5
$\begingroup$

Let $R$ be a Poisson algebra (over $\mathbb C$, say) with Poisson center $Z = \{c \in R : \{c,R\} = 0\}$ and consider two types of ideals $I \leq R$:

  1. $I = \langle (c_i) \rangle$ is generated by Casimirs $c_i$, i.e. each $c_i\in Z$

  2. $I$ is a Poisson ideal, i.e. $\{I,R\} \leq I$.

I'm having trouble distinguishing these geometrically: both seem to be about subschemes of $Spec\ R$ that are unions of symplectic leaves. But that seems too good to be true (or I would hope someone would have told me already). Still, I will ask:

If $I$ is a Poisson ideal, is $\sqrt{I} = \sqrt{ \langle I \cap Z \rangle}?$

If so, what is a reference? If not, what is a counterexample?

Improve this question
$\endgroup$
1
  • 2
    $\begingroup$ Nicola's answer below certainly gives the desired counterexample. I just wanted to mention that if $I$ is already radical then there is a simple obstruction to equality of $I$ and $\langle I\cap Z \rangle$. Namely, the Poisson bracket descends to a Lie algebra structure on $I/I^2$ (the "co-normal Lie algebra"). If $I$ is generated by elements of $Z$ then this Lie bracket will be abelian, but most conormal algebras are not. For example, for the maximal ideal $\mathfrak{m}$ defining the origin in the dual of a Lie algebra $\mathfrak{g}$, we have $\mathfrak{m}/\mathfrak{m}^2\cong\mathfrak{g}$ $\endgroup$
    – Brent Pym
    Commented Jul 29, 2014 at 14:29

2 Answers 2

Reset to default
5
$\begingroup$

I am not that familiar with the algebraic approach but the intersection on the rhs seems quite hard to me. Take the Poisson structure on the plane given by $\{x,y\}=xy$. Then $Z$ contains only constant functions and therefore for any Poisson ideal $I$ you have $I\cap Z$ equal to $Z$ or $0$ depending whether the ideal contains constants or not. Take $I=\langle x,y\rangle$ and it looks to me your condition is not fulfilled...

To put it another way there seems to me to be many situations in which the Poisson center is given only by constants and still the symplectic foliation is quite rich, thus many Poisson ideals.

Improve this answer
$\endgroup$
8
  • $\begingroup$ Being a Poisson ideal, $I$ is also a associative ideal. So to make things interesting, it should not contain the constants. I think your example completely settles the question (in the negative). $\endgroup$
    – Stefan Waldmann
    Commented Jul 29, 2014 at 8:27
  • 2
    $\begingroup$ To put it another way: being a symplectic leaf is still quite far from being a level set of Casimir functions. There are many examples of algebraic Poisson manifolds with dense leaves, or leaves which are dense inside closed submanifolds. $\endgroup$
    – Nicola Ciccoli
    Commented Jul 29, 2014 at 10:38
  • 1
    $\begingroup$ Okay, so the group-action analogue is that a $G$-invariant subvariety may fail to be the intersection of a number of $G$-invariant hypersurfaces. $\endgroup$
    – Allen Knutson
    Commented Jul 29, 2014 at 18:51
  • $\begingroup$ Maybe more than an analogue; what is around here is the infinitesimal action of all the Hamiltonian vector fields... $\endgroup$
    – Nicola Ciccoli
    Commented Jul 30, 2014 at 6:37
  • $\begingroup$ You also asked for references, I guess that here mathnet.or.kr/mathnet/paper_file/glasgow/Gordon/poisord.pdf you may find some pieces fo the general theory... $\endgroup$
    – Nicola Ciccoli
    Commented Jul 30, 2014 at 8:25
0
$\begingroup$

Maybe, the following references can be of help.

  • MR1764436 (2001d:53093) Reviewed Grabowski, Janusz(PL-WASW-IM) Isomorphisms of Poisson and Jacobi brackets. (English summary) Poisson geometry (Warsaw, 1998), 79–85, Banach Center Publ., 51, Polish Acad. Sci., Warsaw, 2000.

  • MR2027202 (2004k:17042) Reviewed Grabowski, J.(PL-PAN); Poncin, N.(LUX-CUL-DM) Automorphisms of quantum and classical Poisson algebras. (English summary) Compos. Math. 140 (2004), no. 2, 511–527.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .