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.
figure 0.0.2:A non-admissible Heegaard diagram If the diagram is chosen so that \(\alpha, \beta\) are disjoint, then the Lagrangian intersection Floer cohomology \(\HF(\alpha, \beta)\) vanishes.
figure 0.0.3:An admissible Heegaard diagram 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.