Constructing a connection $1$-form from local forms.

Question

I am following Section $10.1.3$ of Geometry, Topology and Physics by Nakahara, and have ran in to an issue regarding local connection forms.

Consider a principal $G$-bundle, $P(M,G)$, and an open cover $\{U_{i}\}$ for $M$. Let $\mathcal{A}_{i}\in\mathfrak{g}\otimes\Omega^{1}(U_{i})$ be a Lie algebra valued $1$-form on $U_{i}$, and let $\sigma_{i}:U_{i}\to\pi^{-1}(U_{i})$ be a local section of $P$. I wish to show that there exists a connection $1$-form $\omega$ such that $\mathcal{A}_{i}=\sigma^{*}_{i}\omega$.

The candidate form is given locally by: $$ \omega_{i}=g_{i}^{-1}\pi^{*}\mathcal{A}_{i}g_{i}+g_{i}^{-1}d_{P}g_{i} $$ where $g_{i}$ is the canonical local trivialisation, define by $\phi_{i}^{-1}(u)=(p,g_{i})$ for $u=\sigma_{i}(p)g_{i}$.

I have been able to show that indeed $\mathcal{A}_{i}=\sigma_{i}^{*}\omega_{i}$, but for this to be a connection $1$-form, I must additionally show that: $$ \omega_{i}(A^{\#})=A $$ where $A\in\mathfrak{g}$, and $A^{\#}$ is the fundamental vector field, defined by: $$ A^{\#}f(u) = \frac{d}{dt}f(ue^{tA})|_{t=0} $$ for $f:P\to\mathbb{R}$ and $u\in P$.

I was able to show that indeed $(g_{i}^{-1}\pi^{*}\mathcal{A}_{i}g_{i})(A^{\#})=0$, but I am struggling with the second term. The calculation in Nakahara goes like: $$ (g_{i}^{-1}d_{P}g_{i})(A^{\#}) = g_{i}^{-1}(u) \frac{dg(ue^{tA})}{dt}\biggr|_{t=0} = g_{i}^{-1}(u)g_{i}(u)\left(\frac{d}{dt}e^{tA}\right)\biggr|_{t=0} = A $$ but I am unable to follow. I am very new to this topic, and any help would be much appreciated.

Answer 1

After thinking about this over the weekend, I think I have a solution to my problem.

Essentially by definition, we have that: $$ d_{P}g_{i}(A^{\#}) = \frac{d}{dt}\left(g_{i}(\gamma(t))\right)\biggr|_{t=0} $$ where $\gamma(t)$ is a curve in $P$ passing through $u$ with tangent vector $A^{\#}$ at $u$. But we already have one such curve, again essentially by definition. Namely $c(t):=ue^{tA}$. This gives the first equality: $$ d_{P}g_{i}(A^{\#}) = \frac{dg_{i}(ue^{tA})}{dt}\biggr|_{t=0} $$ For the next equality, we use the canonical local trivialisation condition and the fact that the right action of $G$ on $P$ is associative to deduce that: $$ ue^{tA}=\sigma_{i}(g_{i}(u)e^{tA}) $$ On the other hand, using that $\pi(ue^{tA})=\pi(u)=p$, we have: $$ ue^{tA}=\sigma_{i}(p)g_{i}(ue^{tA}) $$ Using the uniqueness of the canonical local trivialisation, we then have that: $$ g_{i}(ue^{tA})=g_{i}(u)e^{tA} $$ From here, the result follows by smoothness of the group multiplication.

I will leave this up for a while without accepting the answer in case I've made any obvious mistakes.

Answer 2

Define $\mathfrak{g}$ valued 1-form $\omega$ on $P$ \begin{equation} \omega_i \equiv g_i^{-1} \pi^* A_i g_i + g_i^{-1} d_P g_i \end{equation}

Then for the $d_P$ exterior derivative on $P$

$g_i$ canonical local trivialization $\Phi_i^{-1}(u) = (p,g_i)$ , $u = \sigma_i(p)g_i $

then for $X \in T_pM$,

\begin{gathered} \sigma_i^*\omega_i(X) = \omega_i(\sigma_{i*}X) = g_i^{-1} \pi^* A_i g_i(\sigma_{i*} X) + g_i^{-1} d_P g_i(\sigma_{i*}X) = \\ = \pi^*A_i(\sigma_{i*} X) + d_Pg_i(\sigma_{i*}X) = A_i(\pi_* \sigma_{i*}X) + d_Pg_i(\sigma_{i*} X)\quad \quad \quad \text{Q.E.D} \end{gathered}