Example of calculating Lie derivative

Question

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 $v_t(m) = {\frac{d}{ds}}_{|d=t} (f_s \circ f_t^{-1})(m)$ ?

Answer 1

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

1. It holds when $\eta$ is a $0$-form, i.e., a function on $M$.
2. It holds for exact $1$-forms.
3. 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 $\eta$.

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1. 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 $X_{t_0} \circ f_{t_0} = \left.\frac{d}{dt}\right|_{t=t_0}f_t$.

2. 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.)

3. 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.

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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.)