1
$\begingroup$

Given a Hilbert scheme of curves $\mathrm{Hilb}^P(X)$, with $X$ being smooth irreducible scheme. If $h^0(C, \mathcal{N}_{C/X})$ and $h^1(C, \mathcal{N}_{C/X})$ are constant for all curves $C$ parametrized by the scheme, where $\mathcal{N}_{C/X}$ is the normal sheaf of the curve $C$, can one deduce that $\mathrm{Hilb}^P(X)$ is locally a complete intersection?

Intuitively, since $\mathrm{Hilb}^P(X)$ is projective, it seems that it should be possible to use the corresponding embedding in the $\mathbb{P}^n$ and then use the information on the tangent spaces given by the dimension of cohomology of the normal sheaf to get the result we want. The problem is that I don't know how to make it precise.

Thanks in advance.

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ No, that is not true without further hypotheses. It is true if also the Hilbert scheme has dimension equal to $h^0(C,\mathcal{N}_C)-h^1(C,\mathcal{N}_C).$ $\endgroup$ Commented Jan 24 at 10:39
  • $\begingroup$ Could you give me a hint on how to prove it when the Hilbert scheme has dimension $h^0(C,\mathcal{N}_C)-h^1(C,\mathcal{N}_C)$? Indeed, what I do know is the the dimension of the Hilbert scheme is $h^0(C,\mathcal{N}_C)$, even though $h^1(C,\mathcal{N}_C) \neq 0$. $\endgroup$ Commented Jan 24 at 11:11
  • 1
    $\begingroup$ Every connected component of the Hilbert scheme that has dimension $h^0(C,\mathcal{N}_C)$ is smooth. $\endgroup$ Commented Jan 24 at 12:22

0

Reset to default

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.