Equilateral Sets in ℓₚⁿ Beyond n + 1: Linear Excess and Dimension Cost

1. Introduction

How many points can an equilateral set in ${\ell }_p^n$ have as p and n vary? Write $e\left({\ell }_p^n\right)$ for this maximum. For every finite $p>1$, one has $e\left({\ell }_p^n\right)\ge n+1$. In Guy’s 1983 problem list, Kusner conjectured that $e\left({\ell }_p^n\right)=n+1$ for every $n\ge 1$ and $p>2$ [1]. Swanepoel established the case $p=4$; Ge, Xu and Zhou later proved the full interval $2\le p\le 4$ [2,3]. Chalmers then disproved Kusner’s conjecture as stated by Guy, constructing 58 equilateral points in ${\ell }_5^{56}$ [4]. The full $p>4$ range was subsequently covered [5].

The failure of the $n+1$ bound raises two questions. At a fixed $p>4$, how does the number of extra points grow with n? At a fixed dimension n, what happens as p grows?

At every fixed $p>4$, the excess over $n+1$ grows at least linearly with n.

Theorem 1 

(Linear excess). For every real $p>4$ there are constants $c_p>0$ and $n_p$ such that

$$e\left({\ell }_p^n\right)\ge (1+c_p)n\mathrm{for}\mathrm{every}\mathrm{integer}n\ge n_p.$$

Theorem 1 follows by joining copies of a spherical seed. If $S\subset {\ell }_p^d$ has $d+2$ points of common norm-power N and pairwise distance-power $D>2N$, then Lemma 1 gives $r(d+2)$ equilateral points in ${\ell }_p^{r(d+1)-1}$ for every $r\ge 1$. The excess is therefore r. Proposition 1 constructs such a seed for every $p>4$.

At each fixed dimension, sufficiently large finite p forces the maximum back to $n+1$. At $p=\infty$, the $2^n$ vertices of the cube are equilateral.

Theorem 2 

(High-exponent exclusion). If $n\ge 2$ is an integer and $p\ge \max \{6,2n{\left(\log n\right)}^2\}$ is real, then

$$e\left({\ell }_p^n\right)=n+1.$$

The first two theorems bound the dimension needed to retain extra points as p grows. For an integer $k\ge 1$, define

$$D_k\left(p\right):=\min \left\{n\ge 1:e\left({\ell }_p^n\right)\ge n+1+k\right\}.$$

The construction in Theorem 1 makes $D_k\left(p\right)$ finite for $p>4$ and supplies an upper bound. Theorem 2 supplies a lower bound when p is large.

Theorem 3 

(Dimension cost of excess). For every integer $k\ge 1$ and real $p\ge 6$,

$${\displaystyle \frac{p}{2{\left(\log p\right)}^2}}<D_k\left(p\right)\le {\displaystyle \frac{256}{9}}kp.$$

Consequently, $D_k\left(p\right)=p^{1+o\left(1\right)}$ as $p\to \infty$ for each fixed k.

Our seed has $32m$ points in $32m-2$ coordinates. For $p\ge 6$ we can take $m\le 8p/9$. Theorem 2 uses a rank obstruction for an approximate distance matrix. An earlier estimate gave $e\left({\ell }_p^n\right)\le 2(p+1)n$ for $p\ge c{\left(n\log n\right)}^2$ with an absolute constant $c>0$ [6].

Swanepoel proved linear excess for $1<p<2$ [2]. Together with Theorem 1, this gives linear excess for every finite $p\in (1,2)\cup (4,\infty )$; for $2\le p\le 4$, the maximum is $n+1$. The exact values of $e\left({\ell }_p^n\right)$ for general pairs $(p,n)$ and the exact first dimensions $D_k\left(p\right)$ remain open.

2. Construction and Linear Excess

For $x\in {\mathbb{R}}^d$ write ${\parallel x\parallel }_p^p={\sum }_{j=1}^d{\left|x_j\right|}^p$. Joining r copies of a spherical seed with $32m$ points in $32m-2$ coordinates gives $32mr$ points in $r(32m-1)-1$ coordinates, an excess of r over dimension plus one. The joining lemma is related to [2].

