Integer Carath´eodory results with bounded multiplicity

. The integer Carath´eodory rank of a pointed rational cone C is the smallest number k such that every integer vector contained in C is an integral non-negative combination of at most k Hilbert basis elements. We investigate the integer Carath´eodory rank of simplicial cones with respect to their multiplicity, i.e., the determinant of the integral generators of the cone. One of the main results states that simplicial cones with multiplicity bounded by ﬁve have the integral Carath´eodory property, that is, the integer Carath´eodory rank equals the dimension. Furthermore, we present a novel upper bound on the integer Carath´eodory rank which depends on the dimension and the multiplicity. This bound improves upon the best known upper bound on the integer Carath´eodory rank if the dimension exceeds the multiplicity. At last, we present special cones which have the integral Carath´eodory property such as certain dual cones of Gorenstein cones.


Introduction
Given a full-dimensional pointed rational cone C ⊆ R n , every vector contained in C is a non-negative combination of the generators of C, i.e., vectors which lie on extreme rays of C. A classical result by Carathéodory states that each x ∈ C is a non-negative combination of at most n generators.In this manner, an integral analogue of Carathéodory's theorem is to ask whether every integral vector z ∈ C ∩ Z n can be expressed as the non-negative integral combination of at most n elements from the unique minimal generating set of C ∩ Z n , the Hilbert basis of C; see Subsection 2.1 for more details concerning Hilbert bases.
This was proven to be false by Bruns et al. [1999].They present cones C n which require at least 7 6 n elements from the Hilbert basis to express every integral vector in C n for each n ≥ 1.Interestingly, their lower bound is still the largest known lower bound.This example raises the question of what the correct upper bound on the smallest number k is such that every integer vector in C is an integral non-negative combination of at most k Hilbert basis elements.We write ICR(C) for this number k and refer to it as the integer Carathéodory rank of C; see Subsection 2.1 for a precise definition.The best known upper bound on ICR(C) is due to Sebő: Theorem A ([Sebő, 1990, Theorem 2.1]).Let C ⊆ R n be a full-dimensional pointed rational cone.Then Similar to the lower bound by Bruns et al., Sebő's upper bound remains the best known bound since decades.Another challenge concerning the integer Carathéodory rank is to classify and identify cones which admit ICR(C) = n, the minimal possible integer Carathéodory rank for full-dimensional cones.Throughout this paper, we refer to a cone C with ICR(C) = n that C has the integral Carathéodory property and abbreviate this by (ICP).One of the important results concerning the (ICP) is again due to Sebő: Theorem B ([Sebő, 1990, Theorem 2.2]).Let C ⊆ R n be a full-dimensional pointed rational cone with n ≤ 3. Then From the counterexamples in [Bruns et al., 1999] it is known that cones with n ≥ 6 do not have the (ICP), in general.It is open whether cones in dimension n ∈ {4, 5} admit the (ICP).To prove both theorems, Theorem A and Theorem B, Sebő relied on simplicial cones, i.e., full-dimensional cones which have precisely n extreme rays; see Subsection 2.1 for a more general definition.In particular, he employed the following high-level strategy: 1. Reduce from the general cone C to a special simplicial subcone C ′ ⊆ C.
2. Exploit the simplicial structure of C ′ .
In fact, this strategy is commonly utilized when proving results concerning ICR(C) or other variants of this problem; cf. for instance with [Bruns and Gubeladze, 2009, Chapter 3].This leads to the following open intriguing question: Does every simplicial cone have the (ICP)?
An affirmative answer has far-reaching consequences: It implies that the zonotope spanned by the integral generators of a cone contains a set of integral vectors, not necessarily the Hilbert basis, which admits the (ICP).
We contribute to the question of every simplicial cones having the (ICP) by investigating simplicial cones with respect to their multiplicity.In the following, we make precise what we mean by multiplicity in the full-dimensional case; see Subsection 2.1 for a more general definition.Let r 1 , . . ., r n ∈ Z n be linearly independent vectors.Then pos{r 1 , . . ., r n } is a full-dimensional simplicial cone, where pos X denotes the set of all finite non-negative combinations of elements in X ⊆ R n .The multiplicity of pos{r 1 , . . ., r n } is given by det(r 1 , . . ., r n ) .This quantity has various interpretations, for instance, it is known that det(r 1 , . . ., r n ) coincides with the number of integer vectors contained in the half-open parallelepiped spanned by r 1 , . . ., r n ; see, e.g., [Sebő, 1990, Lemma 2] for a proof.
