6
$\begingroup$

Is Heegaard Floer homology the Lagrangian Floer homology of $M=\text{Sym}^g(\Sigma), L_0=\mathbb{T}_{\alpha},L_1=\mathbb{T}_{\beta}$? I am interested in the relationship between the theories

  1. Lagrangian Floer homology, as described in this paper, and
  2. Heegaard Floer homology, as described in this paper.

In Heegaard-Floer, from a Heegaard diagram for a closed oriented three-manifold $Y$, one forms the smooth manifold $\text{Sym}^g(\Sigma)$ and subspaces $\mathbb{T}_{\alpha},\mathbb{T}_{\beta}$.

I know that $\text{Sym}^g(\Sigma)$ is a symplectic manifold, of which $\mathbb{T}_{\alpha}$ and $\mathbb{T}_{\beta}$ are Lagrangian submanifolds. Hence, one can take Lagrangian Floer homology. For some reason, it was my understanding that Heegaard-Floer Homology is precisely the Lagrangian Floer Homology of $M=\text{Sym}^g(\Sigma), L_0=\mathbb{T}_{\alpha},L_1=\mathbb{T}_{\beta}$.

But I'm not sure about this, since the authors of paper 2. seem to construct their own chain complex and differential from scratch using several new ideas (albiet the story is much in the same vein as how one constructs Lagrangian Floer). In particular, nowhere in the paper is there any mention of the action functional.

My question:

  • Is Heegaard-Floer homology precisely the Lagrangian Floer homology of $M=\text{Sym}^g(\Sigma), L_0=\mathbb{T}_{\alpha},L_1=\mathbb{T}_{\beta}$?
  • If so, is this supposed to be apparent from paper 2. or is there another paper where this is proven?
  • If not, what would taking the Lagrangian Floer homology of $M=\text{Sym}^g(\Sigma), L_0=\mathbb{T}_{\alpha},L_1=\mathbb{T}_{\beta}$ give you? In what ways specifically is this different from the graded vector spaces one gets from Heegaard-Floer?
$\endgroup$
2
  • 6
    $\begingroup$ It’s not quite that straightforward, you need to remove a base point in the complement of the Heegaard curves that the pseudoholomorphic disks can’t go through. $\endgroup$
    – Ian Agol
    Commented Jun 27, 2023 at 4:17
  • $\begingroup$ After base-point considerations as Ian says, I think work of Perutz shows that you can choose all the auxiliary data you need (symplectic forms, etc.) so that HF^ agrees with HF. $\endgroup$ Commented Aug 23, 2023 at 21:30

0

You must log in to answer this question.