Proposition

If $F$ is a vector field on $\mathbb{R}^3$, define the forms

\[\begin{align*} \omega^1_F &= F^1dx + F^2dy + F^3dx \\ \omega^2_F &= F^1dy \wedge dx + F^2dz \wedge dx + F^3 dx \wedge dy \end{align*}\]

Prove that
\[\begin{align*} df &= \omega^1_{\grad f}, \\ d(\omega^1_p) &= \omega^2_{\curl F}, \\ d(\omega^2_p) &= (\div F) dx \wedge dy \wedge dz. \end{align*}\]
Use (1) to prove that
\[\begin{align*} \curl \grad f &= 0, \\ \div \curl f &= 0. \end{align*}\]
If $F$ is a vector field on a star-shaped open set $A$ and $\curl F = 0$, show that $F = \grad f$ for some function $f: A \rightarrow R$. Similarly, if $\div F = 0$, show that $F = \curl G$ for some vector field $G$ on $A$.

Solution

1

Let $f: \mathbb{R}^3 \rightarrow \mathbb{R}$.

\[\begin{align*} df &= D_1fdx^1 + D_2fdx^2 + D_3fdx^3 \\ &= \omega^1_{\grad f} \end{align*}\]

where $\grad f = D_1f \cdot e_1 + D_2f \cdot e_2 + D_3f \cdot e_3$.

\[\begin{align*} d\omega^1_F &= d(F^1dx + F^2dy + F^3dz) \\ &= (D_2F^3 - D_3F^2)dy \wedge dz + (D_1F^3 - D_3F^1)dx \wedge dz + (D_1F^2 - D_2F^1)dy \wedge dz \\ &= \omega^2_{\curl F} \end{align*}\]

where $\curl F = (D_2F^3 - D_3F^2)e_1 + (D_1F^3 - D_3F^1)e_2 + (D_1F^2 - D_2F^1)e_3$.

\[\begin{align*} d(\omega^2_F) &= d(F^1 dy \wedge dz + F^2 dz \wedge dx + F^3 dx \wedge dy) \\ &= D_1F^1 dx \wedge dy \wedge dz + D_2F^2 dy \wedge dz \wedge dx + D_3F^3 dz \wedge dx \wedge dy \\ &= (D_1F^1 + D_2F^2 + D_3F^3) dx \wedge dy \wedge dz \\ &= (\div F) dx \wedge dy \wedge dz. \end{align*}\]

2

\[\begin{align*} \omega^2_{\curl(\grad F)} &= d(\omega^1_{\grad F}) \\ &= d(dF) \\ &= d^2F = 0. \\ \div(\curl F) dx \wedge dy \wedge dz &= d(\omega^2_{\curl F}) \\ &= d(d(\omega^1_F)) \\ &= d^2(\omega^1_F) \\ &= 0. \end{align*}\]

3

Let $F$ be a vector field such that $\curl F = 0$. Then $d(\omega^{1}_{F}) = \omega^2_{\curl F} = \omega^2_0 = 0$. Therefore, $\omega^1_{F}$ is closed. By Poincare Lemma[Theoem 4-11, Spivak], $\omega^1_{F}$ is exact. Let $f: A \rightarrow \mathbb{R}$ be a 0-form such that $df = \omega^{1}_F$. In Part 1, we showed that $df = \omega^{1}_{\grad f}$. Therefore, $F = \grad f$.

Similarly, suppose $\div F = 0$. Then $d(\omega^2_{F}) = (\div F)dx \wedge dy \wedge dz = 0$. Therefore, $\omega^2_F$ is a closed 2-form. By Poincare Lemma, there exists a 1-form $G^1dx + G^2dy + G^3dz$ such that $d(G^1dx + \cdots + G^3dz) = \omega^2_F$. Let $G$ be a vector field such that $G(p) = (G^1(p)e^1 + G^2(p)e^2 + G^3(p)e^3)_{(p)}$. Then $\omega^2_F = d(G^1dx + \cdots + G^3dz) = d(\omega^1_G) = \omega^2_{\curl G}$. Therefore, we found a vector field $G$ such that $F = \curl G$.