why does Lie bracket of two coordinate vector fields always vanish?

This is really puzzling me. Say we are dealing with a Riemannian manifold $(M,g)$. Suppose $\nabla$ is the unique torsion free connection on $M$ that is compatible with $g$. Suppose we are in a neighbourhood $U$ with coordinate map $(x^1,\cdots, x^m )$. Since the connection is torsion free, $$[\frac{\partial}{\partial x_i},\frac{\partial}{\partial x_j}]=\nabla_{\frac{\partial}{\partial x_i}}{\frac{\partial}{\partial x_j}}-\nabla_{\frac{\partial}{\partial x_j}}{\frac{\partial}{\partial x_i}}.$$

And since the $\Gamma_{i,j}^k$ is symmetric on $i,j$, the right hand side of the above equation will vanish. So the Lie bracket will be 0. Now here is my confusion. If I start out with $m$ linearly independent vector fields $Y_1, \cdots, Y_m$, then I can find a coordinate system $(y_1,\cdots, y_m)$ such that $Y_i = \frac{\partial }{\partial y_i}$ (Correct me if I am wrong, because I am not sure about this). Then arguing as above, I can show that the Lie bracket of $Y_i$ and $Y_j$ vanishes. I know Lie bracket shouldn't vanish on any two random vector fields I pick. So there must be something wrong with my argument here. Thank you in advance!

• you are right about being wrong. That is exactly where it goes wrong. It is not possible to find a coordinate system on which a neighboorhood reproduces a given set of vector fields as coordinate derivations. At a point, yes. In an open set on the manifold, no. – James S. Cook Jun 24 '15 at 14:53
• @JamesS.Cook, so this in a sense tells me if I start out with two vector fields such that the Lie bracket doesn't vanish, then there is no way I could find a coordinate system such that they happen to be coordinate derivations. Now, is the reverse true? That is, if I start out with two vector fields such that the Lie bracket DOES vanish, is it true that I can find a coordinate system with them being coordinate derivations? Thanks! – henryforever14 Jun 24 '15 at 14:57
• Let's see, if the commutator is nontrivial then I don't think that means it is not possible. I certainly can find vector fields on the plane which have nontrivial Lie Bracket. Converse direction, well, if the bracket is zero on an open set and the vector fields are nonzero on that set, I think that suffices to find a coordinate system for which they appear as derivations. – James S. Cook Jun 24 '15 at 15:02
• some terminology en.wikipedia.org/wiki/Holonomic_basis – James S. Cook Jun 24 '15 at 15:03
• @JamesS.Cook, "Let's see, if the commutator is nontrivial then I don't think that means it is not possible. I certainly can find vector fields on the plane which have nontrivial Lie Bracket." I meant vector fields with nontrivial commutator can't be coordinate derivations. – henryforever14 Jun 24 '15 at 15:20

In $\mathbb{R}^n$ we know that for any smooth function $f$ we have$$\frac{\partial^2f}{\partial x_1\partial x_2}=\frac{\partial^2f}{\partial x_2\partial x_1}.$$In other words:$$\left[\frac{\partial}{\partial x_1},\frac{\partial}{\partial x_2}\right]=0.$$Hence, whenever $X_1,\ldots,X_n$, are local vector fields on a manifold induced by a parametrization, we necessarily have $[X_i,X_j]=0$ for $i,j=1,\ldots,n$.