How does this Lie derivative axiom follow from my definition of the Lie derivative?

Question

$\newcommand{\L}{\mathcal{L}}$ $\newcommand{\der}[2][]{\frac{d#1}{d#2}}$ $\newcommand{\pder}[2][]{\frac{\partial#1}{\partial#2}}$ The definition of the Lie derivative which I'm starting with is $$ \L_v(\alpha) := \der{t} (\phi_t^*(\alpha))|_{t=0}$$ where $X$ is a manifold, $\alpha$ is some tensor, $v \in \Gamma(TX)$ is a vector field and $\phi : \mathbb{R} \times X \rightarrow X$ (writing $\phi_t(x) = \phi(t, x)$) is the 1-parameter group of diffeomorphisms on $X$.

Wikipedia says that this definition follows from four axioms, the first one being $$ \L_v(f) = v(f)$$ for $f$ a smooth function and the second one being $$ \L_v(S \otimes T) = \L_v(S) \otimes T + S \otimes \L_v(T) $$ (plus two more).

I'd like to know how to show the first one I've written. Referring back to the definition I started with, I know that if $\alpha = f$ a smooth function then $\phi_t^*(f) = f \circ \phi_t$. So using the chain rule $$ \begin{align} \L_v(f) &= \der{t}(f \circ \phi_t)|_{t=0} \\ &= \sum_{k=1}^n\pder[f]{x_k} \der[x_k]{t}|_{t=0} \\ &= \sum_{k=1}^n\pder[f]{x_k} v^k \end{align} $$ Here $(x_1, \dots, x_n)$ are local coordinates on $X$ and I've written $v^k = \der[x_k]{t}|_{t=0}$ since these are real numbers (coefficients). But then I think that $\sum_{k=1}^n\pder[f]{x_k} v^k$ is what we get if we "apply $v$ the vector field to $f$", i.e. $$ \L_v(f) = \sum_{k=1}^n\pder[f]{x_k} v^k = v(f) $$ Is this the correct way to do it? Thank you.

Answer 1

Yes, your approach is essentially correct. However, it's not entirely clear just from what you've written that $\sum_k \frac{\partial f}{\partial x_k} v^k = v(f)$. One way to make it more explicit is the following:

Note that $\phi_t(p)=\gamma_p(t)$, where $\gamma_p$ is the integral curve of the vector field $v$ satisfying the initial condition $\gamma_p(0)=p$, so $\gamma_p'(t)=v_{\gamma_p(t)}$ and in particular $\gamma_p'(0)=v_p$. Then

$$(\mathcal{L}_v(f))(p) = \frac{d}{dt} \biggr|_{t=0} f(\phi_t(p)) = \frac{d}{dt} \biggr|_{t=0} (f \circ \gamma_p)(t) = v_p(f)=(v(f))(p)$$

Where the second to last equality is the definition of $v_p(f)$.

Answer 2

Yes, it is the correct way to do it show also it coincides on $1$-tensors (you need for this the axiom 4 in wiki) that your definition verifies the second axioms.

This implies that your definition coincide with the definition of wiki on functions and $1$-forms, the second axiom implies that recursively one can show that it coincides on every tensors.