Hello again :) I asked today a question whether for an arbitrary algebra $(A, \cdot)$ we can define exterior product $\wedge$ by
$$v_1\wedge \ldots \wedge v_k:=\frac{1}{k!}\sum_{\sigma\in S_k}\mathrm{sgn}(\sigma)v_{\sigma(1)}\cdot \ldots \cdot v_{\sigma(k)}$$
and wether it will be an/the exterior algebra.
Commentators pointed out problems such as:
(1). The characteristic of the field must be at least $k+1$ for it and actually $n+1$, since I want such products till $v_1\wedge \ldots \wedge v_n$.
(2) Dimension of such algebra will be wrong and it won't be isomorphic to the exterior algebra.
(3) There is also a question for what linear space $V$ it would be the exterior algebra $\Lambda(V)$ (if for any)?
And I even started to ask these questions myself after publishing the topic. But I asked it because they seem to do apparently something like that at Wikipedia in the article about Geometric algebra. They seem to define the wedge product of geometric algebra as $$a\wedge b:=\frac{1}{2}(ab-ba).$$ But this is likely possible because the geometric algebra/Clifford algebra $Cl(V,Q)$ already exists over some linear space $V$. Ok, but what about definition of $k$-th exterior power $V^{k\wedge}$ as a linear space generated by products $$v_1\wedge \ldots \wedge v_k:=\frac{1}{k!}\sum_{\sigma\in S_k}\mathrm{sgn}(\sigma)v_{\sigma(1)}\otimes \ldots \otimes v_{\sigma(k)}$$ and then defining $$\Lambda(V):=\bigoplus_{k=0}^\infty V^{\wedge k} $$ when my linear space $V$ is over $K=\mathbb R$ or $K=\mathbb C$? Is such definition correct in this particular case? I've seen they define antisimetric Fock space in quantum mechanics thus (but take closures in Hilbert spaces) at Wikipedia, unfortunately they don't get footnote, despite they give bibliography.