Five-dimensional Perfect Simplices

# Five-dimensional Perfect Simplices

Mikhail Nevskii111Department of Mathematics, P.G. Demidov Yaroslavl State University, Sovetskaya str., 14, Yaroslavl, 150003, Russia orcid.org/0000-0002-6392-7618 mnevsk55@yandex.ru    Alexey Ukhalov 222 Department of Mathematics, P.G. Demidov Yaroslavl State University, Sovetskaya str., 14, Yaroslavl, 150003, Russia orcid.org/0000-0001-6551-5118 alex-uhalov@yandex.ru
September 18, 2017
###### Abstract

Let be the unit cube in , . For a nondegenerate simplex , consider the value . Here is a homothetic image of with homothety center at the center of gravity of and coefficient of homothety . Let us introduce the value . We call a perfect simplex if and is inscribed into the simplex . It is known that such simplices exist for and . The exact values of are known for and in the case when there exist an Hadamard matrix of order ; in the latter situation . In this paper we show that and . We also describe infinite families of simplices such that for . The main result of the paper is the existence of perfect simplices in .

Keywords: simplex, cube, homothety, axial diameter, Hadamard matrix

## 1 Introduction

Let us introduce the basic definitions. We always assume that . Element we present in component form as . Denote , .

For a convex body , by we mean the result of homothety of with center of homothety at the center of gravity of and coefficient .

If is a convex polyhedron, then means the set of vertices of . We say that -dimensional simplex is circumscribed around a convex body if and each -dimensional face of contains a point of . Convex polyhedron is inscribed into if each vertex of this polyhedron belongs to the boundary of .

For a convex body , by we denote the length of a longest segment in parallel to the th coordinate axis. The value we call the th axial diameter of . The notion of axial diameter was introduced by P. Scott in [11], [12].

For nondegenerate simplex and convex body in , we consider the value

 ξ(C;S):=min{σ≥1:C⊂σS}.

Denote The equality is equivalent to the inclusion . For convex bodies , we define as the minimal such that belongs to a translate of . Denote . In this paper we compute the values and for simplices We also consider the value

 ξn:=min{ξ(S):S --- n-dimensional simplex,S⊂Qn,vol(S)≠0}.

Let us introduce basic Lagrange polynomials of -dimensional simplex. Let be the set of polynomials in real variables of degree . Consider a nondegenerate simplex in . Denote vertices of by and build the matrix

 A:=⎛⎜ ⎜ ⎜ ⎜ ⎜ ⎜⎝x(1)1…x(1)n1x(2)1…x(2)n1⋮⋮⋮⋮x(n+1)1…x(n+1)n1⎞⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎠.

Let , then . Denote by the determinant obtained from by changing the th row with the row Polynomials from have the property , where is the Kronecker delta. Coefficients of form the th column of . In the following we write , i. e.,

Each polynomial can be represented in the form

 p(x)=n+1∑j=1p(x(j))λj(x).

We call basic Lagrange polynomials related to . By taking , we obtain

 n+1∑j=1λj(x)x(j)=x,n+1∑j=1λj(x)=1. (1)

Therefore, the numbers are barycentric coordinates of with respect to . Simplex can be determined by each of the systems of inequalities or .

It is proved in [1] that for th axial diameter of simplex holds

 1di(S)=12n+1∑j=1∣∣lij∣∣. (2)

There exist exactly one line segment in with the length parallel to the -axis. The center of this segment coincides with the point

 y(i)=n+1∑j=1mijx(j),mij:=∣∣lij∣∣n+1∑k=1|lik|. (3)

Each -face of contains at least one of the endpoints of this segment. These results were generalized to the maximum line segment in parallel to an arbitrary vector . In [4] were obtained the formulae for the length and endpoints of such a segment via coordinates of vertices of and coordinates of .

From equality (2) and properties of (see [3], Chapter 1) it follows that the value is equal to the sum of the positive elements of the th row of and simultaniously is equal to the sum of the absolute values of the negative elements of this row.

