Let $G$ be a matrix Lie group with Lie algebra $\mathfrak{g} = T_e G$. Let $P \to X$ be a $G$-principal bundle with a connection 1-form $A \in \Omega^1(P,\mathfrak{g})$. Suppose we have a loop $\gamma\colon [0,1] \to X$ based at $x \in X$, the parallel transport map $\mathrm{tra}_{\gamma}^{A} \colon P_x \to P_x$ is $G$-equivariant, i.e. $\mathrm{tra}_{\gamma}^{A} \circ R_g = R_g \circ \mathrm{tra}_{\gamma}^{A}$. Therefore, for every loop $\gamma$ there exists a unique $g \in G$ such that $\mathrm{tra}_{\gamma}^{A} = R_g$. In particular, $g \in \operatorname{Hol}_x(A)$ is an element of the holonomy.
Now, if $\gamma$ lies in a subset $U \subset X$ on which $P\vert_U \to U$ is trivializable via a section $s_1 \colon U \to P$. Then we can compute this element via the path ordered exponential
$$ g = \operatorname{\mathcal{P}\mathrm{exp}}\Big(-\int_{\gamma} s_1^{\ast} A \Big) \,.$$
Now comes my confusion. If $s_2 \colon U \to P$ is another section, then there exists $g_{12} \colon U \to G$ such that $s_2 = s_1 \cdot g_{12}$. I believe that in this case
$$ \operatorname{\mathcal{P}\mathrm{exp}}\Big(-\int_{\gamma} s_2^{\ast} A \Big) = g_{12}^{-1} \operatorname{\mathcal{P}\mathrm{exp}}\Big(-\int_{\gamma} s_1^{\ast} A \Big) g_{12} \,,$$
but also $\operatorname{\mathcal{P}\mathrm{exp}}\big(-\int_{\gamma} s_2^{\ast} A \big) = g = \operatorname{\mathcal{P}\mathrm{exp}}\big(-\int_{\gamma} s_1^{\ast} A \big)$ by the uniqueness of the transport map. So either the holonomy lies in the center of $G$ (which it shouldn't) or I have got something wrong. Can someone correct me?
Best Answer
I think the mistake lies in the conclusion
Instead, the choice of some $s\in P_x$ induces a (right) $G$-equivariant identification $h\mapsto s.h$ of $G$ with $P_x$ and the (right) $G$-equivariant maps on $G$ are precisely given by the $\textbf{left}$-multiplications.
If a $G$-equivariant map $\varphi$ on $P_x$ is given by $R_g$ for some $g\in G$, then for fixed $s\in P_x$ and all $h\in G$
$$ s.(hg)=(s.h).g=\varphi(s.h)=\varphi(s).h=(s.g).h=s.(gh) $$
and hence $hg=gh$ since the action is free. So indeed the implication
is correct, although in general it might not need to be true that $\mathrm{tra}_{\gamma}^{A} = R_g$ for some $g\in G$.