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[This question was asked on MSE, but got no answers, I thought it could be more appropriate here]

Let $M$ be a parallelizable manifold.

  1. Is there always a global frame $(X_i)$ such that $[X_i,X_j]=0$ for all $i,j$ ?

  2. 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 ?

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  • $\begingroup$ The maximum number of complete, commuting vector fields on a manifold is sometimes called the rank of the manifold. Generally this number is pretty small. $\endgroup$ – Ryan Budney May 17 '16 at 19:28
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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.

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  • $\begingroup$ Thank you. In the case of non-compact, can we hope to have a similar answer (like : if $M$ is noncompact parallelizable satisfying this relation, then $M$ is a product of a torus with $R^n$) ? $\endgroup$ – Michael May 17 '16 at 18:43
  • $\begingroup$ @Michael: no. For example, if you remove a point from $R^n$ it's no longer diffeomorphic to a product of a torus and $R^n$, I suppose with the exception of $n=1$ and $n=2$. You need to further add that the flows of the vector fields are complete, i.e. exist for all time. $\endgroup$ – Ryan Budney May 17 '16 at 19:27
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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.

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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.

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  • $\begingroup$ Isn't "Open subset of $\mathbb R^n$ modulo discrete group of translation symmetries" a classification? $\endgroup$ – Will Sawin May 17 '16 at 19:52
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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.

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  • $\begingroup$ This answers the non-compact case. Just so that I understand the argument, why noncompactness connectedness and parallelizability implies that we can immerse $M$ in $\mathbb{R}^m$ where $m$ is the dimension of $M$, and why do we get such a framing ? $\endgroup$ – Michael May 17 '16 at 20:42
  • $\begingroup$ The main theorem of immersion theory says that $M$ can be immersed in $N$ if there is a map $M\to N$ covered by a map of tangent bundles $TM\to TN$ that is injective (and linear) on each fiber. The only exception is if $M$ and $N$ have the same dimension and $M$ has a component that is compact and without boundary. In the case when $N=\mathbb R^m$, this means if $M$ is parallelizable. $\endgroup$ – Tom Goodwillie May 17 '16 at 21:18
  • $\begingroup$ When $M$ is immersed in a manifold $N$ of the same dimension (i.e. when some map $M\to N$ is locally a diffeomorphism), then for every tangent vector field on $N$ there is a unique such field on $M$ related to it by that map. This mapping from the set of tangent fields of $N$ to the set of tangent fields of $M$ preserves linear independence and preserves Lie brackets. $\endgroup$ – Tom Goodwillie May 17 '16 at 21:22
  • $\begingroup$ Thank you for you anwers. I don't understand the last sentence of your first comment : "this means if $M$ is parallelizable". $\endgroup$ – Michael May 17 '16 at 23:59
  • $\begingroup$ Sorry, the last sentence is related to the first sentence, not the second, and also badly expressed. $\endgroup$ – Tom Goodwillie May 18 '16 at 11:08

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