Lines on a Quintic Threefold

Question

We work over an algebraically closed field $k$.

I've been given the exercise of showing (using only the technology introduced in the first chapter of Shafarevich's Basic Algebraic Geometry) that every quintic threefold in $\mathbb{P}^4$ contains at least one line. Let $N = \nu_{5,5} - 1$. I've made an argument using the dimensions of the fibers of $\Gamma = \{(F,\ell) \in \mathbb{P}^N \times \text{Gr}(2,5)\mid \ell \subseteq V(F)\}$ over $\mathbb{P^N}$ and $\text{Gr}(2,5)$ which reduces the problem to finding a single quintic hypersurface which contains only finitely many lines, at which point I find I'm stuck. My thought process has been as follows:

A similar argument reduces the problem of showing that every cubic surface in $\mathbb{P}^3$ contains a line to that of finding such a cubic surface which contains finitely many lines. At this point some case analysis shows that the surface given in affine coordinates $T_1T_2T_3 = 1$ can only contain lines at infinity, and that any such line is the intersection of a coordinate plane with the plane at infinity.

This lower-dimensional example doesn't seem to generalize immediately, and all of the examples I've tried have either failed to be examples, or the case analysis in enumerating the lines on a particular surface has grown very unwieldy.

Here are some attempts I've tried:

• Obvious reducible quintics like $x_0^5$ or $x_0x_1x_2x_3x_4$ don't work since they contain hyperplanes.
• $x_0^5 + x_1^5 + x_2^5 + x_3^5 + x_4^5$ contains infinitely many lines. I later found out this is known as the Fermat quintic threefold.
• $x_0^5 = (x_1 + x_2 + x_3 + x_4)x_1x_2x_3x_4$ seems like it might work, but I don't know how to go about trying to prove that.
• This isn't an explicit example, but an idea for generating one that I don't know how to concretize. Let $Y \subset \mathbb{P}^3 \subset \mathbb{P}^4$ be a surface (possibly of degree less than 5) which contains only finitely many lines. Taking the cone over $Y$ in $\mathbb{P}^4$ won't work, but maybe we could somehow take a "twisted" version of a cone over $Y$ in $\mathbb{P}^4$ to obtain a quintic hypersurface, without adding too many new lines. I have no idea whether this is actually possible, but it seems very appealing geometrically.

I'm sure I must be missing something: is there a quintic threefold which can easily be shown to contain only finitely many lines?

Answer 1

This is just a long comment. Maybe some mirror symmetry experts could say something more concrete and useful about this question.

First comment. The moduli space $M_d$ of rational curves of degree $d$ on any quintic threefold is a scheme of finite type. Therefore, if none of the objects it parametrizes have infinitesimal deformations, then $M_d$ is actually finite. Talking about lines, if all of them are infinitesimally rigid in the quintic, $M_1$ will consist of exactly $2875$ reduced points. Because we are on a Calabi-Yau threefold, the only possible normal bundle for an infinitesimally rigid rational curve $C\subset X$ is $$N=\mathscr O_C(-1)\oplus\mathscr O_C(-1).$$ So if you can prove that all lines in $X=V(F)\subset\mathbb P^4$ have normal bundle equal to $N$, then you have your example.

Sheldon Katz has a nice discussion on this in one of his papers about the subject: given a line $L=\{x_0=x_1=0\}\subset V(F)$, he shows that, setting $$F=x_2f_2+x_3f_3+x_4f_4,$$ the line $L$ is infinitesimally rigid if and only if there are no quartic or quintic relations among the $f_i$'s. This may give you a clue on how to find a suitable $F$.

Second comment. One can usually make statements (and conjectures, like Clemens' Conjecture) containing the word generic, like for instance $$\textrm{The generic quintic }X\subset \mathbb P^4\textrm{ contains finitely many lines.}$$ But from there, selecting a special $X_0$ makes it not generic anymore, and so one needs some luck!

I can only give you the completely uninformative recipe to see what you need to check in order to get a finite answer. This will be only slightly better than the empty recipe: "take $X$ generic".

Let $X=V(F)\subset\mathbb P^4$ be a quintic threefold. So what does it mean that a line $L\subset \mathbb P^4$ is contained in $X$? It means that the equation defining $X$ (the single homogeneous form $F$) vanishes on $L$. In fancy terms, it means that the restriction map $$\rho_L:H^0(\mathbb P^4,\mathscr O_{\mathbb P^4}(5))\to H^0(L,\mathscr O_L(5))\qquad\textrm{sends}\,\,F\mapsto \rho_L(F)=0.$$ Now, the ($6$-dimensional) vector spaces $H^0(L,\mathscr O_L(5))$ form a bundle $\mathscr E$ on the ($6$-dimensional) Grassmannian $G(2,5)$ of lines in $\mathbb P^4$. Formally, this bundle is $$\mathscr E=\pi_\ast f^\ast \mathscr O_{\mathbb P^4}(5)=\textrm{Sym}^5(\mathscr Q^\vee),$$ where $\pi:\mathcal U\to G(2,5)$ is the universal line, $f:\mathcal U\to \mathbb P^4$ is the natural projection (and $\mathscr Q$ is the universal quotient bundle). Moreover, the maps $\rho_L$ can be assembled together to form a section $s_F\in H^0(G(2,5),\mathscr E)$. The upshot is: $$F\textrm{ generic }\Rightarrow s_F\textrm{ regular}\Rightarrow \{s_F=0\}\textrm{ finite}.$$ The last implication simply uses $\textrm{rank }\mathscr E=\dim G(2,5)$. Now, the zero locus of $s_F$ is exactly what we are interested in. And we see that all we care about is $s_F$ being a regular section. I agree this is not a big improvement! Of course, checking that $$\textrm{the equation }s_F=0 \textrm{ has finitely many solutions}$$ is not the most pleasant "exercise" one can think of. But certainly you would be able to exclude all forms $F$ such that the equation $\rho_L(F)=0$ has at least $2876$ solutions!