# When do 27 lines lie on a cubic surface?

Consider $27$ (pairwise distinct!) lines in $\mathbb{P}^3$ whose intersection graph is that expected¹ of the $27$ lines on a smooth cubic surface. Question: Is there a simple necessary and sufficient condition for these $27$ lines to indeed lie on a smooth cubic surface?

For a long time I thought this was always the case, but there is at least one obvious necessary condition²:

(T) Whenever three lines pairwise intersect (i.e., are pairwise coplanar), all three lie on a common plane.

(Because the plane through two mutually intersecting lines on a cubic surface cuts the surface as the union of three distinct lines. There are $45$ such tritangent planes on the cubic surface.)

Is this condition (T) sufficient?

Further comment and bonus question: The locally closed subvariety of $\mathrm{Gr}(2,4)^{27}$ (or $\mathrm{Sym}^{27}(\mathrm{Gr}(2,4))$) consisting of configurations of $27$ lines satisfying the incidence conditions (made explicit in note (1) below) is not irreducible (because of the first sentence of note (2) below). What are its irreducible components?

Notes:

1. I.e., we can label the lines as $a_1,\ldots,a_6$, $b_1,\ldots,b_6$ and $c_{12},\ldots,c_{56}$ (the latter indexed by the $15$ unordered pairs in $\{1,\ldots,6\}$) such that the $a_i$ mutually don't intersect, the $b_i$ mutually don't intersect, each $a_i$ intersects each $b_j$ except exactly when $i=j$, the $c_{ij}$ intersect exactly when their index pairs are disjoint, and $c_{ij}$ intersects exactly $a_i,a_j,b_i,b_j$ among the $a_k$ and $b_k$. Equivalently, the intersection graph is the complement of the vertex graph of the Gosset $2_{21}$ polytope. Equivalently, the faithful transitive action of $W(E_6)$ on $27$ elements where the incidence relation is given by the orbit on the unordered pairs such that every line is incident to exactly $10$ others.

2. Three distinct mutually intersecting lines in $\mathbb{P}^3$ either lie on a common plane or, dually, meet at a common point. (Apologies for pointing out something so obvious, but I'm sure I'm not the only one who might have missed this trivial fact.) If we take the $27$ lines on a smooth cubic surface $X$ and dualize (i.e., replace them by their polars w.r.t. some fixed nondegenerate quadric), we get another configuration of $27$ lines (lying on the projective dual $X^\vee$ to the cubic surface) which have $45$ points of intersection of triples of lines, but in general (unless $X$ had some Eckardt points) no planes in which three lines lie; so this configuration does not lie on a cubic surface (and indeed, $X^\vee$ is not a cubic surface).

• there is something I don't understand in the last paragraph: in the configuration of the 27 lines on the cubic there are just 45 points of (triple) intersection, not 135. – Dima Pasechnik Dec 16 '16 at 0:19
• I agree with Dima Pasechnik: there are 45 2-planes that intersect the cubic surface in a "triangle", not 135. – Jason Starr Dec 16 '16 at 12:38
• Indeed, I miscounted (135 is the number of pairs of intersecting lines, so it counts every tritangent plane three times). Fixed. – Gro-Tsen Dec 19 '16 at 16:08

It turns out that condition (T) is, indeed, sufficient for the $27$ lines (distinct and intersecting as expected) to lie on a cubic surface.

To see this, consider the lines $a_1,a_2,a_3,a_4,a_5$ and $b_6$, where the labeling is as in note (2) of the question: $a_1$ through $a_5$ are pairwise skew, and $b_6$ intersects all of them. Choose $4$ distinct points on $b_6$, and $3$ distinct points on each of $a_1$ through $a_5$ also distinct from the intersection point with $b_6$: this makes $4+5\times3=19$ points in total; since there are $20$ coefficients in a cubic form on $4$ variables, there exists a cubic surface $S$ passing through these $19$ points. Since $S$ contains four distinct points on $b_6$, it contains $b_6$ entirely, and since it contains four points (viz., the intersection with $b_6$ and the three additional chosen points) on each of $a_1$ through $a_5$, it contains these also.

