Complex Vectors in Geometric Algebra

[mnb96]

Hello,
I have recently started to study some Geometric Algebra.
I was wondering how should I interpret complex-vectors in [tex]\mathcal{C}^n[/tex] in the framework of Geometric Algebra.
I understand already that a complex-scalar should be interpreted as an entity of the kind:

[tex]z = x + y (\textbf{e}_1 \wedge \textbf{e}_2)[/tex]

where the imaginary-unit is instead the unit bi-vector in [tex]\mathcal{R}^2[/tex]. Now for a real-vector in [tex]\mathcal{R}^n[/tex] one would obviously have:

[tex]\textbf{x} = x_1 \textbf{e}_1 + \ldots + x_n \textbf{e}_n[/tex]

But what would be the equivalent if x were instead complex?

 

[Peeter]

hout4 in the following touches on higher degree complex vector representations

http://www.mrao.cam.ac.uk/~clifford/ptIIIcourse/GeometricAlgebraCourseMateriasl.zip

but you'll probably also be able to find a lot of info under the heading 'complex clifford algebras'

 

[mnb96]

Hi Peeter!
It's always nice to see your replies as soon as GA pops up :)

As far as I understood the trick to handle complex vectors is quite straightforward. They simply consider N extra orthogonal basis versors [tex]\mathbf{f}_i[/tex]. The imaginary-unit for each component of the vector would then be the bi-vector [tex](\mathbf{e}_{i} \wedge \mathbf{f}_{i})[/tex].

However, when I was trying to figure out this by myself I came up to something similar, but I don't know if it's correct. My solution uses only 2 extra dimensions (instead of N). I'll explain how:

[tex]\mathbf{z} = (a_1 + ib_1)\mathbf{e}_{1}+\ldots+ (a_n + ib_n)\mathbf{e}_{n}[/tex]

[tex]\mathbf{z} = a_1\mathbf{e}_{1}+\ldots+a_n\mathbf{e}_{n} \\ + \\ i(b_1\mathbf{e}_{1}+\ldots+b_n\mathbf{e}_{n})[/tex]

At this point I set [tex]i=\mathbf{e}_{n+1}\mathbf{e}_{n+2}[/tex].
Obviously, [tex]i^2=(\mathbf{e}_{n+1}\mathbf{e}_{n+2})^2=-1[/tex]
So, each imaginary component becomes [tex]b_i(\mathbf{e}_{n+1}\mathbf{e}_{n+2}\mathbf{e}_{i}) = b_i(\mathbf{e}_{n+1} \wedge \mathbf{e}_{n+2} \wedge \mathbf{e}_{i}) [/tex]

The difference is that before the bases for complex elements were simply 2-blades, but now they are unit 3-blades.
Can all this work too?

 

[Peeter]

How would you define the inner product with this representation?

 

[mnb96]

That's a good question, in fact, I don't know. But I found a paper which apparently follows the same method I was using and develops also the Hermitian Inner Product.
You can find it here:
http://www.springerlink.com/content/u304122440q01714/
If you don't have access to it, just tell me.


By the way, there is still something that I don't get about these representations of complex-vectors in GA:

We all agree that if u have two vectors (or even two blades), it is possible to give a perfect geometrical meaning to the scalar-product (or contraction) which gives information about the angle of incidence between the two subspaces (vectors, planes, volumes...)

However, when we take the inner-product between two complex-vectors we get the cosine between what? actually it is even a complex scalar! ... is it possible to assign a geometrical interpretation to inner-products of complex vectors in terms of angles between blades?

 

[Peeter]

Do you have a geometric interpretation for the inner product of two plain old complex numbers? I'm not sure what I'd call that entity in general (clearer is the case where the complex numbers are of unit length), and would want to figure that out before moving on to try to give a geometric meaning to the complex vector inner product.

Perhaps dropping a formal attempt at geometrical interpretation is best. When you calculate a Fourier series (which is inherently an inner product based decomposition) do you try to call the Fourier components geometric entities? Calling them projections is about as far as a geometric analogy takes you.

re, the paper. I've been out of school for 10 years, and have no access to private journals.