# A key step in Del Busto's effective Matsusaka theorem

I'm reading a proof of del Busto (http://arxiv.org/pdf/alg-geom/9410018v1.pdf) and am confused at one step. I'm sure its quite easy, but the key trick hasn't come to me yet.

The context here is that $X$ is a nonsingular (complex) projective surface and A an ample divisor. For $\kappa_1,\ldots,\kappa_r$ positive integers $\{x_1,\ldots,x_r\}$ a collection of $r$ distinct points on $X$ and $Z = \sum \kappa_i x_i$, let $f \colon Y \to X$ be the blow up of $X$ at $\{x_1,\ldots,x_r\}$. Denote by $E_1,\ldots,E_r \subset Y$ the corresponding exceptional divisors. He reduces the problem to the vanishing

$$V(k) := H^1\left(Y, \mathcal{O}_Y \left( f^*(kA) - \sum_{i=1}^r \kappa_i E_i \right)\right) = 0.$$

Set $B = f^*(kA - K_X) - \sum_{i=1}^r (\kappa_i + 1)E_i$. For $n > 0$ set $nB = M_n + F_n$ its Zariski decomposition. For $k$ larger than his bound, $\frac{1}{n} F_n \cdot f^* A < 1$ so we can write $\lfloor \frac{1}{n} F_n \rfloor = \sum_{i=1}^r \eta_i E_i$ with $\eta_i \geq 0$. He next shows as a consequence of Kawamata-Viehweg that

$$W(k) := H^1\left(Y, f^* \mathcal{O}_X\left( (k+1) A \right) \otimes \mathcal{O}_Y \left( - \sum_{i=1}^r (\kappa_i + \eta_i) E_i \right) \right) = 0$$ and deduces the result from the claimed implication $W(k) = 0 \Rightarrow V(k+1) = 0$. The two sheaves are very similar, in particular off by $- \sum_{i=1}^r \eta_i E_i$. However, I don't easily see this implication.

• It seems to me that $[(1/n)F_n]$ is exc. and there is a ses inducing $W(k)\to V(k+1)\to H^1((-\sum k_iE_i)|_{\sum \eta _iE_i})$ which has positive degree on a bunch of (non-reduced $\mathbb P^1$'s and so vanishes. If each $\eta _i\in \{0,1\}$, this is clear, if not proceed by induction via the ses of the form $0\to (-k-(\eta-1))E)|_E\to (-kE)|_{\eta E)\to (-kE)|_{(\eta -1)E}\to 0$ (hopefully I got the algebra right) – Hacon Jun 4 '15 at 21:19
• @Hacon Thanks a bunch! That's exactly what I hoping to see. I think the induction when eta_i > 1 is what was tricky to see, but as you noted, you can still induce and so your argument I think works. Also, in case anyone else is interest, there is a subtle change of setting in del Busto's proof. In particular, he really shows that a different lower bound on k implies being (k+1)-jet ample, but his theorem stated on the bottom of page 4 changes everything (hypothesis and conclusion) back to the k-jet ample! – lemiller Jun 4 '15 at 21:27
• @Hacon also it seems your equation array has a markdown error. If you type your comment as an answer I can accept it. – lemiller Jun 4 '15 at 21:31
• OK. Happy that it seems to help. – Hacon Jun 4 '15 at 21:42

It seems to me that $[\frac 1nF_n]$ is exc. and there is a short exact sequence inducing $$W(k)→V(k+1)→H^1((−∑k_iE_i)|_{\sum \eta _iE_i})$$ which has positive degree on a bunch of (non-reduced $\mathbb P^1$'s and so vanishes. If each $\eta _i\in \{0,1\}$, this is clear. If not proceed by induction via the short exact sequences of the form $$0\to (-k-(\eta-1))E)|_E\to (-kE)|_{\eta E}\to (-kE)|_{(\eta -1)E}\to 0.$$