# Is the 'massive' Calogero-Moser system still integrable?

### Background

The (rational) Calogero-Moser system is the dynamical system which describes the evolution of $n$ particles on the line $\mathbb{C}$ which repel each other with force proportional to the cube of their distance. If the particles have (distinct!) position $q_i$ and momentum $p_i$, then the Hamiltonian which describes this system is $$H=\sum_i p_i^2+\sum_{i\neq k}\frac{1}{(x_i-x_k)^2}$$ There are many interesting properties of this system, but one of the first interesting properties is that it is `completely integrable'. This means that solving it explicitly amounts to solving a series of straight-forward integrals.

The integrability can most easily be shown by showing that the phase space for this system includes into a symplectic reduction of a certain matrix space, and then noticing that the above Hamiltonian is a restriction of a integrable Hamiltonian on the whole space. This is done by assigning to any ensemble of points $q_i$ and momenta $q_i$ a pair of $n\times n$ matrices $X$ and $Y$, where $X$ is the diagonal matrix with $q_i$ on the diagonal entries, while $Y$ is given by $$Y_{ii}=p_i, \; Y_{ik}=(x_i-x_k)^{-1}, \; i\neq k$$ This matrix assignment defines a map from the configuration space $CM_n$ of the CM system to the space of pairs of matrices. The space of pairs of matrices $(X,Y)$ is naturally a symplectic space from the bilinear form $(X,Y)\cdot (X',Y')=Tr(XY')-Tr(X'Y)$, and the action of $GL_n$ by simultaneous conjugation naturally has a moment map. Therefore, we sympletically reduce the space of pairs of matrices at a specific coadjoint orbit (not the origin) and get a new symplectic space $\overline{CM}_n$.

Composing the above matrix assignment with symplectic reduction, we get a map $CM_n\rightarrow \overline{CM}_n$. This map turns out to be a symplectic inclusion which has dense image. We also notice that the functions $Tr(Y^i)$, as $i$ goes from $1$ to $n$, descend to a Poisson-commuting family of functions on $\overline{CM}_n$, and because $\overline{CM}_n$ is $2n$ dimensional, each of the functions $Tr(Y^i)$ gives an integrable flow on $\overline{CM}_n$. Finally, we notice that $Tr(Y^2)$ restricts to $H$ on $CM_n$.

### The Massive Version of the CM System

Now, make the following change to the system. To every particle, assign a number $m_i$ (the mass), which can be in $\mathbb{C}$, but I am interested in the case where the $m_i$ are positive integers. Define a the massive CM Hamiltonian as $$H_m=\sum_i\frac{p_i^2}{m_i}+\sum_{i\neq k}\frac{m_im_k}{(x_i-x_k)^2}$$ The physical meaning of this equation is that particles still have force proportional to the inverse of the cube of their distance, but the force is proportional to the mass of that particle; also, particles resist acceleration proportional to their mass. If the force were to drop off proportional to the inverse square of their distance, and attract instead of repel, this would model how massive particles move under the influence of gravity.

### Questions

1. Is this system integrable?
2. Can it be realized in a similar matrix form?
3. Does it have any interesting or new behavior than the usual CM system?

### An Idea

It is almost possible to realize this Hamiltonian in a simple modification of the previous approach. Let $M$ denote the diagonal matrix with the $m_i$s on the diagonal. Then $$Tr(MYMY)=\sum_im_i^2p_i^2+\sum_{i\neq k}\frac{m_im_k}{(x_i-x_k)^2}$$ The functions $Tr( (MY)^i)$ should again be a Poisson commutative family. Rescaling the $p_i$ by $m_i^{3/2}$ gives the massive Hamiltonian $H_m$; however, this rescaling is not symplectic, and so it won't preserve the flows.

