# Root lattices and (resolutions of) singular cubic surfaces

(Cross-posted from math.SE since I'm not sure what platform is suitable -- see https://math.stackexchange.com/questions/3331104/root-lattices-and-resolutions-of-singular-cubic-surfaces)

Given a smooth cubic surface $$X$$ (say over $$\mathbb{C}$$) considered as a blowup of $$\mathbb{P}^2$$ at $$6$$ points) with Neron-Severi group $$\operatorname{NS}(X)$$ and canonical divisor $$K_X$$, the subset $$R = \{ \alpha \in \operatorname{NS}(X) : \alpha\cdot K_X = 0, \alpha\cdot\alpha = -2\}$$ is the root system of the Weyl group of $$E_6$$. I understand that there are previous questions on MSE related to this lattice (e.g. https://math.stackexchange.com/questions/1477220/27-lines-on-a-cubic-surface and https://math.stackexchange.com/questions/82199/automorphism-group-of-the-configuration-of-lines-on-a-cubic-surface-and-quadrati), but I haven't been able to find an explanation about what happens for singular cubic surfaces (on this site or other sources).

More specifically, suppose that $$\varphi : X' \longrightarrow X$$ is a "minimal" resolution of singularities of a singular cubic surface $$X$$ with rational double points as singularities. Suppose that $$X \subset \mathbb{P}^3$$ has rational double points as singularities. For simplicity, we take $$X$$ to have only one singular point (which we take to be $$(1 : 0: 0 : 0)$$) with singularity type $$A_1$$ or $$A_2$$. Then, what sublattice of the root system described above for $$X'$$ does the components of the exceptional divisor of this resolution generate?

Under the assumptions above, we can write $$X = (f = 0)$$ with $$f(t_0, t_1, t_2, t_3) = t_0 g_2(t_1, t_2, t_3) + g_3(t_1, t_2, t_3)$$ for some homogeneous polynomials $$g_2$$ and $$g_3$$ of degrees $$2$$ and $$3$$ respectively (listed in many sources, such as section 9.2 of Dolgachev - Classical algebraic geometry).

In the $$A_2$$ case, I've tried looking at $$(-2)$$-curves coming from two consecutive blowups of three collinear points (blow up one point, and then some point on the exceptional divisor in the next blowup). We repeat this for two collinear triples of points. It isn't clear to me how to obtain a sublattice of $$R$$ (constructed for $$X'$$) that keeps track of the information above. I think this is what I need to better understand "usual" generators for the root lattice in the smooth case. For example, what do classes in $$X'$$ coming from classes of lines passing through the singular point in $$X$$ look like? Are there any suggestions on how to proceed in this situation (or even in the $$A_1$$ case or other singularity types)?

• Please include a link to the question you posted to m.se, and include a link to this question there if you haven't done that already. Aug 23 '19 at 12:07

## 1 Answer

I think you will find everything you want in Demazure, Pinkham, Teissier (eds.), Séminaire sur les singularités des surfaces, Springer LNM 777. Over non-closed fields there's also an article by Coray and Tsfasman, "Arithmetic on singular del Pezzo surfaces", Proc. LMS 57 (1988).

Briefly, the answer is as nice as you might hope. The resolution is what's called a generalised del Pezzo surface of degree 3, that is, a smooth surface obtained by blowing up the projective plane in 6 points in "almost general position". Its Picard lattice is the same as that of a smooth cubic surface; what differs is which classes are effective. It turns out that the effective $$-2$$-curves form a set of positive roots in a sub-root system of $$R$$. The possible configurations of rational singularities on a cubic surface correspond exactly to the sub-root systems of $$E_6$$.

(I believe this all holds more generally for del Pezzo surfaces of degree at least 3; there's an article by Urabe showing that in degrees 1 and 2 there are a few root systems that aren't realised by del Pezzo surfaces.)

• Dear Martin, thank you for the nice answer. I'm wondering do you know of any study on the "root system" on the singular cubic surface $X$? For example, there is still an intersection pairing on Weil divisors as long as $X$ is normal (mathoverflow.net/q/90372). So it still makes sense to talk about the set of $(-2)$-classes on $X$. I'm thinking that since the root system of $X'$ is still $E_6$, could the "root system" of $X$ by any chance is some complement of the sub-root systems corresponding to the rational singularities? Mar 9 at 9:14
• @AGlearner, I don't know of this being studied specifically, but you can look at it like this: the Weil divisors are unaffected by removing the singular locus of X (of codimension 2), so Cl(X) = Pic(U) where U is the complement of the singular locus, isomorphic to the complement of the exceptional divisors in X'. So Cl(X) is a quotient of Pic(X'), and presumably this respects the intersection pairings. Mar 9 at 11:07
• Dear Martin, thank you for the comment. The surjectivity $\pi:Pic(X')\to Cl(X)$ is a very useful observation. This means that the root system on $X$ should be a quotient of the 72 roots on $X'$. However $\pi$ does not seem to preserve the intersection pairing, for example. If $l_1,l_2$ are two lines through an $A_1$ signularity, then $l_1\cdot l_2=\frac12$ is a fraction, according to the formula in Sakai's paper, but their liftings should have self-intersection an integer. Mar 12 at 8:23