Main results

In this paper, we use these two constructions to produce the first examples of empty simplices of width larger than their dimension. The following table summarizes our findings in small dimensions. In it, and elsewhere, volume is normalized to the lattice; unimodular simplices have volume one.

Theorem 1.1. There are empty simplices of the following dimensions and widths:

$$\begin{align*}\begin{array}{c|c|cc} \text{Empty simplex}& \text{Volume} & \text{Dimension} & \text{Width}\\ \hline \operatorname{Cycl}(4, \ 101) = S(4,2) &101 & 4 & 4\\ \operatorname{Cycl}(6, \ 6\,301) = S(6,3) &6\,301 & 6 & 6\\ S(8,4) & 719\,761 & 8 & 8\\ \operatorname{Cycl}(10, \ 582\,595\,883) &582\,595\,883 & 10 & \mathbf{11}\\ S(16,9) &36\,373\,816\,216\,801\,891 & 16 & \mathbf{18}\\ S(30,17) & \simeq 2.9{\cdot} 10^{38}& 30 & \mathbf{34}\\ S(46,26) & \simeq 6.6{\cdot} 10^{66}& 46 & \mathbf{52}\\ S(60,34) & \simeq 5.4{\cdot} 10^{93}& 60 & \mathbf{68}\\ \end{array} \end{align*}$$

The first entry in the table was known; it is the empty $4$ -simplex of width four found by Haase and Ziegler [Reference Haase and Ziegler9], and shown by Iglesias and Santos [Reference Iglesias-Valiño and Santos14] to be the unique empty $4$ -simplex of width larger than three. We include it in the table to emphasize that our constructions reflect two ways of understanding it and of generalizing the unimodular symmetry that acts cyclically on its vertices.

But even our other simplices of width equal to their dimension are new. It is very easy to construct hollow lattice d-simplices of width d (the d-th dilation of a unimodular d-simplex is one example), but for empty simplices, the same is not true (e.g., all empty 3-simplices have width one). In fact, the first proof that there are empty simplices of arbitrarily large width (in growing dimension) was nonconstructive, by Kantor [Reference Kantor17], and the widest empty simplices before this paper in every dimension $d>4$ are those constructed by Sebő [Reference Sebő24], of width $d-1$ for even dimension d, and width $d-2$ when d is odd.

Let us denote by $\operatorname {Flt}_e(d)$ the maximum lattice width among empty d-simplices, which must be attained since it is a positive integer bounded by $\operatorname {Flt}(d)$ . What we know about $\operatorname {Flt}_e(d)$ can be summarized as follows:

Proof. For part (1), let $\Delta =\operatorname {conv}(0, v_1,\dots ,v_d)$ be an empty simplex of width $\operatorname {Flt}_e(d)$ , which we assume to have a vertex at the origin without loss of generality. Let $f\in ( \mathbb {R}^d)^*$ be the linear functional with $f(v_i)=1$ for all i and let $v_{d+1}$ be a lattice point minimizing f among the lattice points in the interior of the cone $\operatorname {cone}(v_1,\dots ,v_d)$ generated by the vertices of $\Delta $ .

Since $\operatorname {conv}(0, v_1,\dots ,v_d, v_{d+1})$ is contained in

$$\begin{align*}\operatorname{cone}(v_1,\dots,v_d) \cap f^{-1}([0,f(v_{d+1})]), \end{align*}$$

minimality of $v_{d+1}$ implies that the polytope $\operatorname {conv}(0, v_1,\dots ,v_d, v_{d+1})$ has no lattice points other than its vertices. In particular, for any choice of $h\in \mathbb {Z}\setminus \{0\}$ , we have that

$$\begin{align*}\tilde \Delta(h) :=\operatorname{conv}( (0,0), (v_1,0),\dots,(v_{d}, 0), (v_{d+1},h)) \end{align*}$$

is an empty $(d+1)$ -simplex. If we take $h\ge \operatorname {Flt}_e(d)$ , we have that $\tilde \Delta (h)$ has width at least $\operatorname {Flt}_e(d)$ because

• ◦ the width of $\tilde \Delta (h)$ with respect to any lattice functional not constant on $ \mathbb {R}^d\times \{0\}$ is at least the width of $\tilde \Delta (h) \cap ( \mathbb {R}^d\times \{0\})$ , which equals $\operatorname {Flt}_e(d)$ .

• ◦ the width of $\tilde \Delta (h)$ with respect to any lattice functional g constant on $ \mathbb {R}^d\times \{0\}$ equals $g(h)$ , which is a positive multiple of h.

Part (2) is proved as Theorem 1.8 in [Reference Codenotti and Santos5], with a construction related to Sebő’s. Part (3) is also in [Reference Codenotti and Santos5, Theorem 1.8], except that there a $\limsup $ appears instead of $\lim $ . But the monotonicity in part (1) allows an easy modification of that proof to give a $\lim $ .

Our constructions imply that the $\lim _{d\to \infty } \frac {\operatorname {Flt}_e(d)}{d} $ in the previous theorem is at least 1.1346 (see Theorem 1.3), in particular disproving the following guess from [Reference Sebő24, p. 403]: ‘it seems to be reasonable to think that the maximum width of an empty integer simplex in $ \mathbb {R}^n$ is n + constant’.

Cyclotomic and circulant simplices

As suggested by our notation for the simplices in Theorem 1.1, we have constructions of two different types:

In Section 3, we introduce cyclotomic simplices: lattice simplices of prime volume N constructed using primitive roots of unity modulo N. There is one cyclotomic simplex $\operatorname {Cycl}(d,N)$ for each choice of dimension d and prime volume N, as long as $d+1$ divides $N-1$ (so that $(d+1)$ -th roots of unity modulo N exist). In order for them to have large width, $d+1$ has to be prime too (see Corollary 3.2). We have computationally enumerated empty cyclotomic simplices of dimensions 4, 6 and 10 and of volume up to $2^{31}$ and found that in dimension 10 there are five of width 11. See a more detailed discussion in Section 3.

