# Lattice Reduced and Complete Convex Bodies

Giulia Codenotti and Ansgar Freyer

(Date: June 2023)

###### Abstract.

Following the established terms in Euclidean geometry, we define reduced and
complete convex bodies $C\subset{\mathbb{R}}^{d}$ when the width and diameter
are measured with respect to a lattice $\Lambda\subset{\mathbb{R}}^{d}$. We
prove that the reduced (complete) bodies in this theory are polytopes with a
bounded number of vertices (facets) and that their width (diameter) is
realized by at least $O(\log d)$ independent directions. Moreover, we present
various methods of constructing such polytopes and relate lattice reduced
simplices to the study of the flatness constant in integer programming.

##  1\. Introduction

A _convex body_ $C\subset{\mathbb{R}}^{d}$ is a convex compact set with non-
empty interior. The _Euclidean diameter_ $\operatorname{diam}_{\mathbb{R}}(C)$
is defined as the length of a longest line segment contained in $C$ and the
_Euclidean width_ $\operatorname{width}_{\mathbb{R}}(C)$ is defined as the
minimum distance between two parallel supporting hyperplanes of $C$, i.e.,

(1.1)
$\operatorname{width}_{\mathbb{R}}(C)=\min_{u\in\mathbb{S}^{d-1}}\max_{a,b\in
C}u\cdot(a-b),$

where $\mathbb{S}^{d-1}$ denotes the Euclidean unit sphere centered at the
origin in ${\mathbb{R}}^{d}$.

In this setting, $C$ is called _complete_ if there is no convex body
$C^{\prime}\supsetneq C$ with
$\operatorname{diam}_{\mathbb{R}}(C^{\prime})=\operatorname{diam}_{\mathbb{R}}(C)$.
Similarly, $C$ is called _reduced_ if there is no convex body
$C^{\prime}\subsetneq C$ with
$\operatorname{width}_{\mathbb{R}}(C^{\prime})=\operatorname{width}_{\mathbb{R}}(C)$.
It is a well-known fact that $C$ is complete if and only if $C$ is of constant
width, i.e., the map

$\mathbb{S}^{d-1}\to{\mathbb{R}},\quad u\mapsto\max_{a,b\in C}u\cdot(a-b)$

is constant. This implies that a complete convex body is in particular
reduced. We refer to the survey [34] for a modern overview on reduced convex
bodies in the Euclidean setting and to the book [10] for extensive background
on complete convex bodies. The study of complete and reduced convex bodies is
interesting in particular because these bodies occur as extremal constructions
in geometric problems. For instance, the Pál-Kakeya problem asks for the
maximal Euclidean width among convex bodies $C\subset{\mathbb{R}}^{d}$ with
$\operatorname{vol}(C)\leq 1$. Since the volume functional is strictly
monotonous, it is clear that the bodies attaining the maximum width in this
problem are reduced. It was shown by Pál [43] in 1921 that for $d=2$, the
maximum is achieved by the regular triangle. In higher dimensions, however,
the problem is still open for more than a century. The fact that the natural
generalization of the regular triangle, the regular simplex, is not reduced
for $d\geq 3$ [40] shows that higher dimensions require a different approach.

In recent years there has been a lot of research on reduced and complete
convex bodies also in non-Euclidean settings, such as spherical geometry and
hyperbolic geometry (see, e.g., [32, 31, 7, 36, 4]), or Minkowski spaces [33,
41, 6]. The purpose of this article is to introduce and study a discrete
interpretation of these notions, in which the diameter and width are measured
with respect to a lattice $\Lambda\subset{\mathbb{R}}^{d}$.

A _lattice_ $\Lambda\subset{\mathbb{R}}^{d}$ is a discrete subgroup of
${\mathbb{R}}^{d}$, which we shall assume to be full-dimensional. The _dual
lattice_ of $\Lambda$ is then defined as
$\Lambda^{\star}=\\{y\in{\mathbb{R}}^{d}:x\cdot y\in{\mathbb{Z}},~{}\forall
x\in\Lambda\\}$. The _lattice width_ of a convex body $C$ with respect to the
lattice $\Lambda$ is defined in analogy to (1.1) as

(1.2)
$\operatorname{width}_{\Lambda}(C)=\min_{y\in\Lambda^{\star}\setminus\\{0\\}}\max_{a,b\in
C}y\cdot(a-b).$

The lattice width is an important parameter in integer linear programming;
Khinchin’s flatness theorem, which states that the lattice width of a convex
body $C$ without interior lattice points is bounded by a constant independent
of $C$ [26], enabled Lenstra to solve integer linear programs in fixed
dimensions in polynomial time [37]. This discovery initiated the search for
the best-possible upper bound $\mathrm{Flt}(d)$ such that any convex body
without interior lattice points in $\Lambda$ satisfies
$\operatorname{width}_{\Lambda}(C)\leq\mathrm{Flt}(d)$. It is conjectured that
$\mathrm{Flt}(d)$ is linear in $d$, but to this day, the best bounds known are
$\mathrm{Flt}(d)\geq 2d-o(d)$ [42] and $\mathrm{Flt}(d)\leq c\cdot
d\log(d)^{8}$, where $c>0$ is a universal constant. The latter goes back on an
argument by Banaszczyk using methods from functional analysis for the
symmetric case [3], and a recent result of Reis and Rothvoss [45] who apply
the reverse Minkowski theorem by Regev and Stephens-Davidowitz to obtain the
inequality for general convex bodies [44].

In order to define the lattice diameter of a convex body $C$, we measure the
length of a segment with respect to a lattice $\Lambda$. To this end, we call
a segment $I=[a,b]\subset{\mathbb{R}}^{d}$ a _lattice segment_ with respect to
$\Lambda$, if $b-a\in\operatorname{span}\\{x\\}$, for some $x\in\Lambda$. If
$I$ is a lattice segment, then there exists a unique vector $v_{I}\in\Lambda$,
which generates $\Lambda\cap\operatorname{span}\\{b-a\\}$ and is a positive
multiple of $b-a$. We call $v_{I}$ the primitive direction of $I$ in
$\Lambda$. The lattice length of $I$ is now defined as

$\operatorname{Vol}_{1}(I)=\frac{|b-a|}{|v_{I}|}.$

Here, $|v|$ denotes the standard Euclidean norm of a vector
$v\in{\mathbb{R}}^{d}$. So in other words, $\operatorname{Vol}_{1}(I)$ is the
Euclidean length of the segment $I$ normalized by the length of its primitive
direction. The _lattice diameter_ of a convex body $C$ with respect to
$\Lambda$ can now be defined as

(1.3) $\operatorname{diam}_{\Lambda}(C)=\max\\{\operatorname{Vol}_{1}(I)\colon
I\subset C\text{ lattice segment}\\}.$

The quantity $\operatorname{diam}_{\Lambda}(C)$ has been used in [11] to study
generalizations of the flatness constant in the plane and in [24] to classify
4-simplices without interior lattice points. Moreover, the lattice diameter
$\operatorname{diam}_{\Lambda}(C)$ also occurs in [25], where it is expressed
in terms of the first minimum of $C-C$ (cf. Section 2).

With the notions of lattice width and diameter at hand, the definitions of
lattice reduced and lattice complete bodies are analogous to the Euclidean
case.

######  Definition 1.1.

A convex body $C\subset{\mathbb{R}}^{d}$ is called _lattice reduced_ with
respect to $\Lambda$ if there is no $C^{\prime}\subsetneq C$ such that
$\operatorname{width}_{\Lambda}(C^{\prime})=\operatorname{width}_{\Lambda}(C)$.
The body $C$ is called _lattice complete_ with respect to $\Lambda$, if there
is no $C^{\prime}\supsetneq C$ with
$\operatorname{diam}_{\Lambda}(C^{\prime})=\operatorname{diam}_{\Lambda}(C)$.

In Figure 1, these definitions are illustrated in the plane. Note that the
square in Figure 1 (a) shows that, in contrast to the Euclidean setting,
completeness does not imply reducedness.

(a) The square is not reduced with respect to ${\mathbb{Z}}^{2}$, since it
contains the dashed diamond, which has the same lattice width. The diamond is
lattice reduced.

(b) The diamond is not complete with respect to ${\mathbb{Z}}^{d}$, since it
is contained in the dashed square, which has the same lattice diameter. The
square is lattice complete.

(c) The “skew square” is simultaneously reduced and complete.

Figure 1. Three examples of reduced or complete quadrilaterals in the plane
(cf. Example 3.10)

An example of a complete convex body with respect to a general $d$-lattice
$\Lambda\subset{\mathbb{R}}^{d}$ is the Voronoi cell

(1.4) $V_{\Lambda}=\\{x\in{\mathbb{R}}^{d}:|x|\leq|x-a|,~{}\forall
a\in\Lambda\\}.$

Exploiting duality we show that the polar body $(V_{\Lambda})^{\star}$ of the
Voronoi cell is reduced with respect to $\Lambda^{\star}$ (cf. Proposition
3.11). Here, the _polar body_ of a convex body $C\subset{\mathbb{R}}^{d}$ is
defined as $C^{\star}=\\{y\in{\mathbb{R}}^{d}\colon x\cdot y\leq 1,~{}\forall
x\in C\\}$.

In the non-symmetric case, the simplex
$S_{d}=\operatorname{conv}\\{-\mathbbm{1},e_{1},...,e_{d}\\}$, where
$\mathbbm{1}$ denotes the all-1-vector, serves as an example of a convex body
which is both reduced and complete (cf. Proposition 3.12). More examples and
constructions of complete and reduced convex bodies are given in Section 3.

Our main results in this paper concern the structure of lattice reduced and
complete convex bodies.

######  Theorem 1.2.

Let $\Lambda\subset{\mathbb{R}}^{d}$ be a $d$-dimensional lattice and let
$C\subset{\mathbb{R}}^{d}$ be a convex body.

  1. (1)

If $C$ is lattice reduced with respect to $\Lambda$, then $C$ is a polytope
with at most $2^{d+1}-2$ vertices.

  2. (2)

If $C$ is lattice complete with respect to $\Lambda$, then $C$ is a polytope
with at most $2^{d+1}-2$ facets.

In both statements, the bound $2^{d+1}-2$ cannot be improved.

This results contrasts the Euclidean situation: A Euclidean complete convex
body is always strictly convex and thus never a polytope [27, Proposition
2.1]. Euclidean reduced polygons exist for $d=2$, for instance in the form of
regular $k$-gons, where $k$ is odd. For $d>2$, constructing a Euclidean
reduced polytope is a non-trivial problem. It was solved affirmatively by
González Merino et al. in [15] for $d=3$, but it is open for $d>3$.

Moreover, we give a sharp lower bound on the dimension of the directions that
realize the width (diameter) of a lattice reduced (complete) convex body.

######  Theorem 1.3.

Let $\Lambda\subset{\mathbb{R}}^{d}$ be a $d$-dimensional lattice and let
$C\subset{\mathbb{R}}^{d}$ be a convex body.

  1. (1)

If $C$ is lattice reduced with respect to $\Lambda$, then the vectors
$y\in\Lambda^{\star}$ that realize the minimum
$\operatorname{width}_{\Lambda}(C)$ span a linear space of dimension at least
$c\cdot\log(d)$, where $c>0$ is a universal constant.

  2. (2)

If $C$ is lattice complete with respect to $\Lambda$, then the vectors
$v\in\Lambda$ for which there exists a lattice segment parallel to $v$
achieving $\operatorname{diam}_{\Lambda}(C)$ span a linear space of dimension
at least $c\cdot\log(d)$, where $c>0$ is a universal constant.

In both statements, the order of magnitude $\log(d)$ is best-possible.

Theorems 1.2 and 1.3 suggest that there might be a duality between reduced and
complete convex bodies in the lattice setting. It is true that an origin-
symmetric convex body $C$ is lattice reduced with respect to some lattice
$\Lambda$ if and only if its polar body $C^{\star}$ is lattice complete with
respect to $\Lambda^{\star}$ (cf. Theorem 3.8). However, there is most likely
no such correspondence in the general case (cf. Remarks 3.13, 4.4 and 4.11).

As a curious consequence of the complete classification of lattice reduced and
complete triangles in the plane up to unimodular equivalence (cf. Section
4.2), we obtain the following theorem.

######  Theorem 1.4.

Let $T\subset{\mathbb{R}}^{2}$ be a triangle and let
$\Lambda\subset{\mathbb{R}}^{2}$ be a 2-dimensional lattice. If $T$ is lattice
complete with respect to $\Lambda$, then $T$ is lattice reduced with respect
to $\Lambda$. The converse does not hold.

This result echoes the fact that the class of Euclidean complete bodies is
properly contained in the class of Euclidean reduced bodies. It further
supports the idea that there is no duality between arbitrary lattice reduced
and lattice complete convex bodies. However, Theorem 1.4 neither extends to
higher-dimensional simplices (cf. Remark 4.11), nor to general 2-dimensional
convex bodies (cf. Example 3.10).

Finally, we apply our theory to the study of the flatness constant
$\mathrm{Flt}(d)$, which may also be regarded as a discrete analogue of the
Pál-Kakeya problem in Euclidean geometry; The side condition
$\operatorname{vol}(C)\leq 1$ is replaced by
$|\operatorname{int}(C)\cap\Lambda|=0$ ($\operatorname{int}C$ denotes the
interior of $C$). In the appendix to this paper, we provide an argument that
shows that the supremum $\mathrm{Flt}(d)$ is in fact achieved by a convex
body, i.e., there exists $C\subset{\mathbb{R}}^{d}$ such that
$\operatorname{width}_{\Lambda}(C)=\mathrm{Flt}(d)$ for any $d$-dimensional
lattice $\Lambda\subset{\mathbb{R}}^{d}$.

Since the interior lattice point enumerator
$C\mapsto|\operatorname{int}(C)\cap\Lambda|$ is not strictly monotonous, it is
not immediate that the flatness constant is achieved by lattice reduced convex
bodies. To prove that this does hold, we first have to show that any convex
body contains a lattice reduced convex body of the same lattice width (cf.
Theorem 3.20). Even so, it is possible that the flatness constant is also
realized by non-reduced convex bodies. For simplices, we can exclude this
possibility. We call a convex body $C$ _hollow_ with respect to a lattice
$\Lambda$ if $C$ contains no lattice points of $\Lambda$ in its interior.

######  Theorem 1.5.

Let $\Lambda\subset{\mathbb{R}}^{d}$ be a $d$-dimensional lattice and let $S$
be a local maximum of $\operatorname{width}_{\Lambda}$ on the class of hollow
$d$-simplices. Then, $S$ is lattice reduced with respect to $\Lambda$.

Here, the term “local” refers to the Hausdorff-distance between compact
subsets of ${\mathbb{R}}^{d}$.

In addition to the search for the realizers of the flatness constant, there is
also a more homogeneous lattice variant of the Pál-Kakeya problem, in which
lattice reduced convex bodies could play an important role. Here one replaces
the condition $\operatorname{vol}(K)\leq 1$ by
$\operatorname{Vol}_{\Lambda}(K):=\tfrac{\operatorname{vol}(K)}{\det\Lambda}\leq
1$, where $\operatorname{Vol}_{\Lambda}$ is the _lattice-normalized volume_.
So the task is to find the maximum lattice width
$\operatorname{width}_{\Lambda}(K)$ among convex bodies $K$ with
$\operatorname{Vol}_{\Lambda}(K)\leq 1$. Makai conjectured that the maximum is
$\tfrac{2^{d}d!}{d+1}$ and that it is attained only by simplices [39]. For
instance, the simplex
$S_{d}=\operatorname{conv}\\{-\mathbbm{1},e_{1},\dots,e_{d}\\}$ has width
exactly the conjectured maximum, if it is dilated so that its volume is 1. We
refer to [16, 20] and the references therein for more information on Makai’s
conjecture. Since the lattice normalized volume is a strictly monotonous
functional, all the extremal cases in Makai’s conjecture are necessarily
reduced.

The paper is organized as follows: In Section 2, we introduce the necessary
tools from convex geometry and the geometry of numbers that are needed for our
investigations. In Section 3, we study lattice reduced and complete convex
bodies in general. This will lead to the proofs of Theorems 1.2 and 1.3, as
well as to several examples of reduced and complete convex bodies. Moreover,
we discuss the special role of origin-symmetric convex bodies and study the
behaviour of reduced and complete convex bodies with respect to standard
operations. We finally show that any convex body $C\subset{\mathbb{R}}^{d}$
contains (is contained in) a lattice reduced (complete) body of the same
lattice width (diameter). This fact follows rather easily in the Euclidean
setting with the help of Zorn’s lemma, but in the discrete setting, more care
is needed. In Section 4, we turn our eye on reduced and complete simplices,
proving Theorems 1.4 and 1.5. We then construct a family of complete
tetrahedra in Section 4 that serves as an interesting source of
(counter-)examples. To conclude, in Section 5, we consider lattice analogs of
various theorems and conjectures on reduced and complete bodies in the
Euclidean theory. The appendix contains a proof for the existence of hollow
convex bodies with maximal lattice width.

For the sake of brevity, in the remainder of the paper, the terms “width”,
“diameter”, “reduced” and “complete” are to be understood with respect to a
given lattice instead of the Euclidean interpretation.

###### Contents

  1. 1 Introduction
  2. 2 Preliminaries
    1. 2.1 Convex Geometry
    2. 2.2 Geometry of Numbers
  3. 3 Results and constructions for general convex bodies
    1. 3.1 Reduced Bodies
    2. 3.2 Complete Bodies
    3. 3.3 The Origin-Symmetric Case
    4. 3.4 Constructions and Examples
    5. 3.5 Proofs of Theorems 1.2 and 1.3
    6. 3.6 Reductions and Completions
  4. 4 Simplices and Triangles
    1. 4.1 Simplices
    2. 4.2 Classification of Triangles in the Plane
    3. 4.3 A Family of Complete Tetrahedra
  5. 5 Comparison to the Euclidean Case
    1. 5.1 Relations between diameter and width
    2. 5.2 Volumetric inequalities
    3. 5.3 Bodies that are both reduced and complete
  6. A Existence of realizers of $\mathrm{Flt}(d)$

##  2\. Preliminaries

For a vector $x\in{\mathbb{R}}^{d}\setminus\\{0\\}$ we denote by $x^{\perp}$
the hyperplane orthogonal to $x$ passing through the origin and by $|x|$ its
Euclidean length. For two vectors $x,y\in{\mathbb{R}}^{d}$, we let $x\cdot y$
be the standard inner product. The Minkowski sum of two non-empty sets
$A,B\subset{\mathbb{R}}^{d}$ is denoted by $A+B=\\{a+b:a\in A,b\in B\\}$ and
for a number $\lambda\in{\mathbb{R}}$, we write $\lambda A=\\{\lambda a:a\in
A\\}$ and $-A=(-1)A$. The convex hull of a set $A\subset{\mathbb{R}}^{d}$ is
denoted by $\operatorname{conv}A$. If $A=\\{x,y\\}$, we write
$[x,y]=\operatorname{conv}A$ for the segment joining $x$ and $y$. The affine
hull of $A$ is denoted by $\operatorname{aff}A$. The interior, the closure and
the boundary of $A$ with respect to the standard topology in
${\mathbb{R}}^{d}$ are denoted by $\operatorname{int}A$, $\operatorname{cl}A$
and $\operatorname{bd}A$ respectively and we write, for instance,
$\operatorname{relint}A$ for the interior of $A$ with respect to the trace
topology on $\operatorname{aff}A$. Moreover, for $n\in{\mathbb{N}}$, we write
$[n]=\\{1,...,n\\}$. For $i\in[d]$, let $e_{i}$ denote the $i$th standard
basis vector of ${\mathbb{Z}}^{d}$ and we let $C_{d}=[-1,1]^{d}$ be the
standard origin-symmetric cube. Finally, we write
$\mathbbm{1}=e_{1}+\cdots+e_{d}$ for the all-1-vector.

