# How can the commutator of vector fields be a genuine vector field?

I want to look at the transformation law for the commutator of two basis vectors. I understand completely how this works for a holonomic frame: The basis vector fields act on functions as the partial derivative operators wrt to the appropriate coordinates, so the commutator vanishes. But when we transform to an anholonomic frame, the basis vector (fields) aren’t going to be the partial derivatives for some coordinate system, so the resulting components will be nonvanishing.

This transformation law seems to me to reek of inhomogeneity, and I can’t seem to reconcile it with what I’ve been taught, which is that the commutator of vector fields is itself a vector field.

UPDATE:

Here’s my problem. The object I have here should be a tensor (right?), and it vanishes in the holonomic frame. But now, in the anholonomic frame, I get nonzero terms. This should not be how a tensor transforms. What am I doing wrong here?

• Well, you have an explicit formula that, in particular, gives you that vector field you get when you take the commutator of arbitrary vector fields (just because the frame is holonomic doesn't mean that the commutator of general vector fields will be $0$). If you have taken a course in smooth manifolds, you know that linearity over $C^\infty$ functions establishes that you have a tensor field, and that is the case with the bracket of two vector fields. Sep 23 '20 at 20:08
• Thanks for your edit (although we'd prefer such equations to be typeset), which proves $e_a=\Lambda_a^\mu\partial_\mu,\,e_b=\Lambda_b^\nu\partial_\nu\implies[e_a,\,e_b]=(e_a\Lambda_b^\mu-e_n\Lambda_a^\mu)\partial_\mu$. This computes the commutator of two components of the same tensor. My calculations were of the commutator of two true vectors, not individual components.
– J.G.
Sep 24 '20 at 17:50
• @J.G. I think you meant ‘b’ instead of ‘n’ for the index on the last e... And I still don’t get why my object goes from being zero to the expression I got below if it is truly a vector. Sep 24 '20 at 17:59
• Sorry, yes, I did. What I'm saying is if the quantities whose commutator you compute have uncontracted indices, the remainder will have twice as many, so won't transform the same way. (My original answer worked without uncontracted indices; the current version is different in that the way it defines a commutator automatically adds a contraction.)
– J.G.
Sep 24 '20 at 18:04
• @J.G. I think I’m starting to understand... But how does the method in your answer work if the basis used is not a coordinate basis? Sep 24 '20 at 18:12

Since vectors transforms as $$V^a=x^a_AV^A$$ between two coordinate systems with $$x^a_A:=\frac{\partial x^a}{\partial x^A}$$,\begin{align}U^a\partial_aV^b&=U^Ax^a_A\partial_a(x^b_BV^B)\\&=U^A\partial_A(x^b_BV^B)\\&=U^A\partial_AV^Bx^b_B+U^AV^B\underbrace{x^b_{AB}}_{\tfrac{\partial^2x^a}{\partial x^A\partial x^B}},\end{align}so $$U^a\partial_aV^b-V^a\partial_aU^b=x^b_B(U^A\partial_AV^B-V^A\partial_AU^B)$$, i.e. $$£_UV^b=x^b_B£_UV^B$$. This is the usual transformation law for a vector.

• @excitonfield Perhaps you instead wanted a proof the Lie derivative transforms as a vector. (Two commutator definitions are common.) I'll let you know when I've updated my answer.
– J.G.
Sep 24 '20 at 5:03
• @excitonfield OK, done.
– J.G.
Sep 24 '20 at 5:24
• I just don’t understand how this reconciles with the whole inhomogeneity issue I had... I don’t understand how we can expect the commutator in my initial question to be well-defined if it vanishes in one frame, but not in another... Sep 24 '20 at 16:35
• @excitonfield I think you need to update your question to show a putative example of a vector that's $0$ in one frame but not another because, as I've shown, the Lie derivative transforms tensorally.
– J.G.
Sep 24 '20 at 16:46
• Just put one in :) Sep 24 '20 at 17:35