6
$\begingroup$

I am interested in the difference between block bundle and fiber bundle.

Let $K$ be a simplicial complex and $p: E\to |K|$ be a continuous map.

A block diffeomorphism of $\Delta^p\times M$ is a diffeomorphism $\Delta^p\times M\to \Delta^p\times M$ which for each face $\sigma \subset \Delta^p$ restricts to a diffeomorphism of $\sigma\times M$.

A block chart for $E$ over a simplex $\sigma\in K$ is a homeomorphism $h_{\sigma}:p^{-1}(\sigma)\to \sigma\times M$ which for every face $\tau$ restricts to a homeomorphism $p^{-1}(\tau)\to \tau\times M$.

A block atlas is a set $\mathcal{A}$ of block charts, at least one over each simplex of K, such that if $h_{\sigma_i}:p^{-1}(\sigma_i)\to \sigma_i\times M$ for $i=0,1$ are two elements of $\mathcal{A}$ then the composition $h_{\sigma_1}\circ h_{\sigma_0}^{-1}$ from $(\sigma_0\cap\sigma_1)\times M$ to itself is a block diffeomorphism.

A block bundle structure is a maximal block atlas. The resulting structure is a block bundle.

This notion is very close to fiber bundle.

I am wondering if there exists a block bundle s.t. both fiber and base are manifolds but it does not admit fiber bundle structure.Is every $S^3$ block bundle over $S^4$ a fiber bundle?

(This may be reduced to a lifting problem,since the fiber bundle has classifying space $BO(4)$ and the concordance class of such block bundle has classifying space $B\widetilde{Cat}(S^3)$.some knowledge about the homotopy group of $B\widetilde{Cat}(S^3)$ and $\widetilde{Cat}(S^3)/Cat(S^3)$ would surely be helpful here.$Cat=Diff,Top,PL$)

$\endgroup$
  • $\begingroup$ A minor corrections: the group of diffeomorphisms of $S^3$ is not homotopy equivalent to $B\text{SO}(4)$---the group of orientation preserving ones is, if you use the Smale conjecture. $\endgroup$ – John Klein Jan 19 '15 at 15:17
3
$\begingroup$

In my paper ``Generalised Miller--Morita--Mumford classes for block bundles and topological bundles" with Johannes Ebert, we construct a block $\mathbf{HP}^2$-bundle $\pi: E^{20} \to S^{12}$ which cannot admit a fibre bundle structure (even up to concordance).

After constructing $\pi$, the property that guarantees that it cannot be a fibre bundle is that a certain Miller--Morita--Mumford class does not vanish, but it must vanish for trivial reasons on any fibre bundle.

$\endgroup$
0
$\begingroup$

I checked some more reference and come up with the following idea,this is too long for a comment,so i present it as an "answer":

(as is pointed out,there is a mistake in my original argument,where i used $S^{diff}(S^n)$ is trivial,which is not true.while the original problem still makes sense in the Top category,where we have $S^{Top}(S^n)$ is trivial and $TOP(S^3)\simeq O(4)$)

The obstruction to the lifting is sitting in $H^4(S^4,\pi_3(\widetilde{TOP}(S^3)/TOP(S^3)))$.

Considere the homotopy exact sequence of the fibration $$\widetilde{TOP}(S^3)/TOP(S^3)\to BTOP(S^3)\to B \widetilde{TOP}(S^3)$$

We have $$\cdots\to\pi_3(O(4))\xrightarrow{p} \pi_3(\widetilde{TOP}(S^3))\to \pi_3(\widetilde{TOP}(S^3)/TOP(S^3))\to \pi_2(O(4))\cdots$$

We know $\pi_2(O(4))=0$ and to compute $\pi_3(\widetilde{TOP}(S^3))$,we need another fibration

$$S(M)\to B\widetilde{TOP}(M)\to BG(M)$$

where $S(M)$ is the structure set of $M$ and $BG(M)$ is the classifying space of the monoid of self homotopy equivalence of $M$.since sphere is topologically rigid,we know $S(M)$ is trivial,hence $$\pi_i(B\widetilde{TOP}(S^3))\cong \pi_i(BG(S^3))$$

If $p$ could be identified with the $J$-homomorphism (not very sure at this time),then it is a surjective homomorphism.This,together with the fact that $\pi_2(O(4))=0$ would imply $\pi_3(\widetilde{TOP}(S^3)/TOP(S^3))=0$,hence no obstruction to the lifting.i.e.every $S^3$ block bundle over $S^4$ admits a topological fiber bundle structure.

could this homomorphism $p$ really be identified with the $J$-homomorphism? why or why not?

$\endgroup$
  • $\begingroup$ well,even if this is true,i guess what this argument proved is:For every $S^3$ block bundle over $S^4$,there exists a concordant block bundle which admits fiber bundle structure.This is weaker than the property appeared in the the original problem. $\endgroup$ – student Jan 21 '15 at 21:32
  • $\begingroup$ I don't think this argument can be right: why should $S(M)$ be contractible? (Which is what I suppose you mean by "trivial".) $\endgroup$ – Oscar Randal-Williams Jan 21 '15 at 22:06
  • $\begingroup$ oop!I see your point.$S^n$ is just Topologically rigid,but not smoothly rigid.Now I want to switch from the category from Diff to Top,and the problem still makes sense. $\endgroup$ – student Jan 21 '15 at 22:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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