Smoothness of the associated G-valued map of a bundle automorphism of a principal G-bundle

Question

Let $\pi: P\rightarrow M$ be a principal $G$-bundle and $f:P\to P$ be a bundle automorphism. Define $\sigma_f: P\to G$ by $f(p)=p\cdot \sigma_f(p)$. It is claimed that $ \sigma_f$ is a G-valued map, i.e. a smooth map and $\sigma_f(p\cdot g)=g^{-1}\sigma_f(p)g$. I can prove the second condition, but why is $\sigma_f$ smooth?

Answer 1

Let $\phi: P_U \to U\times G$ be a local trivialization associated with a local gauge (section) $s: U \to P$, i.e. $\phi(s(x)\cdot g) = (x,g)$. Then $\phi$ can be written as $\phi(p) = (\pi(p),\psi(p))$ where $\psi: P_U \to G$ is smooth.

Therefore, any $p\in P_U$ can be written as $p=s(\pi(p))\cdot \psi(p)$, and $f(p)=s(\pi(f(p)))\cdot \psi(f(p))=s(\pi(p))\cdot \psi(f(p))$ as $f$ preserves fibres.

By the properties of free and transitive actions, and since $f(p)=p\cdot \sigma_f(p)$, we have $\sigma_f(p)=(\psi(p))^{-1}\psi(f(p))$. As the multiplication map, inverse map, $\psi$ and f are all smooth, so is $\sigma_f$ on $P_U$. As smoothness is a local property, $\sigma_f$ is smooth on the whole of $P$.