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Let $E$ be the total space of the sphere bundle $S^k\to E\to M$, is it true that there exists a disk bundle $D^{k+1}\to N\to M$ such that $E=\partial N$? (where $D^{k+1}$ is the unit disk in $\mathbb R^n$)

edited Sep 7 2011 at 21:56

Oscar Randal-Williams
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asked Sep 7 2011 at 15:00

unknown (google)
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With which structure group? – Oscar Randal-Williams Sep 7 2011 at 15:22

orientation preserving Diffeomorphism ($S^k$) – unknown (google) Sep 7 2011 at 17:17

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It is true if you replace smooth by piecewise linear (Alexander trick). – Johannes Ebert Sep 7 2011 at 18:38

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Could someone give a reference for Johannes's comment? – Peter Samuelson Sep 7 2011 at 19:43

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There is a restriction maps $r: Diff(D^{k+1})\to Diff(S^k)$, which induces a map on classifying spaces $Br: BDiff(D^{k+1})\to BDiff(S^k)$. Smooth bundles over $M$ are homotopy classes of mapps into the suitable classifying space. What you are asking is whether any map from $M$ to $BDiff(S^k)$ is homotopic to a map that can be lifted is the image of $Br$. In the PL category the map $r$ has a section given by Alexander trick, as mentioned in the comment above. I do not know the answer in the smooth category, but I suspect it should be "no". – Igor Belegradek Sep 7 2011 at 22:04

Oscar Randal-Williams · Accepted Answer · 2011-09-07 21:55:06Z UTC

If you don't specify what structure group you want the disc bundle $D^{k+1} \to N \to M$ to have, then it is always true: you just take the fibrewise cone on the original family.

If you want to know whether smooth $S^k$-bundles always bound smooth disc bundles, this is true iff $$O(k+1) \to \mathrm{Diff}(S^k)$$ is a homotopy equivalence (known as the Smale conjecture). This is known to be true if $k=0, 1$ (classical), $2$ (Smale) or $3$ (Hatcher). I don't think it is known in any other dimension, and is definitely false in general. In fact, it is false on $\pi_0$ in general, due to the existence of exotic spheres and hence also exotic self-diffeomoprhisms of spheres.

Ryan Budney · Answer 2 · 2011-09-07 23:39:33Z UTC

The question of whether or not a smooth sphere bundle fibrewise extends to a smooth disc bundle over a space $X$ boils down to whether or not the classifying map

$$ X \to BDiff(S^n) $$

lifts up

$$ BDiff(D^{n+1}) \to BDiff(S^n)$$

Where $Diff(D^{n+1})$ is the group of diffeomorphisms of the disc.

The map $Diff(D^{n+1}) \to Diff(S^n)$ splits as a product:

$$O_{n+1} \times PDiff(D^{n}) \to O_{n+1} \times Diff(D^n rel \partial)$$ where $PDiff(D^{n})$ is the group of pseudo-isotopy diffeomorphisms of $D^{n}$. These are diffeomorphisms of $D^{n} \times [0,1]$ that are the identity on $(D^{n} \times {0}) \cup (S^{n-1} \times [0,1])$.

There is a fibre-bundle:

$$Diff(D^{n+1} rel \partial) \to PDiff(D^{n}) \to Diff(D^n rel \partial)$$

so basically this is asking whether or not this bundle has a section. I think it can't have a section, in particular the map $\pi_1 Diff(D^n rel \partial) \to \pi_0 Diff(D^{n+1} rel \partial)$ is epic by Cerf's Pseudoisotopy theorem.

Okay, so this is now an answer. So this is saying that there are sphere bundles over $S^2$ which do not extend to smooth disc bundles over $S^2$.

Igor Rivin · Answer 3 · 2011-09-07 21:39:35Z UTC

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No, it seems that the "fibered cobordism group" is not even necessarily finitely generated, much less trivial: "Some fibred cobordism groups are not finitely generated", by L. Astey, 1988.

EDIT (or, duh) As pointed out by @jc in his comment, the reference indicates that the answer is YES, not NO.

edited Sep 7 2011 at 21:39

answered Sep 7 2011 at 15:19

Igor Rivin
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How does this rule out the possibility that sphere fibrations are nonetheless null-cobordant? – jc Sep 7 2011 at 20:53

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Indeed, after Theorem 1 of your reference, the authors state: "Observe that any $S(\gamma)$ represents the zero element in $N^{S^{4n}}_{4n−2}$, being the boundary of the disc bundle $D(\gamma)$." (here $S(\gamma)$ is some sphere bundle of a 4(n-1) vector bundle $\gamma$, $N^X_p$ is the fiber cobordism group of fibrations over X with fiber p-dimensional manifolds.) – jc Sep 7 2011 at 21:29

Not every sphere bundle arises from a vector bundle since $O(n)\rightarrow Homeo(S^{n-1})$ is not a homotopy equivalence: math.stackexchange.com/questions/40478/… . So unless I'm missing something, this reference doesn't resolve the question... – jc Sep 7 2011 at 21:48

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

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