TITLE: Hamiltonian displaceability of tori in symplectic balls
QUESTION [12 upvotes]: 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.

REPLY [5 votes]: 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