I tried the following symplectic geometry exercise and wanted to make sure it was correct
Let (M,ω) be a closed symplectic manifold, $f,g ∈ C∞(M)$ two smooth functions and $\phi_t,\psi_t ∈ Ham(M,ω)$ the (autonomous) Hamiltonian flows generated by $X_f$ and $X_g$. Then the flows commute, ∀t ∈ [0,1], if and only of the Poisson bracket ${f, g} ≡ ω(Xf , Xg) ∈ C^∞(M)$ is identically zero.
Let's first suppose that the poisson bracket is identically zero. Then we will have that $H:=\omega(X_g,X_f)=-\omega(X_f,X_g)=0$. Then using the fact that $ i([X_f,X_g]) \omega = dH = 0$ we get that $i([X_f,X_g]) \omega =0$ and since $\omega$ is non-degenerate we will get that $[X_f,X_g]=0$, and now using a standard differential geometry fact we get that the flows commute
Now let's assume that $\phi_t \circ \psi_t = \psi_t \circ \phi_t, \forall t\in [0,1]$.It's a well know fact of geometry that this implies that $[X_f,X_g]=0$. Then we know that $i([X_f,X_g]) \omega = dH$ where $H=\omega(X_g,X_f)$. And so this tells us that $d\omega(X_f,X_g)=0$. Now another way to see this is that in each connected component $U$ the function will be constant,i.e. , $\omega(X_f,X_g)=a_U$ for some $a_U\in \mathbb{R}$ and so this tells us that $df(X_g)=a_U$ and so we get that for any $p\in U, \frac{d}{dt}f(\psi_t(p))=a_U*t+b$ for some $b\in \mathbb{R}$. Now if we assume that $a_U \neq 0$ we will get a contradiction with the fact that $M$ is compact. Take the sequence $\{\psi_{n}(p)\}_{n=-\infty}^{n=\infty}$, since $M$ is compact we can take a convergent subsequence $\psi_{n_k}(p)\rightarrow a$ with $n_k\rightarrow \infty$.Now since the functions involved are continuous we must have that $f(\psi_{n_k}(p))=a_U*(n_k)+b\rightarrow f(a)$, but the left side is going to infinity while the right side is a finite number and so we obtain a contradiction,and we must have that $a_U=0$. Since the choice of $U$ was arbitrary we get the desired result.
Any input is appreciated. Thanks in advance.
Best Answer
Yes, this looks correct to me. I'd suggest two simplifications:
(i) the identity $i_{[X_f,X_g]}\omega = d(\omega(X_g,X_f))$ says that $[X_f,X_g] = X_{-\{ f,g\}}$, so $[X_f,X_g] =0 \iff X_{-\{ f,g\}}=0\iff \{f,g\}=$ constant $\iff df(X_g)=$ constant.
(ii) Completing the $\implies$ direction: since $M$ is compact, $f$ will achieve its maximum on $M$. At this point, $df=0$, and so $df(X_g)=0$. Hence the constant will be zero.