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Norms of Minimal Projections

arXiv:math/9211211v1 [math.FA] 17 Nov 1992Norms of Minimal ProjectionsHermann K¨onig1Mathematisches SeminarUniversit¨at KielKiel, GermanyNicole Tomczak-Jaegermann2Department of MathematicsUniversity of AlbertaEdmonton, Alberta, CanadaAbstractIt is proved that the projection constants of two- and three-dimensio-nal spaces are bounded by 4/3 and (1 +√5)/2, respectively. Thesebounds are attained precisely by the spaces whose unit balls are theregular hexagon and dodecahedron.

In fact, a general inequality forthe projection constant of a real or complex n-dimensional space isobtained and the question of equality therein is discussed.1,2 During the work on this paper both authors were partially supported by NATOCollaborative Research Grant CRG 920047.

1Introduction and the main resultsIn this paper we prove results on upper estimates for the norms of mini-mal projections onto finite-dimensional subspaces of Banach spaces, whichare optimal in general. By Kadec–Snobar [KS], onto n-dimensional spacesthere are always projections of norm smaller than or equal to √n.

Gen-eral bounds for these so-called projection constants were further studied byvarious authors, including Chalmers, Garling, Gordon, Gr¨unbaum, K¨onig,Lewis and Tomczak-Jaegermann ([GG], [G], [KLL], [KT], [L], [T]). Someother aspects of minimal projections, like the existence or norm estimatesfor concrete spaces, were investigated by many authors, among them e.g.,Chalmers, Cheney, Franchetti ([CP], [IS], [FV]).In [KT] a very tight formula for the projection constants of spaces withenough symmetries was shown.

We now prove that this formula holds forarbitrary spaces, and study cases of equality.The formula yields, in particular, that the projection constant of any real(resp. complex) 2-dimensional space is bounded by 4/3 (resp.

(1 +√3)/2).Up to isometry, there is just one space (in each case) attaining the bound.The values for 3-dimensional spaces are (1 +√5)/2 (resp. 5/3).

In the realcase, the unique extremal spaces are those whose unit balls are the regularhexagon and the regular dodecahedron. The 4/3-result solves a problem ofGr¨unbaum [G].

A proof of this fact has also been announced by Chalmers etal. [CMSS]; it is our understanding that their argument is incomplete as ofnow.The authors would like to thank to J. J. Seidel for valuable remarksconcerning equiangular lines.We use standard Banach space notation, see e.g., [T.2].

By IK we denotethe scalar field, either the real numbers IR or the complex numbers C . Therelative projection constant of a (closed) subspace E of a Banach space X isdefined byλ(E, X) := {∥P∥| P : X →E ⊂X is a linear projection onto E},the (absolute) projection constant of E is given byλ(E) := {λ(E, X) | X is a Banach space containing E as a subspace}.

(1.1)2

Any separable Banach space E can be embedded isometrically into l∞.For any such embedding, λ(E) = λ(E, l∞), i.e., the supremum in (1.1) isattained. We can therefore restrict our attention to finite-dimensional sub-spaces E ⊂l∞.

Also note that λ(ln∞) = λ(l∞) = 1.Let n ∈IN be a positive integer, ⟨·, ·⟩denote the standard scalar productin IK n and let ∥· ∥2 =q⟨·, ·⟩. For N ∈IN, vectors x1, .

. .

, xN ∈IK n spanninglines in IK n are called equiangular provided that there is 0 ≤α < 1 such that∥xi∥2 = 1 and |⟨xi, xj⟩| = α for i ̸= j, i, j = 1, . .

. , N.PutN(n) :=(n(n + 1)/2if IK = IRn2if IK = C .

(1.2)By Lemmens–Seidel [LS] and Gerzon, in IK n there are at most N(n) equian-gular vectors. (Indeed, the hermitian rank 1 operators xi ⊗xi are linearlyindependent in a suitable real linear space of operators.) This bound is at-tained for n = 2, 3, 7, 23 if IK = IR and for n = 2, 3 if IK = C .

If the bound isattained, necessarily α = 1/√n + 2 if IK = IR and α = 1/√n + 1 if IK = C .Our main result isTheorem 1.1 (a) The projection constant of any n-dimensional normedspace En is bounded byλ(En) ≤(2 + (n −1)√n + 2)/(n + 1)in the real case,(1 + (n −1)√n + 1)/nin the complex case. (1.3)(b) Given IK and n ∈IN, there exist n-dimensional spaces En for whichthe bound is attained if and only if there exist N(n) equiangular vectors inIK n. In this case, such a space En can be realized as an isometric subspaceof lN(n)∞, and the orthogonal projection is a minimal projection onto En.

(c) For IK = IR and n = 2, 3, 7, 23, there are unique spaces En (up toisometry) attaining the bound (1.3); for IK = C and n = 2, 3 such spacesalso exist. For IK = IR and n = 2, 3 the unit balls of En are the regularhexagon and the regular dodecahedron, respectively.Remarks(i) The right hand side of (1.3) equals the bound f(n, N(n))derived in [KLL] for the relative projection constant of an n-dimensionalspace in an N(n)-dimensional superspace.3

(ii) The bounds in (1.3) are of the order √n −1/√n + 2/n if IK = IRand √n −1/2√n + 1/n if IK = C , for large n ∈IN.To prove just (1.3) it would suffice (by approximation) to consider poly-hedral spaces E ⊂lN∞for an arbitrary N ∈IN (the “finite” case); in whichcase the proofs of most of the results which follow can be simplified. Forthe examination of the equality in (1.3) and the uniqueness we need, how-ever, the general (“infinite”) case of E ⊂l∞as well, even though the spacesattaining the bound (1.3) turn out in the end to be polyhedral.To unify the notation in the finite and infinite case which we would liketo discuss simultaneously, we set T = {1, .

. .

, N}, for some N ∈IN, in thefinite case and T = IN, in the infinite case. In particular, l∞(T) denotes lN∞in the former case and l∞in the latter case.If µ = (µt)t∈T is a probablity measure on T, and 1 ≤p < ∞, we letlp(T, µ) := {(ξt)t∈T | ∥(ξt)∥p,µ = (Xt∈T|ξt|pµt)1/p < ∞}.For a subspace E ⊂l∞(T), we denote by Ep,µ the same space E consideredas a subspace of lp(T, µ), via the embedding l∞(T) →lp(T, µ).Finally, for N ∈IN, by RN : lp →lNp we denote the projection onto thefirst N coordinates, acting in an appropriate sequence space (1 ≤p ≤∞).Let n ∈IN.

