# Does pseudo-holomorphic *submanifolds* satisfy unique continuation?

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?

• It looks like Theorem 2.83 in Chris Wendl's book arxiv.org/abs/1011.1690v2 might be what I'm looking for. – John Pardon Sep 4 '15 at 12:55
• 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 ? – BS. Sep 5 '15 at 19:38

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 $\partial D^2(r)$.)

• Indeed, the question was sort of trivial as I originally phrased it. I've now modified it to be nontrivial. – John Pardon Sep 4 '15 at 8:59
• 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. – Nikolaki Sep 4 '15 at 9:11
• 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). – John Pardon Sep 4 '15 at 9:17
• Of course, you are right. Let me try to fix it by applying the application I had in mind, hopefully it will work... – Nikolaki Sep 4 '15 at 9:25
• 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. – John Pardon Sep 4 '15 at 10:06