Scalar Triple Product of Two Polar Vectors and a Pseudovector So I have recently been trying to use methods of exterior algebra to solve quantum mechanics problems, and I still cannot seem to come to a good conclusion about the cross product of two vectors dotted with a pseudovector.
We know that the cross product and wedge product are related according to $$\vec{a} \times\vec{b}=\star(\vec{a}\wedge\vec{b}) $$
$$\vec{a}\wedge\vec{b}=I \vec{a}\times\vec{b} $$
where $I\equiv e_1\wedge e_2\wedge e_3$ is the unit pseudoscalar. Furthermore, we know that in general, a scalar triple product is a pseudoscalar (although I have yet to see a proof for why, my assumption is that somehow dotting the Hodge star of the wedge between a and b with some vector c lies within the $k$th (top) power of $\Lambda^k\mathbb{R}$).
The problem at hand is to classify whether or not the cross product of two momenta operators dotted with the spin operator
$$(\hat{\vec{p}}_1\times\hat{\vec{p}}_2)\cdot\hat{\vec{S}}$$
is a pseudoscalar or scalar. From an exterior algebra approach, we could perhaps express this as
$$\star(\hat{\vec{p}}_1\wedge\hat{\vec{p}}_2)\cdot \hat{\vec{S}}$$
but in this regard, I am uncertain of how to proceed as we cannot specify if spin is orthogonal to the cross product of the momenta here. I presume this is a pseudoscalar simply because it is a scalar triple product.
On the other hand, in quantum mechanics without representation theory, we are told that scalars and pseudovectors under parity have eigenvalue +1, whereas vectors and pseudoscalars have eigenvalue -1 under parity. That is, with the momentum operator
$$\hat{\Pi}^\dagger\hat{\vec{p}}\hat{\Pi}=-\hat{\vec{p}}$$
Applying this to the above problem, we would have
$$\hat{\Pi}^\dagger[(\hat{\vec{p}}_1\times\hat{\vec{p}}_2)\cdot\hat{\vec{S}}]\hat{\Pi}$$
$$=(-\hat{\vec{p}}_1\times-\hat{\vec{p}}_2)\cdot\hat{\vec{S}}\hat{\Pi}=[(\hat{\vec{p}}_1\times\hat{\vec{p}}_2)\cdot\hat{\vec{S}}]\hat{\Pi}$$
Here I am a bit confused what to do with the parity operator not acting on anything, but I assume if I can move it to the beginning of the expression, then it should produce the same result, as spin would not be flipped and since the two momenta would be negative and under a cross product, they would become positive.
So my main two questions are about how to proceed using the exterior algebra method, and if my approach with parity is correct. Thank you
 A: $
\newcommand\rev\widetilde
\newcommand\form[1]{\langle#1\rangle}
$You really want to be using geometric algebra here, which it looks like maybe you are based on your second equation? We can also write $\star(a\wedge b) = -I(a\wedge b)$ in your first equation.
Geometric algebra can be thought of as a product on the exterior algebra induced by a metric, called the geometric or Clifford product. We will use the convention that the geometric product binds weaker than any other product so that e.g.
$$
  \star(a\wedge b) = -I\,a\wedge b.
$$
This product is associative, and it's fundamental property is that $v^2 = v\cdot v$ for all vectors $v$. If $w$ is another vector (and $v^2 \not= 0$—not a problem in Euclidean space), then $-v^{-1}wv$ is exactly the reflection of $w$ through the plane orthogonal to $v$ (and $v^{-1} = v/v^2$). Since all orthogonal transformations are compositions of reflections, geometric algebra represents all orthogonal tranformations as products of vectors. In particular, you'll find that $w \mapsto (-1)^n\rev IwI$ is the parity transformation for $n$-dimensional Euclidean space where $I^{-1} = \rev I = \rev{e_1e_2e_3} = e_3e_2e_1$ is the reversal of $I = e_1e_2e_3 = e_1\wedge e_2\wedge e_3$. When $n = 3$ this looks likes $w \mapsto -e_3e_2e_1we_1e_2e_3$. You will find that this simplifies to $w \mapsto -w$ as it should.
Notice that orientation-preserving transformations are products of an even number of vectors, and orientation-reversing transformations are products of an odd number of vectors. Let us call these even and odd transformations, and let $\iota = 0$ if $T$ is even and $\iota = 1$ if $T$ is odd. In general, we apply $T$ to a vector $w$ via $w \mapsto (-1)^\iota T^{-1}wT$. However, we also want to be mindful of the grade of the object we're transforming: we want a bivector like $a\wedge b$ to transform like
$$
  \bigl[(-1)^\iota T^{-1}aT\bigr]\wedge\bigl[(-1)^\iota T^{-1}bT\bigr]
$$ because we want to transform the individual vectors in the plane spanned by $a$ and $b$. What this amounts to is that if $X$ has grade $k$ then it transforms as
$$
  X \mapsto (-1)^{\iota k}T^{-1}XT.
$$
In particular, the parity transformation of a bivector $X$ is
$$
  X \mapsto I^{-1}XI = X
$$
since $I$ always commutes with bivectors.
The transformations $X \mapsto T^{-1}XT$ are automorphisms of the geometric algebra since they are simply a conjugation action. This means that these transformations are homomorphic over any other product derived from the geometric product. For instance, if $X$ is a $k$-vector and $Y$ is an $l$-vector then
$$
  X\wedge Y = \form{XY}_{k+l}
$$
where $\form{\cdot}_{k+l}$ selects the $k+l$ grade component of a multivector. This means that
$$
  T^{-1}\,X\mkern1mu{\wedge}\mkern1muY\,T = (T^{-1}XT)\wedge(T^{-1}YT).
$$
Similarly, we can extend the metric onto any multivectors $X, Y$ by taking the scalar part:
$$
  X*Y = \form{XY}_0,
$$
and in particular $v\cdot w = \form{vw}_0$ for vectors $v, w$. (It can be argued that $\form{\rev XY}_0$ is more natural, but this is unimportant here.)

$a\times b$ doesn't tell you anything about how it transforms; it is a vector, so if we apply a parity transformation naively we get $-a\times b$. What we really want to transform is the associated bivector $a\wedge b$, so we need to account for the $I$ in $a\times b = -I(a\wedge b)$. For brevity, let us write conjugation by $I$ as $X' = I^{-1}XI$. Then the parity transformation of $(a\times b)\cdot S$ is
$$
  [(a\times b)\cdot S]' = [-I'(a\wedge b)']\cdot S'.
$$
We want to ignore $I'$, and $(a\wedge b)' = a\wedge b$ and $S' = -S$. All together we get a $-$, so $(a\times b)\cdot S$ is a pseudoscalar.
Alternatively, we can consider the fact that in geometric algebra
$$
  (a\times b)\cdot S = (-I\,a\wedge b)\cdot S = -I\,a\wedge b\wedge S.
$$
Since $I$ is invertible, we lose no information by representing this quantity as $a\wedge b\wedge S$. This not only transforms like a pseudoscalar but is a pseudoscalar.
