Let $\mathrm{S}$ be a smooth toric surface. The Hilbert scheme of $n$ points $\mathrm{Hilb}^n(\mathrm{S})$ inherits a torus action, but need not admit the structure of a toric variety itself. For instance, if $\mathrm{S}$ is the projective plane, then $\mathrm{Hilb}^2(\mathrm{S})$ is the $\mathbf{P}^2$-bundle $\mathbf{P}(\mathrm{Sym}^2\Omega(1))$ over the dual projective plane. (If this were a toric variety, then $\mathrm{Sym}^2\Omega(1)$ would split.) Nonetheless, the $\mathrm{Hilb}^n(\mathbf{P}^2)$ share quite a few properties with toric varieties (e.g., they are Mori dream spaces, the Hodge numbers satisfy $h^{p,q}=0$ for $p\neq q$).

On the other hand, if $\mathrm{S}$ is the affine plane, then $\mathrm{Hilb}^2(\mathrm{S})$ is toric, since it is isomorphic to the product of the affine plane with the blow up of a quadric cone at its vertex.

For which toric surfaces $\mathrm{S}$ and $n\geqslant 2$ does $\mathrm{Hilb}^n(\mathrm{S})$ admit the structure of a toric variety?

  • 3
    $\begingroup$ When $S$ is smooth projective, then the fact that $\mathrm{h}^0(S,\mathrm{T}_S)=\mathrm{h}^0(\mathrm{Hilb}^nS,\mathrm{T}_{\mathrm{Hilb}^nS})$ and the lower bound given by the dimension $2n$ on the latter in case it were toric should rule out all but some small cases if I'm not mistaken. $\endgroup$
    – pbelmans
    Jan 24, 2019 at 21:06
  • 3
    $\begingroup$ If $S$ is the affine plane, the natural map from $Hilb^n(S)$ to the symmetric product of $S$ is defined intrinsically, so the action of a torus on the Hilbert scheme would descend to the symmetric product. For $n \geq 3$, the nature of the singularities of the symmetric product (they are non-abelian quotient singularities) rules out its being toric. $\endgroup$
    – naf
    Jan 25, 2019 at 4:45
  • $\begingroup$ Thanks for your comments @pbelmans and @ulrich! So it seems that $\mathrm{Hilb}^n(\mathrm{S})$ is only very rarely toric, which is expected I think. $\endgroup$
    – ssx
    Jan 25, 2019 at 11:08


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