Parity operator and tensor products

Question

When we consider usual vector spaces like $\mathbb{C}^n$, it is simple to consider tensor product operations, like $\mathbb{C}^n\otimes \mathbb{C}^m \cong \mathbb{C}^{n\cdot m}$, or $\mathbb{C}^n\wedge\mathbb{C}^n \cong \mathbb{C}^{{n \choose 2}}$. What happens when we have superspaces? That is, what is $ \mathbb{C}^{n|m}\otimes \mathbb{C}^{p|q} $ and $ \mathbb{C}^{n|m}\wedge \mathbb{C}^{n|m} $? where $ \mathbb{C}^{n|m} = \mathbb{C}^{n}\oplus\Pi\mathbb{C}^{m}$ where $\Pi$ is change of parity operator

Answer 1

Tensor product distributes over direct sum and parity is additive with respect to tensor product: so

$$\text{even} \otimes \text{even} = \text{even}, \text{even} \otimes \text{odd} = \text{odd}, \text{odd} \otimes \text{odd} = \text{even}.$$

This gives

$$\mathbb{C}^{n \mid m} \otimes \mathbb{C}^{p \mid q} \cong \mathbb{C}^{(np + mq) \mid mp + nq}.$$

For exterior powers the situation is more complicated, because it turns out that taking exterior powers of an odd vector space corresponds to taking symmetric powers. It's easier to describe what happens here in a coordinate-free way. If $V, W$ are (even) vector spaces and $\Pi W$ denotes change of parity (so that $\Pi W$ is odd) then taking exterior algebras converts direct sums into (graded) tensor products, which gives

$$\Lambda^{\bullet}(V \oplus \Pi W) \cong \Lambda^{\bullet}(V) \otimes \Lambda^{\bullet}(\Pi W)$$

where $\Lambda^{\bullet}(\Pi W)$ is actually isomorphic as an algebra to the symmetric algebra $S^{\bullet}(W)$, with the elements of $W$ having odd parity. Taking the degree $k$ part of both sides (with respect to the exterior product grading) gives

$$\Lambda^k(V \oplus \Pi W) \cong \bigoplus_{i+j=k} \Lambda^i(V) \otimes \Lambda^j(\Pi W)$$

where $\Lambda^j(\Pi W) \cong S^j(W)$ with the elements of $W$ having odd parity. So the even part of this is the sum over $j$ even and the odd part of this is the sum over $j$ odd.