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This article is cited in 1 scientific paper (total in 1 paper) Diophantine exponents of lattices and the growth of higher-dimensional analogues of partial quotients E. R. Bigushevab, O. N. Germanab a Faculty of Mechanics and Mathematics, Lomonosov Moscow State University, Moscow, Russia b Moscow Center of Fundamental and Applied Mathematics, Moscow, Russia
Russian version:
Matematicheskii Sbornik, 2023, Volume 214, Number 3, DOI: https://doi.org/10.4213/sm9746 
Bibliographic databases:
    § 1. IntroductionIrrationality exponentLet $\theta$ be a real number. By the irrationality exponent of this number we mean the quantity
$$
\begin{equation}
\mu(\theta)=\sup\biggl\{\gamma\in\mathbb{R}\biggm|\biggl|\theta -\frac pq\biggr|\leqslant|q|^{-\gamma}\text{ has $\infty$ solutions in }(q,p)\in\mathbb{Z}^2 \biggr\}.
\end{equation}
\tag{1.1}
$$
For rational numbers the irrationality exponent is obviously equal to $\infty$; for irrational ones it is at least $2$ in view of Dirichlet’s theorem on the approximation of a real number by rational ones. If $\theta$ is irrational, then it can be expanded in an infinite (simple) continued fraction $\theta=[a_0;a_1,a_2,\dots]$. The following inequalities hold for every convergent $p_n/q_n=[a_0;a_1,\dots,a_n]$ (see [1]–[3]):
$$
\begin{equation*}
\frac{1}{q_n^2(a_{n+1}+2)} <\biggl|\theta-\frac{p_n}{q_n}\biggr|\leqslant\frac{1}{q_n^2a_{n+1}}.
\end{equation*}
\notag
$$
Hence, taking into account that for irrational $\theta$ the definition (1.1) is equivalent to
$$
\begin{equation}
\mu(\theta)=\sup\biggl\{\gamma\in\mathbb{R}\biggm|\forall\, n_0>0\ \exists\,n>n_0\colon \biggl|\theta-\frac{p_n}{q_n}\biggr|\leqslant|q_n|^{-\gamma} \biggr\},
\end{equation}
\tag{1.2}
$$
we obtain the classical relation connecting $\mu(\theta)$ with the growth of the partial quotients,
$$
\begin{equation}
\mu(\theta)=2+\limsup_{n\to\infty}\frac{\log a_{n+1}}{\log q_n}.
\end{equation}
\tag{1.3}
$$
Here $q_0,q_1,q_2,\dots$ is the sequence of denominators of the convergents of $\theta$. Many assertions concerning continued fractions have multidimensional analogues. In this paper we consider relation (1.3) from this point of view, interpret it geometrically and prove a three-dimensional analogue of the inequality
$$
\begin{equation*}
\mu(\theta)\leqslant 2+\limsup_{n\to\infty}\frac{\log a_{n+1}}{\log q_n}.
\end{equation*}
\notag
$$
As a multidimensional generalization of continued fractions, we take Klein polyhedra. In the two-dimensional case, which corresponds to ordinary continued fractions, Klein polyhedra are usually called Klein polygons. Klein polygonsContinued fractions have a very elegant geometric interpretation going back to Klein [4]. We work with a natural ‘extension’ of this interpretation, requiring two numbers. This approach makes it possible to use efficiently methods of the geometry of numbers and linear algebra. Let $\theta_1$ and $\theta_2$ be distinct real numbers. Consider linear forms $L_1$ and $L_2$ in two variables with coefficients written as rows of the matrix
$$
\begin{equation}
A=\begin{pmatrix} \theta_1 & -1 \\ \theta_2 & -1 \end{pmatrix}.
\end{equation}
\tag{1.4}
$$
Consider the convex hulls
$$
\begin{equation*}
\mathcal{K}_1=\operatorname{conv}\bigl(\bigl\{\mathbf z\in\mathbb{Z}^2 \setminus\{\mathbf 0\}\bigm|L_1(\mathbf z)\geqslant0,\ L_2(\mathbf z)\leqslant0 \bigr\}\bigr)
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\mathcal{K}_2=\operatorname{conv}\bigl(\bigl\{\mathbf z\in\mathbb{Z}^2 \setminus\{\mathbf 0\}\bigm|L_1(\mathbf z)\leqslant0,\ L_2(\mathbf z)\leqslant0 \bigr\}\bigr).
