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

Question

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, $$\left[\frac{\partial}{\partial x_i},\frac{\partial}{\partial x_j}\right]=\nabla_{\frac{\partial}{\partial x_i}}{\frac{\partial}{\partial x_j}}-\nabla_{\frac{\partial}{\partial x_j}}{\frac{\partial}{\partial x_i}}.$$

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!

Answer 1

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).

Answer 2

What you write before "Correct me if I'm wrong" is indeed wrong.

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 $i,j=1,\ldots,n$.

Answer 3

Here's my attempt at this question. My proof is not super rigorous, but hopefully provides some intuition.

Regarding

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 $Y_i=\frac{\partial}{\partial y_i}$

I think the problem is you can't always find such a coordinate system.

I had the same question and I used the following example to gain intuition. Let $\vec{e_{x}}$, $\vec{e_y}$ be the Cartesian coordinate basis vectors of a 2-D flat plane. Set vector fields $\vec{u}=\vec{e_x}$ and $\vec{v}=x \vec{e_y}$. The Lie bracket $[\vec{u},\vec{v}]=\vec{e_y}$ is not zero.

Let's say you do find a coordinate system such that $\vec{e_u}=\vec{u}$ and $\vec{e_v}=\vec{v}$ everywhere. (Note that $\vec{e_u}$ and $\vec{u}$ are not just single vectors, they are vector fields. The same goes for $\vec{e_v}$ and $\vec{v}$.) Let's say that the plane is described by $R(u,v)=u \vec{e_u} + v\vec{e_v}$. We know that $\frac{\partial{R}}{\partial{u}}=\vec{e_u}=\vec{e_x}$, $\frac{\partial{R}}{\partial{v}}=\vec{e_v}=x\vec{e_y}=u\vec{e_y}$. You can't find such a $R$ that satisfies both partial derivatives. So there doesn't exist such a coordinate system.

More deeply, any coordinate system we can find is describing the plane by the $R(u,v)=u\vec{e_u}+v\vec{e_v}$ function. From calculus, we have $\frac{\partial R}{\partial u \partial v} = \frac{\partial R}{\partial v \partial u}$, so $\frac{\partial}{\partial u}\vec{e_v} = \frac{\partial}{\partial v}\vec{e_u}$, so the Lie bracket $[\vec{e_u},\vec{e_v}]$ is guaranteed to be 0.