My question is in relation to the derivation of the formal adjoint for a connection $D:\Omega^{p-1}(\text{Ad}E)\rightarrow \Omega^p(\text{Ad}E)$ - I am reading through this derivation in Jost's Riemannian Geometry and Geometric Analysis and I have a few questions.
The formal adjoint is defined as the differential operator $D^*:\Omega^{p}(\text{Ad}E)\rightarrow\Omega^{p-1}(\text{Ad}E)$ satisfying
\begin{align*} (D\mu,\nu)=\int_M\langle D\mu,\nu\rangle d\text{vol}&=\int_M\langle u,D^*\nu\rangle d\text{vol}=(u,D^*\nu) \end{align*}
Where $\mu\in \Omega^{p-1}(\text{Ad}E)$ and $\nu\in\Omega^{p}(\text{Ad}E)$.
We write our connection $D$ as $D=d+A$ where $d$ is the exterior derivative, $A\in\Omega^1(\text{Ad}E)$. The author writes \begin{align*} (\nu,D\mu)=(\nu,(d+A)\mu)&=(\nu,d\mu+A_idx^i\wedge \mu)=(d^*\nu)-(A_i\nu,dx^i\wedge \mu). \end{align*} I have some questions about how Jost arrives at the above equation.
First, given that $\mu\in \Omega^{p-1}(\text{Ad}E)=\Gamma(\text{Ad}E)\otimes \Omega^{p-1}(M)$, this means we would write $\mu$ as \begin{align*} \mu=a^j\mu_j\otimes b_I dx^I, \end{align*} for a basis of sections $(u_j)$ of Ad$E$ and increasing multiindices $I$. Then, $D\mu$ is defined as \begin{align*} D\mu=D(a^j\mu_j\otimes b_Idx^I)&=D(a^j\mu_j)\wedge b_Idx^I+a^j\mu_j\otimes d(b_Idx^I)\\ &=d(a^j)\mu_j\wedge b_I dx^I+a^j\mu_kA^k_{ji}dx^i\wedge b_Idx^I+a^j\mu_j\otimes d(b_Idx^I)\\ &=d\mu+A_idx^I\wedge\mu+a^j\mu_j\otimes d(b_Idx^I) \end{align*} But this does not agree with what Jost says $D\mu$ should be which is just $d\mu+A_idx^i\wedge \mu$. That is, I have the extra term on the end. However, surely this term must be there since this is how a connection is extended from the operator $\Gamma(\text{Ad}E)\rightarrow \Gamma(\text{Ad}E)\otimes\Gamma(T^*M)$ to the operator $\Omega^{p-1}(\text{Ad}E)\rightarrow\Omega^{p}(\text{Ad}E)$?
Can anyone explain what is going on here?
The problem is how you're treating $d\mu$ in the calculation. Since everything's local, we have a trivial bundle, and need to use the notion of exterior derivative of a vector-valued form. See this link https://en.wikipedia.org/wiki/Vector-valued_differential_form#Exterior_derivative for discussion.
Basically you fix the basis $\mu_j$ and think of the other stuff as coefficients and do $d$ on that. Which means the calculation should of $d\mu$ should look like this:
$d\mu=d(a^j\mu_j\otimes b_I dx^I)=d(\mu_j\otimes a^jb_Idx^I)=\mu_j\otimes d(a^jb_Idx^I)=\mu_j\otimes d(a^j)\wedge b_Idx^I+\mu_j\otimes a^jd(b_Idx^I)=d(a^j)\mu_j\wedge b_Idx^I+a^j\mu_j\otimes d(b_Idx^I)$
This corroborates Jost's calculation.