Re: involution has fixed point

In article <20875368.1133428489785.JavaMail.jakarta@xxxxxxxxxxxxxxxxxxxxxx>,
Koen Thas <kthas@xxxxxxxxxxxxx> wrote:
>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 $f$.

You seem to be assuming that f is a linear (or at least affine) map.
The real question is for nonlinear f.

Robert Israel israel@xxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
.

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