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.


2 Answers 2


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.

  • $\begingroup$ So do you allow minimal surfaces have finitely many singular points? I want to be sure that the Monotonicity lemma still holds in that case. $\endgroup$
    – Hwang
    Sep 19, 2012 at 3:40
  • $\begingroup$ The monotonicity lemma does not require any further conditions, so it works. $\endgroup$ Sep 19, 2012 at 6:10
  • $\begingroup$ I hope so, but for example see Theorem 2.15 in page 24 in the link. Proper immersion is assumed and they didn't mention about branched minimal suraces. ugr.es/~jperez/papers/bamsJan11.pdf $\endgroup$
    – Hwang
    Sep 19, 2012 at 7:01
  • 1
    $\begingroup$ There's different versions I guess. But the lemma can be stated as "If $\Sigma$ is a surface in $B(r)\subset\mathbb{C}^n$ with boundary in $\partial B(r)$ and is area-minimizing for this boundary condition and $0\in\Sigma$, then $area(\Sigma)\ge\pi r^2$." In particular, our $C$ is minimal (as it is holomorphic) and satisfies the rest of the hypotheses of the Monotonicity lemma by construction, and so we are done. $\endgroup$ Sep 19, 2012 at 7:50
  • $\begingroup$ @ChrisGerig I'm very sorry for resurrecting this very old post, can you send me a reference for that version of the monotonicity lemma? I cannot prove it on my own and I want to understand it $\endgroup$
    – Riccardo
    Apr 6, 2019 at 20:15

I think what you are missing is that any region of a J-holomorphic curve in a symplectic manifold $(M,\omega)$ is always absolutely area minimizing in its homotopy class relative the boundary of the region, with respect to the compatible metric $g_J$. In particular if it is a closed curve it is minimizing in its homology class. This is explained for example in McDuff-Salomon's "J-Holomorphic curves in symplectic topology", in one of the first sections. But is actually a fairly simple calculation.

I don't think we need to assume immersed for this, the singularities will be isolated so that the argument goes through.

This basic phenomenon is generalized by calibration geometry of Lawson in higher dimensions. http://en.wikipedia.org/wiki/Calibrated_geometry

  • $\begingroup$ I'm not sure this works, because our curve $C$ which is pulled-back from the J-holomorphic sphere is not a priori closed, and in particular will have boundary lying on $\partial B(r)$. $\endgroup$ Sep 19, 2012 at 3:32
  • 1
    $\begingroup$ Well I didn't say it has to be closed. The argument just shows any (topologically reasonable) region of the curve will be minimizing relative boundary of the region. It's a Stokes theorem argument. I think this means in particular it is a minimal surface, as in being a critical point for the area functional. $\endgroup$
    – yasha
    Sep 19, 2012 at 4:02
  • $\begingroup$ Isolated singularities should not matter as you said, but I need some time to convince myself. Thank you. $\endgroup$
    – Hwang
    Sep 19, 2012 at 4:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.