The set Fn of all n-dimensional spaces, equipped with the(logarithm of the) Banach–Mazur distance, is a compact metric space, cf.e.g., [T.2]. The projection constant λ, as a function λ : Fn →IR +, is contin-uous with respect to this metric, and hence the supremum supE∈Fn λ(E) isattained: there is F ∈Fn withλ(F) = sup{λ(E) | E ∈Fn}.

(1.4)The proof of the bound (1.3) is based upon an estimate in terms of or-thonormal systems, which, in fact, is a characterization of the maximal pro-jection constant, and it seems to be of independent interest.Theorem 1.2 Let n ∈IN. ThenmaxE∈Fn λ(E) = supµ sup{fj}Xs,t∈IN|nXj=1fj(s)fj(t)|µsµt,(1.5)4

where the outside supremum runs over the set of all discrete probability mea-sures µ = (µt)t on IN and the inside supremum runs over all orthonormal sys-tems {fj} in l2(IN, µ). The double supremum in (1.5) is attained for some µand {fj} ⊂l2(IN, µ) ∩l∞.

In this case, the space E = span {f1, . .

. , fn} ⊂l∞has maximal projection constant.

The square function (Pnj=1 |fj(s)|2)1/2 isconstant µ-a.e. in the extremal case.In the extremal case the support of µ can be finite; and in dimensionsn = 2, 3 it is actually so.

The upper estimate in (1.5) relies on an idea ofLewis [L]. To prove Theorem 1.1, we then have to find an upper estimate forthe right hand side of (1.5).

In certain dimensions (n = 2, 3, and in the realcase additionally n = 7, 23), we find the exact value of (1.5); for other n ∈IN,the expression in (1.5) might be possibly used to slightly improve (1.3).2Projection constants and trace dualityIn this section, we prove the upper bound for max{λ(E) | E ∈Fn} in (1.5).The argument is based on trace duality. For the convenience of the generalreader, we try to use only basic Banach space theory.

The first lemma issimilar to Lemma 1 of [KLL].Lemma 2.1 Let E ⊂l∞(T) be a finite-dimensional subspace, where T ={1, . .

. , N}, or T = IN.

There exists a map u : l∞(T) →l∞(T) with u(E) ⊂Esuch thatλ(E) = tr (u : E →E)andXt∈T∥uet∥∞= 1.Here (et)t∈T denotes the standard unit vector basis in l∞(T).In fact, for any map u with u(E) ⊂E and Pt∈T ∥uet∥∞= 1 one hastr (u : E →E) ≤λ(E), see (??) below.Proof Since E is finite-dimensional, there exists a minimal projection ontoE, say P0 : l∞(T) →E ⊂l∞(T) with ∥P0∥= λ = λ(E) < ∞(cf.

[BC],[IS]). Let F(l∞, l∞) denote the space of finite-rank operators on l∞= l∞(T),equipped with the operator norm.

The setsA = {S ∈F(l∞, l∞) | ∥S∥< λ}5

andB={P ∈F(l∞, l∞) | P = P0 +mXi=1x∗i ⊗xifor some x1, . .

. , xm ∈E, x∗1, .

. .

, x∗m ∈E⊥⊂l∗∞, m ∈IN}are convex and disjoint, since B consists of projections onto E and ∥P∥≥λfor every projection P. Since A is open, by the Hahn–Banach theorem thereis a functional ϕ ∈F(l∞, l∞)∗of norm ∥ϕ∥= 1 such that ϕ(P0) ∈IR andfor S ∈A and P ∈B we haveRe ϕ(S) < λ ≤Re ϕ(P).By the trace duality, ϕ is represented by a map v defined on l∞(T).In the case T = {1, . .

. , N}, the operator norm of w ∈F(l∞, l∞) is justsupt∈T ∥w∗et∥1, so the dual norm isPt∈T ∥vet∥∞.

If T = IN, define a linearoperator v : l∞→l∗∗∞by ⟨v(x), x∗⟩= ϕ(x∗⊗x) for x ∈l∞and x∗∈l∗∞.Writing any S ∈F(l∞, l∞) as S = Pmi=1 x∗⊗xi, one finds that ϕ(S) = tr (v S),with the integral norm i(v) equal toi(v) =supS∈F(l∞,l∞)tr (v S)/∥S∥= 1.Let x∗∈E⊥, x ∈E. Then λ ≤Re ϕ(P0+x∗⊗x) = λ+Re tr (v (x∗⊗x)).Hence Re ⟨vx, x∗⟩≥0 for all x∗∈E⊥, x ∈E, which implies ⟨vx, x∗⟩= 0.Thus v(E) ⊂E⊥⊥= E ⊂l∞, in view of dim E < ∞.

Let Q : l∗∗∞→l∞be thecanonical projection onto l∞with ∥Q∥= 1. Let u := Q v : l∞→l∞.

Thenu(E) ⊂E and, since Qx = x for x ∈E, we have u P0 = v P0. Furthermore,i(u) ≤∥Q∥i(v) = 1 andλ(E) = λ = ϕ(P0) = tr (u P0) = tr (u : E →E).Let RN : l∞→lN∞be the natural projection and let uN := RNu : l∞→lN∞.

Then i(uN) ≤1 and, similar as in the case of T = {1, . .

. , N} discussedabove, this norm, being dual to the operator norm on F(lN∞, l∞), is equal toi(uN) = Pi∈T ∥uNet∥∞≤1.

Taking the limit as N →∞(first for finite sumsin t) we get that Pi∈T ∥uet∥∞≤1. In fact we have the equality.✷The following upper estimate is a consequence of Lemma 2.1 and reliesessentially on an idea of Lewis [L].6

Proposition 2.2 Let E ⊂l∞(T) be an n-dimensional subspace, where T ={1, . .

. , N}, or T = IN.

There is a discrete probability measure µ = (µt)t∈Ton T, ∥µ∥1 = 1 such that for any orthonormal basis (fj)nj=1 in E2,µ we haveλ(E) ≤Xs,t∈T|nXj=1fj(s)fj(t)|µsµt.Note that the double sum in this proposition is finite, since fj ∈l2(T, µ).Proof Let u : l∞(T) →l∞(T) be as in Lemma 2.1 and put µt = ∥uet∥∞, fort ∈T. Then µ is a probability measure on T,Pt∈T µt = 1.

