Towards a classification of incomplete Gabor POVMs in $\mathbb{C}^d$

TOWARDS A CLASSIFICATION OF INCOMPLETE GABOR POVMS IN Cd arXiv:2106.01509v1 [math.FA] 2 Jun 2021 ASSAF GOLDBERGER, SHUJIE KANG, AND KASSO A. OKOUDJOU Abstract. Every (full) finite Gabor system generated by a unit-norm vector g ∈ Cd is a finite unit-norm tight frame (FUNTF), and can thus be associated with a (Gabor) positive operator valued measure (POVM). Such a POVM is informationally complete if the d2 corresponding rank one matrices form a basis for the space of d × d matrices. A sufficient condition for this to happen is that the POVM is symmetric, which is equivalent to the fact that the associated Gabor frame is an equiangular tight frame (ETF). The existence of Gabor ETF is an important special case of the Zauner conjecture. It is known that generically all Gabor FUNTFs lead to informationally complete POVMs. In this paper, we initiate a classification of non-complete Gabor POVMs. In the process we establish some seemingly simple facts about the eigenvalues of the Gram matrix of the rank one matrices generated by a finite Gabor frame. We also use these results to construct some sets of d2 unit vectors in Cd with a relatively smaller number of distinct inner products. Contents 1. Introduction and background 2. Informationally Complete POVMs 2.1. Spectrum of the Gramian G(g) 2.2. Algebraic structure of Gabor POVMs 3. Invariance properties of the rank of Gabor POVMs 3.1. Characterization of rank d Gabor POVMs 3.2. Examples of automorphisms of two types of Gabor systems Acknowledgements References 1. Introduction and background Let M and T denote the modulation and translation operators defined by  M g = (ω n g )d−1 n n=0 T g = (gn−1 )d−1 n=0 d 2πi/d is a dth root of unity. for g = (gn )d−1 n=0 ∈ C , and where ω = e Date: June 4, 2021. 2000 Mathematics Subject Classification. Primary 42C15 Secondary 65T50, 81R05, Key words and phrases. Finite Gabor systems, Equiangular Tight Frames, k-distant sets. 1 1 3 4 8 11 12 13 20 20 2 ASSAF GOLDBERGER, SHUJIE KANG, AND KASSO A. OKOUDJOU The (finite) Gabor system G(g) := G(g, Zd × Zd ) = {M k T ℓ g}d−1 k,ℓ=0 generated by a unit vector d g ∈ C is a Finite Unit Norm Tight Frame (FUNTF), that is d−1 X k,ℓ=0 k ℓ 2 3 2 |hx, M T gi| = d kxk ⇐⇒ x = 1 d3 d−1 X k,ℓ=0 hx, M k T ℓ giM k T ℓ g for each x ∈ Cd , see [8, 23]. It follows that G(g) can be cononically associated with a positive operator valued measure (POVM). By definition, a (Gabor) POVM is informationally complete (IC) if the d2 viewed as the space of d × d (d2 ) rank-one matrices {M k T ℓ gg ∗ T −ℓ M −k }d−1 k,ℓ=0 form a basis for C matrices. If in addition, ′ ′ ′ ′ ′ ′ |hM k T ℓ gg ∗ T −ℓ M −k , M k T ℓ gg ∗ T −ℓ M −k iHS | = |hM k T ℓ g, M k T ℓ gi|2 is constant for (k, ℓ) 6= (k ′ , ℓ′ ), then we say that the Gabor POVM is symmetric (S). Here and in the 2 sequel, we use hA, BiHS = trace(AB ∗ ) to denote the Hilbert-Schmidt inner product on Cd . It is a simple fact to check that every symmetric Gabor POVM is also informationally complete, but the converse is not necessarily true. In the sequel, a (Gabor) FUNTF G(g) will be called a SIC-POVM if the corresponding (Gabor) POVM is symmetric and informationally complete. We refer to [3, 20] for more on POVMs. It was conjectured by Zauner [24] that for each d ≥ 1 there exists a unit vector g ∈ Cd such that the Gabor system G(g) is a SIC-POVM. This conjecture is usually stated in the following form: Conjecture. [24] In any Cd , d > 2, there exists Gabor SIC-POVM generated by a single unit norm vector g under the orbit of Heisenberg group. In particular, for any k, ℓ ∈ Z/dZ \ {(0, 0)}, 1 , |hg, M k T ℓ gi| = √d+1 The conjecture remains open, and the search for the generating vector (or fiducial vector) has been resolved in dimensions 2 − 16, 19, 24, 28, 35, 48, 124, and 323. Moreover, numerical solutions in dimensions up to 67 have also been established. We refer to [4, 15, 24] and the references therein for more on the Zauner conjecture. We note that the (full) Gabor frame G(g) always generates a POVM, so the Zauner conjecture asserts that one can always find a generator so that this POVM is also symmetric, and hence informationally complete. The question of completeness of the Gabor POVM had long been investigated and it is known ([17, 12]) that G(g) is an informationally complete if and only if (1) hg, M k T ℓ gi = 6 0 ∀ (k, ℓ) ∈ Zd × Zd . It is worth observing that the completeness of Gabor POVMs is central in the theory of phase retrieval [5, 6, 10]. In particular, in [10] it was proved that the injectivity of the phase retrieval map generated by a Gabor system G(g) is equivalent to (1). Furthermore, this condition is generic in the sense that (1) holds for “almost all” unit-norm vectors g. In this paper, we offer a new proof of the completeness of Gabor POVMs. Our proof is based on the spectral analysis of the d2 ×d2 Gram matrix associated to the rank-one matrices {M k T ℓ gg ∗ T −ℓ M −k }d−1 k,ℓ=0 . 2 k ℓ 2 In particular, we prove that the d eigenvalues of this Gram matrix are λk,ℓ = d|hg, M T gi| where k, ℓ = 0, 1, . . . , d − 1. To the best of our knowledge this simple fact has never been stated in this manner before, though one of the results in [10, Theorem 2.3] asserts that the spectrum of this Gram TOWARDS A CLASSIFICATION OF INCOMPLETE GABOR POVMS IN Cd 3 matrix lies in the interval [mink,ℓ d|hg, M k T ℓ gi|2 , maxk,ℓ d|hg, M k T ℓ gi|2 ]. After re-deriving (1), we (re) prove that this condition is generic in the sense that the set of all generators of such systems is open dense in the unit sphere of Cd . We then focus on classifying non-IC Gabor POVMs. This is an interesting problem in its own right and is the primary motivation for this paper. We use some of our classification results to construct k− distance sets of d2 elements in Cd , where k is proportional to d. The rest of the paper is organized as follows. In Section 2 we give two different characterization of Gabor IC-POVMs. Both characterizations are based on the spectral analysis of the Gramian of the rank-one matrices associated to the Gabor system. The first approach which was mentioned above is based on explicit formulaes for the eigenvalues of the Gram matrix. The second approach is achieved by formulating the problem using an algebraic framework that we hope to use to investigate other POVMs. The algebraic approach also allows us to investigate the rank of the span of the rank-one matrices associated to the non IC Gabor POVMs. In Section 3 we discuss some invariance of the rank of these POVMs. We refer to [18] for a related work dealing with POVMs generated by rank 2 positive semidefinite matrices. 