Frames over finite fields: Basic theory and equiangular lines in unitary geometry
Gary R. W. Greaves, Joseph W. Iverson, John Jasper, Dustin G. Mixon
aa r X i v : . [ m a t h . M G ] D ec FRAMES OVER FINITE FIELDS: BASIC THEORY ANDEQUIANGULAR LINES IN UNITARY GEOMETRY
GARY R. W. GREAVES, JOSEPH W. IVERSON, JOHN JASPER, AND DUSTIN G. MIXON
Abstract.
We introduce the study of frames and equiangular lines in classicalgeometries over finite fields. After developing the basic theory, we give sev-eral examples and demonstrate finite field analogs of equiangular tight frames(ETFs) produced by modular difference sets, and by translation and modu-lation operators. Using the latter, we prove that Gerzon’s bound is attainedin each unitary geometry of dimension d = 2 l +1 over the field F . We alsoinvestigate interactions between complex ETFs and those in finite unitary ge-ometries, and we show that every complex ETF implies the existence of ETFswith the same size over infinitely many finite fields. Introduction
Among open problems in finite frame theory, perhaps none are more challengingthan Zauner’s conjecture and the existence of equiangular tight frames. To elabo-rate, choose a sequence { ϕ j } j ∈ [ n ] of nonzero vectors in C d , expressed as a matrixΦ = (cid:2) ϕ · · · ϕ n (cid:3) ∈ C d × n , where [ n ] = { , . . . , n } . Then Φ is said to represent:(a) equiangular lines if its columns have equal norm and constant pairwise innerproduct absolute value, (b) a tight frame if its rows are mutually orthogonal andhave equal norm, and (c) an equiangular tight frame (ETF) if both (a) and (b)occur. ETFs arise naturally as optimal solutions to several problems of interest,and they have applications in areas such as wireless communication [48], compressedsensing [3], digital fingerprinting [43], and quantum information theory [46]. Thelatter provided the original setting for Zauner’s conjecture : Conjecture . For every d ≥
1, there exists a complex d × d ETF.Zauner’s conjecture has been the subject of intense interest since its introduc-tion over 20 years ago, and recently a monetary prize was announced for its reso-lution [30]. Nevertheless the problem remains wide open, and at present there areonly finitely many dimensions d for which a complex d × d ETF is known to exist.Beyond Zauner’s conjecture we have the more general (and similarly formidable)existence problem of complex ETFs, for which cash prizes are also available [41, 42].
Problem . For which pairs ( d, n ) does a complex d × n ETF exist?A variety of complex ETF constructions are known, including many infinitefamilies of sizes ( d, n ) with n = d [19, 10, 17, 18, 20, 21, 32, 33]. However verylittle is understood about nonexistence, and the known list of necessary conditionsis as short as this: n ≤ d , n ≤ ( n − d ) , and ( d, n ) / ∈ { (3 , , (5 , } . The first twoconditions amount to Gerzon’s bound [38], and they only fuel interest in the caseof equality predicted by Zauner’s conjecture. Beyond Gerzon, 3 × × by means of a Gr¨obner basis calculation that is not feasible for larger sizes [50].This state of affairs is particularly grim when compared with the case of real ETFs,for which a great many nonexistence results are known, a success that may be dueto the fundamentally discrete nature of the sign pattern in the Gram matrix Φ ⊤ Φof a real ETF [49, 38, 19].Faced with slow progress on problems of import, we take here the advice of Polyaand vary the problem [45]. In particular, we observe that each property of interest(equiangular lines, tight frame, ETF) may be expressed using only polynomials andan order-2 field automorphism (complex conjugation). From this light our problemsare fundamentally algebraic, and they readily admit generalizations over any field,or even ∗ -ring.In this article and its companion [24] we focus on ETFs over finite fields. Herewe join a long tradition of investigating finite field analogs of problems originallyposed over real and complex numbers. Among the many examples, we mention thelocal/global principle of number theory [40], the Ax–Grothendieck theorem [25],sum–product estimates [7], the Erd˝os–Falconer distance problem [31], the Kakeyaproblem [15], and Roth’s theorem on arithmetic progressions [11, 16]. For each ofthese examples the generalization to finite fields proved to be fruitful, either byproducing insight or traction for the original problem, or by creating an alternatearena that demonstrated interest in its own right. We also find a precursor withinframe theory, in the work of Bodmann et al. on frames over F and their relationshipwith coding theory (a subject and viewpoint not represented here, since we do notconcern ourselves with Hamming distance) [5].Generally speaking, when varying a problem one hopes to balance two competinggoals. On the one hand, the new problem should have enough features in commonthat it could reasonably provide some insight for the original one. On the otherhand, the new problem should have some strikingly different features that providenew openings for attack.Both of these objectives are met for ETFs over finite fields. The new problemhas the same linear algebraic expression as the old one, and we are able to derive abasic theory that includes parallels of many of the most important results, includ-ing factorization of Gram matrices (Theorems 3.12 and 3.14) and Gerzon’s bound(Theorem 4.2). Furthermore, we show that the existence of a d × n complex ETFimplies the existence of d × n ETFs over infinitely many finite fields (Theorem 7.1);in this sense the finite field problem is properly a generalization of the complexone. In the case of finite orthogonal geometries, ETFs over finite fields sometimesimply the existence of real ETFs [24]. Meanwhile, the finite field setting providesnew features and new tools, most notably the existence of isotropy (nonzero vectorsbeing orthogonal to themselves) and the absence of norm positivity. Furthermore,the finite vector spaces under consideration provide opportunities for new compu-tational methods (such as exhaustive search). Finally, the discrete nature of finitefields suggests comparison with real ETFs, which are understood much better thantheir complex counterparts.Thanks to these differences, we are able to prove for finite fields what has eludedresearchers in the complex setting for more than 20 years: we prove that Gerzon’sbound is attained in infinitely many dimensions. Specifically, for unitary geometriesover F we use translations and modulations to produce a d × d ETF whenever d = 2 l +1 (Theorem 6.4). Our construction may be viewed as generalizing Hoggar’s RAMES OVER FINITE FIELDS 3 lines to an infinite family over a finite field (Remark 6.6). A similar phenomenonappears in the companion paper [24], where we prove that Gerzon’s bound is at-tained in a finite orthogonal geometry of dimension d whenever | d − | is not apower of 2. Overall, finite fields provide a fruitful environment in which to studyETFs, and due to their rich theory they appear to be worthy of study in their ownright.This paper is organized in two parts. Part 1 develops the basic theory neces-sary for a rigorous investigation in the sequel and in the companion paper [24].Throughout Part 1 we treat both orthogonal and unitary geometries over finitefields, corresponding to real and complex Hilbert spaces, respectively. We beginwith a review of the theory of forms in Section 2, followed by an exposition offrame theory over finite fields in Section 3. Section 4 treats equiangular lines andculminates in Gerzon’s bound (Theorem 4.2).Part 2 focuses on ETFs in unitary geometries, which are the finite field analog ofcomplex Hilbert spaces. In Section 5, we give our first examples of such ETFs, andwe show that a generalization of difference sets creates ETFs just as in the complexsetting (Theorem 5.7). Section 6 develops the theory of Gabor frames over finitefields. After proving that every Gabor frame is tight (Proposition 6.1), we explicitlydescribe fiducial vectors that create d × d ETFs in infinitely many dimensions over F (Theorem 6.4). Finally, Section 7 demonstrates connections with the complexsetting. We first prove that when a complex d × n ETF exists, there is also acomplex d × n ETF with algebraic entries (Theorem 7.3). Then we show thatevery algebraic ETF projects into infinitely many finite fields (Theorem 7.5), sothat every complex ETF implies a multitude of finite field ETFs having the samesize (Theorem 7.1). The paper ends with some open problems for future research.(Additional open problems are scattered throughout.)
Part Basic theory Preliminaries
For the sake of accessibility, we begin with a short review of the theory of forms.Standard references include [26, 52, 2].
Assumptions . Throughout Part 1, we fix a field F and a field automorphism σ : F → F (possibly the identity) that satisfies σ = 1. The subfield fixed by σ is denoted F = { a ∈ F : a σ = a } . We also fix a vector space V over F of finitedimension d = dim V , and we assume V is equipped with a form h· , ·i : V × V → F that satisfies the following conditions:(F1) for every u ∈ V , the induced mapping h u, ·i : V → F is linear,(F2) h u, v i = h v, u i σ for every u, v ∈ V ,(F3) if u ∈ V satisfies h u, v i for every v ∈ V , then u = 0.We write Q : V → F for the associated quadratic form given by Q ( v ) = h v, v i .By (F1) and (F2), h· , ·i is linear in the second variable and σ -linear in the first.Assumptions (F1)–(F3) may also be expressed in terms of matrices. Specifically,for a matrix A = (cid:2) a ij (cid:3) ∈ F m × n we denote A ∗ = (cid:2) a σji (cid:3) ∈ F n × m . Choose anybasis e , . . . , e d ∈ V with coordinate transformation T : V → F d . Then conditions(F1)–(F3) are equivalent to the existence of a Gram matrix M ∈ F d × d satisfying:(F1’) h u, v i = ( T u ) ∗ M ( T v ) for every u, v ∈ V , GARY R. W. GREAVES, JOSEPH W. IVERSON, JOHN JASPER, AND DUSTIN G. MIXON (F2’) M = M ∗ ,(F3’) M is invertible.Explicitly, M = (cid:2) h e i , e j i (cid:3) ∈ F d × d .By (F3’), det M belongs to the multiplicative group F × of F . Throughout thepaper, we write F × ≤ F × for the subgroup of quadratic residues, and in thisnotation the discriminant of V is defined to bediscr( V ) = (det M ) F × ∈ F × / F × . It is an invariant of h· , ·i and independent of the choice of basis behind M .The following space plays the same role as ℓ in classical frame theory. Definition . We write ( · , · ) for the form on F n with Gram matrix M = I , i.e.,( x, y ) = x ∗ y = X i ∈ [ n ] x σi y i for x = { x i } i ∈ [ n ] , y = { y i } i ∈ [ n ] ∈ F n . (Notice that we take F n to consist of column vectors, and that ( · , · ) is conjugate-linear in the first variable.) If σ = 1, then we refer to F n equipped with ( · , · ) as a real model , since it is reminiscent of the real Hilbert space R n . If σ = 1, then wecall it a complex model .Next, let W ≤ V be a subspace. Its orthogonal complement is the subspace W ⊥ = { u ∈ V : h w, u i = 0 for every w ∈ W } , which satisfies dim W + dim W ⊥ = V . We say W is nondegenerate if W ∩ W ⊥ = { } ; equivalently, the restriction of h· , ·i to W again satisfies (F1)–(F3). At theother extreme, W is called totally isotropic if W ≤ W ⊥ , that is, h u, v i = 0 forevery u, v ∈ W . In that case, dim W ≤ dim V .Now let W be another vector space over F with a form satisfying (F1)–(F3).Every linear map A : W → V has a unique adjoint A † : V → W satisfying h Au, v i = h u, A † v i for every u ∈ W and v ∈ V . Its kernel is Ker A † = (Im A ) ⊥ , and inparticular, A † is one-to-one if and only if A is onto (and vice versa). Adjointsare related to but distinct from conjugate transpose matrices, and a matrix for A † can be found as follows. After choosing bases, we may assume there are matrices M and N such that V = F m has form h u, v i = u ∗ M v and W = F n has form h x, y i = x ∗ N y . If we identify operators between F n and F m with matrices, thenthe adjoint of A ∈ F m × n is A † = N − A ∗ M ∈ F n × m .We define the isometry group of V to be the group I( V ) ≤ GL( V ) of operators A : V → V satisfying A † A = I . More generally, ∆( V ) ≤ GL( V ) is the group ofall A satisfying A † A = cI for c ∈ F × . The distinction between the two groupsis necessary: every scalar multiple of an isometry belongs to ∆( V ), but they mayform a proper subgroup.We assume no more than (F1)–(F3) in general, but we will be especially inter-ested in the following special cases, which represent classical geometries over finitefields. Definition . We say we are in
