Maximal integral simplices with no interior integer points

# Maximal integral simplices with no interior integer points

Kent Andersen    Christian Wagner    Robert Weismantel
###### Abstract

In this paper, we consider integral maximal lattice-free simplices. Such simplices have integer vertices and contain integer points in the relative interior of each of their facets, but no integer point is allowed in the full interior. In dimension three, we show that any integral maximal lattice-free simplex is equivalent to one of seven simplices up to unimodular transformation. For higher dimensions, we demonstrate that the set of integral maximal lattice-free simplices with vertices lying on the coordinate axes is finite. This gives rise to a conjecture that the total number of integral maximal lattice-free simplices is finite for any dimension.

## 1 Introduction

In dimension , a simplex is defined to be the convex hull of affinely independent points. It is called integral if its vertices have integer coordinates. Integral simplices have been studied in several contexts. In particular, the literature is rich in investigations of integral simplices that contain no other lattice points besides the vertices - neither on the boundary nor in the interior (see e.g. Reznick [8], Scarf [9], Sebö [11]). This notion of lattice-freeness is an interesting concept which has proven to be valuable for some problems in integer programming and combinatorics. Most notably, this notion is used for primal integer programming and the study of neighbors of the origin (see Scarf [9]).

In this paper, we consider a different notion of lattice-freeness. The application we have in mind is to use integral simplices as a tool to generate cutting planes for (mixed) integer linear problems (see e.g. [1, 2, 3, 4, 5, 13]). For this purpose, we employ the notion of lattice-freeness introduced by Lovász [7].

###### Definition 1.1

A simplex is called lattice-free if .

To obtain deep cutting planes, we look for integral lattice-free simplices which are maximal with respect to inclusion and call them integral maximal lattice-free simplices. A well-known result of Lovász [7] is that for such simplices each facet contains an integer point in its relative interior.

The application to cutting plane generation requires however to have an explicit list of integral maximal lattice-free simplices available. The partial knowledge about structural properties of such bodies is definitely not enough.

In dimension two, it can easily be verified that any integral maximal lattice-free simplex can be unimodularly transformed to . However, to the best of our knowledge, a characterization of integral maximal lattice-free simplices in higher dimensions is not known. Moreover, it is not known if their number is finite.

We note that a recent paper of Treutlein [12] shows finiteness of integral maximal lattice-free simplices in dimension three. In this paper, we extend on this result: we completely characterize integral maximal lattice-free simplices in dimension three and show that - up to unimodular transformation - only seven different simplices exist. Furthermore, for a special class of integral maximal lattice-free simplices, namely simplices with vertices on the coordinate axes, we argue that their number is finite for any dimension . This gives rise to the conjecture that the total number of integral maximal lattice-free simplices is finite, in general.

Section 2 is dedicated to the analysis of the three dimensional case. Extensions are considered in Section 3. Sections 4 and 5 contain details of our proof technique.

## 2 Simplices in dimension three

Let be an integral maximal lattice-free simplex. We assume that , where , and are integer vectors. By computing the Hermite normal form of the matrix , it can be assumed that , , and such that all coefficients , and are integer and, in addition, , , , and hold (see Schrijver [10]). Furthermore, we can assume that and since for it follows and thus the facet spanned by the three points does not contain an interior integer point and therefore is not maximal lattice-free. On the other hand, for , is contained in the split which is a contradiction to its maximality. Hence, we have . In the remainder of this paper we work with the following inequality representation of :

 ⎛⎜ ⎜ ⎜ ⎜⎝00−10−fe−cfbfcd−becff(a−b)e(b−a)+c(a−d)⎞⎟ ⎟ ⎟ ⎟⎠⎛⎜⎝x1x2x3⎞⎟⎠≤⎛⎜ ⎜ ⎜⎝000acf⎞⎟ ⎟ ⎟⎠. (1)

Our proof strategy is based on partitioning the set of potential integral maximal lattice-free simplices according to relations among the unknowns and . For each of the subcases we then manage to compute upper bounds on , and . Once this has been established, the integral maximal lattice-free simplices can be computed by enumeration. The enumeration provides a list of simplices which must then be checked for unimodular equivalence.

###### Definition 2.1

Two sets are unimodularly equivalent if there exist a unimodular matrix and a vector such that , where .

For integral maximal lattice-free simplices and in it follows that they are unimodularly equivalent if there exist a matrix with and a vector such that for all , where is a permutation.

