Smooth section in principal bundle

[sperber]

Hi!

I have the following problem: Let P be a principal bundle over a manifold M, p: P -> M. Let G be the lie group acting on P from the right. Now let U be an open set on M and

s : U -> P

a smooth section. Now it has been said that we can define a local trivialization of P by

t : p^{-1}(U) -> U x G, t( s(x)*g ) = (x,g)

It is clear to me that t is well-defined, bijective and that the inverse t^{-1} given by

t^{-1}(x,g) = s(x)*g

is smooth. What I do not understand is why t itself is smooth. Can somebody help?

Thanks, Sperber

 

[jvicens]

Hello Sperber,

Thank you for your question. The smoothness of t can be understood by examining its local representation. Let p^{-1}(U) be a local trivialization of P over U. Then, for any point x in U, we can choose a local frame e_1, ..., e_n for the fiber of P at x. This frame can be extended to a global frame on P by using the right action of G. This means that for any point y in P, we can write y = s(x)*g for some g in G. Then, the frame at y is given by e_1*g, ..., e_n*g.

Now, let's consider the local representation of t at y. We have t(y) = (x,g) = (x, e_1*g, ..., e_n*g). This is a smooth map from p^{-1}(U) to U x G, since each component is smooth. Therefore, t is a smooth map.

I hope this helps clarify the smoothness of t. Let me know if you have any further questions.