MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Question. Is there any example of a path connected symplectic manifold $(M,\omega)$ that has infinite volume, but which cannot be packed by an infinite number of symplectic balls of a fixed radius $r$, for any $r > 0$?

Note that on such a manifold we would be able to prove the "Poincaré recurrence" of every symplectomorphism:

Given a symplectomorphism $T: M \rightarrow M$ and an open set $U \subset M$ there exists an integer number $k > 0$ such that $T^k(U)$ intersects $U$.

Added and edited: It seems to me that the examples suggested in the comments assume the validity of the following (to my knowledge unproved) statement:

Let $(M,\omega)$ be the symplectic manifold obtained by taking $B(R)$, the symplectic ball of radius $R$ and dimension 2n, and attaching a long thin cylinder with very large volume and very small capacity. Let $r < R$ and consider the ball $B(r)$. There exists constant $c > 0$ such that any symplectic embedding of $B(r)$ into $M$ intersects the ball $B(R) \subset M$ in a set whose volume is at least $c$.

Note that this is not quite the intuition of non-squeezing theorem which would just say that you cannot embedd $B(r)$ into $M$ if $r > R$.

Added remark (10/10/2013). I just had a conversation around this problem with Leonid Polterovich. He tells me that in the early nineties Hofer had asked him whether symplectic geometry could perhaps be used to sharpen the Poincaré recurrence theorem. This question was part of the motivation for his work with McDuff on packing obstructions.

share|cite|improve this question
My intuition is that Gromov non-squeezing might make the following example work: Start with the stupid example of a sequence of disjoint balls with radius converging to 0, but with infinite volume. Then connect these by very thin tubes (thin in the sense that Gromov's non-squeezing theorem doesn't allow large symplectic balls to pass through them.) – Brett Parker Oct 4 '13 at 5:17
Can I not take the infinite "cylinder" whose metric has decreasing ends as $\sim\frac{1}{ln(r)}$? (and then apply Gromov nonsqueezing) – Chris Gerig Oct 4 '13 at 6:04
@BrettParker and Chris Gerig, I think you are implicitly assuming that because one cannot symplectically displace a ball through a thin tube, one cannot displace a large part of its volume through it. – alvarezpaiva Oct 5 '13 at 21:02
I agree that my suggestion relies on this statement about not being able to pass a large volume of a ball through a thin tube. To me, this statement does not seem to follow from the usual argument for Gromov non-squeezing, and after thinking for a little while, I'm inclined to think that it is false. – Brett Parker Oct 6 '13 at 1:10

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.