# Is every $S^3$ block bundle over $S^4$ a fiber bundle?

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.$Cat=Diff,Top,PL$)

• 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. – John Klein Jan 19 '15 at 15:17

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.

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?

• 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. – student Jan 21 '15 at 21:32
• I don't think this argument can be right: why should $S(M)$ be contractible? (Which is what I suppose you mean by "trivial".) – Oscar Randal-Williams Jan 21 '15 at 22:06
• 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. – student Jan 21 '15 at 22:11