In Section 4, we present another construction, using simplices with explicit coordinates that are symmetric under cyclic permutation. We call them circulant simplices since their matrix of vertex coordinates is a circulant matrix. Our exploration of circulant simplices starts with the observation that some of the widest simplices known, including the cyclotomic simplices $\operatorname {Cycl}(4, 101)$ and $\operatorname {Cycl}(6, 6\,301)$ , are circulant of a very specific form. Let $v=(1,m,0,\cdots ,0, -m)\in \mathbb {R}^{d+1}$ , where m is a positive integer and v has $d-2$ zero entries. Let $v^{(i)}$ be the i-th cyclic shift of v. We define

$$\begin{align*}S(d,m):= \operatorname{conv}(v^{(0)}, \dots,v^{(d)}), \end{align*}$$

which is a d-simplex in the hyperplane $\sum _i x_i=1$ in $ \mathbb {R}^{d+1}$ . For odd d, this simplex has width one, but for even d, the situation is quite different. The following statement summarizes the main results of Section 4. The $m_0$ in the statement is defined as the unique positive solution of the polynomial equation

$$\begin{align*}m^{d-1} = \sum_{i=0}^{d/2-1} \binom{d-1-i}{i}m^{2i}. \end{align*}$$

Equivalently, it equals $ \frac {1}{2\sinh \alpha _0}, $ where $\alpha _0=\alpha _0(d)$ is the unique solution of $\cosh \alpha _0 = \sinh (d\alpha _0)$ . See details in Section 4.5.

Theorem 1.3. In each even dimension $d\ge 4$ and for every $m\le \lfloor m_0(d)\rfloor $ , the circulant d-simplex $S(d, m)$ is empty and has width $2 m$ . Hence,

$$\begin{align*}\lim_{d\to \infty} \frac{\operatorname{Flt}_e(d)}{d} \ge \lim_{d\to \infty} \frac{2m_0(d)}{d} = \frac{1}{\operatorname{\mathrm{arcsinh}}(1)} = \frac{1}{\ln(1+\sqrt2)} \simeq 1.1346. \end{align*}$$

The dimensions $16$ , $30$ , $46$ and $60$ that appear in Theorem 1.1 are the smallest values of d for which $2m_0(d)$ exceeds, respectively, $d+2$ , $d+4$ , $d+6$ and $d+8$ .

Discussion and questions

Organization of the paper

Section 2 recalls some results about cyclic lattice simplices. Section 3 presents our investigations on empty cyclotomic lattice simplices. Finally, Section 4 contains the construction of empty circulant lattice simplices.

Algorithmic remarks

We have computationally explored cyclotomic d-simplices for $d=4, 6, 10$ . Our method is the following:

1. 1. We go through all the primes N with $N-1$ a multiple of $d+1$ , in a range that we set appropriately in each case.

2. 2. For each such N, we find the $(d+1)$ -th roots of unity modulo N and construct the cyclotomic simplex $\operatorname {Cycl}(d,N)$ .

3. 3. We check whether $\operatorname {Cycl}(d,N)$ is empty and compute its width.

Steps (1) and (2) are computationally straightforward. (In step (2), we can use brute force since step (3) is much more expensive). Let us discuss how we check emptiness and compute width of a cyclic simplex:

Experimental results

We have explored cyclotomic simplices for $d=4,6,10$ with prime N up to $2^{31} \simeq 2.14{\cdot } 10^9$ , finding the following:

• ◦ For $d=4$ , there are only four empty cyclotomic simplices: the two from Example 3.4 plus another two, of volumes 11 and 61. They are described in Table 1.

  Table 1 The four empty cyclotomic 4-simplices. $\operatorname {Cycl}(4, 101)$ is the unique empty $4$ -simplex of maximum volume (see [Reference Iglesias-Valiño and Santos14]).

  

• ◦ For $d=6$ , there are 88 empty cyclotomic simplices. Their volumes range from 29 to 17683. Their maximum lattice width is 6, attained by the six simplices in Table 2. Table 3 shows additional statistics of cyclotomic $6$ -simplices, both empty and nonempty.

  Table 2 The six empty cyclotomic 6-simplices of width 6. As mentioned in Remark 3.9, extending the search to nonprime volumes, we have found a seventh empty $6$ -simplex of width six, with volume $6\,931$ .

  

  Table 3 Statistics on cyclotomic 6-simplices. Left: number of empty and nonempty cyclotomic simplices for each interval of volume. Right: smallest cyclotomic simplex for each width.

  

• ◦ For $d=10$ , we have found $218\,075$ empty cyclotomic simplices. Their number decreases very fast with volume, as seen in Figure 1. For example, there are only 30 of volume above $10^9$ and only one above $15{\cdot }10^8$ , of volume $1\,757\,211\,061$ and width 10. There are none of width larger than 11, and exactly the following five of width 11:

  $$\begin{align*}\begin{array}{cc} \operatorname{Cycl}(10, \ 582\, 595\, 883), & \operatorname{Cycl}(10, \ 728\, 807\, 201), \\ \operatorname{Cycl}(10, \ 976\, 965\, 023), & \operatorname{Cycl}(10, \ 1\, 066\, 142\, 419), \\ \operatorname{Cycl}(10, \ 1\, 113\, 718\, 783). \end{array} \end{align*}$$
  Table 4 shows the smallest empty cyclotomic $10$ -simplex of width w for each $w\in 1,\dots ,11$ .

  Table 4 Smallest cyclotomic 10-simplex for each width from 1 to 11. The smallest cyclotomic simplex of each width is also empty for widths up to 6 and equal to 8, but not for widths $7,9,10,11$ . For the latter we list both the absolute smallest ‘(n.e.)’ and the smallest empty one ‘(e.)’.

