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Theorem 20.3.27CD874C73EC9

Theorem 20.3.2.7CD874C73EC9 For every divisor $E$ over $Y$, the limit

66AFB7F7397B 20.3.3
\begin{equation} \tau _{Y,u}(E) :=\lim _{t\to \infty }\frac{\operatorname {ord}_ E\mathfrak a_ t(u;Y)}{t} =\sup _{t>0}\frac{\operatorname {ord}_ E\mathfrak a_ t(u;Y)}{t} \end{equation}

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

\[ \mathfrak a_{s+t}\subseteq \mathfrak a_ s\mathfrak a_ t, \qquad b(s+t)\geq b(s)+b(t). \]

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

\[ \liminf _{t\to \infty }\frac{b(t)}t\geq \frac{b(s)}s. \]

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

\[ \frac{b(m)}m =\nu \bigl ((u_ m|_ Y)\circ \pi ,E\bigr ). \]

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. $\square$

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