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.



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


