Let $E$ be a vector bundle over a smooth manifold $M$, equipped with a metric $\eta$ and a metric-compatible connection $\nabla$.
Denote by $\delta_{\nabla^E}:\Omega^1(M,E) \to \Omega^{0}(M,E)=\Gamma(E)$ the adjoint of the connection $${\nabla^E}: \Gamma(E)\to \Omega^1(M,E).$$
Note that $\delta_{\nabla^E}\circ{\nabla^E}: \Gamma(E) \to \Gamma(E)$.
Question: Is $\text{ker} (\delta_{\nabla^E}\circ{\nabla^E})$ always non-zero?
While it is known that $\text{ker} (\delta_{\nabla^E})$ is infinite-dimensional, it is not clear to me that there are non-zero elements in the $\text{ker} (\delta_{\nabla^E})$ which are in the image of $\nabla^E$.
(Note: for a generic connection $\nabla^E$, $\text{ker}(\nabla^E)=0$.
My motivation is trying to understand things about the minimizing properties of harmonic maps.
This answer is pretty much an expansion of Anthony Carapetis's comment together with a reference. The operator $\delta_{\nabla^E} \circ \nabla^E$ (more often denoted by $\nabla^{*} \nabla$) is called the connection Laplacian or the covariant Laplacian (see Chapter 7, Section 3.2 of Petersen). I'll assume that $(M,g)$ is compact and oriented and then by definition of the adjoint operator we have
$$ \int_{M} \left< \nabla s, \nabla s \right> \, d\operatorname{Vol}_g = \int_{M} \left< s, \nabla^{*} \nabla s \right> \, d\operatorname{Vol}_g $$
so any section in the kernel of $\nabla^{*} \nabla$ must be parallel ($\nabla s = 0$).
Since a "generic connection" admits no non-trivial parallel sections, most of the time the kernel of $\nabla^{*} \nabla$ is zero-dimensional.