One of our main results states that simplicial cones with small multiplicity have the (ICP).
We also provide a novel bound on the integer Carathéodory rank which improves upon Sebő's bound, Theorem A, if the multiplicity is bounded by the dimension, that is, det(r 1 , . . ., We allude that these statements resemble recent results on the integer Carathéodory rank and the (ICP) when the simplicial cone is given as a polyhedral representation [Aliev et al., 2022, Theorem 4].Although these results seem to be related, their proofs differ significantly.Also, we remark that it will become evident from the proofs that both results remain valid for k-dimensional simplicial cones in R n with the respective notion of multiplicity; see Subsection 2.1 for an introduction of these concepts.
An immediate follow-up question is to ask whether these bounds generalize to non-simplicial cones.We emphasize that this is not the case for Theorem 1.1: A natural generalization of the multiplicity for a full-dimensional non-simplicial cone pos{r 1 , . . ., r t } given by r 1 , . . ., r t ∈ Z n is to define δ(r 1 , . . ., r t ) as the largest multiplicity of any simplicial subcone formed by r 1 , . . ., r t , that is, Observe that cones with δ(r 1 , . . ., r t ) = 1 have the (ICP) since any triangulation of pos{r 1 , . . ., r t } using r 1 , . . ., r t gives a unimodular triangulation of the cone.However, in contrast to Theorem 1.1, one example given in [Bruns et al., 1999] admits a representation only using 0/1-vectors which satisfies δ(r 1 , . . ., r t ) = 3.So already cones with δ(r 1 , . . ., r t ) = 3 do not have the (ICP).This demonstrates that the simplicial case contains more structure, which makes the question of every simplicial cone having the (ICP) more intriguing.We emphasize that it is open whether cones with δ(r 1 , . . ., r t ) = 2 have the (ICP).
The key technique when proving Theorems 1.1 and 1.2 is a projection argument, Lemma 3.1, which we devote an extra section, Section 3. Afterwards, we prove our main results in Section 4. As an application of Lemma 3.1, one can collect various well-known cones which admit the (ICP); see Corollary 4.2 and Section 5.These include among others special instances of dual cones of simplicial Gorenstein cones.

Related work
Despite the existence of counterexamples to the question whether every cone has the (ICP), it remains an active research area to bound the integer Carathéodory rank and identify cones with minimal integer Carathéodory rank.In [Gijswijt and Regts, 2012], the authors show that cones which arise from polyhedra with additional properties, e.g., polyhedra defined by a totally unimodular constraint matrix, or (poly)matroid base polytopes have the (ICP).Another result related to matroids is ICR(C) ≤ n + r(M ) − 1, where C is given by the incidence vectors of bases of a matroid M and r(M ) denotes the rank of M ; see [de Pina and Soares, 2003].More recent bounds on the integer Carathéodory rank and novel instances which admit the (ICP) are given in [Aliev et al., 2022, Theorem 4].For those results, the authors assume that the cone is represented as the intersection of half-spaces.
Sebő introduced stronger versions of the (ICP) [Sebő, 1990].He asked whether the Hilbert basis gives a unimodular covering of the cone or, even stronger, a unimodular triangulation.Both questions turned out to be false since they would imply (ICP).Nevertheless, this led to the search of quantitative upper bounds on the size of a set which admits a unimodular cover or triangulation; see [Bruns and Gubeladze, 2009, Chapter 3B and 3C] or [Bruns and von Thaden, 2017;von Thaden, 2021] for some more recent work.This work relies on carefully analyzing the simplicial case.
Bruns and Gubeladze introduced a notion of an asymptotic integer Carathéodory rank.They proved an upper bound of 2n − 3 on this asymptotic version [1999].Recently, this was improved to ⌊ 3n 2 ⌋ in [Aliev et al., 2022], which is the first improvement on any version of the integer Carathéodory rank which does not have the factor 2 in front of n.In [2023], Gubeladze conjectured that the 2 in front of the n in Theorem A can not be improved for the integer Carathéodory rank of so-called normal polytopes.
