Example - Differentiation of a Differential Form

Differentiation of a Differential Form

We saw in constants_existence_for_1-forms_multiplication that, if $d\omega$ is a $2-$form in ${\mathbb{R}}^3$, then there exists $a,b,c:{\mathbb{R}}^3\to \mathbb{R}$ such that

$d\omega =a(x,y,z)dx\wedge dy+b(x,y,z)dy\wedge dz+c(x,y,z)dx\wedge dz.$

To figure what $a$ is, for example, we need to determine what $d\omega$ does to the vectors $⟨1,0,0{⟩}_{(x,y,z)}$ and $⟨0,1,0{⟩}_{(x,y,z)}$. Let us compute $d\omega$ assuming that

$\omega =f(x,y,z)dx+g(x,y,z)dy+h(x,y,z)dz.$

\begin{eqnarray} d\omega(\langle1,0,0\rangle, \langle0,1,0\rangle)&=&\dfrac{\partial\omega}{\partial\langle1,0,0\rangle}(\langle0,1,0\rangle)- \dfrac{\partial\omega}{\partial\langle0,1,0\rangle}(\langle1,0,0\rangle)\\ &=&\left\langle\dfrac{\partial g}{\partial x},\dfrac{\partial g}{\partial y},\dfrac{\partial g}{\partial z}\right\rangle\cdot\langle1,0,0\rangle-\left\langle\dfrac{\partial f}{\partial x},\dfrac{\partial f}{\partial y},\dfrac{\partial f}{\partial z}\right\rangle\cdot\langle0,1,0\rangle\\ &=&\dfrac{\partial f}{\partial x}-\dfrac{\partial f}{\partial y}\;. \end{eqnarray}

Analogously, we have

\begin{eqnarray} d\omega(\langle0,1,0\rangle, \langle0,0,1\rangle)&=&\dfrac{\partial h}{\partial y}-\dfrac{\partial g}{\partial z}\\ d\omega(\langle1,0,0\rangle, \langle0,0,1\rangle)&=&\dfrac{\partial h}{\partial x}-\dfrac{\partial f}{\partial z}\;. \end{eqnarray}

Hence,

$d\omega =\left(\frac{\partial f}{\partial x}-\frac{\partial f}{\partial y}\right)dx\wedge dy+\left(\frac{\partial h}{\partial y}-\frac{\partial g}{\partial z}\right)dy\wedge dz+\left(\frac{\partial h}{\partial x}-\frac{\partial f}{\partial z}\right)dx\wedge dz.$

References

@Bac12, pp. 72