A MENGER REDUX: EMBEDDING METRIC SPACES ISOMETRICALLY IN

JOHN C. BOWERS AND PHILIP L. BOWERS

Abstract. We present geometric proofs of Menger’s results on isomet- rically embedding metric spaces in Euclidean space. In 1928, Karl Menger [6] published the proof of a beautiful characteriza- tion of those metric spaces that are isometrically embeddable in the n- dimensional Euclidean space En.WhileavisitoratHarvardUniversityand the Rice Institute in Houston during the 1930-31 academic year, Menger gave courses on metric in which he “considerably shortened and revised [his] original proofs and generalized the formulation.”1 The new proofs of the 1930-31 academic year appear in the English language article New foundation of Euclidean geometry [7] published in the American Journal of in 1931. Leo Liberti and Carlile Lavor in their beautifully written article Six mathematical gems from the history of distance geometry remark on their review of a part of Menger’s characterizationthat“itis remarkable that almost none of the results below offers an intuitive geomet- rical grasp, such as the proofs of Heron’s formula and the Cayley Theorem do. As formal mathematics has it, part of the beauty in Menger’s work con- sists in turning the ‘visual’ geometrical proofs based on intuition into formal symbolic arguments based on sets and relations.”2 Part of the reason that Menger’s results fail to offer an “intuitive geometric grasp” is that Menger offers his results in the general setting of an abstract congruence system and semi-metric spaces. We believe that Menger’s characterization deserves a wider circulation among the mathematics community. The aim of this article is toexpli- cate Menger’s characterization in the category of metric spaces rather than in his original setting of congruence systems and their modelsemi-metric spaces, and in doing so to give a mostly3 self-contained treatment with straightforward geometric proofs of his characterization.Weprovideallthe background material on the geometry of Euclidean space that is needed to prove Menger’s results so that the proofs are accessible to any undergradu- ate who has mastered the basics of real linear algebra and realinnerproduct

1[7], p. 721. 2[4], p. 12. 3See footnote 5. 1 2JOHNC.BOWERSANDPHILIPL.BOWERS spaces.4 The full characterization as presented in Menger [7] may be parsed in three acts. Act 1 reduces the problem of isometrically embedding a X into En to isometrically embedding finite subsets of X into En.Act 2characterizesthosemetricspacesthatfailtoembedinEn isometrically though each subset with n+2 elements does so embed. Acts 1 and 2 together reduce the problem of embedding X isometrically into En to the problem of embedding each subset of X with n +2elementsisometrically.Finally, Act 3 gives an algebraic condition on the distances among the points of a set with n +2elementsthatguaranteesanisometricembeddingintoEn. We give complete geometric proofs of Menger’s results of Acts1and2 and the necessity part of Act 3.5 The approach of this article to Menger’s results offers geometric insights not immediately grasped in Menger’s orig- inal approach. In particular, the construction and characterization of the metric spaces in Act 2 that fail to embed isometrically thougheachsubset with n + 2 elements does embed so, offer details of the structure of these spaces as well as a geometric understanding that are opaque inMenger’s treatment. This geometric approach leads to a characterization of all such spaces with a concrete construction for all of them. As part ofanunder- graduate research project under the direction of the first author, Brittany Braswell has constructed a webpage, Menger Embeddings in Euclidean Space [2], that includes applets that allow for an interactive exploration of these spaces.

Overture: Background on Euclidean Space. We take as our model of Euclidean n-dimensional space, denoted as n E ,thesetoforderedn-tuples x =(x1,...,xn)ofrealnumberswiththe usual real vector space operations and the Euclidean inner product x, y = ⟨ ⟩ x1y1 + + xnyn.ThisgivesrisetotheEuclideannorm x = x, x and the Euclidean··· distance xy = x y .Inthisoverture,wecollectthebasic| | ⟨ ⟩ ! geometric facts about Euclidean| − space| that will be used in theproofsof Menger’s characterization. The proofs of these basic facts use only elemen- tary linear algebra and are left to the reader. The reader seasoned in the geometry of Euclidean space may skip this overture safely andheaddirectly to Act 1, referring only to the statements of the lemmas as theyareused. n Acollection v0,...,vm of m +1 pointsin E is affinely independent { } if the collection v1 v0,...,vm v0 is linearly independent. Note that { − − } 4Mastered being the key word in this sentence. The authors believe that the assignment of this article for careful study could be an appropriate and effective undergraduate project for helping a student master the geometry of Euclidean space using linear algebra and the Euclidean inner product. As of this writing, a student of the first author is the first test case of this belief. 5The sufficiency part of Act 3 is primarily algebraic in nature and depends on a detailed algebraic analysis of certain determinants. As our interestisinexplicatingthegeometry that lies behind Menger’s results, we refer the interested reader to Menger’s original proof in [7] for the sufficiency. AMENGERREDUX 3 this implies that m n.Affineindependenceof v0,...,vm is equivalent ≤ { m } to the condition that, for real coefficients a0,...,am,if k=0 akvk = 0 and m =0 ak =0,thena0 = = am =0.Incase v0,...,vm is affinely k ··· { " } independent, the linear span of the m linearly independent vectors v1 " n { − v0,...,vm v0 is an m-dimensional linear subspace of E that may be expressed as− } m m L =span[v1 v0,...,vm v0]= akvk : ak =0 . − − # =0 =0 % $k $k It follows from this that the affine span of v0,...,vm defined as { } m m A = affspan[v0 ...,vm]= akvk : ak =1 #k=0 k=0 % $ $ is the m-dimensional affine plane obtained by translating the linear subspace L by any element of A so that, in particular, A = v0 + L.Eachelementv of the affine plane A has a unique set of affine coordinates a0,...,am with m respect to the affine basis v0,...,vm for A,withv = k=0 akvk where m { } k=0 ak =1.Noteinparticularthattheaffinebasisitself,V ,iscontained in A and that if x is any element of En not contained in"A,thentheset "V x is affinely independent. ∪Asubset{ } C of En is convex if, for each pair x, y of points in C,theline segment [x, y]= tx +(1 t)y :0 t 1 is contained in C.Theconvex hull of a set B in{En is the− intersection≤ ≤ of all} convex sets containing B,and is the smallest convex set containing B. n The convex hull of some affinely independent set V = v0,...,vm in E { } is called the m- spanned by V ,isdenotedas[v0,...,vm], and has the description m m [v0,...,vm]= akvk : ak =1,ak 0fork =1,...,m . # =0 =0 ≥ % $k $k Each vk is a vertex of [v0,...,vm]. The affine span of V vk has dimension m 1, is called the kth support plane of the simplex,−{ and} is said to be − opposite to the vertex vk,whilethesimplexspannedbyV vk is the kth −{ } face of the simplex. The simplex is regular if all the side-lengths vivj,for 0 i = j m,arethesame.Obviouslyanm-simplex is contained in the ≤ ̸ ≤ affine span of its vertices. It is not difficult to prove that if [v0,...,vn]isan n n-simplex in E and uk is a non-zero vector orthogonal to the kth support n 6 plane for k =0,...n,thenthevectorsu0,...un span E . The most important property of an affine basis for the proof of Menger’s characterization is the following elementary result. Let v0,...,vm be an affine basis for the m-dimensional affine plane A in En{.Theresultsays} that an element a of A is uniqiuely determined by the m +1 distances 6 In fact, the collection {u0,...un} is affinely independent and any n of the vectors n from the list form a vector space basis for E . 4JOHNC.BOWERSANDPHILIPL.BOWERS av0,...,avm.Thisimpliesinparticularthatthem+1 known distances avk for k =0,...,m uniquely determine the m +1affinecoordinatesa0,...am of a with respect to the affine basis v0,...,vm . { } Lemma A. Let a be a point in the affine span A En of the affine basis n ⊂ v0,...,vm for A.Letb E . If avk = bvk for k =0,...,m,thena = b. { } ∈ n Proof. Let a, b E with a A and suppose that avk = bvk for k = 0,...,m.Ifa =∈b,thentheset∈ M = x En : ax = bx of points of En equidistant from̸ both a and b is an ({n ∈1)-dimensional} affine plane, the n − perpendicular bisector of a and b in E .Fork =0,...,m,sinceavk = bvk, vk is a point of M,andthisimpliesthatA M.Sincea A,thena M, which implies that ba = aa =0.Therefore⊂a = b. ∈ ∈ ! Another straightforward argument using perpendicular bisectors provides aproofofthenextresult.

