Chern classes of symplectic manifolds

I have seen that people assign chern classes to the tangent bundle of symplectic manifolds. This confuses me, because to my knowledge chern classes detect differences in the complex structures of vector bundles.

I know that there is a canonical way to assign almost complex structures $$J$$ to symplectic manifolds $$(M,\omega)$$. However, this mechnism seems to depend on a choice of metric $$g$$.

(This is because locally there exists a matrix $$A$$ such that $$\omega(v,w)=g(Av,w)$$ and we can define a complex structure $$J=Q^{-1} A$$ where $$Q^2=-A^2$$.)

So why is this well defined?

The inclusion $$U(n) \to Sp(2n,\mathbb R)$$ is a maximal compact subgroup, hence a homotopy equivalence. This means principal $$U(n)$$-bundles are equivalent to principal $$Sp(2n,\mathbb R)$$-bundles. (Here is a reference, also see nLab page for symplectic group.)

As for your question about the metric - any two compatible almost complex structures lead to isomorphic complex vector bundles, because the space of CACS is path-connected, while the space of topological vector bundles on a manifold is discrete.

In more detail, given two CACS $$J_0$$ and $$J_1$$, you can build a path of almost complex structures $$J_t$$ (cf. Auroux's notes for path-connectedness of space of CACS), which leads to a structure of complex vector bundle on $$p_M^*(TM) \to M \times I$$. Then use the fact that any family of (topological) vector bundles over a (paracompact) manifold is constant (cf. Prop 1.7 of Hatcher's K-theory).

• So you are saying that if I take two different metrics and construct the associated almost complex structure, then the resulting bundles are isomorphic, because the almost complex structures are both compatible with $\omega$? Commented Jan 24, 2019 at 15:26
• @klirk Sorry - I'm not carefully analyzing your description of how to produce the almost complex structure from the metric. However, if it's really canonical then it will be the same canonical assignment as mentioned in Auroux's note I linked to, which does produce a compatible (see the proposition there) ACS. I'll leave it to you to make sure that's really the construction you are talking about.
– Ben
Commented Jan 24, 2019 at 15:33
• thx for taking your time Commented Jan 24, 2019 at 15:44
• @klirk No problem. You might want to see the previous lecture in Auroux's course (ocw.mit.edu/courses/mathematics/…) for more notes about the polar decomposition. Actually if you follow the notes there, he uses pd to construct $\sqrt{-A^2}$ just as you are doing, so I'm pretty sure this is the construction you are referring to.
– Ben
Commented Jan 24, 2019 at 15:52
• @klirk Yes that's right.
– Ben
Commented Jan 24, 2019 at 16:02

This is equivalent to Ben's answer but let me write me thoughts (which is most likely a rephrasing).

The point is that given a symplectic form $$\omega$$ on your manifold, the space of tamed/compatible almost complex structures is $$\textbf{contractible}$$ and, in particular, path connected. Thus any such complex structure will give you the same homotopic invariants and, in particular, the same chern classes.

This is a usual phenomenon when you want to construct topological invariants in differential geometry. Think of of the Chern-Weil construction for the Chern classes. You have to choose various auxiliary data (like a metric) for your bundle and you use this data to construct your class $$\textbf{but}$$ the crucial step is showing that the topological information you get is independent of the choices you made.

• Thank you for your remark. When asking the question, I was probably unaware of the fact that the space of compatible complex structures is contractible or path connected. I still think however, that a more constructive proof would be insightful. Commented Oct 17, 2021 at 21:40