I am having trouble in understanding the section of this paper http://www-math.mit.edu/~etingof/zlecnew.pdf where the author introduces the Calogero-Moser system as the reduction of a manifold $M$ on which is acting a group $G$ along the coadjoint orbit $\mathcal{O}$ of $G$:$\mathcal{C}_n \doteq R(M,G,\mathcal{O})$.

We have the manifold $M = T^\ast\text{Mat}_n(\mathbb{C})$ which, I suppose, is the cotangent fiber of the set of all possible $\mathbb{C}$-valued matrices. Then we define a symplectic structure on it which is "the usual trace form": $$\omega =\text{Tr}\ ( dX \wedge dY).$$ Now, as I see it, this must is some sense be similar to what we do when we define the usual cotangent coordinates $q_i$ and $p_i$ on a cotangent fiber and get our canonical form $\sum_i dq_i\wedge dp_i$. However I do not quite understand what is the exterior derivative of a matrix $X$ or $Y$ and how such an $\omega$ actually defines a symplectic form.

Given all this, we can identify $$ \mathfrak{g}^\ast \simeq \mathfrak{g}$$ $$ M \simeq \text{Mat}_n(\mathbb{C}) \oplus \text{Mat}_n(\mathbb{C}).$$ How...?

Finally we have some functions $$ H_i = \text{Tr}(Y^i), \qquad i=1,\ldots , n$$ which are claimed to be in involution with each other $\{H_i, H_j\}=0$; but to verify that one would have to compute the corresponding hamiltonian vectors $X_i$ and $X_j$ and check that $\omega(X_i, X_j)=0.$ How do we do this?

I don't ask for the whole solution: just for a few good hints or, in alternative, for a good reference where I should look this stuff up.

(My background: basics on symplectic geometry such as Moment Maps, Coadjoint Representation, Reduction and so on)


1 Answer 1


I believe you should interpret $dX\wedge dY$ as follows: $X$ is a matrix valued function on the space of pairs of matrices (which just takes the first one); its entries are honest functions $x_{i,j}$. Thus, the exterior derivative $dX$ is a matrix whose entries are the 1-forms $dx_{i,j}$. The same is true of $dY$. The exterior derivative $dX\wedge dY$ should be like the usual product of matrices, but with $\wedge$ used when you would multiply entries. This is all a clever way of saying the symplectic form is $\omega=\sum_{i,j}dx_{i,j}\wedge dy_{j,i}$.

As for why the functions $\mathrm{Tr}(Y^i)$ Poisson commute: computing the Hamiltonian vector fields is a waste of your time. We have that $\{y_{i,j},y_{k,\ell}\}=0$ for all $i,j,k,\ell$ (since the Hamiltonian vector fields are $\pm \frac{d}{dx_{j,i}}$, where the sign depends on your conventions). Thus, any pair of functions that depend only on the $y_{i,j}$'s commute (this is why Etingof says it is obvious).

  • $\begingroup$ Thanks @Ben Webster, your answer was definitely clarifying. I'm kind of new to this kind of papers, and it seems to me that Etingof's lectures are quite synthetic, when not obscure, in many passages: as a pre-study to Calogero-Moser systems I studied Ana Cannas da Silva's lectures on symplectic geometry; do you think that's enough and I can to tackle Etingof's work or would you suggest some further preliminary reading? Thanks in advance. $\endgroup$
    – Brightsun
    Commented Apr 11, 2014 at 18:55
  • 1
    $\begingroup$ @Brightsun It depends what you want to get out, of course. I don't think you necessarily need more symplectic geometry background, but Etingof's perspective is not one of pure symplectic geometry, so background in other areas like representation theory could also be useful. In terms of symplectic geometry, you might try some references that emphasize the Poisson side of things more, since that seemed to be what was throwing you off. $\endgroup$
    – Ben Webster
    Commented Apr 11, 2014 at 21:18
  • $\begingroup$ Well, I'd like to understand a few initial chapters, not more than that. You are right, I'm not familiar enough with the Poisson stuff, it seems. Thanks again. $\endgroup$
    – Brightsun
    Commented Apr 11, 2014 at 21:20

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