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 $\mathbb R^n$)
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4$\begingroup$ With which structure group? $\endgroup$– Oscar Randal-WilliamsCommented Sep 7, 2011 at 15:22
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$\begingroup$ orientation preserving Diffeomorphism ($S^k$) $\endgroup$– user16750Commented Sep 7, 2011 at 17:17
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7$\begingroup$ It is true if you replace smooth by piecewise linear (Alexander trick). $\endgroup$– Johannes EbertCommented Sep 7, 2011 at 18:38
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2$\begingroup$ Could someone give a reference for Johannes's comment? $\endgroup$– Peter SamuelsonCommented Sep 7, 2011 at 19:43
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2$\begingroup$ 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". $\endgroup$– Igor BelegradekCommented Sep 7, 2011 at 22:04
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3 Answers
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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.
answered Sep 7, 2011 at 21:55
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3$\begingroup$ If any sphere bundle bounds a disc bundle, then the restriction map $Diff(D^k+1) \to Diff(S^k)$ has a section after taking classifying spaces. Why does this imply the Smale conjecture? $\endgroup$ Commented Sep 7, 2011 at 22:18
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$\begingroup$ Why does the existence of exotic spheres imply the existence of exotic self-diffeomorphisms of the standard sphere? $\endgroup$ Commented Sep 7, 2011 at 22:18
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2$\begingroup$ @Greg: the "twist sphere" construction, i.e. gluing two discs together along a diffeo of their boundary gives a short exact sequence $0 \to \mathbb Z_2 \to \pi_0 Diff(S^n) \to \theta_{n+1} \to 0$ where $\theta_{n+1}$ is the group of homotopy $(n+1)$-spheres. This applies only for $n \geq 5$, and is an h-cobordism theorem + Cerf's pseudoisotopy theorem argument. $\endgroup$ Commented Sep 7, 2011 at 22:31
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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 $S^2$.
answered Sep 7, 2011 at 22:41
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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.
answered Sep 7, 2011 at 15:19
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$\begingroup$ How does this rule out the possibility that sphere fibrations are nonetheless null-cobordant? $\endgroup$– j.c.Commented Sep 7, 2011 at 20:53
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1$\begingroup$ 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.) $\endgroup$– j.c.Commented Sep 7, 2011 at 21:29
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1$\begingroup$ 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... $\endgroup$– j.c.Commented Sep 7, 2011 at 21:48