# Spacetime and Geometry: Vanishing commutators#2

Gold Member
This is a refinement of a previous thread (here). I hope I am following correct protocol.

## Homework Statement

I am studying Spacetime and Geometry : An Introduction to General Relativity by Sean M Carroll and have a question about commutators of vector fields. A vector field on a manifold can be thought of as differential operator which transforms smooth functions to smooth functions on the manifold. For a vector field X and a function f(xi) we write

X(f) = g, where g is another function. We then define the commutator of two fields X and Y as

[X,Y](f) = X(Y(f)) - Y(X(f)

In the exercise I am working on, we are asked to find two vector fields whose commutator does not vanish. An important step is to show that if the commutator vanishes for one non-trivial function f, it vanishes for all non-trivial functions. This is implied by the question but not proven. (f = a constant everywhere would be trivial)

See above

## The Attempt at a Solution

I proved it this way using 'Reductio ad absurdum'.

Our starting point is f is non-trivial and [X,Y](f) = 0. We have another function g is non-trivial and [X,Y](g) ≠ 0.

We already know that commutators are linear (from the previous exercise), so

[X,Y](f + g) = [X,Y](f) + [X,Y](g)
or
[X,Y](f + g) = [X,Y](g)

Therefore f is trivial, which breaks our starting assumption, with which there must be some error. The only non trivial possibility is that [X,Y](g) = 0. QED?

Is my solution correct?
Is there a more obvious solution? (I.e. Am I missing something? See below)
I would like a good definition of 'non-trivial'.

In another thread on Commutator of two vector fields (here), I read "[X,Y] describes how far the endpoints of a rectangle vary if you go along X followed by Y or the other way around. Commuting vector fields mean the two path end at the same point". This does not mention a particular function. I think it supports my conclusion.

stevendaryl
Staff Emeritus
There was no reason to create a new thread just to make a second post.

Our starting point is f is non-trivial and [X,Y](f) = 0. We have another function g is non-trivial and [X,Y](g) ≠ 0.

We already know that commutators are linear (from the previous exercise), so

[X,Y](f + g) = [X,Y](f) + [X,Y](g)
or
[X,Y](f + g) = [X,Y](g)

Therefore f is trivial, which breaks our starting assumption, with which there must be some error. The only non trivial possibility is that [X,Y](g) = 0. QED?
I think you're going down a wrong path. ##f## being a constant is not the only possibility for ##[X,Y](f) = 0##. But in any case, how does ##[X,Y](f+g) = [X,Y](g)## imply that ##[X,Y](g) = 0##?

I think that you need to go back to what a vector field is, and what it means to apply it to a function. If you pick a particular coordinate system ##x^1, x^2, ...##, then you can write a vector ##X## in terms of components ##X^1, X^2, ...## and then ##X(f) = X^1 \frac{\partial f}{\partial x^1} + X^2 \frac{\partial f}{\partial x^2} + ...##. Applying ##X## to ##f## means applying the "directional derivative". So what happens when you compute ##X(Y(f))##? (In general, the components ##X^1, X^2, ...## are not constants).

George Keeling
Gold Member
Thanks. You are right. As you have noticed I am still struggling with what X(f) really means. And thanks fort the advice on threads.

stevendaryl
Staff Emeritus
$$X(f) = X^i \frac{\partial f}{\partial x^i}$$ from there I can get pretty quickly to the components of ##[X,Y]##
$$[X,Y]^i = X^j \frac {\partial Y^i}{\partial x^j} - Y^j \frac {\partial X^i}{\partial x^j}$$ That also gives ##[X,Y]##