Thursday, 6 September 2018

Question: Reductio ad absurdum on commutators?

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 (2.05), 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 function f, it vanishes for all functions. This is implied by the question but not proven.

I proved it this way using 'Reductio ad absurdum'. Is this correct?

Our starting point is f  ≠ 0 and [X,Y](f) = 0. We have another function g ≠ 0 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 = 0, which breaks our starting assumption, with which there must be some error. The only non trivial possibility is that [X,Y](g) = 0. QED?
The question is at
https://www.physicsforums.com/threads/spacetime-and-geometry-vanishing-commutators.954767/
I then searched for Commutator and found this
Commutator of two vector fields:
https://www.physicsforums.com/threads/commutator-of-two-vector-fields.950661/
One answer contained
"[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;"

Perhaps I should have put my question in  the physics / Special and General Relativity forum.

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