13
$\begingroup$

For a genus $g$ surface $\Sigma_g$, the mapping class group $\mathrm{Mod}(\Sigma_g)$ acts on the curve complex $\mathcal C(\Sigma_g)$: vertices being isotopy classes of essential, nonseparating, simple, closed curves in $\Sigma_g$ with an edge connecting two vertices if the curves can be made disjoint. In fact, there are many cousins of $\mathcal C(\Sigma_g)$ that have been extensively studied.

In one dimension higher, we have the symplectic mapping class group $\mathrm{SMod}(X)$ of some symplectic manifold $X$. We can build a complex $\mathcal L(X)$ in a similar way: the vertices of $\mathcal L(X)$ are the isotopy classes of Lagrangian spheres in $X$ with an edge connecting two if they are disjoint. Is anything known about this complex and/or it's relatives?

$\endgroup$
5
$\begingroup$

The short answer seems to be: "NO." The long answer seems to be (stolen from Dennis Sullivan): This is a good question - you should think about it!

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

I only just noticed this question, so maybe it's too late, but here's an answer.

Note that some symplectic manifolds (like $\mathbf{CP}^2$) contain no Lagrangian spheres, so this complex is then empty. Let's ignore that and focus on the other examples.

For certain simple symplectic 4-manifolds (small blow-ups of the projective plane or a quadric) this complex should be known explicitly (I can tell you the vertices and tell you what you need to check to get the edges). Let $X_k$ be the $k$-point blow-up of $\mathbf{CP}^2$ with its monotone symplectic form.

  • e.g. $S^2\times S^2$ with its monotone symplectic form has precisely one Lagrangian sphere up to Hamiltonian isotopy (due to Hind https://arxiv.org/abs/math/0311092), so your complex is just a point. Same for $X_2$ (https://arxiv.org/abs/0902.0540).
  • e.g. For $X_3$ and $X_4$, Lagrangian spheres are uniquely determined up to isotopy by their homology classes (https://arxiv.org/abs/0902.0540 again), which gives:
    • $X_3$: the classes $L_{12}=E_1-E_2$, $L_{13}=E_1-E_3$, $L_{23}=E_2-E_3$, $L_{123}=H-E_1-E_2-E_3$ (I'm counting the two orientations of $L$ as the same sphere, so $[L]$ and $-[L]$ give the same sphere). Just for homology reasons, $L_{ij}$ must intersect $L_{jk}$ for $i\neq k$. The only other thing to check is whether $L_{123}\cap L_{ij}$ can be made empty (no homological obstruction). If so then your complex is a $D_4$ Dynkin graph: otherwise it's a disconnected set of 4 points (I didn't think about which case holds, sorry).
    • $X_4$: similarly we get $L_{ij}=E_i-E_j$ and $L_{ijk}=H-E_i-E_j-E_k$. The possible edges are $L_{ij}$ to $L_{k\ell}$ when $\{i,j\}\cap\{k,\ell\}=\emptyset$ and $L_{ijk}$ to $L_{ij}$. If all these edges exist then your complex is obtained from a tetrahedron by putting $L_{ijk}$ at the four vertices and $L_{ij}$ on the edge between $L_{ijk}$ and $L_{ij\ell}$, then connecting opposite edges.
  • e.g. for $X_5$ you have similar homological data, but you can also focus on a single homology class and get infinitely many knotted spheres (as observed by Seidel https://arxiv.org/abs/math/0309012). These are nonetheless classified (Borman-Li-Wu, https://arxiv.org/abs/1211.5952): the symplectic Torelli group acts transitively on them, and in this case the symplectic Torelli group is the pure spherical 5-strand braid group (https://arxiv.org/abs/0909.5622). I haven't thought about how the Lagrangian "curve complex" would work in this example, but my guess is it would be closely related to the curve complex on the punctured sphere, in the following way. All the spheres are vanishing cycles for algebraic degenerations of $X_5$ coming from moving the 5 blow-up points. As three blow-up points become collinear, the blow-up develops a holomorphic $-2$-curve (proper transform of the line through these three points). If you take the anticanonical model then you get a nodal variety (and the Lagrangian sphere is the vanishing cycle of this node). To get 5 points on $S^2$ from this, look at the conic through the 5 blow-up points. You get a holomorphic sphere with 5 marked points. As the node forms, two of these marked points collide, and you can associate to the Lagrangian $S^2$ the simple closed curve which encircles these two marked points. If you can form two nodes simultaneously then these pairs of points collide simultaneously so the simple closed curves will be disjoint (as will the vanishing cycles).

In higher dimensions, little is known. However, by focusing attention on your favourite collection of spheres and computing Floer cohomology you should be able to figure out subcomplexes. Perhaps a more natural/manageable/useful construction here replaces vertices of the complex with quasiequivalence classes of objects (maybe spherical objects if you want to specify that) in the Fukaya category, joined by an edge when their Floer cohomology vanishes. This still admits an action of the symplectic mapping class group (though factoring through the autoequivalence group of the Fukaya category) which is what you'd probably want to use this complex for anyway.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks! I was actually able to find some of your lecture notes describing exactly this, after asking this question. $\endgroup$ – Reid Harris Jul 7 at 19:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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