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Metric thickenings, Borsuk–Ulam theorems, and orbitopes

Henry Adams Department of Mathematics, Colorado State University, Fort Collins, CO 80523, United States adams@math.colostate.edu Johnathan Bush Department of Mathematics, Colorado State University, Fort Collins, CO 80523, United States bush@math.colostate.edu  and  Florian Frick Department of Mathematical Sciences, Carnegie Mellon University, Pittsburgh, PA 15213, United States frick@cmu.edu
(Date: July 28, 2025)
Abstract.

Thickenings of a metric space capture local geometric properties of the space. Here we exhibit applications of lower bounding the topology of thickenings of the circle and more generally the sphere. We explain interconnections with the geometry of circle actions on Euclidean space, the structure of zeros of trigonometric polynomials, and theorems of Borsuk–Ulam type. We use the combinatorial and geometric structure of the convex hull of orbits of circle actions on Euclidean space to give geometric proofs of the homotopy type of metric thickenings of the circle.

Homotopical connectivity bounds of thickenings of the sphere allow us to prove that a weighted average of function values of odd maps Snn+2S^{n}\to\mathbb{R}^{n+2} on a small diameter set is zero. We prove an additional generalization of the Borsuk–Ulam theorem for odd maps S2n12kn+2n1S^{2n-1}\to\mathbb{R}^{2kn+2n-1}. We prove such results for odd maps from the circle to any Euclidean space with optimal quantitative bounds. This in turn implies that any raked homogeneous trigonometric polynomial has a zero on a subset of the circle of a specific diameter; these results are optimal.

Key words and phrases:
Metric thickening, Vietoris–Rips complexes, Borsuk–Ulam theorems, orbitopes, trigonometric polynomials, moment curves, optimal transport
The research of HA and FF was supported through the program “Research in Pairs” by the Mathematisches Forschungsinstitut Oberwolfach in 2019. The research of FF was supported in part by NSF grant DMS-1855591.

1. Introduction

A compact metric space XX admits a canonical isometric embedding into C(X)C(X)^{*}, the dual space of real-valued continuous functions on XX. If C(X)C(X)^{*} is equipped with a Wasserstein metric, then convex combinations of nearby points of XX in C(X)C(X)^{*} give a canonical thickening of the space XX that exhibits local connectivity properties of XX. In particular, if XX is a sufficiently dense sample of points in an ambient space YY, and YY satisfies additional conditions such as being a closed Riemannian manifold with certain curvature bounds, then these metric thickenings of XX recover the homotopy type of YY at small scale parameters [30, 3].

In the present manuscript we relate metric thickenings of the circle S1S^{1} (and more generally the nn-sphere SnS^{n}) to convexity properties of orbits of circle actions on Euclidean space, to Borsuk–Ulam type theorems, and to the structure of zeros of trigonometric polynomials with a prescribed spectrum. We will briefly introduce these notions here and state our main results.

Borsuk–Ulam theorems for higher-dimensional codomains.

The classical Borsuk–Ulam theorem states that any continuous map f:Snnf\colon S^{n}\to\mathbb{R}^{n} identifies some point with its antipode: f(x)=f(x)f(x)=f(-x) for some xSnx\in S^{n}. Equivalently, any odd map f:Snnf\colon S^{n}\to\mathbb{R}^{n}, namely a map satisfying f(x)=f(x)f(-x)=-f(x) for all xSnx\in S^{n}, must have a zero: f(x)=0f(x)=\vec{0} for some xSnx\in S^{n}. For lower-dimensional codomains, Gromov’s “waist of the sphere” theorem gives quantitative bounds for size of the preimage of some point: for any map f:Snkf\colon S^{n}\to\mathbb{R}^{k} with knk\leq n, there is a point yky\in\mathbb{R}^{k} such that ε\varepsilon-neighborhoods of f1(y)f^{-1}(y) have volume bounded below by the volume of the ε\varepsilon-neighborhood of an (nk)(n-k)-dimensional equator of SnS^{n} [22, 23, 35]. Here we investigate quantified generalizations of the Borsuk–Ulam theorem for maps to Euclidean space of dimension greater than nn; see [33] for a different generalization. While a generic odd map f:Snkf\colon S^{n}\to\mathbb{R}^{k} for k>nk>n does not have a zero, we will show that a convex combination of function values must achieve zero for points contained in a set of diameter strictly less than π\pi. Further, the diameter bounds obtained are sharp for maps of the form S1kS^{1}\to\mathbb{R}^{k}, Snn+1S^{n}\to\mathbb{R}^{n+1}, and Snn+2S^{n}\to\mathbb{R}^{n+2}.

In the following, rnr_{n} denotes the diameter of a regular (n+1)(n+1)-simplex inscribed into SnS^{n}, where SnS^{n} carries the standard spherical metric and where each great circle has length 2π2\pi.

Theorem 1.

If f:S12k+1f\colon S^{1}\to\mathbb{R}^{2k+1} is odd and continuous, then there is a subset XS1X\subseteq S^{1} of diameter at most 2πk2k+1\frac{2\pi k}{2k+1} such that conv(f(X))\mathrm{conv}(f(X)) contains the origin.

Refer to caption
Figure 1. The map f:S13f\colon S^{1}\to\mathbb{R}^{3} defined by f(t)=(cos(t),sin(t),cos(3t))f(t)=(\cos(t),\sin(t),\cos(3t)) is odd; it contains (two) subsets XS1X\subseteq S^{1} of three equally-spaced points of diameter 2π3\frac{2\pi}{3} with 0conv(f(X))\vec{0}\in\mathrm{conv}(f(X)).

This result generalizes, with the same diameter bound, to odd maps f:S2n12kn+2n1f\colon S^{2n-1}\to\mathbb{R}^{2kn+2n-1}. When k=0k=0, we recover the classic Borsuk–Ulam theorem.

Theorem 2.

If f:S2n12kn+2n1f\colon S^{2n-1}\to\mathbb{R}^{2kn+2n-1} is odd and continuous, then there is a subset XS2n1X\subseteq S^{2n-1} of diameter at most 2πk2k+1\frac{2\pi k}{2k+1} such that conv(f(X))\mathrm{conv}(f(X)) contains the origin.

We remark that the diameter bound obtained in Theorem 2 also applies to odd and continuous maps f:S2n2kn+2n1f\colon S^{2n}\to\mathbb{R}^{2kn+2n-1}; indeed, ff restricted to the equator S2n1S2nS^{2n-1}\hookrightarrow S^{2n} is odd and continuous with the same codomain.

Theorem 3.

If f:Snn+2f\colon S^{n}\to\mathbb{R}^{n+2} is odd and continuous, then there is a subset XSnX\subseteq S^{n} of diameter at most rnr_{n} such that conv(f(X))\mathrm{conv}(f(X)) contains the origin.

The diameter bounds in Theorems 1 and 3 are optimal. In the case of the circle, there exist odd maps f:S12k2k+1f\colon S^{1}\to\mathbb{R}^{2k}\subseteq\mathbb{R}^{2k+1} already into one dimension lower such that conv(f(X))\mathrm{conv}(f(X)) misses the origin for any set XX of diameter less than 2πk2k+1\frac{2\pi k}{2k+1}. Constructing such an example map ff also proves a result about the structure of zeros of raked trigonometric polynomials, which we explain next.

The structure of zeros of raked trigonometric polynomials.

A trigonometric polynomial is an expression of the form p(t)=c+k=1nakcos(kt)+bksin(kt)p(t)=c+\sum_{k=1}^{n}a_{k}\cos(kt)+b_{k}\sin(kt), inducing a map S1S^{1}\to\mathbb{R}. In the case that c=0,c=0, we call pp a homogeneous trigonometric polynomial. The set S{1,,n}S\subseteq\{1,\dots,n\} of integers kk with ak0a_{k}\neq 0 or bk0b_{k}\neq 0 is called the spectrum of pp, and the largest integer in SS is the degree of pp. The spectrum of pp constrains the set of roots of pp; for example, if pp is homogeneous of degree nn then it has a root on any closed circular arc of length 2πnn+1\frac{2\pi n}{n+1}; see [9, 20]. Kozma and Oravecz in [29] give upper bounds on the length of an arc where a trigonometric polynomial with spectrum bounded away from zero (that is, S[k,n]S\subseteq[k,n]) is non-zero. If the spectrum of pp consists only of odd integers, then pp is called a raked trigonometric polynomial. We show the following structural result about the roots of raked trigonometric polynomials:

Theorem 4.

Let XS1X\subseteq S^{1} be such that diam(X)<2πk2k+1\mathrm{diam}(X)<\frac{2\pi k}{2k+1}. Then there is a raked homogeneous trigonometric polynomial of degree 2k12k-1 that is positive on XX. Moreover, there is a set XS1X\subseteq S^{1} of diameter 2πk2k+1\frac{2\pi k}{2k+1} such that no raked homogeneous trigonometric polynomial of degree 2k12k-1 is positive on XX.

The proof of the first part of Theorem 4 can be used to imply that the quantitative bound on the diameter in Theorem 1 is tight, while the second part of this theorem is a corollary of Theorem 1.

The symmetric moment curve and the Barvinok–Novik orbitope.

The relation between Theorems 1 and 4 is explained by choosing f:S12k2k+1f\colon S^{1}\to\mathbb{R}^{2k}\subseteq\mathbb{R}^{2k+1} to be the symmetric moment curve

SM2k(t)=(cost,sint,cos3t,sin3t,,cos(2k1)t,sin(2k1)t),\mathrm{SM}_{2k}(t)=\bigl{(}\cos t,\sin t,\cos 3t,\sin 3t,\ldots,\cos(2k-1)t,\sin(2k-1)t\bigr{)},

which is an odd function. The convex hull of the curve SM2k\mathrm{SM}_{2k} is referred to as the Barvinok–Novik orbitope 2k\mathcal{B}_{2k}. Now, Theorem 1 implies that there is a set XS1X\subseteq S^{1} of diameter 2πk2k+1\frac{2\pi k}{2k+1} such that conv(SM2k(X))\mathrm{conv}(\mathrm{SM}_{2k}(X)) captures the origin. Thus, for any given z2k{0}z\in\mathbb{R}^{2k}\setminus\{0\}, the inner product z,SM2k(X)\langle z,\mathrm{SM}_{2k}(X)\rangle changes sign in XX since no hyperplane can separate SM(X)\mathrm{SM}(X) from the origin. For varying zz, these inner products range over all possible raked homogeneous trigonometric polynomials, giving the second part of Theorem 4.

The geometry of the Barvinok–Novik orbitope also shows that the bound 2πk2k+1\frac{2\pi k}{2k+1} in Theorem 1 is optimal:

Theorem 5.

Let XS1X\subseteq S^{1} be such that diam(X)<𝒞\mathrm{diam}(X)<\mathcal{C}. Then the convex hull conv(SM2k(X))\mathrm{conv}(\mathrm{SM}_{2k}(X)) does not contain the origin 02k\vec{0}\in\mathbb{R}^{2k} if 𝒞=2πk2k+1\mathcal{C}=\frac{2\pi k}{2k+1}, and this bound is sharp.

This also shows that, given XS1X\subseteq S^{1} with diam(X)<2πk2k+1\mathrm{diam}(X)<\frac{2\pi k}{2k+1}, there is a raked trigonometric polynomial of degree 2k12k-1 that is positive on XX.

Metric thickenings of the circle.

For any metric space XX, we get a continuum spectrum of metric spaces VRm(X;r)\mathrm{VR}^{m}(X;r), r>0r>0, of Vietoris–Rips metric thickenings that capture the local connectivity of XX [3]. Note that the superscript mm is used here to denote the metric thickening of the Vietoris–Rips complex, as opposed to the geometric realization, and is not a numerical parameter. A point in VRm(X;r)\mathrm{VR}^{m}(X;r) is a probability measure on XX with finite support of diameter at most rr. Recently, Vietoris–Rips thickenings have been used in topological data analysis and persistent homology [15, 16]; specifically, they allow for a growing filtration of topological spaces associated to a finite collection or sampling of data. Conjecturally, for XX the circle S1S^{1}, this spectrum ranges over all odd-dimensional spheres S1,S3,S5,,S^{1},S^{3},S^{5},\dots, until eventually becoming contractible.111These homotopy types are known for the Vietoris–Rips simplicial complexes VR(S1;r)\mathrm{VR}(S^{1};r) [2], but not yet for the more natural Vietoris–Rips thickenings VRm(S1;r)\mathrm{VR}^{m}(S^{1};r).

Conjecture 6.

For 2π(k1)2k1r<2πk2k+1\frac{2\pi(k-1)}{2k-1}\leq r<\frac{2\pi k}{2k+1}, the metric thickening VRm(S1;r)\mathrm{VR}^{m}(S^{1};r) is homotopy equivalent to the boundary of the Barvinok–Novik orbitope 2k\mathcal{B}_{2k}, i.e. to the odd-dimensional sphere S2k1S^{2k-1}.

As partial evidence towards this conjecture, we explain how Theorem 5 implies that, for scale parameter rr in this range, the (2k1)(2k-1)-dimensional homology, cohomology, and homotopy groups of VRm(S1;r)\mathrm{VR}^{m}(S^{1};r) are nontrivial.

In Section 5, we show that Conjecture 6 is true up to r=2π3r=\frac{2\pi}{3}, the side-length of an inscribed equilateral triangle, where VRm(S1;r)S3\mathrm{VR}^{m}(S^{1};r)\simeq S^{3}. More importantly, we provide a geometric picture of why this homotopy equivalence is plausible for r0r\geq 0, as follows. For 2π(k1)2k1r<2πk2k+1\frac{2\pi(k-1)}{2k-1}\leq r<\frac{2\pi k}{2k+1} we define a continuous map VRm(S1;r)2k{0}\mathrm{VR}^{m}(S^{1};r)\to\mathbb{R}^{2k}\setminus\{\vec{0}\} via the centrally symmetric moment curve (cost,sint,cos3t,sin3t,)(\cos t,\sin t,\cos 3t,\sin 3t,\ldots). We relate VRm(S1;r)\mathrm{VR}^{m}(S^{1};r) to the facial structure of the Barvinok–Novik orbitope by composing with the radial projection map 2k{0}2k\mathbb{R}^{2k}\setminus\{\vec{0}\}\to\partial\mathcal{B}_{2k}. Finally, for r=2π3r=\frac{2\pi}{3} we obtain the homotopy equivalence VRm(S1;r)4S3\mathrm{VR}^{m}(S^{1};r)\simeq\partial\mathcal{B}_{4}\cong S^{3} via a linear homotopy. This is the only step that we are currently unable to extend to large rr and kk; the missing ingredient is a “diameter non-increasing” property for higher-dimensional Barvinok–Novik orbitopes (Conjecture 22).

To our knowledge, this is the first approach to determine the homotopy type of a Vietoris–Rips thickening by mapping the underlying metric space into a higher-dimensional Euclidean space. This technique is analogous to the “kernel trick” of machine learning, in which data is mapped into a higher dimensional space to illuminate the underlying structure of the data.

As a step towards understanding the relationship between metric thickenings of the circle and the Barvinok–Novik orbitopes, we show that given arbitrary t1,,t2k1S1t_{1},\dots,t_{2k-1}\in S^{1}, there exists a raked homogeneous trigonometric polynomial ff of degree 2k12k-1 with a root at each tit_{i} and its antipode. Further, the polynomial ff alternates signs between these roots, has no other roots in S1S^{1}, and may be written down explicitly in terms of the parameters t1,,t2k1t_{1},\dots,t_{2k-1} (Theorem 26).

We remark that there is an analogous connection between the Čech thickenings of the circle and the Carathéodory orbitopes, i.e. the convex hull of the curve (cost,sint,cos2t,sin2t,,coskt,sinkt)(\cos t,\sin t,\cos 2t,\sin 2t,\ldots,\cos kt,\sin kt\bigr{)} [38].

A preliminary version of several results in this paper appeared in the second author’s master’s thesis [14].

