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Let $({M},g)$ be a connected and non-compact Riemannian manifold without boundary. If $L:\Gamma^{\infty}(E)\to \Gamma^{\infty}(E)$ is a linear second order elliptic operator on some smooth $\mathbb{R}$-bundle $E$ over ${M}$, is it then true that

$$Lu=0$$

for $u\in\Gamma^{\infty}_{c}(E)$ implies that $u=0$, or in other words, there are no compactly-supported smooth homogeneous solutions? I think such a result could be proven via some ``unique continuation property of elliptic system''. However, while looking in the literature, I didn't find a suitable version for this situation.

Examples of such operator $L$ I have in mind is for example the connection Laplacian $$\Delta_{C}:=g^{ij}\nabla_{i}\nabla_{j}:\Gamma^{\infty}(T^{\ast}{M}^{\otimes k})\to\Gamma^{\infty}(T^{\ast}{M}^{\otimes k})$$ with Levi-Civita connection $\nabla$ or closely related, the Hodge-de Rham Laplacian $$\Delta_{H}:=\mathrm{d}\delta+\delta\mathrm{d}:\Omega^{k}({M})\to\Omega^{k}({M})$$ with exterior derivative $\mathrm{d}$ and codifferential $\delta$.

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2 Answers 2

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The unique continuation is valid for generalized Laplacians. This follows from Hörmander's result in

Hörmander, Lars, Uniqueness theorems for second order elliptic differential equations, Commun. Partial Differ. Equations 8, 21-64 (1983). ZBL0546.35023 MR686819

The operators you mentioned are such generalized Laplacians. Here are a few details.

Suppose that $E$ is a metric vector bundle over the Riemann manifold $(M,g)$. Denote by $\Delta_M$ the scalar Laplacian determined by the metric $g$. Fix a connection on $E$ compatible with the metric on $E$ and set $\Delta_E=\nabla^*\nabla: C^\infty(E)\to C^\infty(E)$.

For any $u\in C^\infty(E)$ we have

$$\Delta_M |u(x)|_E^2=2\big\langle \Delta_E u(x),u(x)\big\rangle_E-2\vert \nabla u(x)\vert_E^2.$$

A second order partial differential operator $L$ is called a generalized Laplacian if its principal symbol satisfies

$$ \sigma_L(\xi)=-|\xi|_g^2\cdot \mathbf{1}_{E_x},\;\;\forall x\in M,\;\;\forall \xi\in T^*_xM. $$

Any generalized Laplacian $L:C^\infty(E)\to C^\infty(E)$ has the form

$$ L=\nabla^*\nabla+ T=\Delta_E+T,$$

for some a connection $\nabla$ on $E$ compatible with the metric on $E$ and an endomorphism $T$ of the bundle $E$; see Proposition 10.1.34 here.

If $Lu=0$, then $\Delta_E u=-Tu$ and we deduce

$$\Delta_M |u(x)|_E^2=-2\big\langle Tu(x),u(x)\big\rangle_E-2\vert \nabla u(x)\vert_E^2.$$

At this point you can invoke the above results of Hörmander for the scalar function $|u(x)|_E^2$ to obtain the unique continuation.

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    $\begingroup$ Looking in from the outside, it's unfortunate that (to my knowledge) there doesn't appear to be a handy reference for (strong) unique continuation results for elliptic systems. If one doesn't know this trick of reducing to a scalar equation, or if it's not obviously applicable, the best options seems to be to start climbing up and down the citation tree from (Hile, Protter 1976) to see if the desired class of systems has been studied. $\endgroup$ Apr 29, 2023 at 12:41
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    $\begingroup$ That is true. The only bit of good news for me at least is that in most geometric applications the operators involved are either of Dirac type or Laplacian type and unique continuation results are available in such cases. $\endgroup$ May 1, 2023 at 10:44
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    $\begingroup$ Maybe this is simple but don't you need to be able to show that $|\nabla u|_E^2 \leq C |x|^{-1+\epsilon}|\nabla_M |u|^2|$ for some constants $C,\epsilon>0$ to apply Hormanders result? $\endgroup$
    – RBega2
    May 1, 2023 at 17:15
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    $\begingroup$ @LiviuNicolaescu The result of Hormander requires an estimate on the absolute value of the Laplacian what you wrote only gives a one sided bound. $\endgroup$
    – RBega2
    May 2, 2023 at 22:12
  • $\begingroup$ @RBega2 You are correct and I have deleted that embarrassing comment. The ad-hoc approach indicated by Rafe Mazzeo works. Aronszajn method works in this case. Rafe Mazzeo proves a Carleman inequality for generalized Laplacians in his paper "Unique continuation..." Amer. J. Math. 1991. This is all you need. $\endgroup$ May 4, 2023 at 20:27
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H"ormander's result works perfectly well for any second order system with real diagonal principal part. The last few answers indicate steps to prove this, but it's more direct to just look at the proof and see that all the steps just work directly for systems of this type. It's very common that the hypothesis for such unique continuation results is simply some sort of differential inequality like this, with this sort of extension to bundles being straightforward.

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