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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.

Matias
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    We can't tell you if it is exact or not if you don't tell us what the middle two maps are. I'm guessing the first is inclusion of the constant functions ($M$ is presumably connected) and the second is $f \mapsto X_f$? – Henry T. Horton Jun 30 '13 at 04:15
  • Yes, exactly. But why $M$ should be connected? – Matias Jun 30 '13 at 09:16
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    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, $X_f(g)={f,g}=\omega(X_f,X_g)=0$ for all $g$ implies $f$ constant. You can express $X_f$, $X_g$ in local coordinates an arrive at the result. – Avitus Jun 30 '13 at 09:49

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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.

Avitus
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  • Avitus@ Thanks for the nice answer – Matias Jun 30 '13 at 10:04
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    @h.jolany You are welcome. $M$ is supposed to be connected. This implies that a function with all vanishing derivatives is constant (i.e. each function specifies a unique real number $\mathbb R$). If $M$ admitted multiple connected components then a function with vanishing derivatives would be constant on each connected component, with possibly different values on each of them, though. – Avitus Jun 30 '13 at 10:09
  • ohhh, Yes, thank for this precision and accuracy – Matias Jun 30 '13 at 10:11