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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?

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1 Answer 1

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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.

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  • $\begingroup$ 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? $\endgroup$
    – Yeah
    Commented Jan 23, 2020 at 2:49
  • $\begingroup$ My guess is that the answer is yes, but I don't know for sure. $\endgroup$ Commented Jan 23, 2020 at 5:52
  • $\begingroup$ 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. $\endgroup$
    – Yeah
    Commented Jan 27, 2020 at 3:36

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