Lemma B. Let v0,...,vm be an affine basis for the m-dimensional affine plane A in En,where{ m

Lemma C. Let V = v0,...,vm and W = w0,...,wm be two affinely { n } { } independent subsets of E with vjvk = wjwk for 0 j, k m. Then there n ≤ ≤ is an isometry ρ of E with ρ(vk)=wk for k =0,...,m. The proof of this lemma is an easy induction on m using Lemma B and the observation that any isometry between any two affine planes in En extends to an isometry of En.WemaydescribetheisometrybetweenA and B,the respective affine spans of the sets V and W of the lemma, as follows. Each point a of A has a unique set of affine coordinates a0,...,am with respect m to the affine basis V .Definethefunctionλ : A B by λ(a)= k=0 akwk. Since both V and W are affine bases for the respective→ m-dimensional affine planes A and B, λ is a bijection of A onto B,andLemmaAimpliesthat" λ is an isometric mapping. Let Sn 1(c,r)= x En : xc = r ,the(n 1)- in En centered − { ∈ } − at c of radius r>0. Obviously the vertices of a 1-simplex [v0, v1]liein 1 0 aunique0-sphereinE ,namelyS (c,r)= v0, v1 ,thesphereofradius { 1 } r = v0 v1 /2centeredatc =(v0 + v1)/2inE .Thisisthebasisofan easy| inductive− | argument using elementary Euclidean geometry that proves the next lemma. AMENGERREDUX 5

n Lemma D. Let [v0,...,vn] be an n-simplex in E . There is a unique n 1 sphere S − (c,r) that passes through the n +1 affinely independent vertices v0,...,vn of [v0,...,vn].

This sphere is the circumscribed sphere of the simplex [v0,...,vn]whilethe inscribed sphere is the (n 1)-sphere tangent to each face, i.e., that meets each face of the simplex in− a single point. The final geometric facts that we need concern the inradius and circum- n radius of a regular n-simplex Σε of sidelength ε.Thein-andcircumradii are the respective radii of the inscribed and circumscribed , which n have a common center since Σε is regular. The calculations of these two radii constitute a nice exercise in the use of affine coordinates. Exercise. n e e En+1 Let Σ√2 =[0,,..., n] be the regular n-simplex in with vertices the standard unit coordinate vectors. Then the center of the in- n 1 n and circumscribed spheres of Σ√2 is cn = n+1 k=0 ek. It follows that the inradius δ and the circumradius ζ are given as n n " 1 δn = cn cn 1 = , | − − | n(n +1) !n ζn = cn en = , | − | n +1 & n and those of Σε are obtained from these formulæ by multiplying by the scale factor ε/√2.

Act 1: A reduction to embedding finitely many points isometrically. We are now in a position to state and easily prove the first result of Menger that reduces the question of whether a given metric space embeds isometrically in En to the question of whether each of its finite subsets with at most n+3 elementssoembeds. Ratherthanexplicitly namingthemetric of a metric space, we shall use the convention that the juxtaposition of two points indicates the distance between the points. So, if x and y are points of the metric space X,theexpressionxy means the distance in X between the two points. This of course is the notation we use in the Overture for the distances between points of En.WhenY is a subset of X,wealwayswill view Y as a metric subspace, its metric inherited from that of X,andwhen the metric spaces X and Y are isometric, we will write X ∼= Y . Theorem 1. Let X be a metric space. Then X embeds isometrically in the Euclidean space En if and only if each of its finite subsets with at most n +3 elements embeds isometrically in En. Proof. Suppose each metric subspace of X with at most n +3 elements embeds isometrically in En.Amongallisometricembeddingsofsubsetsof X with at most n +1 elementsontoaffinelyindependentsubsetsof En, 6JOHNC.BOWERSANDPHILIPL.BOWERS