For every N ∈IN,let uN = RNu : l∞→lN∞. Then we have ∥uN : l1(T, µ) →l∞(T)∥≤1.Indeed, the extreme points of the unit ball in l1(T, µ) are, up to a multipleof modulus 1, of the form et/µt for t ∈T and we have ∥uN(et/µt)∥∞=∥uN(et)∥∞/µt ≤1.Now consider E2,µ ⊂l2(T, µ) and fix an arbitrary orthonormal basis{fj}nj=1 in E2,µ.

Thenλ(E)=tr (u : E →E) = tr (u : E2,µ →E2,µ)=nXj=1⟨ufj, fj⟩l2(µ) = limN→∞nXj=1⟨uNfj, fj⟩l2(µ).The second equality is purely algebraic; for the last one use the fact that⟨RNg, h⟩tends to ⟨g, h⟩as N →∞, for all g, h ∈l2(T, µ). Hence, by Lewis’idea of how to use the bound for the norm of uN considered above, we haveλ(E)≤lim supN→∞Xt∈T|nXj=1uNfj(t)fj(t)|µt≤lim supN→∞Xt∈T∥uN(nXj=1fj(t)fj)∥∞µt≤Xs,t∈T|nXj=1fj(s)fj(t)|µsµt,as required.✷7

3Square function in an extremal caseAs a consequence of Proposition 2.2, given a space E ⊂l∞(T), an upperbound for λ(E) would follow from an upper estimate for the quantityφ(n, T) = supµ∈M sup{fj}Xs,t∈T|nXj=1fj(s)fj(t)|µsµt,(3.1)where the outside supremum runs over the set M of all discrete probabilitymeasures µ on T and the inside supremum runs over all orthonormal bases{fj} in E2,µ ⊂l2(T, µ).To estimate (3.1), we first show, using Lagrange multipliers, that thesquare function of an extremal system {f 0j } is constant µ-a.e.We will be mainly concerned with the situation when φ really increasesat the dimension n,φ(n1, T1) < φ(n, T)whenever n1 < n and T1 ⊂T. (3.2)Proposition 3.1 Let n ∈IN and let T = {1, .

. .

, N}, or T = IN satisfy(3.2). Assume that µ0 ∈M and an orthonormal system {f 0j }nj=1 in l2(T, µ)attains the supremumXs,t∈T|nXj=1f 0j (s)f 0j (t)|µ0sµ0t = φ(n, T).

(3.3)Then the square function f 0 is constant µ-a.e.,f 0(s) := nXj=1|f 0j (s)|21/2 =( √nif µ0s ̸= 00if µ0s = 0First notice that if µ0s = 0 for some s ∈T then f 0j (s) = 0 for j = 1, . .

. , n,hence also f 0(s) = 0.

Indeed, otherwise decreasing |f 0j (s)| would allow us tomultiply all the remaining |f 0j (t)| for t ̸= s, by ξ > 1, thus increasing the vof the sum in (3.3).Condition (3.2) implies that the matrix (f 0j (s)) does not split into a non-trivial block diagonal sum of smaller submatrices.8

Lemma 3.2 For all l, m = 1, . .

. , n we have∃l = l0, .

. .

, lρ = m ∀1 ≤r ≤ρ ∃s ∈T, µ0s ̸= 0f 0lr−1(s)f 0lr(s) ̸= 0. (3.4)ProofFor 0 < τ ≤1, by Mτ denote the set of all discrete measures µ onT such that µ(T) = τ.

By φ(n, T, τ) denote the corresponding supremum,analogous to (3.1), so that φ(n, T) = φ(n, T, 1).It is easy to check that φ(n, T, τ) = τφ(n, T, 1). Moreover, φ(n1, T1, 1) ≤φ(n, T, 1) if n1 ≤n and T1 ⊂T.Let J1 ⊂{1, .

. .

, n} be a maximal set such that (3.4) is satisfied for alll, m ∈J1 and let J2 = {1, . .

. , n}\J1 be the complement of J1.

Clearly, J1is non-empty. Let T1 ⊂T be the set of all s such that f 0j (s) ̸= 0 for somej ∈J1, let T2 = T\T1.

By the maximality of J1 and the definition of T1 wehavef 0j (s) = 0whenever(s, j) ∈(T2 × J1) ∪(T1 × J2).Denote by Φ the function whose supremum is taken in (3.1), and by Φ1and Φ2 the functions given by the analogous formulas, with the summationextended over s, t ∈T1 and j ∈J1 for Φ1, and over s, t ∈T2 and j ∈J2 forΦ2. We have Φ = Φ1 + Φ2.

Moreover, as the functions Φi involve only setsJi and Ti, then Φi(zsj, λs) ≤φ(ni, Ti, τi), where ni = |Ji| and τi = Ps∈Ti λ2s,for i = 1, 2. Thusφ(n, T, 1) = F(zsj, λs) = F1(zsj, λs) + F2(zsj, λs)(3.5)≤φ(n1, T1, τ1) + φ(n2, T2, τ2) = τ1φ(n1, T1, 1) + τ2φ(n2, T2, 1).Since τ1+τ2 = 1 and n1 > 0, the assumption (3.2) implies that the inequalityin (3.5) is not possible unless τ2 = 0 and n1 = n. Thus J1 = {1, .

. .

, n} andhence (3.4) holds for all l and m, as required.✷To simplify the orthogonality conditions, we letZsj = fj(s)√µsandΛs = √µsfor s ∈T, j = 1, . .

. , n.(3.6)Given the matrix (Zsj)s∈T,1≤j≤n we consider “short” vectors Zs = (Zsj)j ∈IK n, s ∈T, and “long” vectors eZj = (Zsj)s∈T ∈l2, j = 1, .

. .

, n. The naturalscalar product both in IK n and in l2 will be denoted by ⟨·, ·⟩.9

We work with the function F(Zsj, Λs) defined byF(Zsj, Λs) =Xs,t∈T|⟨Zs, Zt⟩|ΛsΛt =Xs,t∈T|nXj=1ZsjZtj|ΛsΛt. (3.7)Proof of Proposition 3.1 (a) First let IK = IR and T = {1, .

. .

, N}. Weuse Lagrange multipliers.

Clearly, the supremum φ(n, N) described in (3.3)is equal to the maximum of F on the surface given by the conditionsGlm(Zsj, Λs):=⟨eZl, eZm⟩−δlm =Xs∈TZslZsm −δlm = 0for 1 ≤l ≤m ≤n(3.8)G0(Zsj, Λs):=⟨eΛ, eΛ⟩−1 =Xs∈TΛ2s −1 = 0for s ∈T. (3.9)The supremum is attained for a sequence of non-negative Λs; if we set zsj :=f 0j (s)qµ0s and λs :=qµ0s for s ∈T, j = 1, .