Case O if F = F q is a finite field of odd order q ,and σ = 1. On the other hand, Case U occurs when F = F q is finite (possibly ofeven order) and σ is given by α σ = α q .For Case O, we choose to specify that q is odd since the theory of quadraticforms in even characteristic is fundamentally different. In reference to Case U, we RAMES OVER FINITE FIELDS 5 remark that α σ = α q describes the only nontrivial field automorphism on F q thatsatisfies σ = 1.Suppose for the moment that Case O occurs. Then F = F , and h· , ·i is anondegenerate symmetric bilinear form. We call V a quadratic space and sayit has an orthogonal geometry . Since q is odd, h· , ·i can be recovered from Q through the polarization identity h u, v i = (cid:2) Q ( u + v ) − Q ( u − v ) (cid:3) , for u, v ∈ V. Up to isomorphism there are exactly two possibilities for h· , ·i , corresponding towhether or not discr( V ) ∈ F × / F × is trivial. Given another quadratic space W over F q , there exists an isomorphism V → W that preserves the form if and only ifdim V = dim W and discr( V ) = discr( W ). In particular, given an invertible matrix M ∈ F d × d , there exists a basis for V having matrix M as a Gram matrix if andonly if (det M ) F × = discr( V ). Thus, V admits an orthonormal basis if and only ifdiscr( V ) is trivial, if and only if V is isometrically isomorphic with the real modelon F dq . The isometry group is denoted I( V ) = O( V ) since it is an example of aclassical orthogonal group . If d = dim V is even, then O( V ) × F × q is a propersubgroup of ∆( V ) since there exist operators satisfying A † A = αI , where α ∈ F × is not a quadratic residue. If d is odd then the isomorphism type of O( V ) does notdepend on discr( V ), and ∆( V ) = O( V ) × F × q consists of nonzero scalar multiplesof orthogonal matrices.Now suppose we are in Case U. Then h· , ·i is a nondegenerate Hermitian form,and ( V, h· , ·i ) is isometrically isomorphic with the complex model on F dq . We call V a unitary space and say it has a unitary geometry . The subfield fixed by σ is F = F q ≤ F q . We denote T q = { α ∈ F q : α q α = 1 } ≤ F × q , with | T q | = q + 1. If ω ∈ T q is a generator then we have the polarization identity h u, v i = 1 q + 1 q +1 X k =1 ω − k Q ( u + ω k v ) , for u, v ∈ V. An orthonormal basis always exists, so (F1)–(F3) determine h· , ·i uniquely up toisometric isomorphism. The isometry group is denoted I ( V ) = U( V ) since it is anexample of a classical unitary group . For the special case of V = F dq equippedwith ( · , · ) (the complex model) we also write I ( V ) = U( d, q ). Here, U( d, q ) consistsprecisely of unitaries , i.e., matrices U ∈ F d × dq that satisfy U ∗ U = I . Notice thatU(1 , q ) = T q , which justifies the latter notation. Here ∆( V ) = U( V ) × F × q consistsof nonzero scalar multiples of unitary operators. Remark . The reader is warned that the usual spectral theorem fails over finitefields. In Case U, if A is a square matrix that commutes with A ∗ , it does not follow that A is diagonalizable. On the contrary, every B ∈ F n × nq is similar tosome A ∈ F n × nq satisfying A = A ∗ [27]. Many standard techniques in frame theoryhave the spectral theorem at their core, and alternative methods are needed whenworking over finite fields. GARY R. W. GREAVES, JOSEPH W. IVERSON, JOHN JASPER, AND DUSTIN G. MIXON Frame theory
This section develops the basics of frame theory over arbitrary fields, with specialemphasis placed on finite fields. Most of the results have known parallels for realand complex frames [9, 53]. Over finite fields the main differences are as follows:the quadratic form Q ( v ) = h v, v i satisfies no condition akin to positivity, there areusually nonzero vectors v ∈ V with Q ( v ) = 0, there are usually tight frames withframe constant zero, and there are two types of orthogonal geometries in Case O.Each of these differences has repercussions for the basic theory outlined below.3.1. Finite frames.
Definition . Throughout the paper we abuse notation by identifying a finitesequence Φ = { ϕ j } j ∈ [ n ] in V with its synthesis operator Φ : F n → V given byΦ { x i } i ∈ [ n ] = n X k =1 x k ϕ k . Its adjoint with respect to ( · , · ) is the analysis operator Φ † : V → F n given byΦ † v = {h ϕ i , v i} i ∈ [ n ] , and its frame operator is ΦΦ † : V → V , whereΦΦ † v = n X k =1 h ϕ k , v i ϕ k . Multiplying in the other direction gives the
Gramian Φ † Φ : F n → F n , whose rep-resentation in the standard basis is the Gram matrix (cid:2) h ϕ i , ϕ j i (cid:3) of Φ.We call Φ a frame if its vectors span V . It is nondegenerate if its frameoperator is invertible. Since d = dim V , a frame Φ with n vectors is said to have size d × n (matching the size of a matrix for its synthesis operator). A tight frame for V is a frame Φ that satisfies ΦΦ † = cI for c ∈ F . Here c is the frame constant ,and we also call Φ a c -tight frame . If c = 1 the frame is called Parseval . If c = 0it is totally isotropic . We emphasize that the vectors of a totally isotropic tightframe must span V , and it is not sufficient that ΦΦ † = 0.If V = F d then we perform a further abuse by identifying Φ with the matrixof its synthesis operator, which is the d × n matrix whose j -th column is ϕ j . Wealso identify the analysis operator with its matrix, and if the form on V is given by h u, v i = u ∗ M v then Φ † = Φ ∗ M . Example . Totally isotropic (hence degenerate) frames exist, as shown by thesimple example Φ = (cid:2) (cid:3) in the real model over F . Example . Let q be an odd prime power, and consider F q in the real model. If3 ∈ F × q then we can choose β ∈ F × q satisfying β = 3 to obtain the Mercedes–Benzframe Φ = (cid:20) − β β − − (cid:21) . It is tight with constant c = 3 /
2, and its Gram matrix is(3.1) Φ ∗ Φ = − − − − − − . RAMES OVER FINITE FIELDS 7
Conversely, if 3 / ∈ F × q then there does not exist a frame Φ ∈ F × q in the real modelhaving Gram matrix (3.1). This is a consequence of Theorem 3.14 below.The complex model on F q is more permissive, as a tight frame Φ ∈ F × q withGram matrix (3.1) exists provided 3 ∤ q . This follows from Theorem 3.12 below,and an explicit example appears in Example 3.7.As in the real and complex settings, frames are characterized by an expansionproperty involving a dual frame (Ψ below). If Φ is a c -tight frame and c = 0, thenΨ = c Φ is a dual frame. There does not appear to be such a canonical choice ofdual for 0-tight frames.
Proposition 3.4.
A sequence
Φ = { ϕ j } j ∈ [ n ] in V is a frame if and only if thereis another sequence Ψ = { ψ j } j ∈ [ n ] such that ΦΨ † = I , that is, X j ∈ [ n ] h ψ j , v i ϕ j = v for every v ∈ V. Proof.
We may assume V = F d with form h u, v i = u ∗ M v . The reverse implicationis clear. Conversely, if Φ ∈ F d × n is a frame then there exists A ∈ F n × d such thatΦ A = I . Put Ψ = ( AM − ) ∗ to obtain ΦΨ † = I . (cid:3) The characterization below can be proved with standard techniques.
Proposition 3.5. If Φ = { ϕ j } j ∈ [ n ] is a frame for V , then the following are equiv-alent for any choice of c ∈ F : (i) Φ is a c -tight frame, (ii) (Φ † Φ) = c (Φ † Φ) , (iii) (Φ † u, Φ † v ) = c h u, v i for every u, v ∈ V .Moreover, if V = F d and h· , ·i = ( · , · ) , then (i)—(iii) are equivalent to: (iv) the rows ψ , . . . , ψ d ∈ F n of Φ satisfy ( ψ i , ψ j ) = cδ i,j for every i, j ∈ [ d ] .(Here and throughout, δ ij is the Kronecker delta function.)Remark . In Case O or Case U (Definition 2.3) we may add another equivalentcondition to Proposition 3.5:(v) (Φ † v, Φ † v ) = c · Q ( v ) for every v ∈ V .Indeed, (v) is equivalent to (iii) since h· , ·i and ( · , · ) admit polarization identities. Example . Assume Case U (Definition 2.3). Let U = (cid:2) u ij (cid:3) i,j ∈ [ n ] be a unitarymatrix, and choose d rows labeled by J ⊆ [ n ]. Then Proposition 3.5(iv) implies thatthe submatrix Φ = (cid:2) u ij (cid:3) i ∈ J,j ∈ [ n ] is a d × n Parseval frame. This gives a large supplyof tight frames, and in fact every Parseval frame arises this way, as a consequenceof Proposition 3.21 below.As a concrete example, consider the complex model over F . Let α ∈ F × be aprimitive element, and put ω = α ∈ T . Then ω = 1, and U = α (cid:2) ω ij (cid:3) i,j ∈ Z / Z isunitary. Rescaling the rows labeled by J = { , } gives the tight frameΦ = α (cid:20) ω ω ω ω (cid:21) with Gram matrix (3.1). This is the Mercedes–Benz frame over F , cf. Example 3.3.Theorem 3.14 below implies that there does not exist any unitary W ∈ U(2 ,
5) forwhich W Φ has entries in the subfield F . In other words we cannot “rotate” Φ toobtain a version of the Mercedes–Benz frame in the real model over F . GARY R. W. GREAVES, JOSEPH W. IVERSON, JOHN JASPER, AND DUSTIN G. MIXON
Corollary 3.8.
A frame
Φ = { ϕ j } j ∈ [ n ] for V is totally isotropic if and only if Im Φ † is a totally isotropic subspace of F n . Hence, V admits a -tight frame of n vectors only if n ≥ d . The identity (Φ † u, Φ † v ) = h ΦΦ † u, v i easily implies the following parallel charac-terization of nondegenerate frames. Proposition 3.9.
A frame
Φ = { ϕ j } j ∈ [ n ] for V is nondegenerate if and only if Im Φ † is a nondegenerate subspace of F n . Frames from Gram matrices.
Uniqueness.
Proposition 3.10. If Φ is a frame, then Ker Φ † Φ = Ker Φ and
Im Φ † Φ = Im Φ † .Proof. We show Ker Φ † Φ = Ker Φ, and Im Φ † Φ = Im Φ † follows by taking orthogo-nal complements. Given x ∈ F n , we have Φ † Φ x = 0 if and only if h Φ x, Φ y i = 0 forevery y ∈ F n . Since Φ is a frame, this happens if and only if Φ x = 0. (cid:3) Proposition 3.11.
Let Φ and Ψ be frames for V with the same number of vectors.Given c ∈ F × , we have Ψ † Ψ = c Φ † Φ if and only if Ψ = A Φ for unique A ∈ ∆( V ) satisfying A † A = cI .Proof. The reverse implication is trivial. Conversely, if Ψ † Ψ = c Φ † Φ then the ex-pression A ( P j ∈ [ n ] x j ϕ j ) = P j ∈ [ n ] x j ψ j gives a well-defined linear operator A : V → V , and it is clear that h Au, Av i = c h u, v i for every u, v ∈ V . The choice of A isunique since Φ and Ψ are frames. (cid:3) Existence.
