Consider a symplectic manifold $(M, \omega)$. Let us define a concept of a complete set of observables:
A set of functions $f_i : M \to \mathbb R$ form a complete set of observables if any function which Poisson commutes (has vanishing Poisson brackets) with all of them is a constant.
That is the set above can feel the behaviour of a given function in all the directions.
I wonder whether specifying a values of all of $f_i$ will define a unique point on a manifold? That is do such functions form coordinates? If this is false, I'd like to see a simple counterexample. Also if generally the statement is false, is there probably a large class of situations when it is true?
This question is closely related to my question at Physics.SE.
Per @user8268's example. Let $M = \mathbb R^2$ with coordinates $x$, $y$ and the symplectic from $\omega = dx \land dy$. Let's consider functions $x^2$ and $y$ which clearly are not coordinates and show the they satisfy the condition.
Poisson brackets:
$$[f,g] = df \; I(dg)$$
where $I : T^*M \to TM$ is induced by $\omega$ and $I = \frac{\partial}{\partial y} \land \frac{\partial}{\partial x}$. Then:
$$ \begin{cases} I d(x^2) \; df = 0, \\ I d(y) \; df = 0 \end{cases} $$ $$ \begin{cases} -2x \frac{\partial}{\partial y} \; \left(\frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy \right) = 0, \\ \frac{\partial}{\partial x} \; \left(\frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy \right) = 0 \end{cases} $$
or
$$ \begin{cases} -2x \frac{\partial f}{\partial y} = 0, \\ \frac{\partial f}{\partial x}= 0 \end{cases} $$
Thus $f$ is constant.