# Hilbert squares of higher-dimensional hyperkähler varieties are not hyperkähler

Of course, the statement in the title should not come as a surprise, otherwise they would already be listed among the extremely short list of known hyperkähler varieties. Recall that a *hyperkähler* variety is a variety with a holomorphic symplectic form, and they are one of three building blocks (the others are abelian varieties and Calabi–Yau varieties) of varieties with trivial canonical bundle. So far, the list of known hyperkähler varieties is rather short:

- (deformations of) Hilbert schemes of points on K3 surfaces (of dimension $2n$)
- (deformations of) generalised Kummer varieties of abelian surfaces (of dimension $2n$)
- O'Grady's family of 6-dimensional examples
- O'Grady's family of 10-dimensional examples

Inspired by the first type, I was wondering on my walk home yesterday evening whether the Hilbert square $X^{[2]}$, where $X$ is a higher-dimensional hyperkähler variety, would be again hyperkähler. After all, it is true for K3 surfaces. The restriction to Hilbert squares (and Hilbert cubes) is quite natural here: without restricting the number of points, the Hilbert scheme of points on a higher-dimensional variety becomes (very) singular.

But there is a silly and roundabout argument to rule this out, based on the derived category of the Hilbert square (an object I've been thinking about recently, in a different setting). Of course, a more direct analysis of the geometry of the Hilbert square would be more appropriate (the expected geometric argument probably says that the resolution is no longer crepant).

**Proof.** Assume it were hyperkähler. Then its canonical bundle is necessarily trivial. By Bridgeland's indecomposability for derived categories with trivial canonical bundle, we know that its derived category is indecomposable.

On the other hand, in Krug–Ploog–Sosna: Derived categories of resolutions of cyclic quotient singularities a semiorthogonal decomposition of the form \begin{equation*} \mathbf{D}^{\mathrm{b}}(X^{[2]})=\left\langle \mathbf{D}_{\mathbb{Z}/2\mathbb{Z}}^{\mathrm{b}}(X^2),\mathbf{D}^{\mathrm{b}}(X),\ldots,\mathbf{D}^{\mathrm{b}}(X)\right\rangle \end{equation*} with the first component the equivariant derived category, and $\mathop{\rm dim}X-2$ copies of $\mathbf{D}^{\mathrm{b}}(X)$ is given. This contradicts the indecomposability for $\mathop{\rm dim}X\geq 3$.$\blacksquare$

The case $\mathop{\rm dim}X=2$ is an instance of the Bridgeland–King–Reid equivalence by the way.