. .

, n, then F attains its maximumat (zsj, λs).Consider the Lagrange function L defined by2L(Zsj, Λs) = F(Zsj, Λs) −Xl≤meγlmGlm(Zsj, Λs) −βG0(Zsj, Λs).Assume that (zsj, λs) is a point where F attains a local maximum subjectto (3.8) and (3.9). If for some 1 ≤s, t ≤N we had ⟨zs, zt⟩= 0, we wouldleave this term out from the sum defining F. This would lead to a new sum,defining the new function F1.

Clearly, F1 ≤F and max F1 = max F. Themaximum is attained at the same point (zsj, λs) and the function F1 is C2in the neighborhood of this point. Moreover, by setting sgn 0 = 0, in theformulas for derivatives which follow we will still be able to extend the sumsover all indices s, t.To use standard necessary conditions for Lagrange multipliers we firstcheck that the point (zsj, λs) is regular.

This means that the gradients ∇G0and ∇Glm for 1 ≤l ≤m ≤n, are linearly independent vectors in IR N(n+1).Denoting vectors (zsj)Ns=1 by ezj for j = 1, . .

. , N and (λs)Ns=1 by eλ, bya straigtforward differentiation with respect to Zsj and Λs we get, for 1 ≤10

l, m ≤n and l < m,∇G0 =0...02eλ,∇Glm =0...ezm...ezl...00,∇Gmm =0...2ezm...00. (3.10)In the formula for ∇Glm, with l < m, ezm stays on the lth place and ezlstays on the mth place; and in the formula for ∇Gmm, 2ezm stays on the mplace.

Since λs ̸= 0 for s = 1, . .

. , N, the linear independence of the gradientvectors (3.10) follows directly from the linear independence of the vectorsezl ∈IR N, for l = 1, .

. .

, n; the latter fact is an immediate consequence of theorthogonality, hence linear independence, of the system {f 0l }nl=1.Now, the first order condition for Lagrange multipliers states that thereexist multipliers eγlm and β such that after settingγlm = 12eγlmif l < meγmlif m < l2eγllif m = lwe have∂L∂Zsl=Xt∈Tsgn ⟨zs, zt⟩ztlλsλt −nXm=1γlmzsm = 0for s ∈T, l = 1, . .

. , n(3.11)∂L∂Λs=Xt∈T|⟨zs, zt⟩|λt −βλs = 0for s ∈T.

(3.12)First we simplify (3.11) by a suitable orthogonal transformation. Definetwo N × N matrices A and B byA = (sgn ⟨zs, zt⟩λsλt)s,t∈T ,B = (|⟨zs, zt⟩|)s,t∈T .

(3.13)11

Then the conditions (3.11) and (3.12) can be rewritten asAezl=nXm=1γlmezmforl = 1, . .

. , n(3.14)Beλ=βeλ.

(3.15)Let g = (gkl)k,l be an n × n orthogonal matrix which diagonalizes thehermitian n×n matrix Γ = (γlm)l,m, that is, g Γg∗= Dα is a diagonal matrixwith diagonal entries α1, . .

. , αn.For k = 1, .

. .

, n set ez′k =Pnl=1 gklezl. Then ezm =Pnl=1 glmez′l, for m =1, .

. .

, n. We have ⟨z′s, z′t⟩= ⟨zs, zt⟩for 1 ≤s, t ≤N. Thus the function Fand the matrices A and B do not change if we pass from variables inducedby the ezm’s to the variables induced by the ez′k’s.

Similarly, the ez′k’s satisfythe constraints (3.8) and (3.9). Thus the point (z′sk, λs) again gives a localextremum of F, but with a new set of multipliers.Expressing (3.14) in terms of primed vectors ez′k’s we get the n eigenvalueequationsAez′k = αk ez′kfor k = 1, .

. .

, n.(3.16)The last two conditions mean that the multipliers corresponding to (z′sk, λs)are just α1, . .

. , αn, β, with the off-diagonal ones equal to 0.Notice that if {f ′0k} is related to {ez′k} by (3.6), then {f ′0k} is an orthonor-mal basis in span [f 0m]; in particular the new square function f ′0 is equal tof 0.

Thus, without loss of generality, we can and will work with these new“primed” vectors, rather than with the original ones; we will leave howeverthe “primes” out, for clarity of notation. In other words, we will assume that(zsk, λs) satisfies (3.8), (3.9) and (3.15), (3.16).We want to show that all αk’s are equal.

To do so, we use the well-knownsecond order conditions for a relative maximum [H]: the Hessian matrix H,H =∂2L/∂Zpj∂Zqk∂2L/∂Zpj∂Λq∂2L/∂Λp∂Zqk∂2L/∂Λp∂Λq,evaluated at the point (zpj, λp), needs to be negative semi-definite on thetangent space to the surface of constraints at that point.12

We have∂2L∂Zpj∂Zqk=sgn ⟨zp, zq⟩λpλqδjk −αkδpqδjk(3.17)∂2L∂Λp∂Λq=|⟨zp, zq⟩| −βδpq(3.18)∂2L∂Λp∂Zqk=sgn ⟨zp, zq⟩λqzpk(1 + δpq). (3.19)So H is an N(n + 1) × N(n + 1) matrix of the formH =A −α1I.

. .0Ct1.........0. .

.A −αnICtnC1. .

.CnB −βI,(3.20)where A and B are defined in (3.13), I is the identity matrix, and Ck is theN × N matrix Ck = (∂2L/∂Λp∂Zqk)Np,q=1 for k = 1, . .

. , n.The tangent space T to the surface of constraints described by (3.8) and(3.9) consists of all vectors eew ∈IR N(n+1),eew = wpjνp!=ew1...ewneν(3.21)orthogonal to all gradients ∇G0 and ∇Glm for 1 ≤l ≤m ≤n, evaluated at(zpj, λp).

Then necessarily ⟨H eew, eew⟩≤0 for all eew ∈T .¿From (3.10) it follows that eew of the form (3.21) is in the tangent spaceT if and only if it satisfies the following equations:⟨∇(Zpj,Λp)Glmeez, eew⟩=Xp∈T(zplwpm + zpmwpl) = ⟨ezl, ewm⟩+ ⟨ezm, ewl⟩= 0for 1 ≤l ≤m ≤n(3.22)⟨∇(Zpj,Λp)G0eez, eew⟩=2Xp∈Tλpνp = 2⟨eλ, eν⟩= 0. (3.23)We will now show that for any 1 ≤l ̸= m ≤n and any s ∈T(αl −αm)zslzsm = 0.

