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Action of Lie Brackets on vector fields multiplied by functions |
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| Jun14-11, 03:56 PM | #1 |
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Action of Lie Brackets on vector fields multiplied by functions
Hi,
Is there a specific product rule or something one must follow when applying the lie bracket/ commutator to two vector fields such that one of them is multiplied by a function and added to another vector field? This is the expression given in my textbook but I don't see how: [fX+Z,Y] = f[X,Y] + [Z,Y] - (Yf)Xg I don't see where the third term on the right hand side comes from. I'd really appreciate some help on this because I'm self-learning differential geometry for a research project and almost all my doubts revolve around my not understanding how lie brackets work. So any help will be appreciated. Thanks! |
| Jun15-11, 08:11 AM | #2 |
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Mentor
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Let's go back a couple of steps. If [itex]X[/itex] is a vector field and [itex]f[/itex] and [itex]g[/itex] are smooth functions, then both [itex]Xf[/itex] and [itex]Xg[/itex] are functions. Because [itex]f[/itex] and [itex]g[/itex] are both functions, the product [itex]fg[/itex] is also a function on which [itex]X[/itex] can act. Consquently, [itex]X \left(fg\right)[/itex] is a function. [itex]X[/itex] acts like a derivative (is a derivation) on the set (ring) of smooth functions, i.e., [tex]X \left(fg\right) = g Xf + f Xg.[/tex] Now, use the above and expand [tex]\left[ fX,Y \right]g .[/tex] |
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