# $27$ lines on a cubic with multiplicity. Contradiction?

A previous poster asked a question whether there are $27$ lines on every cubic surface allowing for multiplicities. I started to think about this and now I am really confused as I have both a proof and a counterexample.

First, there are cubic surfaces with infinitely many lines, that is fine, but if there are finitely many lines how many are there ?

There is a correspondence (not 1-1)

\left\{\begin{aligned} &\text{Cubic Surfaces}\\ &\text{ in \mathbb{P}^3 }\\ \end{aligned}\right\}\leftrightarrow \left\{\begin{aligned} &\text{Lines }\\ &\text{ in \mathbb{P}^3 }\\ \end{aligned}\right\} Where each surface corresponds to those lines which line on the surface, and the lines to those surfaces which contain it. This is an irreducible correspondence, the variety on the left is just $\mathbb{P}^{20}$, so it is irreducible with no singular points, the variety on the right is the Plucker quadratic in $\mathbb{P}^5$.

Now the general cubic surface corresponds to $27$ lines, and by specialization to any specific surface (which is a simple point in $\mathbb{P}^{20}$) these $27$ lines will specialize and we get $27$ lines counted with multiplicity. (It is a theorem that this is well defined if the point in $\mathbb{P}^{20}$ is simple, which it is.)

On the other hand here is my counterexample, (due to van der Waerden) Let $$f_3(x,y,z)+f_2(x,y,z)=0$$ be a cubic surface in $3$ space (written in affine coordinates), where $f_3$ and $f_2$ are homogeneous polynomials of degree $3$ and $2$ respectively.
And assume that the surface has a double point at the origin. Then for lines through the origin, $x=ta,y=tb,z=tc$ we must have

$$f_3(a,b,c)=0$$ and $$f_2(a,b,c)=0$$ this is the intersection of a cubic curve and a quadratic curve, so $6$ points in common, which we assume to be distinct. This gives $6$ lines through the origin. For a line not through the origin, take the plane through this line and the origin, it intersects the cubic in the line and another conic which must decompose into two lines, since the origin is a double point. The number of pairs of lines through the origin is $\binom{6}{2}=15$ and this gives $15$ lines not through the origin. Thus $6+15=21$ lines in total.

Now if the first argument is correct then there must be some multiplicities in these lines. The lines through the origin are the simple intersections of a cubic and a conic, so cannot be multiple ? And in any case no two of them can be multiple since the plane through then would contain $4$ lines. Further the lines not through the origin cannot be multiple since the plane through them and the origin have three distinct lines.

What is the truth here ? How to resolve this conflict ?

• Just spitballing, but did you take into account the lines at infinity? It looks to me like you only considered lines in an affine chart. – Andrew Mar 22 '18 at 18:25
• @Andrew Yes, it really is a projective surface, I just supressed the fourth variable for convenience, but you can add it in. The arguments, and the use of Bezout's theorem require projective space. – Rene Schipperus Mar 22 '18 at 18:29

To see this, note that, like a smooth cubic surface is $\mathbb{P}^2$ blown up in 6 points in general position, the surface you consider is the blow of 6 points lying on a conic in $\mathbb{P}^2$ followed by the contraction of the conic. Indeed, such a 6-tuple is an intersection of a conic $f_2$ and a cubic $f_3$, and then $$(x:y:z) \mapsto (xf_2/f_3,yf_2/f_3,zf_2/f_3)$$ gives an isomorphism onto the singular cubic surface. The inverse map is given by the projection from the singular point.