Let $f,g:(D^2,j_\mathrm{std})\to(B^{2n}(1),J)$ be two pseudo-holomorphic maps. The following unique continuation result is well-known (it may be proved using either Aronszajn's Lemma or the Carleman similarity principle as in Floer--Hofer--Salamon):

Lemma: Suppose $f=g$ over a neighborhood of $0\in D^2$. Then $f=g$ everywhere.

Is there a similar unique continuation result for pseudo-holomorphic submanifolds (instead of maps)? Here is a precise formulation of what I expect might be true:

Conjecture: Let $U,V\subseteq(D^2)^\circ$ be two closed topological disks with smooth boundary, and let $\phi:U^\circ\to V^\circ$ be a biholomorphism (which necessarily extends continuously to a homeomorphism $U\to V$). Suppose $f=g\circ\phi$ over $U$. Then $\phi$ extends holomorphically to an open neighborhood of $U$ (and hence $f=g\circ\phi$ over this neighborhood by the unique continuation result for maps).

Is some statement along these lines known?

  • $\begingroup$ It looks like Theorem 2.83 in Chris Wendl's book arxiv.org/abs/1011.1690v2 might be what I'm looking for. $\endgroup$ – John Pardon Sep 4 '15 at 12:55
  • $\begingroup$ It seems that you have found the answer, but I thought of a simpler approach (maybe wrong): what about the subset of $D^2\times D^2$ defined by $g(x)=f(y)$ ? It should be complex analytic (for the standard complex structure), and your assumption amounts to say that it contains the graph of your $\phi:U\to V$. Maybe the assumption of smoothness of boundaries of $U,V$ and local structure of complex analytic curves would suffice to conclude ? $\endgroup$ – BS. Sep 5 '15 at 19:38

A partial answer is yes.

Given that the domain has a real analytic boundary the construction is standard. A Schwarz reflection can be performed locally near each boundary point, which together with normal analytic continuation produces the sought extension.

In general, I propose the following approach. Assume that $f(D^2) \subset g(D^2)$. The critical points of $g$ (and $f$) form a discrete set by the aforementioned similarity principle. Away from the critical points $\phi^{-1}(\mathrm{Crit}(g))$, we can write $\phi=g^{-1} \circ f$ and in this way obtain a unique continuation along small punctured discs covering the boundary of $U$. However, since $\phi$ is bounded along the boundary by assumption, the removal of singularities theorem can be applied to complete $\phi$ over the punctures.

Remark. At least in real dimension four, the assumption $f(D^2) \subset g(D^2)$ is not too severe since two unparametrised pseudo-holomorphic curves intersect in a discrete set, as was show in [McDuff; The Local Behaviour of J-holomorphic Curves in Almost Complex 4-manifolds].

(As a particular case, which however is no longer relevant given the new general formulation of the question: Every biholomorphism $\phi$ of $D^2(r)^o \subset \mathbb{C}$ which has a continuous extension to $D^2(r)$ extends to a biholomorphism of the Riemann sphere. Here we use a Schwarz reflection along $\partial D^2(r)$.)

  • $\begingroup$ Indeed, the question was sort of trivial as I originally phrased it. I've now modified it to be nontrivial. $\endgroup$ – John Pardon Sep 4 '15 at 8:59
  • $\begingroup$ I agree. Hopefully the new answer gives the complete picture for your question. For the applications, I would imagine that you can get away by assuming a real analytic boundary (cover your Riemann surface with small analytic discs). But OK, here I'm just guessing what applications you have in mind. $\endgroup$ – Nikolaki Sep 4 '15 at 9:11
  • $\begingroup$ Why is the answer no in general? I think you are missing the hypothesis that $f=g\circ\phi$ (which is really the crux of the question). $\endgroup$ – John Pardon Sep 4 '15 at 9:17
  • $\begingroup$ Of course, you are right. Let me try to fix it by applying the application I had in mind, hopefully it will work... $\endgroup$ – Nikolaki Sep 4 '15 at 9:25
  • $\begingroup$ I don't understand. As you remarked before, the answer is clearly "no" if we omit the hypothesis $f=g\circ\phi$. Now you claim the answer is "yes", but your argument does not use the fact that $f=g\circ\phi$. Note that $\phi$ need not send your disks with analytic boundary to other disks with analytic boundary. $\endgroup$ – John Pardon Sep 4 '15 at 10:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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