I'm self-studying The Geometry of Physics, Third Edition, by Frankel, and the book's two equations defining the Lie derivative of a form, equations 4.16 on page 132, don't seem like they're consistent with each other.
Let $X$ be a vector field with local flow $\phi(t)$, let $x$ be a particular point, and let $\alpha$ be a $p$-form field, with $\alpha_x$ denoting $\alpha$ at point $x$. The book then defines
$$\begin{align}
\mathcal{L}_X \alpha&=\frac{d}{dt}[\phi_t^* \alpha]_{t=0}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)\cr
&=\lim_{t\to 0}\frac{\phi_t^* \alpha_{\phi_t x}-\alpha_x}{t}\ \ \ \ \ (2)
\end{align}$$
Equation (2) makes sense to me. What I don't understand is how equation (1) is equivalent to it. Given that we can write
$$\lim_{t\to 0}\frac{\phi_t^* \alpha_{\phi_t x}-\alpha_x}{t}=\lim_{t\to 0}\frac{\phi_t^* \alpha_{\phi_t x}-\phi_0^* \alpha_{\phi_0 x}}{t}\ ,$$
we can define a function
$$f(\lambda)= \phi_{\lambda}^* \alpha_{\phi_{\lambda} x}$$
in terms of which
$$\mathcal{L}_X \alpha=\lim_{t\to 0}\frac{f(t)-f(0)}{t} = \frac{d}{dt}[f(t)]_{t=0}= \frac{d}{dt}[\phi_{t}^* \alpha_{\phi_t x}]_{t=0}\ ,$$
which is different from equation (1), because the $\alpha$ in equation (1) presumably means $\alpha_x$, not $\alpha_{\phi_t x}$.
Why does the $\phi_t$ disappear? I'm guessing there's some reason pertaining to limits that makes it possible to drop the $\phi_t$, but I'm just not seeing it.
Best Answer
This is a typo.
$$(\mathcal{L}_X\alpha)_p=\lim_{t\rightarrow 0}\frac{1}{t}((\phi_t^*\alpha)_p-\alpha_p).$$
This definition appears on page 150 of Spivak's A Comprehensive Introduction to Differential Geometry, Volume 1, 3rd edition as $$(\mathcal{L}_X\omega)(p)=\lim_{h\rightarrow 0}\frac{1}{h}((\phi_h^*\omega)(p)-\omega(p)).$$
This can also be found here (link to a popular math blog).