# Covariant derivative induced by pullback connection under an automorphism

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?

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.