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Consider the Hilbert scheme parametrizing the curves of degree $d$ and arithmetic genus $g$ in $\mathbf{P}^n$. Is there a formula for its dimension in terms of $g,n,d$? Is there a bound on its number of irreducible components?

flag	asked Jul 19 at 12:45 Alex 227●3

	Are you asking about the Hilbert scheme parameterizing closed subschemes on $\mathbb{P}^n$ with Hilbert polynomial $f(t) = dt+1-g$? – Jason Starr Jul 19 at 13:42
	(Please forgive the self-advertising!) In an old paper [B. Fantechi, R. Pardini, On the Hilbert scheme of curves in higher-dimensional projective space, Manuscripta Math. 90 (1996), 1-15.] B. Fantechi and I showed that for every $n≥3$ there exist smooth projective curves $C_r\subset P^r$ lying on exactly $n$ components of the Hibert scheme, for infinitely many values of $r$. I don't know whether this partially answers your question. – rita Jul 19 at 16:19

Francesco Polizzi · Answer 1 · 2012-07-19 14:35:34Z UTC

It is well known that if $C \subset \mathbf{P}^n$ has degree $d$, arithmetic genus $g$ and is a locally complete intersection, then the dimension at $C$ of the Hilbert scheme $\mathscr{H}=\mathscr{H}^n_{f(t)}$, with $f(t)=dt-g+1$, satisfies $$\dim _C \mathscr{H} \geq h^0(C, \mathscr{N}_C)-h^1(C, \mathscr{N}_C), \quad (*)$$ where $\mathscr{N}_C$ is the normal sheaf of $C$ in $\mathbf{P}^n$.

When $C$ is a smooth and irreducible, by using Riemann-Roch one checks that the right hand side of $(*)$ equals $p(n,d,g):=(n+1)d+ (n-3)(1-g).$

A component of $\mathscr{H}$ of dimension exactly $p(n,d,g)$ is called regular, whereas a component of dimension strictly bigger that $p(n,d,g)$ is called superabundant.

For instance, it is known that every complete intersection curve $C$ belongs to a regular component of $\mathscr{H}$.

Dimension of the Hilbert scheme of curves

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