MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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:


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.

share|cite|improve this question
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
up vote 3 down vote accepted

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 $C$.

share|cite|improve this answer
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 ( 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 $n \ge 6$.

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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.