# Equilateral but not cospherical

In Euclidean spaces, any equilateral set of points (that is, a set of points all of whose distances are equal) must form an equilateral triangle, regular tetrahedron, or higher-dimensional regular simplex. The centroid of the points is at equal distance from all of them, though it is closer than the interpoint distance. The points must all lie on a common sphere centered at the centroid.

In an \( L_{\infty} \) space, too, any equilateral set has a center that is equidistant from all the points. The same is true more generally in any hyperconvex metric space: if the points of an equilateral set are all at distance \( d \) from each other, then by hyperconvexity there is another point at distance \( d/2 \) from them all. So, again, the equilateral points all lie on a common sphere, of radius \( d/2 \).

In an \( L_1 \) space, any three equilateral points are always equidistant from a fourth point. Every \( L_1 \) space forms a median algebra, in which the median of any three points is computed coordinatewise; if three points are at distance \( d \) from each other then they must all be at distance \( d/2 \) from their median.

What about larger numbers of points in \( L_1 \) spaces? Must every equilateral set of four or more points in an \( L_1 \) space be cospherical?

No!

Consider the following five points in four-dimensional \( L_1 \) space: \( (1,1,1,0) \), \( (1,-1,-1,0) \), \( (-1,1,-1,0) \), \( (-1,-1,1,0) \), \( (0,0,0,1) \). To be equidistant from the first four points, any sixth point must belong to the line \( (0,0,0,\ast) \). But, within that line, every point is closer to \( (0,0,0,1) \) than to the other four points by at least two units. So there can be no sixth point equidistant from the previous five, or equivalently the five given points are not cospherical. It doesn't even help to add more dimensions: the same five points remain non-cospherical in any higher dimensional \( L_1 \)space into which this four-dimensional space is embedded.

Another way of looking at this is in terms of extending point sets by adding one or more points. For Euclidean spaces, if \( S \) is a subset of \( T \) as an abstract metric space, and \( S \) and \( T \) both have realizations as Euclidean point sets, then every Euclidean realization of \( S \) can be extended to a Euclidean realization of \( T \). By the injectivity of tight spans in \( L_{\infty} \) spaces, it is also true that every \( L_{\infty} \) realization of \( S \) can be extended to an \( L_{\infty} \) realization of \( T \). But if \( S \) and \( T \) can both be realized as point sets in an \( L_1 \) space, it may not be true that every \( L_1 \) realization of \( S \) can be extended to an \( L_1 \) realization of \( T \). The five points above form an \( L_1 \) realization that cannot be extended to six equilateral points, even though incompatible \( L_1 \) realizations of the same six-point metric space are easy to find (just use the vertices of a regular octahedron in three-dimensional space).

So, for the purposes of extending realizations of finite metric spaces to larger superspaces, in the Euclidean and \( L_{\infty} \) cases it is only the metric that matters. But in \( L_1 \) spaces, there is some property of the realization that is not captured by its intrinsic metric: the same finite metric can be embedded in \( L_1 \) spaces in different incompatible ways. Perhaps this has something to do with the reason that Kusner's conjecture on the maximum size of an equilateral set seems to be so much harder in the \( L_1 \) case than in the Euclidean and \( L_{\infty} \) cases.

Edited 2010-08-18 to replace a more complicated five-dimensional example by a nicer and more symmetric four-dimensional example. The new example also shows that it is possible for a point in an equilateral set to not have any coordinate direction in which it is the unique extreme point.