2. Preliminaries and related work

In this section we review notation and related work on topology, Vietoris–Rips simplicial complexes, metric thickenings, convex geometry, moment curves, and orbitopes.

Topological and metric spaces

We say two continuous maps f,g:XYf,g\colon X\to Y are homotopic, written fgf\simeq g, if there exists a continuous map H:X×[0,1]YH\colon X\times[0,1]\to Y such that H(x,0)=f(x)H(x,0)=f(x) and H(x,1)=g(x)H(x,1)=g(x) for all xXx\in X [24]. Such a map HH is called a homotopy. We say XX and YY are homotopy equivalent, denoted XYX\simeq Y, if there exist continuous maps f:XYf\colon X\to Y and g:YXg\colon Y\to X such that gfidXg\circ f\simeq\mathrm{id}_{X} and fgidYf\circ g\simeq\mathrm{id}_{Y}. We furthermore write XYX\cong Y if spaces XX and YY are homeomorphic.

Given a set of points SXS\subseteq X in a metric space (X,d)(X,d), let the diameter of SS be diam(S)=sup{d(x,y)|x,yS}\mathrm{diam}(S)=\sup\{d(x,y)~|~x,y\in S\}; this value may be infinite.

Conventions regarding S1S^{1}

We equip S1S^{1} with the geodesic metric (of total circumference 2π2\pi), though our results also hold when S1S^{1} is instead equipped with the restriction of the Euclidean metric on 2\mathbb{R}^{2}. Unless otherwise stated, we will always take a representative tS1=/2πt\in S^{1}=\mathbb{R}/2\pi\mathbb{Z} as belonging to [0,2π)[0,2\pi). Let a,bS1a,b\in S^{1}, where aba\neq b, and where aa and bb are each identified with a point in [0,2π)[0,2\pi). Define the open arc (a,b)S1(a,b)_{S^{1}} as

(a,b)S1={{tS1a<t<b}if a<b{tS1a<t<b+2π}if a>b.(a,b)_{S^{1}}=\begin{cases}\{t\in S^{1}\mid a<t<b\}&\text{if }a<b\\ \{t\in S^{1}\mid a<t<b+2\pi\}&\text{if }a>b.\end{cases}

Define the closed arc [a,b]S1[a,b]_{S^{1}} similarly.

Vietoris–Rips simplicial complexes

We identify an abstract simplicial complex with its geometric realization, which is a topological space.

Definition 7.

Let XX be a metric space and fix r0r\geq 0. The Vietoris–Rips simplicial complex of XX with scale parameter rr, denoted VR(X;r)\mathrm{VR}(X;r), has XX as its vertex set and a finite subset σX\sigma\subseteq X as a simplex whenever diam(σ)r\mathrm{diam}(\sigma)\leq r.

A point in VR(X;r)\mathrm{VR}(X;r) can be written in barycentric coordinates as i=0kλixi\sum_{i=0}^{k}\lambda_{i}x_{i}, with diam({x0,,xk})r\mathrm{diam}(\{x_{0},\ldots,x_{k}\})\leq r. We emphasize that in this paper we are using the \leq convention instead of the << convention.

While the theorems of [25, 30] describe conditions under which the homotopy type of a manifold is recoverable from a Vietoris–Rips complex for sufficiently small r0r\geq 0, much less is known about the topological behavior of these constructions for large values of rr, even though large values of rr commonly arise in applications of persistent homology [16]. However, more is known in the specific case when the underlying manifold is the circle. The following theorem from [2] is based on [1, 4].

Theorem 8.

Let 0r<π0\leq r<\pi. There are homotopy equivalences

VR(S1;r)\displaystyle\mathrm{VR}(S^{1};r) {S2k1if 2π(k1)2k1<r<2πk2k+1𝔠S2kif r=2πk2k+1,\displaystyle\simeq\begin{cases}S^{2k-1}&\text{if }\,\frac{2\pi(k-1)}{2k-1}<r<\frac{2\pi k}{2k+1}\\ \bigvee^{\mathfrak{c}}S^{2k}&\text{if }\,r=\frac{2\pi k}{2k+1},\end{cases}

where k=0,1,2,k=0,1,2,\ldots, and where 𝔠\mathfrak{c} denotes the cardinality of the continuum.

Related papers include [19] which studies the 1-dimensional persistence of Čech and Vietoris–Rips complexes of metric graphs, [46] which extends this to geodesic spaces, [47] which studies approximations of Vietoris–Rips complexes by finite samples even at higher scale parameters, and [49] which applies Bestvina–Brady discrete Morse theory to Vietoris–Rips complexes.

Metric thickenings and optimal transport

When a metric space XX is not finite, it is often impossible222A simplicial complex (for example VR(X;r)\mathrm{VR}(X;r)) is metrizable if and only if it is locally finite [37, Proposition 4.2.16(2)]. to equip VR(X;r)\mathrm{VR}(X;r) with a metric without changing the homeomorphism type. In such instances the simplicial complex VR(X;r)\mathrm{VR}(X;r) destroys the metric information about the underlying space XX. This motivates the consideration of the Vietoris–Rips metric thickening, VRm(X;r)\mathrm{VR}^{m}(X;r), which preserves metric information.

Let δx\delta_{x} denote the Dirac delta mass at a point xXx\in X.

Definition 9 ([3]).

Let XX be a metric space and let r0r\geq 0. The Vietoris–Rips thickening is the set

VRm(X;r)={i=0kλiδxi|k,xiX,diam({x0,,xk})r,λi0,λi=1},\mathrm{VR}^{m}(X;r)=\left\{\sum_{i=0}^{k}\lambda_{i}\delta_{x_{i}}~\bigg{|}~k\in\mathbb{N},\ x_{i}\in X,\ \mathrm{diam}(\{x_{0},\dots,x_{k}\})\leq r,\ \lambda_{i}\geq 0,\ \sum\lambda_{i}=1\right\},

equipped with the 11-Wasserstein metric.

This metric is also called the Kantorovich, optimal transport, or earth mover’s metric [42, 43, 44]; it provides a notion of distance between probability measures defined on a metric space. Although it exists much more generally [17, 27, 28], the 11-Wasserstein metric on VRm(X;r)\mathrm{VR}^{m}(X;r) can be defined as follows. Given μ,μVRm(X;r)\mu,\mu^{\prime}\in\mathrm{VR}^{m}(X;r) with μ=i=0kλiδxi\mu=\sum_{i=0}^{k}\lambda_{i}\delta_{x_{i}} and μ=j=0kλjδxj\mu^{\prime}=\sum_{j=0}^{k^{\prime}}\lambda_{j}^{\prime}\delta_{x_{j}^{\prime}}, define a matching pp between μ\mu and μ\mu^{\prime} to be any collection of non-negative real numbers {pi,j}i,j\{p_{i,j}\}_{i,j} such that j=0kpi,j=λi\sum_{j=0}^{k^{\prime}}p_{i,j}=\lambda_{i} and i=0kpi,j=λj\sum_{i=0}^{k}p_{i,j}=\lambda_{j}^{\prime}. Define the cost of the matching pp to be i,jpi,jd(xi,xj)\sum_{i,j}p_{i,j}d(x_{i},x_{j}^{\prime}). The 11-Wasserstein distance between μ,μVRm(X;r)\mu,\mu^{\prime}\in\mathrm{VR}^{m}(X;r) is then the infimum, varying over all matchings pp between μ\mu and μ\mu^{\prime}, of the cost of pp.

Note that VRm(X;0)\mathrm{VR}^{m}(X;0) is isometric to XX. Contrary to the situation for an arbitrary Vietoris–Rips complex, the embedding XVRm(X;r)X\to\mathrm{VR}^{m}(X;r) into the Vietoris–Rips metric thickening given by xδxx\mapsto\delta_{x} is continuous. In fact, more is true: VRm(X;r)\mathrm{VR}^{m}(X;r) is an rr-thickening of XX [21, 3]. For this reason, we identify xXx\in X with the measure δxVRm(X;r)\delta_{x}\in\mathrm{VR}^{m}(X;r) in the image of this embedding.

If MM is a complete Riemannian manifold with curvature bounded from above and below, then VRm(M;r)\mathrm{VR}^{m}(M;r) is homotopy equivalent to MM for rr sufficiently small [3, 6]. This property provides an analogue of Hausmann’s theorem [25] for metric thickenings.

Given a measure μ=i=0kλiδxi\mu=\sum_{i=0}^{k}\lambda_{i}\delta_{x_{i}} with λi>0\lambda_{i}>0, we denote the support of μ\mu by supp(μ)={x0,,xk}\mathrm{supp}(\mu)=\{x_{0},\ldots,x_{k}\}.

Convex geometry

Convex geometry is the study of convex sets, especially polytopes and their facial structures [50]. Given an arbitrary subset YnY\subseteq\mathbb{R}^{n}, we let

conv(Y)={i=1kλivi|k,viY,λi0,i=1kλi=1}\mathrm{conv}(Y)=\left\{\sum_{i=1}^{k}\lambda_{i}v_{i}~\bigg{|}~k\in\mathbb{N},\ v_{i}\in Y,\ \lambda_{i}\geq 0,\ \sum_{i=1}^{k}\lambda_{i}=1\right\}

denote the convex hull of YY. For example, Figure 1 shows the convex hull of the image of the map f:S13f\colon S^{1}\to\mathbb{R}^{3} defined by f(t)=(cos(t),sin(t),cos(3t))f(t)=(\cos(t),\sin(t),\cos(3t)). Similarly, the conical hull of YY is

cone(Y)={i=1kλivi|k,viY,λi0}.\mathrm{cone}(Y)=\left\{\sum_{i=1}^{k}\lambda_{i}v_{i}~\bigg{|}~k\in\mathbb{N},\ v_{i}\in Y,\ \lambda_{i}\geq 0\right\}.

Let YnY\subseteq\mathbb{R}^{n} be convex. Define a face of YY to be any convex set FYF\subseteq Y such that, given xFx\in F, if x=λy+(1λ)zx=\lambda y+(1-\lambda)z for some 0<λ<10<\lambda<1 and y,zYy,z\in Y, then y,zFy,z\in F.

The centrally symmetric trigonometric moment curve

The centrally symmetric moment curve is analogous to the trigonometric moment curve, with the additional property that it is symmetric under reflecting through the origin.

Definition 10.

For kk\in\mathbb{N}, the centrally symmetric moment curve SM2k:S12k\mathrm{SM}_{2k}\colon S^{1}\to\mathbb{R}^{2k} is defined by

SM2k(t)=(cost,sint,cos3t,sin3t,,cos(2k1)t,sin(2k1)t).\mathrm{SM}_{2k}(t)=\bigl{(}\cos t,\sin t,\cos 3t,\sin 3t,\ldots,\cos(2k-1)t,\sin(2k-1)t\bigr{)}.

Here we identify the domain S1S^{1} with /2π\mathbb{R}/2\pi\mathbb{Z}. Since SM2k(t+π)=SM2k(t)\mathrm{SM}_{2k}(t+\pi)=-\mathrm{SM}_{2k}(t), we say that SM2k\mathrm{SM}_{2k} is centrally symmetric about the origin. Interestingly, this curve is closely related to the multidimensional scaling (MDS) embedding S12kS^{1}\hookrightarrow\mathbb{R}^{2k} of the geodesic circle [5, 12, 26, 48]; multidimensional scaling is a way to map a metric space into Euclidean space in a way that distorts the metric (in some sense) as little as possible.

Barvinok–Novik orbitopes

The Barvinok–Novik orbitope is defined by 2k=conv(SM2k(S1))2k\mathcal{B}_{2k}=\mathrm{conv}(\mathrm{SM}_{2k}(S^{1}))\subseteq\mathbb{R}^{2k} [11]. This convex body is not the convex hull of a finite set of points; it is an orbitope instead of a polytope [38].

The faces of 2k\mathcal{B}_{2k} are known for k=2k=2; a subset of these faces are visible in Figure 1 (which is in 3\mathbb{R}^{3} instead of 4\mathbb{R}^{4}).

Theorem 11 ([11, 40]).

The proper faces of 4\mathcal{B}_{4} are

  • the 0-dimensional faces (vertices) SM4(t)\mathrm{SM}_{4}(t) for tS1t\in S^{1},

  • the 1-dimensional faces (edges) conv(SM4({t1,t2}))\mathrm{conv}(\mathrm{SM}_{4}(\{t_{1},t_{2}\})) where t1t2t_{1}\neq t_{2} are the edges of an arc of S1S^{1} of length at most 2π3\frac{2\pi}{3}, and

  • the 2-dimensional faces (triangles) conv(SM4({t,t+2π3,t+4π3}))\mathrm{conv}(\mathrm{SM}_{4}(\{t,t+\frac{2\pi}{3},t+\frac{4\pi}{3}\})) for tS1t\in S^{1}.

Though the facial structure of the Barvinok–Novik orbitopes 2k\mathcal{B}_{2k} is not known for k>2k>2, certain neighborliness results have been established [10]. Sinn has shown that the orbitopes are simplicial [39]. Additionally, Vinzant proved that the edges of 2k\partial\mathcal{B}_{2k} consist of all line segments conv(SM2k({t0,t1}))\mathrm{conv}\left(\mathrm{SM}_{2k}(\{t_{0},t_{1}\})\right) with |t0t1|2π(k1)2k1|t_{0}-t_{1}|\leq\frac{2\pi(k-1)}{2k-1} [45]. In other words, the edges of 2k\mathcal{B}_{2k} are the same as the edges of VR(S1;2π(k1)2k1)\mathrm{VR}(S^{1};\frac{2\pi(k-1)}{2k-1}). The following is an immediate corollary of the work of Sinn and Vinzant.

Corollary 12 ([39, 45]).

Every face of the Barvinok–Novik orbitope 2k\mathcal{B}_{2k} is a simplex whose diameter in S1S^{1} (not in 2k\mathbb{R}^{2k}) is at most 2π(k1)2k1\frac{2\pi(k-1)}{2k-1}.

3. A generalization of the Borsuk–Ulam theorem

The Borsuk–Ulam states that if f:Snnf\colon S^{n}\to\mathbb{R}^{n} is continuous, then there exists a point xSnx\in S^{n} with f(x)=f(x)f(x)=f(-x) [34]. For maps into lower-dimensional Euclidean space, there is a generalization due to Gromov called the “waist of the sphere” theorem [22, 23, 35]. The theorem says that if f:Snkf\colon S^{n}\to\mathbb{R}^{k} is continuous with knk\leq n, then there is some point yky\in\mathbb{R}^{k} such that the ε\varepsilon-neighborhoods of f1(y)f^{-1}(y) have volume at least as large as the volume of the ε\varepsilon-neighborhood of an (nk)(n-k)-dimensional equator of SnS^{n}. There are also versions in which the size of a preimage is measured by its diameter; this is called the Urysohn width [7, 32, 41]. In this section, we ask: what can be said for maps f:Snkf\colon S^{n}\to\mathbb{R}^{k} with knk\geq n?

We say a map f:Snkf\colon S^{n}\to\mathbb{R}^{k} is odd or centrally-symmetric if f(x)=f(x)f(-x)=-f(x) for all xSnx\in S^{n}. An equivalent formulation of the Borsuk–Ulam states that if f:Snnf\colon S^{n}\to\mathbb{R}^{n} is continuous and odd, then there exists a point xSnx\in S^{n} with f(x)=0f(x)=\vec{0} [34, Theorem 2.1.1].

More generally, given topological spaces XX and YY equipped with /2\mathbb{Z}/2\mathbb{Z}-actions μ\mu and ν\nu respectively, we say a map f:XYf\colon X\to Y is odd or /2\mathbb{Z}/2\mathbb{Z}-equivariant if fμ=νff\circ\mu=\nu\circ f. Additionally, we always equip n\mathbb{R}^{n} with the standard antipodal /2\mathbb{Z}/2\mathbb{Z}-action specified by xxx\mapsto-x. For our purposes, we consider another useful reformulation of the Borsuk–Ulam theorem as follows. Given an (n1)(n-1)-connected space XX[34, Proposition 5.3.2(iv)] states that there is no /2\mathbb{Z}/2\mathbb{Z}-equivariant map from XX into Sn1S^{n-1}. Hence, any odd map f:Xnf\colon X\to\mathbb{R}^{n} must hit the origin, because otherwise we would obtain an odd map f|f|:XSn1\frac{f}{|f|}\colon X\to S^{n-1}.

