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To continue with the topic of Plane sextics admitting a conic with 6 double-contact points, I would like to understand the variety of conics touching a given (irreducible) projective plane sextic curve $\{P=0\}$ in 6 points (a.k.a. double contact conics). That is, given $P$, we are interested in the set $\mathcal{Q}$ of conics $\{Q=0\}$ such that $P=-R^2+AQ$, for a cubic $R$ and a quartic $R$.

The obvious cases are $\mathcal{Q}$ empty, and $\mathcal{Q}$ consisting of just one element. C.M.Jessop in Chapter~I.3 of his book Quartic surfaces with singular points, Cambridge: University Press, pp. XXXV+197 (1916) (JFM 46.1501.03) provides a proof that if $|\mathcal{Q}|>1$ then $\mathcal{Q}$ is infinite. Perhaps 110 years later one knows more about this, or understands it better, but I cannot find anything in the modern literature.

Jessop argues roughly as follows (I don't understand some details, in part due to old terminology used).

$P=-R^2+AQ=-R'^2+A'Q'$, with $Q,Q'\in\mathcal{Q}$, specifies the discriminant of $F(u)=Aw^2+2Rw+Q$. Then $\{P=0\}$ is the equation of the tangent cone to the nodal quartic surface $\{F=0\}$ at the node.

Write $AF=(Aw+R)^2+AQ-R^2=(Aw+R)^2+A'Q'-R'^2$, and consider the intersection of surfaces $\{AF=0\}$ and $\{A'=0\}$. It splits into the curve $c_8$ defined by $F=A'=0$ and in 4 lines given by $A=A'=0$ (it's 4 lines, as two plane conics intersect in 4 points).

Then he notes that at $A'=0$ one can write $(Aw+R)^2+A'Q'-R'^2$ as $(Aw+R-R')(Aw+R+R')$ and says that thus $c_8$ splits into two quartic components $c_+$ and $c_-$, each lying in the nodal cubic surface defined by $S_\pm=Aw+R\pm R'$. Then he says that each of $c_+$ and $c_-$ is "the partial intersection of $\{A'=0\}$ with a cubic surface which contains also two generators of $\{A'=0\}$". (In the latter fragment in "" I don't understand terms on italic font - italic added by me. As well, I don't understand which cubic surface is referred to). Then he says that $c_+$ and $c_-$ are therefore quadri-quartic, i.e. having "the type of twisted quartic through which an infinite number of quadrics pass" (Again, I don't understand the latter fragment. What is a twisted quadric, and why $c_+$ and $c_-$ are of this type?)

Then he says "Hence the surface (I suppose, the nodal quartic $\{F=0\}$) contains an infinite number of quadri-quartic curves which are projected from the node into quartic curves which touch the sextic $\{P=0\}$ at each point of intersection" (here in the text there is a footnote saying "For such a point of intersection $p$ is the projection of an actual intersection $q$ of the quadri-quartic and the curve of contact of the tangent cone, and the tangents to these curves at $q$ lie in the tangent plane of the surface".

Finally, he states that the latter implies the claim that if $|\mathcal{Q}|>1$ then $|\mathcal{Q}|=\infty$.

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I don't understand either Jessop's argument, but here is a more modern point of view. Denote by $H$ the divisor of a line on the sextic $C$. Let $D$ be a degree 6 effective divisor on $C$; saying that $D$ is formed by the 6 contact points of a conic with $C$ is equivalent to say that $2D\equiv 2H$ (linear equivalence), that is, $D\equiv H+\alpha $ where $\alpha $ is a divisor class such that $2\alpha \equiv 0$. Given such $\alpha $, either $h^0(H+\alpha )=1$, so $\lvert H+\alpha \rvert=D$ and there is only one contact conic corresponding to $\alpha $; or $h^0(H+\alpha )=2$, and there is a pencil of such conics (one can show that $h^0(H+\alpha )=3$ is impossible).

There are finitely many classes $\alpha\ $ ($2^{20}-1$ to be precise), so your moduli space consists of a finite set plus a number of $\mathbb{P}^1$. That if there is no $\alpha$ with $h^0(H+\alpha )=2$ there could be only one $\alpha $, as Jessop claims, looks very dubious to me.

Edit: Actually the case $h^0(H+\alpha )=2$ does not occur. This follows from the results of Namba (Families of meromorphic functions on compact Riemann surfaces, LN 767, Thm. 2.3.1 and Prop 2.3.6): the only divisor classes $D$ of degree 6 with $h^0(D)\geq 22$ are of the form $H+p-q$, with $p,q\in C$. But $H+\alpha \equiv H+p-q$ would imply $2p\equiv 2q$, hence $p=q$ since $C$ is not hyperelliptic, hence $\alpha =0$.

In conclusion, there are only finitely contact conics to $C$.

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    $\begingroup$ The number of 2-torsion classes $\alpha$ is $2^{20}-1$. $\endgroup$ Commented 20 hours ago
  • $\begingroup$ I am not sure I understand the conclusion - are you saying that such a conic is never unique? $\endgroup$ Commented 11 hours ago
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    $\begingroup$ No. I just quoted Jessop, who says that if the contact conics form a finite set, then this set has only one element. I see no reason why this should be true. $\endgroup$ Commented 11 hours ago
  • $\begingroup$ @abx Which of the cases for $H^0(H+\alpha)$ is "generic", dimension 1 or dimension 2? $\endgroup$ Commented 10 hours ago
  • $\begingroup$ Well, the generic case is $H^0(H+\alpha )=0$ for all $\alpha $, so no contact conic... Then, among plane sextics admitting a contact conic, the generic case will be $H^0(H+\alpha )=1$ for just one $\alpha $. Having $H^0(H+\alpha )=2$ is more special. $\endgroup$ Commented 9 hours ago

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