## 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.