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Let $M,M'$ be homeomorphic smooth, closed, simply connected 4-manifolds. Is it necessarily true that $w_2(TM)=w_2(TM')$ and $p_1(TM)=p_1(TM')$? If so, the comment on this post, shows that $TM$ and $TM'$ are topologically isomorphic as vector bundles.

If the above is false, how does the statement fail, i.e. do we have $w_2(TM)\neq w_2(TM')$, or $p_1(TM)\neq p_1(TM')$, or both?

rpf
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1 Answers1

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Yes, at least in the orientable case. First, the Stiefel-Whitney classes only depend on the Wu classes, which are homotopy invariants (they only depend on the cup product + Steenrod operations), and hence so are the Stiefel-Whitney classes, somewhat surprisingly. Second, by the Hirzebruch signature theorem, $p_1$ is determined by the signature (and an orientation), which is determined by oriented homotopy type, and so $p_1$ is a homotopy invariant in this case as well (the dependence on orientation cancels out).

I'm not sure what happens to $p_1$ in the nonorientable case. Maybe we should pass to the orientable double cover.

Qiaochu Yuan
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    On a non-oriented closed 4-manifold the map in top degree $H^4(M;\Bbb Z) \to H^4(M;\Bbb Z/2)$ is an isomorphism. So you only need to know what the mod 2 reduction of the Pontryagin class is. For this, use the formula $w_2^2(E) = p_1(E) \pmod 2$. Again one finds that applied to $TM$, these are homotopy invariants. –  Oct 22 '18 at 23:33