Lemma 19.5.34.627CFC8D8850 Let $T\in \mathcal{Z}_+(X,\alpha )$ be a Kähler current. Assume that $\nu (T,Y_1)=0$. Then
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19.5.35
\begin{equation} \left\{ y\in \mathbb {R}^{n-1}: (0,y)\in \Delta _{Y_{\bullet }}(T)\right\} =\Delta _{Y_1\supseteq \cdots \supseteq Y_ n}\left(\operatorname{Tr}^{\alpha |_{Y_1}}_{Y_1}(T)\right). \end{equation}
More generally, if $T\in \mathcal{Z}_+(X,\alpha )$ and $\nu (T,Y_1)=0$, suppose in addition that $\operatorname{Tr}^{\alpha |_{Y_1}}_{Y_1}(T)$ is defined, then (19.5.35) still holds.
Proof.
Let $\omega $ be a Kähler form on $X$. The last assertion follows from the first by perturbing $\theta $ to $\theta +\epsilon \omega $.
Step 1. We first handle the case where $T$ has analytic singularities. Let $\pi \colon Z\rightarrow X$ be a modification such that
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$Y_{\bullet }$ admits a lifting $(W_{\bullet },g)$, and
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$\pi ^*T=[D]+R$, where $D$ is an effective $\mathbb {Q}$-divisor on $Z$ and $R$ is a closed positive $(1,1)$-current with bounded potential.
This is possible by theorem 5.18.8 and theorem 19.3.19.
By 20.2.24,
\[ \Pi ^*\operatorname{Tr}_{Y_1}(T)\sim _ P \operatorname{Tr}_{W_1}(\pi ^*T), \]
where $\Pi \colon W_1\rightarrow Y_1$ is the restriction of $\pi $. It follows from theorem 19.5.26 that
\[ \begin{split} \Delta _{W_1\supseteq \cdots \supseteq W_ n}(\operatorname{Tr}_{W_1}(\pi ^*T))=& \Delta _{Y_1\supseteq \cdots \supseteq Y_ n}(\operatorname{Tr}_{Y_1}(T))\operatorname{cor}(Y_1\supseteq \cdots \supseteq Y_ n,\Pi ),\\ \Delta _{W_{\bullet }}(\pi ^*T)=& \Delta _{Y_{\bullet }}(T)g. \end{split} \]
Taking (19.3.13) into account, we find that it suffices to show that
\[ \left\{ y\in \mathbb {R}^{n-1}:(0,y)\in \Delta _{W_{\bullet }}(\pi ^*T) \right\} =\Delta _{W_1\supseteq \cdots \supseteq W_ n}(\operatorname{Tr}_{W_1}(\pi ^*T)). \]
We may assume that $\pi $ is the identity map. Since $\nu (T,Y_1)=0$, the divisor $D$ does not contain $Y_1$. Hence we have
\[ T=[D]+R,\quad T|_{Y_1}=[D]|_{Y_1}+R|_{Y_1}. \]
Note that $[D]|_{Y_1}$ is the current of integration along an effective $\mathbb {Q}$-divisor on $Y_1$.
In particular,
\[ \begin{split} \Delta _{Y_{\bullet }}(T)=& \Delta _{Y_{\bullet }}(\{ R\} )+\nu _{Y_{\bullet }}([D]),\\ \Delta _{Y_1\supseteq \cdots \supseteq Y_ n}(T|_{Y_1})=& \Delta _{Y_1\supseteq \cdots \supseteq Y_ n}(\{ R\} |_{Y_1})+\nu _{Y_1\supseteq \cdots \supseteq Y_ n}([D]|_{Y_1}). \end{split} \]
The first coordinate of $\nu _{Y_{\bullet }}([D])$ is $0$, and the remaining coordinates are exactly $\nu _{Y_1\supseteq \cdots \supseteq Y_ n}([D]|_{Y_1})$. Thus it suffices to show that
\[ \left\{ y\in \mathbb {R}^{n-1}: (0,y)\in \Delta _{Y_{\bullet }}(\{ R\} )\right\} =\Delta _{Y_1\supseteq \cdots \supseteq Y_ n}(\{ R\} |_{Y_1}). \]
Since $R$ has bounded potential, its cohomology class is nef; it is big because $T$ is a Kähler current. Thus the last identity is exactly 19.5.32.
Step 2. Next we consider the case where $T$ is a Kähler current. Take a quasi-equisingular approximation $(T_ j)_ j$ of $T$ in $\mathcal{Z}_+(X,\alpha )$. From Step 1, we know that for large $j\geq 1$,
\[ \left\{ y\in \mathbb {R}^{n-1}: (0,y)\in \Delta _{Y_{\bullet }}(T_ j)\right\} =\Delta _{Y_1\supseteq \cdots \supseteq Y_ n}(\operatorname{Tr}_{Y_1}(T_ j)). \]
Letting $j\to \infty $ and applying theorem 19.5.26 and proposition 20.2.22, we conclude (19.5.35).
$\square$
Pome
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