This is in response to John's addendum. As I understand it, one has the following hierarchy:

- Any Poincare complex $X$ has a Spivak normal spherical fibration $S$.
- If $X$ carries a topological manifold structure then $S$ has a microbundle reduction.
- If $X$ carries a smooth manifold structure then $S$ has a vector bundle reduction refining the microbundle reduction.

I'm going to concentrate on simply connected Poincare 4-complexes $X$ with even intersection form. These have Kirby-Siebenmann smoothing obstruction $ks\in H^4(X;\mathbb{Z}/2)=\mathbb{Z}/2$ equal to $\sigma(X)/8$ mod 2, where $\sigma$ is the signature. This is just the obstruction coming from Rochlin's theorem: $\sigma$ is divisible by $16$ if $X$ is smoothable.

Freedman tells us that $X$ has a unique topological manifold structure, and hence $S$ has a canonical microbundle structure. So, to ask whether there is a vector bundle reduction of the microbundle is the same as asking whether $S$ has a vector bundle reduction.

Let $BG$ be the classifying space for stable spherical fibrations. To solve the obstruction-theory problem of lifting $X\to BG$ to a map $X\to BO$, we need to know the low-dimensional homotopy groups of $BO$ and $BG$ - specifically, whether $\pi_i(BO)\to \pi_i(BG)$ is surjective. I read off from a table in Ranicki's book "Algebraic and geometric surgery" that this is so for $i=1$ and $2$, but that $\pi_3(BG)=\mathbb{Z}/2$ whereas $\pi_3(BO)=0$. So there is an obstruction $o\in H^4(X;\mathbb{Z}/2)$ to finding a vector bundle reduction.

I'm a bit nervous of $ks$ due to my ignorance of topological manifold theory, but I think it should then be the case that $o=ks$ (they seem to be similar beasts; I'm thinking of $ks$ as coming from $\pi_3 (BTOP)$, where $o$ comes from $\pi_3(BG)$). What I actually want to use is the corollary, which if true should have a direct proof - that $o=\sigma/8$ mod 2. Anyone?

Given any unimodular matrix $Q$, I can build a Poincare 4-complex with $Q$ as its intersection matrix (plumb together disc-bundles over $S^2$ according to $Q$, cone off the homology 3-sphere boundary). If it's correct that $o=\sigma/8$, then when $Q=E_8$, I get a complex with no tangent bundle, whereas when $Q=E_8\oplus E_8$ I get a complex which has a tangent bundle but which is not smoothable by Donaldson's diagonalizability theorem.