Re: left/right symmetry, manifest or not

["John S. Denker" <jsd@MONMOUTH.COM>]

Jack Uretsky wrote:
>         The truth is that the RH Rule is built into the Clifford algebra.
> An easy example is with the Pauli matrices, usually defined as:
>         (read s_{x} as "sigma sub-x")
>
> s_{x}s_{y}=is_{z}   s_{i} squared =1, s_{i}s_{j}=-s_{j}s_{i},i not equal j
>
>         This corresponds to a RH- coordinate system.  One could just as
> well have written:
> s_{y}s_{x}=is_{z}
> which would correspond to a LH-coordinate system.

1) I wouldn't have said that.  If you want to say that the RH
rule is built into the Pauli algebra, fine.  But it is not
built into the Clifford Algebra.

2) The axioms of Clifford Algebra sometimes permit but
never require the construction of chiral critters such
as the unit pseudoscalar (i).

The axioms state that each of the basis vectors anticommutes
with each of the others, and that's all that need be said.

Indeed the axioms don't even say how many basis vectors
there are.  You therefore can't take for granted that it is
even possible to construct chiral objects.  If you doubt
me, just try it in D=2 for starters.

3) In particular, you can do all of classical electrodynamics
using Clifford Algebra without mentioning (i) or any other
chiral object, and without imposing any ordering on the
basis vectors.

4) Presumably you need to specify the handedness of the basis
vectors in order to describe weak nuclear interactions.

5) If you insist on expressing the magnetic field as a
pseudovector, you will need (i) or some similar chiral
combination of basis vectors in order to undo the mischief
and convert the magnetic field to the desired bivector
representation, for instance in the expression
        F = (E + i B) gamma_0
but this doesn't repeal item (3) ... you can (and IMHO
should) describe the magnetic field as a bivector from
the beginning, to save yourself a lot of trouble, and to
make manifest the left/right symmetry of the physical laws.