# Isomorphism between strong accessibility algebra and observation space of a Hamiltonian system

Consider a Hamiltonian system $\dot{x} = X_H(x) + X_C(x) u$ and $y = C(x)$ where $u$ is a scalar and $x = (q,p) \in \mathbb R^{2n}$ and $H$ and $C$ are some functions mapping $\mathbb R^{2n} \to \mathbb R$. The vector field $X_F$ is defined as $\dot{q_i} = F_{p_i}$ and $\dot{p_i} = -F_{q_i}$.

My professor claimed that there exists an isomorphism between the {strong accessibility algebra under the Lie-bracket $[\cdot,\cdot]$} and the {observation space under the Poission bracket $\{\cdot, \cdot\}$}.

Question: I am struggling to find this isomorphism however. Can anyone help?

My attempt: The strong accessibility algebra is given by $$\mathcal{C}_{sa} = \operatorname{span}\{ X_C, [X_H,X_C], [X_H,[X_H,X_C]],[X_C,[X_H,X_C]], \text{ higher order Lie-brackets} \},$$ while the observation space is defined as $$\mathcal{O}_s = \operatorname{span} \{ C, L_{X_H}C,L_{X_C}L_{X_H}C, L_{X_C}L_{X_H}C, \text{higher order Lie-derivatives} \}$$ Now the isomorphism seems obvious. Let $\phi : \mathcal{C}_{sa} \to \mathcal{O}_s$ via $$\phi([X_1,[X_2,\ldots ,[X_H, X_C]\ldots]) = L_{X_1}L_{X_2} \ldots L_{X_H} C$$ where $X_i \in \{ X_H, X_C\}$. But I do not know how to show that this is an isomorphism. Thanks in advance for any help!

## 1 Answer

It is a standard result that $$L_X Y = [X, Y]$$. There is a proof of this fact in Lee's Introduction to Smooth Manifolds. With this result in hand, we see that the two algebras are not just isomorphic, but identical. The only difference between the two algebras is which notation we use for the Lie derivative: $$L_X$$ or $$[X, \cdot]$$.