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.

  • $\begingroup$ Why doesn't $H^\*(X,\mathcal L)$ make sense? Look at $\mathcal L$ as a sheaf... $\endgroup$ – Mariano Suárez-Álvarez Jul 9 '11 at 20:28
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    $\begingroup$ There is no natural good sheafification of $\mathcal L$. If I use smooth sections of $\mathcal L$, then the higher cohomology vanishes since it's a fine sheaf. To get twisted (co)homology in the context of sheaf cohomology, one needs the sheaf of locally constant sections of your bundle. I don't have a flat connection on $\mathcal L$, there's no notion of a "locally constant" section of $\mathcal L$. $\endgroup$ – John Pardon Jul 9 '11 at 20:41
  • $\begingroup$ Well, then your «doesn't make sense» is more of a «does not help at all»... That was my point :) $\endgroup$ – Mariano Suárez-Álvarez Jul 9 '11 at 20:49
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    $\begingroup$ If I recall correctly, in the special case when $M$ is the character variety of $SU(2)$ representations of a surface groups and the $L_i$ come from a Heegaard splitting of a 3-manifold this pairing was considered in some notes of Dennis Johnson, perhaps also by Kevin Walker in his extension of Casson's invariant. But I dont think the issue of whether this comes from an intersection pairing on some cohomology theory is considered. $\endgroup$ – Paul Jul 9 '11 at 22:34
  • $\begingroup$ It's true that you can twist ordinary cohomology by a line bundle with flat connection. That's one equivalent way to understand $H^*(M, L)$. If memory serves you can twist 2-periodic cohomology (i.e. add all the even/odd parts together) by a line bundle with a connection which is not necessarily flat. Maybe your $(L,s)$ pairs represent cycles in this twisted cohomology? $\endgroup$ – Chris Schommer-Pries Jul 9 '11 at 23:12

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)


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