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Let $N$ be a symplectic submanifold of $M$. Symplectic blow up of $M$ along $N$ is an operation replacing a tubular neighborhood of $N$ with the projectivization of that neighborhood. So it decreases the volume. I have a question on the change of symplectic capacities.

A symplectic capacity $c$ is a function from the set of symplectic manifolds to $[0, \infty]$ satisfying

  1. $c(M_1) \leq c(M_2)$ if we can embed $M_1$ into $M_2$ symplectically,

  2. $c(M, k\omega) = |k| c(M, \omega)$ for $k \neq 0$, and

  3. $c(B^{2n}(r)) = c (B^2(r) \times \mathbb{R}^{2n-2}) = \pi r^2$, where $B^{2n}(r)$ is a $2n$-dimensional ball of radius $r$.

Symplectic capacities may not change after symplectic blow ups. But it seems to me that it is impossible that symplectic blow ups increase symplectic capacities. I couldn't prove this. Can symplectic blow up increase symplectic capacities?

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Neat question. Seems plausible for Gromov width. What is your motivation? –  Nate Bottman Oct 27 '13 at 16:57
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1 Answer

up vote 1 down vote accepted

The answer is yes.

Let $c$ be the Gromov width except we put $c(M)=\infty$ if $M$ admits an embedding of $B^{2n}(r)$ with $0$ blown up for some $r$.

Using that Gromov width is a capacity it is easy to check 1-3 above, and blowing up $B^{2n}(1)$ at 0 changes this capacity from 1 to $\infty$.

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