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

  • Aaron LinEmail author
Open Access
Original Article


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.


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 \(e(X) \geqslant n + 1\).

We wish to verify this conjecture for the Cartesian product \(\mathbb {R}^a \times \mathbb {R}^b\), equipped with the norm \(||\cdot ||\) given by
$$\begin{aligned} ||(x, y) ||= ||x ||_2 + ||y ||_2, \end{aligned}$$
where \(x \in \mathbb {R}^a\), \(y \in \mathbb {R}^b\), and \(||\cdot ||_2\) denotes the Euclidean norm. We denote this space by \(E^a \oplus _1 E^b\), and refer to it as the \(\ell _1\) sum of the Euclidean spaces \(E^a\) and \(E^b\). This was considered originally by Roman Karasev of the Moscow Institute of Physics and Technology, as a possible counterexample to Conjecture 1. See Swanepoel (2016, Section 3) for more background on equilateral sets.

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.

Let \(d_n\) denote the circumradius of a regular n-simplex (\(n \geqslant 1\)) with unit side length. Note that
$$\begin{aligned} d_n = \left( \sqrt{2+\frac{2}{n}} \right) ^{-1} \end{aligned}$$
is a strictly increasing function of n, and we have \(1/2 \leqslant d_n < 1/\sqrt{2}\).

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

Proposition 2

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


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

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


We first describe an equilateral set of size 2a 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 2a vectors \(\pm \frac{1}{2} e_i\) for \(i = 1, \cdots , a\) form a cross-polytope in \(E^a\), centred on the origin.

We now want to add a point of the form (xo) 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
$$\begin{aligned} \frac{1}{2}> \frac{1}{\sqrt{2}} \left( 1 - \frac{1}{\sqrt{2}} \right) > \left( 1 - \frac{1}{\sqrt{2}} \right) d_{a-1} \end{aligned}$$
for all \(a \geqslant 2\). \(\square \)

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

Let \(b > a \geqslant 2\). Write \(b = (c-1)(a+1) + \beta = c(a+1) - \alpha \) with \(\beta \in \{ 0, \cdots , a\}\) and \(\alpha \in \{1, \cdots , a+1\}\). If either of the conditions \(\beta = 0\), \(\beta = 1\), \(\beta = a\) is satisfied, or the inequality
$$\begin{aligned} \frac{\alpha - 1}{2\alpha } \left( 1 - \sqrt{\frac{c-1}{c}} \right) ^2 + \frac{\beta - 1}{2\beta } \left( 1 - \sqrt{\frac{c}{c+1}} \right) ^2 \leqslant \left( 1 - \sqrt{\frac{1}{2} \left( \frac{c-1}{c} + \frac{c}{c+1} \right) } \right) ^2 \end{aligned}$$
holds, then \(e(E^a \oplus _1 E^b) \geqslant a + b + 1\).

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

We are going to describe an equilateral set of size \(a + b + 1\) with unit distances between points. Noting that \(\alpha \cdot (c-1) + \beta \cdot c = b\), consider the following decomposition of \(E^b\) into pairwise orthogonal subspaces:
$$\begin{aligned} E^b = U_1 \oplus \cdots U_\alpha \oplus V_1 \oplus \cdots \oplus V_\beta , \end{aligned}$$
where \(\dim U_i = c - 1\) for \(i = 1, \cdots , \alpha \) and \(\dim V_j = c\) for \(j = 1, \cdots , \beta \). Let \(u_1^{(i)}, \cdots , u_c^{(i)}\) be the vertices of a regular \((c-1)\)-simplex with unit side length centred on the origin in \(U_i\), and let \(v_1^{(j)}, \cdots , v_{c+1}^{(j)}\) be the vertices of a regular c-simplex with unit side length centred on the origin in \(V_j\).
The \(a + b + 1\) points of our equilateral set will be
$$\begin{aligned} \left\{ \left( w_i, u_k^{(i)} \right) : 1 \leqslant i \leqslant \alpha , 1 \leqslant k \leqslant c \right\} \cup \left\{ \left( z_j, v_\ell ^{(j)} \right) : 1 \leqslant j \leqslant \beta , 1 \leqslant \ell \leqslant c + 1 \right\} . \end{aligned}$$
Note here that \(\alpha \cdot c + \beta \cdot (c+1) = a + b + 1\), and we have \(||u_k^{(i)} - u_{k'}^{(i)} ||_2 = ||v_\ell ^{(j)} - u_{\ell '}^{(j)} ||_2 = 1\) for \(k \ne k'\) and \(\ell \ne \ell '\). All that remains is then to calculate how far apart the \(w_i\)’s and \(z_j\)’s should be in \(E^a\), and see if such a configuration is realisable.
We only have three non-trivial distances to calculate:
  • 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
    $$\begin{aligned} ||w_i - z_{j} ||_2 = 1 - \sqrt{d_{c-1}^2 + d_c^2} = 1 - \sqrt{\frac{1}{2} \left( \frac{c-1}{c} + \frac{c}{c+1} \right) } =: g(c). \end{aligned}$$
What we need in \(E^a\) is thus a regular \((\alpha - 1)\)-simplex with side length \(f(c-1)\) and a regular \((\beta - 1)\)-simplex with side length 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 \(E^a\).

