Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $X$ be an affine holomorphic symplectic variety of dimension $2n$, with the associated Poisson bracket { , }. Let's say it's an integrable system when there are $n$ algebraically independent holomorphic functions $I_i$ ($i=1,\ldots,n$) on $X$ such that they Poisson-commute: $\{I_i,I_j\}=0$.

Pick a simple Lie algebra $\mathfrak{g}$, and pick a coadjoint orbit $O\subset \mathfrak{g}^*$. As is well-known, this is naturally a holomorphic symplectic variety: $x,y\in \mathfrak{g}$ gives a function on $O$, and its Poisson bracket is then given by the Lie bracket:


I wonder which coadjoint orbit is integrable, in the sense given above.

For example, any regular orbit $O$ of $\mathfrak{g}=\mathfrak{gl}(n)$ is integrable: let $X:\mathfrak{g}^*\to\mathfrak{g}$ be the identification via the invariant inner product, choose a generic element $a\in \mathfrak{g}$, and consider a function on $O$ given by

$P_k(t)= \mathrm{tr} (X+ta)^k$

where $t$ is a complex number. (I'm sorry for a slightly confusing notation, but $a$ here is a constant function on $O$ taking value in $\mathfrak{g}$.)

After some manipulation, $P_k(t)$ and $P_{k'}(t')$ are seen to Poisson-commute. Therefore, coefficients $P_{k,i}$ of $t^i$ of $P_k(t)$ all Poisson-commute. Note that $P_{k,k}$ is a constant function on $O$, because $O$ is a coadjoint orbit. Then there are in total $1+2+\cdots+(n-1) = (\dim \mathfrak{gl}(n)-\mathrm{rank} \mathfrak{gl}(n))/2$ independent commuting operators.

Let me conjecture that all coadjoint orbits are integrable.

Another large class of affine holomorphic symplectic varieties are Nakajima's quiver varieties, which include all the coadjoint orbits of $\mathfrak{gl}(n)$. A similar question can be asked: which quiver variety is integrable in this sense?


Thanks to the answers so far, I could track down references, see e.g. the end of Sec. 4 of this paper, showing that any coadjoint orbit of real compact Lie algebras is integrable. I guess the proof should carry over to the semisimple orbits (and presumably nilpotent Richardson orbits) of complex semisimple Lie algebras, in the holomorphic sense. So the question now is: how about nilpotent orbits in $\mathfrak{g}_{\mathbb{C}}$?

Update 2

Indeed, A. Joseph says in this article that he essentially proved that Richardson orbits are integrable in this article (in the setting of enveloping algebras, not the associated graded.) It would be interesting e.g. non-special orbits are not integrable... but I have no idea.

share|cite|improve this question
Some comments. Actually Gelfand-Ceitlin system can be obtained as a limit of $tr(X+ta)^k$ taking $a=diag(u^n, u^{n-1}...u)$ and u-> 0. This some idea result by E. Vinberg, On some commutative subalgebras in universal enveloping algebra, Izv. AN USSR, Ser. Mat., 1990, vol. 54 N 1 pp 3-25 –  Alexander Chervov Jan 16 '12 at 19:11
From the Gaudin's point view matrix "a" is "constant magnetic field" acting on ALL spins. There is GL(n)-GL(m) duality (Howe's duality) which actually interchanges the matrix "a" and "Gaudin's marked point $z_i$" - it is presumable related to duality between monopoles - that "n-monopoles of charge m" is the same as "m-monom with charge n". Being more precise for Gaudin such duality is present only when we have n representations $S(C^m)$, so we have $S(C^m)\otimes ... S(C^m)= S(C^n\otimes C^m)= S(C^m)\otimes ... S(C^m) $ ... see next comment –  Alexander Chervov Jan 16 '12 at 19:19
It is rather easy to check that GL(n) -Gaudin and GL(m) Gaudin defines the same algebra of commutative hamiltonians it is basically matrix identity $det(t+ A + X(z-B)^{-1} P ) = det( z-B + X(t+A)^{-1}P)$ - so you see that role of matrices "A" and "B" interchanged. This actually can be quantized - and this is Capelli identity A generalization of the Capelli identity E. Mukhin, V. Tarasov, A. Varchenko –  Alexander Chervov Jan 16 '12 at 19:23
About relation to monopole duality see [GGM97] section 5.4 and especially formula 5.27 page 23) About integrability point of view [AHH90] MR Adams, J Harnad, J Hurtubise Dual moment maps into loop algebras, Lett.Math.Phys. 20,(1990),299 We also discussed these things but quite shortly in Our point of view I tried to describe in… –  Alexander Chervov Jan 16 '12 at 19:30
Our point of view that we may take Vinberg's limit $a=diag(u^n,u^{n−1}...u)$ and $u-> 0$ on both sides of GLn-GLm duality and we get Gelfand-Ceitlin dual to "bending flows", and it is more or less obvious that bending flows are the same as Jacys-Murphy elements. –  Alexander Chervov Jan 16 '12 at 19:32

3 Answers 3

up vote 7 down vote accepted

Here is answer (YES) from Alexey Bolsinov who is one the main experts in these questions.

"The answer is YES

There is a very general construction allowing to construct an integrable system on more or less any coadjoint orbit for an arbitrary Lie algebra (non necessarily semisimple). This is a recent paper by Vinberg and Yakimova available in arxiv Complete families of commuting functions for coisotropic Hamiltonian actions

In the particular case you are talking about (SEMI-SIMPLE g) the positive answer follows from 2 results:

1) the so-called shifts of polynomial invariants give a completely integrable system on a singular adjoint orbit O(b) in a semi simple Lie algebra G if and only if

the index of the centralizer of b coincides with the index of G

(my paper in Izvestija AN SSSR, 1991 and Acta Appl. Math. 1991), both available on my home page

Bolsinov A.V. Commutative families of functions related to consistent Poisson brackets// Acta Appl. Math., 24(1991), pp. 253-274.

