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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.

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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.

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