Exterior covariant derivative

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Template:Short description Script error: No such module "Labelled list hatnote". In the mathematical field of differential geometry, the exterior covariant derivative is an extension of the notion of exterior derivative to the setting of a differentiable principal bundle or vector bundle with 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 TuP=HuVu of each tangent space into the horizontal and vertical subspaces. Let h:TuPHu 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

Dϕ(v0,v1,,vk)=dϕ(hv0,hv1,,hvk)

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

Rg*ϕ=ρ(g)1ϕ

where Rg(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:

ρ:𝔤𝔤𝔩(V).

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

Dϕ=dϕ+ρ(ω)ϕ,[1]

where, following the notation in Template:Section link, we wrote

(ρ(ω)ϕ)(v1,,vk+1)=1(1+k)!σsgn(σ)ρ(ω(vσ(1)))ϕ(vσ(2),,vσ(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,

D2ϕ=Fϕ.[2]

where F = ρ(Ω) is the representationScript error: No such module "Unsubst". in 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

ρ(Ω)=F=Fijeji

where eij is the matrix with 1 at the (i, j)-th entry and zero on the other entries. The matrix Fij whose entries are 2-forms on P is called the curvature matrix.

For vector bundles

Given a smooth real vector bundle EMScript error: No such module "Check for unknown parameters". with a connection Script error: No such module "Check for unknown parameters". and rank Template:Mvar, the exterior covariant derivative is a real-linear map on the vector-valued differential forms that are valued in Template:Mvar:

d:Ωk(M,E)Ωk+1(M,E).

The covariant derivative is such a map for k = 0Script error: No such module "Check for unknown parameters".. The exterior covariant derivatives extends this map to general Template:Mvar. There are several equivalent ways to define this object:

  • Template:Sfnm Suppose that a vector-valued differential 2-form is regarded as assigning to each Template:Mvar a multilinear map sp: TpM × TpMEpScript error: No such module "Check for unknown parameters". which is completely anti-symmetric. Then the exterior covariant derivative d sScript error: No such module "Check for unknown parameters". assigns to each Template:Mvar a multilinear map TpM × TpM × TpMEpScript error: No such module "Check for unknown parameters". given by the formula
x1(s(X2,X3))x2(s(X1,X3))+x3(s(X1,X2))s([X1,X2],x3)+s([X1,X3],x2)s([X2,X3],x1).
where x1, x2, x3Script error: No such module "Check for unknown parameters". are arbitrary tangent vectors at Template:Mvar which are extended to smooth locally-defined vector fields X1, X2 X3Script error: No such module "Check for unknown parameters".. The legitimacy of this definition depends on the fact that the above expression depends only on x1, x2, x3Script error: No such module "Check for unknown parameters"., and not on the choice of extension. This can be verified by the Leibniz rule for covariant differentiation and for the Lie bracket of vector fields. The pattern established in the above formula in the case k = 2Script error: No such module "Check for unknown parameters". can be directly extended to define the exterior covariant derivative for arbitrary Template:Mvar.
  • Template:Sfnm The exterior covariant derivative may be characterized by the axiomatic property of defining for each Template:Mvar a real-linear map Ωk(M, E) → Ωk + 1(M, E)Script error: No such module "Check for unknown parameters". which for k = 0Script error: No such module "Check for unknown parameters". is the covariant derivative and in general satisfies the Leibniz rule
d(ωs)=(dω)s+(1)kω(ds)
for any differential Template:Mvar-form ωScript error: No such module "Check for unknown parameters". and any vector-valued form Template:Mvar. This may also be viewed as a direct inductive definition. For instance, for any vector-valued differential 1-form Template:Mvar and any local frame e1, ..., erScript error: No such module "Check for unknown parameters". of the vector bundle, the coordinates of Template:Mvar are locally-defined differential 1-forms ω1, ..., ωrScript error: No such module "Check for unknown parameters".. The above inductive formula then says thatTemplate:Sfnm
ds=d(ω1e1++ωrer)=(dω1)e1++(dωr)erω1e1ωrer.
In order for this to be a legitimate definition of dsScript error: No such module "Check for unknown parameters"., it must be verified that the choice of local frame is irrelevant. This can be checked by considering a second local frame obtained by an arbitrary change-of-basis matrix; the inverse matrix provides the change-of-basis matrix for the 1-forms ω1, ..., ωrScript error: No such module "Check for unknown parameters".. When substituted into the above formula, the Leibniz rule as applied for the standard exterior derivative and for the covariant derivative Script error: No such module "Check for unknown parameters". cancel out the arbitrary choice.
  • Template:Sfnm A vector-valued differential 2-form Template:Mvar may be regarded as a certain collection of functions sαijScript error: No such module "Check for unknown parameters". assigned to an arbitrary local frame of Template:Mvar over a local coordinate chart of Template:Mvar. The exterior covariant derivative is then defined as being given by the functions
(ds)αijk=isαjkjsαik+ksαij.
The fact that this defines a tensor field valued in Template:Mvar is a direct consequence of the same fact for the covariant derivative. The further fact that it is a differential 3-form valued in Template:Mvar asserts the full anti-symmetry in i, j, kScript error: No such module "Check for unknown parameters". and is directly verified from the above formula and the contextual assumption that Template:Mvar is a vector-valued differential 2-form, so that sαij = −sαjiScript error: No such module "Check for unknown parameters".. The pattern in this definition of the exterior covariant derivative for k = 2Script error: No such module "Check for unknown parameters". can be directly extended to larger values of Template:Mvar.
This definition may alternatively be expressed in terms of an arbitrary local frame of Template:Mvar but without considering coordinates on Template:Mvar. Then a vector-valued differential 2-form is expressed by differential 2-forms s1, ..., srScript error: No such module "Check for unknown parameters". and the connection is expressed by the connection 1-forms, a skew-symmetric r × rScript error: No such module "Check for unknown parameters". matrix of differential 1-forms θαβScript error: No such module "Check for unknown parameters".. The exterior covariant derivative of Template:Mvar, as a vector-valued differential 3-form, is expressed relative to the local frame by Template:Mvar many differential 3-forms, defined by
(ds)α=d(sα)+θβαsβ.

