# If a compact real submanifold of $\mathbb{CP}^n$ is approximable by complex algebraic curves, is it algebraic?

To make this into a separate question:

If the supports of a sequence of complex algebraic curves in $\mathbb{CP}^n$ (images of non-constant holomorphic maps from compact Riemann surfaces) converge to a real compact immersed submanifold $M \subset \mathbb{CP}^n$, must this submanifold $M$ be complex (algebraic)?

For some motivation for this problem, see here:

If there is a dense geodesic, are almost all geodesics equidistributed? Dense?

Note that if $M$ is complex algebraic, there is certainly a sequence of complex algebraic curves with supports converging to $M$. (In fact, they could be taken lie in $M$). What I ask about is the converse. Unless it has an easy counterexample, this question is somewhat in the spirit of a few famous problems about special subvarieties in diophantine geometry, such as the Manin-Mumford and Andre-Oort questions, or Lang's algebro-geometric conjecture.

• If the convergence is in the Hausdorff topology for compact sets then M is complex algebraic if there is a bound on the degrees of the sequence .In this case M only needs to be a nonempty closed set .This follows from Bishop's convergence theorem for analytic sets along with Chow's theorem .This also works if you have a sequence of pure p dimensional algebraic sets . – Mohan Ramachandran Jul 27 '14 at 18:08
• Thank you for noting this! In this case of bounded degrees, the conclusion is thus that $M$ is in fact an algebraic curve (resp. algebraic set of dimension $p$). I am interested, though, primarily in the case where $M$ is higher dimensional, with the curves wrapping around it and therefore - necessarily - having degrees going to infinity (which is indeed easily seen to be possible if $M$ is complex algebraic). – Vesselin Dimitrov Jul 27 '14 at 20:46
• And yes, I do mean convergence in the Hausdorff topology for compact sets. By the way, I would also like to know whether the submanifold $M$ must be compact - even if we assume from the outset that the limit $M$ is a complex submanifold (or the closure of a complex submanifold). – Vesselin Dimitrov Jul 27 '14 at 21:53
• Yes M will be compact.One does not need any additional assumptions . – Mohan Ramachandran Jul 28 '14 at 14:53
• Commenting to point out that there is a literature studying a different notion of convergence: Thinking of $\frac{1}{\deg Z_i} \int_{Z_i}$ as a linear functional on smooth $(1,1)$-forms and looking at the limit in the space of linear functionals on $\Omega^{(1,1)}$ with the weak topology. See ams.org/mathscinet-getitem?mr=679762 and the papers that cite it. But I don't see how to make this relevant to Hausdorff convergence. – David E Speyer Feb 19 '15 at 19:01