The following was observed in Lemma 2.3 of [27]. Our proof belowis more direct and gives an explicit algorithm.
Theorem 3.12.
Suppose we are in Case U (Definition 2.3). Then G ∈ F n × nq isthe Gram matrix of a frame for V if and only if G = G ∗ and rank G = dim V . Notice that there is no condition akin to positive semidefiniteness in Theo-rem 3.12. This is a significant departure from the real and complex settings.
Proof.
We may assume V = F dq in the complex model. The forward implica-tion is clear from Proposition 3.10. For the converse, we first construct Ψ = (cid:2) ψ · · · ψ n (cid:3) ∈ F n × nq such that Ψ ∗ Ψ = G . It will take the form Ψ = (cid:20) AB (cid:21) ,with A ∈ F n × nq upper triangular and B ∈ F n × nq a diagonal matrix. Later, we willtransform Ψ to produce a frame Φ ∈ F d × nq .We first construct A = (cid:2) a · · · a n (cid:3) in such a way that ( a i , a j ) = G ij for every i = j . We define the entries of A recursively, going down the columns from left toright. To begin, set a to be the first column of the n × n identity matrix. Nowsuppose we have constructed the first j − a j = (cid:2) a kj (cid:3) k ∈ [ n ] so that for any i < j , G ij = ( a i , a j ) = n X k =1 a qki a kj = a ij + i − X k =1 a qki a kj . RAMES OVER FINITE FIELDS 9
To accomplish this, we first set a j = G j . After the first i − a j aredetermined and i < j , we define a ij = G ij − P i − k =1 a qki a kj . Finally, we set a jj = 1and a ij = 0 for i > j . Continuing in this way, we eventually obtain A ∈ F n × nq withthe desired structure.Having built A , we next define the diagonal matrix B = (cid:2) b · · · b n (cid:3) ∈ F n × nq with columns b j = (cid:2) b ij (cid:3) i ∈ [ n ] in such a way that the vectors ψ j = (cid:20) a j b j (cid:21) satisfy G ij = ( ψ i , ψ j ) = ( a i , a j ) + ( b i , b j ) = ( ( a i , a j ) , if i = j ;( a i , a i ) + b q +1 ii , if i = j. It suffices to choose b ii such that b q +1 ii = G ii − ( a i , a i ), and this is possible since G ii − ( a i , a i ) ∈ F q . Make any valid choice to complete the construction of B , henceof Ψ ∈ F n × nq satisfying Ψ ∗ Ψ = G .It remains to transform Ψ ∈ F n × nq into Φ ∈ F d × nq . Set W = Im Ψ ≤ F nq , andconsider its radical rad W := { x ∈ W : h x, y i = 0 for every y ∈ W } ≤ W. Choose any algebraic complement U ≤ W for rad W , that is, W = U ⊕ rad W asvector spaces. Then U ≤ F nq is nondegenerate. For each j , write ψ j = x j + y j with x j ∈ U and y j ∈ rad W . Then x , . . . , x n span U and satisfy ( x i , x j ) = ( ψ i , ψ j ) = G ij for every i, j .Finally, we construct the matrix Φ. Set m = dim U . Since U is nondegenerate,there is an isometric isomorphism T : U → F mq . Define ϕ j = T ( x j ) for every j . Then Φ = (cid:2) ϕ · · · ϕ n (cid:3) ∈ F m × nq satisfies rank Φ = m and Φ ∗ Φ = G . Inparticular, m = rank G = d . This completes the proof. (cid:3) Definition . Given a square matrix M = (cid:2) M ij (cid:3) i,j ∈ [ n ] = (cid:2) m · · · m n (cid:3) , selectcolumns { m j } j ∈ I that form a basis for Im M . We refer to M b = (cid:2) M ij (cid:3) i,j ∈ I as a basic submatrix of M . Theorem 3.14.
Assume Case O (Definition 2.3). Choose a matrix G ∈ F n × nq andbasic submatrix G b . Then, G is the Gram matrix of a frame for V if and only ifthe following hold: (i) G = G ⊤ , (ii) rank G = dim V , (iii) (det G b ) F × q = discr( V ) .Consequently, every symmetric matrix G ∈ F n × nq occurs as the Gram matrix ofa frame for a quadratic space over F q , namely one in dimension rank G whosediscriminant matches the determinant of a basic submatrix of G .Proof. If G is the Gram matrix of a frame Φ = { ϕ j } j ∈ [ n ] , then (i) holds trivially and(ii) follows from Proposition 3.10. For (iii), let g j ∈ F nq denote the j -th column of G and observe that A ( P j ∈ [ n ] x j ϕ j ) = P j ∈ [ n ] x j g j gives a well-defined isomorphism A : V → Im G . If the basic submatrix G b arises from a basis { g j } j ∈ [ n ] for Im G ,then the corresponding vectors { ϕ j } j ∈ I provide a basis for V with Gram matrix G b . This proves (iii).For the converse, we proceed as in the proof of Theorem 3.12. The same ar-gument given there constructs a matrix A = (cid:2) a · · · a n (cid:3) ∈ F n × nq such that ( a i , a j ) = g ij for every i = j . Next, we want to build a block-diagonal matrix B = (cid:2) b · · · b n (cid:3) ∈ F n × nq in such a way that the vectors ψ j = (cid:20) a j b j (cid:21) satisfy g ij = ( ψ i , ψ j ) = ( a i , a j ) + ( b i , b j ) . In other words, we want to arrange so that ( b i , b j ) = 0 for i = j , while ( b j , b j ) = g jj − ( a j , a j ). By Proposition 4.8 of [26], there exist vectors x j ∈ F q such that ( x j , x j ) = g jj − ( a j , a j ). We let B be the block-diagonal matrix B = diag( x , . . . , x n ) ∈ F n × nq .Then Φ = (cid:20) AB (cid:21) = (cid:2) ψ . . . ψ n (cid:3) ∈ F n × nq satisfies Ψ ⊤ Ψ = G .As in the proof of Theorem 3.12, we consider W = Im Ψ and an algebraiccomplement U ≤ W for its radicalrad W := { x ∈ W : h x, y i = 0 for every y ∈ W } ≤ W. For each j we decompose ψ j = x j + y j with x j ∈ U and y j ∈ rad W . Then x , . . . , x n form a frame for U having Gram matrix G . By the forward implication alreadyproved, U = rank G = V and discr( U ) = (det G b ) F × q = discr( V ). Hence there isan isometric isomorphism U → V , and the images of x , . . . , x n provide a frame for V having Gram matrix G . (cid:3) Remark . Suppose we are in Case O (Definition 2.3) and d = dim V is odd.If G ∈ F n × nq is symmetric of rank d , then some scalar multiple cG occurs as theGram matrix of a frame for V . Indeed, if G b is a basic submatrix for G and c = 0,then cG b is a basic submatrix for cG with determinant c d (det G b ). Since d is oddwe can choose c to ensure that (det cG b ) F × q = discr( V ). On the other hand, if d iseven then rescaling G does not alter the existence of a frame for V having G as aGram matrix.3.3. Nondegenerate tight frames, projections, and subspaces.
The isome-try group I ( V ) has a natural action on the space of frames for V , and its orbitsbreak the set of all frames into equivalence classes. As expected, the classes fortight frames correspond with certain projections and subspaces. Proposition 3.16.
Assume Case U (Definition 2.3). Choose c ∈ F × q and n ≥ d =dim V . Then all of the following sets have the same cardinality: (i) F = { unitary equivalence classes of c -tight frames for V with n vectors } , (ii) P = { self-adjoint rank- d matrices G ∈ F n × nq satisfying G = cG } , (iii) S = { nondegenerate d -dimensional subspaces of F nq } .Specifically, the functions f : F → P , g : F → S , and h : P → S given by f ([Φ]) = Φ † Φ , g ([Φ]) = Im Φ † , h ( G ) = Im G are well-defined bijections that satisfy g = h ◦ f .Proof. To see that f is a bijection, consider the related mapping Φ Φ † Φ ofa c -tight frame Φ to its Gramian Φ † Φ. Its fibers consist of unitary equivalenceclasses of c -tight frames, by Proposition 3.11. On the other hand, Proposition 3.5and Theorem 3.12 imply that its range is precisely P . Factoring out equivalenceclasses creates a bijection f : F → P .Next we consider h . Given G ∈ P , set W = Im G and P = c − G . Then P = P = P ∗ and Im P = W . In particular, W ⊥ = (Im P ∗ ) ⊥ = Ker P . Forany x ∈ W , the identity P = P implies that P x = x , and for any x ∈ W ⊥ RAMES OVER FINITE FIELDS 11 we have
P x = 0. It follows that W ∩ W ⊥ = { } . In other words, the mapping G h ( G ) = Im G sends P into S . Furthermore, h : P → S is injective since F nq = W ⊕ W ⊥ and we have determined the action of P = c − G on both spaces.Finally, if we are given W ∈ S we may choose an orthonormal basis w , . . . , w d for W and define G = c P j ∈ [ d ] w j w ∗ j . Then G ∈ P has h ( G ) = Im G = W . Hence h : P → S is a bijection, and so is g = h ◦ f . (cid:3) Lemma 3.17.
Assume Case O (Definition 2.3). If Φ is a nondegenerate frame for V then discr(Im Φ † ) = det(ΦΦ † ) · discr( V ) . Proof.
We may assume V = F dq has form h x, y i = x ∗ M y . Representing Φ as amatrix, we have det(ΦΦ † ) = det(ΦΦ ∗ ) det( M ). Since Φ is nondegenerate, ΦΦ ∗ isa Gram matrix for Im Φ ∗ = Im Φ † . Consequently, det(ΦΦ † ) F × = discr(Im Φ † ) · discr( V ). (cid:3) Proposition 3.18.
Assume Case O (Definition 2.3). Choose c ∈ F × q and n ≥ d =dim V . Then all of the following sets have the same cardinality: (i) F , the set of all O( V ) -equivalence classes of c -tight frames for V with n vectors, (ii) P , the set of all symmetric rank- d matrices G ∈ F n × nq satisfying G = cG ,such that a basic submatrix G b satisfies (det G b ) F × q = discr( V ) , (iii) S , the set of all nondegenerate d -dimensional subspaces W ≤ F nq with discr( W ) = c d · discr( V ) .Specifically, the functions f : F → P , g : F → S , and h : P → S given by f ([Φ]) = Φ † Φ , g ([Φ]) = Im Φ † , h ( G ) = Im G are well-defined bijections that satisfy g = h ◦ f . As a consequence of Proposition 3.18, when d is odd every nondegenerate sub-space W ≤ F nq corresponds with a tight frame for V , but the frame constant c mustsatisfy c F × q = discr( W ) / discr( V ). On the other hand, when d is even, a nondegen-erate subspace W ≤ F nq yields a tight frame for V if and only if discr( W ) = discr( V ).Here the frame constant may be arbitrary. Proof.