Instead of analyzing the integer Carathéodory rank with respect to the Hilbert basis of a cone, it is also of interest to study an integer Carathéodory rank for arbitrary generating sets with respect to fixed elements; see [Eisenbrand and Shmonin, 2006].In addition to this, the authors provide a link to problems in integer programming such as the cutting stock problem.

Notation and definitions
In this section, we introduce additional notation and definitions which we use throughout the paper.In particular, we define the notion of Hilbert bases, integer Carathéodory rank, and the multiplicity.Moreover, we introduce lattices and discuss a projected version of the integer Carathéodory rank.
We abbreviate [m] := {1, . . ., m}.By int X we denote the interior of a set X and lin X is the linear hull of X.The standard unit vectors of R n are denoted by e 1 , . . ., e n .Let r 1 , . . ., r k ∈ Z n .We abbreviate the linear hull lin{r 1 , . . ., r k } by lin R for R = (r 1 , . . ., r k ) ∈ Z n×k .Similarly, we write pos R for pos{r 1 , . . ., r k }.

Hilbert bases, integer Carathéodory rank, and multiplicity
A cone C is pointed if 0 is the vertex of C. The Hilbert basis of a pointed rational cone C with respect to Z n is the minimal set of integral vectors in C such that every element in C ∩ Z n is a non-negative integral combination of the elements in the set.We denote the Hilbert basis of C by H(C).The integer vectors in the Hilbert basis are called Hilbert basis elements.It is known that the Hilbert basis is finite and unique for each pointed rational cone C; see [van der Corput, 1931].Moreover, a Hilbert basis element can not be written as the sum of two non-zero integral vectors in C ∩ Z n .The integer Carathéodory rank of C is defined as Let r 1 , . . ., r k ∈ Z n be linearly independent and R = (r 1 , . . ., r k ).Then the kdimensional cone pos R is referred to as simplicial.We define by par{r 1 , . . ., the set of integer vectors in the k-dimensional half-open parallelepiped spanned by r 1 , . . ., r k .The multiplicity of pos{r 1 , . . ., r k } is given by the number of integer vectors contained in the half-open parallelepiped spanned by r 1 , . . ., r k , that is, par{r 1 , . . ., r k } .We write ∆(r 1 , . . ., r k ) for the multiplicity.Observe that ∆(r 1 , . . ., r n ) = det(r 1 , . . ., r n ) when k = n; see, e.g., [Sebő, 1990, Lemma 2] for a proof.If ∆(r 1 , . . ., r k ) = 1, we refer to pos R as unimodular.Moreover, if C is a collection of unimodular cones such that pos R ⊆ C∈C C, we refer to the unimodular cones in C as a unimodular cover of pos R. According to the previously introduced notation, we abbreviate par{r 1 , . . ., r k } by par R.

Lattices
We introduce lattices and present several useful properties which we need throughout the paper; see [Lekkerkerker and Gruber, 1987] for an introduction to lattices.
The set Λ * := {x ∈ lin Λ : y T x ∈ Z for all y ∈ Λ} is a lattice and referred to as the dual lattice of Λ.Note that (b 1 ) * , . . ., In the case k = n, the dual basis of B = (b 1 , . . ., b n ) is given by the columns of B −T .

Projections and the integer Carathéodory rank
Let M ⊆ R n be a linear subspace.For a set X ⊆ R n , we denote by X|M the orthogonal projection of X onto M .If X = {x}, we simply write x|M and treat x|M as a vector.We denote by r ⊥ the orthogonal complement of the linear space spanned by the vector r ∈ R n \{0}.Let r 1 , . . ., r k be linearly independent with R = (r 1 , . . ., r k ) and M = (r k ) ⊥ .Observe that pos R|M is a (k − 1)-dimensional pointed cone given by the generators r 1 |M, . . ., r k−1 |M ; see also the first part of Lemma 3.3.We define ICR(pos R|M ) to be the integer Carathéodory rank of pos R|M with respect to the projected lattice (lin R ∩ Z n )|M .Indeed, this is equivalent to our setting, where we consider the integer lattice, due to the following transformation: ) is invertible.We return to the integer lattice by transforming everything with B −1 .So we obtain the cone This does not alter the integer Carathéodory rank.