Note the following formulae for introduced numerical characteristics. Let be a nondegenerate simplex and be a convex body in It was shown in [3] (see § 1.3) that in the case we have

 ξ(C;S)=(n+1)max1≤k≤n+1maxx∈C(−λk(x))+1. (4)

The condition

 maxx∈C(−λ1(x))=…=maxx∈C(−λn+1(x)) (5)

is equivalent to the fact that simplex is circumscribed around . When is a cube in , equality (4) can be written in the form

 ξ(S)=(n+1)max1≤k≤n+1maxx∈ver(C)(−λk(x))+1, (6)

and (5) can be replaced by the condition

 maxx∈ver(C)(−λ1(x))=…=maxx∈ver(C)(−λn+1(x)). (7)

For formula (6) was proved in [1]. The statement that (7) holds if and only if is circumscribed around follows from the results of [1] and [10].

It was proved in [3] (see § 1.4) that, for arbitrary convex body and nondegenerate simplex in , holds

 α(C;S)=n+1∑j=1maxx∈C(−λj(x))+1. (8)

If , then (8) is equivalent to

 α(S)=n∑i=11di(S). (9)

Equality (9) was established in [10]. There are several interesting corollaries of this result. For instance, we present here the formula for via coefficients of

 α(S)=12n∑i=1n+1∑j=1|lij|. (10)

From (10) and properties of , it follows that is equal to the sum of the positive elements of upper rows of and simultaneously is equal to the sum of the absolute values of the negative elements of these rows.

It is obvious that for convex body and simplex holds . The equality takes place only when simplex is circumscribed around .

If , then . Applying (9) for this case we have

 ξ(S)≥α(S)≥n. (11)

Consequently, By 2009, the first author obtained that

 ξ1=1,ξ2=3√55+1=2.34…,ξ3=3,4≤ξ4≤133=4.33…,
 5≤ξ5≤5.5,6≤ξ6≤6.6,ξ7=7.

If , then

 ξn≤n2−3n−1. (12)

Hence, for any , we have . Inequality (12) was established by calculations for the simplex with vertices , , , , (see [2], [3], § 3.2). If , then which gives (12). If , then has the following property (see [7], Lemma 3.3): replacement of an arbitrary vertex of by any point of decreases the volume of the simplex. For (and only in these cases), is a simplex of maximum volume in If , then , therefore If , then ; for we have .

If is an Hadamard number, i.e., there exist an Hadamard matrix of the order , then (see section 4). In this and only this case there exist a regular simplex inscribed into such that vertices of coincide with vertices of ([7], Theorem 4.5). For such a simplex, we have . It follows from (11) that , and (9) gives .

Since 2011, the first author supposed that the equality holds only if is an Hadamard number. In this paper we will demonstrate that it is not so. The smallest such that is not an Hadamard number, while , is equal to ; the next number with such a property is . Below we study simplices such that , for odd , . Since , there no exist such a simplex in the case . For even numbers , the problem of existence of simplices with the condition is unsolved. In [5] the authors proved that , and conjectured that .

A nondegenerate simplex in we will call a perfect simplex with respect to an -dimensional cube if and cube is inscribed into simplex , i. e., the boundary of contains all the vertices of . If simplex is perfect with respect of the cube , then we shortly call such a simplex perfect.

By the moment of submitting the paper, only three values of , such that perfect simplices in exist, are known to the authors. These numbers are . In all these cases we have . The case is trivial. The unique (up to similarity) three-dimensional perfect simplex is described in section 5. The main result of this paper is the existence of perfect simpices in , see section 6. Moreother, in section 7 we describe the whole family of -dimensional perfect simplices.

## 2 Supporting Vertices of the Cube

Let be an -dimensional nondegenerate simplex with vertices and let , , be basic Lagrange polynomials of . We define the th facet of as its -dimensional face which does not contain the vertex . In other words, th facet of is an -face cointained in hyperplane . By the th facet of simplex , we mean the facet parallel to the th facet of .

