From Wikipedia:
A multilinear map of the form $f\colon V^n \to W$ is said to be alternating if it satisfies any of the following equivalent conditions:
- whenever there exists $1 \leq i \leq n-1$ such that $x_i = x_{i+1}$ then $f(x_1,\ldots,x_n) = 0$.
- whenever there exists $1 \leq i \neq j \leq n$ such that $x_i = x_j$ then $f(x_1,\ldots,x_n) = 0$.
- if $x_1,\ldots,x_n$ are linearly dependent then $f(x_1,\ldots,x_n) = 0$.
I want to extend this definition to the general case and prove equivalence between the three conditions above. $f\colon V^X \to W$, where $X$ is a possibly infinite set. I can prove equivalence of the three definitions in the finite case:
- (1) ⇒ (2): The property is tautological in the base case $ j = i + 1$. For the induction case, use the lemma that if adjacent elements are swapped $(\forall i\in X,i \notin \{m,n\}\colon b_i = a_i) \cap b_m = a_n \cap b_n = a_m$ then $c_x = a_x + b_x \Rightarrow c_m = c_n \Rightarrow f(c) = 0 \Rightarrow f(a) + f(b) = 0 \Rightarrow f(a) = -f(b)$ the function becomes negative.
- (2) ⇒ (3): Let there be a linear combination $v \in X^n \cap s \in F^n \cap n \in \mathbb N$ so that $\sum_{k=1}^n s_k a_{v_k} = 0$. Replace the value at one of the indices in the linear combination with the full combination and expand using linearity.
In the infinite case I don't know how to prove it. I think $X$ must be well-ordered and maybe transfinite induction is helpful here but I don't know how.