\end{equation*}
\notag
$$
These convex hulls (Figure 1), as well as the sets $-\mathcal{K}_1$ and $-\mathcal{K}_2$, are called Klein polygons. The integer-combinatorial structure of the boundaries $\partial\mathcal{K}_1$ and $\partial\mathcal{K}_2$ is closely related to continued fractions of the numbers $\theta_1$ and $\theta_2$. A detailed description of this relationship can be found, for example, in [5] and [6] (also see [7]). Here we only point out the following crucial fact. Assume that This is the case shown in Figure 1. Then the coordinates of vertices of the polygons $\mathcal{K}_1$ and $\mathcal{K}_2$ are equal (up to signs) to the denominators and numerators of the convergents of $\theta_1$ and $\theta_2$, and the integer lengths of edges are equal to the corresponding partial quotients of these numbers. Recall that by the integer length of an integer segment (that is, the line segment whose endpoints have integer coordinates) we mean the number of empty integer segments contained in it. Thus, to every edge of the Klein polygon we assign the corresponding partial quotient. Moreover, if $\mathbf v=(q_n,p_n)$, where $p_n/q_n$ is a convergent of $\theta_1$, then the edges meeting at this vertex have integer lengths equal to the partial quotients $a_n$ and $a_{n+2}$ of $\theta_1$. Some partial quotients can also be assigned to all vertices. The reason is that there is a bijection (Figure 2) between the vertices of $\mathcal{K}_1$ and the edges of $\mathcal{K}_2$, under which a vertex $\mathbf v$ corresponds to an edge of integer length
$$
\begin{equation}
\operatorname{ang}(\mathbf v)=|{\det(\mathbf r_1,\mathbf r_2)}|,
\end{equation}
\tag{1.6}
$$
where $\mathbf r_1$ and $\mathbf r_2$ are the primitive integer vectors parallel to the edges incident to $\mathbf v$. In Figure 2 we have $\mathbf r_1=\mathbf w-\mathbf v$ and $\mathbf r_2=\mathbf u-\mathbf v$. The quantity $\operatorname{ang}(\mathbf v)$ is called the integer angle at the vertex $\mathbf v$. Now, if $\mathbf v=(q_n,p_n)$, then $\operatorname{ang}(\mathbf v)=a_{n+1}$ (see Figure 1). Thus, Klein polygons equipped with integer edge lengths and with integer angles at their vertices can be regarded as a geometric interpretation of continued fractions (more precisely, of pairs of continued fractions). Under assumption (1.5) the continued fraction of $\theta_1$ corresponds to the part of the above construction lying in the first quadrant. If (1.5) is not satisfied, then the above correspondence between the vertices and convergents can fail, but only in a (possibly large) neighbourhood of the origin. For instance, if $\theta_1>1$ and $0<\theta_2<\theta_1$, then we consider an arbitrary number $\theta'_2$ satisfying $-1<\theta'_2<0$ and the Klein polygons $\mathcal{K}'_1$ and $\mathcal{K}'_2$ corresponding to $\theta_1$ and $\theta'_2$. Then the sets $\partial\mathcal{K}_1\cap\partial\mathcal{K}'_1$ and $\partial\mathcal{K}_2\cap\partial\mathcal{K}'_2$ are polygonal lines infinite in one direction, on which the sequence of partial quotients of $\theta_1$ is written starting from some index. The vertices of these polygonal lines correspond to the convergents of $\theta_1$ with indices greater than some bound. Thus, if $\mathbf v$ is a sufficiently distant vertex of one of the Klein polygons, then $\mathbf v=(\pm q_n,\pm p_n)$, where $p_n/q_n$ is a convergent of $\theta_1$ or $\theta_2$. Also, $\operatorname{ang}(\mathbf v)=a_{n+1}$ is a partial quotient of the same number. We obtain
$$
\begin{equation*}
\frac{\log a_{n+1}}{\log q_n}\asymp\frac{\log(\operatorname{ang}(\mathbf v))}{\log|\mathbf v|} \quad\text{as } n\to\infty.
\end{equation*}
\notag
$$
Therefore, if we denote by $\mathcal{V}=\mathcal{V}(\pm\mathcal{K}_1,\pm\mathcal{K}_2)$ the set of vertices of the Klein polygons $\mathcal{K}_1$, $-\mathcal{K}_1$, $\mathcal{K}_2$, and $-\mathcal{K}_2$, then equality (1.3) (more precisely, its analogue for two numbers) can be written as follows:
$$
\begin{equation}
\max\bigl(\mu(\theta_1),\mu(\theta_2)\bigr) =2+\displaystyle\limsup_{\substack{ \mathbf v\in \mathcal{V},\ |\mathbf v|>1 \\ |\mathbf v|\to\infty }} \frac{\log(\operatorname{ang}(\mathbf v))}{\log|\mathbf v|}.
\end{equation}
\tag{1.7}
$$
Here and throughout, $|\cdot|$ denotes the sup-norm. The Diophantine exponent of a latticeFor lattices in an arbitrary dimension the notion of Diophantine exponent is defined. For every $\mathbf x=(x_1,\dots,x_n)\in\mathbb{R}^n$ set Definition 1. Let $\Lambda$ be an arbitrary lattice of complete rank in $\mathbb{R}^n$. By the Diophantine exponent of this lattice we mean the quantity It follows from Minkowski’s convex body theorem that $\omega(\Lambda)\geqslant0$ (also see [8] and [9]). We turn back to the continued fractions of $\theta_1$ and $\theta_2$ and the Klein polygons $\mathcal{K}_1$ and $\mathcal{K}_2$. As above, let $\mathcal{V}$ be the set of vertices of $\pm\mathcal{K}_1$ and $\pm\mathcal{K}_2$. Let the matrix $A$ be given by (1.4). Consider the lattice
$$
\begin{equation*}
\Lambda=A\mathbb{Z}^2=\bigl\{(L_1(\mathbf z), L_2(\mathbf z))\bigm|\mathbf z\in\mathbb{Z}^2 \bigr\}.
\end{equation*}
\notag
$$
For every $\gamma\geqslant0$ the set
$$
\begin{equation}
\bigl\{\mathbf x\in\mathbb{R}^2\bigm|\Pi(\mathbf x)\geqslant|\mathbf x|^{-\gamma} \bigr\}
\end{equation}
\tag{1.9}
$$
is the union of four convex sets. If $\omega(\Lambda)=\infty$, then it does not contain $A(\mathcal{K}_1\,{\cup}\,\mathcal{K}_2)$ for any $\gamma\in\mathbb{R}$. If $\omega(\Lambda)\,{<}\,\infty$, then
$$
\begin{equation*}
\omega(\Lambda)=\inf\bigl\{\gamma\in\mathbb{R}\bigm| A(\mathcal{K}_1\cup\mathcal{K}_2)\text{ is contained in the set (1.9)} \bigr\}.
\end{equation*}
\notag
$$
Since every polygon $\mathcal{K}_i$ is a convex hull of its vertices, if $\Pi(\mathbf x)\geqslant|\mathbf x|^{-\gamma}$ for all points in $A(\mathcal{V})$, then this condition also holds for all nonzero points of $\Lambda$. Therefore, in the definition of $\omega(\Lambda)$ (see (1.8)) the set of all points of $\Lambda$ can be replaced by $A(\mathcal{V})$:
$$
\begin{equation}
\omega(\Lambda)=\sup\bigl\{\gamma\in\mathbb{R}\bigm|\exists\,\infty\ \mathbf x\in A(\mathcal{V})\colon \Pi(\mathbf x)\leqslant|\mathbf x|^{-\gamma} \bigr\}.