2. Informationally Complete POVMs The goal of this section is to re-derive the characterization of IC Gabor POVMs. In the sequel, for every vector g we define the support of g to be the set supp(g) = {i| gi 6= 0}. In addition, the cardinality of supp(g) will be denoted by kgk0 . Given a unit vector g ∈ Cd , the collection G(g) = {gk,ℓ := M k T ℓ g}(k,ℓ)∈Zd ×Zd will denote the (Gabor) FUNTF generated by g. We will view G(g) as an ordered multiset, with respect to the lexicographical order of Zd × Zd , and we shall abuse notation and denote again by G(g) the d × d2 matrix whose columns are the vectors of this frame given in this order. To each vector gk,ℓ we associate a rank-one matrix given by ∗ Πk,ℓ := Πk,ℓ (g) = gk,ℓ ⊗ gk,ℓ = gk,ℓ gk,ℓ . d−1 We seek a characterisation of all g for which the rank-one matrices {Πk,ℓ }k,ℓ=0 span the space of all d × d matrices. This is the case if and only if the Gram matrix G(g) := (hΠk,ℓ , Πk′ ,ℓ′ i) = (tr(Π∗k,ℓ Πk′ ,ℓ′ )) of this set of matrices if full rank. This is also equivalent to the information completeness of the associated Gabor POVM. Thus we focus on analyzing the rank of G(g) by giving a full descriptions of its spectrum. The completness of Gabor POVM was also obtained in [17, Theorem 15; Section IV-A] and in [12, Section 4.1]. Nonetheless, we point out that our result also gives the rank of the Gram matrix G(g). The main result of this section is the following theorem that gives the rank of the Gram matrix G(g). 4 ASSAF GOLDBERGER, SHUJIE KANG, AND KASSO A. OKOUDJOU Theorem 1. Let g ∈ Cd be a unit vector. The rank of the Gramian G(g) equals the number of nonzero entries in the multiset {hg, M k T ℓ gi}(k,ℓ)∈Zd ×Zd . Consequently, the Gabor POVM is informationally complete if and only if hg, M k T ℓ gi = 6 0 for all k, ℓ. In Section 2.1 we compute all the eigenvalues of G(g) obtaining its rank as a consequence. We note that [10, Theorem 2.3] shows that (1) implies that G(g) has full rank. Their proof consists of showing that (1) implies that the lowest eigenvalue of G(g) is positive. Our result shows that (1) is also a necessary for G(g) to be full rank. In Section 2.2, we use an algebraic framework to give a second proof of Theorem 1. We then use this framework in Section 3 to identify some invariants of the rank of G(g). 2.1. Spectrum of the Gramian G(g). In this section, we prove that the Gram matrix G(g) is block circulant with circulant blocks. This structure allows us to compute the eigenvalues of G(g) since it is diagonalizable by an appropriate DFT matrix. In particular, the first part of result is an extension of a well-known fact about block circulant matrices [22, 11]. To the best of our knowledge, the remaining parts are new, though they seem quite simple. Let F := Fd denote the d × d DFT √ matrix. In particular, the (k, ℓ) entry of F is Fk,ℓ = ω kℓ / d, where ω = e2πi/d . Theorem 2. Let g be a unit vector in Cd , and G(g) = (hΠk,ℓ , Πk′ ,ℓ′ i) be the Gram matrix of the rank-one projectors {Πk,ℓ = M k T ℓ g ⊗ M k T ℓ g}d−1 k,ℓ=0 . The following statements hold. (i) G(g) is a block circulant matrix with circulant blocks. In particular, the ((k, ℓ), (k ′ , ℓ′ ))-th entry of G is ′ ′ G((k,ℓ),(k′ ,ℓ′ )) ≡ hΠk,ℓ , Πk′ ,ℓ′ i = |hg, M k −k T ℓ −ℓ gi|2 , where k, k ′ , ℓ, ℓ′ ∈ {0, · · · , d − 1}. (ii) G(g) can be diagonalized by F ⊗ F . In particular, the eigenvalues {λa,b }d−1 a,b=0 of G(g) are given by λa,b = d−1 X k,l=0 ω ak+bl |hg, M k T l gi|2 = d d−1 X 2 ∗ gk gk+a ω bk = d|hg, M b T a gi|2 k=0 where a, b ∈ Zd . In particular, up to a factor of d, the entries of G(g) are the same as its eigenvalues. Proof. Let g be a unit vector in Cd . ′ ′ ′ ′ (i) G((k,ℓ),(k′ ,ℓ′ )) (g) = |hgk,ℓ , gk′ ,ℓ′ i|2 = |hM k T ℓ g, M k T ℓ gi|2 = |hg, M k −k T ℓ −ℓ gi|2 . To prove that G(g) is block circulant matrix with circulant blocks, we decompose G(g) into d2 blocks, each of the size d × d as following:   A0,0 A0,1 A0,2 · · · A0,d−1   A1,1 A1,2 · · · A1,d−1   A1,0   G(g) =  . .. .. .. ..  . . . .  ..  Ad−1,0 Ad−1,1 Ad−1,2 · · · Ad−1,d−1 The (ℓ, ℓ′ )-th entry in block Ak,k′ is then G((k,ℓ),(k′ ,ℓ′ )) (g). TOWARDS A CLASSIFICATION OF INCOMPLETE GABOR POVMS IN Cd 5 First we show that G is block circulant, that is, Ak,k′ = Ak+1,k′ +1 for any k, k ′ ∈ Zd . For any ℓ, ℓ′ ∈ Zd the (ℓ, ℓ′ )-entry in Ak+1,k′ +1 is ′ ′ ′ ′ G((k+1,ℓ),(k′ +1,ℓ′ )) = |hM k+1 T ℓ g, M k +1 T ℓ gi|2 = |hg, M k −k T ℓ −ℓ gi|2 . But this is exactly the (ℓ, ℓ′ )-entry in Ak,k′ . This shows that G(g) is a block circulant matrix. Furthermore we have Ak,k′ = A0,k′ −k . For simplicity, we denote Ak ≡ A0,k , and note that   A0 A1 A2 . . . Ad−1   Ad−1 A0 A1 . . . Ad−2   G(g) =  . ..  .. .. .. . . .  . . . .  A1 A2 A3 . . . A0 We next prove that each block Ak is also a circulant matrix. Without loss of generality, we focus on the first row of the blocks. For any k, ℓ, ℓ′ ∈ Zd , the (ℓ + 1, ℓ′ + 1)-th entry of Ak is equal to the (ℓ, ℓ′ )-th entry of Ak , since ′ ′ |hT ℓ g, M k T ℓ gi|2 = |hg, M k T ℓ −ℓ gi|2 . This concludes the proof of the first part of the Theorem. 2 (ii) For any a ∈ Zd , consider the functions ha : Cd → Cd defined as following   v  ρ v   a   2   ha (v) =   ρa v  ,  ..   .  ρd−1 a v where ρa = ω a . We claim that for any eigenvector w of G, w ∈ Range(ha ) for some a ∈ Zd . Denote Ha = A0 + A1 ρa + A2 ρ2a + · · · + Ad−1 ρd−1 a . By part (i), each Hi is a circulant matrix, thus can be diagonalized by the d × d DFT matrix F . Suppose v is an eigenvector of Ha and Ha v = λv. Then      A0 A1 A2 . . . Ad−1 v Ha v    Ad−1 A0 A1 . . . Ad−2   ρa v   ρa Hi v      = Gha (v) =  . = λha (v).  . . . . .  .   ···  . . . . .  . . . .  .   . ρd−1 A1 A2 A3 . . . A0 ρd−1 a Hi v a v So the columns of F ⊗ F are eigenvectors of G(g). Since F ⊗ F is invertible, its columns account for all d2 eigenvectors of G(g). We now find the spectrum of G(g) which consists of the collections of eigenvalues of n {Ha }d−1 a=0 . Denoting the (0, n) in Ha as Ha , n ∈ {0, 1, · · · , d − 1}. Then the d eigenvalues of Ha are given by Had−1 λa,b = Ha0 + ρb Ha1 + ρ2b Ha2 + · · · + ρd−1 b = d−1 X ℓ=0 ρℓb Haℓ = d−1 X ℓ=0 ρℓb ( d−1 X k=0 ρka |hg, M k T ℓ gi|2 ) = d−1 X d−1 X ℓ=0 k=0 ρkb ρℓa |hg, M k T ℓ gi|2 6 ASSAF GOLDBERGER, SHUJIE KANG, AND KASSO A. OKOUDJOU = d−1 X k,ℓ=0 ω ak+bℓ |hg, M k T ℓ gi|2 Writing, |hg, M k T ℓ gi|2 = Pd−1 λa,b = ∗ ∗ k(n−m) , n,m=0 gn gm gm−ℓ gn−ℓ ω d−1 X we see that ∗ ∗ gm−ℓ gn−ℓ ω k(n−m+a)+bℓ . gn gm n,m,k,ℓ=0 Changing variables r = n − ℓ, we obtain λa,b = d−1 X ∗ gn gm gr+m−n gr∗ ω k(a+n−m) ω b(n−r) . n,m,k,r=0 Summing first over k, we may replace m = n + a we obtain that λa,b = d d−1 X Pd−1 k=0 ω k(a+n−m) ∗ gn gn+a gr+a gr∗ ω b(n−r) = d n,r=0 d−1 X by dδ0,a+n−m . Then substituting 2 ∗ gn gn+a ω bn , n=0 which is (ii).  We can now give a proof of Theorem 1. Proof of Theorem 1. Theorem 1 follows directly from Theorem 2.  Pd−1 k d Remark. (1) If we encode g by a polynomial fg (X) = k=0 gk X ∈ C[X]/(X − 1), then Theorem 1 can be rephrased as: The rank of G(g) is the total number of zero coefficients in the collection of polynomials n o fg (ω −ℓ X)fg (X d−1 ) mod (X d − 1) | ℓ = 0, 1, . . . , d − 1 . √ (2) When d = 2, the condition for a vector g = (ceiθ1 , 1 − c2 eiθ2 ) to generate an informationally complete POVM is c2 ∈ / {0, 1, 21 }. (3) λa,b = λd−a,d−b . (4) Denote the matrix Λ with entries Λa,b = λa,b , the matrix Ξ with entries Ξk,l = |hg, M k T l gi|2 . Then Λ = F ΞF ∗ . The next result shows that Gabor informationally complete POVMs are generic in the sense that the set of all such POVMs is an open dense in the set of all FUNTFs of d2 elements in Cd . Proposition 1. For each d ≥ 2 there exists a complete Gabor POVM in Cd . Moreover, the set of all normalized vectors g such that rank(G(g)) = d2 is open dense in the unit sphere in Cd . Proof. Consider the unit sphere S 2d−1 ⊂ Cd as an algebraic subvariety of the affine 2d space with coordinates Re(gi ) and Im(gi ) for 1 ≤ i ≤ d. The condition rank(G(g)) = d2 is given by the P polynomial conditions λk,ℓ (g) = d| gi ḡi+k ω −iℓ |2 6= 0, for all (k, ℓ). For each pair (k, ℓ) with TOWARDS A CLASSIFICATION OF INCOMPLETE GABOR POVMS IN Cd 7 √ k 6= 0 take g = (1, 0, . . . , 1, 0, . . .)/ 2, with supp(g) = {1, 1 + k}, so we have λk,ℓ (g) = d/4 6= 0. Likewise, for k = 0 we take g = (1, 0, . . . , 0), and again λ0,ℓ (g) 6= 0. Thus for each (k, ℓ) the set Sk,ℓ := {g ∈ S 2d−1 | λk,ℓ (g) 6= 0} is nonempty open dense in the Zariski topology, and so is T  S = k,ℓ Sk,ℓ . [1, 21] Clearly, if the support of g is the set {1, 2, . . . , s} for s ≤ d/2, then rank(G(g)) < d2 , regardless of what is g, as we have gi ḡi+s+1 = 0 for all i. However, we can refine Proposition 1 for supports of cardinality > d/2. Proposition 2. Let S ⊆ {1, . . . , d} be a subset of cardinality > d/2. Then there exists a unit vectors g with supp(g) = S and rank(G(g)) = d2 . Proof. Like in the proof of Proposition 1 we consider the algebraic subvariety VS ⊆ S 2d−1 of all vectors g with supp(g) = S. It in enough to find for each pair (k, ℓ) a vector g ∈ VS such that λk,ℓ (g) 6= 0. For each k, the intersection T := S ∩(k+S) 6= ∅, where we define k+S = {k+i | i ∈ S}. Pick up j ∈ T and let g ′ be the vector such that gi′ = 0 if i ∈ / S, gi′ = 1 if i ∈ S and i ∈ / {j, k + j}, ′ ′ ′ ′ ¯ and gi = d if i ∈ {j, k + j}. Let g = g /||g ||. Then the vector wk defined by (wk )i = gi g ′ i+k satisfies (wk )j = d2 , (wk )i = 0 if i ∈ / T , (wk )i = 1 if i ∈ T and i ∈ / {j, j + k, j − k} and (wk )i = d otherwise. P iℓ 2 ′ 4 2 Therefore λk,ℓ (g) = d| i∈T (wk )i ω | /||g || ≥ d|d − (#T − 1) · d|2 /||g ′ ||4 > 0.  In the next result we collect a number of properties about the rank of G(g). In particular, we give all possible values of this rank when g is a unit vector in Cd whose support has size at most 2. In Theorem 4 we will prove a more general result than part (iii). Proposition 3. For a unit vector g ∈ Cd , we denote by kgk0 the number of nonzero entries in g. The following statements hold. (i) (ii) (iii) (iv) rank(G(g)) ≥ d. If d is odd, the rank of the Gramian G(g) is also an odd number. Suppose that d ≥ 2 and kgk0 = 1. Then rank(G(g)) = d. Suppose that kgk0 = 2. Then rank(G(g)) = 3d, 3d − 1, 2d, 23 d or d. d Proof. Let g ∈ Cd be a unit vector. For each ℓ ∈ Zd , we let wℓ = (gi gi+ℓ )d−1 i=0 ∈ C . (i) By part (ii) of Theorem 2 we see that λ0,0 = d. In addition, note that since the diagonal entries of G(g) are |hM k T ℓ g, M k T ℓ gi|2 = 1 where (k, ℓ) ∈ Zd × Zd , we have tr(G(g)) = d2 . If rank(G(g)) < d, then tr(G(g)) < d2 . This is a contradiction. (ii) Suppose that d is odd. Recall that rank(G(g)) equals to the number of nonzero eigenvalues of G, and that λ0,0 = d > 0. When (a, b) 6= (0, 0), then (a, b) 6= (d−a, d−b) and λa,b = λd−a,d−b . P Consequently, rank(G(g)) = 1 + 2 (a,b)∈S sgn(λa,b ), where S = {(a, b) |0 ≤ a ≤ d − 1, 0 ≤ b ≤ d−1 2 , (a, b) 6= (0, 0)}. (iii) Without loss of generality we assume that g0 = 1, and gk = 0 for all k 6= 0. Then kŵ0 k0 = d and kŵℓ k0 = 0 for all ℓ 6= 0. (iv) Without loss of generality we assume that g0 , gκ 6= 0 for some κ 6= 0. The proof is divided into two cases. 