Equilateral sets in the $$\ell _1$$ sum of Euclidean spaces

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Original Article

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

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

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

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 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}
(1)
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}
(2)
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.

Proof

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 $$E^{29} \oplus _1 E^{39}$$.

References

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