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19.5.14 The case of Kähler currents8B697E317627

Definition 19.5.15.24F2B706F011 Let $T\in \mathcal{Z}_+(X,\alpha )$ be a Kähler current. Take a quasi-equisingular approximation $(T_ j)_ j$ of $T$ in $\mathcal{Z}_+(X,\alpha )$. Then we define

\[ \Delta _{Y_{\bullet }}(T)\coloneqq \bigcap _{j=1}^{\infty }\Delta _{Y_{\bullet }}(T_ j). \]

Lemma 19.5.16.08C830D7FB38 The convex body $\Delta _{Y_{\bullet }}(T)$ in definition 19.5.15 is independent of the choices of the $T_ j$'s.

In particular, if $T$ also has analytic singularities, then the $\Delta _{Y_{\bullet }}(T)$'s defined in definition 19.5.15 and in definition 19.5.7 coincide.

Proof. It is more convenient to use the language of $\theta $-psh functions at this point. Let $\psi _ k$ (resp. $\varphi _ k$) denote the potentials in $\mathrm{PSH}(X,\theta )$ corresponding to $S_ k$ (resp. $T_ k$) for each $k\geq 1$. Note that $\psi _ k$ and $\varphi _ k$ are unique up to additive constants. By proposition 5.18.12, for any small rational $\epsilon >0$, $j>0$, we can find $k>0$ so that

\[ \psi _ k\preceq (1-\epsilon )\varphi _ j. \]

By 19.5.12,

\[ \bigcap _{k=1}^{\infty }\Delta _{Y_{\bullet }}(\theta ,\psi _ k)\subseteq \Delta _{Y_{\bullet }}(\theta ,(1-\epsilon )\varphi _ j). \]

On the other hand, observe that

\[ \bigcap _{\epsilon \in \mathbb {Q}_{>0}\text{ small enough}}\Delta _{Y_{\bullet }}(\theta ,(1-\epsilon )\varphi _ j)=\Delta _{Y_{\bullet }}(\theta ,\varphi _ j). \]

In fact, the $\supseteq $ direction follows from 19.5.12, so it suffices to show that the two sides have the same volume, which follows from (19.5.13).

It follows that

\[ \bigcap _{k=1}^{\infty }\Delta _{Y_{\bullet }}(\theta ,\psi _ k)\subseteq \bigcap _{j=1}^{\infty }\Delta _{Y_{\bullet }}(\theta ,\varphi _ j). \]

The other inclusion follows by symmetry. $\square$

The same argument shows that

Corollary 19.5.17.A07A6093F1F3 Suppose that $T,S\in \mathcal{Z}_+(X,\alpha )$ are two Kähler currents satisfying $T \preceq _{\mathcal{I}} S$. Then

\[ \Delta _{Y_{\bullet }}(T)\subseteq \Delta _{Y_{\bullet }}(S)\subseteq \Delta _{Y_{\bullet }}(\alpha ). \]

Proposition 19.5.18.F190AF72550B Let $T\in \mathcal{Z}_+(X,\alpha )$ be a Kähler current. Then

4DF6CA86E7CD 19.5.19
\begin{equation} \operatorname{vol}\Delta _{Y_{\bullet }}(T)=\frac{1}{n!}\operatorname{vol}T. \end{equation}

Proof. Take a quasi-equisingular approximation $(T_ j)_ j$ of $T$ in $\mathcal{Z}_+(X,\alpha )$. Note that $\Delta _{Y_{\bullet }}(T_ j)$ is decreasing in $j$, as follows from 19.5.12. Our assertion follows from (19.5.13) and theorem 16.9.57. $\square$

Lemma 19.5.20.1493F0A1407E Let $T\in \mathcal{Z}_+(X,\alpha )$ be a Kähler current and $\omega $ be a Kähler form on $X$. Then

7D755FB49272 19.5.21
\begin{equation} \Delta _{Y_{\bullet }}(T)\subseteq \Delta _{Y_{\bullet }}(T+\omega ). \end{equation}

Moreover,

2F017272546D 19.5.22
\begin{equation} \Delta _{Y_{\bullet }}(T)=\bigcap _{\epsilon >0}\Delta _{Y_{\bullet }}(T+\epsilon \omega ). \end{equation}

Proof. We first prove (19.5.21). Taking quasi-equisingular approximations, we reduce immediately to the case where $T$ has analytic singularities. By 19.5.11, we may assume that $T$ has log singularities. Take $D$ and $R$ as in (19.5.8). By definition again, it suffices to show that

\[ \Delta _{Y_{\bullet }}(\{ R\} )\subseteq \Delta _{Y_{\bullet }}(\{ R\} +\{ \omega \} ), \]

which is clear by definition.

Next we prove (19.5.22). Thanks to (19.5.21), it remains to prove that both sides have the same volume:

\[ \lim _{\epsilon \to 0+}\operatorname{vol}(T+\epsilon \omega )=\operatorname{vol}T. \]

This is proved in proposition 18.1.27. $\square$

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