Cubic surfaces with 27 rational lines

The simplest way to describe these seems to be as follows. Choose three skew lines L_{1}, L_{2}, L_{3} on the cubic surface V and let M_{1}, M_{2}, M_{3} be their three transversals on V; then there is a quadric which contains these six lines. We can take this quadric to be

X_{0}X_{1} = X_{2}X_{3}

and the lines to be

\begin{matrix}L_{1} : X_{0} = X_{2} = 0, & M_{1} : X_{0} = X_{3} = 0, \\ L_{2} : X_{1} = X_{3} = 0, & M_{2} : X_{1} = X_{2} = 0, \\ L_{3} : X_{0} = X_{3}, X_{1} = X_{2}, & M_{3} : X_{0} = X_{2}, X_{1} = X_{3}. \end{matrix}

The general cubic surface containing these six lines has the form

\begin{matrix}c_{4}(X_{0} + X_{1} − X_{2} − X_{3})(X_{0}X_{1} + X_{2}X_{3}) \\ +(c_{0}X_{0} + c_{1}X_{1} + c_{2}X_{2} + c_{3}X_{3})(X_{0}X_{1} − X_{2}X_{3}) = 0. \end{matrix}

This surface necessarily contains a further nine rational lines, each line being the residual intersection of V with the plane containing L_{i} and M_{j} for some i, j; we shall call this line L_{ij}.

The remaining lines on V fall into six pairs, a typical pair being the two lines N, N′ on V which meet L_{1} and L_{2} but not L_{3}. In general, the individual lines of such a pair are only defined over a quadratic extension; what is crucial for us is that this quadratic extension is the same for each pair, because each line in one pair meets just one of the two lines in any other pair. We have now two ways to proceed: we can choose one line in the ambient space which meets L_{1} and L_{2} and impose the two linear conditions on the c_{i} which make this line lie in V, or we can choose two such lines and solve the resulting four linear equations in the c_{i}.

In either case, the first step is to write down the conditions that the line

X_{0} = uX_{2}, X_{1} = vX_{3}

lies in V. These are

\begin{matrix}c_{4}(u − 1)(uv + 1) + (uc_{0} + c_{2})(uv − 1) = 0, \\ c_{4}(v − 1)(uv + 1) + (vc_{1} + c_{3})(uv − 1) = 0. \end{matrix}

The simplicity of these equations leads me to prefer the second alternative above. We therefore ask what is the cubic surface of our type which also contains the two lines

\begin{matrix}N : X_{0} = u_{1}X_{2}, X_{1} = v_{1}X_{3}, \\ N′ : X_{0} = u_{2}X_{2}, X_{1} = v_{2}X_{3}; \end{matrix}

and we find that the values of the c_{i} are

c_{0} = u_{1}u_{2}v_{1}v_{2} − 1 − (u_{1}v_{1} − u_{2}v_{2})(u_{1} + u_{2} − 2)/(u_{1} − u_{2}),
c_{1} = u_{1}u_{2}v_{1}v_{2} − 1 − (u_{1}v_{1} − u_{2}v_{2})(v_{1} + v_{2} − 2)/(v_{1} − v_{2}),
c_{2} = −u_{1}u_{2}v_{1}v_{2} + 1 + (u_{1}v_{1} − u_{2}v_{2})(2u_{1}u_{2} − u_{1} − u_{2})/(u_{1} − u_{2}),
c_{3} = −u_{1}u_{2}v_{1}v_{2} + 1 + (u_{1}v_{1} − u_{2}v_{2})(2v_{1}v_{2} − v_{1} − v_{2})/(v_{1} − v_{2}),
c_{4} = −(u_{1}v_{1} − 1)(u_{2}v_{2} − 1).

This is almost certainly the simplest formulation possible.

If you need explicit equations for each of the 27 lines, you should proceed as follows. You already have the equations of the L_{i} and M_{i}, together with an easy recipe for the equations of the N_{ij}. You also have the equations of N and N′. Four more lines are the residual intersections of V with the four planes containing either L_{1} or L_{2} and either N or N′. But N and N′ also meet each N_{3i}, and the final six lines are the residual intersections of V with the six planes containing N or N′ together with one of the N_{3i}.

Peter Swinnerton-Dyer 15 April 1998