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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.$Cat=Diff,Top,PL$)

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  • $\begingroup$ 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, 2015 at 15:17

2 Answers 2

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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.

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  • $\begingroup$ Thanks for this paper Oscar; it has the clearest introduction to block bundles I've come across. Here is a simple question I haven't been able to puzzle out: are the total spaces of two concordant block bundles homotopy equivalent? $\endgroup$ Nov 6, 2020 at 14:42
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    $\begingroup$ @Connor: Yes they are. One way to see this is to show they are both homotopy equivalent to the total space of the concordance, and for that you can use that the fibre and homotopy fibre of a block bundle over any vertex are homotopy equivalent, which is proved in this paper. $\endgroup$ Nov 6, 2020 at 20:02
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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?

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  • $\begingroup$ 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, 2015 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".) $\endgroup$ Jan 21, 2015 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, 2015 at 22:11
  • $\begingroup$ It is true that $p$ is the j homomorphism. $Bp$ classifies taking a vector bundle to its sphere bundle thought of as a block bundle, while $Bj$ classifies taking a vector bundle to its sphere bundle. The equivalence you describe $B\widetilde{TOP}(S^3) \rightarrow BG(S^3)$ classifies taking a spherical block bundle to an equivalent fibration (which is a spherical fibration). So the evident diagram commutes showing p is the J-homomorphism. $\endgroup$ Aug 10, 2020 at 22:22

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