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).

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.