\end{equation}
\tag{1.10}
$$
Further, if $\mathbf v=(q,p)$ is a point in $\mathcal{V}$, then for sufficiently large $|q|$ and $\mathbf w=A\mathbf v=\bigl(L_1(\mathbf v),L_2(\mathbf v)\bigr)$ we have
$$
\begin{equation*}
\begin{aligned} \, \Pi(\mathbf w)^2 &=|q\theta_1-p|\cdot|q\theta_2-p| \\ &\asymp|q|\min\bigl(|q\theta_1-p|,|q\theta_2-p|\bigr) =q^2\min\biggl(\biggl|\theta_1-\frac pq\biggr|,\biggl|\theta_2-\frac pq\biggr|\biggr) \end{aligned}
\end{equation*}
\notag
$$
and Therefore, for sufficiently large $|q|$,
$$
\begin{equation}
\Pi(\mathbf w)\asymp|\mathbf w|^{-\gamma} \quad\Longleftrightarrow\quad \min\biggl(\biggl|\theta_1-\frac pq\biggr|, \biggl|\theta_2-\frac pq\biggr|\biggr)\asymp q^{-2-2\gamma}.
\end{equation}
\tag{1.11}
$$
The equivalence (1.11), in view of (1.10) and (1.2), results in the equality
$$
\begin{equation}
\max\bigl(\mu(\theta_1),\mu(\theta_2)\bigr)=2+2\omega(\Lambda).
\end{equation}
\tag{1.12}
$$
Combining (1.7) and (1.12) we obtain a geometric interpretation of (1.3):
$$
\begin{equation*}
\omega(\Lambda)= \frac12\limsup_{\substack{ \mathbf v\in \mathcal{V},\ |\mathbf v|>1 \\ |\mathbf v|\to\infty }}\frac{\log(\operatorname{ang}(\mathbf v))}{\log|\mathbf v|}.
\end{equation*}
\notag
$$
Correspondingly, the aim of this paper is to generalize the inequality
$$
\begin{equation*}
\omega(\Lambda)\leqslant\frac12\limsup_{\substack{ \mathbf v\in \mathcal{V},\ |\mathbf v|>1 \\ |\mathbf v|\to\infty }} \frac{\log(\operatorname{ang}(\mathbf v))}{\log|\mathbf v|}
\end{equation*}
\notag
$$
to the three-dimensional case.
§ 2. Statement of the main resultLet $L_1$, $L_2$ and $L_3$ be linearly independent linear forms in three variables. The zero subspaces of $L_1$, $L_2$ and $L_3$ partition $\mathbb{R}^3$ into eight simplicial (closed) cones. In each of them consider the convex hull of the nonzero integer points. We denote these hulls by $\mathcal{K}_i$, $i=1,\dots,8$ (in any order). They are called the Klein polyhedra (corresponding to the linear forms $L_1$, $L_2$ and $L_3$). If the forms $L_1$, $L_2$ and $L_3$ do not vanish at nonzero integer points, then, as shown in [10], all the $\mathcal{K}_i$ are generalized polyhedra (that is, their intersections with compact polyhedra are polyhedra too). In particular, in this case a finite number of edges meet at every vertex of the Klein polyhedron. Therefore, for the edge star of a vertex of the Klein polyhedron we can define a multidimensional analogue of integer angle. Generally speaking, such a multidimensional analogue can be defined in various ways. We use the concept of the determinant of an edge star, which was introduced in [11] and [12]. Let $\mathcal{K}$ be one of the $\mathcal{K}_i$, and let $\mathbf v$ be a vertex of $\mathcal{K}$. Let $\mathbf r_1,\dots,\mathbf r_k$ be primitive integer vectors parallel to the edges of $\mathcal{K}$ incident to $\mathbf v$. We denote the edge star of the vertex $\mathbf v$ by $\operatorname{St}_{\mathbf v}$. Definition 2. By the determinant of the edge star $\operatorname{St}_{\mathbf v}$ we mean the quantity Since the vectors $\mathbf r_1,\dots,\mathbf r_k$ have integer coordinates and all terms in (2.1) are nonzero, $\det\operatorname{St}_{\mathbf v}$ is a positive integer. It is not difficult to prove that $\det\operatorname{St}_{\mathbf v}$ defined in this way is equal to the volume of the Minkowski sum of the line segments $[\mathbf 0,\mathbf r_1],\dots,[\mathbf 0,\mathbf r_k]$. It is also clear that in the two-dimensional case the analogous quantity coincides exactly with the integer angle at a vertex (see (1.6)). The following assertion is the main result of our paper. Theorem. Consider a lattice Assume that the forms $L_1$, $L_2$ and $L_3$ do not vanish at nonzero integer points. Let $\mathcal{K}_1,\dots,\mathcal{K}_8$ be the Klein polyhedra corresponding to these linear forms, and let $\mathcal{V}$ be the set of all their vertices. Then § 3. Proof of the theoremBy analogy with the two-dimensional case denote the matrix of size $3\times3$ whose rows consist of the coefficients of the linear forms $L_1$, $L_2$ and $L_3$ by $A$. Without loss of generality we can assume that $\det A=1$, since both sides of (2.2) do not change under a homothety with centre at the origin. Indeed, if the inequality $\Pi(\mathbf x)\leqslant|\mathbf x|^{-\gamma}$ has infinitely many solutions $\mathbf x\,{\in}\,\Lambda$, then for any fixed positive $\lambda$ and $\varepsilon$ the inequality $\Pi(\lambda\mathbf x)\leqslant|\lambda\mathbf x|^{-\gamma+\varepsilon}$ also has infinitely many solutions $\mathbf x\in\Lambda$, because $\Pi(\lambda\mathbf x)=\lambda\Pi(\mathbf x)$. Similarly, if the inequality $\det\operatorname{St}_{\mathbf v}\geqslant|\mathbf v|^{-\gamma}$ has infinitely many solutions $\mathbf v\in\mathcal{V}$, then for any fixed positive $\lambda$ and $\varepsilon$ the inequality $\lambda^3\det\operatorname{St}_{\mathbf v}\geqslant|\lambda\mathbf v|^{-\gamma-\varepsilon}$ also has infinitely many solutions. Thus, we assume that $\det A=1$. Then
$$
\begin{equation*}
\Lambda=A\mathbb{Z}^3\quad\text{and} \quad\det\Lambda=1.