En let Y = y0,...,ym ∼= y0,...,ym ,wherem n,beonewitha maximum{ number of} elements.{ Here} of⊂ course our notation≤ suggests that yk yk for k =0,...,munder the isometric embedding so that yjyk = yjyk -→ for 0 j, k m.LetA be the affine span of y0,...,ym ,anm-dimensional affine≤ plane≤ in En.WeshallshowthatX embeds{ isometrically} into A. n Define ρ : X A E as follows. First, ρ(yk)=yk for k =0,...,m. → ⊂ Now let x X Y and note that the set Yx = Y x has m +2

Act 2: Characterizing certain finite sets that fail to embed isometrically. In some respects, the results of this section offer the most surprising aspects of Menger’s characterization. We begin by asking whether we may improve upon Theorem 1 by reducing the number n +3. Specializingto the case n = 1 offers a simple, but somewhat surprising glimpse into the AMENGERREDUX 7 general results of this section. A metric space that isometrically embeds in the Euclidean line E1 is said to be linear.Inthiscase,thequestionofthe linearity of three points of X is just the question of whether the triangle inequality among an appropriate ordering of the three pointsisinfactan equality. Indeed, for three pairwise distinct points x, y,andz of the metric space X,wesaythaty is between x and z provided xz = xy + yz,whichwe denote by xyz.Obviously,thetripleofpoints x, y, z is linear if and only if xyz,oroneofitspermutations,holds.Theorem1saysthatifeve{ } ry three- and four-point subset of X is linear, then X is linear. Can we do better and say that X is linear whenever each triple of its points is linear? The answer is ‘no’ with a family of similar couterexamples provided by the four-point metric spaces whose elements are the vertices of Euclidean rectangles, with distances the lengths of the shortest edge paths in the rectangles connecting any two. Now the really interesting fact is that these counterexamples are, up to isometry, the only counterexamples! What we mean is this. Let X be ametricspaceinwhicheverytripleofitspointsislinear.Then either X is linear or X consists of precisely four distinct points w, x, y,andz,andwxy, xyz, yzw, zwx, wx = yz,andxy = zw.Inparticular,ifX has more than four points, then X is linear if and only if any three of its points is linear, improving upon Theorem 1. All of these statements about the n =1case can be proved by chasing around betweeness relations among four points of X.Theaimofthissectionistoprovethatthisspecialcaseofn =1is representative of the general case. First we address the general question of whether there are anymetric spaces with n +3 elementsthatfailtoembedin En isometrically, even though every one of its subsets with at most n +2 elementsdoessoembed. This is answered by the next result, rather surprising to us (especially the uniqueness assertion), which is used later to describe all such examples. To n set up the result, let V = v0,...,vn E be affinely independent, and therefore an affine basis for{ En,andlet}⊂a be any element of En.Foreach j =0,...,n,letAj be the jth support plane of the n-simplex [v0,...,vn] opposite vj that is affinely spanned by Vj = V vj .Letaj be the isometric reflection in En of the point a through− the{ (n} 1)-dimensional − support plane Aj.Wesaythata is a polar point for [v0,...,vn]iftheset a0,...,an is a set of n +1 affinelyindependentpoints,whichtherefore are{ the vertices} of an n-simplex. In this case, Lemma D of the Overture guarantees the existence of the unique circumscribing sphere of the n-simplex [a0,...,an], say with center a∗ and radius r = r(a, v0,...,vn). The point a∗ is said to be antipodal to a,theradiusr is the polar diameter of the set a, a ,andtheorderedpair(a, a )issaidtobeantipodal with respect to { ∗} ∗ the n-simplex [v0,...,vn]; see Fig. 1. It very well may be that a is self- antipodal,meaningthata = a∗;infact,a = a∗ if and only if a is equidistant from all the support planes A0,...,An of [v0,...,vn]. This includes the incenter of the simplex [v0,...,vn], the center of its inscribed sphere, as well as, for each k =0,...,n,apointintheintersectionofthehalfspaces 8JOHNC.BOWERSANDPHILIPL.BOWERS

(a) a is not a polar point for T even (b) a is a polar point for T with a∗ an- though it does not lie on any support tipodal to a,anda∗ is polar for T with line of T . a∗∗ = a.

Figure 1. Polar points for a given simplex T =[v0, v1, v2].

H0 Hk 1 Hk Hk+1 Hn,whereHj is the halfspace bordered ∩···∩ − ∩ ∩ ∩···∩ by Aj that contains vj and Hj is the halfspace bordered by Aj opposite vj. Note that if [v0,...,'vn]isaregularsimplex,theincenterof[v0,...,vn]isthe only one of these self-antipodal' points that is equidistant from the vertices of [v0,...,vn]. We will use this fact in the proof of Theorem 3. A metric space X is an extension of the metric space Y if Y embeds isometrically in X. n Theorem 2. Let V = v0,...,vn be an affine basis for E and let a be n { } any element of E other than one of v0,...,vn. Then there is an extension of V a to a metric space X with n +3 elements that fails to embed in En isometrically∪{ } while every one of its subsets with at most n +2 elements does so embed if and only if a is a polar point for [v0,...,vn] that does not lie in any of the n +1 support planes Aj,forj =0,...,n;moreover,X is unique up to isometry.

Proof. We make the following preliminary claim. If a is a polar point for [v0,...,vn]withantipodalpair(a, a∗), then a lies on the support plane Aj, for some j =0,...,n,ifandonlyifa∗ = vj.Indeed,observethata lies on Aj if and only if aj = a.Thisimplies,sincetheset a0,...,an has exactly n +1points,thata lies on at most one support plane.{ Assume} then that a lies on Aj for some j =0,...,n,sothataj = a.Thenforanyindexk = j, ̸ ak = a,andthisimpliesthatthesupportplaneAk is the perpendicular ̸ bisector of the two distinct points a and ak.Sinceaa∗ = aja∗ = r = aka∗, we have a Ak.Hence ∗ ∈ a∗ Ak : k =0,...,j 1,j +1,...,n = vj . ∈ { − } { } ( AMENGERREDUX 9

