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Re: A question about mirrors

Hi all-
David has correctly picked a nit. The subset with det=1 is a
(proper) subgp., the subset with det=-1 is not a subgp.
Regards,
Jack

Adam was by constitution and proclivity a scientist; I was the same, and
we loved to call ourselves by that great name...Our first memorable
scientific discovery was the law that water and like fluids run downhill,
not up.
Mark Twain, <Extract from Eve's Autobiography>

On Mon, 7 Aug 2000, David Bowman wrote:

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