BC31851ECB21
Fix $n\in \mathbb {N}$. Let $\Delta ,\Delta '\subseteq \mathbb {R}^ n$ be convex bodies with positive volumes. The standard Lebesgue measure on $\mathbb {R}^ n$ is denoted by $\operatorname{vol}$.
Recall that $\mathcal{K}_ n$ denotes the set of convex bodies in $\mathbb {R}^ n$ and $d_{\mathrm{Haus}}$ denotes the Hausdorff metric. We refer to section 19.7 for the basic properties of these objects.
We study the notion of Okounkov test curves in this section; this leads to the definition of the Duistermaat–Heckman measure of a non-Archimedean metric in subsection 22.1.54 below. This section can be skipped on a first reading.
Definition 19.6.1.ACD722F45231 An Okounkov test curve relative to $\Delta $ consists of
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a number $\Delta _{\max }\in \mathbb {R}$ and
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an assignment $(-\infty ,\Delta _{\max })\ni \tau \mapsto \Delta _{\tau }\in \mathcal{K}_ n$ satisfying
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the assignment $\tau \mapsto \Delta _{\tau }$ is decreasing and concave;
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we have $\Delta _{\tau }\xrightarrow {d_{\mathrm{Haus}}}\Delta $ as $\tau \to -\infty $.
The set of Okounkov test curves relative to $\Delta $ is denoted by $\mathrm{TC}(\Delta )$.
An Okounkov test curve $\Delta _{\bullet }$ relative to $\Delta $ is bounded if $\Delta _{\tau }=\Delta $ when $\tau $ is small enough. The subset of bounded Okounkov test curves is denoted by $\mathrm{TC}^{\infty }(\Delta )$.
An Okounkov test curve $\Delta _{\bullet }$ relative to $\Delta $ is said to have finite energy if
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19.6.2
\begin{equation} \mathbf{E}(\Delta _{\bullet })\coloneqq n!\Delta _{\max } \operatorname{vol}\Delta +n!\int _{-\infty }^{\Delta _{\max }} \left(\operatorname{vol}\Delta _{\tau }-\operatorname{vol}\Delta \right)\, \mathrm{d}\tau >-\infty . \end{equation}
The subset of Okounkov test curves with finite energy is denoted by $\mathrm{TC}^{1}(\Delta )$.
Given $\Delta _{\bullet }\in \mathrm{TC}(\Delta )$ and $\Delta '_{\bullet }\in \mathrm{TC}(\Delta ')$, we say $\Delta _{\bullet }\leq \Delta '_{\bullet }$ if $\Delta _{\max }\leq \Delta '_{\max }$ and for any $\tau <\Delta _{\max }$, we have $\Delta _{\tau }\subseteq \Delta '_{\tau }$.
Sometimes it is convenient to introduce
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19.6.3
\begin{equation} \Delta _{\Delta _{\max }}=\bigcap _{\tau <\Delta _{\max }}\Delta _{\tau }\in \mathcal{K}_ n. \end{equation}
We shall always make this extension in the sequel when we talk about $\Delta _{\Delta _{\max }}$. Observe that $(-\infty ,\Delta _{\max }]\ni \tau \mapsto \Delta _{\tau }$ is still concave.
Proposition 19.6.4.29A7DDA38CA2 Any Okounkov test curve $(\Delta _{\tau })_{\tau < \Delta _{\max }}$ relative to $\Delta $ is continuous in $\tau $. Moreover, $\operatorname{vol}\Delta _{\tau }>0$ for all $\tau <\Delta _{\max }$.
Proof.