Now $b_1$ intersects the four lines $a_2$ through $a_5$ in four distinct points since $a_2$ through $a_5$ are pairwise skew; and since these four points lie on $S$, it follows that $b_1$ lies entirely on $S$; similarly, $b_2$ through $b_5$, and finally $a_6$ (which intersects $b_1$ through $b_5$ in distinct points), all lie on $S$. So $S$ contains the "double-six" $\{a_1,\ldots,a_6,b_1,\ldots,b_6\}$.

So far, property (T) has not been used, only the incidence relation. Now it remains to see that the $c_{ij}$ lie on $S$. Consider the intersection line $\ell$ of the planes $a_i\vee b_j$ (spanned by $a_i$ and $b_j$) and $a_j\vee b_i$ (spanned by $a_j$ and $b_i$; these planes are well-defined since $a_i$ meets $b_j$ and $a_j$ meets $b_i$, and they are distinct since $a_i$ and $a_j$ are skew): this line $\ell$ must be equal to the $c_{ij}$ of the given configuration, because $c_{ij}$ intersects both $a_i$ and $b_j$ so property (T) implies that it lies in the plane $a_i\vee b_j$, and similarly it lies in the plane $a_j\vee b_i$. But $\ell$ also lies on $S$ for similar reasons¹, in other words, $c_{ij}$ lies on $S$, and all the given lines lie on $S$.

Finally, $S$ must be smooth because it contains the configuration of $27$ lines expected of a smooth cubic surface (it is easy to rule out the case where $S$ is a cubic cone, a reducible surface or a scroll by considering the intersecting and skew lines in the double-six; and every configuration where $S$ has double point singularities has fewer than $27$ lines).

1. Let me be very precise here, because at this stage we don't know whether $S$ is smooth (so we can't invoke (T) directly on $S$). We have four distinct lines $a_i,a_j,b_i,b_j$ on $S$ such that $a_i$ meets $b_j$ and $a_j$ meets $b_i$ and all other pairs are skew. Call $\pi := a_i\vee b_j$ and $\pi' := a_j \vee b_i$ the planes generated by the two pairs of concurrent lines, and $\ell := \pi\wedge\pi'$ their intersection. We want to show that $\ell$ lies on the surface $S$. If $\pi$ or $\pi'$ is contained in $S$ (reducible) then the conclusion is trivial, so we can assume this is not the case. So the (schematic) intersection of $S$ with the plane $\pi$ is a cubic curve containing two distinct lines ($a_i$ and $b_j$), so it is the union of three lines: $a_i$, $b_j$ and a third line $m$ (a priori possibly equal to one of the former). Consider the intersection point of $a_j$ and $\pi$ (which is well-defined since $a_j$ is skew with $a_i$ so does not lie in $\pi$): it lies on $\ell$ because it is on both $\pi$ and $\pi'$; and it must also lie on $m$ since it is on $\pi$ but neither on $a_i$ nor on $b_j$ (as the two are skew with $a_j$); similarly, the intersection point $b_i\wedge\pi$ is well-defined and lies on both $\ell$ and $m$; so $\ell=m$ lies on $S$ (and we are finished) unless perhaps the two intersections considered are equal, i.e., $a_j,b_i,\pi$ concur at a point $P$, necessarily on $m$. Assume the latter case: $S$ must be singular at $P$ because the line $m$ through $P$ does not lie on the plane $\pi'$ generated by two lines ($a_j,b_i$) through $P$ (i.e., we have three non-coplanar tangent directions at $P$). Now symmetrically, if we call $m'$ the third line of the intersection of $S$ with $\pi'$ (besides $a_j$ and $b_i$), we are done unless $a_i,b_j,\pi'$ concur at a point $P'$, necessarily on $m'$ and necessarily singular on $S$. The points $P$ and $P'$ are distinct because $a_i$ and $a_j$ are skew; and they are on $\ell$ because they are on $\pi$ and $\pi'$; and the line joining two singular points on a cubic surface lies on the surface, so $\ell = P\vee P'$ lies on $S$ in any case. (Phew!)

What I still don't know is whether there are configurations of $27$ distinct lines with the expected incidence relations and which satisfy neither condition (T) nor its dual (viz., whenever three lines pairwise meet, all three meet at a common point), and in particular, what are the irreducible components of the space of configurations. (I also don't know if there is a way to substantially simplify the tedious argument given in note (1) above.)