Coordinate curves and the Lie bracket on a differentiable manifold

On a differentiable manifold, let $$x^i$$ be the coordinates of points on the manifold in a local parametrization. Then it is straightforward to show that the Lie bracket between vector fields defined by the coordinate curve vanishes, i.e., $$\left[\frac{\partial}{\partial x^i},\frac{\partial}{\partial x^j}\right] = 0$$. I, however, find this very strange. It seems that given any linearly independent vector fields, $$X, Y,\ldots$$, one can always choose a local parametrization such that the vector fields $$X, Y,\ldots$$ are the associated basis of the parametrization. But that would mean that any two vector fields would have zero Lie brackets, which is clearly wrong.

So where is wrong with this reasoning? (It seems it has to be that not all linearly independent vector fields can be made into associated basis of a local parametrization. But why not? What is the obstruction?)

Thanks!

• "one can always choose a local parametrization such that the vector fields $X,Y,...$ are the associated basis of the parametrization". No. Precisely because of what you said. This is not true even if you have a single vector field $X$ with $X_p=0$ for some $p$. Look up the Frobenius' theorem. May 22 '20 at 2:28
• Yes, it seems that your basic logic is messed up. May 22 '20 at 2:31
• @IvoTerek Thanks very much for the answer. I was not aware of Frobenius' theorem before, so was very confused. Now that I do, I'm much happier. May 22 '20 at 16:00
• @TedShifrin The missing link in my knowledge was the relation between Lie brackets and integrability. May 22 '20 at 16:01

In fact, if $$X_1,\ldots,X_n$$ are linearly independant vector fields, there exist coordinates $$x^i$$ such that $$X_i = \partial/\partial x^i$$ if and only if $$[X_i,X_j]=0$$. A nice generalisation is Frobenius's theorem : there exists a submanifold tangent to your linearly independant vector fields if and only if their bracket always vanish.