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I have the following question. Any input would be appreciated. If you have a counterexample, I would be very interested to know.

Question. Let $(X, \omega, J)$ be a Kahler manifold of complex dimension $2$ of general type, with $\omega$ a symplectic form and $J$ an integrable complex structure compatible with $\omega$. Let $L$ be an ample line bundle on $X$. Assume the divisor $D$ corresponding to $L$ is connected, symplectic curve with a positive self-intersection. Is there is a surjection $\pi_{1}(D) \rightarrow \pi_{1}(X)$?

Remark: Note that here I don't assume that the divisor $D$ is holomorphic. In this case the answer to my question is yes by Nori's result. I am asking if Nori's result holds for symplectic submanifold.

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    $\begingroup$ Could you please define what you mean by a pesudo-holomorphic curve in your case? In order to do this, you need to define an almost complex structure on your Kahler surface, but have not explained this. Note also, that any Kahler manifold with an ample bundle is projective thanks to Kodaira embedding theorem. $\endgroup$
    – Dmitri Panov
    Commented Jul 29, 2012 at 11:31
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    $\begingroup$ @kla: "Dear Jason: In my question above, I assume that the manifold X is a complex surface of general type." In that case, what is your distinction between "pseudo-holomorphic" and "holomorphic"? A pseudo-holomorphic curve in a <complex> manifold is holomorphic. Once again, I think you should tell us what reference you are using, since you seem to be confused about basic definitions. $\endgroup$
    – Jason Starr
    Commented Jul 29, 2012 at 15:49
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    $\begingroup$ I think the question is badly phrased. Maybe the OP asks if a connected symplectic surface $D$ in $X$ whch is Poincaré dual to $c_1(L)$ with ample $L$ and $c_1(L)^2>0$ still verifies Nori's theorem. Am I right ? $\endgroup$
    – BS.
    Commented Jul 29, 2012 at 17:21
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    $\begingroup$ You should edit your question, then. $\endgroup$
    – BS.
    Commented Jul 30, 2012 at 9:30
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    $\begingroup$ ... In particular, remove integrability assumption and Kahler assumption. I am also unsure what "general type" would mean in the setting suggested by BS. $\endgroup$
    – Misha
    Commented Jul 30, 2012 at 11:04

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