Let $M:=U(n)/T^n$ be a complete flag variety, where $U(n)$ is an unitary group and $T^n \simeq (S^1)^n$ consists of its diagonal matrices. I have heard the following construction of a symplectic moment map $\mu: M \to t^*$. Let $B=\operatorname(b_1, \ldots, b_n)$ be a diagonal matrix with different $b_i \in \mathbb R$. Then $M$ may be identified with a $U(n)$orbit $M_B$ of $B$ in hermitian matrices $Herm_n$ under conjugation, namely $$i: M \to M_B \subset Herm_n, \quad X \mapsto XBX^{1},$$ and one can see that the stabilizer of $B$ is exactly $T^n$. It is said that there is a symplectic form on $M_B$ such that a natural action of $T^n$ by conjugation is Hamiltonian with a moment map $$\mu: M_B \to t^*, \quad Y \mapsto (y_{11}, \ldots, y_{nn}),$$ that is it maps matrix to its diagonal. Could you help me to see this symplectic form?
1 Answer
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The form on the orbit of coadjoint action is called KirillovKostantSouriau form (thank to Friedrich Knop for reference!), it descends from a form $\omega_\alpha: T^*_\alpha \mathfrak g^* \times T^*_\alpha \mathfrak g^*=\mathfrak g \times \mathfrak g \to \mathbb C$ that in a point $\alpha \in \mathfrak g^*$ is given by $\omega_\alpha(x, y)=\alpha([x, y])$.
The details are, for example, in Chriss, Ginzburg Representation Theory and Complex Geometry, proposition 1.1.5 and claim 1.1.6.

1$\begingroup$ The definition of the KKS form is actually $$ \omega_\alpha(\mathrm{ad}_x^* \alpha,\mathrm{ad}_y^*\alpha) = \alpha([x,y]).$$ $\endgroup$ Commented Jan 5, 2017 at 19:44