7
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ 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? $\endgroup$
    – Qmechanic
    Nov 30, 2013 at 21:37
  • $\begingroup$ 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. $\endgroup$
    – Austen
    Dec 2, 2013 at 16:32
  • $\begingroup$ Thanks, found them both. $\endgroup$
    – Qmechanic
    Dec 2, 2013 at 18:54

1 Answer 1

3
$\begingroup$

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:

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

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.