- #1

George Keeling

Gold Member

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*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(x

^{i}) 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)

## Homework Equations

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.