8 ASSAF GOLDBERGER, SHUJIE KANG, AND KASSO A. OKOUDJOU (a) Suppose that d is odd.    kw0 k0 = 2,    kwκ k0 = kw−κ k0 = 1, kwℓ k0 = 0 for ℓ 6= 0, κ, −κ.  kwˆ k = kŵ k = kŵ k = d, 0 0 κ 0 −κ 0 =⇒ kŵℓ k0 = 0 for ℓ 6= 0, κ, −κ. It follows that, rank(G(g)) = 3d. (b) Suppose now that d is even. If κ 6= d2 and |g0 | 6= |gκ |, then by the same arguments we have rank(G(g)) = 3d. Suppose κ 6= d2 and |g0 | = |gκ |. Then kŵ0 k0 = d − 1 when there exists an integer c such that ω cκ = −1, otherwise kŵ0 k0 = d. So rank(G(g)) = 3d − 1 or 3d. Next, suppose that κ = d2 , then wκ = w−κ . It follows that  kw k = kw k = 2, κ 0 0 0 kwℓ k0 = 0 for ℓ 6= 0, κ. Since ω κ = −1, we conclude that    d |g | = |g | d 0 κ kwˆ0 k0 = 2 and kŵκ k0 = d |g0 | = d 6 |gκ | 2 g0 gk 6= ±gk g0 g0 gk = ±gk g0 Consequently, rank(G) = 3d, 3d − 1, 2d, 3d 2 or d.  2.2. Algebraic structure of Gabor POVMs. In this section we give another proof of Theorem 1 from an algebraic point of view. In particular, this approach allows us to list several actions that leave invariant the rank of G(g). The Weyl-Heisenberg group is the subgroup W ⊆ Ud (C) generated by M and T . It has order 3 d and every element of W can be written uniquely as ω a M b T c for integers 0 ≤ a, b, c < d. The elements of W are monomial matrices. A monomial matrix is a square matrix with a unique nonzero entry in each row and column, which is a phase. A matrix X is monomial, if and only if it can be written (uniquely) as a product DP , where D is diagonal with diagonal entries of modulus 1, and P is a permutation matrix. Gabor systems give rise to a monomial representation of W on the vector space C[G(g)], g is a symbolic vector of indeterminates. Here C[S] denotes the vector space of formal complex linear combinations of the set S. Explicitly, w = ω a M b T c acts on basis elements gk,ℓ = M k T ℓ g by ω a M b T c gk,ℓ = ω a−ck gk+b,ℓ+c . We introduce a monomial matrix M(w) of size d2 to denote this action. This matrix is indexed by (t, s) ∈ Zd ⊕ Zd and its ((t, s), (t′ , s′ )) entry is given by  ω a−ct′ if (t, s) = (t′ + b, s′ + c) M(w)((t,s),(t′ ,s′ )) = 0 else. TOWARDS A CLASSIFICATION OF INCOMPLETE GABOR POVMS IN Cd 9 It is clear by the construction that M is a group homomorphism: M(ww′ ) = M(w)M(w′ ). We define another monomial representation, by M′ (w) := |M(w)|, the entrywise absolute-value. Consider the Gram matrix H(g) := G(g)∗ G(g), for a unit vector g. Note that this is the Gram matrix of the Gabor frame G(g), and is different from the Gram matrix G(g) of the rank-one matrices we have discussed thus far. In particular, for every d ≥ 2, H(g) is a d2 × d2 matrix, and its eigenvalues are only 0 and d. Moreover, √1d H(g) is a self adjoint idempotent of rank d. The matrix H(g) is invariant under the monomial action M: M(w)H(g)M(w)∗ = H(g), ∀w ∈ W. The collection A = A(M) := {X ∈ Md2 (C) | ∀w ∈ W, M(w)XM(w)∗ = X} is a matrix algebra, closed under conjugate transpose. Similarly we have the matrix algebra A′ = A(M′ ) = {X ∈ Md2 (C) | ∀w ∈ W, M′ (w)XM′ (w)∗ = X}. Both algebras have dimension d2 . In the following theorem, if Z is a group, let C[Z] denote the group algebra over C. In addition, we shall make use of the C-algebra isomorphism (2) θ: given by X 7→ (ω i )0≤i<d . C[X] → Cd , Xd − 1 Theorem 3. There is an isomorphism of algebras over C, ∼ = ε:A− → Md (C). (3) There is a sequence of isomorphisms of algebras over C ∼ = ∼ = ∼ = (4) A′ − → C[Zd ⊕ Zd ] − → C[Zd ] ⊗C C[Zd ] − → C[X]/(X d − 1) ⊗C C[X]/(X d − 1) 2 θ⊗θ ∼ = −−−−−→ Cd ⊗C Cd ∼ = Cd . ∼ = 2 The first isomorphism respects the conjugate-transpose. Let µ : A′ − → Cd denote the composition. 2 Under µ the conjugate transpose becomes complex conjugation on Cd . Proof. Let us construct first the inverse map ε−1 : Md (C) → A. The complex vector space Md (C) has the special basis {M i T j }0≤i,j<d . It is enough to define ε−1 on this basis, and show that it is an algebra homomorphism. For each (i, j), we first construct a matrix E(i, j) ∈ A, which satisfies E(i, j)(p,q),(0,0) = δ(i,j),(p,q) . The matrix E(i, j) is uniquely determined by these properties. Indeed, by the invariance under the monomial matrices M(w), E(i, j)(a+i,b+j),(a,b) = M(M a T b )(a+i,b+j),(i,j) E(i, j)(i,j),(0,0) M(T −b M −a )(0,0),(a,b) = ω −bi · 1 · M(ω −ab M −a T −b )(0,0),(a,b) = ω −bi · 1 · ω −ab+ab = ω −bi . 10 ASSAF GOLDBERGER, SHUJIE KANG, AND KASSO A. OKOUDJOU A similar computation shows that E(i, j)(a′ ,b′ ),(a,b) = 0 if (a′ , b′ ) 6= (a + i, b + j). This shows the uniqueness of E(i, j). On the other hand, it is easy to check that if we define E(i, j) by these formulae, then E(i, j) ∈ A. Note that E(i, j) is monomial and that |E(i, j)| = |M(M i T j )|. Next we compare E(i + s, j + t) with E(i, j)E(s, t). We have (E(i, j)E(s, t))(a+i+s,b+j+t),(a,b) = E(i, j)(a+i+s,b+j+t),(a+s,b+t) E(s, t)(a+s,b+t),(a,b) = ω −(b+t)i ω −bs = ω −b(i+s) ω −it . We also have E(i + s, j + t)(a+i+s,b+j+t),(a,b) = ω −b(i+s) . Hence E(i, j)E(s, t) = ω −it E(i + s, j + t). We try to define (5) ε−1 (M i T j ) = ω f (i,j) E(i, j) for some f (i, j) ∈ Zd , and we extend ε−1 by linearity to a vector-space homomorphism. We wish to find an f , such that ε−1 will be actually an algebra homomorphism. A necessary and sufficient condition for this is that ε−1 (M i T j · M s T t ) = ε−1 (M i T j ) · ε−1 (M s T t ). Then using (5) and equality M i T j · M s T t = ω −js M i+s,j+t , we must have f (i, j)f (s, t) = ω −js−it f (i + s, j + t). This condition is satisfied by the choice f (i, j) = ω ij , so (6) ε−1 (M i T j ) := ω ij E(i, j) extends to a C-algebra homomorphism Mn (C) → A. For (a, b) 6= (c, e) modulo d, ε−1 (M a T b ) and ε−1 (M c T e ) have disjoint supports, hence ε−1 is injective. By comparing dimensions we conclude that ε−1 is an algebra isomorphism. Since both M i T j and ε−1 (M i T j ) are monomial matrices, then their conjugate-transpose is equal to their inverse, and ε−1 being a ring isomorphism respects matrix inverses. Hence ε commutes with the conjugate-transpose. This completes the first part. The map α : Zd ⊕ Zd → GLd2 (C) given by α(s, t) = M′ (M a T b ) is a group homomorphism. Then we can extend α to a C-algebra map ε′ : C[Zd ⊕ Zd ] → Md2 (C). This map is injective since as above {M′ (M a T b )}(a,b) have disjoint supports. Also, as ww′ and w′ w differ only by a phase, M′ (w′ w) = M′ (ww′ ) = M′ (w)M′ (w′ ), and in particular M′ (w′ ) is stable under conjugation with M′ (w). It follows that the image of ε′ is in A′ . Comparing dimensions, we obtain the first isomorphism in (4). The other isomorphisms are well known and natural. They are given by the maps [(a, b)] 7→ [a] ⊗ [b] 7→ X a ⊗ X b 7→ [1, ω a , ω 2a , . . . ω (d−1)a ] ⊗ [1, ω b , ω 2b , . . . ω (d−1)b ] 7→ (ω ia ω jb )i,j . Under those maps, the conjugate-transpose in A′ is compatible with the negation map [(a, b)] 7→ [(−a, −b)], which translates into complex conjugation at the rightmost term.  