I About the maximal extension of local charts on a manifold

[cianfa72]

TL;DR

About the limits in the extension of a local chart on a smooth manifold

I'm a bit confused by the conditions on the existence of coordinate basis given by Frobenius's theorem.

Namely, let's take a n-dimensional smooth manifold and a set of n smooth vector fields defined on it. Suppose they are pointwise linearly independent and do commute each other (i.e. zero commutator/Lie bracket).

That means they span the entire tangent space at any point and since commute, they define a local coordinate basis. What does this mean? Well, starting from any point on the manifold, there is a neighborhood such that the integral curves of those vector fields define a local coordinate chart. So far so good.

Also, what is the maximum extension of such local charts, can any of them cover the entire manifold ?

Take a 2-sphere as opposed to an infinite cylinder. On the sphere, by the hairball theorem, there are no two smooth vector fields pointwise linearly independent. However parallels and meridians define two commuting vector fields excluding both poles, thus defining local charts based on them (excluding the poles).

On the other hand, on the cylinder, there exist two pointwise linearly independent smooth vector fields that commute. Therefore starting from any point they define a local coordinate chart, yet neither of them can be extended globally for topological reasons. Indeed, trying to extend any of them becomes problematic: the same point would have multiple coordinate values.

Thus, in the latter case, although the two smooth vector fields define a constant rank bundle and do commute, the maximal extension of any local chart based on them is actually limited.

Does the above make sense ?

 

[wrobel]

TL;DR: About the limits in the extension of a local chart on a smooth manifold

Also, what is the maximum extension of such local charts, can any of them cover the entire manifold ?

you can cover a manifold with a single chart if only this manifold is diffeomorphic to $\mathbb{R}^m$. I do not understand other questions

 

[wrobel]

O, it seems I understand the question. Consider a torus $\mathbb{T}^2$ with angle coordinates $(\psi_1,\psi_2),\quad \psi_i\pmod{2\pi}$. And take two linearly independent vector fields $(\omega_1,\omega_2)$ and $(\omega'_1,\omega'_2)$. These vector fields commute. What will you get if for example the numbers $\omega_1,\omega_2$ are rationally independent? etc.

 

[cianfa72]

you can cover a manifold with a single chart if only this manifold is diffeomorphic to $\mathbb{R}^m$. I do not understand other questions

Yes. Better, I'd say since a smooth manifold is first and foremost a topological manifold, we can cover it iff, as topological manifold, is homeomorphic to $\mathbb R^m$ as topological space.

My point was about the topological "obstructions" that can prevent from extending a local chart based on some set of pointwise linearly independent and commuting smooth vector fields. That is the case of the cylinder, for instance.

 

[cianfa72]

O, it seems I understand the question. Consider a torus $\mathbb{T}^2$ with angle coordinates $(\psi_1,\psi_2),\quad \psi_i\pmod{2\pi}$. And take two linearly independent vector fields $(\omega_1,\omega_2)$ and $(\omega'_1,\omega'_2)$. These vector fields commute. What will you get if for example the numbers $\omega_1,\omega_2$ are rationally independent? etc.

Sorry, not sure to understand which are the two your linearly independent vector fields $(\omega_1,\omega_2)$ and $(\omega'_1,\omega'_2)$.

 

[wrobel]

for example $(1,2)$ and $(3,4)$ then
$(1,\sqrt 2)$ and $(\sqrt 2,1)$

 

[cianfa72]

for example $(1,2)$ and $(3,4)$ then
$(1,\sqrt 2)$ and $(\sqrt 2,1)$

Sorry, is $(1,2)$ or $(3,4)$ a vector field defined on the torus $\mathbb{T}^2$ ?