Let $\pi:X \to U$ be a projective morphism, and $(X, \Delta = A + B)$ be a KLT pair, where $A$ is a general ample divisor and $B$ is effective.

Suppose $K_X + \Delta$ is not nef (over $U$) and there exists a nef divisor $C$ such that $K_X + \Delta + C$ is nef. Then there exists an extremal ray $R$ which is $(K_X + \Delta)$-negative and there exists $\lambda \in (0,1]$ such that $K_X + \Delta + \lambda C$ is nef but trivial on $R$. Now, we run MMP with scaling $C$, my questions are the following:

Suppose $f: X \to Z$ is the contraction of the extremal ray $R$, if it is a divisoral contraction, in order to run MMP, we need to use $(Z, \Delta')$ replace $(X, \Delta)$ where $\Delta' = f_* \Delta$, why in this case $K_Z + \Delta' + \lambda C'$ is nef ( here $C'$ is $f_*C$)?

Similarly, if $f$ is a small contraction, and suppose the flip $X^+$ exists, why $K_{X^+}+ \Delta^+ + \lambda C^+$ is nef?


We have $K_X+\Delta +\lambda C=f^*(K_Z+\Delta '+\lambda C')$ by the Base Point Free Thm (3.3 and 3.7(4) in Koll\'ar-Mori 1998). Clearly $K_Z+\Delta '+\lambda C'$ is nef. If $f^+:X^+\to Z $ is the flip, then $K_{X^+}+\Delta^+ +\lambda C^+={f^+}^*(K_Z+\Delta '+\lambda C')$ where $\Delta^+ $ and $C^+$ are the strict transforms of $\Delta$ and $C$. Thus $K_{X^+}+\Delta^+ +\lambda C^+$ is nef (as it is the pull back of a nef $\mathbb Q$-divisor).

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  • $\begingroup$ Dear Prof. Hacon, thank you very much for your answer! I have a little concern about the formula $K_X + \Delta + \lambda C = f^*(K_Z + \Delta' + \lambda C')$. When $f: X \to Z$ is a small contraction, could $K_Z + \Delta' + \lambda C'$ no longer $\mathbb{Q}$-Cartier? $\endgroup$ – Li Yutong Oct 21 '14 at 15:22
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    $\begingroup$ That is a valid concern; the point is that since $K_X+\Delta +\lambda C$ is $f$-trivial, there is a chance that it is actually a pull-back from $Z$. To check this is not easy; it is essentially a consequence of the BPF thm; see 3.7(4) in Koll\'ar-Mori 1998 $\endgroup$ – Hacon Oct 21 '14 at 15:29

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