Interpretation of "27" lines for cubic surface with rational double points

Question

It is well known that a smooth cubic surface has $27$ distinct lines. Explicitly, if we choose a planar representation, i.e., blowup $\mathbb P^2$ at $6$ general points $p_1,...,p_6$, the $27$ lines are (1) $E_i$, $1\le i\le 6$, the exceptional divisors, (2) $F_{ij}$, $1\le i<j\le 6$, the proper transform of lines joining $p_i$ and $p_j$, and (3) $Q_i$, $1\le i\le 6$, the proper transform of conics passing $5$ points except $p_i$.

When a cubic surface acquires with one node ($A_1$ singularity), it has $21$ lines. One can think this happens in a specialization as the $6$ points become to lie on a single conic, and the line $E_i$ and $Q_i$ coincide in the limit as a double line, for $i=1,...,6$, while the rest of the $15$ lines $F_{ij}$ stays simple. So $27$ is interpreted as $2\times 6+15$.

What happens in general? My understanding is that, since the number $27$ (or $2875$ for quintic threefolds) is calculated via the intersection theory, it should be interpreted as the length of the Hilbert scheme of lines, especially when the cubic surface is not too singular and the number of lines is still finite.

According to Dolgachev's book section 9.2.2, all cubic surfaces with at worst rational double point singularities have finitely many lines. (e.g., a cubic surface with an $A_2$ singularity has $15$ lines; a cubic surface with an $E_6$ singularity has only $1$ line.)

So my question is, is there work been done to describe the Hilbert scheme of lines for cubic surfaces with rational double point singularities, or is there a geometric interpretation of how the number $27$ are attributed to the multiplicities of geometric lines in those cubic surfaces?

Answer 1

As mentioned in Dolgachev's book, Schläfli classified cubic surfaces according to their singularities. In A Memoir on Cubic Surface Cayley tabulates for each type of singular cubic surface the number of distinct lines and their multiplicity. The multiplicity of a line in the Hilbert scheme of lines depends on whether it passes through a singularity and the type of that singularity. I'll illustrate this in some examples.

(II) In the case you mention (one $\mathrm{A}_1$-singularity $p$) $15$ lines don't pass through $p$ (which have therefore multiplicity 1), and $6$ do (and each has multiplicity $2$).

(IV) If you consider a cubic surface with two $\mathrm{A}_1$-singularities $p$ and $q$, then $7$ lines miss both $p$ and $q$, $8$ lines pass through one of $p,q$, and exactly one line passes through both $p$ and $q$ (which has multiplicity $2\times 2=4$).

(III) If a cubic surface has just one $\mathrm{A}_2$-singularity $p$, then $9$ lines miss $p$, and the $6$ lines who pass through $p$ have multiplicity $3$.

(XXI) As mentioned by Balazs in the comments, the case XXI of three $\mathrm{A}_2$-singularities is particularly nice. The singularities form the vertices of a triangle, whose edges are the three lines in the cubic surface, each of which has multiplicity $3\times 3=9$ in the Hilbert scheme. In this case it is particularly simple to write down the equations cutting out the Hilbert scheme as a subscheme of the Grassmannian $\mathrm{Gr}(2,4)$; one obtains that the Hilbert scheme of lines is the spectrum of three copies of $\mathbf{C}[x,y]/(x^3,y^3)$ (which confirms that each line has multiplicity $9$).

(Note the following consequence: as the universal Hilbert scheme of lines is flat over the locus of cubic surfaces which contain finitely many lines, and since the Hilbert polynomial is constant in flat families, this computation shows that if a cubic surface has finitely many lines, then the number of lines must be 27, counted with multiplicity of course.)

I guess you could wonder which finite $\mathbf{C}$-algebras occur as rings of functions of Hilbert schemes of lines of singular cubic surfaces; I don't think Cayley tabulated these.