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Not thinking about $h$-cobordism, one usually defines a cobordism between manifolds, realizes it is an equivalence relation, chooses an appropriate class of structured manifolds (framed, unoriented, spin, ...), and dreams about the group of cobordism classes of these manifolds.

Thinking about $h$-cobordism, whose definition I learned today, one seems to not think this way. At least, I've glanced through the Wikipedia page and some notes/articles about $h$-cobordism and there is no explicit mention of 'the set of $h$-cobordism classes of $n$-manifolds (possibly with structure)' or its (group) structure.

(Taking after Will Sawin and Connor Malin's comments)

There is not an interesting group of $h$-cobordism classes of manifolds in general (see comments.) But there is, for instance, interesting groups of $h$-cobordism classes of homotopy spheres.

Is there any interpretation of these groups of a similar spirit (e.g., using spectra) to the relationship between cobordism and Thom spectra?

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  • $\begingroup$ The whole point of the group structure in manifold cobordism is that the disjoint union of two manifolds can be cobordant to one manifold (e.g. their connected sum). But in $h$-cobordism, that is impossible, since a space with two components can't be homotopic to a space with one component. So why would one expect there to be a useful group structure? $\endgroup$
    – Will Sawin
    Commented Jul 12, 2022 at 1:26
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    $\begingroup$ @WillSawin There is a particularly famous h-cobordism group of homotopy spheres studied by Kervaire and Milnor... $\endgroup$
    – Connor Malin
    Commented Jul 12, 2022 at 1:29
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    $\begingroup$ @WillSawin Am I misremembering? I thought the elements were h-cobordism classes of homotopy spheres. Of course this turns out to be diffeomorphism classes, but that doesn't take away from the fact they defined them as a type of h-cobordism group. $\endgroup$
    – Connor Malin
    Commented Jul 12, 2022 at 1:34
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    $\begingroup$ @ConnorMalin and Will Sawin - I've edited my question a good bit after seeing these comments. $\endgroup$
    – Matthew Niemiro
    Commented Jul 12, 2022 at 1:43
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    $\begingroup$ Whitehead groups are parameterized by fundamental groups, not integers. In the $h$-cobordism theorem (and Whitehead's theorem), the Whitehead group is an obstruction to uniqueness. The corresponding obstruction to existence is Siebenmann's end theorem, using Wall's finiteness invariant. These are the cases $n=1,0$ of a spectrum related to $K$-theory whose homotopy groups are called higher Whitehead torsion or (stable) parameterized $h$-cobordism groups. But this introduces a new direction of variation, not weaving together very similar objects like in the Thom construction. $\endgroup$
    – Ben Wieland
    Commented Jul 12, 2022 at 2:28

1 Answer 1

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This topic has a very different flavor from what is usually meant by cobordism. No, the correspondence between cobordism classes and homotopy classes (of a Thom space or Thom spectrum) has no analogue here; when manifolds, or cobordisms between manifolds, are created by transversality, we don't have much control over their homotopy types.

I would say that the main point of considering h-cobordisms is that surgery theory (in the sense of Wall, Browder, Novikov, ...) is pretty good at answering questions like "How many manifolds, if any, have a given homotopy type?" but where "how many" means up to the equivalence relation of h-cobordism. So, to complete the picture we need to know something about the distinction between this equivalence relation and the one that we originally cared about.

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  • $\begingroup$ In response to some comments I have edited my question since they clarified some misunderstanding and led me to be a bit more specific. I think this answer works regardless, though. $\endgroup$
    – Matthew Niemiro
    Commented Jul 12, 2022 at 1:45

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