###  2.1. Convex Geometry

Here we recall the basic terms and facts from convex geometry that are
necessary for the understanding of the following sections. For a thorough
read, we refer to the books [17, 46], as well as [47] for polytope theory.

A _convex body_ $C\subset{\mathbb{R}}^{d}$ means a convex compact set with
non-empty interior. A _face_ $F\subset C$ is a convex subset of $C$ such that
for any $x,y\in C$ with $[x,y]\cap F\neq\emptyset$, we have $x,y\in F$. A face
is called _proper_ , if neither $F$, nor $C\setminus F$ are empty. An _extreme
point_ is a 0-dimensional face, the set of extreme points of $C$ is denoted by
$\operatorname{ext}(C)$.

A hyperplane $H\subset{\mathbb{R}}^{d}$ is said to _support_ a convex body
$C$, if $C\cap H\neq\emptyset$ and $C$ is contained in one of the two closed
halfspaces determined by $H$. If $H$ is a supporting hyperplane of $C$, the
set $F=C\cap H$ is called an _exposed face_ of $C$. Any exposed face is indeed
a face, but the converse is not true. A 0-dimensional exposed face is called
_exposed point_ and the set of exposed points is denoted by
$\operatorname{expo}(C)$. We have
$\operatorname{cl}(\operatorname{expo}(C))=\operatorname{ext}C$.

For a vector $v\in{\mathbb{R}}^{d}$, the _support function_ of $C$ is defined
as $h(C;v)=\max_{v\in C}x\cdot v.$ In fact, if $v\neq 0$, the hyperplane
$\\{x\in{\mathbb{R}}^{d}\colon x\cdot v=h(C;v)\\}$ is a supporting hyperplane
of $C$.

For a convex set $K\subset{\mathbb{R}}^{d}$, its _polar set_ is defined as

$K^{\star}=\\{y\in{\mathbb{R}}^{d}\colon x\cdot y\leq 1,~{}\forall x\in K\\}.$

There is a variety of nice properties for this polarity operation [46, Section
1.6.1]. Here we want to highlight that $K^{\star}$ is bounded if and only if
$0\in\operatorname{int}K$. In particular, for a convex body $C$ which contains
the origin as an interior point, $C^{\star}$ is a convex body with
$0\in\operatorname{int}C^{\star}$.

We measure the distance between two convex bodies with respect to the
_Hausdorff distance_ , i.e., we have

$d(C,C^{\prime})=\min\\{r\geq 0\colon C\subset C^{\prime}+rB\text{ and
}C^{\prime}\subset C+rB\\},$

for any $C,C^{\prime}\subset{\mathbb{R}}^{d}$, where
$B\subset{\mathbb{R}}^{d}$ denotes the Euclidean unit ball. Indeed, $d$ turns
the set of convex bodies in ${\mathbb{R}}^{d}$ into a metric space.

A _polytope_ $P\subset{\mathbb{R}}^{d}$ is the convex hull of a finite set
$A\subset{\mathbb{R}}^{d}$, or, equivalently, a bounded intersection of closed
half-spaces. The extreme points of a polytope are commonly called _vertices_
and its $(\dim(P)-1)$-dimensional faces are called _facets_. For a face
$F\subset P$, the _normal cone_ of $F$ is defined as

$\operatorname{ncone}(F;P)=\\{y\in{\mathbb{R}}^{d}\colon\forall x\in
F,x^{\prime}\in P.~{}x\cdot y\geq x^{\prime}\cdot y\\}.$

The _normal fan_ $\mathrm{N}(P)$ of $P$ is the collection of all normal cones
of $P$. $\mathrm{N}(P)$ is a complete fan, i.e., the union of all normal cones
of $P$ is ${\mathbb{R}}^{d}$, any face of a normal cone is again a normal cone
and any two normal cones intersect in a normal cone.

If $0\in\operatorname{int}P$, then $P^{\star}$ is again a polytope and there
is a 1-1 correspondence between $k$-faces of $P$ and $(n-1-k)$-faces of
$P^{\star}$; For a $k$-face $F\subset P$, the set

$F^{\diamond}=\\{y\in P^{\star}\colon x\cdot y=1,~{}\forall x\in F\\}$

is a $(n-k-1)$-face of $P^{\star}$, the so-called _polar face_ of $F$. Another
fan associated to a polytope $P$ with $0\in\operatorname{int}P$ is the face
fan. For a face $F\subset P$ one considers the positive hull

$\operatorname{cone}(F)=\\{\lambda x\colon x\in F,~{}\lambda\geq 0\\}.$

The collection of these cones is the _face fan_ $\mathrm{F}(P)$. The face fan
is connected to the normal fan via polarity; we have
$\operatorname{ncone}(F;P)=\operatorname{cone}(F^{\diamond})$ and so
$\mathrm{N}(P)=\mathrm{F}(P^{\star})$.

###  2.2. Geometry of Numbers

In this subsection we summarize tools we need from the geometry of numbers
that help describe the interplay of convex bodies and lattices in
${\mathbb{R}}^{d}$. Further background on the geometry of numbers can be found
in the books [18, 9].

A _lattice_ $\Lambda\subset{\mathbb{R}}^{d}$ is by definition a discrete
subgroup of ${\mathbb{R}}^{d}$. Equivalently, we may express $\Lambda$ as the
integer span of linearly independent vectors
$b_{1},\dots,b_{k}\in{\mathbb{R}}^{d}$, i.e.,

$\Lambda=\operatorname{span}_{{\mathbb{Z}}}\\{b_{1},\dots,b_{k}\\}=\Big{\\{}\sum_{i=0}^{k}z_{i}b_{i}\colon
z_{1},\dots,z_{k}\in{\mathbb{Z}}\Big{\\}}.$

The vectors $b_{1},\dots,b_{k}$ are then called a _basis_ of $\Lambda$. Two
bases $B=(b_{1},\dots,b_{k})$ and
$B^{\prime}=(b_{1}^{\prime},\dots,b_{k}^{\prime})$ of $\Lambda$ differ only by
a _unimodular transformation_ , i.e., $B^{\prime}=BU$, for a matrix
$U\in{\mathbb{Z}}^{k,k}$ with $\det U=1$. The group of unimodular
transformations on ${\mathbb{R}}^{d}$ is denoted by
$\mathrm{GL}_{d}({\mathbb{Z}})$. The _determinant_ of a lattice $\Lambda$ can
now be defined as $\det\Lambda=\sqrt{\det(B^{T}B)}$, where $B$ is a basis of
$\Lambda$.

Subgroups of $\Lambda$ are called _sublattices_. The _dual lattice_ of
$\Lambda$ is defined as

$\Lambda^{\star}=\\{y\in\operatorname{span}\Lambda\colon x\cdot
y\in{\mathbb{Z}},~{}\forall x\in\Lambda\\}.$

A vector in $x\in\Lambda$ is called _primitive_ , if $x\neq 0$ and the segment
$[0,x]$ contains no lattice points of $\Lambda$ in its relative interior. More
generally, a sublattice $\Lambda^{\prime}\subset\Lambda$ is called primitive,
if
$\Lambda^{\prime}=\operatorname{span}_{\mathbb{R}}(\Lambda^{\prime})\cap\Lambda$.
The primitive vectors $y\in\Lambda^{\star}$ are in 2-1 correspondence to the
primitive $(\dim\Lambda-1)$-dimensional sublattices of $\Lambda$ via the map
$y\mapsto\Lambda\cap y^{\perp}$. The vectors $y$ and $-y$ both lead to the
same sublattice.

Three standard examples of lattices that will be of interest in the paper are
the integer lattice ${\mathbb{Z}}^{d}$ and the root lattice
$A_{d}=\\{x\in{\mathbb{Z}}^{d+1}\colon x\cdot\mathbbm{1}=0\\}$, as well as its
dual $A_{d}^{\star}$. We refer to the book [13] for a detailed treatment of
special lattices.

When working with width and diameter, it will be convenient to write
$\operatorname{width}(C;v)=h(C-C;v)$ for the width of $C$ in direction
$v\in{\mathbb{R}}^{d}\setminus\\{0\\}$. This way, we have
$\operatorname{width}_{\Lambda}(C)=\min_{y}\operatorname{width}(C;y)$, where
$y$ ranges over $\Lambda^{\star}\setminus\\{0\\}$. Similarly, we write for
$v\in{\mathbb{R}}^{d}\setminus\\{0\\}$

$\operatorname{diam}(C;v)=\max\Big{\\{}\frac{\operatorname{vol}_{1}(I)}{|v|}\colon
I\subset C\text{ segment parallel to }v\Big{\\}}$

and find $\operatorname{diam}_{\Lambda}(C)=\max_{v}\operatorname{diam}(C;v)$,
where $v$ ranges over $\Lambda\setminus\\{0\\}$.

An important parameter in the geometry of numbers is the first successive
minimum of an origin-symmetric convex body $C\subset{\mathbb{R}}^{d}$ with
respect to a lattice. It is given by

$\lambda_{1}(C;\Lambda)=\min\big{\\{}\lambda\geq 0\colon(\lambda
C)\cap(\Lambda\setminus\\{0\\})\neq\emptyset\big{\\}}=\max\big{\\{}\lambda\geq
0\colon\operatorname{int}(\lambda C)\cap\Lambda=\\{0\\}\big{\\}}.$

It is well-known that the width and diameter of a convex body $C$ can be
expressed in terms of the first minimum as stated in the following lemma (see,
e.g., [25, Section 1] and [24, Section 2.1]). Its proof is a good exercise for
anyone that wants to familiarize themselves with the definitions.

######  Lemma 2.1.

Let $C\subset{\mathbb{R}}^{d}$ be a convex body and let
$\Lambda\subset{\mathbb{R}}^{d}$ be a $d$-lattice. Then we have

  1. (1)

$\operatorname{width}_{\Lambda}(C)=\lambda_{1}((C-C)^{*};\Lambda)$ and

  2. (2)

$\operatorname{diam}_{\Lambda}(C)=\lambda_{1}(C-C;\Lambda)^{-1}$.

Moreover, the directions where the width (resp. diameter) of the convex body
is acheived correspond to the non-trivial lattice points in
$\lambda_{1}((C-C)^{*};\Lambda)(C-C)^{\star}$ (resp.
$\lambda_{1}(C-C;\Lambda)(C-C)$).

We call these directions where the width (diameter) is achieved, i.e. lattice
points in $\lambda_{1}((C-C)^{*};\Lambda)(C-C)^{\star}$ (resp.
$\lambda_{1}(C-C;\Lambda)(C-C)$) simply _width (diameter) directions_. Since
the first successive minimum is continuous, it follows immediately that the
width and diameter directions of a convex body are stable with respect to
small perturbations of $C$. More precisely, the following Lemma holds.

######  Lemma 2.2.

Let $C\subset{\mathbb{R}}^{d}$ be a convex body. Then there exists
$\varepsilon>0$ such that for any convex body
$C^{\prime}\subset{\mathbb{R}}^{d}$ with $d(C,C^{\prime})<\varepsilon$ the
width (diameter) directions of $C^{\prime}$ are contained in the width
(diameter) directions of $C$.

We finish the section by turning to the main focus of the paper, the new
definitions of reduced and complete convex bodies, and collecting some first
useful properties of these.

######  Lemma 2.3.

Let $C\subset{\mathbb{R}}^{d}$ be a convex body and let
$\Lambda\subset{\mathbb{R}}^{d}$ be a $d$-lattice.

  1. (1)

If $C$ is reduced (complete) with respect to $\Lambda$, then $\lambda C+t$ is
reduced (complete) with respect to $\Lambda$, for any
$\lambda\in{\mathbb{R}}\setminus\\{0\\}$ and $t\in{\mathbb{R}}^{d}$.

  2. (2)

If $C$ is reduced (complete) with respect to $\Lambda$, then $AC$ is reduced
(complete) with respect to $A\Lambda$, for any
$A\in\mathrm{GL}_{d}({\mathbb{R}})$.

  3. (3)

If $C$ is reduced (complete) with respect to ${\mathbb{Z}}^{d}$, then $UC$ is
reduced (complete) with respect to ${\mathbb{Z}}^{d}$, for any
$U\in\mathrm{GL}_{d}({\mathbb{Z}})$.

All statements follow directly from the properties of the successive minima.
The lemma shows that there is no restriction in considering only the integer
lattice $\Lambda={\mathbb{Z}}^{d}$. However, when constructing a reduced or
complete convex body it is oftentimes more convenient to modify the lattice
rather than the convex body, which is why we formulate our results for
arbitrary lattices.

##  3\. Results and constructions for general convex bodies

###  3.1. Reduced Bodies

First we show that a convex body is reduced if and only if every exposed point
is selected by a width direction. An analog of this statement is also known in
the Euclidean case [34, Theorem 1].

######  Proposition 3.1.

Let $C\subset{\mathbb{R}}^{d}$ be a convex body and let
$\Lambda\subset{\mathbb{R}}^{d}$ be a $d$-lattice. The following are
equivalent:

  1. (1)

$C$ is reduced with respect to $\Lambda$.

  2. (2)

For every exposed point $p\in C$, there exists a width direction
$y\in\Lambda^{\star}$ such that $y\cdot p>y\cdot x$, for all $x\in
C\setminus\\{p\\}$.

###### Proof.

If $C$ is not reduced, there exists a convex body $C^{\prime}\subsetneq C$
such that
$\operatorname{width}_{\Lambda}(C^{\prime})=\operatorname{width}_{\Lambda}(C)$.
Since $C^{\prime}$ is closed and strictly contained in $C$, there exists a
point $p\in\operatorname{expo}(C)$ which is not in $C^{\prime}$. Thus if
condition (2) would hold, then there would exist a width direction
$y\in\Lambda^{\star}$ such that
$\operatorname{width}(C^{\prime};y)<\operatorname{width}(C;y)=\operatorname{width}_{\Lambda}(C)$,
a contradiction since $C^{\prime}$ and $C$ have the same width.

For the converse, let $p\in\operatorname{expo}(C)$ be an exposed point and let
$H=\\{x:c\cdot x=b\\}$, $c\in{\mathbb{R}}^{d}\setminus\\{0\\}$, be the
hyperplane selecting $p$, that is, $H\cap C=\\{p\\}$ and $C\subset
H^{-}=\\{x:c\cdot x\leq b\\}$. We define $C_{\varepsilon}=C\cap
H_{\varepsilon}^{-},$ where $H_{\varepsilon}^{-}=\\{x:c\cdot x\leq
b-\varepsilon\\}$. Moreover, let
$\\{y_{1},\dots,y_{k}\\}=\operatorname{width}_{\Lambda}(C)(C-C)^{\star}\cap\Lambda^{\star}\setminus\\{0\\}$
be the width directions of $C$, which are finite (cf. Lemma 2.1).

Suppose now that condition (2) does not hold for $p$, that is, for each of the
width directions $y_{i}$, there exists some $q_{i}\in C\setminus\\{p\\}$ such
that $y_{i}\cdot q_{i}\geq y_{i}\cdot p$. We want to show that for
$\varepsilon>0$ small enough we have
$\operatorname{width}_{\Lambda}(C_{\varepsilon})=\operatorname{width}_{\Lambda}(C)$.
First observe that, if $\varepsilon$ is chosen sufficiently small,
$H_{\varepsilon}^{-}$ will still contain $q_{1},\dots,q_{k}$, which implies
$q_{i}\in C_{\varepsilon}$ and thus
$\operatorname{width}(C_{\varepsilon};y_{i})=\operatorname{width}(C;y_{i})$.

Now we recall from Lemma 2.2 that for $\varepsilon>0$ small enough, the width
directions of $C_{\varepsilon}$ are among the width direction
$\\{y_{1},\dots,y_{k}\\}$ of $C$. In conclusion, for $\varepsilon>0$ small
enough it holds that
$\operatorname{width}_{\Lambda}(C_{\varepsilon})=\operatorname{width}_{\Lambda}(C)$,
and thus $C$ is not lattice reduced. ∎

The next theorem entails the first statements of Theorems 1.2 and 1.3.

######  Theorem 3.2.

Let $C\subset{\mathbb{R}}^{d}$ be a convex body, reduced with respect to a
$d$-lattice $\Lambda$. Let $k=\dim\\{y\in\Lambda^{\star}\colon y\text{ is a
width direction of }C\\}$. Then, $C$ is a polytope with at most $2^{k+1}-2$
vertices.

Before we come to the proof, we need the following lemma.

######  Lemma 3.3.

Let $C\subset{\mathbb{R}}^{d}$ be a convex body with
$\operatorname{int}C\cap\Lambda=\\{0\\}$. Let
$A\subseteq\operatorname{bd}(C)\cap\Lambda$ such that no two points in $A$ lie
in a common facet of $C$. Then, $|A|\leq 2^{k+1}-2$, where $k$ is the
dimension of $\operatorname{span}A$.

###### Proof.

First we assume that $k=d$. Suppose there are points $a,b\in A$ with $a\neq b$
and $a\equiv b\mod 2\Lambda$. Then $(a+b)/2$ is a lattice point in
$\operatorname{int}C$, because otherwise, $a$ and $b$ had to be contained in a
common face of $C$. Consequently, $-a=b$. Thus, there cannot be more than 2
points of $A$ within one coset of $\Lambda/2\Lambda$. Further, there is no
point $a\in A\cap 2\Lambda$, since then, $a/2$ would be a non-zero point in
$\operatorname{int}C$. As there are $2^{d}$ cosets, the statement follows.

