\( \def\CC{{\mathbb C}} \def\RR{{\mathbb R}} \def\NN{{\mathbb N}} \def\ZZ{{\mathbb Z}} \def\TT{{\mathbb T}} \def\CF{{\operatorname{CF}^\bullet}} \def\HF{{\operatorname{HF}^\bullet}} \def\SH{{\operatorname{SH}^\bullet}} \def\ot{{\leftarrow}} \def\st{\;:\;} \def\Fuk{{\operatorname{Fuk}}} \def\emprod{m} \def\cone{\operatorname{Cone}} \def\Flux{\operatorname{Flux}} \def\li{i} \def\ev{\operatorname{ev}} \def\id{\operatorname{id}} \def\grad{\operatorname{grad}} \def\ind{\operatorname{ind}} \def\weight{\operatorname{wt}} \def\Sym{\operatorname{Sym}} \def\HeF{\widehat{CHF}^\bullet} \def\HHeF{\widehat{HHF}^\bullet} \def\Spinc{\operatorname{Spin}^c} \def\min{\operatorname{min}} \def\div{\operatorname{div}} \def\SH{{\operatorname{SH}^\bullet}} \def\CF{{\operatorname{CF}^\bullet}} \def\Tw{{\operatorname{Tw}}} \def\Log{{\operatorname{Log}}} \def\TropB{{\operatorname{TropB}}} \def\wt{{\operatorname{wt}}} \def\Span{{\operatorname{span}}} \def\Crit{\operatorname{Crit}} \def\CritVal{\operatorname{CritVal}} \def\FS{\operatorname{FS}} \def\Sing{\operatorname{Sing}} \def\Coh{\operatorname{Coh}} \def\Vect{\operatorname{Vect}} \def\into{\hookrightarrow} \def\tensor{\otimes} \def\CP{\mathbb{CP}} \def\eps{\varepsilon} \) SympSnip:

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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.
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.