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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.

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    $\begingroup$ 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. $\endgroup$ – Jason Starr May 16 '16 at 11:57
  • $\begingroup$ 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? $\endgroup$ – Francesco Polizzi May 16 '16 at 12:07
  • $\begingroup$ 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? $\endgroup$ – Jason Starr May 16 '16 at 12:45
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    $\begingroup$ 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. $\endgroup$ – Francesco Polizzi May 16 '16 at 13:00
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    $\begingroup$ 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/… $\endgroup$ – Francesco Polizzi May 16 '16 at 13:14
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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 $S_3$.

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  • $\begingroup$ 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 $$z^3+bz+c, \quad b=-f_2/\sqrt[3]{4}, \, c=f_3/ \sqrt{-27},$$ but is it immediate that there are no others? $\endgroup$ – Francesco Polizzi May 17 '16 at 6:50
  • $\begingroup$ @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. $\endgroup$ – Alex Degtyarev May 17 '16 at 7:44
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    $\begingroup$ @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. $\endgroup$ – Alex Degtyarev May 17 '16 at 23:08
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    $\begingroup$ @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 $S_3$. $\endgroup$ – Alex Degtyarev May 18 '16 at 10:45
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    $\begingroup$ @FrancescoPolizzi: This must all be very classical, but being ignorant, I can only refer to my paper (see S 4): Alex Degtyarev. Oka's conjecture on irreducible plane sextics. J. London Math. Soc., 78:2 (2008), 329--351 doi:10.1112/jlms/jdn029. Lots of other stuff on sextics in general and torus type in particular is found on my web page :) $\endgroup$ – Alex Degtyarev May 18 '16 at 10:47

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