# volume of big line bundles under finite morphisms

Let $X$, $Y$ be complex projective varieties of dimension $n$, let $f:X \rightarrow Y$ be a surjective finite morphism of degree $d$ and let $B$ be a big line bundle on $Y$.

Is that true that vol($f^*B$)=d $\cdot$ vol($B$)?

(I know that if $B$ is not only big but also nef then the formula is true by Lazarsfeld's Positivity in Algebraic geometry I, remark 1.1.13, using the well-known fact that if $B$ is nef then vol($B$)=$B^n$).

Yes, this is true, even in a slightly more general context (the varieties can be defined over any field $k$, and the morphism only needs to be generically finite and surjective).

The main parts of the argument are given in the books of Lazardfeld (Positivity in Algebraic geometry) and Debarre (Higher-dimensional Algebraic Geometry), although this specific property is never stated there (as far as I know).

By the projection formula, we have $H^0(X,f^* B^{\otimes m}) \cong H^0(Y,f_* \mathcal O_X \otimes B^{\otimes m}),$ so we can restrict our attention to $f_* \mathcal O_X$ and its twists by $B$. There is an open dense subset $U$ of $Y$ such that $f_* \mathcal O_X$ is free of rank $d = \deg(f)$, so $(f_* \mathcal O_X)|_U \simeq \mathcal O_U^d$. This isomorphism gives an injection $f_* \mathcal O_X \hookrightarrow \mathcal K_Y^d$, where $\mathcal K_Y$ is the sheaf of total quotient rings of $\mathcal O_Y$. Set $\mathcal G = f_* \mathcal O_X \cap \mathcal O_Y^d$ and define $\mathcal G_1$ and $\mathcal G_2$ by the exact sequences of sheaves

$0 \to \mathcal G \to f_* \mathcal O_X \to \mathcal G_1 \to 0$ ,

$0 \to \mathcal G \to \mathcal O_Y^d\; \to \mathcal G_2 \to 0.$

The supports of $\mathcal G_1$ and $\mathcal G_2$ do not meet $U$, hence have dimension less than $n$. Therefore,

$h^0(Y, \mathcal G_i\otimes B^{\otimes m}) := \dim_k H^0(Y, \mathcal G_i\otimes B^{\otimes m}) = O(m^{n-1})$

for $i=1,2$ (see e.g. Proposition 1.31 in Debarre's book). Using the long exact cohomology sequence for the two short exact sequences above twisted by $B^{\otimes m}$, this implies

$h^0(X,f^* B^{\otimes m}) = h^0(Y,(f_*\mathcal O_X)\otimes B^{\otimes m})= d \cdot h^0(Y, B^{\otimes m}) + O(m^{n-1}),$

from which the assertion follows.