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($B$)=$B^n$). 



volume of big line bundles under finite morphisms
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.


See also Proposition 4.1 in [this arXiv paper](http://arxiv.org/PS_cache/arxiv/pdf/1006/1006.2340v1.pdf).


