Example of calculating Lie derivative Let $(M, \omega)$ be a symplectic manifold. Let $f_t \in Diff (M) $ be a smooth family of diffeomorphisms on $M$, $t \in \mathbb{R}, $such that $f_0 = id_M$.
Why do we have this equality:
$$\frac{d}{dt}f_t^* \omega = f_t^*(L(v_t) \omega),$$
Where $L(v_t)$ denotes the Lie derivative by the vector field $v_t$ which is given  by $v_t(m) = {\frac{d}{ds}}_{|d=t} (f_s \circ f_t^{-1})(m)$  ?
 A: I'll write $\mathcal{L}_X$ for the Lie derivative with respect to $X$. We will prove that, for any $k$-form $\eta$ on $M$,
$$
\frac{d}{dt} f_t^*\eta = f_t^*\mathcal{L}_{X_t}\eta.
$$
In particular, it holds for the symplectic $2$-form $\omega$, which is your question. To do this, we will prove that

*

*It holds when $\eta$ is a $0$-form, i.e., a function on $M$.

*It holds for exact $1$-forms.

*If it holds for forms $\alpha$ and $\beta$, then it holds for $\alpha\wedge\beta$.

Then, since the algebra of differential forms on $M$ is (locally) generated by sums of wedge products of functions and exact $1$-forms, the result holds for an arbitrary $\eta$.


*

*If $\eta\colon M \to \mathbb R$ is a smooth function on $M$, then

$$
\left.\frac{d}{dt}\right|_{t=t_0} f_t^*\eta = \left.\frac{d}{dt}\right|_{t=t_0}\eta \circ f_t = (X_{t_0}\eta) \circ f_{t_0} = f_{t_0}^*(\mathcal{L}_{X_{t_0}}\eta), 
$$
where in the second equality we used $X_{t_0} \circ f_{t_0} = \left.\frac{d}{dt}\right|_{t=t_0}f_t$.


*For $\eta$ as in (1),

$$
\left.\frac{d}{dt}\right|_{t=t_0}f_t^*(d\eta) = \left.\frac{d}{dt}\right|_{t=t_0}d(f_t^*\eta) = d\left(\left.\frac{d}{dt}\right|_{t=t_0}f_t^*\eta\right) = d(f_{t_0}^*(\mathcal{L}_{X_{t_0}}\eta)) = f_{t_0}^*(\mathcal{L}_{X_{t_0}}(d\eta)),
$$
where in the third equality, $d$ commutes with the time derivative by equality of mixed partials. (One can see this by writing it out in local coordinates.)


*Finally, if it holds for $\alpha$ and $\beta$, then

\begin{align*}
\left.\frac{d}{dt}\right|_{t=t_0} f_t^*(\alpha\wedge\beta) &= \left.\frac{d}{dt}\right|_{t=t_0} (f_t^*\alpha)\wedge(f_t^*\beta) \\
&= \left(\left.\frac{d}{dt}\right|_{t=t_0}f_t^*\alpha\right)\wedge(f_{t_0}^*\beta) + (f_{t_0}^*\alpha)\wedge\left(\left.\frac{d}{dt}\right|_{t=t_0}f_t^*\beta\right) \\
&= (f_{t_0}^*(\mathcal{L}_{X_{t_0}}\alpha)) \wedge (f_{t_0}^*\beta) + (f_{t_0}^*\alpha)\wedge(f_{t_0}^*(\mathcal{L}_{X_{t_0}}\beta)) \\
&= f_{t_0}^*\left( (\mathcal{L}_{X_{t_0}}\alpha)\wedge\beta + \alpha\wedge(\mathcal{L}_{X_{t_0}}\beta) \right) \\
&= f_{t_0}^*(\mathcal{L}_{X_{t_0}}(\alpha\wedge\beta)).
\end{align*}
The product rule in the second equality can be proven using the exact same method as the single-variable calculus product rule: write out the limit definition of the derivative, and add and subtract a suitable term.

I'd like to remark that, following the exact same outline, one gets the following formula for a time-dependent family $\eta_t$ of differential forms:
$$
\frac{d}{dt} f_t^*\eta_t = f_t^*\left(\mathcal{L}_{X_t}\eta_t + \frac{d}{dt}\eta_t\right).
$$
Step 1 becomes a tiny bit trickier. Steps 2 and 3 don't really change, other than becoming a little messier in notation. This kind of formula shows up when you're using a Moser-type argument, such as in a proof of Darboux's theorem. (This is why I wanted to bring it up.)
