# Ample divisors on blown-up projective space

Let $\mathbb{P}=\mathrm{Proj}(\mathbb{C}[x_0,\ldots,x_n])$ be complex projective $n$-space. Assume I have linear subvarieties $L_1,\ldots,L_k\in\mathbb{P}$ of codimension $r_i\ge 2$, respectively. Let $\pi:X\to\mathbb{P}$ be the composition of blowing up in $L_1$, then blowing up in the strict transform of $L_2$, and so on.

Let $E_i\:=\pi^{-1}(L_i)$ and $E=E_1+\cdots+E_k$ the exceptional divisor. Now, any divisor on $X$ is of the form

$H=\pi^\ast(D) + \sum_{i=1}^k a_i E_i$

I am wondering when $H$ is ample - I am very much willing to assume that $D$ is ample, and I am looking for a condition that depends mostly on the $a_i$. If this is still too general, I would like to know if an anticanonical divisor

$-K_X=-K_{\mathbb{P}} - \sum_{i=1}^k (r_i-1) E_i$

on $X$ is ample.

The above is the least general scenario that I am willing to study - more generally, what are the "best" sufficient conditions for ampleness of a divisor on a nonsingular blow-up?

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I might be wrong, but if $L_i$ are line and two of those lines have non-empty intersection, consecutive blow-up of those two is not equal to blowing up the union at the same time.(This is was just a general comment and has nothing to do with your question) – Mohammad F. Tehrani Oct 9 '11 at 14:42
Hm, I have to admit I haven't thought about that very thoroughly. I most definitely want to blow up $L_1$, then $\tilde L_2$, and so on. I will edit my post. – Jesko Hüttenhain Oct 9 '11 at 15:19

Your question is actually far too general, so let me assume $n=2$. Also in this case, there are only partial results.

In the case where all $a_i$ are equal to $1$, Kurchle and (independently) Xu showed that $$H=\pi^*(dL) + \sum_{i=1}^r E_i$$ (where $L$ is the class of a line) is ample provided that $H^2 > 0$ and $d \geq 3$.

Later, Szemberg and Tutaj-Gasinska, in their paper General blow-ups of the projective plane (Proceedings of the Amer. Math. Soc. 130, 2002), proved the following (non optimal) result:

Theorem. Let $X$ the blow-up of $\mathbb{P}^2$ at $r$ general points and let $k\geq 2$ and $r$ be integers such that $d \geq 3k+1$. If $r \leq \frac{d^2}{k^2} -1$, then the line bundle $$H=\pi^*(dL) + \sum_{i=1}^r kE_i$$ is ample.

I refer you to Szemberg-Tutaj-Gasinska paper for more details on this problem and on its relations with the so-called Nagata Conjecture.

In higher dimensions, there is a paper by Angelini that generalizes the result of Xu for $n=3$, in the case where all blown-up subvarieties are points. See Ample divisors on the blow up of $\mathbb{P}^3$ at points , Manuscripta Mathematica 93 (1997).

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Thanks a bunch, I also corrected my question. Is there anything known for $n=3$? – Jesko Hüttenhain Oct 9 '11 at 6:01
You are welcome. I added a reference to a result concerning the case $n=3$. – Francesco Polizzi Oct 9 '11 at 8:22

An example in dimension $3$ where $-K_X$ is not ample. Take $X$ the blow-up of $\mathbb{P}^3$ at six general points $p_1,...,p_6$. The anticanonical divisor is $$-K_X = 4H-2E_1-...-2E_6 = 2(2H-E_1-...-E_6).$$ Therefore for any strict transform $\widetilde{L}_{i,j}$ line $L_{i,j} = \left\langle p_i,p_j\right\rangle$ we have $-K_X\cdot\widetilde{L}_{i,j} = 0$. Furthermore, if $\widetilde{C}$ is the strict transform of the twisted cubic $C$ through $p_1,...,p_6$ we have $-K_X\cdot\widetilde{C} = 0$ as well. Therefore $-K_X$ is not ample. However, since the base locus of $|-K_X|$ is zero dimensional $-K_X$ is nef.

On the other hand if we take $p_1,...,p_k$ general points, with $2\leq k\leq 4$ then the $\widetilde{L}_{i,j}$ still have intersection zero with $-K_X$. However, $X$ is toric and in particular log Fano. This means that there exists a reduced simple normal crossing divisor $D$ on $X$ such that $-(K_X+D)$ is ample and $(X,D)$ is klt.

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Why do you need $6$ points? If $X$ is the blow-up of $k$ points in $\mathbb{P}^n$ with $n\ge 3$ and $k\ge 2$, then $-K_X$ is not ample, by intersecting a line through the points with it. – Jérémy Blanc Mar 20 '14 at 13:24
Of course, I just wanted to provide a case where a curve which is not the strict transform of a line has zero intersection with the anticanonical. – S_Z_S Mar 20 '14 at 13:55
Ah OK. A simple remark: the twisted cubic through $6$ points are equivalent to the lines (sum of three). – Jérémy Blanc Mar 20 '14 at 14:00