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Let $U \subseteq\mathbb{R}^n$ be open, and let $\omega$ be an $(n-1)$-form that's given by

$$\omega = \sum_{i=1}^n (-1)^{i-1} F_i\, dx_1 \wedge\dots\wedge dx_{i-1} \wedge dx_{i+1} \wedge\dots\wedge dx_n\qquad(1) $$

where the $F_i$ are continuously differentiable functions $F_i: U \to \mathbb{R}$.

I now want to show that the (exterior) derivative of $\omega$ is given by $d \omega = \text{div}F \cdot dx_1 \wedge\dots\wedge dx_n$ where $F$ is the vector field $F(x) := (F_1(x), \dots, F_n(x))$, and $\operatorname{div}F$ the divergence of $F$ (i.e. $\operatorname{div}F = \sum_{i=1}^n \frac{\partial F_i}{\partial x_i} $).

Using that, I want to conclude that if the $F_i \in C^2$ (i.e. the $F_i$ are not only once but twice continuously differentiable), then we have $d^2 \omega := d(d \omega) = 0$.

(My definition of the exterior derivative of a $k$-form was as follows: Given any differential $\mu$ form in base representation $\mu = \sum_{1 \leq i_1 < \dots < i_k \leq n} a_{i_1 \dots i_k} dx_{i_1} \wedge\dots\wedge dx_{i_k}$ with the $a_{i_1 \dots i_k} \in C^1$, i.e. continuously differentiable, the derivative is defined as $d\mu = \sum_{1 \leq i_1 < \dots < i_k \leq n} da_{i_1 \dots i_k} \wedge dx_{i_1} \wedge\dots\wedge dx_{i_k}$.)

I must admit that I didn't know how to get started. I don't really know what I can do to get form the formula (1) to the divergence of $F$, and how to "get" the missing $dx_i$ in each of the sums. Working with $k$-forms and the exterior derivative is still very new to me, and I know I'm still lacking the proper intuition for these kind of constructions.

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Note that

$$d\omega = \sum_{i=1}^n(-1)^{i-1}dF_i\wedge dx_1\wedge\dots\wedge dx_{i-1}\wedge dx_{i+1}\wedge\dots\wedge dx_n.$$

and $dF_i = \sum_{j=1}^n\frac{\partial F_i}{\partial x_j}dx_j$. As wedge product is skew-symmetric, if some $dx_k$ appears twice in a term, that term must be zero. So we see that

\begin{align*} &\ (-1)^{i-1} dF_i\wedge dx_1\wedge\dots\wedge dx_{i-1}\wedge dx_{i+1}\wedge\dots\wedge dx_n\\ =&\ (-1)^{i-1}\left(\sum_{j=1}^n\frac{\partial F_i}{\partial x_j}dx_j\right)\wedge dx_1\wedge\dots\wedge dx_{i-1}\wedge dx_{i+1}\wedge\dots\wedge dx_n\\ =&\ (-1)^{i-1}\sum_{j=1}^n\frac{\partial F_i}{\partial x_j}dx_j\wedge dx_1\wedge\dots\wedge dx_{i-1}\wedge dx_{i+1}\wedge\dots\wedge dx_n\\ =&\ (-1)^{i-1}\frac{\partial F_i}{\partial x_i}dx_i\wedge dx_1\wedge\dots\wedge dx_{i-1}\wedge dx_{i+1}\wedge\dots\wedge dx_n\\ =&\ \frac{\partial F_i}{\partial x_i} dx_1\wedge\dots\wedge dx_{i-1}\wedge dx_i\wedge dx_{i+1}\wedge\dots\wedge dx_n\\ =&\ \frac{\partial F_i}{\partial x_i} dx_1\wedge\dots\wedge dx_n. \end{align*}

So we see that

\begin{align*} d\omega &= \sum_{i=1}^n(-1)^{i-1}dF_i\wedge dx_1\wedge\dots\wedge dx_{i-1}\wedge dx_{i+1}\wedge\dots\wedge dx_n\\ &= \sum_{i=1}^n\frac{\partial F_i}{\partial x_i} dx_1\wedge\dots\wedge dx_n\\ &= \left(\sum_{i=1}^n\frac{\partial F_i}{\partial x_i}\right) dx_1\wedge\dots\wedge dx_n\\ &= \operatorname{div}F\, dx_1\wedge\dots\wedge dx_n. \end{align*}

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