Let $(P,\pi,M;G)$ be a principal bundle with connection form $A\in\mathcal{C}(P)$ and
let $\rho:G\rightarrow\mathrm{GL}(V)$ be a representation of $G$ on some finite-dimensional vector space $V$. From these data we can construct an associated vector bundle
$E=P\times_\rho V$ with typical fibre $V$. Using the parallel transport induced by $A$ on
$E$ we can define a covariant derivative
$$\nabla^A:\Gamma(E)\longrightarrow\Omega^1(M,E)$$
on $E$. There is also another (equivalent) way to introduce this covariant derivative: start with equation (1) (derived in the following) and show its invariance under a change of local section. $\textbf{This is where I am stuck}$. See below for details.
Let $s:U\rightarrow P$ be a local section of $P$ and $\Phi:U\rightarrow E$ a local section of $E$. Then we can find a smooth map $\phi:U\rightarrow V$ such that $\Phi(x)=[s(x),\phi(x)]$ on $U$. We can also pull back the connection $A$ to $A_s=s^\ast A\in\Omega^1(U,\mathfrak{g})$. The covariant derivative of $\Phi$ can then be written as
\begin{equation}
(\nabla^A_X\Phi)(x)=[s(x),\mathrm{d}\phi(X(x))+\rho_\ast(A_s(X(x))\phi(x)],
\qquad\qquad\qquad\qquad (1)
\end{equation}
where $X\in\mathfrak{X}(U)$ is a vector field.
There is another way to introduce the covariant derivative which is maybe more familiar to physicists: we start with the local formula (1) and show its covariance under a change of the section $s$. But this calculation is where I am stuck. Let $s':U'\rightarrow P$ be a nother section such that $U\cap U'\neq\emptyset$. Then there is a transition function $g:U\cap U'\rightarrow G$ of $P$ such that $s=s'\cdot g$. We also find another smooth map $\phi':U'\rightarrow V$ such that $\Phi=[s',\phi']$. By the definition of $E$ it then follows that $[s,\phi]=[s',\phi']$ if and only if $\phi=\rho(g)^{-1}\phi'$ on $U\cap U'$. We calculate
$$(\nabla^{A_s}_X\phi)(x)=\mathrm{d}(\rho(g(x))^{-1}\phi')(X(x))
+\rho_\ast\left(\mathrm{Ad}(g(x)^{-1})A_{s'}(X(x))+g^\ast\mu_G(X(x))\right)\rho(g(x))^{-1}\phi'(x)$$
I assume this has to be equal to $\rho(g(x))^{-1}(\nabla^{A_{s'}}_X\phi')(x)$. For the first differential I get
$$\mathrm{d}(\rho(g(x))^{-1}\phi')(X(x))=\rho(g(x))^{-1}\mathrm{d}\phi'(X(x))-\rho_\ast\left(g^\ast\mu_G(X(x))\right)\rho(g(x))^{-1}\phi'(x),$$
so that the second term cancels the last term in the equation before. The thing that bothers me is the adjoint representation in the remaining argument of $\rho_\ast$. How do I get rid of that? Or is my calculation of the differential wrong?
Best Answer
With the help of @ahersh23's answer I think I got it now:
It is $$\mathrm{d}\left(\rho(g(x))^{-1}\phi'\right)(X(x)) =\rho(g(x))^{-1}\mathrm{d}\phi'(X(x)) -\rho_\ast\left((g^\ast\mu_G)(X(x))\right)\rho(g(x))^{-1}\phi'(x),$$ so that we find $$\left(\nabla^{A_{s'}}_X\phi'\right)(x)=\rho(g(x))^{-1}\mathrm{d}\phi'(X(x))+\rho_\ast\left(\mathrm{Ad}\left(g(x)^{-1}\right)A_{s'}(X(x))\right)\rho(g(x))^{-1}\phi'(x).$$ Now the adjoint representation of $G$ is defined by $$\mathrm{Ad}:G\rightarrow\mathrm{GL}(\mathfrak{g}),\,g\mapsto\left(L_g\circ R_{g^{-1}}\right)_\ast.$$ With this we find for $Y\in\mathfrak{g}$ and $h\in G$ $$\rho_\ast\left(\mathrm{Ad}(h^{-1})Y\right) =\left(\rho\circ L_{h^{-1}}\circ R_h\right)_\ast(Y) =\left(\rho(h^{-1})\rho\circ R_h\right)_\ast(Y) =\rho(h^{-1})\left(\rho\circ R_h\right)_\ast(Y) =\rho(h^{-1})\rho_\ast(Y)\rho(h),$$ where the second equal sign follows from the fact that $(\rho\circ L_{h^{-1}})(g)=\rho(h^{-1}g)=\rho(h^{-1})\rho(g)$ and the last equal sign follows analogously.