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Suppose $(X,I)$ is an almost-complex real analytic manifold where $I$ is a real-analytic almost complex structure. Suppose there exists an $I$-almost complex submanifold $M\subset X$ where this means that $M$ is a real analytic submanifold such that given a tangent vector $v$ to $M,$ then $I(v)$ is also tangent to $M.$ Suppose that the restriction of $I$ to $M$ is integrable. The question is: are there conditions under which this implies that $I$ is integrable on all of $X?$ I would be supremely surprised if the condition was empty, namely that this always implied that $I$ was integrable, although I wouldn't look a gift horse in the mouth. Nevertheless, a condition such as the restriction of $I$ to $M$ being Stein, or something of such a global nature, doesn't sound quite so ridiculous.

I should add that in the exact situation I'm considering, $X$ is non-compact (it's topologically a ball) and $M$ is non-compact (also topologically a ball) and the restriction of $I$ to $M$ is Stein, and lots of other nice things you might dream it to be as a complex manifold.

Edit: I've just realized that I need to add that $M$ is not of complex dimension one, otherwise the integrability of $I$ restricted to $M$ is a trivial consequence of there being no $(2,0)$-forms on $M.$

Thanks for any comments or ideas you might have.

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  • $\begingroup$ I think it would be messy to write down an example, but my intuition says that there should be lots of examples. Since a ball is Stein, we just take the usual complex structure on the ball $\mathbb{B}^3 \subset \mathbb{C}^3$, and perturb it as an almost complex structure away from a complex linear subspace $\mathbb{C}^2$. I think we will get no other complex surfaces in the resulting $X$ other than $M=\mathbb{B}^2 \subset \mathbb{C}^2$. $\endgroup$
    – Ben McKay
    Nov 23, 2014 at 10:05
  • $\begingroup$ Also if you start with $M\subset X$ of complex dimension $1$, then $M\times\mathbb{C}\subset X\times\mathbb{C}$ will be of complex dimension $2$ and it will be Stein if $M$ is. You still won't be able to 'extend' integrability in such a case. Starting with $X$ having complex dimension $2$, you can see that you can have $M\times\mathbb{C}^k\subset X\times\mathbb{C}^k$ for any $k$, so even codimension $1$ won't help you without some further hypotheses. $\endgroup$ Nov 23, 2014 at 10:05
  • $\begingroup$ For better or worse, I think these comments close the question. I'm not sure how to proceed, but thanks. $\endgroup$ Nov 23, 2014 at 14:19

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