Let us work over $\mathbb{C}$. It is a classic result that if $S$ is a smooth cubic surface, then there are 27 lines contained in $S$.
My question is: can we compute the number of points which are intersection of these lines? Does the resulting number depend on the choice of the cubic $S$? In particular, I'm looking for a surface $S$ with 216 of such points.
Any reference in this direction would be much appreciated.
Best Answer
There can be at most 135 such points on a smooth cubic surface $S$.
In the standard proof of the fact that the number of lines is 27, one ingredient is the lemma that every line on $S$ meets exactly 10 other lines.
So the number of pairs $(L_i,p_{ij})$ where $L_i$ is a line on $S$ and $p_{ij}$ is the intersection point of $L_i$ with $L_j$ is $27 \times 10 = 270$.
But in the computation each intersection point $p_{ij}$ appears at least twice. If we assume that no point lies on more than 2 lines, then each of the $p_{ij}$ appears exactly twice, so the actual number of intersection points is $270/2=135$.
As Angina Seng says in the comments, some cubic surfaces have so-called Eckardt points, where 3 lines meet. An Eckardt point will be counted ${3 \choose 2} = 3$ times in the above calculation, so the the number of intersection points is $135-2e$ where $e$ is the number of Eckardt points. The possible numbers of Eckardt points on a smooth cubic surface are
$$0, 1, 2, 3, 4, 6, 9, 10, 18$$ so the possible numbers of intersection points of 2 lines are $$135, 133, 131, 129, 127, 123, 117, 115, 99.$$