If $k<d$, we consider the convex body $C^{\prime}=C\cap L$, where
$L=\operatorname{span}A$, and the lattice $\Lambda^{\prime}=\Lambda\cap L$.
Then, $C^{\prime}$ and $\Lambda^{\prime}$ are $k$-dimensional in $L$. So the
statement follows by applying the full-dimensional case in $L$. ∎

###### Proof of Theorem 3.2.

Since there are only finitely many width directions of $C$, it follows from
Proposition 3.1 that $C$ has only finitely many exposed points and is, thus, a
polytope.

Let $v_{1},\dots,v_{n}\in C$ be the vertices of $C$ and let
$y_{i}\in\Lambda^{\star}$ be the width direction selecting $v_{i}$ uniquely
guaranteed by Proposition 3.1, $1\leq i\leq n$. Then, $y_{i}$ is contained in
the interior of the normal cone
$\operatorname{ncone}(v_{i};C)\in\mathrm{N}(C)$. Since the normal fan of $C-C$
is the common refinement of $\mathrm{N}(C)$ and $\mathrm{N}(-C)$ [47,
Proposition 7.12], no two distinct vectors $y_{i}$ and $y_{j}$ fall in the
same normal cone of $C-C$. By duality, this implies that no two of the vectors
$y_{i}$ and $y_{j}$ are contained in a common face cone in
$\mathrm{F}((C-C)^{\star})$. In particular, it follows that there is no
dilation of $(C-C)^{\star}$ such that $y_{i}$ and $y_{j}$ lie in a common
facet of this dilation.

Now consider the polytopes
$P=\lambda_{1}((C-C)^{\star};\Lambda)(C-C)^{\star}$. From the definition of
the successive minima we have that $P$ contains the origin as its unique
interior lattice point. Since the $y_{i}$ are width directions of $C$, they
are contained in $\operatorname{bd}(P)$. Since no two of the vectors $y_{i}$
and $y_{j}$ are contained in a common facet of $P$, we can apply Lemma 3.3 to
$P$ and $A=\\{y_{1},\dots,y_{n}\\}$ to obtain $n\leq 2^{k+1}-2$ as desired. ∎

###  3.2. Complete Bodies

Here we prove a similar statement to Theorem 3.2 for complete convex bodies.

######  Theorem 3.4.

Let $C\subset{\mathbb{R}}^{d}$ be a convex body, complete with respect to a
$d$-lattice $\Lambda$. Let $k=\dim\\{y\in\Lambda^{\star}\colon y\text{ is a
diameter direction of }C\\}$. Then, $C$ is a polytope with at most $2^{k+1}-2$
facets.

Although Theorem 3.4 is exactly the dual statement to Theorem 3.2 in the
origin-symmetric case, we are not aware of an argument that allows to derive
one of the theorems from the other.

Again, we need some preliminary results in order to prove Theorem 3.4.

######  Lemma 3.5.

Let $C$ be a convex body and consider $[a,b],[c,d]\subset C$ such that $b-a$
and $d-c$ are in a common face of $C-C$. Then there exist opposite exposed
faces $F_{1},F_{2}\subset C$ of $C$ with $a,b\in F_{1}$ and $c,d\in F_{2}$.
Here, $F_{1}$ and $F_{2}$ are called _opposite_ , if there exist parallel
supporting hyperplanes $H_{1},H_{2}\subset{\mathbb{R}}^{d}$ such that
$F_{i}=C\cap H_{i}$, $i\in\\{1,2\\}$.

###### Proof.

Since any face of $C-C$ is contained in an exposed face, there exists a vector
$v\in{\mathbb{R}}^{n}\setminus\\{0\\}$ such that

$v\cdot(b-a)=v\cdot(d-c)=h(C-C;v)=h(C;v)+h(-C;v).$

It follows that

$v\cdot b=v\cdot d=h(C;v)\text{\quad and \quad}v\cdot(-c)=v\cdot(-a)=h(-C;v),$

which proves the lemma. ∎

From this it follows immediately that for a fixed direction $v$, the segments
of maximal length in a convex body that are parallel to $v$ pass between two
fixed opposite facets.

######  Lemma 3.6.

Let $v\subset{\mathbb{R}}^{d}\setminus\\{0\\}$ and let $S,S^{\prime}\subset C$
be two segments parallel to $v$ of maximal length in $C$. Then there exists a
pair of opposite exposed faces $F,G$ such that $S$ and $S^{\prime}$ have an
endpoint in $F$ and $G$ each.

We call _diameter segments_ those segments contained in the convex body which
have maximum lattice length. The following proposition gives a helpful
characterization of complete polytopes.

######  Proposition 3.7.

A $d$-polytope $P\subset{\mathbb{R}}^{d}$ is complete with respect to a
$d$-lattice $\Lambda\subset{\mathbb{R}}^{d}$ if and only if for each facet
$F\subset P$, there is a diameter segment $I_{F}\subset P$ with an endpoint in
$\operatorname{relint}F$.

###### Proof.

If there were a facet not satisfying the condition above, then stacking that
facet by adding a point outside the polytope but very close to the interior of
the facet would give a polytope with the same diameter, since this operation
will not change diameter directions and does not lengthen existing diameter
segments (the only maximal segments that can be extended in the stacking have
an endpoint in $\operatorname{relint}F$).

If $P$ were not complete but satisfies the condition of the proposition, let
$Q$ be a polytope strictly containing it of the same diameter. If $q$ is a
point of $Q\setminus P$, then it must lie beyond some facet of $P$, say $F$.
Then the segment $I_{F}$ of the condition of the proposition can be extended
to a longer segment in $Q$, contradicting the diameter condition on $Q$. ∎

Now we are well-equipped for the proof of Theorem 3.4.

###### Proof of Theorem 3.4.

Without loss of generality, suppose $\operatorname{diam}_{\Lambda}(C)=1$ (cf.
Lemma 2.3). In view of Lemma 2.1, the diameter of $C$ is attained in finitely
many directions. By Lemma 3.6, for each such diameter direction $v$, there
exist two opposite exposed faces that contain the endpoints of all diameter
segments parallel to $v$. So we have finitely many exposed faces
$F_{1},...,F_{s}\subset C$ that contain the endpoints of all diameter
segments. Let $u_{1},...,u_{s}$ be the outer unit normal vectors of $C$ that
realize the $F_{i}$’s, that is, $F_{i}=\\{x\in C\colon x\cdot
u_{i}=h(C;u_{i})\\}$.

Suppose that $C$ is strictly contained in the intersection of the half-spaces
$\\{x\in{\mathbb{R}}^{d}\colon u_{i}\cdot x\leq h(C;u_{i})\\}$. Then there is
a point $x\in\operatorname{bd}(C)$ with $u_{i}\cdot x<h(C;u_{i})$, for all
$1\leq i\leq s$. Let $\varepsilon>0$ be such that
$\varepsilon<h(C;u_{i})-u_{i}\cdot x$, for all $1\leq i\leq s$, and define
$x_{\varepsilon}=x+\varepsilon v\not\in C$, for some outer normal vector $v$
of $C$ at $x$. We claim that $C^{\prime}=\operatorname{conv}(C\cup\\{x\\})$
has the same diameter as $C$.

If $\varepsilon$ is small enough, we can assume that the lattice diameter of
$C^{\prime}$ is attained by one of the diameter directions of $C$ (cf. Lemma
2.2). Denote this direction by $y$ and let $F_{1}$ and $F_{2}$, say, be the
faces of $C$ that contain the endpoints of diameter segments in this
direction. By the choice of $\varepsilon$, we have

$x_{\varepsilon}\cdot u_{i}=x\cdot u_{i}+\varepsilon v\cdot u_{i}\leq x\cdot
u_{i}+\varepsilon<h(u_{i};C),$

for all $i$. Thus, the hyperplanes $H_{i}=\\{x\in{\mathbb{R}}^{d}:u_{i}\cdot
x=h(u_{i};C)\\}$ are also supporting hyperplanes of $C^{\prime}$. Therefore, a
segment in $C^{\prime}$ parallel to $y$ cannot be longer than a segment
passing between $F_{1}$ and $F_{2}$. So we obtain
$\operatorname{diam}_{\Lambda}(C^{\prime})=\operatorname{diam}_{\Lambda}(C)$,
contradicting the completeness of $C$. It follows that $C$ is indeed the
intersection of the halfspaces $\\{x\in{\mathbb{R}}^{d}\colon u_{i}\cdot x\leq
h(u_{i};C)\\}$ and as such a polytope.

In order to see that $C$ has at most $2^{k+1}-2$ facets, we recall from
Proposition 3.7, that for each facet $F\subset C$, there exists a diameter
segment $[x_{F},y_{F}]\subset C$ such that $x_{F}\in\operatorname{relint}F$.
For each facet $F\subset C$, let $v_{F}=x_{F}-y_{F}$. Since
$\operatorname{diam}_{\Lambda}(C)=1$, these are boundary lattice points of
$C-C$. Suppose $v_{F}$ and $v_{G}$ are in a common facet of $C-C$. By Lemma
3.5, we have that $x_{F}$ and $x_{G}$ are in a common facet of $C$. By
construction, we must have $F=G$. Thus, no two distinct $v_{F}$’s share a
facet. Since, due to $\operatorname{diam}_{\Lambda}(C)=1$, we have
$\operatorname{int}(C-C)\cap\Lambda=\\{0\\}$, the inequality follows from
applying Lemma 3.3 to the $v_{F}$’s and $C-C$. ∎

###  3.3. The Origin-Symmetric Case

For origin-symmetric convex bodies, there is a duality between reducedness and
completeness.

######  Theorem 3.8.

Let $C\subset{\mathbb{R}}^{d}$ be an origin-symmetric $d$-polytope and let
$\Lambda\subset{\mathbb{R}}^{d}$ be a $d$-lattice.

  1. (1)

The width directions of $C$ with respect to $\Lambda$ are the diameter
directions of $C^{\star}$ with respect to $\Lambda^{\star}$.

  2. (2)

$C$ is reduced with respect to $\Lambda$, if and only if $C^{\star}$ is
complete with respect to $\Lambda^{\star}$.

###### Proof.

For (1), we recall from Lemma 2.1 that the width directions of $C$ with
respect to $\Lambda^{\star}$ are given by

$W=\lambda_{1}((C-C)^{*};\Lambda^{\star})(C-C)^{\star}\cap\Lambda^{\star}\setminus\\{0\\}.$

Since $-C=C$, we have

$\begin{split}\lambda_{1}((C-C)^{*};\Lambda^{\star})(C-C)^{\star}&=\lambda_{1}\Big{(}\frac{1}{2}C^{\star};\Lambda^{\star}\Big{)}\frac{1}{2}C^{\star}\\\
&=\lambda_{1}(2C^{\star};\Lambda^{\star})2C^{\star}=\lambda_{1}(C^{\star}-C^{\star};\Lambda^{\star})(C^{\star}-C^{\star}),\end{split}$

where we used that the successive minima are $(-1)$-homogeneous in the first
argument and that $C^{\star}$ is origin-symmetric as well. It follows that

$W=\lambda_{1}(C^{\star}-C^{\star};\Lambda^{\star})(C^{\star}-C^{\star})\cap\Lambda^{\star}\setminus\\{0\\},$

which proves the claim.

For (2), we may assume that $C$ is a polytope. Moreover, we can assume that
$\operatorname{diam}_{\Lambda^{\star}}(C^{\star})=2$ (cf. Lemma 2.3). In this
case the width directions of $C$ and the diameter directions of $C^{\star}$
are given by $W=C^{\star}\cap\Lambda^{\star}\setminus\\{0\\}$.

First, let $C^{\star}$ be complete with respect to $\Lambda^{\star}$. Consider
a vertex $v\in C$ of $C$ and its polar facet $v^{\diamond}\subset C^{\star}$.
Since $C^{\star}$ is complete, there exists a diameter direction $y\in W$ and
a segment $I=[a,a+2y]\subset C^{\star}$ such that
$a+2y\in\operatorname{relint}v^{\diamond}$ (cf. Proposition 3.7). Since $C$ is
origin-symmetric, the segment $-I\subset C$ is also diameter realizing and by
Lemma 3.5 we have $-a\in v^{\diamond}$. It follows that
$y=\tfrac{1}{2}(a+2y-a)\in\operatorname{relint}v^{\diamond}$.

This shows that
$y\in\operatorname{int}(\operatorname{cone}{v^{\diamond}})=\operatorname{int}(\operatorname{ncone}(v;C))$,
which means that $y\cdot v>y\cdot x$ for all $x\in C\setminus\\{v\\}$. Thus,
$C$ is reduced with respect to $\Lambda$ by (1) and Proposition 3.1. The
reverse implication is proved by similar means. ∎

######  Remark 3.9.

There is no direct generalization of Theorem 3.8 to arbitrary convex bodies.
In fact, if $C$ is not origin-symmetric, $C^{\star}$ might even be unbounded.
Since width and diameter are invariant with respect to translations one could,
however, consider $(C-x_{0})^{\star}$ instead, where $x_{0}$ is a point in
$\operatorname{int}C$. A natural choice for $x_{0}$ might be the centroid
$\operatorname{g}(C)=\operatorname{vol}(C)^{-1}\int_{C}x\,\mathrm{d}x$. But as
we will see in Remarks 3.13 and 4.11, we have

(3.1) $C\text{ is reduced w.r.t.\
$\Lambda$}\centernot\implies~{}(C-\operatorname{g}(C))^{\star}\text{ is
complete w.r.t.\ }\Lambda^{\star}$

and

(3.2) $C\text{ is complete w.r.t.\
$\Lambda$}\centernot\implies~{}(C-\operatorname{g}(C))^{\star}\text{ is
reduced w.r.t.\ $\Lambda^{\star}$}.$

###  3.4. Constructions and Examples

We start by considering 2-dimensional examples.

######  Example 3.10.

In this example, we consider the lattice $\Lambda={\mathbb{Z}}^{2}$.

  1. (1)

The triangle $T_{\mathrm{orth}}=\operatorname{conv}\\{0,e_{1},e_{2}\\}$ is
reduced, but not complete. Its width is attained by the directions $\pm
e_{1}$, $\pm e_{2}$ and $\pm\mathbbm{1}$. Each of these directions cut out one
of the three vertices of $T_{\mathrm{orth}}$. The diameter of
$T_{\mathrm{orth}}$ is attained only by its edges. Thus, $T_{\mathrm{orth}}$
is not complete by Proposition 3.7.

  2. (2)

The square $C_{2}=[-1,1]^{2}$ is complete since it is a multiple of the
Voronoi cell of ${\mathbb{Z}}^{2}$ (cf. Proposition 3.11). It is not reduced
since its width is attained by $\pm e_{1}$ and $\pm e_{2}$, but these
directions support $C_{2}$ at its edges, not at vertices, thus violating
Proposition 3.1. By polarity, $C_{2}^{\star}=\operatorname{conv}\\{\pm
e_{1},\pm e_{2}\\}$ is reduced, but not complete.

  3. (3)

Interestingly, the “twisted squares”
$Q_{x}=\operatorname{conv}\\{\pm\tbinom{1}{x},\pm\tbinom{x}{-1}\\}$ are
simultaneously reduced and complete for $x\in(0,1)$; Indeed, both width and
diameter are attained by the directions $\pm e_{1},\pm e_{2}$ and, due to the
choice of $x$, the corresponding lines support $Q_{x}$ at its vertices and the
corresponding segments pass between two opposite edges (cf. Figure 1).

  4. (4)

The pentagon
$\operatorname{conv}\\{\tbinom{-1}{1},\tbinom{1}{2},\tbinom{2}{-2},\tbinom{-2}{-1}\\}$
is reduced and not complete.

  5. (5)

The hexagon
$H=\operatorname{conv}\\{\pm\tbinom{2}{-1},\pm\tbinom{-1}{2},\pm\tbinom{1}{1}\\}$
is simultaneously reduced and complete.

(4) and (5) can be verified by means of elementary, but rather lengthy,
computations.

We continue with our “standard examples” for complete and reduced polytopes.

######  Proposition 3.11.

Let $\Lambda\subset{\mathbb{R}}^{d}$ be a $d$-lattice and let $V_{\Lambda}$ be
its Voronoi cell as defined in (1.4). Then, $V_{\Lambda}$ is complete with
respect to $\Lambda$ and $(V_{\Lambda})^{\star}$ is reduced with respect to
$\Lambda^{\star}$.

Before we come to the proof, we note that the Voronoi cell of $\Lambda$ may
equivalently be represented as

(3.3) $V_{\Lambda}=\\{x\in{\mathbb{R}}^{d}\colon v\cdot
x\leq|v|^{2}/2,~{}\forall v\in\Lambda\\}$

Since $\Lambda$ is a discrete set, it follows that $V_{\Lambda}$ is a
polytope. The vectors $v\in\Lambda$ such that the inequality $v\cdot
x\leq|v|^{2}/2$ contributes a facet to $V_{\Lambda}$ are called _Voronoi
relevant_.

###### Proof.

It is well known (see for instance [21, Section 4]) that $2V_{\Lambda}$
contains each Voronoi relevant vector $v_{F}$ in the relative interior of its
corresponding facet $F$ and moreover, we have
$\operatorname{int}(2V_{\Lambda})\cap\Lambda=\\{0\\}$. Since $V_{\Lambda}$ is
origin-symmetric, we have
$\lambda_{1}(V_{\Lambda}-V_{\Lambda};\Lambda)=\lambda_{1}(2V_{\Lambda};\Lambda)=1$
and, thus, $\operatorname{diam}_{\Lambda}(V_{\Lambda})=1$. Again by symmetry
around the origin, the diameter is attained by the segments $[v_{F},-v_{F}]$.
Hence, Proposition 3.7 applies and $V_{\Lambda}$ is indeed complete. The
statement for $(V_{\Lambda})^{\star}$ follows directly from Proposition 3.8. ∎

An interesting non-symmetric example is given by the simplices

$S_{d}=\operatorname{conv}\\{-\mathbbm{1},e_{1},\dots,e_{d}\\}\subset{\mathbb{R}}^{d}.$

######  Proposition 3.12.

The simplices $S_{d}\subset{\mathbb{R}}^{d}$ are simultaneously reduced and
complete with respect to ${\mathbb{Z}}^{d}$.

