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Let $G$ be a Lie group and $\pi:P\to M$ a smooth principal $G$-bundle. Let $\omega$ be a connection on $P$; it is a $\mathfrak{g}$-valued 1-form on $P$ where $\mathfrak{g}$ is the Lie algebra of $G$. (Recall that $\ker(\omega)$ gives a $G$-equivariant horizontal distribution of $P$.) The connection defines a covariant derivative $D_\omega: \Omega^k(P,\mathfrak{g})\to \Omega^{k+1}(P,\mathfrak{g})$ defined by $D_\omega \alpha (X_0,\dots,X_k)= d\alpha ((X_0)_{\text{hor}},\dots,(X_k)_{\text{hor}}).$

Consider an associated vector bundle $E=P\times_\rho V$ for some represenation $\rho:G\to GL(V)$. There is a canonical isomorphism between the space $\Omega^k_{\rho,h}(P,\mathfrak{g})$ of horizontal $\rho$-equivariant forms and the space $\Omega^k(M,E)$ of $E$-valued forms on $M$. Using this isomorphism, $D_\omega$ defines a covariant derviative $d_\omega:\Omega^k(M,E)\to \Omega^{k+1}(M,E)$, and in particular for $k=0$ the map $d_\omega:\Omega^0(M,E)\to \Omega^1(M,E)$ is denoted by $\nabla^\omega$. Following the definitions it can be shown that for a given $f\in \Omega^0(M,E)$ (which is just a function $M\to E$), $x\in M$, and $v\in T_xM$, we have $(\nabla^\omega f)_x(v)=[p,(D_\omega\tilde{f})_p(\tilde{v})]$, where $\tilde{f}:P\to V$ is determined by $f(\pi(p))=[p,\tilde{f}(p)]$, $p\in P_x$, and $\tilde{v}\in T_pP$ is a lift of $v$.

Now, suppose that $\Theta:P\to P$ is a bundle automorphism. It is given by $\Theta(p)=p\cdot u(p)$ for some well-defined map $u:P\to G$. Also the pullback $\Theta^*\omega$ is a connection 1-form on $P$, and as above it defines a covariant derivative $\nabla^{\Theta^*\omega}$. I want to find a relation between $\nabla^{\omega}$ and $\nabla^{\Theta^*\omega}$. It seems that it is given by $\theta((\nabla^{\Theta^*\omega} f)_x(v))=(\nabla^\omega (\theta f))_x(v)$ for $x\in M$ and $v\in T_xM$, where $\theta:E\to E$ is the automorphism $[p,v]\mapsto [\Theta(p),v]$ induced by $\Theta$. But I can't see how to prove this.

For some $p\in P_x$ and lift $\tilde{v}\in T_pP$ of $v$, I've got $[\Theta(p), (\Theta^* D_\omega(\tilde{f}\circ \Theta^{-1}))_p(\tilde{v})]$ for the right-hand-side, and $[p, D_\omega(\tilde{f}\circ \Theta^{-1})_p(\tilde{v})]$ for the left-hand-side. But are these two the same?

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They are the same. The reason is that $$ (\Theta^*D_\omega(\tilde{f}\circ\Theta^{-1}))_p(\tilde{v}) = \Theta^*(D_\omega(\tilde{f}\circ\Theta^{-1})(\Theta_*\tilde{v}))(p) $$ where $(\Theta_*\tilde{v})_p:= T_{\Theta^{-1}(p)}\Theta(\tilde{v}_{\Theta^{-1}(p)})$. But $\Theta_*\tilde{v}$ is just another lift of $v$, since $$ T_p\pi(\Theta_*\tilde{v})_p = T_{\Theta^{-1}(p)}(\pi\circ\Theta)(\tilde{v}_{\Theta^{-1}(v)}) = T_{\Theta^{-1}(p)}\pi(\tilde{v}_{\Theta^{-1}(p)}) = v_x, $$ where $x=\pi(p)$, so the above expression is equivalent to $$ \Theta^*(D_\omega(\tilde{f}\circ\Theta^{-1})(\tilde{v}))(p). $$ Now just use the fact that for any $\tilde{g}\in\Omega^1(P,\mathfrak{g})$ (in particular, for $\tilde{g} = D_\omega(\tilde{f}\circ\Theta^{-1})(\tilde{v})$) $$ [\Theta(p), (\Theta^*\tilde{g})(p)] = [p,\tilde{g}(p)]. $$

In proving the result, I personally find the notation a bit easier if one uses the equivalent definition $$ \widetilde{\nabla_{v_x}f}(p) = v_p^{\operatorname{hor}_\omega}\tilde{f} $$ where $v_p^{\operatorname{hor}_\omega}$ denotes the horizontal lift of $v_x$ with respect to $\omega$, plus the easily proved fact that $$ T_p\Theta(v_p^{\operatorname{hor}_{\Theta^*\omega}}) = v_{\Theta(p)}^{\operatorname{hor}_\omega}. $$ It follows that $$ v_p^{\operatorname{hor}_{\Theta^*\omega}}\tilde{f} = v_{\Theta(p)}^{\operatorname{hor}_\omega}(\tilde{f}\circ\Theta^{-1}) \quad\text{i.e.}\quad \widetilde{\nabla^{\Theta^*\omega}_{v_x}f} = \widetilde{\nabla^\omega_{v_x}(\theta f)}\circ\Theta = \widetilde{\theta^{-1}\nabla^\omega_{v_x}(\theta f)}, $$ using $\tilde{g}\circ\Theta^{-1} = \widetilde{\theta g}$ for $g\in\Omega^0(M,E)$ twice. Then just drop the tildes and the $v_x$ to get your result.

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