TOWARDS A CLASSIFICATION OF INCOMPLETE GABOR POVMS IN Cd 11 We remark that the isomorphism (2) θ : C[X]/(X d − 1) → Cd is given by the Chinese Remainder Q Theorem for polynomials. That is, since X d − 1 = i (X − ω i ), then M C[X]/(X d − 1) ∼ C[X]/(X − ω i ) ∼ = = Cd . i The map θ takes a polynomial f to the vector (f (ω i ))i . Then this linear operation is nothing Pd−1 i d but the discrete Fourier transform. If we write f (X) = i=0 fi X ∈ C[X]/(X − 1) we have √ θ(f ) = dF · [f0 , f1 , . . . , fd−1 ]T ∈ Cd . Given an element X ∈ A, the matrix X (2) defined by (2) Xi,j := |Xi,j |2 is a member of A′ . In the spacial case X = H(g), writing G(g) = {g0,0 , . . . , gd,d } and (2) ∗ , then G(g) := X ∗ ′ letting Πi,j = gi,j gi,j i,j = T r(Πi,j Πi,j ) ∈ A , and rank G(g) is the dimension of the complex vector space spanned by the Πi,j . We have 2 Lemma 1. Under the isomorphism (4) µ : A′ → Cd , the rank of X ∈ A′ equals the number of nonzero entries in µ(X). Proof. By the isomorphism µ, the algebra A′ contains a list of d2 nontrivial idempotents e1 , . . . , en2 , 2 summing up to 1 and satisfying ei ej = 0 for i 6= j. On letting A′ act on a the vector space V = Cd , L we have V = j ej V , and this decomposition respects the action of A′ . As A′ acts faithfully on V , and ek A′ acts as zero on ej V for j 6= k, then ej V 6= 0 for all j, and by equating dimensions we conclude that each ej V is 1-dimensional. Hence µ is nothing but a simultaneous diagonalization of the algebra A′ and the lemma follows.  3. Invariance properties of the rank of Gabor POVMs In this section we identify a number of transformations leaving the rank of G(g) invariant, and moreover leave invariant the multiset of the internal angles between the vectors of G(g). We exploit this to first classify all unit-norm vectors g for which the rank of G(g) is d. Next, we introduce a notion of an automorphism group of Gabor frames, and construct examples of m-distance sets with small m. Finally, we prove that rank(G(g)) cannot take a value strictly between d and 2d when d > 2 is a prime. For a nonzero vector g = (g0 , g1 , . . . , gd−1 )T ∈ Cd , we let X ak,ℓ = ak,ℓ (g) := gn ḡn+k ω nℓ . n We recall that the k-translation k + S of a subset S ⊆ Zd , is the set {k + s| s ∈ S}. We shall √ −1 ∈ Z× make the convention that if r is a real number, then ω r := e2 −1rπ/d . For m ∈ Z× d , let m d denote its group inverse. Given g ∈ Cd , it is not difficult to prove that each of the following transformations of g resulting in a vector h preserves the ranks rank(G(h)) = rank(G(g)) and the following transformation rules: 1. Phase: Let h = cg for |c| = 1. Then ak,ℓ (h) = ak,ℓ (g). 2. Additive translation: Let hi = gi+t . We have (7) ak,ℓ (h) = ω −tℓ akℓ (g). 12 ASSAF GOLDBERGER, SHUJIE KANG, AND KASSO A. OKOUDJOU 3. Multiplicative: Let hi = gmi , where gcd(m, d) = 1. We have (8) ak,ℓ (h) = amk,m−1 ℓ (g). i 4. (Phase) Quadratic: Let hi = gi ω a(2)+bi+c , wherec is a real number, a is an integer, and b is an integer of a half integer such that a(d − 1)/2 + b is an integer. We have k ak,ℓ (h) = ak,ℓ−ak (g)ω −bk−a(2) . (9) We remark that the condition on b implies that the transformation is well-defined if we consider the index i as real integer. 4’. (Phase) Quadratic: Suppose that supp(g) ⊂ κZd , for some positive integer κ|d, and let a i h = g ω κ (2)+bi+c for i ∈ supp(g) and h = 0 otherwise, where c is a real number, a is an i i i integer and b is an integer of a half integer such that a(d − 1)/2 + b is an integer. We have for every integer s (10) a sκ asκ,ℓ (h) = asκ,l−as (g)ω −bsκ− κ ( 2 ) . 3.1. Characterization of rank d Gabor POVMs. In this section give a complete characterization of all vectors g leading to rank d Gabor POVMs. We note that this generalizes part (iii) of Proposition 3. Theorem 4. Let r be a divisor of d, and define a vector g := g(r, d) ∈ Cd by  1 i ≡ 0 mod r g(r, d)i = 0 otherwise. Then the rank of G(g) is d, and G(g) is a 2-distance set. Conversely, up to translation and phase quadratic transformations, a normalized vector g has rank d, if and only if g = √1 g(κ, d) for some d/κ κ|d. P Proof. For g defined as above, we see that ak,ℓ (g) = n gn ḡn+k ω nℓ = 0 if k is not a multiple of r. P P Otherwise atr,ℓ = n=rj grj ḡr(j+t) ω nℓ = n=rj ω nℓ . This quantity vanishes if and only if ω rℓ = 1. Equivalently ℓ is a multiple of d/r. Hence atr,sd/r (g) are precisely the ones that do not vanish and there are d of them. It remains to prove the only if part. Assume that rank(G(g)) = d for a unit vector g. One P P 2 2 eigenvalue of G(g) is d|a0,0 (g)|2 = d i |gi ||gi+k | ≤ kgk = 1 i |gi | = d. Notice that |ak,ℓ (g)| ≤ by the Cauchy-Schwartz inequality, hence d is the largest eigenvalue. As tr(G(g)) = d2 and the rank is d, we must conclude that d appears with multiplicity d, and all other eigenvalues are 0. Let S = supp(g), and let κ be the smallest positive integer such that i, κ + i ∈ S. Then the vector (gi gi+κ )i 6= 0, and (aκ,ℓ )ℓ is its discrete Fourier transform. It follows that aκ,ℓ 6= 0 for some ℓ. If κ = d then S is a singleton and g = g(d, d) up to phase and translation. So we shall assume from now that κ < d. TOWARDS A CLASSIFICATION OF INCOMPLETE GABOR POVMS IN Cd Now |aκ,ℓ (g)| = |a0,0 (g)| = 1 implies that 1=| X i∈S,i+κ∈S gi ḡi+κ ω iℓ | ≤ X i∈S,i+κ∈S  |gi gi+κ | ≤  X i∈S, i+κ∈S 1/2  |gi |2   X i∈S, i+κ∈S 13 1/2 |gi+κ |2  , where we applied the Cauchy-Schwartz inequality once more. This can happen if and only if S ∩ (κ + S) = S, so S must be κ-periodic. In particular κ must divide d. By applying translation, we may assume without loss of generality that S is the subsets of multiples of κ, and for a given k there exists an ℓ such that ak,ℓ 6= 0, if and only if k = tκ. Take t = 1. By applying a phase quadratic transformation (type 4’), we may assume that |aκ,0 | = 1. By Cauchy-Schwartz we conclude that the vectors (gi ) and gκ+i are proportional, so gi = αgκ+i for all p p i, with |α| = 1. Modifying by a phase we may assume that g0 = 1/ d/κ hence gκj = αj / d/κ. In particular for j = d/κ we obtain αd/κ = 1. Modifying by linear phase (type 4) we obtain p g = g(κ, d)/ d/κ.  