When are Hilbert schemes smooth? - MathOverflow

When are Hilbert schemes smooth?

2009-10-09T22:43:45Z

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?

Answer by Ben Webster for When are Hilbert schemes smooth?

2009-10-10T01:35:18Z

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."

Answer by David Speyer for When are Hilbert schemes smooth?

2009-10-10T03:48:37Z

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.

Answer by Daniel Erman for When are Hilbert schemes smooth?

2009-10-11T18:39:24Z

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).

Answer by mdeland for When are Hilbert schemes smooth?

2009-10-17T17:27:14Z

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.