  
  

  Figure 1 The counts of empty cyclotomic 10-simplices with volumes up to $2^{31}$ . Each bar counts the simplices whose volume falls within a block of 10 million. The vertical scale is logarithmic.

  Remark 3.9. For $d=4,6,10$ , we have also explored what happens if we allow N not to be prime. This expanded search is significantly more computationally expensive because, as mentioned in Remark 3.6, roots of unity may come in several orbits for each N, and they all have to be considered; in particular, for cyclotomic d simplices with nonprime volume N, we use the notation $\operatorname {Cycl}(d,N,r)$ , where r is the root of unity producing that simplex. Yet, we have the following necessary condition for emptiness: for some $\operatorname {Cycl}(d,N,r)$ to be empty, one needs $\operatorname {Cycl}(d,p)$ empty for all prime divisors p of N.In dimension four, no extra empty cyclotomic simplices arise, but in dimension six, we get 28 new ones, besides the 88 simplices of prime volume. Their volumes range from 841 to 13 021 and are always the product of two primes from $\{29, 43, 71, 113, 197, 211, 239$ , $337, 449\}$ . Among them there is one of width 6, with volume $N= 6931$ . Nonprime volumes often produce several orbits of 7th roots of unity. For example, for 6931, there are six orbits, which produce three nonempty simplices, two empty simplices of width 4 and one empty simplex of width 6.In dimension ten, we only considered simplices of volume less than 10¹⁰ and with two factors in their volume. This search resulted in many additional empty simplices, with volume ranging between 529 and 3 637 493 861. In this set, three additional empty simplices were found with width 11:

  $$ \begin{align*} \operatorname{Cycl}(10, 476\, 164\, 613, 403\, 772\, 212), \\ \operatorname{Cycl}(10, 702\, 431\, 819, 305\, 486\, 391), \\ \operatorname{Cycl}(10, 1\, 419\, 547\, 823, 822\, 028\, 992). \end{align*} $$

  Observe that these include 10-simplices of width 11 both with smaller and bigger volumes than all the ones obtained with prime volumes.

4.1 Circulant simplices, and our specific family $S(d,m)$

A second way of guaranteeing that a d-simplex has a unimodular cyclic symmetry acting on its vertices is by permuting coordinates. Fix a dimension d, and let $v=(v_0,\dots ,v_d)\in \mathbb {R}^{d+1}$ be any real vector. For each $i\in \mathbb {N}$ , let $v^{(i)}$ be the vector obtained from v by a cyclic shift (to the right) of i units in its coordinates. That is, $v^{(d+1)} = v^{(0)}=v$ , and for $i =1, \ldots , d$ ,

$$\begin{align*}v^{(i)}:=(v_{d+1-i}, \dots,v_d,v_0,\dots,v_{d-i}) \end{align*}$$

Definition 4.1. If the $v^{(i)}$ ’s defined above are linearly independent, we call

$$\begin{align*}\operatorname{conv}(v^{(0)},\dots, v^{(d)}) \end{align*}$$

the circulant simplex (of dimension d) with generator v.

The name comes from the fact that square matrices in which all columns are obtained from the first one by cyclic shifts are called circulant, and the matrix $(v^{(0)} \dots v^{(d)})$ with columns the vertex coordinates of a circulant simplex is circulant. Linear independence requires that $\sum _i v_i \ne 0$ . In fact, since the circulant simplex with generator v affinely spans the hyperplane $\sum _i x_i = \sum _i v_i$ , its volume equals

(1) $$ \begin{align} \operatorname{Vol}(\operatorname{conv}(v^{(0)} \dots v^{(d)}))= \left| \frac{\det(v^{(0)} \dots v^{(d)})}{ \sum_{i=0}^d v_i} \right|. \end{align} $$

The example of interest to us is the following. For each $m\in \mathbb {R}$ and $d\in \mathbb {N}$ , we denote $S(d,m)$ the circulant simplex of dimension d and generator $(1,m,0,\dots ,0,-m)$ . That is, the vertices of $S(d,m)$ are the columns of the following $(d+1)\times (d+1)$ circulant matrix:

$$\begin{align*}M(d,m):= \begin{pmatrix} 1 & -m & 0 & \ldots & \ldots & 0 & m \\ m & 1 & -m & \ddots & & & 0 \\ 0 & m & 1 & \ddots & \ddots & & \vdots \\ \vdots & \ddots & \ddots & \ddots & \ddots & \ddots & \vdots \\ \vdots & & \ddots & \ddots & 1 & -m & 0 \\ 0 & & & \ddots & m & 1 & -m \\ -m & 0 & \ldots & \ldots & 0 & m & 1 \end{pmatrix}. \end{align*}$$

For odd d, the width of $S(d,m)$ is $1$ with respect to the functional $\sum _{i=0}^{(d-1)/2} x_{2i}$ , which alternates values 0 and 1 on vertices. Thus, we will always assume d even. Since we are interested in lattice simplices, we will sometimes assume m to be an integer, but in some proofs, we need to regard m as a continuous parameter.

Remark 4.2. Although this family looks too specific to deserve attention per se, we were led to studying it from the (not at all obvious) realization that the unique empty $4$ -simplex of width 4 is unimodularly equivalent to $S(4,2)$ . Moreover, $S(6,3)$ is also an empty $6$ -simplex of width 6 (in fact, it coincides with the smallest cyclotomic empty simplex of that dimension). That is,

$$\begin{align*}S(4,2) = \operatorname{Cycl}(4,101), \qquad S(6,3) = \operatorname{Cycl}(6,6301). \end{align*}$$

An explicit formula for this circulant determinant is easy to compute.