We first claim that $\operatorname{vol}\Delta _{\tau '}>0$ for all $\tau '<\Delta _{\max }$. By Condition (2b) in definition 19.6.1 and theorem 19.7.5, we know that $\operatorname{vol}\Delta _{\tau ''}>0$ when $\tau ''$ is small enough. Fix one such $\tau ''$. We may assume that $\tau ''\leq \tau '$ since otherwise there is nothing to prove. Next take $\tau '''\in (\tau ',\Delta _{\max })$. Take $t\in (0,1)$ such that $\tau '=t\tau '''+(1-t)\tau ''$. It follows that
\[ \operatorname{vol}\Delta _{\tau '}\geq \operatorname{vol}\left( t\Delta _{\tau '''}+(1-t)\Delta _{\tau ''}\right)\geq (1-t)^ n\operatorname{vol}\Delta _{\tau ''}>0. \]
Next we claim that $\operatorname{vol}\Delta _{\tau }$ is continuous for $\tau < \Delta _{\max }$. In fact, it follows from theorem 19.7.14 that $(-\infty ,\Delta _{\max }) \ni \tau \mapsto \log \operatorname{vol}\Delta _{\tau }$ is concave, but we already know that it is finite, hence the continuity follows.
Next we show that
\[ \Delta _{\tau }=\bigcap _{\tau '<\tau }\Delta _{\tau '}. \]
The $\subseteq $ direction is obvious. By the continuity of the volume, both sides have the same volume and the volume is positive, we therefore obtain the equality.
Similarly, we have
\[ \Delta _{\tau }=\overline{\bigcup _{\tau '>\tau }\Delta _{\tau '}}. \]
The continuity of $\Delta _{\tau }$ at $\tau <\Delta _{\max }$ is proved.
$\square$
Definition 19.6.5.F0D4BAA63D8A A test function on $\Delta $ is a function $G\colon \Delta \rightarrow [-\infty ,\infty )$ such that
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$G$ is concave,
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$G$ is finite on $\operatorname{Int}\Delta $, and
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$G$ is upper semicontinuous.
A test function $G$ is bounded if $G$ is bounded from below.
A test function $G$ has finite energy if
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19.6.6
\begin{equation} \mathbf{E}(G)\coloneqq n!\int _{\Delta } G\, \mathrm{d}\lambda >-\infty . \end{equation}
Definition 19.6.7.7B20D9977149 Let $\Delta _{\bullet }\in \mathrm{TC}(\Delta )$. We define its Legendre transform as
\[ G[\Delta _{\bullet }] \colon \Delta \rightarrow [-\infty ,\infty ),\quad a\mapsto \sup \left\{ \tau <\Delta _{\max }:a\in \Delta _{\tau }\right\} . \]
Given a test function $G\colon \Delta \rightarrow [-\infty ,\infty )$, we define its inverse Legendre transform $\Delta [G]_{\bullet }$ as the Okounkov test curve relative to $\Delta $ defined as follows:
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$\Delta [G]_{\max }=\sup _{\Delta } G$, and
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for each $\tau <\sup _{\Delta } G$, we set
\[ \Delta [G]_{\tau }=\{ x\in \Delta :G\geq \tau \} . \]
We observe that
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19.6.8
\begin{equation} G[\Delta _{\bullet }](a)=\max \left\{ \tau \leq \Delta _{\max }: a\in \Delta _{\tau }\right\} ,\textup{ if } G[\Delta _{\bullet }](a)>-\infty . \end{equation}
Lemma 19.6.9.21BC38898B4A Let $\Delta _{\bullet }\in \mathrm{TC}(\Delta )$. Then $G[\Delta _{\bullet }]$ defined in definition 19.6.7 is a test function.
Similarly, if $G\colon \Delta \rightarrow [-\infty ,\infty )$ is a test function, then $\Delta [G]_{\bullet }$ is an Okounkov test curve.
Proof.
First suppose that $\Delta _{\bullet }\in \mathrm{TC}(\Delta )$. We want to verify that $G[\Delta _{\bullet }]$ satisfies the conditions in definition 19.6.5.
