Re: [Phys-L] closed vectors

[John Denker <jsd@av8n.com>]

On 10/06/2014 11:45 AM, Jeffrey Schnick wrote:
> It would be better to say that being closed under multiplications
> means that the cross product of any two vectors, each of which is an
> element of the set, is also an element of the set.

Right.

I would add that contrary to the wording of the subject
line, no vector is "closed".  It is the /set/ that is
closed (with respect to a given operation).

Closure is a concept that comes up a lot in /group theory/.
However, in the context of the cross-product operator, a 
closed set of vectors does not form a group.  The group 
axioms require an identity element, which is lacking here.

==============

Whenever you see a cross product, you should ask yourself
if the physics would be better expressed in terms of a 
/wedge/ product.  The answer is almost always "yes".

If you do that, you can set up a system that is not only
closed but forms a group.  In three dimensions, a basis
for the group will contain
  -- one scalar                 grade 0
  -- three vectors              grade 1
  -- three pseudovectors        grade 2
  -- one pseudoscalar           grade 3

for a total of 2^3 blades, which is the right answer. 

To this way of looking at it, a set consisting of grade=1 
vectors is /never/ closed under the wedge product operation, 
because the wedge product of two vectors is a grade=2
pseudovector.  On the other side of the same coin, the 
set of grade=1 vectors is never closed under the dot 
product operation either, because the dot product of two 
vectors is a grade=0 scalar.

If you include all 8 blades, the set is closed, and forms
a group.  Forsooth it's more than a group, it's an /algebra/ 
i.e. Clifford algebra.

===========

Very unlike the cross product, the wedge product generalizes
to any number of dimensions, not just 3 dimensions.  There
are tremendous applications in d=2 and d=4.  For a discussion
of why this is useful, see
  https://www.av8n.com/physics/clifford-intro.htm