As in the proof of Proposition 3.16, the mapping Φ Φ † Φ induces a bijec-tion F → P . Next we consider h . Given G ∈ P , Theorem 3.12 provides a c -tightframe Φ for V such that G = Φ † Φ. Then Lemma 3.17 implies that Im G = Im Φ † has discriminant c d · discr V . In other words, the mapping G h ( G ) = Im G sends P into S . The same argument of Proposition 3.16 shows that h : P → S isinjective, and it remains only to prove it is sujective.Choose any subspace W ≤ F nq in S . To show that W = Im G for some G ∈ P ,it suffices to construct a c -tight frame Φ for V having Im Φ † = Im Φ † Φ = W . Fixa basis v , . . . , v d for V , and let M = (cid:2) h v i , v j i (cid:3) ∈ F d × dq be its Gram matrix. Sincedet( cM ) F × q = c d · det( M ) F × q = c d · discr( V ) = discr( W ) , there is a basis w , . . . , w d for W satisfying ( w i , w j ) = c h v i , v j i for every i, j ∈ [ d ].Let A : V → F n × nq be the unique linear operator with Av i = w i for every i ∈ [ d ]. Itis clearly injective, so Φ := A † : F nq → V is the synthesis operator of a frame for V . Furthermore, we have arranged so that (Φ † u, Φ † v ) = c h u, v i for every u, v ∈ V , soΦ is a c -tight frame. The proof is complete since Im Φ † = W and g = h ◦ f . (cid:3) Remark . In either Case U or Case O (Definition 2.3) we can find the number ofI( V )-equivalence classes of nondegenerate tight frames by counting subspaces of F n .This can be done with an orbit-stabilizer argument for the action of an isometrygroup, as described on page 148 of [2]. We omit details but report the results.In Case U, choose n ≥ d = dim V and c ∈ F × q . Then the number of equivalenceclasses of c -tight frames for V with n vectors equals | U( n, q ) || U( d, q ) | · | U( n − d, q ) | , where | U( n, q ) | can be found in Theorem 11.28 of [26].A similar formula applies in Case O, but now discriminants play a role alongwith the frame constant c ∈ F × q . For n ≥ d = dim V the number of equivalenceclasses of c -tight frames for V with n vectors equals | O( F nq ) || O( W ) | · | O( U ) | , where F nq refers to the real model and W and U are quadratic spaces over F q suchthat dim W = d , dim U = n − d , discr W = c d · discr V , and discr U = c d F × q . Theorders of the orthogonal groups are given in Theorem 9.11 of [26]. Remark . In contrast with nondegenerate tight frames, I ( V )-equivalence classesof 0-tight frames may not be identified with the range of a corresponding analysisoperator. For example, suppose Φ ∈ F d × nq is a 0-tight frame under the complexmodel, and choose any invertible operator A ∈ GL( d, q ). Then Ψ := A Φ ∈ F d × nq is a frame, and Im Ψ † = Im Φ † . By Corollary 3.8, Ψ is a 0-tight frame. However, A may be chosen so that Ψ † Ψ = Φ † ( A † A )Φ = Φ † Φ, in which case Ψ and Φ areunitarily inequivalent by Proposition 3.11.3.4.
Naimark complements.Proposition 3.21.
Assume Case U (Definition 2.3). Let Φ be a c -tight frame( c = 0 ) for V with Gram matrix G ∈ F n × nq . Then cI − G is the Gram matrix of a c -tight frame Ψ for a unitary space W of dimension n − d , such that Im Ψ † = (Im Φ † ) ⊥ .Proof. Since G = Φ † Φ satisfies G = cG , the kernel of H := cI − G coincideswith the range of G . As H = H ∗ it follows that Im H = (Im G ) ⊥ , and rank H = n − d . Apply Theorem 3.12 to write H = Ψ † Ψ, where Ψ is a d × n frame in aunitary geometry. Then Ψ is a c -tight frame since H = cH , and Im Ψ † = Im H =(Im Φ † ) ⊥ . (cid:3) Proposition 3.22.
Assume Case O (Definition 2.3). Let Φ be a c -tight frame( c = 0 ) for V with Gram matrix G ∈ F n × nq , and choose a ∈ F × q . Then a ( cI − G ) is the Gram matrix of an ac -tight frame Ψ for a quadratic space W of dimension n − d such that discr( W ) = a n − d c n · discr( V ) . Here Im Ψ † = (Im Φ † ) ⊥ .Proof. The same argument of Proposition 3.21 applies, and we need only computethe discriminant of W . Put H = a ( cI − G ), and let W be a quadratic space ofdimension n − d admitting a frame Ψ with Gram matrix H . As Φ is nondegenerate, F nq = Im G ⊕ Im H . Therefore discr(Im G ) · discr(Im H ) = discr( F nq ) is trivial, RAMES OVER FINITE FIELDS 13 i.e., discr(Im G ) = discr(Im H ). The latter are related to discr( V ) and discr( W ) asin Proposition 3.18, and, in particular, c d · discr( V ) = ( ac ) n − d · discr( W ). (cid:3) In Proposition 3.22, the choice of a ∈ F × q makes a difference for discr( W ) if andonly if n − d is odd. For both Proposition 3.21 and Proposition 3.22 it is essentialthat c = 0, since a 0-tight frame of size d × n exists only if n ≥ d .3.5. Equal norm tight frames.
Definition . Let Φ = { ϕ j } j ∈ [ n ] be a c -tight frame for V . If there is a constant a ∈ F (possibly zero) such that h ϕ j , ϕ j i = a for every j ∈ [ n ], then we call Φ an( a, c )- equal norm tight frame , or ( a, c )-NTF.NTFs are generalizations of unit norm tight frames that allow arbitrary norms. Ifthere exists nonzero α ∈ F such that αα σ = a , then we may rescale an ( a, c )-NTF toobtain a (1 , c/a )-NTF, with unit norm. However this is not always possible. Someauthors use the abbreviation ENTF instead of NTF. We eschew this terminology inorder to avoid confusion with the stronger notion of equiangular tight frame (ETF).If Φ is an ( a, c )-NTF with n vectors, then the traces of Φ † Φ and ΦΦ † give(3.2) na = dc, where d = dim V as usual. For example, an ( a, a = 0 orchar F | n . Example . There are finite field versions of harmonic frames, which provide alarge supply of NTFs in Case U (Definition 2.3). If m | q + 1 then a primitive m -th root of unity ω ∈ F × q satisfies ω q +1 = 1, and we may create the matrix F = (cid:2) ω ij (cid:3) ij ∈ F m × mq . It is a Hadamard matrix of order m since F ∗ F = mI and every entry of F is unimodular. By taking tensor powers, we may create aHadamard matrix H of order n whenever every prime factor of n divides q + 1.Then we may select any d ≤ n rows of H to produce a ( d, n )-NTF Φ ∈ F d × nq . Remark . Assume q is odd in Case U (Definition 2.3). By modifying the“spectral tetris” construction of [8], one may create an ( a, c )-NTF of size d × n whenever (3.2) holds and either c = 0 or a = c = 0. We omit details, and leaveopen the general problem of characterizing NTF existence. Example . In the complex model, choose any d > F dq . Then Φ ∈ F d × nq is a (1 , n = q d − q d +( − d +1 q +1 vectors, as we now explain.Let ω ∈ F × q be a generator for the subgroup T q ≤ F × q of unimodular scalars,and consider the unit sphere of F dq expressed as columns of the matrixΨ = (cid:2) Φ ω Φ · · · ω q Φ (cid:3) ∈ F d × n ( q +1) q . By an inductive argument Ψ has q d − [ q d + ( − d +1 ] = n ( q + 1) columns, whichgives the formula for n . Furthermore, the columns of Ψ (hence also of Φ) span F dq since the former contain the standard basis, and ΨΨ ∗ = ( q + 1)ΦΦ ∗ = ΦΦ ∗ .Therefore it suffices to show ΨΨ ∗ = 0.Any U ∈ U( d, q ) permutes the unit sphere, so it commutes with ΨΨ ∗ . Takingpermutation matrices for U we see that ΨΨ ∗ has the form cI + bJ for c, b ∈ F q . Then bJ also commutes with every U ∈ U( d, q ). By an application of the WittExtension Theorem (Theorem 10.12 of [26]), it follows that b = 0 and ΨΨ ∗ = cI .To get c = 0, it suffices to show the last row v ∈ F nq of Ψ satisfies ( v, v ) = 0. Thecolumns x = (cid:20) yβ (cid:21) of Ψ are in bijection with pairs ( y, β ) ∈ F d − q × F q such that( y, y ) = 1 − β q +1 . As such ( v, v ) = P b ∈ F × q n b b , where n b is the number of y ∈ F d − q with ( y, y ) = 1 − b . For b = 1 this is the size of the unit sphere in F d − q , or n b = q d − [ q d − + ( − d ]. For b = 1 subtraction gives n = q d − + ( − d − ( q d − − q d − ).If d > n b is divisible by q , so that ( v, v ) = 0. On the other hand if d = 2 then q divides every n b −
1, so that ( v, v ) = P b ∈ F × q b = 0. Example . In the real model, choose a ∈ F × q and let Φ ∈ F d × nq consist of onevector from each pair { x, − x } in F dq such that ( x, x ) = a . Then Φ is an NTF, byan argument similar to that of Example 3.26. The frame constant may or may notbe zero depending on q, a, d . 4. Equiangular lines
Next we develop the basic theory of equiangular lines over arbitrary fields. Inthe real and complex case this is just the theory of equiangular lines. Our mainresult (Theorem 4.2) is a generalization of Gerzon’s bound.Recall that F ≤ F is the subfield fixed by σ . In Case O we have F = F q = F ,and in Case U it is F = F q ≤ F q = F . Definition . Given a, b ∈ F , we say Φ = { ϕ j } j ∈ [ n ] forms an ( a, b ) -equiangularsystem in V if the following hold:(i) h ϕ j , ϕ j i = a for every j ∈ [ n ],(ii) h ϕ i , ϕ j ih ϕ j , ϕ i i = b for every i = j in [ n ].If this holds and ϕ j = 0 for every j ∈ [ n ], then L = { span ϕ j } j ∈ [ n ] forms a sequenceof equiangular lines . For c ∈ F , Φ is an ( a, b, c ) -equiangular tight frame , or( a, b, c ) -ETF , if the following hold in addition to (i)–(ii):(iii) span Φ = V ,(iv) ΦΦ † = cI .In other words, an ( a, b, c )-ETF is an ( a, b )-equiangular system that is also a c -tightframe. Equivalently, it is an ( a, c )-NTF for which (ii) holds.In our general setting, we have the following version of Gerzon’s bound [38]. Ouroverall method of proof is the usual one, but the abstract setting presents a fewsubtleties to address. Theorem 4.2 (Gerzon’s bound) . Denote k = dim F F ∈ { , } , depending onwhether or not σ is trivial. Suppose a = b . Then there exists an ( a, b ) -equiangularsystem Φ of n vectors in V only if n ≤ d + k ( d − d ) . If equality holds, then ΦΦ † = cI for some c ∈ F , where c = 0 if a = 0 . If equality holds in Case O orCase U (Definition 2.3), then Φ is an ( a, b, c ) -ETF. The case a = b is exceptional, and in the real or complex case it may only de-scribe vectors chosen repeatedly from a single line, as a consequence of the conditionfor equality in Cauchy–Schwarz. Stranger behavior can occur over finite fields, asdemonstrated by Example 4.3 further below. RAMES OVER FINITE FIELDS 15
Proof.
