8
$\begingroup$

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?

$\endgroup$
2
  • 2
    $\begingroup$ For surfaces, $\textrm{Hilb}^[n]$ is smooth for any $n$ (this is a classical result of Fogarty) $\endgroup$ Sep 5, 2012 at 11:48
  • 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. $\endgroup$
    – naf
    Sep 5, 2012 at 13:18

1 Answer 1

9
$\begingroup$

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.

  1. Every length 3 scheme is abstractly isomorphic to a subscheme of the plane.
  2. 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!

$\endgroup$
1
  • 1
    $\begingroup$ Thank you for your detailed answer. I am curious about the global description too. $\endgroup$
    – Allen
    Sep 5, 2012 at 12:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.