For the reverse direction, assume that a∗ = vj for some j =0,...,n and that a Aj.Thena Ak for any index k = j,andthisimpliesthat ̸∈ ∗ ∈ ̸ a a = a ak = r = a aj,Sincea Aj,thesupportplaneAj is the perpen- ∗ ∗ ∗ ̸∈ dicular bisector of the distinct points a and aj,andsincea∗a = r = a∗aj, n we conclude that a Aj.Butthena Ak = ,acontradiction. ∗ ∈ ∗ ∈∩k=0 ∅ Therefore a Aj.Thisprovesthepreliminaryclaim. ∈ ( )Assumethata is a polar point for [v0,...,vn]thatdoesnotliein ⇐ any of the n +1 supportplanes Aj,forj =0,...,n.Asapointset,let X be any set with precisely n +3elementsandnamethedistinctelements n as y0,...,yn, a, a .LetY = y0,...,yn .Definethefunctionρ : X E ∗ { } → by ρ(yk)=vk, ρ(a)=a,andρ(a∗)=a∗.Sincea does not lie on any support plane Aj,neithera nor a∗ is one of the vertices of V .Itfollows that ρ is bijective between Y a and V a and between Y a and ∪{ } ∪{ } ∪{ ∗} V a∗ .DefinedistancesamongthepointsofX by xy = ρ(x)ρ(y), the Euclidean∪{ } distance between ρ(x)andρ(y), provided x, y = a, a ,and { }̸ { ∗} aa∗ = r = r(a, v0,...,vn). We will show that (i)everysubsetofX with n +2 elementsembedsin En preserving the distances defined among the points of X,and(ii) X does not embed isometrically in En.Notethatitem (i)impliesthatthedistancesassignedonX define a metric. For (i), first note that the restrictions of ρ to both the sets Y a and to n ∪{ } Y a∗ are embeddings into E that preserve the distances defined among the∪ points{ } of X.Considerthensubsetswithn +2pointsthatcontainboth a and a∗.Foranyj =0,...,n,therestrictionofthedistancefunctionto the set Xj = X yj is just the pull back to Xj of the Euclidean metric −{ } on Vj a , aj .Tobemorespecific,letf be the mapping on Xj defined ∪{ ∗ } by f(x)=ρ(x)forx Yj a∗ and f(a)=aj.AnapplicationofLemma Bandthedefinitionofthedistancesdefinedamongthepointso∈ ∪{ } f X imply that xy = f(x)f(y)forallx, y Xj. We now verify (ii). Suppose∈ that g is an isometric embedding of X into En.BycomposingwithanisometryofEn if necessary, Lemma C implies that we may assume that g(yk)=vk for k =0,...,n.Twoapplications of Lemma A imply that g(a)=a and g(a∗)=a∗.Thisimplies,forevery k =0,...,n,thataka = r = aa = g(a)g(a )=aa ,sothata Ak ∗ ∗ ∗ ∗ ∗ ∈ for each k =0,...,n,sinceAk is the perpendicular bisector of the distinct n points ak and a.Butthisisimpossiblesince k=0Ak = . As Y a is isometric with V a ,wehaveshownthat∩ ∅ X is an extension of V ∪a{ with} the desired properties.∪{ } ( ∪)Let{ } X be a metric extension of V a with n +3 elementsthatfails to embed⇒ in En isometrically while every∪ one{ of} its subsets with at most n+2 elements does so embed. As above, let Y = y0,...,yn , X = Y a, b ,and ρ : Y a V a an isometric bijection{ with ρ(Y )=} V .ByLemmasA∪{ } and C∪ and{ } the→ fact∪{ that} Y b admits an isometric embedding in En,there is a unique point b En such∪{ that} the extension of ρ to X via ρ(b)=b is an ∈ isometry on Y b .Weclaimthata is a polar point for [v0,...,vn]andthat ∪{ } b = a∗ is antipodal to a.Byhypothesis,ρ is not an isometry and this implies 10 JOHNC.BOWERSANDPHILIPL.BOWERS that ab = ab.LemmasA,B,andCusedasintheprecedingparagraphs ̸ imply that, for each j =0,...,n,themapgj : Vj a, b defined by gj = ρ ∪{ } on Vj b while gj(a)=aj is an isometry. It follows that ajb = ab r for ∪{ } ≡ j =0,...,n,sothatthepointsa0,...,an all lie on the sphere centered at b of radius r.Iftheset a0,...,an is affinely independent, then the claim { } that a is polar and a = b is confirmed. Assume then that a0,...,an fails ∗ { } to be affinely independent and let H be the affine span of a0,...,an ,an { } affine plane of dimension at most n 1. Note that the vector uj = aj a − − is a normal vector to the support plane Aj of the n-simplex [v0,...,vn], n and therefore the vectors u0,...,un span E .Nowifa were an element of H,then u0,...,un would be contained in the linear subspace H a { } − of dimension less than n,contradictingthefactthatthevectorsu0,...,un n span E .Weconcludethata H,andhence,sincethepointsa0,...,an all lie on a sphere in H and a ̸∈H,thereisasphereSn 1(c,R)inEn that ̸∈ − contains the point a as well as the points a0,...,an.Itnowfollowsthat, since each support plane Aj is the perpendicular bisector of the pair a and aj,thecenterc of this sphere is an element of each Aj,butthisisimpossible n as Ak = .Weconcludethenthat a0,...,an is affinely independent, ∩k=0 ∅ { } a is a polar point for [v0,...,vn], and b = a∗ is antipodal to a.Nowwe may observe that a is not in any support plane Aj,forotherwisethefirst paragraph of this proof would imply that b = a∗ = vj,contradictingthatρ is bijective between Y b and V b . The uniqueness claim∪{ follows} easily∪{ since} the proof thus far implies that any metric space extending V a as required must have its distances ∪{ } determined by the pair (a, a∗)asdescribedintheprecedingparagraph. ! Remark. It is a consequence of the proof of the ( )caseofTheorem2 ⇒ that when a is a polar point for [v0,...,vn]thatdoesnotlieonanyofthe n +1supportplanesAj,forj =0,...,n,thena∗ is also a polar point for [v0,...,vn]thatliesonnosupportplaneanda∗∗ = a.Inthiscase,wesay that a and a∗ form a polar pair for, and are antipodal with respect to, the 7 simplex [v0,...,vn]. We are now in a position to improve upon Theorem 1. Theorem 3. Let X be a metric space for which every subset with at most n +2 elements isometrically embeds in En. Then either X isometrically embeds in En,orX has precisely n +3 elements and X contains a subset with n +1 elements that embeds isometrically onto an affinely independet subset of En. Proof. We assume that X fails to embed isometrically in En even though all of its subsets with at most n +2 elementsdoessoembed. Itfollowsthat X has at least n +3 elements. Exactlyasinthefirsttwoparagraphsofthe n proof of Theorem 1, let Y = y0,...,ym = y0,...,ym = V E ,where { } ∼ { } ⊂ 7See the webpage [2] for interactive applets that manipulate polar pairs for n-simplices n in E ,forn =1, 2, 3. AMENGERREDUX 11 m n,beasubsetofX with a maximum number of elements that embeds isometrically≤ onto an affinely independent subset of En,letA be the affine n span of V ,anddefinethefunctionρ : X A E so that ρ(yk)=yk for → ⊂ k =0,...,m and ρ(x)=ρx(x)forx X Y .Thoughρ fails to be an ∈ − isometry, for each x X Y ,therestrictionofρ to Yx = Y x ,viz., ∈ − ∪{ } ρx : Yx A,isanisometry. We now→ prove that m = n.Ifnot,thenm

8We will show in the next paragraph that the self-antipodal point x must be the incenter of the n-simplex [y0,...,yn]. 12 JOHNC.BOWERSANDPHILIPL.BOWERS follows that the (n 2)-dimensional affine plane Ai,j and the 2-dimensional − affine plane Bi,j,theaffinespanofa, b,andc,meetatasinglepoint n .NowAi,j may be described as z E : az = bz = cz ,andthis implies,O since a, b,andc are the vertices{ ∈ of an equilateral triangle,} that the (n 2)-dimensional affine plane Ai,j and the 2-dimensional affine plane Bi,j are− orthogonal with the single point of intersection ,whichistheincenter of the 2-simplex [a, b, c]. This says that En decomposesO as an orthogonal n direct sum E = Ai,j Bi,j,whichimpliesthatVi along with at least one of a, b,andc is affinely⊥ independent. In turn this implies that, if the set n Yi a, b, c were to embed isometrically into E ,thenusingLemmaC,τi,j ∪{ } could be extended to an isometric embedding of Yi a, b, c by τi,j(yj)=yj. ∪{ } The fact that ayj = byj = cyj then would imply that yj Ai,j.Thiswould ∈ contradict the fact that the collection Vi,beinganaffinelyindependentset of n elements, spans an (n 1)-dimensional affine plane and cannot be − contained in the (n 2)-dimensional affine plane Ai,j.ThusYi a, b, c , − n ∪{ } and similarly Yj a, b, c ,cannotembedisometricallyinE ,andtherefore ∪{ } both are metric extensions of Vi,j a, b, c that fail to embed isometrically in En.TheuniquenessofTheorem2nowappliestoensurethatforan∪{ } y index k not equal to i nor j, yiyk = yjyk,andalsoayi = ayj, byi = byj,and cyi = cyj.Becausei and j are arbitrary indices, we conclude that all the distances among the n +1pointsof Y are equal to a common positive value n En ε,sothatthesetV is the vertex set of a regular n-simplex Σε in ,and that all the distances xy for x a, b, c and y Y have a common value ζ.Sincetheincenterofaregularsimplexistheonlyself-ant∈{ } ∈ ipodal point that is equidistant from the vertices of [y0,...,yn], we conclude that x is n the incenter of the regular simplex Σε and hence ζ is the circumradius of n Σε while the polar diameter r = ab = bc = ca is 2δ,whereδ is the inradius n of Σε . The preceding paragraphs have uncovered the metric structure of Y n ∪ a, b, c . Y is isometric to the vertex set of a regular n-simplex Σε of side- length{ }ε, a, b, c is isometric to the vertices of a regular 2-simplex (equilat- { 2} n eral triangle) Σ2δ of side-length 2δ,whereδ is the inradius of Σε ,andthe distance from any point of Y to any of a, b,orc is the distance ζ from the in- n n center of Σε to any of its vertices, the circumradius of Σε .Armedwiththese facts, we are ready to derive a contradiction using the embedding τi,j of the immediately preceding paragraph. Specifically, set i =0andj =1.Theset n Y0,1 a, b, c = y2,...,yn, a, b, c embeds isometrically in E via τ0,1 with ∪{ } { } image y2,...,yn, a, b, c .RecallthattheaffinespansA0,1 of y2,...,yn { } { } and B0,1 of a, b, c are orthogonal with a single point of intersection ,the incenter of the{ 2-simplex} [a, b, c]. From the Exercise of the Overture, theO cir- cumradius of [a, b, c]iss =(2δ/√2)ζ2 = εδnζ2,sinceδ =(ε/√2)δn.Since the distance from a to any point of y2,...,yn is constant, the points { } y2,...,yn lie on a sphere in A0,1 centered at .Butsincethepoints { } O y2,...,yn are the vertices of a regular (n 2)-simplex of side-length ε, { } − this sphere must be the circumscribing sphere of radius R =(ε/√2)ζn 2. − AMENGERREDUX 13