(3.24)13

This will follow from the negative-definiteness of the matrix H, by evaluating⟨H eew, eew⟩on suitable vectors eew.Consider the vector few0 of the form (3.21) with ewl = ezm, ewm = −ezl andewk = 0 otherwise, and eν ∈IR N arbitrary satisfying (3.23). The orthogonalityof the vectors ezk ensured by (3.8) implies that few0 ∈T.Let fH denote the Nn×Nn matrix which appears in the upper left cornerof (3.20) and let eC be the N × Nn matrix from the bottom left corner.A simple calculation using (3.16) shows that⟨ fH000!few0, few0⟩= ⟨ (αm −αl)ezl(αm −αl)ezm!, ezl−ezm!⟩= 0.Thus⟨H few0, few0⟩=⟨ fHeCteCB −βI!few0, few0⟩=⟨(B −βI)eν, eν⟩+nXk=1⟨Ck ewk, eν⟩+nXk=1⟨ewk, Ctkeν⟩=⟨(B −βI)eν, eν⟩+ 2⟨Clezm −Cmezl, eν⟩≤0.

(3.25)Replacing eν by εeν and taking ε →0 we get that the first term in thefinal sum (which is of the second order in ε) can be disregarded from theestimate. In the inequality obtained this way eν can be replaced by −eν, hence⟨Clezm −Cmezl, eν⟩= 0.

Since this equality holds for an arbitrary eν orthogonalto eλ, we conclude that there is a constant γ such that Clezm −Cmezl = γeλ.Equivalently, the definition of Cl and (3.19) yield that for p ∈T,γλp=zplXq∈Tsgn ⟨zp, zq⟩λqzqm(1 + δpq) −zpmXq∈Tsgn ⟨zp, zq⟩λqzql(1 + δpq)=zplXq∈Tsgn ⟨zp, zq⟩λqzqm −zpmXq∈Tsgn ⟨zp, zq⟩λqzql. (3.26)Observe that by (3.13) and (3.16) we have, for p ∈T,Xq∈Tsgn ⟨zp, zq⟩λpλqzqm = (Aezm)p = αmzpmand an analogous equality holds for the second term in (3.26).

This impliesthatγλ2p = (αm −αl)zpmzplforp ∈T. (3.27)14

Summation over p yields by (3.8) and (3.9) thatγ = (αm −αl)⟨ezm, ezl⟩= 0.Hence (3.24) holds. It obviously follows from Lemma 3.2 thatαm = αl =: αfor all 1 ≤l, m ≤n.

(3.28)Expressing (3.16) coordinatewise we haveXt∈Tsgn ⟨zs, zt⟩ztlλsλt −αzsl = 0for l = 1, . .

. , n, s ∈T.Multiplying by zsl, summing up over l and using (3.15), we find, for s ∈T,0 =Xt∈T|⟨zs, zt⟩|λsλt −αnXl=1z2sl = βλ2s −αnXl=1z2sl.In terms of µs and f 0 this means that βµs = αf 0(s)µs, i.e., f 0(s) = β/αis constant for all s ∈T with µs ̸= 0.

If µs = 0, then f 0(s) = 0, as mentionedalready. Since Ps∈T |f 0(s)|2µs = n = (β/α)2, we conclude that β/α = √n,completing the proof in the case (a).

(b) We now consider the infinite case T = IN for IK = IR . Assume thatthe function F given by (3.7) attains a relative maximum subject to theconstraints (3.8) and (3.9) at the point (zsj, λs), where zsj = f 0j (s)qµ0s andλs =qµ0s for s ∈T, j = 1, .

. .

, n.Note that ezj = (zsj)s∈T ∈l2, forj = 1, . .

. , n, and eλ = (λs)s∈T ∈l2.

For N ∈IN and an arbitrary vector ez ∈l2set ezN = RN ez ∈lN2 .Fix N ∈IN sufficiently large so that ezN1 , . .

. , ezNn are linearly independent.In (3.7)–(3.9) fix the variables for s > N by puttingZsj = zsj, Λs = λsfor s > N, j = 1, .

. .

, n.Relative to the new constraints, F as a function in the variables (Zsj, Λs),with s = 1, . .

. , N, j = 1, .

. .

, n, attains a relative maximum at (zsj, λs), withs = 1, . .

. , N, j = 1, .

. .

, n. The first order Lagrange multiplier conditions(3.11) and (3.12) now take the formXt∈Tsgn ⟨zs, zt⟩ztlλsλt−nXm=1γlmzsm = 0for s = 1, . .

. , N, l = 1, .

. .

, n(3.29)Xt∈T|⟨zs, zt⟩|λt−βλs = 0for s = 1, . .

. , N.(3.30)15

For a fixed s = 1, . .

. , N, the first sum in (3.29) is independent of N. SinceezN1 , .

. .

, ezNn are linearly independent, this uniquely determines the matrixΓ = (γlm)nl,m=1, which is then independent of N. Thus (3.29) and (3.30) holdfor all s ∈IN. Again, we diagonalize the n × n matrix Γ and we introduceez′1, .

. .

, ez′n satisfying the eigenvalue equations Aez′k = αk ez′k for k = 1, . .

. , n.Moreover, Beλ = βeλ.Note that A and B, formally given by (3.13), arenow infinite Hilbert–Schmidt matrices.

Again, in what follows, we leave the“primes” out and write simply ezk.Now let T be the space of (infinite) vectors eew in the direct sum L l2 ofn + 1 copies of l2, which are of the form (3.21) and satisfy (3.22) and (3.23)(with T = IN). Let H be the (infinite) matrix of the form (3.20), with A, Band Ck being infinite as well.

To conclude the same proof as in part (a), itsuffices to show that ⟨H eew, eew⟩≤0 for all eew ∈T .To this end, denote by HN and AN, BN, CNk the restricted matrices oforder N(n + 1) × N(n + 1) and N × N respectively.The constraints for the restricted problem in the variables (Zsj, Λs), withs = 1, . .

. , N, j = 1, .

. .

, n, are still of the formGNlm(Zsj, Λs)=NXs=1ZslZsm −dlm = 0GN0 (Zsj, Λs)=NXs=1Λ2s −d = 0,for some d, dlm ∈IR .This implies that the corresponding tangent spaceT N ⊂IR N(n+1) of vectors eewN of the form (3.21) is defined by the equationsNXp=1(zplwpm + zpmwpl)=0for 1 ≤l ≤m ≤n(3.31)NXp=1λpνp=0. (3.32)Hence, in general, the projection eewN = RN(n+1) eew of eew onto IR N(n+1) is notin T N, since ⟨ezl, ewm⟩+ ⟨ezm, ewl⟩= 0 for ≤l ≤m ≤n does not imply⟨ezNl , ewNm⟩+ ⟨ezNm, ewNl ⟩= 0 for ≤l ≤m ≤n.