In Theorems 12, and 3 we prove the generalizations of the Borsuk–Ulam thoerem for maps SnkS^{n}\to\mathbb{R}^{k} with knk\geq n. The first result is for n=1n=1.

Theorem 1.

If f:S12k+1f\colon S^{1}\to\mathbb{R}^{2k+1} is odd and continuous, then there is a subset XS1X\subseteq S^{1} of diameter at most 2πk2k+1\frac{2\pi k}{2k+1} such that conv(f(X))\mathrm{conv}(f(X)) contains the origin.

Equivalently, if f:S12k+1f\colon S^{1}\to\mathbb{R}^{2k+1} is continuous, then there exists a subset {x1,,xm}S1\{x_{1},\ldots,x_{m}\}\subseteq S^{1} of diameter at most 2πk2k+1\frac{2\pi k}{2k+1} such that i=1mλif(xi)=i=1mλif(xi)\sum_{i=1}^{m}\lambda_{i}f(x_{i})=\sum_{i=1}^{m}\lambda_{i}f(-x_{i}), for some choice of convex coefficients λi\lambda_{i}.

For example, if f=SM2k:S12k2k+1f=\mathrm{SM}_{2k}\colon S^{1}\to\mathbb{R}^{2k}\subseteq\mathbb{R}^{2k+1}, then this set XX is easy to find: we can let XX be 2k+12k+1 evenly-spaced points on the circle. Theorem 5 shows that the above diameter bound is sharp, both for maps S12k+1S^{1}\to\mathbb{R}^{2k+1} and for maps S12kS^{1}\to\mathbb{R}^{2k}. Indeed, SM2k:S12k2k+1\mathrm{SM}_{2k}\colon S^{1}\to\mathbb{R}^{2k}\subseteq\mathbb{R}^{2k+1} is an odd map in which the convex hull of the image of every set of diameter strictly less than 2πk2k+1\frac{2\pi k}{2k+1} misses the origin.

Proof.

Fix 0<ε<2π(2k+1)(2k+3)0<\varepsilon<\frac{2\pi}{(2k+1)(2k+3)}; this ensures 2πk2k+1+ε<2π(k+1)2k+3\frac{2\pi k}{2k+1}+\varepsilon<\frac{2\pi(k+1)}{2k+3}. The induced map f:VR(S1;2πk2k+1+ε)2k+1f\colon\mathrm{VR}(S^{1};\frac{2\pi k}{2k+1}+\varepsilon)\to\mathbb{R}^{2k+1} given by f(iλixi)=iλif(xi)f(\sum_{i}\lambda_{i}x_{i})=\sum_{i}\lambda_{i}f(x_{i}) is odd with domain VR(S1;k2k+1+ε)S2k+1\mathrm{VR}(S^{1};\frac{k}{2k+1}+\varepsilon)\simeq S^{2k+1} by Theorem 8. By the Borsuk–Ulam theorem, this map has a zero, giving a subset XX of diameter at most 2πkk+1+ε\frac{2\pi k}{k+1}+\varepsilon with conv(f(X))\mathrm{conv}(f(X)) containing the origin. Furthermore, by Carathéodory’s theorem, we can take the size of XX to be at most 2k+22k+2. It remains to reduce the diameter to 2πkk+1\frac{2\pi k}{k+1} by a compactness argument.

For each integer n1n\geq 1, we obtain a subset XnS1X_{n}\subseteq S^{1} of diameter bounded above by 2πk2k+1+εn\frac{2\pi k}{2k+1}+\frac{\varepsilon}{n}, of size |Xn|2k+2|X_{n}|\leq 2k+2, and with 0conv(f(Xn))\vec{0}\in\mathrm{conv}(f(X_{n})). If |Xn|<2k+2|X_{n}|<2k+2, then duplicate an arbitrary point in XnX_{n} to obtain a multi-set of size exactly 2k+22k+2. Arbitrarily order these points so that XnX_{n} can be thought of as a point in the torus (S1)2k+2(S^{1})^{2k+2}. By compactness of the torus, the sequence {Xn}\{X_{n}\} has a subsequence converging to a limit configuration X(S1)2k+2X\in(S^{1})^{2k+2} of diameter at most 2πk2k+1\frac{2\pi k}{2k+1} and with 0conv(f(X))\vec{0}\in\mathrm{conv}(f(X)). Removing duplicate points (and ignoring the ordering) gives us the desired subset XS1X\subseteq S^{1}. ∎

We remark that Theorem 1 may also be proven using the Barvinok–Novik orbitopes and Corollary 12, though this perspective does not generalize as nicely for the proofs of Theorems 2 and 3. In particular, an odd and continuous map f:S12k+1f\colon S^{1}\to\mathbb{R}^{2k+1} defines a continuous map F:VRm(S1;2πk2k+1)2k+1F\colon\mathrm{VR}^{m}(S^{1};\frac{2\pi k}{2k+1})\to\mathbb{R}^{2k+1} obtained by extending linearly to convex combinations of points in S1S^{1}. Then, FF induces an odd and continuous map F~:2k+22k+1\tilde{F}\colon\partial\mathcal{B}_{2k+2}\to\mathbb{R}^{2k+1} factoring through the well-defined inclusion ι:2k+2VRm(S1;2πk2k+1)\iota\colon\partial\mathcal{B}_{2k+2}\hookrightarrow{}\mathrm{VR}^{m}(S^{1};\frac{2\pi k}{2k+1}) (this inclusion exists by Corollary 12). Since 2k+2\partial\mathcal{B}_{2k+2} is homeomorphic to a (2k+1)(2k+1)-sphere and hence 2k2k-connected, we may apply [34, Proposition 5.3.2(iv)] to obtain the result. This approach depends crucially on known properties of the boundary of the Barvinok–Novik orbitope 2k+2\mathcal{B}_{2k+2} and does not immediately generalize to maps from higher-dimensional spheres.

Corollary 13.

Fix a list of odd continuous functions fi(t):S1f_{i}(t)\colon S^{1}\to\mathbb{R} for 1i2k+11\leq i\leq 2k+1. Let PP be the set of functions of the form p:S1p\colon S^{1}\to\mathbb{R} defined by p(t)=j=12k+1zjfj(t)p(t)=\sum_{j=1}^{2k+1}z_{j}f_{j}(t) with zjz_{j}\in\mathbb{R}. Then there is a subset XS1X\subseteq S^{1} of diameter at most 2πk2k+1\frac{2\pi k}{2k+1} such that no function in PP is positive on XX.

Proof.

Consider the odd map f:S12k+1f\colon S^{1}\to\mathbb{R}^{2k+1} given by f(t)=(f1(t),,f2k+1(t))f(t)=(f_{1}(t),\ldots,f_{2k+1}(t)). Note that each function pPp\in P is specified by a vector z2k+1z\in\mathbb{R}^{2k+1}, in the sense that p(t)=zf(t)p(t)=z^{\intercal}f(t) for all tS1t\in S^{1}. By Theorem 1, there exists a subset XS1X\subseteq S^{1} of diameter at most 2πk2k+1\frac{2\pi k}{2k+1} such that conv(f(X))\mathrm{conv}(f(X)) contains the origin. Hence, if we write X={x1,,xm}X=\{x_{1},\ldots,x_{m}\} with i=1mλif(xi)=0\sum_{i=1}^{m}\lambda_{i}f(x_{i})=\vec{0} for λi0\lambda_{i}\geq 0, then i=1mλip(xi)=i=1mλizf(xi)=zi=1mλif(xi)=z0=0\sum_{i=1}^{m}\lambda_{i}p(x_{i})=\sum_{i=1}^{m}\lambda_{i}z^{\intercal}f(x_{i})=z^{\intercal}\sum_{i=1}^{m}\lambda_{i}f(x_{i})=z^{\intercal}\vec{0}=0. In particular, p(xi)p(x_{i}) must be non-positive for at least some ii. ∎

The next corollary follows immediately from Corollary 13, and proves the second part of Theorem 4.

Corollary 14.

Fix a list of odd degrees did_{i} for 1i2k+11\leq i\leq 2k+1, and fix a list of trigonometric functions fi(t)=sin(t)f_{i}(t)=\sin(t) or fi(t)=cos(t)f_{i}(t)=\cos(t). Let PP be the set of all polynomials of the form p(t)=j=12k+1zjfj(djt)p(t)=\sum_{j=1}^{2k+1}z_{j}f_{j}(d_{j}t) with zjz_{j}\in\mathbb{R}. Then there is a subset XS1X\subseteq S^{1} of diameter at most 2πk2k+1\frac{2\pi k}{2k+1} such that no polynomial in PP is positive on XX.

For example, the above corollary applies if PP is the set of all raked homogeneous trigonometric polynomials of degree at most 2k12k-1, namely

p(t)=j=1kajcos((2j1)t)+j=1kbjsin((2j1)t),p(t)=\sum_{j=1}^{k}a_{j}\cos\bigl{(}(2j-1)t\bigr{)}+\sum_{j=1}^{k}b_{j}\sin\bigl{(}(2j-1)t\bigr{)},

after noting that we are considering the special case in which one of the constants zjz_{j} defining p(t)=j=12k+1zjfj(djt)p(t)=\sum_{j=1}^{2k+1}z_{j}f_{j}(d_{j}t) is zero.

We also show the sharpness of the above result: the number of summands defining the trigonometric polynomial cannot be increased, and the upper bound on the diameter of XX can not be decreased, giving the first part of Theorem 4.

Corollary 15.

Given a subset XS1X\subseteq S^{1} of diameter less than 2πk2k+1\frac{2\pi k}{2k+1}, there exists a raked homogeneous trigonometric polynomial of degree 2k12k-1 that is positive on all of the points in XX.

Proof.

This result is a corollary of Theorem 5 in Section 4, which says that the convex hull conv(SM2k(X))\mathrm{conv}(\mathrm{SM}_{2k}(X)) does not contain the origin. Hence, there is a separating hyperplane HzH_{z} with orthogonal vector z2kz\in\mathbb{R}^{2k} given by Hz={x2k|zx>0}H_{z}=\{x\in\mathbb{R}^{2k}~|~z^{\intercal}x>0\} such that SM2k(X)Hz\mathrm{SM}_{2k}(X)\subseteq H_{z}. Therefore, the raked homogeneous trigonometric polynomial of degree 2k12k-1 given by pz(x)=zSM2k(x)p_{z}(x)=z^{\intercal}\mathrm{SM}_{2k}(x) is positive on all of the points in XX. ∎

We give two versions of the Borsuk–Ulam theorem for maps SnkS^{n}\to\mathbb{R}^{k} with knk\geq n, now also with n2n\geq 2. Our results would be strengthened if we better understood the homotopy types of Vietoris–Rips thickenings of nn-spheres for n2n\geq 2 at all scale parameters.

The first version generalizes Theorem 1 (take n=1n=1) to maps from odd-dimensional spheres into Euclidean spaces of higher dimensions.

Theorem 2.

If f:S2n12kn+2n1f\colon S^{2n-1}\to\mathbb{R}^{2kn+2n-1} is odd and continuous, then there is a subset XS2n1X\subseteq S^{2n-1} of diameter at most 2πk2k+1\frac{2\pi k}{2k+1} such that conv(f(X))\mathrm{conv}(f(X)) contains the origin.

Proof.

The case of k=0k=0 follows from the standard Borsuk–Ulam theorem.

For k1k\geq 1, we will think of S2n1S^{2n-1} as a join of nn circles (S1)n(S^{1})^{*n}. Explicitly, if S2n1S^{2n-1} is viewed as the unit sphere in 2n\mathbb{R}^{2n}, then the subset of S2n1S^{2n-1} with all coordinates zero, with the (possible) exception of coordinates 2i12i-1 and 2i2i, is a circle. The distance between any two points in distinct such circles is π2\frac{\pi}{2} in the geodesic metric. Let 2πk2k+1<r<2π(k+1)2k+3\frac{2\pi k}{2k+1}<r<\frac{2\pi(k+1)}{2k+3}. Since k1k\geq 1 implies r>π2r>\frac{\pi}{2}, this will allow us to construct a /2\mathbb{Z}/2\mathbb{Z}-equivariant embedding of (VR(S1;r))n(\mathrm{VR}(S^{1};r))^{*n} into VR(S2n1;r)\mathrm{VR}(S^{2n-1};r). In barycentric coordinates, a point in VR(S1;r)\mathrm{VR}(S^{1};r) can be written as xXλxx\sum_{x\in X}\lambda_{x}x, where the vertex set XX of the simplex containing this point has diameter at most rr and where the positive coefficients λx\lambda_{x} sum to one. Hence, a point in (VR(S1;r))n(\mathrm{VR}(S^{1};r))^{*n} consists of nn collections of such points xXiλxx\sum_{x\in X_{i}}\lambda_{x}x for 1in1\leq i\leq n, along with non-negative numbers κ1,,κn\kappa_{1},\ldots,\kappa_{n} that add up to one. We map the points in XiX_{i} to the ii-th copy of S1S^{1} in S2n1=(S1)nS^{2n-1}=(S^{1})^{*n}, and we multiply their weights by κi\kappa_{i}. This gives the barycentric coordinates of a well-defined point in VRm(S2n1;r)\mathrm{VR}^{m}(S^{2n-1};r); the diameter of the supporting simplex is at most rr since r>π2r>\frac{\pi}{2}. Furthermore, this map respects the antipodal /2\mathbb{Z}/2\mathbb{Z}-actions on (VR(S1;r))n(\mathrm{VR}(S^{1};r))^{*n} and VR(S2n1;r)\mathrm{VR}(S^{2n-1};r); these actions are free since antipodal points are at distance r<πr<\pi apart. By Theorem 8 we have

(VR(S1;r))n(S2k+1)n=S(2k+1)n+n1=S2kn+2n1.(\mathrm{VR}(S^{1};r))^{*n}\simeq(S^{2k+1})^{*n}=S^{(2k+1)n+n-1}=S^{2kn+2n-1}.

It follows from [34, Proposition 5.3.2(iv)] that any odd map from (VR(S1;r))n(\mathrm{VR}(S^{1};r))^{*n}, and hence also from VR(S2n1;r)\mathrm{VR}(S^{2n-1};r), into 2kn+2n1\mathbb{R}^{2kn+2n-1} hits the origin. This gives a subset XS2n1X\subseteq S^{2n-1} of diameter at most r=2πk2k+1+εr=\frac{2\pi k}{2k+1}+\varepsilon such that conv(f(X))\mathrm{conv}(f(X)) contains the origin. By a compactness argument as in the proof of Theorem 1, we can reduce this diameter to exactly 2πk2k+1\frac{2\pi k}{2k+1}. ∎

In the following theorem, rnr_{n} is the diameter of an inscribed regular (n+1)(n+1)-simplex in SnS^{n}.

Theorem 3.

If f:Snn+2f\colon S^{n}\to\mathbb{R}^{n+2} is odd and continuous, then there is a subset XSnX\subseteq S^{n} of diameter at most rnr_{n} such that conv(f(X))\mathrm{conv}(f(X)) contains the origin.

This diameter bound is sharp, both for maps Snn+2S^{n}\to\mathbb{R}^{n+2} and maps Snn+1S^{n}\to\mathbb{R}^{n+1}. Indeed, the standard inclusion f:Snn+1n+2f\colon S^{n}\hookrightarrow\mathbb{R}^{n+1}\subseteq\mathbb{R}^{n+2} is an odd map that satisfies 0conv(f(X))\vec{0}\notin\mathrm{conv}(f(X)) for all XSnX\subseteq S^{n} of diameter less than rnr_{n} [31, Proof of Lemma 3].

Proof.

