Suppose that a compact Kähler manifold $(X,\omega)$ has a real torus acting on it by symplectomorphisms in a Hamiltonian way (the torus is not necessarily of maximal rank). Then for any smooth function from $PSH_{tor}(X,\omega) := \{\phi \in PSH(X,\omega)\;|\; \phi\text{ - invariant}\}$ the form $\omega_\phi := \omega + i\partial\bar{\partial}\phi$ is another invariant symplectic form in the same cohomology class, thus it has the same moment polytope.

Now the question is: if $J$ is the moment map for $\omega$ and $J_\phi$ is the moment map for $\omega_\phi$, do they produce the same Duistermaat-Heckman measure on the polytope? I suppose they do, but why?


We just need to show that $(X,\omega)$ and $(X,\omega_\phi)$ are $T$-equivariant symplectomorphic, then there is a commutative diagram relating the moment maps $J$ and $J_\phi$ over the same polytope through the equivariant symplectomorphism, and this directly implies their induced Duistermaat-Heckman measure agree.

We apply Moser's argument. Let $\omega_s=(1-s)\omega+s\omega_\phi$. Then $\omega_s-\omega=sd(\Re i\overline{\partial}\phi)$. Let $Y_s$ be the vector field satisfying $\iota_{Y_s}\omega_s=-\Re i\overline{\partial}\phi$, and let $\varphi_Y^s$ be the time $s$ flow for $Y_s$. Then we have $$\frac{d}{ds}(\varphi_Y^s)^*\omega_s=(\varphi_Y^s)^*(\mathcal{L}_{Y_S}\omega_s+\frac{d}{ds}\omega_s)=(\varphi_Y^s)^*(d\iota_{Y_s}\omega_s+d\Re i\overline{\partial}\phi)=0.$$ Hence $(\varphi_Y^1)^*\omega_\phi=\omega$. Since $\varphi_Y^1$ commutes with the $T$-action (because by definition $Y_s$ is $T$-invariant), it gives a $T$-equivariant symplectomorphism between $(X,\omega)$ and $(X,\omega_\phi)$.

  • $\begingroup$ One more question: how do we know that the symplectic volumes of reduced $J^{-1}(t)$ and $J_\phi^{-1}(t)$ are equal? In general those will be different sets. $\endgroup$ – Nikodem Dyzma May 16 at 8:12
  • $\begingroup$ @NikodemDyzma: Thanks for pointing that out. I've edited the answer. $\endgroup$ – Wonderfield May 17 at 2:27

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.