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 P → M 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 to the horizontal subspace.

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 v_i are tangent vectors to P at u.

Suppose V is a representation of G; i.e., there is a Lie group homomorphism ρ : G → GL(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, ..., v_k) = ψ(hv₀, ..., hv_k).)

Example: if ω is the connection form on P, then Ω = Dω is called the curvature form of ω. Bianchi's second identity says the exterior covariant derivative of Ω is zero; i.e., DΩ = 0.

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

(\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)}).

Example: Bianchi's second identity (DΩ = 0) can be stated as: $d\Omega + \operatorname{ad}(\omega) \cdot \Omega = d\Omega + [\omega \wedge \Omega] = 0$ .

Unlike the usual exterior derivative, which squares to 0, we have: with F = ρ(Ω), for a tensorial zero-form ϕ,

D^2\phi=F \cdot \phi.

^[2]

In particular D² vanishes for a flat connection (i.e., 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 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

Notes

↑ 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 tangent vector, then $D \phi(X) = d\phi(X)$ and $\omega(X) = 0$ . For the general case, let X_i's be tangent vectors to P at some point such that some of X_i's are horizontal and the rest vertical. If X_i is vertical, we think of it as a Lie algebra element and then identify it with the fundamental vector field generated by it. If X_i is horizontal, we replace it with the horizontal lift of the vector field extending the pushforward πX_i. This way, we have extended X_i's to vector fields. Note the extension is such that we have: [X_i, X_j] = 0 if X_i is horizontal and X_j is vertical. Finally, by the invariant formula for exterior derivative, we have:
$D \phi(X_0, \dots, X_k) - d \phi(X_0, \dots, X_k) = {1 \over k+1} \sum_0^k (-1)^i \rho(\omega(X_i)) \phi(X_0, \dots, \widehat{X_i}, \dots, X_k)$ ,
which is $(\rho(\omega) \cdot \phi)(X_0, \cdots, X_k)$ .
↑ Proof: Since ρ acts on the constant part of ω, it commutes with d and thus
$d(\rho(\omega) \cdot \phi) = d(\rho(\omega)) \cdot \phi - \rho(\omega) \cdot d\phi = \rho(d \omega) \cdot \phi - \rho(\omega) \cdot d\phi$ .
Then, according to the example at Lie algebra-valued differential form § Operations,
$D^2 \phi = \rho(d \omega) \cdot \phi + \rho(\omega) \cdot (\rho(\omega) \cdot \phi) = \rho(d \omega) \cdot \phi + {1 \over 2} \rho([\omega \wedge \omega]) \cdot \phi,$
which is $\rho(\Omega) \cdot \phi$ by E. Cartan's structure equation.

References

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

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Exterior covariant derivative

Definition

Exterior covariant derivative for vector bundles

See also

Notes

References