\end{equation*}
\notag
$$
We write Then $\mathcal{K}_1',\dots,\mathcal{K}_8'$ are the convex hulls of the nonzero points of the lattice $\Lambda$ lying in each of the eight orthants. They are called the Klein polyhedra of the lattice $\Lambda$. Consider $\mathcal{W}=A(\mathcal{V})$, the set of all of their vertices. As in the two-dimensional case, we use the fact that points in $\mathcal{W}$ are local minima of the function $\Pi(\mathbf x)$ (considered on the union of the polyhedra $\mathcal{K}_1',\dots, \mathcal{K}_8'$). Then
$$
\begin{equation}
\omega(\Lambda) =\limsup_{\substack{\mathbf x\in\Lambda,\ |\mathbf x|>1 \\ |\mathbf x|\to\infty}}\frac{\log\bigl(\Pi(\mathbf x)^{-1}\bigr)}{\log|\mathbf x|} =\limsup_{\substack{\mathbf w\in\mathcal{W},\ |\mathbf w|>1 \\ |\mathbf w|\to\infty}}\frac{\log\bigl(\Pi(\mathbf w)^{-1}\bigr)}{\log|\mathbf w|}.
\end{equation}
\tag{3.1}
$$
For every point $\mathbf w\in\mathcal{W}$ we denote by $\operatorname{St}_{\mathbf w}$ the edge star of $\mathbf w$, that is, the union of the edges of $\mathcal{K}_i'$ incident to this point. The determinant $\det\operatorname{St}_{\mathbf w}$ of this edge star is defined similarly to (2.1). To every point $\mathbf w\in\mathcal{W}$ there corresponds $\mathbf v\in\mathcal{V}$ such that $\mathbf w=A\mathbf v$. Under this correspondence $\operatorname{St}_{\mathbf w}=A(\operatorname{St}_{\mathbf v})$, that is,
$$
\begin{equation*}
\det\operatorname{St}_{\mathbf w}=\det\operatorname{St}_{\mathbf v}\quad\text{and} \quad |\mathbf w|\asymp|\mathbf v| \quad\text{as } |\mathbf v|\to\infty.
\end{equation*}
\notag
$$
We note that $\det\operatorname{St}_{\mathbf w}$, as well as $\det\operatorname{St}_{\mathbf v}$, is a positive integer. Thus, in view of (3.1), to prove (2.2), it suffices to show that
$$
\begin{equation}
\frac{\log\bigl(\Pi(\mathbf w)^{-1}\bigr)}{\log|\mathbf w|} \leqslant\frac23\cdot\frac{\log(\det\operatorname{St}_{\mathbf w})}{\log|\mathbf w|}+o(1)
\end{equation}
\tag{3.2}
$$
for $\mathbf w\in\mathcal{W}$ as $|\mathbf w|\to\infty$. In fact, we claim slightly more. The following local assertion is true. Proof. We set $\varepsilon$ to be equal to an arbitrary sufficiently small positive number, for example, If $\Pi(\mathbf w)\geqslant\varepsilon$, then for $c_1=\varepsilon^{3/2}$ we have $\det\operatorname{St}_{\mathbf w}\geqslant1\geqslant c_1\Pi(\mathbf w)^{-3/2}$. We also assume that
We split our argument into several steps. Step 1: a hyperbolic rotation. Let $\mathbf w=(w_1,w_2,w_3)$. Consider the diagonal operator
$$
\begin{equation*}
D=\operatorname{diag}\biggl(\frac{\Pi(\mathbf w)}{w_1}, \frac{\Pi(\mathbf w)}{w_2},\frac{\Pi(\mathbf w)}{w_3}\biggr).
\end{equation*}
\notag
$$
We write
$$
\begin{equation*}
\mathbf u=D\mathbf w, \qquad \Lambda_{\mathbf w}=D\Lambda\quad\text{and} \quad \mathcal{K}_{\mathbf w}=D\mathcal{K}_i',
\end{equation*}
\notag
$$
where $\mathcal{K}_i'$ is the polyhedron of which $\mathbf w$ is a vertex. Then $\mathcal{K}_{\mathbf w}$ is the Klein polyhedron of the lattice $\Lambda_{\mathbf w}$ corresponding to the positive orthant (the cone of points with nonnegative coordinates), and we have
$$
\begin{equation*}
\det\Lambda_{\mathbf w}=\det\Lambda=1 \quad\text{and} \quad \det\operatorname{St}_{\mathbf u}=\det\operatorname{St}_{\mathbf w}
\end{equation*}
\notag
$$
since $|\det D|=1$. Here all coordinates of the point $\mathbf u$ are equal to $\Pi(\mathbf w)$. In particular,
$$
\begin{equation*}
|\mathbf u|=\Pi(\mathbf u)=\Pi(\mathbf w)<\varepsilon.
