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Let $X$ be a smooth complex algebraic variety (or, better, complex analytic manifold). Let $\{C_i\}$ be a sequence of compact algebraic subvarieties (resp. analytic reduced subspaces) which converges to a compact reduced subspace $C$ in the sense of the analytic topology on complex points of the Hilbert scheme (resp. Douady space).

Question. Is it true that $\{C_i\}$ converges to $C$ in the sense of currents, namely for any smooth differential form $\omega$ on $X$ one has $$\lim_{i\to\infty}\int_{C_i}\omega=\int_C\omega?$$ A reference would be helpful.

Added: In the case when $C$ is not reduced, the right hand side in the last equality should probably be multiplied by an appropriate multiplicity.

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    $\begingroup$ this is true for Barlet space the analog of chow variety.See A Fujiki Publications RIMS vol 14 (1978) 1-52 proposition 2.3 $\endgroup$ – Mohan Ramachandran Sep 15 '14 at 16:41
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    $\begingroup$ See paper of Jon Magnusson Math Scandinavica vol 101 (2007) 19-28 A global morphism from Douady space to the cycle space which combined with Fujiki's paper answers your question . $\endgroup$ – Mohan Ramachandran Nov 3 '14 at 19:09
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For Kahler manifold this is true. First, notice that limits in the sense of Barlett or Douady clearly (up to multiplicity) coincide with the limits in Hausdorff topology on the set of compact subsets. This can be proven using the following argument: the Douady moduli space $D$ with the usual topology is Hausdorff, compact, and the map to the same space $H$ with the Hausdorff topology is continuous. By Bishop's theorem, $H$ is also compact. Bijective continuous maps between compacts are homeomorphisms.

Now, the map from $H$ to currents is continuous in currents topology (also by construction, or, if you like, by semicontinuity of Lelong numbers). The same argument proves that it is also a homeomorphisms (currents are compact for obvious reasons).

The same argument for complex manifold would work if the volume of the subvarieties is bounded, because this is the condition under which the Bishop's theorem remains true.

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  • $\begingroup$ Thank you. It will take me some time to digest your answer. $\endgroup$ – makt Sep 18 '14 at 16:17

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