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 $\mathrm{d}, \mathrm{d}^\nabla$, or just $\nabla$, 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.

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, let $(\omega^1,\ldots,\omega^n)$ be an orthonormal frame of $T^*M$ in a neighbourhood about $p$, and let $(s_1,\ldots, s_r)$ be an orthonormal frame of $E$ in a neighbourhood about $p$.

Claim: The adjoint of the exterior derivative in the $L^2$ norm is given by