# How can I see the “missing” Poisson center when the rank of the Poisson structure drops?

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?

-
Incidentally, there are other ways to see the leaves with extra codimension. For example, the Poisson bivector determines a differential on the graded algebra of antisymmetric multivector fields (the differential is given by the Schouten-Nijenhuis bracket with the Poisson bivector; that it squares to zero is equivalent to the Jacobi identity); this complex computes Poisson cohomology, and is related to the BRST-BV complex that physicists use. The homology of this complex can (sometimes) see the missing leaves, e.g. I think I can see the (locally) closed leaves by studying $H^0$. – Theo Johnson-Freyd Aug 17 '11 at 14:18

I do not have a definite answer but rather some reflections I've being doing myself and with a colleague I'll mention later, recently, on the subject. Sure Poisson cohomology can help you in detecting the"missing" leaves, my favourite example being the triple of bivectors $\partial_x\wedge\partial_y$ (trivial 1-Poisson cohomology), $(x^2+y^2)\partial_x\wedge\partial_y$ (1-dimensional 1-Poisson cohomology), $(x^2+y^2)^2\partial_x\wedge\partial_y$ (infinite-dimensional 1-Poisson cohomology). In a sense $1$-dimensional Poisson cohomology represents the tangent space to the set of leaves (which is a badly behaved non Hausdorff space and may therefore have no nonconstant functions defined on it) and is therefore slightly more sensible.

Now, you may know that Poisson cohmology may be defined also with values in a Poisson module. If $T$ is a distribution (I mean a linear continuous functional on $\cal C_0^\infty(M)$ - compactly supported smooth functions) then letting $\left\{f,T\right\} (g)=T(\{f,g\})$ one gets that $\{f,T\}$ is still a distribution. One can show that in this way distributions form a Poisson module over $\cal C_0^\infty(M)$. One can of course look for the annihilator of such module, i.e. Casimir distributions defined by $\left\{f,T\right\}=0$ for every $f$. This space enlarges naturally the space of Casimir functions. On the quadratic singular Poisson structure I was referring to above it is possible to show that the vector space of Casimir distributions is $5$-dimensional generated by the $\delta$ function of the origin, its first derivatives and its mixed second derivative.

One could consider this space of Casimir distributions, or more generally the Poisson cohomology with coefficients in this Poisson module. This I've never attempted to compute, even in easy examples.

Examples of Poisson Modules, I, Rendiconti del Circolo Matematico di Palermo(2) 52 (2003), 419-452

Examples of Poisson Modules, II, Rendiconti del Circolo Matematico di Palermo(2) 53 (2004), 23-60.

The examples above are all $2$-dimensional. It is important to recall that every $2$-dimensional bivector is Poisson and determined by a smooth functions on the plane. We are therefore looking for something that should be, in this specific case, distinguish singularities of smooth functions, quite subtle, therefore.

Something easier, though not so capable of detailed analysis is to consider a maximal subalgebra of continous functions to which the Poisson bracket may be extended. This was done, for example, in an old paper by Albert Sheu on the quantization of the Podles sphere (the one with an appendix by Lu-Weinstein, I do not have a chance to get the exact reference now) where he quantizes the algebra of all smooth functions on the sphere minus the North Pole that can be continously extended to the North Pole. The reason why this is much less sensible is that as long as you have quadratic singularities in the bivector you can had square root singularities in the functions, which immediatly throws in all continous functions.

Hope this may help.

-
I wonder whether there is a way to merge this user which I wrongly created for this answer, with the user writing this comment (i.e. merging me into me... ) – Nicola Ciccoli Oct 15 '11 at 8:42
Ask on tea.mathoverflow.net or flag your answer for moderator attention. – Mariano Suárez-Alvarez Oct 15 '11 at 8:47
One point about which I was not clear. Casimirs are indeed $0$-Poisson cohomology (and Casimir distributions are $0$-cohomology with coefficients in the Poisson module of distributions). But to "see the singular leaves" one may need to go to $1$-Poisson cohomology. – Nicola Ciccoli Oct 15 '11 at 9:09
I have made no edits except to fix some TeX. As always, there is a problem that Markdown sometimes strips backslashes off of special characters. I have fixed this by adding <p> tags around the problematic paragraph — Markdown is instructed to do no processing to things inside such tags. – Theo Johnson-Freyd Oct 15 '11 at 18:36
Great. So my comment to my question was of course nonsense — $H^0$ is precisely the space of Casimirs. (Somehow I had gotten myself confused, and thought I had a homology theory, so that $H_0$ was "all functions modulo the vanishing ideal of the Poisson bivector". I do get something like this if I move to $H^1$.) If I think the answer is to use distributions, then I can formalize them as a Poisson module. Indeed, it is the module dual to the module of smooth functions. (Of course, "module" really means "sheaf of modules over the corresponding algebroid.) – Theo Johnson-Freyd Oct 15 '11 at 18:41