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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? :)

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

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