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!
Best Answer
Ultimately, the vanishing is due the fact that partial derivatives commute. The reason we cannot realize a given set of vector fields as coordinate derivations is ultimately due to non-trivial curvature on the manifold.
On the other hand, if a given set of nontrivial vector fields have vanishing brackets, or more generally Lie brackets which close on the span of the vector fields then we say such a set of vector fields is involutive. This means there exists a submanifold of the given manifold which takes the given set of vector fields as tangents. See Frobenius Theorem or this related MSE question for example.
I should also mention, it is always possible to take one nontrivial vector field and make it a coordinate derivation see straightening theorem (which is a trivial case of the more general result of Frobenius).