Lemma 4.3. For every even $d\ge 2$ , and every $m\in \mathbb {R}$ , we have

$$ \begin{align*} \operatorname{Vol}(S(d,m)) =\det(M(d,m)) & = \sum_{i=0}^{ d/2} \frac{d+1}{d+1-i}\binom{d+1-i}{i} m^{2i}\\ & = 1 + \sum_{i=1}^{ d/2} \frac{d+1}{i}\binom{d-i}{i-1} m^{2i}. \end{align*} $$

Remark 4.4. The same proof gives that the circulant determinant with generator $v=(a,b,c,0,\dots ,0)$ equals

$$\begin{align*}a^{d+1} + c^{d+1} +(-1)^d \sum_{i=0}^{\lceil d/2\rceil} \frac{d+1}{d+1-i}\binom{d+1-i}{i} (-ac)^i b^{d+1-2i}. \end{align*}$$

4.2 Width of $S(d,m)$

Proof. If d is odd (that is, if the number of coordinates is even), then the sum of the even coordinates is a lattice functional giving width one to $S(d,m)$ .

So, for the rest of the proof, we assume d even. The width of $S(d,m)$ is clearly $\le 2m$ , attained by any of the coordinate functionals. Let $w \in \mathbb {Z}^{d+1}$ be a nonconstant vector, which we regard as a functional, and let us show that it gives width at least $2m$ to $S(d,m)$ .

Recall that $v^{(i)}$ denotes the i-th column of $M{d}{m}$ . Indices i go from $0$ to d and are considered modulo $d+1$ . Observe that

$$\begin{align*}\langle w_,v^{(i)}\rangle =m(w_{i+1}-w_{i-1}) +w_i. \end{align*}$$

Since $d+1$ is odd, we can cyclically go through all the i’s in steps of length two (that is, skipping every second element). Thus, there exist indices $i_0$ and $j_0$ with $w_{2i_0}= \min (w)$ and

$$ \begin{align*} w_{2(i_0-1)}> w_{2i_0} = w_{2(i_0+1)} = \dots = w_{2j_0} < w_{2(j_0+1)}. \end{align*} $$

(Observe that it could well be that $i_0=j_0$ , but this is not a problem for us). We distinguish two cases:

• Case 1: $w_{2i_0-1} \leq w_{2j_0+1}$ In this case, we compute

  $$ \begin{align*} \langle w, v^{(2j_0+1)}\rangle - \langle w, v^{(2i_0-1)}\rangle &= m\underbrace{(w_{2j_0+2} - w_{2j_0})}_{\geq 1} \\ &\quad-m\underbrace{(w_{2i_0} - w_{2i_0-2})}_{\leq -1} \\ &\quad+ \underbrace{w_{2j_0+1} - w_{2i_0-1}}_{\geq 0} \qquad\qquad \geq 2m. \end{align*} $$

• Case 2: $w_{2i_0-1}> w_{2j_0+1}$ In this case, there exists a $k_0$ in $[i_0, j_0]$ (considered cyclically) such that $w_{2k_0-1}> w_{2k_0+1}$ and (by construction) $w_{2k_0} = \min (w)$ . We compute that

  $$ \begin{align*} \langle w, v^{(2j_0+1)}\rangle - \langle w, v^{(2k_0)}\rangle &= m\underbrace{(w_{2j_0+2} - w_{2j_0})}_{\geq 1} \\ &\quad- m\underbrace{(w_{2k_0+1} - w_{2k_0-1})}_{\leq -1} \\ &\quad+ \underbrace{w_{2j_0+1} - w_{2k_0}}_{\geq 0 } \qquad\qquad \geq 2m. \\[-57pt] \end{align*} $$

Example 4.6. The situation for odd d generalizes as follows to any d such that $d+1$ is not prime: let $d+1=ab$ be a nontrivial factorization and consider the circulant d simplex $S^{a}(d,m)$ generated by

$$\begin{align*}v^{(0)} = (1,0,\dots,0,m,0,\dots,0,-m), \end{align*}$$

where the first block of zeroes has length $a-2$ . Observe that $S^{2}(d,m)=S(d,m)$ .

Let $f_i: \mathbb {R}^{d+1} \to \mathbb {R}$ , for $i\in \{0,\dots ,a-1\}$ , be the functional that adds the coordinates with indices equal to $i \mod a$ . The simplex $S^{a}(d,m)$ has width one with respect to the functional $f_0$ , and it is an empty simplex by the following argument. Let $F_a=(f_0,\dots ,f_{a-1}): \mathbb {R}^{d+1} \to \mathbb {R}^a$ . Let $F_b: \mathbb {R}^{d+1} \to \mathbb {R}^b$ be the projection that forgets all coordinates exept those with indices multiple of a. Then, $F_a(S^{a}(d,m))$ is the standard unimodular simplex, and the fiber containing $v^{(0)}$ projects by $F_b$ isomorphically to the standard simplex. Since all vertex-fibers are isomorphic, $F_a$ sends $S^{a}(d,m)$ to a unimodular simplex, and all of its vertex fibers are unimodular. This implies that $S^{a}(d,m)$ is empty.

Hence, if $d+1$ is not prime, then there are infinitely many empty circulant simplices of dimension d.

