Difference Between Families of Weakly and Strongly Maximal Integral Lattice-Free Polytopes

Abstract

A d-dimensional closed convex set K in $\mathbb {R}^d$ is said to be lattice-free if the interior of K is disjoint with $\mathbb {Z}^d$. We consider the following two families of lattice-free polytopes: the family $\mathcal {L}^d$ of integral lattice-free polytopes in $\mathbb {R}^d$ that are not properly contained in another integral lattice-free polytope and its subfamily $\mathcal {M}^d$ consisting of integral lattice-free polytopes in $\mathbb {R}^d$ which are not properly contained in another lattice-free set. It is known that $\mathcal {M}^d = \mathcal {L}^d$ holds for $d \le 3$ and, for each $d \ge 4$, $\mathcal {M}^d$ is a proper subfamily of $\mathcal {L}^d$. We derive a super-exponential lower bound on the number of polytopes in $\mathcal {L}^d {\setminus } \mathcal {M}^d$ (with standard identification of integral polytopes up to affine unimodular transformations).

1.1 Introduction

By |X| we denote the cardinality of a finite set X. Let $\mathbb {N}$ be the set of all positive integers and let $d \in \mathbb {N}$ be the dimension. Elements of $\mathbb {Z}^d$ are called integral points or integral vectors. We call a polyhedron $P \subseteq \mathbb {R}^d$ integral if P is the convex hull of $P \cap \mathbb {Z}^d$. Let ${{\,\mathrm{Aff}\,}}(\mathbb {Z}^d)$ be the group of affine transformations $A: \mathbb {R}^d \rightarrow \mathbb {R}^d$ satisfying $A(\mathbb {Z}^d) = \mathbb {Z}^d$. We call elements of ${{\,\mathrm{Aff}\,}}(\mathbb {Z}^d)$ affine unimodular transformations. For a family $\mathcal {X}$ of subsets of $\mathbb {R}^d$, we consider the family of equivalence classes

$$ \mathcal {X}/ {{\,\mathrm{Aff}\,}}(\mathbb {Z}^d) := \left\{ \left\{ A (X) \, : \, A \in {{\,\mathrm{Aff}\,}}(\mathbb {Z}^d) \right\} \, : \, X \in \mathcal {X} \right\} $$

with respect to identification of the elements of $\mathcal {X}$ up to affine unimodular transformations. A subset K of $\mathbb {R}^d$ is called lattice-free if K is closed, convex, d-dimensional and the interior of K contains no points from $\mathbb {Z}^d$. A set K is called maximal lattice-free if K is lattice-free and is not a proper subset of another lattice-free set.

Our objective is to study the relationship between the following two families of integral lattice-free polytopes:

• 1. The family $\mathcal {L}^d$ of integral lattice-free polytopes P in $\mathbb {R}^d$ such that there exists no integral lattice-free polytope properly containing P. We call elements of $\mathcal {L}^d$ weakly maximal integral lattice-free polytopes.

• 2. The family $\mathcal {M}^d$ of integral lattice-free polytopes P in $\mathbb {R}^d$ such that there exists no lattice-free set properly containing P. We call the elements of $\mathcal {L}^d$ strongly maximal integral lattice-free polytopes.

The family $\mathcal {L}^d$ has applications in mixed-integer optimization, algebra and algebraic geometry; see [1, 3, 4, 13], respectively. In [2, 11] it was shown that $\mathcal {L}^d$ is finite up to affine unimodular transformations:

Theorem 1

([2, Theorem 2.1], [11, Corollary 1.3]) $\mathcal {L}^d / {{\,\mathrm{Aff}\,}}(\mathbb {Z}^d)$ is finite.

