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This is a question about the relationship between two ways of viewing the Calogero-Moser system.

$$\ddot x_i=2\sum_{j\neq i}\frac{1}{(x_i-x_j)^3}\qquad i=1,\ldots N$$

  • By introducing the $N$ variables $y_i$ ("dual variables" for want of a better term), one has the equivalent first order system

$$i\dot x_j=\sum_{k\neq j}\frac{1}{x_k-x_j}-\sum_{k=1}^N\frac{1}{y_k-x_j}\\ i\dot y_j=\sum_{k\neq j}\frac{1}{y_j-y_k}-\sum_{k=1}^N\frac{1}{y_j-x_k}\\$$

I am not sure who to attribute these equations to, but for the case where the $x$'s and $y$'s are complex conjugate, they appear in KM Case, PNAS, 75, 3562 (1978) in the context of pole dynamics for the Benjamin-Ono equation.

  • Alternatively, the construction of Kazhdan, Kostant, Sternberg (KKS) views the CM system as (to cut a long story short) the dynamics of the eigenvalues of the matrix $X+Pt$, where $P_{ij}=i/(x_i-x_j)$ for $i \neq j$ and the diagonal elements give the initial velocities. This is as an example of Hamiltonian reduction.

My question is: what is the relationship (if any) between these two viewpoints? Specifically, to what do the $y_i$ correspond in the matrix formulation?

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Comment to the question (v3): That PNAS reference is quite short. Do you have another reference where the above coupled first order system is mentioned? –  Qmechanic Nov 30 '13 at 21:37
    
Yes, it appears in: Alexander G Abanov et al 2009 J. Phys. A: Math. Theor. 42 135201 (for the case of periodic boundary conditions) and Michael Stone et al 2008 J. Phys. A: Math. Theor. 41 275401. –  Austen Dec 2 '13 at 16:32
    
Thanks, found them both. –  Qmechanic Dec 2 '13 at 18:54

1 Answer 1

I) Let us introduce a collective notation $z_i$, $i\in I$, for OP's $x_i$'s and $y_i$'s (which by the way do not have to be equal in numbers). Here $I$ is a finite index set. We assume that the map $z:I\to \mathbb{C}$ is injective. Also let us introduce a parity $\sigma: I\to \{\pm 1\}$, which is $+1$ for an $x_i$ and $-1$ for a $y_i$. Define a bi-linear skew-symmetric bracket$^1$

$$\tag{1} \{z_i,z_j\}~:=~ \left\{ \begin{array}{ccl} \frac{1}{z_i-z_j} & \text{for} & i\neq j, \\ 0& \text{for} & i= j. \end{array} \right.$$

Then OP's coupled first-order system can be written in Hamiltonian form

$$\tag{2} \mathrm{i}\dot{z}_j ~=~\sum_{i\in I\backslash\{j\}}\frac{\sigma_i}{z_i-z_j} ~\stackrel{(1)+(3)}{=}~\{z_j,H\},$$

with Hamiltonian

$$\tag{3} H~:=~ -\sum_{i\in I} \sigma_i z_i. $$

More generally, for a function $f=f(z)$, the time evolution is given as

$$\tag{4} \mathrm{i}\frac{d f}{dt} ~=~\{f,H\}+ \mathrm{i}\frac{\partial f}{\partial t}. $$

II) The corresponding second-order system is the Calogero-Moser equations$^2$

$$\tag{5} -\ddot{z}_j ~=~\sum_{i\in I\backslash\{j\}}\frac{1+\sigma_i\sigma_j}{(z_i-z_j)^3}. $$

It is a major point that the sum on the rhs. of eq. (5) only runs over elements of the same kind, i.e. if ${z}_j$ on the lhs. is an $x_j$, then the non-zero terms in the sum on the rhs. is only over the $x_i$'s, i.e. independent of the $y_i$'s. Hence the evolution of $x_j$ only depend on the $y_i$'s via their initial conditions [1]. And vice-versa with the roles $x_i \leftrightarrow y_i$ exchanged.

III) The Calogero-Moser Hamiltonian in Darboux coordinates $(z_i,p_i)$ reads

$$ \tag{6}H_{CM}~=~\frac{1}{2}\sum_{i\in I}p_i^2 +\frac{1}{4}\sum_{i,j\in I}^{i\neq j}\frac{1+\sigma_i\sigma_j}{(z_i-z_j)^2}. $$

IV) In the spirit of the Kazhdan-Kostant-Sternberg (KKS) construction [2,3], let us define a position matrix

$$ \tag{7} Z~:=~\text{diag}(z_i),$$

a parity matrix

$$ \tag{8} \Sigma~:=~\text{diag}(\sigma_i), \qquad \Sigma^2 = {\bf 1},$$

and a momentum matrix

$$ \tag{9} P_{ij}~:=~\left\{ \begin{array}{ccl} \frac{\mathrm{i}}{z_j-z_i} & \text{for} & i\neq j, \\ p_i& \text{for} & i= j. \end{array} \right.$$

The diagonal elements $p_i$ of the momentum matrix (9) have a physical interpretation as initial velocities. The momentum matrix (9) satisfies the following Canonical Commutation Relation (CCR) for finite-dimensional matrices:

$$ \tag{10} [Z,P]-\mathrm{i}{\bf 1}~=~\text{rank-one matrix}.$$

The Calogero-Moser Hamiltonian (6) can then be written as

$$ \tag{11} H_{CM}~=~\frac{1}{4}{\rm Tr}(P^2+P\Sigma P\Sigma).$$

The flow of $z_i(t)$ is given by the eigenvalues of the matrix

$$\tag{12} Z(t=0)+t\frac{P(t=0)+\Sigma P(t=0)\Sigma}{2}. $$

The matrix (12) is block diagonal consisting of two blocks. Each block is just the standard KKS construction. In particular, the eigenvalues $x_i(t)$ and $y_i(t)$ do only talk to each other via the initial conditions.

References:

  1. M. Stone, I. Anduaga, and L. Xing, The classical hydrodynamics of the Calogero–Sutherland model, J. Phys. A: Math. Theor. 41 (2008) 275401, arXiv:0803.3735.

  2. D. Kazhdan, B. Kostant, and S. Sternberg, Hamiltonian group actions and dynamical systems of Calogero type, Comm. Pure Appl. Math. 31 (1978) 481.

  3. P. Etingof, Lectures on Calogero-Moser systems, arXiv:math/0606233.

--

$^1$ Note that the bracket (1) does not satisfy the Jacobi identity. We suspect that the bracket (1) can be extended to a homotopy hierarchy of higher brackets, although we did not pursuit the matter, partly because the bracket (1) is not important for the rest of the answer.

$^2$ To prove Eq. (5) from eq. (2), the following identity is helpful:

$$\mathrm{i}(\dot{z}_i -\dot{z}_j) ~=~\{z_i-z_j,H\}$$ $$\tag{13} ~\stackrel{(1)+(3)}{=}~-\frac{\sigma_i+\sigma_j}{z_i-z_j} +(z_i-z_j)\sum_{k\in I\backslash\{i,j\}}\frac{\sigma_k}{(z_i-z_k)(z_j-z_k)}.$$

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