Cohomology theory for symplectic manifolds

2011-07-09T17:06:45Z

Suppose I have a symplectic manifold $(M,\omega)$ and a line bundle $\mathcal L$ with a connection with curvature $\omega$ (or perhaps it's more standard to say $\frac i{2\pi}\omega$; anyway, the constant factor doesn't make any difference here).

Now I'd like to consider the following operation, and I'm hoping that it can be interpreted in some cohomology theory.

Consider objects of the form $(L,s)$, where $L\subset M$ is a lagrangian submanifold, and $s$ is a flat section of $\mathcal L|_L$ (note that since $\omega|_L=0$, the connection on $\mathcal L|_L$ is flat, so such sections exist at least locally).

Consider dual objects of the form $(L,s)$, where $L\subset M$ is a lagrangian submanifold, and $s$ is a flat section of $\mathcal L^\ast|_L$ (i.e. the dual bundle, with the induced connection).

Now if we have $(L_1,s_1)$ and $(L_2,s_2)$ (an object and a dual object), where $L_1$ and $L_2$ intersect transversally, then of course we can take the following quantity as their "pairing":

$$\langle(L_1,s_1),(L_2,s_2)\rangle:=\sum_{x\in L_1\cap L_2}\operatorname{sign}(x)\langle s_1(x),s_2(x)\rangle$$

Of course, this looks awfully similar to the intersection pairing (cup product) on $H^n(M)$ (say $\dim M=2n$). My question is: is there a (co)homology theory in which what I've written above is an honest intersection pairing (cup product)?

Of course, a natural thing to try is $H^n(M,\mathcal L)$ for the first type of object and $H^n(M,\mathcal L^\ast)$ for the dual object (singular homology with twisted coefficients). Then naturally the cup product goes to $H^{2n}(M,\mathcal L\otimes\mathcal L^\ast)=H^{2n}(M,\mathbb C)=\mathbb C$. But of course this is nonsense since $H^\ast(M,\mathcal L)$ doesn't make sense unless we specify a flat connection on $\mathcal L$, and our natural connection has curvature! Perhaps there's an easy fix that I'm missing.

Answer by Eigenbunny for Cohomology theory for symplectic manifolds

2011-07-09T19:36:22Z

How about looking at the homology of the homotopy fibre of the map $M \rightarrow K(\mathbb R,2)$ which represents your cohomology class? At least, any Lagrangian submanifold $L$ has the property that $L \rightarrow M \rightarrow K(\mathbb R,2)$ is (canonically) nullhomotopic, so gives rise to a homology class in the homotopy fibre. Absolutely no warranty that this is helpful ...

(... in fact, maybe you'd better look in the direction of Chern-Simons differential cocycles)

Cohomology theory for symplectic manifolds - MathOverflow

Cohomology theory for symplectic manifolds

Answer by Eigenbunny for Cohomology theory for symplectic manifolds