Let $E \to M$ be a vector bundle over some Riemannian metric $(M, g)$ and endow it with some fibre metric. Assume that covariant derivative $\nabla$ is compatible with the metric. 


It is essentially an application of Stokes theorem to derive the following identity
$$\nabla^\*\_X = - \nabla\_X - div(X)$$
for the formal adjoint of $\nabla\_X$ with respect to the usual $L^2$-scalarproduct. 


*Question 1:* Is there a closed expression for the formal adjoint of $\nabla$ regarded as a map $\Gamma^\infty(E) \to \Gamma^\infty(T^\*M \otimes E)$, i.e. $\nabla^\*: \Gamma^\infty(T^\*M \otimes E) \to \Gamma^\infty(E)$.


I'm especially interested in the case where one drops the compatibility condition on $\nabla$ and the metric. The background for this question lies in the equation $\nabla^\* \nabla \phi = - \text{Tr}\_g (\nabla^2\_{\cdot, \cdot} \phi)$ for the Bochner Laplacian. I can see where it steams from for metric connections, but not for arbitary $\nabla$. Is this forumla just taken as a generalization of the compatible case?


*Question 2:* The word 'formal' refers to the fact that the adjoint is only a 'true' adjoint operator after extension to appropriate Hilbert spaces. Exists a definition of adjointness on Fréchet spaces such that one can view $\nabla^\*$ as a true adjoint operator? (Assume that $M$ is compact so that the section spaces are Fréchet spaces). I had the idea to represent $\Gamma^\infty(E)$ as the projective limit of Hilbert spaces and then piece the the adjoints on the building blocks together, but failed working out the details.



Formal adjoint of the covariant derivative
**Ad 1:** Yes, there is. The formula is
$$\nabla^\*(X^\flat \otimes u) = - \nabla\_X u -\mathrm{div}(X) \cdot u,$$
as can easily seen by local computation. Here, $X$ is a vector field and $X^\flat$ is the dual one form w.r.t. the metric.


Note that unless you have a scalar product on the bundle $E$, too, the dual operator will be an operator $\Gamma^\infty(T^\*M \otimes E^\*) \longrightarrow \Gamma^\infty(E^\*)$. The above formula holds in both cases (depending on which case you are in, $u$ will be a section of $E$ or $E^\*$, and the connection on the right will be either your given connection on $E$ or the corresponding dual connection on $E^\*$ (this connection is always well-defined by the formula $(\nabla u)(s) = d(u(s)) - u(\nabla(s))$).


A similar formula holds even in the case that you don't have a metric on $M$, in which case the dual operator will send sections of $TM \otimes E^\* \otimes |\Lambda|$ to sections of $E^\* \otimes |\Lambda|$, where $|\Lambda|$ is the density bundle.


**Ad 2:** Let me first say that in my opinion, one uses the word "formal" only to indicate that one doesn't bother about any functional analytic meaning of the word adjoint whatsoever, and one does not automatically want to say that there has to be any setting where there is an "actual" adjoint.


However, to give a more constructive answer: Yes, this operator is always an adjoint operator in the functional analytic sense.


First, this is true for elliptic differential operators on manifolds. On compact manifolds, they have a unique closed extension, whose adjoint operator is given on smooth functions by the formal adjoint.


In general, consider for example the space $\mathscr{D}=\mathscr{D}(M, E)$ of test sections (which is not a Fréchet space unless $M$ is compact!). It can be embedded into its dual $\mathscr{D^\prime}(M, E^\*)$. Now any differential operator $A$ is a continuous operator on $\mathscr{D}$, and its adjoint $A^\prime$ is an operator on $\mathscr{D}^\prime$. However, on the elements $\mathscr{D}\subset \mathscr{D}^\prime$ $A^\prime$ acts just as the formal adjoint of $A$. In this sense, the formal adjoint is also an "actual" adjoint in some sense.


However, this is quite formal and not at all deep. It follows at once from the definitions of distributions etc.


