Recall that a Poisson algebra is a commutative algebra $A$ along with a bracket $\lbrace,\rbrace: A^{\otimes 2} \to A$ which is a Lie bracket and which is also a derivation in each variable. The Poisson center of $A$ is the subalgebra of those $f\in A$ such that $\lbrace f,\rbrace : A \to A$ is the $0$ derivation. Elements of the Poisson center are generally called Casimirs. Let me write $Z(A)$ for the Poisson center of $A$.
The most important case is when $A = \mathcal C^\infty(M)$ for a Poisson manifold $M$. Then $M$ is foliated by symplectic leaves, which are the orbits for the Lie algebra action of $A$ on $M$. It follows that every Casimir is constant on each symplectic leaf. Symplectic leaves can have interesting macroscopic topology — for example, they can wrap around $M$ "irrationally" — so I would rather think locally on $M$, and replace $A$ by the corresponding sheaf of Poisson structures. Then there is a sheaf whose sections are local Casimirs; this sheaf might not have very many global sections. Anyway, one generally tries to believe that at least locally the symplectic leaves are precisely the common level sets of the Casimirs.
But even this fails locally when the Poisson bivector drops in rank. As an easy example, consider $\mathbb R^2$ with coordinates $x$ and $y$ and Poisson bivector $x \frac{\partial}{\partial x} \wedge \frac{\partial}{\partial y}$. There are two 2-dimensional symplectic leaves, namely $\lbrace (x,y) \text{ s.t. } x>0\rbrace$ and $\lbrace (x,y) \text{ s.t. } x<0\rbrace$, and uncountably many $0$-dimensional symplectic leaves, namely the points $\lbrace (0,y)\rbrace$ for each $y$. But any Casimir is locally constant on each of the 2-dimensional leaves, and if it extends to the $y$-axis it must continue to be constant.
So, far from $\operatorname{spec}(Z(A))$ being the set of symplectic leaves, $\operatorname{spec}(Z(A))$ is more accurately thought of as the "GIT quotient" of $\operatorname{spec}(A) = M$ under the Poisson action.
But there is a qualitative difference between, say, $\mathcal C^\infty(\mathbb R^2)$ with bracket $x \frac{\partial}{\partial x} \wedge \frac{\partial}{\partial y}$ and the same algebra with the nondegenerate bracket $\frac{\partial}{\partial x} \wedge \frac{\partial}{\partial y}$. This difference is not detected by the algebraic Poisson center, but in some sense this is because $\mathcal C^\infty$ isn't quite the right type of function. Indeed, suppose that I had some type of "delta functions"; then there would be functions of the form $\delta(x)f(y)$ in $Z\bigl( \mathcal C^\infty(\mathbb R^2), x \frac{\partial}{\partial x} \wedge \frac{\partial}{\partial y}\bigr)$, and the collection of all such functions would correctly cut out the symplectic leaves as the common level sets.
Hence, my question:
Can I always find the symplectic leaves of a Poisson manifold as the common level sets of the Casimirs if I allow some sort of "generalized function"? If so, how precisely should I define such generalized functions?
As a step towards the second question (assuming the answer to the first is "yes"), let me describe the approach I've been imagining. I do not have $\delta$-functions in $\mathcal C^\infty$, but I do have sequences of functions that approach $\delta$-functions for some norm. I'm not very good at analysis, so I don't have intuition of which is the correct norm to use, but let's suppose I've picked such a norm. Then I might say that $f\in \mathcal C^\infty$ is almost central or an almost Casimir if the operator norm of $\lbrace f,\rbrace$ is small. One should then expect that functions approaching $\delta(x)f(y)$ are almost Casimirs for $x \frac{\partial}{\partial x} \wedge \frac{\partial}{\partial y}$, whereas some version of Heisenberg Uncertainty says that there are no almost Casimirs for the nondegnerate bracket $\frac{\partial}{\partial x} \wedge \frac{\partial}{\partial y}$. So perhaps every Poisson manifold does have (locally) an "almost Poisson center", and this almost center is enough to detect all the symplectic leaves?
