SympSnip:

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_{Σ_{1}} D^{2} \times S^{1}

. The Heegaard diagram

(Σ_{1}, α, β)

consists of a torus with two meridional cycles.

If the diagram is chosen so that

α, β

are disjoint, then the Lagrangian intersection Floer cohomology

{HF}^{∙} (α, β)

vanishes.

However, if the diagram is chosen so that

α, β^{'}

intersect transversely, the Lagrangian intersection Floer cohomology (with

Z / 2 Z

coefficients) is

Z / 2 Z \oplus Z / 2 Z

. Note that

β^{'}

can be chosen so that it is Hamiltonian isotopic to

β

. The discrepancy between these two answers comes from the non-convergence of the homotopy between the composition of continuation maps

f \circ g : {CF}^{∙} (α, β^{'}) \to {CF}^{∙} (α, β) \to {CF}^{∙} (α, β^{'})

id : {CF}^{∙} (α, β^{'}) \to {CF}^{∙} (α, β^{'})

over

Z / 2 Z

coefficients. The presence of an annulus between

α, β

is the culprit for the non-convergence. One can make the quantities converge by using Novikov coefficients instead of

Z / 2 Z

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

α, β

are Hamiltonian isotopic.