Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 0 down vote accepted

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:

$$ \begin{cases} I d(x^2) \; df = 0, \\ I d(y) \; df = 0 \end{cases} $$ $$ \begin{cases} -2x \frac{\partial}{\partial y} \; \left(\frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy \right) = 0, \\ \frac{\partial}{\partial x} \; \left(\frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy \right) = 0 \end{cases} $$


$$ \begin{cases} -2x \frac{\partial f}{\partial y} = 0, \\ \frac{\partial f}{\partial x}= 0 \end{cases} $$

Thus $f$ is constant.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.