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Action of Lie Brackets on vector fields multiplied by functions

by tut_einstein
Tags: action, brackets, fields, functions, multiplied, vector
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tut_einstein
#1
Jun14-11, 03:56 PM
P: 31
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!
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George Jones
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Jun15-11, 08:11 AM
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Quote Quote by tut_einstein View Post
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
Yes.
Quote Quote by tut_einstein View Post
This is the expression given in my textbook
Which book?

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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