Vector-valued differential form: Difference between revisions

Revision as of 11:18, 26 January 2021

Revision as of 22:50, 21 September 2021

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Here ''R''_''g'' denotes the right action of ''G'' on ''P'' for some ''g'' ∈ ''G''. Note that for 0-forms the second condition is [[vacuously true]].

*Example: If ρ is the [[adjoint representation]] of ''G'' on the Lie algebra, then the connection form ω satisfies the first condition (but not the second). The associated [[curvature form]] Ω satisfies both; hence Ω is a tensorial form of adjoint type. The "difference" of two connection forms is a tensorial form.

Given ''P'' and ''ρ'' as above one can construct the [[associated vector bundle]] ''E'' = ''P'' &times;_''ρ'' ''V''. Tensorial ''q''-forms on ''P'' are in a natural one-to-one correspondence with ''E''-valued ''q''-forms on ''M''. As in the case of the principal bundle F(''E'') above, given a ''q''-form <math>\overline{\phi}</math> on ''M'' with values in ''E'', define φ on ''P'' fiberwise by, say at ''u'',

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:<math>\Gamma(M, E) \simeq \{ f: P \to V | f(ug) = \rho(g)^{-1}f(u) \}, \, \overline{f} \leftrightarrow f</math>.

*Example: Let ''E'' be the tangent bundle of ''M''. Then identity bundle map id_''E'': ''E'' →''E'' is an ''E''-valued one form on ''M''. The [[tautological one-form]] is a unique one-form on the frame bundle of ''E'' that corresponds to id_''E''. Denoted by θ, it is a tensorial form of standard type.<!--Mention this somewhere else: The [[exterior covariant derivative]] of θ, Θ = ''D''θ is called a [[torsion form]].-->

Now, suppose there is a connection on ''P'' so that there is an [[exterior covariant differentiation]] ''D'' on (various) vector-valued forms on ''P''. Through the above correspondence, ''D'' also acts on ''E''-valued forms: define ∇ by

:<math>\nabla \overline{\phi} = \overline{D \phi}.</math>

In particular for zero-forms,

:<math>\nabla: \Gamma(M, E) \to \Gamma(M, T^*M \otimes E)</math>.

This is exactly the [[covariant derivative]] for the [[connection (vector bundle)|connection on the vector bundle ''E'']].<ref>Proof: <math>D (f\phi) = Df \otimes \phi + f D\phi</math> for any scalar-valued tensorial zero-form ''f'' and any tensorial zero-form φ of type ρ, and ''Df'' = ''df'' since ''f'' descends to a function on ''M''; cf. this [[Chern–Weil homomorphism#Definition of the homomorphism|Lemma 2]].</ref>