Cohomology theory for symplectic manifolds - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T06:17:52Z http://mathoverflow.net/feeds/question/69888 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/69888/cohomology-theory-for-symplectic-manifolds Cohomology theory for symplectic manifolds unknown (google) 2011-07-09T17:06:45Z 2011-07-09T21:36:31Z <p>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).</p> <p>Now I'd like to consider the following operation, and I'm hoping that it can be interpreted in some cohomology theory.</p> <p>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).</p> <p>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).</p> <p>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":</p> <p>$$\langle(L_1,s_1),(L_2,s_2)\rangle:=\sum_{x\in L_1\cap L_2}\operatorname{sign}(x)\langle s_1(x),s_2(x)\rangle$$</p> <p>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)?</p> <p>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.</p> http://mathoverflow.net/questions/69888/cohomology-theory-for-symplectic-manifolds/69903#69903 Answer by Eigenbunny for Cohomology theory for symplectic manifolds Eigenbunny 2011-07-09T19:36:22Z 2011-07-09T19:36:22Z <p>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 ...</p> <p>(... in fact, maybe you'd better look in the direction of Chern-Simons differential cocycles)</p>