Why is the pseudoscalar called pseudoscalar in Geometric Algebra

Question

It makes sense to call it a pseudoscalar in odd dimensions, because it commutes with all other objects. But in even dimensions it anticommutes, why is it still called pseudoscalar?

Further I don't get the relations with complex numbers and the notation $I$ (analog to the unit imaginary) for the pseudoscalar. In $\mathbb{G_{3}}$ the bivectors generate rotations in their planes so it makes sense to denote them by $i_{1}$,$i_{2}$,$i_{3}$ (also analog to the unit imaginary). But the pseudoscalar $I$ doesn't rotate vektors, it generates duality transformations, again why is it denoted by $I$? On the other hand, the pseudoscalar in $\mathbb{G_{2}}$ generates duality transformations (scalar-> pseudoscalar) AND rotations. So in $\mathbb{G_{2}}$ the bivectors have similar properties like the pseudoscalar and in $\mathbb{G_{3}}$ bivectors and the pseudoscalar have different properties. And all of these objects square to -1. Is this just a notational issue or what is going on?

Answer 1

I have trouble with physicists' terminology, so I will use the standard ones in the theory of quadratic forms. Let $q$ be a quadratic form on a vector space $V$ of dimension $n$ over a field $K$, and let $C$ be the Clifford algebra of $q$ (so $C$ has dimension $2^n$). We also note $C_0$ the even Clifford algebra, ie the subalgebra of $C$ generated by products of an even number of elements of $V$ (so the subalgebra generated by scalars and what you call bivectors).

Then there are two cases :

1. if $n$ is odd, then the center of $C$ (the elements that commute with everyone) is $Z(C)=K(I)$ following your notation and calling $I=v_1v_2\cdots v_n$ where $(v_i)$ is an orthogonal basis of $V$. So $Z(C)$ is a quadratic extension of $K$ (which is reminiscent of $\mathbb{C}$ and $\mathbb{R}$, hence the notation $I$, I guess). Actually, $I^2=\delta$ where $\delta$ is the (signed) discriminant of the quadratic form $q$, so $Z(C)=K(\sqrt{\delta})$. On the other hand, $C_0$ is a central simple algebra over $K$, so $Z(C_0)=K$ (indeed, $I$ is not in $C_0$).

2. if $n$ is even, then $Z(C)=K$ ; $C$ is a central simple algebra over $K$. On the other hand, $Z(C_0)=K(I)$ and $C_0$ is a central simple algebra over $K(I)$. So in this case $I$ commutes with all even elements of $C$ (and anti-commutes with homogeneous odd elements), and you still get $I^2 = \delta$.

So although in the even-dimensional case your $I$ does not commute with everyone, it still commutes with all the even elements, so I guess that's where your terminology of "pseudoscalar" comes from. Also, its square is a scalar in $K$, which is reminiscent of $i\in \mathbb{C}$ (and this square has a strong signification for the quadratic form $q$).

Answer 2

The multiples of $I$ form a one-dimensional vector space, just as multiples of $1$ do. Nevertheless, the two can be distinguished by how they transform under reflections: pseudoscalars change sign while scalars do not (by definition).

Hence, while the set of pseudoscalars has the same dimensionality as the set of scalars, they do not have the same transformation laws. They are not scalars. They're close, though. They're... pseudo-scalars.