# Triple covers of $\mathbb{P}^2$ with fixed branch locus

Let us consider a smooth (complex) cubic surface $X \subset \mathbb{P}^3$ and a general point $p \notin X$. Then it is classically well-known that linear the projection $$\pi_p \colon X \longrightarrow \mathbb{P}^2$$ is a triple cover whose branch locus $B$ is a sextic plane curve with six cusps lying on a conic. Conversely, given such a sextic $B \subset \mathbb{P}^2$ there exists a triple cover as above.

Question. Let $B \subset \mathbb{P}^2$ be a fixed plane sextic with six cusps on a conic. What is the dimension of the space of cubics surfaces $X \subset \mathbb{P}^3$ that admit a projection branched over $B$?

I'm rather sure that this must be written somewhere in Zariski's work, but I was not able to find it. Anyway, any reference is appreciated.

• Are you asking about the dimension in the $\mathbb{P}^{19}$ of all cubic surfaces, or in the $4$-dimensional quotient by the action of $\textbf{PGL}_4$? I would have guessed that in the $4$-dimensional quotient you get a discrete set of such cubics, since the topological type is determined by the monodromy action, and then the complex structure away from the ramification locus is uniquely determined as well. May 16, 2016 at 11:57
• Right, I would like to know the dimension of the "moduli space", namely the dimension of the quotient by the $\mathbf{PGL}_4$-action. So, if I understand correctly, you say that it should be zero by Riemann Extension Theorem. Is this 0-dimensional space a finite set? A single point? May 16, 2016 at 12:07
• There is a naive bound on the size of the set. The open complement of the sextic plane curve has finitely presented fundamental group. The set of homomorphisms of that group into $S_3$ (up to conjugation) is an upper bound. But it is probably an overestimate: most topological coverings of a quasi-projective variedy do not "close up" to a proper analytic space (the one-dimensional case is an exception). Even if there is a proper covering, why should it be a cubic surface? May 16, 2016 at 12:45
• If the cover $X$ is a proper analytic space, then it is algebraic by Riemann and Grauert-Remmert Extension Theorems (Serre, Topics in Galois Theory, Thm. 6.1.4), so a projective surface. Then it is no too difficult to show that it is smooth and actually a cubic surface, see arxiv.org/abs/1605.02102, Prop. 3.3. May 16, 2016 at 13:00
• Furthermore, if I'm not mistaken, it seems to me that, since the branch locus $B \subset \mathbb{P}^2$ is an algebraic curve, then the cover is necessarily projective-algebraic. This should follow again by Grauert-Remmert applied to the unramified cover over the complement $\mathbb{P}^2-B$. See mathoverflow.net/questions/40791/… May 16, 2016 at 13:14

A sextic with 6 cusps on a conic has a unique (up to obvious equivalence) torus structure, i.e., a representation of the equation in the form $f_2^3+f_3^2=0$, where $\deg f_i=i$. (Note $\{f_2=0\}$ is necessarily the conic and, in fact, the six cusps are the intersection of the two curves $\{f_i=0\}$.) Then, it is more or less obvious that this polynomial is the discriminant (with respect to an extra variable) of a unique cubic polynomial. Thus, the cubic is unique.
Of course, over $\mathbb{C}$ the same conclusion follows from the fact that $\pi_1$ of the complement, which is the modular group $\langle u,v\,|\,u^2=v^3=1\rangle$, has a unique epimorphism to $S_3$.
• Thanks. I agree with your second proof (over $\mathbb{C}$). Why do you claim that it is "more or less" obvious that the polynomial is the discriminant of a unique cubic polynomial (up to multiplicative scalars)? I clearly see one of these polynomials, namely $$z^3+bz+c, \quad b=-f_2/\sqrt{4}, \, c=f_3/ \sqrt{-27},$$ but is it immediate that there are no others? May 17, 2016 at 6:50
• @FrancescoPolizzi: Yes. Any polynomial $z^3+bz+c$ gives rise to a torus structure, and the latter is unique. (Extra roots of unity are compensated by coordinate changes.) For the uniqueness, the conic $f_2=0$ passes through all cusps, and the cubic $f_3=0$ is tangent to the curve at each cusp, so both are overdetermined (or, one can use Bezout's theorem). BTW, it may happen that one or both split; they should just be transversal at the intersection points. May 17, 2016 at 7:44
• @FrancescoPolizzi Yes, of course on a conic. I just wanted to warn you that there may be more degenerate singularities "on a conic", either due to a singular cubic or non-generic projection. In most cases, a torus structure is unique, but there are a few exceptions (e.g., nine cusps automatically constitute 12 sextuples lying on 12 conics; this is also the only case when there is an extra covering that is not a cubic). As above, torus structures $\leftrightarrow$ cubics is a bijection. May 17, 2016 at 23:08
• @FrancescoPolizzi: He claims there is one of degree 3; the others are of degree 4. According to his definition of "generic", he counts just the one that is not a cubic; each of the the 12 cubics has three cusps (the three cusps that are not on the chosen conic). The maximal dihedral quotient of $\pi_1$ in this case is $\mathbb{D}[\mathbb{Z}_3^3]$, so there are 13 quotients to $S_3$. May 18, 2016 at 10:45