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? :)

 

[fzero]

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.

 

[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 $\xi|_p(\alpha) = \dot{\gamma}|_p (\alpha) = \frac{d}{dt}(\alpha \circ \gamma)|_{t = 0} = \frac{d}{dt}\{(\alpha \circ \phi_t)(p)\}|_{t = 0}$.