# 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; this yields a natural direct sum decomposition $T_{u}P=H_{u}\oplus V_{u}$  of each tangent space into the horizontal and vertical subspaces. Let $h:T_{u}P\to H_{u}$  be the projection to the horizontal subspace.

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

$D\phi (v_{0},v_{1},\dots ,v_{k})=d\phi (hv_{0},hv_{1},\dots ,hv_{k})$

where vi are tangent vectors to P at u.

Suppose that ρ : G → GL(V) is a representation of G on a vector space V. If ϕ is equivariant in the sense that

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

where $R_{g}(u)=ug$ , then 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).)

By abuse of notation, the differential of ρ at the identity element may again be denoted by ρ:

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

Let $\omega$  be the connection one-form and $\rho (\omega )$  the representation of the connection in ${\mathfrak {gl}}(V).$  That is, $\rho (\omega )$  is a ${\mathfrak {gl}}(V)$ -valued form, vanishing on the horizontal subspace. If ϕ is a tensorial k-form of type ρ, then

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

where, following the notation in Lie algebra-valued differential form § Operations, we wrote

$(\rho (\omega )\cdot \phi )(v_{1},\dots ,v_{k+1})={1 \over (1+k)!}\sum _{\sigma }\operatorname {sgn} (\sigma )\rho (\omega (v_{\sigma (1)}))\phi (v_{\sigma (2)},\dots ,v_{\sigma (k+1)}).$

Unlike the usual exterior derivative, which squares to 0, the exterior covariant derivative does not. In general, one has, for a tensorial zero-form ϕ,

$D^{2}\phi =F\cdot \phi .$ 

where F = ρ(Ω) is the representation[clarification needed] in ${\mathfrak {gl}}(V)$  of the curvature two-form Ω. The form F is sometimes referred to as the field strength tensor, in analogy to the role it plays in electromagnetism. Note that D2 vanishes for a flat connection (i.e. when Ω = 0).

If ρ : G → GL(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 : G → GL(V) is a representation, one can form the associated bundle E = P ×ρ V. Then the exterior covariant derivative D given by a connection on P induces an exterior covariant derivative (sometimes called the exterior connection) on the associated bundle, this time using the nabla symbol:

$\nabla :\Gamma (M,E)\to \Gamma (M,T^{*}M\otimes E)$

Here, Γ denotes the space of local sections of the vector bundle. The extension is made 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 form; thus, it is given on decomposable elements of the space $\Omega ^{k}(M;E)=\Gamma (\Lambda ^{k}(T^{*}M)\otimes E)$  of $E$ -valued k-forms by

$\nabla (\omega \otimes s)=(d\omega )\otimes s+(-1)^{k}\omega \wedge \nabla s\in \Omega ^{k+1}(M;E)$ .

For a section s of E, we also set

$\nabla _{X}s=i_{X}\nabla s$

where $i_{X}$  is the contraction by X.

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

$-2F(X,Y)s=\left([\nabla _{X},\nabla _{Y}]-\nabla _{[X,Y]}\right)s$

which can be easily recognized as the definition of the Riemann curvature tensor on Riemannian manifolds.

## Examples

• If ω is the connection form on P, then Ω = is called the curvature form of ω.
• Bianchi's second identity, which says that the exterior covariant derivative of Ω is zero (that is, DΩ = 0) can be stated as: $d\Omega +\operatorname {ad} (\omega )\cdot \Omega =d\Omega +[\omega \wedge \Omega ]=0$ .