Existence of a 1 - Form With Differential Equal to a Given 2 - Form
If\[\omega^2\]
is a 2 - form in \[\mathbb{R}^3\]
then we can write\[\omega^2 = f_1 dx_2 \wedge dx_3 +f_2 dx_3 \wedge dx_1 +f_3 dx_1 \wedge dx_2\]
If
\[omega^1\]
is a 1 - form in \[\mathbb{R}^3\]
then we can write\[ \omega^1 =g_1 dx_1 + g_2 dx_2 + g_3 dx_3 \rightarrow d \omega^1 = (\frac{\partial g_3}{\partial x_2} - \frac{\partial g_2}{\partial x_3}) dx_2 \wedge dx_3 + (\frac{\partial g_1}{\partial x_3} - \frac{\partial g_3}{\partial x_1}) dx_3 \wedge dx_1 +(\frac{\partial g_2}{\partial x_1} - \frac{\partial g_1}{\partial x_1}) dx_1 \wedge dx_2\]
Equating
\[\omega^2 , \: d \omega^1\]
gives\[f_1 =(\frac{\partial g_3}{\partial x_2} - \frac{\partial g_2}{\partial x_3}) , \: f_2 = (\frac{\partial g_1}{\partial x_3} - \frac{\partial g_3}{\partial x_1}) , \; f_3 = (\frac{\partial g_2}{\partial x_1} - \frac{\partial g_1}{\partial x_1})\]
or with \[\mathbf{f} =(f_1 ,f_2,f_3)^T , \: \mathbf{g}=(g_1,g_2 g_3)^T\]
we have \[ \mathbf{f} = \mathbf{\nabla} \times \mathbf{g} \]
.Hence such a 1 - form does exist.