Let $G$ be a Lie group, $M$ be a smooth manifold and $\pi:P\rightarrow M$ be a principal $G$-bundle. Given a connection $1$-form $A:P\rightarrow \Lambda^1_{\mathfrak{g}}T^*P$ on the principal bundle, fixing a point $x\in M$ we have the notion of holonomy map $\Omega(M,x)\rightarrow \text{Aut}(\text{fiber of P at x})$
Recall : Given a loop (path) $\gamma$ in $M$, based at $x$, and a point $u\in \pi^{-1}(x)$, the connection gives a path $\gamma^*_u$ in $P$ starting at $u$.Varying $u$ over $\pi^{-1}(x)$, we get paths $\{\gamma^*_u:[0,1]\rightarrow P|u\in \pi^{-1}(x)\}$. This gives a map $\Phi_x:\pi^{-1}(x)\rightarrow \pi^{-1}(x)$ defined as $u\mapsto \gamma^*_u(1)$. I was wondering if $\Phi_x$ is a smooth map (recall that $\pi^{-1}(x)$ is an embedded submanifold of $P$).
Is the parallel transport/displacement map $\Phi_x:\pi^{-1}(x)\rightarrow \pi^{-1}(x)$ a smooth map for each $x\in M$? I am sure this is true but could not prove it now. Any help is appreciated.
Best Answer
As I said in a comment, this holds even for general smooth bundles (which need not be not $G$-bundles). You can find a proof for instance in Theorem 9.8, page 80, in:
I.Kollar, P.Michor, J.Slovak, "Natural operators in differential geometry", Springer-Verlag, 1993.
But the key is a local theorem from ODEs (proof of which you can find in any graduate-level textbook on ODEs):
Assuming that $\Phi$ is a vector field on an open subset $U$ of $R^n$, the Cauchy problem of the form $$ x'(t)= \Phi(x), x(t_0)=v\in R^n $$ has unique smooth short-term solutions on relatively compact open subsets in $U$. See for instance Appendix I of Kobayashi-Nomizu "Foundations of Differential Geometry".