We first verify the concavity. Take $a,b\in \Delta $. We want to prove that for any $t\in (0,1)$,
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19.6.10
\begin{equation} G[\Delta _{\bullet }](ta+(1-t)b)\geq tG[\Delta _{\bullet }](a)+(1-t)G[\Delta _{\bullet }](b). \end{equation}
There is nothing to prove if $G[\Delta _{\bullet }](a)$ or $G[\Delta _{\bullet }](b)$ is $-\infty $. So we assume that both are finite. In this case, by (19.6.8),
\[ a\in \Delta _{G[\Delta _{\bullet }](a)},\quad b\in \Delta _{G[\Delta _{\bullet }](b)}. \]
Thus,
\[ ta+(1-t)b\in t \Delta _{G[\Delta _{\bullet }](a)}+(1-t)\Delta _{G[\Delta _{\bullet }](b)}\subseteq \Delta _{tG[\Delta _{\bullet }](a)+(1-t)G[\Delta _{\bullet }](b)}. \]
We deduce that
\[ G[\Delta _{\bullet }](ta+(1-t)b)\geq tG[\Delta _{\bullet }](a)+(1-t)G[\Delta _{\bullet }](b). \]
Therefore, (19.6.10) follows.
It is clear that $G[\Delta _{\bullet }]$ is finite on the interior of $\Delta $. It remains to argue that $G[\Delta _{\bullet }]$ is upper semicontinuous.
Let $(a_ i)_{i\geq 1}$ be a sequence in $\Delta $ with limit $a\in \Delta $. Define $\tau _ i=G[\Delta _{\bullet }](a_ i)$. Let $\tau =\varlimsup _ i \tau _ i$. We need to show that
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19.6.11
\begin{equation} G[\Delta _{\bullet }](a)\geq \tau . \end{equation}
There is nothing to prove if $\tau =-\infty $. We assume that this is not the case. Passing to a subsequence, we may assume that $\tau _ i\to \tau $. In particular, we can assume that $\tau _ i\neq -\infty $ for all $i\geq 1$. It follows from (19.6.8) that $a_ i\in \Delta _{\tau _ i}$ for all $i\geq 1$. If $\tau <\Delta _{\max }$, then $\Delta _{\tau _ i}\xrightarrow {d_{\mathrm{Haus}}}\Delta _{\tau }$, so theorem 19.7.6 implies that $a\in \Delta _{\tau }$. If $\tau =\Delta _{\max }$, then the same conclusion follows from the definition $\Delta _{\Delta _{\max }}=\bigcap _{\tau '<\Delta _{\max }}\Delta _{\tau '}$. Thus, (19.6.11) follows.
Conversely, suppose that $G\colon \Delta \rightarrow [-\infty ,\infty )$ is a test function. We argue that $\Delta [G]_{\bullet }$ is an Okounkov test curve. We verify the conditions in definition 19.6.1.
Firstly, for each $\tau < \sup _{\Delta }G$, the set $\Delta [G]_{\tau }$ is a convex body as $G$ is concave and usc. Moreover, $\Delta [G]_{\tau }$ is clearly decreasing in $\tau $.
Secondly, for each $a\in \Delta $, we can write $a=\lim _{i}a_ i$ with $a_ i\in \operatorname{Int}\Delta $. By assumption, $G$ is finite at $a_ i$. Thus,
\[ a\in \overline{\{ G>-\infty \} }=\overline{\bigcup _{\tau <\sup _{\Delta }G}\Delta [G]_{\tau }}. \]
By theorem 19.7.6, $\Delta [G]_{\tau }\xrightarrow {d_{\mathrm{Haus}}}\Delta $ as $\tau \to -\infty $.
Thirdly, $\Delta [G]$ is concave. To see this, take $\tau ,\tau '< \Delta _{\max }$. We need to prove that for any $t\in (0,1)$,
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19.6.12
\begin{equation} \Delta [G]_{t\tau +(1-t)\tau '}\supseteq t\Delta [G]_{\tau }+(1-t)\Delta [G]_{\tau '}. \end{equation}
Let $a\in \Delta [G]_{\tau }$ and $b\in \Delta [G]_{\tau '}$. We have $G(a)\geq \tau $ and $G(b)\geq \tau '$. As $G$ is concave, we have $G(ta+(1-t)b)\geq t\tau +(1-t)\tau '$. Thus,
\[ ta+(1-t)b\in \Delta [G]_{t\tau +(1-t)\tau '} \]
and (19.6.12) follows.