###### Theorem 1.

Let . The vertex of belongs to the th facet of simplex if and only if

 −λj(v)=max1≤k≤n+1,x∈ver(Qn)(−λk(x)). (13)
###### Proof.

Denote . First note that the equation of hyperplane containing the th facet of simplex can be written in the form

 λj(x)=1−ξn+1. (14)

Indeed, this equation has a form . Remind that is a linear function of . Since the center of gravity of is contained in the hyperplane and the th facet of belongs to the hyperplane , we have

 a−1n+10−1n+1=ξ.

Hence, .

Let us give another proof of this fact. Let be the center of gravity of , be the vertex of not coinciding with , and be the homothetic image of the point with coefficient of homothety and center of homothety at . Then and . The functional is a linear (homogeneous and additive) functional on . Since and we have

 λj(z)−λj(0)=μj(z)=μj(ξy+(1−ξ)c)=ξμj(y)+(1−ξ)μj(c)=
 =ξ[λj(y)−λj(0)]+(1−ξ)[λj(c)−λj(0)]=1−ξn+1−λj(0).

This implies . The point belongs to the th facet of simplex , therefore, the equation of hyperplane cointaining this facet has the form (14).

Now assume that satisfy the equation (14). Since

 ξ=(n+1)max1≤k≤n+1,x∈ver(Qn)(−λk(x))+1, (15)

see (6), the right part of (13) is equal to . Therefore, belongs to the th facet of . On the other hand, if belongs to the th facet of , then . By (15), this is equivalent to (13). This completes the proof. ∎

## 3 On a Simplex Satisfying the Condition S⊂Q⊂nS

In this section we denote by the center of gravity of a convex body .

###### Theorem 2.

Let be a nondegenerate parallelotope and is a nondegenerate simplex in . If , then .

###### Proof.

Any two nondegenerate parallelotopes in are affine equivalent. Corresponding nondegenerate affine transformation maps a simplex to a simplex. This transformation maps the center of gravity of each polyhedron into the center of gravity of the image of this polyhedron. Therefore, it is enough to prove the theorem in the case . Namely, we will show that if and conditions of the theorem are satisfied, then .

Let be vertices of , and be basic Lagrange polynomials of .

Since , we have . For any simplex , holds . Hence, . It follows that simplex is circumscribed around . So, the following equalities hold

 maxx∈ver(Q′n)(−λ1(x))=…=maxx∈ver(Q′n)(−λn+1(x)).

In other words, does not depend on . From the equality

 n=ξ(Q′n;S)=(n+1)max1≤j≤n+1,x∈ver(Q′n)(−λj(x))+1

we obtain, for any ,

 maxx∈ver(Q′n)(−λj(x))=n−1n+1. (16)

Let be the coefficients of , i. e., . Then

 maxx∈ver(Q′n)(−λj(x))=n∑i=1|lij|−ln+1,j. (17)

Consider the value . It is easy to see that

 maxx∈Q′nλj(x)=n∑i=1|lij|+ln+1,j. (18)

Since , we have

 maxx∈Q′nλj(x)≥λj(x(j))=1. (19)

Using (16)–(19), we obtain

 2⋅ln+1,j=[n∑i=1|lij|+ln+1,j]−[n∑i=1|lij|−ln+1,j]=
 =maxx∈Q′nλj(x)−maxx∈ver(Q′n)(−λj(x))≥1−n−1n+1=2n+1.

Consequently, for any , the following inequality holds:

 λj(0)=ln+1,j≥1n+1. (20)

The numbers are the barycentric coordinates of the point (see (1)), hence,

 n+1∑j=1λj(0)=1. (21)

If, for some , inequality (20) is strict, then the left part of (21) is strictly greater than the right part. This is not possible, therefore,

 λj(0)=1n+1,1≤j≤n+1. (22)

Thus, we have

 c(S)=1n+1n+1∑j=1x(j)=n+1∑j=1λj(0)x(j)=0.

