Covariant derivative on an associated representation bundle

On a principal chart, each $p\in {\pi }^{-1}(U_i)$ has a unique expression $p=s_i(x)h$. The associated-bundle chart therefore assigns $[s_i(x)h,v]$ the vector coordinate $\rho (h)v$. In particular $[s_i(x),{\psi }_i(x)]$ determines a unique smooth component ${\psi }_i$. If $s_j=s_i{g}_{ij}$, then $[s_j,{\psi }_j]=[s_i,\rho (g_{ij}){\psi }_j]$; equality with $[s_i,{\psi }_i]$ gives ${\psi }_j=\rho (g_{ij}^{-1}){\psi }_i$.

In a finite basis of $V$, ${\mathrm{GL}}_{\mathbb{C}}(V)$ is the open subset where the determinant is nonzero, so its tangent space at the identity is ${\mathrm{End}}_{\mathbb{C}}(V)$. Thus the differential ${\rho }_{\ast }$ is real-linear by [F6]. For fixed $g$, [F9] gives $\rho \circ C_{g^{-1}}=C_{\rho (g^{-1})}\circ \rho$. The derivative of target conjugation is the linear map $T\mapsto \rho (g^{-1})T\rho (g)$, while the derivative of source conjugation is ${\mathrm{Ad}}_{g^{-1}}$ by [F7]. The chain rule [F6] therefore gives ${\rho }_{\ast }({\mathrm{Ad}}_{g^{-1}}Z)=\rho (g^{-1}){\rho }_{\ast }(Z)\rho (g)$. For a smooth transition $g_{ij}:U_i\cap U_j\to G$, differentiation of $s_j=s_i{g}_{ij}$ decomposes a tangent vector into the right translate of the tangent from $s_i$ and the fundamental vertical vector with generator ${\theta }_L(g_{ij})(X)$, by [F5]. Reproduction and equivariance in [F4] yield $A_j={\mathrm{Ad}}_{g_{ij}^{-1}}{A}_i+{\theta }_L(g_{ij})$.

Put $g=g_{ij}(x)$, $Z={\theta }_L(g)(X)$, and $R=\rho (g^{-1})$. Let $c$ be a smooth curve through $x$ with $c^{\mathrm{\prime }}(0)=X$, and write $h(t)=g_{ij}(c(t))$. By [F5], $h(0)=g$ and $h^{\mathrm{\prime }}(0)=d(L_g{)}_{e}Z$. For each $t\in \mathbb{R}$, $s\mapsto {\gamma }_Z(ts)$ is a one-parameter subgroup with initial tangent $tZ$, so uniqueness in [F8] and the exponential definition give ${\exp }_G(tZ)={\gamma }_Z(t)$. Hence $k(t):=g{\exp }_G(tZ)=g{\gamma }_Z(t)$ has the same initial value and tangent as $h$. By [F6], differentiating the smooth map $q\mapsto \rho (q^{-1})$ along these curves gives the same derivative at zero. Since $k(t{)}^{-1}={\gamma }_Z(-t)g^{-1}$, the homomorphism law [F9] and ${\gamma }_Z^{\mathrm{\prime }}(0)=Z$ yield $dR(X)=-{\rho }_{\ast }(Z)R$. Using step 1.1, step 2.1, and the adjoint-intertwining identity in step 2.1, $d{\psi }_j(X)+{\rho }_{\ast }(A_j(X)){\psi }_j=-{\rho }_{\ast }(Z)R{\psi }_i+Rd{\psi }_i(X)+{\rho }_{\ast }({\mathrm{Ad}}_{g^{-1}}{A}_i(X))R{\psi }_i+{\rho }_{\ast }(Z)R{\psi }_i=R(d{\psi }_i(X)+{\rho }_{\ast }(A_i(X)){\psi }_i)$. The two Maurer–Cartan terms cancel, and the remaining equality uses $\rho (g)R={\mathrm{i}\mathrm{d}}_V$. This is precisely the associated-bundle coordinate transition for an $E$-valued 1-form, so the local expressions define one global smooth $E$-valued 1-form and do not depend on the selected sections.

In any local chart, multiplication by a scalar function obeys $d(f{\psi }_i)=df\otimes {\psi }_i+fd{\psi }_i$; the endomorphism term obeys ${\rho }_{\ast }(A_i)(f{\psi }_i)=f{\rho }_{\ast }(A_i){\psi }_i$ by pointwise linearity. Adding these identities gives $({\mathrm{\nabla }}^{\omega }(f\sigma ){)}_i=df\otimes {\psi }_i+f({\mathrm{\nabla }}^{\omega }\sigma {)}_i$, which is the asserted Leibniz rule in the associated bundle. If $V=0$, every term is zero and the same formulas hold; the argument is pointwise and also applies on an empty base.