Timeline for If a compact real submanifold of $\mathbb{CP}^n$ is approximable by complex algebraic curves, is it algebraic?
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Feb 20, 2015 at 1:23 | comment | added | Vesselin Dimitrov | @DavidSpeyer: Taking $Z_d$ to be the degree $d$ Fermat curve in $\mathbb{CP}^2$, as in Demailly's paper, the supports converge in the Hausdorff topology to the set $S$ consisting of those $[z_0:z_1:z_2]$ having $|z_i| \leq |z_k| = |z_l|$ for some permutation $i,j,k$ of $0,1,2$. This is not exactly a real manifold, but it comes close; and it is not algebraic. This suggests that the answer to my question is negative. | |
| Feb 20, 2015 at 0:09 | comment | added | Vesselin Dimitrov | @DavidSpeyer: This means looking at the $Z_i$ as defining closed, positive $(1,1)$-currents of integration (normalized to have total mass one). Then their limit, if it exists, is also closed and positive as $(1,1)$-current, but it need not be an analytic cycle. (I think a counterexample is contained in the paper you pointed to?) Even if the limit is an analytic cycle $Z$, though, this $Z$ is certainly of complex dimension one, whereas the problem with the Hausdorff notion of convergence is most interesting when the $M$ has $\dim_{\mathbb{R}}{M} > 2$, with the complex curves wrapping around it. | |
| Feb 19, 2015 at 19:01 | comment | added | David E Speyer | 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. | |
| Jul 28, 2014 at 14:53 | comment | added | Mohan Ramachandran | Yes M will be compact.One does not need any additional assumptions . | |
| Jul 27, 2014 at 21:53 | comment | added | Vesselin Dimitrov | 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). | |
| Jul 27, 2014 at 20:46 | comment | added | Vesselin Dimitrov | 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). | |
| Jul 27, 2014 at 18:08 | comment | added | Mohan Ramachandran | 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 . | |
| Jul 25, 2014 at 7:56 | history | asked | Vesselin Dimitrov | CC BY-SA 3.0 |