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Question. Let $S$ be a non-singular complex projective surface and let $S^{[n]}$ be its Hilbert scheme of $n$ points. Is there a natural way to associate to a Kähler metric $\omega$ on $S$ a Kähler metric $\omega^{[n]}$ on $S^{[n]}$?

More generally, if $S$ is any (not necessarily projective) Kähler surface, is $S^{[n]}$ Kähler? In [1], Beauville writes "J'ignore si c'est toujours vrai" (translation: "I don't know if it's always true") in reference to this question, but that's 37 years ago, so maybe this is known now.

In case $S$ is projective, the Hilbert scheme $S^{[n]}$ is also projective and hence Kähler, but I was wondering if there is a more intrinsic and explicit construction. Naively, one can look at the $\mathfrak{S}_n$-invariant Kähler metric on $S^n$ induced by $\omega$ and ask if the corresponding Kähler metric on the smooth locus of $S^{(n)} = S^n / \mathfrak{S}_n$ extends to $S^{[n]}$.

Added. According to David E Speyer's comment below, $S^{[n]}$ is the blow up of $S^n$ along the reduced union of codimension 2 subvarieties. The blow up of a Kähler manifold is Kähler, but I am not aware of a way to construct a Kähler metric canonically. The only argument that I am aware of uses partitions of unitity, as in, e.g., Voisin [2, Proposition 3.24].

References.

[1] Beauville, A. Variétés Kähleriennes dont la première classe de Chern est nulle. J. Differential Geom. 18 (1983), no. 4, 755–782 (1984).

[2] Voisin, C. Hodge theory and complex algebraic geometry. I. Translated from the French original by Leila Schneps. Cambridge Studies in Advanced Mathematics, 76. Cambridge University Press, Cambridge, 2002.

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  • $\begingroup$ "...and ask if the corresponding Kähler metric on the smooth locus of $S^{[n]}$ extends". I do not understand: $S^{[n]}$ is smooth. Maybe you are confusing it with the symmetric product, of which it is a desingularization? $\endgroup$
    – Francesco Polizzi
    Commented Oct 5, 2021 at 14:29
  • $\begingroup$ @FrancescoPolizzi Yes, I meant $S^{(n)} = S^n / \mathfrak{S}_n$ (whose smooth locus is an open dense subset of $S^{[n]}$). $\endgroup$
    – Jost Schultze
    Commented Oct 5, 2021 at 15:38
  • $\begingroup$ I believe it follows from results of Haiman that $S^{[n]}$ is the blow up of $S^n$ along the reduced union of the $\binom{n}{2}$ codimension $2$ subvarieties $D_{ij} := \{ (x_1, \ldots, x_n) \in S^n : x_i = x_j \}$. Does that imply an answer? $\endgroup$
    – David E Speyer
    Commented Oct 5, 2021 at 17:34
  • $\begingroup$ $H^2$ of $S^{[n]}$ is one dimensional larger than $H^2$ of $S$. The Kähler metric on $S^{(n)}$ induced from one on $S$ is at the boundary of the Kähler cone because of this extra dimension. One should at least specify the Kähler class in the extra dimension. $\endgroup$
    – Hiraku Nakajima
    Commented Nov 20, 2021 at 0:48

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