Pseudovectors in Geometric / Clifford Algebra

Question

I am studying geometric algebra and I am confused about why pseudovectors are written as single vectors with an $i$ in front. In other words, for basis vectors $\gamma_{\mu}$ where $\mu = 1,2,3,4$, there are 4 pseudovectors, written as $i \gamma_{\mu}$.

I know that in general, for a space of dimension $N$, pseudovectors are elements of the set of $(N-1)$ fold wedge products, namely $\wedge^{N-1}\mathcal{C}_N$ where $\mathcal{C}$ denotes a Clifford algebra from dimension $N$.

So from this, for a 4 dimensional space, the pseudovectors are trivectors, for example $(\gamma_1\wedge \gamma_2 \wedge \gamma_3)$. Where does the $i$ come in? I know that the pseudoscalar is written as an $N$ fold wedge product, so in this case would be a quad-vector, but I don't see how that can be worked in.

I can't seem to find any literature that explains this directly. I did find something that says there is a one-to-one mapping between $(N-1)$ fold wedge products, and single vectors, but it didn't go much into exactly how this mapping works.

Answer 1

The pseudoscalar for an N=4 dimensional algebra is $I=\gamma_1\gamma_2\gamma_3\gamma_4$, so if you multiply $I$ by any vector you get a grade 3 object (trivector). For example $I\gamma_4=\gamma_1\gamma_2\gamma_3\gamma_4\gamma_4=\gamma_1\gamma_2\gamma_3$ (assuming each basis vector $\gamma_\mu$ squares to +1). I think this is the type of calculation you were looking for.

Answer 2

I believe I have gathered enough information from these replies here, other sources, as well as from the replies here:

Hodge star operator

I believe this aggregation of information warrants me answering my own question here:

For a particular 'c-number' from a Clifford Algebra, it can be decomposed into a scalar part, a 1 form part, 2 form part, and so on until you get to an N-form part, which is the pseudoscalar $i$.

The Hodge Star operation is critical in this context it seems. In Clifford Algebra, the Hodge Star is captured by multiplying by $i$. This multiplication by $i$ maps a $k$ form into an $N-k$ form by the same kind of steps done here by foghorn (and the steps done in the link above). The important aspect here is that this mapping is one-to-one, meaning there is no ambiguity in transformations between these two regimes.

In 4 dimensions, a c-number has a scalar, vector, bivector, trivector, and pseudoscalar (quad-vector) parts. A trivector is an 3-form ($k=3$). Taking the Hodge dual of this (see foghorn above), yields a 1-form (vector) ($N-k = 4-3 = 1$). Taking the hodge dual again would again yield back the same 3-form you started with, for example:

$$i e^0 = e^0 e^0 e^1 e^2 e^3 = e^1 e^2 e^3$$

(assuming all basis vectors norm to +1). Because this mapping is one-to-one, one could instead represent the 3-forms as a vector multiplied by $i$ as shown above. This allows one to treat the object as if it were a vector, but with a sign change upon conjugation (which is why its called a pseudovector).

Answer 3

"I know that in general, for a space of dimension N, pseudovectors are elements of the set of (N−1) fold wedge products."

I think that this is wrong: it should be N fold wedge products. Thus in a plane pseudovectors are bivectors, wedge products of 2 vectors.

"Aren't bivectors, like torque, pseudovectors in 3D though?" No.