Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Forums
Homework Help
Advanced Physics Homework Help
Spacetime and Geometry: Vanishing commutators#2
Reply to thread
Message
[QUOTE="George Keeling, post: 6052701, member: 647945"] [I]This is a refinement of a previous thread ([URL='https://www.physicsforums.com/threads/spacetime-and-geometry-vanishing-commutators.954767/#post-6052150']here[/URL]). I hope I am following correct protocol.[/I] [h2]Homework Statement [/h2] 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[SUP]i[/SUP]) 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) [h2]Homework Equations[/h2] See above [h2]The Attempt at a Solution[/h2] 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 [I]Commutator of two vector fields[/I] ([URL='https://www.physicsforums.com/threads/commutator-of-two-vector-fields.950661/']here[/URL]), 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. [/QUOTE]
Insert quotes…
Post reply
Forums
Homework Help
Advanced Physics Homework Help
Spacetime and Geometry: Vanishing commutators#2
Back
Top