# Exterior covariant derivative

In mathematics, the exterior covariant derivative is an analog of an exterior derivative that takes into account the presence of a connection.

## Definition

Let G be a Lie group and PM be a principal G-bundle on a smooth manifold M. Suppose there is a connection on P so that it gives a natural direct sum decomposition of each tangent space $T_u P = H_u \oplus V_u$ into the horizontal and vertical subspaces. Let $h: T_u P \to H_u$ be the projection.

If ϕ is a k-form on P with values in a vector space V, then its exterior covariant derivative Dϕ is a form defined by

$D\phi(v_0, v_1,\dots, v_k)= d \phi(h v_0 ,h v_1,\dots, h v_k)$

where vi are tangent vectors to P at u.

Suppose V is a representation of G; i.e., there is a Lie group homomorphism ρ: GGL(V). If φ is equivariant in the sense:

$R_g^* \phi = \rho(g)^{-1}\phi$

where $R_g(u) = ug$, then Dϕ is a tensorial (k + 1)-form on P of the type ρ: it is equivariant and horizontal (a form ψ is horizontal if ψ(v0, …, vk) = ψ(hv0, …, hvk).)

We also denote the differential of ρ at the identity element by ρ:

$\rho: \mathfrak{g} \to \mathfrak{gl}(V).$

If φ is a tensorial k-form of type ρ, then

$D \phi = d \phi + \rho(\omega) \cdot \phi,$[1]

where $\rho(\omega)$ is a $\mathfrak{gl}(V)$-valued form, and

$(\rho(\omega) \cdot \phi)(v_1, \dots, v_{k+1}) = 1/{(k+1)}! \sum_{\sigma} \operatorname{sgn}(\sigma)\rho(\omega(v_{\sigma(1)})) \phi(v_{\sigma(2)}, \dots, v_{\sigma(k+1)}).$
• Example: Bianchi's second identity (DΩ = 0) can be stated as: $d\Omega + \operatorname{ad}(\omega) \cdot \Omega = 0$.

Unlike the usual exterior derivative, which squares to 0 (that is d2 = 0), we have:

$D^2\phi=F \cdot \phi,$[2]

where F = ρ(Ω). In particular D2 vanishes for a flat connection (i.e., Ω = 0).

If ρ: GGL(Rn), then one can write

$\rho(\Omega) = F = \sum {F^i}_j {e^j}_i$

where ${e^i}_j$ is the matrix with 1 at the (i, j)-th entry and zero on the other entries. The matrix ${F^i}_j$ whose entries are 2-forms on P is called the curvature matrix.

## Exterior covariant derivative for vector bundles

When ρ: GGL(V) is a representation, one can form the associated bundle E = Pρ V. Then the exterior covariant differentiation D given by a connection on P defines

$\nabla: \Gamma(M, E) \to \Gamma(M, TM \otimes E)$

through the correspondence between E-valued forms and tensorial forms of type ρ (see tensorial forms on principal bundles.) Requiring ∇ to satisfy Leibniz's rule, ∇ also acts on any E-valued forms. This ∇ is called the exterior covariant differentiation on E. One also sets: for a section s of E,

$\nabla_X s = i_X \nabla s$

where $i_X$ is the contraction by X. Explicitly,

$\nabla_X s = (hX) s$

since $\nabla_X \overline{\phi} = \overline{D \phi (X)} = \overline{d \phi (hX)} = (hX)s$ when $s = \overline{\phi}$.

Conversely, given a vector bundle E, one can take its frame bundle, which is a principal bundle, and so gets an exterior covariant differentiation on E (depending on a connection). Identifying tensorial forms and E-valued forms, there is, for example,

$2F(X, Y) s = (-[\nabla_X, \nabla_Y] + \nabla_{[X, Y]}) s$.

1. ^ If k = 0, then, writing $X^{\#}$ for the fundamental vector field (i.e., vertical vector field) generated by X in $\mathfrak{g}$ on P, we have:
$d \phi(X^{\#}_u) = {d \over dt}|_0 \phi(u \operatorname{exp}(tX)) = -\rho(X)\phi(u) = -\rho(\omega(X^{\#}_u))\phi(u)$,
since φ(gu) = ρ(g-1)φ(u). On the other hand, Dφ(X#) = 0. If X is a horizontal vector field, then $D \phi(X) = d\phi(X)$ and $\omega(X) = 0$. In general, by the invariant formula for exterior derivative, we have: for any vector fields Xi's, since φ takes the same values at hXi's and Xi's,
\begin{align} &D \phi(X_0, \dots, X_k) - d \phi(X_0, \dots, X_k) = {1 \over k+1} \sum_0^k (-1)^i \rho(\omega(X_i)) \phi(X_0, \dots, \widehat{X_i}, \dots, X_k) \\ &= {1 \over (k+1)!} \sum_0^k (-1)^i \rho(\omega(X_i)) \sum_{\sigma: \{ 0, \dots, \widehat{i}, \dots, k \}} \operatorname{sgn}(\sigma) \phi(X_{\sigma(0)}, \dots, \widehat{X_{\sigma(i)}}, \dots, X_{\sigma(k)}) \end{align}
where the hat means the term is omitted. This equals $(\rho(\omega) \cdot \phi)(X_0, \cdots, X_k)$.
$D^2 \phi = \rho(d \omega) \cdot \phi + \rho(\omega) \cdot (\rho(\omega) \cdot \phi) = \rho(d \omega) \cdot \phi + {1 \over 2} \rho([\omega \wedge \omega]) \cdot \phi,$
(cf. the example at Lie algebra-valued differential form#Operations), which is $\rho(\Omega) \cdot \phi$ by E. Cartan's structure equation.