Consider the hamiltonian system $(M, \omega, H)$ with hamiltonian vector field $X$ defined by $$ \tag{1}\label{1} \iota_{X}\omega = -dH $$ It is a simple matter of applying definitions to show that a smooth function $f$ on $M$ is an integral of motion $$\mathscr{L}_X f = 0 \tag{2} \label{2}$$ iff its hamiltonian vector field $Z$, defined by $$\iota_Z \omega = -df \tag{3}\label{3}$$ preserves the Hamiltonian $H$, that is $H$ is pull-back invariant along the flow of $Z$, or $$\mathscr{L}_Z H = 0 \tag{4}\label{4}$$

Indeed: \begin{split} \mathscr{L}_X f = X(f) & = df(X) \\ & = - \iota_Z \omega \, (X) = -\omega(Z,X) = \omega(X,Z) = \iota_X\omega \, (Z) \\ & = -dH(Z) = -Z(H) = - \mathscr{L}_Z H = 0 \quad \square \end{split} where \eqref{3} is used at the first line, \eqref{1} at the second and \eqref{4} at the last to conclude \eqref{2}.

Since exact forms are closed, hamiltonian vector fields are symplectic by Cartan's magic formuala: $$\mathscr{L}_Z\omega = \iota_Zd\omega + d\iota_Z\omega = 0$$ where the first terms vanishes being $\omega$ closed per definition, and the second being $\iota_Z \omega$ exact, so closed, by \eqref{3}.

There is thus a 1-to-1 correspondence between integrals of motions and hamiltonian (in particular symplectic), Hamiltonian-preserving vector fields.

Can I weaken this assumption? Is it enough to have a Hamiltonian preserving, symplectic vector field $$\mathscr{L}_Z H = 0, \quad \mathscr{L}_Z \omega = 0 $$ to obtain a constant of motion? In other words, is this enough to make $Z$ hamiltonian? Of course the answer is yes if the first de Rham cohomology group is trivial, since then 1-forms are closed iff exact and vector fields are symplectic iff hamiltonian; but what if this is not the case?

I'm pretty sure the answer lies in the moment map, but I was trying to come up with an intuitive answer before studying it.


No, it's not. For example, take $M = \mathbb{R}^2\times(S^1\times\mathbb{R})$ with the symplectic form $\omega = dx\wedge dy + \alpha\wedge dz$, $H=H(x,y)$, and $Z=\frac{\partial}{\partial z}$. Here $\alpha$ is the closed 1-form that takes value 1 on $\frac{\partial}{\partial\theta}$: informally, $\alpha=d\theta$, although this is only true in a coordinate patch. Then $\mathscr{L}_ZH=0$ and $\mathscr{L}_Z\omega=0$. But $i_Z\omega = -\alpha$, and $\alpha$ is not actually an exact form (since its integral around $S^1$ is 1). So there's no corresponding globally defined integral of motion. Locally, we can say the integral of motion is $\theta$. However this can't be extended to a globally defined function.

  • $\begingroup$ Thanks. I understand this amounts to saying that closed forms are locally exact, and I was indeed interested in the global obstruction. $\endgroup$
    – DavideL
    Mar 9 '20 at 11:36
  • $\begingroup$ Yesh, pretty much (that's my understanding anyway). $\endgroup$
    – user17945
    Mar 9 '20 at 15:40

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