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I am trying to see if there is a way to translate the computation of the index of the Floer operator for Hamiltonian Floer to Lagrangian Floer. Hamiltonian Floer homology is a theory that counts (perturbed-)J-holomorphic cylinder connecting fixed points of Hamiltonian symplectomorphism. Let $$\mathcal F(u)=u_s+Ju_t-JX$$ where $X=X_t$ is a (time-dependent) Hamiltonian vector field, and

$$u=u(s,t):\mathbb R\times S^1\to W$$

is a cylinder in the symplectic manifold $W$. When we linearize $\mathcal F$, we get

$$d\mathcal F_u(Y)=\bar\partial Y+S(s,t)Y$$

where $S(s,t)$ is a zeroth order operator and $Y\in \Gamma(u^*TW)$. We can compute and show that $S^{\pm}(t):=S(\pm\infty,t)$ is symmetric and there is a way to create a homotopy of $S$ so that $\bar\partial Y+SY$ stays Fredholm and it turns $S$ into diagonal matrices. From that we can compute the Fredholm index directly and it equals to the Conley-Zehnder index. All the treatments are given in the book of Audin and Damian's Morse Theory and Floer Homology.

Is there a way to carry out a similar treatment for the Lagrangian Floer Homology? The theory counts J-holomorphic strip between two Lagrangians, i.e. loosely speaking,

$$\mathcal F(u)=u_s+Ju_t$$ where $$u=u(s,t):\mathbb R\times [0,1]\to W$$ $$u(s,0)\in L_1,$$ $$u(s,1)\in L_2,$$ $$u(\pm\infty,t)\in L_1\cap L_2.$$

Is there a way to rewrite this setup to the previous case, where we compute the index of Hamiltonian Floer operator?

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    $\begingroup$ mathoverflow.net/questions/43338/… (but otherwise in the reverse direction the issue is the Lagrangian boundary conditions). $\endgroup$ Commented Apr 5, 2023 at 23:01
  • $\begingroup$ In their example, the two Lagrangians are given by the diagonal of $W\times W$ and a graph of Hamiltonian. How does it generalize to arbitrary (transversal) pair of Lagrangians in $W$? Also, in their definition of Hamiltonian Floer, their equation is $u_s+J_tu_t$ and requires only $u(s,t+1)=f(u(s,t))$, instead of $u_s+Ju_t+JX_t$ with $u(s,t+1)=u(s,t)$. Are they equivalent? $\endgroup$
    – Y.H. Chan
    Commented Apr 7, 2023 at 17:37

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