In the case of the trivial real line bundle ℝ × MMScript error: No such module "Check for unknown parameters". with its standard connection, vector-valued differential forms and differential forms can be naturally identified with one another, and each of the above definitions coincides with the standard exterior derivative.

Given a principal bundle, any linear representation of the structure group defines an associated bundle, and any connection on the principal bundle induces a connection on the associated vector bundle. Differential forms valued in the vector bundle may be naturally identified with fully anti-symmetric tensorial forms on the total space of the principal bundle. Under this identification, the notions of exterior covariant derivative for the principal bundle and for the vector bundle coincide with one another.Template:Sfnm

The curvature of a connection on a vector bundle may be defined as the composition of the two exterior covariant derivatives Ω0(M, E) → Ω1(M, E)Script error: No such module "Check for unknown parameters". and Ω1(M, E) → Ω2(M, E)Script error: No such module "Check for unknown parameters"., so that it is defined as a real-linear map F: Ω0(M, E) → Ω2(M, E)Script error: No such module "Check for unknown parameters".. It is a fundamental but not immediately apparent fact that F(s)p: TpM × TpMEpScript error: No such module "Check for unknown parameters". only depends on s(p)Script error: No such module "Check for unknown parameters"., and does so linearly. As such, the curvature may be regarded as an element of Ω2(M, End(E))Script error: No such module "Check for unknown parameters".. Depending on how the exterior covariant derivative is formulated, various alternative but equivalent definitions of curvature (some without the language of exterior differentiation) can be obtained.

It is a well-known fact that the composition of the standard exterior derivative with itself is zero: d(dω) = 0Script error: No such module "Check for unknown parameters".. In the present context, this can be regarded as saying that the standard connection on the trivial line bundle ℝ × MMScript error: No such module "Check for unknown parameters". has zero curvature.

Some equations involving covariant derivative can be locally solved using Chen's iterated integrals[3] or using approach based on linear homotopy operator.[4]

Example

  • Bianchi's second identity, which says that the exterior covariant derivative of Ω is zero (that is, DΩ = 0) can be stated as: dΩ+ad(ω)Ω=dΩ+[ωΩ]=0.

Notes

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  1. If k = 0, then, writing X# for the fundamental vector field (i.e., vertical vector field) generated by X in 𝔤 on P, we have:
    dϕ(Xu#)=ddt|0ϕ(uexp(tX))=ρ(X)ϕ(u)=ρ(ω(Xu#))ϕ(u),
    since ϕ(gu) = ρ(g−1)ϕ(u). On the other hand, (X#) = 0. If X is a horizontal tangent vector, then Dϕ(X)=dϕ(X) and ω(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:
    Dϕ(X0,,Xk)dϕ(X0,,Xk)=1k+10k(1)iρ(ω(Xi))ϕ(X0,,Xi^,,Xk),
    which is (ρ(ω)ϕ)(X0,,Xk).
  2. Proof: Since ρ acts on the constant part of ω, it commutes with d and thus
    d(ρ(ω)ϕ)=d(ρ(ω))ϕρ(ω)dϕ=ρ(dω)ϕρ(ω)dϕ.
    Then, according to the example at Template:Section link,
    D2ϕ=ρ(dω)ϕ+ρ(ω)(ρ(ω)ϕ)=ρ(dω)ϕ+12ρ([ωω])ϕ,
    which is ρ(Ω)ϕ by E. Cartan's structure equation.
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References

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