0
$\begingroup$

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 $\mathbb{P}^3$.

Improve this question
$\endgroup$
5
  • 3
    $\begingroup$ 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. $\endgroup$
    – Jason Starr
    Commented Jun 22, 2012 at 17:18
  • $\begingroup$ Is it obvious that not every quartic surface contains a plane conic? $\endgroup$
    – Will Sawin
    Commented Jun 22, 2012 at 17:33
  • 3
    $\begingroup$ @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 $\mathbb{P}^3$. $\endgroup$
    – Jason Starr
    Commented Jun 22, 2012 at 17:36
  • $\begingroup$ 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 $X$. $\endgroup$
    – Arend Bayer
    Commented Jun 23, 2012 at 9:10
  • 1
    $\begingroup$ @Arend: The OP says "quartics", not "plane quartics". Indeed if you interpret the question with "plane quartics", it is absurd. $\endgroup$
    – Jason Starr
    Commented Jun 24, 2012 at 12:10

1 Answer 1

Reset to default
1
$\begingroup$

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.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .