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

center o bass

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? :)

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fzero

Homework Helper
Gold Member
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.

WannabeNewton

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

center o bass

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.

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WannabeNewton

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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