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?


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    $\begingroup$ The degree of the surface equals $320$, so I believe $n_L$ equals $80$. $\endgroup$ Mar 5, 2020 at 17:37
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    $\begingroup$ The standard reference is "Tennison, B. R. On the quartic threefold. Proc. London Math. Soc. (3) 29 (1974), 714–734." $\endgroup$
    – Sasha
    Mar 5, 2020 at 19:34
  • $\begingroup$ @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? $\endgroup$
    – AG learner
    Mar 6, 2020 at 22:16
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    $\begingroup$ @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 $n_L$. $\endgroup$
    – Sasha
    Mar 7, 2020 at 8:19
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    $\begingroup$ @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. $\endgroup$
    – Sasha
    Mar 7, 2020 at 8:21

1 Answer 1


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 $n_L=L\cdot C-1=81$.


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