Re: A question about mirrors

[David Bowman <David_Bowman@GEORGETOWNCOLLEGE.EDU>]

I had planned on sitting out the discussion about mirrors & rotations but
could not help picking a nit in a comment by Jack U.

In particular, regarding:
> ...
>        The uniqueness of the mirror image in classical physics,
> then, is represented by the fact that the isometric transformations
> in 3-space divide into two "discrete" subgroups that cannot be
> connected by a rotation.  ...
>  ...      The succinct answer to your question is that there are
> only 2 discrete subgroups, because there are only two real solutions
> to the equation x^2 = 1.

As much as I appreciate how much Jack's case is nicely argued, I (fussy
fellow that I am) still wish to point out that these two disjoint
subsets of O(3) (i.e. the 3x3 orthogonal matrices with determinants being
+1 and -1, respectively) are not both subgroups.  Only the subset (called
O(3)^+) of the orthogonal matrices with determinant of +1 form a
subgroup.  The other set of those orthogonal matrices with determinant -1
do *not* form a subgroup.  The reason that they are not a subgroup is
that they do not form a group among themselves since they do not contain
the identity, nor are they closed under multiplication (the product of two
-1 determinant matrices is a +1 determinant matrix).  The elements of
O(3) that have determinant -1 are called a *coset* (or coset 'space' since
they all happen to be continuously connected to each other), *not* a
subgroup.

David Bowman
David_Bowman@georgetowncollege.edu