###### Proof.

We start with the reducedness of $S_{d}$. Since $S_{d}$ has integer vertices,
we have $\operatorname{width}_{{\mathbb{Z}}^{d}}(S_{d})\in{\mathbb{Z}}_{>0}$.
The vectors $e_{i}$, $1\leq i\leq d$, show that
$\operatorname{width}_{{\mathbb{Z}}^{d}}(S_{d})\leq 2$ and the fact that the
origin is an interior lattice point of $S_{d}$ gives
$\operatorname{width}_{{\mathbb{Z}}^{d}}(S_{d})>1$. Hence,
$\operatorname{width}_{{\mathbb{Z}}^{d}}(S_{d})=2$, realized by the standard
basis vectors of ${\mathbb{Z}}^{d}$. The linear functions $x\mapsto x\cdot
e_{i}$ are uniquely maximized by the vertices $e_{i}\in S_{d}$ and uniquely
minimized by the vertex $-\mathbbm{1}\in S_{d}$. Therefore, $S_{d}$ is reduced
by Proposition 3.1.

For the completeness of $S_{d}$ we observe that the facets of $C=S_{d}-S_{d}$
are of the form $F=G-G^{\prime}$, where $G\subset S_{d}$ is a proper face of
$S_{d}$ and $G^{\prime}$ is its complementary face, i.e., the convex hull of
the vertices of $S_{d}$ which are not in $F$. So the vertices of $G$ and
$G^{\prime}$ partition $\\{-\mathbbm{1},e_{1},\dots,e_{d}\\}$ and we shall
assume that $-\mathbbm{1}\in G^{\prime}$. Let $J\subseteq[d]$ be the set of
indices $j$ such that $e_{j}\in G$ . A simple computation then shows that

$\begin{split}\operatorname{aff}F&=\operatorname{aff}\\{e_{j}-e_{i},e_{j}+\mathbbm{1}\colon
j\in J,~{}i\not\in J\\}\\\
&=\left\\{x\in{\mathbb{R}}^{d}\colon\frac{d-|J|+1}{d+1}\sum_{j\in
J}x_{j}-\frac{|J|}{d+1}\sum_{j\not\in J}x_{j}=1\right\\}\end{split}$

and, thus,

(3.4) $C=\left\\{x\in{\mathbb{R}}^{d}\colon\left|\frac{d-|J|+1}{d+1}\sum_{j\in
J}x_{j}-\frac{|J|}{d+1}\sum_{j\not\in J}x_{j}\right|\leq 1,~{}\forall
J\subseteq[d],J\neq\emptyset\right\\}.$

Now we claim that
$\operatorname{diam}_{{\mathbb{Z}}^{d}}(S_{d})=1+\tfrac{1}{d}$. First, we
observe that the segment $I_{0}=[-\mathbbm{1},\tfrac{1}{d}\mathbbm{1}]\subset
S_{d}$ has lattice length $1+\tfrac{1}{d}$, which shows that
$\operatorname{diam}_{{\mathbb{Z}}^{d}}(S_{d})\geq 1+\tfrac{1}{d}$, or, in
other words, $\lambda_{1}(C;{\mathbb{Z}}^{d})\leq\tfrac{d}{d+1}$. It, thus,
follows from $S_{d}\subset[-1,1]^{d}$ that
$\operatorname{diam}_{{\mathbb{Z}}^{d}}(S_{d})$ is attained by a non-zero
vector $v\in\tfrac{d}{d+1}C\cap{\mathbb{Z}}^{d}\subset[-1,1]^{d}$. We may
assume that $J=\\{j\in[d]\colon v_{j}=1\\}$ is non-empty (otherwise, we
consider $-v$ instead). For $j\not\in J$, we have $v_{j}\leq 0$ and we obtain

$\frac{d-|J|+1}{d+1}\sum_{j\in J}v_{j}-\frac{|J|}{d+1}\sum_{j\not\in
J}v_{j}\geq|J|\frac{d-|J|+1}{d+1}\geq\frac{d}{d+1}.$

So the only non-zero integer points in $\tfrac{d}{d+1}C$ lie in the boundary.
This shows $\lambda_{1}(C;{\mathbb{Z}}^{d})=\frac{d}{d+1}$ and thus
$\operatorname{diam}_{{\mathbb{Z}}^{d}}(S_{d})=1+\tfrac{1}{d}$.

Now it suffices to observe that for any $i\in[d]$, the segments
$I_{i}=[e_{i},-\tfrac{1}{d}e_{i}]$ have lattice length $1+\tfrac{1}{d}$ and
are contained in $S_{d}$; indeed,

$-\frac{1}{d}e_{i}=\frac{1}{d}(-\mathbbm{1}+\sum_{j\neq
i}e_{j})\in\operatorname{relint}F_{i},$

where $F_{i}\subset S_{d}$ denotes the facet opposite to $e_{i}$. Thus,
$S_{d}$ is complete according to Proposition 3.7. ∎

######  Remark 3.13.

The simplex $S_{d}$ proves (3.1). Its centroid is the origin and its polar
$S_{d}^{\star}$ is equivalent to the orthogonal simplex
$\operatorname{conv}\\{0,e_{1},\dots,e_{n}\\}$, which is not complete.

The following classical constructions preserve completeness and reducedness
respectively.

######  Lemma 3.14.

Let $P\subset{\mathbb{R}}^{k}$ and $Q\subset{\mathbb{R}}^{l}$ be full-
dimensional polytopes.

  1. (1)

If $P$ and $Q$ are complete with respect to ${\mathbb{Z}}^{k}$ and
${\mathbb{Z}}^{l}$ respectively and if
$\operatorname{diam}_{{\mathbb{Z}}^{k}}(P)=\operatorname{diam}_{{\mathbb{Z}}^{l}}(Q)$,
then their Cartesian product $P\times Q$ is complete with respect to
${\mathbb{Z}}^{k+l}$.

  2. (2)

If $0\in\operatorname{int}P$, $0\in\operatorname{int}Q$, and if $P$ and $Q$
are reduced with respect to ${\mathbb{Z}}^{k}$ and ${\mathbb{Z}}^{l}$
respectively, such that
$\operatorname{width}_{{\mathbb{Z}}^{k}}(P)=\operatorname{width}_{{\mathbb{Z}}^{l}}(Q)$,
then their free sum

$P\oplus
Q=\operatorname{conv}\big{(}(P\times\\{0\\}^{l})\cup(\\{0\\}^{k}\times
Q)\big{)}\subset{\mathbb{R}}^{k+l}$

is reduced with respect to ${\mathbb{Z}}^{k+l}$

  3. (3)

Under the same conditions of (2), the join of height $h$

$P\star
Q=\operatorname{conv}\big{(}(P\times\\{0\\}^{l+1})\cup(\\{0\\}^{k}\times
Q\times\\{h\\})\big{)}\subset{\mathbb{R}}^{k+l+1},$

where $|h|\geq\operatorname{width}_{{\mathbb{Z}}^{k}}(P)$, is reduced with
respect to ${\mathbb{Z}}^{k+l+1}$.

###### Proof.

In the following, let
$\operatorname{diam}=\operatorname{diam}_{{\mathbb{Z}}^{n}}$ and
$\operatorname{width}=\operatorname{width}_{{\mathbb{Z}}^{n}}$, where the
ambient dimension $n$ will be clear from the context.

For (1), we start by proving $\operatorname{diam}(P\times
Q)=\operatorname{diam}(P)$. Clearly, any lattice segment in $P$ is also a
lattice segment in $P\times Q$ with the same length and so we have
$\operatorname{diam}(P\times Q)\geq\operatorname{diam}(P)$. Conversely, let
$I=[(a,a^{\prime}),(b,b^{\prime})]\subset P\times Q$ be a lattice segment,
where $a,b\in P$ and $a^{\prime},b^{\prime}\in Q$. If $I$ is orthogonal to
$\\{0\\}^{k}\times{\mathbb{R}}^{l}$, i.e., if $a^{\prime}=b^{\prime}$, then
$\operatorname{Vol}_{1}(I)=\operatorname{Vol}_{1}([a,b])\leq\operatorname{diam}(P)$.
If $I$ is not orthogonal to $\\{0\\}^{k}\times{\mathbb{R}}^{l}$, then the
projection

$\Pi:P\times Q\to Q,~{}(x,y)\mapsto y$

restricted to the segment $I$ is injective and maps lattice points to lattice
points. It follows that

$\operatorname{Vol}_{1}(I)\leq\operatorname{Vol}_{1}(\Pi(I))\leq\operatorname{diam}(Q)=\operatorname{diam}(P).$

This proves $\operatorname{diam}(P\times Q)=\operatorname{diam}(P)$. In order
to see that $P\times Q$ is complete, we note that any facet of $P\times Q$ is
of the form $F\times Q$ or $P\times G$, where $F\subset P$ and $G\subset Q$
are facets of $P$ and $Q$, respectively. For a facet of the form $F\times Q$,
there exists a point $x\in\operatorname{relint}F$ and a diameter segment
$I_{F}=[x,x^{\prime}]\subset P$, for some $x^{\prime}\in P$, since $P$ is
complete. Let $y\in\operatorname{int}Q$ be any interior point of $Q$. The
segment $[(x,y),(x^{\prime},y)]\subset P\times Q$ is diameter realizing for
$P\times Q$ and the point $(x,y)$ is in the relative interior of $F\times Q$.
A facet of the type $P\times G$ is treated analogously.

For (2), we use [12, Theorem 2.2, (3)], which shows that
$\operatorname{width}(P\oplus Q)=\operatorname{width}(P)$. In order to see
that $P\oplus Q$ is reduced, we note that the vertices of $P\oplus Q$ are of
the form $(v,0)$, or $(0,w)$, where $v\in P$ and $w\in Q$ are vertices of $P$
and $Q$, respectively. A width realizing lattice direction
$y\in{\mathbb{Z}}^{k}$, which is uniquely maximized at $v$ leads to a width
realizing direction $(y,0)\in{\mathbb{Z}}^{k+l}$, which is uniquely maximized
by $(v,0)$. Thus, $P\oplus Q$ is reduced.

For (3), we first consider a vector $z\in{\mathbb{Z}}^{k+l}\times\\{0\\}$. By
projecting onto ${\mathbb{R}}^{k+l}\times\\{0\\}$, we see that
$\operatorname{width}(P\star Q;z)=\operatorname{width}(P\oplus
Q;z)\geq\operatorname{width}(P)$, with equality, if $z$ is the canonical
embedding of one of the width direction of $P$ or $Q$. If
$z\in{\mathbb{Z}}^{k+l+1}$ with $z_{k+l+1}\neq 0$, we have

(3.5) $\operatorname{width}(P\star Q;z)\geq|z\cdot((0,h)-(0,0))|\geq
h\geq\operatorname{width}(P).$

Thus, $\operatorname{width}(P\star Q)=\operatorname{width}(P)$, which is
attained by the canonical embedding of the width directions of $P$ and $Q$.
These vectors also testify that $P\star Q$ is reduced. ∎

Lemma 3.14 and its proof lead us to a first example of a reduced 3-polytope
whose width directions do not span its ambient space.

######  Example 3.15.

The simplex
$S=\operatorname{conv}\\{(-1,0,0)^{T},(1,0,0)^{T},(0,1,3)^{T},(0,-1,3)^{T}$ is
reduced since it arises as a join as in Lemma 3.14, where $P=Q=[-1,1]$ and
$h=3$. As we saw in the proof, the width directions of $S$, $\pm e_{1}$ and
$\pm e_{2}$, are given by the canonical embedding of the width directions of
$P$. Since $h>2$, the inequality in (3.5) is strict, so we have no width
direction outside $\operatorname{span}\\{e_{1},e_{2}\\}$.

The remainder of this subsection is devoted to a lifting construction which
allows us to construct origin-symmetric complete (reduced) polytopes whose
diameter (width) directions are not full-dimensional. The basic idea of the
construction is to lift the normal vectors of a polytope
$P\subset{\mathbb{R}}^{d}$ into ${\mathbb{R}}^{d+1}$ in such a way that $P\cap
e_{d+1}^{\perp}=P$ and then adjust the lattice
$\Lambda\subset{\mathbb{R}}^{d+1}$ afterwards such that the lifted polytope is
indeed complete.

The following lemma makes this geometric idea precise.

######  Lemma 3.16.

Let $P=\\{x\in{\mathbb{R}}^{d}\colon|x\cdot a_{i}|\leq 1,~{}1\leq i\leq m\\}$
be an origin-symmetric $d$-polytope. Suppose that $m>d$ and each $a_{i}$
contributes a pair of facets to $P$. Then there exist numbers
$\eta_{1},\dots\eta_{m}\in{\mathbb{R}}$ such that the polyhedron

$\overline{P}=\\{y\in{\mathbb{R}}^{d+1}\colon|y\cdot\overline{a_{i}}|\leq
1,~{}1\leq i\leq m\\},$

where $\overline{a_{i}}=(a_{i},\eta_{i})^{T}\in{\mathbb{R}}^{d+1}$, is a
bounded origin-symmetric $(d+1)$-polytope with $2m$ facets and
$\overline{P}\cap e_{d+1}^{\perp}=P$.

###### Proof.

It is clear that $\overline{P}\cap e_{d+1}^{\perp}=P$ holds for any choice of
$\eta_{1},...\eta_{m}\in{\mathbb{R}}$. By polarity, it suffices to prove that
we can choose $\eta_{1},...,\eta_{m}$ such that
$\overline{P}^{\star}=\operatorname{conv}\\{\pm\overline{a_{1}},\dots,\pm\overline{a_{m}}\\}$
contains the origin in its interior and has all $a_{i}$’s as vertices. Again
by polarity, we have that
$0\in\operatorname{int}P^{\star}=\operatorname{conv}\\{\pm a_{1},\dots,\pm
a_{m}\\}$. Similarly to the proof of Steinitz’ theorem [46, Theorem 1.3.10],
we may consider an arbitrary line $\ell$ passing through the origin and apply
Caratheodory’s theorem in the two opposite facets of $P^{\star}$ intersecting
$\ell$ in order to choose $d$ linearly independent vectors
$a_{i_{1}},\dots,a_{i_{d}}$ of the $a_{i}$’s such that the origin is contained
in the interior of $\operatorname{conv}\\{\pm a_{i_{1}},\dots,\pm
a_{i_{d}}\\}$. Since $m>d$, we may assume that $a_{m}$ is not among the
vectors $a_{i_{1}},\dots,a_{i_{d}}$. We set $\eta_{1}=\cdots=\eta_{m-1}=0$ and
$\eta_{m}=1$. This way, $\overline{P}^{\star}$ is a double pyramid over
$\operatorname{conv}\\{\pm a_{1},\dots,\pm a_{m-1}\\}$, which contains the
origin in its interior and the vectors $\overline{a_{i}}$, $1\leq i\leq m$, as
vertices. ∎

If we perform a lifting of the normal vectors of $P$ as in the above Lemma and
extend a lattice $\Lambda\subset{\mathbb{R}}^{d}$ to a $(d+1)$-dimensional
lattice $\overline{\Lambda}\subset{\mathbb{R}}^{d+1}$ by adding a vector
“sufficiently far away from $\operatorname{span}\Lambda$” we can lift a
complete polytope $P\subset{\mathbb{R}}^{d}$ to a complete polytope
$\overline{P}\subset{\mathbb{R}}^{d+1}$ (cf. Figure 2). This is formalized in
the following proposition.

Figure 2. Lifting the normal vectors of the complete hexagon $P$ into
${\mathbb{R}}^{3}$ yields a parallelotope $\overline{P}$ that is complete with
respect to the 2-lattice $\Lambda$. The lattice $\Lambda$ can easily be
extended to a 3-lattice $\overline{\Lambda}$ to which $\overline{P}$ is
complete by choosing a plane parallel to $\Lambda$ which does not intersect
$\overline{P}$.

######  Proposition 3.17.

Let $P\subset{\mathbb{R}}^{d}$ be a complete origin-symmetric $d$-polytope
with respect to some $d$-dimensional lattice $\Lambda$. If $P$ has more than
$2d$ facets, there exists an origin-symmetric polytope
$\overline{P}\subset{\mathbb{R}}^{d+1}$ and a $(d+1)$-dimensional lattice
$\overline{\Lambda}\subset{\mathbb{R}}^{d}$ such that the following holds:

  1. (1)

$\overline{P}$ arises from $P$ via a lifting of the normals as in Lemma 3.16.

  2. (2)

$\overline{\Lambda}$ contains $\Lambda$ as a primitive sub-lattice.

  3. (3)

$\overline{P}$ is a complete polytope w.r.t. $\overline{\Lambda}$.

  4. (4)

The diameter directions of $\overline{P}$ are the same than the ones of $P$
(when the latter is embedded in ${\mathbb{R}}^{d+1}$).

###### Proof.

We apply Lemma 3.16 to $P$ and obtain an origin-symmetric polytope
$\overline{P}$ with the same number of facets and $\overline{P}\cap
e_{d+1}^{\perp}=P$. We choose a vector $t\in{\mathbb{R}}^{d+1}$ such that
$\overline{P}\cap(t+\operatorname{span}\Lambda)=\emptyset$. Then,
$\overline{P}$ contains no lattice points of
$\overline{\Lambda}=(\Lambda\times\\{0\\})\oplus{\mathbb{Z}}t$ other than
those in $P\cap\Lambda$. Since $\overline{P}$ is symmetric, its diameter is
then attained by the same directions as for $P$ (cf. Lemma 2.1). It remains to
check that $\overline{P}$ is complete. To this end, it is enough to note that
for any boundary point $x\in\operatorname{bd}(P)$, we have $x\cdot(\pm
a_{i})=1$ if and only if $(x,0)^{T}\cdot(\pm\overline{a_{i}})=1$. Hence, the
segments that realized the diameter of $P$ intersect the facets of
$\overline{P}$ in their interior and so $\overline{P}$ is complete. ∎

###  3.5. Proofs of Theorems 1.2 and 1.3

We want to use our construction methods in order to prove the main theorems
1.2 and 1.3.

###### Proof of Theorem 1.2.

(1) is a direct consequence of Theorem 3.2, since $k\leq d$. (2) follows from
Theorem 3.4 in the same way. In order to see that the number $2^{d+1}-2$ is
best possible, we recall that the Voronoi cell of $A_{d}^{\star}$ is a
permutohedron [13, Ch. 4, Sec. 6.6], so it has $2^{d+1}-2$ facets. Since
$V_{A_{d}^{\star}}$ is complete with respect to $A_{d}^{\star}$ (cf.
Proposition 3.11), the number $2^{d+1}-2$ is optimal with respect to
$A_{d}^{\star}$. But in view of Lemma 2.3 (2), it is also optimal with respect
to any other lattice. For reduced polytopes, the optimality is proven in the
same way, since $(V_{A_{d}^{\star}})^{\star}$ is reduced with respect to
$A_{d}$. ∎

###### Proof of Theorem 1.3.

(1) can be derived from Theorem 3.2, since a $d$-polytope has at least $d+1$
vertices. So it follows that $d+1\leq 2^{k+1}-2$, where $k$ is the dimension
of the linear space spanned by the diameter realizing segments. The claim
follows by rearranging this inequality.