$\square$
Theorem 19.6.13.2584C29ED70F The Legendre transform and inverse Legendre transform are inverse to each other, defining a bijection between $\mathrm{TC}(\Delta )$ and the set of test functions on $\Delta $.
Under this bijection, $\mathrm{TC}^1(\Delta )$ corresponds to test functions on $\Delta $ with finite energy and $\mathrm{TC}^{\infty }(\Delta )$ corresponds to bounded test functions on $\Delta $.
Proof.
Thanks to 19.6.9, in order to prove the first assertion, it only remains to see that the Legendre transform and the inverse Legendre transform are inverse to each other, which is immediate by definition.
It is obvious that $\mathrm{TC}^{\infty }(\Delta )$ corresponds to bounded test functions. Moreover, a direct computation shows that if $\Delta _{\bullet }\in \mathrm{TC}(\Delta )$, then
\[ \mathbf{E}(\Delta _{\bullet })=\mathbf{E}(G[\Delta _{\bullet }]), \]
concluding the $\mathrm{TC}^{1}(\Delta )$ case.
$\square$
Proposition 19.6.14.7B4552909292 Let $(\Delta ^ i)_{i\in I}$ be a decreasing net in $\mathcal{K}_ n$ such that $\Delta \subseteq \bigcap _{i\in I}\Delta ^ i$. Consider a decreasing net $(\Delta ^ i_{\bullet })_{i\in I}$ with $\Delta ^ i_{\bullet }\in \mathrm{TC}(\Delta ^ i)$ for all $i\in I$ such that there is $\Delta _{\bullet }\in \mathrm{TC}(\Delta )$ satisfying the following properties:
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$\Delta _{\max }= \lim _{i\in I}\Delta ^ i_{\max }$;
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for any $\tau <\Delta _{\max }$, we have $\Delta ^ i_{\tau }\xrightarrow {d_{\mathrm{Haus}}}\Delta _{\tau }$.
Then for any $a\in \Delta $, we have
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19.6.15
\begin{equation} \lim _{i\in I} G[\Delta ^ i_{\bullet }](a)=G[\Delta _{\bullet }](a). \end{equation}
Note that in general,
\[ \Delta \subsetneq \bigcap _{i\in I}\Delta ^ i. \]
Proof.
Fix $a\in \Delta $. It follows immediately from the definition of $G$ that the net $(G[\Delta ^ i_{\bullet }](a))_{i\in I}$ is decreasing and the $\geq $ direction in (19.6.15) holds. Let us prove the reverse inequality. Let $\tau $ denote the left-hand side of (19.6.15) for the moment. By definition, for any $\epsilon >0$ and any $i\in I$, we have $a\in \Delta ^ i_{\tau -\epsilon }$. It follows that
\[ a\in \Delta _{\tau -\epsilon }. \]
Therefore,
\[ \tau \leq G[\Delta _{\bullet }](a). \]
This completes the proof.
$\square$
Similarly, for increasing nets, we have:
Proposition 19.6.16.D70EFA3100C9 Let $(\Delta ^ i)_{i\in I}$ be an increasing net in $\mathcal{K}_ n$ with Hausdorff limit $\Delta $ such that $\operatorname{vol}\Delta ^ i>0$ for all $i\in I$. Consider an increasing net $(\Delta ^ i_{\bullet })_{i\in I}$ with $\Delta ^ i_{\bullet }\in \mathrm{TC}(\Delta ^ i)$ for all $i\in I$. Let $\Delta _{\max }=\lim _{i\in I}\Delta ^ i_{\max }$. For any $\tau <\Delta _{\max }$, let $\Delta _{\tau }$ be the Hausdorff limit of $\Delta ^ i_{\tau }$. Then $\Delta _{\bullet }\in \mathrm{TC}(\Delta )$ and
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19.6.17
\begin{equation} \lim _{i\in I}G[\Delta ^ i_{\bullet }](a)=G[\Delta _{\bullet }](a) \end{equation}
for any $a\in \operatorname{Int}\Delta $.