Several groups of researchers are interested in enumeration of $\mathcal {L}^d$, up to affine unimodular transformations, in fixed dimensions. This requires understanding geometric properties of $\mathcal {L}^d$. Currently, no explicit description of $\mathcal {L}^d$ is available for dimensions $d \ge 4$ and, moreover, it is even extremely hard to decide if a given polytope belongs to $\mathcal {L}^d$. A brute-force algorithm based on volume bounds for $\mathcal {L}^d$ (provided in [11]) would have doubly exponential running time in d. In contrast to $\mathcal {L}^d$, its subfamily $\mathcal {M}^d$ is easier to deal with. Lovász’s characterization [9, Proposition 3.3] of maximal lattice-free sets leads to a straightforward geometric description of polytopes belonging to $\mathcal {M}^d$. This characterization can be used to decide whether a given polytope is an element of $\mathcal {M}^d$ in only exponential time in d. Thus, while enumeration of $\mathcal {M}^d$ in fixed dimensions is a hard task, too, enumeration of $\mathcal {L}^d$ is even more challenging.

For a given dimension d, it is a priori not clear whether or not $\mathcal {M}^d$ is a proper subset of $\mathcal {L}^d$. Recently, it has been shown that the inequality $\mathcal {M}^d = \mathcal {L}^d$ holds if and only if $d \le 3$. The equality $\mathcal {M}^d = \mathcal {L}^d$ is rather obvious for $d \in \{1,2\}$, as it is not hard to enumerate $\mathcal {L}^d$ in these very small dimensions and to check that every element of $\mathcal {L}^d$ belongs to $\mathcal {M}^d$. Starting from dimension three, the problem gets very difficult. Results in [1, 2] establish the equality $\mathcal {M}^3 = \mathcal {L}^3$ and enumerate $\mathcal {L}^3$, up to affine unimodular transformations. As a complement, in [11, Theorem 1.4] it was shown that for all $d \ge 4$ there exists a polytope belonging to $\mathcal {L}^d$ but not to $\mathcal {M}^d$.

While Theorem 1.4 in [11] shows that $\mathcal {L}^d$ and $\mathcal {M}^d$ are two different families, it does not provide information on the number of polytopes in $\mathcal {L}^d$ that do not belong to $\mathcal {M}^d$. Relying on a result of Konyagin [6], we will show that, asymptotically, the gap between $\mathcal {L}^d$ and $\mathcal {M}^d$ is very large.

For $a_1,\ldots ,a_d >0$, we introduce

$$ \kappa (a):=\kappa (a_1,\ldots ,a_d) = \frac{1}{a_1} + \cdots + \frac{1}{a_d}. $$

Reciprocals of positive integers are sometimes called Egyptian fractions. Thus, if $a \in \mathbb {N}^d$, then $\kappa (a)$ is a sum of d Egyptian fractions. We consider the set

$$ \mathcal {A}_d := \left\{ (a_1,\ldots ,a_d) \in \mathbb {N}^d \, : \, a_1 \le \cdots \le a_d, \ \kappa (a_1,\ldots ,a_d)=1 \right\} $$

of all different solutions of the Diophantine equation

$$ \kappa (x_1,\ldots ,x_d)=1 $$

in the unknowns $x_1,\ldots ,x_d \in \mathbb {N}$. The set $\mathcal {A}_d$ represents possible ways to write 1 as a sum of d Egyptian fractions. It is known that $\mathcal {A}_d$ is finite. Our main result allows is a lower bound on the cardinality of $(\mathcal {L}^d{\setminus } \mathcal {M}^d) / {{\,\mathrm{Aff}\,}}(\mathbb {Z}^d)$:

Theorem 2

$\bigl |(\mathcal {L}^{d+5}{\setminus } \mathcal {M}^{d+5}) / {{\,\mathrm{Aff}\,}}(\mathbb {Z}^{d+5})\bigr | \ge \bigl |\mathcal {A}_d\bigr |.$

