We know that the intersection of a smooth cubic surface in $\mathbb{P}^3$ with a plane is a cubic, a line+a conic or three disjoint lines. Can this intersection consist of a conic and its tangent line?

  • $\begingroup$ Three disjoint lines does not occur as a hyperplane section is connected. What can happen though is that the intersection is three lines in a triangle configuration (tritangent trios, there's 45 of these), or for special cubics there are also three lines passing through a point (Eckardt point). $\endgroup$ – Frank Dec 26 '18 at 13:37

I guess so. Try the cubic surface whose equation is $$x(xy-z^2)-y^3+w^3=0.$$

If my computations are correct, it is smooth and the intersection with any of the three planes $y- \eta w=0$ (where $\eta$ is a third root of unity) is isomorphic to $x(xy-z^2)=0$, which is the union of a smooth conic and one of its tangent lines.

  • $\begingroup$ Your computations are correct. :) $\endgroup$ – diverietti Jun 25 '12 at 10:07
  • $\begingroup$ Yes, they are. Sorry for the stupid post then... $\endgroup$ – mathdonk Jun 25 '12 at 10:11

In fact, the result is more general. Let $S$ be any smooth cubic in $\mathbb{P}^3$ and let $L\subset S$ be any line (there are $27$ on it). Then, there exist exactly two conics $C_1,C_2$ in $S$ which are tangent to $L$ and such that the union of $L$ with $C_i$ is the intersection of $S$ with a plane.

To show this, let $\pi\colon S\to \mathbb{P}^1$ be the map given by the projection from the line (for example if $L$ has equation $x=y=0$, take $\pi\colon (w:x:y:z)\mapsto (x:y)$). One can see that the map $\pi$ is a morphism (for $x=y=0$, the equation of $S$ is $xL=yM$ where $L,M$ are two polynomials of degree $2$ and you send $(w:x:y:z)$ on $(M:L)$). The fibres of $\pi$ correspond to the conics of $S$ such that the union with $L$ is the intersection of $S$ with a plane. In general, such a conic intersects $L$ into $2$ points, so $\pi$ induces a $2:1$ map from $L$ to $\mathbb{P}^1$. By Hurwitz-formula, the map has exactly two ramification points, which are the two conics meeting $L$ in only one point, i.e. with a tangence.

PS: Cutting $S$ with a plane, it is also possible to get a singular irreducible cubic, which can be cuspidal or nodal. Or three lines. A similar argument shows that you can choose any $L$ and find five pairs of $2$ lines associated to it.

PS2: It was not a stupid post. Even if the answer is easy, it gives nice questions of elementary geometry.

  • 1
    $\begingroup$ So another question arises. Let $S$ be a smooth cubic surface in $\mathbb{P}^3$. We can assume that the point $p=[0:0:0:1]$ and the line $z=w=0$ are contained in $S$ and $p$ is not lying on any of the $27$ lines. It means that the polynomial $F$ defining $S$ has the form $P(x,y,z)+Q(x,y,z)w+R(x,y,z)w^2$. We can now consider the projection $[x:y:z:w]\mapsto [x:y:z]$ which makes $S$ a double cover of $\mathbb{P}^2$. Its ramification locus is the set of the points where this cover is one-to-one, given by the homogenous equation $Q^2-4PR=0$ (if I'm not wrong). $\endgroup$ – mathdonk Jun 25 '12 at 15:49
  • $\begingroup$ The images of these $27$ lines should be bitangent to this plane quartic (I tjink it doesn't have to be smooth) but I can write an equation where the image is tangent only in one point. What's wrong here? $\endgroup$ – mathdonk Jun 25 '12 at 15:51
  • $\begingroup$ The double cover to the plane is from the blow-up of $S$ at a point, not from $S$ (otherwise it is not a morphism). The images of the $27$ lines actually are bitangent to the discriminant quartic; the missing bitangent is the image of the exceptional divisor. $\endgroup$ – Francesco Polizzi Jun 25 '12 at 16:05
  • $\begingroup$ Yes, yes, I know it should be like that, but I have some problems with that. If you take the line $z=w=0$ and project it, you obtain the line z=0. If it meets the quartic $Q^2−4PR$ it means that $Q=0$ in that point (since $P$ is zero for $z=0$). It gives some information about the coefficients of $Q$ (and only $Q$!) and I can't find any contradiction if it's actually a tangent, not bitangent. $\endgroup$ – mathdonk Jun 25 '12 at 16:08
  • $\begingroup$ The geometric explanation of Francesco is correct. If you take the projection from a point $p$ outside of the $27$ lines, the map is a morphism from the blow-up of $p$ (del Pezzo surface of degree $2$) to $\mathbb{P}^2$ and is a double covering ramified over a SMOOTH quartic (and all smooth quartics are obtained this way). I think the problems you have are in your choice of coordinates. If you take order $(w:x:y:z)$, then your point does not belong to the line $L$, but your equation is false. If your order is $(x:y:z:w)$, the equation is OK but the line passes through the point. $\endgroup$ – Jérémy Blanc Jun 25 '12 at 21:47

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