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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.