From: sim_stef@yahoo.com (Simeon Stefanov)
Subject: Re: f(f(x,y)) = (x,y) => f(a,b)=(a,b)
Date: 15 Nov 2001 10:55:38 -0800
Newsgroups: sci.math
Summary: Homeomorphisms of the plane, of finite order, have fixed points

> OK, but don't all these transversality results depend on having
> a smooth structure on the manifold M? If f is smooth one acquires
> such a structure from f, butif not then maybe one has to appeal
> to the big theorem that every 2-manifold is smoothable.
> See my correspondence with Lee Rudolph elsewhere in the thread.
> 
> Robin Chapman

     It turns out there is no need to go into smoothness and
trasversality theorems. Here is a short proof based on the Borsuk-Ulam
theorem.
     Let f be a continuous involution in R^2. We shall prove that it
has a fixed point. Since the plane is connected and 1-connected, one
constructs a continuous map
                   h: S^2 -> R^2
sending antipodal points into involutive ones:
                    h(-x)=fh(x).
     Then by Borsuk-Ulam's theorem there exists a point x in S^2 such
that h(-x)=h(x), so fh(x)=h(x) and h(x) is a fixed point of f.

Simeon
==============================================================================

From: sim_stef@yahoo.com (Simeon Stefanov)
Subject: Re: f(f(x,y)) = (x,y) => f(a,b)=(a,b)
Date: 16 Nov 2001 00:58:13 -0800
Newsgroups: sci.math

> >      Let f be a continuous involution in R^2. We shall prove that it
> > has a fixed point. Since the plane is connected and 1-connected, one
> > constructs a continuous map
> >                    h: S^2 -> R^2
> > sending antipodal points into involutive ones:
> >                     h(-x)=fh(x).
> 
> How?
> 
>  >      Then by Borsuk-Ulam's theorem there exists a point x in S^2 such
>  > that h(-x)=h(x), so fh(x)=h(x) and h(x) is a fixed point of f.
> 
> Robin Chapman


     Set first h(A)=B, where A and B are arbitrary points in S^2 and
R^2 respectively. Then set h(-A)=f(B). Let L be a half great circle
with ends A and -A. Extend h to h: L -> R^2. Now extend it to LU(-L)
setting for x in -L
                             h(x)=fh(-x).
     Let W be one of the (closed) semispheres with boundary LU(-L).
Since the plane is 1-connected (in fact contractible) h may be
continuously extended to h: W -> R^2. Now setting as above h(x)=fh(-x)
for x in -W, we get continuous h: S^2 -> R^2 with the desired
property.

Simeon
==============================================================================

From: sim_stef@yahoo.com (Simeon Stefanov)
Subject: Re: f(f(x,y)) = (x,y) => f(a,b)=(a,b)
Date: 19 Nov 2001 07:06:50 -0800
Newsgroups: sci.math

> Is there a similar argument for period p for odd primes p?
> 
> Robin Chapman

     Yes, for prime periodic map in R^n  there is a similar proof
which makes use of Smith-Yang index theory. I thing there was a
counterexample for composite period.
     Maybe there is an elementary proof that every periodic map in the
plane has a fixed point.

Simeon