Let $D$ be a $\mathbb{Q}$-divisor in a smooth variety $X$. In Lazarsfeld book "Positivity in Algebraic Geometry 2" I found Proposition 9.5.13 saying that if for any $x\in D$ we have $mult_xD < 1$ then the pair $(X,D)$ is klt.

I am wondering if it possible to check that a pair is klt by compunting the discrepancies of a partial resolution and apply the fact above. I will be more precise.

Assume we have a birational divisor contraction $f:Y\rightarrow X$ such that $f^{-1}D\cup Exc(f)$ is simple normal crossing and the discrepancies of $(X,D)$ with respect any exceptional divisor are gretaer than $-1$. Let $Z = f(Exc(f))\subset X$. If for any $x\in D\setminus (D\cap Z)$ we have $mult_xD< 1$ is it true that $(X,D)$ is klt?


1 Answer 1


If I interpreted correctly your question I guess the answer is positive. Here is the statement:

Let $f:Y\rightarrow X$ be a proper birational morphism between normal $\mathbb{Q}$-factorial varieties, and let $D$ be an effective $\mathbb{Q}$-divisor in $X$. Let us write $$K_Y = f^{*}(K_X+D)+\sum_{i}a_iE_i-\widetilde{D}$$ where $E_i$ are $f$-exceptional divisors and $\widetilde{D}$ is the strict transform of $D$. If $a_i \leq 0$ for any $i$, and $mult_y(\widetilde{D}-\sum_{i}a_iE_i) < 1$ for any $y\in Y$ , then the pair $(X,D)$ is klt.

To prove this you can proceed as follows. We have $\widetilde{D}-\sum_{i}a_iE_i = \widetilde{D}+\sum_{i}b_iE_i$ with $b_i = -a_i \geq 0$. Then, $\widetilde{D}+\sum_{i}b_iE_i$ is effective. Since $mult_y(\widetilde{D}+\sum_{i}b_iE_i) < 1$ for any $y\in Y$, by Proposition 9.5.13 of "Positivity in Algebraic Geometry 2" we have that the multiplier ideal $\mathcal{I}(\widetilde{D}+\sum_{i}b_iE_i)$ is trivial. Then, by Section 9.3.B of "Positivity in Algebraic Geometry 2" we get that $(Y,\widetilde{D}+\sum_{i}b_iE_i)$ is klt. Now, we may write $$K_Y+\widetilde{D}+\sum_{i}b_iE_i = K_Y+\widetilde{D}-\sum_{i}a_iE_i = f^{*}(K_X+D).$$ Finally, since $(Y,\widetilde{D}+\sum_{i}b_iE_i)$ is klt, by Lemma 3.10 in http://arxiv.org/abs/alg-geom/9601026 the pair $(X,D)$ is klt as well.


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