Remark. Choosing r = d, we recover part (iii) of Proposition 3. 3.2. Examples of automorphisms of two types of Gabor systems. The list of transformations 1–4’ above gives rise to a notion of an automorphism group of a Gabor systems. We define the group Gd as the group of all maps Cd → Cd generated by the transformations 1–4’. For every unit vector g, let Aut(g) ⊆ Gd be the subgroup fixing g. We call this the Automorphism group of g. We can easily create vectors g with nontrivial automorphisms. For example, if gi+δ = gi for all i and δ|d, then g is a periodic vector having nontrivial translations as automorphisms. Likewise if gmi = gi for all i, gcd(m, d) = 1, then g has multiplicative automorphisms. A more interesting i example will be a vector g, such that for all i, gi+δ = gi ω α(2)+βi+γ , for fixed α, β, γ, δ ∈ Zd . So any such g has a nontrivial automorphism, which is a composition of phase-quadratic and a translation transformations. If gcd(δ, d) = 1, then by iterating this relation, it is easy to show that (11) i i φ gi = √ ω a(3)+b(2)+ci , d |φ| = 1, where a, b, c ∈ Zd solve uniquely the linear system   α  aδ =
δ a 2 + bδ = β 

 a 3δ + b 2δ + cδ = γ. Conversely, every vector of the form (11) has this specific automorphism. A vector g = (gi ) satisfying (11) is what is known in the literature as an Alltop sequence. Such sequences were constructed by Alltop [2] for applications of spread spectrum radars and communication. By the transformation rule of a phase quadratic symmetries, we have that |ak,ℓ (g)| = |ak,ℓ−αk (g)|, which means that zero eigenvalues may be duplicated by the automorphism, giving some limitations on rank(G(g)). 14 ASSAF GOLDBERGER, SHUJIE KANG, AND KASSO A. OKOUDJOU Assume now that d is prime and 1 ≤ a < d. Then the above symmetry implies that |ak,ℓ (g)| is independent of ℓ. In fact we can show Proposition 4. (See [16, 19]) If 1 ≤ a < d, d is an odd prime, and g is as in (11), then   √1 k > 0 d |ak,ℓ (g)|2 = . 0 k = 0, ℓ > 0 d−1 In particular rank(G(g)) = d2 − d + 1. Moreover, the Gabor System {T ℓ M k g}ℓ,k=0 , together with the standard basis is a maximal MUB of d + 1 bases. Proof. We compute directly for k 6= 0: |ak,ℓ (g)| = X gi ḡi+k ω iℓ = i i = |φ|2 X a( i )+b( i )+ci −a(i+k)+b(i+k)+c(i+k) iℓ 2 3 2 ω ω ω 3 d 1 X iℓ−ak( i )−ai(k)−bki −a(k)−b(k)−ck 1 X −ak( i )−ai(k)−bki+iℓ 2 2 2 2 2 = ω ω 3 ω d d i i = √ 1 X − ak (i− 1 − 1 (ℓ+bk+a(k)))2 1 1 X − ak i2 1 2 2 ak = = | ± ±d| = √ ω 2 ω 2 d d d d i i In this computation we have interpreted a fraction m/n as the unique integer f such that m ≡ nf √ P 2 mod d. We have used the fact that for an odd prime number d, and gcd(δ, d) = 1, i ω αi = ± ±d. This is a variant of the well-known Gauss sums [16, Theorem 1]. P When k = 0, ak,ℓ (g) = d1 i ω iℓ = 0 if ℓ 6= 0. This in particular means that g is orthogonal to M i g for all i, and moreover B0 := {M j g}j is an orthonormal basis. It follows that for every i, Bi = {T i M j g}i is an orthonormal basis, and that the Gabor system is a MUB of d bases. Since all entries of G(g) are of the same modulus, we can adjoin the standard basis to obtain a maximal MUB.  We next construct another family of vectors g having a nontrivial automorphism group, and leading to a non complete Gabor POVM with few distinct inner products. We consider the subspace of Cd given by (12) i V(a, b, c, κ) = {g ∈ Cd : gκi = ω a(2)+bi+c gi } for some integers a, b, c ∈ Zd and κ ∈ Z× d. The group (Zd )× acts on the set Zd by multiplication. The subgroup hκi generated by κ yields a disjoint decomposition of hκi- orbits G Zd = Qr . r For each r fix a point ir ∈ Qr . Then the value gir determines uniquely the values of gi for all i ∈ Qr . Namely, if i = κm ir , then by iterating the condition in (12) we get (13) gi = ω Pm−1 j=0 j a(κ 2ir )+bκj ir +c g ir . TOWARDS A CLASSIFICATION OF INCOMPLETE GABOR POVMS IN Cd 15 Notice however, that this must apply to any index s such that κs ir ≡ ir mod d. Hence if gir 6= 0, the following condition must hold X j  a + bj + c ≡ 0 mod d. (14) 2 j∈Qr In this case, we say that Qr is an orientable hκi-orbit for the triple (a, b, c). Otherwise it is nonorientable. We have Proposition 5. Using the above notations and definitions, the following statements hold. (a) The dimension over C of the space V(a, b, c, κ) is the number of orientable hκi-orbits for the triple (a, b, c). (b) If d is odd and gcd(d, κ2 − 1) = 1, then all the hκi-orbits are orientable for (a, b, c). Proof. Let Q be the set of all hκi-orbits in Zd , and denote by N the number of orientable hκi-orbits. (a) Pick an index ir ∈ Qr for any orbit Qr ∈ Q. We define a linear map E : V(a, b, c, κ) → CQ by sending g to the vector (gir )r . The map E is injective, because by (13) g is determined uniquely by the collection {gir }. Also, when (14) is not satisfied, then gir = 0, so we conclude that dim V(a, b, c, κ) = dim Image(E) ≤ N . To prove the equality, for every orientable orbit Qr we will construct a vector g ∈ V(a, b, c, κ) with gir = 1 and gis = 0 for all s 6= i. We define gi for i ∈ Qr by equation (13) taken with gir = 1, and for i ∈ / Qr we set gi = 0 for. Then condition (14) guarantees that g is well-defined, regardless of the choice of a lifting m ∈ Z, and it is easy to check now that the condition in (12) is satisfied. Hence g ∈ V(a, b, c, κ), and we are done proving (a). (b) We must check condition (14). Since d is odd and hence 2 is invertible modulo d, it is enough P P to prove that for each orbit Qr , j∈Qr j 2 ≡ j∈Qr j ≡ 0 mod d. Since Qr = {κt ir | 0 ≤ t < s}, where κs ir = ir , then using gcd(d, κ2 − 1) = 1: s−1 X t=0 (κt ir )2 = κ2s − 1 2 ir − ir i = 2 ir = 0, κ2 − 1 r κ −1 where the quantities are considered as elements of Zd . A similar argument proves that P j∈Qr j ≡ 0 mod d.  