In a symplectic manifold $(M, \omega)$, $J$-holomorphic curves are special minimal surfaces if $J$ is compatible with the symplectic structure and the metric is induced by $\omega$ and $J$.

Minimal surfaces satisfy the monotonicity property, i.e., if $S$ is a minimal surface passing through $p$, then for $r$ small enough, the area of $S \cap B_r(p)$ is at least $c r^2$, for some constant $c>0$ depending on the metric. This property can be used to prove, for example, the removal of singularity of $J$-holomorphic curves, and also, the diameter of a $J$-holomorphic curve cannot be arbitrarily large if the energy (area) is bounded.

In Gromov-Witten theory, one also consider perturbations of the Cauchy-Riemann equation, as $\overline\partial u = \nu(u)$, for other reasons. If just to prove removal of singularity, we can view the graph of a solution as a holomorphic curve in $\Sigma \times M$ (if domain is $\Sigma$), with respect to some almost complex structure on $\Sigma \times M$ depending on $\nu$.

But my question is, if $\nu$ is sufficiently small (in $C^0$, for example), can we prove a monotonicity result for a solution $u$ to $\overline\partial u = \nu(u)$ (but not its graph)? I suspect this holds because for $\nu$ small, a solution should be close to a real $J$-holomorphic curve.


1 Answer 1


I think that no monotonicity theorem of the above form holds.

For example, consider a constant holomorphic curve, then perturb this to a nonconstant map $u$ with 0 area, and small $\bar \partial u$. This curve will violate your monotonicity inequality.

That said, this rules out only the most mindless attempt to generalize the monotonicity inequality for holomorphic curves. Maybe some inequality involving area + the integral of $\bar \partial u$ still applies?

  • $\begingroup$ Yes, I see what the problem is. There could be very "thin" solutions. $\endgroup$
    – Guangbo Xu
    Sep 20, 2013 at 16:10

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