Throughout the proof we work in the F -space S of linear operators A : V → V satisfying A † = A , and we equip S with the (possibly degenerate) symmetric F -bilinear form h A, B i F = tr( AB ). For any choice of basis e , . . . , e d of V , themapping T : S → F d × d given by T ( A ) = (cid:2) h e i , Ae j i (cid:3) is easily seen to be an F -linear isomorphism of S onto the space of self-adjoint d × d matrices. By countingentries on and above the diagonal, we deduce that dim F S = d + k ( d − d ). Thisgives an upper bound on the size of a linearly independent set in S , which we willuse to prove the theorem.Let Φ = { ϕ j } j ∈ [ n ] be an ( a, b )-equiangular system in V . For each j ∈ [ n ] definethe outer product A j = ϕ j ϕ † j ∈ S by A j v = h ϕ j , v i ϕ j , and let A = { A j } j ∈ [ n ] .We start by identifying linear dependencies in A from its Gram matrix. For any i, j ∈ [ n ], h A i , A j i F = tr( ϕ † i ϕ j ϕ † j ϕ i ) = h ϕ i , ϕ j ih ϕ j , ϕ i i = ( a , if i = j ; b, if i = j. Therefore A has Gram matrix G := (cid:2) h A i , A j i F (cid:3) = bJ n + ( a − b ) I n ∈ F n × n , where J n is the all-ones matrix. Since h· , ·i F may be degenerate, we cannot always factor G = A † A (in particular, A † may not be well defined); however it is still true thatKer A ≤ Ker G . Furthermore, since a − b = 0 we have Ker G ≤ Im J n = span { n } ,where n ∈ F n is the all-ones vector. Equality holds only if nb + ( a − b ) = 0.Therefore, Ker A = { } when a = − ( n − b , and Ker A ≤ span { n } generally.First assume a = 0. Since A maps F n into S , and since Ker A ≤ span { n } ,rank-nullity gives the bounddim F S ≥ dim F Im A = n − dim F Ker A ≥ n − , that is, n ≤ d + k ( d − d ) + 1. Equality holds only if a = − ( n − b and Ker A =span { n } , that is, P j ∈ [ n ] ϕ j ϕ † j = 0. We claim this cannot happen. Otherwise, therelation tr(Φ † Φ) = tr(ΦΦ † ) says that na = 0, while char F cannot divide n since b = a = − ( n − b . Therefore a = 0, contrary to assumption. It follows that n ≤ d + k ( d − d ).Now assume a = 0. Then tr( A j ) = tr( ϕ † j ϕ j ) = a = 0 for each j , so that A liesin the subspace S = { A ∈ S : tr( A ) = 0 } ≤ S of traceless self-adjoint operators. Notice that dim F S = dim F S − S linearly onto F . Proceeding as before, we find thatdim F S − F S ≥ dim F Im A = n − dim F Ker A ≥ n − , i.e., n ≤ dim F S . Equality holds only if 0 = P j ∈ [ n ] ϕ j ϕ † j = ΦΦ † .Next consider the case of equality, n = dim F S , with a = 0. Define A n +1 = I d and B = { A j } j ∈ [ n +1] , which equals A appended by the identity matrix. For any j ≤ n we have h A j , A n +1 i F = tr( ϕ i ϕ † i I ) = a , so B has Gram matrix H := (cid:2) h A i , A j i F (cid:3) i,j ∈ [ n +1] = (cid:20) bJ n + ( a − b ) I n a n a ⊤ n d (cid:21) ∈ F ( n +1) × ( n +1)0 . Considering that B has n + 1 > dim F S vectors, we conclude it is linearly de-pendent. Choose any nonzero x = { x i } i ∈ [ n +1] ∈ Ker B . As above, we have x ∈ Ker H . For any choice of i, j ≤ n , expanding matrix products in the equation (cid:2) Hx (cid:3) i = 0 = (cid:2) Hx (cid:3) j shows that a x i + bx j = bx i + a x j , or ( a − b ) x i = ( a − b ) x j . Since a = b it follows that x i = x j , and x = (cid:2) α ⊤ n β (cid:3) ⊤ for some α, β ∈ F .Furthermore, α = 0 since (cid:2) Hx (cid:3) = 0 and a = 0. Defining c = − β/α , we conclude0 = α − X j ∈ [ n ] x j A j + α − x n +1 I d = X j ∈ [ n ] ϕ j ϕ † j − cI d . Therefore, ΦΦ † = cI d as desired.Finally, suppose that the bound is attained with n = d + k ( d − d ) in eitherCase O or Case U. Then(4.1) d ≥ rank Φ ≥ rank Φ † Φ =: d ′ . By Theorem 3.14 (Case O) or Theorem 3.12 (Case U), there is a frame Ψ of n vectors in an orthogonal (Case O) or unitary (Case U) geometry on F d ′ , suchthat Ψ † Ψ = Φ † Φ. Then Ψ is an ( a, b )-equiangular system in dimension d ′ , andconsidering the bound proved above we must have d ′ = d . Therefore equality holdsthroughout (4.1), and Φ is a frame. By the above it is an ETF. (cid:3) We now demonstrate the pathology of the case a = b for Gerzon’s bound. Example . For d ≥ f ( d ) on the size of anequiangular system in any space of dimension d over any field. We cannot evenbound the number of distinct lines spanned by vectors in an equiangular systemwithout accounting for the base field. For example, a sequence of vectors formsa (0 , d ≥ q is allowed to vary, the number N ( d, q )of distinct lines in a maximal totally isotropic subspace of F dq under the complexmodel grows to infinity with q . Hence there exist arbitrarily large (0 , d -dimensional spaces (over various fields). Example . More generally, let a ∈ F be such that there exists ϕ ∈ V with h ϕ , ϕ i = a . Choose a totally isotropic subspace W ≤ span { ϕ } ⊥ . For w ∈ W define ϕ w = ϕ + w . Then Φ = { ϕ w } w ∈ W is an ( a, a )-equiangular system. Thisgives very large examples. Example . Not every ( a, a )-equiangular system takes the form of the last exam-ple. To see this, choose d ≥ q , and consider the complexmodel on V = F dq . For any isotropic x, y ∈ F d − q satisfying ( x, y ) = − (cid:20) x y (cid:21) is a (1 , a, b )-equiangular system in Case O or Case U thatbeats Gerzon when we know the values of a, b . We leave this as an open problem. Problem . In Case O and Case U (Definition 2.3), find a relative bound on thesize of an ( a, b )-equiangular system that outperforms Gerzon (Theorem 4.2).Despite not yet having a relative bound, we can relate the parameters of an( a, b, c )-ETF, as in Welch [54].
Proposition 4.7. If V admits an ( a, b, c ) -ETF of n vectors, then a ( c − a ) = ( n − b .When char F fails to divide both d and n − , it follows that b = ( n − d ) d ( n − a . RAMES OVER FINITE FIELDS 17
Proof.
Let Φ = { ϕ j } j ∈ [ n ] be an ( a, b, c )-ETF in V , and consider the matrix A =Φ † Φ − aI , whose diagonal is zero. Expanding (Φ † Φ − aI ) with the relation (Φ † Φ) = c Φ † Φ shows that A = ( c − a ) A + a ( c − a ) I , and in particular, ( A ) ii = a ( c − a )for every i ∈ [ n ]. On the other hand, when we compute the matrix product wefind that ( A ) ii = P j ∈ [ n ] A ij A ji = ( n − b for every i ∈ [ n ]. Comparing theseexpressions shows that a ( c − a ) = ( n − b . Finally, when char F fails to divide d and n −
1, we can solve (3.2) to find c = nd a , so that b = ( n − d ) d ( n − a . (cid:3) As in the real and complex settings, ETFs over finite fields often come in Naimarkcomplementary pairs. More precisely, we have the following consequence of Propo-sition 3.21 and Proposition 3.22.
Proposition 4.8 (Naimark complements of ETFs) . (a) If there exists an ( a, b, c ) -ETF of n vectors in a unitary geometry on F dq and c = 0 , then there exists a ( c − a, b, c ) -ETF of n vectors in a unitarygeometry on F n − dq . (b) If there exists an ( a, b, c ) -ETF of n vectors in an orthogonal geometry on F dq and c = 0 , then there exists a ( c − a, b, c ) -ETF of n vectors in an orthogonalgeometry on F n − dq . We emphasize that c = 0 in Proposition 4.8, and that the orthogonal geometriesin Proposition 4.8(b) may have different discriminants. Part ETFs in unitary geometry
For the remainder of the paper, we focus on ETFs in finite unitary geometries.ETFs in finite orthogonal geometries are the subject of the companion paper [24].5.
First examples
In this section we demonstrate some constructions of ETFs in unitary geome-tries, focusing especially on those derived from modular difference sets. We have notinvestigated finite field analogs of other sources of complex ETFs, such as Steinersystems [21], hyperovals [20], graph coverings [10, 18, 34], the Tremain construc-tion [17], association schemes [12, 33], or Gelfand pairs [32]. We leave these topicsfor future research.
Example . Every ETF in a finite orthogonal geometry produces one in a finiteunitary geometry, as we now explain. If Φ is an ( a, b, c )-ETF of n vectors in anorthogonal geometry on F dq , then its Gram matrix G may be viewed as an elementof F n × nq with rank F q G = rank F q G = d . By Theorem 3.12, there is a frame Ψ of n vectors in a unitary geometry on F dq having G as its Gram matrix. Consideringthe entries of G and the fact that G = cG , we conclude that Ψ is an ( a, b, c )-ETFin a unitary geometry. This creates a large number of examples, which are exploredmore fully in [24].Next we show three ETFs in unitary geometries having unusual sizes. Theauthors discovered Examples 5.2, 5.3, 5.4 while searching for large (0 , vertices are isotropic vectors in F dq under the complex model, with vertices u and v adjacent precisely when ( u, v ) q +1 = 1. Example . Take q = 3 and ζ ∈ F × as a primitive element. Then the followingis a (0 , , ×
16 in the complex model on F :Φ = ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ . Example . Take q = 3 and ζ ∈ F × as a primitive element. Then the followingis a (0 , , ×
28 in the complex model on F : Φ = ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ . Example . Take q = 2 and ζ ∈ F × as a primitive element. Then the followingis a (0 , , ×
27 in the complex model on F : Φ = ζ ζ ζ ζ ζ ζ ζ ζζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ . ETFs from modular difference sets.
Next we show how ETFs in the com-plex model can be constructed from the following generalization of difference sets.
Definition . Let k, n ∈ N . A set D ⊆ Z /n Z is called a k -modulardifference set if the function c : Z /n Z → Z /k Z given by c ( g ) = |{ ( a, b ) ∈ D : a − b = g }| mod k is constant on ( Z /n Z ) \ { } .In order to convert modular difference sets to ETFs, we now define the discreteFourier transform matrix over a finite field. For simplicity we will only considerthe Fourier transform over finite cyclic groups, rather than the more general caseof finite abelian groups. Definition . Let q be a power of the prime p , and let α be a generator of themultiplicative group of F q . Given n ∈ N such that n | q + 1, set ω = α ( q − /n .Define the n × n discrete Fourier transform (DFT) matrix over F q by F = (cid:2) ω ij (cid:3) i,j ∈ Z /n Z . For D ⊂ Z /n Z we define the | D | × n submatrix F D = (cid:2) ω ij (cid:3) i ∈ D,j ∈ Z /n Z . RAMES OVER FINITE FIELDS 19
Using the notation of Definition 5.6, the next theorem shows that the matrix F D is an ETF if and only if D is a p -modular difference set. However, we wish toemphasize that the assumption n | q + 1 is essential. Indeed, there are p -modulardifference sets that do not give rise to ETFs over a finite field because they fail tosatisfy this condition. The following theorem generalizes a construction of complexETFs due to Strohmer and Heath [48, 55, 13]. Theorem 5.7.
