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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,

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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).

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  • $\begingroup$ Hmm, yes you are right! Thanks for this. I realize now I was vague in the question as I was rather interested in just the possibility of this extension in a small neighborhood of M, that is to say given this one form can we just extend it to a slightly larger manifold $\endgroup$
    – Ali
    Commented Feb 16, 2019 at 15:03
  • $\begingroup$ @Ali **The answer is still no.**My example has singularity at P=(0,0). Take my example and place the singularity on the boundary of your manifold. You cannot extend this form to a closed form past the point P no matter how small neighborhood you want to consider. $\endgroup$
    – Piotr Hajlasz
    Commented Feb 16, 2019 at 21:55

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