Everyone of us had sometimes this awful feeling that some sign is lost in a calculation and that this sign is perturbing some fundamental understanding of what is going on. I feel the same has happened for me today and I can't figure this sign problem out, so I count on you.

A Calogero-Moser system is defined as a Hamiltonian system with a Hamiltonian $$H=\sum p_i^2 + \sum_{i \neq k} \frac{1}{(x_i-x_j)^2}$$ It is widely known that this system is completely integrable.

I am trying to understand this widely known fact.

One of the proofs relies on the relation of the system with a linear flow in the space of matrices, the relationship is nicely explained in this MathOverflow entry: Is the 'massive' Calogero-Moser system still integrable?

The question that I already asked as a comment there is the following: the standard proof of the integrability rewrites $H$ as a restriction of some other function on the space of matrices which is actually $\mathrm{Tr }Y^2$ for a matrix $Y$ defined by $$ Y_{ii}=p_i, \; Y_{ik}=(x_i-x_k)^{-1}, \; i\neq k $$. A simple calculation will give us not $H$ but

$$H^-=\sum p_i^2 - \sum_{i \neq k} \frac{1}{(x_i-x_j)^2}$$

This is exactly the expression Etingof obtains in his Lectures on Calogero-Moser systems, http://www-math.mit.edu/~etingof/zlecnew.pdf. Etingof starts from the dynamics on matrix space and defines CM system as its symplectic reduction. So no problems for him.

But for a system of the particles on the real line, I feel lost. How one can prove integrability? And also, $H^-$ is giving the trajectories that would collapse.

  • 1
    $\begingroup$ What's wrong with the $i*x$ change of variables? $\endgroup$
    – AHusain
    Commented Jan 21, 2016 at 19:53
  • 1
    $\begingroup$ Hi Olga ! I think that even though the CM system is real, you can consider it coming from complex (hermitian) matrices, whose (real) eigenvalues should be the positions of the "particles". Hence it would be legit to take $Y_{jk}=\sqrt{-1}/(x_j-x_k)$... It's sometimes hard to guess something from its shadow! $\endgroup$
    – BS.
    Commented Jan 21, 2016 at 20:39
  • $\begingroup$ The condition $\mathrm{rk}([X,Y]+1)=1$ should still hold and by changing non-diagonal values of $Y$ like this I think we will get out of this condition. We can change both matrices by conjugating them by some matrix from $PGL_n(\mathbb{C})$. $X$ is diagonal. We would love to get $Y_{j k}$ as you write them but this is not possible to pass from a matrix [ M= \begin{bmatrix} p_1 & \frac{1}{x_1-x_2} \\ \frac{1}{x_2-x_1} & p_2 \end{bmatrix} ] to [ M= \begin{bmatrix} p_1 & \frac{i}{x_1-x_2} \\ \frac{i}{x_2-x_1} & p_2 \end{bmatrix} ] by (even complex) conjugacy $\endgroup$
    – Olga
    Commented Jan 22, 2016 at 10:52

1 Answer 1


The answer to my question (provided by BS) is the following:

We have to change the action by looking at the group $G=U(n, \mathbb{C})$ and its action by conjugacy on pairs of Hermitian matrices. The space of pairs of such matrices can be identified with $T^* (Lie U(n, \mathbb{C})^*)$ because $Lie U(n, \mathbb{C})$ consists of antihermitian matrices (so we should only divide by $i$ to obtain Hermitian matrices).

The coadjoint orbit is an orbit of a matrix with ones everywhere except for the diagonal (where it has zeroes). Its orbit is all Hermitian matrices $T$ such that $rk (T+\mathrm{Id})=1$. Moment map is $J(X,Y)=-i[X,Y]$. This subtle change in moment map will permit us to change the entries in $Y$ matrix.

And then a representative of each element in the orbit can be chosen in a form $(X,Y)$ where $X$ is a diagonal matrix and $Y_{j k }= \frac{i}{x_j-x_k}$.

So the idea of AHusain to multiply by $i$ was a good one -- but one has to change the action...

Note that this proof (for $H$) is quite the same as a proof for the potential $H^-$: it is related to the fact that they both come from the group $SL(n, \mathbb{C})$:this group has (among others) two real parts: $SL(n, \mathbb{R})$ and $SU(n, \mathbb{R})$. The first part corresponds to $H^-$ and the second to $H^+=H$.


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