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Re: mirrors: two or more?

Subject: Re: mirrors: two or more?
Date: Sun, 6 Aug 2000 11:56:56 +0530
From: Abhishek Roy <fingerslip@YAHOO.COM>

I follow your argument below but you assume the property of
involution (A*A = I) which John Denker has chosen to ignore.

Not so. I don't assume R*R=I, but, rather, I prove that, given the
definition of R as x -> -x, it *must* have this property. I then
prove by contradiction that there are no reflection operators which
have the properties Decker 1 and Decker 2.

The physical
mirror with which we are all familiar has this property of course, but
needless to say the whole exercise is precisely a hypothesis on the
(mathematical) existence of others.

Of "other" what? Mirrors? Perhaps, I still don't understand what
some people are talking about when they use the terms "mirror" and
"reflection." If we are not referring to the transformation x -> -x,
then *what* are we talking about?

Enantiomorph prop. 1 : Enantiomorphs in n dimensions are are identical in
n+1 dimensions. Please show exactly how this is dependent on mirrors. As I
said in my previous post, IMO an independent definition is possible, because
of statements like the one above.

Given that R transforms as x > -x, then a 180 degree rotation in n+1
space is equivalent R. Consider the letter "L." It has handedness in
2 dimensions, but we can transform it into its mirror image by
rotating it out of its plane 180 degrees in the third dimension. The
result is exactly what R gives. The same goes for 3-d objects rotated
in 4-d, and n-d objects in (n+1)-d. Rotation by 180 degrees in n+1
dimensions is equivalent to reflection in n dimensions; they are the
same thing.

Enantiomorphs are *defined* by their behavior under the reflection
transformation, i.e, an enantiomorph is an object which cannot be
superimposed on its mirror image. An equivalent definition is an
enantiomorph is an n-dimensional object which cannot be superimposed
on itself when rotated through 180 degrees in n+1 dimensions.

The behavior of enantiomorphs under the reflection operator is a
*necessary* property of enantiomorphs. I don't think you can define
enantiomorphs any other way, but, if you could, it would still have to
have the same property under reflection. If not, it would not be an
enantiomorph.

Glenn A. Carlson, P.E.
St. Charles County Community College
St. Peters, MO
gcarlson@mail.win.org