Re: [Phys-l] constructing perpendiculars in D dimensions

[John Denker <jsd@av8n.com>]

On 11/30/2009 08:29 AM, David Bowman wrote:

> Considering John's repeated lobbying for exterior/geometric algebra
> I suspect that he would be expecting an answer somewhat along the
> lines of:
>
> Take the Hodge dual of the (D-1)-form exterior product of the D-1
> (covariant) basis vectors that are already mutually orthogonal.

That'll do it.

I might have expressed the idea slightly differently:
if you're a geometric algebra maven you can get the 
job done even if you've never heard of a Hodge dual
... not that I'm recommending reinventing wheels or
complaining about knowing more than the bare minimum,
just that pedagogically speaking introducing the 
Hodge dual seems unnecessarily complicated.

Here's an easy and compact way of doing the job:
Take all D of your basis vectors and multiply them
together to form a top-grade element e.  (This is
often called a "pseudoscalar" even though it really
works as a pseudoscalar only when D is odd.)

Multiplying by e is very nearly the same as taking
the Hodge dual.  (In a non-Euclidean space e.g.
Minkowski space you might need do to a teeny bit
more work to account for the signature of the
metric.)

Specifically, wedge-multiply together the given D-1
vectors, then multiply by e, and you're done.

The wikipedia article
  http://en.wikipedia.org/wiki/Hodge_dual
seems to know what it's talking about, although
I haven't reviewed it super-carefully.  The final
section on writing div, grad, and curl in terms 
of exterior derivatives and Hodge stars shows 
why the idea is useful ... but also supports the
idea that Clifford algebra allows you to do the
same things more easily.  You can write div as
(* d *) but in Clifford algebra you can just
write it as (∇ •) in the usual way.  Or maybe
I'm just missing something;  I don't pretend to
be a big expert in this area.

In any case it is interesting to reflect on the
isomorphism between objects of grade k and objects
of grade (D-k).

===

The Gram-Schmidt idea also works.  You would 
have to iterate over the D basis vectors to
find one that is linearly independent of the
D-1 given vectors, then use Gram-Schmidt to
orthogonalize it.

===

A third way of getting the job done is a D
dimensional generalization of the tableau
used to compute cross products.  Write out
the components of the D-1 given vectors 
using D columns and D-1 rows, then fill in
the Dth row with the D known basis vectors.
Consider this tableau to be the elements of 
a DxD matrix and take the determinant.

========

Amusing note:  Using any of these methods,
you discover that the same method works for
an even more general problem:  The D-1 given
vectors do not even need to be orthogonal,
as long as they are linearly independent.
The constructed vector will be orthogonal
to each of the D-1 given vectors even if 
they are not orthogonal to each other.

=========

Another thing I noticed:  By way of background
recall that the usual rule for using Clifford
algebra to reflect a vector v is:

       mirror(v) = r v r        [2]

which reflects v in the plane (or hyperplane)
perpendicular to r, for any unit vector r.

However, it seems much cleverer to write it as
 
                  r v r
     mirror(v) = -------           [3]
                   r•r

               = (1/r) v (r)

because equation [3] works even if r is not normalized,
... and works even when r is timelike i.e. r•r < 0.