Moment map of isometries on Kähler manifolds Let us assume we are given a Kähler manifold $M$, equipped with its metric $g_{\imath\bar\jmath}$ and with the associated symplectic form
$$
\Omega = i\, g_{\imath \bar \jmath}dz^\imath \wedge d\bar z^{\bar\jmath},
$$
which satisfies $d\Omega=0$.
Consider now the isometry algebra $\mathfrak g$ of this manifold, i.e. the Lie algebra of the $d$-dimensional isometry group $G$; these infinitesimal isometries are described on $M$ by those vector fields $\xi_A$ for $A=1,\ldots,d$ satisfying
$$
\xi_A^\imath = -i g^{\imath\bar\jmath}\partial_{\bar\jmath}P_A\\
\bar \xi_A^{\bar\imath}=i g^{\imath\bar\jmath}\partial_{\imath}P_A,
$$
for some functions $P_A$ defined on $M$. In particular, each $\xi_A$ corresponds to the infinitesimal action on $M$ of a generator $X_A$ of $\mathfrak g$.
I am trying to show that the map $P:M \to \mathfrak g^\ast$, from the manifold to the dual of the Lie algebra, defined by $P_A = \langle P, X_A\rangle$, is the moment map of the $G$-action on $M$. Here the angular brackets are used to indicate contraction of the Lie algebra with its dual:
$$
\langle\cdot\,,\cdot\rangle:\mathfrak g^\ast\times\mathfrak g \rightarrow \mathbb R.
$$
Indeed 
$$
dP_A = \partial_\imath P_A dz^\imath + \partial_{\bar\jmath}P_A d\bar z^{\bar\jmath}
=-ig_{\imath\bar\jmath}\bar \xi_A^{\bar \jmath}dz^{\imath} + i g_{\imath\bar\jmath} \xi_A^\imath d\bar z^{\bar\jmath} = \Omega(\xi_A,\cdot) 
$$
by the above relations for $X_A$: this means that $P_A$ is a Hamiltonian for (the homotopy generated by) $\xi_A$, as is part of the definition of moment map.
However I cannot verify the property of equivariance, i.e. that 
$$
P\circ \psi_g = \mathrm{Ad}_g^\ast P,
$$
where $g\in G$ and $\psi$ denotes the group action.
My try:
At the infinitesimal level, equivariance is equivalent to
$
\mathcal L_{\xi_A} P = \mathrm{ad}_{X_A}^\ast P,
$
for $X\in\mathfrak g$.
Applying this to some $X_B\in\mathfrak g$, the right-hand side then gives 
$$
\langle \mathrm{ad}_{X_A}^\ast P,X_B\rangle =
-\langle P, \mathrm{ad}_{X_A}X_B\rangle = -\langle P, [X_A,X_B]\rangle = -f^C_{AB}P_C
$$
where $f^{C}_{AB}$ are the structure constants of $\mathfrak g$, but on the other hand
$$
\langle \mathcal L_{\xi_A} P, \xi_B\rangle = \mathcal L_{\xi_A}P_B= dP_B(\xi_A)= - \Omega(\xi_A, \xi_B) 
$$
because, by our first computation, $P_B$ is a Hamiltonian for $\xi_B$.
I have two hypotheses:


*

*the computation is wrong somewhere (but I can't see where!)

*it is actually true that
$
\Omega(\xi_A, \xi_B) = f^{C}_{AB}P_C
$
(but I can't see why!).


Please, any help is appreciated.
 A: It turns out that this condition is not true in general. Consider for instance the plane, with complex coordinates $z$, $\bar z$ and the standard Euclidean metric. In particular we have two isometries given by the translations ($a$ and $b$ are real constants)
$$
\xi_1 = a (\partial_z+ \bar\partial_{\bar z})\,,\qquad
\xi_2 = i b (\partial_z - \bar \partial_{\bar z})\,.
$$
The corresponding moment maps can be taken to be 
$$
P_1 = \frac{a}{2i} (z -\bar z)\,,\qquad
P_2 = \frac{b}{2}(z+\bar z)\,.
$$
However, clearly $[\xi_1,\xi_2]=0$, whereas
$$
\Omega(\xi_1, \xi_2) = -2 ab\,.
$$
So, the equivariance property has to be considered case by case.
More in general, we may use the fact that the Hamilton function of $[\xi_A, \xi_B]$ is $\Omega(\xi_A, \xi_B)$ up to some constant $c_{AB}$, namely
$$
P_{[X_A,X_B]} = \Omega(\xi_A, \xi_B)+c_{AB}\,,
$$
where $c_{AB}$ has to satisfy the two-cocycle condition with respect to the structure constants. The equivariance condition thus reduces to checking that $c_{AB}$ can be consistently set to zero.
