# Pseudovectors in Geometric / Clifford Algebra

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.

• Just to get clarity on your question, are you asking what $i$ (perhaps more usually $I$) means in this context? Are you asking for a derivation that $I$ times a vector is a pseudovector? Or both? (Or neither?) Mar 5, 2021 at 16:42
• More towards the latter, but both would be good! Mar 5, 2021 at 16:47
• $I$ is generally the unit pseudoscalar; and multiplication by $I$ is the dual (Hodge star) of an element of a geometric algebra. This duality is the reason behind the term "pseudo", and the link between the $k$ and $n-k$ grade objects. Mar 5, 2021 at 19:19

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.

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).

"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.

• Aren't bivectors, like torque, pseudovectors in 3D though ? Mar 5, 2021 at 19:09
• Wouldn’t an N fold wedge be the pseudo scalar? In 3D, an oriented plane is a two-fold wedge product, namely N-1. Also can you explain more about why something like torque isn’t a pseudo vector? This directly goes against many articles that I have read recently. Mar 7, 2021 at 18:09