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edited Oct 1 2011 at 14:15 |
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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edited Oct 1 2011 at 3:41 |
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: Let $e_1, \ldots, e_N$ be unit vectors along the axes in $\mathbb{R}^N$. For each $j$, $1 \leq j \leq k$, let $v_j := (a_{j1}, \sum a_{ji}e_i$, 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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edited Oct 1 2011 at 3:34 |
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 $dist(v, 0) > c$ v$ from both the origin and the boundary of $dist(v, \partial P) > c$, nP$ is greater than $c$, then $v$ is in fact an non-negative integral linear combination of $v_1, \ldots, v_k$. Here $dist$ means the Euclidean distance and $\partial P$ is the topological boundary of $P$.
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: Let $e_1, \ldots, e_N$ be unit vectors along the axes in $\mathbb{R}^N$. For each $j$, $1 \leq j \leq k$, let $v_j := \sum a_{ji}e_i$, 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 sum 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 $P$ mP$ and the distance of $[mv] - v_0$ from the origin and the boundary of $P$ 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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edited Oct 1 2011 at 3:04 |
Edit 13: I was stupid and, as J.C. Ottem pointed out, ignored the "very" of "very ample". I was also confused, and, as Artie Prendergast-Smith pointed out, thought of the operation taking place in the Néron-Severi group. So let me try again This is my 3rd attempt to give an elementary proof. It is essentially the same proof as in Edits 1 and 2, under a certain assumption (but with some corrections, and hopefully will be clearer. I do not know how restrictive hope you see that the idea is very simple and geometrically almost obvious. If it seems complicated, then the fault is ) (see Edit 2 below)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 $D_j = \sum_i sum_{i=1}^N a_{ji} C_i$, for irreducible divisors $C_i$. Extra Assumption: The $\mathbb{Z}$-span of $D_j$'s equals the $\mathbb{Z}$-span of C_i$ and integers $C_i$'s. Added later: this assumption actually can always made a_{ji}$. We want to be satisfied - see Edit 2 below. Under this extra assumption I show that $[mD]$ is very ample for all sufficiently large $m$.I 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 $r$ c$ such that for all $n \geq 1$, if $v \in nP \cap \mathbb{Z}^N$ such that $dist(v, 0) > r$ c$ and $dist(v, \partial P) > r$c$, then $v$ is in fact an non-negative integral linear combination of $v_1, \ldots, v_k$. Here $dist$ means the Euclidean distance and $\partial P$ is the topological boundary of $P$. Now Here starts the proof: Step 1: Without loss of generality we may assume that $[mD]$ \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: At first note Let $e_1, \ldots, e_N$ be unit vectors along the axes in $\mathbb{R}^N$. For each $j$, $1 \leq j \leq k$, let $v_j := \sum a_{ji}e_i$, 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 sum big multiples 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 $m_0$ m_j$ such that $m_0D_j + sD_j$ mD_j$ is very ample for all non-negative integer $s$ 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 $j$. It follows 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 $s_1, \ldots, s_k$. FinallyIndeed, note 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 for if $m$ is sufficiently large, then $m$, [mv] - v_0$ is in the interior of $P$ and the distance of $[mv] - v_0$ from the origin and the boundary of $P$ is bigger than $c$. Therefore, Khovanskii's lemma implies that $[mD]$ is a positive integral linear combination of $D_1, [mv] - v_0 = \ldots, D_k$sum a_j v_j$ for non-negative integers $a_j$. Consequently, and thereforeif $m$ is sufficiently large, we may assume then $[mD] $[mD] = m_0(D_1 + \cdots + D_k) D_0) + \sum s_jD_j$ a_j D_j$$ for some non-negative integers $s_1, a_1, \ldots, s_k$a_k$. Consequently, Step 4 then tells that $[mD]$ is ample! Edit 2: In fact the Extra Assumption above can always be satisfied: by Kleiman's condition, we may add to our collection of $D_j$'s very ampledivisors of the form $D_j + \sum \epsilon_i C_i$ as long as $\epsilon_i$'s are sufficiently small rational numbers. After choosing sufficiently many new $D_j$'s of this form we can make sure that the Set Up and Extra Assumption are both satisfied! This completes the proof in the general case.
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edited Sep 30 2011 at 23:26 |
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 1: I was stupid and, as J.C. Ottem pointed out, ignored the "very" of "very ample". I was also confused, and, as Artie Prendergast-Smith pointed out, thought of the operation taking place in the Néron-Severi group. So let me try again to give an elementary proof, under a certain assumption (I do not know how restrictive it is)is) (see Edit 2 below).
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 $D_j = \sum_i a_{ji} C_i$, for irreducible divisors $C_i$.
Extra Assumption: The $\mathbb{Z}$-span of $D_j$'s equals the $\mathbb{Z}$-span of $C_i$'s. Added later: this assumption actually can always made to be satisfied - see Edit 2 below.
