i'm reading the famous article "Varietes kahleriennes dont la premiere classe de chern est nulle" by Beauville, in particular proposition 6, which characterizes the second cohomology group for the hilbert scheme of points on a k3 surface, i.e. $H^2(S^{[r]},\mathbb{C})=i(H^2(S),\mathbb{C})\oplus \mathbb{C}[E]$, where $E$ is the preimage of the singular locus of $Sym^{r}(S)$ through the Hilbert-Chow map $\rho:S^{[r]}\rightarrow Sym^{r}(S)$.

Given $\Delta\subset S^r$ the diagonal in which at least two elements are equal and $\Delta_{ij}\subset S^r$ the diagonal composed by the elements $(x_1,\ldots,x_r)$ with $x_i=x_j$ and all the other elements not equal between them, we set $S^r_*:=(S^r\setminus \Delta)\cup \cup_{i>j}\Delta_{ij}$ and as $S^{[r]}_*$ the preimage through the Hilbert-chow map of the quotient of $S^r_*$ by the action of the symmetric group on $r$ elements.

Beauville proves this formula computing $H^2(S^{[r]}_*,\mathbb{C})$ as the part of the cohomology $\oplus_i pr_i^*H^2(S,\mathbb{C})\oplus_{i>j}\mathbb{C}E_{ij}$ which is invariant for the action of the symmetric group on $r$ elements.

Then he remarks that one can prove also $H^2(S^{[r]},\mathbb{Z})=i(H^2(S),\mathbb{Z})\oplus \mathbb{Z}\delta$ with $2\delta=E$, but "the proof is more delicate".. do you have any idea how it goes? also from where comes the equality $2\delta=E$? Thank you very much