When does $\nabla\times(\nabla\times F)=0$ imply $\nabla \times F=0$ On a (simply connected) domain $\Omega$ for a smooth vector field $F\colon \Omega \to \mathbb{R}^3$, when does $\nabla\times(\nabla\times F)=0$ imply $\nabla \times F=0$. I know that $n\cdot(\nabla\times F)=0$ on $\partial\Omega$ is sufficient, and also $t\cdot(\nabla\times F)=0$ is sufficient ($t$ the tangential). Is there a weaker condition?
 A: Here are some basic thoughts. Let $G$ be a vector field which is of the form $\nabla \times F$, and also obeys $\nabla \times G = 0$.
Since $\nabla \times G = 0$, the vector field $G$ is locally of the form $\nabla h$ for some scalar valued function $h$. The condition that $G = \nabla \times F$ imples that $\nabla \cdot G=0$ or, in other words, $\nabla^2(h)=0$. This says that $h$ is harmonic. So, locally, the condition is equivalent to $G$ being the gradient of a harmonic function. Globally, if $\Omega$ is not simply connected, then traveling around a loop in $\Omega$ may change $h$ by a constant.
From this, we can see that, if $G$ is compactly supported, it is zero: If $G$ is $0$ outside a ball of radius $R$, then $h$ is constant outside that ball, and thus $h$ is constant everywhere. 
We can also trot out our favorite conditions to ensure that a harmonic function is constant. For example, if $\Omega = \mathbb{R}^3$, and $G(x) \to 0$ as $|x| \to \infty$, then $|h(x)| = o(x)$ and a variant of Liouville's theorem tells us that $h$ is constant and $G=0$.
The OP seems to be interested in conditions on the flux of $G$ across $\partial \Omega$. In order to make this make sense, I am going to assume that the set up is that $\overline{\Omega}$ is a compact manifold with boundary.
Let $R$ be the flux of $G$ across $\partial \Omega$. As the OP already knows, if $R=0$ then $G=0$. And the assignment of $G \mapsto R$ is linear, so this shows that $R$ determines $G$. Let $\mathcal{V}$ be the vector space of functions on $\partial \Omega$ which can occur as such an $R$. 
The OP asks for conditions which force $G$ to be zero. In other words, he wants a vector space $\mathcal{W}$ of functions on $\partial \Omega$ which is transverse to $\mathcal{V}$. I find that an odd way to think about the question -- better to just characterize $\mathcal{V}$! 
Now, the obvious observation is that $\int_{\partial \Omega} R=0$ for any $R$ in $\mathcal{V}$, since $\nabla \cdot G=0$. It would be really cool if that were a precise characterization of $\mathcal{V}$. But I don't think it is. Let $\Omega$ be a solid cylinder, and let $R$ be zero on the sides of $\Omega$ and a hat function on top and bottom, dying out smoothly as it approaches the sides. Is $R$ the flux of some $G$? I couldn't figure out how to make it be.
A: For some scalar function g: $ rot\  F=grad\  g$. Necessary and sufficient!