However, since the limit of (3.31)and (3.32), as N →∞, coincides with (3.22) and (3.23) (for T = IN), it is16

clear that for any eew ∈T , there is a sequence ( eewN)∞N=1, witheewN = ( ewN)leνN!∈T N,such that eewN →eew in the L l2-norm.Since ⟨HN eewN, eewN⟩≤0 for all N ∈IN, it suffices to show thatlimN→∞⟨HN eewN, eewN⟩= ⟨H eew, eew⟩.This is shown term by term. A typical case islimN→∞⟨(AN −αlI)( ewN)l, ( ewN)l⟩= ⟨A ewl −αl ewl, ewl⟩,which reduces to⟨( ewN)l, ( ewN)l⟩→⟨ewl, ewl⟩and ⟨AN( ewN)l, ( ewN)l⟩→⟨A ewl, ewl⟩.

(3.33)But (3.33) follows from limN→∞( ewN)l = ewl in the l2-norm and the fact thatmatrices AN converge to A in the Hilbert–Schmidt norm.As before, we find that α1 = . .

. = αn =: α and βµs = αf 0(s)µs.

Wethen complete the proof as in case (a). (c) Finally, we indicate the necessary changes in the proof of the complexcase, IK = C .

We assume for simplicity that T = {1, . .

. , N}.

The functionF, as defined by (3.7), is now a function of the complex variables Zsj =Xsj + iYsj and the real variables Λs. We consider F as a function of realvariables (Xsj, Ysj, Λs).

There are now n2 real constraints for 1 ≤l ≤m ≤n,G(1)lm(Xsj, Ysj, Λs):=Re Glm(Zsj, Λs) = 0for l < mG(2)lm(Xsj, Ysj, Λs):=Im Glm(Zsj, Λs) = 0for l < m(3.34)Gll(Xsj, Ysj, Λs):=Gll(Zsj, Λs) = 0for l = m,as well as (3.9).Consider the Lagrange function L defined by2L := F −Xl

If F attains the extremum subject to conditions (3.9) and (3.34) at (zsj =xsj + iysj, λs), then a calculation shows that the first order conditions can bewritten in the following complex form∂L∂Xsl+ i ∂L∂Ysl=Xt∈Tsgn ⟨zs, zt⟩ztlλsλt −nXm=1γlmzsm = 0,(3.35)for s ∈T, l = 1, . .

. , n. Here sgn w = w/|w| for w ∈C , w ̸= 0 and sgn 0 = 0.Moreover,γlm = 12eγ(1)lm −ieγ(2)lmif l < meγ(1)ml + ieγ(2)mlif m < l2eγllif m = ldefine an n × n hermitian complex matrix Γ.

Thus we can again diagonalizeΓ and rewrite (3.35) and (3.12) asAezk = αk ezkfor k = 1, . .

. , nand Beλ = βeλ,(3.36)where A and B are formally defined as in (3.13).The tangent space T to the surface of constraints (3.34) and (3.9) nowconsists of vectors eew ∈C N(n+1), whose complex form is formally describedby (3.21) and whose real form iseew =eu1ev1...eunevneν∈IR N(2n+1)(3.37)where ewl = eul + evl for l = 1, .

. .

, n. The equations defining T can be writtenin the following (complex) formXp∈T(zplwpm + wplzpm)=⟨ezl, ewm⟩+ ⟨ewl, ezm⟩= 0for 1 ≤l ≤m ≤n,(3.38)⟨eλ, eν⟩=0. (3.39)18

The Hessian matrix H in the real form has now size N(2n+1)×N(2n+1).In particular, the matrix C in (3.25) consists of 2n real matricesC(1)l= ∂2L∂Λp∂Xql!Np,q=1,C(2)l= ∂2L∂Λp∂Yql!Np,q=1for l = 1, . .

. , nof size N × N, evaluated at (xsj, ysj, λs).

The condition ⟨H eew, eew⟩≤0 trans-lates intonXl=1⟨C(1)leul + C(2)levl, eν⟩= 0,(3.40)for all eν satisfying (3.39). ThusPnl=1(C(1)leul + C(2)levl) is a multiple of eλ, forall eew ∈T of the (complex) form (3.21) satisfying (3.38) and (3.39).For 1 ≤l ̸= m ≤n we pick two different types of vectors in T .

In thecomplex form (3.21), the first vector eew looks as before, that is,ewl = ezm,ewm = −ezl and ewk = 0 otherwise, and eν ∈IR N arbitrary satisfying (3.39).The second type, eew′, is defined similarly by setting ew′l = iezm, ew′m = iezl andewk = 0 otherwise. The real form of these vectors is the following, writing thenon-zero terms only,eew =exmeym−exl−eyleνandeew′ =−eymexm−eylexleν.Both vectors satisfy (3.38) and (3.39).

Calculating (3.40) for eew and eew′ andusing the eigenvalue equations (3.36) we find, in an analogous way as weobtained (3.27) in case (a), that there are γ1 and γ2 such thatλp(C, 0) eewp=Re (αm −αl)zpmzpl = γ1λ2pλp(C, 0) eew′p=Im (αm −αl)zpmzpl = γ2λ2pfor all p ∈T. Summing over p and using (3.34) and (3.9) we infer thatγ1 = γ2 = 0, thus(αm −αl)zpmzpl = 0.Just as in case (a), the last equality implies α1 = .

. .

= αn =: α and βµs =αf 0(s)µs for all s ∈T, which completes the proof of (c).✷19

4The estimate for the projection constantWe start by a simple but useful lemma of Sidelnikov [Si] and Goethals andSeidel [GS.1]. It gives a lower bound for expressions related to those appear-ing in the definition (3.1) of φ.

Since the bound is essential for our estimate,we include its proof.Lemma 4.1 Let T = {1, . .

. , N}, or T = IN and let (µs)s∈T be a probabilitymeasure on T. Let (zs)s∈T ∈IK n with ∥zs∥2 = 1.

Let ω be the normalizedrotation-invariant measure on Sn−1 = Sn−1(IK ). Then for every even naturalnumber k ∈2IN,Xs,t∈T|⟨zs, zt⟩|kµsµt ≥ZSn−1ZSn−1 |⟨z, w⟩|k dω(z)dω(w).