Consider now the triangle with vertices a, yn,and :ithasarightangleat O and side-lengths a = s, yn = R,andayn = ζ =(ε/√2)ζn.Anappli- O O O 2 2 2 2 2 2 2 cation of the Pythagorean Theorem gives s +R = ζ ,or2δnζ2 +ζn 2 = ζn. The Exercise of the Overture then implies that − 4 1 n 2 n + − = . 3 n(n +1) n 1 n +1 − Straightforward algebra applies to show that this latter equation has a unique solution, n = 2, and this contradicts that the dimension n 1 of Euclidean space En −satisfies the equation. ≥ Case 2: The set Y b, c does embed isometrically in En. In this case the restriction of ρ to∪Y{ } b, c is an isometric embedding. Let a = ρ(a), b = ρ(b), and c = ρ(c). If∪Y{ a,} c were to fail to embed isometrically, then ∪{ } the uniqueness of Theorem 2 would imply that ykb = ykc for k =0,...,n, which would imply by Lemma A that ρ(b)=b = c = ρ(c), contradicting that ρ isometrically embeds b, c .Thusρ isometrically embeds both sets Y a, c and Y b, c .Wewillshowthatthisfact—thatboththesesets{ } with∪{n +3} elementsdoembedisometrically—impliesthat∪{ } Y a, b, c nec- essarily embeds in En isometrically, a contradiction that finishes∪{ the proof.} For the moment we make two assumptions that are addressed in the next paragraph: (i)the(n +1)-elementset y1 ...,yn, c is affinely inde- { } pendent, and (ii)then-element set y3,...,yn, b, c is affinely indepen- dent and, hence, spans an (n 1)-dimensional{ affine} plane A.Byhy- − n pothesis, the (n +1)-elementset S = y3,...,yn, a, b, c embeds in E isometrically, and by Lemma C we may{ assume the isometric} embedding µ : S En satisfies µ = ρ on S a with µ(a)=a .Leta be the → −{ } ′ ′′ isometric reflection of a′ through A.ByhypothesisandLemmaC,the three (n +2)-elementsets Sk = yk,y3,...,yn, a, b, c , k =0, 1, 2, embed { } isometrically onto the respective sets yk, y3,...,yn,αk, b, c , k =0, 1, 2, n { } for some αk E .ByLemmaB,since y3,...,yn, b, c is affinely indepen- ∈ { } dent, each αk is either a′ or a′′.Itfollowsthatforatleasttwovaluesof k =0, 1, 2, the αk’s agree; i.e., without loss of generality, we may assume that α1 = α2 = a ,andthisimpliesthat y1,...,yn, a, b, c embeds isometrically ′ { } onto y1,...,yn, a , b, c . By Lemma A, since ρ restricts to Y a, c as { ′ } ∪{ } an isometric embedding and y1 ...,yn, c is affinely independent, the dis- { } tances of a to the points of y1,...,yn,c uniquely determine ρ(a)=a.But { } we have seen that y1,...,yn, a, b, c is isometric with y1,...,yn, a , b, c , { } { ′ } so the distances of a to the points of y1,...,yn,c uniquely determine the point a .Thusa = a ,andthenρ isometrically{ embeds} Y a, b, c since ′ ′ ∪{ } then ab = a′b = ab.ThisofcoursecontradictsthatY a, b fails to embed in En isometrically. ∪{ } We now show that the two assumptions of the preceding paragraph al- ways hold for an appropriate ordering of the vertices of the n-simplex σ = n [y0,...,yn]. Indeed, since k=0Ak = ,thereisatleastonesupportplane of σ that does not contain∩ the point ∅c,andwithoutlossofgeneralitywe 14 JOHNC.BOWERSANDPHILIPL.BOWERS may assume that c A0.Assumption(i)followsimmediately.Nowsince n ̸∈ Ak = y0 and c = y0,thereisanothersupportplanenotcontaining ∩k=1 { } ̸ c,andwithoutlossofgeneralitywemayassumethatc A1.LetE be the ̸∈ affine span of the set y2,...,yn, c ,an(n 1)-dimensional affine plane. If { } − b E,thenassumption(ii)holdssincethenthelargerset y2,...,yn, b, c is̸∈ affinely independent. If b E,thenthesameargumentappliedabove{ } to c and the n-simplex σ in E∈n applied now to b and the (n 1)-simplex − τ =[y2 ...,yn, c]inE implies that there are at least two support planes of τ in the (n 1)-dimensional Euclidean space E that do not contain b. These two affine− planes are opposite two of the vertices of τ,andthisimplies that one of these vertices must be from the list y2,...,yn,sowithoutloss of generality we may assume that b does not lie on the support plane of τ opposite y2.Assumption(ii)isnowautomatic. !

Theorem 2 offers in its corollary below the claimed improvementupon Theorem 1 as long as X is not a metric space of cardinality n +3. Corollary. Let X be a metric space whose cardinality is not equal to n +3. Then X embeds isometrically in the Euclidean space En if and only if each of its finite subsets with at most n +2 elements embeds isometrically in En. We summarize in the next corollary facts uncovered in this section about all the finite metric spaces with precisely n +3elementsthatfailtoembed isometrically in En,eventhougheverysubsetwithatmostn +2elements does so embed. This provides a characterization up to isometry of all such metric spaces. Indeed, all such spaces up to isometry are obtained from the vertex set of an n-simplex in En and a polar pair. All distances are the Euclidean ones, except for that between the points of the polar pair, and that distance is the polar diameter of the pair. Corollary. Let X be a metric space that fails to embed isometrically in En though all of its subsets with at most n +2 elements does so embed. Then X has precisely n +3 elements and any subset of X of cardinality n +1 isometrically embeds in En as an affine basis. Moreover, for any two points a and b of X,thereexistsamappingρ : X En that restricts to isometric embeddings of X a and X b ,andthen→ ρ(a) and ρ(b) form a polar pair for the n-simplex−{ }σ spanned− by{ } the points of ρ(X a, b ). The points ρ(a) and ρ(b) do not lie on any of the n+1 support planes−{ of σ}. The distance ab = r,thepolardiameteroftheset ρ(a),ρ(a)∗ ,whereρ(a)∗ = ρ(b),while the remaining distances in X are xy{= ρ(x)ρ(y)},when x, y = a, b .For { }̸ { } any j =0,...,n,thepolardiameterisr = ρ(a)jρ(a)∗ = ρ(a)j ρ(b),where ρ(a)j is the isometric reflection of ρ(a) through the (n 1)-dimensional − support plane Aj of the n-simplex σ. Now that we have characterized those spaces X with the desired embed- ding properties into En,howcommonarethey?Theansweriscontainedin the next proposition, whose proof is left to the reader. AMENGERREDUX 15

n Proposition. Let [v0,...,vn] be an n-simplex in E . The set of points a n that are polar to [v0,...,vn] is a dense open subset of E whose complement has measure zero.

Act 3: Cayley-Menger determinants and isometric embedding. Acts 1 and 2 have reduced the question of the isometric embeddability into En of the metric space X with more than n +3elementstothequestionof the isometric embeddability of each subset of X with n+2 elements. Menger in his 1931 paper [7] gave a purely algebraic invariant for an (n +2)-element metric space that encodes isometric embeddability into En.Todescribethis we need the classical formula for the volume of a simplex.

