Let $V$ be a vector bundle over a smooth, complete Riemannian manifold $M$. In general, the manifold is not compact.
Further, denote by $g$ the metric on $M$, the volume measure by $d\mu$ and $h$ a metric on $V$. Without hopefully causing too much confusion, let $\nabla$ denote both the Levi-Cevita connection on $M$ and a metric compatible connection on $V$. Let the $L^2(V)$ be denoted by $u:M \to V$ such that $u(x) \in V_x$ where $V_x$ is the fibre over $x \in M$ and $$ \int_{M} h_x(u(x),u(x))\ d\mu < \infty.$$
Now, a few facts. First, it is possible to prove that $\nabla$ is a closeable, densely defined operator on $L^2(V)$. Thus, identify it with its closure and let $\rm{div}=-\nabla^\ast$. Since we assumed that $h$ and $\nabla$ are compatible, it is possible to show that for all compactly supported sections $S \in C_c^\infty(T^\ast M \otimes V)$ and $T \in C_c^\infty(V)$ that
$$(\nabla T, S) = (T, \rm{tr}\nabla S).$$
Indeed, the $(.,.)$ are exactly what you expect them to be - on the left it is the inner product over ${L^2(T^\ast M \otimes V)}$ and on the right, the inner product over ${L^2(V)}$.
It is also possible to show that $\rm{tr}\nabla$ is a closeable, densely defined operator. Thus, I can define $\tilde{\nabla} = -(\rm{tr}\nabla)^\ast$. We immediately find that: $$\nabla \subset \tilde{\nabla}\quad\text{and}\quad \rm{tr}\nabla \subset \rm{div}.$$
What I'm interested in is knowing when the closure $\rm{tr}\nabla = \rm{div}$. Equivalently, this amounts to showing that $\tilde{\nabla} = \nabla$.
Does anyone know if there are answers to this? It would be nice enough if this was true when $V$ are the $(p,q)$ tensors.
