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.



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> **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 $B$?
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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. 



Triple covers of $\mathbb{P}^2$ with fixed branch locus
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 $S\_3$. 


