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Spacetime and Geometry: Vanishing commutators#2
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[QUOTE="stevendaryl, post: 6052715, member: 372855"] There was no reason to create a new thread just to make a second post. 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). [/QUOTE]
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Spacetime and Geometry: Vanishing commutators#2
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