TITLE: On a parallelizable manifold, is there always a frame satisfying $[X_i,X_j]=0$?
QUESTION [5 upvotes]: [This question was asked on MSE, but got no answers, I thought it could be more appropriate here]
Let $M$ be a parallelizable manifold.

Is there always a global frame $(X_i)$ such that $[X_i,X_j]=0$ for all $i,j$ ?
If the answer is no, what kind of obstruction there is to find such a frame ? and what kind of general (topological ?) condition on $M$ makes it possible to find such a frame ?

REPLY [10 votes]: Suppose that $M$ is compact, such a frame exists implies that the commutative group $R^n$ acts transitively on it, this implies that $M$ is a torus. But a compact Lie group is parallelizable, so that is not always possible.

REPLY [4 votes]: On a compact manifold, you have a global frame such that  $[X_i, X_j]=0$ if and only if your manifold is the torus. Starting from dimension 3,   there are parallelisable manifolds different from the torus,  possibly the simplest  example is  $S^3$. 
Indeed, the flows of the vector fields generate an action of $R^n$ on the manifold, the manifold is therefore $R^n$ quotient by  the stabilisor,  and the  stabilisor is a discrete subgroup of $R^n$ which preserves the vector fields and is therefore a lattice.

REPLY [3 votes]: Every Lie group is parallelizable, but a compact manifold with such a global frame is an abelian Lie group, so a torus. Hence lots of counterexamples, but I don't know the classification. For the noncompact, you can take out a Cantor set, or something worse, from Euclidean space or from a torus, so there is no classification possible.

REPLY [3 votes]: If $M$ is connected and non-compact of dimension $m$, then parallelizability implies that $M$ can be immersed in $\mathbb R^m$, and this implies existence of such a framing.