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8 added 1 characters in body

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.

7 deleted 46 characters in body

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. 6 added 39 characters in body; deleted 40 characters in body 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.

5 Fixed some gaps