4.3 The inequality description of $S(d,m)$

Let $u^{(j)} \in \mathbb {Z}^{d+1}$ be the j-th row of the matrix (where we use indices $j=0,\ldots ,d$ )

$$\begin{align*}\det(M(d,m)) \cdot M(d,m)^{-1}. \end{align*}$$

The factor $\det (M(d,m))$ makes any $u^{(j)}$ an integer vector whenever m is an integer. Clearly, the inequality description of $S(d,m)$ is

$$\begin{align*}S(d,m) = \left\{x \in \mathbb{R}^{d+1} \,:\, \sum_{i=0}^{d} x_i = 1,\, \text{ and } \langle u^{(j)}, x \rangle \ge 0 \ \forall j=0, \ldots, d\right\}. \end{align*}$$

In order to explicitly compute the vectors $u{{}^{(j)}}$ , let us define the following continuant matrices of size $k\times k$ , for each $k\in \mathbb {N}$ .

$$\begin{align*}C(k,m) := \begin{pmatrix} 1 & -m & 0 & \ldots & \ldots & 0 & 0 \\ m & 1 & -m & \ddots & & & 0 \\ 0 & m & 1 & \ddots & \ddots & & \vdots \\ \vdots & \ddots & \ddots & \ddots & \ddots & \ddots & \vdots \\ \vdots & &\ddots & \ddots & 1 & -m & 0 \\ 0 & & & \ddots & m & 1 & -m \\ 0 & 0 & \ldots & \ldots & 0 & m & 1 \end{pmatrix} \end{align*}$$

and the following continuant numbers

$$\begin{align*}c_k := c_k(m):=\det(C(k,m)).\end{align*}$$

We additionally define $c_{-1} := 0$ , $c_{0} := 1$ , and observe that $c_1 = 1$ and $c_2 = 1+m^2$ . We have the following:

It follows from the lemma that $c_k=c_k(m)$ is a polynomial in m of degree k or $k-1$ (depending on the parity of k) with constant coefficient $1$ . In fact, we later relate them to the well-studied Fibonacci polynomials.

Since $M(d,m)$ is circulant, its inverse is circulant too. Thus, each $u^{(j)}$ is indeed the jth cyclic shift of the first row $u^{(0)}$ of $\det (M(d,m)) \cdot M(d,m)^{-1}$ . Using the continuant numbers, we can give an explicit formula for $u^{(0)}$ :

For the rest of this section, let $u =(u_0, \ldots , u_{d})$ denote the first row of the matrix $\det (M(d,m)) \cdot M(d,m)^{-1}$ . Analyzing the entries of u is crucial in what follows.

Proposition 4.8.

$$\begin{align*}u_k = c_{d-k} \, m^{k} + (-1)^{d+k-1} c_{k-1}\, m^{d+1-k}, \qquad \forall\ k \in \{0, \ldots, d\}. \end{align*}$$

In particular, if d is even, we get the following:

1. 1. $u_k>0$ for every odd $k=1,3,\dots ,d-1$ .

2. 2. $u_k> u_{k+2}$ for every even $k=0,2,\dots ,d-2$ .

Proof. For a matrix M, denote by $M_{\setminus (i,j)}$ the matrix M with the ith row and the jth column removed (with indices in $\{0, \ldots , d\}$ ). For the first formula, observe that $u_k$ equals $(-1)^k$ times the determinant of $M(d,m)_{\setminus (k,0)}$ . If $k=0$ , we immediately see that $\det (M(d,m)_{\setminus (0,0)}) = c_d$ . So, let $k> 0$ . Now, we use Laplacian expansion along the first row of $\det (M(d,m)_{\setminus (k,0)})$ . We get two summands (corresponding to the first and the last entry of the first row). The first one equals $(-m)$ times the determinant of a block tridiagonal matrix that is easily seen to be equal to $(-m)^{k-1} \cdot c_{d-k}$ . The second summand equals $(-1)^{d+1} m$ times the determinant of a block tridiagonal matrix that evaluates to $c_{k-1} \cdot m^{d-k}$ . Together, we get that $u_k$ equals

$$\begin{align*}(-1)^k ((-m) \cdot (-m)^{k-1} \cdot c_{d-k} + (-1)^{d+1} m \cdot c_{k-1} \cdot m^{d-k}), \end{align*}$$

as desired.

For the second part, positivity for odd k is obvious, since in this case, $u_k$ is a positive combination of two of the $c_i$ ’s. To deal with even k, observe that denoting $\tilde c_k = m^{d-k}c_k$ , the recurrence in Lemma 4.7 becomes

$$\begin{align*}\tilde c_k = \frac{\tilde c_{k-1}}m + \tilde c_{k-2}. \end{align*}$$

Hence, both the even and the odd subsequences of $\tilde c_k$ are strictly increasing. This implies that $u_k$ decreases with k for even k, since for even k, we have

$$\begin{align*}u_k = c_{d-k} \, m^{k} - c_{k-1}\, m^{d+1-k}= \tilde c_{d-k} - \tilde c_{k-1}. \\[-38pt] \end{align*}$$

4.4 Emptiness of $S(d,m)$

In this section, we treat $m\in [0,\infty )$ as a continuous parameter. For positive small m, the standard basis vectors lie outside $S(d,m)$ ; for example, the functional $w=(1, 0,\dots ,0,\epsilon )$ separates $e_1$ from all the $v^{(i)}$ , whenever $\epsilon < 1-m$ . In the other extreme, for $m\to \infty $ , the $e_i$ ’s lie inside $S(d,m)$ . With the above computations, it is now straightforward to see where the limit between these two situations is. The nonobvious and maybe surprising part of the following result is that containing the standard basis is the only obstruction for $S(d,m)$ to be empty.

To show that $S(d,m)$ is empty whenever $u_d<0$ , we introduce the following notation:

Let $e_0, \ldots , e_d$ denote the standard basis of $ \mathbb {R}^{d+1}$ . We will again use indices modulo $d+1$ . For $i=0, \ldots , d$ , let $a_i = e_{i+1} - e_{i-1}$ , so that the vertices of $S(d,m)$ are the points $e_i + m a_i$ . Let A be the convex hull of all the $a_i$ ’s, which is a cyclically symmetric d-simplex in the hyperplane $\{\sum _i x_i =0\} \subset \mathbb {R}^{d+1}$ that contains the origin in its relative interior. Let $F_i$ be the facet of A opposite to the vertex $a_i$ and let $A_i$ be the convex hull of the origin and $F_i$ . Observe that the simplex A is barycentrically subdivided into the $d+1$ simplices $A_i$ . Finally, let $\Delta = \operatorname {conv} (e_0,\dots , e_d)$ denote the standard unimodular simplex in the hyperplane $\{\sum _{i=0}^d x_i =1\}$ .