Theorem 4. The n-dimensional volume vn of an n-simplex [v0,...,vn] in En is 1 v = v ([v0,...,v ]) = det V , n n n n! where, for k =0,...,n,thekth row of the (n +1)) )(n +1) square matrix ) ×) V consists of the n coordinates of the vertex v)k augmented) by a ‘1’inthe (n +1)st position. The reader may find many references to this in the literature and perhaps one of the shortest proofs is given by P. Stein [10]. Since our desire is to give proofs of Menger’s results using geometric arguments at the undergraduate level, we give a quick geometric proof, even shorter than Stein’s. Proof. First we derive the n-dimensional volume of a cone C over an (n 1)- n− dimensional base B.Lete1,...,en be the standard unit basis of E and n 1 A = hen + E − ,the(n 1)-dimensional affine plane at height h 0above − n 1 ≥ the coordinate hyperplane E − =span[e1,...,en 1]. Let B be a subset − of A with (n 1)-dimensional volume vn 1 = vn 1(B), and let C = 0 B, the cone over−B with vertex 0,theoriginof− En.Now− C is just the union∗ of the line segments in En between the origin and the points of B;precisely, C = tb :0 t 1, b B = 0 t 1tB,wheretB = tb : b B . For each{ 0 ≤t 1,≤ since∈ all distances} ∪ ≤ ≤ are dilated by a factor{ of t by∈ the} ≤ ≤ n 1 dilation x tx,wehavevn 1(tB)=t − vn 1(B). The ‘volume by slicing’ technology-→ of elementary calculus− implies that− h h n 1 x x − 1 vn(C)= vn 1 B dx = vn 1(B) n 1 dx = vn 1(B)h. 0 − h − 0 h − n − * + , * The classical formula for the volume of the n-simplex now follows by induction. Indeed, the basis of the induction, n =1,holdstrivially.Assume the formula holds for any (n 1)-simplex in En 1 and let h be the distance − − from v0 to the 0th support plane A0 of σ =[v0,...,vn], an n-simplex in n E .Letσ0 =[v1,...,vn]bethefaceofσ opposite v0.Firstmakethe simplifying assumption that A0 =span[e2,...,en]andv0 = he1.ThenV, the matrix in the statement of the theorem, has h, 0,...,0, 1asitsfirstrow and h, 0,...,0asitsfirstcolumn,andthen n minor V0 of V obtained by × 16 JOHNC.BOWERSANDPHILIPL.BOWERS striking out the first row and column is the matrix in the statement of the theorem for the simplex σ0 in the (n 1)-dimensional Euclidean space A0. − It follows from this and the inductive hypothesis that det V = h det V0 = (n 1)!hvn 1(σ0). Since σ = v0 σ0,theresultoftheprecedingparagraph − − 1 ∗ 1 implies that vn(σ)=n− vn 1(σ0)h =(n!)− det V,verifyingtheinductive step. − n To finish the proof, since any n simplex [v0,...,vn]inE may be moved so that A0 =span[e2,...,en]andv0 = he1 by a rigid motion, all we need do is verify that det V is invariant under any translation of the vertices by avectora and under the application of any n n orthogonal matrix On. This is left as a nice exercise in the manipulation× of determinants, using that det On =1,detAB =detA det B,anddetA =0whenonecolumnis amultipleofanother. · ! Menger’s clever idea is to manipulate the (n+1) (n+1) square matrix V to obtain an associated (n+2) (n+2)squarematrix× D whose determinant × expresses the squared volume of the n-simplex [v0,...,vn]entirelyinterms of dij = vivj,0 i, j n,thedistancesamongtheverticesofthesimplex. Noting that Cayley≤ in≤ his first paper used this same determinant in dis- cussing finite subsets of E3,Blumenthal[1]termeddetD the Cayley-Menger determinant.Firstconsiderthetwomatrices

v01 ... v0n 10 v01 ... v0n 01 ...... A = ⎛ . . . . .⎞ and B = ⎛ . . . . .⎞ , vn1 ... vnn 10 vn1 ... vnn 01 ⎜ ⎟ ⎜ ⎟ ⎜ 0 ... 001⎟ ⎜ 0 ... 010⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ where vk =(vk1,...,vkn)fork =0,...,n,andnotethatdetA =detV = det B.TheproductofA with the transpose of B is − 2 2 v0, v0 ... v0, vn 1 ⟨ . ⟩ . ⟨ . ⟩ . ABtr ⎛ . .. . .⎞ = 2 2 vn, v0 ... vn, vn 1 ⎜⟨ ⟩ ⟨ ⟩ ⎟ ⎜ 1 ... 10⎟ ⎜ ⎟ and the Cayley-Menger matrix⎝ is ⎠ 2 2 d00 ... d0n 1 ...... D = D(v0,...,vn)=⎛ ⎞ . d2 ... d2 1 ⎜ n0 nn ⎟ ⎜ 1 ... 10⎟ ⎜ ⎟ ⎝ ⎠2 In the Cayley-Menger determinant, det D,substitutingd = vi vj, vi ij ⟨ − − vj = vi, vi + vj, vj 2 vi, vj ,thenfori, j =0,...,n,multiplying ⟩ ⟨ ⟩ ⟨ ⟩− ⟨ ⟩ the last row by vi, vi and subtracting from the ith row, then multipying ⟨ ⟩ the last column by vj, vj and subtracting from the jth column, gives ⟨tr ⟩ +1 2 det D =( 2)n det AB =( 1)n 2n(det V) .CouplingthiswithTheorem4 − − AMENGERREDUX 17

