Pome65B836952506The restriction of a multiplier ideal to $Y$ records the vanishing of its local generators along $Y$. Its vanishing order along a divisor on a modification of $Y$ can grow with the weight, even when the ideal is generically trivial on $Y$.
For $t>0$ put
Here $\mathcal I(tu)\mathcal O_ Y$ denotes the image of $\mathcal I(tu)$ under the restriction map $\mathcal O_ X\to \mathcal O_ Y$. This is a coherent ideal sheaf on $Y$, and it is nonzero: since $\nu (u,Y)=0$, there is $y\in Y$ with $\nu (u,y)<1/t$, and Skoda's integrability theorem gives $\mathcal I(tu)_ y=\mathcal O_{X,y}$.
In this chapter, a divisor over $Y$ is a prime divisor $E$ on a connected smooth compact modification $\pi :Z\to Y$. For a nonzero coherent ideal $\mathfrak a$ on $Y$, let $\operatorname {ord}_ E(\mathfrak a)$ be the generic vanishing order along $E$ of the ideal generated by its pullback to $Z$; it is a nonnegative integer.
Theorem 20.3.2.7CD874C73EC9 For every divisor $E$ over $Y$, the limit
66AFB7F7397B
20.3.3
exists and is finite, and it equals the limit along positive integers. These numbers depend only on the multiplier-ideal equivalence class of $u$ on $X$.
Proof. Fix the modification $\pi :Z\to Y$ carrying $E$, and write $b(t)=\operatorname {ord}_ E\mathfrak a_ t(u;Y)$ for $t>0$. For $s,t>0$, the subadditivity of multiplier ideals (the second inclusion in (5.37.6)), followed by restriction to $Y$, gives
Moreover, $b(t)\geq 0$. For fixed $s>0$ and $t=ks+r$ with $k$ a nonnegative integer and $0\leq r<s$, repeated superadditivity gives $b(t)\geq k b(s)$. Hence
Taking the supremum over $s$ proves the existence of the limit and the supremum formula, with the value $+\infty $ allowed at this stage. In particular, the limit along positive integers is the same.
For every sufficiently large integer $m$, the restriction $u_ m|_ Y$ has analytic type $(1/m,\mathfrak a_ m(u;Y))$ by section 5.34, and therefore
These numbers are uniformly bounded by corollary 5.43.26, so the limit along integers, and hence the supremum, is finite. Finally, the ideals $\mathfrak a_ t(u;Y)$ are unchanged when $u$ is replaced by a potential with the same multiplier ideals, which proves the last assertion. $\square$
We call the function $E\mapsto \tau _{Y,u}(E)$, defined on all divisors over $Y$, the divisorial trace data of $u$ along $Y$.
Inclusions of ideals give monotonicity and homogeneity of the trace data directly. For sums, subadditivity of multiplier ideals gives one inequality; for the reverse inequality, an analytic approximant of the sum controls sufficiently fine approximants of both summands.
Proposition 20.3.4.89D22AB4CBD8 Let $v\in \mathrm{QPSH}(X)$ also have zero generic Lelong number along $Y$, and recall that $u\preceq _{\mathcal I}v$ means $\mathcal I(ku)\subseteq \mathcal I(kv)$ for every real $k>0$ (section 5.33). Then, as functions on divisors over $Y$,
$u\preceq _{\mathcal I}v$ implies $\tau _{Y,u}\geq \tau _{Y,v}$;
$\tau _{Y,au}=a\tau _{Y,u}$ for every $a>0$;
$\tau _{Y,u+v}=\tau _{Y,u}+\tau _{Y,v}$;
$\tau _{Y,\max (u,v)}\leq \min \{ \tau _{Y,u},\tau _{Y,v}\} $.
Proof. The inclusions of ideals defining $\preceq _{\mathcal I}$ persist after restriction and reverse the vanishing orders, which proves (1). Replacing $t$ by $at$ in (20.3.3) proves (2). Moreover,
Since the order of a sum of ideals is the minimum of their orders, restriction and passage to the limit prove (4); note that $\max (u,v)$ also has zero generic Lelong number along $Y$. Generic Lelong numbers of sums add, so $u+v$ has zero generic Lelong number along $Y$ as well.
For (3), subadditivity of multiplier ideals gives $\tau _{Y,u+v}\geq \tau _{Y,u}+\tau _{Y,v}$. For the reverse inequality, fix a sufficiently large integer $k$ and the analytic approximant $S_ k$ of $u+v$ given by theorem 5.36.1, of analytic type $(1/k,\mathcal I(k(u+v)))$; that theorem gives $u+v\leq S_ k$. Let $U_ m,V_ m$ be the corresponding approximants of $u$ and $v$. Lemma 5.38.12 gives
Here both $\delta _ m$ and the bounded error may depend on the fixed $k$. The restrictions of $U_ m$, $V_ m$ and $S_ k$ to $Y$ are not identically $-\infty $ by corollary 5.43.26. Restricting, pulling back to a fixed modification and taking the generic Lelong number along a divisor $E$ on it, we obtain
Letting first $m\to \infty $, with $k$ fixed, and then $k\to \infty $, the finite limits of theorem 20.3.2 give the reverse inequality. $\square$
Comments
Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.