Re: involution has fixed point

Hi,
there is an easy solution, relying on the case $n = 2$.

Let $L$ be any line of $R^n$, $n > 2$, and let $f$ be the involution. If $L^f$ meets $L$ in a point, and $L^f$ is not $L$, the intersection point is fixed, and we are done.
So for any line $L$, $L$ and $L^f$ generate a $3$-subspace $R^3$ which is fixed (globally) by $f$.
Take any point $x$ in this $R^3$. We suppose $x^f$ is not $x$, so that $x$ and $x^f$ generate a line fixed by $f$. So on each point there is a unique line fixed by $f$.
Now it is easy to see that $f$ must have fixed planes in $R^3$, and the case $n = 2$ can be used.
.