Let $\Pi:=\mathbb{P}(H^0(\mathbb{P}^5,\mathcal{O}_{\mathbb{P}}(3)))$ be the space of cubic fourfolds in $\mathbb{P}^5$. It is well-known that those cubics which are pfaffian, i.e. defined by the pfaffian of a 6-by-6 skew-symmetric matrix of linear forms, form a hypersurface in $\Pi$. Does anyone know the degree of that hypersurface?
$\begingroup$
$\endgroup$
2
-
2$\begingroup$ You are looking for the degree of $\mathcal{C}_{14}$ in Hasset's notation, right? $\endgroup$– Francesco PolizziCommented Feb 13, 2019 at 11:22
-
$\begingroup$ @Francesco Polizzi: Yes. $\endgroup$– abxCommented Feb 13, 2019 at 12:38
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
1
According to Example 3, p. 319 of
Li, Zhiyuan; Zhang, Letao: Modular forms and special cubic fourfolds, Adv. Math. 245, 315-326 (2013). ZBL1290.11077,
the degree of $\mathcal{C}_{14}$ should be 915678.
answered Feb 13, 2019 at 12:38
-
1$\begingroup$ Very interesting, thanks! I was totally unaware of that paper. $\endgroup$– abxCommented Feb 13, 2019 at 20:10
$\begingroup$
$\endgroup$
1
Just a remark: the Pfaffian locus is not quite the $\mathcal{C}_{14}$ divisor, but rather a constructible dense subset of $\mathcal{C}_{14}$, see
https://link.springer.com/article/10.1007/s00208-018-1707-7.
The divisor $\mathcal{C}_{14}$ is rather the locus of cubic fourfolds containing a quartic rational normal scroll $T$ (or a deformation of such $T$ with $T\cdot H^2=4$ and $T\cdot T=10$).
-
2$\begingroup$ Right. Of course the hypersurface I had in mind is the closure of the Pfaffian locus. $\endgroup$– abxCommented Feb 13, 2019 at 20:12