# Moment map of isometries on Kähler mainfolds

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 infinitesimal isometries $\mathfrak g$ of this manifold, i.e. the Lie algebra of the $d$-dimensional isometry group $G$; this infinitesimal isometries are given by those vector fields $X_A$ for $A=1,\ldots,d$ satisfying $$X_A^\imath = -i g^{\imath\bar\jmath}\partial_{\bar\jmath}P_A\\ \bar X_A^{\bar\imath}=i g^{\imath\bar\jmath}\partial_{\imath}P_A,$$ for some functions $P_A$ defined on $M$.

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 $$dP_A = \partial_\imath P_A dz^\imath + \partial_{\bar\jmath}P_A d\bar z^{\bar\jmath} =-ig_{\imath\bar\jmath}\bar X_A^{\bar \jmath}dz^{\imath} + i g_{\imath\bar\jmath}X_A^\imath d\bar z^{\bar\jmath} = \Omega(X_A,\cdot)$$ as it should, by the above relations defining $X_A$: this means that $P_A$ is the Hamiltonian for the homotopy generated by $X_A$, 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 $\mathcal L_X P = \mathrm{ad}_X^\ast P,$ 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 $f^{ABC}$ are the structure constants of $\mathfrak g$, but on the other hand $$\langle \mathcal L_{X_A} P,X_B\rangle = \mathcal L_{X_A}P_B= dP_B(X_A)= - \Omega(X_A, X_B)$$ because, by our first computation, $dP_B = \Omega(X_B, \cdot)$.

I have two hypotheses:

• the computation is wrong somewhere (but I can't see where!)
• it is actually true that $\Omega(X_A, X_B) = f^{C}_{AB}P_C$ (but I can't see why!).

Please, any help is appreciated.