Question in pseudovectors

Question

I learnt that under parity transformation a vector $\vec{A}$ <---(Parity)------> $-\vec{A}$

and a pseudovector can be written as $\vec{c}=\vec{A} $ $\times$ $\vec{B}$ and since A goes negative A and B goes negative $ \vec{c}$ is invariant under parity transformation

But I can define vector A as an arbitrary cross product of $$ \vec{A}=\vec{z} \times \vec{s}$$ Then z and s changes values and A is a pseudovector , now This is inconsistent with our original statement Is there a better way to define pseudovectors ??

Answer 1

Why have you assumed that $\vec{c}$ and $\vec{B}$ also change sign under a parity transformation?

Note that the cross-product also changes sign under parity transformation! Thus, if you change the handedness of your coordinate system you get $\vec{c} = -(-\vec{A})\times\vec{B} = \vec{A} \times \vec{B}$. So actually, both $\vec{c}$ and $\vec{B}$ should be invariant under parity transformation.

Similarly, since the cross product changes sign under parity transformation, you automatically see that $\vec{A} = \vec{z} \times \vec{s}$ behaves as a pseudovector when both $\vec{z},\vec{s}$ are invariant under parity transform.

Answer 2

Short answer. A pseudo-vector can be written as the cross product of two (non-pseudo-) vectors and, conversely, the cross product of two pseudo-vectors produces a vector.

More generally, if $P_A$ and $P_B$ are the parities of $\vec{A}$ and $\vec{B}$ respectively, then the parity of $\vec{A} \times \vec{B}$ is given by $P_{A\times B} = -P_A \cdot P_B$.

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Long answer. In fact, what is usually called pseudovectors in physics corresponds to bivectors in mathematics, which are themselves the second-grade elements of the exterior algebra of $\mathbb{R}^3$, usually denoted by $\Lambda^2(\mathbb{R}^3)$ (see e.g. here, here or here).

Now, it can be shown that the dimension of the space of $k$-vectors $\Lambda^k(\mathbb{R}^n)$ is given by the binomial coefficient $\binom{n}{k}$. In the case of $\mathbb{R}^3$, i.e. with $n = 3$, its exterior algebra $\Lambda(\mathbb{R}^3) = \Lambda^0(\mathbb{R}^3) \oplus \Lambda^1(\mathbb{R}^3) \oplus \Lambda^2(\mathbb{R}^3) \oplus \Lambda^3(\mathbb{R}^3)$, where

1. $\Lambda^0(\mathbb{R}^3) = \mathbb{R}$ are the scalars,
2. $\Lambda^1(\mathbb{R}^3) = \mathbb{R}^3$ are the vectors,
3. $\Lambda^2(\mathbb{R}^3)$ are the bivectors and
4. $\Lambda^3(\mathbb{R}^3)$ are the trivectors.

Since $\binom{n}{k} = \binom{n}{n-k}$, there exists some isomorphisms such that $\Lambda^2(\mathbb{R}^3) \approx \Lambda^1(\mathbb{R}^3) = \mathbb{R}^3$ and $\Lambda^3(\mathbb{R}^3) \approx \Lambda^0(\mathbb{R}^3) = \mathbb{R}$; in other words, bivectors are associated to vectors and trivectors to scalars by isomorphism.

Actually, the bivectors are constructed as products of vectors, more precisely with the help of the exterior product, which is nothing else than the antisymmetric cross product; in the same way, the trivectors are composed of a triple exterior product. However, the properties of the exterior product induce a switch in parity, so that the bivectors stay the same under a parity transformation and the trivectors are reversed.

To conclude, bivectors are called pseudovectors and trivectors pseudoscalars, because they are identified respectively to vectors and scalars through isomorphism, but with a reversed parity due to the exterior product from which they are constructed.