From: Allan Adler <ara@zohar.ai.mit.edu>
Subject: Re: Name the cubic surface! (Repost)
Date: 29 Mar 1999 11:01:17 -0500
Newsgroups: sci.math.research

p_mclean@postoffice.utas.edu.au (Patrick McLean) writes:

> 
> Does anyone know anything about this cubic equation and the surface it
> describes?
> 
> x^2 + y^2 + z^2 - 2xyz =1
> 
> Someone previously suggested that Newton classified all cubic surfaces,
> does anyone know of a reference to this result.

I don't really know, but my vague recollection of things I've heard
is that Newton classified cubic curves, not surfaces.

If you write your cubic homogeneously as wx^2+wy^2+wz^2-2xyz=w^3 and
set the first partials equal to 0, you find just 4 common solutions,
namely those where w is nonzero (hence we can assume it is 1)
and x^2=y^2=z^2=xyz=1. So this is a cubic surface with 4 singularities
and it is easy to check that they are double points. There are
old books on cubic surfaces, e.g. The 27 Lines On A Cubic Surface,
The Nonsingular Cubic Surfaces, etc., in which one can look up
classical descriptions of cubic surfaces with various kinds of
singularities. You might also look in Salmon's Analytic Geometry
of Three Dimensions, which has a chapter devoted to cubic surfaces.
For example, section 523 of the Chelsea edition has a table of 23
kinds of cubic surfaces, classified according to their singularities
and giving the class and number of lines on the surface in each case.

There are also recent results on moduli of cubic surfaces and 
their connection with automorphic functions.

In this case, however, it might be possible to improvise. First of
all, if a cubic has two singularities P,Q then the line L joining
P,Q meets the cubic in 4 points since, P and Q being double points,
each counts as two intersections. Therefore, L must lie entirely in
the cubic surface. The 4 singularities I mentioned above are linearly
independent, hence are the vertices of a tetrahedron and we conclude
that the cubic surface contains the edges of a tetrahedron. A tetrahedron
has 6 edges, so that gives us at least 6 lines in the cubic and the
plane W=0 contains none of these edges. However, if we set W=0, we
get XYZ=0, which is satisfied by setting any one of X,Y or Z to 0.
Therefore, we have three more lines, i.e. W=X=0, W=Y=0 and W=Z=0,
giving us 9 lines in the cubic. Messages from the past will tell
us below that there are exactly 9 lines.

We can always change coordinates so that the tetrahedron becomes
the standard one spanned by 
E1=[1,0,0,0], E2=[0,1,0,0], E3=[0,0,1,0], E4=[0,0,0,1]. What is
the most general cubic surface containing the edges of this
standard tetrahedron? The edges of this tetrahedron consist of
all points with at least 2 vanishing coordinates. Therefore,
any monomial with at least 3 different variables will vanish
automatically. If L is one of the edges, say the one defines by
X=Y=0, then the restriction of a cubic f(W,X,Y,Z) to L is f(0,0,Y,Z)
and this must be identically 0. Therefore there are no monomials
in f involving just two variables. Likewise, since f vanishes at
the 4 points E1,E2,E3,E4, there are no terms W^3, X^3, Y^3, Z^3.
Therefore, the most general cubic vanishing on this tetrahedron
is: aXYZ  + b WYZ + c WXZ + d WXY. If we divide through by
WXYZ, this becomes a/W + b/X + c/Y + d/Z, so such a cubic is
the image of a plane in 3-space under the transformation
[W,X,Y,Z] -> [1/W, 1/X, 1/Y, 1/Z]. (Salmon uses the terminology
"reciprocal surface" but is not referring to this transformation;
instead, he means what we would call the dual surface.)

If one of the coefficients a,b,c,d vanishes, say a, then the cubic
is reducible and has a line of singularlties at least. This is not
the case for your cubic so we can assume abcd is nonzero. Now, the
quadruples (p,q,r,s) with pqrs nonzero form a group under coordinatewise
multiplication. The mapping (p,q,r,s) -> (qrs, prs, pqs, qrs) defines a
homomorphism from that group to itself which one easily shows to be
surjective. Therefore, we can choose (p,q,r,s) which maps to (a,b,c,d).
If we then replace W,X,Y,Z by W/p, X/q, Y/r, Z/s respectively,
the equation of the cubic becomes

XYZ + WYZ + WXZ + WXY =0.

Such a surface looks like it ought to be pretty famous, doesn't it?

From the table in Salmon I mentioned above, guessing at the notation
for singularities since I never studied it, it appears that the
only entry corresponding to a cubic with 4 double points is 4 C_2.
Assuming I have guessed correctly, the class of this surface should
be 4 and it should have exactly 9 lines on it, according to the
table. On the following page, Salmon says explicitly that a cubic
having the vertices of a pyramid for double points has a dual
surface of the 4th degree, which he then shows how to compute. The
answer is: the plane pW + qX + rY + sZ=0 is tangent to this cubic
iff sqrt(p)+sqrt(q)+sqrt(r)+sqrt(s)=0, which he tells us becomes
a quartic after one eliminates the radicals. He even gives the explicit
formula for the result, which is:
(w^2+x^2+y^2+z^2-2wx-2wy-2wz-2xy-2yz-2xz)^2-64wxyz=0.

He then tells us that this quartic surface is commonly known as
Steiner's quartic. He remarks that it has 3 double lines meeting
in a point and every tangent plane cuts it in two conics. He refers us
to section 554a for more information about it. There we learn that
conversely, every quartic with 3 nodal lines meeting at a point
is the dual of a 4-nodal cubic surface. In the course of proving
this, he mentions that the equation of the quartic can be brought
to the nicer form y^2 z^2 + z^2 x^2 + x^2 y^2 - 2wxyz by a suitable
linear change of coordinates.

The 4 double points of the cubic correspond to tangent planes of the
quartic that happen to be tangent along a conic instead of cutting
it in two distinct conics (if I am reading Salmon correctly).

After worrying about what happens when two of the three nodal lines
of the Steiner quartic are allowed to coincide, Salmon then states the
following theorem of Darboux, as generalized by Castelnuovo:
"Apart from ruled surfaces, Steiner's quartic is the only surface
which is cut in non-proper curves by all planes of a dobuly infinite
series." Until I study it more closely I will have to guess at his
meaning, but I think he means that if there is a two dimensional
family of planes, each of which cuts the surface in a reducible curve,
then the surface is either ruled or else it is a Steiner surface.

The fun never ends.

Allan Adler
ara@altdorf.ai.mit.edu