example 0.0.1
We look at the example of \(M=S^2\times S^1\). Observe that \(S^2=D^2\cup_{S^1} D^2\), so we can write \(M=D^2\times S^1 \cup_{\Sigma_1} D^2\times S^1\). The Heegaard diagram \((\Sigma_1, \alpha, \beta)\) consists of a torus with two meridional cycles. If the diagram is chosen so that \(\alpha, \beta\) are disjoint, then the Lagrangian intersection Floer cohomology \(\HF(\alpha, \beta)\) vanishes. However, if the diagram is chosen so that \(\alpha, \beta'\) intersect transversely, the Lagrangian intersection Floer cohomology (with \(\ZZ/2\ZZ\) coefficients) is \(\ZZ/2\ZZ\oplus \ZZ/2\ZZ\). Note that \(\beta'\) can be chosen so that it is Hamiltonian isotopic to \(\beta\). The discrepancy between these two answers comes from the non-convergence of the homotopy between the composition of continuation maps \[f\circ g:\CF(\alpha, \beta')\to \CF(\alpha, \beta) \to \CF(\alpha, \beta')\] \[\id: \CF(\alpha, \beta')\to \CF(\alpha, \beta')\] over \(\ZZ/2\ZZ\) coefficients. The presence of an annulus between \(\alpha, \beta\) is the culprit for the non-convergence. One can make the quantities converge by using Novikov coefficients instead of \(\ZZ/2\ZZ\) coefficients. In that setting, the differential in figure 0.0.2 will be exact unless the areas of the two strips agree --- that is, the Lagrangians \(\alpha, \beta\) are Hamiltonian isotopic.definition 0.0.4
A pointed Heegaard diagram \((\Sigma_g, \underline \alpha, \underline \beta, z)\) is a Heegaard diagram \((\Sigma_g, \underline \alpha, \underline \beta)\) along with a choice of point \(z\) disjoint from the cycles \(\underline \alpha, \underline \beta\).- Gromov-Compactness: Observe that this is defined over \(\ZZ/2\ZZ\) coefficients. In general, when we look at the strips connecting two points \(x, y\), we are only guaranteed compactness of the moduli space of pseudoholomorphic strips which have bounded energy. To get around this problem --- when strips from \(x\) to \(y\) have possibly unbounded energy --- we would employ Novikov coefficients in Lagrangian intersection Floer cohomology. Here, we do not use such a coefficient ring.
- If we treat \(X\) as a symplectic manifold, the Lagrangian submanifolds \(L_{\underline \alpha}, L_{\underline \beta}\) are not tautologically unobstructed (i.e. \(\omega(\pi_2(X, L))\neq 0\)). We therefore must rule out disk and sphere bubbling to show that the differential squares to zero.
lemma 0.0.5
Suppose that \(g>2\). Then \(\pi_2(\Sym^g(\Sigma_g))=\ZZ\). The generator of this group intersects \(Y_z\).lemma 0.0.6
Suppose that \(u: D^2\to L_\alpha\) is a holomorphic disk. Then the image of \(u\) intersects \(Y_z\).References
[Per08] | Timothy Perutz. Hamiltonian handleslides for Heegaard Floer homology, 2008. |