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Lie Derivatives

  1. Jan 6, 2007 #1
    Let [itex]M[/itex] be a diff. manifold, [itex]X[/itex] a complete vectorfield on [itex]M[/itex] generating the 1-parameter group of diffeomorphisms [itex]\phi_t[/itex]. If I now define the Lie Derivative of a real-valued function [itex]f[/itex] on [itex]M[/itex] by

    [tex]\mathscr{L}_Xf=\lim_{t\rightarrow 0}\left(\frac{\phi_t^*f-f}{t}\right)=\frac{d}{dt}\phi_{t}^{*}f |_{t=0}[/tex]

    (where [itex]...^{*}[/itex] denotes the pull-back by ...) that's mere notation, right? I.e. the limit is not a functional limit, right? This simply defines how I should evaluate the Lie Derivative, true?

    Now, if I know that

    (1) [tex]\phi_t^*\theta\cdot\phi_{-t*}Y\s=\s\phi_t^*(\theta\cdot Y)[/tex]

    where [itex]Y[/itex] is another vectorfield on [itex]M[/itex], how could I prove that

    (2) [tex](\mathscr{L}_X\Theta)\cdot Y + \Theta\cdot (\mathscr{L}_XY)=\mathscr{L}_X(\Theta\cdot Y)[/tex]?

    (Here, the Lie Derivative is defined correspondingly.) If I take the time derivative at [itex]t=0[/itex] of both sides in (1), I can't apply the standard (i.e. banach space) product rule, because the constituents are not real functions! What can I do?

    Thanks in advance. Best regards...Cliowa
  2. jcsd
  3. Jan 10, 2007 #2


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    Well, you could approximate the flow as
    \phi_t \simeq 1 + t \phi
    power expand in t, and prove it that way.

    Or you could be even more of a physicist and write it out in components. ;)
  4. Feb 12, 2008 #3
    Lie derivatives for vector fields and tensor fields are not so tricky as Lie derivatives for spinor fields. I would like to contact somebody who is expert in this area. Please, let me know if you are such a person.
  5. Feb 12, 2008 #4

    George Jones

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    Have you looked at chapter 7, Differentiation of Spinor Fields, from of the book Geometry, Spinors, and Applications by Donal J. Hurley and Michael A. Vandyck?
  6. Feb 12, 2008 #5
    you define the operator by what it do on a function f, so if we take a point m in the manifold

    [tex]\mathscr{L}_Xf(m)=\lim_{t\rightarrow 0}\left(\frac{\phi_t^*f(m)-f(m)}{t}\right) [/tex]

    defines a map g(t) from R to R, so this is just the usual derivative, so you have defined an operator by what it does on a function.

    a bit like you could say that you can define a function on R by saying what it should do on a real number fx.

    fx = x^2

    or as we usually write

    f(x) = x^2

    so you see, this is what you usually does, but now it is not from R but some space of functions

    you could write

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