Showing that the Lie derivative of a function is the directional deriv

  • #1
Hi!

To boost my understanding of the mathematics in relation to general relativity, I'm reading about Lie derivatives in Sean Carroll's "Spacetime and geometry". Here he defines the Lie derivative of a (k,l) tensor at the point p along the vectorfield V as

$$\mathcal{L}_V T^{\mu_1 \cdots \mu_k}_{\ \ \ \nu_1 \cdots \nu_l} = \lim_{t \to 0} \frac{\Delta_t T^{\mu_1 \cdots \mu_k}_{\ \ \ \nu_1 \cdots \nu_l}}{t} $$

where

$$\Delta_t T^{\mu_1 \cdots \mu_k}_{\ \ \ \nu_1 \cdots \nu_l}(p) = \phi^*_t[T^{\mu_1 \cdots \mu_k}_{\ \ \ \nu_1 \cdots \nu_l}(\phi_t(p))] - T^{\mu_1 \cdots \mu_k}_{\ \ \ \nu_1 \cdots \nu_l}(p)$$

and ##\phi_t## is a family of diffeomorphisms parametrised by ##t## generated by the vector field V. The * here denotes the pullback operation. In the book at page 432 he states that one should after a moments reflection be able to be convinced that the Lie derivative of an ordinary function is given by

$$\mathcal{L}_V f = V(f)$$

i.e. that it is just given by the action of the vector V on the scalar (the directional derivative). However I do not see how this follows from the above definition. If I just plug in a (0,0)-tensor (i.e. a scalar) I get

$$\mathcal{L}_V(p) = \lim_{t\to 0} \frac{\phi^*_t[f (\phi_t(p))] - f(p)}{t}$$

and I do not see how this is supposed to reduce to the directional derivative. Can someone enlighten me? :)
 

Answers and Replies

  • #2
fzero
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Let the manifold be ##M## and the vector field be expressed in local coordinates as

$$ V = \sum_\mu v^\mu \partial_\mu.$$

The family of diffeomorphisms generated by ##V## is the map ##\phi_t:\mathbb{R}\times M \rightarrow M##, where

$$\phi_t: (t, p^1 , \ldots ,p^n) \mapsto (v^1 t + p^1 , \ldots, v^n t + p^n).$$

If you now apply this to your formula (by Taylor expanding ##f(\phi_t)## or something more rigorous), you should be able to arrive at the result.
 
  • #3
WannabeNewton
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There's a very easy way to see it. Let ##\xi## be a vector field and ##\alpha## a scalar field. Taking ##p\in M##, and ##\phi_t## the one-parameter group of diffeomorphisms generated by ##\xi## on a neighborhood of ##p##, we have ##\xi|_p (\alpha) = \frac{\mathrm{d} }{\mathrm{d} t}\{(\alpha \circ \phi_t)(p)\}|_{t = 0}##.

Now ##\phi_t^{*}(\alpha) = \alpha \circ \phi_t## hence
##(\mathcal{L}_{\xi}\alpha)(p) = \lim_{t\rightarrow 0}\frac{(\phi_t^{*}(\alpha))(p) - \alpha(p)}{t}\\ = \lim_{t\rightarrow 0}\frac{(\alpha \circ \phi_t)(p) - (\alpha \circ \phi_0)(p)}{t}\\ = \frac{\mathrm{d} }{\mathrm{d} t}\{(\alpha \circ \phi_t)(p)\}|_{t = 0}\\ = \xi|_{p}(\alpha)##.
 
  • #4
There's a very easy way to see it. Let ##\xi## be a vector field and ##\alpha## a scalar field. Taking ##p\in M##, and ##\phi_t## the one-parameter group of diffeomorphisms generated by ##\xi## on a neighborhood of ##p##, we have ##\xi|_p (\alpha) = \frac{\mathrm{d} }{\mathrm{d} t}\{(\alpha \circ \phi_t)(p)\}|_{t = 0}##.

Now ##\phi_t^{*}(\alpha) = \alpha \circ \phi_t## hence
##(\mathcal{L}_{\xi}\alpha)(p) = \lim_{t\rightarrow 0}\frac{(\phi_t^{*}(\alpha))(p) - \alpha(p)}{t}\\ = \lim_{t\rightarrow 0}\frac{(\alpha \circ \phi_t)(p) - (\alpha \circ \phi_0)(p)}{t}\\ = \frac{\mathrm{d} }{\mathrm{d} t}\{(\alpha \circ \phi_t)(p)\}|_{t = 0}\\ = \xi|_{p}(\alpha)##.

Thanks! The only think I'm not sure if I understand is why

##\xi|_p (\alpha) = \frac{\mathrm{d} }{\mathrm{d} t}\{(\alpha \circ \phi_t)(p)\}|_{t = 0}##.

I might not have the definition of the generating vector field straight. Equation (B.3) in Carroll does not help that much.
 
Last edited:
  • #5
WannabeNewton
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It's because ##\gamma(t) = \phi_t(p)## is an integral curve of ##\xi## that starts at ##p## (by definition of what it means for ##\phi_t## to be the one-parameter group of diffeomorphisms generated by ##\xi##) so ##\xi|_p(\alpha) = \dot{\gamma}|_p (\alpha) = \frac{d}{dt}(\alpha \circ \gamma)|_{t = 0} = \frac{d}{dt}\{(\alpha \circ \phi_t)(p)\}|_{t = 0}##.
 

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