n gives the squared-volume of the n-simplex [v0,...,vn]inE entirely in terms of the distances among its vertices as n+1 2 2 ( 1) v = vn([v0,...,vn]) = − det D. n 2n(n!)2 2 D n+1 Since vn 0, the sign of the determinant det is ( 1) .Noticethatvn = ≥ n − 0=detD for the points v0,...,vn in E if and only if the set v0,...,vn fails to be affinely independent. { } If X is any metric space of cardinality r,theCayley-Menger determinant for X is 2 2 d11 ... d1r 1 . . . . D D ) . .. . .) det =det (X)=) 2 2 ) , )d 1 ... d 1) ) r rr ) ) 1 ... 10) ) ) ) ) where x1,...,xr is some ordering of the) points of X and) dij = xixj.Note that det D is independent of the ordering) of the points of )X and, obviously, is an isometric invariant of finite metric spaces. This implies the following. If X is a metric space that isometrically embeds in En via the isometry ρ,then for any finite subset Y X with, say, r distinct points, the Cayley-Menger determinant det D(Y )eithervanishesorhasthesign(⊂ 1)r.Moreover,the determinant vanishes if and only if ρ(Y )isaffinelydependent,whichalways− occurs when r>n+1.Thisgivesanecessaryconditionforn +2pointsof X to embed isometrically in En,andMengershowedittobesufficient. Theorem 5. The (n +2)-point metric space X embeds isometrically in En if and only if, for each subset Y X, the Cayley-Menger determinant ⊂ Y det D(Y ) either vanishes or has the sign ( 1)| |,where Y is the cardinality of Y ,anddet D(X)=0. − | | Necessity has been demonstrated. The proof of sufficency is by induction and involves a detailed algebraic analysis. The authors offer no geometric proof of sufficiency and refer the reader to Menger’s original argument in [7]. We close this act by condensing all three fundamental resultsofMenger into a characterization of those metric spaces isometrically embeddable in Euclidean space, as he did in [7]. Metrical Characterization of Euclidean Sets. AmetricspaceX em- beds isometrically in En if and only if, when X contains more than n +3 points, then (i) for every Y X with precisely r n +1 points, the Cayley-Menger determinant⊂det D(Y ) either vanishes≤ or has the sign ( 1)r,and (ii) the determinant associated to each n+2 distinct points of−X vanishes; and when X contains exactly n +3 points, in addition to these conditions, (iii) det D(X)=0. 18 JOHNC.BOWERSANDPHILIPL.BOWERS

Postlude: Distance and Metric Geometry. Menger’s 1931 results may be counted as the modern starting point for the discipline of distance geometry,9 adisciplinethatstudiessetsbased on given distances between pairs of its points. Special interest rests on the relation of general metric spaces to the metric spaces of the classical —spherical and elliptic, Euclidean, and hyperbolic spaces. More recently distance geometry has found applications to a greatvarietyoffields, including protein structure, molecular structure, wireless sensor networks, multidimensional scaling, and graph rigidity. Liberti et al. [5] is a fantas- tic reference source for these and other applications. Distance geometry has close affinity to and not insignificant overlap with the discipline of metric ge- ometry,10 which studies the metric relationships that characterize segments and lines and and surfaces, and today finds expression in the study of metric and length spaces, with applications in geometric theory and . The classic reference for distance geometry is Leonard Blumenthal’s 1953 treatise Theory and Applications of Distance Geometry [1] and a modern reference for the still-developing discipline of metric geometry is Dmitri Burago, Yuri Burago, and Sergei Ivanov’s 2001 treatise A Course in Metric Geometry [3]. Aside from the original articles [6] and [7], detailed proofsofMenger’s results appear in Chapter IV of Blumenthal [1], whose approach is that of Menger’s with proofs that are more algebraic and combinatorial than ours. Blumenthal goes much further in that he develops the corresponding iso- metric embedding results for both spherical and hyperbolic spaces, as well as Hilbert spaces. Of interest also is C.L. Morgan [8], where the problem of embedding metric spaces into Euclidean spaces is addressed entirely al- gebraically. There Morgan gives determinant conditions, similar in flavor to Menger’s use of the Cayley-Menger determinant, to give necessary and sufficient algebraic conditions based on the distances among points for a metric space to admit a Euclidean embedding. This is a little different from Schoenberg’s [9] earlier algebraic approach for finite metric spaces that gives a condition on the the matrix of squared distances that is necessary and sufficient for embedding into Euclidean space.

9Ancient Greek mathematics made modest contributions to the discipline, but the subject lay dormant until Arthur Cayley’s contributions in the nineteenth century. Menger is the successor of Cayley in the development of the discipline and his work inspired agenerationofgeometerstodevelopthetheorytomaturityinthemiddleofthelast century. 10The early pioneers in this development, no doubt influenced byMenger,werethe German-American mathematician Herbert Busemann and the Russian mathematician Aleksandr Aleksandrov, working in the 1940’s and 1950’s. Mikhail Gromov, along with Eliyahu Rips, revived interest in their work in the 1980’s in the early development of geometric group theory,oneofthemostsignificantdevelopmentsingeometrictopology of the last half century, and one that enjoys intense researchactivityamonggeometric topologists today. AMENGERREDUX 19

References [1] L. M. Blumenthal, Theory and Applications of Distance Geometry,OxfordUniv.Press, Oxford, 1953. [2] B. Braswell, J. C. Bowers, Menger embeddings in Euclideanspace,(2016), https://w3.cs.jmu.edu/bowersjc/page/menger/. [3] D. Burago, Y. Burago, S. Ivanov, ACourseinMetricGeometry,GraduateStudiesin Mathematics, Vol. 33, Amer. Math. Soc., Providence, 2001. [4] L. Liberti, C Lavor, Six mathematical gems from the history of distance geometry, In- ternational Trans. in Oper. Res. (Special Issue on Distance Geometry) 30 (Apr. 2015), DOI: 10.1111/itor.12170. [5] L. Liberti, C. Lavor, N. Maculan, A. Mucherino, Euclideandistancegeometryand applications, SIAM Review 56 no. 1 (2014), 3–69. [6] K. Menger, Untersuchungen ¨uber allgeneine Metrik?, Math. Ann. 100 (1928), 75–163. [7] K. Menger, New foundation of Euclidean geometry, Amer. J. of Math. 53(4) (1931), 721–745. [8] C. L. Morgan, Embedding metric spaces in Euclidean space, J. Geometry 5(1) (1974), 101–107. [9] I. J. Schoenberg, Sur la d´efinition axiomatique d’une classe d’espace distancis vecto- riellement applicable sur l’espace de Hilbert, Ann. of Math. 36 (1935), 724–732. [10] P. Stein, A note on the volume of a simplex, Amer. Math. Monthly 73 no. 3 (1966), 299-301.