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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.

Francesco Polizzi
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    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. – Jason Starr May 16 '16 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? – Francesco Polizzi May 16 '16 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? – Jason Starr May 16 '16 at 12:45
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    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 http://arxiv.org/abs/1605.02102, Prop. 3.3. – Francesco Polizzi May 16 '16 at 13:00
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    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 http://mathoverflow.net/questions/40791/finite-covers-of-complex-varieties-all-but-two-questions-answered – Francesco Polizzi May 16 '16 at 13:14
  • Let $U $ be a quasi-projective scheme over $\mathbb C$, e.g. , $U= \mathbb P^2 - B$. Then the topological fundamental group of $U$ is finitely generated; see SGA7, tome I, Expose II, Theoreme 2.3.1. In particular, for all $d\geq 1$, the set of finite etale covers of $U$ of degree $d$ is finite (up to equivalence). Therefore, $\mathbb P^2 -B$ has only finitely many finite etale covers of degree $3$. Normalization now tells us that the category of finite covers of degree three of $\mathbb P^3$ ramified precisely along $B$ is equivalent to... – Ariyan Javanpeykar May 16 '16 at 15:29
  • ...the category of finite etale covers of $\mathbb P^2 -B$ of degree $3$. It follows that, as Jason Starr mentioned, the set of cubic surfaces with the desired property is finite. – Ariyan Javanpeykar May 16 '16 at 15:29
  • Thanks. And from this + Grauert-Remmert extension it follows that every such a cover is algebraic (hence projective), right? – Francesco Polizzi May 16 '16 at 17:10
  • Yes, although in your set-up Riemann's existence theorem (Theorem 5.1 in SGA 1, Expose XII) might already be enough to conclude. (Note that to prove Grauert-Remmert one uses Riemann's existence theorem; see Theorem 5.4 in loc . cit..) – Ariyan Javanpeykar May 16 '16 at 19:17

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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 $S_3$.

Alex Degtyarev
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  • 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? – Francesco Polizzi May 17 '16 at 06:50
  • @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. – Alex Degtyarev May 17 '16 at 07:44
  • @FrancescoPolizzi Of course, this polynomial is unique up to coordinate changes: one should first bring it to the reduced form $z^3+bz+c$. In particular, we assume that the characteristic is not $3$ or $2$ (just in case), in case you care about positive characteristic. – Alex Degtyarev May 17 '16 at 07:47
  • @FrancescoPolizzi I guess I should also mention that all this applies to a sextic with exactly 6 cusps. This works for most other sextics of torus type that are not too degenerate. However, a sextic with 8 cusps has 4 torus structure (each selecting 6 out of the 8 cusps). Most interesting are 9-cuspidal sextics: each admits 13 coverings, of which only 12 are cubics! – Alex Degtyarev May 17 '16 at 20:09
  • When you write "I guess I should also mention that all this applies to a sextic with exactly 6 cusps" I guess you mean "with exactly six cusps on a conic", right? – Francesco Polizzi May 17 '16 at 22:59
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    @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. – Alex Degtyarev May 17 '16 at 23:08
  • Thank you again. A sextic with 9 cusps is the dual of a smooth cubic, right? If so, you say that there are 13 non-equivalent coverings over such a sextic, but In this paper by Kulikov it is claimed instead that there are four. http://arxiv.org/pdf/math/9803144v1.pdf Am I missing something? – Francesco Polizzi May 18 '16 at 10:04
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    @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 $S_3$. – Alex Degtyarev May 18 '16 at 10:45
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    @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 :) – Alex Degtyarev May 18 '16 at 10:47