dg.differential geometry - tangent bundle of a fiber bundle

I just wanted to elaborate on Benoît Kloeckner's answer, so if you like what I say, please upvote his answer.

By a frame, I mean a basis of the tangent space at a point on a smooth
manifold $M$. The space $F$ of all possible frames, called the frame
bundle, is a principal $GL(n)$-bundle over the manifold, $n$ is the
dimension of the manifold. A point in $F$ is given by $(x, e)$, where
$x in M$, $e = (e_1, dots, e_n)$, and $e_i in T_xM$. Associated
with each point is the dual frame $omega^1, dots, omega^n in
T_x^*M$. Let $pi: F rightarrow M$, $pi(x,e) = x$, denote the
natural projection.

There is a natural set of $n$ $1$-forms $hatomega^1, dots,
hatomega^n$ on $F$, which are called either "tautological" or
"semi-basic" and act as follows: If $v in T_{(x,e)}F$, then
$
langle hatomega^i,vrangle = langleomega^i,pi_* v rangle,
$
where $omega^1, dots, omega^n in T^*_xM$ form a dual basis to the basis
$e_1, dots, e_n in T_xM$. These forms have the universal property
that given any section $s: M rightarrow F$, $s^*baromega^i$ are
$1$-forms on $M$ dual to the moving frame given by the $e_i$.

You can check that any connection $nabla$ on $T_*M$ determines a set
of global $1$-forms $hatomega^i_j$ on $F$, such given any section
$s = (s_1, dots, s_n): M rightarrow F$, $nabla s_j =
s_is^*hatomega^i_j$. Therefore, a connection on $F$ gives a set of global
$1$-forms $hatomega^1, dots, hatomega^n, hatomega^1_1, dots, hatomega^n_n$
that trivialize $T^*F$. The dual vector fields
trivialize $T_*F$.

Since there always exists a connection on $T_*M $, this shows that $F$
has a parallelizable tangent bundle. The same argument can be extended
to any principal $G$-bundle of tangent frames. As observed by
Hoeckner, the case $G = O(n)$ corresponds to a Riemannian structure.

This, of course, does not answer the original question, but it is a
important case where the answer is yes. These global $1$-forms are
extremely useful in many contexts; the work of Robert Bryant
illustrates this.