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I am a bit struggling with the following exercise.

Let Q be a non-degenerate quadric in $\mathbb{RP}^3$, $l \subset \mathbb{RP}^3$ a line and l' $\subset \mathbb{RP}^3$ its polar. Show that all polar planes of points on l pass throught l'.

I tried to prove this algebraically. Let X=[x] be a point on l and let b be the symmetric bilinearform describing Q. By defnition of the polar, for each point P=[p] on l' it follows that b(x,p)=0. This implies that P is contained in the polar plane $P_X$ of X.

So for each X $\in$ l it follows l' $\subset P_X$. But this does not agree with my sketches where the polar l' is orthogonal to l and intersects the polar planes in exactly one point. So is my proof wrong or is my understanding what the polar line is flawed?

Parcly Taxel
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Polymorph
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  • Out of curiosity, what’s your definition of the polar line to a line? This proposition might simply be dual to the definition. – amd Dec 09 '17 at 22:10

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Let us set apart the fact that we are in projective space $\mathbb{RP}^3$ because the issue can be set in ordinary Euclidean space $\mathbb{R}^3$.

The point where I do not agree with you is that the polar line of a given line is not orthogonal in general to the initial line.

Let us look at an example. Take for our quadric the ellipsoid with equation :

$$x^2+4y^2+z^2=1$$

Besides, let us take $\ell$ with parametric equations :

$$\begin{cases}x=2t+1\\y=t\\z=3t-1\end{cases}$$

the polar line $\ell^{\circ}$ of $\ell$ is defined as the set of points $(x,y,z)$ such that:

$$\forall t \in \mathbb{R}, \ \ \ \ (2t+1)x+4ty+(3t-1)z-1=0.$$

Collecting constant terms and coefficients of $t$, we get the equivalent system:

$$\begin{cases}x-z-1=0\\ 2x+4y+3z=0\end{cases}$$

whose solution is straight line $\ell^{\circ}$ with equations:

$$\begin{cases}x=z+1\\y=\dfrac{-5z-2}{4}\\z=1z+0\end{cases}$$

In particular, a directing vector of this straight line $\ell^{\circ}$ is $(1,-5/4,1)$ which is not orthogonal in the Euclidean sense with respect to the directing vector : $(2,1,3)$ of $\ell.$

Nevertheless, these two former vectors can be considered as orthogonal with respect to the quadratic form induced by the ellipsoid:

$$\begin{pmatrix}1 & -5/4 & 1\end{pmatrix}\begin{pmatrix}1 & 0 & 0 \\ 0 & 4 & 0\\0& 0& 1\end{pmatrix}\begin{pmatrix}2\\1\\3\end{pmatrix}=0.$$

Jean Marie
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    In the literature, the two vectors are often called conjugate w/r the ellipsoid. – amd Dec 09 '17 at 02:15