differential form identity and permutations

Question

If $t^1,...,t^k$ are the coordinates of a k-cube.

Then apparently

$$dt^{\sigma(1)} \wedge \ldots \wedge dt^{\sigma(k)}= (\operatorname{sgn} (\sigma)) dt^1 \wedge dt^k $$

I cannot see how this proved, nor does it seem intuitively true.

Answer 1

If you consider $dt^1\wedge\cdots\wedge dt^k$ as a wedge product of $k$ 1-forms, the conclusion can't be more obvious.

Recall that the wedge product of $k$ $1$-forms: $$ dt^1\wedge\cdots\wedge dt^k\triangleq\frac{k!}{1!\cdots 1!}\mathcal A(dt^1\otimes\cdots\otimes dt^k). $$ where $\mathcal A$ is the alternating operator: $$ \mathcal A(\omega)(v_1,\dots,v_k)=\frac{1}{k!}\sum_{\tau\in S_k}\text{sgn}(\tau)\omega(v_{\tau(1)},\dots,v_{\tau(k)}),\forall\text{ vector fields }v_1,\dots,v_k. $$ As a result of the definition: $$ \begin{split} dt^{\sigma(1)}\wedge\cdots\wedge dt^{\sigma(k)}(v_1,\dots,v_k)&=\sum_{\tau\in S_k}\text{sgn}(\tau)dt^{\sigma(1)}\otimes\cdots\otimes dt^{\sigma(k)}(v_{\tau(1)},\dots,v_{\tau(k)})\\ &=\sum_{\tau\in S_k}\text{sgn}(\tau)\text{sgn}(\sigma)^2dt^{1}\otimes\cdots\otimes dt^{k}(v_{\sigma\tau(1)},\dots,v_{\sigma\tau(k)})\\ &=\text{sgn}(\sigma)\sum_{\tau\in S_k}\text{sgn}(\sigma\tau)dt^{1}\otimes\cdots\otimes dt^{k}(v_{\sigma\tau(1)},\dots,v_{\sigma\tau(k)})\\ &=^*\text{sgn}(\sigma)\sum_{\sigma\tau\in S_k}\text{sgn}(\sigma\tau)dt^{1}\otimes\cdots\otimes dt^{k}(v_{\sigma\tau(1)},\dots,v_{\sigma\tau(k)})\\ &\triangleq\text{sgn}(\sigma)dt^1\wedge\cdots\wedge dt^k(v_1,\dots,v_k) \end{split} $$ The $=^*$ is true since $\tau\mapsto\sigma\tau$ is a bijective auto map on the permutation group $S_k$ of $k$ elements.