Concrete Constructions of Real Equiangular Line Sets
We give some concrete constructions of real equiangular line sets. The emphasis here is on {\em building blocks} for certain angles which are then used to build up larger equiangular line sets. We concentrate on angles greater than or equal to 1/7.
Authors: Janet C. Tremain
Concrete Constructions of Real Equiangular Line Sets Janet C. T remai n ∗ Department of Mathemat ics Universit y of Mi sso uri Columbia, MO 6521 1-410 0 jane t@ma th.mis souri.edu No vem ber 4 , 2018 Abstract W e giv e some concrete constructions of real equiangular line s ets. The emphasis here is on b uilding blo cks for certain angles which are th en used to bu ild up larger equiangular line sets. W e concentrat e on angles g reater than or equal to 1 / 7. 0 Math Subje ct Classi fic ation. P rimary: 51M04 ∗ The author was suppo rted by NSF DMS 070 4216 1 Con ten ts 1 Index 4 2 In tro duction 6 3 M = 3 vectors in R 2 at angle 1 2 8 4 M = 6 vectors in R 3 at angle 1 √ 5 8 5 Angle 1 / 3 9 5.1 M = 4 v ectors in R 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 5.2 M = 6 v ectors in R 4 at angle 1 / 3 . . . . . . . . . . . . . . . . . . . . . . . . 9 5.3 M = 8 v ectors in R 5 at angle 1 / 3 . . . . . . . . . . . . . . . . . . . . . . . . 9 5.4 M = 8 v ectors in R 7 at angle 1 / 3 . . . . . . . . . . . . . . . . . . . . . . . . 9 5.5 M = 6 v ectors in R 4 at angle 1 / 3 . . . . . . . . . . . . . . . . . . . . . . . . 10 5.6 M = 8 v ectors in R 5 at angle 1 / 3 . . . . . . . . . . . . . . . . . . . . . . . . 11 5.7 M = 8 v ectors in R 5 at angle 1 / 3 . . . . . . . . . . . . . . . . . . . . . . . . 11 5.8 M = 10 v ectors in R 5 at angle 1 / 3 . . . . . . . . . . . . . . . . . . . . . . . 1 1 5.9 M = 12 v ectors in R 6 at angle 1 / 3 . . . . . . . . . . . . . . . . . . . . . . . 1 2 5.10 M = 12 vec tors in R 6 at angle 1 / 3 - another example . . . . . . . . . . . . . 12 5.11 M = 16 vec tors in R 6 at angle 1 / 3 . . . . . . . . . . . . . . . . . . . . . . . 1 3 5.12 M = 16 vec tors in R 6 at angle 1 / 3 fro m building blo ck s . . . . . . . . . . . . 14 5.13 M = 28 vec tors in R 7 at angle 1 / 3 . . . . . . . . . . . . . . . . . . . . . . . 1 5 5.14 M = 28 vec tors in R 7 at angle 1 / 3 - simpler represen tation . . . . . . . . . . 16 5.15 M = 28 vec tors in R 7 at angle 1 / 3 - Differen t format . . . . . . . . . . . . . 17 5.16 M = 28 vec tors in R 7 at angle 1 / 3 - second example . . . . . . . . . . . . . . 18 6 M = 2( N − 1) v ectors in R N at angle 1 / 3 19 7 Angle 1 / 5 20 7.1 M = 4 v ectors in R 4 at angle 1 / 5 . . . . . . . . . . . . . . . . . . . . . . . . 20 7.2 M = 5 v ectors in R 5 at angle 1 / 5 . . . . . . . . . . . . . . . . . . . . . . . . 20 7.3 M = 5 v ectors in R 5 at angle 1 / 5 and is circulan t . . . . . . . . . . . . . . . 20 7.4 M = 5 v ectors in R 5 at angle 1 / 5 and is circulan t . . . . . . . . . . . . . . . 20 7.5 M = 4 v ectors in R 7 at angle 1 / 5 . . . . . . . . . . . . . . . . . . . . . . . . 21 7.6 M = 8 v ectors in R 7 at angle 1 / 5 . . . . . . . . . . . . . . . . . . . . . . . . 21 7.7 M = 8 v ectors in R 8 at angle 1 / 5 . . . . . . . . . . . . . . . . . . . . . . . . 