# Independence of Duistermaat-Heckman measure

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?

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Nikodem Dyzma is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.

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 $$(X,\omega_\phi)$$.
• 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. – Nikodem Dyzma May 16 at 8:12