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Nov 8, 2016 at 19:01 comment added Hacon That's right. We use the LC condition for adjunction purposes (in the cutting down the LC centers process), but once the co-support is o an isolated point, we no longer need this.
Nov 8, 2016 at 14:34 comment added Li Yutong @Hacon By the way, it seems that as long as the singularity at $x$ is not klt (i.e. it is not necessarily to be lc), one will have such lifting property by the same argument (with $\mathcal{O}_X/\mathcal J = \mathcal{O}_X/{m^l_x} \oplus \mathcal{O}_Z$).
Nov 8, 2016 at 12:12 comment added Chen Jiang $x$ is general, so may assume that $x$ is not contained in $B$ nor $D$, so adding them does not affect singularities around $x$
Nov 8, 2016 at 7:04 comment added Li Yutong Dear Prof. Hacon, this is exactly what I am asking and I see your point! Thank you very much! But I still have a technical detail: I understand that $(X, \Delta)$ is klt in a punctured neighborhood of $x$ with $\{x\}$ as non-klt centre. But the multiplier ideal is $\mathcal{J}(\Delta+ \epsilon B + \lceil D\rceil -D)$. How could I know the non-klt centre for $(X, \Delta+ \epsilon B + \lceil D\rceil -D)$ is still $\{x\}$. If the non-klt centre for the above is strictly larger than $\{x\}$, then one may not have direct sum $\mathcal{O}_x \oplus \mathcal{O}_Z$. Where did I got wrong?
Nov 8, 2016 at 3:48 comment added Hacon The point is that $O_X/\mathcal J =\mathcal O _x\oplus \mathcal O _Z$ where $y\in Z$. By the above vanishing, considering the long exact sequence in cohomology, we can lift the section $1\oplus 0$ from $H^0(\mathcal O _x)\oplus H^0(\mathcal O _Z)$ to a section $\sigma \in H^0(K_X+\lceil D\rceil)$ such that $s(x)=1$ and $s(y))=0$. Does this answer your question or did you mean something else?
Nov 7, 2016 at 9:32 history asked Li Yutong CC BY-SA 3.0