Let $\textbf{F}\in \mathbb{R}^3$ be a smooth vector field for all space. It is well known using [Helmholtz decomposition](https://en.wikipedia.org/wiki/Helmholtz_decomposition) that we can decompose $\textbf{F}$ into two vector fields in $V$: $$\textbf{F} = \nabla \sigma + \nabla \times \Gamma,$$ where $\nabla \sigma$ is called the irrotational part and $\nabla \times \Gamma$ the solenoidal part of $\textbf{F}$. If both these fields are non-zero, is it true that $\nabla \sigma$ and $\nabla \times \Gamma$ are linearly independent?



Are the irrotational and solenoidal parts of a smooth vector field linearly independent?
$\newcommand\R{\mathbb R}\newcommand\na{\nabla}\newcommand\om{\boldsymbol{\omega}}\newcommand\si{\sigma}\newcommand\Ga{\Gamma}\newcommand\F{\mathbf F}\newcommand\x{\mathbf x}\newcommand\0{\mathbf 0}$The answer is no, in general. E.g., take any nonzero $\om\in\R^3$ and let
$$\si(\x):=\om\cdot\x\quad\text{and}\quad\Ga(\x):=\om\times\x$$
for $\x\in\R^3$, with $\F:=\na\si+\na\times\Ga$. Then $\na\si(\x)=\om$ and $\na\times\Ga(\x)=2\om$ for all $\x\in\R^3$, so that $\na\si$ and $\na\times\Ga$ are nonzero but linearly dependent.




---


On a positive note, suppose that $\na\si$ and $\na\times\Ga$ are nonzero and linearly dependent, so that $\na\si=c\na\times\Ga$ for some nonzero real $c$. Then $\na^2\si=\na\cdot(\na\si)=c\na\cdot(\na\times\Ga)=\0$, so that $\si$ is a harmonic scalar field. If we assume that $\si$ is bounded above or below, then [Liouville's theorem](https://en.wikipedia.org/wiki/Harmonic_function#Liouville%27s_theorem) will imply that $\si$ is constant. [Moreover](https://math.stackexchange.com/a/1685017/96609), if $\si(\x)=o(|\x|)$ as $|\x|\to\infty$, then $\si$ is constant. Furthermore, similarly it is easy to show that, if $\si$ is bounded above or below by a function $\si\_\*$ such that $\si\_\*(\x)=o(|\x|)$ as $|\x|\to\infty$, then $\si$ is constant -- and hence the fields $\na\si$ and $\na\times\Ga$ are zero, a contradiction.


Thus, the answer to your question becomes yes given that $\si$ is bounded above or below by a function $\si\_\*$ such that $\si\_\*(\x)=o(|\x|)$ as $|\x|\to\infty$.


On the other hand, the counterexample in the first part of the answer shows that the sublinear growth condition, $\si\_\*(\x)=o(|\x|)$ as $|\x|\to\infty$, cannot be relaxed to the condition that $\si\_\*(\x)=O(|\x|)$ as $|\x|\to\infty$.


