Reply: does your teacher know what you are doing?

Subject: Re: Q: Geometric Algebra instead of Vector Algebra??
From: john baez <baez@math.ucr.edu>
Date: 1996/09/26
Message- Id: <52f22c$m5s@pulp.ucs.ualberta.ca>
Newsgroups: sci.physics.research
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In article <524snh$at4@agate.berkeley.edu>,
Doug B Sweetser <sweetser@alum.mit.edu> wrote:

>Physicist always acknowledge the importance of fields.

Yes, but usually they mean something different than "commutative
ring with all nonzero elements invertible" when they say "field".
Usually they mean something like the electromagnetic or gravitational
field. :-)

>When describing a
>vector space or a particular algebra, the clause, "over the field of real
>(or complex) numbers" is added. That way calculus can be done! It is
>quite the exception for one to read "over the field of quaternions."
>There is a good reason for this: quaternions don't commute.

Indeed. Most of the time people don't refer to noncommutative
rings with all nonzero elements invertible as "fields". The
term "skew field" is more common for these. Folks also use
"division ring" to mean any ring for which all nonzero elements
have inverses.

>While this
>may be "the work of the devil" (Lord Kelvin) from a mathematical
>viewpoint, it is quite common in quantum mechanics.

Noncommutativity is typical in quantum mechanics, but most noncommutative
operator algebras aren't fields or even skew fields; the only algebras
over the reals for which all nonzero elements are invertible are the
reals, the complexes, and the quaternions. (Here of course by "algebra"
I mean "associative algebra", as usual --- which rules out the octonions.)

>One technical issue I have with Hestenes approach is that I'm not sure
>his Geometric Algebra is always a field. I know for example that
>quaternions with complex values are no longer a field.

Right. Hestenes' "geometric algebras" are, as far as I can tell, what most
folks call "Clifford algebras". There are lots of interesting Clifford
algebras, of which the "complex quaternions" are one, but the only ones
that are division rings are the reals, complexes and quaternions.

>To lose a
>founding property of calculus is too high a price for me to pay!

Why? It's not as if it's an all-or-nothing matter: either to use
Clifford algebras for *everything* or *nothing*. Rings of all sorts
are interesting in physics: tensor algebras, Grassman algebras,
matrix algebras, Clifford algebras... and most of these aren't
division rings! Surely you don't refuse to study matrices on
the grounds that not all nonzero matrices have inverses?

>If quaternions are a powerful tool, then they should be useful for
>solving problems. To test that hypothesis, I am taking a undergraduate
>class in special relativity at MIT and doing all the problem sets using
>quaternions instead of the Lorentz transformation!

Bold fellow!

>I have made it through
>problems dealing with time dilation, length contraction and asynchronous
>moving clocks, now on to velocity addition with quaternions. This has
>been an immense amount of fun, a real challenge to develop the tools in
>Mathematica and then apply them (and my problem set look sooo neat typed!).

Neat. Does your teacher understand what you are doing?