MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
This paper: 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
up vote 8 down vote accepted

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?

share|cite|improve this answer
@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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.