(2) follows in the same fashion from Theorem 3.4, since a $d$-polytope also
has at least $d+1$ facets.

In order to see that the logarithmic dependence on $d$ is optimal, we apply
Proposition 3.17 to the permutohedron $P=V_{A_{k}^{\star}}$, which is complete
with respect to $\Lambda=A_{k}^{\star}$ (Here we think of
$\operatorname{span}(A_{k}^{\star})$ as ${\mathbb{R}}^{k}$). Since $P$ has
$2^{k}-1>k$ pairs of opposite facets, we can apply Proposition 3.17 multiple
times until we obtain an origin-symmetric polytope $Q$ of dimension $2^{k}-1$
which is complete with respect to some $(2^{k}-1)$-dimensional lattice
$\Lambda\supset A_{k}^{\star}$ and whose diameter directions are exactly the
ones of $P$. Thus, the diameter directions of $Q\subset{\mathbb{R}}^{2^{k}-1}$
form a $k$-dimensional subspace, which proves the optimality of the
logarithmic order in the complete case. For the reduced case it is again
enough to use the symmetry of $Q$ in conjunction with Proposition 3.8. ∎

###  3.6. Reductions and Completions

In the Euclidean case, it follows from Zorn’s lemma that any convex body
$C\subset{\mathbb{R}}^{d}$ has a completion and a reduction. These are defined
as complete (reduced) convex bodies that are supersets (subsets) of $C$ with
the same Euclidean diameter (width) as $C$.

In the discrete setting, the terms “completion” and “reduction” are defined in
the same way with the appropriate definitions of diameter and width. We want
to argue the existence of completions and reductions in a similar way to the
Euclidean setting. However, the argument is more involved since the Euclidean
diameter cannot be bounded by the lattice diameter. Therefore, it is not clear
that the completion is indeed a bounded set. Likewise, when using Zorn’s lemma
to find a reduction, we have to make sure that it has non-empty interior to be
a convex body in the sense of our definition. To this end we use the following
two Lemmas.

######  Lemma 3.18.

Let $K\subset{\mathbb{R}}^{d}$ be an origin-symmetric $d$-dimensional convex
set and let $\Lambda\subset{\mathbb{R}}^{d}$ be a $d$-dimensional lattice such
that $\operatorname{int}K\cap\Lambda=\\{0\\}$. Then, $K$ is bounded.

###### Proof.

Suppose $K$ is unbounded. Then there exists a sufficiently large $R>0$ such
that $\operatorname{vol}(K\cap[-R,R]^{d})>2^{d}\det\Lambda$. It follows from
Minkowski’s first theorem [18, Theorem 2.5.1] that $K\cap[-R,R]^{d}$ contains
a non-trivial interior lattice point of $\Lambda$, a contradiction. ∎

On the polar side, we have the following statement, where for an arbitrary
bounded set $A$, the lattice width $\operatorname{width}_{\Lambda}(A)$ is
defined similarly to (1.2) as

$\operatorname{width}_{\Lambda}(A)=\inf_{y\in\Lambda^{\star}}\sup_{a,b\in
A}y\cdot(a-b).$

######  Lemma 3.19.

Let $K\subset{\mathbb{R}}^{d}$ be a compact convex set and let
$\Lambda\subset{\mathbb{R}}^{d}$ be a $d$-dimensional lattice. If
$\operatorname{width}_{\Lambda}(K)>0$, then $K$ has non-empty interior.

###### Proof.

Let $w=\operatorname{width}_{\Lambda}(K)$ and consider the origin-symmetric
convex set $M=w(K-K)^{\star}$. Then $M$ contains no interior lattice points of
$\Lambda^{\star}$: Suppose to the contrary that there exists
$y\in\operatorname{int}M\cap\Lambda^{\star}\setminus\\{0\\}$. Then we have
$y\cdot(a-b)\leq w$ for all $a,b\in K$. But as $K$ is compact, there exist
$a_{0},b_{0}\in K$ such that
$\operatorname{width}(K;y)=y\cdot(a_{0}-b_{0})<w,$ a contradiction. Moreover,
since $K$ is bounded, $M$ is $d$-dimensional. Thus, Lemma 3.18 applies and we
obtain that $M$ is bounded. Consequently $K-K$, and thus also $K$, have non-
empty interior. ∎

######  Theorem 3.20.

Let $C\subset{\mathbb{R}}^{d}$ be a convex body and let
$\Lambda\subset{\mathbb{R}}^{d}$ be a $d$-lattice.

  1. (1)

There exists a complete convex body $K\supset C$ with
$\operatorname{diam}_{\Lambda}(K)=\operatorname{diam}_{\Lambda}(C)$.

  2. (2)

There exists a reduced convex body $L\subset C$ with
$\operatorname{width}_{\Lambda}(L)=\operatorname{width}_{\Lambda}(C)$.

###### Proof.

Without loss of generality, we assume that $0\in\operatorname{int}C$. For (1),
we consider the set

$\mathcal{P}=\\{K\supset C\text{ convex
body}\colon\operatorname{diam}_{\Lambda}(K)=\operatorname{diam}_{\Lambda}(C)\\},$

partially ordered by inclusion. Since a completion of $C$ is a maximal element
of $\mathcal{P}$, we aim to use Zorn’s lemma in order to find such a maximal
element. To this end, we consider a chain $\mathcal{C}\subset\mathcal{P}$ and
we define $M^{\prime}=\bigcup\mathcal{C}$ as the union of all convex bodies in
$\mathcal{C}$, as well as $M=\operatorname{cl}M^{\prime}$. If
$M\in\mathcal{P}$, then Zorn’s lemma applies and the proof is finished. It,
thus, remains to show that $M$ is a convex body with
$\operatorname{diam}_{\Lambda}(M)=\operatorname{diam}_{\Lambda}(C)$.

Clearly, $M$ has non-empty interior, since $C\subset M$. Since $\mathcal{C}$
is a chain, $M^{\prime}$ is convex, which implies that $M$ is convex as well.
Moreover, the convexity of $M^{\prime}$ yields $\operatorname{int}M\subset
M^{\prime}$; Otherwise, there was a point $x\in\operatorname{int}(M)\setminus
M^{\prime}$, which could be separated from $M^{\prime}$ by a hyperplane and,
since $x\in\operatorname{int}(M)$, we may assume that the separation is
strict. In this case, however, $x\not\in\operatorname{cl}M^{\prime}=M$, a
contradiction.

Now consider a primitive vector $v\in\Lambda\setminus\\{0\\}$ and let
$I=[a,b]\subset M$ be a segment of maximal length in $M$ in direction $v$.
Towards a contradiction, assume that $M$ is unbounded in direction $v$, i.e.,
$\operatorname{diam}(M;v)=\infty$. Since
$0\in\operatorname{int}C\subset\operatorname{int}M$, it follows that
$M\cap\operatorname{span}\\{v\\}$ is unbounded. Since
$\operatorname{int}M\subset M^{\prime}$, we find that
$M^{\prime}\cap\operatorname{span}\\{v\\}$ is also unbounded. Hence, there
exists a sequence of points $(a_{n})_{n\in{\mathbb{N}}}\subset
M^{\prime}\cap\operatorname{span}\\{v\\}$ with
$\lim_{n\to\infty}|a_{n}|=\infty$. In particular, there exists
$n\in{\mathbb{N}}$ with $|a_{n}|>\operatorname{diam}_{\Lambda}(C)\cdot|v|$.
Consider $A\in\mathcal{C}$ with $a_{n}\in A$. We have

$\operatorname{diam}_{\Lambda}(A)\geq\operatorname{Vol}_{1}([0,a_{n}])=\frac{|a_{n}|}{|v|}>\operatorname{diam}_{\Lambda}(C),$

a contradiction to $A\in\mathcal{P}$. Hence, the longest segment $I\subset M$
parallel to a given primitive vector $v\in\Lambda$ is of finite length.
Moreover, if $I=[a,b]$, we have, since $0\in\operatorname{int}M$, that

$I_{n}=\left(1-\frac{1}{n}\right)I\subset\operatorname{int}M\subset
M^{\prime}.$

It follows from the definition of $M^{\prime}$ that there exist convex bodies
$A,B\in\mathcal{C}$ with $(1-1/n)a\in A$ and $(1-1/n)b\in B$. Since
$\mathcal{C}$ is a chain, we can assume that $A\subset B$ and, thus,
$I_{n}\subset B$. Hence,
$\operatorname{Vol}_{1}(I_{n})\leq\operatorname{diam}_{\Lambda}(B)=\operatorname{diam}_{\Lambda}(C)$
and, taking the limit $n\to\infty$,
$\operatorname{Vol}_{1}(I)\leq\operatorname{diam}_{\Lambda}(C)$. Since
$C\subset M$, we have
$\operatorname{diam}_{\Lambda}(M)=\operatorname{diam}_{\Lambda}(C)$. It
follows that the convex set

$D=\frac{1}{\operatorname{diam}_{\Lambda}(C)}(M-M)$

contains no non-trivial interior lattice points of $\Lambda$. From Lemma 3.18
we have that $D$ is bounded. This implies that $M$ is bounded as well.

In summary, we saw that $M\in\mathcal{P}$ is a valid upper bound of
$\mathcal{C}$ and thus, by Zorn’s lemma, the completion of $C$ exists.

For (2), we consider the set

(3.6) $\mathcal{Q}=\\{K\subset C\text{ convex
body}\colon\operatorname{width}_{\Lambda}(K)=\operatorname{width}_{\Lambda}(C)\\},$

partially ordered by inclusion. Again let $\mathcal{C}\subset\mathcal{Q}$ be a
chain in $\mathcal{Q}$. As in the proof of (1) it suffices to show that the
intersection $M=\bigcap\mathcal{C}$ of all convex bodies in $\mathcal{C}$ is
an element of $\mathcal{Q}$.

It is clear, that $M$ is a compact convex set. We have to show that
$\operatorname{int}M\neq\emptyset$ and
$\operatorname{width}_{\Lambda}(M)=\operatorname{width}_{\Lambda}(C)$. To this
end, we consider a primitive direction $y\in\Lambda^{\star}$ and, towards a
contradiction, we assume that
$\operatorname{width}(M;y)<\operatorname{width}_{\Lambda}(C)$. For a scalar
$\alpha\in{\mathbb{R}}$, let $H_{\alpha}=\\{x\in{\mathbb{R}}^{d}\colon x\cdot
y=\alpha\\}$ and let $H_{\alpha}^{-}$ and $H_{\alpha}^{+}$ be the closed half-
spaces defined by $H_{\alpha}$ depending on whether $x\cdot y$ is less than or
greater than $\alpha$ respectively. Due to our assumption on
$\operatorname{width}(M;y)$, there exist $\alpha<\beta$ with $M\subset
H_{\alpha}^{+}\cap H_{\beta}^{-}$ and
$\beta-\alpha<\operatorname{width}_{\Lambda}(C)$.

Let $\alpha^{\prime}=\alpha-\varepsilon$ and
$\beta^{\prime}=\beta+\varepsilon$, where $\varepsilon>0$ is such that
$\beta^{\prime}-\alpha^{\prime}<\operatorname{width}_{\Lambda}(C)$. Suppose
that each $A\in\mathcal{C}$ intersects $H_{\alpha^{\prime}}^{-}$. Then we
consider the family of non-empty convex bodies

$\mathcal{C}^{\prime}=\\{A\cap H_{\alpha^{\prime}}^{-}\colon
A\in\mathcal{C}\\}.$

Since $\mathcal{C}$ is a chain, so is $\mathcal{C}^{\prime}$. In particular,
since all $A^{\prime}\in\mathcal{C^{\prime}}$ are non-empty,
$\mathcal{C}^{\prime}$ satisfies the finite intersection property, i.e., for
any finite collection
$\\{A_{1}^{\prime},\dots,A_{n}^{\prime}\\}\subset\mathcal{C}^{\prime}$, we
have $\bigcap_{i=1}^{n}A_{i}^{\prime}\neq\emptyset$. Since any
$A^{\prime}\in\mathcal{C}^{\prime}$ is contained in the compact set $C$, it
follows that $M\cap
H_{\alpha^{\prime}}^{-}=\bigcap\mathcal{C}^{\prime}\neq\emptyset$, which
contradicts $M\subset H_{\alpha}^{+}$. Hence, there is a convex body
$A\in\mathcal{C}$ with $A\subset H_{\alpha^{\prime}}^{+}$ and, likewise, a
convex body $B\in\mathcal{C}$ with $B\subset H_{\beta^{\prime}}^{-}$. We may
assume that $A\subset B$, because $\mathcal{C}$ is a chain. It follows that

$\operatorname{width}(A;y)\leq\beta^{\prime}-\alpha^{\prime}<\operatorname{width}_{\Lambda}(C),$

a contradiction to $A\in\mathcal{Q}$. Thus, we have
$\operatorname{width}(M;y)\geq\operatorname{width}_{\Lambda}(C)>0$, for all
$y\in\Lambda^{\star}\setminus\\{0\\}$. Lemma 3.19 gives that $M$ has non-empty
interior. Also, since $M\subset C$, it follows that
$\operatorname{width}_{\Lambda}(M)=\operatorname{width}_{\Lambda}(C)$ as
desired. So $M\in\mathcal{Q}$ is a minimal element of $\mathcal{C}$ and Zorn’s
lemma gives the existence of a minimal element $L\in\mathcal{Q}$, which is a
reduction of $C$. ∎

######  Remark 3.21.

  1. (1)

If the convex body $C$ in Theorem 3.20 is symmetric, then its reduction and
completion may also be chosen to be symmetric. To see this, we start with the
reduction of $C$. Replacing the partially ordered set $\mathcal{Q}$ in (3.6)
by

$\mathcal{Q}_{s}=\\{K\subset C\text{ convex
body}\colon\operatorname{width}_{\Lambda}(K)=\operatorname{width}_{\Lambda}(C),~{}K\text{
is origin-symmetric}\\},$

Zorn’s Lemma yields the existence of a minimal element $L\in\mathcal{Q}_{s}$.
Indeed, $L$ is reduced. Otherwise, there would be an exposed point $p\in L$ at
which the width is not uniquely attained (cf. Proposition 3.1). The same is
then true for $-p\in L$. Applying the truncation described in the proof of
Proposition 3.1 to $p$ and $-p$ simultaneously yields an origin-symmetric
convex body $L^{\prime}\subsetneq L$ with
$\operatorname{width}_{\Lambda}(L^{\prime})=\operatorname{width}_{\Lambda}(L)$,
a contradiction to the minimality of $L$ in $\mathcal{Q}_{s}$. The existence
of an origin-symmetric completion can now be deduced immediately from Theorem
3.8.

  2. (2)

The reduction and completion of $C$ are in general not unique. For instance,
each $Q_{x}$, $x\in(0,1)$ in Example 3.10 constitutes a symmetric reduction of
$[-1,1]^{2}$. Moreover, the right triangle
$\operatorname{conv}\\{\tbinom{1}{1},\tbinom{1}{-1},\tbinom{-1}{-1}\\}$ is a
non-symmetric reduction.

  3. (3)

In [25, Lemma 1.2] it is shown that the product of the successive minima of
$C$ and its polar is bounded from above by $c\cdot d$, where $c>0$ is an
absolute constant, i.e., we have

(3.7)
$\frac{\operatorname{width}_{\Lambda}(C)}{\operatorname{diam}_{\Lambda}(C)}=\lambda_{1}(C-C;\Lambda)\,\lambda_{1}((C-C)^{\star};\Lambda^{\star})\leq
c\cdot d,$

for any convex body $C\subset{\mathbb{R}}^{d}$. Conway and Thompson showed
that the linear order of this upper bound is achieved by certain ellipsoids
[23, Theorem 9.5], i.e., there exists another universal constant $c^{\prime}$
such that for each $d\in{\mathbb{N}}$ there exists an ellipsoid
$\mathcal{E}\subset{\mathbb{R}}^{d}$ such that

$\frac{\operatorname{width}_{\Lambda}(\mathcal{E})}{\operatorname{diam}_{\Lambda}(\mathcal{E})}\geq
c^{\prime}\,d.$

Considering a reduction (completion) of $\mathcal{E}$ shows that the linear
order in the inequality is also achieved by polytopes with at most $2^{d+1}-2$
vertices (facets).

For polytopes we can construct a reduction explicitly without the help of
Zorn’s lemma. Moreover, this reduction does not increase the number of
vertices of the polytope.

######  Proposition 3.22.

Let $P\subset{\mathbb{R}}^{d}$ be a $d$-polytope and let
$\Lambda\subset{\mathbb{R}}^{d}$ be a $d$-lattice. Then there exists a reduced
$d$-polytope $Q\subset P$ such that
$\operatorname{width}_{\Lambda}(P)=\operatorname{width}_{\Lambda}(Q)$ and
$|\operatorname{ext}(Q)|\leq|\operatorname{ext}(P)|$.

In particular, any simplex has a reduction which is also a simplex. The idea
of the construction is to push the vertices at which the width is not uniquely
achieved inside the polytope. This is formalized in the following lemma.

######  Lemma 3.23.

Let $P\subset{\mathbb{R}}^{d}$ be a $d$-polytope and let
$\Lambda\subset{\mathbb{R}}^{d}$ be a $d$-lattice. Let
$W\subset\Lambda^{\star}$ be the set of width directions of $P$ and let
$v\in\operatorname{ext}(P)$ be a vertex at which
$\operatorname{width}_{\Lambda}(P)$ is not uniquely attained, i.e.,
$\operatorname{int}(\operatorname{ncone}(v;P))\cap W=\emptyset$. Then, for any
$x_{0}\in\operatorname{int}P$, there exists a number $\lambda_{0}\in[0,1)$
such that

(3.8)
$P_{\lambda}=\operatorname{conv}\big{(}(\operatorname{ext}(P)\setminus\\{v\\})\cup\\{\lambda
v+(1-\lambda)x_{0}\\}\big{)}\subset P$

satisfies
$\operatorname{width}_{\Lambda}(P_{\lambda})=\operatorname{width}_{\Lambda}(P)$,
for all $\lambda\in[\lambda_{0},1]$.

