Equilateral dimension

Max number of equidistant points in a metric space
Regular simplexes of dimensions 0 through 3. The vertices of these shapes give the largest possible equally-spaced point sets for the Euclidean distances in those dimensions

In mathematics, the equilateral dimension of a metric space is the maximum size of any subset of the space whose points are all at equal distances to each other.[1] Equilateral dimension has also been called "metric dimension", but the term "metric dimension" also has many other inequivalent usages.[1] The equilateral dimension of d {\displaystyle d} -dimensional Euclidean space is d + 1 {\displaystyle d+1} , achieved by a regular simplex, and the equilateral dimension of a d {\displaystyle d} -dimensional vector space with the Chebyshev distance ( L {\displaystyle L^{\infty }} norm) is 2 d {\displaystyle 2^{d}} , achieved by a hypercube. However, the equilateral dimension of a space with the Manhattan distance ( L 1 {\displaystyle L^{1}} norm) is not known. Kusner's conjecture, named after Robert B. Kusner, states that it is exactly 2 d {\displaystyle 2d} , achieved by a cross polytope.[2]

Lebesgue spaces

The equilateral dimension has been particularly studied for Lebesgue spaces, finite-dimensional normed vector spaces with the L p {\displaystyle L^{p}} norm

  x p = ( | x 1 | p + | x 2 | p + + | x d | p ) 1 / p . {\displaystyle \ \|x\|_{p}={\bigl (}|x_{1}|^{p}+|x_{2}|^{p}+\cdots +|x_{d}|^{p}{\bigr )}^{1/p}.}

The equilateral dimension of L p {\displaystyle L^{p}} spaces of dimension d {\displaystyle d} behaves differently depending on the value of p {\displaystyle p} :

Unsolved problem in mathematics:

How many equidistant points exist in spaces with Manhattan distance?

  • For p = 1 {\displaystyle p=1} , the L p {\displaystyle L^{p}} norm gives rise to Manhattan distance. In this case, it is possible to find 2 d {\displaystyle 2d} equidistant points, the vertices of an axis-aligned cross polytope. The equilateral dimension is known to be exactly 2 d {\displaystyle 2d} for d 4 {\displaystyle d\leq 4} ,[3] and to be upper bounded by O ( d log d ) {\displaystyle O(d\log d)} for all d {\displaystyle d} .[4] Robert B. Kusner suggested in 1983 that the equilateral dimension for this case should be exactly 2 d {\displaystyle 2d} ;[5] this suggestion (together with a related suggestion for the equilateral dimension when p > 2 {\displaystyle p>2} ) has come to be known as Kusner's conjecture.
  • For 1 < p < 2 {\displaystyle 1<p<2} , the equilateral dimension is at least ( 1 + ε ) d {\displaystyle (1+\varepsilon )d} where ε {\displaystyle \varepsilon } is a constant that depends on p {\displaystyle p} .[6]
  • For p = 2 {\displaystyle p=2} , the L p {\displaystyle L^{p}} norm is the familiar Euclidean distance. The equilateral dimension of d {\displaystyle d} -dimensional Euclidean space is d + 1 {\displaystyle d+1} : the d + 1 {\displaystyle d+1} vertices of an equilateral triangle, regular tetrahedron, or higher-dimensional regular simplex form an equilateral set, and every equilateral set must have this form.[5]
  • For 2 < p < {\displaystyle 2<p<\infty } , the equilateral dimension is at least d + 1 {\displaystyle d+1} : for instance the d {\displaystyle d} basis vectors of the vector space together with another vector of the form ( x , x , ) {\displaystyle (-x,-x,\dots )} for a suitable choice of x {\displaystyle x} form an equilateral set. Kusner's conjecture states that in these cases the equilateral dimension is exactly d + 1 {\displaystyle d+1} . Kusner's conjecture has been proven for the special case that p = 4 {\displaystyle p=4} .[6] When p {\displaystyle p} is an odd integer the equilateral dimension is upper bounded by O ( d log d ) {\displaystyle O(d\log d)} .[4]
  • For p = {\displaystyle p=\infty } (the limiting case of the L p {\displaystyle L^{p}} norm for finite values of p {\displaystyle p} , in the limit as p {\displaystyle p} grows to infinity) the L p {\displaystyle L^{p}} norm becomes the Chebyshev distance, the maximum absolute value of the differences of the coordinates. For a d {\displaystyle d} -dimensional vector space with the Chebyshev distance, the equilateral dimension is 2 d {\displaystyle 2^{d}} : the 2 d {\displaystyle 2^{d}} vertices of an axis-aligned hypercube are at equal distances from each other, and no larger equilateral set is possible.[5]

Normed vector spaces

Equilateral dimension has also been considered for normed vector spaces with norms other than the L p {\displaystyle L^{p}} norms. The problem of determining the equilateral dimension for a given norm is closely related to the kissing number problem: the kissing number in a normed space is the maximum number of disjoint translates of a unit ball that can all touch a single central ball, whereas the equilateral dimension is the maximum number of disjoint translates that can all touch each other.

For a normed vector space of dimension d {\displaystyle d} , the equilateral dimension is at most 2 d {\displaystyle 2^{d}} ; that is, the L {\displaystyle L^{\infty }} norm has the highest equilateral dimension among all normed spaces.[7] Petty (1971) asked whether every normed vector space of dimension d {\displaystyle d} has equilateral dimension at least d + 1 {\displaystyle d+1} , but this remains unknown. There exist normed spaces in any dimension for which certain sets of four equilateral points cannot be extended to any larger equilateral set[7] but these spaces may have larger equilateral sets that do not include these four points. For norms that are sufficiently close in Banach–Mazur distance to an L p {\displaystyle L^{p}} norm, Petty's question has a positive answer: the equilateral dimension is at least d + 1 {\displaystyle d+1} .[8]

It is not possible for high-dimensional spaces to have bounded equilateral dimension: for any integer k {\displaystyle k} , all normed vector spaces of sufficiently high dimension have equilateral dimension at least k {\displaystyle k} .[9] more specifically, according to a variation of Dvoretzky's theorem by Alon & Milman (1983), every d {\displaystyle d} -dimensional normed space has a k {\displaystyle k} -dimensional subspace that is close either to a Euclidean space or to a Chebyshev space, where

k exp ( c log d ) {\displaystyle k\geq \exp(c{\sqrt {\log d}})}
for some constant c {\displaystyle c} . Because it is close to a Lebesgue space, this subspace and therefore also the whole space contains an equilateral set of at least k + 1 {\displaystyle k+1} points. Therefore, the same superlogarithmic dependence on d {\displaystyle d} holds for the lower bound on the equilateral dimension of d {\displaystyle d} -dimensional space.[8]

Riemannian manifolds

For any d {\displaystyle d} -dimensional Riemannian manifold the equilateral dimension is at least d + 1 {\displaystyle d+1} .[5] For a d {\displaystyle d} -dimensional sphere, the equilateral dimension is d + 2 {\displaystyle d+2} , the same as for a Euclidean space of one higher dimension into which the sphere can be embedded.[5] At the same time as he posed Kusner's conjecture, Kusner asked whether there exist Riemannian metrics with bounded dimension as a manifold but arbitrarily high equilateral dimension.[5]

Notes

References