(4.1)(In the complex case, express the integrand in the real variables andintegrate over Sn−1(C ) = S2n−1(IR ). )Proof Let n ∈IN and k = 2m ∈2IN.

For z ∈IK n, let z⊗j = z⊗. .

.⊗z ∈IK njdenote the j-fold tensor product of z with itself, for j = 1, 2, . .

.. Scalarproducts in IK nj will be denoted by ⟨·, ·⟩j, and for j = 1 just by ⟨·, ·⟩. Thenfor any z, w ∈IK n and j = 1, 2, .

. .

we have⟨z⊗j, w⊗j⟩j = ⟨z, w⟩j,and⟨z⊗m ⊗z⊗m, w⊗m ⊗w⊗m⟩k = ⟨z, w⟩m ⟨z, w⟩m = |⟨z, w⟩|k.Considerξ :=Xs∈T(z⊗ms⊗z⊗ms)µs −ZSn−1 (z⊗m ⊗z⊗m) dω(z) ∈IK nk.By the rotation invariance of ω, integrals of the formRSn−1 |⟨e, w⟩|k dω(w) donot depend on e ∈Sn−1. This allows to evaluate ⟨ξ, ξ⟩k as follows:0 ≤⟨ξ, ξ⟩k=Xs,t∈T|⟨zs, zt⟩|kµsµt +ZSn−1ZSn−1 |⟨z, w⟩|k dω(z)dω(w)−2Xs∈TµsZSn−1 |⟨zs, w⟩|k dω(w)=Xs,t∈T|⟨zs, zt⟩|kµsµt −ZSn−1ZSn−1 |⟨z, w⟩|k dω(z)dω(w),20

which proves the lemma.✷Now we are ready for the proof of Theorem 1.1.Proof of Theorem 1.1 (a) Let n ∈IN and let G(n) denote the right handside of (1.3). We have to show that for any n-dimensional space E we haveλ(E) ≤G(n).

By Proposition 2.2, λ(E) ≤φ(n, T), where T = {1, . .

. , N}, ifE ⊂lN∞, and T = IN, if E ⊂l∞.

By Lemma ? ?

it suffices to show thatφ(n, T) ≤G(n)for T = {1, . .

. , N}.

(4.2)Given n and T we may assume that φ(n1, T1) < φ(n, T) for all n1 < nand all T1 ⊂T; otherwise the proof which follows would be applied to theminimal n1 with φ(n1, T) = φ(n, T), to show that φ(n, T) ≤G(n1) < G(n).The double supremum in the definition (3.1) of φ(n, T) is attained forsome probablity measure µ = (µs)s∈T on T and some orthonormal systemfj = (fj(s))s∈T ∈l2(T, µ), j = 1, . .

. , n.By Proposition 3.1, the squarefunction f := (Pnj=1 |fj|2)1/2 equals √n for all s where µs ̸= 0, and equals to0 otherwise.

The span of the fj’s is therefore supported by S := suppµ ⊂T.For s ∈S, let zs := n−1/2(fj(s))nj=1 ∈ln2. Then ∥zs∥2 = 1 andφ(n, T) = nXs,t∈S|⟨zs, zt⟩|µsµt.

(4.3)Define α and β byα; =(1/√n + 2IK = IR1/√n + 1IK = C,β :=(3/(n + 2)IK = IR2/(n + 1)IK = C.Then for u ∈[−1, 1] we have(|u| −α)2 =(u2 −α2)/(|u| + α)2 ≥(u2 −α2)2/(1 + α)2.This implies|u| ≤γ0 + γ2u2 −γ4u4for u ∈[−1, 1],(4.4)whereγ0 = α2 −α32(1 + α)2, γ2 = 12α +α(1 + α)2, γ4 =12α(1 + α)2(4.5)21

are non-negative. Equality in (4.4) occurs for u ∈[−1, 1] if and only if |u|equals to 1 or α.

(The right hand side of (4.4) touches |u| at ±α and intersects|u| at ±1. )Using (4.4) and (4.1) we can estimate (4.3).φ(n, T)≤nXs,t∈Tγ0 + γ2|⟨zs, zt⟩|2 −γ4|⟨zs, zt⟩|4µsµt≤nγ0 + γ2/n −γ4ZSn−1ZSn−1 |⟨z, w⟩|4 dω(z)dω(w)=nγ0 + γ2 −γ4β = G(n).

(4.6)Here we used the orthonormality of the fj’s to evaluate the double sumXs,t∈T|⟨zs, zt⟩|2µsµt = 1/nand the fact that for any e ∈Sn−1,I :=ZSn−1 |⟨e, w⟩|4 dω(w) = β/n,since e.g., in the real case,I =Z 1−1 t4(1 −t2)(n−3)/2 dt/Z 1−1(1 −t2)(n−3)/2 dt = 3/(n(n + 2));in the complex case the calculation yields I = 2/(n(n + 1)).The last equality in (4.6) is established by a direct calculation using (4.5). (b) and (c) We now assume that E is an n-dimensional space attain-ing the extremal bound, λ(E) = G(n).

By Proposition 2.2, there is T ={1, . .

. , N} or T = IN, a probability measure µ = (µs)s∈T on T and an or-thonormal basis (fj)nj=1 in E2,µ such thatλ(E) ≤Xs,t∈T|nXj=1fj(s)fj(t)|µsµt.

(4.7)For all n1 < n and all T1 ⊂T we have φ(n1, T1) < φ(n, T); otherwise, forsome n1 < n and some T1 ⊂T we would have, by part (a),G(n) = λ(E) ≤φ(n, T) ≤φ(n1, T1) ≤G(n1) < G(n).22

Therefore, by Proposition 3.1, on the support S ⊂T of µ, the square functionf = (Pnj=1 |fj|2)1/2 is equal to √n. For s ∈S consider again the short vectorszs = (fj(s))nj=1/√n.

Hence ∥zs∥2 = 1 for s ∈S. We may and will furtherassume that S is minimal in the sense that for s ̸= t we have zs ̸= θzt with|θ| = 1.Otherwise, we could replace the short vectors zs and zt by onevector zs, assigning to it the measure µs + µt; the orthogonality and thenormalization of the corresponding long vectors and the double sum in (4.7)would remain unchanged.