Lemma 1 

(Spherical amplification). Let $p>1$ and let $S\subset {\ell }_p^d$ have M points, common distance-power $D>0$, and common norm-power N with $2N<D$. For every integer $r\ge 1$ there is an equilateral set of $rM$ points in ${\ell }_p^{rd+r-1}$ with distance-power D.

Proof. 

The case $r=1$ is S. For $r\ge 2$, let $b\in (0,1)$ solve

$${(1+b)}^p+(r-2)b^p=2.$$

The left-hand side is strictly increasing, starts at 1, and exceeds 2 at $b=1$. Thus the points

$$e_1,\dots ,e_{r-1},-b(1,\dots ,1)\in {\mathbb{R}}^{r-1}$$

have common distance-power 2. Scale these tags to distance-power $D-2N$. Place r copies of S in disjoint d-coordinate blocks, appending the corresponding tag to each point of a copy. Distances within a copy retain power D; cross-copy distances have power $N+N+(D-2N)=D$.    □

To construct a seed for Lemma 1, define sixteen type vectors from a Walsh matrix. Index the rows and columns of the symmetric 8-by-8 Walsh matrix $H_8$ by ${\mathbb{F}}_2^3$, with ${\left(H_8\right)}_{ij}={(-1)}^{i·j}$. For $t>1$ define sixteen vectors in ${\mathbb{R}}^{16}$ by

$$q_i=({\left(H_8\right)}_{i,*},t{e}_i),q_{8+i}=(t{e}_i,-{\left(H_8\right)}_{*,i}),i\in {\mathbb{F}}_2^3.$$

Since $H_8$ is symmetric, either row or column notation may be used in the second expression. Put

$$\begin{array}{cccccc}\hfill R& =8+t^p,\hfill & \hfill A& =4·2^p+2t^p,\hfill & \hfill B& ={(t+1)}^p+{(t-1)}^p+14,\hfill \end{array}$$

(1)

$$\begin{array}{cccccc}\hfill L& =2^{p-1}R,\hfill & \hfill T& =(1+m^{-1})L,\hfill & \hfill F_{p,m}\left(t\right)& =T+7A-8B.\hfill \end{array}$$

(2)

Here and below m is a power of two. Each $q_i$ has norm-power R. Moreover, for distinct $i,j$ and any signs $a,b\in \{-1,1\}$,

$$\parallel a{q}_i-b{q}_j{\parallel }_p^p=\left\{\begin{array}{cc}A,\hfill & i,j\mathrm{belong}\mathrm{to}\mathrm{the}\mathrm{same}\mathrm{half}\mathrm{of}\{0,\dots ,15\},\hfill \\ B,\hfill & i,j\mathrm{belong}\mathrm{to}\mathrm{different}\mathrm{halves}.\hfill \end{array}\right.$$

(3)

Within one half, two signed Walsh rows or columns differ in exactly four places, and the two spikes contribute $2t^p$. Across halves, fourteen coordinates contribute 1 each; the two remaining terms are ${(t+1)}^p$ and ${(t-1)}^p$. This proves (3) for both sign choices.

Proposition 1 

(Spherical seed above four). For every real $p>4$ there is a power of two m and an equilateral set of $32m$ points in ${\ell }_p^{32m-2}$. Its points have common norm-power N, its pairwise distances have common power D, and $2N<D$. For $p\ge 6$ one may take m to be the least power of two strictly greater than $4p/9$.

Proof. 