Proof. Part (1) is obvious, since the vertices of $S(d,m)$ are the points $e_i +m a_i$ .

Part (3) uses the description of the vector u. The vertices of $e_d + m A_0$ are the points $e_d +m a_i$ for $i\ne 0$ , together with $e_d$ . We have $\langle u, e_d \rangle = u_d < 0$ , and for the other vertices of $e_d + m A_0$ , we write

$$\begin{align*}\langle u, e_d +m a_i\rangle = \langle u, e_d -e_i\rangle + \langle u, e_i +m a_i\rangle. \end{align*}$$

The second summand equals zero since $e_i +m a_i$ (for $i\ne 0$ ) is a vertex of the facet of $S(d,m)$ given by $\langle u , x \rangle =0$ . The first summand equals $u_d-u_i$ , which is negative except for $i=d$ , by parts (1) and (2) in Proposition 4.8 (assuming d even).

It only remains, then, to prove part (2). Let $C=\frac 1{d+1}(1,\dots ,1)$ be the ‘center’ of the hyperplane $\{\sum _i x_i =1\}$ – that is, the barycenter of both $S(d,m)$ and $\Delta $ .

Let $\Sigma _i = \operatorname {cone}(A_i)$ be the ith facet-cone of A, and consider the complete fan centered at C given by the affine cones $C+\Sigma _i $ , $i=0,\dots ,d$ . We will represent points in the ∑ _i=0 ^d x _i = 1 hyperplane in the form

(2) $$ \begin{align} e_d + \sum_{i=1}^d \lambda_i a_i. \end{align} $$

Part (2) is equivalent to the claim that all the lattice points in $(C+\Sigma _i ) \cap (\Delta + m A)$ lie in $e_{i-1} + m A_i$ . We are going to prove the claim for i = 0; that is, we show that the lattice points in $(C+\Sigma _0 ) \cap (\Delta + m A)$ are contained in $e_{d} + m A_0$ . The claim for all i follows cyclically.

Figure 2 Illustration of parts (1) and (2) of Lemma 4.10. Left: the simplex $m\,A$ and its decomposition into $d+1$ dilated unimodular simplices $m\,A_i$ . Center: our circulant simplex $S(d,m)$ , contained in the Minkowski sum $\Delta + m\,A$ . Right: the simplices $e_{i-1} + m A_{i}$ contain all the lattice points in $\Delta + m\,A$ . In the picture, we have $m=2.6$ , which is bigger than the emptiness threshold in dimension two, $m_0(2)=1$ .

Figure 3 Illustration of part (3) of Lemma 4.10. In this picture, $m=0.8$ , below the threshold $m_0(2)=1$ . The simplex $e_{d} + m A_{0}$ lies completely in the open half-space $\{u <0\}$ , except for its vertex $e_{d} + m a_{d}$ , common to $S(d,m)$ .

The cone $\Sigma _0$ , generated by $a_1,\dots , a_{d}$ , is unimodular. Thus, its lattice points are the semigroup Z generated by $a_1,\dots , a_{d}$ :

$$\begin{align*}Z:=\left\{\sum_{i=1}^d \lambda_i a_i: \lambda_i \in \mathbb{Z}_{\ge0} \ \forall i\right\}. \end{align*}$$

This implies that the lattice points in the translated cone $C+\Sigma _0 $ are the set $p+Z$ , where p is the unique lattice point in the translated half-open parallelepiped

$$\begin{align*}C +\left\{\sum_{i=1}^d \lambda_i a_i: \lambda_i \in [0,1) \ \forall i\right\}. \end{align*}$$

This unique point p is $e_d$ , since

$$\begin{align*}e_d = C + \sum_{i=1}^d \frac{i}{d+1} a_{2i}. \end{align*}$$

(Remember that indices are regarded modulo $d+1$ , so $a_{2i}$ for $i> d/2$ means $a_{2i-d-1}$ ). Summing up, we have so far proved that

(3) $$ \begin{align} \mathbb{Z}^d \cap (C+\Sigma_0) = e_d + Z. \end{align} $$

Recall that we denote $F_0$ the facet of A opposite to $a_0$ , which equals the facet of $A_0$ opposite to C. Let $f_0$ be its outer normal vector and let $b_0$ be the value of $f_0(x)$ on $F_0$ . We have that $f_0( a_j)=b_0$ for all $j\ne 0$ and $f_0( a_0) < b_0$ .

The facet $F^{\prime }_0$ of Δ + mA with this same normal vector $f_0$ equals e _i + mF ₀, where e _i is the vertex of $\Delta $ maximizing $f_0 (e_i)$ . Since

$$\begin{align*}f_0(a_j) = f_0(e_{j+1}-e_{j-1})= f_0(e_{j+1}) - f_0(e_{j-1})>0 \quad \forall j \ne 0, \end{align*}$$

the $e_i$ maximizing $f_0$ can only be $e_{-1} =e_d$ . Thus, F ₀ ′ equals e _d + mF ₀, and its facet inequality is $f_0(x) \le f_0(e_d) + m b_0$ . In particular,

(4) $$ \begin{align} \Delta +m A \subset\{f_0(x) \le f_0(e_d) + m b_0\}. \end{align} $$

Let now p be a lattice point in $(C + \Sigma _0) \cap (\Delta +m A)$ . By (3), we have that $p=e_d +\sum _{i=1}^d\lambda _i a_i$ for some $\lambda _1,\dots , \lambda _d\in \mathbb {Z}_\ge 0$ . This leads to

$$\begin{align*}f_0 (p) = f_0(e_d) + b \sum_{i=1}^d\lambda_i \end{align*}$$

and (4) implies $\sum _{i=1}^d\lambda _i\le m$ . Thus, $p\in e_d +mA_0$ , as we wanted to show.

