The following is a part of the proof of Gromov nonsqueezing theorem.
The existence of a $J$-holomorphic curve gives an upper bound for the radius of a symplectically embedded ball.
Let $\psi: B(r) \rightarrow M$ be a symplectic embedding of a ball (of dimension equal to dim $M$) of radius $r$ centered at the origin into a symplectic manifold $(M, \omega)$.
An $\omega$-compatible almost complex structure $J$ on $M$ is chosen so that, on the image of $\psi$, it coincides with the pushforward of the standard complex structure by $\psi$.
Let $u: S^2 \rightarrow M$ be a $J$-holomorphic sphere passing through the point $\psi(0)$.
Then the preimage $C$ of this holomorphic sphere by $\psi$ is a minimal surface in $B(r)$ with boundary in $\partial B(r)$.
By the monotonicity formula, the area of $C$ is at least $\pi r^2$, which is the area of the flat $2$-disk.
This gives an upper bound on $r$.
Now I have a question. How do we consider $C$ as a minimal surface in $B(r)$? A minimal surface is a surface with mean curvature zero, and the mean curvature is defined on the image of an immersion. But a $J$-holomorphic curve is not an immersion in general. How do we deal with critical points?
One possible option seems to be taking critial points off. But then the immersion is not proper and the proof of the monotonicity formula seems to use properness. For example, the proof in the book "holomorphic curves in symplectic geometry" uses a compactly supported vector field.
I am not familiar with minimal surfaces, so I must be missing something. Any comment is appreciated.
Best Answer
So the key is that we are taking the standard complex structure on $\mathbb{C}^n$ (hence the ball in it). Then the pseudo-holomorphic preimage $C$ is actually a holomorphic curve, and such surfaces are minimal. Then you apply the Monotonicity lemma to get your bound.