The space VRm(Sn;rn)\mathrm{VR}^{m}(S^{n};r_{n}) has a free /2\mathbb{Z}/2\mathbb{Z}-action that maps the convex combination i=1kλiδxi\sum_{i=1}^{k}\lambda_{i}\delta_{x_{i}} of Dirac measures for points x1,,xkx_{1},\dots,x_{k} on SnS^{n} to i=1kλiδxi\sum_{i=1}^{k}\lambda_{i}\delta_{-x_{i}}, that is, to the measure that is supported on the antipodal point set with the same weights λi\lambda_{i}. This action is free since antipodal points on SnS^{n} are farther than rnr_{n} apart.

Let f:Snn+2f\colon S^{n}\to\mathbb{R}^{n+2} be odd and continuous. Because ff is bounded, [3, Lemma 5.2] implies that ff induces a continuous map F:VRm(Sn;rn)n+2F\colon\mathrm{VR}^{m}(S^{n};r_{n})\to\mathbb{R}^{n+2} defined by F(i=1kλiδxi)=i=1kλif(xi)F(\sum_{i=1}^{k}\lambda_{i}\delta_{x_{i}})=\sum_{i=1}^{k}\lambda_{i}f(x_{i}). Notice that FF commutes with the antipodal action on VRm(Sn;rn)\mathrm{VR}^{m}(S^{n};r_{n}) and SnS^{n}:

F(i=1kλiδxi)=i=1kλif(xi)=i=1kλif(xi)=F(i=1kλiδxi).F\left(\sum_{i=1}^{k}\lambda_{i}\delta_{-x_{i}}\right)=\sum_{i=1}^{k}\lambda_{i}f(-x_{i})=-\sum_{i=1}^{k}\lambda_{i}f(x_{i})=-F\left(\sum_{i=1}^{k}\lambda_{i}\delta_{x_{i}}\right).

Next, fix a regular (n+1)(n+1)-simplex Δ\Delta inscribed in SnS^{n}, and let An+2A_{n+2} denote the group of rotational symmetries of Δ\Delta, that is, the alternating group on n+2n+2 elements. In a noncanonical fashion, we may identify An+2A_{n+2} as a subgroup of SO(n+1)\mathrm{SO}(n+1) by associating to each gAn+2g\in A_{n+2} the matrix MgSO(n+1)M_{g}\in\mathrm{SO}(n+1) such that Mgv=gvM_{g}\cdot v=g\cdot v for each vertex vv of Δ\Delta. In this way, we obtain the orbit space SO(n+1)An+2\frac{\mathrm{SO}(n+1)}{A_{n+2}} of SO(n+1)\mathrm{SO}(n+1) under the action of An+2A_{n+2} by left multiplication. Theorem 5.4 of [3] states that the homotopy type of VRm(Sn;rn)\mathrm{VR}^{m}(S^{n};r_{n}) is Σn+1SO(n+1)An+2\Sigma^{n+1}\frac{\mathrm{SO}(n+1)}{A_{n+2}}, and because SO(n+1)An+2\frac{\mathrm{SO}(n+1)}{A_{n+2}} is connected, its (n+1)(n+1)-fold suspension Σn+1SO(n+1)An+2VRm(Sn;rn)\Sigma^{n+1}\frac{\mathrm{SO}(n+1)}{A_{n+2}}\simeq\mathrm{VR}^{m}(S^{n};r_{n}) is (n+1)(n+1)-connected. Thus, the map FF, as a /2\mathbb{Z}/2\mathbb{Z}-equivariant map from an (n+1)(n+1)-connected space to n+2\mathbb{R}^{n+2}, has a zero [34, Proposition 5.3.2(iv)]. That is, there are points x1,,xmSnx_{1},\dots,x_{m}\in S^{n} that are pairwise at distance at most rnr_{n} and such that i=1mλif(xi)=0\sum_{i=1}^{m}\lambda_{i}f(x_{i})=\vec{0} for some λ1,,λm0\lambda_{1},\dots,\lambda_{m}\geq 0 with i=1mλi=1\sum_{i=1}^{m}\lambda_{i}=1. ∎

4. Diameter bound for Carathéodory sets on the symmetric moment curve

Let YkY\subseteq\mathbb{R}^{k} be a set in Euclidean space. Carathéodory’s theorem states that if the convex hull of YY contains the origin, then there is a subset of YY of at most k+1k+1 points whose convex hull also contains the origin. We say that YYY^{\prime}\subseteq Y is a Carathéodory subset of YY if the convex hull of YY^{\prime} contains the origin. The following theorem gives a lower bound on the diameter of the preimage of any Carathéodory subset of the symmetric moment curve in 2k\mathbb{R}^{2k}. Here, the circle S1S^{1} is equipped with the geodesic metric of total circumference 2π2\pi.

Theorem 5.

Let XS1X\subseteq S^{1} be such that diam(X)<𝒞\mathrm{diam}(X)<\mathcal{C}. Then the convex hull conv(SM2k(X))\mathrm{conv}(\mathrm{SM}_{2k}(X)) does not contain the origin 02k\vec{0}\in\mathbb{R}^{2k} if 𝒞=2πk2k+1\mathcal{C}=\frac{2\pi k}{2k+1}, and this bound is sharp.

To prove Theorem 5, we may restrict attention to subsets of SM2k(S1)\mathrm{SM}_{2k}(S^{1}) of size at most 2k+12k+1 by Carathéodory’s theorem. Suppose X={t0,,t2k}S1X=\{t_{0},\dots,t_{2k}\}\subseteq S^{1} is such that the origin is contained in the convex hull of {SM2k(t0),,SM2k(t2k)}\{\mathrm{SM}_{{2k}}(t_{0}),\dots,\mathrm{SM}_{2k}(t_{2k})\}. Then, there exist scalars λi0\lambda_{i}\geq 0 such that 0=i=02kλiSM2k(ti)\vec{0}=\sum_{i=0}^{2k}\lambda_{i}\mathrm{SM}_{2k}(t_{i}) and i=02kλi=1\sum_{i=0}^{2k}\lambda_{i}=1. In this way, we obtain a system of 2k2k equations

i=02kλicos(nti)=0andi=02kλisin(nti)=0forn=1,3,,2k1.\sum_{i=0}^{2k}\lambda_{i}\cos(nt_{i})=0\quad\text{and}\quad\sum_{i=0}^{2k}\lambda_{i}\sin(nt_{i})=0\quad\text{for}\quad n=1,3,\dots,2k-1.

We therefore let M2kM_{2k} be the 2k×(2k+1)2k\times(2k+1) matrix

M2k=(cos(t0)cos(t1)cos(t2k)sin(t0)sin(t1)sin(t2k)cos(3t0)cos(3t1)cos(3t2k)sin(3t0)sin(3t1)sin(3t2k)cos((2k1)t0)cos((2k1)t1)cos((2k1)t2k)sin((2k1)t0)sin((2k1)t1)sin((2k1)t2k)),M_{2k}=\begin{pmatrix}\cos(t_{0})&\cos(t_{1})&\ldots&\cos(t_{2k})\\ \sin(t_{0})&\sin(t_{1})&\ldots&\sin(t_{2k})\\ \cos(3t_{0})&\cos(3t_{1})&\ldots&\cos(3t_{2k})\\ \sin(3t_{0})&\sin(3t_{1})&\ldots&\sin(3t_{2k})\\ \vdots&\vdots&\ddots&\vdots\\ \cos((2k-1)t_{0})&\cos((2k-1)t_{1})&\ldots&\cos((2k-1)t_{2k})\\ \sin((2k-1)t_{0})&\sin((2k-1)t_{1})&\ldots&\sin((2k-1)t_{2k})\\ \end{pmatrix},

and consider the vector equation M2kλ=0M_{2k}\vec{\lambda}=\vec{0}. To prove Theorem 5, we build towards describing the nullspace of M2kM_{2k}, which we complete in Lemma 18.

Lemma 16.

Let AA denote the 2k×2k2k\times 2k matrix whose columns are SM2k(t1),SM2k(t2),,SM2k(t2k)\mathrm{SM}_{2k}(t_{1}),\mathrm{SM}_{2k}(t_{2}),\ldots,\mathrm{SM}_{2k}(t_{2k}). Then det(A)=κ1j<l2ksin(tltj)\det(A)=\kappa\prod_{1\leq j<l\leq 2k}\sin(t_{l}-t_{j}) for some nonzero constant κ\kappa depending only on kk.

We would like to thank Harrison Chapman for the insights behind the proof of Lemma 16. The main idea of the proof is to perform elementary row and column operations to AA to obtain a Vandermonde matrix. In addition to the general case, the simpler case k=2k=2 of this proof is written out in more detail in [14]. The determinant of a related matrix is given in [18].

A Vandermonde matrix is an n×nn\times n matrix of the form

V=(1a1a12a1n11a2a22a2n11anan2ann1).V=\begin{pmatrix}1&a_{1}&a_{1}^{2}&\cdots&a_{1}^{n-1}\\ 1&a_{2}&a_{2}^{2}&\cdots&a_{2}^{n-1}\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 1&a_{n}&a_{n}^{2}&\cdots&a_{n}^{n-1}\\ \end{pmatrix}.

Its determinant is det(V)=1i<jn(ajai)\det(V)=\prod_{1\leq i<j\leq n}(a_{j}-a_{i}); see for example [36, Section 2.8.1].

Proof of Lemma 16.

We will perform elementary row and column operations to AA to obtain a Vandermonde matrix. Given f:f\colon\mathbb{R}\to\mathbb{C}, define the function f:2k2kf\colon\mathbb{R}^{2k}\to\mathbb{C}^{2k} via f(t¯)=(f(t1),f(t2),,f(t2k))Tf(\underline{t})=(f(t_{1}),f(t_{2}),\ldots,f(t_{2k}))^{\text{T}} for t¯=(t1,t2,,t2k)T\underline{t}=(t_{1},t_{2},\ldots,t_{2k})^{\text{T}}. Since

A=(cos(t1)cos(t2)cos(t2k)sin(t1)sin(t2)sin(t2k)cos(3t1)cos(3t2)cos(3t2k)sin(3t1)sin(3t2)sin(3t2k)cos((2k1)t1)cos((2k1)t2)cos((2k1)t2k)sin((2k1)t1)sin((2k1)t2)sin((2k1)t2k)),A=\begin{pmatrix}\cos(t_{1})&\cos(t_{2})&\ldots&\cos(t_{2k})\\ \sin(t_{1})&\sin(t_{2})&\ldots&\sin(t_{2k})\\ \cos(3t_{1})&\cos(3t_{2})&\ldots&\cos(3t_{2k})\\ \sin(3t_{1})&\sin(3t_{2})&\ldots&\sin(3t_{2k})\\ \vdots&\vdots&\ddots&\vdots\\ \cos((2k-1)t_{1})&\cos((2k-1)t_{2})&\ldots&\cos((2k-1)t_{2k})\\ \sin((2k-1)t_{1})&\sin((2k-1)t_{2})&\ldots&\sin((2k-1)t_{2k})\\ \end{pmatrix},

we have

det(A)\displaystyle\det(A) =det(AT)=det(cos(t¯)sin(t¯)cos(3t¯)sin(3t¯)cos((2k1)t¯)sin((2k1)t¯))\displaystyle=\det\left(A^{\text{T}}\right)=\det\begin{pmatrix}\cos(\underline{t})&\sin(\underline{t})&\cos(3\underline{t})&\sin(3\underline{t})&\cdots&\cos((2k-1)\underline{t})&\sin((2k-1)\underline{t})\end{pmatrix}
=det(eit¯+eit¯2eit¯eit¯2ie3it¯+e3it¯2e3it¯e3it¯2ie(2k1)it¯+e(2k1)it¯2e(2k1)it¯e(2k1)it¯2i)\displaystyle=\det\begin{pmatrix}\frac{e^{i\underline{t}}+e^{-i\underline{t}}}{2}&\frac{e^{i\underline{t}}-e^{-i\underline{t}}}{2i}&\frac{e^{3i\underline{t}}+e^{-3i\underline{t}}}{2}&\frac{e^{3i\underline{t}}-e^{-3i\underline{t}}}{2i}&\cdots&\frac{e^{(2k-1)i\underline{t}}+e^{-(2k-1)i\underline{t}}}{2}&\frac{e^{(2k-1)i\underline{t}}-e^{-(2k-1)i\underline{t}}}{2i}\end{pmatrix}
=122k(i)kdet(eit¯+eit¯eit¯eit¯e(2k1)it¯+e(2k1)it¯e(2k1)it¯e(2k1)it¯).\displaystyle=\frac{1}{2^{2k}}(-i)^{k}\det\begin{pmatrix}e^{i\underline{t}}+e^{-i\underline{t}}&e^{i\underline{t}}-e^{-i\underline{t}}&\cdots&e^{(2k-1)i\underline{t}}+e^{-(2k-1)i\underline{t}}&e^{(2k-1)i\underline{t}}-e^{-(2k-1)i\underline{t}}\end{pmatrix}.

Next, let CjC_{j} denote the jj-th column of the above matrix. For j=1,3,,2k1j=1,3,\dots,2k-1, perform the column operations CjCj+Cj+1C_{j}\mapsto C_{j}+C_{j+1}, and then after each CjC_{j} has been updated, perform the column operations Cj+1Cj+112CjC_{j+1}\mapsto C_{j+1}-\frac{1}{2}C_{j}. It follows that

det(A)\displaystyle\det(A) =122k(i)kdet(2eit¯eit¯2e3it¯e3it¯2e(2k1)it¯e(2k1)it¯)\displaystyle=\frac{1}{2^{2k}}(-i)^{k}\det\begin{pmatrix}2e^{i\underline{t}}&-e^{-i\underline{t}}&2e^{3i\underline{t}}&-e^{-3i\underline{t}}&\cdots&2e^{(2k-1)i\underline{t}}&-e^{-(2k-1)i\underline{t}}\end{pmatrix}
=ik2kdet(eit¯eit¯e3it¯e3it¯e(2k1)it¯e(2k1)it¯)\displaystyle=\frac{i^{k}}{2^{k}}\det\begin{pmatrix}e^{i\underline{t}}&e^{-i\underline{t}}&e^{3i\underline{t}}&e^{-3i\underline{t}}&\cdots&e^{(2k-1)i\underline{t}}&e^{-(2k-1)i\underline{t}}\end{pmatrix}

by factoring out column multiples. Letting ω=e(2k1)i(t1+t2++t2k)\omega=e^{-(2k-1)i(t_{1}+t_{2}+\dots+t_{2k})}, we may factor e(2k1)itje^{-(2k-1)it_{j}} from row jj to obtain

det(A)\displaystyle\det(A) =ik2kωdet(e((2k1)+1)it¯e((2k1)1)it¯e((2k1)+(2k1))it¯e((2k1)(2k1))it¯)\displaystyle=\frac{i^{k}}{2^{k}}\omega\det\begin{pmatrix}e^{((2k-1)+1)i\underline{t}}&e^{((2k-1)-1)i\underline{t}}&\cdots&e^{((2k-1)+(2k-1))i\underline{t}}&e^{((2k-1)-(2k-1))i\underline{t}}\end{pmatrix}
=ik2kωdet(e2kit¯e(2k2)it¯e(2k+2)it¯e(2k4)it¯e2(2k1)it¯1¯),\displaystyle=\frac{i^{k}}{2^{k}}\omega\det\begin{pmatrix}e^{2ki\underline{t}}&e^{(2k-2)i\underline{t}}&e^{(2k+2)i\underline{t}}&e^{(2k-4)i\underline{t}}&\cdots&e^{2(2k-1)i\underline{t}}&\underline{1}\end{pmatrix},

where 1¯\underline{1} is the column of all 11’s. After re-ordering rows by a permutation σ\sigma and taking the determinant of the resulting Vandermonde matrix, we have

det(A)=sign(σ)ik2kωdet(1¯e2it¯e4it¯e(2(2k1))it¯)=sign(σ)ik2kω1j<l2k(e2itle2itj).\det(A)=\mathrm{sign}(\sigma)\frac{i^{k}}{2^{k}}\omega\det\begin{pmatrix}\underline{1}&e^{2i\underline{t}}&e^{4i\underline{t}}&\cdots&e^{(2(2k-1))i\underline{t}}\end{pmatrix}=\mathrm{sign}(\sigma)\frac{i^{k}}{2^{k}}\omega\prod_{1\leq j<l\leq 2k}\left(e^{2it_{l}}-e^{2it_{j}}\right).