The proof of Theorem 2 is constructive. This means that, for every $a \in \mathcal {A}_d$, we generate an element in $P_a \in \mathcal {L}^{d+5} {\setminus } \mathcal {M}^{d+5}$ such that for two different elements a and b of $\mathcal {A}_d$, the respective polytopes $P_a$ and $P_b$ do not coincide up to affine unimodular transformations. The proof of Theorem 2 is inspired by the construction in [11]. Using lower bounds on $\bigl |\mathcal {A}_d\bigr |$ from [6], we obtain the following asymptotic estimate:

Corollary 3

$\ln \ln \bigl |\bigl ( \mathcal {L}^d {\setminus } \mathcal {M}^d \bigr ) / {{\,\mathrm{Aff}\,}}(\mathbb {Z}^d)\bigr | = \Omega \left( \frac{d}{\ln d} \right)$, as $d \rightarrow \infty$.

Note 4

We view the elements of $\mathbb {R}^d$ as columns. By o we denote the zero vector and by $e_1,\ldots ,e_d$ the standard basis of $\mathbb {R}^d$. If $x \in \mathbb {R}^d$ and $i \in \{1,\ldots ,d\}$, then $x_i$ denotes the i-th component of x. The relation $a \le b$ for $a, b \in \mathbb {R}^d$ means $a_i \le b_i$ for every $i \in \{1,\ldots ,d\}$. The relations $\ge ,>$ and < on $\mathbb {R}^d$ are introduced analogously. The abbreviations ${{\,\mathrm{aff}\,}}, {{\,\mathrm{conv}\,}}, {{\,\mathrm{int}\,}}$ and ${{\,\mathrm{relint}\,}}$ stand for the affine hull, convex hull, interior and relative interior, respectively.

1.2 An Approach to Construction of Polytopes in $\mathcal {L}^d {\setminus } \mathcal {M}^d$

We will present a systematic approach to construction of polytopes in $\mathcal {L}^d {\setminus } \mathcal {M}^d$, but first we discuss general maximal lattice-free sets.

Definition 5

Let P be a lattice-free polyhedron in $\mathbb {R}^d$. We say that a facet F of P is blocked if the relative interior of F contains an integral point.

Maximal lattice-free sets can be characterized as follows:

Proposition 6

([9, Proposition 3.3]) Let K be a d-dimensional closed convex subset of $\mathbb {R}^d$. Then the following conditions are equivalent:

• 1. K is maximal lattice-free;

• 2. K is a lattice-free polyhedron such that every facet of K is blocked.

It can happen that some facets of a maximal lattice-free polyhedron are more than just blocked. We introduce a respective notion. Recall that the integer hull ${K}_I$ of a compact convex set K in $\mathbb {R}^d$ is defined by

$$ {K}_I:= {{\,\mathrm{conv}\,}}( K \cap \mathbb {Z}^d). $$

Definition 7

Let P be a d-dimensional lattice-free polyhedron in $\mathbb {R}^d$. A facet F of P is called strongly blocked if ${F}_I$ is $(d-1)$-dimensional and $\mathbb {Z}^d \cap {{\,\mathrm{relint}\,}}{F}_I \ne \emptyset$. The polyhedron P is called strongly blocked if all facets of P are strongly blocked.

The following proposition extracts the geometric principle behind the construction from [11, Sect. 3]. (Note that arguments in [11, Sect. 3] use an algebraic language.)

Proposition 8

Let P be a strongly blocked lattice-free polytope in $\mathbb {R}^d$. Then ${P}_I\in \mathcal {L}^d$. Furthermore, if P is not integral, then ${P}_I \not \in \mathcal {M}^d$.

Proof

In order to show ${P}_I \in \mathcal {L}^d$ it suffices to verify that, for every $z \in \mathbb {Z}^d$ such that ${{\,\mathrm{conv}\,}}({P}_I \cup \{z\})$ is lattice-free, one necessarily has $z \in {P}_I$. If $z \not \in {P}_I$, then $z \not \in P$ and so, for some facet F of P, the point z and the polytope P lie on different sides of the hyperplane ${{\,\mathrm{aff}\,}}F$. Then $\emptyset \ne \mathbb {Z}^d \cap {{\,\mathrm{relint}\,}}{F}_I \subseteq {{\,\mathrm{int}\,}}( {{\,\mathrm{conv}\,}}(P \cup \{z\}))$, yielding a contradiction to the choice of z. Thus, for every facet F of P, z and P lie on the same side of ${{\,\mathrm{aff}\,}}F$. It follows $z \in P$. Hence $z \in P \cap \mathbb {Z}^d \subseteq {P}_I$.

