4
$\begingroup$

Let $X$ be a smooth and projective variety of dimension $d>1$. Let $X^{[2]}$ denote the Hilbert scheme of length two subschemes of $X$. Let $X^{(2)}:=X\times X/\mathbb{Z}_2$, where $\mathbb{Z}_2$ acts by $(x,y)\mapsto (y,x)$. Then there is a birational map $X^{[2]}\to X^{(2)}$. Let $E$ denote the exceptional divisor if this map. Or $E$ can be described as the divisor whose locus is the set of non-reduced subschemes. Can someone point a nice reference where it is explained that there is a line bundle, whose square is the line bundle corresponding to $E$.

Alternatively, the question can be posed as follows. Let $Y$ denote the blow up of the diagonal of $X\times X$. Then the action of $\mathbb{Z}_2$ extends to $Y$, (I think this action is trivial when restricted to the exceptional divisor). The quotient is $\pi:Y\to X^{[2]}$. How do I see that there is a divisor $F$ on $X^{[2]}$ such that $2F=\pi_*(E)$?

I am looking for a reference or explanation of this fact.

$\endgroup$
3
  • 1
    $\begingroup$ I would imagine the original reference for this is one of Fogarty's Hilbert scheme papers, or Beauville's article (in French) which constructs higher dimensional examples of hyperkähler manifolds. You can find a more modern account in O'Grady's notes here irma.math.unistra.fr/~pacienza/notes-ogrady.pdf (see 2.1.22). $\endgroup$
    – Frank
    Jun 9, 2018 at 21:16
  • $\begingroup$ @Frank: Thanks. I think equation (2.1.22), and Proposition 2.2 might be relevant to my question. $\endgroup$
    – Rex
    Jun 9, 2018 at 23:44
  • 2
    $\begingroup$ If $\pi :Y\rightarrow Z$ is a double covering of smooth varieties, we have $\pi _*\mathscr{O}_Y\cong \mathscr{O}_Z\oplus L^{-1}$ and the multiplication $L^{-1}\otimes L^{-1}\rightarrow \mathscr{O}_Z$ is given by a section of $L^2$, whose divisor is the branch locus. $\endgroup$
    – abx
    Jul 24, 2020 at 12:22

1 Answer 1

1
$\begingroup$

Note that $\mathrm{E}$ is the branch divisor of the covering $p:\mathrm{Z}\rightarrow\mathrm{X}^{[n]}$, where $\mathrm{Z}\subset\mathrm{X}\times\mathrm{X}^{[n]}$ is the universal subscheme. Hence $-2c_1(p_\ast\mathscr{O}_{\mathrm{Z}})$ is linearly equivalent to $\mathrm{E}$. Now $\mathscr{O}^{[n]}=p_\ast\mathscr{O}_{\mathrm{Z}}$ is locally free of rank $n$ on $\mathrm{X}^{[n]}$, and so $\det(\mathscr{O}^{[n]})^\vee$ the invertible sheaf you are after.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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