# Fix points and invariant planes of projectivity not using coordinates

Let $A,B,C,D \in \mathbb{RP}^2$ be a quadrivertex. Let $f: \mathbb{RP}^2 \to \mathbb{RP}^2, A\mapsto B, B \mapsto C, C \mapsto D, D \mapsto A$.

Find the fix points and invariant planes of $f$ and relate these results to the constraints without using coordinates.

Using a coordinate frame, $f$ would be determined by the four points, thus the eigenvectors of its matrix and those of its dual (inverse) could be calculated determining both the fix point and the invariant planes. Unfortunately i lack the understanding of how to do this without using explicit coordinates. Any ideas? Thanks already for your time!

One fixed point is easy. The line $AC$ gets mapped to $BD$ by the transformation $f$. Conversely $BD$ gets mapped to $CA$. So the intersection of these two lines must remain fixed.
Assume your four points form a regular square. Then the operation $f$ is a rotation by $90°$. A rotation is an Euclidean transformation, i.e. an isometry, and as such preserves the ideal circle points $[1:\pm i:0]$. For a generic quadrilateral, those points will likely have different coordinates, but it is important to take possibly complex positions into account. Any other quadrilateral is equivalent to a square under real projective transformations. Which means that the second and third fixed point will always be a pair of mutually conjugate complex points.
Now let's look at the fixed lines, which I assume is what you mean by invariant planes. One is again fairly easy. Since line $AB$ gets mapped to $BC$, that to $CD$ and that to $DA$, you can do with these lines what we did for the points. The intersection of $AB$ with $CD$ will trade places with that of $BC$ with $DA$. Which means the line joining them will be fixed.