# Number of lines incident to a fixed line on quartic threefolds

Let $$X$$ be a general quartic threefold over $$\mathbb C$$. It is known that the Fano scheme of lines $$F$$ is a smooth curve and every line $$L\in F$$ has normal bundle $$N_{L|X}\cong \mathcal{O}+\mathcal{O}(-1)$$. (See Collino's Lines on Quartic Threefolds.) Therefore if two lines are infinitesimally closed in $$F$$, they are disjoint.

Let $$n_L$$ be the number of lines on $$X$$ incident to the line $$L$$. I'd like to know

(1) If $$n_L$$ is a constant number?

(2) If so, how could we calculate this number?

Thanks!

• The degree of the surface equals $320$, so I believe $n_L$ equals $80$. Mar 5, 2020 at 17:37
• The standard reference is "Tennison, B. R. On the quartic threefold. Proc. London Math. Soc. (3) 29 (1974), 714–734." Mar 5, 2020 at 19:34
• @Sasha Thanks for the nice reference! Now I am thinking the following way to compute: Let $G=Gr(2,5)$, and $I\subset G\times G$ be the correspondence that parameterizes pair of incident lines. Let $p,q$ be coordinate projection. I think $n_L$ should be $p^*F\cdot I\cdot q^*F$. Is that right? Project it to the first coordinate, it should be $F\cdot p_*(I\cdot q^*F)$. I know $F=32\sigma_1\sigma_{1,1}(3\sigma_1^2+4\sigma_{1,1})$ from Tennison's paper, but I don't know how to do the rest into Schubert cycles. Could you help me on this? Mar 6, 2020 at 22:16
• @AGlearner: You can also look into "Markushevich, D. G. Numerical invariants of families of lines on some Fano varieties. Mathematics of the USSR-Sbornik, 1983, 44:2, 239–260", where similar computations with more explanations for other Fano threefolds are performed. In partiuclar, Lemma 2.6 explains how to go from the number 320 to the number $n_L$. Mar 7, 2020 at 8:19
• @JasonStarr: In fact, when you count $n_L$ you should also count $L$ with multiplicity $-1$, so the actual number of lines intersecting $L$ and distinct from $L$ is 81. Mar 7, 2020 at 8:21

The answer is that $$n_L=81$$. The proof was originally due to Fano and the literature can be found in page 40 from Tyurin's Five lectures on three-dimensional varieties, 1972. For convenience of others and for my own benefit, I will rewrite the proof below. The notation is the same as above.

I. Degree of the Scroll: Let $$I$$ be the incidence variety of the pair $$(L,x)$$ such that $$x\in L$$ $$\require{AMScd}$$ $$\begin{CD} I @>{\varphi}>> X\\ @V{p}VV \\ F \end{CD}$$

with $$p:I\to F$$ is a $$\mathbb P^1$$-bundle over the curve $$F$$, and $$\varphi(I)$$ is the scroll of the surface swept out by lines in quartic threefold $$X$$, which is singular at points where lines are incident to each other. According to a standard Schubert calculus computation in Tennison's On the quartic threefold, 1974, the degree of surface $$\varphi(I)$$ is $$320$$, or equivalently, $$\varphi(I)=80H\tag{1}\label{1}$$ in $$A^1(X)$$, where $$H$$ is a hyperplane section.

II. $$n_L$$ as Intersection Number: Let $$\Gamma, L\subset I$$ be a section and a fiber of $$p:I\to F$$, respectively, then among the points of the intersection $$\varphi(L)\cdot \varphi(\Gamma)\tag{2}\label{2}$$ is one point of the intersection of $$L$$ with $$\Gamma$$ and the remaining points are intersection of $$L$$ with other lines in the family. Therefore it suffices to find a special section such that the intersection number $$(\ref{2})$$ is easy to compute.

Let $$H'$$ be a general hyperplane containing a line $$L$$ in $$X$$, then $$S=H'\cap X$$ is a smooth surface with $$L$$ as the unique line. So in the Chow ring $$A^*(X)$$, $$\varphi(I)\cdot S=L+C\tag{3}\label{3}$$ with $$C$$ the residue curve which does not contain a line. Therefore $$\varphi^{-1}(\varphi(I)\cdot S)=L+\varphi^{-1}(C)$$, and

Proposition: $$\varphi^{-1}(C)$$ is a section of the bundle $$p:I\to F$$. It follows that $$n_L=L\cdot C-1.$$

III. Final Computation: Intersect with $$L$$ on both side of $$(\ref{3})$$, and use $$(\ref{1})$$, we have

$$80(L+C')\cdot L=80(L^2+3)=L^2+L\cdot C$$

where $$C'$$ is a plane cubic intersecting $$L$$ transversely at $$3$$ points. Now use the fact that $$S$$ has trivial canonical bundle, so $$L$$ has self-intersection $$(-2)$$ in $$S$$, which leads to $$n_L=L\cdot C-1=81$$.