Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
    
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 '12 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 '12 at 16:19
add comment

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 $f(t)=dt-g+1$, 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 $\mathscr{H}$.

share|improve this answer
2  
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. 33-35 of Koll\'ar's "Rational Curves on Algebraic Varieties". –  Jason Starr Jul 19 '12 at 14:24
    
You are right. I corrected the answer, thank you. –  Francesco Polizzi Jul 19 '12 at 14:36
    
I am aware of this inequality. I wonder if something more precise is known (or at least conjectured). I am mainly interested in smooth $C$. –  Alex Jul 19 '12 at 16:05
    
@Alex: "I am mainly interested in smooth $C$". Are you interested in "general" or "generic" $C$? If so, then Brill-Noether theory essentially gives a complete answer. –  Jason Starr Jul 19 '12 at 20:05
    
I am interested in general $g,n,r$, for generic $C$ their values are restricted by the Brill-Noether number being nonnegative. –  Alex Jul 20 '12 at 11:32
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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