Does duality mapping preserve cross ratio?

Question

I'm new to projective geometry.

I learned the definition of cross ratio of 4 collinear points and that of 4 concurrent lines in $\mathit{P}\mathbb{R}^{2}$.

The question is, by duality we can map 4 collinear points into 4 concurrent lines. Does such maps always keep the cross ratio of the points and that of their images (cross ratio of the lines) the same?

By duality mapping, I mean a map $f$, satisfying:

1. if $P$ is a point, then $f(P)$ is a line.
2. if $l$ is a line, then $f(l)$ is a point.
3. $P$ lies on $l$ $\Longleftrightarrow$ $f(l)$ lies on $f(P)$.

And the question can be described as:

Is it always the case $(AB;CD)=(f(A)f(B);f(C)f(D))$, no matter how we choose the mapping $f$, where $(AB;CD)$ is the cross ratio $\frac{AC}{BC}:\frac{AD}{BD}$.

I think it's true if we construct the mapping in regard of homogeneous coordinates.

If $A,B,C,D$ are 4 collinear points, we can pick their homogeneous coordinates to be $\vec{a},\vec{b},\vec{c}= \vec{a}+\vec{b},\vec{d}= k\vec{a}+\vec{b}$ then $(AB;CD) = k$

Let $f(\vec{a})=\vec{a}$, that is, the dual of the point $A = \vec{a}=(P,Q,R)$ is the line $Px+Qy+Rz = 0$. So the dual of the point has the same coordinate of its preimage. It's just a matter of whether we interpret the coordinate as a point or a line.

If so, we have $(f(A)f(B);f(C)f(D))=k$, too.

However, as far as I know there are more than one way to construct duality mappings (pole and polar, etc.). And so comes the question.

Thank you for any response!

Answer 1

If you have two "dualities" $f$ and $f'$, then $f'\circ f^{-1}$ is a self-equivalence of the projective plane (it sends points to points and preserves alignment). This means that this is a projective transformation (this is not a completely trivial statement!), which are known to preserve the cross-ratio.

Try to see that this means that is one duality preserves the cross-ratio, then they all do.