We first give the coordinates for any m and $t>1$ satisfying $F_{p,m}\left(t\right)=0$; the existence of these parameters is proved in Lemmas 2 and 4 below. Set

$$W={\displaystyle \frac{T-A}{8}},{\alpha }^p={\displaystyle \frac{R}{16}},{\beta }^p={\displaystyle \frac{mW}{2^p}},$$

choosing the positive roots. These choices are valid because

$$T-A=(2^{p-1}-2)t^p+{\displaystyle \frac{L}{m}}>0.$$

The equation $F_{p,m}\left(t\right)=0$ gives

$$A+8W=B+7W=T.$$

(4)

Write $m=2^d$, index the rows and columns of $H_m$ by ${\mathbb{F}}_2^d$, and set $H_m(s,r)={(-1)}^{s·r}$. Identify the type indices $0,\dots ,15$ with ${\mathbb{F}}_2^4$ and write ${\chi }_i\left(c\right)={(-1)}^{i·c}$. Let $C={\mathbb{F}}_2^4\setminus \{0,8\}$, where 8 denotes the first coordinate character. For each $(\sigma ,i,s)\in \{-1,1\}\times {\mathbb{F}}_2^4\times {\mathbb{F}}_2^d$ form a point $x_{\sigma ,i,s}$ by concatenating

$$\begin{array}{cccc}\hfill & \sigma H_m(s,r)q_i\left[a\right]\hfill & \hfill & (r\in {\mathbb{F}}_2^d,a=0,\dots ,15),\hfill \\ \hfill & \alpha {\chi }_i\left(c\right)H_m(s,u)\hfill & \hfill & (c\in {\mathbb{F}}_2^4,u\in {\mathbb{F}}_2^d\setminus \left\{0\right\}),\hfill \\ \hfill & \beta {\chi }_i\left(c\right)\hfill & \hfill & (c\in C).\hfill \end{array}$$

There are $32m$ labels and $16m+16(m-1)+14=32m-2$ coordinates.

Put $D=2^{p}mR$. If $i,s$ agree and only $\sigma$ changes, the front block contributes D. If i agrees and s changes, the balance of distinct Walsh rows gives $2^{p-1}mR$ from the front block and the same amount from the ordinary character block. If $i\ne j$, the front contributes $mA$ or $mB$ according to the halves containing $i,j$. The ordinary block contributes $2^{p-1}(m-1)R$: for each nonzero Walsh column, exactly eight of the sixteen character signs differ. Of the fourteen retained characters, eight differ when $i,j$ are in the same half and seven differ when they are in different halves. Thus the retained contribution is $8mW$ or $7mW$, respectively. Equation (4) makes the total in either case

$$mT+2^{p-1}(m-1)R=2^{p}mR=D.$$

Since $D>0$, the labels give distinct points.

Every point has norm-power

$$N=mR+16(m-1){\alpha }^p+14{\beta }^p=(2m-1)R+{\displaystyle \frac{14mW}{2^p}}.$$

As $A>0$, one has $W<T/8$, so

$$N<{\displaystyle \frac{23m-1}{8}}R<2^{p-1}mR={\displaystyle \frac{D}{2}}$$

for $p>4$. The root lemmas complete the proof.    □

Lemma 2 

(Scalar root for $p\ge 6$). If $p\ge 6$ and m is the least power of two strictly greater than $4p/9$, the equation $F_{p,m}\left(t\right)=0$ has a root $t\in (1,{64}^{1/p})$, and $m\le 8p/9$.

Proof. 

Direct substitution gives

$$F_{p,m}\left(1\right)=49(2^{p-1}-2)+{\displaystyle \frac{92^{p-1}}{m}}>0.$$

Set $t_0={64}^{1/p}$, $a=\log 64$ and $y=a/\left(2p\right)$. With $F_{p,\infty }=L+7A-8B$, write $\epsilon =-F_{p,\infty }\left(t_0\right)/L\left(t_0\right)$. Expanding (1) at $t_0$ yields

$$\epsilon ={\displaystyle \frac{16}{9}}\left(\cosh {\left(y\right)}^p-1\right)+{\displaystyle \frac{2}{9}}{\left({\displaystyle \frac{t_0-1}{2}}\right)}^p-{\displaystyle \frac{196}{9·2^p}}.$$