Suppose q is a prime power, and n ∈ N satisfies n | q + 1 . Given D ⊆ Z /n Z , the matrix F D of Definition 5.6 is an ETF if and only if D is a p -modular difference set.Proof. Let D = { a , a , . . . , a d } ⊆ Z /n Z . As in Definition 5.5, we let c : Z /n Z → F q be defined by c ( g ) = |{ ( a, b ) ∈ D : a − b = g }| mod p, and we also consider c in vector form as c = (cid:2) c (0) · · · c ( n − (cid:3) ⊤ ∈ F nq . Foreach j ∈ Z /n Z , let ϕ j denote the j th column of F D , that is, ϕ j = (cid:2) ω a j ω a j ω a j · · · ω a d j (cid:3) ⊤ . A simple calculation shows that F D F ∗ D = nI. Since n | q + 1, we see that p ∤ n and hence n = 0. This implies that F D is an n -tight frame which is not totally isotropic. Since the entries in F D are unimodular,we see that ( ϕ i , ϕ i ) = d = | D | . Thus, for any set D ⊆ Z /n Z , the matrix F D is a( | D | , n )-NTF.Next, note that( ϕ i , ϕ j )( ϕ j , ϕ i ) = tr( ϕ ∗ i ϕ j ϕ ∗ j ϕ i ) = tr( ϕ i ϕ ∗ i ϕ j ϕ ∗ j ) = d X k =1 d X l =1 ω ( j − i )( a k − a l ) = X g ∈ Z /n Z c ( g ) ω ( j − i ) g = [ F c ] j − i . (5.1)Assume D is a p -modular difference set. That is, there is a number λ ∈{ , . . . , p − } such that c ( g ) = λ (mod p ) for all g ∈ ( Z /n Z ) \ { } . Now, for i = j , from (5.1) we see that( ϕ i , ϕ j )( ϕ j , ϕ i ) = X g ∈ ( Z /n Z ) \{ } c ( g ) ω ( j − i ) g + d = λ X g ∈ ( Z /n Z ) \{ } ω ( j − i ) g + d = − λ + d. Therefore, { ϕ i } i ∈ Z /n Z is a ( | D | , | D | − λ, n )-ETF.Finally, we assume F D is an ETF. From (5.1) we can deduce that there is aconstant α ∈ F q such that F c = (cid:2) n α α · · · α (cid:3) ⊤ . Since F ∗ F = nI = 0, we see that c = n − F ∗ F c = n − (cid:2) n − ( n − α n − α n − α · · · n − α (cid:3) ⊤ . Thus, D is a p -modular difference set. (cid:3) Example . For n = 14, the reader can check that D = { , , , , , , } is a3-modular difference set. Since n |
27 + 1, this produces a 7 ×
14 ETF in a unitarygeometry on F . A complex ETF of this size is known to exist, but it cannot beharmonic [19]. Indeed, there is no difference set of d = 7 elements in a group of n = 14 elements since n − d ( d −
1) [37].
Example . Let k ∈ N such that p = 3 k − q = p . Let H be a subgroup of Z / k Z such that | H | = 3 k . We claim that theset D = H ∪ { } is a p -modular difference set. Indeed, consider the function c fromDefinition 5.5. One can easily deduce that c ( g ) = ( k, if g ∈ H \ { } ;1 , if g ∈ ( Z / k Z ) \ H. Since 3 k = 1 (mod p ), this shows that D is a p -modular difference set. Additionally,we see that q + 1 = (3 k − + 1 = 0 (mod 9 k ), and hence 9 k | q + 1 . By the abovetheorem, the matrix F D is a (3 k + 1) × k ETF over F q . Since there are infinitelymany k such that 3 k − k = 2 this construction produces a 7 ×
18 ETF over F . Notethat there is no known construction of a complex ETF of this size, and there cannotbe a 7 ×
18 complex harmonic ETF for the same reason as in Example 5.8 [19].6.
ETFs from translation and modulation
In this section we introduce translation and modulation operators over finitefields, and show that they can be used to create NTFs just as in the complexsetting. By identifying an appropriate fiducial vector, we show that Gerzon’s boundis attained in unitary geometries of every dimension d = 2 l +1 over the field F .6.1. NTFs from translation and modulation.
Fix a prime power q , and chooseintegers d , . . . , d m ≥ q + 1. In this section we consider a finite fieldversion of the Heisenberg group over G := Q mk =1 Z /d k Z , where | G | = Q mk =1 d k =: d .This presents a finite model in which to investigate Zauner’s conjecture. We workin the unitary space V = F Gq of functions ϕ : G → F q , equipped with the form h ϕ, ψ i = X x ∈ G ϕ ( x ) q ψ ( x ) . Then V has orthonormal basis { δ x } x ∈ G , where δ x ∈ F dq is the indicator functionof a point, with δ x ( x ) = 1 and δ x ( y ) = 0 for y = x .In order to define modulation, we first introduce notation for T q -valued charac-ters on G . For each k ∈ [ m ], we fix a generator ω k ∈ T q ≤ F × q for the unique sub-group of order d k . Given x k , y k ∈ Z /d k Z we denote ˆ y k ( x k ) = ω x k y k k = ˆ x k ( y k ), andfor x = ( x , . . . , x m ) and y = ( y , . . . , y m ) in G we define ˆ y ( x ) = Q mk =1 ˆ y k ( x k ) ∈ T q .Then(6.1) ˆ y ( x ) = ˆ x ( y ) and ( x + y )ˆ( z ) = ˆ x ( z )ˆ y ( z ) , for x, y, z ∈ G, and y = 0 if and only if ˆ y ( x ) = 0 for every x ∈ G . Furthermore, for any x ∈ G ˆ y ( x ) X z ∈ G ˆ y ( z ) = X z ∈ G ˆ y ( x + z ) = X z ∈ G ˆ y ( z ) , RAMES OVER FINITE FIELDS 21 that is, (cid:0) ˆ y ( x ) − (cid:1) P z ∈ G ˆ y ( z ) = 0. It follows that(6.2) X z ∈ G ˆ y ( z ) = ( d, if y = 0;0 , otherwise.In this notation, each y ∈ G determines a modulation operator M y ∈ U( F Gq )and a translation operator T y ∈ U( F Gq ) given by( M y ϕ )( x ) = ˆ y ( x ) ϕ ( x ) and ( T y ϕ )( x ) = ϕ ( x − y ) , for ϕ ∈ F Gq , x ∈ G. It is straightforward to verify the usual relations(6.3) T x T y = T x + y , M x M y = M x + y , M y T x = ˆ y ( x ) T x M y , for x, y ∈ G. Our immediate goal is the following.
Proposition 6.1.
For any nonzero ϕ ∈ F Gq , the collection Φ = { T x M y ϕ } x,y ∈ G isan ( a, da ) -NTF for F Gq , where a = h ϕ, ϕ i . The NTF Φ of Proposition 6.1 is known as a
Gabor frame , and ϕ is its fiducialvector . If Φ happens to be an ETF then it is known as a Gabor ETF .In order to prove Proposition 6.1 we leverage a unitary geometry on the spaceHS( F Gq ) of all linear operators on F Gq . Given two such operators A, B we define h A, B i F = tr( A † B ) = X z ∈ G h Aδ z , Bδ z i . If A and B are represented by their matrices (cid:2) A xy (cid:3) and (cid:2) B xy (cid:3) over the standardbasis { δ x } x ∈ G , then h A, B i F = P x,y ∈ G A qxy B xy . It follows easily that h· , ·i F is anondegenerate form, and HS( F Gq ) is a unitary space. Lemma 6.2.
The collection { T x M y } x,y ∈ G is an orthogonal basis for HS( F Gq ) , andany A ∈ HS( F Gq ) satisfies (6.4) X x,y ∈ G h T x M y , A i F T x M y = dA. Proof.
For any x, y ∈ G we computetr( T x M y ) = X z ∈ G h δ z , T x M y δ z i = X z ∈ G ( T x M y δ z )( z ) = X z ∈ G ˆ y ( z − x ) δ z ( z − x ) . This is clearly 0 if x = 0, and otherwise it equals P z ∈ G ˆ y ( z ). By (6.2)tr( T x M y ) = ( d, if x = y = 0;0 , otherwise . For any x, y, s, t ∈ G the relations (6.3) and (6.1) now imply(6.5) h T x M y , T s M t i F = tr( M − y T − x T s M t ) = ˆ y ( x − s ) tr( T s − x M t − y ) = dδ x,s δ y,t . Furthermore, h T x M y , T x M y i = d d = Q mk =1 d k and each d k is coprimewith q . It follows easily that { T x M y } x,y ∈ G is an orthogonal basis for its span,which must equal HS( F Gq ) by dimension count. Finally, for any A ∈ HS( F Gq ) we can expand A = P x,y ∈ G c xy T x M y with c xy ∈ F q , and for any s, t ∈ G theidentity (6.5) produces h T s M t , A i F = X x,y ∈ G c xy h T s M t , T x M y i F = dc st . This implies (6.4). (cid:3)
Proof of Proposition 6.1.
Fix nonzero ϕ ∈ F Gq . Given ψ ∈ F Gq we denote ψϕ † ∈ HS( F Gq ) for the operator ( ψϕ † )( ϑ ) = h ϕ, ϑ i ψ , ϑ ∈ F Gq . Then for any choice of A ∈ HS( F Gq ) h A, ψϕ † i F = X x ∈ G h Aδ x , ψϕ † δ x i = X x ∈ G D Aδ x , h ϕ, δ x i ψ E = D A X x ∈ G h δ x , ϕ i δ x , ψ E . That is, h A, ψϕ † i F = h Aϕ, ψ i . Applying this identity and (6.4) we find X x,y ∈ G h T x M y ϕ, ψ i T x M y ϕ = X x,y ∈ G h T x M y , ψϕ † i F T x M y ϕ = d ( ψϕ † )( ϕ ) = daψ. Therefore ΦΦ † = daI . Furthermore, if ψ = 0 then ψϕ † = 0 since there ex-ists ϑ ∈ F Gq with h ϕ, ϑ i 6 = 0. Consequently, there exist x, y ∈ G with 0 = h T x M y , ψϕ † i F = h T x M y ϕ, ψ i . Since this holds for every ψ = 0, we conclude thatspan { T x M y ϕ } x,y ∈ G = F Gq . Finally, we have h T x M y ϕ, T x M y ϕ i = h ϕ, ϕ i = a forevery x, y ∈ G , since T x and M y are unitary. Therefore Φ is an ( a, da )-NTF. (cid:3) Gerzon equality in finite unitary geometries.
As in the complex setting,translations and modulations can be used to create ETFs that achieve equalityin Gerzon’s bound. The difficulty (as ever) lies in identifying a suitable fiducialvector ϕ . The next example shows how finite fields can simplify this problem bypresenting a finite search space that retains many salient features of the complexsetting. (See Theorem 6.4 and Example 7.9 further below for more examples ofETFs achieving Gerzon’s bound.) Example . We produce a 4 ×
16 Gabor ETF over a finite field. Take m = 1, d = d = 4, and q = 31. Let ζ ∈ F × be a primitive element, and define µ = ζ and ω = µ . The latter generate the unique subgroups of T with orders 8 and 4, re-spectively. Define R ∈ F × to be the diagonal matrix with entries R jj = µ j ( j +4) for j ∈ Z / Z , and let F = (cid:2) ω ij (cid:3) ∈ F × be the DFT matrix of Definition 5.6. (Through-out this example we use the ordering Z / Z = { , , , } .) Then Z := R F ∈ F × isakin to (a scalar multiple of) Zauner’s 4 × ×
16 ETF [56]. A similar phenomenon occursover F , where the finite search space makes it easier to identify an appropriatefiducial vector. Specifically, Z has exactly three eigenvalues: λ = ζ , λ = ζ ,and λ = ζ with respective geometric multiplicities 2 , ,
1. The two-dimensionaleigenspace for λ contains exactly 924 one-dimensional subspaces. Checking onerepresentative from each line, we find (up to rescaling) exactly four fiducial vectorsin the λ -eigenspace that generate Gabor ETFs, namely ϕ = ζζ ζ , ϕ = ζ ζ ζ , ϕ = ζ ζ ζ , ϕ = ζ ζ ζ . RAMES OVER FINITE FIELDS 23
We now identify fiducial vectors for infinitely many Gabor ETFs. Theorem 6.4proves that Gerzon’s bound is attained in unitary geometries of infinitely manydimensions over the field F = F . This is the first field for which this phenomenonis known to occur. (In a companion paper we perform a similar feat for Gerzon’sbound in orthogonal geometries [24].) Theorem 6.4.
