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?



Can symplectic blow up increase symplectic capacities?
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 $\infty$.


