Reduction along an Orbit for C.-M. systems

Question

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)

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).