Take q = 3 , m to be odd, and d k = 2 for ≤ k ≤ m , so that G = ( Z / Z ) m and d = 2 m . Let ζ ∈ F be a primitive element, and define ϕ ∈ F d by ϕ ( x ) = ( − − ζ , if x = 0;1 , otherwise . Then
Φ = { T x M y ϕ } x,y ∈ G is a (0 , , -ETF. In particular, Gerzon’s bound is at-tained in a unitary space over F whenever its dimension is twice a power of .Proof. Observe that x = − x in G and 2 ≡ − F . In particular, d = 2 m ≡ − ζ ) = − ζ ) = − ζ .Let ∈ F G be the function that is constantly 1, so ϕ = + (1 − ζ ) δ . Herethe coefficient on δ satisfies (1 − ζ ) = 1 + ζ and (1 − ζ ) = 1 − ( ζ ) = − x, y ∈ G we expand to find h ϕ, T x M y ϕ i = h , T x M y i +(1 − ζ ) h , T x M y δ i +(1+ ζ ) h δ , T x M y i−h δ , T x M y δ i . Given z ∈ G we have ( T x M y )( z ) = ˆ y ( z − x ) and ( T x M y δ )( z ) = δ x ( z ). Thus h , T x M y i = X z ∈ G ( T x M y )( z ) = X z ∈ G ˆ y ( z − x ) = X z ∈ G ˆ y ( z ) = dδ y, = − δ y, and h , T x M y δ i = X z ∈ G ( T x M y δ )( z ) = 1 , while h δ , T x M y i = ( T x M y )(0) = ˆ y ( − x ) = ˆ y ( x )and h δ , T x M y δ i = ( T x M y δ )(0) = δ x, . Therefore h ϕ, T x M y ϕ i = − δ y, + 1 − ζ + (1 + ζ )ˆ y ( x ) − δ x, . Since d k = 2 for every k we must have ˆ y ( x ) ∈ {± } . Simplifying above, we find h ϕ, T x M y ϕ i = , if x = y = 0;1 , if { } ( { x, y } ; − , if 0 / ∈ { x, y } and ˆ y ( x ) = 1; ζ , otherwise . In particular, h ϕ, T x M y ϕ i = 1 whenever { x, y } 6 = { } . Now for any s, t ∈ G with ( s, t ) = ( x, y ) the relations (6.3) produce h T x M y ϕ, T s M t ϕ i = h ϕ, M − y T − x T s M t ϕ i = ˆ y ( x − s ) h ϕ, T s − x M t − y ϕ i , so that h T x M y ϕ, T s M t ϕ i = 1. Hence Φ = { T x M y ϕ } x,y ∈ G is a (0 , , , (cid:3) Remark . Over the complex numbers an ETF that attains equality in Gerzon’sbound is also known as a symmetric informationally complete positive oper-ator valued measure , or a
SIC-POVM . This terminology comes from quantuminformation theory, where such ETFs are important partly because their outer prod-ucts ϕϕ † provide a basis for operator space that consists of rank-one projectionssumming to a nonzero multiple of the identity [46].We intentionally avoid this terminology in Theorem 6.4 since the ETFs it pro-duces consist of isotropic vectors, and the frame constant is zero. Consequentlythe outer products all have trace zero, and no linear combination of them recreatesthe identity operator. However, as shown in the proof of Gerzon’s bound (Theo-rem 4.2) the outer products span the codimension-one space of traceless operators,and they form the finite field equivalent of a simplex in that space. In doing so theyproduce the largest possible collection of traceless rank-one self-adjoint operatorswith constant pairwise value of h· , ·i F . Remark . The ETFs created in Theorem 6.4 may be seen as an infinite familythat generalizes Hoggar’s lines [28, 29]. The latter refers to an 8 ×
64 complex ETFwith entries in the ring Z [ i ] of Gaussian integers. (Here we have in mind the versiongiven by Jedwab and Wiebe [36].) Specifically, let G = ( Z / Z ) , and consider thetranslation and modulation operations on C G given by( ˜ T y ψ )( x ) = ψ ( x − y ) and ( ˜ M y ψ )( x ) = ( − x · y for ψ ∈ C G , x, y ∈ G, where x · y denotes the dot product. Define ψ ∈ C G by ψ ( x ) = ( − i, if x = 0;1 , otherwise.Then the system Ψ = { ˜ T x ˜ M y ψ } x,y ∈ G is an 8 ×
64 complex ETF [36, 51]. To relatethis with an ETF given by Theorem 6.4, let ζ ∈ F × be a primitive element, and let f : Z [ i ] → F be the unique ring homomorphism given by f ( a + bi ) = a + bζ . (Thisis well defined since Z [ i ] ∼ = Z [ x ] / ( x + 1) and ζ = − f ( z ) = f ( z ) for every z ∈ Z [ i ]. It follows that f maps the ETF Ψ ∈ Z [ i ] × to an equiangularsystem f (Ψ) ∈ F × . (Here we choose orderings on G and G × G to identify C G with C , and so on. We also extend f to a mapping on matrices by entrywiseapplication.) In fact, f (Ψ) = Φ is exactly the ETF of Theorem 6.4 when m = 3,where f ( ψ ) = ϕ is the given fiducial vector in that case.Furthermore, it is possible to recover Hoggar’s lines from the finite field ETF Φwhen m = 3. Explicitly, let H ∈ F × q be the Gram matrix of Φ. As shown inthe proof of Theorem 6.4, H has zeros on the diagonal, with off-diagonal entriesin T = {± , ± ζ } . Let g : { } ∪ T → C be the multiplicative character given by g (0) = 0 and g ( ζ l ) = i l . Extending g to a mapping on matrices, it turns out that S := g ( H ) ∈ C × has exactly two eigenvalues. Adding an appropriate amountof identity, we find that S + 3 I is the Gram matrix of a complex 8 ×
64 ETF. Infact S + 3 I = Ψ ∗ Ψ is the Gram matrix of Hoggar’s lines.Sadly, this procedure does not produce a complex ETF of size 32 × m = 5 in Theorem 6.4, where we found that the corresponding matrix S ∈ C × has three eigenvalues. This is not surprising, since Godsil and Royhave shown that the group G = ( Z / Z ) m generates a complex 2 m × m GaborETF only if m ∈ { , } [22]. RAMES OVER FINITE FIELDS 25
Finally, we remark that Hoggar’s lines are highly symmetric and have a doublytransitive automorphism group [57, 34]. We have not investigated the symmetriesof the ETFs given by Theorem 6.4, and we leave this problem for future study.
Problem . For Φ as in Theorem 6.4, determine the group Aut Φ of all permu-tations µ ∈ S ( G × G ) having components µ ( x, y ) =: (cid:0) µ ( x, y ) , µ ( x, y ) (cid:1) for whichthere exist scalars c µ ( x, y ) ∈ F × q such that T µ ( x,y ) M µ ( x,y ) ϕ = c µ ( x, y ) T x M y ϕ forevery x, y ∈ G . 7. Finite field ETFs from complex ETFs
In this section we prove that every complex ETF produces ETFs in infinitelymany finite fields. In the process we show that the existence of a d × n complexETF implies that of a d × n ETF with algebraic entries. Our main result is thefollowing.
Theorem 7.1.
Suppose there is a d × n complex ETF. Then, for infinitely manypairwise coprime q , there is an ETF of n vectors in a unitary geometry on F dq . We proceed in two steps. First we nudge the complex ETF to have algebraicentries, then we map the algebraic ETF into infinitely many finite fields.7.1.
Preliminaries.
First we recall some basic number theory, where standardreferences include [14, 35, 44]. An algebraic number is a zero of a polynomialwith rational coefficients; an algebraic integer is a zero of a monic polynomialwith integer coefficients. The algebraic integers form a ring, and every algebraicnumber is a ratio of algebraic integers. A number field E is a field with Q ≤ E ≤ C and dim Q E < ∞ . We write O E for the ring of algebraic integers contained in E .Its ideals have the following properties. Proposition 7.2. If E is a number field, then the following hold for any propernonzero ideal a ⊂ O E : (a) O E / a is finite, (b) if a is prime then it is maximal, (c) only finitely many prime ideals of O E contain a . If E is a number field, then there exists α ∈ E such that E = Q ( α ), and E = { f ( α ) : f ∈ Q [ x ] } . The minimal polynomial of α is the monic polynomial m α ∈ Q [ x ] of lowest degree such that m α ( α ) = 0. We say E is Galois if it containsevery root of m α . In any case, there exists a Galois number field containing E .If E is Galois then both E and O E are closed under complex conjugation. Thisgives each the structure of a ∗ -ring , that is, a ring equipped with an involutoryring automorphism. A ∗ -ring homomorphism is a ring homomorphism between ∗ -rings that preserves the involution. This completes our brief review.7.2. Complex ETFs with algebraic entries.Theorem 7.3.
If there is a d × n complex ETF, then there is a d × n complex ETFwith algebraic entries. To prove Theorem 7.3, consider the algebra of sets generated by sets of the form { x ∈ R n : f ( x ) ≥ } , f ∈ Z [ x , . . . , x n ] . We refer to members of this algebra as integral semialgebraic sets . For example,by taking real and imaginary parts of matrix entries, the set of d × n complex ETFsmay be viewed as an integral semialgebraic subset of R dn . As such, Theorem 7.3is a special case of the following. Lemma 7.4.
Every nonempty closed integral semialgebraic set contains a pointwhose coordinates are all real algebraic numbers.Proof.
Let S ⊆ R n be nonempty, closed, and integral semialgebraic. By nonemp-tyness, there exists an integer k such that S := S ∩ { x : k x k ≤ k } is nonempty.For each i ∈ [ n ], let π i : R n → R denote projection onto the i th coordinate. We williteratively take z i := sup π i ( S i − ) , S i := S i − ∩ { x ∈ R n : π i ( x ) = z i } . We claim that the following hold for each i ∈ [ n ]:(i) z i is a real algebraic number,(ii) S i is a nonempty compact integral semialgebraic subset of R n .Considering S ⊇ S n = { ( z , . . . , z n ) } , the result follows from this claim.Observe that S is a nonempty compact integral semialgebraic subset of R n . Wewill show that (i) and (ii) together follow from S i − being a nonempty compactintegral semialgebraic subset of R n , and then the claim follows by induction. Since S i − is nonempty, compact, and integral semialgebraic, it follows from Tarski–Seidenberg (Theorem 1.4.2 of [4]) that π i ( S i − ) is also nonempty, compact, andintegral semialgebraic. As such, π i ( S i − ) is the disjoint union of finitely manycompact intervals with real algebraic endpoints. It follows that (i) holds. Since z i ∈ π i ( S i − ), it follows that S i is nonempty. To finish the proof of (ii), consideringour hypothesis on S i − , it suffices to demonstrate that { x : x i = z i } is closed andintegral semialgebraic. It is closed since π i is continuous. To see it is integralsemialgebraic, consider any polynomial f ∈ Z [ x ] for which f ( z i ) = 0, and select a, b ∈ Q such that z i is the only root of f in [ a, b ]. Then { x : x i = z i } = { x : f ( x i ) = 0 } ∩ { x : x i ≥ a } ∩ { x : x i ≤ b } is integral semialgebraic, as desired. (cid:3) Projecting complex ETFs with algebraic entries.Theorem 7.5.
Suppose Φ ∈ C d × n is an ETF with entries in O E , where E is aGalois number field. Then, for infinitely many pairwise coprime q there exists a ∗ -ring homomorphism f : O E → F q such that f (Φ) ∈ F d × nq is a d × n ETF in thecomplex model.
Our proof of Theorem 7.5 uses the following.
Lemma 7.6.
