Let $(E, \nabla)$ be a vector bundle with connection over a manifold $M$. Denote by $\Omega^k(E)$ the set of $k$-forms valued in $E$: that is, \[ \Omega^k(E) = \Gamma(\Lambda^k T^* M \otimes E) = \Omega^k(M) \otimes \Gamma(E). \] (Note the inconsistency in notation: $M$ is a rank zero vector bundle over $M$, but $\Omega^k(M)$ does not mean the $M$-valued $k$-forms on $M$; it means the standard $k$-forms, and one really ought to write $\Omega^k(\mathbb{R} \times M)$. But this abuse is standard.) The last equality is not obvious. See section isomorphisms for more information.

The connection $\nabla$ can be thought of as a map $\Omega^0(E) \to \Omega^1(E)$. The exterior derivative associated to $\nabla$, written often as $\mathrm{d}$ or $\mathrm{d}^\nabla$ is a map $\Omega^k(E) \to \Omega^{k+1}(E)$ for each non-negative integer $k$ characterized by the following properties:

1. Restricted to $\Omega^0(E)$, one has $\mathrm{d}^\nabla = \nabla$
2. For a $k$-form $\omega$ and a section $\sigma$ of $E$, a Leibniz rule holds:

\[ \mathrm{d}^\nabla (\omega \otimes \sigma) = \mathrm{d}\omega \otimes \sigma + (-1)^k \omega \otimes \mathrm{d}^\nabla \sigma. \]

In this Leibniz rule, $\mathrm{d}\omega$ is the usual exterior derivative on forms.

In a local orthonormal frame $(e_i)_{i=1}^n$ for $TM$, the exterior derivative takes the following form: \[ \mathrm{d}\alpha = e^i \wedge \nabla_{e_i} \alpha. \] The crux of the proof is that $\nabla$ is torsion-free; one can first argue that this identity holds in a coordinate frame (so, for the Levi-Civita connection, $\nabla_{e_i} \nabla_{e_j} = -\nabla_{e_j} \nabla_{e_i}$) and then extend by $C^\infty(M)$-linearity as appropriate.

Suppose that $M$ is closed and equipped with a Riemannian metric $g$, and that $E$ is equipped with a compatible fibre metric $h$. Then there is an induced metric on each fibre of each bundle $\Lambda^k T^*M \otimes E$, and by integrating there is an induced $L^2$ metric on each $\Omega^k(E)$. Let $p \in M$ be any point, and $(e_i)_{i=1}^n$ be an orthonormal frame of $TM$ in a neighbourhood about $p$. Let $(\omega^i)_{i=1}^n$ be the dual frame of $T^*M$ about $p$.

Claim: The adjoint to the exterior derivative is given by \[ \mathrm{d}^* = -\omega(e_i)^* \nabla_{e_i}. \]

Proof: Take $\sigma \in \Omega^{k+1}(E)$ and $\tau \in \Omega^k(E)$. We compute: \begin{align*} (\mathrm{d}\tau, \sigma)_{L^2} &= (e(\omega^i) \nabla_{e_i} \tau, \sigma)_{L^2} \\ &= (\nabla_{e_i} \tau, e(\omega^i)^* \sigma)_{L^2} \end{align*} On the other hand, \begin{align*} (\tau, -\omega(e_i)^* \nabla_{e_i} \sigma)_{L^2} &= -(e(\omega^i) \tau, \nabla_{e_i} \sigma)_{L^2} \\ \end{align*} Now recall that by metric compatibility, \begin{align*} \nabla_{e_i} \langle e(\omega^i)\tau, \sigma\rangle &= \langle \nabla_{e_i} e(\omega^i)\tau, \sigma\rangle + \langle e(\omega^i)\tau, \nabla_{e_i}\sigma\rangle \\ &= \langle (\nabla_{e_i} \omega^i) \wedge \tau + e(\omega^i) \nabla_{e_i} \tau, \sigma\rangle + \langle e(\omega^i)\tau, \nabla_{e_i}\sigma\rangle. \end{align*} The left-hand side is a divergence; integrating (using round brackets everywhere for the $L^2$ norm, over a compact neighbourhood and multiplying by a suitable bump function to remove boundary terms), we have on a neighbourhood about $p$ that \begin{align*} ( e(\omega^i)\tau, \nabla_{e_i}\sigma) &= -((\nabla_{e_i} \omega^i) \wedge \tau,\sigma) - ( e(\omega^i) \nabla_{e_i} \tau, \sigma) \\ &= -( (\nabla_{e_i} \omega^i) \wedge \tau,\sigma)- ( \nabla_{e_i} \tau, e(\omega^i)^* \sigma), \end{align*}