Why is Heegaard Floer Homology defined in terms of Sym$^g\Sigma_g$ instead of Pic$^g\Sigma_g$? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T12:37:33Z http://mathoverflow.net/feeds/question/100263 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/100263/why-is-heegaard-floer-homology-defined-in-terms-of-symg-sigma-g-instead-of-pic Why is Heegaard Floer Homology defined in terms of Sym$^g\Sigma_g$ instead of Pic$^g\Sigma_g$? unknown (google) 2012-06-21T17:57:03Z 2012-07-13T01:47:48Z <p>Recall the definition of Heegaard Floer homology: $\Sigma_g$ is a closed surface, and <code>$\{\alpha_1,\ldots,\alpha_g\}$</code> and <code>$\{\beta_1,\ldots,\beta_g\}$</code> 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 $\operatorname{Sym}^g\Sigma_g$.</p> <p>Now if we think of $\Sigma_g$ as a complex curve, then there is a birational map $\phi:\operatorname{Sym}^g\Sigma_g\to\operatorname{Pic}^g\Sigma_g$.</p> <blockquote> <p>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).</p> </blockquote> <p>There is at least one concrete reason (and one philosophical reason) why one might try this definition instead of the original:</p> <ol> <li><p>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.</p></li> <li><p>$\operatorname{Pic}^g\Sigma_g$ 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.</p></li> <li><p>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.</p></li> </ol> http://mathoverflow.net/questions/100263/why-is-heegaard-floer-homology-defined-in-terms-of-symg-sigma-g-instead-of-pic/102089#102089 Answer by Tim Perutz for Why is Heegaard Floer Homology defined in terms of Sym$^g\Sigma_g$ instead of Pic$^g\Sigma_g$? Tim Perutz 2012-07-13T01:47:48Z 2012-07-13T01:47:48Z <p>There is a tacit assumption behind this question, which I don't think is justified: that the Abel-Jacobi images of the Heegaard tori <code>$\mathbb{T}_{\alpha}$</code> and <code>$\mathbb{T}_{\beta}$</code> are Lagrangian with respect to some reasonable symplectic form on the Jacobian torus. </p> <p>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?</p>