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Spacetime and Geometry: Vanishing commutators

  • #1
George Keeling
Gold Member
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10

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

Homework Equations


Is my solution below correct?
Is there a more obvious solution? (I.e. Am I missing something?)

The Attempt at a Solution


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

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?
 

Answers and Replies

  • #2
stevendaryl
Staff Emeritus
Science Advisor
Insights Author
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It's not true that " if the commutator vanishes for one function f, it vanishes for all functions". A commutator produces another vector field, and a vector field can yield zero when applied to one function, and something nonzero when applied to another. For example, in one dimension, let ##f## be the function that is equal to 1 at every point. Then ##X(f) = 0## for any vector field ##X##.
 

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