Leibniz rule for induced connections on associated bundles

The induced covariant derivative on an associated vector bundle is a derivation with respect to multiplying sections by functions.
Leibniz rule for induced connections on associated bundles

Let π:PM\pi:P\to M be a with structure group GG, and let ω\omega be a on PP. Given a representation ρ:GGL(V)\rho:G\to \mathrm{GL}(V), form the associated vector bundle E:=P×ρVME:=P\times_\rho V\to M. The connection ω\omega induces a (covariant derivative)

:Γ(E)Ω1(M;E), \nabla:\Gamma(E)\to \Omega^1(M;E),

characterized by the usual horizontal-lift construction (or equivalently by local connection 1-forms obtained from ω\omega in trivializations).

Proposition (Leibniz rule)

For every smooth function fC(M)f\in C^\infty(M) and every section sΓ(E)s\in \Gamma(E), the induced covariant derivative satisfies

(fs)  =  dfs  +  fs. \nabla(fs) \;=\; df\otimes s \;+\; f\,\nabla s.

Equivalently, for every XX on MM,

X(fs)  =  X(f)s  +  fXs. \nabla_X(fs) \;=\; X(f)\,s \;+\; f\,\nabla_X s.

In particular, the induced \nabla is a bona fide connection on EE in the standard sense: it is R\mathbb R-linear in ss and is a first-order differential operator over multiplication by functions.

Examples

  1. Trivial bundle with matrix-valued 1-form. If P=M×GP=M\times G and E=M×VE=M\times V, the induced connection can be written as =d+ρ(A)\nabla = d + \rho_*(A) for a g\mathfrak g-valued 1-form AA. Then (fs)=d(fs)+ρ(A)fs=(df)s+f(ds+ρ(A)s), \nabla(fs)=d(fs)+\rho_*(A)\,fs=(df)s+f(ds+\rho_*(A)s), which is exactly the Leibniz rule.
  2. Associated line bundle. For a principal U(1)U(1)-bundle and the standard 1-dimensional representation, \nabla is the usual connection on a complex line bundle; the Leibniz identity reduces to the familiar product rule for covariant differentiation of functions times sections.
  3. Tangent bundle from the frame bundle. If PP is the frame bundle of TMTM and ω\omega is a principal connection on PP, the associated bundle for the defining representation is TMTM. The resulting \nabla on TMTM satisfies X(fY)=X(f)Y+fXY\nabla_X(fY)=X(f)Y+f\nabla_XY.