Parametrising the Clebsch cubic

Question

The Clebsch cubic surface is a famous cubic surface whose 27 lines are all defined over the real numbers.

It can be described as the solution set of the equation (in homogeneous coordinates)

$$x^3+y^3+z^3+w^3 =(x+y+z+w)^3 \quad (\ast)$$

On the other hand, like any cubic surface, it is parametrised by the four-dimensional space of cubic curves passing through six given points in the plane. In this special case the points are the vertices of a pentagon together with its centre. Choosing a basis $(f_1,f_2,f_3,f_4)$ for that space gives a rational map $\mathbf P^2 \dashrightarrow \mathbf P^3$ sending $x$ to $(f_1(x), f_2(x), f_3(x), f_4(x))$, and blowing up the six base points makes this an isomorphism onto its image. For an arbitrary choice of basis of this space, the image will be projectively equivalent to the Clebsch cubic, but it won't be exactly the solution set of the above equation.

Question: How do we choose $(f_1,f_2,f_3,f_4)$ so that the image of this map is exactly the solution set of the equation $(\ast)$ above?

Note that intersecting the Clebsch surface with any of the coordinate planes in $\mathbf P^3$, we get a cubic curve which is a union of three lines. But this curve is the same as the solution set of $f_i=0$, unless it contains one of the exceptional lines. So each of the $f_i$ should be a product of 3 linear forms, or the union of a conic through 5 of the 6 points and a line through 2 of the 6 points, only one of which lies on the conic.

I guess there is a simple brute-force method to answer this question, as follows: pick any basis for the space of cubic curves, find the degree-3 equation they satisfy, find the necessary linear transformation to put that in Clebsch form, then apply that transformation to the elements of the original basis. But there is something repugnant about that way of solving it: the setup of the problem is so beautiful and symmetric that one ought to be able to find the answer "by pure thought".

However, I have failed to do that so far. Please enlighten me!

Answer 1

In section 9.5.4 of Dolgachev's Classical Algebraic Geometry: A Modern View here, Dolgachev refers to Klein's book on the icosahedron Lectures on the Icosahedron and the Solution of Equations of the Fifth Degree here, to give $5$ polynomials $F_i$ that pass through $6$ fundamental points and the image of the rational map defined by these cubic polynomials is the Clebsch diagonal surface.

• I. Dolgachev. Classical algebraic geometry: a modern view. Cambridge University Press, Cambridge, 2012. xii+639 pp.

• F. Klein. Lectures on the icosahedron and the solution of equations of the fifth degree. Translated into English by George Gavin Morrice. Second and revised edition. Dover Publications, Inc., New York, N.Y., 1956. xvi+289 pp.

However, Dolgachev was incorrect to say that the image is given in the standard form of the surface. It is, in fact given by $s_1 = s_3 = 0$, where $s_i$ are elementary symmetric polynomials in $5$ variables. So what you need is to make a linear transformation to replace $s_3$ with the sum of cubes. I think these transformations must be something of the form$$G_i = \sum e^i F_i, \text{ where }e^5 = 1, \text{ }i = 0, 1, 2, 3, 4.$$Let me know if it works.

Dear Erin, thanks for your answer. As you say, Dolgachev's presentation gives a parametrisation of the Clebsch surface as a surface in $\mathbb{P}^4$. That is one standard way to present this surface, so I wouldn't say it is incorrect, just not what I was looking for. And indeed, one should be able to go from these $F_i$ to the ones I want by an appropriate substitution, but that seems a bit messy.

However, one can find the $F_i$ I am looking for directly by taking appropriate products of the linear forms defining lines joining two of the basepoints. I will write the details as an answer if I get a chance.

The intersection with the hyperplane $w = 0$ is given by the equation, say$$x^3 + y^3 + z^3 = (x + y + z)^3.$$It is an irreducible cubic curve. This agrees with that $F_i$ is an irreducible cubic.

Yes, there is no mistake in Dolgachev's book. I forgot that$$s_3(F_1, \ldots, F_5) = s_1(F_1, \ldots, F_3) \iff \sum F_i^3 = \sum F_i = 0.$$It happens.