# Equilateral sets in the \(\ell _1\) sum of Euclidean spaces

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## Abstract

Let \(E^n\) denote the (real) *n*-dimensional Euclidean space. It is not known whether an equilateral set in the \(\ell _1\) sum of \(E^a\) and \(E^b\), denoted here as \(E^a \oplus _1 E^b\), has maximum size at least \(\dim (E^a \oplus _1 E^b) + 1 = a + b + 1\) for all pairs of *a* and *b*. We show, via some explicit constructions of equilateral sets, that this holds for all \(a \leqslant 27\), as well as some other instances.

## Keywords

Equilateral sets Normed spaces Regular simplices## Mathematics Subject Classification

46B20 52A21 52C10## 1 The problem

An equilateral set in a normed space \((X, ||\cdot ||)\) is a subset \(S \subset X\) such that for all distinct \(x,y \in S\), we have \(||x - y ||= \lambda \) for some fixed \(\lambda \). Since *X* is a normed space, the maximum size of an equilateral set in *X* is independent of \(\lambda \), and we denote it by *e*(*X*). When \(\dim (X) = n\), we have the tight upper bound \(e(X) \leqslant 2^n\), proved by Petty (1971) nearly 50 years ago. However, the following conjecture concerning a lower bound on *e*(*X*), formulated also by Petty (amongst others), remains open for \(n \geqslant 5\). (The \(n=2\) case is easy; see Petty (1971) and Väisälä (2012) for the \(n = 3\) case, and Makeev (2005) for the \(n = 4\) case.)

### Conjecture 1

Let *X* be an *n*-dimensional normed space. Then \(e(X) \geqslant n + 1\).

## 2 The results

Observe that we need only construct \(a + b + 1\) points in \(E^a \oplus _1 E^b\) which form an equilateral set to show that \(e(E^a \oplus _1 E^b) \geqslant \dim (E^a \oplus _1 E^b) + 1 = a + b + 1\). We will work with these points in the form \((x_i, y_i) \in \mathbb {R}^a \times \mathbb {R}^b\), since we can then examine the \(x_i\)’s and \(y_i\)’s separately when necessary. By abuse of notation, we will denote the origin of any Euclidean space by *o*.

*n*-simplex (\(n \geqslant 1\)) with unit side length. Note that

*n*, and we have \(1/2 \leqslant d_n < 1/\sqrt{2}\).

The \(a = 1\) case is easy.

### Proposition 2

\(e(E^1 \oplus _1 E^b) \geqslant b + 2\).

### Proof

Let \(y_1, \cdots , y_{b+1}\) be the vertices of a regular *b*-simplex with unit side length centred on the origin. Then the points \((o, y_1), \cdots , (o, y_{b+1}), (1-d_b, o)\) are pairwise equidistant. \(\square \)

We next deal with the case where \(b = a\).

### Proposition 3

\(e(E^a \oplus _1 E^a) \geqslant 2a + 1\).

### Proof

We first describe an equilateral set of size 2*a* in \(E^a \oplus _1 E^a\): consider the set of points \(\{ (v_i, \frac{1}{2}e_i) : i = 1, \cdots , a\} \cup \{ (v_i, -\frac{1}{2}e_i) : i = 1, \cdots , a\}\), where \(v_1, \cdots , v_a\) are the vertices of a regular simplex of codimension one, centred on the origin with side length \(1 - 1/\sqrt{2}\), and \(e_1, \cdots , e_a\) are the standard basis vectors. Note that the 2*a* vectors \(\pm \frac{1}{2} e_i\) for \(i = 1, \cdots , a\) form a cross-polytope in \(E^a\), centred on the origin.

*x*,

*o*) to the above set, a unit distance away from every other point. Note that we must have \(||x - v_i ||_2 = 1/2\) for \(i = 1, \cdots , a\), and

*x*must lie on the one-dimensional subspace orthogonal to the \((a-1)\)-dimensional subspace spanned by the \(v_i\)’s. This is realisable if \(||x - v_i ||_2 \geqslant (1 - 1/\sqrt{2})d_{a-1}\) (note that the \((a-1)\)-simplex formed by the \(v_i\)’s has side length \(1 - 1/\sqrt{2}\)), in which case we have an equilateral set of size \(2a + 1\) in \(E^a \oplus _1 E^a\). But we have

In the remaining case and our main result, we have \(b > a \geqslant 2\), and we find sufficient conditions for an equilateral set of size \(a + b + 1\) to exist in \(E^a \oplus _1 E^b\).

### Theorem 4

Note that if inequality (1) is satisfied by all pairs of *a* and *b* with \(b > a \geqslant 2\) and \(b \ne 0\), 1, or \(a \pmod {a+1}\), then Proposition 2, Proposition 3, and Theorem 4 cover all possible cases, as \(E^a \oplus _1 E^b\) is isometrically isomorphic to \(E^b \oplus _1 E^a\). Unfortunately, this is not true, and we explore its limitations after the proof of Theorem 4.