21 7.8 M = 28 v ectors in R 14 at angle 1 / 5 . . . . . . . . . . . . . . . . . . . . . . . 22 7.9 M = 30 in R 15 at angle 1 / 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 7.10 M = 36 vec tors in R 15 at angle 1 / 5 . . . . . . . . . . . . . . . . . . . . . . . 24 7.11 M = 40 vec tors in R 16 at angle 1 / 5 . . . . . . . . . . . . . . . . . . . . . . . 25 7.12 M = 60 vec tors in R 19 at angle 1 / 5 . . . . . . . . . . . . . . . . . . . . . . . 27 7.13 M = 45 vec tors in R 21 at angle 1 / 5 . . . . . . . . . . . . . . . . . . . . . . . 28 7.14 M = 45 vec tors in R 21 at angle 1 / 5 . . . . . . . . . . . . . . . . . . . . . . . 29 7.15 N − 1 ve ctors in R N at angle 1 / 5 . . . . . . . . . . . . . . . . . . . . . . . . 30 2 7.16 N − 1 ve ctors in R N at angle 1 / 5 . . . . . . . . . . . . . . . . . . . . . . . . 30 7.17 M = 4 N ve ctors in R 3 N +1 at angle 1 / 5 . . . . . . . . . . . . . . . . . . . . . 31 8 Angle 1 / 7 32 8.1 M = 7 v ectors in R 7 at angle 1 / 7 . . . . . . . . . . . . . . . . . . . . . . . . 32 8.2 M = 8 v ectors in R 7 at angle 1 / 7 . . . . . . . . . . . . . . . . . . . . . . . . 32 9 M = 2 N v ect ors in R N 33 9.1 M = 6 v ectors in R 3 at angle 1 / √ 5 . . . . . . . . . . . . . . . . . . . . . . . 33 9.2 M = 10 v ectors in R 5 at angle 1 / 3 . . . . . . . . . . . . . . . . . . . . . . . 3 3 9.3 M = 12 v ectors in R 6 at angle 1 / 3 . . . . . . . . . . . . . . . . . . . . . . . 3 4 9.4 M = 28 v ectors in R 14 at angle 1 / 3 . . . . . . . . . . . . . . . . . . . . . . . 34 9.5 M = 28 v ectors in R 14 at angle 1 / 5 . . . . . . . . . . . . . . . . . . . . . . . 35 10 M = N + 1 vectors in R N 36 11 Unitary Matrices 37 11.1 Circulant self-adjoint Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 11.2 Circulant self-adjoint Unitary Matrix . . . . . . . . . . . . . . . . . . . . . . 37 11.3 Another example of circulant matrices . . . . . . . . . . . . . . . . . . . . . 37 12 Multiple angles 38 12.1 M = 2 N ve ctors at t w o a ngles: 1 / √ 5 , 0 . . . . . . . . . . . . . . . . . . . . . 38 3 1 Index In these notes, w e are c onstructing M equ iangular lines in R N at angle 1 /α . (I) N=2 , M=3. (I I) N=3, M=6. (I I I) Angle 1 / 3. (A) B uilding Blo cks 1. N= 3 , M=4. 2. N= 4 , M=6. 3. N= 5 , M=8. 4. N= 7 , M=8. (B) Spec ific Dimensions 1. N= 4 , M=6. 2. N= 5 , M=8 3. N= 5 , M=8 4. N= 5 , M=10. 5. N= 6 , M=12. 6. N= 6 , M=12 (Another example). 7. N= 6 , M=16 (All entries ± 1 / √ 6). 8. N= 6 , M=16 (F rom building blo c ks). 9. N= 7 , M=28. 10. N=7, M=28 (Simpler F ormat). 11. N=7, M=28 (Differen t F ormat) . 12. N=7, M=28 (Another example). (C) General Sets 1. R N alw a ys has M = 2( N − 1). (IV) A ngle 1 / 5. (A) B uilding Blo cks 1. N= 4 , M=4. 2. N= 5 , M=5 (Circulan t con t a ining 4 × 4 orthogo nal). 3. N= 5 , M=5 (Circulan t) 4. N= 5 , M=5 (Circulan t) 5. N= 7 , M=4. 6. N= 7 , M=8. 7. N= 8 , M=8. 4 (B) Spec ific Dimensions 1. N= 1 4, M=28. 2. N= 1 5, M=30. 3. N= 1 5, M=36. 4. N= 1 6, M=40. 5. N= 1 9, M= 60. 6. N= 2 1, M=45. 7. N= 2 1, M= 45 (Using building blo c ks) (C) General sets 1. R N alw a ys has M = N − 1 v ectors at ang le 1 / 5 . 