Example of a Hamiltonian Lie group action I was wondering why the following Lie group action is Hamiltonian. 
Equip $\mathbb{C}^{k\times n}\cong\mathbb{R}^{2kn}$ with the canonical symplectic form $\omega_0$ on $\mathbb{R}^{2kn}$. We have an action by the Lie group $G=U(k)$ on $\mathbb{C}^{k\times n}$ by matrix multiplication, which is of course smooth. However, I can't see why this action is symplectic, because I don't really see how the form $\omega_0$ works on $\mathbb{C}^{k\times n}$.
The momentum map for this action is supposed to be the map $\mu:\mathbb{C}^{k\times n}\to\mathfrak{u}(k)^*$, $\mu(A)\xi=\frac{i}{2}\text{Tr}(AA^*\xi).$ This map is well-defined, but how do I compute $d\langle\mu,\xi\rangle$, and show that this is equal to $\omega_0(\xi_M,\cdot)$ where $\xi_M$ is the fundamental vector field associated to the action. I have computed $\xi_M(x)=\xi x$, but how do I work with $\omega_0$ on $\mathbb{C}^{k\times n}?$
 A: On $\mathbb{C}$, the standard symplectic form is given by 
$\omega_0(v,w) = \text{Im}(\overline v w),$
for $v,w\in\mathbb{C}$ (the area form). In terms of the coordinate $z$ on $\mathbb{C}$, we have: 
$\omega_0 = \frac{d\overline z\wedge d z}{2i} = \frac{d(\overline z dz)}{2i}$,
So that $\omega_0$ is exact, namely for $\lambda_z(v) := \frac{\overline z v}{2i}$, we have $d\lambda = \omega_0$.
For a general complex vector space, $\mathbb{C}^N$, one does this on each factor:
$\omega_0((v_1,...,v_N), (w_1,...,w_N)) := \text{Im}(\sum \overline v_j w_j) = \text{Im} \langle v, w\rangle$.
Where $\langle\cdot,\cdot\rangle$ is the standard Hermitian inner product on $\mathbb{C}^N$. Moreover, $\omega_0$ is also exact, with $\omega_0 = d\lambda$ for
$\lambda_z(v) = \frac{\langle z, v\rangle}{2i}$
For your case, the standard Hermitian inner product on $\mathbb{C}^{kn}$ thought of as complex linear maps is given by the trace norm:
$\langle\langle V, W\rangle\rangle := Tr(V^*W)$.
So you have:
$\omega_0(V, W) = \text{Im} \langle\langle V, W\rangle\rangle$
Now, it is clear that the $U(k)$ action is symplectic, since $Tr((UV)^*(UW)) = Tr(V^*W)$ for $U\in U(k)$.
As for the momentum map, the fact that $\omega_0$ is exact makes things easier. We have $\omega_0 = d\lambda$ where
$\lambda_A(V) = \frac{1}{2i} \langle\langle A, V\rangle\rangle$
By the Cartan formula, and since the flow of $\xi_M$ is symplectic, we may take:
$\mu(A)\xi = -\lambda_A(\xi_M(A)) = - \frac{1}{2i} \langle\langle A, \xi A\rangle\rangle = -\frac{1}{2i} Tr(A^*\xi A) = \frac{i}{2}Tr(AA^*\xi)$.
