Independence of Duistermaat-Heckman measure

Question

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?

Answer 1

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 $(X,\omega_\phi)$.