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 $\hat\omega^1, \dots,
\hat\omega^n$ on $F$, which are called either "tautological" or
"semi-basic" and act as follows: If $v \in T_{(x,e)}F$, then
$
\langle \hat\omega^i,v\rangle = \langle\omega^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 $s: M \rightarrow F$, $s^*\bar\omega^i$ 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 $\hat\omega^i_j$ on $F$, such given any section
$s = (s_1, \dots, s_n): M \rightarrow F$, $\nabla s_j =
s_is^*\hat\omega^i_j$. Therefore, a connection on $F$ gives a set of global
$1$-forms $\hat\omega^1, \dots, \hat\omega^n, \hat\omega^1_1, \dots, \hat\omega^n_n$
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.