RE: [Axiom-developer] RE: learning in public

[Page, Bill]

Bertfried on Wednesday, June 02, 2004 12:55 PM you wrote:
> 
> On Tue, 1 Jun 2004, Page, Bill wrote:
> 
> > Bertfried and Tim,
> 
> > Formal definition
> >
> > Let V be a vector space over a field k, and q : V -> k
> > a quadratic form on V. The Clifford Algebra C(q) is a
> > unital associative algebra over k together with a linear
> > map i : V -> C(q) defined by the following universal
> > property:
> >
> > for every associative algebra A over k with a linear map
> > j : V -> A such that for every v in V we have
> > j(v)^2 = q(v)1 (where 1 denotes the multiplicative
> > identity of A), there is a unique algebra homomorphism
> >
> >     f : C(q) -> A
> >
> > such that the following diagram commutes:
> >
> >                V ----> C(q)
> >                |     /
> >                |    / Exists and is unique
> >                |   /
> >                v  v
> >                A
> >
> > i.e. such that fi = j.
> 
> Of course, this is standard! But....

It is standard and it is categorical but even in the
case of fields and vector spaces the constructive
definition (in terms of a particular quotient over the
the tensor algebra) is not at all "obvious" to me.

> 
> 1) Usually you will need Clifford algebras over rings,
>  not fields, like the ring of smooth functions -> tempered
>  distributions
> 
> 2) The form should be a bilinear form, not only a quadratic
> form. See
> 
>    symmetric forms =  bilinear forms mod alternating forms
> 
> iff char of the base field/ring is not 2. The above 
> construction is then no longer so obvious.

I do not understand why you say these generalizations
make the constructive definition any less obvious. It
seems to me that we can have tensor algebra over a
module

  http://planetmath.org/encyclopedia/TensorAlgebra.html

and the analogous construction at

  http://planetmath.org/encyclopedia/CliffordAlgebra2.html

"Let M be a (left-)module over a ring R, and B:VxV->k a
symmetric bilinear form. Then the Clifford algebra Cliff(M,B)
is the quotient of the tensor algebra T(M) by the relations
  v (x) w + w (x) v = -2 B(v,w)"

where (x) denotes tensor product, makes good sense to me.
I don't see any problem if B(v,w) is degenerate. The
Z_2-grading on the tensor algebra is still inherited in
the same way.

> 
> 3) ANY base introduces a filtration. Physics is _sensitive_ to
> this filtration, even if the algebras are _isomorphic_ and
> hence mathematically indistinguishable. This stems from additional
> features employed in physics. Hence it is _tremendously dangerous_
> to introduce bases. However, certain physical applications
> _need_ such a reference base, alas I am confused.

I do not think one needs to make assumptions about a basis
in order to prove that the Clifford algebra is filtered

  http://planetmath.org/encyclopedia/FilteredAlgebra.html

and (most of?) the construction of the exterior algebra

  http://planetmath.org/encyclopedia/ExteriorAlgebra.html

goes through.

> 
> 4) From a technical point of view, the most easy construction of a
> Clifford algebra (and that one which is capable of arbitrary even
> degenerated bilinear forms) is that of Chevalley, which unfortunately
> assumes a grading, which is _not_ inherent in a Clifford structure.
> 
> Let x,y in V, u,v,w in V^, the exterior (Grassmann) algebra 
> over V (unique up to isomorphism) define the Clifford product
> recursively as:
> 
> i) x _| y = B(x,y) = y |_ x    (bilinear for acting on VxV )
> 
> ii) x _| (u /\ v) = (x _| u) /\ v + ßhat{u} /\ (x _| v)   (derivation)
> 
> iii) u _| ( v _| w) = (u /\ v) _| w  (left action on the module V^)
> 
> definition:  the Clifford multiplication wrt the bilinear 
> form B (used in the contraction _|) is defined as
> 
>   x ° u = x _| u + x /\ u
> 
> a general element v can be multiplied by recursive use of i) 
> ii) and iii)
> 
> This is not the most efficient algorithm, but a plain approach.

Chevalley's recursive construction is interesting since it
simultaneously builds the graded structure and the quotient.

> A much better approach uses Hopf algebras and defines the
> Clifford product as a twist by a (Laplace) 2-cocycle on the
> Grassmann Hopf algebra.

It is not immediately clear to me how this approach is
reflected in your definition below. What part is the Hopf
algebra and what part is the twist?

> 
> Hence the way to go is:
> 
> Define a category "graded module"
> Define the categories symmetric and exterior algebras over 
> such a module (this amounts to introduce super symmetric
> multilinear algebra)

I think that this is equivalent to the tensor algebra
defined above, isn't it?

> Define the category of reflexive spaces with inner product.

I presume you mean something more general than the
inner product?

> Define the Clifford algebra as a Functor which assigns to 
> every reflexive module a Clifford algebra.

Aw, now that's the hard part isn't it? But I think it is
the same problem as in the less general case. We need
(somehow) to efficiently construct the appropriate quotient
of the tensor algebra.

> 
> Much better is to make this that way that an iteration can
> be done.
>

I think I *might* know how to do this an incremental fashion
in terms of a map (i.e. a "remember table") that defines the
partition

  T(M) ->> Cliff(M,B)

given by the bilinear relations.
 
> It is mandatory that such modules may be direct sum of graded 
> modules and that one can somehow manage the grading information,
> eg we should think of modules composes from symmetric and
> exterior parts
> 
>  M = Sym(V) \oplus Alt(W)
> 
> this is needed for more advanced Cliffordizations.

Isn't this logically part of in the tensor algebra and the
Z_2-grading?

> 
> hope this helps
> ciao
> BF.
> 

Yes, I think it does. Does what I say above make sense
to you?

Regards,
Bill Page.