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Let $\Delta$ be a skew-symmetric $n$-linear map. I have the following in my notes and I am having trouble seeing how it follows: $$ \Delta\left(\sum_{i=1}^n{e_i}, \sum_{i=1}^n{(e_i)} -e_2, \sum_{i=1}^n{(e_i)} -e_3, \dots, \sum_{i=1}^n{(e_i)} -e_n\right) = \Delta\left(\sum_{i=1}^n{e_i}, -e_2, -e_3, \dots, -e_n\right). $$

I am having trouble seeing how the $\displaystyle\sum_{i=1}^n{e_i}$ terms vanish.

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The fact that it is a sum of vectors is not relevant. Let $v = \sum_{i=1}^ne_i$, then the left hand side becomes $\Delta(v, v - e_2, v - e_3, \dots, v - e_n)$. By linearity in the second argument, we have

$$\Delta(v, v - e_2, v - e_3, \dots, v - e_n) = \Delta(v, v, v - e_3, \dots, v - e_n) + \Delta(v, - e_2, v - e_3, \dots, v - e_n)$$

As $\Delta$ is skew-symmetric, if two of its arguments are equal, then its value is zero (assuming $\Delta$ does not take values in a ring of characteristic two). Therefore, the first term on the right hand side is zero, so

$$\Delta(v, v - e_2, v - e_3, \dots, v - e_n) = \Delta(v, - e_2, v - e_3, \dots, v - e_n).$$

Repeating the same argument, but for the third argument, we find that

$$\Delta(v, - e_2, v - e_3, \dots, v - e_n) = \Delta(v, -e_2, v, \dots, v - e_n) + \Delta(v, - e_2, - e_3, \dots, v - e_n)$$

and again, the first term on the right hand side is zero by the skew-symmetry of $\Delta$. Repeating for each argument of $\Delta$, we see that

$$\Delta(v, v - e_2, v - e_3, \dots, v - e_n) = \Delta(v, -e_2, -e_3, \dots, -e_n).$$


Here's another way to see it. By multilinearity, we can expand $\Delta(v, v - e_2, v - e_3, \dots, v - e_n)$ as a sum of $2^{n-1}$ terms which are precisely all the possible terms of the form $\Delta(v, a_2, \dots, a_n)$ where $a_i \in \{v, -e_i\}$. Note that if a term has two $v$'s (i.e. there is some $i$ such that $v = a_i$), then that term is zero by skew-symmetry. So the only terms which aren't automatically zero are the ones with $a_i \neq v$ for every $i$, that is, $a_i = - e_i$. There is only one such term, namely $\Delta(v, -e_2, -e_3, \dots, -e_n)$ so as before, we see that

$$\Delta(v, v - e_2, v - e_3, \dots, v - e_n) = \Delta(v, -e_2, -e_3, \dots, -e_n).$$

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