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Let $G$ be an ample $\mathbb{Q}$-divisor on a smooth variety $X$. Let $D$ be a $\mathbb{Q}$-divisor linearly equivalent to $G$. Let $f: Y\to X$ be a common log resolution of $G$ and $D$. We define the multiplier ideal of a divisor $G$ as $I(G)=f_*O_Y(K_{Y/X}-[f^*G])$. Are the multiplier ideals $I(G)$ and $I((1-t)G+tD)$ the same for sufficiently small $t>0$?

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Oops, thank you quim. –  Fei YE May 30 '11 at 21:58
    
You are welcome! –  quim May 31 '11 at 5:32
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3 Answers

up vote 3 down vote accepted

I don't think so. Take for example $X=\mathbb{P}^2$, and $G$ and $D$ to be distinct lines. then $I(G)=\mathcal{O}_X(-G)$ while $I((1-t)g+tD)=\mathcal{O}_X$ for every small $t>0$. Maybe you might want to look at the multiplier ideal associated to the linear series $|G|$.

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That's right. I was thinking too general. Thanks! –  Fei YE May 29 '11 at 23:53
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Think about it this way: The co-support of the multiplier ideal of $G$ is the locus where the pair $(X,G)$ has worse than klt singularities. If $(X,G)$ has non-klt singularities, say because some of its coefficients are at least $1$, and $(X,D)$ is dlt, then some of the small perturbations will likely be still klt, so you get a jump in the multiplier ideal. A simple concrete example for this is given by Gianni Bello in his answer.

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By co-support do you mean the complement of the scheme defined by $I(G)$. Can you give a little bit more detail? Thanks a lot. –  Fei YE May 30 '11 at 0:07
    
Yes, I meant the support of $\mathscr{O/I}$. –  Sándor Kovács May 30 '11 at 7:21
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However, I should point out that one always has the multiplier ideal containment $$I((1−t)G+tD) \supseteq I(G)$$ for $t > 0$ sufficiently small.

In particular, the singularities of $(X, (1-t)G + tD)$ can only be epsilon more severe than the singularities of the pair $(X, G)$, thus the multiplier ideal won't get smaller, at least for sufficiently small $t$. Furthermore, they might even be better (in other words, less severe) as Gianni's example shows.

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