### Another Idea

In the case of integer $m_i$, one possibility is to work with $N\times N$ matrices rather than $n\times n$ matrices, where $N=\sum m_i$. Then it is possible to construct a matrix $X$ with eigenvalues $q_i$, each occuring with multiplicity $m_i$, as well as a matrix $Y$ such that $(X,Y)$ defines a point in $\overline{CM}_N$. The Hamiltonian $Tr(Y^2)$ even restricts to the correct 'massive' Hamiltonian $H_m$. However, the flow described by this Hamiltonian on $\overline{CM}_n$ will in almost all cases immediately separate eigenvalues that started together, which we don't want. If we restrict the Hamiltonian to the closed subspace where the eigenvalues are required to stay together, then this gives the desired flow. Unfortunately, restricting to a closed subvariety doesn't preserve a Hamiltonian being integrable.

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Is there a Lax formulation for the massive model? – José Figueroa-O'Farrill Feb 11 '10 at 21:16
My colleague Harry Braden, who apparently has no time to answer himself, claims that the case of $N=3$ is still integrable, but not for higher $N$. I don't have a reference, whence I am writing this as a comment, since I don't think that "hearsay" should be considered an answer :) – José Figueroa-O'Farrill Feb 12 '10 at 14:24
An even more general case $H=\sum_i\frac{p_i^2}{m_i}+\sum_{i\neq k}\frac{g_{ik}}{(x_i-x_k)^2}$ also is apparently nonintegrable as stated at Scholarpedia (but there's no direct proof there either, that's why this is a comment): scholarpedia.org/article/Calogero-Moser_system – mathphysicist Feb 12 '10 at 16:10
Hmm, its interesting that $N=3$ is a special case, since that was the number of CM particles I could get to stay together, in the manner I describe in the 'Another Idea' section. – Greg Muller Feb 12 '10 at 16:25
Since the particles lie in C, shouldn't expression in the Hamiltonian be |x_i - x_j|² and not (x_i - x_j)² ? – Tom LaGatta Feb 15 '10 at 21:53

The paper "Meromorphic Parametric Non-Integrability, the Inverse Square Potential" by E. J. Tosel, proves almost what was claimed in the comments. Except for Jacobi's theorem:

The 3-body problem on a line with arbitrary masses and inverse square potential is completely integrable with rational first integrals.

and Calogero-Moser’s Theorem:

The n-body problem with equal masses on a line with an inverse square potential is completely integrable. More precisely, there exists a complete family of commuting first integrals which are rational in $(Q,P)$.

all other cases are non-integrable. The main theorem is:

Theorem 3 (Non-integrability meromorphic in linear momenta and masses, rational in positions).

(i) For $n = 4$, the $n$-body problem on a line with an inverse square potential does not have a complete system of generically independent first integrals which are rational with respect to $Q$ and meromorphic with respect to $P$ and $(m _i) _{1\le i\le n}$

(ii) For $n = 3$ and $p \geq 2$, the $n$-body problem in $\mathbb{R} ^p$ with an inverse square potential does not have a complete system of generically independent first integrals which are rational with respect to $Q$ and meromorphic with respect to $P$ and $(m _i) _{1\le i\le n}$

A corollary of this theorem is that Calogero-Moser’s theorem deals with an exceptional case: there cannot exist a Lax pair $(L,B)$ which would depend meromorphically on the masses for $n$ bodies on a line ($n \geq 4$).

There is a remark there that the rationality condition in the case of the line needs only be checked in $n-4$ positions.

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Dear Gjerrji, thanks for this answer ! Can you provide some bibliography on Theorem 3? – Olga Mar 8 at 11:16

Just to add to Gjergji Zaimi's answer, Harry Braden has sent me the expressions for the conserved charges responsible for the integrability of the $N=3$ model:

• The total momentum $P = p_1 + p_2 + p_3$

• The hamiltonian $H$

• $Q = 2 H \sum_{i=1}^3 m_i x_i^2 -(\sum_{i=1}^3 x_i p_i )^2$

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Do you happen to know where the third integral of motion comes from? For my purposes, I don't really need a complete set of $n$ integrals of motion, I really only need the natural analog of the third integral of motion, after momentum and energy. My guess is it has the above form even when there are more particles. – Greg Muller Feb 15 '10 at 17:09
My guess -- I'll try to confirm -- is that this is probably the trace of the square of the Lax operator for this model. – José Figueroa-O'Farrill Feb 15 '10 at 17:57