Permutations of basis elements in Clifford Algebras

In summary: Clifford algebra CL(2,0) has a mapping f that permutes the coordinates of vectors in a way that preserves algebraic relations between generators.
  • #1
mnb96
715
5
Hello,
let's consider, for example, the Clifford algebra CL(2,0) and the following mapping f for an arbitrary multivector:

[tex]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[/tex]

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 [itex]f:\mathcal{C}\ell_{2,0} \rightarrow \mathcal{C}\ell_{2,0}[/itex] ?

Thanks.
 
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  • #2
What you have described is essentially the Hodge * operator - that is up to a sign equivalent to multiplication by the volume element [tex]e_{12}[/tex] in this case. Its general definition is

[tex]x\wedge \star y=(x,y)e[/tex]

where [tex]e=e_1\ldots e_n[/tex]

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.
 
  • #3
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 [itex]2^n[/itex] 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
[tex] a\mathbf{e_{12}}+b\mathbf{e_{23}}+c\mathbf{e_{32}} \longmapsto a + b\mathbf{e_{23}}+c\mathbf{e_{32}}
[/tex]

we would have that [itex]f(e_{12})=1[/itex], but [itex]f(e_{12}e_{12})\neq f(e_{12})f(e_{12})[/itex] so that mapping is not a morphism.
 
  • #4
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.
 
  • #5
Thanks!
now everything is more clear.
 

Related to Permutations of basis elements in Clifford Algebras

1. What is a Clifford algebra?

A Clifford algebra is a mathematical structure that generalizes the properties of vectors in three-dimensional Euclidean space. It is based on a set of basis elements and follows specific rules for multiplication and addition.

2. What are basis elements in a Clifford algebra?

Basis elements in a Clifford algebra are the building blocks of the algebra. They are typically represented by vectors or matrices and are used to create all other elements in the algebra through multiplication and addition.

3. What are permutations of basis elements?

Permutations of basis elements in a Clifford algebra refer to rearranging the order of the basis elements in a given product. This can result in different values for the product and is an important concept for understanding the algebra's properties.

4. How do permutations of basis elements affect the properties of a Clifford algebra?

The permutations of basis elements in a Clifford algebra can affect the properties of the algebra in various ways. For example, they can change the values of products, affect the commutativity and associativity of the algebra, and impact the representation of geometric transformations.

5. What are some practical applications of permutations of basis elements in Clifford algebras?

Permutations of basis elements in Clifford algebras have applications in many fields, including physics, engineering, and computer science. They can be used to study geometric transformations, solve problems in quantum mechanics, and develop efficient algorithms for data processing and analysis.

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