Local one sections and connections $1$-forms?

Question

Let $\pi:P\longrightarrow M$ be a $G$-principal bundle endowed with a connection $1$-form $\omega\in \Omega^1(P; \mathfrak{g})$ where $\mathfrak{g}$ is the Lie algebra of $G$.

Let $\{U_i\}_{i\in I}$ be an open cover of $M$ and for every $j\in I$ choose a local section $s_j:U_j\longrightarrow \pi^{-1}(U_i)$. Then one might define $$\omega_j:=s_j^*(\omega)\in \Omega^1(U_j; \mathfrak{g}).$$

Since $G$ acts freely and transitively on the fibers one might find a function $g:U_i\cap U_j\longrightarrow G$ such that $$s_j(p)=s_i(p)\cdot g(p)\quad (p\in U_i\cap U_j).$$ Using this data I'd like to obtain a relation between $\omega_i$ and $\omega_j$ on $U_i\cap U_j$, can anyone help me?

Thanks

Answer 1

The sought relation is $$ \omega_j|_p = \operatorname{Ad}_{g(p)^{-1}}\omega_i|_p + (g^*\theta)_p, $$ where $\operatorname{Ad}: G\rightarrow \operatorname{Aut}(\mathfrak{g})$ is the adjoint representation of $G$ and $\theta\in\Omega^1(G,\mathfrak{g})$ is the Maurer-Cartan form on $G$.

I think it is fruitful to prove a more general result. As a motivation consider this:

Since $G$ acts freely and transitively on $P$, there is a vertical automorphism $\phi$ of $P$ such that we have on $U_i \cap U_j$: $$ s_j = \phi \circ s_i $$ $\phi$ is given by the right action on $P$: $\phi(q)=R_{g\circ\pi(q)}(q)$ for $q\in\pi^{-1}(U_i\cap U_j)$. Hence we have $$ \omega_j = s_j^*\omega=s_i^*(\phi^*\omega) $$

Claim: The pullback of $\omega$ under some vertical automorphism $\phi$ (meaning $\phi$ commutes with the right action on $P$ and $\pi\circ\phi=\pi$) is given by: $$ (\phi^*\omega)_q = \operatorname{Ad}_{h(q)^{-1}}\omega_q + (h^*\theta)_q, $$ where $q\in P$ and $h: P\rightarrow G$ is such that $\phi(q)=R_{h(q)}(q)$.

Note: Such a map obviously exists. In the above setting $h=g\circ\pi$.

Proof

1. Verify that for $X\in T_q P$: $$ \phi_*X(f)=(R_\bullet(q)\circ h)_*X(f) + (R_{h(q)})_*X(f), $$ thus: $$ (\phi^*\omega)(X)=\omega((R_{h(q)})_*X)+\omega((R_\bullet(q)\circ h)_*X)=\operatorname{Ad}_{h(q)^{-1}}\omega(X)+\omega((R_\bullet(q)\circ h)_*X) $$

2. Now it holds ($L$ is the left action by multiplication on $G$): $$ (h^*\theta)_qX=\theta_{h(q)}(h_*X)=(L_{h(q)^{-1}})_*(h_*X)=:B\in\mathfrak{g}. $$ Then, for any $x\in G$: $$ R_\bullet(q)\circ L_{h(q)}(x) = R_\bullet(q)(h(q)\cdot x)=R_x\circ R_{h(q)}\in P(q). $$ By choosing $x=\exp(tB)$ we thus get a curve $\gamma$ in $P$ with $\gamma(0)=q$. Its tangent vector at $t=0$ is just $(R_\bullet(q)\circ h)_*X$. On the other hand, $\gamma(t)=R_{\exp(tB)}(\tilde q)$ with $\tilde q=R_{h(q)}(q)$, which is a integral curve through $\tilde q$ of the flow generated by $B$ on $P$. Therefore, its tangent vector field is induced by a fundamental vector field $B^\sharp$, especially we have $$ B^\sharp_{\tilde q} = (R_\bullet(q)\circ h)_*X $$ and hence $$ \omega((R_\bullet(q)\circ h)_*X)=\omega(B^\sharp_{\tilde q})=B=(h^*\theta)_qX, $$ which then shows: $$ (\phi^*\omega)_q = \operatorname{Ad}_{h(q)^{-1}}\omega_q + (h^*\theta)_q. $$

Back to sections We apply $s_i^*$ on both sides ($p=\pi(q)$): $$ \omega_j|_p = s_i^*(\operatorname{Ad}_{g\circ\pi(q)^{-1}}\omega_q) + s_i^*((g\circ\pi)^*\theta)_q=\operatorname{Ad}_{g(p)^{-1}}\omega_i|_p + (g^*\theta)_p, $$ since $\pi\circ s_i=id$.