We distinguish our analysis into the two major cases and . For there are two integer points which play a key role in our subcase analysis. In the following let

 k:=⌈ac+b−ac⌉−1.

The distinctions in our subcase analysis for are based on the locations of the points and relative to . Geometrically, this can be interpreted as follows: Firstly, we investigate simplices where the point either lies on or violates the fourth facet of (1). Afterwards, we consider the opposite case and divide, secondly, into simplices where the point either lies on or violates the third facet of (1) and simplices where this is not the case.

Here is the structure of the case distinction for with the corresponding bounds on , and :

1.       ( means: either lies on or violates the fourth facet of (1) )

1. no integral maximal lattice-free simplex possible

2.       ( means: strictly satisfies the fourth facet of (1) )

1.       ( means: either lies on or violates the third facet of (1) )

2.       ( means: strictly satisfies the third facet of (1) )

1. no integral maximal lattice-free simplex possible

The analysis of the subcases is technical and tedious, but not complicated, in principle. The complete analysis is given in Section 4. In summary, the analysis shows that any integral maximal lattice-free simplex in with satisfies , , and .

The case where must be treated differently. Here, the integer point and the unknown parameter play a key role. The structure of the case distinction for with the corresponding bounds on and is shown below.

1.       ( means: either lies on or violates the fourth facet of (1) )
no integral maximal lattice-free simplex possible

2.       ( means: strictly satisfies the fourth facet of (1) )

1. reducible to the case

The complete subcase analysis for is given in Section 5. It shows that any integral maximal lattice-free simplex in with satisfies and or is unimodularly transformable to a simplex with . Thus, in both cases and there is only a finite number of potential simplices that need to be checked. After ruling out simplices which are equivalent by unimodular transformation only seven different simplices remain.

###### Theorem 2.1

Any integral maximal lattice-free simplex in dimension three can be brought by a unimodular transformation into one of the simplices - .

The convex hulls of the columns of the seven matrices listed below represent these integral maximal lattice-free simplices. We remark that six of these simplices are given as examples in [12].

1. Simplices with vertices on the coordinate axes:

2. Other simplices:

## 3 Simplices in higher dimensions

The case distinctions in Sections 4 and 5 are specialized to the three dimensional geometry. In order to provide a characterization of integral maximal lattice-free simplices in higher dimensions, it seems unavoidable to develop a general proof technique. Although we do not have this machinery at hand today, we believe that the number of integral maximal lattice-free simplices is finite for any . As a first indication for the correctness of this conjecture we will show finiteness for a special class of simplices, namely those that have vertices on the coordinate axes.

Let be an integral maximal lattice-free simplex with one vertex being and the other vertices lying on the coordinate axes. Without loss of generality we assume that , where for all and denotes the -th unit vector in . We further assume that . In particular, we have , otherwise there exists a facet of which does not contain an interior integer point. The inequality representation of is given by the inequalities , , and an additional inequality of the form

 α1x1+⋯+αdxd≤r, (2)

where for all and . It may be checked that the following inequalities together with nonnegativity define integral simplices in dimension four that are maximal lattice-free.

 21 x1 + 14 x2 + 6 x3 + x4 ≤ 4 2, (T1) 15 x1 + 10 x2 + 3 x3 + 2 x4 ≤ 3 0, (T2) 12 x1 + 8 x2 + 3 x3 + x4 ≤ 2 4, (T3) 10 x1 + 5 x2 + 4 x3 + x4 ≤ 2 0, (T4) 9 x1 + 6 x2 + 2 x3 + x4 ≤ 1 8, (T5) 6 x1 + 4 x2 + x3 + x4 ≤ 1 2, (T6) 6 x1 + 3 x2 + 2 x3 + x4 ≤ 1 2, (T7) 5 x1 + 2 x2 + 2 x3 + x4 ≤ 1 0, (T8) 4 x1 + 4 x2 + 3 x3 + x4 ≤ 1 2, (T9) 4 x1 + 3 x2 + 3 x3 + 2 x4 ≤ 1 2, (T10) 4 x1 + 2 x2 + x3 + x4 ≤ 8, (T11) 3 x1 + x2 + x3 + x4 ≤ 6, (T12) 2 x1 + 2 x2 + x3 + x4 ≤ 6, (T13) x1 + x2 + x3 + x4 ≤ 4. (T14)
###### Theorem 3.1

Let be integers and .

