Name this Vector Calculus Theorem 
There is an important theorem in vector calculus that says $\boldsymbol{\nabla}\boldsymbol{\cdot}\mathbf{G}\boldsymbol{=}0$
(where $\mathbf{G}$ is some differentiable vector function) implies and is implied by $\mathbf{G}\boldsymbol{=}\boldsymbol{\nabla}\boldsymbol{\times}\mathbf{H}$ (where $\mathbf{H}$ is another differentiable function).

Does anyone know if there is a name for this vector calculus theorem this description is referring to?
 A: $\nabla \cdot \mathbf{G} = 0 \implies  \mathbf{G} = \nabla \times \mathbf{H}$ is part of the Helmholz theorem (you also need to make some assumptions about the asymptotic behavior of $G$).
$\nabla \cdot \nabla \times \mathbf{H} = 0$ is a relatively straightforward identity in vector calculus, an easy proof is just to write out the components in Cartesian coordinates and use the fact that partial derivatives commute.
One observation that can make these facts seem more intuitive, is that if you go to Fourier space and work with a plane wave with wavevector $\mathbf{k}$, then the statements become
\begin{equation}
\mathbf{k} \cdot \tilde{\mathbf{G}} = 0 \implies \tilde{\mathbf{G}} = i \mathbf{k} \times \tilde{\mathbf{H}}
\end{equation}
and
\begin{equation}
\mathbf{k} \cdot (\mathbf{k} \times \mathbf{\tilde{H}} ) = 0
\end{equation}
where $\mathbf{k}$ is just an ordinary 3-vector, and $\tilde{\mathbf{G}}, \tilde{\mathbf{H}}$ are the spatial Fourier transforms of $\mathbf{G}$ and $\mathbf{H}$ (which are also just ordinary vectors, just on k space).
A: Exactly what Andrew said.
If you've taken EM or electrodynamics you also know that $\vec{B} = \vec{\nabla}\times\vec{A}$. 
It is a gauge freedom that exists due to the fact the the magnetic field is sourceless or, in other words, $\vec{\nabla}\cdot\vec{B}=0$.
 This is a classic use of the theorem in physics.
Note that if $\vec{\nabla}\cdot\vec{G}=0$ you could also claim that $\vec{G}=\vec{\nabla}\times\left(\vec{H}+\vec{\nabla}\phi\right)$ where $\phi$ is a scalar function. It is true since the curl of a gradient field is always 0.
$\vec{\nabla}\times\vec{\nabla}\phi=0$ 
You can visit https://en.wikipedia.org/wiki/Gauge_fixing for more gauge freedoms that arise from similar truths in vector calculus.