Let g be a vector satisfying (12). By using (8) and (9) we have |aκi,κ−1 ℓ | = |ai,ℓ−ai | for all (i, ℓ), or equivalently (15) |ai,ℓ | = |aκ−1 i,κℓ−κ−1 ai |. Assume now the conditions of Proposition 5. Our next goal is to estimate the number distinct angles associated with the Gabor system generated by a vector g that satisfies (12). Towards this we transform coordinates on (Zd )2 by i′ = i and ℓ′ = ℓ − 2aκℓ/(κ2 − 1). Write a′i′ ,ℓ′ (g) = a′i′ ,ℓ′ = ai,ℓ (g). Then (15) transforms to the simpler form (16) |a′i,ℓ | = |a′κ−1 i,κℓ |. 16 ASSAF GOLDBERGER, SHUJIE KANG, AND KASSO A. OKOUDJOU Recall that we also have the conjugacy symmetry |ai,ℓ | = |a−i,−l |. Equivalently, |a′i,ℓ | = |a′−i,−l |. (17) Let Bκ be the subgroup of Z× d generated by κ and −1. Then (16)-(17) are equivalent to |a′i,ℓ | = |a′t−1 i,tℓ |, for all t ∈ Bκ . (18) The group Bκ satisfies Bκ = hκi or [Bκ : hκi] = 2. Theorem 5. Suppose that d is odd and that gcd(κ2 − 1, d) = 1. Then the Gabor system G(g) is m-angular where X gcd(δ(d1 ), δ(d2 ))  d   d  (19) m≤ ϕ . ϕ r d1 d2 d1 ,d2 |d Here r = |Bκ |, δ(di ) is the order of the image of Bκ in Z× di , and ϕ is the Euler totient function. Proof. The number of angles in this Gabor system is the number of distinct |ai,ℓ (g)|, which is the number of distinct |a′i,ℓ (g)|. The group Bκ acts on (Zd )2 via (i, ℓ) 7→ (t−1 i, tℓ) for all t ∈ Bκ , hence by (18) |a′i,ℓ | are constant along the orbits. We will be done if we show that the right hand side of (19) is the number of Bκ -orbits. We note that if C is a finite group acting on a finite set S, and Fix(s) is the cardinality of the P stabilizer of the point s ∈ S, then the number of orbits is s∈S 1/Fix(s). Let κ0 be a generator for the cyclic group Bκ , and consider a point (i, ℓ) ∈ (Zd )2 . Let d1 = gcd(i, d) and d2 = gcd(ℓ, d). Then δ(d ) κ0 has multiplicative order δ(d1 ) modulo d1 and δ(d2 ) modulo d2 . Thus hκ0 1 i is the stabilizer of δ(d ) i, and hκ0 2 i is the stabilizer of ℓ. It follows that Fix((i, ℓ)) = r/gcd(δ(d1 ), δ(d2 )). The number of the pairs (i, ℓ) with gcd(i, d) = d1 and gcd(ℓ, d) = d2 is ϕ(d/d1 )ϕ(d/d2 ). The theorem follows.  Corollary 1. Suppose that d is prime and κ generates (Zd )× , or that d ≡ 3 mod 4 and κ has order (d − 1)/2. Then, we have  m ≤ d + 2 rank(G(g)) ≡ d2 mod (d − 1). Proof. In the both cases Bκ = Z× d and r = d−1. The pair (d1 , d2 ) takes the values (1, 1), (d, 1), (1, d), (d, d). We have δ(1) = 1 and δ(d) = d − 1. The right hand side of (19) is d + 2. The second equation follows from the fact that all the non-trivial Bκ -orbits of Z2d are of order d − 1.  We illustrate Corollary 1 with an example, namely the family of Gabor frames generated by the
Björck sequences. Suppose d is an odd prime, and denote χ[k] ≡ kd the Legndre symbol. We can define a vector in Cd accordingly as following: • When d is prime and d ≡ 1 mod 4, 1 gk = √ eiθχ[k] , where θ = arccos d for all k ∈ Zd .   1 √ , 1+ d TOWARDS A CLASSIFICATION OF INCOMPLETE GABOR POVMS IN Cd • When d is prime and d ≡ 3 mod 4,   √1 eiφ d gk =  √1 d 17 if k ∈ QC ⊆ (Zd )× , otherwise, 1−d ), and QC is the preimage of −1 under χ. for all k ∈ Zd . Where φ = arccos( 1+d Björck sequences, which were constructed in [9] are CAZAC (Constant Amplitude Zero Auto Correlation) sequences, meaning that hg, T ℓ gi = 0 for all ℓ ∈ Z× d . For this reason, they provide d examples of vectors g ∈ C with kgk0 = d and do not generate IC-POVMs. In fact, the number of different values in the Gramian G(g) is relatively small. When d ≡ 3 mod 4, we know that rank(G(g)) ≤ d2 − 2d + 2 by the result in [7] and the following observation. Proposition 6. Suppose d is prime and d ≡ 3 mod 4 and g is a unit vector in Cd , then |hg, M k T ℓ gi| ′ ′ take d + 1 different values. Furthermore, |hg, M k T ℓ gi| = |hg, M k T ℓ gi| if kℓ ≡ k ′ ℓ′ mod d and (k, ℓ), (k ′ , ℓ′ ) 6= (0, 0). Note that by Corollary 1, the upper bound of the number of angles is d + 2. Under the specific choice of of the phase φ, the values of a0,ℓ and aℓ,0 , ℓ 6= 0 all degenerate to 0. General choices of φ will result in d + 2 angles. Remark. We provide an example illustrating Proposition 6 with the Björck sequence of length d = 7 is  3  √ + i 41 k = 3, 5, 6. gk = 4 7  √1 k = 0, 1, 2, 4. 7 Figure 1 shows the different values of |hg, M k T ℓ gi|. Figure 1. Heatmap of |hg, M k T ℓ gi|, where g is the Björck sequence of length 7. We conclude this section by showing that rank(G(g)) cannot lie in the interval (d, 2d) when d is odd and prime. We contrast this with Proposition 3 where we gave an example of g leading to rank(G(g)) = 3d/2 for d even. But first, we need the following Lemma. 18 ASSAF GOLDBERGER, SHUJIE KANG, AND KASSO A. OKOUDJOU Lemma 2. Assume that d is an odd prime. If kgk0 > d/2 and rank(G(g)) 6= d, then rank(G(g)) ≥ 2d. Proof. The assumption kgk0 > d/2 implies that the vector wk := (gi ḡi+k )i is not the zero vector for all k. Hence for each k, ak,ℓ 6= 0 for at least one ℓ. Suppose by contradiction that rank(G(g)) < 2d. Therefore, we must have some value of k for which ak,ℓ 6= 0 for exactly one ℓ. In particular wk is proportional to (1, ω ℓ , ω 2ℓ , . . . , ω (d−1)ℓ ). There are two cases to consider: √ Case I: k = 0. Then w0 = (1, 1, . . . , 1)/ d. There must be some k ′ 6= 0 for which ak′ ,ℓ 6= 0 for at most two values of ℓ. Thus wk′ equals to a vector (aω im +bω in )i , for some constants a, b. But √ as |(wk′ )i | = 1/ d for all i, |aω im + bω in | is independent of i, which implies a = 0 or b = 0. WLOG b = 0. Thus gi /gi+k′ = daω im . Modifying g by multiplicative transformation, and using gcd(k ′ , d) = 1, we may assume that k ′ = 1, which implies that gi = g0 (da)i ω mi(i−1)/2 . Substituting i = d we obtain (da)d = 1, hence da = ω r for some r. Thus g is a quadratic √ transformation of (1, 1, . . . , 1)/ d and rank(G(g)) = d. A contradiction. Case II: k 6= 0. Since gcd(k, d) = 1, then by modifying g be a multiplicative transformation we may assume that k = 1, and gi ḡi+1 = cω ℓi . By a phase quadratic transformation, we may reduce to the case ℓ = 0, so gi ḡi+1 = c for all i. Dividing this by ḡi+1 gi+2 = c̄ we obtain √ gi /gi+2 = c/c̄. This implies that |gi | = |gi+2 | for all i and as d is odd, |gi | = 1/ d for all i and from this point the proof is identical to case I.  