# Pseudo-holomorphic disk which is constant along boundary

Let $$(M,J,\omega)$$ be a symplectic manifold with a compatible almost complex structure, $$D$$ be the closed unit disk in $$\mathbb{C}$$, and $$u:(D,i)\to (M,J)$$ be a $$(J,i)$$-holomorphic map.

Question: Assume $$u|_{\partial D}$$ is constant, does this imply $$u$$ is a constant map?

## 1 Answer

Extend $$u$$ to get a $$C^1$$ pseudoholpmorphic map defined on $$\mathbb{C}$$ by setting $$u$$ constant outside the unit disc. It's $$C^1$$ because you know the derivative of $$u$$ along the unit circle vanishes (by assumption), so the Cauchy-Riemann equations satisfied by $$u$$ on the disc tell you that $$du$$ vanishes along the unit circle; clearly $$du$$ continues to vanish outside the disc, hence it's $$C^1$$. It's pseudoholomorphic because this is a pointwise condition on derivatives which clearly holds piecewise for this map. Now by unique continuation, $$u$$ is constant.

• Thank you Jonny! I now understand. Actually I am thinking about a seemingly harder question: if $u:D\to M$ is nonconstant pseudoholomorphic, then does $u|_{\partial D}$ have only (finitely many) isolated critical points? May I ask if you have an idea about this? – Yeah Jan 23 '20 at 2:49
• My guess is that the answer is yes, but I don't know for sure. – Jonny Evans Jan 23 '20 at 5:52
• I realized that boundary critical points are isolated if we add Lagrangian/ totally real boundary condition. This is due to a similar similarity principle as interior points. Since I only care about holomorphic disks bounded by a Lagrangian submanifold, it’s all done. – Yeah Jan 27 '20 at 3:36