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Let D be an ample R-divisor, is the round down [mD] very ample for any sufficiently divisible number m?

I think it's true. But I do not know how to arrange an argument.

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Are you willing to assume that the ambient space is smooth? Otherwise what if $\lfloor mD \rfloor$ isn't even $\mathbb{Q}$-Cartier? –  Karl Schwede Sep 28 '11 at 20:13
    
Yes. The ambient space is assumed to be smooth. –  Zhengyu Hu Sep 29 '11 at 0:58
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2 Answers 2

up vote 5 down vote accepted

I am not sure if this is the shortest proof, but I think that it is a proof.

Let $A=$ very ample line bundle. After replacing D by a multiple, you may assume that $$C=D - K_X - (n+1) A$$ is ample where $n=\dim X$.

By Angehrn and Siu, we know that $$K_X+(n+1)A + \text{(ample line bundle)}$$ is very ample.

Now $$[mD] = K_X+ (n+1) A + C + [mD] - D$$

and you just need to make sure that $[mD]+C-D$ is ample if $m\gg 0$. This is easy to check by Diophantine approximation.

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Thank you. I think it is correct... –  Zhengyu Hu Oct 1 '11 at 1:10
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At first a question: If I understand correctly, then the rounding down operation depends on your choice of basis for the Néron-Severi group, right?

So I am assuming you fix a basis $D_1, \ldots, D_k$ of $NS(X) \otimes_\mathbb{Z} \mathbb{R}$ and for each $D := \sum_j r_jD_j$, you define $[mD] := \sum [mr_j]D_j$.

If this is true, then doesn't your assertion follow from the following geometric fact?

Let $C$ be a full dimensional cone in $\mathbb{R}^k$ and $K$ be the standard cube of length $2$ in $\mathbb{R}^k$ centered at the origin, i.e.

$K := \lbrace\sum_{j=1}^k s_je_j: -1 \leq s_j \leq 1$ for all $j$, $1 \leq j \leq k \rbrace$,

where $e_1, \ldots, e_k$ are unit vectors along the axes. If $v$ belongs to the interior of a full dimensional cone $C$ in $\mathbb{R}^k$, then $mv + K$ also lies in the interior of $C$ for all sufficiently large $m$.

If as your basis you choose ample divisors, then $K$ can be replaced by a cube of length one.

Edit 3: This is my 3rd attempt to give an elementary proof. It is essentially the same proof as in Edits 1 and 2, but with some corrections, and hopefully will be clearer. I hope you see that the idea is very simple and geometrically almost obvious. If it seems complicated, then the fault is in my exposition.

Set Up: Let $D_1, \ldots, D_k$ be ample divisors and $D := \sum_j r_jD_j$ for positive real numbers $r_1, \ldots, r_k$. Also, let $D_j = \sum_{i=1}^N a_{ji} C_i$, for irreducible divisors $C_i$ and integers $a_{ji}$. We want to show that $[mD]$ is very ample for large $m$.

In the proof we will use the following fact about finite sums of integral points in a lattice:

Lemma: Let $v_1, \ldots, v_k \in \mathbb{Z}^N$ such that $\mathbb{Z}$-span of $v_j$'s equals $\mathbb{Z}^N$. Let $P$ be the convex hull (over $\mathbb{R}$) of $\lbrace 0, v_1, \ldots, v_k \rbrace$. Then there exists a positive real number $c$ such that for all $n \geq 1$, if $v \in nP \cap \mathbb{Z}^N$ such that the (Euclidean) distance of $v$ from both the origin and the boundary of $nP$ is greater than $c$, then $v$ is in fact an non-negative integral linear combination of $v_1, \ldots, v_k$.

The above statement (actually a more precise formulation of it) is due to Khovanskii. The proof is very elementary and beautiful, and is in Proposition 2 of this article.