The aim of this section is to prove Lemma 3.1.To do so, we need two necessary results which we state and prove below.
We prove the converse direction.At first, we claim that the Hilbert basis of pos R is contained in the parallelepiped spanned by r 1 , . . ., r k .Let y ∈ pos R ∩ Z n be such that y = k i=1 λ i r i and λ j > 1 for some j ∈ [k], i.e., y is not contained in the parallelepiped spanned by r 1 , . . ., r k .However, then we can decompose y into two integral vectors such that Both vectors are integral, contained in pos R, and not 0. Therefore, y is not a Hilbert basis element.
Thus, we get that the Hilbert basis is contained in par R ∪ {r 1 , . . ., r k }.So the elements in par R combined with r 1 , . . ., r k generate pos R ∩ Z n with integral and non-negative coefficients.We claim that they integrally generate L ∩ Z n as well.In order to see this, take some w ∈ L ∩ Z n .We have where the first term is an integer vector and, thus, the second term is contained in pos R ∩ Z n .Hence, the second term is a (non-negative) integral combination of the Hilbert basis.We obtain that the set par R ∪ {r 1 , . . ., r k } is a generating set of the lattice L ∩ Z n .
Since ((r i ) * −(r j ) * ) T y = 0 ∈ Z for all y ∈ par R and (( The proof of Lemma 3.1 consists of a projection argument.In the following, we show that pos R and par R behave well under certain orthogonal projections. Lemma 3.3.Let r 1 , . . ., r k ∈ Z n be linearly independent, R = (r 1 , . . ., r k ), and L = lin R. Then we have Proof.The linearity of the orthogonal projection onto (r i ) ⊥ and the fact that r i |(r i ) ⊥ = 0 imply the first claim and the inclusion " ⊆ " in the second claim.Without loss of generality let i = 1.Given y ∈ par{r 2 |(r 1 ) ⊥ , . . ., r k |(r 1 ) ⊥ }, there exist λ 2 , . . ., λ k ∈ [0, 1) such that y = k j=2 λ j (r j |(r 1 ) ⊥ ).We know there exists λ ∈ R such that k j=2 λ j r j + λr The claim follows from the observation that ỹ|(r 1 ) ⊥ = y.
We are in the position to prove the main statement of this section.
For the first part of the statement, we observe that This implies already the first statement.
For the second part of the statement, we assume without loss of generality that µ i ≥ µ j .There are with s ≤ ICR(pos R|(r i ) ⊥ ) and y 1 , . . ., y s ∈ par R ∪ {r 1 , . . ., r k } by Lemma 3.3.Hence, there exists some σ ∈ R which satisfies Our goal is to show that σ ∈ N which then implies the second claim.There are where τ t ∈ R n with τ t j = τ t j for all j ∈ [n] and t ∈ [s].This yields Further, we obtain τ t i = τ t j if y t ∈ par R from Lemma 3.2.This implies that the difference of the two sums above is 0 provided that y t ∈ par R ∪ {r 1 , . . .r k }\{r i , r j } for all t ∈ [s].In this special case, we get σ ∈ N as So we are left with the case y l = r i or y l = r j for some l ∈ [s].Note that y l = r i is not possible since this would mean y l |(r i ) ⊥ = 0. Hence, we assume y l = r j .Here, the difference of the two sums above is −σ l .As σ l ∈ N by construction, we get σ ∈ Z by the same argument as before.Furthermore, we have 0 ≤ µ i −µ j ≤ σ.We conclude that σ ∈ N which proves the second part of the lemma.
4 Proofs of Theorems 1.1 and 1.2 To prove Theorem 1.1, Theorem 1.2, and the Corollary 4.2 below, we employ the following strategy: First, we observe that we are in the situation of the first or second case of Lemma 3.1.After applying the lemma once, we argue that the multiplicity and coset structure of the elements in par R behave well.Afterwards, we transform everything back to integer lattice via the transformation outlined in Subsection 2.3 if necessary and repeat this procedure.
Before we begin to prove the two main statements, we have to ensure that the second step above is well-defined.
