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In Kollar and Mori "Birational geometry of algebraic varieties" discrepancy is defined as following way.

Let X be a normal variety and $D = \sum_i a_i D_i$ be a $\mathbb{Q}$ divisor. Assume that $m(\mathrm{K}_X + D ) $ is a Cartier divisor for some integer $m >0$. Let $f \colon Y \rightarrow X$ be a birational morphism from the normal variety $Y$ and $E$ be the exceptional locus of $f$. Then there exist rational numbers $a(E_i,X,D)$ such that $m a(E_i ,X,D)$ is an integer and $ m(K_Y + f^{-1}_* D)$ is linearly equivalent to $f^{*}(K_X + D) + \sum_i ma(E_i,X,D)E_i $ where $E_i \subset E$ is an exceptional divisor for all $i$.

My question: Are the numbers $ma(E_i,X,D)$ independent of the choice of representative of the linear equivalence class of $K_Y$?

If there is no principal divisor $ \operatorname{div}h = \sum_k a_k D_k$, $h \in K(Y) $ satisfying $D_k \subset E = \operatorname{Ex}(f)$, the answer of above question is yes.

When $f \colon Y \rightarrow X$ is proper, I can show there is no principal divisor $ \operatorname{div}h = \sum_k a_k D_k$, $h \in K(Y) $ satisfying $D_k \subset E = \operatorname{Ex}(f)$ as follows:

By valuative criterion, every discrete valuation which has center at X has center at Y. Hence $\operatorname{div}h = 0$ as divisor on $X$.Then $h \in \mathcal{O}^{*}_X$. We have $h \in \mathcal{O}^{*}_Y$ by morphism $\mathcal{O}_X \rightarrow f_{*}\mathcal{O}_Y $.

Above proof does not work when $f$ is not proper. So I have trouble when $f$ is not proper.

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    $\begingroup$ Assume that $f_*(K_Y)=K_X$, then $K_Y+f^{-1}_* D-f^*(K_X+D)\equiv \sum a_iE_i$ has a unique solution by the negativity lemma (if there is another solution $ \sum a'_iE_i$, then $\sum (a_i'-a_i)E_i$ is numerically equivalent to 0 and exceptional so in fact equal to 0). $\endgroup$
    – Hacon
    Commented Oct 3, 2019 at 21:24
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    $\begingroup$ @Hacon Negativity lemma holds when $f$ is not proper? If it holds, I want reference. Thank you. $\endgroup$
    – tukudani
    Commented Oct 4, 2019 at 1:28
  • $\begingroup$ Can you clarify your original question? When you say "independent of linear equivalence", what divisor are you talking about? $D$? $\endgroup$ Commented Oct 4, 2019 at 2:37
  • $\begingroup$ @ Sándor Kovács I'm talking about linear equivalece class of $K_Y$. $\endgroup$
    – tukudani
    Commented Oct 4, 2019 at 3:28
  • $\begingroup$ @tukudani Do you finally solve this problem? $\endgroup$
    – Hydrogen
    Commented Jul 20, 2022 at 21:09

1 Answer 1

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If the morphism is not proper then an exceptional divisor can be linearly equivalent to zero. Indeed locally near a point of any exceptional divisor this must be true. Then localising gives such a birational morphism. But then for any value of the coefficient of such a divisor, we get the desired equality.

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