# Why is Heegaard Floer Homology defined in terms of Sym$^g\Sigma_g$ instead of Pic$^g\Sigma_g$?

Recall the definition of Heegaard Floer homology: $\Sigma_g$ is a closed surface, and $\{\alpha_1,\ldots,\alpha_g\}$ and $\{\beta_1,\ldots,\beta_g\}$ are sets of attaching circles. Then Heegaard Floer homology is (more or less) the Lagrangian intersection Floer homology of $\mathbb T_\alpha=\prod_{i=1}^g\alpha_i$ and $\mathbb T_\beta=\prod_{i=1}^g\beta_i$ in $\operatorname{Sym}^g\Sigma_g$.

Now if we think of $\Sigma_g$ as a complex curve, then there is a birational map $\phi:\operatorname{Sym}^g\Sigma_g\to\operatorname{Pic}^g\Sigma_g$.

What happens if instead we consider the Lagrangian intersection Floer homology of $\phi(\mathbb T_\alpha)$ and $\phi(\mathbb T_\beta)$ inside $\operatorname{Pic}^g\Sigma_g$? Are the resulting groups trivially the same, trivially different, or at least interesting? (if they're not the same, then I guess there may be no good reason why they would even be invariants of the underlying three-manifold).

There is at least one concrete reason (and one philosophical reason) why one might try this definition instead of the original:

1. There are no holomorphic spheres in $\operatorname{Pic}^g\Sigma_g$ (because it is an abelian variety; in fact the map $\phi$ is exactly contracting all the embedded $\mathbb P^n$'s in the symmetric product). This means we don't have to worry about some types of bubbling.

2. $\operatorname{Pic}^g\Sigma_g$ is a complex torus; in particular its topology is very concrete and easy to understand. Also it is perhaps algebrogeometrically more natural than the symmetric product.

3. I could imagine that maybe there is some general statement whereby blowing down all the $\mathbb P^n$'s always does something understandable (perhaps nothing) to the Lagrangian intersection Floer homology.

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This paper: arxiv.org/abs/math/0006192 of Ozsvath-Szabo might be relevant as I seem to remember it does something along these lines (a precursor to HF). – Jonny Evans Jun 21 '12 at 18:37

## 1 Answer

There is a tacit assumption behind this question, which I don't think is justified: that the Abel-Jacobi images of the Heegaard tori $\mathbb{T}_{\alpha}$ and $\mathbb{T}_{\beta}$ are Lagrangian with respect to some reasonable symplectic form on the Jacobian torus.

One can make the Heegaard tori Lagrangian by using a Kaehler form on the symmetric product that is product-like outside some neighbourhood of the diagonal. And one can probably find symplectic forms for which Abel-Jacobi is a symplectomorphism outside a neighbourhood of the theta-divisor (this is certainly true in the genus 2 case). Doing both of these things at once would suffice to make the images Lagrangian, but this might be tricky to achieve - and it's perhaps not very natural?

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@Perutz: Why is this true for $g=2$? – Chris Gerig Jul 14 '12 at 1:03
Chris: what I had in mind is that in that case, the Abel-Jacobi map simply blows down a (-1)-sphere, and one could use a standard symplectic blow-down construction. – Tim Perutz Jul 14 '12 at 14:52