In addition, note that for any we have . Indeed, from (17), (18), and (22) it follows

 maxx∈Q′nλj(x)=n∑i=1|lij|+ln+1,j=maxx∈ver(Q′n)(−λj(x))+2ln+1,j=
 =n−1n+1+2n+1=1.

Hence, if , then the cube lies in the intersection of half-spaces . The theorem is proved. ∎

## 4 The Case when n+1 is an Hadamard Number

A nondegenerate -matrix is called an Hadamard matrix of order if its elements are equal to or and

 H−1m=1mHTm.

Some information on Hadamard matrices can be found in [6]. It is known that if exists, then or is a multiple of . It is established that exists for infinite set of numbers of the form , including powers of two . By 2008, the smallest number for which it is unknown whether there is an Hadamard matrix of order was equal to . We call a natural number an Hadamard number if there exist an Hadamard matrix of order .

The regular simplex , inscribed into in such a way that vertices of are situated in vertices of , exists if and only if is an Hadamard number (see [7], Theorem 4.5). If is an Hadamard number, then and, therefore, . The latter statement was proved by two different methods in the paper [2] and in the book [3], § 3.2.

Here we give yet another proof of this fact. This new proof differs from the proofs given in [5] and [6] and directly utilizes properties of Hadamard matrices. We also establish some other properties of regular simplex .

###### Theorem 3.

Let be an Hadamard number and be a regular simplex inscribed in . Then .

###### Proof.

Taking into consideration similarity, we can prove the statement for the cube . Since is an Hadamard number, there exist a normalized Hadamard matrix of order , i. e., such an Hadamard matrix that its first row and first column consist of ’s (see [9], Chapter 14). Let us write rows of this matrix in inverse order:

 H=⎛⎜ ⎜ ⎜ ⎜ ⎜ ⎜⎝111…1…………1…………1………………………1⎞⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎠.

The obtained matrix H also is an Hadamard matrix of order . Consider the simplex with vertices formed by first numbers in rows of .

It is clear that all vertices of are also the vertices of , therefore, the simplex is inscribed into the cube. Let us show that is a regular simplex and the lengths of its edges are equal to . Let , be two different rows of . All elements of are , hence, we have . Rows of an Hadamard matrix are mutually orthogonal, therefore,

 ∥a−b∥2=(a−b,a−b)=∥a∥2+∥b∥2−2(a,b)=2(n+1).

Denote by and the vertices of obtained from and respectively by removing the last component. This last component is equal to 1. It follows that -dimensional length of the vector is equal to -dimensional length of the vector , i. e., .

Denote by basic Lagrange polynomials of . Since , the coefficients of are situated in rows of . All constant terms of these polynomials stand in the last column of . Consequently, they are equal to . It means that constant terms of polynomials are equal to . By this reason, for any ,

 (n+1)maxx∈ver(Q′n)(−λj(x))=n−1.

The coefficients of polynomials are equal to , therefore, for any , the vertex of , such that , is unique. Indeed, is defined by equalities , where are the coefficients of .

Now let us find using formula (6) for . We have

 ξ(Q′n;S′)=(n+1)max1≤j≤n+1maxx∈ver(Q′n)(−λj(x))+1=n−1+1=n.

Consider the similarity transformation which maps into . This transformation also maps into a simplex inscribed into . Denote by the image of . Obviously, . It follows that, if is an Hadamard number, then . As we know, for any , (see (11)). Hence, Since the coefficient of similarity for mapping to is equal to , the length of any edge of is equal to .

From the condition

 maxx∈ver(Q′n)(−λ1(x))=…=maxx∈ver(Q′n)(−λn+1(x))=n−1n+1,

it follows that simplex is circumscribed around the cube . From the said above it also follows that each -face of contains only one vertex of . This means that simplex is circumscribed around the cube in the same way. Since is circumscribed around , we have and . These equalities could be also obtained from inequalities and equality .

Equalities and can be also derived in another way. Note that the regular simplex inscribed into has the maximum volume among all the simplices in , see Theorem 2.4 in [7]. Additionally, for any simplex of maximum volume in , all the axial diameters are equal to . The latter property of a maximum volume simplex in was established by Lassak in [8]. This fact also can be obtained from (9) (see [10] and [3], § 1.6]).

The theorem is proved. ∎

## 5 Perfect Simplices in R1 and R3

The case is very simple. For the segment , we have . Therefore, and is the unique perfect simplex. The equality could be also obtained from Theorem 3.

Consider the case . The conditions of Theorem 3 are satisfied, hence, . There exist a regular simplex inscribed into such that . As an example, we consider the simplex with vertices , , , . Matrices and for have the form

 A=⎛⎜ ⎜ ⎜⎝0001110110110111⎞⎟ ⎟ ⎟⎠,A−1=12⎛⎜ ⎜ ⎜⎝−111−1−11−11−1−1112000⎞⎟ ⎟ ⎟⎠.

Using (2), we get , i. e., . It follows from (9) that . Basic Lagrange polynomials of are

 λ1(x) =12(−x1−x2−x3+2), λ2(x) =12(x1+x2−x3), λ3(x) =12(x1−x2+x3), λ4(x) =12(−x1+x2+x3).

By (6), we have

 ξ(S1)=4max1≤k≤4maxx∈ver(Q3)(−λk(x))+1.

Substituting vertices of into basic Lagrange polynomials we obtain

 max1≤k≤4maxx∈ver(Q3)(−λk(x))=12. (23)

Hence, . This implies . For each , maximum in (23) is achieved only for one vertex of :

 −λ1(1,1,1)=−λ2(0,0,1)=−λ3(0,1,0)=−λ4(1,0,0)=12. (24)

It follows from Theorem 1 and (24) that the extremal vertices of the cube, i. e., the vertices , , , belong to the facets of the simplex . Every facet of contains only one vertex of . The rest vertices of the cube belong to the interior of . Thus, though for we have , the simplex is not perfect.

From (3), we obtain that three maximum segments in parallel to coordinate axes are intersected at the center of the cube.

Now consider the simplex with vertices , , , . For this simplex,

 A=⎛⎜ ⎜ ⎜ ⎜ ⎜ ⎜⎝12001121010121111211⎞⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎠,A−1=12⎛⎜ ⎜ ⎜⎝00−22−2200−1−111201−1⎞⎟ ⎟ ⎟⎠.

We have . Consecuently and . Basic Lagrange polynomials of are

 λ1(x) =−x2−12x3+1, λ2(x) =x2−12x3, λ3(x) =12(−2x1+x3+1), λ4(x) =12(2x1+x3−1).

The formula (6) takes the form

 ξ(S2)=4max1≤k≤4maxx∈ver(Q3)(−λk(x))+1.

Substituting vertices of into polynomials we obtain

 −λ1(0,1,1)=−λ1(1,1,1)=−λ2(0,0,1)=−λ2(1,0,1)=−λ3(1,0,0)=−λ3(1,1,0)=−λ4(0,0,0)=−λ4(0,1,0)=12. (25)

Hence, and . In this case all the vertices of are extremal. It follows from Theorem 1 and (25) that each vertex of belongs to the boundary of . Therefore, , unlike , is a perfect simplex.

It is proved in [3] that and are the only simplices in (up to similarity) with property . Each perfect simplex in is similar to .

The simplices , , , are shown in Fig. 1 and Fig. 2. Vertices of situated on the boundary of and are marked with bold points. Bold lines mark the segments corresponding to the axial diameters.

## 6 The Exact Value of ξ5

###### Theorem 4.

There exist simplex such that simplex is circumscribed around and boudary of contains all vertices of the cube.

###### Proof.

Consider the simplex with vertices , , , , , . Obviously, . Matrices and for the simplex have the form

 A=⎛⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜⎝12113111120131111212130111212013