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Let $S$ be a generic cubic surface and let $C$ be its intersection with a generic quadric surface. In the linear system of hyperplane sections of $S$, how many points represent the planes $H$ tangent to $C$ at two points, such that the curve $H \cap S$ is singular?

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OK, I'll try to answer your corrected question : you are looking at planes which are bitangent to $C$, and tangent to $S$ somewhere. Now these are two independent properties, so in the dual $(\mathbb{P}^3)^*$ your planes are the intersection points of the dual surface $S^*$ and the curve of bitangent planes, say $\Gamma $.

The degree of $S^*$ is $3.2^2=12$. To compute the degree of $\Gamma $ one counts its intersection with a general plane in $(\mathbb{P}^3)^*$, that is, bitangent planes to $C$ passing through a general point $p$ of $\mathbb{P}^3$. From $p$ $C$ projects to a plane sextic $C'$ with 6 nodes, and the bitangent planes through $p$ to bitangent lines to $C'$. Their number is given by the Plücker formula; if I am not mistaken it is 12. So the answer is $12^2=144$.

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  • $\begingroup$ Ok, thank you. About the degree of $\Gamma$, using Plucker formulas, I obtain a different $b$ (probably I did a mistake). I have $d^*=18$ and $2b+3f=300$, $132=b+f$. What do you obtain? $\endgroup$
    – sqrt2sqrt2
    Dec 18, 2013 at 15:47
  • $\begingroup$ Sorry, I forgot a term : the number of bitangents is actually 72. I use the formula in Salmon (Higher plane curves), for a curve of degree $d$ with $n$ nodes: the number of bitangents is $\frac{1}{2}d(d-2)(d^2-9)-2n(d^2-d-6)+2n(n-1) $. $\endgroup$
    – abx
    Dec 18, 2013 at 16:01
  • $\begingroup$ You mean 96 (your formula and mine give the same result). $\endgroup$
    – sqrt2sqrt2
    Dec 18, 2013 at 16:07
  • $\begingroup$ One last thing: in the intersection locus of $\Gamma$ and $S^*$ there are two kind of points: those representing planes $H$ bitangent to $C$ and such that $S \cap H$ is singular in another point; and those (with multiplicity 2 in $\Gamma \cap S^*$, I would say) representing planes $H$ bitangent to $C$ and such that $S \cap H$ is singular at one of the two tangency points. How to count the first set? $\endgroup$
    – sqrt2sqrt2
    Dec 18, 2013 at 16:19
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Zero. The plane cubic $H\cap S$ must have two singular points $p,q$; this implies that the line $\ell:=\langle p,q\rangle$ is contained in $S$. For each point $r$ of $\ell$ the tangent plane $T_r(S)$ cuts $S$ along $\ell$ and a conic $c $; if $\ell\cap c =\{r,r'\} $, we have $T_r(S)=T_{r'}(S)$, and the map $r\mapsto r'$ is an involution of $\ell$. The pairs $(r,r')$ form a curve in $\mathrm{Sym^2}(\ell)$, say $\Gamma _{\ell}$.

Now we want $r,r'$ in a quadric $Q$. This quadric will meet $\ell$ in two points $s,s'$ which are arbitrary; if we choose $Q$ general enough, we will have $(s,s')\notin \Gamma _{\ell}$, and this will hold for all of the 27 lines in $S$. Thus none of the pairs $(r,r')$ with $T_r(S)=T_{r'}(S)$ lies in $C\times C$.

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  • $\begingroup$ Sorry, I forgot a comma in the sentence: I mean "how many points represent the planes H tangent to C at two points, such that the curve H∩S is singular?" So I don't require that the tangency points are singular points of $H \cap S$. The singular point is another point. $\endgroup$
    – sqrt2sqrt2
    Dec 15, 2013 at 16:05
  • $\begingroup$ Indeed this makes a completely different question. $\endgroup$
    – abx
    Dec 15, 2013 at 16:20

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