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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?

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  • $\begingroup$ Nice question. I don't know an easy example offhand. Can you tell us which examples you have tried? $\endgroup$ – Nefertiti Mar 30 '16 at 9:30
  • $\begingroup$ Thanks, I've edited the question to give more details. $\endgroup$ – Sebastian Conybeare Mar 30 '16 at 9:47
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    $\begingroup$ By the way, just in case you don't know, Macaulay 2 has a command $\texttt{Fano(k,I)}$ which will calculate the ideal of the so-called Fano variety of $k$-planes inside the projective variety defined by the homogenous ideal $I$. That gives a quick way to check whether a candidate quintic is going to work or not; of course, you then have to prove the result to your own satisfaction, but at least you know you're not barking up the wrong tree. $\endgroup$ – Nefertiti Mar 30 '16 at 11:11
  • $\begingroup$ I see that you still have no answer to this question. Did Katz give you the exercise? If so, have you asked him how to solve it? If and when you find an answer, it would be great if you would post it here. Alternatively, you could try this question on MO instead --- there are people there (e.g. Borisov) who may well know such an example $\endgroup$ – Nefertiti Apr 7 '16 at 8:37
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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!

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