Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $\mathbb R^n$)

share|improve this question
4  
With which structure group? –  Oscar Randal-Williams Sep 7 '11 at 15:22
    
orientation preserving Diffeomorphism ($S^k$) –  user16750 Sep 7 '11 at 17:17
6  
It is true if you replace smooth by piecewise linear (Alexander trick). –  Johannes Ebert Sep 7 '11 at 18:38
1  
Could someone give a reference for Johannes's comment? –  Peter Samuelson Sep 7 '11 at 19:43
2  
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 '11 at 22:04

3 Answers 3

up vote 9 down vote accepted

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.

share|improve this answer
1  
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? –  Johannes Ebert Sep 7 '11 at 22:18
    
Why does the existence of exotic spheres imply the existence of exotic self-diffeomorphisms of the standard sphere? –  Greg Friedman Sep 7 '11 at 22:18
2  
@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. –  Ryan Budney Sep 7 '11 at 22:31

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 $S^2$.

share|improve this answer

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.

share|improve this answer
    
How does this rule out the possibility that sphere fibrations are nonetheless null-cobordant? –  j.c. Sep 7 '11 at 20:53
1  
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.) –  j.c. Sep 7 '11 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... –  j.c. Sep 7 '11 at 21:48

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.