I may be totally wrong here, but after a bit of studying I've reach the conclusion of the title. My reasoning is as follows:
For a principal bundle $P(M,G),$ and $VP$ is vertical bundle, let $\sigma : \mathfrak g \to \mathfrak X(P)$ be the fundamental vector field Lie algebra morphism. Since the bundle is principal, fundamental vector fields are all vertical vector fields. Now, fix a basis $\{e_1,\dotsc. e_k\}$ of the Lie algebra $\mathfrak g, $ where $k$ is its dimension. The vector fields $E_i = \sigma(e_i), \ i = 1\dotsc, k$ are smooth vector fields which form a basis of $V_p$ at each point $p \in P,$ since at each point $p,$ the mapping $\sigma_p : X \in \mathfrak g \mapsto (\sigma(X))_p\in V_p $ is an isomorphism. So we have a global smooth frame for $VP,$ which means $VP$ is a trivial vector bundle.
I'm a bit cautious about this affirmation since I haven't seen any comments about it on lectures notes nor books I've looked so far, but I can't see any issues in my argument.
Am I doing something wrong here? Thanks.
Best Answer
Yes, this is correct. You can think of this as a generalization of the fact that the tangent bundle of $G$ is trivial. Indeed, when the bundle $P\cong M\times G$ is trivial, $VP$ can be identified with the pullback of the tangent bundle of $G$. The trivialization of $VP$ thus obtained will actually be preserved by the transitions between different local trivializations of $P$, since those transitions correspond to multiplying by elements of $G$ on each fiber, and our trivialization of $TG$ is translation-invariant. So, the local trivializations of $VP$ combine to give a global trivialization, even if $P$ is not trivial.