A 1 - Form That is Not The Differential of a 0 - Form

If  
\[\omega^1\]
  is a 1 - form in  
\[\mathbb{R}^3\]
  then we can write
\[\omega^1 = f_1 dx_1 +f_2 dx_2 +f_3 dx_3\]

If  
\[omega^0\]
  is a 0 - form in  
\[\mathbb{R}^3\]
  then we can write
\[ \omega^0 =g(x_1,x_2,x_3) \rightarrow d \omega^0 = \frac{\partial g}{\partial x_1} dx_1 + \frac{\partial g}{\partial x_2} dx_2 + \frac{\partial g}{\partial x_3} dx_3\]

Equating  
\[\omega^1 , \: d \omega^0\]
  gives
\[f_1 = \frac{\partial g}{\partial x_1} , \: f_2 = \frac{\partial g}{\partial x_2} , \; f_3 = \frac{\partial g}{\partial x_3}\]
  or with  
\[\mathbf{f} =(f_1 ,f_2,f_3)^T , \: \mathbf{f} = \mathbf{\nabla} g\]
.
The existence of a scalar function  
\[g\]
  such that for a vector field  
\[\mathbf{f}\]
  we have  
\[\mathbf{f} = \mathbf{\nabla} g\]
  is one of the properties of a conservative vector field, or equivalently  
\[\mathbf{\nabla} \times \mathbf{f} =0\]

Take  
\[f_1 =x_2 , \: f_2=-x_1 \: f_3=0\]
  then
\[(\mathbf{\nabla} \times \mathbf{f} = (\frac{\partial}{\partial x_1} , \frac{\partial}{\partial x_2} , \frac{\partial}{\partial x_3} )^T \times (x_2, -x_1,0)^T = -2 \]

Hence no such function  
\[g\]
  exists.

You have no rights to post comments