From this lemma, we quickly obtain that (2) ⟹ (4) in Theorem 4.9. Indeed, by part (1) of the lemma, all lattice points of S(d, m) lie in $\Delta + mA$ , and by part (2), they also lie in $e_{i-1} + m A_{i}$ for some i. Part (3) shows that the only point in both $e_{d} + m A_{0}$ and $S(d,m)$ is the vertex $e_d+ma_ d$ . Hence, by cyclic symmetry, the only lattice points that S(d, m) contains are its vertices, the definition of empty.

Observe that the equation $m^{d-1} = c_{d-1}$ in part (1) of Theorem 4.9 has a unique solution with $m>0$ , since the right-hand side is a polynomial of degree $d-2$ with all coefficients nonnegative (and not all zero). Thus, the following is a more explicit version of Theorem 4.9:

Corollary 4.11. Let d be even and let $m_0(d)$ be the unique positive solution of $u_{d}(m)=0$ . Equivalently, of

$$\begin{align*}m^{d-1} = c_{d-1}. \end{align*}$$

If $m <m_0(d)$ , then $S(d,m)$ is empty.

We have computationally verified emptiness up to $d=22$ .

4.5 Asymptotics of $m_0(d)$ , via Fibonacci polynomials and hyperbolic functions

We now investigate the asymptotics of m ₀(d) to understand how wide an empty circulant simplex S(d, m) may be. We continue to use the base assumption that d is even, since otherwise, the width is 1.

The continuant numbers and thus the coefficients of the u-vector have many interesting interpretations in terms of special functions (e.g., generalized Binomial polynomials, hypergeometric functions, etc. [Reference Graham, Knuth, Patashnik and Mathematics8, p.203,204,213]). Here, we stress the relation to the following sequence of polynomials, one for each degree h. Following [Reference Hoggatt and Bicknell11, Reference Hoggatt and Long12], we call them Fibonacci polynomials:Footnote ²

$$\begin{align*}F_h(x):=\sum_{j=0}^{\infty} \binom{h-1-j}{j} x^{h-1-2j}. \end{align*}$$

As usual, in the definition, we take the natural convention that $\binom {a}{b}$ is well-defined for every $a\in \mathbb {Z}$ and $b\in \mathbb {N}$ , with $\binom {a}{b}=0$ if $a <b$ . From Lemma 4.7, we see that

$$\begin{align*}c_{k} = \sum_{i=0}^{\infty} \binom{k-i}{i} m^{2i}=m^{k}F_{k+1}\left(\frac1{m}\right), \end{align*}$$

and from Proposition 4.8, we derive that for $k \in \{0, \ldots , d\}$

(5) $$ \begin{align} u_{k} = \left(F_{d+1-k}\left(\frac{1}{m}\right)+(-1)^{k-1} F_{k}\left(\frac{1}{m}\right)\right) \cdot m^d. \end{align} $$

The property of $F_h$ that we need is the following [Reference Hoggatt and Bicknell11]:

(6) $$ \begin{align} F_{2n}(2\sinh z) = \frac{\sinh (2nz)}{\cosh(z)}, \quad F_{2n+1}(2\sinh z) = \frac{\cosh ((2n+1)z)}{\cosh(z)}. \end{align} $$

Let us define

$$\begin{align*}\alpha = \alpha(m) := \operatorname{\mathrm{arcsinh}}\left(\frac{1}{2m}\right),\end{align*}$$

so that $\frac 1m=2\sinh \alpha $ . Note that $\cosh (\alpha )=\cosh (\operatorname {\mathrm {arcsinh}}\left (\frac {1}{2m}\right )) = \sqrt {\frac {1}{4m^2}+1}$ .

Using Equations (5) and (6), we get that, for even d,

$$\begin{align*}\frac{u_d}{m^d} = F_1\left(\frac1m\right) - F_d\left(\frac1m\right) =1 - \frac{\sinh (d \alpha)}{\cosh(\alpha)}. \end{align*}$$

Since $m_0(d)$ is the unique solution to $u_d=0$ , we obtain that $\alpha _0(d):=\alpha (m_0(d))$ is the unique solution to $\cosh \left (\alpha \right ) - \sinh \left (d \alpha \right ) =0$ .

Theorem 4.12.

$$\begin{align*}\lim_{d\to \infty} \frac{2m_0(d)}{d} = \frac{1}{\operatorname{\mathrm{arcsinh}}(1)} \simeq 1.1346. \end{align*}$$

Proof. The equation

$$\begin{align*}\sinh (d\alpha_0)= \cosh (\alpha_0) \end{align*}$$

clearly implies $\lim _{d\to \infty } \alpha _0(d)=0$ . Thus, we can approximate $\sinh (\alpha _0)\sim \alpha _0$ and $\cosh (\alpha _0)\sim 1$ in the standard sense that $a\sim b$ means $\lim _{d\to \infty } a/b =1$ . Hence,

$$\begin{align*}\sinh(d\alpha_0) \sim 1 \ \ \Rightarrow \ \ d\alpha_0 \sim \operatorname{\mathrm{arcsinh}}(1) \ \ \Rightarrow \ \ \frac1{2m_0(d)} = \sinh(\alpha_0) \sim \alpha_0 \sim \frac{\operatorname{\mathrm{arcsinh}}(1)}d.\\[-44pt] \end{align*}$$

Remark 4.13. This proof also implies that $2m_0(d)$ is always smaller than the asymptotic value $\frac {d}{\operatorname {\mathrm {arcsinh}}(1)} \simeq 1.1346 d$ , because all the ‘ $\sim $ ’ in the proof are in fact ‘ $>$ ’. But (experimentally) the convergence is fairly quick. Up to at least $d=1000$ , we have

$$\begin{align*}\lfloor m_0(d) \rfloor = \left\lfloor\frac{d}{2\operatorname{\mathrm{arcsinh}}(1)}\right\rfloor. \end{align*}$$