If P is not integral, then ${P}_I \not \in \mathcal {M}^d$ since ${P}_I \varsubsetneq P$ and P is lattice-free.   $\square$

1.3 Lattice-Free Axis-Aligned Simplices

For $a \in \mathbb {R}_{>0}^d$, the d-dimensional simplex

$$ T(a):= {{\,\mathrm{conv}\,}}\{ o, a_1 e_1, \ldots , a_d e_d \}. $$

is called axis-aligned. The proof of the following proposition is straightforward.

Proposition 9

For $a \in \mathbb {R}_{>0}^d$, the following statements hold:

• 1. the simplex T(a) is a lattice-free set if and only if $\kappa (a) \ge 1$;

• 2. the simplex T(a) is a maximal lattice-free set if and only if $\kappa (a)=1$.

We introduce transformations which preserve the values of $\kappa$. The transformations arise from the following trivial identities for $t>0$:

$$\begin{aligned} \frac{1}{t}&= \frac{1}{t+1} + \frac{1}{t(t+1)}, \end{aligned}$$

(1.1)

$$\begin{aligned} \frac{1}{t}&= \frac{1}{t+2} + \frac{1}{t(t+2)} + \frac{1}{t(t+2)}, \end{aligned}$$

(1.2)

$$\begin{aligned} \frac{1}{t}&= \frac{2}{3 t} + \frac{1}{3 t}. \end{aligned}$$

(1.3)

Consider a vector $a \in \mathbb {R}_{>0}^d$. By (1.1), if t is a component of a, we can replace this component with two new components $t+1$ and $t(t+1)$ to generate a vector $b \in \mathbb {R}_{>0}^{d+1}$ satisfying $\kappa (b)=\kappa (a)$. Identities (1.2) and (1.3) can be applied in a similar fashion. For every $d \in \mathbb {N}$, with the help of (1.1)–(1.3), we introduce the following maps:

$$\begin{aligned} \phi _d&:\mathbb {R}_{>0}^d \rightarrow \mathbb {R}_{>0}^{d+1},&\phi _d(a)&:= \begin{pmatrix} a_1 \\ \vdots \\ a_{d-1} \\ a_d+1 \\ a_d(a_d+1) \end{pmatrix}, \end{aligned}$$

(1.4)

$$\begin{aligned} \psi _d&: \mathbb {R}_{>0}^d \rightarrow \mathbb {R}_{>0}^{d+3},&\psi _d(a)&: = \begin{pmatrix} a_1 \\ \vdots \\ a_{d-1} \\ a_d+3 \\ a_d(a_d+1) \\ (a_d+1)(a_d+3) \\ (a_d+1)(a_d+3) \end{pmatrix} , \nonumber \\ \xi _d&: \mathbb {R}_{>0}^d \rightarrow \mathbb {R}_{>0}^{d+1}&\xi _d(a)&:= \begin{pmatrix} a_1 \\ \vdots \\ a_{d-1} \\ \frac{3}{2} a_d \\ 3a_d \end{pmatrix}. \end{aligned}$$

(1.5)

The map $\phi _d$ replaces the component $a_d$ by two other components based on (1.1), while $\xi _d$ replaces $a_d$ based on (1.3). The map $\psi _d$ acts by replacing the component $a_d$ based on (1.1) and then replacing the component $a_d+1$ based on (1.2). Identities (1.1)–(1.3) imply

$$\begin{aligned} \kappa (\phi _d(a)) = \kappa (\psi _d(a))=\kappa (\xi _d(a)))= \kappa (a). \end{aligned}$$

(1.6)

Lemma 10

Let $P=T(\xi _d(a))$, where $a \in \mathcal {A}_d$ and $d \ge 2$. Then P is a strongly blocked lattice-free $(d+1)$-dimensional polytope. Furthermore, if $a_d$ is odd, P is not integral.

