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Put $V= \mathbb{C}^3$. Let $D \subset V$ be an isolated singularity and
$\mu: \tilde{V} \rightarrow V$ be a log resolution of the pair $(V,D)$ whose exceptional locus $E$ and the strict transform $\tilde{D}$ satisfies that $\tilde{D} \cup E$ has a normal crossing support. We can define $c_j \in \mathbb{Z}$ such that $K_{\tilde{V}} + \tilde{D} = \mu^* (K_V +D)+ \sum c_j E_j $ where $E = \bigcup E_j$ is the irreducible decomposition.

Question Is there $\mu$ such that $c_j \le 0$ for all $j$?

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1 Answer 1

The answer is surely no.

If $D$ itself does not have a minimal log smooth resolution, then certainly $(V,D)$ couldn't have such a log resolution you need. On the other hand, there are bunch of isolated surface singularities whose minimal resolution is not log smooth, e.g., the log canonical surface singularity whose minimal log resolution has its exceptional locus an nodal rational curve.

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Thank you for the comment. I think that every normal surface singularity $D$ has a log resolution such that each discrepancy is non-positive. It is called an essential resolution, e.g. in Ishii's paper. Actually, if I blow up the node of the nodal curve in your example, the coefficient of the $-1$-curve is 0. –  tarosano Jan 14 '12 at 18:04
    
I see. You are right. We also need to consider the curves of discrepancy 0 which doesn't necessarily appear in the minimal resolution of the surface itself. –  CYXU Jan 16 '12 at 0:58
    
Just for safety, maybe the discrepancy of the $-1$-curve is $-1$. –  tarosano Mar 4 '13 at 11:24
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