# Clifford code

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### Bertfried Fauser

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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

### Waldek Hebisch

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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 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

### Bertfried Fauser

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

### Martin Baker

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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.

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

### Bertfried Fauser

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