###### Proof.

We consider
$\lambda_{0}=\inf\\{\lambda\in[0,1]\colon\operatorname{width}_{\Lambda}(P_{\lambda})=\operatorname{width}_{\Lambda}(P)\\}$.
Since $\operatorname{width}_{\Lambda}$ is continuous, the infimum
$\lambda_{0}$ is indeed attained. It remains to show that $\lambda_{0}<1$. To
this end, we recall that for $\varepsilon>0$ sufficiently small, the width
directions of $P_{1-\varepsilon}$ are given by $W$, the set of width
directions of $P$ (cf. Lemma 2.2). But since for any $y\in W$,
$\operatorname{width}(P;y)$ is attained by two vertices
$a,b\in\operatorname{ext}(P)\setminus\\{v\\}$, which are still present in
$P_{1-\varepsilon}$, it follows that
$\operatorname{width}(P_{1-\varepsilon};y)=\operatorname{width}(P;y)$ and,
thus,
$\operatorname{width}_{\Lambda}(P_{1-\varepsilon})=\operatorname{width}_{\Lambda}(P)$.
This shows $\lambda_{0}<1$ and finishes the proof. ∎

###### Proof of Proposition 3.22.

Let $V=\operatorname{ext}(P)$ be the set of vertices of $P$. We call a vertex
$v\in V$ reduced if there exists a width realizing direction
$y\in\Lambda^{\star}\setminus\\{0\\}$ with $v\cdot y>x\cdot y$, for all $x\in
P\setminus\\{v\\}$. We can thus rephrase Proposition 3.1 as follows: $P$ is
reduced if and only if all of its vertices are reduced.

So assume that there exists a vertex $v\in V$ which is not reduced. We
distinguish two cases.

Case 1:
$\operatorname{width}_{\Lambda}(\operatorname{conv}(V\setminus\\{v\\}))=\operatorname{width}_{\Lambda}(P)$.
We set $P^{\prime}=\operatorname{conv}(V\setminus\\{v\\})\subset P$ and obtain
a polytope inside $P$ that has the same width as $P$ and one non-reduced
vertex less than $P$. The polytope $P^{\prime}$ is $d$-dimensional in view of
Lemma 3.19.

Case 2:
$\operatorname{width}_{\Lambda}(\operatorname{conv}(V\setminus\\{v\\}))<\operatorname{width}_{\Lambda}(P)$.
We fix some
$x_{0}\in\operatorname{relint}\operatorname{conv}(V\setminus\\{v\\})$ and let
$x_{\lambda}=\lambda v+(1-\lambda)x_{0}\in\operatorname{int}P$, for
$\lambda\in[0,1]$. In view of Lemma 3.23, there exists a minimal $\lambda_{0}$
such that
$\operatorname{width}_{\Lambda}(\operatorname{conv}(V\setminus\\{v\\}\cup\\{x_{\lambda_{0}}\\})=\operatorname{width}_{\Lambda}(P)$.
We let $v^{\prime}=x_{\lambda_{0}}$ and
$P^{\prime}=\operatorname{conv}(V\setminus\\{v\\}\cup\\{v^{\prime}\\})\subset
P$. By construction, we have
$\operatorname{width}_{\Lambda}(P^{\prime})=\operatorname{width}_{\Lambda}(P)$.
We claim that the $v^{\prime}$ is a reduced vertex of $P^{\prime}$.

Clearly, $v^{\prime}$ is a vertex of $P^{\prime}$ to begin with. Otherwise, it
would not be necessary for the convex hull description of $P^{\prime}$ and we
find ourselves in Case 1. For the same reason (and by the choice of $x_{0}$),
we have $\lambda_{0}>0$. Let $W\subset\Lambda^{\star}$ be the set of width
directions of $P^{\prime}$, i.e.,
$W=(\Lambda^{\star}\setminus\\{0\\})\cap\operatorname{width}_{\Lambda}(P)(P^{\prime}-P^{\prime})^{\star}$.
If $v^{\prime}$ is not a reduced vertex of $P^{\prime}$, then for each $y\in
W$, there exist $u,w\in V\setminus\\{v\\}$ with
$\operatorname{width}_{\Lambda}(P)=(u-w)\cdot y$. If we replace
$v^{\prime}=x_{\lambda}$ by $x_{\lambda-\varepsilon}$, for $\varepsilon>0$
sufficiently small, we obtain a polytope $P^{\prime\prime}$ whose width
directions are still among the vectors $W$ (cf. Lemma 2.2). But this implies
$\operatorname{width}_{\Lambda}(P^{\prime\prime})=\operatorname{width}_{\Lambda}(P)$,
contradicting the minimality of $\lambda$. Hence, $v^{\prime}$ is indeed a
reduced vertex of $P^{\prime}\subset P$. It is also clear by construction that
$P^{\prime}$ is a $d$-polytope.

In both cases, we constructed a $d$-polytope $P^{\prime}\subset P$ such that
$\operatorname{width}_{\Lambda}(P^{\prime})=\operatorname{width}_{\Lambda}(P)$
and $P^{\prime}$ has one non-reduced vertex less than $P$. Also, the total
number of vertices of $P^{\prime}$ is not higher than the number of vertices
in $P$. Thus, after finitely many steps, we obtain a reduced polytope with the
desired properties. ∎

Similar to Remark 3.21 i), if $P$ is origin-symmetric, a slight adaption of
the proof of Proposition 3.22 yields an origin-symmetric reduction of $P$; we
simply have to treat the opposite vertices of $P$ simultaneously. Using the
polarity in the symmetric case, we can find a completion of $P$, which does
not rely on Zorn’s Lemma and does not posses more facets than $P$. It is
obtained by dualizing the construction of the reduction of $P$, i.e., pairs of
facets of $P$ are pushed outward until they either vanish or contain a lattice
point in their interiors.

However, in the non-symmetric case this process will not yield a completion.
In fact, if $P$ is a simplex, it produces merely homothetic copies of $P$ that
have a larger width. Therefore, the following problem remains open.

######  Question 3.24.

Is there an explicit way of constructing a completion of a (non-symmetric)
polytope $P\subset{\mathbb{R}}^{d}$ with respect to a given lattice
$\Lambda\subset{\mathbb{R}}^{d}$?

##  4\. Simplices and Triangles

###  4.1. Simplices

We begin with the proof of Theorem 1.5.

###### Proof of Theorem 1.5.

Towards a contradiction, we assume that
$S=\operatorname{conv}\\{v_{0},\dots,v_{d}\\}\subset{\mathbb{R}}^{d}$ is a
local maximum of $\operatorname{width}_{\Lambda}$ on the class of hollow
$d$-simplices, which is not reduced. Our goal is to find for any
$\varepsilon>0$ a hollow simplex $S^{\prime}\subset{\mathbb{R}}^{d}$ with
$d(S,S^{\prime})<\varepsilon$ and
$\operatorname{width}_{\Lambda}(S)<\operatorname{width}_{\Lambda}(S^{\prime})$.

If $S$ is not reduced, then there exists a vertex $v_{0}$, say, of $S$ such
that $\operatorname{int}(\operatorname{ncone}(v_{0};S))\cap W=\emptyset$,
where $W$ denotes the set of width directions of $S$. Using Lemma 3.23, we can
replace this vertex by a new vertex $v_{0}^{\prime}\in\operatorname{int}S$ and
obtain a simplex
$T=\operatorname{conv}\\{v_{0}^{\prime},v_{1},\dots,v_{d}\\}\subset S$ with
$d(S,T)<\tfrac{\varepsilon}{2}$ and
$\operatorname{width}_{\Lambda}(S)=\operatorname{width}_{\Lambda}(T)$. We
consider an inequality description

(4.1) $T=\\{x\in{\mathbb{R}}^{d}\colon a_{i}\cdot x\leq b_{i},~{}0\leq i\leq
d\\},$

for certain $a_{i}\in{\mathbb{R}}^{d}\setminus\\{0\\}$ and
$b_{i}\in{\mathbb{R}}$. We assume that the new vertex $v_{0}^{\prime}$ is
contained in the facets $F_{i}=\\{x\in T\colon a_{i}\cdot x=b_{i}\\}$, for
$1\leq i\leq d$. Since $v_{0}\in\operatorname{int}S$, we have
$\operatorname{relint}F_{d}\subset\operatorname{int}S$. Since $S$ is hollow,
it follows that $T$ is hollow as well and
$\operatorname{relint}(F_{d})\cap\Lambda=\emptyset$. Therefore, there exists a
number $\delta>0$ such that

$S^{\prime}=\\{x\in{\mathbb{R}}^{d}\colon a_{i}\cdot x\leq b_{i},~{}0\leq
i<d,~{}a_{d}\cdot x\leq b_{d}+\delta\\}$

is hollow and $d(T,S^{\prime})<\tfrac{\varepsilon}{2}$. So we have
$d(S,S^{\prime})<\varepsilon$. Moreover, since $T$ is a simplex, $S^{\prime}$
is a translation of $\mu T$, for some scale factor $\mu>1$. This implies
$\operatorname{width}_{\Lambda}(S^{\prime})>\operatorname{width}_{\Lambda}(T)=\operatorname{width}_{\Lambda}(S)$.
∎

######  Remark 4.1.

In the plane, Hurkens found a triangle $T_{2}\subset{\mathbb{R}}^{2}$ that
realizes $\mathrm{Flt}(2)$ [22]. In higher dimensions, the realizers of
$\mathrm{Flt}(d)$ are unknown, but local maximizers of
$\operatorname{width}_{\Lambda}$ within the class of hollow simplices have
been found. In [12] a tetrahedron $T_{3}\subset{\mathbb{R}}^{3}$ that locally
maximizes the width has been obtained, and in [42] the authors construct a
4-dimensional local maximum $T_{4}\subset{\mathbb{R}}^{4}$, as well as a
5-dimensional local maximum $T_{5}\subset{\mathbb{R}}^{5}$. By Theorem 1.5,
these simplices are reduced. Moreover, a sagemath computation shows that
$T_{3}$ and $T_{4}$ are complete, while $T_{5}$ is not complete. The script
that we used to check the reducedness and completeness of polytopes can be
accessed under the following URL:
https://github.com/AnsgarFreyer/lattice_reduced_code

It remains open whether Theorem 1.5 holds true for arbitrary convex bodies. We
call a convex body $C\subset{\mathbb{R}}^{d}$ a _realizer_ of
$\mathrm{Flt}(d)$ if $C$ is hollow and
$\operatorname{width}_{\Lambda}(C)=\mathrm{Flt}(d)$.

######  Question 4.2.

Is any realizer of $\mathrm{Flt}(d)$ is reduced?

Note that Question 4.2 is equivalent to the question whether any realizer $C$
of $\mathrm{Flt}(d)$ is _inclusion-maximal hollow_ , i.e., $C$ is hollow,
while any $C^{\prime}\supsetneq C$ is not hollow. Due to a result Lovász,
inclusion-maximal hollow polytopes have at most $2^{d}$ facets [38], see also
[2].

We finish this subsection with an interesting observation concerning complete
simplices.

######  Proposition 4.3.

Let $S\subset{\mathbb{R}}^{d}$ be a complete $d$-simplex with respect to some
$d$-lattice $\Lambda$. Then, the diameter directions of $S$ span
${\mathbb{R}}^{d}$.

###### Proof.

According to Proposition 3.7, for every facet $F\subset S$, there exists a
diameter segment $I_{F}\subset S$ with an endpoint
$x_{F}\in\operatorname{relint}F$. By Lemma 3.6, the other endpoint of $I_{F}$
lies inside the face opposite of $F$. Since $S$ is a simplex, this means that
the other endpoint is the unique vertex $v_{F}\in S$ opposite to $F$.

Towards a contradiction, assume that the directions $x_{F}-v_{F}$, where $F$
ranges over all facets of $S$, are contained in a $(d-1)$-dimensional space
$H\subset{\mathbb{R}}^{d}$. Let $t\in{\mathbb{R}}^{d}$ such that $H+t$
supports $S$. Then, $S\cap(H+t)$ contains a vertex $v$ of $S$. Let $F$ be the
facet opposite to $v$. Then, $\operatorname{relint}F$ is disjoint from $H+t$;
otherwise $F\subset H+t$ and the entire simplex would be lower-dimensional.
But now the segment $I_{F}$ is not parallel to $H$, a contradiction. ∎

######  Remark 4.4.

As we saw in Example 3.15, there exist reduced simplices
$S\subset{\mathbb{R}}^{d}$ whose width directions $W$ do not span
${\mathbb{R}}^{d}$. Consider any point of $x_{0}\in\operatorname{int}S$. Since
the polar $(S-x_{0})^{\star}$ is again a simplex, the above proposition shows
that $(S-x_{0})^{\star}$ cannot be complete with the same diameter directions
$W$, as it is the case for origin-symmetric convex bodies when polarized with
respect to the origin (cf. Theorem 3.8).

It would be interesting to find more classes of complete (or reduced) convex
bodies for which the diameter (width) directions span the entire ambient
space. One such class apart from simplices are the Voronoi cells from
Proposition 3.11, since any Voronoi relevant vector of $\Lambda$ is a diameter
direction. By duality the width directions of $(V_{\Lambda})^{\star}$ are
full-dimensional.

###  4.2. Classification of Triangles in the Plane

In this subsection we give a complete classification of reduced triangles and
triangles that are simultaneously reduced and complete. Finally, we show that
every complete triangle is among the reduced and complete triangles, proving
Theorem 1.4.

For convenience, in this subsection we only consider the lattice
$\Lambda={\mathbb{Z}}^{2}$ and we write $\operatorname{diam}(C)$ for
$\operatorname{diam}_{{\mathbb{Z}}^{2}}(C)$, as well as
$\operatorname{width}(C)$ for $\operatorname{width}_{{\mathbb{Z}}^{2}}(C)$. We
call two polygons equivalent, if they are equivalent up to translations,
dilations and unimodular transformations.

If $P\subset{\mathbb{R}}^{2}$ is a reduced polygon, then the primitive vectors
in ${\mathbb{Z}}^{2}$ that realize $\operatorname{width}(P)$ must be
2-dimensional. Moreover, they contain a basis of ${\mathbb{Z}}^{2}$ (cf. [18,
Ch.1, Theorems 3.4 and 3.7]). Thus, we have the following lemma.

######  Lemma 4.5.

Let $P\subset{\mathbb{R}}^{2}$ be reduced w.r.t. ${\mathbb{Z}}^{2}$. Then $P$
is equivalent to a polygon $Q\subset[-1,1]^{2}$, such that each edge of
$[-1,1]^{2}$ contains a vertex of $Q$ and $\operatorname{width}(Q)=2$.

First, consider a reduced triangle $T\subset[-1,1]^{2}$ with
$\operatorname{width}(T)=2$ such that every edge of $[-1,1]^{2}$ contains a
vertex of $T$. Clearly, one of the vertices of $T$ is then a vertex of
$[-1,1]^{2}$ as well. Without loss of generality, let $(-1,-1)^{T}$ be a
vertex of $T$. The other two vertices are then given by $(x,1)^{T}$ and
$(1,y)^{T}$ for certain $x,y\in[-1,1]$.

Since $T$ is reduced, we must have $x+y\leq 0$; otherwise the vector
$(-1,1)^{T}$ would achieve a smaller width. Indeed, the condition $x+y\leq 0$
is also sufficient for $T$ to be reduced:

######  Proposition 4.6.

A triangle $T\subset{\mathbb{R}}^{2}$ is reduced with respect to
${\mathbb{Z}}^{2}$ to if and only if it is equivalent to a triangle of the
form

(4.2)
$T_{xy}=\operatorname{conv}\left\\{\binom{-1}{-1},\binom{x}{1},\binom{1}{y}\right\\},$

where $x+y\leq 0$ and $-1\leq x,y\leq 1$.

###### Proof.

It only remains to prove the sufficiency, i.e., we aim to prove that $T_{xy}$
is reduced for any $x,y\in[-1,1]$ with $x+y\leq 0$. Since
$\operatorname{width}(T_{xy};e_{i})=2$, $i=1,2$, we have
$\operatorname{width}(T_{xy})\leq 2$ and if equality holds, then $T_{xy}$ is
reduced by Proposition 3.1.

So we consider a vector $v=(a,b)^{T}\in{\mathbb{Z}}^{2}\setminus\\{0\\}$ with
$a\neq 0$ and $b\neq 0$ and we distinguish two cases (note that the width does
not depend on the orientation of $v$).

Case 1: $a,b>0$.
We consider the vertices $(x,1)^{T}$ and $(-1,-1)^{T}$. We have

$\operatorname{width}(T_{xy};v)\geq\left|\binom{a}{b}\cdot\left(\binom{x}{1}-\binom{-1}{-1}\right)\right|=a(x+1)+2b\geq
2b\geq 2,$

where we used that $x+1\geq 0$ and that $b$ is an integer.

Case 2: $a>0>b$.
We consider the vertices $(x,1)^{T}$ and $(1,y)^{T}$. We have

$\begin{split}\operatorname{width}(T_{xy};v)\geq\left|\binom{a}{b}\cdot\left(\binom{1}{y}-\binom{x}{1}\right)\right|&=a(1-x)+b(y-1)\geq
1-x-(y-1)\\\ &=2-(x+y)\geq 2\end{split}$

where we used that $1-x\geq 0$, $y-1\leq 0$ and $a,b\in{\mathbb{Z}}$ in the
first line, as well as $x+y\leq 0$ in the second line. ∎

Now suppose that $T_{xy}$ as in (4.2) is also complete. From Proposition 4.3,
we see that each vertex of $T_{xy}$ needs to be the endpoint of a diameter
segment. The diameter directions of $T_{xy}$ are the lattice points in
$\lambda_{1}(C;{\mathbb{Z}}^{2})C$, where $C=T_{xy}-T_{xy}\subset[-2,2]^{2}$.
Since the origin is an interior point of $T_{xy}$ (unless $x=y=-1$ in which
case $T_{xy}$ is not complete), the diameter of $T_{xy}$ is greater than 1.
Thus, it is realized by lattice points in
$\operatorname{int}C\cap[-2,2]^{2}\subset[-1,1]^{2}$.