Finally, note ω=1j<l2kei(tl+tj)\omega=\prod_{1\leq j<l\leq 2k}e^{-i(t_{l}+t_{j})}, and multiply each term (e2itle2itj)\left(e^{2it_{l}}-e^{2it_{j}}\right) above by the factor ei(tl+tj)e^{-i(t_{l}+t_{j})} extracted from ω\omega to obtain

det(A)\displaystyle\det(A) =sign(σ)ik2k1j<l2k(ei(tltj)ei(tltj))\displaystyle=\mathrm{sign}(\sigma)\frac{i^{k}}{2^{k}}\prod_{1\leq j<l\leq 2k}\left(e^{i(t_{l}-t_{j})}-e^{-i(t_{l}-t_{j})}\right)
=sign(σ)ik2k1j<l2k2isin(tltj)=κ1j<l2ksin(tltj)\displaystyle=\mathrm{sign}(\sigma)\frac{i^{k}}{2^{k}}\prod_{1\leq j<l\leq 2k}2i\sin(t_{l}-t_{j})=\kappa\prod_{1\leq j<l\leq 2k}\sin(t_{l}-t_{j})

where κ=sign(σ)ik2k(2i)2k2k=sign(σ)i2k222k(k1)=sign(σ)22k(k1)\kappa=\mathrm{sign}(\sigma)\tfrac{i^{k}}{2^{k}}(2i)^{2k^{2}-k}=\mathrm{sign}(\sigma)i^{2k^{2}}2^{2k(k-1)}=\mathrm{sign}(\sigma)2^{2k(k-1)}. ∎

The following corollary is immediate.

Corollary 17.

For 0i2k0\leq i\leq 2k, let M2k,iM_{2k,i} denote the 2k×2k2k\times 2k matrix obtained by removing the ii-th column of M2kM_{2k}. Then

det(M2k,i)=κ0j<l2kj,lisin(tltj),\det(M_{2k,i})=\kappa\prod_{\begin{subarray}{c}0\leq j<l\leq 2k\\ j,l\neq i\end{subarray}}\sin(t_{l}-t_{j}),

for some nonzero constant κ\kappa depending only on kk.

Lemma 18.

If no two points t0,t1,,t2kS1t_{0},t_{1},\ldots,t_{2k}\in S^{1} are equal or antipodal, then the nullspace of M2kM_{2k} is one-dimensional and is spanned by λ=(λ0,λ1,,λ2k)\vec{\lambda}=(\lambda_{0},\lambda_{1},\ldots,\lambda_{2k})^{\intercal}, where

λi=(1)i0j<l2kj,lisin(tltj).\lambda_{i}=(-1)^{i}\prod_{\begin{subarray}{c}0\leq j<l\leq 2k\\ j,l\neq i\end{subarray}}\sin(t_{l}-t_{j}).
Proof.

Because M2kM_{2k} has 2k2k rows and 2k+12k+1 columns, it has nullity at least one. Further, by Corollary 17, observe that M2k,0M_{2k,0} is invertible if and only if no two points tl,tjS1t_{l},t_{j}\in S^{1} are equal or antipodal. Hence, M2kM_{2k} contains 2k2k linearly independent columns and has nullity exactly one.

Next, we prove λ\vec{\lambda} is contained in the nullspace of M2kM_{2k}. To ease notation, write

M2kλ=(C1S1C3S3C2k1S2k1)T.M_{2k}\vec{\lambda}=\begin{pmatrix}C_{1}&S_{1}&C_{3}&S_{3}&\cdots&C_{2k-1}&S_{2k-1}\end{pmatrix}^{\text{T}}.

Note λi=(1)i1κdet(M2k,i)\lambda_{i}=(-1)^{i}\frac{1}{\kappa}\det(M_{2k,i}), and hence for n=1,3,5,,2k1n=1,3,5,\dots,2k-1 we have

Cn=i=02kcos(nti)λi=1κi=02k(1)icos(nti)det(M2k,i).C_{n}=\sum_{i=0}^{2k}\cos(nt_{i})\lambda_{i}=\frac{1}{\kappa}\sum_{i=0}^{2k}(-1)^{i}\cos(nt_{i})\det(M_{2k,i}).

Therefore, CnC_{n} is equal to 1κ\frac{1}{\kappa} times the determinant of the matrix

(cos(nt0)cos(nt1)cos(nt2k)cos(t0)cos(t1)cos(t2k)sin(t0)sin(t1)sin(t2k)cos(3t0)cos(3t1)cos(3t2k)sin(3t0)sin(3t1)sin(3t2k)cos((2k1)t0)cos((2k1)t1)cos((2k1)t2k)sin((2k1)t0)sin((2k1)t1)sin((2k1)t2k)).\begin{pmatrix}\cos(nt_{0})&\cos(nt_{1})&\ldots&\cos(nt_{2k})\\ \cos(t_{0})&\cos(t_{1})&\ldots&\cos(t_{2k})\\ \sin(t_{0})&\sin(t_{1})&\ldots&\sin(t_{2k})\\ \cos(3t_{0})&\cos(3t_{1})&\ldots&\cos(3t_{2k})\\ \sin(3t_{0})&\sin(3t_{1})&\ldots&\sin(3t_{2k})\\ \vdots&\vdots&\ddots&\vdots\\ \cos((2k-1)t_{0})&\cos((2k-1)t_{1})&\ldots&\cos((2k-1)t_{2k})\\ \sin((2k-1)t_{0})&\sin((2k-1)t_{1})&\ldots&\sin((2k-1)t_{2k})\end{pmatrix}.

Since n=2j1n=2j-1 for some 1jk1\leq j\leq k, the first row of this matrix is equal to one of the other rows. Hence, the matrix is singular, giving that Cn=0C_{n}=0.

Similarly, it follows that SnS_{n} is equal to 1κ\frac{1}{\kappa} times the determinant of the same matrix, except with the first row replaced by (sin(nt0),sin(nt1),,sin(nt2k))(\sin(nt_{0}),\sin(nt_{1}),\ldots,\sin(nt_{2k})). For the same reasons as before, it follows that Sn=0S_{n}=0. ∎

For convenience, we rescale λ\vec{\lambda} by γ:=0j<l2k1sin(tltj)\gamma:=\prod_{0\leq j<l\leq 2k}\frac{1}{\sin(t_{l}-t_{j})} (which is well-defined for t1,,t2kt_{1},\ldots,t_{2k} distinct) to obtain

γλ=(1α0(t0,,t2k),,1α2k(t0,,t2k)),whereαi(t0,,t2k)=0j2kjisin(tjti).\gamma\vec{\lambda}=\left(\frac{1}{\alpha_{0}(t_{0},\dots,t_{2k})},\dots,\frac{1}{\alpha_{2k}(t_{0},\dots,t_{2k})}\right)^{\intercal},\quad\text{where}\quad\alpha_{i}(t_{0},\dots,t_{2k})=\prod_{\begin{subarray}{c}0\leq j\leq 2k\\ j\neq i\end{subarray}}\sin(t_{j}-t_{i}).

Recall that entries of λ\vec{\lambda} correspond to coefficients in the linear combination 0=i=02kλiSM2k(ti)\vec{0}=\sum_{i=0}^{2k}\lambda_{i}\mathrm{SM}_{2k}(t_{i}). In particular, we are concerned only with convex linear combinations. Hence, after normalizing λ\vec{\lambda} (and potentially rescaling by 1-1), it is necessary that each entry λi\lambda_{i} is positive. In other words, the origin may be contained in the convex hull of {SM2k(t0),,SM2k(t2k)}\{\mathrm{SM}_{2k}(t_{0}),\dots,\mathrm{SM}_{2k}(t_{2k})\} only in the case that the terms αi(t0,,t2k)\alpha_{i}(t_{0},\dots,t_{2k}) share the same sign. We next relate the sign of each term αi(t0,,t2k)\alpha_{i}(t_{0},\dots,t_{2k}) to the configuration of points t0,,t2kS1t_{0},\dots,t_{2k}\in S^{1}.

Lemma 19.

Let t0,,t2kS1t_{0},\dots,t_{2k}\in S^{1}, with no two points equal or antipodal. Then, the numbers αi(t0,,t2k)\alpha_{i}(t_{0},\dots,t_{2k}) have the same sign for all 0i2k0\leq i\leq 2k if and only if χ(ti):=#{tjtj(ti+π,ti)S1}=k\chi(t_{i}):=\#\{t_{j}\mid t_{j}\in(t_{i}+\pi,t_{i})_{S^{1}}\}=k for all ii.

Proof.

Throughout, we assume that the points t0,,t2kS1t_{0},\dots,t_{2k}\in S^{1} are distinct, with no two points antipodal, and furthermore that they are ordered by index with a counterclockwise orientation. Observe that sign(αi(t0,,t2k))=(1)χ(ti)\mathrm{sign}(\alpha_{i}(t_{0},\dots,t_{2k}))=(-1)^{\chi(t_{i})}.

We first prove two preliminary properties.

  • (i)

    i=02kχ(ti)=k(2k+1)\sum_{i=0}^{2k}\chi(t_{i})=k(2k+1).

  • (ii)

    If t0,,t2kt_{0},\dots,t_{2k} are not all contained in a semicircle, then 1χ(ti+1)χ(ti)1\geq\chi(t_{i+1})-\chi(t_{i}) for 0i2k0\leq i\leq 2k, where we set t2k+1=t0t_{2k+1}=t_{0}.

For (i), note that since no two points are equal or antipodal, we have that tj(ti+π,ti)S1t_{j}\in(t_{i}+\pi,t_{i})_{S^{1}} if and only if ti(tj+π,tj)S1t_{i}\notin(t_{j}+\pi,t_{j})_{S^{1}}. Therefore i=02kχ(ti)=(2k+12)=k(2k+1)\sum_{i=0}^{2k}\chi(t_{i})=\binom{2k+1}{2}=k(2k+1).

For (ii), observe that the open arc (ti+1+π,ti)S1(t_{i+1}+\pi,t_{i})_{S^{1}} contains exactly χ(ti+1)1\chi(t_{i+1})-1 points. Indeed, (ti+1+π,ti)S1(t_{i+1}+\pi,t_{i})_{S^{1}} contains exactly χ(ti+1)1\chi(t_{i+1})-1 points for all ii if and only if ti(ti+1+π,ti+1)S1t_{i}\in(t_{i+1}+\pi,t_{i+1})_{S^{1}} for all ii, which is true if and only if the points are not contained in a semicircle. Hence, (ti+π,ti+1+π)S1(t_{i}+\pi,t_{i+1}+\pi)_{S^{1}} must contain exactly χ(ti)(χ(ti+1)1)\chi(t_{i})-(\chi(t_{i+1})-1) points. Because this number is non-negative, it follows that 1χ(ti+1)χ(ti)1\geq\chi(t_{i+1})-\chi(t_{i}).

We now prove Lemma 19. In the case that χ(ti)=k\chi(t_{i})=k for all ii, we see that the numbers αi(t0,,t2k)\alpha_{i}(t_{0},\dots,t_{2k}) are all positive or are all negative.

Conversely, suppose the numbers αi(t0,,t2k)\alpha_{i}(t_{0},\dots,t_{2k}) have the same sign. Since sign(αi(t0,,t2k))=(1)χ(ti)\mathrm{sign}(\alpha_{i}(t_{0},\dots,t_{2k}))=(-1)^{\chi(t_{i})}, the numbers χ(ti)\chi(t_{i}) have the same parity. Further, in the case kk is odd (resp. even), (i) implies each χ(ti)\chi(t_{i}) is odd (resp. even). Therefore, in either case, we may write χ(ti)=k+2ni\chi(t_{i})=k+2n_{i} for some integer nin_{i}\in\mathbb{Z}. Note that (i) implies

k(2k+1)=i=02kχ(ti)=i=02k(k+2ni)=k(2k+1)+2i=02kni,k(2k+1)=\sum_{i=0}^{2k}\chi(t_{i})=\sum_{i=0}^{2k}(k+2n_{i})=k(2k+1)+2\sum_{i=0}^{2k}n_{i},

giving i=02kni=0\sum_{i=0}^{2k}n_{i}=0. Therefore, it is sufficient to prove that ni=njn_{i}=n_{j} for all i,ji,j. Toward that end, define t2k+1=t0t_{2k+1}=t_{0} and n2k+1=n0n_{2k+1}=n_{0}, and observe

0=i=02kni+1=i=02k(ni+1+(ni+ni))=i=02k((ni+1ni)+ni)=i=02k(ni+1ni)+i=02kni=i=02k(ni+1ni).\displaystyle 0=\sum_{i=0}^{2k}n_{i+1}=\sum_{i=0}^{2k}(n_{i+1}+(-n_{i}+n_{i}))=\sum_{i=0}^{2k}((n_{i+1}-n_{i})+n_{i})=\sum_{i=0}^{2k}(n_{i+1}-n_{i})+\sum_{i=0}^{2k}n_{i}=\sum_{i=0}^{2k}(n_{i+1}-n_{i}).

It cannot be the case that all of the points tit_{i} are contained in a semicircle, since then χ(ti)\chi(t_{i}) would obtain all of the values 0,1,,2k0,1,\dots,2k, contradicting the fact that these values have the same parity. Therefore, we may apply (ii) to obtain

1(k+2ni+1)(k+2ni)=2(ni+1ni),1\geq(k+2n_{i+1})-(k+2n_{i})=2(n_{i+1}-n_{i}),

which implies 0ni+1ni0\geq n_{i+1}-n_{i} for all ii. Together with i=02k(ni+1ni)=0\sum_{i=0}^{2k}(n_{i+1}-n_{i})=0, this gives ni+1=nin_{i+1}=n_{i} for all ii. ∎

We are now prepared to prove Theorem 5.

Proof of Theorem 5.

Let distinct t0,,t2kS1t_{0},\dots,t_{2k}\in S^{1} be given in counterclockwise order, and define D=diam({t0,,t2k})D=\mathrm{diam}(\{t_{0},\dots,t_{2k}\}). We claim that if χ(ti):=#{tjtj(ti+π,ti)S1}=k\chi(t_{i}):=\#\{t_{j}\mid t_{j}\in(t_{i}+\pi,t_{i})_{S^{1}}\}=k for all ii, then D2πk2k+1D\geq\frac{2\pi k}{2k+1}. Indeed, define t2k+1=t0,t_{2k+1}=t_{0}, and let i\ell_{i} be the length of (ti,ti+1)S1(t_{i},t_{i+1})_{S^{1}} for all ii. Because χ(ti)=k=χ(ti+1)\chi(t_{i})=k=\chi(t_{i+1}), it follows that there exists exactly one point tjt_{j} in the arc (ti+π,ti+1+π)S1.(t_{i}+\pi,t_{i+1}+\pi)_{S^{1}}. Further, because the function f:(ti+π,ti+1+π)S1f\colon(t_{i}+\pi,t_{i+1}+\pi)_{S^{1}}\to\mathbb{R} defined by f(t)=max{dS1(t,ti),dS1(t,ti+1)}f(t)=\max\{d_{S^{1}}(t,t_{i}),d_{S^{1}}(t,t_{i+1})\} is minimized at the midpoint of (ti+π,ti+1+π)S1(t_{i}+\pi,t_{i+1}+\pi)_{S^{1}}, it follows that Dπi2D\geq\pi-\frac{\ell_{i}}{2}. On the other hand, because there are 2k+12k+1 consecutive pairs of points ti,ti+1t_{i},t_{i+1}, we must have j2π2k+1\ell_{j}\leq\frac{2\pi}{2k+1} for some 0j2k0\leq j\leq 2k. Hence Dππ2k+1=2πk2k+1D\geq\pi-\frac{\pi}{2k+1}=\frac{2\pi k}{2k+1}.