\end{equation*}
\notag
$$
Step 2: short and long vectors. Let us show that $\mathbf u$ is the shortest vector of the lattice $\Lambda_{\mathbf w}$ in the sup-norm. We consider the cube $\mathcal{B}$, which is a closed ball of radius $|\mathbf u|$ in the sup-norm. It has the centre at the origin $\mathbf 0$ and edge length $2|\mathbf u|$. Since $\mathbf u$ is a vertex of $\mathcal{K}_{\mathbf w}$, there is a support hyperplane to $\mathcal{K}_{\mathbf w}$ intersecting $\mathcal{K}_{\mathbf w}$ at $\mathbf u$. This hyperplane divides the cube $\mathbf u+\mathcal{B}$ into two parts, which are symmetric with respect to $\mathbf u$. In the part containing $\mathbf 0$ there are no points of $\Lambda_{\mathbf w}$ distinct from $\mathbf 0$ and $\mathbf u$. Therefore, there are no lattice points distinct from $\mathbf 0$, $\mathbf u$ and $2\mathbf u$ in the whole cube $\mathbf u+\mathcal{B}$. Hence $\mathcal{B}$ does not contain points of $\Lambda_{\mathbf w}$ other than $\mathbf 0$, $\mathbf u$ and $-\mathbf u$ either. Thus, $\mathbf u$ is indeed the shortest lattice vector of $\Lambda_{\mathbf w}$ in the sup-norm. Let $\mathbf p_1,\dots,\mathbf p_k$ be the primitive vectors of $\Lambda_{\mathbf w}$ that are parallel to the edges of $\mathcal{K}_{\mathbf w}$ incident to $\mathbf u$. These vectors are the images of the vectors $\mathbf r_1,\dots,\mathbf r_k$ in Definition 2 under the action of the operator $DA$, and
$$
\begin{equation*}
\det\operatorname{St}_{\mathbf u} =\sum_{1\leqslant i_1<i_2<i_3\leqslant k}|{\det(\mathbf p_{i_1},\mathbf p_{i_2},\mathbf p_{i_3})}|.
\end{equation*}
\notag
$$
We order $\mathbf p_1,\dots,\mathbf p_k$ in the ascending order of their Euclidean norms, for which we will use the notation $|\,{\cdot}\,|_2$. For any $\mathbf p_i$ such that $\mathbf u$, $\mathbf p_1$ and $\mathbf p_i$ are linearly independent we have $|\mathbf u|_2\,|\mathbf p_i|_2^2\,{\geqslant}\,|\mathbf u|_2\,|\mathbf p_1|_2|\mathbf p_i|_2\,{\geqslant}\, |{\det(\mathbf u,\mathbf p_1,\mathbf p_i)}|\,{\geqslant}\det\Lambda_{\mathbf w}\,{=}\,1$, that is, $|\mathbf p_i|_2\geqslant|\mathbf u|_2^{-1/2}$, from which we obtain
$$
\begin{equation}
|\mathbf p_i|\geqslant \frac{|\mathbf p_i|_2}{\sqrt3} \geqslant\frac{|\mathbf u|_2^{-1/2}}{\sqrt3}\geqslant\frac{(|\mathbf u|\sqrt3\,)^{-1/2}}{\sqrt3} =\frac{|\mathbf u|}{3^{3/4}|\mathbf u|^{3/2}} >\frac{|\mathbf u|}{3^{3/4}\varepsilon^{3/2}} >\frac{4|\mathbf u|}{\varepsilon}.
\end{equation}
\tag{3.4}
$$
Note for the future that for this $\mathbf p_i$ we have
$$
\begin{equation}
\mathbf u+\frac{4|\mathbf u|}{|\mathbf p_i|}\mathbf p_i\in(1-\varepsilon)\mathbf u +\mathcal{O}.
\end{equation}
\tag{3.5}
$$
Indeed, since $\mathbf p_i\in-\mathbf u+\mathcal{O}$, it follows that
$$
\begin{equation*}
\frac{4|\mathbf u|}{|\mathbf p_i|}\mathbf p_i \in-\frac{4|\mathbf u|}{|\mathbf p_i|}\mathbf u+\mathcal{O}\subset -\varepsilon\mathbf u+\mathcal{O},
\end{equation*}
\notag
$$
which is equivalent to (3.5).
Step 3: a Klein polygon. Consider the two-dimensional subspace $\pi$ spanned by $\mathbf u$ and $\mathbf p_1$ and the lattice $\Gamma=\pi\cap\Lambda_{\mathbf w}$. Since the triangle with vertices $\mathbf 0$, $\mathbf u$ and $\mathbf u+\mathbf p_1$ does not contain points of $\Lambda_{\mathbf w}$ other than its vertices, the vectors $\mathbf u$ and $\mathbf p_1$ form a basis of the lattice $\Gamma$. We denote by $\mathcal{C}$ the intersection of $\pi$ with the positive orthant and consider $\mathcal{K}_\pi=\operatorname{conv}(\mathcal{C}\cap\Gamma\setminus\{\mathbf 0\})$, the Klein polygon of the lattice $\Gamma$ corresponding to the cone $\mathcal{C}$. Then $\mathbf p_1$ is a primitive vector of the lattice $\Gamma$ parallel to an edge of $\mathcal{K}_\pi$ incident to $\mathbf u$. We denote the primitive vector of $\Gamma$ parallel to the other edge by $\mathbf p_1'$. We note that, generally speaking, $\mathbf p_1'$ is not necessarily one of the vectors $\mathbf p_2,\dots,\mathbf p_k$. The vectors $\mathbf u$ and $\mathbf p_1'$ also form a basis of $\Gamma$. Therefore, the points $\mathbf p_1$ and $\mathbf p_1'$ lie at the same distance from the line spanned by $\mathbf u$ (but on different sides of it). Hence
$$
\begin{equation}
\mathbf p_1+\mathbf p_1'=t\mathbf u, \qquad t\in\mathbb{N}
\end{equation}
\tag{3.6}
$$
(cf. Figure 2). Thus, if for $\mathbf x=(x_1,x_2,x_3)$ we set then $M(\mathbf p_1+\mathbf p_1')\geqslant M(\mathbf u)=3|\mathbf u|$ by (3.6), and we see that either $M(\mathbf u+\mathbf p_1)\geqslant\frac92|\mathbf u|$ or $M(\mathbf u+\mathbf p_1')\geqslant\frac92|\mathbf u|$.