Proof

In this proof, we use the all-ones vector

$$ \mathbbm {1}_{d} := \begin{pmatrix} 1 \\ \vdots \\ 1 \end{pmatrix} \in \mathbb {R}^d. $$

For the sake of brevity we introduce the notation $t:=a_d$. One has $1= \kappa (a) = \sum _{i=1}^d \frac{1}{a_i} \ge \sum _{i=1}^d \frac{1}{t} = \frac{d}{t}$, which implies $t \ge d \ge 2$. By (1.6), one has $\kappa (\xi _d(a))=1$ and so, by Proposition 9, P is maximal lattice-free.

If t is even, the polytope P is integral and hence every facet of P is integral, too. In view of Proposition 6, integral maximal lattice-free polytopes are strongly blocked, and so we conclude that P is strongly blocked.

Assume that t is odd, then the polytope P has one non-integral vertex. In this case, we need to look at facets of P more closely, to verify that P is strongly blocked. We consider all facets of P.

• 1. The facet $F={{\,\mathrm{conv}\,}}\bigl \{o,a_1 e_1,\ldots ,a_{d-1} e_{d-1}, 3 t e_{d+1}\bigr \}$ is a d-dimensional integral integral axis-aligned simplex. Since

  $$ \kappa (a_1,\ldots ,a_{d-1},3t) < 1, $$

  the integral point $e_1 + \cdots + e_{d-1} + e_{d+1}$ is in the relative interior of F. Hence, F is strongly blocked.

• 2. The facet $F = {{\,\mathrm{conv}\,}}\Bigl \{o,a_1 e_1,\ldots ,a_{d-1} e_{d-1}, \frac{3}{2} t e_d\Bigr \}$ contains the d-dimensional integral axis-aligned simplex

  $$ G:={{\,\mathrm{conv}\,}}\Bigl \{o, a_1 e_1,\ldots ,a_{d-1} e_{d-1}, \frac{3t-1}{2} e_d\Bigr \}, $$

  as a subset. In view of $t \ge 2$, we have

  $$ \kappa \Bigl (a_1,\ldots ,a_{d-1},\frac{3t-1}{2}\Bigr ) < 1, $$

  which implies that the integral point $e_1 + \cdots + e_d$ is in the relative interior of G. It follows that F is strongly blocked.

• 3. The facet $F:= {{\,\mathrm{conv}\,}}\Bigl \{a_1e_1,\ldots ,a_{d-1} e_{d-1}, \frac{3}{2} t e_d, 3 t e_{d+1} \Bigr \}$ contains the integral d-dimensional simplex

  $$ G: = {{\,\mathrm{conv}\,}}\Bigl \{a_1 e_1,\ldots ,a_{d-1} e_{d-1}, \frac{3 t-1}{2} e_d + e_{d+1}, 3 t e_{d+1}\Bigr \}. $$

  as a subset. It turns out that $\mathbbm {1}_{d+1}$ is the relative interior of G, because $\mathbbm {1}_{d+1}$ is a convex combination of the vertices of ${{\,\mathrm{relint}\,}}G$, with positive coefficients. Indeed, the equality

  $$ \mathbbm {1}_{d+1} = \sum _{i=1}^{d-1} \frac{1}{a_i} \bigl (a_i e_i\bigr ) + \lambda \Bigl (\frac{3t-1}{2} e_d + e_{d+1}\Bigr ) + \mu \bigl (3 t e_{d+1}\bigr ) $$