So the only possible direction of a segment $I_{1}$ that connects
$(-1,-1)^{T}$ with its opposite edge and that realizes the diameter of
$T_{xy}$ is $(1,1)^{T}$. A routine computation shows that the lattice length
of $I_{1}$ is

$\operatorname{Vol}(I_{1})=\frac{3-xy-(x+y)}{2-(x+y)}.$

At the vertex $(x,1)^{T}$, the diameter could be obtained in the directions
$(0,-1)^{T}$ and $(1,-1)^{T}$. But $(0,-1)^{T}$ clearly gives the longer
segment $I_{2}$ and is therefore diameter realizing. The length of $I_{2}$ is
given by

$\operatorname{Vol}_{1}(I_{2})=\frac{3-y}{2}-\frac{1+y}{2}x.$

Note that, by exchanging the roles of $x$ and $y$ the third segment $I_{3}$ in
direction $(-1,0)^{T}$ that connects the vertex $(1,y)^{T}$ to its opposite
edge is of length

$\operatorname{Vol}_{1}(I_{3})=\frac{3-x}{2}-\frac{1+x}{2}y=\frac{3-y}{2}-\frac{1+y}{2}x=\operatorname{Vol}_{1}(I_{2}).$

Combining these equations we obtain, that $T_{xy}$ is complete, only if the
following equation is fulfilled:

$\frac{3-xy-(x+y)}{2-(x+y)}=\frac{3-y}{2}-\frac{1+y}{2}x$

After multiplying with $2-(x+y)$, we obtain an equation which is quadratic in
$x$. Its solutions are $x=\frac{3-y}{y+1}$ and $x=-y$. The first solution is
not in the interval $[-1,1]$ for any value of $y\in(-1,1)$. So we obtain that
$T_{xy}$ is complete, only if $x=-y$. On the other hand, for $|x|<1$ and
$x=-y$, the three segments $I_{1}$, $I_{2}$ and $I_{3}$ pass from their
respective vertices to the interior of the opposite edge of that vertex and
are all of the same length. The only direction for a longer lattice segment
would be $(1,-1)^{T}$, but we already saw that a segment in this direction has
shorter lattice length than $I_{2}$ (and $I_{3}$). Since for $|x|=1$, the
triangle $T_{-xx}$ is not complete, we can now classify the triangles that are
both complete and reduced as follows:

######  Proposition 4.7.

A triangle $T\subset{\mathbb{R}}^{2}$ is simultaneously reduced and complete
with respect to ${\mathbb{Z}}^{2}$, if and only if it is equivalent to a
triangle of the form

$T_{x}=\operatorname{conv}\left\\{\binom{-1}{-1},\binom{x}{1},\binom{1}{-x}\right\\},$

where $0\leq x<1$.

Next, we show that the triangles obtained in Proposition 4.7 are indeed the
only complete triangles up to equivalence. For this, we need two elementary
lemmas.

######  Lemma 4.8.

Let $C\subset{\mathbb{R}}^{2}$ be an origin-symmetric convex body such that
$\operatorname{int}C\cap{\mathbb{Z}}^{2}=\\{0\\}$ and $\\{\pm e_{1},\pm
e_{2}\\}\subset C$. If $C$ contains a lattice point other than $0$, $\pm
e_{1}$, or $\pm e_{2}$, it also contains one of the points $(1,1)^{T}$ and
$(1,-1)^{T}$.

###### Proof.

Let $x=(a,b)\neq\pm e_{i}$ be a non-zero lattice point in $C$. As the standard
basis vectors are on the boundary of $C$, $x$ must be in the interior of one
of the four quadrants. Without loss of generality, we can assume that
$x\in{\mathbb{Z}}_{>0}^{2}$. Also, we can assume that $a\geq b$. We may write

$\binom{1}{1}=\frac{1}{a}x+\left(1-\frac{b}{a}\right)e_{2}.$

Since $a$ and $b$ are integers with $0<b\leq a$, it follows that
$1-b/a=(a-b)/a\in[0,1]$ and $1/a+(1-b/a)=(1-b+a)/a\leq 1$. Hence,
$(1,1)^{T}\in\operatorname{conv}\\{0,e_{2},x\\}\subset C$. ∎

######  Lemma 4.9.

Consider two disjoint segments $I_{i}=[v_{i},w_{i}]$. $i=1,2$, and define the
point $v_{3}$ as the intersection of the lines
$\operatorname{aff}\\{v_{1},w_{2}\\}$ and
$\operatorname{aff}\\{v_{2},w_{1}\\}$ (if any). Then the triangle
$T=\operatorname{conv}\\{v_{1},v_{2},v_{3}\\}$ does not contain both of the
segments $I_{1}$ and $I_{2}$.

###### Proof.

Let $\ell_{1}=\operatorname{aff}\\{v_{1},w_{2}\\}$ and
$\ell_{2}=\operatorname{aff}\\{v_{2},w_{1}\\}$ and suppose that $I_{1}\subset
T$, i.e., $T$ contains the triangle
$T^{\prime}=\operatorname{conv}\\{v_{1},v_{2},w_{1}\\}$. Since $v_{1}$ and
$w_{1}$ are on the two edges of $T$ adjacent to $v_{2}$, any segment $I\subset
T$ that connects $v_{2}$ to the line $\ell_{1}$ must intersect
$[v_{1},w_{1}]=I_{1}$. But $I_{1}$ and $I_{2}$ are disjoint, so $I_{2}$ cannot
be contained in $T$. ∎

###### Proof of Theorem 1.4.

Let $T$ be a complete triangle. After applying a unimodular transformation and
a dilation we may assume that $\operatorname{diam}(T)=1$ and that the standard
basis vectors are among the diameter realizing directions of $T$. Thus, there
is a horizontal segment $I_{1}\subset T$ and a vertical segment $I_{2}\subset
T$ that both realize the diameter of $T$. By Proposition 4.3, $I_{i}$ connects
a vertex $v_{i}$ to a point $w_{i}$ in the relative interior of the edge
$E_{i}$ opposite to $v_{i}$, $i=1,2$. Again, after applying a unimodular
transformation if necessary, we can assume that the right endpoint of $I_{1}$
is $v_{1}$ and that the upper endpoint of $I_{2}$ is $v_{2}$. The triangle $T$
is now uniquely described; the third vertex $v_{3}$ has to be the intersection
of the lines $\operatorname{aff}\\{v_{1},w_{2}\\}$ and
$\operatorname{aff}\\{v_{2},w_{1}\\}$ (in particular, these lines are not
parallel).

In view of Lemma 4.9, $I_{1}$ and $I_{2}$ intersect. After a translation, we
may assume that $I_{1}=[(x-1,y)^{T},(x,y)^{T}]$ and $I_{2}=[0,e_{2}]$, for
certain $0<x,y<1$ (Note that we cannot have equality in one of the constraints
on $x$ and $y$, since then, one of the segments $I_{i}$ would be an edge of
$T$). In order for the other endpoints $0$ and $(x-1,y)^{T}$ to be part of the
triangle $\operatorname{conv}\\{v_{1},v_{2},v_{3}\\}$ it is necessary that
$x>y$.

By Proposition 4.3, there exists a third pair of lattice directions in which
the lattice diameter of $T$ is attained. Since by assumption,
$\operatorname{diam}(T)=1$, the difference body $C=T-T$ fulfills the
assumptions of Lemma 4.8. So the diameter of $T$ has to be attained in
direction $(1,-1)^{T}$ or $(1,1)^{T}$ as well. The direction $(1,-1)^{T}$ can
be ruled out; If it was diameter realizing, the corresponding segment in $T$
had to include $v_{1}$ as a vertex. But then the segment would have a shorter
lattice length than $I_{1}$. So there is a diameter realizing segment $I_{3}$
in direction $(1,1)^{T}$. The only vertex that this segment can contain is
$v_{3}$. So we have $I_{3}=[v_{3},w_{3}]$, where $w_{3}$ is the intersection
of $v_{3}+\operatorname{span}\\{(1,1)^{T}\\}$ with the edge
$E_{3}=[v_{1},v_{2}]$.

We saw that $T$ is complete, if and only if it is equivalent to a triangle of
the form $\operatorname{conv}\\{e_{2},(x,y)^{T},v_{3}\\}$, where $0<y<x<1$ and
$\operatorname{Vol}_{1}(I_{3})=1$ (with $v_{3}$ and $I_{3}$ as above). In the
the next step we fix the value of $x\in(0,1)$ and show that there is exactly
one value of $y\in(0,x)$ such that $\operatorname{Vol}_{1}(I_{3})=1$. In this
setting the points $v_{3}$ and $w_{3}$ depend on $y$, which is why we denote
them by $v_{3}(y)$ and $w_{3}(y)$. Computing the point of intersection of the
lines
$\operatorname{aff}\\{v_{1},w_{2}\\}=\operatorname{aff}\\{(x,y)^{T},(0,0)^{T}\\}$
and
$\operatorname{aff}\\{v_{2},w_{1}\\}=\operatorname{aff}\\{(0,1)^{T},(x-1,y)^{T}\\}$
shows that

$v_{3}(y)=\begin{pmatrix}\frac{x(x-1)}{x-y}\\\
\frac{y(x-1)}{x-y}\end{pmatrix}.$

In particular, as $y$ increases, $v_{3}(y)$ moves strictly monotonously
downward on the diagonal line $\ell=\\{(a,b)^{T}:a-b=x-1\\}$. The degenerate
cases $y\to 0$ and $y\to x$ correspond to $(x-1,0)$ and the “point at
infinity” of $\ell$. Since $w_{3}(y)$ is by definition the intersection of
$v_{3}(y)+\operatorname{span}\\{(1,1)^{T}\\}$ with $[e_{1},(x,y)^{T}]$, it
follows from the fact that all $v_{3}(y)$ lie on the same diagonal $\ell$,
that $w_{3}(y)$ is the intersection of $\ell$ and $[e_{1},(x,y)^{T}]$. Since
$\ell$ is independent of $y$, the point $w_{3}(y)$ moves strictly monotonously
upward on $\ell$ as $y$ increases. In total, we find that
$\operatorname{Vol}_{1}(I_{3}(y))$ is a strictly monotonous function in
$y\in(0,x)$ (see also Figure 3). By considering, e.g., a projection on a
horizontal line, we see that $\lim_{y\to
0}\operatorname{Vol}_{1}(I_{3}(y))<1$. On the other hand, since $v_{3}(y)$ is
unbounded for $y\to x$, we have $\lim_{y\to
x}\operatorname{Vol}_{1}(I_{3}(y))=\infty$. Consequently, for any $x\in(0,1)$,
there is exactly one $y_{x}\in(0,x)$ such that the triangle
$\Delta_{x}=\operatorname{conv}\\{e_{2},(x,y_{x})^{T},v_{3}(y_{x})\\}$ is
complete.

To conclude, we recall the triangles
$T_{x}=\operatorname{conv}\\{(-1,-1)^{T},(x,1),(1,-x)^{T}\\}$ from Proposition
4.7. Let $r\in(0,1)$ be the fraction of the horizontal diameter realizing
segment that is to the right of the vertical diameter segment. By considering
the limits $x\to\pm 1$, we see that $r$ can attain any value between 0 and 1.
Thus, the triangle $\Delta_{r}$ is in fact equivalent to one of the triangles
$T_{x}$, which finishes the proof.

Figure 3. Illustration of the monotonicity of
$y\mapsto\operatorname{Vol}_{1}(I_{3}(y))$. As $y$ increases, the black vertex
$v_{3}(y)$ moves downward on the diagonal dashed line $\ell$, while the other
endpoint $w_{3}(y)$ of $I_{3}(y)$ moves upward on $\ell$.

∎

At this point, it is natural to ask whether Theorem 1.4 extends to arbitrary
$d$-simplices. The following construction, however, leads to counterexamples
for $d=3$ (Remark 4.11 (3)).

###  4.3. A Family of Complete Tetrahedra

We start with the regular tetrahedron

(4.3) $\Delta=\operatorname{conv}\\{0,e_{i}+e_{j}\colon 1\leq i<j\leq
3\\}\subset[0,1]^{3}$

and we let

$D=\Delta-\Delta=\operatorname{conv}\\{\pm e_{i}\pm e_{j}\colon 1\leq i<j\leq
3\\}\subset C_{3}$

be its difference body. Note that $D$ is the convex hull of the midpoints of
the edges of the cube $C_{3}$. Thus, $D$ has 6 quadrilateral facets
corresponding to the original facets of $C_{3}$ and 8 triangular facets
corresponding to the 8 vertices of $C_{3}$. Moreover, each quadrilateral facet
is of the form $e-f$, where $e,f\subset\Delta$ are opposite edges of $\Delta$,
and each triangular facet is of the form $\pm(F-v)$, where $F\subset\Delta$ is
a facet of $\Delta$ and $v\in\Delta$ is the vertex opposite to it.

Note that a segment $I=[a,b]\subset\Delta$ passes from a vertex $v\in\Delta$
to the interior of the facet opposite to $v$, if and only if, $b-a$ is in the
interior of one of the triangular facets of $D$. Hence, $\Delta$ is complete
with respect to a 3-lattice $\Lambda\subset{\mathbb{R}}^{3}$, if and only if
for some $\lambda>0$, we have $\operatorname{int}(\lambda
D)\cap\Lambda=\\{0\\}$ and $\lambda D$ possesses a lattice point of $\Lambda$
in the relative interior of each of its triangular facets.

For each choice of signs $\sigma\in\\{\pm 1\\}^{3}$, there is a triangular
facet

$F_{\sigma}=\\{x\in C_{3}\colon\sigma\cdot x=2\\}\subset D$

of $D$. We start with a point
$(\alpha,\beta,\gamma)^{T}\in\operatorname{int}F_{(1,1,1)}$, i.e., we have
$\alpha+\beta+\gamma=2$ and $\alpha,\beta,\gamma<1$. Moreover, we want to
require that $\alpha,\beta,\gamma\geq\frac{1}{2}$. Now let

(4.4)
$\Lambda=\operatorname{span}_{{\mathbb{Z}}}\left\\{\begin{pmatrix}-\alpha\\\
\beta\\\ \gamma\end{pmatrix},\begin{pmatrix}\alpha\\\ -\beta\\\
\gamma\end{pmatrix},\begin{pmatrix}\alpha\\\ \beta\\\
-\gamma\end{pmatrix}\right\\}.$

Since

$\begin{pmatrix}\alpha\\\ \beta\\\
\gamma\end{pmatrix}=\begin{pmatrix}-\alpha\\\ \beta\\\
\gamma\end{pmatrix}+\begin{pmatrix}\alpha\\\ -\beta\\\
\gamma\end{pmatrix}+\begin{pmatrix}\alpha\\\ \beta\\\ -\gamma\end{pmatrix},$

the difference body $D$ contains a lattice point of $\Lambda$ in each of the
triangular facets $F_{\sigma}$, $\sigma\in\\{\pm 1\\}^{3}$. It remains to
check that $\operatorname{int}D\cap\Lambda=\\{0\\}$. To this end, we first
consider a point of the form $x=(0,b,c)\in\operatorname{int}D\cap\Delta$.
There exist $z_{1},z_{2},z_{3}\in{\mathbb{Z}}$ with

$\begin{pmatrix}0\\\ b\\\ c\end{pmatrix}=z_{1}\begin{pmatrix}-\alpha\\\
\beta\\\ \gamma\end{pmatrix}+z_{2}\begin{pmatrix}\alpha\\\ -\beta\\\
\gamma\end{pmatrix}+z_{3}\begin{pmatrix}\alpha\\\ \beta\\\
-\gamma\end{pmatrix}.$

It follows from the first equation that $z_{1}=z_{2}+z_{3}$. Thus, the lattice
point $x$ is of the form

$x=\begin{pmatrix}0\\\ 2z_{3}\beta\\\ 2z_{2}\gamma\\\ \end{pmatrix}.$

If either $z_{3}$ or $z_{2}$ is non-zero, then, since
$\beta,\gamma\geq\frac{1}{2}$, the corresponding entry has an absolute value
of at least 1. But then $x$ is on the boundary of $C_{3}$. Hence,
$z_{2}=z_{3}=0$, which implies $x=0$.

By symmetry, we can conclude that a non-zero point in
$x\in\operatorname{int}D\cap\Lambda$ cannot have a zero coordinate. So,
towards a contradiction, let $x=(a,b,c)^{T}\in\operatorname{int}D\cap\Lambda$,
where $a,b,c\in{\mathbb{R}}\setminus\\{0\\}$. Since both $D$ and $\Lambda$ are
symmetric with respect to the coordinate hyperplanes, may even assume that
$a,b,c>0$. Since $x\in\Lambda$, there are positive integers
$z_{1},z_{2},z_{3}\geq 1$ such that
$x=(z_{1}\alpha,z_{2}\beta,z_{3}\gamma)^{T}$. It now follows from
$x\in\operatorname{int}D$ and the symmetry properties of $D$ that
$(\alpha,\beta,\gamma)^{T}\in\operatorname{int}D$, a contradiction to
$(\alpha,\beta,\gamma)^{T}\in F_{(1,1,1)}\subset\operatorname{bd}D$.

So we have established that $\operatorname{int}D\cap\Lambda=\\{0\\}$. This
proves the following proposition.

######  Proposition 4.10.

The regular simplex $\Delta$ defined in (4.3) is complete with respect to the
lattice $\Lambda$ as defined in (4.4) for any choice of

$\begin{pmatrix}\alpha\\\ \beta\\\
\gamma\end{pmatrix}\in\left\\{\begin{pmatrix}a\\\ b\\\
c\end{pmatrix}\colon\frac{1}{2}\leq a,b,c<1,~{}a+b+c=2\right\\}.$

######  Remark 4.11.

  1. (1)

There is no restriction in considering the regular tetrahedron $\Delta$; Since
all simplices are affinely equivalent, finding a lattice with respect to which
a fixed simplex is complete is the same as finding a simplex which is complete
with respect to a fixed lattice.

  2. (2)

If one of the parameters ($\alpha$, say) is equal to $\frac{1}{2}$, the
difference body $D$ contains the point

$\begin{pmatrix}1\\\ 0\\\ 0\end{pmatrix}=\begin{pmatrix}\frac{1}{2}\\\
\beta\\\ \gamma\end{pmatrix}+\begin{pmatrix}\frac{1}{2}\\\ -\beta\\\
-\gamma\end{pmatrix}\in\Lambda,$

which is then a boundary point in one of the quadrilateral facets of $D$. For
the tetrahedron $\Delta$, this implies that the diameter is not only attained
by the four segments connecting the facets to their opposite vertices as
guaranteed by Proposition 4.3, but also by an additional segment between two
opposite edges.

  3. (3)

For a generic choice of the parameters, such as $(0.65,0.65,0.7)$, the
tetrahedron $\Delta$ is not reduced with respect to $\Lambda$, as can be
checked using sagemath (cf. Remark 4.1). This shows that Theorem 1.4 does not
hold in dimension 3.

  4. (4)

The same tetrahedron has the property that also
$(\Delta-\operatorname{g}(\Delta))^{\star}$ is not reduced with respect to
$\Lambda^{\star}$. Thus, it confirms (3.2) from Remark 3.9.

##  5\. Comparison to the Euclidean Case

We want to finish the paper by discussing discrete analogs of classical
properties and conjectures about reduced and complete convex bodies.

