The cross product is alive and well and living with Clifford (was: left/right symmetry, manifest or not)

[Jack Uretsky <jlu@HEP.ANL.GOV>]

Hi all-
        In an effort to demistify some of this geometric algebra stuff, I
will make two points.  The Harke paper
                               http://www.harke.org/ps/intro.ps.gz

 is directed to more profound
problems than are normally faced in undergraduate mechanics, and the
ordinary vector analysis that you have long known and loved is ("is"="may
be described as") a Clifford Algebra.
        As to the first point, I quote from Harke's paper, p.3:
"These techniques will enable us to carry
out many operations of linear algebra in a coordinate free manner. Finally
I will discuss the spacetime algebra, a particular geometric algebra of four
dimensions based upon the Minkowski metric. All of these methods play an
essential role in the development of Gauge Theory Gravity and, I believe,
give it an elegant expression."
        So,IMO, foisting this stuff on helpless undergraduates (and their
teachers) qualifies as corruption of youth in the worst sense of the
excesses wrought by the "new math" of the 'sixties.
        As to the second point, I will present a proof that the positive
subalgebra (to be defined) of the Clifford Algebra of V(4,(4)) is just the
ordinary vector algebra that you have all used since childhood.  You will,
I think, find the proof entertaining and instructive.  I presented the
axioms of Clifford Algebra in yesterday's posting.
        V(4,(4)) is venerated by the basis vectors e_i, 1=1,..,4 which
have the property that (e_i)^2=1 and the "dot product" (e_i|e_j) =delta(i,j).
That is, the vectors form an orthonormal set.  The "dot product" is
defined by the relation
 (e_i|e_j) = 1/2(e_ie_j + e_je_i) which I abbreviate as {e_i,e_j}/2.

        The ordered products of basis vectors (like 12, 13, 14, 123, etc)
are the elements of the Clifford algebra.  The positive subalgebra has as
its elements the products composed of an even number of e_i's.  There are
7 such elements.  I define them as follows:
        a1=e_1e_2, a2=e_1e_3, a3=e_1e_4, a4=e_2e_3, a5=e_2e_4, a6=e_3e_4,
and a7=e_1e_2e_3e_4.

1) First note that (a7)^2=1, and a7 commutes with each of the other 6
elements of the algebra.  By a well-known theorem (it's called Schur's
Lemma), a7 is proportional to the identity vector of the algebra.  So I
can write a7 as either +I or -I, since its square is 1.
2) Next, it is easy to show that
        a6= -a1a7
        a5=  a2a7
        a4= -a3a7
so that there are only three independent elements of the algebra which we
take as a1, a2, a3.  Their products are:
        a1a2=a3, a2a3=a1, a3a1=a2, and they all satisfy aiaj=-ajai for i
not equal to j.
3.  The three elements a1, a2, a3 therefore are the basis elements of a
vector space over the real number.  The algebra of that vector space
admits two products of elements u,v:  The ordinary product, which in your
former life you wrote as the cross-product, where the basis elements
satisfy a1a2=a3, a2a3=a1, a3a1=a2; and the "dot product" which may also be
written as u dot v ={u,v}/2.
4.  Finally, this all looks more familiar if we write:
        a1 = i
        a2 = j
        a3 = k
and put little hats over the i,j, k.
                                        Q.E.D.
                        Regards,
                                        Jack



On Wed, 28 Aug 2002, Robert Cohen wrote:

> I wrote:
> > > Now, suppose we can agree on a coordinate system.  In other
> > words, suppose I
> > > give you enough information so that you can reproduce the
> > coordinate system
> > > when you arrive in my room.
> > >
> > > Is it now possible to describe, via words, which way it is
> > spinning?  If so,
> > > how?
> to which John S. Denker responded:
> > If you say the object's orientation is first north, then east, then
> > south, then west, then north again, there's no doubt about the
> > direction of rotation.
> >
> > Does that answer the question?
>
> Yes, it does.  Thank you.
>
> Now, suppose I define a particular vector product such that its magnitude
> represents a "circulation" in the "direction of A then in the direction of
> B".  In what way is this equivalent to the wedge product?  How about the
> cross product?
>
> ____________________________________________
> Robert Cohen; rcohen@po-box.esu.edu; 570-422-3428; http://www.esu.edu/~bbq
> Physics, East Stroudsburg Univ., E. Stroudsburg, PA 18301

--
"But as much as I love and respect you, I will beat you and I will kill
you, because that is what I must do.  Tonight it is only you and me, fish.
It is your strength against my intelligence.  It is a veritable potpourri
of metaphor, every nuance of which is fraught with meaning."
                     Greg Nagan from "The Old Man and the Sea" in
                     <The 5-MINUTE ILIAD and Other Classics>