Covariant derivative on associated vector bundle under change of section 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?
 A: I'm still new to this topic, and I'm having a little trouble figuring out everything you're doing in the second half of your question, but it may be helpful to look at how the connection coefficients transform under a change of section.
Given a trivialization, we can write our connections as follows:
$$\omega=g^{-1}dg+g^{-1}A_\mu dx^\mu g$$
where $g^{-1}dg$ is the maurer cartan form, and $A_\mu dx^\mu$ is a lie algebra valued one form on the base. We can think of $A$ as the pullback of $\omega$ with respect to a section corresponding to the trivialization:
$$A=s^*\omega$$
We can go from the trivialization corresponding to the section $s$ to a trivialization corresponding to a different section $s'$ by multiplying the section $s$ by some lie group valued function on the base $h(x)$:
$$s'=s(x)h(x)$$
The trivialization corresponding to $s'$ is then obtained by noting that:
$$(x,s'(x)g)=(x,s(x)h(x)g)$$
So the change of trivialization is given by $g\rightarrow h(x)g$. Under this transformation, $\omega$ becomes:
$$\omega^h=(h(x)g)^{-1}d(h(x)(g))+(h(x)g)^{-1}A_udx^u(h(x)g)$$
$$\omega^h=g^{-1}dg+g^{-1}h^{-1}(x)dh(x)g+g^{-1}h^{-1}(x)A_udx^uh(x)g$$
Thus we obtain:
$$A_udx^u\longmapsto h^{-1}dh(x)+h^{-1}A_udx^uh(x)$$
The Lie algebra valued one form $A$ are (as far as I have understood) the connection coefficients in the associated vector bundle, so under a change of section, which amounts to a change of trivialization, I believe that you would expect a term that looks like $\text{Ad}_{h^{-1}}(A)$. My apologies if this was unhelpful, or if I misunderstood what you are getting at.
Edit: You seem to be following Hamiltons Gauge Theory text, Im going to go look at the connections chapter and see if I can update this answer more.
Edit: I see you have already figured this out. For the sake of posterity here is my full solution to your question, sorry it took me so long to get around to this.
Fixing a connection $A$ on $P$ we have that:
$$\nabla^A_X\Phi=[s,d\phi(X_x)+\rho_*(A_s(X_x))\phi] $$
where $\phi$ is a map $U\rightarrow V$ and $A_s=s^*A$. For some map $g:U\rightarrow G$, let $s'=s\cdot g$ and $
\phi'=\rho(g^{-1})\phi$, under this change of section we have:
$$A'=\text{Ad}_{g^{-1}}\circ A+g^*\mu_G$$
We have that:
$$d\phi'(X_x)=\rho(g^{-1})d\phi+\left(D_x\rho(g^{-1})(X_x)\right)\phi$$
and that:
$$\rho_*(A'(X_x))\phi'=\rho_*(\text{Ad}_{g^{-1}}A_s(X_x))\rho(g^{-1})\phi+\rho_*(\mu_G(D_xg(X_x)))\rho(g^{-1})\phi$$
Note that by chain rule:
$$D_x\rho(g^{-1})(X_x)=D_{g^{-1}}\rho\circ D_gi\circ D_x g(X_x)$$
where $i$ denotes the inversion map. Denote $D_x g(X_x)$ by $\tilde{Z}\in T_gG$, we then see that for any $g\in G$:
$$D_gi(\tilde{Z})=
\lim_{t\rightarrow 0}\frac{d}{dt}g\exp(\mu_G(\tilde{Z})t)^{-1}$$
$$=\lim_{t\rightarrow 0}\frac{d}{dt}\exp(-\mu_G(\tilde{Z})t)g^{-1}$$
$$=-\mu_G(\tilde{Z})g^{-1}$$
Thus we obtain:
$$D_x\rho(g^{-1})(X_x)=-D_g^{-1}\rho\left(\mu_G(\tilde{Z})g^{-1}\right)$$
$$=-D_{g^{-1}}\rho\circ D_e R_{g^{-1}}\left(\mu_G(\tilde{Z})\right)$$
$$=-D_{\rho(e)}R_{\rho(g^{-1})}\circ D_e\rho\left(\mu_G(\tilde{Z})\right)$$
$$=-\left(D_e\rho(\mu_G(\tilde{Z})\right)\rho(g^{-1})$$
$$=-\rho_*(\mu_G(D_xg(X_x)))\rho(g^{-1})$$
Therefore we have that:
$$d\phi'=\rho(g^{-1})d\phi-\rho_*(\mu_G(D_xg(X_x)))\rho(g^{-1})\phi$$
Now examine the following:
$$\rho_*(\text{Ad}_{g^{-1}}\circ A_s(X_x))=\rho_*(C_{g^{-1}*}\circ A_s(X_x))$$
$$=\rho_*\circ C_{g^{-1}*}\circ A_s(X_x)$$
$$=\left(\rho\circ C_{g^{-1}}\right)_*\circ A_s(X_x)$$
$$=\left(\rho\circ R_g\circ L_{g^{-1}}\right)_*A_s(X_x)$$
$$=\left(R_{\rho(g)}\circ L_{\rho(g^{-1})}\circ \rho\right)_*\circ A_s(X_x)$$
$$=\rho(g^{-1})\rho_*(A_s(X_x))\rho(g)$$
Thus we have that:
$$\rho_*(A'_s(X_x))\phi'=\rho(g^{-1})\rho_*(A_s(X_x))\phi+\rho_*(\mu_G(D_xg(X_x)))\rho(g^{-1})\phi$$
Therefore:
$$d\phi'(X_x)+\rho_*(A'_s(X_x))\phi'=\rho(g^{-1})d\phi(X_x)+\rho(g^{-1})\rho_*(A_s(X_x))\phi$$
Putting it all together we obtain:
$$\nabla^A_X\Phi'=[s\cdot g,\rho(g^{-1})d\phi(X_x)+\rho(g^{-1})\rho_*(A_s(X_x))\phi $$
$$=[s,d\phi(X_x)+\rho_*(A_s(X_x))\phi]$$
as desired.
A: 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.
A: I've checked over your calculation for $d(\rho(g(x)^{-1}\phi')$and I think it is correct. The trick to pulling out the adjoint from inside $\rho_*$ is to remember that $\rho$ is a Lie Gp. homomorphism, hence $\rho(Ad(h))(x) = Ad(\rho(h))\rho(x)$. Let $Y=A_s'(X)$. Passing to differentials we have:
$ \rho_*(ad(g(x)^{-1})Y)(\rho(g(x)^{-1})\phi') = \rho_*(Ad_{g(x)^{-1}})_*(Y)(\rho(g(x)^{-1})\phi') = Ad_{\rho(g(x)^{-1})}(\rho_*(Y))(\rho(g(x)^{-1})\phi') = \rho(g(x)^{-1})\rho_*(Y)\phi' =\rho(g(x)^{-1})\rho_*(A_s')\phi'$
which is what you want.
