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I know that Hilbert schemes can be very singular. But are there any interesting and nontrivial Hilbert schemes that are smooth? Are there any necessary conditions or sufficient conditions for a Hilbert scheme to be smooth?

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up vote 5 down vote accepted

I don't know of many global conditions for a Hilbert scheme to be smooth/singular. Ben's answer probably gives the most interesting example of a smooth Hilbert scheme, namely the Hilbert scheme of n points on a smooth surface.

Here are two more examples of smooth Hilbert schemes.

1) The Hilbert scheme of hypersurfaces of degree d in PP^n. Such hypersurfaces are parametrized by homogeneous degree d polynomials in n+1 variables, and hence this Hilbert scheme is a projective space of dimension n+d choose d.

2) The Hilbert scheme of linear subpsace of dimension d of PP^n. This is just the Grassmanian Gr(d+1,n+1).

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in the same line as example (1): if $X$ is a projective surface and $\rm{H}^1(\mathcal{O}_X = 0$ then its Hilbert schemes of curves are smooth. So, the family of surfaces with $b_1(X) = 0$ yields examples of smooth Hilbert schemes. – James O Feb 11 '10 at 0:28

A very well-known condition is that the Hilbert scheme of a smooth surface is smooth. As David pointed out below, the Hilbert scheme of a smooth curve is smooth and equal to the symmetric product (since k[t] has only one finite dimension quotient of each dimension).

I don't know of any other examples, but one of the versions of Murphy's Law in algebraic geometry is roughly "if you don't have a good reason for a Hilbert scheme to not be horrible, it will be as horrible as you can possibly imagine."

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Yes, the Hilbert scheme of n points on a smooth curve is also smooth (and it equals the n-fold symmetric product of the curve). – David Rydh Oct 10 '09 at 19:07

Here is yet another example of a smooth Hilbert scheme. Let $X$ be a smooth degree 3 hypersurface in projective space of dimension $n \geq 3$ (say, over an algebraically closed field), and let $H$ be the Hilbert scheme of lines on $X$ (i.e., corresponding to Hilbert polynomial $t + 1$).

The tangent space to $H$ at a point $[L]$ (corresponding to a line $L$ in $X$) is $H^0(L, N)$ where $N$ is the normal bundle of $L$ in $X$. The rank of $N$ is $n - 2$ and the degree of $N$ is $2n - 6$ (you can see this by looking at the standard tangent bundle and normal bundle sequences). Every vector bundle on $L = \mathbb{P}^1$ splits into the direct sum of line bundles. Then the degree of each rank 1 summand of $N$ is at most 1 ($N$ injects into the normal bundle of $L$ in $\mathbb P^n$) and then you can show that no piece can have degree less than $-1$. This allows us to conclude that $H^1(L, N) = 0$. This means that $H$ is smooth at the point $[L]$ (see for example Kollár's book Rational Curves on Algebraic Varieties, Chapter 1, where he explains the infinitesimal behavior of the Hilbert scheme). Since this is true for any line $L$ in $X$, the Hilbert scheme is smooth.

The same argument works for lines on a smooth Quadric. In the same book, Kollár proves that for a general degree $d$ hypersurface $X$ in $\mathbb P^n$, the Hilbert scheme of lines on $X$ is smooth.

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Another reference, arguably more readable, is Higher-Dimensional Algebraic Geometry, by Olivier Debarre. – Charles Staats Jun 28 '12 at 20:11

The Hilbert scheme of n points on a 3-fold is not smooth for n sufficiently large, but the exact value of sufficiently large is unknown. See the chapter on Hilbert schemes in "Combinatorial Commutative Algebra", by Miller and Sturmfels.

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I think that David's answer is not correct. The Hilbert scheme of 4 points on a 3-fold is not smooth. The singular point is defined by the ideal (x,y,z)^2. Most likely, David is alluding to the fact that the Hilbert shceme of n points on a 3-fold is REDUCIBLE for n>>0, and the precise value of n is unknown. – Daniel Erman Oct 11 '09 at 18:29
3-folds shouldn't be smooth, you can compute tangent space I think. – Ilya Nikokoshev Oct 11 '09 at 20:17

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