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(adsbygoogle = window.adsbygoogle || []).push({});

[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

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Lie Derivatives

**Physics Forums | Science Articles, Homework Help, Discussion**