4.6 The boundary of $S(d,m)$

We here again assume m to be an integer, so that $S(d,m)$ is a lattice simplex. It is easy to show that it is not always cyclic:

In what follows, we show that the values of d and m in this proposition are the only ones for which $S(d,m)$ is not cyclic (assuming d even) and that when this happens, the quotient group $G_{S(d,m)}$ is isomorphic to $ \mathbb {Z}_{N/2} \oplus \mathbb {Z}_2$ . The key fact is that the volume of each facet of $S(d,m)$ , which does not depend on the facet, by cyclic symmetry, equals the $\gcd $ of the vector u from the previous sections. (See the proof of Theorem 4.16 for an explanation). Our next result computes this $\gcd $ :

Proof. (1) $\Rightarrow $ (2) $\Rightarrow $ (3) are clear. In the following, we will often use Lemma 4.7 and Proposition 4.8 without further mentioning.

To show (3) $\Rightarrow $ (4), if p is a prime that divides $u_0,u_2,u_{d}$ , then $u_0 = c_d=c_{d-1}+m^2 c_{d-2} \equiv 0 \ \pmod p$ . As $c_d \equiv 1 \ \pmod m$ , p does not divide m, and

(7) $$ \begin{align} c_{d-1} \equiv -m^2c_{d-2} \ \pmod p. \end{align} $$

Next, $u_2 = c_{d-2}m^2-m^{d-1} \equiv 0 \ \pmod p$ , hence, as $p \not |m$ ,

(8) $$ \begin{align} c_{d-2} \equiv m^{d-3} \ \pmod p. \end{align} $$

Finally, $u_{d}=m^d - c_ {d-1}m \equiv 0 \ \pmod p$ ; hence, again as $p \not |m$ ,

(9) $$ \begin{align} c_{d-1} \equiv m^{d-1} \quad \pmod p \end{align} $$

From Equations (7),(8),(9) we get $m^{d-1} \equiv -m^{d-1} \ \pmod p$ . As p does not divide m, we see $p=2$ . Now, $c_d \equiv 1 \ \pmod m$ implies m is odd and $c_d=u_0$ is even. This gives $d\equiv 2\ \pmod 3$ since, for odd m, Lemma 4.7(1) gives the Fibonacci recursion

$$\begin{align*}c_k \equiv c_{k-1} + c_{k-2} \quad \pmod 4, \qquad c_{-1}=0, \quad c_0=1, \end{align*}$$

which implies

(10) $$ \begin{align} c_k \text{ is even } \;\Leftrightarrow\; k \equiv 2 \quad \pmod 3. \end{align} $$

Since d is even, $d \equiv 2 \ \pmod 3$ implies $d \equiv 2 \ \pmod 6$ , as desired.

It remains to prove (4) $\Rightarrow $ (1). Let $d \equiv 2\ \pmod 6$ and m be odd. Let $k \in \{0, \ldots , d\}$ . From Proposition 4.8, we have $u_{k} = c_{d-k} \, m^{k} + (-1)^{k-1} c_{k-1}\, m^{d+1-k} \equiv c_{d-k} + c_{k-1} \ \pmod 2$ , as m is odd. Therefore, from the equivalence (10), it is straightforward to verify that $d \equiv 2 \ \pmod 6$ implies that $u_{k}$ is even for every k. So, $2$ divides $\gcd (u_0,\ldots ,u_{d})$ . From the argument in the proof of (3) $\Rightarrow $ (4), we know that $2$ is the only prime divisor of $\gcd (u_0,\ldots ,u_{d})$ . Hence, to show (1), it suffices to show that $4$ does not divide $\gcd (u_0,\ldots ,u_{d})$ . This follows again from the Fibonacci recursion modulo $4$ , which implies

$$\begin{align*}c_k \equiv 0 \quad \pmod 4 \;\Leftrightarrow\; k \equiv 5 \quad \pmod 6.\end{align*}$$

Thus, if $u_0 = c_d \equiv 0 \ \pmod 4$ , then $d \equiv 5 \ \pmod 6$ , a contradiction.

Still, even in the noncyclic case, the facets of $S(d,m)$ (for $d> 2$ ) are empty, no matter how large m is.

Proof. Observe that u is, modulo signs, the vector of $d \times d$ -minors of the matrix obtained by deleting the first column from $M(d,m)$ . Thus, $\gcd (u_0, \ldots , u_d)$ equals the volume of the facet F opposite to the first vertex of $S(d,m)$ . Since, by cyclic symmetry, all facets have the same volume, Lemma 4.15 implies most of the statement: if $d \ne 2 \ \pmod 6$ or m is even, then all facets are unimodular; hence, $S(d,m)$ is cyclic. If $d = 2 \ \pmod 6$ and m is odd, then all facets have volume two, so the exact sequence of Proposition 2.1 becomes

$$\begin{align*}0 \to \mathbb{Z}_2 \to G_{S(d,m)} \to \mathbb{Z}_{N/2} \to 0. \end{align*}$$

Since $G_{S(d,m)}$ is not cyclic (by Proposition 2.3), this implies $G_{S(d,m)}\cong \mathbb {Z}_{N/2}\oplus \mathbb {Z}_2$ .

To show that facets are still empty, observe that $\operatorname {Vol}(F)=2$ implies that every barycentric coordinate of every lattice point in $\operatorname {aff}(F)$ is in $\frac 12 \mathbb {Z}$ . Hence, if F was not empty, then it would contain a lattice point with two barycentric coordinates equal to $\frac 12$ and the rest equal to $0$ . That is to say, the mid-point of an edge would be a lattice point, which clearly is not the case in $S(d,m)$ , when $d\ge 3$ .