Using $\cosh y\ge 1+y^2/2+y^4/24$ and ${(1+z)}^p\ge 1+pz+p(p-1)z^2/2$ for $z\ge 0$ gives

$$p\epsilon \ge {\displaystyle \frac{2a^2}{9}}+{\displaystyle \frac{a^4(3p-2)}{216p^2}}-{\displaystyle \frac{196p}{9·2^p}}.$$

(5)

The bound $a>112/27$ follows from

$$\log 2=2{\int }_0^{1/3}{\displaystyle \frac{dz}{1-z^2}}>2{\int }_0^{1/3}(1+z^2)dz={\displaystyle \frac{56}{81}}.$$

On $6\le p\le 8$, the middle term in (5) is bounded below by its value at 8 and the last term by its value at 6, giving

$$p\epsilon >{\displaystyle \frac{258663895}{114791256}}>{\displaystyle \frac{9}{4}}.$$

For $p\ge 8$, discarding the middle term and bounding the last term at 8 gives

$$p\epsilon >{\displaystyle \frac{164983}{52488}}>{\displaystyle \frac{9}{4}}.$$

These comparisons use that $(3p-2)/p^2$ and $p/2^p$ decrease for $p\ge 6$. Thus $1/m<9/\left(4p\right)<\epsilon$, so $F_{p,m}\left(t_0\right)=L\left(t_0\right)(1/m-\epsilon )<0$. Continuity supplies the root. The least power of two above $4p/9$ is at most $8p/9$.    □

Lemma 3 

(Exponential zero count). If $0<a_1<\dots <a_s$ and each $c_j\ne 0$, then $f\left(x\right)={\sum }_{j=1}^s{c}_j{a}_j^x$ has at most as many real zeros, counted with multiplicity, as the sign sequence $(c_1,\dots ,c_s)$ has changes.

Proof. 

Induct on s. Dividing f by $a_1^x$ preserves its zeros. The derivative of the resulting function is an exponential polynomial with coefficient signs $(c_2,\dots ,c_s)$. By induction it has at most $V(c_2,\dots ,c_s)$ zeros. Rolle’s theorem, including multiplicities, allows at most one additional zero for f. When $c_1$ and $c_2$ have opposite signs, this is exactly the additional sign change. When they have the same sign, the normalized function and its derivative have that sign sufficiently far to the left. Any first zero of the function then requires a preceding zero of its derivative. Together with the zeros supplied by Rolle between subsequent zeros, this removes the additional allowance.    □

Lemma 4 

(Scalar root for $4<p<6$). For every real $4<p<6$ there are a power of two m and $t\in (1,\sqrt{8})$ satisfying $F_{p,m}\left(t\right)=0$.

Proof. 

Put $t_0=\sqrt{8}$ and $f\left(p\right)=F_{p,\infty }\left(t_0\right)$. Algebraic expansion gives

$$f\left(p\right)=-112-8{(t_0-1)}^p+322^p+14t_0^p-8{(t_0+1)}^p+\frac{1}{2}{\left(2t_0\right)}^p.$$

Its bases are strictly increasing and its coefficient signs are $-,-,+,+,-,+$, with three changes. Direct expansion gives, for general t,

$$F_{2,\infty }\left(t\right)=0,F_{4,\infty }\left(t\right)=6{(t^2-8)}^2,F_{6,\infty }\left(t\right)=30(t^2-8)(t^4-8).$$

Thus f has zeros at $2,4,6$. Lemma 3 shows that these are all its real zeros and each is simple. Its largest-base term is positive, hence $f\left(p\right)>0$ for $p>6$ and $f\left(p\right)<0$ for $4<p<6$. Choose a power of two $m>L\left(t_0\right)/[-f\left(p\right)]$. Then $F_{p,m}\left(t_0\right)=f\left(p\right)+L\left(t_0\right)/m<0$, whereas $F_{p,m}\left(1\right)>0$ by the preceding lemma’s direct calculation. Continuity gives the asserted root.    □

