27 lines on a cubic surface help

Question

I am studying the theorem that states that on a non-singular cubic there are 27 lines. I am trying to prove it using elementary projective geometry. I have found that on the cubic surface every line meets other 10 lines, and in particular every line lies, with other two of those ten lines, on a plane. It should be an easy combinatoric exercise to prove that the lines are 27. How can I do that?

Answer 1

I don't know if this is exactly what you want, but let me outline a very elementary argumment starting from your information. (In fact this is more or less exactly the original argument of Salmon, as found in his book A Treatise on the Analytic Geometry of Three Dimensions.)

If I understand correctly you have proved that, given a smooth cubic surface $S$:

(a) there is a plane $P$ such that $P \cap S = L_1 \cup L_2 \cup L_3$, a union of 3 distinct lines;

(b) given any line $L$, there are 10 exactly 10 lines in $S$ that meet $L$.

You need one more fact:

(c) for the lines $L_1, \ L_2, \ L_3$ as in (a), every other line on $S$ meets exactly one of these lines.

(The proof of "at least one" is trivial. The proof of "at most one" is slightly less trivial but still easy. Both involve some amount of geometric input, so this argument is not "purely combinatorial".)

Then you conclude as follows: for each $i$, there are 10 lines meeting $L_i$, namely two in the plane $P$ and 8 others. For $i \neq j$ the sets of 8 lines meeting $L_i$ and $L_j$ are disjoint by (c), and also by (c) every line on $S$ is either one of the $L_i$ or in one of these sets of 8 lines. So altogether we get

$$ 3 + 8 + 8 + 8 = 27$$ lines on $S$.