Theorem 22.1.56.0AD7F80A793D Assume that $X$ admits a smooth flag $Y_{\bullet }$. The Duistermaat–Heckman measure $\operatorname{DH}(\Gamma )$ of $\Gamma \in \mathrm{PSH}^{\mathrm{NA}}(X,\theta )_{>0}$ in definition 22.1.55 is independent of the choice of the smooth flag $Y_{\bullet }$. Furthermore, for any $m\in \mathbb {Z}_{>0}$, the following moment identity holds:
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22.1.57
\begin{equation} \int _{\mathbb {R}}x^ m\operatorname{DH}(\Gamma )(x)=\Gamma _{\max }^ m\operatorname{vol}\Gamma +m\int _{-\infty }^{\Gamma _{\max }}\tau ^{m-1}\Bigl ( \operatorname{vol}(\theta +\mathrm{dd}^{\mathrm{c}}\Gamma _{\tau })-\operatorname{vol}\Gamma \Bigr )\, \mathrm{d}\tau , \end{equation}
and
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22.1.58
\begin{equation} \int _{\mathbb {R}}\operatorname{DH}(\Gamma )=\operatorname{vol}\Gamma . \end{equation}
Proof.
We observe that (22.1.57) holds by the computations in proposition 19.6.19 and theorem 19.5.26 1. It is independent of the choice of the flag. Furthermore, (22.1.58) follows immediately from theorem 19.5.26 1.
Assume first that $\Gamma $ is bounded. Since the Duistermaat–Heckman measure has bounded support in this case (cf. theorem 19.6.13), we conclude that $\operatorname{DH}(\Gamma )$ is uniquely determined.
By translating the parameter, it suffices to treat the case $\Gamma _{\max }=0$. Indeed, if $a=\Gamma _{\max }$ and
\[ \widetilde\Gamma _\tau \coloneqq \Gamma _{\tau +a}, \]
then $\widetilde\Gamma _{\max }=0$, and the associated test function is translated by $a$. Consequently,
\[ \operatorname{DH}(\Gamma )=(x\mapsto x+a)_*\operatorname{DH}(\widetilde\Gamma ). \]
Thus independence of the flag and the moment identity for $\widetilde\Gamma $ imply the corresponding assertions for $\Gamma $.
For each $\epsilon >0$, we define $\Gamma ^{\epsilon }\in \mathrm{PSH}^{\mathrm{NA}}(X,\theta )_{>0}$ as follows:
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Let $\Gamma ^{\epsilon }_{\max }=0$, and
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we set
\[ \Gamma ^{\epsilon }_{\tau }= \begin{cases} \Gamma _{-\infty },& \text{ if }\tau \leq -\epsilon ^{-1},\\ P_{\theta }\left[ (1+\epsilon \tau )\Gamma _{\tau }-\epsilon \tau \Gamma _{-\infty } \right],& \text{ if }\tau \in \left(-\epsilon ^{-1},0\right). \end{cases} \]
Let us check that $\Gamma ^\epsilon $ is indeed an element of $\mathrm{PSH}^{\mathrm{NA}}(X,\theta )_{>0}$. The concavity and monotonicity in $\tau $ follow from proposition 16.4.2. Since the $\Gamma _\tau $'s are $\mathcal I$-model, they are $\mathcal I$-good by example 18.1.11; hence
\[ (1+\epsilon \tau )\Gamma _\tau -\epsilon \tau \Gamma _{-\infty } \]
is $\mathcal I$-good by proposition 18.1.17. Therefore its $P$-envelope is also its $\mathcal I$-envelope, and the corresponding level is $\mathcal I$-model. The value at $-\infty $ is $\Gamma _{-\infty }$. Thus $\Gamma ^\epsilon \in \mathrm{PSH}^{\mathrm{NA}}(X,\theta )_{>0}$.
For each fixed $\tau <0$, the argument of theorem 17.13.21 Step 3.3 gives
\[ \Gamma ^\epsilon _\tau \searrow \Gamma _\tau \quad \textup{as }\epsilon \to 0+. \]
By proposition 16.4.4 and corollary 16.9.14, this convergence is in $d_ S$. Hence theorem 19.5.26 4 gives
\[ \Delta _{Y_\bullet }(\theta ,\Gamma ^\epsilon _\tau )\xrightarrow {d_{\mathrm{Haus}}}\Delta _{Y_\bullet }(\theta ,\Gamma _\tau ) \]
for every $\tau <0$. Since
\[ \Gamma ^\epsilon _{-\infty }=\Gamma _{-\infty } \]
for all $\epsilon $, the underlying convex body is fixed. Therefore 19.6.22 applies and gives
\[ \operatorname{DH}(\Gamma ^\epsilon )\rightharpoonup \operatorname{DH}(\Gamma ). \]
It follows that $\operatorname{DH}(\Gamma )$ is independent of the choice of the flag.
$\square$
Pome
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