Chronology Current Month Current Thread Current Date
[Year List] [Month List (current year)] [Date Index] [Thread Index] [Thread Prev] [Thread Next] [Date Prev] [Date Next]

Re: [Phys-L] closed vectors

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.


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