Let $V$ be a vector space over some field $\mathbb{K}$ and consider its $n$-fold tensor product $V\otimes \cdots \otimes V = \bigotimes^{n}V$. Call a tensor $v_{1}\otimes \cdots v_{n}$ skew-symmetric if exchanging the order of a pair of vectors exchanges its sign, that is:
$$v_{1}\otimes \cdots \otimes v_{i}\otimes \cdots \otimes v_{j}\otimes \cdots \otimes v_{n} = -v_{1}\otimes \cdots \otimes v_{j}\otimes \cdots \otimes v_{i}\otimes \cdots \otimes v_{n}$$
The space of skew-symmetric tensors is a subspace of $V\otimes \cdots \otimes V$, denoted by $S_{n}(V)$.
The usual construction of exterior powers of a vector space $V$ goes like this. The exterior power $\bigwedge^{n}V$ is proved to exist because it can be explicitly constructed by taking the quotient space $\bigwedge^{n}V = \bigotimes^{n}V/S_{n}(V)$.
Fact: There is an isomorphism $\bigwedge^{n}V \cong S_{n}(V)$.
Define a linear function $f: V\times \cdots \times V \to S_{n}(V)$ by setting:
$$f(v_{1},…,v_{n}) = \frac{1}{n!}\sum_{\sigma \in S_{n}}\operatorname{sign}(\sigma)v_{\sigma(1)}\otimes \cdots \otimes v_{\sigma(n)}$$
where the sum is over all permutations $\sigma$ of the set $\{1,…,n\}$.
Question: Let $v_{1}\wedge \cdots \wedge v_{n}$ be an arbitrary element of $\bigwedge^{n}V$. Does the above isomorphism imply that:
$$v_{1}\wedge \cdots \wedge v_{n} = f(v_{1},…,v_{n})$$
for every $v_{1},…,v_{n} \in V$?
Best Answer
First, some notation: the standard term for what you've called skew-symmetric tensors is alternating tensors, and the standard notation is $\text{Alt}^n(V)$.
Your question, as written, does not type check. The LHS $v_1 \wedge \dots \wedge v_n$ is an element of the $n^{th}$ exterior power $\Lambda^n(V)$ (your definition of the exterior power is incorrect, by the way; you should be quotienting by a suitable action of $S_n$, or by a much larger subspace than what you've written), while the RHS $f(v_1, \dots v_n)$ is an alternating tensor in $\text{Alt}^n(V)$. So it doesn't make sense to ask that they're equal.
What is true is that $f$ as you've defined it is an alternating multilinear map so, by the universal property of the exterior power, it canonically factors through $\Lambda^n(V)$. If we call $f' : \Lambda^n(V) \to \text{Alt}^n(V)$ the factored map then it is true that $f'(v_1 \wedge \dots \wedge v_n) = f(v_1, \dots v_n)$; maybe this is what you meant to ask?
The following general discussion may be helpful. Suppose $G$ is a finite group acting on a finite-dimensional vector space $W$ over a field of characteristic not dividing $|G|$. We can consider, on the one hand, the invariant subspace $W^G = \{ w \in W : \forall g \in G, gw = w \}$, and on the other hand the coinvariant subspace $W_G = W/(gw \sim w)$; these are respectively the left and right adjoints of the functor which equips a vector space with the trivial $G$-action. They have the same dimension, namely $\frac{1}{|G|} \sum_{g \in G} \text{tr}_W(g)$, so one might hope that they're even canonically isomorphic, and they are, as follows.
There is a canonical inclusion $W^G \to W$, and also a canonical quotient map $W \to W_G$. So there is a canonical composite map $i : W^G \to W_G$. This map is an isomorphism because it has an inverse $i^{-1} : W_G \to W^G$ which can be written down explicitly. Given $[w] \in W_G$, we want to write down a $G$-invariant vector which maps back down to it. This can be done by taking
$$[w] \mapsto \frac{1}{|G|} \sum_{g \in G} gw$$
where $w$ is any lift of $[w] \in W_G$ to $w \in W$. We can also sum over the orbit of $w$ and then divide by the size of the orbit.
As an application, taking $G = S_n$ and $W = V^{\otimes n}$ with the usual action by permuting coordinates, we conclude that, over a field of characteristic greater than $n$, there is a natural isomorphism between the space of symmetric tensors $\text{Sym}^n(V)$ (the invariant subspace of this action) and the symmetric power $S^n(V)$ (the coinvariant subspace). Tensoring this action with the sign representation of $S_n$, we conclude that there is a natural isomorphism between the space of alternating tensors $\text{Alt}^n(V)$ and the exterior power $\Lambda^n(V)$, whose inverse is exactly what you wrote.
(You only claim that there is an isomorphism but the naturality of the isomorphism is a much stronger statement; "isomorphism" only means they have the same dimension but in fact they are isomorphic as representations of $GL(V)$, and naturality is even stronger than this.)
(This need to divide by $|G|$ in the formulas suggests that we don't have this isomorphism without that assumption on the characteristic, which is true, at least for the symmetric case; I don't remember off the top of my head whether it's true for the alternating case. So in positive characteristic one really needs to be careful to make a distinction.)