Permutations of basis elements in Clifford Algebras

[mnb96]

Hello,
let's consider, for example, the Clifford algebra CL(2,0) and the following mapping f for an arbitrary multivector:

$$a + b\mathbf{e_1}+c\mathbf{e_2}+d\mathbf{e_{12}} \longmapsto a\mathbf{e_{12}} + b\mathbf{e_1}+c\mathbf{e_2}+d$$

For vector spaces R^n we can permute the coordinates of vectors by a linear (and orthogonal) transformation defined as a permutation matrix.
Is it possible to do something similar for multivectors? or should we just say that we are applying a mapping $f:\mathcal{C}\ell_{2,0} \rightarrow \mathcal{C}\ell_{2,0}$ ?

Thanks.

 

[arkajad]

What you have described is essentially the Hodge * operator - that is up to a sign equivalent to multiplication by the volume element $$e_{12}$$ in this case. Its general definition is

$$x\wedge \star y=(x,y)e$$

where $$e=e_1\ldots e_n$$

and x,y are arbitrary elements of the algebra.

Remark: for this you need to extend the scalar product to the whole algebra, but that is canonical.

 

[mnb96]

That is true!
In the particular case of CL(2,0) the Hodge star operator works as expected!

So, if I understood correctly there is no "general" operator that permutes any of the $2^n$ basis into another. I guess we can do it by simply forcing to define a mapping, but then we would probably need to prove that such mapping is a morphism (in order to be useful).

For example if we consider CL(3,0) and the following mapping
$$a\mathbf{e_{12}}+b\mathbf{e_{23}}+c\mathbf{e_{32}} \longmapsto a + b\mathbf{e_{23}}+c\mathbf{e_{32}}$$

we would have that $f(e_{12})=1$, but $f(e_{12}e_{12})\neq f(e_{12})f(e_{12})$ so that mapping is not a morphism.

 

[arkajad]

If you permute the basis in such a way that the algebraic relations between generators are preserved - you get an algebra morphism, otherwise you get just a linear morphism. If you are lucky - your linear morphism may have some additional properties.

 

[mnb96]

Thanks!
now everything is more clear.