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Jan 24, 2010, 1:47:52 PM1/24/10

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Dear Waldek and Martin,

I am currently totally absorbed with other things and will most likely

not have much

(any) time to look in detail into the code. What kind of thing was

that what you did not

understand about the paper (which paper by the way)? Perhaps I can at

least explain

what the different products and the Clifford algebras with arbitrary

bilinear forms.

[I have not invented Clifford algebras of an arbitrary bilinear form,

to my knowledge

Rafal Ablamowicz and Pertti Lounesto did this roughly at the same time

when I started

to use them, but their paper is prior to mine, I even used then their

notation of a 'dotted'

wedge product, see eg. Lounestos book Clifford algebras and spinors, LMS 286.]

There is also a relation between such 'quantum Clifford algebras' (as

they became called)

to star products in star-product deformation quantization. This was

investigated to some

length by Hirshfeld and Henselder (see arXiv) I have not fully

understood their work, they

make some claims which I find odd, but had no time to see if that is

just a matter of

different notation.

I think for geometric reasoning one needs only symmetric bilinear

forms, perhaps unless

one starts to think about Grassmannians and Borel subgroups of the

(projective) GL(n)

(Clifford algebras over V+V* where V* is the dual space and the

bilinear form is given

by the (mutual) evaluation maps). Heinrich Saller called these

algebras 'quantum algebras',

however they occur frequently in the representation theory of rational

(meromorphic)

representations of the GL(n) groups (co and contravariant representations).

Cheers

BF.

--

% PD Dr Bertfried Fauser

% Research Fellow, School of Computer Science, Univ. of Birmingham

% Honorary Associate, University of Tasmania

% Privat Docent: University of Konstanz, Physics Dept

<http://www.uni-konstanz.de>

% contact |-> URL : http://www.cs.bham.ac.uk/~fauserb/

% Phone : +44-121-41-42795

Jan 26, 2010, 3:19:59 PM1/26/10

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Bertfried Fauser wrote:

> I am currently totally absorbed with other things and will most likely

> not have much

> (any) time to look in detail into the code. What kind of thing was

> that what you did not

> understand about the paper (which paper by the way)? Perhaps I can at

> least explain

> what the different products and the Clifford algebras with arbitrary

> bilinear forms.

> I am currently totally absorbed with other things and will most likely

> not have much

> (any) time to look in detail into the code. What kind of thing was

> that what you did not

> understand about the paper (which paper by the way)? Perhaps I can at

> least explain

> what the different products and the Clifford algebras with arbitrary

> bilinear forms.

I mean the paper:

Vertex Normalordering as a Consequence of Nonsymmetric Bilinearforms

in Clifford Algebras

http://arxiv.org/pdf/hep-th/9504055

I the abstract you wrote:

We consider Clifford algebras with nonsymmetric bilinear forms, which are

isomorphic to the standard symmetric ones, but not equal.

Observing, that the content of physical theories is dependent

on the injection \oplus^n /\V^(n) ->CL(V, Q) one has to transform

to the standard construction. The injection

is of course dependent on the antisymmetric part of the bilinear form.

You use the word "injection" but AFAICS the Clifford you get from

non-symmntric form is canonically isomorphic to the the Clifford

algebra you get from symmetric part of the form and in particular

injection of V into CL(V, Q) should agree. More precisly for

me "classical" Clifford algebra is an algebra with mapping

\phi : V -> CL(V, Q) such that:

1) \phi(v)^2 = Q(v)I

2) which is universal with respect to property above, that is

for any mapping \phi' from V into an algebra A which

satisfies \phi'^2 = Q(v)I there is unique mapping from CL(V, Q)

to A such that the following diagram commute:

CL(V, Q)

^ |

/ |

/ |

/ |

/ v

V -------> A

Given the word "isomorphic" I think that your Clifford algebra

is associative algebra which has the same dimension as symmetic

one. My reading of your formulas is that \phi(v)^2 = Q(v)I

is satisfied (note that Q(v) = <Bv, v> depend only on symmetric

part of B) so universal property 2) implies canonical

isomoprphism.

In other words, only extra structure (like dotted wedge and other

products) depend on antysymmetric part. Is this correct?

Next, in 6_11 to 6_14 you wrote:

Clearly, if we had chosen J to be the common use involution on V

...