Proof.Every three-dimensional cone has the (ICP) due to Theorem B. Therefore, we assume that k ≥ 4. The assumption that {(r 1 ) * + (L ∩ Z n ) * , . . ., (r k ) * + (L ∩ Z n ) * } contains at most three non-trivial pairwise different elements implies that there either exists . Hence, we are in one of the cases of Lemma 3.1.
Hence, we can apply Lemma 3.1 as long as the dimension is at least four.In each step, we only add one element to our non-zero integral combination.We repeat this procedure until the dimension equals three.In this case, we utilize Theorem B to finish the proof of the statement.
We are in the position to prove Theorem 1.1 and Theorem 1.2.Here, we apply Lemma 3.1 and Lemma 4.1 in a similar manner as in the proof of Corollary 4.2.The main part of the proof is concerned with constructing a suitable cover by unimodular subcones in the case when |det R| = 5.
Proof of Theorem 1.1.The assumption 1 ≤ |det R| ≤ 5 implies that the number of cosets in Z n /RZ n is bounded by five.Hence, by duality, the number of cosets in (RZ n ) * /Z n equals five as well.Since one of the cosets is trivial, the set {(r 1 ) * + Z n , . . ., (r n ) * + Z n } contains at most four non-trivial pairwise different elements.In particular, if |det R| ≤ 4, the number of non-trivial pairwise different cosets is bounded by three.In this case, the claim follows from Corollary 4.2.
We show the theorem by covering the cone pos R with unimodular subcones formed by the Hilbert basis.The resulting unimodular covering implies the (ICP) since every integer vector in the cone is contained in an unimodular subcone spanned by the Hilbert basis.We proceed as follows: First, we present the construction of the covering in detail.In this process, it is important that the interiors of the subcones do not intersect.For the sake of readability and brevity, we do not formally verify this property.This can be done by formulating for each pair of simplicial subcones a linear program with an arbitrary non-zero linear functional, where the constraints are given by the polyhedral description of the two subcones with strict inequalities.The resulting linear program will turn out to be infeasible as the intersection of the interiors are empty.Alternatively, one can utilize suitable software such as polymake or Sage which is capable of computing the intersection and dimension of polyhedra.Once we have the collection of subcones, we prove that they indeed cover pos R via a volume argument.
To motivate the covering, observe that pos{y 1 , y 2 , y 3 , y 4 } is a three-dimensional cone contained in R 4 ∩ {x ∈ R 4 : (1, −1, −1, 1)R −1 x = 0}.For each one-dimensional face of pos{y 1 , y 2 , y 3 , y 4 }, there exists a unique three-dimensional face of pos R such that the cone spanned by the one-dimensional and three-dimensional face is unimodular.For instance, select y 1 , a generator of a one-dimensional face of pos{y 1 , y 2 , y 3 , y 4 }.Then, the unique three-dimensional face of pos R corresponding to y 1 is pos{r 2 , r 3 , r 4 }.Repeating this procedure for every one-dimensional face of pos{y 1 , y 2 , y 3 , y 4 } yields By computing determinants, one can verify that each cone is indeed a unimodular subcone of pos R.Moreover, it is possible to check that the interiors of these subcones do not intersect.In a similar manner, we fix two-dimensional faces of pos{y 1 , y 2 , y 3 , y 4 }.As earlier, for each of those faces, there exists a unique twodimensional face of pos R such that the resulting subcone is unimodular.This gives us the subcones Again, one can verify that the cones are unimodular and the interiors of all our current subcones do not intersect.We are left with covering the three-dimensional cone pos{y 1 , y 2 , y 3 , y 4 }.In addition to this, we still need to utilize the remaining two-dimensional faces of pos R, which are pos{r 1 , r 4 } and pos{r 2 , r 3 }.We proceed with covering the subcones pos{r 2 , r 3 , y 1 , y 2 , y 3 , y 4 } and pos{r 1 , r 4 , y 1 , y 2 , y 3 , y 4 }.This suffices to cover pos R, which we verify at the end of the proof.By symmetry, we only present a unimodular covering of pos{r 2 , r 3 , y 1 , y 2 , y 3 , y 4 } in detail.The covering of pos{r 1 , r 4 , y 1 , y 2 , y 3 , y 4 } works analogously.We begin with triangulating pos{y 1 , y 2 , y 3 , y 4 } using the two-dimensional subcone pos{y 2 , y 3 }.This yields the unimodular subcones We are left with the unimodular subcones Again, one can verify that the interiors have an empty intersection.When covering the other subcone pos{r 1 , r 4 , y 1 , y 2 , y 3 , y 4 }, one has to choose the two-dimensional subcone pos{y 1 , y 4 } to triangulate pos{y 1 , y 2 , y 3 , y 4 } and select the corresponding subcones as above.
