From: rusin@washington.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: Involutions
Date: 8 May 1995 15:45:07 GMT
In article <3ocvo4$dk5@rzsun02.rrz.uni-hamburg.de>,
Hauke Reddmann wrote:
>Let f1(x)=x,f2(x)=-x,f3(x)=1/x.
>Then for all i,j fi(fi(x))=x and fi(fj(x))=fj(fi(x)).
>Is it possible to add a fourth involution which commutes
>with the others? (Continuous, of course.)
No. Now the lecture.
There seems to be some fascination with finding functions satisfying some
relations with composites. I'll admit, they entertain me too. As
recent examples (including this one) show, one usually discovers that one
must refine the assumptions along the way -- instead of thinking
about "formulas" one must make decisions about domain (and codomain),
continuity, and so on. Perhaps someone can suggest a reference addressing
these concerns generally so we can add something to the FAQ.
The unspecified part of this particular problem is just what the domain
ought to be. I am guessing the poster wants further functions to be
defined on X=R-{0}, just as f2 and f3 are.
In the present case, the poster is inquiring about continuous
involutions, which are necessarily homeomorphisms, once the domain is
specified. I find it helpful to use the perspective of group theory to
answer such questions. We may state this as asking if there is a
larger elementary-abelian 2-group < f2, f3, g > in Homeo(X).
The answer is no.
Observe that f o g = g o f and f(x)=x imply f fixes g(x). Applying
this to f=f3 and x=1 implies g(1) must be either 1 or -1. Replacing
g with f2 o g as necessary, we may assume g(1)=1.
Now, X is the disjoint union of two connected components, so since
g is a homeomorphism, it must permute these. Since g(1)=1, g must
in fact be an involution of each of the two components.
By applying the logarithm homeomorphism, we see g gives rise to
an involution G of the real line which fixes 0, and which commutes
with F3(x) = -x. We can delete this fixed point and see that G
is a homeomorphism of R-{0} (only by accident the same as X, if
you follow me); by the same reasoning as above we conclude that
(composing with F3 as necessary) G is a homeomorphism of order 2
of (0, oo), now with the extra condition that the further value
G(0)=0 make G continuous on [0, oo).
The Intermediate Value Theorem then forces G to be an increasing
function: if 0 G(b) [both must be positive], the IVT
shows G(b) = G(c) for some c in [0,a], which violates the condition
that G be invertible.
It then follows that if for some x we have x < G(x), then also
G(x) < G(G(x)) = x, a contradiction; x > G(x) is likewise excluded.
Consequently, G must be the identity function on [0,oo). Since
G commutes with F3, G(x) =x for all real x, so that g(x) = x for
all positive x. Since g commutes with f2, g(x)=x for all nonzero x.
Thus, an element of order 2 in Homeo(X) which commutes with < f2, f3 >
must actually lie in that subgroup.
Now, if you drop continuity, or no longer require that g be defined on
all of X, then more possibilities arise. For example, you can use
f4(x) = 1-x on (0, 1) and extend to the rest of X'=R-{-1, 0, 1} using
commutativity with f2 and f3; but then the same reasoning as
above suggests there will be no further involution on this X'
commuting with f2, f3, f4. If you delete N points of R, your
homeomorphism group is isomorphic to
Homeo(R) wreath Sym(N+1)
which includes the elementary-abelian 2-subgroups of Sym(N+1) (whose
cardinality and description are known) and the unique (up to conjugation)
order-2 subgroup of Homeo(R) (included diagonally). Unless I'm
missing something that will give you a maximal elementary 2-subgroup.
Other maximal examples will include the homeomorphisms which invert
some of the copies of R and fix others.
dave
References:
that rational-function-of-order-3 problem
that f(x+a sqrt(x)) problem
that f(x)f(y)=f(x+y) problem
etc.