Question

It is well known that a generic hypersurface of degree $2n-3$ in $\mathbb CP^n$ has finite number of lines. I would like to ask a couple of questions about lines on Fermat hypersurfaces and their symmetries:

$$\sum_{i=1}^{n+1}x_i^{2n-3}=0.$$

Fermat hypersurfaces have a group of automorphisms of order $(2n-3)^n(n+1)!$. In the case $n=3$ (the case of cubic) this group is acting transitively on the collection of $27$ lines and this rases some questions.

The first question is pedagogical, I plan to use it for teaching and really want to know the answer.

Question 1. Is there some slick way to give a high-school proof of the fact that there are exactly $27$ lines on Femat cubic in $\mathbb CP^3$ using (or not) the symmetries of the cubic but without using any theory at all?

Further questions are not for teaching, I am just curious about them.

Question 2. Is it known that a Fermat hypersurface of degree $2n-3$ has finite number of lines for any $n$? Is it known that these lines are never multiple?

Question 3. Can one say something about the number of orbits of the action of symmetries on lines on a Fermat hypersurface of degree $2n-3$? For example, what happen in the case of quintic, $n=5$? According to wiki a generic quintic has $2875=125\cdot 23$ lines, so if Fermat quintic is generic, there should be more than one orbit in the action on lines on it. What is the number of orbits?

I would be happy to know the answer on any of these questions.

Answer 1

Regarding question 1, any line in the Fermat cubic $C = \{X_0^3 + X_1^3 + X_2^3 + X_3^3 = 0\}$ must meet the coordinate hyperplane $H_0 = \{X_0 = 0\}$. So which points $x \in (C \cap H_0)$ can lie on lines? If $Y, Z$ are homogenous coordinates on $T_x(C \cap H_0) \cong \mathbb{P}^1$, then the restriction of $X_0^3 + X_1^3 + X_2^3 + X_3^3$ to $T_x C$ is of the form $X_0^3 + F(Y,Z)$ for a homogeneous cubic $F$. For $x$ to lie on a line, $X_0^3 + F$ must factorise, so $F$ is a cube. This means that $x$ is an inflection point of the plane cubic curve $C \cap H_0 = \{X_1^3 + X_2^3 + X_3^3 = 0\}$. The inflection points are given by intersection with the zero set of the Hessian determinant $216X_1X_2X_3$. Hence the intersection of any line in $C$ with any coordinate hyperplane must actually have two corrdinates equal to 0, and it follows that the lines consist of $\{X_0^3 + X_1^3 = X_2^3 + X_3^3 = 0\}$ and its two images under permutating the coordinates (9 lines in each).

P.S. Here is a related exercise I like. Once one has identified the 27 lines in the Fermat cubic $C$, one can use the symmetries of $C$ to guess how to arrange 6 points in $\mathbb{P}^2$ so that the blow-up is isomorphic to $C$, and then write down an explicit rational map $\mathbb{P}^2 \dashrightarrow \mathbb{P}^3$ that maps birationally onto $C$.

Answer 2

Assume for example that $n = 2k + 1$ is odd. Let $\xi^{2n-3} = -1$. Then for any $(y_0,y_1,\dots,y_k) \in \mathbb{CP}^k$ the point $(y_0,\xi y_0,y_1,\xi y_1,\dots,y_k, \xi y_k)$ is on the Fermat hypersurface. So, it contains $\mathbb{CP}^k$. In particular, if $k \ge 2$ (and so $n \ge 5$) the number of lines is infinite. A similar argument works for even $n \ge 6$.