Let C be the collection of the subcones above, including the unmentioned subcones in the covering of pos{r 1 , r 4 , y 1 , y 2 , y 3 , y 4 }.We claim that these subcones cover pos R. To show this, let H = {x ∈ R 4 : (1, 1, 1, 1)R −1 x ≤ 2}.Since it is possible to verfiy that the pairwise intersection of each subcone in C has empty interior, it suffices to argue that vol where vol(•) denotes the four-dimensional Lebesgue measure.In fact, we will show that equality holds in the inequality above.This implies that the simplices C ∩ H cover the simplex pos R ∩ H which proves the covering property of the cone by scaling.Observe that we have pos R ∩ H = 2 • conv{0, r 1 , r 2 , r 3 , r 4 }, where conv X denotes the convex hull of the set X ⊆ R n .So we deduce To determine the volume of the simplices corresponding to unimodular subcones in C, we have to take into account that the vertices coming from the generators are scaled to 2r 1 , . . ., 2r 4 .So we begin by counting the number of generators, r 1 , r 2 , r 3 , r 4 , in each subcone.There are four subcones with precisely three generators, ten subcones with two generators, and four subcones with only one generator.As every subcone is unimodular, we calculate vol So we conclude that pos R = C∈C C.
Although the collection of subcones from the previous proof form a unimodular covering and their interiors do not intersect they are not a unimodular triangulation of the cone pos R. The reason for this is that the three-dimensional cone pos{y 1 , y 2 , y 3 , y 4 } is triangulated once using pos{y 2 , y 3 } and once using pos{y 1 , y 4 }.
Proof of Theorem 1.2.Since n ≥ |det R| and (RZ n ) * /Z n has cardinality |det R|, either one of the vectors (r 1 ) * , . . ., (r n ) * is integral or the difference of two pairwise different vectors is integral by the pigeonhole principle.So we are in the situation of Lemma 3.1.Thus, we can reduce the dimension by one.By the first part of Lemma 4.1, the corresponding lower-dimensional volume of the respective parallelepiped does not exceed ∆(r 1 , . . ., r n ) = |det R|.We repeat this procedure until the dimension equals |det R|−1.As |det R|−1 ≥ 5 > 2, we can apply Sebő's bound, Theorem A, and obtain 5 Special cones with the (ICP) Utilizing our methods, we collect special instances of well-known simplicial cones which have the (ICP).This demonstrates the applicability of our developed theory.In fact, it should not be too difficult to obtain more examples using the methods from the earlier chapters.
We proceed by applying the slightly more abstract statement given in Corollary 4.2.We will see that the restriction on the number of cosets naturally translates to restrictions on the representation of our following cones.
Proof.We claim that {R −T e 1 + Z n , . . ., R −T e n + Z n } contains at most three nontrivial cosets in R −T Z n /Z n .Then, the statement follows from Corollary 4.2.
For our second example, we assume that there exists y ∈ int(pos R) ∩ par R with y + (pos R ∩ Z n ) = int(pos R) ∩ Z n .Special cones satisfying this premise are dual cones of simplicial Gorenstein cones.The general class of Gorenstein cones and polytopes plays an important role in the study of toric varieties; see, e.g., [Batyrev, 1994] for the first appearance of Gorenstein cones and their related polyhedra.