Thus, if there exists a distinct involution of period two, we

have the desired extension.

But I did not found if such J always exist and if it exist

there is question of uniqueness. In particular, do the

extra structure depend only on the bilinear form or there

is extra degree of freedom due to choice of J?

Also, formulas (13), (14) and (16) seem to give simple definition

of inner (_|) product, is there equally simple definition of

dotted wedge? (of course, those formula are good definition

only if _| always exists).

> [I have not invented Clifford algebras of an arbitrary bilinear form,

> to my knowledge

> Rafal Ablamowicz and Pertti Lounesto did this roughly at the same time

> when I started

> to use them, but their paper is prior to mine, I even used then their

> notation of a 'dotted'

> wedge product, see eg. Lounestos book Clifford algebras and spinors, LMS 286.]

>

Thanks for this info, on the net your paper showed up and the others

not...

--

Waldek Hebisch

heb...@math.uni.wroc.pl

Feb 1, 2010, 9:42:55 AM2/1/10

to fricas...@googlegroups.com

Dear Waldek,

> Vertex Normalordering as a Consequence of Nonsymmetric Bilinearforms

> in Clifford Algebras

>

> http://arxiv.org/pdf/hep-th/9504055

This is a bad source, it was one of my first papers on that subject

and there are by now much

better descriptions, eg:

arXiv:math-ph/0212032 and arXiv:math-ph/0212031

and more mathematical ones in:

arXiv:math-ph/0208018 and arXiv:math/9911180

showing how the antisymmetric part in the bilinear form changes lots

of things, and providing some

commutative diagrams showing the relations between the structures....

> Given the word "isomorphic" ....

The usage of isomorphism is problematic. It depends _very_ much on

what you consider to be conserved under the map. I can shortly

summarise this as follows:

* given a quadratic space (V,Q), one can construct the universal

Clifford algebra CL(V,Q)

as you did using the commutative diagram. These algebras come with

different _additional_

structures:

-- a Z_2 grading (even and odd parts of the algebra. The even part

is a sub-Clifford-algebra,

the odd part is a module over the even part only)

-- a multivector structure or Z-grading

since Clifford algebras are Z_2 graded algebras, isomorphisms need

_not_ preserve the Z-grading.

The multivector structure (Z-grading) in a CLifford algebra is

inherited from the underlying

Grassmann algebra (module), which in turn comes from the grading of

the tenor algebra.

So concerningthe modules a different notion of isomorphism may apply.

* The Clifford relation v*v=Q(v) shows via polarisation, that the

Clifford algebra depends only on a

symmetric bilinear form 2*G(u,v) = Q(u+v)-Q(u)-Q(v), this is

reflected in the (anti)commutator

relations:

uv+vu = 2G(u,v) for all u,v in V

the symmetrisation prevents G to be non-symmetric.

* Looking at Clifford algebras as (sub) algebras of endormophisms of

the Grassmann algebra

induces a multivector structure. However, the contraction maps _|

need an identification of the

space V and its dual space V* which is not universal (but may be

chosen canonically as is usually

done using the dual basis). This extends to the Grassmann algebras

/\V and /\V* = ~(/\V)*.

The isomorphism is usually taken to be the canonical one, but that

has not to be the madatory

choice another choice can use the nondegenerate bilinrear form b: V -> V*, via

B(u,v) = <b(u), v>, where < , > is the evaluation map : V* x V --> R

* Given an arbitrary bilinear form B it can always be decomposed into

a symmetric part G and

an antisymmetric part F, B=G+F. The Clifford algebras CL(V,B), due

to its construction will

only see the symmetric part G, however the contraction maps will

depend on B and not only

on G. Hence one arrives at the result:

CL(V,B) ~= CL(V,G) as Z_2 graded algebras

However the multivector structures of CL(V,B) and CL(V,G) will be

different if F=\=0 and the

isomorphism is Z-2 graded and not in general Z-graded, but only

Z-filtered. One has:

bilinear forms

quadratic forms ~= symmetric foms = -----------------------------

alternatig forms

* The map J was introduced by Marcel Riesz, in characteristic 0 he

showed that such a map always

exists. C. Chevalley introduced the contraction map to be able to

construct Clifford algebras

in characteristic 2, where the above polarization trick is not

available. I never looked deep into

the characteristic problem.

