My question is very basic (I don't know too much of differential geometry):
given a fiber bundle, is there a necessary and sufficient condition for its tangent bundle to be trivial?


I have some ideas, but submitted to some conditions on the cohomology ring of the bundle.
(I apologize if it is trivial.)



Is there a necessary and sufficient condition for the tangent bundle of a fiber bundle to be trivial?
This is far from being a complete answer, but there is a case when one construct a parallelizable bundle (meaning its total space has trivial tangent bundle) from a given (geometric) bundle. 


The context is that of $G$-structure, which are a formalization of the concept of geometric structures. A $G$-structure is a set of data including a fiber bundle on a smooth manifold, which shall be thought of as the bundle of admissible frames. For example, in the Riemannian case ($G=O(n)$) the bundle is that of orthonormal frames. If the group $G$ has a finite-order rigidity property, then one can construct a sequence of bundles, the total space of each one being the base space of the next one, so that after a finite number of steps one gets a bundle whose total space is parallelizable. This is a tool to prove that the group of automorphisms of the $G$-structure is a Lie group. As an example, if I remember well the total space of the bundle of orthonormal frames on a Riemannian manifold is parallelizable.


All details are available in Kobayashi's *transformation groups in differential geometry*. 


