# Is composition of discrete Hamiltonian flows integrable?

Consider $$\Bbb{R}^2$$ with the usual symplectic form $$\omega = dx \wedge dy$$

For a function $$H \colon \Bbb{R}^2 \to \Bbb{R}$$, let $$X_H$$ be the Hamiltonian vector field. Then the map $$\Bbb{R}^2 \to \Bbb{R}^2$$, defined as the flow of $$X_H$$ for time $$t=1$$, is symplectic, and Liouville integrable (since $$H$$ is invariant).

My question is this: given two functions $$H_1(x,y)$$ and $$H_2(x,y)$$, consider the composition of performing the time-1 flow of $$X_{H_1}$$ followed by the time-1 flow of $$X_{H_2}$$. Is this map Liouville integrable (i.e. does it have a conserved quantity)?

If so, why? Can the conserved quantity be described in terms of $$H_1$$ and $$H_2$$?

If not, is there an easy counter-example?

Note: I asked this same quesiton on math stack exchange (here), but got no answers.

• What about $H_1(x,y) = \omega^2 x^2 /2$ and $H_2(x,y) = y^2/2$? Then the composition is exponentially unstable if $\omega$ is larger than $2$. Dec 10, 2020 at 19:26