Let $X$ be a complete intersection of a quadric and a cubic in $\mathbb{P}^5$. In the smooth case it is a so-called Fano threefold of index one and degree six.

I would like to consider the case when $X$ has only ordinary double points as singularities. The question is: what's maximal number of ordinary double points that $X$ can have?

Here is a lower bound. The threefold $X$ defined by the following two equations $$ x_0^2 + x_1^2 + x_2^2 - x_3^2 - x_4^2 - x_5^2 = (x_0 + x_1 + x_2 - x_3 - x_4 - x_5)^2 $$ $$ x_0^3 + x_1^3 + x_2^3 - x_3^3 - x_4^3 - x_5^3 = (x_0 + x_1 + x_2 - x_3 - x_4 - x_5)^3 $$ has $34$ ordinary double points. (I found these by projecting a $5$-dimensional Segre cubic with $35$ ordinary double points from one of them.) The singularities are located at points $[x_0, ..., x_5]$ such that all $x_i$'s are $0$ or $1$, and the number of $1$'s among $x_0, x_1, x_2$ is either equal to or one more than the number of $1$'s among $x_3, x_4, x_5$.

Can there be more than $34$?

Another question: is $X$ necessarily a rational variety when the number of ordinary double points is large enough?

  • $\begingroup$ @Pop: yes, thank you, I fixed this typo. $\endgroup$ – Evgeny Shinder Sep 3 '19 at 20:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.