Taking (3.4) into account we conclude that the inequality holds for at most one of the points $\mathbf u+\mathbf p_i$, $i=1,\dots,k$, and this is either $\mathbf u+\mathbf p_1$ or ${\mathbf u+\mathbf p_2}$.Step 4: a cross-section and a convex hull. We write
$$
\begin{equation*}
M_1(\mathbf x)=x_1+x_2+\frac94x_3, \qquad M_2(\mathbf x)=x_1+\frac94x_2+x_3\quad\text{and} \quad M_3(\mathbf x)=\frac94x_1+x_2+x_3.
\end{equation*}
\notag
$$
For every point $\mathbf x=(x_1,x_2,x_3)$ satisfying
$$
\begin{equation}
x_1,x_2,x_3\geqslant0\quad\text{and} \quad M_1(\mathbf x),M_2(\mathbf x),M_3(\mathbf x)\leqslant\frac92|\mathbf u|,
\end{equation}
\tag{3.8}
$$
we have $x_1,x_2,x_3\leqslant2|\mathbf u|$. However, as noted above, the cube $\mathbf u+\mathcal{B}$ contains no lattice points distinct from $\mathbf 0$, $\mathbf u$ and $2\mathbf u$. Therefore, none of the points $\mathbf u+\mathbf p_i$, $i=1,\dots,k$, satisfies (3.8). Thus, if there is a point satisfying (3.7) among these points, then it also satisfies at least one of the inequalities
$$
\begin{equation*}
M_1(\mathbf x)>\frac92|\mathbf u|, \qquad M_2(\mathbf x)>\frac92|\mathbf u|\quad\text{and} \quad M_3(\mathbf x)>\frac92|\mathbf u|.
\end{equation*}
\notag
$$
Without loss of generality we can assume that the first inequality holds for this point. Then for every $i=1,\dots,k$ we also have
$$
\begin{equation*}
M_1(\mathbf u+\mathbf p_i)\geqslant\frac92|\mathbf u|,
\end{equation*}
\notag
$$
since for nonnegative $x_1$, $x_2$ and $x_3$ we have $M_1(\mathbf x)\geqslant M(\mathbf x)$.
We denote by $\mathcal{O}$ the positive orthant, that is, the cone consisting of the points with nonnegative coordinates. Consider the plane
$$
\begin{equation*}
\mathcal{M}=\biggl\{\mathbf x\in\mathbb{R}^3\biggm|M_1(\mathbf x)=\frac92|\mathbf u| \biggr\}
\end{equation*}
\notag
$$
and the cross-sections
$$
\begin{equation*}
\mathcal{S}=\mathcal{M}\cap\mathcal{O}, \qquad \mathcal{T}=\mathcal{M}\cap(\mathbf u+\mathcal{O})\quad\text{and} \quad \mathcal{P}=\mathcal{M}\cap\mathcal{K}_{\mathbf w}.
\end{equation*}
\notag
$$
Since $\mathbf u+\mathcal{O}\subset\mathcal{K}_{\mathbf w}\subset\mathcal{O}$, it follows that $\mathcal{T}\subset\mathcal{P}\subset\mathcal{S}$. Here $\mathcal{T}$ is nonempty because $M_1(\mathbf u)=\frac{17}4|\mathbf u|<\frac92|\mathbf u|$. Moreover, $\mathcal{T}$ is a triangle with vertices at the points
$$
\begin{equation*}
\biggl(\frac54|\mathbf u|,|\mathbf u|,|\mathbf u|\biggr), \qquad \biggl(|\mathbf u|,\frac54|\mathbf u|,|\mathbf u|\biggr), \qquad \biggl(|\mathbf u|,|\mathbf u|,\frac{10}9|\mathbf u|\biggr).
\end{equation*}
\notag
$$
The set $\mathcal{S}$ is a triangle with vertices at
$$
\begin{equation*}
\biggl(\frac92|\mathbf u|,0,0\biggr), \qquad \biggl(0,\frac92|\mathbf u|,0\biggr)\quad\text{and} \quad \Bigl(0,0,2|\mathbf u|\Bigr).
\end{equation*}
\notag
$$
The set $\mathcal{P}$ is the polygon with vertices at the points of intersection of the line segments $[\mathbf u,\mathbf u+\mathbf p_i]$, $i=1,\dots,k$, with the plane $\mathcal{M}$. Here (3.5) implies that the vertices of $\mathcal{P}$ corresponding to the $\mathbf p_i$ not lying in the subspace $\pi$ (that is, linearly independent of $\mathbf u$ and $\mathbf p_1$) are contained in the set
$$
\begin{equation*}
\mathcal{T}_\varepsilon=\mathcal{M}\cap\bigl((1-\varepsilon)\mathbf u+\mathcal{O}\bigr).
\end{equation*}
\notag
$$
The vertices corresponding to the points $\mathbf p_i$ lying in $\pi$ are contained in the line segment Thus,
$$
\begin{equation}
\mathcal{T}\subset\mathcal{P} \subset\operatorname{conv}(\mathcal{T}_\varepsilon\cup\mathcal{I}).
