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Re: reflection operators



At 05:04 AM 8/3/00 +0530, Abhishek Roy wrote:
I was looking for some way to describe a reversal of 'handedness' without
using mirrors. For example- two books I looked at started by saying that all
isometries could be descibed as products of reflections. Then you show that
any even number of reflections may be reduced to a rotation and an odd
number of reflections into a (reflection+rotation). Alright so far, but I
don't understand how you now classify the latter as 'orientation-reversing'.

I'm not sure I understand your question(s), but let me take another stab at
answering....

If all you want is a way of distinguishing reflection-like transformations
from rotation-like transformations, then a topological answer leaps to
mind: rotations can be continuously transformed to the identity (just
continuously crank the angle to zero). Reflections cannot; they are
all-or-nothing.

It seemed to me that there must be more fundamental way of describing
orientation (not the specific L or R, just handedness), which you could use
to see that ther were only two forms and *then* apply that to a mirror
transformation.

The just-given criterion won't help you with this second question (which
is, I assume, the main objective).

First of all, I say again, you might be trying to prove something that
isn't true. The transformation of right-handed glove into a left-handed
glove is an isometry. But the transformation into an antimatter glove is
also an isometry. So there might well be more possibilities than just L and R.

It's bad luck to prove things that aren't true.

========

Let me take yet another stab at divining what the intended question is.

The reflection operator is a square root of the identity operator; V^2 =
I. Of course there are other square roots of the identity, including 180
degree rotations (and the zero-degree rotation for that matter).

Perhaps the intended question is something like this: suppose there was an
operator W which was a _cube_ root of unity and which was not a rotation
(i.e. not continuously deformable into the identity). Can we rule this
out? The answer is yes, under mild conditions, we can rule it out.

Obviously the key ingredient in the proof is the requirement that W be an
isometry; |W P| = |P| for all vectors P. So write out this requirement in
terms of the matrix elements of W. You will find lots of squares and
square roots. You can choose the +- roots arbitrarily and still satisfy
the requirement; about half the time such choices will result in
reflection-operators. You won't find any cube roots. I think that's all
there is to the story.