Let $(P,\Omega)$ be a symplectic manifold, and let $[\cdot,\cdot]$ be the natural Poisson bracket. Let $\varphi^h(a)$ be the Hamiltonian flow generated by the smooth function $h:P\rightarrow\mathbb{R}$, and let $\varphi^g(b)$ be the Hamiltonian flow generated by the smooth function $g:P\rightarrow\mathbb{R}$.


Suppose the two flows commute, $\varphi^g(b)\varphi^h(a) = \varphi^h(a)\varphi^g(b)$. Are there interesting circumstances under which it follows that the generators commute as well, $[g,h]=0$?


I know that the flows commuting implies that $[g,h]=k$ for some constant $k$; a reference for the proof of this can be found, for example, in Arnold's (1989) *Mathematical Methods of classical mechanics*, pg. 218 Cor. 9. It is also easy to show that if $[g,h]=0$, then $\varphi^g(b)\varphi^h(a) = \varphi^h(a)\varphi^g(b)$. But under what circumstances does the converse hold?


It's interesting that in quantum mechanics, the analogous relations hold in both directions. That is, if $G$ and $H$ are linear self-adjoint operators on a Hilbert space, and if $e^{ibG}$ and $e^{iaH}$ are the continuous one-parameter unitary groups they generate, then $e^{ibG}e^{iaH}=e^{iaH}e^{ibG}$ if and only if $[H,G]=0$ (where now $[\cdot,\cdot]$ is the commutator bracket).



When do commuting Hamiltonian flows have commuting generators?
This is not so much an answer as a suggestion to change the question. When $(P,\Omega)$ is *prequantizable*, i.e. there exists over $P$ a hermitian line bundle with connection $(L,\nabla)$ having curvature $\Omega$ (see e.g. [Kostant 1970](http://dx.doi.org/10.1007/BFb0079068)), then your hamiltonian vector fields $X^g, X^h$ and their flows $\varphi^g, \varphi^h$ lift canonically to $\nabla$-preserving vector fields $\xi^g, \xi^h$ and flows $\psi^g, \psi^h$ on $L$. ***These*** commute if and only if $[g,h]=0$.


That, I believe, is the correct "classical analogue" of the quantum facts you allude to.


Conversely, the true quantum analogue of looking at $\varphi^g, \varphi^h$ is looking at the action of $e^{ibG}, e^{iaH}$ not on Hilbert space $\mathcal{H}$ but on *projectivized* Hilbert space $\mathbb{P}\mathcal{H}$. ***There*** they commute iff (disregarding the usual domain questions) $[G,H]$ is a constant multiple of the identity.


In fact the analogy is good enough that $\mathbb{P}\mathcal{H}$ is a (usually infinite-dimensional) symplectic manifold, to which the first paragraph above applies, with $g, h$ the expectation values of $G, H$ and $L^\times \to P$ the tautological projection $\mathcal{H}\setminus\lbrace0\rbrace\to \mathbb{P}\mathcal{H}$. Moreover $\xi^h$ and $\psi^h(a)$ are just $H$ and $e^{iaH}$ (acting on $\mathcal{H}\setminus\lbrace0\rbrace$) -- so we've come full circle.


[P.S.: Regarding functions whose Poisson brackets are constant, you might be interested in [this paper](http://www.ams.org/mathscinet-getitem?mr=0289078) of Roels and Weinstein.]




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**Update** regarding your extra question ("Isn't $e^{ibG}e^{iaH}=e^{iaH}e^{ibG}\Leftrightarrow[G,H]=0$ also true on $\mathbb{P}\mathcal{H}$?"): This is a statement about transformations of $\mathcal{H}$, not $\mathbb{P}\mathcal{H}$. Write $\underline{e^{iaH}}$ for the diffeo of $\mathbb{P}\mathcal{H}$ induced by $e^{iaH}\in\mathrm{U}(\mathcal{H})$, and likewise $\underline{iH}$ for the vector field on $\mathbb{P}\mathcal{H}$ induced by $iH\in\mathrm{End}(\mathcal{H})$. Then (exercise!) $e^{iaH}\mapsto\underline{e^{iaH}}$ is a group morphism with kernel the multiples of the identity, and likewise $iH\mapsto\underline{iH}$ is a Lie algebra morphism with kernel the multiples of the identity. Therefore we have


\begin{array}{cccl}
\underline{e^{ibG}}.\underline{e^{iaH}}=\underline{e^{iaH}}.\underline{e^{ibG}} & \Leftrightarrow & [\underline{iG},\underline{iH}]=0 &\text{(actions on }\mathbb{P}\mathcal{H})\\\
\Updownarrow & & \Updownarrow\\\
e^{ibG}e^{iaH}e^{-ibG}e^{-iaH}\in\mathbb{C}\cdot\mathbf{1} & \Leftrightarrow & [iG,iH]\in\mathbb{C}\cdot\mathbf{1} & \text{(actions on }\mathcal{H})
\end{array}


and my claim is that *these*, not $[G,H]=0$, are the quantum analogs of $\varphi^g(b)\varphi^h(a)=\varphi^h(a)\varphi^g(b)$.


