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 $h$. 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 \[ \mathrm{d}^* = -e(\omega^i)^* \nabla_{e_i}. \]

Proof 1: We use Stokes's theorem. First, note that \[ e_i \langle e(\omega^i)\phi, \psi \rangle = \langle \nabla_{e_i} e(\omega^i)\phi, \psi\rangle + \langle e(\omega^i)\phi, \nabla_{e_i} \psi \rangle,\] and moreover that \[ \nabla_{e_i} e(\omega^i) \phi = e(\omega^i) \nabla_{e_i} \phi + e(\nabla_{e_i} \omega^i) \phi = = e(\omega^i) \nabla_{e_i} \phi + e(\nabla_{e_i} e_i) \phi. \] One therefore computes: \begin{align*} \langle \mathrm{d}\phi, \psi\rangle_{L^2} &= \int_M \langle \mathrm{d} \phi, \psi \rangle \mathrm{vol} \\ &= \int_M \langle e(\omega^i) \nabla_{e_i} \phi, \psi \rangle \mathrm{vol} \\ &= \int_M \left(e_i \langle e(\omega^i), \psi\rangle - \langle e(\nabla_{e_i} e_i) \phi, \psi\rangle - \langle e(\omega^i)\phi, \nabla_{e_i} \psi\rangle\right) \mathrm{vol}. \end{align*} The last term is the one we want to survive; so we hope to show that \[ \left(e_i \langle e(\omega^i), \psi\rangle - \langle e(\nabla_{e_i} e_i\rangle\right)\mathrm{vol} \] is exact.