# Lines on degree 2n-3 Fermat hypersufaces

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.

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The Fermat threefolds contain several one-parameter families of lines: partition the variables into two sets of size 2 and 3 and set them separately to zero. You obtain $10d$ families of lines in this way. Of I recall correctly, these are "non-reduced" as soon as $d$ is at least 5. You can find out more in papers by Albano-Katz and more recently Candelas and others. –  M P Jan 4 '13 at 9:11
Thank you MP, so together with Sashas answer this solves all my questions apart from the first one –  aglearner Jan 4 '13 at 9:17

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 $C$.

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In fact I have a couple of questions. Do I understand correctly that by $T_xC\cap H_0$ you mean the tangent projective line to $C\cap H_0$? Why $X_0^3+F$ must factorisу for $x$ to lie on a line? –  aglearner Jan 4 '13 at 19:42
Also, I don't understand why you want to write the restiction as $X_0^3+F(Y,Z)$. Any function on a line can be written as $F(Y,Z)$? (maybe I don't understand what you mean when you say that $Y,Z$ are homogeneous coordinates on the line). But anyway thank you for giving this proof, I got the main idea and the style of the proof is almost as elementary as I would like it to be. –  aglearner Jan 4 '13 at 20:45
In the same order as your questions: Yes, I do. If $x$ lies on a line, then that line is contained in $T_xC$, and its defining equation is a factor in $X_0^3 + F$. $F(Y,Z)$ is indeed intended to mean an arbitrary cubic on the line $T_x(C \cap H_0)$; what I try to emphasise (perhaps unsuccessfully) is that the restriction of equation of $C$ to $T_xC \cong \mathbb{P}^2$, which is a cubic in $X_0$, $Y$ and $Z$, does not contain any cross-terms like $X_0Y^2$. –  Johannes Nordström Jan 4 '13 at 21:33
Johannes, thanks again for the proof. Also, by any chance, do you know a book that contains this exercise with cubic surface? –  aglearner Jan 6 '13 at 20:31
I don't have any references for the specifics of my answer, but I've found the first of Reid's Chapters on Algebraic Surfaces (arxiv.org/abs/alg-geom/9602006) a useful general reference for cubic surfaces. –  Johannes Nordström Jan 6 '13 at 22:14

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 $n \ge 6$.

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Thanks Sasha, this is neat! I still wonder about the cubic and the quintic. –  aglearner Jan 4 '13 at 8:31