\end{equation}
\tag{3.9}
$$
Step 5: an analysis of the cross-section. We denote by $\mathbf a$ the point of intersection of the plane $\mathcal{M}$ with the line spanned by the vector $\mathbf u$. Then all coordinates of $\mathbf a$ are equal, this point belongs to the segment $\mathcal{I}$, and it is the centre of homothety of the triangles $\mathcal{T}$, $\mathcal{T}_\varepsilon$ and $\mathcal{S}$. We distinguish three zones in $\mathcal{S}$ (Figure 3):
$$
\begin{equation*}
\begin{aligned} \, \Omega_1 & =\biggl\{\mathbf x=(x_1,x_2,x_3)\in\mathcal{S}\setminus\mathcal{T} \biggm|x_2<\frac{x_1+x_3}2,\ x_3\leqslant\frac{x_1+x_2}2 \biggr\}, \\ \Omega_2 & =\biggl\{\mathbf x=(x_1,x_2,x_3)\in\mathcal{S}\setminus\mathcal{T} \biggm|x_3<\frac{x_1+x_2}2,\ x_1\leqslant\frac{x_2+x_3}2 \biggr\}, \\ \Omega_3 & =\biggl\{\mathbf x=(x_1,x_2,x_3)\in\mathcal{S}\setminus\mathcal{T} \biggm|x_1<\frac{x_2+x_3}2,\ x_2\leqslant\frac{x_1+x_3}2 \biggr\}. \end{aligned}
\end{equation*}
\notag
$$
The segment $\mathcal{I}$ has a nonempty intersection with exactly one of the sets $\Omega_1$, $\Omega_2$ and $\Omega_3$. Figure 3 shows the case when it intersects $\Omega_3$. Let us examine this case in detail. The other two cases are treated in a similar way.
We denote by $\mathbf b$ the endpoint of $\mathcal{I}$ lying outside $\Omega_3$. Let us draw the straight lines from $\mathbf b$ through the bottom vertices of the triangle $\mathcal{T}$. These lines cut off triangles $\Delta_1$ and $\Delta_2$ from $\mathcal{T}_\varepsilon$. Each of these must contain at least one vertex of the polygon $\mathcal{P}$, because these vertices lie in $\mathcal{T}_\varepsilon\cup\mathcal{I}$. Let $\Delta_1$ contain the vertex corresponding to $\mathbf p_{i_1}$ and $\Delta_2$ contain the vertex corresponding to $\mathbf p_{i_2}$. We claim that $\Omega_3$ also contains a vertex of the polygon $\mathcal{P}$. If this is not the case, then in (3.9) the segment $\mathcal{I}$ can be replaced by $[\mathbf a,\mathbf b]$:
$$
\begin{equation*}
\mathcal{T}\subset\mathcal{P} \subset\operatorname{conv}(\mathcal{T}_\varepsilon\cup[\mathbf a,\mathbf b]).
\end{equation*}
\notag
$$
Then it follows from the assumption that $\Omega_3$ does not contain vertices of $\mathcal{P}$ that the top vertex of $\mathcal{T}$ does not belong to $\mathcal{P}$, which contradicts the inclusion $\mathcal{T}\subset\mathcal{P}$. Therefore, $\Omega_3$ contains a vertex of the polygon $\mathcal{P}$. Assume that it corresponds to $\mathbf p_{i_3}$.
Step 6: a straightforward estimate. Since $\varepsilon$ is small, the vectors $\mathbf p_{i_1}$, $\mathbf p_{i_2}$ and $\mathbf p_{i_3}$ found above satisfy the following two conditions: 1) any two of them are linearly independent of $\mathbf u$; 2) $|{\det(\mathbf p_{i_1},\mathbf p_{i_2}, \mathbf p_{i_3})}|\asymp|\mathbf p_{i_1}|\cdot|\mathbf p_{i_2}|\cdot|\mathbf p_{i_3}|$. Here the constants implied by the symbol $\asymp$ are absolute, because
$$
\begin{equation*}
|\mathbf p_{i_1}|\cdot|\mathbf p_{i_2}|\cdot|\mathbf p_{i_3}|\geqslant |{\det(\mathbf p_{i_1},\mathbf p_{i_2},\mathbf p_{i_3})}|= |\mathbf p_{i_1}\wedge\mathbf p_{i_2}\wedge\mathbf p_{i_3}|\gg_\varepsilon |\mathbf p_{i_1}|\cdot|\mathbf p_{i_2}|\cdot|\mathbf p_{i_3}|.
\end{equation*}
\notag
$$
In view of these properties we obtain
$$
\begin{equation*}
\begin{aligned} \, \Pi(\mathbf u)^3\det(\mathbf p_{i_1},\mathbf p_{i_2},\mathbf p_{i_3})^2 &\asymp |\mathbf u|^3|\mathbf p_{i_1}|^2|\mathbf p_{i_2}|^2|\mathbf p_{i_2}|^2 \\ &\geqslant|{\det(\mathbf u,\mathbf p_{i_1},\mathbf p_{i_2})\cdot \det(\mathbf u,\mathbf p_{i_2},\mathbf p_{i_3})\cdot \det(\mathbf u,\mathbf p_{i_3},\mathbf p_{i_1})}|\geqslant1. \end{aligned}
\end{equation*}
\notag
$$
Thus, there is a positive constant $c_2$ such that
$$
\begin{equation*}
\det\operatorname{St}_{\mathbf w} =\det\operatorname{St}_{\mathbf u}\geqslant|{\det(\mathbf p_{i_1}, \mathbf p_{i_2},\mathbf p_{i_3})}|\geqslant c_2\Pi(\mathbf u)^{-3/2}=c_2\Pi(\mathbf w)^{-3/2}.
\end{equation*}
\notag
$$
Taking into account that we have obtained this inequality under assumption (3.3), it remains to set $c=\min(c_1,c_2)$. The lemma implies (3.2) and therefore (2.2). This completes the proof of the theorem. § 4. On the converse of the lemmaLet us describe an example showing that a verbatim conversion of our lemma is impossible. We take an arbitrary integer $n\geqslant3$ and set
$$
\begin{equation*}
\mathbf e_1=(n,1,0), \qquad \mathbf e_2=(0,n,1), \qquad \mathbf e_3=(1,0,n)\quad\text{and} \quad \mathbf v=(1,1,1).
