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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.

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if you take $M=\mathbb{R}^2$ with coordinates $x$,$y$ then the functions $x^2$,$y$ satisfy your property but are not coordinates –  user8268 Sep 11 '12 at 13:57

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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.

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