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?](https://mathoverflow.net/questions/15031/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.



Sign problem in a Calogero-Moser system: proof of integrability?
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 $H^+=H$.


