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

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!

• 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. Commented Jun 24, 2015 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! Commented Jun 24, 2015 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. Commented Jun 24, 2015 at 15:02
• some terminology en.wikipedia.org/wiki/Holonomic_basis Commented Jun 24, 2015 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. Commented Jun 24, 2015 at 15:20

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).

• If you have a copy, see Chapter 19 in John M. Lee's Introduction to Smooth Manifolds Commented Jun 24, 2015 at 15:13

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

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.