Exterior covariant derivative

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

Definition[edit]

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[edit]

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.

See also[edit]

Notes[edit]

  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).
  2. ^ Proof: We have:
    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.

References[edit]