Remark. It follows that for d = 3 there is no g with rank(G(g)) = 5. If this was not the case, the generator g would satisfy kgk0 = 1, but this in turn implies rank(G(g)) = 3, a contradiction. We need introduce a definition and a preliminary result that is interesting in its own right. Definition 1. The density δ = δ(f ) of a polynomial f (X) ∈ F [X] over a field F is the number of nonzero coefficients in f . We denote µd the group of complex roots of unity of order d. The main part of the proof lies in the following result. Proposition 7. Let d be a prime integer and f (X) ∈ C[X] a polynomial of degree less than d. Then the number of roots of f which are in µd is at most δ(f ) − 1. Proof. The result is an immediate consequence of the fact that all minors of d × d DFT matrix are nonzero when d is prime, see [14, Theorem 6] or [13, Theorem 4]. Nonetheless, we provide here an algebraic number field argument. Suppose, by contradiction, that the polynomial f has at least δ(f ) roots in µd . Pick δ(f ) roots, Pδ(h)−1 αi = ω ei , 0 ≤ i < δ(f ). Let h = j=0 cj X hj be the polynomial of degree less than d, with the smallest possible density (≤ δ(f )) having all the αi as roots. Then the coefficients of h are solution to a linear system of equations given by the vectors δ(f ) vj := (ω ei hj )i ∈ µd , 0 ≤ j ≤ δ(h) − 1. By the assumption on h, this is the minimal linear dependency. The rank of the matrix V = (ω ei hj )i,j is δ(h) − 1, thus there are δ(h) − 1 independent rows, and the linear dependency coefficients can be read from the δ(h) − 1 size minors belonging to TOWARDS A CLASSIFICATION OF INCOMPLETE GABOR POVMS IN Cd 19 the submatrix corresponding to these rows. In particular the coefficients belong to the cyclotomic field K := Q(ω) and h is proportional to a polynomial in K[X]. WLOG we assume that h ∈ K[X]. The maximal order of K is known to be the ring Z[ω], and by again rescaling h we may assume that h ∈ Z[ω][X]. There is a ring homomorphism φ : Z[ω] → Zd . The kernel of φ is the unique prime ideal D above d, which is known to be principal, and generated by ω − 1. We extend φ to a ring homomorphism (denoted again by φ) φ : Z[ω][X] → Zd [X]. Let vD (z) denote the D-adic valuation of z ∈ K, and let vD (h) = mini vD (ci ). Then replacing h by h/(ω − 1)vD (h) , we still have h ∈ Z[ω][X], and at least on ci ∈ / D. In particular φ(h) 6= 0. Q Now, h(X) = h0 (X) i (X − αi ), and still h0 (X) ∈ Z[ω][X]. This can be proved by induction by dividing g(X) successively by each (X − αi ). For example, in the first step write h(X) = h(X − α0 + α0 ) and expand each monomial around X − α0 using the binomial formula. On applying φ we obtain φ(h)(X) = (X − 1)δ(f ) φ(h0 )(X). In particular, all first δ(h) ≤ δ(f ) derivatives of φ(h) vanish at X = 1: X ci hi (hi − 1) · · · (hi − k + 1) = 0 mod d, ∀k < δ(h). φ(h)(k) (X = 1) = i
Hence, since not all ci are 0 mod d, the matrix W = hi (hi −1) · · · (hi −k +1) i,k ∈ (Zd )δ(h),δ(h) has linearly dependent rows. Notice that the kth column of W are the values of polynomial Fk (X) = X(X − 1) · · · (X − k + 1) substituted at X = hi . Thus by performing column elementary operations on W , we may clear the lower terms in Fk (X), and our matrix is Gauss equivalent to the matrix U = (hki )i,k . But 0 ≤ hi < d are distinct and U is the Vandermonde matrix in the field Zd , hence det(W ) = det(U ) 6= 0 in Zd . This is a contradiction, and the theorem is proved.  We are now ready to prove: Theorem 6. For an odd prime d, there is no unit vector g with d < rank(G(g)) < 2d. Proof. Suppose that there is such g. We know by Lemma 2 that kgk0 < d/2. Then for each k, kwk k0 < d/2. Suppose that k is taken such that wk 6= 0. By the above theorem, the DFT w ck can have at most (d − 3)/2 zero entries, hence the number of ℓ such that ak,ℓ (g) 6= 0 is at least (d + 3)/2. This implies that there can be at most 3 values of k such that wk 6= 0. This implies in turn that kgk0 ≤ 2. But in this case the conclusion of the theorem follows from Proposition 3.  We conclude the paper by proving the rank of Gabor POVMs in dimensions 4 and 5 when the generator g does not have full support. Proposition 8. Suppose that g is a unit-norm vector. The following statements hold. (i) If g ∈ C4 , then   if kgk0 = 1  4 rank(G(g)) = 4, 6, 8, 11, or 12 if kgk0 = 2    11, 12, 13, 14, 15, or 16 if kgk0 = 3 20 ASSAF GOLDBERGER, SHUJIE KANG, AND KASSO A. OKOUDJOU (ii) If g ∈ C5 , then   if kgk0 = 1  5 rank(G(g)) = 15 if kgk0 = 2    21, 23, or 25 if kgk0 = 3 Proof. The result for kgk0 = 1, 2 can be obtained form Theorem 4 and Proposition 3. (i) Assume now kgk0 = 3. Without loss of generality, let g0 , g1 , g2 6= 0 and g3 = 0. We have kŵ0 k0 =3 or 4, kŵ1 k0 = kŵ3 k0 =3 or 4, and kŵ2 k0 =2 or 4. • kŵ0 k0 = 3 if |g0 |2 + |g2 |2 6= |g1 |2 , else kŵ0 k0 = 4. • kŵ2 k0 = 2 if g0 ḡ2 = ±g2 ḡ0 , else kŵ2 k0 = 4. • kŵ1 k0 = kŵ3 k0 = 3 if g0 ḡ1 + ω k g1 ḡ2 = 0 for some k ∈ Z4 , else kŵ1 k0 = kŵ3 k0 = 4. Since all possible combinations of (||ŵ0 ||0 , ||ŵ1 ||0 , kŵ2 k0 ) can be obtained, we can conclude that rank(G(g)) can be any integer between 11 and 16. (ii) Next, suppose kgk0 = 3. We assume g0 , g1 , g2 6= 0. All other possibilities can be obtained from additive and multiplicative translation from this vector. Then kw0 k0 = 3, kw1 k0 = kw4 k0 = 2, and kw2 k0 = kw3 k0 = 1. So kŵ0 k0 = 3 or 5; kŵ2 k0 = kŵ3 k0 = 5; and kŵ1 k0 = kŵ4 k0 =4 or 5. • kŵ0 k0 = 3 if w0 = (1, −2 cos(4π/5), 1, 0, 0). Then g is equivalent to p (1, −2 cos(4π/5))eiθ1 , eiθ2 , 0, 0) for θ1 , θ2 ∈ [0, 2π]. Otherwise kŵ0 k0 = 5. 1 j • kŵ1 k0 = kŵ4 k0 = 4 if and only if g is equivalent to a scalar multiple of (1, g1 , −g g1 ω , 0, 0) for some j ∈ Z5 . 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