User armin holschbach - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T10:41:22Z http://mathoverflow.net/feeds/user/8333 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/30299/volume-of-big-line-bundles-under-finite-morphisms/35137#35137 Answer by Armin Holschbach for volume of big line bundles under finite morphisms Armin Holschbach 2010-08-10T16:18:36Z 2010-08-10T16:18:36Z <p>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).</p> <p>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).</p> <p>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</p> <p>$0 \to \mathcal G \to f_* \mathcal O_X \to \mathcal G_1 \to 0$ , </p> <p>$0 \to \mathcal G \to \mathcal O_Y^d\; \to \mathcal G_2 \to 0.$</p> <p>The supports of $\mathcal G_1$ and $\mathcal G_2$ do not meet $U$, hence have dimension less than $n$. Therefore,</p> <p>$h^0(Y, \mathcal G_i\otimes B^{\otimes m}) := \dim_k H^0(Y, \mathcal G_i\otimes B^{\otimes m}) = O(m^{n-1})$ </p> <p>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</p> <p>$h^0(X,f^* B^{\otimes m}) = h^0(Y,(f_*\mathcal O_X)\otimes B^{\otimes m})= d \cdot h^0(Y, B^{\otimes m}) + O(m^{n-1}),$</p> <p>from which the assertion follows.</p> <p>See also Proposition 4.1 in <a href="http://arxiv.org/PS_cache/arxiv/pdf/1006/1006.2340v1.pdf" rel="nofollow">this arXiv paper</a>.</p>