###  5.1. Relations between diameter and width

In the Euclidean setting, it is evident that the diameter is bounded from
below by the width of a convex body, that is, we have
$\operatorname{diam}_{\mathbb{R}}(C)\geq\operatorname{width}_{\mathbb{R}}(C)$
for any convex body $C\subset{\mathbb{R}}^{d}$. The lattice analog of this
fact is inequality (3.7) in Remark 3.21. In general, it is impossible to
reverse these inequalities, even at the cost of a constant depending on the
dimension. However, if $C\subset{\mathbb{R}}^{2}$ is a Euclidean reduced
convex body in the plane, it was shown by Lassak [28] that

(5.1)
$\operatorname{diam}_{\mathbb{R}}(C)\leq\sqrt{2}\operatorname{width}_{\mathbb{R}}(C).$

Also, if $C\subset{\mathbb{R}}^{d}$ is a Euclidean complete convex body, we
readily have
$\operatorname{diam}_{\mathbb{R}}(C)=\operatorname{width}_{\mathbb{R}}(C)$,
since $C$ is of constant width (recall that in the Euclidean setting, complete
is stronger than reduced). In the lattice setting in the plane, the following
analogue to Lassak’s statement holds.

######  Proposition 5.1.

Let $C\subset{\mathbb{R}}^{2}$ be lattice reduced or complete with respect to
some lattice $\Lambda\subset{\mathbb{R}}^{2}$. Then,
$\operatorname{diam}_{\Lambda}(C)\leq\operatorname{width}_{\Lambda}(C)$.

We note that equality holds if $C=\operatorname{conv}\\{\pm e_{1},\pm
e_{2}\\}$ (for reduced bodies) or $C=[-1,1]^{2}$ (for complete bodies).

###### Proof.

Let $C$ be reduced with respect to $\Lambda$. Consider a diameter direction
$v\in\Lambda\setminus\\{0\\}$ of $C$. Since $C$ is 2-dimensional it follows
from Proposition 3.1 that there exist two independent width directions of $C$.
So let $y\in\Lambda^{\star}$ be a width direction with $v\cdot y\neq 0$. Let
$[a,a+\operatorname{diam}_{\Lambda}(C)v]\subset C$ be the diameter realizing
segment in direction of $v$. We have

$\operatorname{width}_{\Lambda}(C)=\operatorname{width}(C;y)\geq|y\cdot(a+\operatorname{diam}_{\Lambda}(C)v-a)|=\operatorname{diam}_{\Lambda}(C)\,|y\cdot
v|\geq\operatorname{diam}_{\Lambda}(C),$

where we used that $|y\cdot v|\geq 1$. The argument for complete bodies is
analogous, the existence of a non-orthogonal pair of diameter and width
directions is guaranteed by Proposition 3.7. ∎

Just like Lassak’s inequality (5.1) does not generalize to higher dimensions
(see [29]), it is impossible to extend Proposition 5.1 to dimensions $d\geq
3$. As a counterexample one can consider the cube $C=[-1,1]^{3}$ together with
the lattice $\Lambda=A_{2}\oplus\alpha{\mathbb{Z}}\mathbbm{1}$, for any
$\alpha\geq 1$. This is a complete convex body (in fact, it is a special case
of Proposition 3.17, see also Figure 2) with
$\operatorname{diam}_{\Lambda}(C)=2$ and
$\operatorname{width}_{\Lambda}(C)=2/\alpha$. Since $\alpha\geq 1$ is
arbitrary, this shows that there cannot be a constant $c>0$ such that
$\operatorname{diam}_{\Lambda}(C)\leq c\cdot\operatorname{width}_{\Lambda}(C)$
holds for any 3-dimensional complete convex body. For reduced bodies, it
suffices to consider $C^{\star}$ and $\Lambda^{\star}$, since $C$ is
symmetric.

###  5.2. Volumetric inequalities

In Euclidean geometry, the isodiametric inequality states that the volume of a
convex body $C\subset{\mathbb{R}}^{d}$ of Euclidean diameter $d$ is bounded
from above by the volume of a Euclidean ball of diameter $d$ [46, Inequality
(7.22)]. Again, it is not possible bound the volume of $C$ from below in terms
of a diameter, if $C$ is arbitrary. For complete convex bodies in the plane,
however, such a lower bound exists: For any Euclidean complete convex body
$C\subseteq{\mathbb{R}}^{2}$ of diameter $d$ we have [35, 5]

(5.2) $\operatorname{vol}(C)\geq\operatorname{vol}(U_{d}),$

where $U_{d}$ is a _Releaux triangle_ of diameter $d$, i.e., the intersection
of three disks of radius $d$ centered at the vertices of a regular triangle of
side length $d$. Generalizing (5.2) to higher dimensions is an open problem,
see for instance [1, 8].

For the Euclidean width, the dual statement to the isodiametric inequality is
the Pál-Kakeya problem, which asks for the optimal lower bound on
$\operatorname{vol}(C)$ in terms of $\operatorname{width}_{\mathbb{R}}(C)$.
While it is not possible to bound $\operatorname{vol}(C)$ from above by
$\operatorname{width}_{\mathbb{R}}(C)$, Lassak conjectured a nice reverse
version of Pál’s inequality in the plane [30]: If $C\subset{\mathbb{R}}^{2}$
is a Euclidean reduced convex body with
$\operatorname{width}_{\mathbb{R}}(C)=w$, let $B_{w}$ be a Euclidean disk with
width $w$. Do we have

(5.3) $\operatorname{vol}(C)\leq\operatorname{vol}(B_{w})?$

Interestingly, the equality cases in (5.3) (if true) are not only the disks of
width $w$, but also the “quarter of a disk” $\\{x\in{\mathbb{R}}_{\geq
0}^{2}\colon|x|\leq w\\}$.

For lattice reduced and lattice complete convex bodies the volumetric analogs
of the isodiametric inequality and the Pál-Kakeya problem are Minkowski’s
first theorem and Makai’s conjecture (cf. Section 1). Moreover, in the plane
we obtain inequalities similar to (5.2) and (5.3) with comparatively little
effort. Recall that
$\operatorname{Vol}_{\Lambda}(C)=\operatorname{vol}(C)\det(\Lambda)^{-1}$
denotes the lattice normalized volume of $C$.

######  Proposition 5.2.

Let $C\subset{\mathbb{R}}^{2}$ be a convex body and let
$\Lambda\subset{\mathbb{R}}^{2}$ be a 2-dimensional lattice.

  1. (1)

If $C$ is reduced with respect to $\Lambda$, then
$\operatorname{Vol}_{\Lambda}(C)\leq\operatorname{width}_{\Lambda}(C)^{2}$.

  2. (2)

If $C$ is complete with respect to $\Lambda$, then
$\operatorname{Vol}_{\Lambda}(C)\geq\tfrac{1}{2}\operatorname{diam}_{\Lambda}(C)^{2}$.

Both inequalities are sharp.

###### Proof.

Without loss of generality, let $\Lambda={\mathbb{Z}}^{2}$ (cf. Lemma 2.3).
For (1), let $y_{1},y_{2}\in{\mathbb{Z}}^{2}$ be two independent width
directions of $C$. After applying a unimodular transformation and a
translation, we obtain a copy of $C$ that is contained in the square
$Q=[0,\operatorname{width}_{{\mathbb{Z}}^{2}}(C)]^{2}$.

For (2), let $S_{1},S_{2}$ be two non-parallel lattice diameter realizing
segments. Again, after applying a unimodular transformation, we can assume
that $S_{i}$ is parallel to $e_{i}$. Let $P=\operatorname{conv}(S_{1}\cup
S_{2})\subseteq C$. If the segments $S_{1}$ and $S_{2}$ intersect, it is clear
that $\operatorname{vol}(P)=\tfrac{1}{2}\operatorname{diam}_{\Lambda}(C)^{2}$.
Otherwise, we have
$\operatorname{vol}(P)>\tfrac{1}{2}\operatorname{diam}_{\Lambda}(C)^{2}$, as
can be seen by first applying a Steiner symmetrization with respect to one of
the affine hulls of the $S_{i}$ (see [19, Remark 1.1] for details). The
inequality now follows from $P\subseteq Q$.

In order to see that the inequalities are sharp, we consider the twisted
squares $Q_{x}$, $x\in(0,1)$, from Example 3.10. $Q_{x}$ is simultaneously
reduced and complete and we have $\operatorname{width}(Q_{x})=2$ and
$\lim_{x\nearrow 1}\operatorname{vol}(Q_{x})=4$, as well as $\lim_{x\searrow
0}\operatorname{diam}(Q_{x})=2$ and $\lim_{x\searrow
0}\operatorname{vol}(Q_{x})=2$. ∎

###  5.3. Bodies that are both reduced and complete

Recall that a convex body is complete in the Euclidean setting if and only if
it is of constant width. Hence, the class of Euclidean complete bodies is
(properly) contained in the class of Euclidean reduced bodies.

On the discrete side we have seen that for a fixed lattice
$\Lambda\subset{\mathbb{R}}^{d}$ the class of complete bodies and the class of
reduced bodies are neither contained in one another, nor dual to one another
(at least not in an obvious way). Nonetheless, it would be interesting to gain
a better understanding on how the notions of completeness and reducedness
interact in this setting:

######  Question 5.3.

What can be said about convex bodies $C\subset{\mathbb{R}}^{d}$ that are
simultaneously reduced and complete? Do there exist origin-symmmetric convex
bodies that are reduced and complete for $d\geq 3$?

Below we list the convex bodies that are both reduced and complete that we
know.

  1. (1)

The simplices $S_{d}\subset{\mathbb{R}}^{d}$ from Proposition 3.12.

  2. (2)

The quadrangles $Q_{x}\subset{\mathbb{R}}^{2}$, $x\in(-1,1)$ and the hexagon
$H$ from Example 3.10.

  3. (3)

The triangles obtained in Proposition 4.7.

  4. (4)

The local maximizers of the lattice width $T_{3}\subset{\mathbb{R}}^{3}$ and
$T_{4}\subset{\mathbb{R}}^{4}$ in dimensions 3 and 4 (cf. Remark 4.1).

  5. (5)

The regular simplex $\Delta$ from (4.3) with respect to the lattice $\Lambda$
as in (4.4), where $\alpha=\beta=\tfrac{3}{4}$ and $\gamma=\tfrac{1}{2}$. This
can be verified using the sagemath script from Remark 4.1. Note that since
$\gamma=\tfrac{1}{2}$, this simplex has an exceptional diameter segment $I$
passing between two of its edges. Curiously, our computation shows that all of
the width directions of $\Delta$ are orthogonal to $I$.

All examples in this list are either 2-dimensional, or simplices. To start
with, it would be interesting to see any reduced and complete polytope in
dimension higher than 2 which is not a simplex.

## Acknowledgements

The authors wish to thank Martina Juhnke-Kubitzke, Thomas Kahle, Raman Sanyal
and Christian Stump for organizing the Combinatorial CoworkSpace 2022, where
this collaboration started. Many thanks go to Gennadiy Averkov, Francisco
Santos, Martin Henk, Bernardo González Merino and Stefan Kuhlmann for helpful
discussions about various sections of the paper. Ansgar Freyer is partially
supported by the Austrian Science Fund (FWF) Project P34446-N.

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##  Appendix A Existence of realizers of $\mathrm{Flt}(d)$

Since we are not aware of an immediate argument or a reference in the
literature that shows that the flatness constant is achieved by a convex body,
we present a proof here in order to keep the paper self-contained.

######  Proposition A.1.

Let $\Lambda\subset{\mathbb{R}}^{d}$ be a $d$-dimensional lattice. There
exists a hollow (with respect to $\Lambda$) convex body
$C\subset{\mathbb{R}}^{d}$ with
$\operatorname{width}_{\Lambda}(C)=\mathrm{Flt}(d)$.

The proof requires the selection theorems of Blaschke and Mahler.

######  Theorem A.2 (Blaschke).

Let $(C_{i})_{i\in{\mathbb{N}}}$ be a sequence of convex bodies in
${\mathbb{R}}^{d}$ such that there exist convex bodies
$K,L\subset{\mathbb{R}}^{d}$ with $K\subset C_{i}\subset L$ for all
$i\in{\mathbb{N}}$. Then, $(C_{i})_{i\in{\mathbb{N}}}$ has a convergent
subsequence (in the Hausdorff metric).

Recall from Section 2 that two bases $A,B\in\mathrm{GL}_{d}({\mathbb{R}})$ of
a $d$-dimensional lattice $\Lambda\subset{\mathbb{R}}^{d}$ differ only by a
right multiplication with a unimodular matrix, i.e., we have $B=AU$, for some
$U\in\mathrm{GL}_{d}({\mathbb{Z}})$. Therefore the $d$-dimensional lattices in
${\mathbb{R}}^{d}$ correspond one-to-one to the elements of
$\mathrm{GL}_{d}({\mathbb{R}})/\mathrm{GL}_{d}({\mathbb{Z}})$. This turns the
set of $d$-dimensional lattices into a (metrizable) topological space by
considering the quotient topology on
$\mathrm{GL}_{d}({\mathbb{R}})/\mathrm{GL}_{d}({\mathbb{Z}})$.

######  Theorem A.3 (Mahler).

Let $(\Lambda_{i})_{i\in{\mathbb{N}}}$ be a sequence of $d$-dimensional
lattices in ${\mathbb{R}}^{d}$ such that there exist constants $c_{1},c_{2}>0$
such that $\det\Lambda_{i}<c_{1}$ and $\lambda_{1}(B_{d},\Lambda_{i})>c_{2}$,
where $B_{d}$ denotes the Euclidean unit ball. Then the sequence
$(\Lambda_{i})_{i\in{\mathbb{N}}}$ has a convergent subsequence

###### Proof of Proposition A.1.

Without loss of generality, let $\Lambda={\mathbb{Z}}^{d}$. Let
$(C_{i})_{i\in{\mathbb{N}}}$ be a sequence of convex bodies such that
$\lim_{i\to\infty}\operatorname{width}_{{\mathbb{Z}}^{d}}(C_{i})=\mathrm{Flt}(d)$.
By John’s theorem [46, Theorem 10.12.2] there exist vectors
$t_{i}\in{\mathbb{R}}^{d}$ and linear transformations
$A_{i}\in\mathrm{GL}_{d}({\mathbb{R}})$ such that

(A.1) $B_{d}\subseteq A_{i}C_{i}+t_{i}\subseteq d\,B_{d}.$

Let $\widetilde{C_{i}}=A_{i}C_{i}+t_{i}$ and
$\Lambda_{i}=A_{i}{\mathbb{Z}}^{d}$. By Theorem A.2, we can assume that that
$(\widetilde{C_{i}})_{i\in{\mathbb{N}}}$ is convergent to a convex body $C$.
Assume that also the sequence $(\Lambda_{i})_{i\in{\mathbb{N}}}$ converges to
a lattice $\Lambda$. Then, by the continuity of the successive minima,

$\operatorname{width}_{\Lambda}(C)=\lim_{i\to\infty}\operatorname{width}_{\Lambda_{i}}(\widetilde{C_{i}})=\lim_{i\to\infty}\operatorname{width}_{{\mathbb{Z}}^{d}}(C_{i})=\mathrm{Flt}(d),$

which proves the claim.

In order to show that $(\Lambda_{i})_{i\in{\mathbb{N}}}$ has a convergent
subsequence, we show that the hypothesis of Theorem A.3 are met. First, we
note that by Minkowski’s first theorem on successive minima combined with the
reverse Blaschke-Santaló inequality, there exists a weak resolution of
Makais’s conjecture, i.e., a constant $c_{1}>0$ depending only on $d$ such
that $\operatorname{vol}(C_{i})\geq
c_{1}\operatorname{width}_{{\mathbb{Z}}^{d}}(C_{i})^{d}$ (see for instance
[14, Lemma 3.2]). Thus,

(A.2) $\det\Lambda_{i}=\det
A_{i}=\frac{\operatorname{vol}{\widetilde{C_{i}}}}{\operatorname{vol}(C_{i})}\leq\frac{d^{d}\operatorname{vol}(B_{d})}{c_{1}\operatorname{width}_{{\mathbb{Z}}^{d}}(C_{i})^{d}},$

where we also used (A.1) to estimate the enumerator. It was shown in [12] that
$\mathrm{Flt}(d)\geq\mathrm{Flt}(d-1)+1$. Since
$\operatorname{width}_{{\mathbb{Z}}^{d}}(C_{i})$ converges to
$\mathrm{Flt}(d)$, we may assume

(A.3)
$\operatorname{width}_{{\mathbb{Z}}^{d}}(C_{i})\geq\mathrm{Flt}(d-1)+\tfrac{1}{2}.$

So we obtain from (A.2) that $\det\Lambda_{i}\leq c_{2}$, for a constant
$c_{2}>0$ depending only on $d$.

In order to treat the successive minima, we first use the inclusion (A.1) that
leads to the estimate

$\lambda_{1}(B_{d},\Lambda_{i})\geq\operatorname{diam}_{\Lambda_{i}}(\widetilde{C_{i}})^{-1}=\operatorname{diam}_{{\mathbb{Z}}^{d}}(C_{i})^{-1}.$

We can finish the proof by showing that

(A.4)
$\operatorname{diam}_{{\mathbb{Z}}^{d}}(C_{i})<\max\\{\mathrm{Flt}(d),2(\mathrm{Flt}(d-1)+1)\\}=:c_{3}.$

Towards a contradiction, assume that
$\operatorname{diam}_{{\mathbb{Z}}^{d}}(C_{i})>c_{3}$. Then there exists a
segment $S=[a,a+c_{3}v]\subset C_{i}$, where
$v\in{\mathbb{Z}}^{d}\setminus\\{0\\}$. Let $y\in{\mathbb{Z}}^{d}$ be a width
direction of $C_{i}$. We have

$\operatorname{width}_{{\mathbb{Z}}^{d}}(C_{i})=\operatorname{width}(C_{i};y)\geq
c_{3}|v\cdot y|.$

Since $v\cdot y\in{\mathbb{Z}}$ and $c_{3}>\mathrm{Flt}(d)$ it follows that
the width directions of $C_{i}$ are orthogonal to $v$. Let $K_{i}$ be the
orthogonal projection of $C_{i}$ on $v^{\perp}$ and let $\Gamma$ be the
lattice obtained by projecting ${\mathbb{Z}}^{d}$ on $v^{\perp}$. Then we have
$\operatorname{width}_{\Gamma}(K_{i})=\operatorname{width}_{{\mathbb{Z}}^{d}}(C_{i})$.

While $K_{i}$ is in general not hollow, we show in the following that it is
“almost hollow”. Let $p\in v^{\perp}$ be the point to which $S$ projects under
the orthogonal projection $\pi_{v}:{\mathbb{R}}^{d}\to v^{\perp}$. For a point
$x\in K_{i}\setminus\\{p\\}$, let