# Pushforward of a log canonical pair

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.

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.
• Why $\phi_*(K_X)=K_Y$? Could you point out some reference? – Li Yutong Jul 30 '14 at 7:29
• @dhy Which form of the theorem do you mean, for the G-R vanishing theorem I check in "Positivity of algebraic geometry" (THM 4.3.9), it only states that $R^i\phi_*(K_X)=0$ for $i>0$. – Li Yutong Jul 30 '14 at 14:09
• $\phi_* K_X = K_Y$ 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 $\phi_* K_X = K_Y$. – Karl Schwede Jul 31 '14 at 5:06
• 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). – Karl Schwede Jul 31 '14 at 5:06