Let E be a Galois number field. Given any choice of nonzero algebraicintegers z , . . . , z n ∈ E , there exist infinitely many ideals p ⊂ O E such that all ofthe following are true: (i) O E / p is a finite field, (ii) p is closed under complex conjugation, (iii) p does not contain any of z , . . . , z n .Additionally, the set of field characteristics { char O E / p : p ⊂ O E is an ideal satisfying (i)–(iii) } RAMES OVER FINITE FIELDS 27 so obtained is infinite.Proof.
Let F = E ∩ R . Then F is the fixed field for the subgroup of Gal( E/ Q )generated by complex conjugation. By the Fundamental Theorem of Galois Theory, E is a Galois extension of F with automorphism group Gal( E/F ) ∼ = Z / Z . Inparticular, E is a cyclic extension of F .As a consequence of the Frobenius density theorem (Cor. 5.4 in [35, Ch. V]),there are infinitely many prime ideals q ⊂ O F for which p = q O E is a prime idealof O E . Furthermore, for each p ⊂ O E there is at most one prime ideal q ⊂ O F such that p = q O E , namely q = p ∩ O F . (Indeed, if q ⊂ O F is prime and p = q O E ,then p ∩ O F contains the maximal ideal q .) Hence there are infinitely many primeideals p ⊂ O E of the form p = q O E , where q is a prime ideal of O F . Moreover,there are only finitely many prime ideals in O E that contain any of the ideals z O E , . . . , z n O E . Overall, there are infinitely many choices of prime ideals p ⊂ O E that avoid z , . . . , z n and take the form p = q O E , where q ⊂ R . Any such p isclosed under complex conjugation since this is true of both q and O E , and O / p isa finite field by Proposition 7.2.To prove the “additionally” statement, we verify that any prime ideal p contains(char O E / p ) O E . By considering the minimal polynomial of a nonzero element of p ,we see that p contains a nonzero integer (namely the opposite of the polynomial’sconstant term). As such, there exists a positive prime p ∈ Z such that p Z = p ∩ Z ,the latter being a nonzero prime ideal of Z . We have p O E ⊂ p , and the containment O E / p ⊃ ( Z + p ) / p ∼ = Z / ( Z ∩ p ) ∼ = Z /p Z , demonstrates that p = char O E / p . This proves the claim. By Proposition 7.2(c),there are only finitely many prime ideals p ⊂ O E associated with any given charac-teristic p = char O E / p . Since infinitely many prime ideals satisfy (i)–(iii), the setof associated field characteristics must also be infinite. (cid:3) Proof of Theorem 7.5.
First observe that the frame constant c for Φ belongs to O E since cI = ΦΦ ∗ . Select an ideal p ⊂ O E as in Lemma 7.6, where p does not contain c .Denote K = O E / p , and let g : O E → K be the quotient mapping. Since p is closedunder complex conjugation, there is a well-defined field automorphism σ of K givenby g ( x ) σ = g ( x ) for x ∈ O E , and σ = 1. By passing to a quadratic extension of K if necessary, we obtain a finite field F q and a ring homomorphism f : O E → F q with kernel p , such that f ( x ) = x q for every x ∈ O E . Then f (Φ) f (Φ) ∗ = f (ΦΦ ∗ ) = f ( c ) I = 0, and so f (Φ) is a tight frame for F dq . The other properties are verifiedsimilarly, and f (Φ) is an ETF. (cid:3) Finally, we obtain our main result as a corollary of Theorem 7.5.
Proof of Theorem 7.1.
Suppose there is an ETF Φ ∈ C d × n . By Theorem 7.3 wemay take Φ to have algebraic entries. After rescaling Φ to clear any algebraic integerdenominators, we may assume its entries are in fact algebraic integers. Now let E beany Galois extension of Q containing the entries of Φ, and apply Theorem 7.5. (cid:3) Remark . Many ETFs Φ ∈ C d × n are constructed by way of their Gram matrices G = Φ ∗ Φ. Often G carries a nice structure while Φ is essentially unknown (yetguaranteed to exist by Cholesky decomposition). In such cases it may be desirableto project G into a finite field instead of Φ itself. The procedure in Theorem 7.5works just as well to map G into finite fields with infinitely many characteristics, provided the entries of G are all algebraic integers. Indeed, let E be a number fieldwhose integer ring O E contains the entries of G as well as its nonzero eigenvalue c .Choose any ideal p ⊂ O E not containing c as in Lemma 7.6, and let f : O E → F q bea ∗ -ring homomorphism with kernel p . Then f ( G ) ∈ F n × nq is self-adjoint, and d ′ :=rank f ( G ) ≤ rank G = d since the rank of a matrix is the largest size of a squaresubmatrix having nonzero determinant. Moreover, applying f to the coefficients ofthe characteristic polynomial of G shows that det (cid:0) xI − f ( G ) (cid:1) = [ x − f ( c )] d x n − d .Considering the Jordan normal form of f ( G ) we conclude that d ′ ≥ d , and equalityholds. Applying Theorem 3.12, we conclude that f ( G ) factors to produce a d × n ETF in a unitary geometry over F q . Example . Let k ≥ α k ∈ C be a primitive k -throot of unity. Suppose Φ ∈ C d × n is an ETF whose frame constant c and Grammatrix entries all lie in Z [ α k ]. Then we can project Φ into the complex model overa finite field as follows. Choose any prime power q such that k divides q + 1, andlet ω k ∈ T q ≤ F × q be a generator for the unique subgroup of order k . Denote the k -th cyclotomic polynomial by m k ( x ) ∈ Z [ x ], and recall that x k − Q j | k m j ( x ).Then Z [ α k ] ∼ = Z [ x ] / (cid:0) m k ( x ) Z [ x ] (cid:1) , and m k ( ω k ) = 0 since ω k is a root of x k − x j − j < k . Consequently, there is a well-defined ring homomorphism f : Z [ α k ] → F q given by f ( g ( α k )) = g ( ω k ) for every g ∈ Z [ x ]. Furthermore, f ( z ) = f ( z ) q for every z ∈ Z [ α k ] since α k = α k − k and ω k − k = ω qk . It followseasily that f (Φ ∗ Φ) ∈ F n × nq is the Gram matrix of an ETF in the complex modelon F d ′ q , where d ′ = rank f (Φ ∗ Φ). We have d ′ ≤ d with equality if f ( c ) = 0, as inRemark 7.7. Finally, if Φ ∈ Z [ α k ] d × n and f ( c ) = 0 then f (Φ) ∈ F d × nq is itself anETF, as in the proof of Theorem 7.5. Example . Table 1 describes some finite unitary geometries that admit GaborETFs, where the abelian group that provides translations and modulations is cyclic.In creating the table we began with a known fiducial vector for a complex GaborETF, and then applied the construction of Theorem 7.5 to map it into a finitefield. The fiducial vector ϕ ∈ C d was taken from one of [23, 47, 1]. In each case theresulting Gabor frame Φ ∈ C d × d had entries in a number field E , and we multipliedby a scalar to clear fractions and put the entries in O E . Defining F = E ∩ R , we thenused Magma [6] to identify a prime ideal q ⊂ O F for which p := q O E remaineda prime ideal of O E . Table 1 gives the size of O E / p ∼ = F q . Here, the quotientmapping produces a ∗ -ring homomorphism f : O E → F q , and f (Φ) ∈ F d × d q is anETF as in Theorem 7.5. Furthermore, considering the images under f of complextranslation and modulation matrices, we see that f (Φ) is itself a Gabor ETF.We emphasize that Table 1 does not describe all finite fields that admit projec-tions of the given complex fiducial vectors. Our method focused on ideals of O E that lie entirely in R , but in many cases there are other prime ideals of O E that areclosed under complex conjugation, and the corresponding quotient mappings yieldGabor ETFs in other finite fields not listed in Table 1.7.4. Open problems.
We end with problems for future investigation. The first isthe finite field analog of Zauner’s conjecture.
RAMES OVER FINITE FIELDS 29 ϕ q
2a 167, 191, 239, 263, 311, 383, 743, 863, 887, 911, 983, 1031, 11033a, 3b, 3c 167, 191, 239, 263, 311, 383, 743, 863, 887, 911, 983, 1031, 11034a 71, 191, 239, 311, 359, 431, 479, 599, 719, 839, 911, 1031, 11515a 179, 239, 359, 419, 479, 599, 659, 719, 839, 1019, 1259, 13196a 47 , 59 , 83 , 131, 167 , 227, 251 , 311 , 383 , 419 , 467, 479 , 47 , 103 , 167, 199 , 223, 271 , 311 , 367 , 383 , 439 , 479
8a 479, 911, 2351, 2399, 2591, 2879, 3119, 5279, 5471, 5711, 69598b 79, 191, 239, 271, 431, 479, 719, 751, 911, 991, 1039, 11519a, 9b 11 , 59 , 71 , 131 , 179 , 191 , 239 , 251 , 311 , 359 , 419 , 43110a 19 , 479 , 1319 , 1559, 1979 , 2939, 2999 , 3659 , 3779 , 4259 , 2879 , 5519 , 10559, 11519 , 12239 , 14159 , 1451911c 167 , 239 , 1223 , 1487 , 2063 , 2111, 2207 , 2543 , 2591 , 2879
12a 263 , 503 , 599 , 647 , 1439 , 1871 , 2063 , 2207, 2447 , 259112b 263, 503, 599, 647, 1439, 1871, 2063, 2207, 2447, 2591, 268713a, 13b 251 , 439 , 1291 , 1559, 2539 , 3631 , 4339 , 5659 , 479 , 1091 , 1151 , 1319 , 1559 , 1811 , 1931 , 1979 , 359 , 719 , 1319 , 2879, 2999 , 3359 , 4799 , 4919, 5039
15d 59 , 179 , 239 , 359 , 419, 479, 659, 719 , 839 , 1019, 1259 Table 1.
The first column describes a known fiducial vector fora complex Gabor ETF, as in [47, 1, 23]. The number d in front ofthe letter gives the dimension of the ETF, and Z /d Z is the abeliangroup that provides translations and modulations. The secondcolumn lists some prime powers q for which the complex ETF maybe projected into F d × d q as a Gabor ETF over a finite field. Thefinite field fiducial vectors are included in an ancillary file with thearXiv version of this paper. Example 7.9 gives our methodology. Conjecture . For every d , there existinfinitely many pairwise coprime q such that a unitary geometry on F q admits anETF of n = d vectors.According to Theorem 7.1, if Zauner’s conjecture holds over the complex numbersthen Conjecture 7.10 is also satisfied. More generally, we pose the following. Problem . For which ( d, q ) does a unitary geometry on F dq admit an ETF of n = d vectors?Next, the converse of Theorem 7.1 remains open. Problem . If there exist d × n ETFs in unitary geometries over infinitely manyfinite fields with distinct characteristics, does there also exist a d × n complex ETF?If the words “unitary” and “complex” are replaced by “orthogonal” and “real”in Problem 7.12, then the resulting question has an affirmative answer. See Propo-sition 3.3 of [24]. Problem . Identify necessary conditions for the existence of d × n ETFs in finiteunitary geometries.
An effective solution to Problem 7.13, when paired with Theorem 7.1, wouldprovide necessary conditions on the existence of complex ETFs.
Acknowledgments
GRWG was partially supported by the Singapore Ministry of Education Aca-demic Research Fund (Tier 1); grant numbers: RG29/18 and RG21/20. JJ was sup-ported by NSF DMS 1830066. DGM was partially supported by AFOSR FA9550-18-1-0107 and NSF DMS 1829955. This project began at the 2018 MFO Mini-Workshop on Algebraic, Geometric, and Combinatorial Methods in Frame Theory.The authors thank the other participants, Matt Fickus, and Steve Flammia forcomments that motivated the authors or provided insight.
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School of Physical and Mathematical Sciences, Nanyang Technological University,Singapore 637371
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