We are given the complex special linear group $SL(2, \mathbb{C})$ which is a Lie group $G$, with its corresponding Lie algebra $sl(2, \mathbb{C})$ denoted by $g$. We are to prove that the natural action of $G$ on $(\mathbb{C}^2,ω=dz\wedge dw)$ is Hamiltonian with the moment map
$µ(z, w) = \frac{1}{2} \begin{pmatrix} zw & w^2 \\ -z^2 & -zw\\ \end{pmatrix}$
and the duality between the dual space of the Lie algebra $g$ and $g^*$ itself is given by the inner product on g defined by $(A, B) := Tr(AB)$.
So we need to prove $3$ things:
$(1)$ The action $\phi: G\to Symp(\mathbb{C}^2)$ is symplectic.
$(2)$ $µ(\phi_g(z,w))=Ad_{g^{-1}}^*(µ(z,w))$
$(3)$ $dµ_X = i_{U_M} ω$ where i. e $µ_X$ is a hamiltonian function for the vector field $U_M$
I have alerady proved part $(1)$ and $(2)$ and I need some help with part $(3)$, can someone help me please? Thank you in advance!
Edit:
for $X\in g$, $µ_X(z,w)$ is defined like $µ_X(z,w)=(µ(z,w),X)=Tr(µ(z,w)X)$. So we have $dµ_X(z,w)(Y)=Y(µ_X(z,w))= \frac{d}{dt}|_{t=0} μ_X(\exp(tY)(z, w))=\frac{d}{dt}|_{t=0}(\exp(tY)(z, w),X)=\frac{d}{dt}|_{t=0}Tr(\exp(tY)(z, w)X)$
on the other hand:
$ω(U_M(z, w), Y(z, w)) = ?$
Here, we have that $U_M$ be the vector field on $M$ generated by the one-parameter subgroup $\{\exp tU | t ∈ \mathbb{R}\} ⊆ G$.
so, can I say that $\frac{d}{dt}|_{t=0}\exp tU =U(\exp tU )$?
And How can I use that to calculate $ω(U_M(z, w), Y(z, w))$?