Suppose $\delta$ is a collineation of a projective plane which is its own inverse (i.e. $\,\delta^2 = i$ is the identity map). Trying to show that $\delta$ must fix more than one point and more than one line.
This problem has me sort of confused and I feel like I'm missing something.
I know:
- If $F$ is a collineation, then the inverse of $F$ is also a collineation.
- If $F$ is a collineation, then $F$ is (naturally induces) a bijection on the set of all lines.
Tried completing by showing:
Given a collineation $\delta$ and a line $l$, we can fix distinct points $p$ and $q$ on $l$, and since $\delta$ is a 1-1 and onto, $p'=\delta'()$ and $q'= \delta'(q)$ are distinct. Let $l′$ be the line through $p'$ and $q'$. Then $\delta[l']$ is a line through $p$ and $q$, so it must be $l$. This then shows that $\delta$ is 1-1 on lines. It is clear that $F$ is also injective on lines.
Now I am a little confused about how to tie this into a collineation which is it's own inverse.
The argument is tricky and you need to pay a lot of attention to your cases. I'll start you off with the part showing that more than one point is fixed (the lines part is just the dual argument): If $\delta$ fixes every line, it also fixes every point, and we're done. If not, let's choose a line $l$ which is not fixed. What can you say about the intersection of $l$ and $\delta(l)$? Is there another line not fixed by $\delta$ and not incident to this intersection?