BREAKING NEWS
Exterior covariant derivative

## Summary

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 ${\displaystyle T_{u}P=H_{u}\oplus V_{u}}$ of each tangent space into the horizontal and vertical subspaces. Let ${\displaystyle h:T_{u}P\to H_{u}}$ 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

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

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

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

where ${\displaystyle R_{g}(u)=ug}$, 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:

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

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

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

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

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

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,

${\displaystyle D^{2}\phi =F\cdot \phi .}$[2]

where F = ρ(Ω) is the representation[clarification needed] in ${\displaystyle {\mathfrak {gl}}(V)}$ 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

${\displaystyle \rho (\Omega )=F=\sum {F^{i}}_{j}{e^{j}}_{i}}$

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

## 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:

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

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 Vector-valued differential forms § Basic or 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 ${\displaystyle \Omega ^{k}(M;E)=\Gamma \left(\Lambda ^{k}(T^{*}M)\otimes E\right)}$ of ${\displaystyle E}$-valued k-forms by

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

For a section s of E, we also set

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

where ${\displaystyle i_{X}}$ 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

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

which can be seen to be a generalization of the Riemann curvature tensor on Riemannian manifolds.

## Example

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

## Notes

1. ^ If k = 0, then, writing ${\displaystyle X^{\#}}$ for the fundamental vector field (i.e., vertical vector field) generated by X in ${\displaystyle {\mathfrak {g}}}$ on P, we have:
${\displaystyle d\phi \left(X_{u}^{\#}\right)=\left.{d \over dt}\right\vert _{0}\phi (u\operatorname {exp} (tX))=-\rho (X)\phi (u)=-\rho \left(\omega \left(X_{u}^{\#}\right)\right)\phi (u)}$,
since ϕ(gu) = ρ(g−1)ϕ(u). On the other hand, (X#) = 0. If X is a horizontal tangent vector, then ${\displaystyle D\phi (X)=d\phi (X)}$ and ${\displaystyle \omega (X)=0}$. For the general case, let Xi's be tangent vectors to P at some point such that some of Xi's are horizontal and the rest vertical. If Xi is vertical, we think of it as a Lie algebra element and then identify it with the fundamental vector field generated by it. If Xi is horizontal, we replace it with the horizontal lift of the vector field extending the pushforward πXi. This way, we have extended Xi's to vector fields. Note the extension is such that we have: [Xi, Xj] = 0 if Xi is horizontal and Xj is vertical. Finally, by the invariant formula for exterior derivative, we have:
${\displaystyle D\phi (X_{0},\dots ,X_{k})-d\phi (X_{0},\dots ,X_{k})={1 \over k+1}\sum _{0}^{k}(-1)^{i}\rho \left(\omega \left(X_{i}\right)\right)\phi \left(X_{0},\dots ,{\widehat {X_{i}}},\dots ,X_{k}\right)}$,
which is ${\displaystyle (\rho (\omega )\cdot \phi )(X_{0},\cdots ,X_{k})}$.
2. ^ Proof: Since ρ acts on the constant part of ω, it commutes with d and thus
${\displaystyle d(\rho (\omega )\cdot \phi )=d(\rho (\omega ))\cdot \phi -\rho (\omega )\cdot d\phi =\rho (d\omega )\cdot \phi -\rho (\omega )\cdot d\phi }$.
Then, according to the example at Lie algebra-valued differential form § Operations,
${\displaystyle D^{2}\phi =\rho (d\omega )\cdot \phi +\rho (\omega )\cdot (\rho (\omega )\cdot \phi )=\rho (d\omega )\cdot \phi +{1 \over 2}\rho ([\omega \wedge \omega ])\cdot \phi ,}$
which is ${\displaystyle \rho (\Omega )\cdot \phi }$ by E. Cartan's structure equation.

## References

• Kobayashi, Shoshichi; Nomizu, Katsumi (1996). Foundations of Differential Geometry, Vol. 1 (New ed.). Wiley-Interscience. ISBN 0-471-15733-3.