Transformation of the local gauge connection from one chart to another

Question

Assume a principal $G$-bundle $P \to B$. Then consider two trivialisations $U_i$ and $U_j$ and the transition functions $\psi$ such that the sections transform as $$ s_i(x) = s_j(x) \cdot \psi_{ji}(x) $$ Of course $\psi_{ji}(x) = \psi_{ij}^{-1}(x)$ with $x\in B$. Now, on $P$ a connection $\omega \in \Omega^1(P,\mathfrak{g})$ transforms at a point $ua \in G$ moved along by the right action of $a \in G$ as $$ \omega_{ua} = \text{adj}_{a^{-1}} \omega_u $$ I want to find how a local representation of the connection changes if we change trivialisation. For sections $s_i, s_j$ in $U_i, U_j$ respectively (with non-empty intersection) we have for the pull-backs $$ s_i^*(x) = [\,s_j(x) \psi_{ji}(x)\,]^* = \psi_{ji}^{*}(x) s_j^{*}(x) $$ The local representation of a connection is $$ \omega_i = s_i^{*}(\omega) $$ Thus what should we get for $$ s_i^{*}(\omega_{ua})=? $$ In Marathe's book "Topics in physical mathematics" the answer is given as $$ \omega_j(x) = \text{adj}_{\psi_{ij}^{-1}}\omega(x) + \psi_{ij}\Theta(x) $$ with $\Theta(x)$ the Maurer-Cartan form. How does he get this equation though remains quite unclear to me. I understand that $a$ should in principle not play a role since the projection of $u$ and $ua$ to $x$ should be the same but I suspect it is somehow related to the Maurer-Cartan form.

Answer 1

We start with an observation regarding tangent vectors in $P$. Let $p\in P$, let $\gamma:(-\epsilon,\epsilon)\to P$ be a path with $\gamma(0)=p$, let $g:(-\epsilon,\epsilon)\to G$ with $g(0)=e$, and let $\gamma':(-\epsilon,\epsilon)\to P$ be given by$$\gamma'(t)=\gamma(t)\cdot g(t).$$

Each one of $\gamma$ and $\gamma'$ represents a tangent vector at $p$, and both paths have the same image in $B$. It follows that$$\dot{\gamma}'(0)-\dot{\gamma}(0)\in V_p,$$where $V_p$ denotes the vertical tangent space at $p$. Since $\omega$ is a connection, it restricts to the Maurer-Cartan form on every vertical space. Thus, we obtain the identity $$\omega\left(\dot{\gamma}'(0)\right)=\omega\left(\dot{\gamma}(0)\right)+\Theta\left(\dot{\gamma}'(0)-\dot{\gamma}(0)\right)=\omega\left(\dot{\gamma}(0)\right)+\dot{g}(0).$$

We continue by slightly generalizing the above observation. Using the same notation, relax the assumption $g(0)=e$. This means that now, $\gamma$ and $\gamma'$ represent vectors in two different tangent spaces. Write $$\dot{\gamma}'(0)=\left.\frac{d}{dt}\right|_{t=0}\gamma(t)\cdot g(t)=\left.\frac{d}{dt}\right|_{t=0}(\gamma(t)\cdot g(0))\cdot\left(g(0)^{-1}g(t)\right).$$

Using $G$-equivariance of $\omega$ and the previous identity, we obtain in this case $$\omega\left(\dot{\gamma}'(0)\right)=\mathrm{adj}_{g(0)}^{-1}\omega\left(\dot{\gamma}(0)\right)+\Theta\left(\dot{g}(0)\right).$$

Now, let $b\in B$, and let $v\in T_bB$ be represented by the path $\alpha:(-\epsilon,\epsilon)\to B$. The desired formula follows from the preceding identity, taking $$\gamma(t):=s_i(\alpha(t)),\quad g(t):=\psi_{ij}(\alpha(t)).$$