* I hit on Clifford algebras with arbitrary bilinear form in Quantum

Field Theory calculations. There

an ordering issue becomes apparent, say time- and normal-odered

fields. One can exploit the

morphism /\V -> /\_FV (with /\_F the 'dotted wedge) to reformulate

the Wick theorem reordering

for operators, the iso is encoded in the multivector structure, see

: arXiv:hep-th/0007032.

From a cohomological point of view this can be seen as a trivial

deformation of the Grassmann

algebra.

Chevalley has the theorem: CL(V,Q) ~= CL(V,-Q)^op

This can be generalized to: CL(V,B) ~= CL(V, -B^t)^op where B^t is

the transposition.

In term of Martin's Clifford package one can conclude:

-- Clifford elements are given in a Grassmann basis

-- There is a second Grassmann basis (dotted wedge) which allows to

'absorb' any antisymmetric

part F.

-- W.r.t. a Grassmann basis the Clifford products cmul_G and cmul_B

are different

-- The geometric content of a Clifford algebra is encoded in its

multivector structure

(and that is what Martin is interested in). Further more, for

geometric applications you will

usually not need any recourse to a general bilinear form, an

arbitrary symmetric bilinear form

will do. A basis of orthonormal vectors of G~=Q is too narrow for

geometric applications and

this assumption was made in the previous Clifford package (by Watt (et al?))

-- One can introduce a Clifford basis (monomials are given by the

Clifford product of elements in V)

such elements are not geometric, but represent much more

'operators'. The Grassmann basis

and Clifford basis are related by an isomorphism.

Does this help?

Cheers

BF.

PS: In my eyes many things get much clearer if Hopf algebras are

considered. At the moment I do not see how to use Franz' and your

tensor package to do this efficiently.

Feb 1, 2010, 10:44:44 AM2/1/10

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On Monday 01 Feb 2010 14:42:55 Bertfried Fauser wrote:

> In term of Martin's Clifford package one can conclude:

> -- Clifford elements are given in a Grassmann basis

> -- There is a second Grassmann basis (dotted wedge) which allows to

> 'absorb' any antisymmetric

> part F.

> -- W.r.t. a Grassmann basis the Clifford products cmul_G and cmul_B

> are different

> -- The geometric content of a Clifford algebra is encoded in its

> multivector structure

> (and that is what Martin is interested in). Further more, for

> geometric applications you will

> usually not need any recourse to a general bilinear form, an

> arbitrary symmetric bilinear form

> will do.

> In term of Martin's Clifford package one can conclude:

> -- Clifford elements are given in a Grassmann basis

> -- There is a second Grassmann basis (dotted wedge) which allows to

> 'absorb' any antisymmetric

> part F.

> -- W.r.t. a Grassmann basis the Clifford products cmul_G and cmul_B

> are different

> -- The geometric content of a Clifford algebra is encoded in its

> multivector structure

> (and that is what Martin is interested in). Further more, for

> geometric applications you will

> usually not need any recourse to a general bilinear form, an

> arbitrary symmetric bilinear form

> will do.

Having said that, why not keep FriCAS as general as possible unless there is a

need to constrain it?

The use of the QuadraticForm domain did not add any functionality to the

CliffordAlgebra domain apart from enforcing the bilinear form to be symmetric.

So is there reason why we would want to re-introduce it? its just more

complexity and more overhead.

Martin Baker

Feb 1, 2010, 10:51:10 AM2/1/10

to fricas...@googlegroups.com

Hi Martin,

> The use of the QuadraticForm domain did not add any functionality to the

> CliffordAlgebra domain apart from enforcing the bilinear form to be symmetric.

> So is there reason why we would want to re-introduce it? its just more

> complexity and more overhead.

If you understood my mail in this direction, that was not my

intention. I need such Clifford

algebras with arbitrary bilinear forms anyhow and am very pleased to

have them. The effort to

make a Clifford product for ageneral symmetric bilinear form and an

arbitrary one are the same

form a computational point of view. Faster code could only be made for

the special case of an orthogonal basis, which sometimes helps.

I will see if I can test the new code, but I cannot promise anything

before March, sorry about

that, I had not even time to do anything for teh Schur functions. I

am, by the way, very pleased

to see what you have already achieved and that the new package is

going to be shiped with

FriCAS.

Cheers

BF.

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