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

Let $P_1, P_2, Q$ denote the Hilbert scheme of a plane conic in $\mathbb{P}^3$, a quartic and a degree $d$ surface in $\mathbb{P}^3$. Then there is a natural inclusion map $i$ from Hilbert flag scheme $\mathrm{Hilb}_{P_1,Q}$ to $\mathrm{Hilb}_{P_2,Q}$ under the map, $(C,X) \mapsto (2C,X)$. Then is the image under the composition of the maps, $i$ with the natural projection map $\mathrm{pr}_2$,

$\mathrm{Hilb}_{P_1,Q} \to\mathrm{Hilb}_{P_2,Q}\to \mathrm{Hilb}_{Q}$

an irreducible component of the image of $\mathrm{pr}_2$? If so can this result be generalized to the case when we can replace plane conic and quartic by curves $C_1, C_2$ such that $rC_1$ has the same Hilbert polynomial as $C_2$?

Note:Hilbert flag scheme $\mathrm{Hilb}_{P_i,Q}$ parametrize pairs of the form $C \subset X$ where $P_i$ is the Hilbert polynomial of $C$ and $X$ is a degree $d$ surface in $\mathbb{P}^3$.

share|improve this question
3  
I disagree that you have such a morphism. You have a rational transformation which is regular on the open subset parameterizing pairs $(C,X)$ such that $C$ is a Cartier divisor in $X$. Also, when $d$ equals $4$, clearly the image of $\text{Hilb}_{P_2,Q}$ equals all of $\text{Hilb}_Q$ (consider hyperplane sections), whereas the image of $\text{Hilb}_{P_1,Q}$ will be a proper closed subset. –  Jason Starr Jun 22 '12 at 17:18
    
Is it obvious that not every quartic surface contains a plane conic? –  Will Sawin Jun 22 '12 at 17:33
3  
@Will -- Yes, it is obvious. It is 9 linear conditions on quartic surfaces to contain a given smooth plane conic, yet there is only an 8-parameter family of plane conics in $\mathbb{P}^3$. –  Jason Starr Jun 22 '12 at 17:36
    
I don't understand the map on Jason's open subset either. I assume $Hilb_{P_2}$ is the Hilbert scheme of plane quartics. If $C \subset X$ is planar, that does not mean that $2C \subset X$ is also. You need to adjust $P_2$ to be the Hilbert polynomial of the first infinitesimal neighborhood of $C \subset X$. You can compute this by determining the degree of the normal bundle of $C$ (via adjunction), but the answer will depend on the degree $d$ of $X$. –  Arend Bayer Jun 23 '12 at 9:10
1  
@Arend: The OP says "quartics", not "plane quartics". Indeed if you interpret the question with "plane quartics", it is absurd. –  Jason Starr Jun 24 '12 at 12:10
add comment

1 Answer 1

up vote 1 down vote accepted

This question has been answered in the comments by Jason Starr. The morphism $i$ is in general not defined and the composition will in general not map to a component of the target. I am reposting this as a CW answer; if it gets upvoted, this question will not reappear on the front page.

share|improve this answer
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.