Exact sequence arising from symplectic manifold

Question

Let $M$ be a symplectic manifold, why is the following sequence exact? $$0\to \mathbb{R} \to C^\infty (M)\to A\to 0$$ Here $A$ is the set of global Hamiltonian vector fields.

Answer 1

Hint: let $\phi:C^{\infty}(M)\rightarrow A$ be $f\mapsto \phi(f)=X_f$. The only non trivial part of your statement to prove is $\ker\phi=\mathbb R$. In other words, $X_f(g)=\{f,g\}=0$ for all $g$ implies $f$ constant. Here we consider $M$ to be connected.

Let $\omega=dq^i\wedge dp_i$ be the symplectic 2-form s and

$$X_f=\frac{\partial f}{\partial p_i}\frac{\partial}{\partial q^i}- \frac{\partial f}{\partial q^i}\frac{\partial}{\partial p_i}$$

be the Hamiltonian vector field $X_f$, both in canonical coordinate.

We want to prove that $X_f(g)=\{f,g\}=0$ for all $g\in C^{\infty}(M)$ implies $f$ constant. Then

$$X_f(g)=\{f,g\}=\frac{\partial f}{\partial p_i}\frac{\partial g}{\partial q^i}- \frac{\partial f}{\partial q^i}\frac{\partial g}{\partial p_i}=0~~ (*) $$ for all $g$ implies

$$\frac{\partial f}{\partial p_i}=\frac{\partial f}{\partial q^i}=0,$$

for all $i=1,\dots,2n$. We chose $g=q^j$ and $g=p_j$ for all $j$'s in (*). The statement follows.