Proposition 19.3.14.D094DD00A592 Let $\pi \colon Z\rightarrow X$ be a proper bimeromorphic morphism, where $Z$ is a Kähler manifold. Let $W_{\bullet }$ be a lifting of $Y_{\bullet }$. Then for any closed positive $(1,1)$-current $T$ on $X$, we have
10.14
\begin{equation} \nu _{W_{\bullet }}(\pi ^*T)= \nu _{Y_{\bullet }}(T)\operatorname{cor}(Y_{\bullet },\pi ). \end{equation}
Proof.
We argue by induction on $n\geq 0$. The case $n=0$ is trivial. In general, assume that $n\geq 1$ and the result is proved in dimension $n-1$.
For simplicity, we write $\nu =\nu _{Y_{\bullet }}$ and $\nu '=\nu _{W_{\bullet }}$. Let $\mu $ (resp. $\mu '$) be the valuation of currents defined by the truncated flag $Y_1\supseteq \cdots \supseteq Y_ n$ (resp. $W_1\supseteq \cdots \supseteq W_ n$). Then we need to show that
579570889FCE
19.3.15
\begin{equation} \begin{split} & \begin{bmatrix} \nu '(\pi ^*T)_1
& \mu '\Bigl ( \operatorname{Tr}_{W_1}\bigl (\pi ^*T-\nu '(\pi ^*T)_1[W_1]\bigr )\Bigr )
\end{bmatrix}\\ =& \begin{bmatrix} \nu (T)_1
& \mu \Bigl (\operatorname{Tr}_{Y_1}\bigl (T-\nu (T)_1[Y_1]\bigr )\Bigr )
\end{bmatrix}\operatorname{cor}(Y_{\bullet },\pi ). \end{split} \end{equation}
By Zariski's main theorem theorem 9.10.5, $W_1\to Y_1$ is bimeromorphic and hence
\[ \nu '(\pi ^*T)_1=\nu (T,Y_1)\eqqcolon c. \]
By the inductive hypothesis, we have
92459CED0BC1
19.3.16
\begin{equation} \mu '\Bigl (\Pi ^{*} \operatorname{Tr}_{Y_1}(T-c[Y_1])\Bigr )=\mu \Bigl (\operatorname{Tr}_{Y_1}(T-c[Y_1])\Bigr )\operatorname{cor}\left(Y_1\supseteq \cdots \supseteq Y_ n,\Pi \right), \end{equation}
where $\Pi \colon W_1\rightarrow Y_1$ is the restriction of $\pi $. Since $\pi ^*[Y_1]=[W_1]+(\pi ^*[Y_1]-[W_1])$, 20.2.24 and proposition 20.2.21 give
\[ \begin{split} \operatorname{Tr}_{W_1}\bigl (\pi ^*T-c[W_1]\bigr )\sim _ P& \operatorname{Tr}_{W_1}\bigl (\pi ^*(T-c[Y_1])\bigr )\\ & +c \operatorname{Tr}_{W_1}\bigl (\pi ^*[Y_1]-[W_1]\bigr )\\ \sim _ P& \Pi ^{*}\operatorname{Tr}_{Y_1}\bigl (T-c[Y_1]\bigr )+c \operatorname{Tr}_{W_1}\bigl (\pi ^*[Y_1]-[W_1]\bigr ). \end{split} \]
So
\[ \mu '\Bigl (\operatorname{Tr}_{W_1}\bigl (\pi ^*T-c[W_1]\bigr )\Bigr )=\mu '\Bigl (\Pi ^{*}\operatorname{Tr}_{Y_1}\bigl (T-c[Y_1]\bigr )\Bigr )+c\mu '\Bigl (\operatorname{Tr}_{W_1}\bigl (\pi ^*[Y_1]-[W_1]\bigr )\Bigr ). \]
Combining the above with (19.3.16), we see that (19.3.15) follows.
$\square$
Pome
Comments
Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.