Re: isometries, and symmetry groups



On Mar 30, 2008 7:43 PM CT, crossedproduct wrote:

If M is a (nonempty) set, we can consider the group
of all bijections M -> M.

Suppose X is a subset of M; does the set of all
bijections which leave X invariant, necessarily form
a group ?
(A bijection f : M -> M which leaves X invariant is
such that f(X) = X).

I'm pretty certain that it does.

I believe that a few more conditions are required to
gaurantee a group structure.

If we require the bijection f to be a homeomorphism, then
given a topological space X, the set of all homeomorphisms
of X onto itself (denoted Homeo(X)) indeed form a group.

How does this differ from the symmetry group of X,
which consists of all *isometries* of M (when M is a
metric space) leaving X fixed ?

It would seem that the difference lies in the fact that
Homeo(X) is a group regardless of the topological space X
being metric, whereas Isom(M) requires M be a metric
space -- it would appear that Homeo(X) is a little more
general than Isom(M).

Of course, that begs the question: Is Homeo(M) =~ Isom(M)
for every metric space M? Intuitively, I would like to
say yes, however, this matter is a little beyond me at
the moment.

In other words, why are isometries important to
studying (or defining, for that matter) groups of
transformations of plane figures, and not just those
bijections which leave them invariant?

I may be mistaken, but if you consider your question a
bit more deeply, then I think it will answer itself. You
are talking about bijections which leave figures
invariant. Well, what is meant by this "invariance" of
figures? Does it not mean that distances between points
on the figure are preserved?

In any case, Homeo(X) and Isom(M) are both very imporant
groups and much is known about them (i.e. algebraic
topology, algebraic geometry, etc. are developed branches
of mathematics).

Regards,
Kyle Czarnecki
.



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