19.3 Flags and valuationsE9A341A325BE

Proof. We will prove both statements at the same time by induction on $n\geq 0$. The cases $n=0,1$ are trivial.

Let us consider the case $n>1$ and assume that the result is known in dimension $n-1$. We first observe that $\nu _{Y_{\bullet }}(T)$ is independent of the choice of the trace operator: Different choices of $\operatorname{Tr}_{Y_1}(T-\nu (T,Y_1)[Y_1])$ are $\mathcal{I}$-equivalent by proposition 20.2.8. Therefore, by induction, its valuation is well-defined.

Next, let $T'$ be another closed positive $(1,1)$-current such that $T\sim _{\mathcal{I}}T'$. Using proposition 16.6.2, we know that $\nu (T,Y_1)=\nu (T',Y_1)$. Therefore,

It follows by induction that

This completes the proof. $\square$

Proof. 1 We make an induction on $n\geq 0$. The case $n=0,1$ is trivial. Assume that $n\geq 2$ and the case $n-1$ is known. Observe that $\nu (T,Y_1)\geq \nu (S,Y_1)$, if the inequality is strict, we are done. So let us assume that $\nu (T,Y_1)=\nu (S,Y_1)$. By proposition 20.2.21, we find that

Hence the same comparison holds for $\preceq _{\mathcal{I}}$ by corollary 16.8.14. By the inductive hypothesis, we conclude (19.3.10).

2 We make an induction on $n\geq 0$. The cases $n=0,1$ are trivial. Assume that $n\geq 2$ and the case $n-1$ is known. By proposition 5.16.8, we have

By proposition 20.2.21, we have

Since $\sim _ P$ implies $\sim _{\mathcal{I}}$ by corollary 16.8.14, the inductive hypothesis and the independence of the valuation from the trace representative conclude (19.3.11). $\square$

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

By Zariski's main theorem theorem 9.10.5, $W_1\to Y_1$ is bimeromorphic and hence

By the inductive hypothesis, we have

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

So

Combining the above with (19.3.16), we see that (19.3.15) follows. $\square$

Proof. We let $\pi '=\pi \circ p$. Then $W'_{\bullet }$ is also the lifting of $W_{\bullet }$ with respect to $p$.

We argue by induction on $n\geq 1$. The case $n=1$ is trivial. Assume that $n\geq 2$ and the case $n-1$ has been solved. Then by (19.3.13), the desired formula (19.3.18) can be reformulated as

By the inductive hypothesis, this is equivalent to

which, thanks to proposition 19.3.14, can be further rewritten as

Indeed, as divisors on $W_1'$,

The desired equality is therefore exactly the additivity of the valuation, proposition 19.3.9. $\square$

Proof. By Hironaka's Chow lemma theorem 9.10.8, we may assume that $\pi $ is a modification.

We begin by setting $W_0=Z$. We will construct $W_ i$ inductively for each $i$. Assume that for $0\leq i<n$ a smooth partial flag $W_0\supseteq \cdots \supseteq W_ i$ has been constructed on a modification $\pi _ i:Z_ i\rightarrow Z$ so that $\pi \circ \pi _ i$ restricts to bimeromorphic morphisms $W_ j\rightarrow Y_ j$ for each $j=0,\ldots ,i$.

By Zariski's main theorem, $W_ i\rightarrow Y_ i$ is an isomorphism outside a codimension $2$ subset of $Y_ i$. We let $W_{i+1}$ be the strict transform of $Y_{i+1}$ in $W_ i$. The problem is that $W_{i+1}$ is not necessarily smooth.

We will further modify $Z_ i$ and lift $W_1,\ldots , W_{i+1}$ in order to make the flag smooth. Take the embedded resolution of $(W_{j},W_{i+1})$, say $W_ j'\rightarrow W_ j$ for each $j=0,\ldots ,i$.

We have canonical embeddings $W_{i}'\hookrightarrow W_{i-1}'\hookrightarrow \cdots \hookrightarrow W_0'$ making the following diagram commutative:

Let $W_{i+1}'$ be the strict transform of $W_{i+1}$ in $W_ i'$. It suffices to define $\pi _{i+1}$ as the morphism $W_0'\rightarrow Z_ i\rightarrow Z$ and replace $W_0\supseteq \cdots \supseteq W_{i+1}$ by $W_0'\supseteq \cdots \supseteq W_{i+1}'$. $\square$

Proof. We first prove independence of the choice of $h$. Let $h_ i\colon \widehat Z_ i\rightarrow Z$, $i=1,2$, be two choices, and let $(W^ i_{\bullet },g_ i)$ be the corresponding liftings, with $g_ i=\operatorname{cor}(Y_{\bullet },\pi \circ h_ i)$. Choose a common modification over $Z$, $q_ i\colon U\rightarrow \widehat Z_ i$ such that $Y_{\bullet }$ admits a lifting $W_{\bullet }$ to $U$. Then $W_{\bullet }$ is also the lifting of $W^ i_{\bullet }$ with respect to $q_ i$. Put

By proposition 19.3.14, proposition 19.3.17,

Since $h_1\circ q_1=h_2\circ q_2$, the left-hand side is independent of $i$. This proves that (19.3.21) is well-defined.

For the birational invariance, choose a proper bimeromorphic morphism $q\colon \widehat Z\rightarrow Z'$ from a compact Kähler manifold such that $Y_{\bullet }$ admits a lifting to $\widehat Z$ with respect to $\pi \circ p\circ q$. Computing $\nu _{Y_{\bullet }}(p^*S)$ using $q$, and computing $\nu _{Y_{\bullet }}(S)$ using $p\circ q$, gives the same expression. $\square$