As is well known, the Hilbert scheme of two points on a given smooth projective variety X are blow up along diagonal of product of X and then quotient the Z2 action. It is smooth. My question is whether Hilbert schemes of 3 points on arbitrary smooth projective varieties are smooth. If so, why and how to describe the geometry of them?
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2$\begingroup$ For surfaces, $\textrm{Hilb}^[n]$ is smooth for any $n$ (this is a classical result of Fogarty) $\endgroup$– Francesco PolizziCommented Sep 5, 2012 at 11:48
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4$\begingroup$ A useful reference might be the article of Fantechi and Göttsche "The cohomology ring of the Hilbert scheme of 3 points on a smooth projective variety." J. Reine Angew. Math. 439 (1993), 147–158. $\endgroup$– nafCommented Sep 5, 2012 at 13:18
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1 Answer
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Yes, the Hilbert scheme of 3 points on a smooth variety is smooth. I don't know of a global description for the resulting Hilbert scheme, but here's the local reason this is true.
- Every length 3 scheme is abstractly isomorphic to a subscheme of the plane.
- For a zero dimensional subscheme $\text{Spec} A$ of a smooth variety $X$, there is a natural functorial map from embedded deformations of $\text{Spec} A\subseteq X$ to abstract deformations of $\text{Spec} A$, and this map is smooth.
Fact 1 is easy. Fact 2 takes more work, but it follows from some elementary arguments about deformation theory for affine schemes. Combining these facts: $\text{Spec} A$ is a smooth point of $\text{Hilb}^3 X$ if and only if the miniversal abstract deformation ring of $\text{Spec} A$ is smooth if and only if, after any reembedding of $\text{Spec} A$ into $\mathbb A^2$ we have that $\text{Spec} A$ is a smooth point of $\text{Hilb}^3 \mathbb A^2$; the last statement is true by Fogarty.
If somebody has a global description of the resulting Hilbert scheme, I would be very curious!
answered Sep 5, 2012 at 12:15
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1$\begingroup$ Thank you for your detailed answer. I am curious about the global description too. $\endgroup$– AllenCommented Sep 5, 2012 at 12:51