Let $f:X \to Y$ be a finite map from a normal projective variety to a smooth projective variety, $D$ be a Cartier divisor on $X$. Do we have any relation between $\kappa(X,D)$ and $\kappa(Y,f_*D)$?


1 Answer 1


There is an obvious relation: the pushforward map $f_*:|mD|\rightarrow |f_*(mD)|=|mf_*D|$ is injective, hence $\kappa (X,D)\leq \kappa (X,f_*D)$. It is easy to see that you cannot get more: for instance, take for $f$ a general projection from a cubic surface $X\subset \mathbb{P}^3$ to $\mathbb{P}^2$, and for $D$ a line in $S$. Then $\kappa (S,D)=0$, but $f_*D$ is a line in $\mathbb{P}^2$, so $\kappa (\mathbb{P}^2,f_*D)= 2$.

  • $\begingroup$ why $f_*$ is injective? Is there a reference? $\endgroup$
    – Hu Zhengyu
    Feb 26, 2020 at 17:05
  • $\begingroup$ I can imagine that $|mD| \to |mf^*f_*D|$ is injective but I am not sure if $f_*$ is injective. $\endgroup$
    – Hu Zhengyu
    Feb 26, 2020 at 17:31
  • $\begingroup$ @Hu Zhengyu: It is finite because $f$ is finite, on the other hand it is linear. $\endgroup$
    – abx
    Feb 26, 2020 at 17:32
  • $\begingroup$ I think $f_*$ is not injective. If you take $D=0$, then $f_*$ would map some nonzero rational function of $X$ to zero since $K(X)$ is obviously bigger than $K(Y)$. You just said $f$ is finite? What is the definition of finite for a linear map? $\endgroup$
    – Hu Zhengyu
    Feb 27, 2020 at 3:01

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