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I saw in the answer of this post:

Is it true that all sphere bundles are boundaries of disk bundles?

that a $S^3$-bundle over $S^4$ bounds a disc bundle over $S^4$ iff $O(4)\rightarrow Diff(S^3)$ is a homotopy equivalence.

Can anyone direct me to a reference that proves it?

I'm studying Milnor's paper "On manifolds homeomorphic to the 7-sphere" and trying to avoid Thom's paper to see that a closed 7-manifold is the boundary of an 8-manifold. More specifically, I have a $SO(4)$-bundle

$$S^3 \rightarrow M^7 \rightarrow S^4$$

and I want to see this bundle as the boundary of some bundle:

$$D^4 \rightarrow N^8 \rightarrow S^4$$

with $\partial N^8 = M^7$

Thanks in advance.

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    $\begingroup$ If you simply want to avoid Thom's cobordism result you can restrict the invariant Milnor defined to those manifolds that are bounded. And if your sphere bundle comes with transition function in $SO(4)$, you can use the same information to construct a $\mathbb{R}^4$ vector bundle and look at its disc bundle. $\endgroup$ Dec 8, 2015 at 4:40

2 Answers 2

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The map $O(4) \to \text{Diff}(S^3)$ being homotopy equivalence is not equivalent to the assertion that every smooth $S^3$-bundle bounds a disk bundle, but the much stronger statement that every smooth $S^3$-bundle can be linearized, i.e. arises as the unit bundle of some four-dimensional vector bundle (and this clearly implies that it is the boundary of a disk bundle).

You can see this by observing that smooth $S^3$-bundles over a space $X$ are in bijective correspondence with homotopy classes $[X,B \text{Diff}(S^3)]$ and the same holds for vector bundles and $[X,B O(4)]$ .

But I think you do not need all this stuff to understand what Milnor did: There we already start with a $S^3$-bundle with structure group $SO(4)$, i.e. a linear sphere bundle.

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The proof that the inclusion $\mathsf O(4)\hookrightarrow \mathrm{Diff}(S^3)$ is a homotopy equivalence (a fact that was known as Smale's conjecture) was obtained by Allen Hatcher in his Annals of Math. paper in 1983.

As pointed out by Igor Belegradek, analogues of this statement are not true in sufficiently high dimension; namely, if $n\geq7$, then $\mathrm{Diff}(S^n)$ does not have the homotopy type of a finite-dimensional Lie group, see P.L. Antonelli, D. Burghelea, P.J. Kahn. However, if $\mathrm{Diff}(S^n)$ is equipped with pointwise $C^k$-topology, then it has a deformation retract to $\mathsf{SO}(n+1)$, see reference [44] in the latter paper. Finally, a $\mathsf G$-equivariant version of the statement remains true, see Grove and Kim's JDG paper in 2004 (Thm B).

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    $\begingroup$ Renato, what you say about higher dimensions is not true. See e.g. [The non-finite homotopy type of some diffeomorphism groups P.L. Antonelli, D. Burghelea, P.J. Kahn] sciencedirect.com/science/article/pii/0040938372900213. The inclusion is not even a rational homotopy equivalence in some dimensions, see [Farrell, F.T. Hsiang, W.C. On the rational homotopy groups of the diffeomorphism groups of discs, spheres and aspherical manifolds. Proc. Symp. Pure. Math. 32 (1977) 403--415. ] $\endgroup$ Dec 8, 2015 at 4:06
  • $\begingroup$ @Igor: Ooops, thanks for that correction! I will edit to reflect that. $\endgroup$ Dec 8, 2015 at 4:14
  • $\begingroup$ Renato, just to avoid confusion: In reference 44 the author uses pointwise convergence $C^k$ topology. This is different from uniform convergence (on compact sets) $C^k$ topology. To my knowledge nobody uses pointwise $C^k$ topology, and it is kind of meaningless geometrically. Thus this is not the issue of $k$ against infinity. $\endgroup$ Dec 8, 2015 at 12:19

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