I also conjectured that

this condition ind Cent (b) = ind G, in fact, holds true for all singular elements b\in G and checked it for G=sl(n) (in particular for all nilpotent)

2) This conjecture (widely known as Elashvili conjecture) has been proved for an arbitrary semi simple Lie algebra and for all elements (in fact the proof is easily reduced to nilpotent elements)

First, Elashvili did it by, in some sense, straightforward computation which in the most difficult case of e_8 involved some computer program (unpublished)

Recently a conceptual proof has been done by Jean-Yves Charbonnel (IMJ), Anne Moreau (available in arxiv) The index of centralizers of elements of reductive Lie algebras

To the best of my knowledge, this is the only known universal way to construct an integrable system on an arbitrary orbit.

Remark: I am talking about classical integrable systems, not quantum. These systems can be quantized too, but this is another story. "

share|cite|improve this answer

Here is a related result of Azad, Ban and Biswas which you may find interesting.

Let $G$ be a complex semisimple Lie group, $\mathfrak{g}=\textrm{Lie}(G)$. Let $c\in \mathfrak{g}$ be a semi-simple element and $\mathcal{O}:=\textrm{Ad}(G)\cdot c$. Let $\Omega$ be the (holomorphic) Kostant-Kirillov symplectic form. Suppose all eigenvalues of $\textrm{ad}(c)$ are real. Then $(\mathcal{O},\textrm{Re }\Omega)\simeq T^\vee (G/P)$, where $P\subset G$ is any parabolic with Levi component $Z(c)$, the centraliser of $c$ in $G$. Here $\simeq$ means ''an isomorphism of real symplectic manifolds'', and $T^\vee (G/P)$ is equipped with the Liouville symplectic form.

A similar result holds if all eigenvalues of $\textrm{ad}(c)$ are purely imaginary, but then you have to take $\textrm{Im }\Omega$.

This generalises a theorem of Arnold for $SL(n,\mathbb{C})$.

share|cite|improve this answer
Thank you; this paper seems to explicitly construct involutive Hamiltonians on $T^*(G/P)$ (I'm confused if I'm talking about hol. sympl. spaces or just sympl. spaces here...) Do you have any comments on nilpotent orbits? –  Yuji Tachikawa Jan 12 '12 at 11:21
Interesting fact. Can we understand it somehow ? Toy model for such considerations - take sl_2 = C^3, regular orbit - is qaudric defined by Casimir = Const. Let me point out that Orbit - affine complex manifold and TP - is NOT, since contains a projective P. So they cannot be isomorphic as complex manifolds. But according to what I heard Orbit - is deformation of TP. –  Alexander Chervov Jan 12 '12 at 12:54
@Alexander: You're right.The orbit is an affine bundle over $G/P$, not a vector bundle. In the toy example you map the orbit to $\mathbb{P}^1$ by sending a (regular) matrix to the flag determined by an eigenvector. The affine bundle doesn't have a holomorphic section, but has a smooth one. –  Peter Dalakov Jan 12 '12 at 15:37
@Alexander: I should've been more explicit about the section, sorry. Use the hermitian metric on $\mathbb{C}^2$: given a point $p\in\mathbb{P}^1$ take a unit vector $v_1\in p$. Then take an orthogonal unit vector, say $v_2$. Let $P$ be the matrix with columns $v_i$. The (smooth) section is $p\mapsto PDP^{-1}$, where $D=\textrm{diag}(\sqrt{\lambda},-\sqrt{\lambda})$. Here $\lambda\in \mathbb{C}^\times$ is the value of the determinant, which is fixed on the orbit. –  Peter Dalakov Jan 12 '12 at 16:37
@Peter Thank You very much ! Is it possible to see the result in this toy model ? –  Alexander Chervov Jan 12 '12 at 20:32

My feeling is that for semi-simple g (at least classical g) this should be known to be true. But I cannot provide a reference now.

I am not sure about details, but the similar sounding conjecture sometimes associated with names of A.S. Mishchenko and A.T. Fomenko (early 80-ies ???). As far as I understand their conjecture is that any (not only semi-simple) g has maximal Poisson-commutative subalgebra in S(g). I am not sure did they conjecture that we can find int. sys. on any orbit. May be not explicitly. There is a bulk of works on this conjecture. Some names - A. Bolsinov, Trofimov, Fomenko himself wrote books on it, many of his students worked on it. I would suggest to look at

The argument shift method and maximal commutative subalgebras of Poisson algebras

Dmitri I. Panyushev, Oksana S. Yakimova

As far as I understand they rise question not only for g, but also for any affine Poisson manifold (I am not sure this work or not - but Yakimova surely discussed it).

About quivers - Nekrasov's old paper contains some examples of int.sys. on quivers . My feeling was that one can see (at least some of) quivers as moduli spaces on vector bundles on very degenerated curves and so these are in a sense Hitchin's system.

share|cite|improve this answer
Thank you very much as always. Indeed, the statement contained in the reference you gave me settles my question positively for regular orbits. Do you know any paper where non-regular orbits are studied? –  Yuji Tachikawa Jan 12 '12 at 8:06
Aaah... so this is the Gelfand-Zeitlin system (for $gl(n)$) I often hear about recently. –  Yuji Tachikawa Jan 12 '12 at 9:10
Thanks for thanks:). I'll try to e-mail to some colleagues in evening - the question must be discussed somewhere. I am also interested in it, but never had a motivations to ask... Since it is interesting not only for me - it gives some motivation. –  Alexander Chervov Jan 12 '12 at 13:44

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.