Proof.
It is obvious that $\Delta _{\bullet }\in \mathrm{TC}(\Delta )$.
Fix $a\in \operatorname{Int}\Delta $. Since $\Delta ^ i\xrightarrow {d_{\mathrm{Haus}}}\Delta $, we have $a\in \Delta ^ i$ for all large enough $i$. By definition, the net $(G[\Delta ^ i_{\bullet }](a))_{i\in I}$ is increasing and the $\leq $ direction in (19.6.17) holds. Let us write $\tau =G[\Delta _{\bullet }](a)$ for the time being. By definition of $G$, for any $\epsilon >0$, we have
\[ a\in \Delta _{\tau -\epsilon /2}. \]
The concavity of $\Delta _{\bullet }$ guarantees that
\[ a\in \operatorname{Int}\Delta _{\tau -\epsilon }. \]
Indeed, choose $\sigma \ll 0$ so that a small ball around $a$ is contained in $\Delta _\sigma $. Since $a\in \Delta _{\tau -\epsilon /2}$ and
\[ \tau -\epsilon =t\sigma +(1-t)(\tau -\epsilon /2) \]
for a suitable $t\in (0,1)$, concavity gives a small ball around $a$ contained in $\Delta _{\tau -\epsilon }$.
Since $\Delta ^ i_{\tau -\epsilon }\xrightarrow {d_{\mathrm{Haus}}}\Delta _{\tau -\epsilon }$ and $a\in \operatorname{Int}\Delta _{\tau -\epsilon }$, it follows that for all large enough $i\in I$,
\[ a\in \Delta ^ i_{\tau -\epsilon }. \]
Therefore,
\[ \tau -\epsilon \leq G[\Delta ^ i_{\bullet }](a) \]
for all large enough $i$. Taking the limit with respect to $i$ and then with respect to $\epsilon $, we conclude the desired inequality.
$\square$
Definition 19.6.18.51940EF88F4F Let $\Delta _{\bullet }$ be an Okounkov test curve relative to $\Delta $. We define the Duistermaat–Heckman measure $\operatorname{DH}(\Delta _{\bullet })$ as
\[ \operatorname{DH}(\Delta _{\bullet })\coloneqq G[\Delta _{\bullet }]_*(\operatorname{vol}). \]
It is a Radon measure on $\mathbb {R}$, since $G[\Delta _{\bullet }]$ is finite on $\operatorname{Int}\Delta $.
In other words, $\operatorname{DH}(\Delta _{\bullet })$ is the distribution of the random variable $G[\Delta _{\bullet }]$.
Proposition 19.6.19.F397AFA38927 Let $\Delta _{\bullet }\in \mathrm{TC}(\Delta )$. Let $m\in \mathbb {Z}_{>0}$. Then the following extended moment identity holds:
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19.6.20
\begin{equation} \int _{\mathbb {R}}x^ m \operatorname{DH}(\Delta _{\bullet })(x)=\Delta _{\max }^ m\operatorname{vol}\Delta +m\int _{-\infty }^{\Delta _{\max }} \tau ^{m-1}(\operatorname{vol}\Delta _{\tau }-\operatorname{vol}\Delta )\, \mathrm{d}\tau \end{equation}
and
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19.6.21
\begin{equation} \int _{\mathbb {R}}\operatorname{DH}(\Delta _{\bullet })=\operatorname{vol}\Delta . \end{equation}
Proof.
