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A homotopy sphere is a topological $n$-manifold $M$ which is homotopy equivalent to $S^n$.

A homology sphere is a topological $n$-manifold $M$ such that $H_i(M) \cong H_i(S^n)$ for all $i$.

Note, by (one version of) Whitehead's Theorem, every simply connected homology sphere is a homotopy sphere. However, there are homology spheres which are not homotopy spheres, for example, the Poincaré homology sphere.

I've seen in some references that homotopy/homology spheres are assumed to be smooth, but of course the smooth structure is not needed to define them. My question is whether or not this assumption is restrictive. That is:

Are there homotopy/homology spheres which admit no smooth structure?

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    $\begingroup$ I believe the answer is no for homotopy spheres, although I won't be able to assemble all the references here. In dimension 3 and below it's classical all manifolds can be smoothed. In dimension 4 it follows by Freedman's work -- the smoothing obstruction requires some homology to exist. In high dimensions it pops out of smoothing theory... but precisely who to credit or where to look I am uncertain. I would strongly suspect Kirby and Siebenmann being responsible for this. For homology spheres I believe the answer is yes, and there are some classical low-dimensional examples. $\endgroup$ – Ryan Budney Apr 12 '16 at 18:47
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    $\begingroup$ @RyanBudney In high dimensions it should be easier than smoothing theory, because homotopy spheres are all topologically standard, no? $\endgroup$ – Mike Miller Apr 12 '16 at 19:16
  • $\begingroup$ I suppose it depends on what you call smoothing theory. The transition from topological to PL or smooth structure is what I'm calling smoothing theory. You don't need the full machine at this step, but you need the basics of it. $\endgroup$ – Ryan Budney Apr 12 '16 at 19:19
  • $\begingroup$ What does "topologically standard" mean? $\endgroup$ – Mariano Suárez-Álvarez Apr 12 '16 at 20:16
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    $\begingroup$ @Mariano: presumably it means homeomorphic to the usual sphere. $\endgroup$ – Qiaochu Yuan Apr 12 '16 at 20:37
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Since the Poincaré conjecture is known in all dimensions any homotopy $n$-sphere is homeomorphic to $S^n$ and hence admits a smooth structure.

Any manifold of dimension $\le 3$ admits a smooth structure.

Kirby-Siebenmann famously showed that a closed manifold $M$ of dimension $>4$ admits a PL structure if and only if a certain element of $H^4(M;\mathbb Z_2)$ vanishes. The element is called the Kirby-Siebenmann class. Thus any homology sphere of dimension $\neq 4$ admits a PL structure.

Kervaire noted on p.71 of Smooth homology spheres and their fundamental groups that any PL homology sphere of dimension $\neq 3$ admits a smooth structure (because it bounds a contractible PL manifold and all such manifolds are smoothable).

Thus the only remaining case is homology $4$-spheres. Some of them have large fundamental groups for which $4$-dimensional surgery is not yet known to work. It seems smoothability of such spheres is open. Note that any closed PL $4$-manifold admits a smooth structure.

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    $\begingroup$ Are there any explicit examples of homology $4$-spheres for which it is not yet known if they admit a smooth structure or not? $\endgroup$ – Michael Albanese Apr 13 '16 at 16:45
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    $\begingroup$ I do not know such examples, but then if you remember Freedman's work in the simply-connected case the non-smoothable examples arise in a roundabout way. It might be that the same will be true for 4-manifolds with large fundamental group, i.e. a smoothable manifold may have a non-smoothable "evil twin" which has the same intersection form and nontrivial Kirby–Siebenmann invariant. $\endgroup$ – Igor Belegradek Apr 13 '16 at 20:54

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