Therefore, if diam({t0,,t2k})<𝒞=2πk2k+1\mathrm{diam}(\{t_{0},\dots,t_{2k}\})<\mathcal{C}=\frac{2\pi k}{2k+1}, then χ(ti)k\chi(t_{i})\neq k for some 0i2k0\leq i\leq 2k. Hence Lemmas 18 and 19 imply that there do not exist positive scalars λi\lambda_{i} with 0=i=02kλiSM2k(ti)\vec{0}=\sum_{i=0}^{2k}\lambda_{i}\mathrm{SM}_{2k}(t_{i}).

To see that this bound is sharp, let tiS1t_{i}\in S^{1} denote the vertices of a regular inscribed (2k+1)(2k+1)-gon. Note that 0=i=02k12k+1SM2k(ti)\vec{0}=\sum_{i=0}^{2k}\frac{1}{2k+1}\mathrm{SM}_{2k}(t_{i}) in this case. ∎

5. A connection between metric thickenings and orbitopes

We now connect the Vietoris–Rips metric thickenings of the circle to the Barvinok–Novik orbitopes. Indeed, we conjecture (Conjecture 6) that for 2π(k1)2k1r<2πk2k+1\frac{2\pi(k-1)}{2k-1}\leq r<\frac{2\pi k}{2k+1}, the metric thickening VRm(S1;r)\mathrm{VR}^{m}(S^{1};r) is homotopy equivalent to the boundary 2k\partial\mathcal{B}_{2k} of the Barvinok–Novik orbitope, i.e. to the odd-dimensional sphere S2k1S^{2k-1}. We are able to show the partial result that the (2k1)(2k-1)-dimensional homology, cohomology, and homotopy groups of VRm(S1;r)\mathrm{VR}^{m}(S^{1};r) are nontrivial; we only obtain the full homotopy type for r2π3r\leq\frac{2\pi}{3}.

Refer to caption

VRm(S1;r)\mathrm{VR}^{m}(S^{1};r)

2k{0}\mathbb{R}^{2k}\setminus\{\vec{0}\}

2k\partial\mathcal{B}_{2k}

Refer to caption

SM2k\mathrm{SM}_{2k}

pp

Refer to caption
Figure 2. The composition of maps VRm(S1;r)SM2k2k{0}𝑝2k\mathrm{VR}^{m}(S^{1};r)\xrightarrow{\mathrm{SM}_{2k}}\mathbb{R}^{2k}\setminus\{\vec{0}\}\xrightarrow{\,p\,}\partial\mathcal{B}_{2k}, drawn in the case k=1k=1.

Towards Conjecture 6, we build the following sequence of maps:

VRm(S1;r)SM2k2k{0}𝑝2k𝜄VRm(S1;r).\mathrm{VR}^{m}(S^{1};r)\xrightarrow{\mathrm{SM}_{2k}}\mathbb{R}^{2k}\setminus\{\vec{0}\}\xrightarrow{\,p\,}\partial\mathcal{B}_{2k}\xrightarrow{\,\iota\,}\mathrm{VR}^{m}(S^{1};r).

This construction will proceed as outlined below.

  1. (1)

    Section 5.1: We define the radial projection map p:2k{0}2kp\colon\mathbb{R}^{2k}\setminus\{\vec{0}\}\to\partial\mathcal{B}_{2k}. We extend the domain of SM2k\mathrm{SM}_{2k} to VRm(S1;r)\mathrm{VR}^{m}(S^{1};r), and note that the composition pSM2kp\circ\mathrm{SM}_{2k} is well-defined.

  2. (2)

    Section 5.2: We define the inclusion ι:2kVRm(S1;r)\iota\colon\partial\mathcal{B}_{2k}\to\mathrm{VR}^{m}(S^{1};r). Since (pSM2k)ι=id2k(p\circ\mathrm{SM}_{2k})\circ\iota=\mathrm{id}_{\partial\mathcal{B}_{2k}}, we obtain that the (2k1)(2k-1)-dimensional homology, cohomology, and homotopy groups of VRm(S1;r)\mathrm{VR}^{m}(S^{1};r) are nontrivial.

  3. (3)

    Section 5.3: We prove that pSM2kp\circ\mathrm{SM}_{2k} and ι\iota are homotopy inverses; this is the step that we can currently only complete for r2π3r\leq\frac{2\pi}{3} (and hence k2k\leq 2).

When r<2π3r<\frac{2\pi}{3} and k=1k=1, this proof is quite easy to interpret. The map SM2\mathrm{SM}_{2} maps the space VRm(S1;r)\mathrm{VR}^{m}(S^{1};r) to an annulus missing the origin in 2\mathbb{R}^{2} (see Figure 2). Map pp radially projects the annulus to its outer circle, and map ι\iota includes the circle back into VRm(S1;r)\mathrm{VR}^{m}(S^{1};r).

As a result of step (3), we obtain VRm(S1;2π3)S3\mathrm{VR}^{m}(S^{1};\frac{2\pi}{3})\simeq S^{3}. Note that VRm(S1;2π3)≄VR(S1;2π3)𝔠S2\mathrm{VR}^{m}(S^{1};\frac{2\pi}{3})\not\simeq\mathrm{VR}(S^{1};\frac{2\pi}{3})\simeq\bigvee^{\mathfrak{c}}S^{2}. We think of the metric thickening VRm(S1;2π3)\mathrm{VR}^{m}(S^{1};\frac{2\pi}{3}) as having the “right” homotopy type, whereas the wild homotopy type of the simplicial complex VR(S1;2π3)\mathrm{VR}(S^{1};\frac{2\pi}{3}) is an artifact of it being equipped with the “wrong” topology333The inclusion S1VR(S1;2π3)S^{1}\hookrightarrow\mathrm{VR}(S^{1};\frac{2\pi}{3}) is not continuous. As further evidence for being more interested in the homotopy type S3S^{3} rather than 𝔠S2\bigvee^{\mathfrak{c}}S^{2}, note that for all 0<ε<2π150<\varepsilon<\frac{2\pi}{15} we have VR(S1;2π3+ε)S3\mathrm{VR}(S^{1};\frac{2\pi}{3}+\varepsilon)\simeq S^{3}..

There is an analogous relationship between the Čech thickenings of the circle and the Carathéodory orbitopes, i.e. the convex hull of the curve (cost,sint,cos2t,sin2t,,coskt,sinkt)(\cos t,\sin t,\cos 2t,\sin 2t,\ldots,\cos kt,\sin kt\bigr{)} [38]. We do not detail that connection here, although some connections between Čech complexes of finite points on the circle and cyclic polytopes (convex hulls of finite sets of points from this trigonometric moment curve) are given in [4].

5.1. Map from the Vietoris–Rips thickening to the Barvinok–Novik orbitope

We first define the radial projection map p:2k{0}2kS2k1p\colon\mathbb{R}^{2k}\setminus\{\vec{0}\}\to\partial\mathcal{B}_{2k}\simeq S^{2k-1}. As 2k\mathcal{B}_{2k} is a convex body containing the origin in its interior, each ray emanating from the origin intersects 2k\partial\mathcal{B}_{2k} exactly once. Hence, pp is well-defined.

We extend SM2k:S12k\mathrm{SM}_{2k}\colon S^{1}\to\mathbb{R}^{2k} to SM2k:VRm(S1;r)2k\mathrm{SM}_{2k}\colon\mathrm{VR}^{m}(S^{1};r)\to\mathbb{R}^{2k} by declaring SM2k(iλiδti)=iλiSM2k(ti)\mathrm{SM}_{2k}\left(\sum_{i}\lambda_{i}\delta_{t_{i}}\right)=\sum_{i}\lambda_{i}\mathrm{SM}_{2k}(t_{i}). Here the sum on the left-hand side defines a measure as a convex sum of Dirac delta functions at the points tiS1t_{i}\in S^{1} (of diameter at most rr), whereas the sum on the right-hand side is a sum of vectors in 2k\mathbb{R}^{2k}. Because SM2k\mathrm{SM}_{2k} restricted to S1S^{1} is continuous and bounded, Lemma 5.2 of [3] proves that this extension to all of VRm(S1;r)\mathrm{VR}^{m}(S^{1};r) is continuous.

Finally, suppose r<2πk2k+1r<\frac{2\pi k}{2k+1}. Then, Theorem 5 implies that the origin 02k\vec{0}\in\mathbb{R}^{2k} is not in the image of the map SM2k:VRm(S1;r)2k\mathrm{SM}_{2k}\colon\mathrm{VR}^{m}(S^{1};r)\to\mathbb{R}^{2k}, and hence the composition pSM2k:VRm(S1;r)2kp\circ\mathrm{SM}_{2k}\colon\mathrm{VR}^{m}(S^{1};r)\to\partial\mathcal{B}_{2k} from the Vietoris–Rips thickening to the boundary of the Barvinok–Novik orbitope is well-defined.

5.2. Inclusion from the Barvinok–Novik orbitope boundary to the Vietoris–Rips thickening

For r2π(k1)2k1r\geq\frac{2\pi(k-1)}{2k-1}, we define the map ι:2kVRm(S1;r)\iota\colon\partial\mathcal{B}_{2k}\to\mathrm{VR}^{m}(S^{1};r) as follows. Given a point iλiSM2k(ti)2k\sum_{i}\lambda_{i}\mathrm{SM}_{2k}(t_{i})\in\partial\mathcal{B}_{2k} with λi>0\lambda_{i}>0 for all ii, let ι(iλiSM2k(ti))=iλiδti\iota\left(\sum_{i}\lambda_{i}\mathrm{SM}_{2k}(t_{i})\right)=\sum_{i}\lambda_{i}\delta_{t_{i}}. Recall that Corollary 12 states every face of 2k\mathcal{B}_{2k} is a simplex whose diameter in S1S^{1} (not in 2k\mathbb{R}^{2k}) is at most 2π(k1)2k1\frac{2\pi(k-1)}{2k-1}, and hence the image of ι\iota indeed lands in VRm(S1;r)\mathrm{VR}^{m}(S^{1};r). We now prove that ι\iota is continuous.

Lemma 20.

Let r2π(k1)2k1r\geq\frac{2\pi(k-1)}{2k-1}. The map ι:2kVRm(S1;r)\iota\colon\partial\mathcal{B}_{2k}\to\mathrm{VR}^{m}(S^{1};r) is continuous.

Proof.

Recall pp is the radial projection to the boundary of 2k\mathcal{B}_{2k}. We will show that

(pSM2k)|ι(2k):ι(2k)2k(p\circ\mathrm{SM}_{2k})|_{\iota(\partial\mathcal{B}_{2k})}\colon\iota(\partial\mathcal{B}_{2k})\to\partial\mathcal{B}_{2k}

is a bijective continuous function from a compact space to a Hausdorff space. It will then follow from [8, Theorem 3.7] that (pSM2k)|ι(2k)(p\circ\mathrm{SM}_{2k})|_{\iota(\partial\mathcal{B}_{2k})} is a homeomorphism, with a continuous inverse ι:2kι(2k)\iota\colon\partial\mathcal{B}_{2k}\to\iota(\partial\mathcal{B}_{2k}). Therefore ι:2kVRm(S1;r)\iota\colon\partial\mathcal{B}_{2k}\to\mathrm{VR}^{m}(S^{1};r) is continuous.

The fact that (pSM2k)|ι(2k)(p\circ\mathrm{SM}_{2k})|_{\iota(\partial\mathcal{B}_{2k})} is a bijective function follows from Corollary 12. The space 2k\partial\mathcal{B}_{2k} is Hausdorff since it inherits the subspace topology from Euclidean space. Finally, to see that ι(2k)\iota(\partial\mathcal{B}_{2k}) is compact, we note that ι(2k)\iota(\partial\mathcal{B}_{2k}) is a closed subset of 𝒫(S1)\mathscr{P}(S^{1}), the space of all Radon probability measures on S1S^{1} equipped with the Wasserstein metric. Since S1S^{1} is compact, it follows that 𝒫(S1)\mathscr{P}(S^{1}) is compact by [44, Remark 6.19], and therefore ι(2k)\iota(\partial\mathcal{B}_{2k}) is compact as a closed subset of a compact space. ∎

We can now give the following corollary of Theorem 5.

Corollary 21.

For 2π(k1)2k1r<2πk2k+1\frac{2\pi(k-1)}{2k-1}\leq r<\frac{2\pi k}{2k+1}, the (2k1)(2k-1)-dimensional homology, cohomology, and homotopy groups of VRm(S1;r)\mathrm{VR}^{m}(S^{1};r) are nontrivial.

Proof.

Theorem 5 implies that for in this range of rr values, the map (pSM2k)ι(p\circ\mathrm{SM}_{2k})\circ\iota is the identity map on 2k\partial\mathcal{B}_{2k}, i.e. that the space 2kS2k1\partial\mathcal{B}_{2k}\cong S^{2k-1} is a retract of VRm(S1;r)\mathrm{VR}^{m}(S^{1};r). ∎

5.3. Show pSM2kp\circ\mathrm{SM}_{2k} and ι\iota are homotopy inverses.

We conjecture that the composition ιpSM2k\iota\circ p\circ\mathrm{SM}_{2k} has a controllable effect on the diameter of any measure in the Vietoris–Rips thickening.

Conjecture 22.

Given 2π(k1)2k1r<2πk2k+1\frac{2\pi(k-1)}{2k-1}\leq r<\frac{2\pi k}{2k+1} and μVRm(S1;r)\mu\in\mathrm{VR}^{m}(S^{1};r), we conjecture

diam(supp(μ))=diam(supp(μ)supp(ιpSM2k(μ))).\mathrm{diam}(\mathrm{supp}(\mu))=\mathrm{diam}(\mathrm{supp}(\mu)\cup\mathrm{supp}(\iota\circ p\circ\mathrm{SM}_{2k}(\mu))).
Theorem 23.

Conjecture 22 would imply Conjecture 6, namely that for 2π(k1)2k1r<2πk2k+1\frac{2\pi(k-1)}{2k-1}\leq r<\frac{2\pi k}{2k+1}, we have

VRm(S1;r)2kS2k1.\mathrm{VR}^{m}(S^{1};r)\simeq\partial\mathcal{B}_{2k}\cong S^{2k-1}.
Proof.

As observed in the proof of Corollary 21, we have that (pSM2k)ι=id2k(p\circ\mathrm{SM}_{2k})\circ\iota=\mathrm{id}_{\partial\mathcal{B}_{2k}}. Hence, it remains only to show that ι(pSM2k)idVRm(S1;r)\iota\circ(p\circ\mathrm{SM}_{2k})\simeq\mathrm{id}_{\mathrm{VR}^{m}(S^{1};r)}. We will do so using the simplest possible homotopy, a linear homotopy. Indeed, consider the linear homotopy H:VRm(S1;r)×IVRm(S1;r)H\colon\mathrm{VR}^{m}(S^{1};r)\times I\to\mathrm{VR}^{m}(S^{1};r) defined by

H(μ,t)=(1t)μ+t[ι(pSM2k)(μ)].H(\mu,t)=(1-t)\mu+t[\iota\circ(p\circ\mathrm{SM}_{2k})(\mu)].

Conjecture 22 would imply that HH is well-defined, and hence also continuous by Lemma 3.8 of [3]. Note H(,0)=idVRm(S1;r)H(-,0)=\mathrm{id}_{\mathrm{VR}^{m}(S^{1};r)} and H(,1)=ι(pSM2k)H(-,1)=\iota\circ(p\circ\mathrm{SM}_{2k}). Hence, this would imply VRm(S1;r)B2kS2k1\mathrm{VR}^{m}(S^{1};r)\simeq\partial B_{2k}\cong S^{2k-1}. ∎

We remark that Conjecture 22 is true for r<2π3r<\frac{2\pi}{3}, giving VRm(S1;r)S1\mathrm{VR}^{m}(S^{1};r)\simeq S^{1} for r<2π3r<\frac{2\pi}{3}.