In fact, (19.6.21) follows immediately from the definition. Let $G=G[\Delta _{\bullet }]$. We first assume that $G$ is bounded from below. Then (19.6.20) follows from a straightforward computation:
\[ \begin{aligned} & \int _{\mathbb {R}}x^ m \operatorname{DH}(\Delta _{\bullet })(x)\\ =& \int _{\Delta } G[\Delta _{\bullet }](a)^ m\, \mathrm{d}\operatorname{vol}(a)\\ =& \int _{\Delta } \left( \Delta _{\max }^ m-\int _{G[\Delta _{\bullet }](a)}^{\Delta _{\max }} m\tau ^{m-1}\, \mathrm{d}\tau \right)\, \mathrm{d}\operatorname{vol}(a)\\ =& \Delta _{\max }^ m\operatorname{vol}\Delta -m\int _{\mathbb {R}}\int _{\Delta } \mathds {1}_{[G[\Delta _{\bullet }](a),\Delta _{\max }]}(\tau )\tau ^{m-1}\, \mathrm{d}\operatorname{vol}(a)\, \mathrm{d}\tau \\ =& \Delta _{\max }^ m\operatorname{vol}\Delta -m\int _{-\infty }^{\Delta _{\max }}\int _{\Delta \setminus \Delta _{\tau }} \tau ^{m-1}\, \mathrm{d}\operatorname{vol}(a)\, \mathrm{d}\tau \\ =& \Delta _{\max }^ m\operatorname{vol}\Delta -m\int _{-\infty }^{\Delta _{\max }} \tau ^{m-1}\left(\operatorname{vol}\Delta -\operatorname{vol}\Delta _{\tau }\right)\, \mathrm{d}\tau . \end{aligned} \]
In general, set $G_ A=G\lor (-A)$. The same pointwise identity applied to $G_ A$ gives
\[ \int _{\Delta }G_ A^ m\, \mathrm{d}\operatorname{vol}=\Delta _{\max }^ m\operatorname{vol}\Delta +m\int _{-A}^{\Delta _{\max }}\tau ^{m-1}(\operatorname{vol}\Delta _{\tau }-\operatorname{vol}\Delta )\, \mathrm{d}\tau . \]
Letting $A\to \infty $ gives (19.6.20): if $m$ is even, this follows from monotone convergence applied to $G_ A^ m\nearrow G^ m$; if $m$ is odd, apply monotone convergence to $\Delta _{\max }^ m-G_ A^ m\nearrow \Delta _{\max }^ m-G^ m$. Thus both sides of (19.6.20) are interpreted in the corresponding extended sense.
$\square$
Lemma 19.6.22.9C05354365F4 Let $(\Delta ^ i)_{i\geq 1}$ be a decreasing sequence in $\mathcal{K}_ n$ with limit $\Delta $. Suppose that $(\Delta _{\bullet }^ i)_{i\geq 1}$ is a decreasing sequence with $\Delta _{\bullet }^ i\in \mathrm{TC}(\Delta ^ i)$. Suppose that there is $\Delta _{\bullet }\in \mathrm{TC}(\Delta )$ such that
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$\Delta _{\max }=\lim _{i\to \infty }\Delta ^ i_{\max }$;
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for any $\tau <\Delta _{\max }$, we have $\Delta ^ i_{\tau }\xrightarrow {d_{\mathrm{Haus}}}\Delta _{\tau }$.
Then $\operatorname{DH}(\Delta _{\bullet }^ i)\rightharpoonup \operatorname{DH}(\Delta _{\bullet })$.
Proof.
Let $f\in C_ c(\mathbb {R})$. It follows from proposition 19.6.14 that
\[ G[\Delta _{\bullet }^ i]\to G[\Delta _{\bullet }] \]
pointwise on $\Delta $. Hence the dominated convergence theorem gives
\[ \int _{\Delta } f(G[\Delta _{\bullet }^ i])\, \mathrm{d}\lambda \to \int _{\Delta } f(G[\Delta _{\bullet }])\, \mathrm{d}\lambda . \]
Since $\Delta ^ i\xrightarrow {d_{\mathrm{Haus}}}\Delta $, theorem 19.7.5 gives $\operatorname{vol}(\Delta ^ i\setminus \Delta )\to 0$. The contribution of $\Delta ^ i\setminus \Delta $ is bounded by $\| f\| _{\infty }\operatorname{vol}(\Delta ^ i\setminus \Delta )$, so
\[ \int _{\Delta ^ i} f(G[\Delta _{\bullet }^ i])\, \mathrm{d}\lambda \to \int _{\Delta } f(G[\Delta _{\bullet }])\, \mathrm{d}\lambda . \]
This is the weak convergence.