Under this extra assumption I show that $[mD]$ is very ample for all sufficiently large $m$. I use the following:
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 $r$ such that for all $n \geq 1$, if $v \in nP \cap \mathbb{Z}^N$ such that $dist(v, 0) > r$ and $dist(v, \partial P) > r$, then $v$ is in fact an non-negative integral linear combination of $v_1, \ldots, v_k$. Here $dist$ means the Euclidean distance and $\partial P$ is the topological boundary of $P$.
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.
Now the proof that $[mD]$ is ample: At first note that there exists $m_0$ such that $m_0D_j + sD_j$ is very ample for all non-negative integer $s$ and all $j$. It follows that $m_0(D_1 + \cdots +D_k) + \sum s_jD_j$ is very ample for all collections of non-negative integers $s_1, \ldots, s_k$. Finally, note that for sufficiently large $m$, Khovanskii's lemma implies that $[mD]$ is a positive integral linear combination of $D_1, \ldots, D_k$, and therefore, we may assume $[mD] = m_0(D_1 + \cdots +D_k) + \sum s_jD_j$ for some non-negative $s_1, \ldots, s_k$. Consequently, $[mD]$ is ample!
Edit 2: In fact the Extra Assumption above can always be satisfied: by Kleiman's condition, we may add to our collection of $D_j$'s ample divisors of the form $D_j + \sum \epsilon_i C_i$ as long as $\epsilon_i$'s are sufficiently small rational numbers. After choosing sufficiently many new $D_j$'s of this form we can make sure that the Set Up and Extra Assumption are both satisfied! This completes the proof in the general case.
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edited Sep 30 2011 at 22:38 |
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.
I was stupid and, as J.C. Ottem pointed out, ignored the "very" of "very ample". I was also confused, and, as Artie Prendergast-Smith pointed out, thought of the operation taking place in the Néron-Severi group. So let me try again to give an elementary proof, under a certain assumption (I do not know how restrictive it is).
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 $D_j = \sum_i a_{ji} C_i$, for irreducible divisors $C_i$.
Extra Assumption: The $\mathbb{Z}$-span of $D_j$'s equals the $\mathbb{Z}$-span of $C_i$'s.
Under this extra assumption I show that $[mD]$ is very ample for all sufficiently large $m$. I use the following:
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 $r$ such that for all $n \geq 1$, if $v \in nP \cap \mathbb{Z}^N$ such that $dist(v, 0) > r$ and $dist(v, \partial P) > r$, then $v$ is in fact an non-negative integral linear combination of $v_1, \ldots, v_k$. Here $dist$ means the Euclidean distance and $\partial P$ is the topological boundary of $P$.
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.
Now the proof that $[mD]$ is ample: At first note that there exists $m_0$ such that $m_0D_j + sD_j$ is very ample for all non-negative integer $s$ and all $j$. It follows that $m_0(D_1 + \cdots +D_k) + \sum s_jD_j$ is very ample for all collections of non-negative integers $s_1, \ldots, s_k$. Finally, note that for sufficiently large $m$, Khovanskii's lemma implies that $[mD]$ is a positive integral linear combination of $D_1, \ldots, D_k$, and therefore, we may assume $[mD] = m_0(D_1 + \cdots +D_k) + \sum s_jD_j$ for some non-negative $s_1, \ldots, s_k$. Consequently, $[mD]$ is ample!
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edited Sep 30 2011 at 22:32 |
I was stupid and, as J.C. Ottem pointed out, ignored the "very" of "very ample". I was also confused, and, as Artie Prendergast-Smith pointed out, thought of the operation taking place in the Néron-Severi group. So let me try again to give an elementary proof, under a certain assumption (I do not know how restrictive it is). 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 $D_j = \sum_i a_{ji} C_i$, for irreducible divisors $C_i$. Extra Assumption: The $\mathbb{Z}$-span of $D_j$'s equals the $\mathbb{Z}$-span of $C_i$'s. Under this extra assumption I show that $[mD]$ is very ample for all sufficiently large $m$. I use the following: 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 $r$ such that for all $n \geq 1$, if $v \in nP \cap \mathbb{Z}^N$ such that $dist(v, 0) > r$ and $dist(v, \partial P) > r$, then $v$ is in fact an non-negative integral linear combination of $v_1, \ldots, v_k$. Here $dist$ means the Euclidean distance and $\partial P$ is the topological boundary of $P$. 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. Now the proof that $[mD]$ is ample: At first note that there exists $m_0$ such that $m_0D_j + sD_j$ is very ample for all non-negative integer $s$ and all $j$. It follows that $m_0(D_1 + \cdots +D_k) + \sum s_jD_j$ is very ample for all collections of non-negative integers $s_1, \ldots, s_k$. Finally, note that for sufficiently large $m$, Khovanskii's lemma implies that $[mD]$ is a positive integral linear combination of $D_1, \ldots, D_k$, and therefore, we may assume $[mD] = m_0(D_1 + \cdots +D_k) + \sum s_jD_j$ for some non-negative $s_1, \ldots, s_k$. Consequently, $[mD]$ is ample!
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answered Sep 30 2011 at 18:28 |
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.
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