1. If is maximal lattice-free, then is bounded by which is a solution to the following recursion:

 λ∗1 =2, λ∗j =1+j−1∏i=1λ∗i,∀ j=2,…,d−1, λ∗d =d−1∏i=1λ∗i.
2. For , every integral maximal lattice-free simplex of the form is defined by nonnegativity and one of the inequalities - .

###### Proof.

(a): First, consider the inequality (2). We show that . Observe that , otherwise the point is in the interior of . For the purpose of deriving a contradiction, assume that . Since is maximal lattice-free there exists an integer point in the relative interior of the facet . In particular, satisfies for all and for at least one . However, since for all this implies which is a contradiction. Thus, we have .

From the definition of , it follows that for all which implies that . Hence, we obtain

 1=1λ1+⋯+1λd. (3)

We look for an upper bound on . Due to (3), the coefficient is the bigger the smaller the other coefficients are. It follows from an inductive argument that, starting with , gradually fixing the ’s at their minimal possible value yields the maximal value for . In this way, we can recursively compute . Again, this recursion can be formally shown using an inductive argument. Thus, is bounded.

(b): For the recursion yields . By enumeration, we obtain the simplices defined by nonnegativity and one of the inequalities - . ∎

The key element of the proof of Theorem 3.1 is a recursive formula. Next we illustrate this recursion for the case of .

###### Example 3.1

First, is fixed at . Then, (3) implies that . Thus, the choice is minimal and substituting in (3) yields . The minimal possible value for is now and from (3) we obtain . The last step is to fix which leads to .

Observe that the above recursion does not only give an upper bound on , but also constructs an integral maximal lattice-free simplex with the largest possible value for . For instance, if , is maximal lattice-free.

As a second indication why in any dimension there should only be a finite number of integral maximal lattice-free simplices, we note that the follwing property holds. For any integral maximal lattice-free simplex , let us denote by its facets. Each facet contains interior integer points. In fact, we can search for a maximal sublattice fully contained in the interior of that is unimodularly transformable to some for .

###### Observation 3.1

We have for at most one .

###### Proof.

For the purpose of deriving a contradiction, assume that for two facets and with . Then, and have each at least integer points of different parity in their interior. If two integer points and on and , respectively, have the same parity, then is an interior point in . Hence, we have integer points with different parity in the interior of . Now, any interior integer point of a facet different from and will lead to a contradiction. ∎

This shows that the possibilities for the sublattice structure in the interior of the facets is somehow limited. Finally, let us remark, that if for all , finiteness follows from a result of Lagarias and Ziegler [6].

## 4 Details on the case distinction for a≥2

Let be an integral maximal lattice-free simplex with given by the following inequality description:

 − x3≤0, (4) − f x2 + e x3≤0, (5) −cf x1 + bf x2 + (cd−be) x3≤0, (6) cf x1 + f(a−b) x2 + (e(b−a)+c(a−d)) x3≤acf. (7)

In this section, we will often state interior integer points as counterexamples and simply prove that for those points all the restrictions (4) - (7) are satisfied with a strict inequality. Recall, that the unknowns , and are integer and satisfy the following properties:

 2 ≤a, 0≤b

We make frequently use of the following simple observation.

###### Observation 4.1
1. Let . Then, for any integer , we have .

2. Let . Then, for any integers and , we have .

3. Let and . Then, for any integers and , we have .

### I) cf+f(a−b)+e(b−a)+c(a−d)≥acf

#### 1) b≥a

In this case, we have which implies that . Since and this inequality is only satisfied for . Substituting this in yields . Since and we obtain and therefore . If then is a contradiction. Hence, we have . However, is now contained in which is a contradiction to its maximality.

#### 2) b<a

##### i) (a,c)=(2,2)

If then there is no interior integer point in the facet . Hence, . Substituting in yields and therefore .

##### ii) (a,c)≠(2,2)

Here, we obtain which implies

 f(ac−a−c)≤ac. (8)

From (8) and Observation 4.1(c), it follows that .

Now assume . Then and . Substituting in implies that . If or it follows and therefore . Hence, the facet does not contain an interior integer point. Thus, let . We must have since otherwise the facet does not contain an interior integer point. This implies and .

Therefore, we can assume that . From (8) and Observation 4.1(c), it follows that and it remains to find an upper bound on . If there is no interior integer point in the facet . On the other hand, if we have by (8) and Observation 4.1(c).

