0
$\begingroup$

Consider the conic bundle of $\mathbb{P}^2(\mathbb R)$with matrix $$A(\lambda,\mu)=\begin{pmatrix} 0 & \mu & \mu \\ \mu & 0 & \lambda \\ \mu & \lambda & 0 \end{pmatrix}$$ This is a tangent conic bundle with base cycle $2A,B,C$ with $$A= \begin{pmatrix}1 \\ 0 \\ 0 \end{pmatrix} B= \begin{pmatrix}0 \\ 1 \\ 0 \end{pmatrix} C= \begin{pmatrix}0 \\ 0 \\ 1 \end{pmatrix}$$ The fixed tangent in $A$ is $r:x_1+x_2=0$. I want to understand which points $P \in \mathbb{P}^2(\mathbb R)$ have the same polar for every conic in the bundle, i.e. for which $p_0,p_1,p_2 \in \mathbb R$ the plucker coordinates $$p=P^TA(\lambda,\mu)=\begin{pmatrix}p_0 &p_1 & p_2 \end{pmatrix} \begin{pmatrix} 0 & \mu & \mu \\ \mu & 0 & \lambda \\ \mu & \lambda & 0 \end{pmatrix}=\begin{pmatrix}\mu(p_1+p_2) & \mu p_0 +\lambda p_2 & \mu p_0 +\lambda p_1\end{pmatrix}$$ of the polar line with pole $P$ do not depend on $\lambda, \mu$. Obviously, one of these points is $A$: I would like to verify if there are other points of this kind.

For the polar to be defined, the conic must be non-degenerate i.e. $\lambda \neq 0$ and $\mu \neq 0$, hence

$$\begin{pmatrix}\mu(p_1+p_2) & \mu p_0 +\lambda p_2 & \mu p_0 +\lambda p_1\end{pmatrix}\stackrel{\text{x}}{=}\begin{pmatrix}p_1+p_2 & p_0 +\frac{\lambda}{\mu} p_2 & p_0 +\frac{\lambda}{\mu} p_1\end{pmatrix}$$ Let $k:=\frac{\lambda}{\mu}$. The polar of $P$ is the same for every conic in the bundle if and only if $$\begin{pmatrix}p_1+p_2 & p_0 +k_1 p_2 & p_0 +k_1 p_1\end{pmatrix}\stackrel{\text{x}}{=}\begin{pmatrix}p_1+p_2 & p_0 +k_2 p_2 & p_0 +k_2 p_1\end{pmatrix}$$ for every $k_1,k_2 \in \mathbb{R}\setminus \{0\}$. If $p_1+p_2 \neq 0$, the two remaining coordinates must be constant, implying $p_1=p_2=0$, contradiction. Such equation is true if and only if $$\begin{cases}p_1+p_2=0 \\(p_0+k_1p_2)(p_0+k_2p_1)=(p_0+k_1p_1)(p_0+k_2p_2) \end{cases}\iff \begin{cases}p_1+p_2=0 \\ p_0(p_2-p_1)(k_1-k_2)=0\end{cases}$$

The two projective points for which these equality hold for every $k_1,k_2$ are

$$A=\begin{pmatrix} 1 \\ 0 \\ 0\end{pmatrix},Q=\begin{pmatrix} 0 \\ 1 \\ -1\end{pmatrix}$$

I cannot figure out geometrically why the point $Q$ has a fixed polar. Can somebody give an intuition as to why this happens?

$\endgroup$
0
$\begingroup$

Further thinking about the problem brought me to this conclusion.

Let $\mathcal C$ be a tangent bundle of conics in $\mathbb{P}^2(\mathbb{R})$ with base cycle $2A,B,C$.Let $a$ be the fixed tangent through $A$. Let $D:=a~ \land(B~ \lor C)$. Since $D\in a$, its polar line $d$ passes through $A$ by reciprocity. Moreover, $X\in d$ because the cross-ratio $(BCDX)$ is harmonic. This implies that the polar $d$ is fixed, because it passes through $X$ and $A$ for every conic in the bundle.

enter image description here

In my problem, $D$ is exactly $\begin{pmatrix} 0 \\ 1 \\ -1\end{pmatrix}.$

$\endgroup$

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.