  holds for $\lambda = \frac{2}{3t-1}$ and $\mu = \frac{t-1}{t(3t-1)}$, where

  $$ \sum _{i=1}^{d-1} \frac{1}{a_i} + \lambda + \mu = 1. $$

• 4. It remains to consider faces F with the vertex set

  $$ \left\{ o, a_1 e_1,\ldots ,a_d e_d, \frac{3}{2} t e_d, 3 t e_{d+1} \right\} {\setminus } \{a_i e_i\}, $$

  where $i \in \{1,\ldots ,d+1\}$. Without loss of generality, let $i=1$ so that

  $$\begin{aligned} F = {{\,\mathrm{conv}\,}}\left\{ o, a_2 e_2,\ldots , \frac{3}{2} t e_d, 3 t e_{d+1} \right\} . \end{aligned}$$

  This facet contains the integral d-dimensional simplex

  $$ G: = {{\,\mathrm{conv}\,}}\Bigl \{o, a_2 a_2,\ldots ,a_{d-1} e_{d-1}, \frac{3 t-1}{2} e_d + e_{d+1}, 3 t e_{d+1}\Bigr \}. $$

  Similarly to the previous case, one can check that $e_2 + \cdots + e_{d+1}$ is an integral point in the relative interior of G. Consequently, F is strongly blocked.   $\square$

1.4 Proof of the Main Result

For $d \ge 4$, Nill and Ziegler [7] construct one vector $a\in \mathbb {R}_{>0}^d$ with ${T(a)}_I \in \mathcal {L}^d {\setminus } \mathcal {M}^d$. We generalize this construction and provide many further vectors a with the above properties. We will also need to verify that for different choices of a, we get essentially different polytopes ${T(a)}_I$.

Lemma 11

Let P and Q be d-dimensional strongly blocked lattice-free polytopes such that for their integral hulls the equality ${Q}_I=A({P}_I)$ holds for some $A \in {{\,\mathrm{Aff}\,}}(\mathbb {Z}^d)$. Then $Q=A(P)$.

Proof

Since A is an affine transformation, we have

$$\begin{aligned} A(P_I) = A({{\,\mathrm{conv}\,}}(P \cap \mathbb {Z}^d)) = {{\,\mathrm{conv}\,}}A(P \cap \mathbb {Z}^d). \end{aligned}$$

Using $A \in {{\,\mathrm{Aff}\,}}(\mathbb {Z}^d)$, it is straightforward to check the equality $A(P \cap \mathbb {Z}^d) = A(P) \cap \mathbb {Z}^d$. We thus conclude that $A({P}_I) = {A(P)}_I$. The assumption ${Q}_I = A({P}_I)$ yields ${Q}_I {=} {A(P)}_I$. Since P is strongly blocked lattice-free, A(P) too is strongly blocked lattice-free. We thus have the equality ${Q}_I={A(P)}_I$ for strongly blocked lattice-free polytopes Q and A(P). To verify the assertion, it suffices to show that a strongly blocked lattice-free polytope Q is uniquely determined by the knowledge of its integer hull ${Q}_I$. This is quite easy to see. For every strongly blocked facet G of ${Q}_I$, the affine hull of G contains a facet of Q. Conversely, if F is an arbitrary facet of Q, then $G= {F}_I$ is a strongly blocked facet of ${Q}_I$. Thus, the knowledge of ${Q}_I$ allows to determine affine hulls of all facets of Q. In other words, ${Q}_I$ uniquely determines a hyperplane description of Q.   $\square$

Lemma 12

Let $a,b \in \mathbb {R}_{>0}^d$ be such that the equality $T(b)= A(T(a))$ holds for some $A \in {{\,\mathrm{Aff}\,}}(\mathbb {Z}^d)$. Then a and b coincide up to permutation of components.

