at the moment I am self-studying differential geometry using the book by Guillemin and Haine. In this book, Exercise 2.6.ix asks me to prove the following:
Let $U \subset \mathbb{R}^n$ open, $f_t: U \to U, t \in \mathbb{R}$, a one-parameter group of diffeomorphisms and (the vector field) $\boldsymbol{v}$ its infinitesimal generator. Given $\omega \in \Omega^k(U)$, show that $$ \left. \frac{\mathrm{d}}{\mathrm{d}t}f_t^\star \omega\right\vert_{t=0} = L_\boldsymbol{v}\omega,\qquad\qquad(*) $$ where $L_\boldsymbol{v}\omega := \mathrm{d}\iota_\boldsymbol{v}\omega + \iota_\boldsymbol{v}\mathrm{d}\omega$ is the Lie derivative and $f_t^\star \omega$ is the pullback of $\omega$ by $f_t$.
Some more context:
They give a sketch of the proof: (1) Show that ($*$) holds for zero-forms. (2) Show that if ($*$) holds for some $k$-form $\omega$, it also holds for $\mathrm d\omega$. (3) Show that if holds for $\omega_1$ and $\omega_2$, it also holds for $\omega_1 \wedge \omega_2$. (4) Show that ($*$) thus holds for general forms, since they are a wedge product of zero-forms and exact one-forms.
(Importantly, in the next Exercise one uses the result ($*$) to show that ($*$) holds also for all $t$, and that the Lie-derivative $L_\boldsymbol{v}$ commutes with $f^\star_t$. I see that in the literature ($*$) for all $t$ is used as a definition of the Lie-derivative. However, for solving the exercise I can only work with what I stated above.)
I have a hard time proving point (2), and would appreciate some feedback to make my prove correct or point me to what I am missing. Judging from the structure of the exercises so far, I expect the proof can be done in a couple of lines.
My attempt at a proof looks like this:
Claim:
If ($*$) holds for $\omega$, then $$ \left.\frac{\mathrm{d}}{\mathrm{d}t}f_t^\star \mathrm{d}\omega\right\vert_{t=0} = L_\boldsymbol{v}\mathrm{d}\omega,\qquad\qquad(2). $$
Proof (attempt):
First, by differentiating $f^\star_t \mathrm{d}\omega = \mathrm{d}f^\star_t\omega$ with respect to $t$, and then evaluating at $t=0$, the LHS of (2) becomes: $$ LHS(2) = \left.\frac{\mathrm{d}}{\mathrm{d}t}\mathrm{d}f^\star_t \omega\right\vert_{t=0}.$$
By using $L_\boldsymbol{v}\mathrm{d}\omega = \mathrm{d}L_\boldsymbol{v}\omega$ (which follows directly from the definition of the Lie derivative) together with the assumption, the RHS becomes $$ RHS(2) = \mathrm{d}\left.\frac{\mathrm{d}}{\mathrm{d}t}f^\star_t \omega\right\vert_{t=0}.$$
So, I am done if I show that $$ \left.\frac{\mathrm{d}}{\mathrm{d}t}\mathrm{d}f^\star_t \omega\right\vert_{t=0} = \mathrm{d}\left.\frac{\mathrm{d}}{\mathrm{d}t}f^\star_t \omega\right\vert_{t=0}.\qquad\qquad(**)$$
So now the wonky bit:
I assumed (maybe wrongly) that for $p \in U$ the pullback $f^\star_t \omega$ has the form $$ (f^\star_t \omega)_{p} = \sum_I (a_I\circ f_t)(p) \mathrm{d}x_I.\qquad\qquad(***)$$.
What I mean by that is that if $\omega = \sum_I \Phi_I \mathrm{d}x_I$, then $$f_t^\star \omega = \sum_I (\Phi_I \circ f_t) \mathrm{d}(x_I \circ f_t) = \sum_I (\Phi_I \circ f_t) \mathrm{d}f_{t,i_{1}} \wedge \dots \wedge \mathrm{d}f_{t,i_{k}} \\ = \sum_I (\Phi_I \circ f_t) \sum_{j_1} (\frac{\partial f_{t,i_{1}}}{\partial x_{j_1}} \circ f_t) \mathrm{d}x_{j_1} \wedge \dots \wedge \sum_{j_k} (\frac{\partial f_{t,i_{k}}}{\partial x_{j_k}} \circ f_t) \mathrm{d}x_{j_k} \\ = \sum_I (\Phi_I \circ f_t) \left(\prod_{l=1}^k \left(\frac{\partial f_{t,i_{k}}}{\partial x_{j_k}} \circ f_t\right)\right) \bigwedge_{l=1}^k \mathrm{d}x_{j_l}\\ =: \sum_I (a_I\circ f_t) \mathrm{d}x_I, $$ where $a_I$ is just some function defined by the last equality. This $a_I$ incorporates the sums and products of $\Phi_I$, the partial derivatives of the $f_{t,i}$ and the orthogonality and redundancy of the basis elements. Since all these functions are $C^\infty$, this combination can be written as a $C^\infty$-function $a_I$.
Then, after some explicit calculation using chain and product rules for functions, and the fact that $\gamma(t) = f_t(p)$ is an integral curve starting at $p$ for $t=0$, and thus $\left(\left.\frac{\mathrm{d}}{\mathrm{d}t}(f_t)\right\vert_{t=0}\right)_i = \left(\left.\frac{\mathrm{d}\gamma(t)}{\mathrm{d}t}\right\vert_{t=0}\right)_i=v_i$, I could explicitly calculate that both sides of ($**$) agree.
However, I still haven't been able to properly show ($***$). Also, I have the feeling that I am overcomplicating things and trying to do too much in coordinates.
Can you help me simplify or complete my proof?