If \(\beta = 1\), in which case \(\alpha = a\), consider the decomposition \(E^a = E^{a-1} \oplus E^1\). Consider the points \((p_1, o), \cdots , (p_a, o)\), where \(p_1, \cdots , p_a\) are the vertices of a regular \((a-1)\)-simplex with side length \(f(c-1)\), centred on the origin in \(E^{a-1}\). We want to add a point \((o, \zeta )\) for some \(\zeta \in E^1\) such that, for any \(i = 1, \cdots , a\), we have
$$\begin{aligned} ||(p_i, o) - (o, \zeta ) ||_2 = g(c), \end{aligned}$$
or equivalently,
$$\begin{aligned} d_{a-1}^2 f(c-1)^2 + \zeta ^2 = g(c)^2. \end{aligned}$$
Noting that \(d_{a-1} < 1/\sqrt{2}\), it suffices to show, for all \(c \geqslant 2\), that
$$\begin{aligned} f(c-1)^2 < 2 g(c)^2. \end{aligned}$$
But this is easily verifiable to be true, and so the desired 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 \(\beta = a\).
Now suppose \(2 \leqslant \beta \leqslant a - 1\) so that \(\alpha , \beta \geqslant 2\). Consider this time, the decomposition \(E^a = E^{\alpha - 1} \oplus E^{\beta - 1} \oplus E^1\), noting that \(\alpha + \beta = a + 1\). Suppose \(p_1, \cdots , p_\alpha \) are the vertices of a regular \((\alpha -1)\)-simplex with side length \(f(c-1)\), centred on the origin in \(E^{\alpha -1}\), and \(q_1, \cdots , q_\beta \) are the vertices of a regular \((\beta -1)\)-simplex with side length 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
$$\begin{aligned} ||(p_i, o, o) - (o, q_j, \zeta ) ||_2 = g(c), \end{aligned}$$
or equivalently
$$\begin{aligned} \left( d_{\alpha -1} f(c-1) \right) ^2 + \left( d_{\beta -1} f(c) \right) ^2 \leqslant g(c)^2. \end{aligned}$$
But this is exactly inequality (1). \(\square \)

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.


Since f(n) is a decreasing function of n, inequality (2) holds if a and b satisfy
$$\begin{aligned} \left( d_{\alpha -1}^2 + d_{\beta -1}^2 \right) f(c-1)^2 < g(c)^2. \end{aligned}$$
Using the fact that \(\alpha = a + 1 - \beta \) implies \(d_{\alpha -1}^2 + d_{\beta -1}^2 \leqslant (a-1)/(a+1)\), we therefore just need a and b to satisfy
$$\begin{aligned} \frac{a-1}{a+1} < \left( \frac{g(c)}{f(c-1)} \right) ^2. \end{aligned}$$
But the latter expression is an increasing function of c, and so if \(c \geqslant a\), or equivalently, when \(b \geqslant a^2 + a\), we need only consider the inequality
$$\begin{aligned} \frac{a-1}{a+1} < \left( \frac{g(a)}{f(a-1)} \right) ^2, \end{aligned}$$
which is then easily verifiable to be true. \(\square \)

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 \(E^{29} \oplus _1 E^{39}\).



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


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  3. Swanepoel, K.: Combinatorial distance geometry in normed spaces. arXiv:1702.00066
  4. Väisälä, J.: Regular simplices in three-dimensional normed spaces. Beitr. Algebra Geom. 53(2), 569–570 (2012)MathSciNetCrossRefzbMATHGoogle Scholar

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Authors and Affiliations

  1. 1.Department of MathematicsThe London School of Economics and Political ScienceLondonUK

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