Proof

We use induction on d. For $d=1$, the assertion is trivial. Let $d \ge 2$. One of the d facets of T(a) containing o is mapped by A to a facet of T(b) that contains o. Without loss of generality we can assume that the facet $T(a_1,\ldots ,a_{d-1}) \times \{0\}$ of T(a) is mapped to the facet $T(b_1,\ldots ,b_{d-1}) \times \{0\}$ of T(b). By the inductive assumption, $(a_1,\ldots ,a_{d-1})$ and $(b_1,\ldots ,b_{d-1})$ coincide up to permutation of components. Since unimodular transformations preserve the volume, T(a) and T(b) have the same volume. This means, $\prod _{i=1}^d a_i = \prod _{i=1}^d b_i$. Consequently, $a_d=b_d$ and we conclude that a and b coincide up to permutation of components.

Proof

(Proof of Theorem 2) For every $a \in \mathcal {A}_d$, we introduce the $(d+5)$-dimensional integral lattice-free polytope

$$ P_a:={T(\eta (a))}_I, $$

where

$$ \eta (x) := \xi _{d+4}(\psi _{d+1}(\phi _d(x))) $$

and the functions $\xi _{d+4}, \psi _{d+1}$ and $\phi _d$ are defined by (1.4)–(1.5).

By (1.6) for each $a \in \mathcal {A}_d$, we have $\kappa (\eta (a))=1$. For $a \in \mathcal {A}_d$ the last component of $\phi _d(a)$ is even. This implies that the last component of $\psi _{d+1} (\phi _d(a))$ is odd. Thus, by Lemma 10, $T(\eta (a))$ is strongly blocked lattice-free polytope which is not integral.

Let $a, b \in \mathcal {A}_d$ be such that the polytopes $P_a$ and $P_b$ coincide up to affine unimodular transformations. Then, by Lemma 11, $T(\eta (a))$ and $T(\eta (b))$ coincide up to affine unimodular transformations. But then, by Lemma 12, $\eta (a)$ and $\eta (b)$ coincide up to permutations. Since the components of a and b are sorted in the ascending order, the components of $\eta (a)$ and $\beta (b)$ too are sorted in the ascending order. Thus, we arrive at the equality $\eta (a) = \eta (b)$, which implies $a=b$.

In view of Proposition 8, each $P_a$ with $a \in \mathcal {A}_d$ belongs to $\mathcal {L}^d$ but not to $\mathcal {M}^d$. Thus, the equivalence classes of the polytopes $P_a$ with $a \in \mathcal {A}_d$ with respect to identification up to affine unimodular transformations form a subset of $(\mathcal {L}^{d+5} {\setminus } \mathcal {M}^{d+5} ) / {{\,\mathrm{Aff}\,}}(\mathbb {Z}^{d+5})$ of cardinality $|\mathcal {A}_d|$. This yields the desired assertion.   $\square$

Proof

(Proof of Corollary 3) The assertion is a direct consequence of Theorem 2 and the asymptotic estimate

$$ \ln \ln |\mathcal {A}_d| = \Omega \left( \frac{d}{\ln d} \right) $$

of Konyagin [6, Theorem 1] (see also [5, Corollary 1.2]).   $\square$

Remark 13

In view of the asymptotic upper bound $\ln \ln |\mathcal {A}_d| = O(d)$, determined with different degrees of precision in [8, 10] and [12, Theorem 2], the lower bound of Konyagin is optimal up to the logarithmic factor in the denominator.

Since all known elements of $\mathcal {L}^d$ are of the form ${P}_I$, for some strongly blocked lattice-free polytope P, we ask the following:

Question 14

Do there exist polytopes $L \in \mathcal {L}^d$ which cannot be represented as $L={P}_I$ for any strongly blocked lattice-free polytope P?

If there is a gap between the families $\mathcal {L}^d$ and the family

$$ \left\{ {P}_I \, : \, P \subseteq \mathbb {R}^d \ \text {strongly blocked lattice-free polytope} \right\} , $$

then it would be interesting to understand how irregular the polytopes from this gap can be. For example, one can ask the following:

Question 15

Do there exist polytopes $L \in \mathcal {L}^d$ with the property that no facet of L is blocked?