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 $\delta$.
Best Answer
The unique continuation is valid for generalized Laplacians. This follows from Hörmander's result in
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.