Proposition 5.2.Let R = (r 1 , . . ., r n ) ∈ Z n×n be such that there exists a vector y ∈ int(pos R) ∩ par R with y + (pos R ∩ Z n ) = int(pos R) ∩ Z n .Further, let |det R| have at most four divisors and let Z n /RZ n be cyclic.Then Proof.Let y = Rλ with λ ∈ 1 |det R| {1, . . ., |det R| − 1} n by Cramer's rule.Observe that λ i = 0 for i ∈ [n] since y is in the interior of the cone.As a first step, we prove that the coefficients of λ have to be small.In order to do so, we apply the transformation R −1 and analyze the elements in R Thus, we assume that µ is contained in the boundary of R n ≥0 , i.e., there exists l ∈ As a next step, we show that λ has at most three different entries.In order to do so, we need that Z n /RZ n is cyclic.This is equivalent to Λ/Z n being cyclic.We want to show that λ + Z n generates Λ/Z n .Since Λ/Z n is cyclic, there exists a µ ∈ [0, 1) n ∩ Λ such that µ + Z n generates Λ/Z n .This means that some entry of µ indexed by j ∈ [n] satisfies µ j = s |det R| for s ∈ {1, . . ., |det R|}, where s and |det R| are coprime.There exists m ∈ {1, . . ., |det R| − 1} with ms ≡ 1 mod |det R| which implies that mµ− ⌊mµ⌋ ∈ [0, 1) n ∩ Λ has j-th entry 1 |det R| .So we obtain λ j = 1 |det R| by (1).This yields that λ + Z n is a generator of Λ/Z n .We argue that λ has at most three different entries.Select i ∈ [n].Let λ i = p i q i for q i ∈ {1, . . ., |det R|} and p i ∈ {1, . . ., q i − 1}.Since λ + Z n generates Λ/Z n , we have q i µ i ∈ N for all µ ∈ [0, 1) n ∩ Λ.By a similar argument as before, there exists m ∈ {1, . . ., q i − 1} such that mp i ≡ 1 mod q i and, thus, mµ − ⌊mµ⌋ ∈ [0, 1) n ∩ Λ with µ i = 1 q i .Again, we conclude that λ i = 1 q i by (1).We know that q i divides |det R| by Cramer's rule.Since there are only four divisors of |det R|, where one of them is 1, we conclude that λ has at most three different entries.
As a final step, we claim that λ k = λ l for k = l implies R −T (e k − e l ) ∈ Z n .This property combined with the fact that λ has at most three different entries and Corollary 4.2 prove the statement.
In the statement of Proposition 5.2 we assume among other things that y ∈ par R. We remark that the statement remains valid if we drop this assumption and suppose that y ∈ int(pos R) ∩ Z n with y + (pos R ∩ Z n ) = int(pos R) ∩ Z n .In this case, it is possible to deduce that a face of pos R satisfies the premises of Proposition 5.2.
We end this section by presenting an example of a cone which meets the assumptions of Proposition 5.2 and does not satisfy the assumptions of Proposition 5.1.Let p, q ∈ N be distinct prime numbers and k ∈ {1, . . ., q − 1} and l ∈ {1, . . ., p − 1} such that kp + lq = pq − 1.The integers k and l always exist.For instance, choose k ∈ {1, . . ., q − 1} such that kp ≡ q − 1 mod q which implies pq − 1 − kp ≡ 0 mod q.So we have pq − 1 − kp = lq for some l ∈ {1, . . ., p − 1}.We define Note that det R = pq has four divisors, 1, p, q, and pq, and observe that yields y = Rλ ∈ Z 4 .So y + RZ 4 generates Z 4 /RZ 4 which implies that Z 4 /RZ 4 is cyclic.If there is some z ∈ int(pos R) ∩ Z 4 with z = Rµ which is not contained in y + (pos R ∩ Z 4 ), then there exists some j ∈ [4] with 0 < µ j < λ j .As λ 1 and λ 2 are already as small as possible, j has to be either three or four.However, 0 < µ j < λ j implies that (Rµ) j / ∈ Z.So we obtain y + (pos R∩ Z 4 ) = int(pos R)∩ Z 4 .Therefore, y meets the assumptions of Proposition 5.2.The matrix R is already in Hermite normal form; see [Schrijver, 1986, Chapter 4] for an introduction to Hermite normal forms and unimodular transformations.As the Hermite normal form is unique, the matrix R can not be transformed into a matrix represented by three standard unit vectors and an additional vector r via an unimodular transformation.We conclude that pos R does not satisfy the assumptions of Proposition 5.1.