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.
$\square$
Pome
Comments
Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.