It's been a while since there was an honest mathematical blogpost, for reasons explained in an earlier post. I should maybe also write something about the preprint we put on the arXiv in April.

But today I will share you a fun computation I did involving hyperkähler varieties, as I expect that some of the code used in the computation will be useful to future me, and therefore maybe also to someone else.

Hyperkähler varieties

There is a curious class of varieties with trivial canonical bundle, which are irreducible holomorphic symplectic varieties, or hyperkähler varieties. For an excellent introduction one is referred to O'Grady's introductory lecture notes.

Up to deformation, we currently know of two infinite families and 2 exceptional (maybe sporadic is a better word? maybe they are not so sporadic too but we just don't know):

  • Hilbert schemes of points on K3 surfaces
  • generalised Kummer varieties
  • O'Grady's 6-dimensional varieties
  • O'Grady's 10-dimensional varieties

Roughly there are 2 types of results on hyperkähler varieties:

  • results proven for an abstract hyperkähler variety
  • results proven by inspecting the known cases explicitly

On $\int\mathrm{td}_X^{1/2}$

A very interesting cohomology class is the Todd class, which first appears when you look at the Hirzebruch–Riemann–Roch formula. It is also possible to take the square root of this class, and Hitchin–Sawon proved that for a $2n$-dimensional hyperkähler variety $X$ \begin{equation} \int\mathrm{td}_X^{1/2} = \frac{1}{(192\pi^2n)^n} \frac{||R||^{2n}}{(\operatorname{vol}X)^{n-1}} \end{equation}

Here $||R||$ is the $\mathcal{L}^2$-norm of the Riemann curvature tensor. I don't claim I understand the right-hand side very well, but even without deep understanding the right-hand side exhibits beauty.

For an abstract hyperkähler variety, the work of Hitchin–Sawon and Jiang shows that \begin{equation} 0 < \int\mathrm{td}_X^{1/2} \leq 1 \end{equation} with equality if and only if $X$ is a K3 surface.

For the two known finite families on the other hand Sawon has given in his PhD thesis an explicit expression, namely \begin{align} \int\mathrm{td}_{\mathrm{K3}^{[n]}}^{1/2}&=\frac{(n+3)^n}{4^n n!}, \\ \int\mathrm{td}_{\mathrm{Kum}^n}^{1/2}&=\frac{(n+1)^{n+1}}{4^n n!}. \end{align}

In the first version of Jiang's preprint an exercise left to the reader was to compute these integrals for O'Grady's exceptional hyperkähler varieties, using the fact that the Chern numbers are known.

Over the weekend I had some fun doing this in Sage, resulting in a short note which could be of interest to someone. The code is available in the pdf, but also as a gist. The numbers that I obtained led Jiang to update his preprint, giving a more conceptual explanation for them.

There is no serious mathematics happening here, but I hope the very explicit nature and available code is of interest. I had fun making my first steps in hyperkähler varieties!

For good measure, here is the output of the Sage code:

O'Grady 6-fold
Defining c2, c4, c6

O'Grady 10-fold
Defining c2, c4, c6, c8, c10

Generalised Kummer variety of dimension 6-fold: computation by hand
Defining c2, c4, c6

I don't have the Chern numbers of the Hilbert scheme of 5 points on a K3 surface, so I couldn't do an explicit check for that case, but the closed formula of course just works too.

And here's a table with some values

$\mathrm{K3}^n$ $\mathrm{Kum}^n$ $\mathrm{OG}6$ $\mathrm{OG}10$
$n=1$ 1 1
$n=2$ $\frac{25}{32}=0.78125$ $\frac{27}{32}=0.84375$
$n=3$ $\frac{9}{16}=0.5625$ $\frac{2}{3}\approx0.66667$ $\frac{2}{3}\approx0.66667$
$n=4$ $\frac{2401}{6144}\approx0.39079$ $\frac{3125}{6144}\approx0.50863$
$n=5$ $\frac{4}{15}\approx0.26667$ $\frac{243}{640}\approx0.37969$ $\frac{4}{15}\approx0.26667$
$n=6$ $\frac{59049}{327680}\approx0.18020$ $\frac{823543}{2949120}\approx0.27925$
$n=7$ $\frac{15625}{129024}\approx0.12110$ $\frac{64}{315}\approx0.20317$
$n=8$ $\frac{214358881}{2642411520}\approx0.081127$ $\frac{43046721}{293601280}\approx0.14662$
$n=9$ $\frac{243}{4480}\approx0.054244$ $\frac{1953125}{18579456}\approx0.10512$
$n=10$ $\frac{137858491849}{3805072588800}\approx0.036233$ $\frac{285311670611}{3805072588800}\approx0.07498$