# If there exists a global section then the principal bundle is trivial - problem with smoothness

Let $\pi \colon P \to B$ be a principal $G$-bundle and let $s \colon B \to P$ be it's smooth section. In order to show that $P \simeq B \times G$ I define the map $\varphi \colon P \to B \times G$ by the following rule: $$\varphi(p) = (\pi(p),g(p))$$ where $g(p) \in G$ is such that $p = s(\pi(p))g(p)$ (it always exists since the right action of $G$ is transitive on each fiber and it is unique since the action is free on each fiber). It is clear that $\varphi(ph) = (\pi(p),g(p)h)$ and that $\pi = \pi_1 \circ \varphi$. Also one can see that $\varphi$ is a bijection. The only problem for me to finish the demonstration is to show that $\varphi$ is smooth. Please, help me with this.

Added: a principal $G$-bundle is given by two smooth manifolds $P$ and $B$ and by a Lie group $G$ together with a smooth submersion $\pi \colon P \to B$ and with a free smooth right action of $G$ on $P$ such that $B \simeq P/G$ as sets and for any $b \in B$ there exists a neighborhood $U$ of $b$ and a diffeomorphism $\varphi \colon \pi^{-1}(U) \to U \times G$, such that $\pi = \pi_1 \circ \varphi$ and $\pi_2 \circ \varphi$ is $G$-equivariant.

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How do you know the bundle is trivial? – Isaac Solomon Dec 17 '13 at 22:54
@IsaacSolomon there is a theorem that a principal $G$-bundle is (isomorphic to) a trivial bundle if and only if it has a smooth global section – Nimza Dec 17 '13 at 22:58
Ah, right. That makes sense. – Isaac Solomon Dec 17 '13 at 22:59

It is simpler to prove smoothness of the inverse map $$G\times B\to P,\quad(g,b)\mapsto s(b)\cdot g$$ The smoothness of the section and of the $G$-action on $P$ imply smoothness, and deduce smoothness of the map you are looking at from the inverse function theorem.
One can also show, using the inverse function theorem, that the canonical map $$P\times_B P\to G, (p,q)\mapsto g$$ (where $q=p\cdot g$) is smooth (using the inverse function theorem) and deduce the result you are interested in from there.
You can use the fact that the map must have constant rank. What's your definition of a principal $G$-bundle? – Olivier Bégassat Dec 18 '13 at 0:05
But could you please explain in more detail the fact about constant rank? The map $(x,g) \mapsto s(x)g$ is a composition of $s \times id_G$ and of the action of $g$. The differential of first map has the maximal rank, I see this – Nimza Dec 18 '13 at 21:23