# Hamiltonian displaceability of tori in symplectic balls

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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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? –  Thomas Kragh May 21 '10 at 13:12
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. –  Tim Perutz May 21 '10 at 16:30

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.