# Exceptional divisor of the Hilbert-Chow morphism of the punctual Hilbert scheme

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.

• 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). Jun 9, 2018 at 21:16
• @Frank: Thanks. I think equation (2.1.22), and Proposition 2.2 might be relevant to my question.
– Rex
Jun 9, 2018 at 23:44
• 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.
– abx
Jul 24, 2020 at 12:22

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.