Let N := |S|. We have to show that N is finite,and, in fact, bounded by N(n) as defined in (1.2).By (4.7), (4.4) and (4.6) we haveλ(E)=nXs,t∈S|⟨zs, zt⟩|µsµt≤nXs,t∈Sγ0 + γ2|⟨zs, zt⟩|2 −γ4|⟨zs, zt⟩|4µsµt≤G(n).Thus, the assumption λ(E) = G(n) implies the equality of all terms.

Theequality in the first inequality requires that |⟨zs, zt⟩| = α or 1 for all s, t ∈S(note that µs ̸= 0 for s ∈S). For s ̸= t, zs ̸= θzt, hence |⟨zs, zt⟩| = α. Recallthat α = 1/√n + 2 in the real case, and α = 1/√n + 1 in the complexcase.

We thus proved that the vectors (zs)s∈S ⊂Sn−1(IK ) are equiangular.Since in IK n there are at most N(n) equiangular vectors, it follows thatN = |S| ≤N(n). Using the Cauchy–Schwartz inequality, we get anotherchain of inequalities which become equalities,G(n)=n Xs,t∈Sµsµtα +Xs∈Sµ2s(1 −α)≥nα + (n/N)(1 −α) ≥nα + (n/N(n))(1 −α) = G(n),where the last equality follows by a direct calculation, inserting the value ofα.

The equality implies, in particular, that N = N(n). Also, all values ofµs have to be equal (µs = 1/N(n)).

Since the vectors fj are all supportedby S, it follows that E is isometric to a subspace of lN(n)∞. The orthogonalprojection, given by the matrix (n/N(n))(⟨zs, zt⟩)s,t, is a minimal projection.Conversely, if in IK n there exist N(n) equiangular vectors (zs), we mayconstruct E = span [f1, .

. .

, fn] ⊂lN(n)∞by letting fj(s) = √n zsj (j =1, . .

. , n, s = 1, .

. .

, N(n)). By [K], the projection constant of E is equal23

to G(n), and the fj’s are orthonormal with respect to the equidistributedprobability measure µ on {1, . .

. , N(n)}.

Moreover, P, given by the matrix(n/N(n))(⟨zs, zt⟩)N(n)s,t=1, and acting as an operator from lN(n)∞to lN(n)∞, is aminimal (and orthogonal) projection onto E with norm G(n).Either way, the norm of a vector Pnj=1 αjfj in lN(n)∞is given by∥nXj=1αjfj∥∞=sup1≤s≤N(n)√n |⟨α, zs⟩|.Thus, given N(n) equiangular vectors (zs) in IK n, we get an n-dimensionalnormed space with the maximal projection constant by setting∥(αj)nj=1∥:=sup1≤s≤N(n)|⟨α, zs⟩|. (4.8)In the real case, N(n) = n(n + 1)/2 equiangular vectors exist in IR n forn = 2, 3, 7, 23 and these systems are unique up to orthogonal transforma-tions.

Hence the real spaces with projection constant G(n) are unique upto isometry if n = 2, 3, 7, 23. For n = 2, the uniqueness (up to orthogonaltransformations) of the three vectors at angle 2π/3 each, is trivial.

For n = 3one considers the 6 × 6 Gram matrix (⟨zs, zt⟩), with |⟨zs, zt⟩| = 1/√5 fors ̸= t. It is easy to see that up to permutations and multiplications of thezs’s by −1, the sign pattern is uniquely determined. The standard paper onthe subject is Lemmens, Seidel [LS]; for the uniqueness for n = 7, 23 we referto Goethals, Seidel [GS.2] and Seidel [S].

For n = 2, (4.8) yields the normwith the (regular) hexagonal unit ball; for n = 3, the extremal ball definedvia (4.8) is the (regular) dodecahedron, since the 6 equiangular vectors inIR 3 are the diagonals of the icosahedron.In the complex case, N(n) = n2 equiangular vectors exist in C n at leastfor n = 2, 3. For n = 2, the system and the extremal space are again uniqueup to isometry.

For n = 3, the system of 9 vectors in C 3 is connected to theHessian polyhedron, cf. Coxeter [C].✷RemarkPart (a) of the previous proof also shows that Theorem 1.2 canbe restated asmaxE∈Fn λ(E) = maxµmaxzsXs,t∈IN|⟨zs, zt⟩|µsµt.

(4.9)24

where the double maximum is taken over all discrete probability measuresµ = (µs)s∈IN and all sets of unit vectors (zs)s∈IN ⊂Sn−1(IK ) such thatIdIK n = nXs∈INµszs ⊗zs.Example In IR 4, consider the following 10 vectors of the formxs =1√123−1−1−1,xs = 1√2sin αsin α−cos α−cos α2,permuting the 3 to all places in the first type of vectors (1 ≤s ≤4) andpermuting the two −cos α in the second type of vectors (5 ≤s ≤10). Seta = −sin 2α + 1/2.

One checks that4 4Xs=1(a/2(1 + 2a))xs ⊗xs +10Xs=5(1/6(1 + 2a))xs ⊗xs= IdIR 4.Hence, letting µs = a/2(1 + 2a) for 1 ≤s ≤4 and µs = 1/6(1 + 2a) for5 ≤s ≤10, we see that the xs’s and µs’s satisfy the constraints in (4.9). Thescalar products |⟨xs, xt⟩| satisfy the following: for 1 ≤s ̸= t ≤4 they areequal to 1/3; for 5 ≤s ̸= t ≤10 they take two values, (1 −sin 2α)/2 appears24 times and | sin 2α|/2 appears 6 times; for 1 ≤s ≤4 and 5 ≤t ≤10, theyare equal to (1/√6)| sin α + cos α|.The maximum of the function10Xs,t=1|⟨xs, xt⟩|µsµtis equal to 1.8494 and it is attained for α = 1.4592.In IR 4, 10 equiangular vectors do note exist.

By Theorem 1.1 and (4.9),the maximal projection constant λ = sup λ(E4), for 4-dimensional real spacesE4 satisfies1.8494 ≤λ < (2 + 3√6)/5 ∼1.8697.The known explicit examples of equiangular lines allow to write down theextremal norms in the cases mentioned above, using (4.8).25

IK n∥(αj)n1∥λ(X)IR 2max(|2α1|, |α1 −√3α2|, |α1 +√3α2|)4/3hexagonIR 3max±(|τα1 ± σα2|, |τα2 ± σα3|, |τα3 ± σα1|)where τ :=q(√5 + 1)/2, σ :=q(√5 −1)/2√5 + 12dodeca-hedronIR 7maxmax1≤i

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Gr¨unbaum, F. A. Sherk (eds. ), Springer Verlag, New York 1982,203–218.[GS.2]J.

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