Suppose $(M,g)$ is a compact Riemannian manifold with smooth boundary and that $M\subset \tilde{M}$ with $(\tilde{M},g)$ also a compact Riemannian manifold with smooth boundary. 
Let us consider a one-form $\alpha \in L^2(M;T^\*M)$ with the additional property that $\nabla \cdot \alpha \in L^2(M)$, that is the divergence of $\alpha$ weakly makes sense as an element of $L^2(M)$. Does $\alpha$ admit an extension to $\tilde{M}$ such that the same exact regularity properties hold in the larger domain?


Thanks,



Sobolev extension operators
**The answer is no**.
Consider the form 
$$
\omega = \frac{-y}{x^2+y^2}\, dx + \frac{x}{x^2+y^2}\, dy
$$
on the annulus $\Omega=\{ (x,y):\, 1<x^2+y^2<2\}$. This form is closed, but not exact, because the integral along a circle around zero equals $2\pi$. 


If you would manage to extend it to a closed form in the disc $\{ (x,y):\, x^2+y^2<2\}$
it would be exact (Poincare lemma).


