# Composition of flow maps of Hamiltonian systems

Given a pair of autonomous Hamiltonian vector fields $$X_H,X_K\in\mathfrak{X}(M)$$, with flow maps which are respectively $$\Phi^t,\Psi^s$$, is $$\Phi^t\circ \Psi^s$$ the $$(t+s)-$$flow map of some Hamiltonian system? I think this is the case, possibly with the final map being the flow of a time-dependent Hamiltonian system.

I know that the composition map is a symplectomorphism. However, since not all the functions in this space are flows of Hamiltonian systems, I am not sure of the answer to this question.

Here is an additional explanation of what I am interested in.

I do not mean that for any $$t$$ and $$s$$, the Hamiltonian of the resulting vector field remains unchanged. More precisely, here are some settings which, for my interest, are not counterexamples: $$\Phi^t\circ \Psi^s = \varphi_{X_L}^{t+s}$$ $$\Psi^t\circ \Phi^s = \alpha_{X_R}^{t+s}$$ $$\Phi^u\circ \Psi^v = \varphi_{X_W}^{u+v},$$ where $$L,R,W$$ are three Hamiltonian functions which possibly do not coincide.

This means that, when we fix $$t$$ and $$s$$, there is a Hamiltonian vector field $$X_{U}$$ so that its $$(t+s)-$$flow is the same map as $$\Phi^t\circ \Psi^s$$.

• If so, wouldn't you have $\Phi^t\circ \Psi^s = \Phi^{t+s} = \Psi^{t+s}$ for all $t$ and $s$ by writing $(t+s) = (t+s)+0 = 0+(t+s)$? Commented Sep 10, 2021 at 11:55
• I am not saying that it is the flow of one of the two hamiltonians $H$ or $K$. I am saying of a new one. Thus in principle there is no relation with $\Phi^{t+s}$ or $\Psi^{t+s}$. @Didier Commented Sep 11, 2021 at 6:45
• @Dadelsam I think you did not understand what I said. Suppose it is the flow of any vector field, say $\forall t, s, \varphi^{t+s} = \Phi^t \circ \Psi^s$. Writing $t+s$ = $(t+s) + 0 = 0 + (t+s)$ yields the equality $\Phi^{t+s}\circ \Psi^0 = \Phi^0 \circ \Psi^{t+s}$. Commented Sep 11, 2021 at 8:06
• @Didier Thanks for the comments. Maybe I have not written enough details in the questions and, indeed, your reasoning is completely clear. I have added some more details to my question. Commented Sep 12, 2021 at 12:47

Firstly, the length of the parametrizing time interval does not matter: if $$\phi_t$$ is the flow of $$X_t$$ for $$t \in [0, T]$$, i.e. $$\frac{d}{dt} \phi_t(x)=X_t(\phi_t(x))$$ then defining $$\psi_s=\phi_{Ts}$$ with $$s\in [0,1]$$ one has $$\frac{d}{ds} \psi_s(x)=TX_{Ts}(\phi_{Ts}(x))$$. This says that the reparameterized $$\psi_s$$ is generated by $$Y_s=TX_{Ts}$$. Of course if $$X_t$$ is Hamiltonian, then so is $$Y_s$$ and vice versa.

Thus we can assume that both $$\Phi$$ and $$\Psi$$ are time 1 maps of Hamiltonian systems, i.e. $$\Phi=\phi_1$$ for a family $$\phi_t$$ with $$t\in[0,1]$$, and $$\Psi=\psi_1$$ for a family $$\psi_t$$ with $$t\in[0,1]$$. Denote the corresponding Hamiltonian vector fields by $$X_t$$ and $$Y_t$$. Now, consider

$$\theta_t=\phi_t\cdot \psi_t$$ so that

$$\Theta=\theta_1=\phi_1\cdot \psi_1=\Phi\cdot\Psi.$$

We just need to show that $$\theta_t$$ is a family of Hamiltonian diffeos. Of course this means we need to compute the generating (time dependent) vector field and see that it's Hamiltonia.

To do this we differentiate

$$\frac{d}{dt}\theta_t(x)=\frac{d}{dt}(\phi_t\cdot \psi_t(x))=$$ $$(\frac{d}{dt}\phi_t)(\psi_t(x))+(\phi_t)_*(\frac{d}{dt}\psi_t(x))=$$ $$X_t(\phi_t(\psi_t(x)))+(\phi_t)_*(Y_t(\psi_t(x)))=X_t(\theta_t(x))+(\phi_t)_*Y_t(\phi_t^{-1}\theta_t(x))$$

so $$\theta_t$$ is generated by the vector field $$Z_t(x)=X_t(x)+(\phi_t)_*Y_t(\phi_t^{-1}(x))$$.

Now all that remains is to show that this is Hamiltonian. What is the corresponding function? Well, if the function for $$X_t$$ is $$H_t$$ and for $$Y_t$$ is $$K_t$$, then I claim that for $$Z_t$$ it's $$H_t+K_t\cdot\phi_t^{-1}$$. Of course it's the $$K_t\cdot\phi_t^{-1}$$ part that's the issue. Naturality of Hamiltonian vector fields under symplectic diffeos means that for any function $$K$$ and any symplectomorphism $$f$$, we have $$f_*(X_{K\cdot f}(p))=X_K(f(p))$$. Applied to $$f=\phi^{-1}$$ this gives what we want.

As a remark, it is in fact the case that we need time-dependent flows. In fact, it follows from the simplicity of the group of Hamiltonian diffeos that every Hamiltonian diffeomorphism is a product of autonomous ones (it's a nice exercise to check that inverse of an autonomous Hamiltonean diffeo is also autonomous, so the set of Hamiltonean diffeos that are a finite product of autonomous ones forms a conjugation invariant subgroup, hence is the whole group), but sometimes you need arbitrarily many of them (see https://arxiv.org/abs/1207.0624 and follow-up work).

• Thank you very much for the nice and clear answer! I will check the mentioned paper...do you have other references where I can find material on this composition of autonomous hamiltonian flows? Commented Sep 13, 2021 at 6:29
• I'm not sure what kind of references you are asking about. The basic (if bit older) reference is a book by Leonid Polterovich "The geometry of the group of symplectic diffeomorphisms". For "autonomous Hamiltonians are not everything" there is the line of work referenced above, and there is also arxiv.org/abs/1412.8277 and extensions (arxiv.org/abs/1605.07594, arxiv.org/abs/1405.7931, arxiv.org/abs/1707.06020 etc).
– Max
Commented Sep 13, 2021 at 15:07
• Great, thank you again. Commented Sep 13, 2021 at 15:46