# Hilbert schemes of points on toric surfaces

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?

• 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. Jan 24, 2019 at 21:06
• 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.
– naf
Jan 25, 2019 at 4:45
• 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.
– ssx
Jan 25, 2019 at 11:08