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?

$\begingroup$ Are you asking about the Hilbert scheme parameterizing closed subschemes on $\mathbb{P}^n$ with Hilbert polynomial $f(t) = dt+1g$? $\endgroup$– Jason StarrJul 19, 2012 at 13:42

2$\begingroup$ (Please forgive the selfadvertising!) In an old paper [B. Fantechi, R. Pardini, On the Hilbert scheme of curves in higherdimensional projective space, Manuscripta Math. 90 (1996), 115.] 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. $\endgroup$– ritaJul 19, 2012 at 16:19
1 Answer
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 $f(t)=dtg+1$, 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 RiemannRoch one checks that the right hand side of $(*)$ equals $p(n,d,g):=(n+1)d+ (n3)(1g).$
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 $\mathscr{H}$.

2$\begingroup$ The dimension inequality you write is true if $C$ is a local complete intersection scheme. In general, you do need some extra conditions to reduce the "natural" obstruction group to $H^1(C,\mathcal{N})$, cf. pp. 3335 of Koll\'ar's "Rational Curves on Algebraic Varieties". $\endgroup$ Jul 19, 2012 at 14:24

$\begingroup$ You are right. I corrected the answer, thank you. $\endgroup$ Jul 19, 2012 at 14:36

$\begingroup$ I am aware of this inequality. I wonder if something more precise is known (or at least conjectured). I am mainly interested in smooth $C$. $\endgroup$– AlexJul 19, 2012 at 16:05

$\begingroup$ @Alex: "I am mainly interested in smooth $C$". Are you interested in "general" or "generic" $C$? If so, then BrillNoether theory essentially gives a complete answer. $\endgroup$ Jul 19, 2012 at 20:05

$\begingroup$ I am interested in general $g,n,r$, for generic $C$ their values are restricted by the BrillNoether number being nonnegative. $\endgroup$– AlexJul 20, 2012 at 11:32