\end{equation*}
\notag
$$
Consider the cone $\mathcal{C}$ with vertex at the origin and edges spanned by the vectors $\mathbf e_1$, $\mathbf e_2$ and $\mathbf e_3$. The Klein polyhedron $\mathcal{K}=\operatorname{conv}(\mathcal{C}\,{\cap}\,\mathbb{Z}^3\setminus\{\mathbf 0\})$ has three bounded faces and three unbounded ones. The bounded faces are the triangles $\mathbf v\mathbf e_1\mathbf e_2$, $\mathbf v\mathbf e_2\mathbf e_3$ and $\mathbf v\mathbf e_3\mathbf e_1$. Indeed, the tetrahedron $\mathbf 0\mathbf v\mathbf e_1\mathbf e_2$ is empty (which means that it does not contain integer points other than vertices) because it is squeezed between the planes $x_3=0$ and $x_3=1$, while the vectors $\mathbf e_1\,{-}\,\mathbf 0$ and $\mathbf e_2\,{-}\,\mathbf v$ are primitive. In a similar way the tetrahedra $\mathbf 0\mathbf v\mathbf e_2\mathbf e_3$ and $\mathbf 0\mathbf v\mathbf e_3\mathbf e_1$ are also empty. Thus, three faces meet at the vertex $\mathbf v$, namely, $\mathbf v\mathbf e_1\mathbf e_2$, $\mathbf v\mathbf e_2\mathbf e_3$ and $\mathbf v\mathbf e_3\mathbf e_1$. The determinant of the edge star $\operatorname{St}_{\mathbf v}$ is
$$
\begin{equation*}
|{\det(\mathbf e_1-\mathbf v,\mathbf e_2-\mathbf v,\mathbf e_3-\mathbf v)}|= \left| \begin{matrix} n-1 & -1 & 0 \\ 0 & n-1 & -1 \\ -1 & 0 & n-1 \end{matrix} \right|= (n-1)^3-1.
\end{equation*}
\notag
$$
We define linear forms $L_1$, $L_2$ and $L_3$ by the rows of the matrix
$$
\begin{equation*}
A=(n^3+1)^{-2/3} \begin{pmatrix} n^2 & 1 & -n \\ -n & n^2 & 1 \\ 1 & -n & n^2 \end{pmatrix}.
\end{equation*}
\notag
$$
Then $L_i(\mathbf e_j)=0$ for $i\neq j$ and $\det A=1$. We write
$$
\begin{equation*}
\Lambda=A\mathbb{Z}^3, \qquad\mathcal{K}'=A\mathcal{K}\quad\text{and} \quad\mathbf w=A\mathbf v.
\end{equation*}
\notag
$$
Then $\mathbf w$ is a vertex of the polyhedron $\mathcal{K}'$,
$$
\begin{equation}
\Pi(\mathbf w)^3=L_1(\mathbf v)L_2(\mathbf v)L_3(\mathbf v) =\frac{(n^2-n+1)^3}{(n^3+1)^2}\asymp1 \quad\text{as } n\to\infty,
\end{equation}
\tag{4.1}
$$
$$
\begin{equation}
\det\operatorname{St}_{\mathbf w}=\det\operatorname{St}_{\mathbf v}=(n-1)^3-1\asymp n^3 \quad\text{as } n\to\infty.
\end{equation}
\tag{4.2}
$$
Of course, $\Lambda$ can be ‘corrected’ slightly so that it has no nonzero points on the coordinate planes. For example, for a small positive $\varepsilon$, instead of $A$ we can take the matrix
$$
\begin{equation*}
A_\varepsilon=A+\begin{pmatrix} 0 & 0 & \varepsilon \\ \varepsilon & 0 & 0 \\ 0 & \varepsilon & 0 \end{pmatrix}.
\end{equation*}
\notag
$$
Then for sufficiently small $\varepsilon$ the point $\mathbf v$ remains a vertex of the corresponding Klein polyhedron, the triangles $\mathbf v\mathbf e_1\mathbf e_2$, $\mathbf v\mathbf e_2\mathbf e_3$ and $\mathbf v\mathbf e_3\mathbf e_1$ remain its faces, and thus the edge star $\operatorname{St}_{\mathbf v}$ is preserved. At the same time, in view of (4.1) and (4.2), $\Pi(\mathbf w)$ is bounded away from zero, while $\det\operatorname{St}_{\mathbf w}$ can be arbitrarily large. Thus, a verbatim conversion of our lemma is impossible. However, it is an appropriate observation that, as proved in [12], $\Pi(\mathbf x)$ is bounded away from from zero at the nonzero points of the lattice $\Lambda$ if and only if the determinants of edge stars and faces of the Klein polyhedron of $\Lambda$ corresponding to the positive orthant are bounded. To prove this one must consider vertices of both $\mathcal{K}$ itself and the adjacent polyhedra corresponding to other orthants. Thus, we arrive at two questions, the answers to which would make it possible to reverse (2.2) (possibly changing the constant). Question 1. Is it true that if $\det\operatorname{St}_{\mathbf v}$ is large, then ‘near’ $\mathbf v$ there is a vertex $\mathbf v'$ of one of the eight Klein polyhedra of the lattice $\Lambda$ such that $\Pi(\mathbf v')$ is small? Question 2. Is there an integer affine invariant of an edge star different from $\det\operatorname{St}_{\mathbf v}$ for which a ‘convertible’ analogue of our lemma holds? AcknowledgementsThe authors would like to thank an anonymous referee for reading the paper carefully and making numerous useful comments, which improved the presentation. 
Citation in format AMSBIB
\Bibitem{BigGer23} |
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