Does a continuous involution of R^n have a fixed point?
Alan D. Burns
Does a continuous involution f: R^n -> R^n have a fixed point?
I expect the answer to this question is well known, but not by me.
The following summarises my limited progress in answering the question.
Easy Proof for R^1.
-------------------
We may assume f is not the identity, so choose distinct points x, x'
which are swapped by f. Simple continuity arguments show that f maps
the closed interval [x,x'] into itself. Hence, f has a fixed point in
[x,x'] by Brouwers fixed point theorem.
Almost (?) a Proof for R^2.
---------------------------
As above, choose distinct points x, x' which are swapped by f.
Let c be a continuous curve joining x, x', and c' its image under f.
The concatenation c//c' is then a closed curve in R^2.
Case 1: c intersects c' in x, x' only.
--------------------------------------
By the Jordan curve theorem, R^2 - (c//c') has two connected components,
the bounded interior I, and unbounded exterior E, of c//c'.
These are either fixed or swapped by f - they must in fact be fixed by
f, as f must preserve compactness of the closure of I, Cl(I).
Hence, f maps Cl(I) into itself, which is topologically a closed disk.
So, the Brouwer fixed point theorem gives a fixed point of f in Cl(I).
Case 2: c intersect c' in a finite number of points.
----------------------------------------------------
f acts as an involution on the points of intersection of c and c'.
So, if c intersects c' in an odd number of points, then one of them
must be fixed by f, that is, the middle point ordered parameterically
along c, so we are done.
If c intersects c' in an even number of points, label these
x=x(1), x(2), ...., x(2n), ordered parametrically along c.
f permutes these pairwise:
x(1) <--> x(2n)
x(2) <--> x(2n-1)
:
x(n) <--> x(n+1)
If we take the segment d of the curve c between and x(n), x(n+1), then
the points x(n), x(n+1) and the curve d between them then satisfy the
conditions of case 1. Hence we are done.
Case 3: c intersects c' in an infinite number of points.
--------------------------------------------------------
This is the part of the 'proof' that I cannot finish.
To complete the proof, we need to either handle case 3, or show that
x, x' and c may be chosen to satisfy case 2. In a counterexample, every
possible x, x', c would have to satisfy case 3. I find it unimaginable
that such a counterexample should exist, but I cannot prove it!
Ideas for R^n
-------------
A proof along the lines of my R^2 attempt may be possible, but it is
likely to be a lot messier.
I think a more fruitful approach is to consider the continuous
extension of f to the one-point compactification S^n of R^n.
It is easy to see that this extension exists, is unique, and satisfies
f(infinity) = infinity. So, the conjecture is equivalent to the
alternative conjecture:
If a continuous involution of S^n has a fixed point, then it has at
least two fixed points.
Is this a well-known result from algebraic topology?
Thanks for all contributions,
Alan D. Burns
Joerg Winkelmann
Alan D. Burns (al...@abnumerics.demon.co.uk) wrote:
: Does a continuous involution f: R^n -> R^n have a fixed point?
:
:
Yes.
One can prove this in the following way:
Assume that there is a fixed-point-free involution.
In this case take the quotient Q of R^n divided by the equivalence
relation given by the involution. Now we got an unramified Galois
covering R^n --> Q and it follows that Q is an Eilenberg-MacLane space
of type K(Z/2Z,1).
Now the infinite-dimensional real projective space
P_infty(R) is also an Eilenberg-Maclane space of type K(Z/2Z,1).
This implies that the cohomology groups of Q and P_infty(R)
coincide.
But this is impossible:
H^k(P_infty(R),Z/2Z) = Z/2Z for all natural numbers k,
while Q is a manifold of dimension n which implies that
H^k(Q,Z/2Z) = 0 for all k> n=dim(Q).
Thus there can not exist such an involution.
Joerg
Cyril.Banderier
You will have a complete prove of this from C to C in a french review :
"Quadrature" and also in "Revue de mathematiques speciales".
I remember that there was also a proof which used Schoenfield's theorem.
Jeremy Rickard
Alan D. Burns wrote:
>
> Does a continuous involution f: R^n -> R^n have a fixed point?
>
Yes. Suppose there is such an involution without a fixed
point, and let G be the cyclic group of order two generated
by this involution. Consider the quotient space X = R^n/G.
Then X is an n-dimensional manifold, and so its integral
homology vanishes in degrees greater than n, but also
X is an Eilenberg-MacLane space K(G,1), and so has integral
homology in all odd degrees.
More generally, no non-trivial finite group can act freely
on R^n.
Jeremy Rickard.
Steve Ferry
In article <D$IhpPA8A...@abnumerics.demon.co.uk>, "Alan D. Burns"
<al...@abnumerics.demon.co.uk> wrote:
> Does a continuous involution f: R^n -> R^n have a fixed point?
>
> I expect the answer to this question is well known, but not by me.
> The following summarises my limited progress in answering the question.
>
For the R^2 case he mentions, you can give a short argument: Take the
quotient space and get a covering projection R^2 -> M^2, where M^2 is a
connected non-compact 2-manifold. By covering space theory, the
fundamental group of M^2 is Z/2Z. But every connected noncompact
n-manifold is homotopy equivalent to an (n-1)-dimensional polyhedron^1, so
the fundamental group of M^2 is the fundamental group of a graph, which is
free.
The general case follows by cohomology of groups, as pointed out by
Rickard. Smith theory, as exposed in Bredon's book on group actions, gives
more information about possible fixed-point sets.
Steve Ferry
st...@math.binghamton.edu
^1 It's easier to see that if pi_1 M^2 contains a torsion element, then
pi_1 K^2 contains a torsion element for some compact connected submanifold
K^2 of M^2. But it's easy to see that compact connected 2-manifolds with
boundary collapse to 1-complexes, giving a contradiction.
--
Steve Ferry
st...@math.binghamton.edu