Re: magnitude of parallelograms and parallelepipeds

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

Bob Sciamanda asked about:
>  interpreting the geometric product of two bivectors.

OK, here goes:

> The definition for vectors: AB =A dot B + A /\ B

I wouldn't call it the definition, but it's a valid
and useful formula.  (Some people, including the Cambridge
mafia, introduce the geometric product this way, but
most people don't, and I don't recommend it.)

I consider the geometric product to be primary and
fundamental.  The dot product and wedge product can
be derived therefrom.  See below.

> is not directly applicable if A and B are bivectors (?)

Right, not applicable at all.

> What do "dot" and /\ mean between bivectors?

I like the interpretation given by equations 18 and
20 in Harke, because it is simultaneously maximally
practical and maximally sophisticated:  If A has grade r
and B has grade s, then
 -- A.B is the low-grade piece of the geometric
    product, having grade |r-s|, assuming nonzero r,s
 -- A/\B is the high-grade piece of the geometric
    product, having grade r+s.

Note that for a pair of vectors, r=s=1, this reduces
to the familiar
        A . B  = (A B + B A)/2
        A /\ B = (A B - B A)/2

while if A is a vector and B is a bivector, it's
just the reverse:
        A /\ B = (A B + B A)/2
        A . B  = (A B - B A)/2

(Compare equations 30 and 31 in Harke.)

And (!) if both A and B are both bivectors, it is not in
general possible to write the geometric product in terms
of the high-grade piece (grade=4, exterior product) and the
low-grade piece (grade=0, inner product) because there could
be a "middle product" (grade=2).

My recommendation:  to evaluate (A/\B)(C/\D), express both
factors in terms of the geometric product,
        (A B - B A)(C D - D C)
and turn the crank.

To evaluate (A/\B).(C/\D), proceed as above, then throw away
all non-scalar terms in the product.

To evaluate (A/\B)/\(C/\D) proceed as above, but throw away
everything except the grade=4 piece.  Note that /\ is
associative, so we can call it A/\B/\C/\D.  (This is guaranteed
to be zero in D=3.)

Reference:
  http://www.harke.org/ps/intro.ps.gz


========================================
Jack Uretsky wrote:
>         Writing AB=AdotB + A/\B is misleading, at best.  It seems to imply
> that the scalar AdotB can be added to the elements of A/\B, which is
> wrong.

I diametrically disagree.  Adding A.B to A/\B is exactly
the right thing to do.  As emphasized by the Cambridge
mafia among others,
  http://www.mrao.cam.ac.uk/~clifford/introduction/intro/node5.html
the whole point of Clifford Algebra is to add scalars
and vectors et cetera.  Yes, this is like adding apples
and oranges.  People add apples and oranges all the time.
(You can't safely _compare_ apples and oranges, but that's
a different issue.)  Most people on this list can cope
with adding a real number plus an imaginary number to form
a complex number;  it shouldn't be much of a stretch to add
scalars, vectors, bivectors etc. to build a Clifford Algebra.

Indeed, talking about "components" at all is retrogressive.
I agree with Hestenes that thinking about vectors primarily
in terms of their components is a harmful _conceptual virus_
  http://modelingnts.la.asu.edu/pdf/OerstedMedalLecture.pdf

Vectors, bivectors, etc. have a physical and geometrical
reality that transcends any particular coordinate system.

There is a pictorial representation that appeals to the
least-sophisticated students.  There is an axiomatic
structure that appeals to the most-sophisticated students.