Given a real vector bundle, say the tangent bundle of a manifold, some obstructions to this being the underlying real bundle of a complex bundle come from characteristic classes. These can prove the lack of complex structure on the underlying manifold. I was wondering if perhaps there was a similar set of classes for symplectic manifolds?

  • $\begingroup$ Given a symplectic manifold, there exists a contractible set of almost complex structures compatible with the symplectic structure. These almost complex structures give a complex vector bundle structure on the tangent bundle. You can define characteristic classes to be the chern classes of this complex vector bundle. $\endgroup$ – Thomas Rot Jun 13 '16 at 8:41

You have classifying spaces $BSp(n)$ and $BGl(2n)$ and from the inclusion $Sp(n)$ into $Gl(2n)$ you get a map $\iota \colon BSp(n) \to BGl(n)$.

A characteristic class for a real vector bundle is nothing else than a cohomology class of $BGl(n)$ (This follows from naturality). And you can define characteristic classes analogously for symplectic bundles by taking cohomology classes for $BSp(n)$. A symplectic structure of your vector bundle now corresponds to a lift of the classifying map to $BGl(n)$ into $BSp(n)$.

But now if you understand $\iota^*$ you can get that certain characteristic classes can't occure if you have such a lift. For example if you were looking at real bundles and asking for a complex structure you would get that every odd Stiefel-Whitney-Class would have to vanish since the complex grassmanian has no cohomology in odd dimensions.

I'm not aware of how the cohomology ring of $BSp(n)$ looks like, but with that information and a little computation of $\iota^*$ you would get explicit classes, that obstruct such a lift.

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    $\begingroup$ I think $BSP(n)=G_n(\mathbb{H}^\infty)=\mathbb{Z}[a_1,\ldots, a_n]$ with degree $a_i=4$, so all stiefel whitney classes in degree $0,1,2,3 \pmod 4$ have to vanish, and probably $c_n=0$ for $n$ odd as well. $\endgroup$ – Thomas Rot Jun 13 '16 at 8:43

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