Exterior covariant derivative on a vector bundle

Let $E\to M$ be a smooth vector bundle with a connection $\nabla$ . The exterior covariant derivative is the unique linear operator of degree one on $E$ -valued forms,

d_\nabla:\Omega^k(M;E)\longrightarrow\Omega^{k+1}(M;E),

whose action on a local scalar $k$ -form $\alpha$ and a local section $s$ is

d_\nabla(\alpha\otimes s) =d\alpha\otimes s+(-1)^k\alpha\wedge\nabla s.

Here $d$ is the exterior derivative, $\nabla s$ is viewed as an $E$ -valued one-form, and the wedge product acts on the scalar-form factors. In degree zero this gives $d_\nabla s=\nabla s$ . Linearity is over $\mathbb R$ , or over $\mathbb C$ for a complex connection.

Well-definedness and local formula

The connection Leibniz identity makes the formula compatible with the equality $(f\alpha)\otimes s=\alpha\otimes(fs)$ . Local frames then give existence and uniqueness, and the locally defined operators agree on overlaps.

If $\nabla=d+A$ in a local frame and $\omega$ is a column of coefficient forms, then

d_\nabla\omega=d\omega+A\wedge\omega.

Consequently $d_\nabla^2\omega=F_\nabla\wedge\omega$ , where $F_\nabla=dA+A\wedge A$ is the curvature matrix. The operator need not square to zero.

Examples and principal bundles

For the trivial real line with its product connection, $d_\nabla=d$ . For a line bundle with local connection one-form $A$ , it is $d+A\wedge\cdot$ .

A principal connection induces this operator on associated vector bundles. Under the correspondence with tensorial forms on the principal bundle, it agrees with the horizontal-projection definition of exterior covariant differentiation.