$\square$
Similarly, we have
Lemma 19.6.23.22B2EF8281A7 Let $(\Delta ^ i)_{i\geq 1}$ be an increasing sequence in $\mathcal{K}_ n$ with Hausdorff limit $\Delta $ such that $\operatorname{vol}\Delta ^ i>0$ for all $i\geq 1$. Consider an increasing sequence $(\Delta ^ i_{\bullet })_{i\geq 1}$ with $\Delta ^ i_{\bullet }\in \mathrm{TC}(\Delta ^ i)$ for all $i\geq 1$. Let $\Delta _{\bullet }\in \mathrm{TC}(\Delta )$ be defined as
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$\Delta _{\max }=\lim _{i\to \infty }\Delta ^ i_{\max }$;
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for any $\tau <\Delta _{\max }$, $\Delta _{\tau }$ is the Hausdorff limit of $\Delta ^ i_{\tau }$.
Then we have
\[ \operatorname{DH}(\Delta _{\bullet }^ i)\rightharpoonup \operatorname{DH}(\Delta _{\bullet }). \]
Proof.
Let $f\in C_ c(\mathbb {R})$. It follows from proposition 19.6.16 that
\[ G[\Delta _{\bullet }^ i]\to G[\Delta _{\bullet }] \]
almost everywhere on $\Delta $. Extending $f(G[\Delta _{\bullet }^ i])$ by $0$ outside $\Delta ^ i$, we also have $\mathds {1}_{\Delta ^ i}\to \mathds {1}_{\Delta }$ almost everywhere on $\Delta $, since the boundary of the convex body $\Delta $ has Lebesgue measure zero. The dominated convergence theorem gives
\[ \int _{\Delta ^ i} f(G[\Delta _{\bullet }^ i])\, \mathrm{d}\lambda \to \int _{\Delta } f(G[\Delta _{\bullet }])\, \mathrm{d}\lambda . \]
This is the weak convergence.
$\square$
The main source of Okounkov test curves is the following:
Theorem 19.6.24.74133E3920A2 Let $X$ be a connected compact Kähler manifold and $\theta $ be a closed smooth real $(1,1)$-form on $X$ representing a big cohomology class $\alpha $. Let $Y_{\bullet }$ be a smooth flag on $X$ and $\Gamma \in \mathrm{TC}(X,\theta )_{>0}$. Then the map
\[ (-\infty ,\Gamma _{\max })\ni \tau \mapsto \Delta _{Y_{\bullet }}(\theta ,\Gamma )_{\tau }\coloneqq \Delta _{Y_{\bullet }}(\theta ,\Gamma _{\tau }) \]
defines an Okounkov test curve relative to $\Delta _{Y_{\bullet }}(\theta ,\Gamma _{-\infty })$.
If furthermore $\Gamma \in \mathrm{TC}^1(X,\theta ;\Gamma _{-\infty })$ (resp. $\mathrm{TC}^{\infty }(X,\theta ;\Gamma _{-\infty })$), then we have $\Delta _{Y_{\bullet }}(\theta ,\Gamma )\in \mathrm{TC}^1(\Delta _{Y_{\bullet }}(\theta ,\Gamma _{-\infty }))$ (resp. $\mathrm{TC}^{\infty }(\Delta _{Y_{\bullet }}(\theta ,\Gamma _{-\infty }))$).
See definition
17.12.1 and definition
17.12.10 for the relevant definitions.
Proof.