### Proof of Theorem 4

*c*-simplex with unit side length centred on the origin in \(V_j\).

- the distance between \(\left( z_j, v_\ell ^{(j)} \right) \) and \(\left( z_{j'}, v_{\ell '}^{(j')} \right) \) for \(j \ne j'\) should be one, and so$$\begin{aligned} ||z_j - z_{j'} ||_2 = 1 - \sqrt{d_{c}^2 + d_{c}^2} = 1 - \sqrt{\frac{c}{c+1}} =: f(c), \end{aligned}$$
- the distance between \(\left( w_i, u_k^{(i)} \right) \) and \(\left( w_{i'}, u_{k'}^{(i')} \right) \) for \(i \ne i'\) should be one, and so$$\begin{aligned} ||w_i - w_{i'} ||_2 = 1 - \sqrt{d_{c-1}^2 + d_{c-1}^2} = 1 - \sqrt{\frac{c-1}{c}} = f(c-1), \end{aligned}$$
- finally, the distance between \(\left( w_i, u_k^{(i)} \right) \) and \(\left( z_{j}, v_{\ell }^{(j)} \right) \) should also be one, and so$$\begin{aligned} ||w_i - z_{j} ||_2 = 1 - \sqrt{d_{c-1}^2 + d_c^2} = 1 - \sqrt{\frac{1}{2} \left( \frac{c-1}{c} + \frac{c}{c+1} \right) } =: g(c). \end{aligned}$$

*f*(

*c*), with the distance between any point from one simplex and any point from the other being

*g*(

*c*). Note that here we consider the \((-1)\)-simplex to be empty. We now show that this configuration is realisable (in \(E^a\)) if the conditions in the statement of the theorem are satisfied.

We first consider the special cases \(\beta = 0\) and \(\beta = 1\) or *a*, and then the main case \(2 \leqslant \beta \leqslant a-1\). It is trivial if \(\beta = 0\): then \(\alpha = a + 1\) and we only need to find a regular *a*-simplex with side length \(f(c-1)\) in \(E^a\).

*a*-simplex exists in \(E^a\). By symmetry and the fact that \(f(c)^2 < f(c-1)^2\), the desired

*a*-simplex also exists if \(\beta = a\).

*f*(

*c*), centred on the origin in \(E^{\beta -1}\). Consider then the set of points \(\{ (p_i, o, o) : i = 1, \cdots , \alpha \} \cup \{ (o, q_j, \zeta ) : j = 1, \cdots , \beta \}\), where \(\zeta \in E^1\) is to be determined. As before, we want a \(\zeta \) such that for all

*i*and

*j*, we have

As mentioned above, inequality (1), and thus inequality (2), does not hold for all pairs of *a* and *b*. However, we have the following result.

### Lemma 5

If \(b \geqslant a ^2 + a\), then inequality (2) holds.

### Proof

*f*(

*n*) is a decreasing function of

*n*, inequality (2) holds if

*a*and

*b*satisfy

*a*and

*b*to satisfy

*c*, and so if \(c \geqslant a\), or equivalently, when \(b \geqslant a^2 + a\), we need only consider the inequality

It can be checked (by computer) that inequality (2) holds for all \(a \leqslant 27\), but does not hold for \(a = 28\) and \(b = 40\), \(a = 29\) and \(39 \leqslant b \leqslant 44\), and \(a = 30\) and \(40 \leqslant b \leqslant 47\). The spaces of smallest dimension where we could not find an equilateral set of size \(a + b + 1\) are \(E^{28} \oplus _1 E^{40}\) and \(E^{29} \oplus _1 E^{39}\).

## Notes

### Acknowledgements

The author would like to thank Konrad Swanepoel for introducing him to this problem, and for the numerous helpful suggestions in writing this up.

## References

- Makeev, V.V.: Equilateral simplices in a four-dimensional normed space. Zap. Nauchn. Sem. St. Peterburg. Otdel. Mat. Inst. Steklov. (POMI) Geom. I Topol
**329**(9), 88–91, 197 (2005)Google Scholar - Petty, C.M.: Equilateral sets in Minkowski spaces. Proc. Am. Math. Soc.
**29**, 369–374 (1971)MathSciNetCrossRefzbMATHGoogle Scholar - Swanepoel, K.: Combinatorial distance geometry in normed spaces. arXiv:1702.00066
- Väisälä, J.: Regular simplices in three-dimensional normed spaces. Beitr. Algebra Geom.
**53**(2), 569–570 (2012)MathSciNetCrossRefzbMATHGoogle Scholar

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