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

MathOverflow Asked on December 27, 2021

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 $X^{(2)}:=Xtimes X/mathbb{Z}_2$, 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 $Xtimes X$. 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 $pi:Yto X^{[2]}$. 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.

One Answer

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

Answered by The Sand Reckoner on December 27, 2021

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