The remainder of this section is devoted to proving that Conjecture 22 is true for r=2π3r=\frac{2\pi}{3}, and hence VRm(S1;2π3)S3\mathrm{VR}^{m}(S^{1};\frac{2\pi}{3})\simeq S^{3}. In order to prove this, we first describe a number of intermediate lemmas.

The first such lemma, Farkas’ Lemma, characterizes when a vector lies in the convex cone generated by a set of vectors. Let +={t|t0}\mathbb{R}^{+}=\{t\in\mathbb{R}~|~t\geq 0\}.

Lemma 24 (Farkas’ Lemma [13]).

Let Am×nA\in\mathbb{R}^{m\times n}, let aima_{i}\in\mathbb{R}^{m} for 1in1\leq i\leq n denote the columns of AA, and let vmv\in\mathbb{R}^{m}. Then, exactly one of the following is true:

  1. (1)

    There exists x(+)nx\in(\mathbb{R}^{+})^{n} such that Ax=vAx=v.

  2. (2)

    There exists ymy\in\mathbb{R}^{m} such that aiy0a_{i}^{\intercal}y\geq 0 for all ii and vy<0v^{\intercal}y<0.

Case (1) above is equivalent to vcone({a1,,an})v\in\mathrm{cone}(\{a_{1},\dots,a_{n}\}), and case (2) is equivalent to vcone({a1,,an})v\notin\mathrm{cone}(\{a_{1},\dots,a_{n}\}). We can use Farkas’ Lemma to study how cones intersect.

Lemma 25.

Let u0,,un,v0,,vkmu_{0},\dots,u_{n},v_{0},\dots,v_{k}\in\mathbb{R}^{m}. If there exists some ymy\in\mathbb{R}^{m} such that uiy0u_{i}^{\intercal}y\geq 0 for 0in0\leq i\leq n and viy<0v_{i}^{\intercal}y<0 for 0ik0\leq i\leq k, then cone({u0,,un})cone({v0,,vk})=0\mathrm{cone}\left(\left\{u_{0},\dots,u_{n}\right\}\right)\cap\mathrm{cone}\left(\left\{v_{0},\dots,v_{k}\right\}\right)=\vec{0}.

Proof.

Suppose such a vector ymy\in\mathbb{R}^{m} exists, and let 0v=i=0kλivicone({v0,,vk})\vec{0}\neq v=\sum_{i=0}^{k}\lambda_{i}v_{i}\in\mathrm{cone}(\{v_{0},\dots,v_{k}\}). Then, because there exists some 0jk0\leq j\leq k with λj>0\lambda_{j}>0, we have vy=i=0kλiviyλjvjy<0v^{\intercal}y=\sum_{i=0}^{k}\lambda_{i}v_{i}^{\intercal}y\leq\lambda_{j}v_{j}^{\intercal}y<0. Hence, by Lemma 24, vv is not contained in the convex cone generated by {u0,,un}\{u_{0},\dots,u_{n}\}. ∎

The following theorem will be used to construct a vector satisfying the hypotheses of Lemma 25, given certain configurations of points along the curve SM2k\mathrm{SM}_{2k}.

Theorem 26.

Fix a positive integer kk and distinct v1,,v2k1S1v_{1},\dots,v_{2k-1}\in S^{1} with no two points antipodal. Let u1,,u4k2u_{1},\dots,u_{4k-2} denote the set of points {v1,,v2k1}{v1+π,,v2k1+π}\{v_{1},\dots,v_{2k-1}\}\cup\{v_{1}+\pi,\dots,v_{2k-1}+\pi\} labeled in counterclockwise order such that u1=v1u_{1}=v_{1}. Then, there exists a raked homogeneous trigonometric polynomial ff of degree 2k12k-1 such that f(ui)=0f(u_{i})=0 for 1i4k21\leq i\leq 4k-2. Further, sign(f(t))=(1)i\mathrm{sign}(f(t))=(-1)^{i} for t(ui,ui+1)S1t\in(u_{i},u_{i+1})_{S^{1}}, where we define u4k1=u1u_{4k-1}=u_{1}.

Proof.

For tS1t\in S^{1}, consider points SM2k(t)2k\mathrm{SM}_{2k}(t)\in\mathbb{R}^{2k} to be written as column vectors and define the 2k×2k2k\times 2k matrix

N(t)=(SM2k(t)SM2k(v1)SM2k(v2)SM2k(v2k2)SM2k(v2k1)).N(t)=\begin{pmatrix}\mathrm{SM}_{2k}(t)&\mathrm{SM}_{2k}(v_{1})&\mathrm{SM}_{2k}(v_{2})&\cdots&\mathrm{SM}_{2k}(v_{2k-2})&\mathrm{SM}_{2k}(v_{2k-1})\end{pmatrix}.

By Lemma 16,

det(N(t))=κ(1j<l2k1sin(vlvj))(1l2k1sin(vlt)),\det(N(t))=\kappa\Biggl{(}\prod_{\begin{subarray}{c}1\leq j<l\leq 2k-1\end{subarray}}\sin(v_{l}-v_{j})\Biggr{)}\Biggl{(}\prod_{\begin{subarray}{c}1\leq l\leq 2k-1\end{subarray}}\sin(v_{l}-t)\Biggr{)},

where κ\kappa is a nonzero constant that depends only on kk. Further, by considering the cofactor expansion of this determinant along the first column of N(t)N(t), observe that det(N(t))\det(N(t)) is a raked homogeneous trigonometric polynomial of degree 2k12k-1. Because no two elements of {v1,,v2k1}\{v_{1},\ldots,v_{2k-1}\} are equal or antipodal, note that 1j<l2k1sin(vlvj)0\prod_{\begin{subarray}{c}1\leq j<l\leq 2k-1\end{subarray}}\sin(v_{l}-v_{j})\neq 0. Hence,

f(t)=1κ(1j<l2k1sin(vlvj))1det(N(t))=1l2k1sin(vlt)f(t)=\frac{1}{\kappa}\Biggl{(}\prod_{\begin{subarray}{c}1\leq j<l\leq 2k-1\end{subarray}}\sin(v_{l}-v_{j})\Biggr{)}^{-1}\det(N(t))=\prod_{\begin{subarray}{c}1\leq l\leq 2k-1\end{subarray}}\sin(v_{l}-t)

is a well-defined raked homogeneous trigonometric polynomial of degree 2k12k-1 with real roots {v1,,v2k1}{v1+π,,v2k1+π}\{v_{1},\dots,v_{2k-1}\}\cup\{v_{1}+\pi,\dots,v_{2k-1}+\pi\}. Finally, observe for t{v1,,v2k1}{v1+π,,v2k1+π}t\notin\{v_{1},\dots,v_{2k-1}\}\cup\{v_{1}+\pi,\dots,v_{2k-1}+\pi\} we have

sign(f(t))=sign(1l2k1sin(vlt))=(1)ρ(t),\mathrm{sign}\left(f(t)\right)=\mathrm{sign}\Biggl{(}\prod_{\begin{subarray}{c}1\leq l\leq 2k-1\end{subarray}}\sin(v_{l}-t)\Biggr{)}=(-1)^{\rho(t)},

where we define ρ(t)=#{vlvl(t+π,t)S1, 1l2k1}\rho(t)=\#\{v_{l}\mid v_{l}\in(t+\pi,t)_{S^{1}},\,1\leq l\leq 2k-1\}. ∎

Remark 27.

In the setting of Theorem 26, there exists a vector y2ky\in\mathbb{R}^{2k} such that (SM2k(ui))y=0\left(\mathrm{SM}_{2k}(u_{i})\right)^{\intercal}y=0 for all ii. Further, sign((SM2k(t))y)=(1)i\mathrm{sign}\left(\left(\mathrm{SM}_{2k}(t)\right)^{\intercal}y\right)=(-1)^{i} for t(ui,ui+1)S1t\in(u_{i},u_{i+1})_{S^{1}}, where we define u4k1=u1u_{4k-1}=u_{1}.

Proposition 28.

Let distinct t1,,tnS1t_{1},\dots,t_{n}\in S^{1} be in counterclockwise order and contained in an arc [t1,tn]S1[t_{1},t_{n}]_{S^{1}} of length at most 2π3\frac{2\pi}{3}. Let distinct s1,,smS1s_{1},\dots,s_{m}\in S^{1} be such that conv(SM4({s1,,sm}))\mathrm{conv}(\mathrm{SM}_{4}(\{s_{1},\dots,s_{m}\})) is a face of 4\mathcal{B}_{4}, and {s1,,sm}[t1,tn]S1\{s_{1},\dots,s_{m}\}\nsubseteq[t_{1},t_{n}]_{S^{1}}. Then

cone(SM4({s1,,sm}))cone(SM4({t1,,tn}))=cone(SM4({s1,,sm}{t1,,tn})).\mathrm{cone}\left(\mathrm{SM}_{4}(\{s_{1},\dots,s_{m}\})\right)\cap\mathrm{cone}\left(\mathrm{SM}_{4}(\{t_{1},\dots,t_{n}\})\right)=\mathrm{cone}\left(\mathrm{SM}_{4}(\{s_{1},\dots,s_{m}\}\cap\{t_{1},\dots,t_{n}\})\right).

For the above proposition we agree cone()=0\mathrm{cone}(\varnothing)=\vec{0}.

Proof.

Throughout, for convenience, consider points SM4(t)4\mathrm{SM}_{4}(t)\in\mathbb{R}^{4} to be written as column vectors. In light of the known facial structure of 4\mathcal{B}_{4} (Theorem 11), it follows that m3m\leq 3. Hence, there are three cases:

  1. (i)

    The sets {s1,,sm}\{s_{1},\dots,s_{m}\} and {t1,,tn}\{t_{1},\dots,t_{n}\} are disjoint.

  2. (ii)

    The sets {s1,,sm}\{s_{1},\dots,s_{m}\} and {t1,,tn}\{t_{1},\dots,t_{n}\} contain one point of intersection. In this case, m{2,3}m\in\{2,3\}, i.e. {s1,,sm}\{s_{1},\dots,s_{m}\} determines an edge or an equilateral triangle in 4\partial\mathcal{B}_{4}.

  3. (iii)

    The sets {s1,,sm}\{s_{1},\dots,s_{m}\} and {t1,,tn}\{t_{1},\dots,t_{n}\} contain two points of intersection. In this case, m=3m=3, the points {s1,s2,s3}\{s_{1},s_{2},s_{3}\} determine an equilateral triangle in 4\partial\mathcal{B}_{4}, {s1,s2,s3}{t1,,tn}={t1,tn}\{s_{1},s_{2},s_{3}\}\cap\{t_{1},\dots,t_{n}\}=\{t_{1},t_{n}\}, and the length of (t1,tn)S1(t_{1},t_{n})_{S^{1}} is 2π3\frac{2\pi}{3}.

The proof will proceed as follows. We will consider first the case that {t1,,tn}\{t_{1},\dots,t_{n}\} and {s1,,sm}\{s_{1},\dots,s_{m}\} are disjoint and apply Lemma 25 to prove that the resulting cones in 4\mathbb{R}^{4} must be disjoint. Then, we will generalize this argument to allow for intersections and consider the remaining two cases.

Toward that end, suppose {s1,,sm}{t1,,tn}=\{s_{1},\dots,s_{m}\}\cap\{t_{1},\dots,t_{n}\}=\varnothing and note, by Lemma 25, that it is sufficient to find y4y\in\mathbb{R}^{4} such that (SM4(ti))y0\left(\mathrm{SM}_{4}(t_{i})\right)^{\intercal}y\geq 0 for 1in1\leq i\leq n and (SM4(si))y<0\left(\mathrm{SM}_{4}(s_{i})\right)^{\intercal}y<0 for 1im1\leq i\leq m. To define such a vector yy, fix points v1,v2,v3S1v_{1},v_{2},v_{3}\in S^{1} as follows. By the assumptions on the configuration of the points {s1,,sm}\{s_{1},\dots,s_{m}\}, observe there must exist an arc Γ=(γ1,γ2)S1\Gamma=(\gamma_{1},\gamma_{2})_{S^{1}} of length π\pi such that

  • [t1,tn]S1Γ[t_{1},t_{n}]_{S^{1}}\subseteq\Gamma,

  • {s1,,sm}{γ1,γ2}=\{s_{1},\dots,s_{m}\}\cap\{\gamma_{1},\gamma_{2}\}=\varnothing, and

  • |{s1,,sm}Γ|=N|\{s_{1},\dots,s_{m}\}\cap\Gamma|=N for N1N\leq 1.

Indeed, to see that we can arrange N2N\leq 2, note that if m=3m=3 then {s1,s2,s3}\{s_{1},s_{2},s_{3}\} are the vertices of an equilateral triangle, and hence not in an arc of length π\pi. To see that we can arrange N1N\leq 1, note that if m=2m=2, then since one of the sis_{i} points is outside [t1,tn]S1[t_{1},t_{n}]_{S^{1}}, we can choose Γ\Gamma so that the same sis_{i} point is also outside Γ\Gamma.

\begin{overpic}[width=346.89731pt]{PointsOnS1_N0.pdf} \put(97.7,50.0){\footnotesize$\displaystyle t_{1}$} \put(94.0,67.5){\footnotesize$\displaystyle t_{2}$} \put(88.0,78.0){\footnotesize$\displaystyle t_{3}$} \put(67.0,94.3){\footnotesize$\displaystyle t_{4}$} \put(60.0,96.5){\footnotesize$\displaystyle t_{5}$} \put(13.3,72.8){\footnotesize$\displaystyle v_{1}$} \put(11.0,69.5){\footnotesize$\displaystyle v_{2}$} \put(10.0,66.0){\footnotesize$\displaystyle v_{3}$} \put(56.0,0.0){\footnotesize$\displaystyle s_{1}$} \put(86.0,17.0){\footnotesize$\displaystyle s_{2}$} \put(95.0,33.3){\footnotesize$\displaystyle\gamma_{1}$} \put(0.5,66.0){\footnotesize$\displaystyle\gamma_{2}$} \end{overpic}
An example of points {t1,,t5}\{t_{1},\dots,t_{5}\} and {s1,s2}\{s_{1},s_{2}\} in S1S^{1} in the case N=0N=0.
\begin{overpic}[width=346.89731pt]{PointsOnS1_N1.pdf} \put(97.7,50.0){\footnotesize$\displaystyle t_{1}$} \put(94.0,67.5){\footnotesize$\displaystyle t_{2}$} \put(88.0,78.0){\footnotesize$\displaystyle t_{3}$} \put(78.0,79.0){\footnotesize$\displaystyle v_{1}$} \put(78.6,87.4){\footnotesize$\displaystyle s_{1}$} \put(71.5,84.2){\footnotesize$\displaystyle v_{2}$} \put(67.0,94.3){\footnotesize$\displaystyle t_{4}$} \put(60.0,96.5){\footnotesize$\displaystyle t_{5}$} \put(10.2,67.0){\footnotesize$\displaystyle v_{3}$} \put(-2.0,56.0){\footnotesize$\displaystyle s_{2}$} \put(67.0,3.7){\footnotesize$\displaystyle s_{3}$} \put(95.0,33.3){\footnotesize$\displaystyle\gamma_{1}$} \put(0.5,66.0){\footnotesize$\displaystyle\gamma_{2}$} \end{overpic}
An example of points {t1,,t5}\{t_{1},\dots,t_{5}\} and {s1,s2,s3}\{s_{1},s_{2},s_{3}\} in S1S^{1} in the case N=1N=1.
Figure 3. In both cases, the points {γ1,γ2}\{\gamma_{1},\gamma_{2}\} and {v1,,v3}\{v_{1},\dots,v_{3}\} are defined in the proof of Proposition 28 and are used to construct a vector satisfying the hypotheses of Lemma 25.

