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Here is my first try at a question, which is a really easy to state question about displaceability:

Let $D$ be the unit disk in the complex plane $D = \{ |z| \leq 1 \}$ equipped with its standard symplectic form and for $r \in (0,1)$ let $S(r) \subset D$ be the Lagrangian circle of radius $r$ centered at $0$, enclosing area $\pi r^2$.

Question: for which $r_1,\ldots, r_n$ is the Lagrangian torus $S(r_1) \times \ldots \times S(r_n)$ Hamiltonian displaceable in $D^n$?

Conjecture: for a given $r_1,\ldots, r_n$, the Lagrangian is non-displaceable iff $r_j^2 \ge 1/2$ for all $j$.

Known: Using McDuff's probes, one can see that if some $r_j^2 < 1/2$, then the Lagrangian is displaceable. I think I can show that if each $r_j^2$ lies in the set $\{ 1/2, 2/3,3/4, \ldots \}$ then the Lagrangian is non-displaceable, by embedding D^n in a product of weighted projective lines and showing the Floer homology of the Lagrangian is non-vanishing. But there must be a way of doing better.

Sub-question: (which came out of a discussion with Abouzaid): is there a Floer-theoretic way of seeing the non-displaceability of $S(r)$ for $r^2 \ge 1/2$? This is easy to prove by elementary means.

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  • $\begingroup$ Is part of your argument: if $L$ is the product of the equators in a product of integer scalings of $S^2$ then $FH(L,L)$ is non-trivial? If so why does this only work for integer scalings? $\endgroup$
    – Thomas Kragh
    Commented May 21, 2010 at 13:12
  • $\begingroup$ Welcome, Chris! Thomas: a product of $S^2$'s is monotone only when the factors have the same area. Without this condition, you have to worry about obstructions to defining Floer homology. I suspect what Chris has in mind is to work in a product of orbifolds (weighted projective lines) in which he can make the Lagrangian monotone, and then check that the bounding holomorphic discs cancel algebraically (as they do for the equator in $S^2$) so that Floer homology is non-zero. $\endgroup$
    – Tim Perutz
    Commented May 21, 2010 at 16:30

1 Answer 1

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I think Thomas is right: taking a different area form on $S^2$, but still taking the Lagrangian to be the equator is equivalent to changing r, since the unit disk still embeds symplectically as long as the two-sphere has area at least \pi. Then non-vanishing Floer homology of the product of equators in the product of $S^2$'s implies non-displaceability in the product of unit disks. Thanks Thomas, I missed that!

Here is a similar problem that this technique solves: take the ball $B_r$ of radius r in $C^2$ centered at zero, and a product of circles $S(r_1) \times S(r_2)$ contained in $B_r$. When is $S(r_1) \times S(r_2)$ displaceable? Thomas' suggestion, applied to the embedding in $CP^2$, gives non-displaceability for $r_1^2 = r_2^2 \ge 1/3$ for $r = 1$. Using probes this is the complete answer.

  • Chris
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  • $\begingroup$ Good point! But I'm trying to understand how one would actually define and compute Floer homology here. Is the idea that the obstruction chains vanish by a symplectic involution argument, and then that the Oh spectral sequence degenerates for much the same reason? I think equivariant regularity should be OK, as there don't seem to be non-trivial discs fixed by the involution. $\endgroup$
    – Tim Perutz
    Commented May 21, 2010 at 19:31
  • $\begingroup$ Hi Tim, Yes, there is potentially an an obstruction here, but the equator is a special case of the Clifford torus handled by Cho in arxiv.org/abs/math/0308224. In general, the obstruction is explicitly computed for Fano toric varieties, by Fukaya et al in arxiv.org/abs/0802.1703, and I had a different approach in arxiv.org/abs/1004.2841 which gives a version of Floer homology in the non-Fano case without vfc's. $\endgroup$
    – Chris Woodward
    Commented May 21, 2010 at 19:54
  • $\begingroup$ (ctd) just to add a little explanation for Tim's question, even though product of equators isn't monotone, but still one can understand the moduli space of Maslov index two disks in this case using that there are no spheres of negative index, and the disks can be explicitly described. So then one can check that the Floer cohomology is still well-defined, and it is non-vanishing exactly for the product of equators. $\endgroup$
    – Chris Woodward
    Commented May 21, 2010 at 21:11
  • $\begingroup$ (ctd) In general, there are spheres of negative index which Fukaya et al have to deal with, but no negative index disks, which is something of a miracle special to the toric case. In my paper I used a different version of Floer cohomology which counts "quasimaps", which then avoids the negative index problem and still obstructs displaceability. $\endgroup$
    – Chris Woodward
    Commented May 21, 2010 at 21:12

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