Alternating tensors vs $p$-vectors

Question

Is there a reason to differentiate between alternating tensors and $p$-vectors? More precisely, is the exterior algebra always isomorphic to the subalgebra of alternating tensors?

Thanks.

Answer 1

This is a special case of a more general phenomenon. Let $G$ be a finite group acting on a vector space $W$ over a characteristic $0$ field. One can define two spaces. The invariants $W^G=\{w\in W\,|\, gw=w\text{ for all }g\in G\}$, and the coinvariants $W_G=W/\langle g\cdot w-w\rangle$. Here $\langle g\cdot w-w\rangle$ is the subspace spanned by vectors of the form $gw-w$ for $g\in G$ and $w\in W$. If $G$ is the symmetric group acting on $W=V^{\otimes n}$ via $\sigma\cdot v_1\otimes\cdots\otimes v_n=\mathrm{sgn}(\sigma) v_{\sigma^{-1}(1)}\otimes\cdots\otimes v_{\sigma^{-1}(n)}$, then $W^G$ and $W_G$ are both isomorphic to the exterior power $\Lambda^n V$. The general phenomenon that I spoke about is the fact that $W_G\cong W^G$, which gives your question as a special case. To prove this we define a map $W_G\to W^G$ by sending $[w]\mapsto\frac{1}{|G|} \sum_{g\in G} g\cdot w$. This is well-defined because it sends $w-g\cdot w\mapsto 0$. Similarly define a map in the opposite direction $W^G\to W_G$ via $w\mapsto [w]$. Then the composition $W_G\to W^G\to W_G$ is the identity $[w]\mapsto [w]$. Similarly the composition $W^G\to W_G\to W^G$ is the identity.

In order to get the is argument to work, I needed to divide by $|G|$, which is why I am assuming characteristic $0$. As a counterexample in the characteristic $2$ case note that $[v\otimes v]\neq 0\in [V^{\otimes 2}]_{\mathbb Z_2}$, whereas the corresponding element of $[V^{\otimes 2}]^{\mathbb Z_2}$ should be $v\otimes v+v\otimes v=2v\otimes v=0$. Here, $\mathbb Z_2$ denotes the field with $2$ elements.