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Let $M$ be a smooth Riemannian manifold and $E$ be a finite-rank vector bundle over $M$ equipped with a bundle metric $\langle\cdot,\cdot\rangle\in\Gamma^{\infty}(E^{\ast}\otimes E^{\ast})$, i.e. $\langle\cdot,\cdot\rangle$ is a positive-definite inner product on each fibre. This induces an inner product on the level of (compactly-supported) sections via $$(\cdot,\cdot):=\int_{M}\langle\cdot,\cdot\rangle\,\mathrm{vol}_{g}.$$

Now, if $M$ is compact and $L:\Gamma^{\infty}(E)\to\Gamma^{\infty}(E)$ a formally self-adjoint elliptic operator, then the famous Fredholm alternatives states that the elliptic equation $$Lu=f$$ for some source $f\in\Gamma^{\infty}(E)$ has a smooth solution $u\in\Gamma^{\infty}(E)$ if and only if $(f,g)=0$ for all $g\in\ker(L)$.

If $M$ is non-compact, is there a similar statement, i.e. if $f\in\Gamma^{\infty}_{c}(E)$ does there exists a compactly-supported smooth solution $u\in\Gamma^{\infty}_{c}(E)$ if and only if $f$ is orthogonal to every element in $\ker(L\vert_{\Gamma^{\infty}_{c}})$?

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    $\begingroup$ This is tricky. There are few results in the noncompact case since noncompactness can manifest itself in so many ways. An old result that covers a special case of non compactness is due to Lockhard and McOwen. numdam.org/item/ASNSP_1985_4_12_3_409_0 $\endgroup$ Commented Jul 21, 2023 at 14:34
  • $\begingroup$ Thanks, I will check it out! Also, I just realized, my precise formulation also makes not a lot of sense since $\mathrm{ker}(L\vert_{c})$ is probably empty for many operators by unique continuation $\endgroup$ Commented Jul 21, 2023 at 14:44
  • $\begingroup$ More precisly, the example I am interested most, is whether there are compactly supported smooth solutions of the equation $\Delta_g f=\delta\omega$ for some compactly supported smooth 1 form omega. $\endgroup$ Commented Jul 21, 2023 at 14:45
  • $\begingroup$ @G.Blaickner in view of your last comment then this might be relevant: mathoverflow.net/questions/442169/poisson-equation-on-manifolds $\endgroup$ Commented Jul 21, 2023 at 14:54
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    $\begingroup$ @G.Blaickner Well in the setting of manifolds with cylindrical ends, under certain conditions, if the source is L^2-orthogonal to the kernel of the formal adjoint of the operator then there is a solution which decays exponentially over the ends but this doesn't mean that the solution will be compactly supported. $\endgroup$ Commented Jul 22, 2023 at 16:41

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