On the transformation for wedge product of covectors

Question

I found this question on the site and the first answer which states that

$$\begin{align*} \beta^1\wedge\cdots\wedge\beta^k&=\left(\sum_{j=1}^ka_j^1\gamma^j\right)\wedge\cdots\wedge\left(\sum_{j=1}^ka_j^k\gamma^j\right)\\ &=\sum_{\sigma\in S_k}a_{\sigma(1)}^1\gamma^{\sigma(1)}\wedge\cdots\wedge a^k_{\sigma(k)}\gamma^{\sigma(k)}\\ &=\sum_{\sigma\in S_k}\left(\prod_{i=1}^ka^i_{\sigma(i)}\right)\gamma^{\sigma(1)}\wedge\cdots\wedge \gamma^{\sigma(k)}\\ &=\sum_{\sigma\in S_k}\text{sgn}\sigma\left(\prod_{i=1}^ka^i_{\sigma(i)}\right)\gamma^{1}\wedge\cdots\wedge \gamma^{k}\\ &=(\det A)\gamma^{1}\wedge\cdots\wedge \gamma^{k} \end{align*}$$

and I'm wondering about the equalities in this. Firstly how is $$\left(\sum_{j=1}^ka_j^1\gamma^j\right)\wedge\cdots\wedge\left(\sum_{j=1}^ka_j^k\gamma^j\right) = \sum_{\sigma\in S_k}a_{\sigma(1)}^1\gamma^{\sigma(1)}\wedge\cdots\wedge a^k_{\sigma(k)}\gamma^{\sigma(k)}$$

and secondly how is

$$\sum_{\sigma\in S_k}\left(\prod_{i=1}^ka^i_{\sigma(i)}\right)\gamma^{\sigma(1)}\wedge\cdots\wedge \gamma^{\sigma(k)} = \sum_{\sigma\in S_k}\text{sgn}\sigma\left(\prod_{i=1}^ka^i_{\sigma(i)}\right)\gamma^{1}\wedge\cdots\wedge \gamma^{k}?$$

There seems to be some usage of some property related to the wedge product here which I'm not familiar with. I know the definition, but this seems to be not using that and instead something else?

Answer 1

1. The first step is actually two steps in one. The wedge product is multi-linear, so we expand $$\left(\sum_{j=1}^ka_j^1\gamma^j\right)\wedge\dotsc\wedge\left(\sum_{j=1}^ka_j^k\gamma^j\right)=\sum_{j_1,\dotsc,j_k=1}^ka^1_{j_1}\gamma^{j_1}\wedge\dotsc\wedge a_{j_n}^k\gamma^{j_k}.$$ Next, the wedge product is alternating, so only the summands for which $j_1,\dotsc,j_k$ are pairwise distinct are non-zero. To say that $j_1,\dotsc,j_k$ are pairwise distinct, however, is just to say that they are a permutation of $\{1,\dotsc,k\}$, so that's how we get $$\sum_{j_1,\dotsc,j_k=1}^ka^1_{j_1}\gamma^{j_1}\wedge\dotsc\wedge a_{j_n}^k\gamma^{j_k}=\sum_{\sigma\in S_k}a_{\sigma(1)}^1\gamma^{\sigma(1)}\wedge\cdots\wedge a^k_{\sigma(k)}\gamma^{\sigma(k)}.$$

2. The wedge product is alternating, hence anti-symmetric, meaning $\gamma^{\sigma(1)}\wedge\dotsc\wedge\gamma^{\sigma(k)}=\mathrm{sgn}(\sigma)\gamma^1\wedge\dotsc\wedge\gamma^k$.

Answer 2

$$\begin{align*} \beta^1\wedge\cdots\wedge\beta^k&=\left(\sum_{j=1}^ka_j^1\gamma^j\right)\wedge\cdots\wedge\left(\sum_{j=1}^ka_j^k\gamma^j\right) \\ &=\sum_{j=1}^k \left(a_j^1\gamma^j\wedge\cdots\wedge\left(\sum_{j=1}^ka_j^k\gamma^j\right)\right) \\ &= \cdots \\ &= \sum_{1 \leq i_1, \ldots, i_k \leq k} \left(a_{i_1}^1\gamma^{i_1}\wedge\cdots\wedge a_{i_k}^k\gamma^{i_k}\right) \\ &=\sum_{\sigma \in S_k} \left(a_{\sigma(1)}^1\gamma^{\sigma(1)}\wedge\cdots\wedge a_{\sigma(k)}^k\gamma^{\sigma(k)}\right), \end{align*} $$ where the last equality comes up because we can't have repeats (by antisymmetry).

Now let $\sigma \in S_k$. If $\sigma = (13)$, then notice that

$$\begin{align*} \gamma^3 \wedge \gamma^2 \wedge \gamma^1 \wedge \cdots \wedge \gamma^k &= -\gamma^3 \wedge \gamma^1 \wedge \gamma^2 \wedge \cdots \wedge \gamma^k \\ &= \gamma^1 \wedge \gamma^3 \wedge \gamma^2 \wedge \cdots \wedge \gamma^k \\ &= -\gamma^1 \wedge \gamma^2 \wedge \gamma^3 \wedge \cdots \wedge \gamma^k .\end{align*}$$

Check that this works for every transposition. Using the definition of $\text{sgn}$, the second equality follows.