Proof 

(Proof of Theorem 1). Fix $p>4$ and take the seed of Proposition 1. Lemma 1 gives, for every $r\ge 1$, $32mr$ equilateral points in dimension $r(32m-1)-1$. Appending zero coordinates preserves all distances, so for every $n\ge 32m-2$,

$$e\left({\ell }_p^n\right)\ge 32m⌊{\displaystyle \frac{n+1}{32m-1}}⌋.$$

(6)

Consequently

$$\underset{n\to \infty }{\mathrm{lim\; inf}}{\displaystyle \frac{e\left({\ell }_p^n\right)}{n}}\ge {\displaystyle \frac{32m}{32m-1}}>1.$$

Choosing any $c_p<1/(32m-1)$ proves the theorem for all sufficiently large n.    □

3. High-Exponent Exclusion and Dimension Cost

For Theorem 2, a matrix of coordinate powers approximates the distance matrix of a putative $(n+2)$-point set. The error bound and rank constraint give a contradiction.

Proof 

(Proof of Theorem 2). Suppose that $n+2$ equilateral points exist in ${\ell }_p^n$, with $n\ge 2$ and $p\ge \max \{6,2n{\left(\log n\right)}^2\}$. A larger set contains such a subset. Rescale the common distance to 1 and translate each coordinate by its minimum over the points. Denote the resulting coordinates by $x_{i\ell }$, where $1\le i\le n+2$ and $1\le \ell \le n$. Every $x_{i\ell }$ belongs to $[0,1]$, and

$${\displaystyle \sum _{\ell =1}^n}{|x_{i\ell }-x_{j\ell }|}^p=1(i\ne j).$$

(7)

Define the $(n+2)$-by-n matrices

$$P_{i\ell }=x_{i\ell }^p,Q_{i\ell }={(1-x_{i\ell })}^p,$$

and put $K=P{Q}^{\mathsf{T}}+Q{P}^{\mathsf{T}}$ and $E=K-(J-I)$, where J is the all-ones matrix.

First, every entry of E is nonnegative. For $0\le x\le y\le 1$, set $d=y-x$ and $t=x(1-y)$. Then $(1-x)y=d+t$, so

$$x^p{(1-y)}^p+{(1-x)}^p{y}^p-d^p={(d+t)}^p-d^p+t^p\ge 0.$$

(8)

Summing (8) over coordinates and using (7) proves $E_{ij}\ge 0$ for $i\ne j$; the diagonal is nonnegative directly.

We bound the off-diagonal error. Put

$$v={\displaystyle \frac{\log n}{p}},u={\displaystyle \frac{{\left(\log n\right)}^2}{p}},\rho =n^{-2/p}=e^{-2v}.$$

The elementary inequality $t\le {(1-d)}^2/4$ gives

$$d+t\le {\left({\displaystyle \frac{1+d}{2}}\right)}^2=d\cosh {\left({\displaystyle \frac{\log (1/d)}{2}}\right)}^2(d>0).$$

(9)

Since $\log \cosh z\le z^2/2$ for $z\ge 0$, coordinates with $d\ge \rho$ satisfy ${(d+t)}^p-d^p\le d^p(e^u-1)$. For $d<\rho$, the first expression in (9) gives

$${(d+t)}^p\le {\left({\displaystyle \frac{1+\rho }{2}}\right)}^{2p}=n^{-2}\cosh {\left(v\right)}^{2p}\le n^{-2}{e}^u.$$

Also $t^p\le 4^{-p}$. Summing the two classes of coordinates and using (7), we obtain

$$0\le E_{ij}\le \eta :=\left(1+{\displaystyle \frac{1}{n}}\right)e^u-1+n{4}^{-p}(i\ne j).$$

(10)

On the diagonal, a separate calculation gives

$$0\le E_{ii}=2{\displaystyle \sum _{\ell =1}^n}{\left[x_{i\ell }(1-x_{i\ell })\right]}^p\le \delta :=2n{4}^{-p}.$$

(11)

