2
$\begingroup$

Let $\mathbb{K}$ be an algebraically closed field. Let $\mathcal{Q}$ be the 6-dimensional vector space of the quadratic (or bilinear symmetric) form over $\mathbb{K}^3$, and $\mathbb{P}(\mathcal{Q})$ the 5-dimensional projective space associated. An element of $\mathbb{P}(\mathcal{Q})$ is a conic section over $\mathbb{P}_2(\mathbb{K})=\mathbb{P}(\mathbb{K^3})$.

A pencil of conics in $\mathbb{P}(\mathcal{Q})$ is a linear system of dimension $1$. My professor states that there are 5 different types of non-degenerate of pencil of conics:

  • there exists 4 points such that the pencil is the pencil of the conics through them;
  • there exists 3 points and a line through one of those points and not containig the other 2 points such that the pencil is the pencil of the conics through the 3 points and having the line as a tangent in the first point;
  • there exists 2 points and 2 lines, where each line contains one of the 2 points and such that the pencil is the pencil of conics passing through the 2 points and in those points tangent to the 2 lines;
  • the pencil is generated by a smooth conic ad a couple of lines such that the one of the 2 lines is tangent to the first conic and the intersection of the 2 lines and the conic is empty;
  • the pencil is generated by a double line and a smooth conic, tangent to each other.

This result is obtained considering that in a pencil of conics, in the condition i described, there exist a smooth and a degenerate conic and treating the possible intersection cases using Bezout theorem. But I am not sure that the cases are all the possible one. In particular is it possible to reduce the following case to one of the cases quoted above: a smooth conic and a couple of lines, where the lines and the conic have a point in common, one line is tangent to the conic and the other intersects the conic at another point? This case is the one where the intersection multiplicity of the 2 conics is divided in one point, where it is 3, and another point where it is 1. The sum is 1 as Bezout says.

If you want you can give me a reference to the classification of the pencil of conics.

Thanks!!

$\endgroup$
  • $\begingroup$ I don't understand the case you are refering to. Your second line should intersect your conic at 2 points (with multiplicity). $\endgroup$ – Carot Aug 23 at 16:33
  • $\begingroup$ Yes, I cannot see the case I explained as one of the 5 mentioned above. I can try explaining it better, if this is the problem. $\endgroup$ – Alessandro Pecile Aug 23 at 19:28
  • $\begingroup$ No I don't understand what case you are refering to. I don't think it is possible. $\endgroup$ – Carot Aug 23 at 21:06
0
$\begingroup$

It is true that there exist 5 types of non-degenerate pencils of conics (in case field $\mathbb K$ is algebraically closed). However, the fourth case you mentioned coincides with the second one. Indeed, considering case 2, denote the base points by $a, b, c$ and let $l$ be the mentioned line, $a \in l$, and let $m$ be the line passing through $b,c$. The pencil contains the conic $l\cup m$, since $l$ is tangent to this conic. Now, choosing any smooth conic $C$ in the pencil, one can see that $C$ does not pass through the point $l\cap m$, because it touches $l$ at point $a$, $a\ne l\cap m$. Thus the pencil is generated by conic $C$ and the couple of lines $m\cup l$. $l$ is tangent to $C$ and the intersection of the $2$ lines and the conic is empty, so one obtains exactly a pencil of the fourth type.

At the same time, if the condition 'the intersection of the 2 lines and the conic is empty' is replaced with its opposite, namely 'the two lines and the conic have a common point', one obtains exactly the case you are referring to. Therefore, you are right and this case must be considered in order to obtain the complete classification of pencils of conics.

As for the other cases, everything is OK.

You can find the classification of non-degenerate pencils of conics here, pages 34-37: http://gorod.bogomolov-lab.ru/ps/stud/projgeom/1718/lec_03.pdf.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.