Here starts the proof:

Step 1: Without loss of generality we may assume that $\mathbb{Z}$-span of $D_j$'s equals the $\mathbb{Z}$-span of $C_i$'s. Indeed, it follows from Kleiman's criterion, and finite dimensionality of $N_1(X)$ that for every $m \gg 1$ and $\epsilon := (\epsilon_1, \ldots, \epsilon_N) \in \lbrace 1, 0, -1 \rbrace^N$, $D_{m,\epsilon} := mD_1 + \sum_{i=1}^N\epsilon_i C_i$ is ample. Choosing different values of $\epsilon$ and $m$ and adding $D_{m,\epsilon}$'s to the collection of $D_j$'s, we may ensure that $\mathbb{Z}$-span of $D_j$'s equals the $\mathbb{Z}$-span of $C_i$'s. Moreover, and this is essential, choosing $D_{m,\epsilon}$'s to be sufficiently close to the ray generated by $D_1$, we may ensure that $D$ still lies in the interior of the cone generated by $D_j$'s, i.e. $D = \sum_{j=1}^k r_jD_j$ with each $r_j$ being a positive real number.

Step 2: For each $j$, $1 \leq j \leq k$, let $v_j := (a_{j1}, \ldots, a_{jN}) \in \mathbb{R}^N$, i.e. $v_j$ is the "coordinate" vector of $D_j$ for each $j$ (and therefore $v_j \in \mathbb{Z}^N$ for each $j$). Adding some big multiples of $D_j$'s to the existing collection of $D_j$'s if necessary, we may assume that $v := \sum r_j v_j$ is in the interior of the convex hull $P$ of $0, v_1, \ldots, v_k$.

Step 3: For each $j$, $1 \leq j \leq k$, there exists a positive integer $m_j$ such that $mD_j$ is very ample for all $m \geq m_j$. Indeed, there is $l_j, n_j$ such that $n_jD_j$ is very ample and $mD_j$ is globally generated for all $m \geq l_j$. Setting $m_j := l_j + n_j$ does the job (due to Exercise II.7.5(d) of Hartshorne).

Step 4: There exists a positive integer $m_0$ such that $m_0(D_1 + \cdots +D_k) + \sum s_jD_j$ is very ample for all collections of non-negative integers $s_1, \ldots, s_k$. Indeed, set $m_0 := \max \lbrace m_1, \ldots, m_k \rbrace$ and apply the same exercise of Hartshorne.

Step 5: Let $v, v_1, \ldots, v_k$ and $P$ be as in Step 2. Let $c$ be the constant we get from applying Khovanskii's lemma to $v_1, \ldots, v_k$. Let $v_0 := m_0(v_1 + \cdots + v_k)$, where $m_0$ is as in Step 4. Since $v$ is in the interior of $P$, it follows that if $m$ is sufficiently large, then $[mv] - v_0$ is in the interior of $mP$ and the distance of $[mv] - v_0$ from the origin and the boundary of $mP$ is bigger than $c$. Therefore, Khovanskii's lemma implies that $[mv] - v_0 = \sum a_j v_j$ for non-negative integers $a_j$. Consequently, if $m$ is sufficiently large, then

$$[mD] = m_0(D_1 + \cdots + D_0) + \sum a_j D_j$$

for non-negative integers $a_1, \ldots, a_k$. Step 4 then tells that $[mD]$ is very ample.

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Where do you show that $[mD]$ is $very$ ample? –  J.C. Ottem Sep 30 '11 at 19:02
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Dear auniket, I am not sure I follow your answer. First, the round-down operation is defined on the level of divisors, not (linear or numerical) equivalence classes. For instance, suppose D is any (integral) divisor, and $\Delta$ is another Weil divisor which is linearly equivalent to $2D$. Then D and $\Delta/2$ are numerically equivalent $\Q$-divisors, but the first one has round-down D, while the second has round-down zero. Second, very ampleness is unfortunately not defined on the level of numerical classes, unlike ampleness. (See Hartshorne Ex. V.1.12.) So I think... –  Artie Prendergast-Smith Sep 30 '11 at 19:03
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...one cannot simply argue on the level of numerical classes. –  Artie Prendergast-Smith Sep 30 '11 at 19:04
    
I am still confused, you seem to show that $[mD]$ is ample for large $m$ (which is obvious right?), but I think the hard part of the problem is the very ampleness. Am I missing something? –  J.C. Ottem Oct 1 '11 at 0:27
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I agree with J.C. I think the crux of the problem is in the statement "At first note that there exists $m_0$ such that $m_0 D_j+sD_j$ is very ample for all non-negative integer s and all j". –  Artie Prendergast-Smith Oct 1 '11 at 1:13
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