This question comes from learning the paper "Existence of minimal models for varieties of log general type", where they define the log terminal model (See Definition 3.6.7). However, the question itself does not need that definition.

Let $\pi: X \to U$ be a projective morphism of normal quasi-projective varieties. Suppose that $K_X + \Delta$ is log canonical and let $\phi: X -\to Y$ be a birational contraction of normal quasi-projective varieties over $U$, where $Y$ is projective over $U$ (here, $\phi$ is a "birational contraction" means $\phi^{-1}$ does not contract any divisor, for example, when $\phi$ is a morphism). Set $\Gamma=\phi_* \Delta$. Then is it true $$\phi_*(K_X + \Delta)=K_Y +\Gamma\quad?$$

I am equally satisfied with an answer in the simplified case as $U$ is a point, or $\phi$ is a morphism.


1 Answer 1


Yes, since $$\phi_*K_X=K_Y.$$ Here is the reason. We take a common resolution of $X$ and $Y$, say $$ p:W\rightarrow X, \\q: W \rightarrow Y. $$ Then we can right $$ K_W-p^*K_X=E,\\ K_W-q^*K_Y=F, $$ where $E$ is exceptional over $X$ (and over $Y$ because "contraction") and $F$ is exceptional over $Y$. If apply $q_*$ to $$ p^*K_X+E=q^*K_Y=F, $$ we get the equality.

  • $\begingroup$ Why $\phi_*(K_X)=K_Y$? Could you point out some reference? $\endgroup$
    – Li Yutong
    Jul 30, 2014 at 7:29
  • $\begingroup$ I think the key words here are "Grauert-Reimenschneider vanishing theorem." $\endgroup$
    – dhy
    Jul 30, 2014 at 8:07
  • 1
    $\begingroup$ $\phi_* K_X = K_Y$ has nothing to do with Grauert-Riemenschneider vanishing I think. The edit to the question gives one way to see it. Here is another. Both $X$ and $Y$ are normal and they coincide on an open set $U$ whose complement has codimension 2 on $Y$. $K_X$ is any divisor such that $O_X(K_X) \cong \omega_X$ and $K_Y$ is any divisor such that $O_Y(K_Y) \cong \omega_Y$. If I have a divisor $K_X$, this gives me a divisor $K_U = K_X|_U$. But since the complement of $U$ has codimension 2 on $Y$, $K_U$ uniquely determines a choice of $K_Y$ such that $\phi_* K_X = K_Y$. $\endgroup$ Jul 31, 2014 at 5:06
  • 3
    $\begingroup$ Of course, there are other choice of $K_Y$ too (although each one comes from a choice of some $K_X$). It is worth emphasizing that just because $\phi_* K_X = K_Y$ does not mean that $\phi_* \omega_X = \omega_Y$ (this latter condition is basically rational singularities). $\endgroup$ Jul 31, 2014 at 5:06
  • 1
    $\begingroup$ I don't think people typically define $\omega_X = \mu_* \O_{X'}(K_{X'})$. That is not a sheaf which corresponds to a divisor, it does not coincide with the $\text{Ext}$ definition. It does satisfy various Kodaira-type vanishing theorems though. I would NOT use that definition as that of the canonical sheaf (it is really a early version of the multiplier ideal). Hopefully this helps. $\endgroup$ Jul 31, 2014 at 12:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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