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1

It is an application of the following fundamental property of the exterior derivative: $d(\alpha \wedge \beta) = d\alpha \wedge \beta + (-1)^{\mathrm{deg} \alpha}(\alpha \wedge d\beta)$. When $\alpha = f$ is a $0$-form (a function), this identity says that $d(f \beta) = df \wedge \beta + f d\beta$. In your situation, $f = x^{i_1}$ and $\beta = ...


1

Here is a slightly more systematic answer than in my comment above. Suppose we have the $(k-1)$-form $\omega=f(\vec{x})\omega_{i_1}$ where $\omega_{i_1}:=dx_{i_2}\wedge dx_{i_3}\wedge\cdots \wedge dx_{x_k}$ has been introduced for convenience. Then the exterior derivative of $\omega$ is $$d\omega=df \omega_{x_i}=\left(\frac{\partial f}{\partial ...


3

I'll write out an answer for a form of three variables. This might yield some intuition. Thing to notice is that $dx^i\wedge dx^i=0$, and that $dx^i\wedge dx^j=-dx^j\wedge dx^i$, and $f_{xy}=f_{yx}$. Thus, because $dx^i\wedge dx^i=0$ $$d^2f=f_{xy}dy\wedge dx+f_{xz}dz\wedge dx+f_{yx}dx\wedge dy+f_{yz}dz\wedge dy+f_{zx}dx\wedge dz+f_{zy}dy\wedge dz $$ ...


1

$$ \begin{split} d(df)&=d(\frac{\partial f}{\partial x^j})\wedge dx^j\\ &=\frac{\partial^2 f}{\partial x^jx^k}dx^k\wedge dx^j\\ &=0 \end{split} $$ since $$ dx^k\wedge dx^j=-dx^j\wedge dx^k $$ and $$ \frac{\partial^2 f}{\partial x^jx^k}=\frac{\partial^2 f}{\partial x^kx^j} $$


0

It might just be easier define $a_{ji} := -a_{ij}$ for $i < j$ and write $w = \tfrac{1}{2}\sum_{i,j} a_{ij} e^i \wedge e^j$, so that $$ w \wedge w = \frac{1}{4} \sum_{i,j,k,l} a_{ij}a_{kl} e^i \wedge e^j \wedge e^k \wedge e^l = \sum_{i<j<k<l} \frac{1}{4}\left(\sum_{\pi \in S_4} (-1)^\pi a_{\pi(i)\pi(j)}a_{\pi(k)\pi(l)} \right)e^i \wedge e^j ...



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