Lie algebra of Hamiltonian functions

Question

Could anyone explain to me how to obtain the Corollaries 2 and 3 in the following paragraph taken by Arnold's book?

The Lie algebra of hamiltonian functions

The hamiltonian vector fields on a symplectic manifold form a subalgebra of the Lie algebra of all fields. The hamiltonian functions also form a Lie algebra: the operation in this algebra is called the Poisson bracket of functions. The first integrals of a hamiltonian phase flow form a subalgebra of the Lie algebra of hamiltonian functions.

A The Poisson bracket of two functions

Let $(M^{2n}, \omega^2)$ be a symplectic manifold. To a given function $H: M^{2n} \to \mathbb{R}$ on the symplectic manifold there corresponds a one-parameter group $g^t_H : M^{2n} \to M^{2n}$ of canonical transformations of $M^{2n}$—the phase flow of the hamiltonian function equal to $H$. Let $F: M^{2n} \to \mathbf{R}$ be another function on $M^{2n}$.

Definition. The Poisson bracket $(F,H)$ of functions $F$ and $H$ given on a symplectic manifold $(M^{2n}, \omega^2)$ is the derivative of the function $F$ in the direction of the phase flow with hamiltonian function $H$: $$ (F,H)(x) = \frac{d}{dt}\Big|_{t=0} F(g^t_H(x)). $$ Thus, the Poisson bracket of two functions on $M$ is again a function on $M$.

Corollary 1. A function $F$ is a first integral of the phase flow with hamiltonian function $H$ if and only if its Poisson bracket with $H$ is identically zero: $(F,H) = 0$.

We can give the definition of Poisson bracket in a slightly different form if we use the isomorphism $I$ between 1-forms and vector fields on a symplectic manifold $(M^{2n},\omega^2)$. This isomorphism is defined by the relation (cf. Section 37 (wherein $\eta$ is a vector in the tangent space and $\omega^1$ is the corresponding 1-form in the cotangent space)) $$ \omega^2(\mathbf{\eta},I \omega^1) = \omega^1(\mathbf{\eta}). $$ The velocity vector of the phase flow $g^t_H$ is $I d H$. This implies

Corollary 2. The Poisson bracket of the functions $F$ and $H$ is equal to the value of the 1-form $dF$ on the velocity vector $I dH$ of the phase flow with hamiltonian function $H$: $$ (F,H) = dF(I dH). $$

Using the preceding formula again, we obtain

Corollary 3. The Poisson bracket of the functions $F$ and $H$ is equal to the "skew scalar product" of the velocity vectors fo the phase flows with hamiltonian functions $H$ and $F$: $$ (F,H) = \omega^2(I dH, I dF).$$

The emphasized explanation of the notation of Sec. 37 was added.

Answer 1

Corollary 2 is just working out the chain rule in the derivative that defines the Poisson bracket. In Sec. 38A, Arnol'd defines the one-parameter group of diffeomorphisms as $$ \frac{d}{dt}\Big|_{t=0} g^t\mathbf{x} = I dH(\mathbf{x}). $$ This is a vector field. In general, the derivative of $F$ evaluated at a point $\mathbf{y}(t)$ with respect to the real parameter $t$ is \begin{align} \frac{d}{dt} F(\mathbf{y}(t)) &= dF(\dot{\mathbf{y}}), \end{align} which is just the exterior calculus version of the chain rule. You can guess its form because it has to be a function from the reals to the reals, and there's no other such function that's linear in $F$ and in $\dot{\mathbf{y}}$ that you can make out of the one-form $dF$ and the vector $\dot{\mathbf{y}}$. Plugging these two equations into each other proves the corollary.

Corollary 3 uses the definition of $I$ backwards, where $dF$ takes the place of $\omega^1$ and $I dH$ takes the place of $\eta$.