# Generalisation of the Poincaré Lemma

Let $\Omega \subset \mathbb{R}^3$ be an open but not simply connected domain and let $v \: \colon \Omega \to \mathbb{R}^3$ be a continuously differentiable vector field. Assume that $\textrm{curl} \, v = 0$ in $\Omega$. As $\Omega$ is not simply connected, we cannot conclude that there is a function $\varphi \in C^2(\Omega, \mathbb{R})$ such that $v = \nabla \varphi$.

However, if we additionally assume that for all smooth loops $\gamma \: \colon [0, 1] \to \Omega$ the line integral $$\int_{\gamma} v(r) \cdot \textrm{d}r = 0, \qquad (1)$$ then we find $\varphi \in C^2(\Omega, \mathbb{R})$ such that $v = \nabla \varphi$.

My question is now:

Let $\Omega \subset \mathbb{R}^3$ be an open but not simply connected domain and given a vector field $v \in L^2(\Omega, \mathbb{R}^3)$ with $\textrm{curl} \, v = 0$ in $\Omega$. Is there a similar condition like $(1)$ that ensures the existence of a function $\varphi \in H^1(\Omega)$ such that $v = \nabla \varphi$?

The general condition is that if for all divergence free compactly supported smooth vector fields, that is $$f \in \mathcal{D}(\Omega,\mathbb{R}^3)$$ with $$div(f)=0$$, $$v \in L^2(\Omega,\mathbb{R}^3)$$ satisfies, $$$$\int_\Omega v.f dx = 0$$$$
then there will exist $$\phi \in H^1(\Omega)$$ such that $$v=\nabla \phi$$. This condition on $$v$$ will imply $$curl (v)=0$$ but not the other way around. This is in fact quite a general result and would work for $$v$$ in more general Sobolev spaces or in the space of distributions (each component of $$v$$ is a distribution).