with rational homology of a sphere and $H_2(W, \Bbb Z) = H_3(W, \Bbb Z)$ — some finite abelian group; denote it by $T$. I think that first example known was Wu manifold $SU(3)/SO(3)$ with $T = \Bbb Z/2$. You can obtain a lot of examples in such way: take 2 degree Moore space of any finite abelian group $T$. It's a simply connected 3-complex, so embeddable in $\Bbb R^6$. Gluing two copies of tubular neighborhoods of this embedding together we obtain spin rational homology sphere with $H_2 = T^2$. Examples are, indeed, good, but classification is better; there's one.


If we have smooth simply connected 5-manifold $M$, there are three most obvious invariants: $H_2(M, \Bbb Z) = \Bbb Z^r \oplus T$, evaluation on second Stiefel-Whitney class $w: H_2 \to \Bbb Z/2$ and torsion pairing $b: T \times T \to \Bbb Q / \Bbb Z$.

What's so good in dimension $5$ is that it's high enough that all pwoerful surgery techniques are applicable, low enough to not yet contain all horrible (but beautiful) things like exotic spheres, and $5$ is odd, so no issues related with pairing on middle homology.

Theorem (Barden, 1965) [B] . Simply connected smooth manifolds are classified by $w$ and $b$ (i. e. every abstract isomorphism between $H_2$'s preserving pairing and $w$ is realized by diffeomorphism).

Proof is done by surgery and pretty easy.

Remark. Skew nondegenerate pairings on torsion groups are classified by homomorphism $w': T \times T \to \Bbb Z/2$, $x \mapsto b(x, x)$ — it's straightforward linear algebra. Also, it is pretty easy to show that $b(x, x) = w(x)$ if we regard $\Bbb Z/2$ as 2-torsion in $\Bbb Q/\Bbb Z$ (AFAIR, it's a theorem of Wall) Putting these together we see that actually we only need $w$ in classification of 5-manifolds, because pairing on homology is recovered from it.

[B] D. Barden, Simply connected five-manifolds, Ann. of Math. (2) 82 (1965).

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    $\begingroup$ Instead of considering $o: W \to K(\Bbb Z, 5)$ to obtain a homology equivalence $o' : W \to S^5$, you could instead use the fact that every closed orientable manifold admits a degree one map to a sphere of the same dimension (this map will be a homology equivalence). $\endgroup$ May 6 at 19:56
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Your first question seems hopeless given Kervaire's result that any finitely presented, perfect group having trivial second (group) homology is the fundamental group of a smooth homology $n$-sphere for $n > 4$.

Your second question kills this construction (since you require $\pi_1$ trivial now). I would imagine you could shift Kervaire's construction up to $\pi_2$ and $\pi_3$ in the same manner as one can use Eilenberg-MacLane to place a group on any $\pi_i$, but I haven't thought any of the details through.


Edit: I imagined wrongly. xsnl's answer explains the situation for simply connected homology 5-spheres. (A skeleton of that answer is also present as comments to this answer.)

Also, I corrected the link to group homology.

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    $\begingroup$ For homology sphere $N$ we have fundamental class $N \to K(\Bbb Z, 5)$. We can assume that image lies in 5-skeleton, so by Whitehead theorem (for simply connected spaces homology equivalence is homotopy equivalence) it's a homotopy equivalence $N \to S^5$. By Poincare theorem, it is diffeo/homeomorphic to a sphere. $\endgroup$
    – xsnl
    Apr 26, 2018 at 16:32
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    $\begingroup$ However, if we require that only $H_*(N, \Bbb Q) = H_*(S^5, \Bbb Q)$, then for every natural number there's a smooth manifold $W_k$ with $|H_2(W_k)| = k^2$. Another example is Wu manifold $SU(3)/SO(3)$ with $H_2 = \Bbb Z/2$. $\endgroup$
    – xsnl
    Apr 26, 2018 at 16:43
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    $\begingroup$ Also there's a full classification of 5-dimensional simply connected manifolds (they are very nice because homotopy, homeo and diffeo classifications all coincide) — D. Barden, Simply connected five-manifolds, Ann. of Math. (2) 82 (1965). $\endgroup$
    – xsnl
    Apr 26, 2018 at 16:52
  • $\begingroup$ @xsni : I think everyone would benefit from you writing an Answer, not just brief comments. $\endgroup$ Apr 27, 2018 at 4:47
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    $\begingroup$ Okay, I'll move this to an answer. $\endgroup$
    – xsnl
    Apr 27, 2018 at 21:53

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