There is a nonzero $c\in {\mathbb{R}}^{n+2}$ satisfying $P^{\mathsf{T}}c=0$ and ${\sum }_i{c}_i=0$, since these are at most $n+1$ homogeneous equations in $n+2$ unknowns. Hence $c^{\mathsf{T}}Kc=0$ and $c^{\mathsf{T}}(J-I)c=-{\parallel c\parallel }_2^2$, so

$$c^{\mathsf{T}}Ec={\parallel c\parallel }_2^2.$$

(12)

Let s be the sum of the positive entries of c; the negative entries sum to $-s$. Nonnegativity of E allows us to discard the terms with $c_i{c}_j<0$ in an upper bound. The same-sign off-diagonal coefficients sum to $2s^2-{\parallel c\parallel }_2^2$. Thus

$$c^{\mathsf{T}}Ec\le \eta (2s^2-{\parallel c\parallel }_2^2{)+\delta \parallel c\parallel }_2^2\le \left({\displaystyle \frac{n}{2}}\eta +\delta \right){\parallel c\parallel }_2^2,$$

where the last step uses $4s^2={\parallel c\parallel }_1^2\le (n+2){\parallel c\parallel }_2^2$.

The coefficient is strictly less than 1. For $n\ge 3$, $u\le 1/\left(2n\right)$ and $e^u\le {(1-u)}^{-1}\le 2n/(2n-1)$. Since $\log n\ge \log 3>1$, the hypothesis implies $p\ge 2n$. Therefore

$${\displaystyle \frac{n}{2}}\eta +\delta \le {\displaystyle \frac{3n}{2(2n-1)}}+\left({\displaystyle \frac{n^2}{2}}+2n\right){16}^{-n}\le {\displaystyle \frac{9}{10}}+{\displaystyle \frac{21}{8192}}<1.$$

The last exponential expression decreases for integer $n\ge 3$. For $n=2$, the bounds $\log 2<3/4$ and $p\ge 6$ give $u<1/8$ and $e^u\le 8/7$, whence

$${\displaystyle \frac{n}{2}}\eta +\delta \le {\displaystyle \frac{5}{7}}+{\displaystyle \frac{6}{4096}}<1.$$

This contradicts (12).

Finally, $n+1$ points always exist. Take $e_1,\dots ,e_n,-a(1,\dots ,1)$, where $a\in (0,1)$ solves ${(1+a)}^p+(n-1)a^p=2$. Their pairwise distance-powers all equal 2. Thus $e\left({\ell }_p^n\right)=n+1$.    □

Proof 

(Proof of Theorem 3). Fix $p\ge 6$ and $k\ge 1$. Let m be the seed order of Lemma 2. Joining k copies of the seed yields $32mk$ points in dimension $n=k(32m-1)-1$. Since $32mk=n+1+k$,

$$D_k\left(p\right)\le k(32m-1)-1\le k\left({\displaystyle \frac{256}{9}}p-1\right)-1<{\displaystyle \frac{256}{9}}kp.$$

Conversely, let $n=D_k\left(p\right)$. Then $n\ge 2$ and ${\ell }_p^n$ contains at least $n+2$ equilateral points. Theorem 2 implies $p<2n{\left(\log n\right)}^2$. If $n\le p$, this gives $p<2n{\left(\log p\right)}^2$. If $n>p$, the desired lower bound holds because $2{\left(\log p\right)}^2>1$. In either case,

$${\displaystyle \frac{p}{2{\left(\log p\right)}^2}}<D_k\left(p\right).$$

For fixed k, taking logarithms of the two bounds gives $\log D_k\left(p\right)/\log p\to 1$, which is $D_k\left(p\right)=p^{1+o\left(1\right)}$.    □

For each fixed $p>4$, the number of equilateral points beyond $n+1$ grows at least linearly with dimension. For each fixed dimension, sufficiently large p restores the maximum $n+1$. Our bounds place the dimension needed to retain any fixed number of extra points at $p^{1+o\left(1\right)}$. The remaining question is whether that dimension must grow linearly with p.

Question. For each fixed integer $k\ge 1$, is there a constant $c_k>0$ such that $D_k\left(p\right)\ge c_{k}p$ for all sufficiently large real p?