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?


  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).

  • 2
    $\begingroup$ 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. $\endgroup$ Dec 16, 2016 at 0:19
  • $\begingroup$ I agree with Dima Pasechnik: there are 45 2-planes that intersect the cubic surface in a "triangle", not 135. $\endgroup$ Dec 16, 2016 at 12:38
  • $\begingroup$ Indeed, I miscounted (135 is the number of pairs of intersecting lines, so it counts every tritangent plane three times). Fixed. $\endgroup$
    – Gro-Tsen
    Dec 19, 2016 at 16:08

1 Answer 1


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.)


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