Is it possible a trivial fiber bundle with nonzero holonomy? Let $P\rightarrow M$ be a principal bundle with structure group $G$. Suppose that the bundle is trivial $M\times G$; is it possible to have a nonzero holonomy along some closed trajectory on $M$ for some connection form?
 A: I believe so, although it's been a while since I've really thought about principal bundle. 
Consider the associated SO(2)-principle bundle to the tangent bundle of the unit sphere in 3-space (alternatively: consider the tangent circle bundle of the sphere). There's clearly holonomy there (think about parallel translation around a non-equatorial latitude line). 
Now remove the two polar regions, and you've got a trivial bundle over a cylinder, but the holonomy is still nonzero around, say, the 30-north latitude line. 
If you don't like the fact that my manifold is open, you can do the same thing with a large torus with a small minor-radius. At some point around the major circumference, inflate things (as if the torus were a snake that swawllowed a beach-ball, and then ate its own tail). In the inflated area, you have my sphere example, while the bundle over the whole torus is still trivial. 
A: Take the trivial circle bundle over the circle with the connection form $iad\theta$ for some $a\in \mathbb R$. Then the  holonomy of going counter-clockwise once around the circle  is $e^{2\pi i a}$. If $a$ is not an integer this is not trivial. 
In this example the connection is flat (vanishing curvature) so the holonomy gives a homomorphism from  the fundamental group of $M$ to $G$. In fact, there is a bijection between such homomorphisms and flat connections on $M\times G\to M$ (with appropriate equivalence relations on both sides of the bijection).  
So to get an example with a simply-connected $M$  the connection cannot be flat. Take for example the trivial circle bundle over  $\mathbb R^2$ with the connection form $ixdy.$ The curvature is $idx\wedge dy$ and so the holonomy along a closed trajectory is $e^{iA}$, where $A$ is the (signed) area enclosed by the trajectory. If $A/2\pi$ is not an integer then  the holonomy is not trivial.   
A: All you need is for the curvature of the connection to not vanish. Try, for example, the connection defined by the standard contact structure on $\mathbb{R}^3$ (this is a principle $\mathbb{R}$-bundle over $\mathbb{R}^2$, where the group action on $\mathbb{R}$ is addition).
