The set $Ham(M,\omega)$ of Hamiltonian diffeomorphisms is a subgroup of the set of symplectic diffeomorphisms $Symp(M\omega)$

Question

The set $Ham(M,\omega)$ of Hamiltonian diffeomorphisms is a subgroup of the set of symplectic diffeomorphisms $Symp(M\omega)$ is proven in the following lectures notes https://jumpshare.com/s/tkHDIaSse4wwcxlG3mW1 (proof of proposition 3.42 -pag 110 (=pag 120 of the pdf))

But I have no idea what is going on in some computations:

What is going in the two highlighted statements?

1. In the first one how do we get a sum? The derivative of a composition should be by the chain rule $\frac{d}{dt}(\varphi_F^t\circ\varphi_G^t)=\frac{d}{dt}(\varphi_F^t)|_{\varphi_G^t}\frac{d}{dt}(\varphi_G^t)$. I don't see how they I getting a + there and why are they getting those terms

2. In the second one, how does 3.33 imply that the flow of $G^t$ is $(\varphi_F^t)^{-1}$? and how is this showing that $\varphi^{-1}\in Ham(M,\omega)$?

edit:

Answer 1

@TedShifrin pointed out the way to go. Let me write it out.

Consider formally the map $(r, s)\to \phi_F^r\circ \phi_G^s$. Then by the definition of flow and the chain rule, \begin{align*} \frac{\partial}{\partial r} (\phi_F^r\circ \phi_G^s) &= X_{F^r}\circ \phi_F^r\circ \phi_G^s,\\ \frac{\partial}{\partial s} (\phi_F^r\circ \phi_G^s) &= D\phi_F^r X_{G^s}\circ \phi_G^s \quad \text{chain rule here for }\phi_F^r\circ. \end{align*} Now consider the map $t\to (r, s)=(t, t)$, then another chain rule for this change of variables gives \begin{align*} \frac{d}{dt} (\phi_F^t\circ \phi_G^t) &= \frac{dr}{dt} \frac{\partial}{\partial r}\Big|_{r=t,s=t} (\phi_F^r\circ \phi_G^s) + \frac{ds}{dt}\frac{\partial}{\partial s}\Big|_{r=t,s=t} (\phi_F^r\circ \phi_G^s)\\ &= X_{F^t}\circ \phi_F^t\circ \phi_G^t + D\phi_F^t X_{G^t}\circ \phi_G^t. \end{align*}

Now the second question mainly follows from the two lines after (3.3). So for the particular choice of $$ G^t = -F^t\circ \phi_F^t, $$ the function $$ F^t + G^t\circ (\phi_F^t)^{-1} = F^t - F^t = 0. $$ Its associated flow is hence the identity, $Id$, but it is also $\phi_F^t\circ \phi_G^t$. So for this particular $G^t$, we have $$ \phi_F^t\circ \phi_G^t=Id. $$ So $(\phi_F^t)^{-1} = \phi_G^t\in \text{Ham}(M, \omega)$, since it is generated by a function $G^t$.