If N=0N=0, define v1=γ2δv_{1}=\gamma_{2}-\delta, with δ>0\delta>0 small enough such that both (v1+π,v1)S1{s1,,sm}=(v_{1}+\pi,v_{1})_{S^{1}}\cap\{s_{1},\dots,s_{m}\}=\varnothing and (v1+π,v1)S1{t1,,tn}={t1,,tn}(v_{1}+\pi,v_{1})_{S^{1}}\cap\{t_{1},\dots,t_{n}\}=\{t_{1},\dots,t_{n}\}. Then, define v2v_{2} and v3v_{3} so that v1v_{1}, v2v_{2}, v3v_{3}, and γ2\gamma_{2} appear in counterclockwise order.

If N=1N=1, assume without loss of generality that Γ{s1,,sm}={s1}\Gamma\cap\{s_{1},\dots,s_{m}\}=\{s_{1}\}. Then, define v1=s1εv_{1}=s_{1}-\varepsilon and v2=s1+εv_{2}=s_{1}+\varepsilon. Choose ε>0\varepsilon>0 small enough such that (v1,v2)S1(v_{1},v_{2})_{S^{1}} does not contain any point in {t1,t2,,tn,γ1,γ2}\{t_{1},t_{2},\dots,t_{n},\gamma_{1},\gamma_{2}\} and furthermore so that (v1+π,v2+π)S1{s1,,sm}=(v_{1}+\pi,v_{2}+\pi)_{S^{1}}\cap\{s_{1},\dots,s_{m}\}=\varnothing. Such points must exist because no two elements of {s1,,sm}\{s_{1},\dots,s_{m}\} are antipodal. Finally, define v3=γ2δv_{3}=\gamma_{2}-\delta, with δ>0\delta>0 small enough such that both (v3+π,v3)S1{s1,,sm}={s1}(v_{3}+\pi,v_{3})_{S^{1}}\cap\{s_{1},\dots,s_{m}\}=\{s_{1}\} and (v3+π,v3)S1{t1,,tn}={t1,,tn}(v_{3}+\pi,v_{3})_{S^{1}}\cap\{t_{1},\dots,t_{n}\}=\{t_{1},\dots,t_{n}\}.

Now, apply Remark 27 to obtain y4y\in\mathbb{R}^{4} such that for t{v1,v2,v3}{v1+π,v2+π,v3+π}t\notin\{v_{1},v_{2},v_{3}\}\cup\{v_{1}+\pi,v_{2}+\pi,v_{3}+\pi\}, we have

sign((SM4(t))y)=sign(1l3sin(vlt))=(1)ρ(t),\mathrm{sign}\left(\left(\mathrm{SM}_{4}(t)\right)^{\intercal}y\right)=\mathrm{sign}\Biggl{(}\prod_{\begin{subarray}{c}1\leq l\leq 3\end{subarray}}\sin(v_{l}-t)\Biggr{)}=(-1)^{\rho(t)},

where ρ(t)=#{vlvl(t+π,t)S1, 1l3}\rho(t)=\#\{v_{l}\mid v_{l}\in(t+\pi,t)_{S^{1}},\,1\leq l\leq 3\}. When we consider the case t=tit=t_{i} for 1in1\leq i\leq n, we note by construction that ρ(ti)\rho(t_{i}) is even for each tit_{i}, and so (SM4(ti))y0\left(\mathrm{SM}_{4}(t_{i})\right)^{\intercal}y\geq 0 for 1in1\leq i\leq n.

On the other hand, in the case N=0N=0, we note that ρ(si)=3\rho(s_{i})=3 and sign((SM2k(si))y)=1\mathrm{sign}\left((\mathrm{SM}_{2k}(s_{i}))^{\intercal}y\right)=-1 for 1im1\leq i\leq m. Finally, in the case N=1N=1, note that ρ(s1)=1\rho(s_{1})=1 and sign((SM2k(s1))y)=1\mathrm{sign}\left((\mathrm{SM}_{2k}(s_{1}))^{\intercal}y\right)=-1. Further, the pair {v1,v2}\{v_{1},v_{2}\} has zero net effect on the parity of ρ(si)\rho(s_{i}) for 2im2\leq i\leq m by the fact that (v1+π,v2+π)S1{s1,,sm}=(v_{1}+\pi,v_{2}+\pi)_{S^{1}}\cap\{s_{1},\dots,s_{m}\}=\varnothing. Hence, sign((SM2k(si))y)=1\mathrm{sign}\left((\mathrm{SM}_{2k}(s_{i}))^{\intercal}y\right)=-1 for 2im2\leq i\leq m.

This concludes the proof of case (i) that cone(SM4({s1,,sm}))cone(SM4({t1,,tn}))={0}\mathrm{cone}\left(\mathrm{SM}_{4}(\{s_{1},\dots,s_{m}\})\right)\cap\mathrm{cone}\left(\mathrm{SM}_{4}(\{t_{1},\dots,t_{n}\})\right)=\{\vec{0}\} when {s1,,sm}{t1,,tn}=.\{s_{1},\dots,s_{m}\}\cap\{t_{1},\dots,t_{n}\}=\varnothing.

Next, consider case (ii). Assume without loss of generality that {s1,,sm}{t1,,tn}={s1},\{s_{1},\dots,s_{m}\}\cap\{t_{1},\dots,t_{n}\}=\{s_{1}\}, and write s1=tαs_{1}=t_{\alpha} for some 1αn1\leq\alpha\leq n. Given ucone(SM4({t1,,tn}))cone(SM4({s1,,sm}))\vec{u}\in\mathrm{cone}\left(\mathrm{SM}_{4}(\{t_{1},\dots,t_{n}\})\right)\cap\mathrm{cone}\left(\mathrm{SM}_{4}(\{s_{1},\dots,s_{m}\})\right), write u=i=1nλiSM4(ti)=j=1mκjSM4(sj)\vec{u}=\sum_{i=1}^{n}\lambda_{i}\mathrm{SM}_{4}(t_{i})=\sum_{j=1}^{m}\kappa_{j}\mathrm{SM}_{4}(s_{j}) for some non-negative scalars λi,κj\lambda_{i},\kappa_{j}. To show ucone(SM4(tα))\vec{u}\in\mathrm{cone}\left(\mathrm{SM}_{4}(t_{\alpha})\right), observe that it is sufficient to prove λi=0\lambda_{i}=0 for all i{1,,n}αi\in\{1,\dots,n\}\setminus\alpha. We consider the possibilities λακ1\lambda_{\alpha}\geq\kappa_{1} and λα<κ1\lambda_{\alpha}<\kappa_{1} separately.

If λακ1\lambda_{\alpha}\geq\kappa_{1}, then

uκ1SM4(s1)=(λακ1)SM4(tα)+i{1,,n}αλiSM4(ti)=j=2mκjSM4(sj).\vec{u}-\kappa_{1}\mathrm{SM}_{4}(s_{1})=(\lambda_{\alpha}-\kappa_{1})\mathrm{SM}_{4}(t_{\alpha})+\sum_{i\in\{1,\dots,n\}\setminus\alpha}\lambda_{i}\mathrm{SM}_{4}(t_{i})=\sum_{j=2}^{m}\kappa_{j}\mathrm{SM}_{4}(s_{j}).

It follows that uκ1SM4(s1)cone(SM4({t1,,tn}))cone(SM4({s2,,,sm}))\vec{u}-\kappa_{1}\mathrm{SM}_{4}(s_{1})\in\mathrm{cone}\left(\mathrm{SM}_{4}(\{t_{1},\dots,t_{n}\})\right)\cap\mathrm{cone}\left(\mathrm{SM}_{4}(\{s_{2},,\ldots,s_{m}\})\right). Hence, because {t1,,tn}{s2,,sm}=\{t_{1},\dots,t_{n}\}\cap\{s_{2},\dots,s_{m}\}=\varnothing, we have obtained a configuration of points satisfying the hypotheses of case (i) of this proof. Therefore, uκ1SM4(s1)=0,\vec{u}-\kappa_{1}\mathrm{SM}_{4}(s_{1})=\vec{0}, and by Corollary 29 of case (i) below, it follows that λα=κ1\lambda_{\alpha}=\kappa_{1} and λi=0\lambda_{i}=0 for all i{1,,n}αi\in\{1,\dots,n\}\setminus\alpha.

If λα<κ1\lambda_{\alpha}<\kappa_{1}, then uλαSM4(tα)=i{1,,n}αλiSM4(ti)=j=2mκjSM4(sj)λαSM4(tα)\vec{u}-\lambda_{\alpha}\mathrm{SM}_{4}(t_{\alpha})=\sum_{i\in\{1,\dots,n\}\setminus\alpha}\lambda_{i}\mathrm{SM}_{4}(t_{i})=\sum_{j=2}^{m}\kappa_{j}\mathrm{SM}_{4}(s_{j})-\lambda_{\alpha}\mathrm{SM}_{4}(t_{\alpha}). That is,

uλαSM4(tα)=(κ1λα)SM4(s1)+j=2mκjSM4(sj).\vec{u}-\lambda_{\alpha}\mathrm{SM}_{4}(t_{\alpha})=(\kappa_{1}-\lambda_{\alpha})\mathrm{SM}_{4}(s_{1})+\sum_{j=2}^{m}\kappa_{j}\mathrm{SM}_{4}(s_{j}).

As before, because ({t1,,tn}{tα}){s1,,sm}=,\left(\{t_{1},\dots,t_{n}\}\setminus\{t_{\alpha}\}\right)\cap\{s_{1},\dots,s_{m}\}=\varnothing, we have obtained a configuration of points satisfying the hypotheses of case (i) of this proof. Hence uλαSM4(tα)=0\vec{u}-\lambda_{\alpha}\mathrm{SM}_{4}(t_{\alpha})=\vec{0}, and by Corollary 29 of case (i), it follows that λi=0\lambda_{i}=0 for all i{1,,n}αi\in\{1,\dots,n\}\setminus\alpha. This concludes the proof for case (ii).

Last, observe that case (iii) follows by a similar trick: by rewriting a vector u\vec{u} contained in the intersection of both cones, we may obtain a configuration of points satisfying the hypotheses of case (i) or case (ii). ∎

We emphasize that the following is a corollary of case (i) only in the proof of Proposition 28; indeed it is used in the proof of case (ii).

Corollary 29.

Let distinct t1,,tnS1t_{1},\dots,t_{n}\in S^{1} be in counterclockwise order and contained in an arc [t1,tn]S1[t_{1},t_{n}]_{S^{1}} of length at most 2π3\frac{2\pi}{3}. If i=1nλiSM4(ti)=0\sum_{i=1}^{n}\lambda_{i}\mathrm{SM}_{4}(t_{i})=\vec{0} with λi0\lambda_{i}\geq 0, then λi=0\lambda_{i}=0 for all 1in1\leq i\leq n.

Proof.

The claim is obvious in the case n=1n=1. Otherwise, because SM4(t)=SM4(t),\mathrm{SM}_{4}(-t)=-\mathrm{SM}_{4}(t), we may write i=1n1λiSM4(ti)=λnSM4(tn)\sum_{i=1}^{n-1}\lambda_{i}\mathrm{SM}_{4}(t_{i})=\lambda_{n}\mathrm{SM}_{4}(-t_{n}), with tn[t1,tn]S1-t_{n}\notin[t_{1},t_{n}]_{S^{1}}. With s1=tn,s_{1}=-t_{n}, observe that the hypotheses of case (i) of Proposition 28 are satisfied, implying

cone(SM4({t1,,tn1}))cone(SM4(tn))=0\mathrm{cone}\left(\mathrm{SM}_{4}(\{t_{1},\ldots,t_{n-1}\})\right)\cap\mathrm{cone}\left(\mathrm{SM}_{4}(-t_{n})\right)=\vec{0}

Since λnSM4(tn)\lambda_{n}\mathrm{SM}_{4}(-t_{n}) is in this intersection of cones, this implies λn=0\lambda_{n}=0. Hence i=1n1λiSM4(ti)=0\sum_{i=1}^{n-1}\lambda_{i}\mathrm{SM}_{4}(t_{i})=\vec{0}, and we may proceed iteratively to conclude λi=0\lambda_{i}=0 for all ii. ∎

We are now ready to prove that the “diameter non-increasing” result in Conjecture 22 is true for r=2π3r=\frac{2\pi}{3}.

Proposition 30.

For μVRm(S1;2π3)\mu\in\mathrm{VR}^{m}(S^{1};\frac{2\pi}{3}), we have diam(supp(μ))=diam(supp(μ)supp(ιpSM4(μ)))\mathrm{diam}(\mathrm{supp}(\mu))=\mathrm{diam}(\mathrm{supp}(\mu)\cup\mathrm{supp}(\iota\circ p\circ\mathrm{SM}_{4}(\mu))).

Proof.

Let μ=i=1nλiδtiVRm(S1;2π3)\mu=\sum_{i=1}^{n}\lambda_{i}\delta_{t_{i}}\in\mathrm{VR}^{m}(S^{1};\frac{2\pi}{3}) for tiS1t_{i}\in S^{1} and λi>0\lambda_{i}>0 with iλi=1\sum_{i}\lambda_{i}=1. There are two cases. If {t1,,tn}\{t_{1},\dots,t_{n}\} are in counterclockwise order and belong to an arc of length at most 2π3\frac{2\pi}{3}, then Proposition 28 implies that supp(ιpSM4(μ))[t1,tn]S1\mathrm{supp}(\iota\circ p\circ\mathrm{SM}_{4}(\mu))\subseteq[t_{1},t_{n}]_{S^{1}}, and hence

diam(supp(μ))=diam(supp(μ)supp(ιpSM4(μ))).\mathrm{diam}(\mathrm{supp}(\mu))=\mathrm{diam}(\mathrm{supp}(\mu)\cup\mathrm{supp}(\iota\circ p\circ\mathrm{SM}_{4}(\mu))).

Otherwise, n=3n=3 and {t1,t2,t3}\{t_{1},t_{2},t_{3}\} form the vertices of an equilateral triangle. In this case, we have ιpSM4(μ)=μ\iota\circ p\circ\mathrm{SM}_{4}(\mu)=\mu in light of Theorem 11. ∎

6. Conclusion

We provide a lower bound on the diameter of a Carathéodory set in the centrally symmetric trigonometric moment curve, i.e., a set whose convex hull contains the origin. As applications, we obtain sharp versions of the Borsuk–Ulam theorem for maps into higher-dimensional codomains, and we gain control over the zeros of raked trigonometric polynomials. Furthermore, we provide a geometric proof (taking advantage of continuous maps afforded by the optimal transport metric) that the Vietoris–Rips metric thickening of the circle achieves the homotopy type of the 3-sphere S3S^{3} at scale parameter r=2π3r=\frac{2\pi}{3}, in contrast to the uncountably infinite wedge-sum of 2-spheres attained by the ordinary Vietoris–Rips complex on the circle. This proof reveals connections between Vietoris–Rips thickenings of the circle and the Barvinok–Novik orbitopes 2k\mathcal{B}_{2k}; analogous connections exist between Čech thickenings of the circle and the Carathéodory orbitopes.

The homotopy types of Vietoris–Rips metric thickenings of the circle VRm(S1;r)\mathrm{VR}^{m}(S^{1};r) are currently unknown for r>2π3r>\frac{2\pi}{3}. To obtain Conjecture 6, that VRm(S1;r)2kS2k1\mathrm{VR}^{m}(S^{1};r)\simeq\partial\mathcal{B}_{2k}\cong S^{2k-1} for 2π(k1)2k1r<2πk2k+1\frac{2\pi(k-1)}{2k-1}\leq r<\frac{2\pi k}{2k+1}, it remains to prove the homotopy equivalence ι(pSM2k)idVRm(S1;r)\iota\circ(p\circ\mathrm{SM}_{2k})\simeq\mathrm{id}_{\mathrm{VR}^{m}(S^{1};r)}, where a linear homotopy may again be well-defined (see Conjecture 22).

7. Acknowledgements

We would like to thank Harrison Chapman for the insights behind the proof of Lemma 16, Michał Adamaszek, Alexander Barvinok, Yuliy Baryshnikov, Isabella Novik, Raman Sanyal, Rainer Sinn, and Cynthia Vinzant for helpful conversations, and Arkadiy Skopenkov for pointing us to [33].

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