Consider $\Gamma \in \mathrm{TC}(X,\theta )_{>0}$. We need to verify that $\Delta _{Y_{\bullet }}(\theta ,\Gamma )$ is an Okounkov test curve relative to $\Delta _{Y_{\bullet }}(\theta ,\Gamma _{-\infty })$.
First observe that $\tau \mapsto \Delta _{Y_{\bullet }}(\theta ,\Gamma _{\tau })$ is concave and decreasing for $\tau <\Gamma _{\max }$. The decreasing property follows from theorem 19.5.26 2. For concavity, use the valuative characterization theorem 19.5.40 together with the additivity of valuations in proposition 19.3.9: if $x\in \Delta _{Y_\bullet }(\theta ,\Gamma _\tau )$ and $y\in \Delta _{Y_\bullet }(\theta ,\Gamma _{\tau '})$, then $tx+(1-t)y$ is obtained from the current $tS+(1-t)S'$, which is more singular than $\Gamma _{t\tau +(1-t)\tau '}$. Hence
\[ t\Delta _{Y_\bullet }\left(\theta ,\Gamma _\tau \right)+(1-t)\Delta _{Y_\bullet }\left(\theta ,\Gamma _{\tau '}\right)\subseteq \Delta _{Y_\bullet }\left(\theta ,\Gamma _{t\tau +(1-t)\tau '}\right). \]
Next we show that as $\tau \to -\infty $, we have
\[ \Delta _{Y_{\bullet }}(\theta ,\Gamma _{\tau })\xrightarrow {d_{\mathrm{Haus}}}\Delta _{Y_{\bullet }}(\theta ,\Gamma _{-\infty }). \]
By theorem 19.5.26 2, the bodies $\Delta _{Y_{\bullet }}(\theta ,\Gamma _{\tau })$ are contained in $\Delta _{Y_{\bullet }}(\theta ,\Gamma _{-\infty })$ and increase as $\tau \to -\infty $. Thus it suffices to compute their volumes:
\[ \begin{split} \lim _{\tau \to -\infty }\operatorname{vol}\Delta _{Y_{\bullet }}(\theta ,\Gamma _{\tau })=\frac{1}{n!}\lim _{\tau \to -\infty }\operatorname{vol}(\theta +\mathrm{dd}^{\mathrm{c}}\Gamma _{\tau })=\frac{1}{n!}\operatorname{vol}(\theta +\mathrm{dd}^{\mathrm{c}}\Gamma _{-\infty }) \\ =\operatorname{vol}\Delta _{Y_{\bullet }}(\theta ,\Gamma _{-\infty }), \end{split} \]
where we applied theorem 19.5.26 and theorem 16.9.57.
When $\Gamma \in \mathrm{TC}^{\infty }(X,\theta ;\Gamma _{-\infty })$, it is clear that $\Delta _{Y_{\bullet }}(\theta ,\Gamma )\in \mathrm{TC}^{\infty }(\Delta _{Y_{\bullet }}(\theta ,\Gamma _{-\infty }))$.
When $\Gamma \in \mathrm{TC}^{1}(X,\theta ;\Gamma _{-\infty })$, by theorem 19.5.26 1, (17.12.12) and (19.6.2), we have
\[ \mathbf{E}^{\Gamma _{-\infty }}(\Gamma )=\mathbf{E}(\Delta _{Y_{\bullet }}(\theta ,\Gamma )). \]
Indeed, by theorem 19.5.26 1,
\[ n!\operatorname{vol}\Delta _{Y_{\bullet }}(\theta ,\Gamma _\tau )=\int _ X\theta _{\Gamma _\tau }^ n \]
for every $\tau <\Gamma _{\max }$, and the same identity holds at $\tau =-\infty $. Comparing (17.12.12) with (19.6.2) gives the asserted equality of energies.
So $\Delta _{Y_{\bullet }}(\theta ,\Gamma )\in \mathrm{TC}^1(\Delta _{Y_{\bullet }}(\theta ,\Gamma _{-\infty }))$.
$\square$
Pome
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