### Ii) cf+f(a−b)+e(b−a)+c(a−d)<acf

For the purpose of deriving a contradiction assume that . In this case, the point is in the interior of as one can easily check by substituting in the inequalities (4)-(7): in all four inequalities, the left hand side is strictly less than the right hand side. We therefore must have

 −cf+bf+cd−be≥0. (9)

We first show that . Note that holds true only for . However, then the facet does not contain an interior integer point. Thus, we have which implies that .

#### 1) −cfk+bf+cd−be≥0

We have already shown that . Assume . Then which is a contradiction. Hence, and it follows

 1

From (10), we obtain which implies by Observation 4.1(a).

Moreover, we have since otherwise is a contradiction. From (10), it follows which means that . From (9), we know that .

Assume . Then, (10) implies which can never hold true for and . Thus, . So assume . Now, (10) implies which is only satisfied for . Substituting this in (9) yields . For the point is in the interior of : obviously, (4) and (5) are strict;

 −2cf+bf+cd−be=f(b−c)+c(d−f)−be <0; 2cf+f(a−b)+e(b−a)+c(a−d) = 4f+2f−2e+6−2d=6f+2(3−d−e)<6f =acf.

So we have which implies and . It remains to consider the case .

##### i) (a,c)=(2,2)

From , it follows that and (9) implies that . First note that since otherwise the point is in the interior of : clearly, (4) and (5) are strict;

 −2cf+bf+cd−be=f(b−c)+c(d−f)−be <0; 2cf+f(a−b)+e(b−a)+c(a−d) = 4f+f−e+4−2d≤4f−2e+4<4f =acf.

Now assume . If holds true, then the point is in the interior of : as above, (4), (5), and (6) are strict;

 2cf+f(a−b)+e(b−a)+c(a−d)=4f+f−e+4−2d<4f =acf.

If , the point is in the interior of : clearly, (4) is strict; (5) is strict since and ;

 −2cf+bf+2cd−2be=−4f+f+4d−2e≤−f+8−4e <0; 2cf+f(a−b)+2e(b−a)+2c(a−d) = 4f+f−2e+8−4d≤4f−f−4e+8<4f =acf.

Hence, we must have .

##### ii) (a,c)≠(2,2)

In this case, we have . If the point is in the interior of : clearly, (4),(5) and (6) are strict (see above);

 2cf+f(a−b)+e(b−a)+c(a−d) = cf+cf−bf−cd+be+2(c+f−e) ≤ cf+2(c+f−e)

Here, the strict inequality in the last row follows from the fact that for and (use Observation 4.1(b)). Therefore, we obtain and it follows .

#### 2) −cfk+bf+cd−be<0

For the purpose of deriving a contradiction assume . Then, the point is in the interior of . One can see this by substituting in the inequalities (4)-(7). Thus, it follows

 cfk+f(a−b)+e(b−a)+c(a−d)≥acf. (11)

We know that . Now assume . From (9) and (11) it follows that . Note that holds true for any feasible triple and would lead to a contradiction in this chain of inequalities. Therefore, we must have . However, in this case is contained in which contradicts the maximality of . Thus, we have .

From (9) it follows that and from (11) it follows that . Thus, we have

 cd−be>0 (12)

and

 e(b−a)+c(a−d)>0. (13)
##### i) e>0

Here, the point is in the interior of : obviously, (4) is strict; (5) is strict since we have ;

 −cfd+bfe+cd−be=(1−f)(cd−be) <0; cfd+f(a−b)e+e(b−a)+c(a−d) = (1−f)(−cd+e(b−a))+ac<(1−f)(−ac)+ac =acf.

The strict inequalities follow from (12) and (13).

##### ii) e=0

By (12) and (13), we obtain . Furthermore, (9) and (11) change to

 −cf+bf+cd≥0 (14)

and

 cfk+f(a−b)+c(a−d)≥acf. (15)
###### A) a=b

In this case we have . Substituting this in (15) yields and since this implies that . If the point is in the interior of : evidently, (4) and (5) are strict;

 −2cf+bf+cd−be=f(b−c)+c(d−f) <0; 2cf+f(a−b)+e(b−a)+c(a−d) = 2cf+c(a−d)<2cf+cf(a−d)=acf+cf(2−d) ≤acf.

So we have . If we have in the interior of : as above, (4), (5), and (6) are strict;

 2cf+f(a−b)+e(b−a)+c(a−d)=2cf+c(a−1)<2cf+cf(a−2) =acf.

The strict inequality follows from Observation 4.1(a): since . Thus, let . The chain implies that . Substituting this in (14) yields .