2. R N alw a ys has M = N − 1 v ectors at ang le 1 / 5 (Another example). 3. R 3 N +1 alw a ys has 4 N v ectors at angle 1 / 5. (V) A ngle 1 / 7. (A) B uilding Blo cks 1. N= 7 , M=7. 2. N= 7 , M=8. (IX) M = 2 N v ectors in R N . 1. N= 3 , M=6, angle 1 / √ 5. 2. N= 5 , M=10, angle 1 / 3. 3. N= 6 , M=12, angle 1 / 3. 3. N= 1 4, M=28, angle 1 / 5. 4. N= 1 5, M=30, angle 1 / 5. (X) M = N + 1 vectors in R N (XI) Unitary Matrices 1. M = N (Circulant self-adjoin t unitary with tw o en tries) 2. M = N (Circulant self-adjoin t unitary with tw o en tries) 3. M = N (Circulant unitary with tw o en tries) (XI I) Multiple Angles 5. M = 2 N , inner pro ducts ± 1 / √ 5 , 0. (Made up of t w o circulan ts). 5 2 In t ro duction A very old pro blem is: Problem 2.1. How many e quiangular line s c an we dr aw thr ough the origin in R N ? This means, if w e c ho ose a set of unit length v ectors { f m } M m =1 , o ne on each line, then there is a constan t c so tha t f or all 1 ≤ m 6 = n ≤ M we ha v e |h f n , f m i| = c. These inner pro ducts represen t the cosine of the acute angle b etw een the lines. The pro blem of constructing an y n umber (esp ecially , the maximal n umber) o f equiangular lines in R N is o ne of the most elemen t a ry and at the same time one of the most difficult pr o blems in mathematics. After sixt y y ears of researc h, the maximal n um b er of equiangular lines in R N is known only for 35 dimensions. F or a slightly more general view of this topic see Benedetto and Kolesar [1]. This line of researc h w as started in 1948 b y Hann tjes [4] in the setting of elliptic geometry where he identifie d the maximal num b er of equiangular lines in R N for n = 2 , 3. Later, V an Lint and Seidel [7] classified the largest num b er of equiangular lines in R N for dimensions N ≤ 7 and at the same t ime emphasized the relations to discrete ma t hematics. In 197 3 , Lemmens and Seidel [6] made a comprehens iv e study o f real equiangular line sets whic h is still to day a fundamen tal piece of w ork. Gerzon [6] gav e an upp er b ound fo r the maximal n um b er of equiangular lines in R N : Theorem 2.2 (Gerzon) . I f we have M e quiangular lines in R N then M ≤ N ( N + 1) 2 . W e will see that in most cases there are man y few er lines t han this b ound giv es. Also, P . Neumann [6] pro duced a fundamen tal result in the area: Theorem 2.3 (P . Neumann) . I f R N has M e quiangular li nes at angle 1 / α a nd M > 2 N , then α is an o dd inte ger. Finally , there is a low er b ound o n the angle formed b y equiangular line sets. Theorem 2.4. If { f m } M m =1 is a s et of norm one ve ctors in R N , then max m 6 = n |h f m , f n i| ≥ s M − N N ( M − 1) . (2.1) Mor e over, we have e quality if and only if { f m } M m =1 is an e q uiangular tight fr ame and in this c ase the tight fr am e b ound is M N . 6 This inequality go es bac k to W elch [9]. Strohmer and Heath [8] and Ho lmes and Paulsen [5] giv e more direct argumen ts whic h also yields the ”moreo v er” part. F or some reason, in the lit era t ure t here is a further assumption added to the ”mo r eo v er” part of Theorem 2.4 that the vec tors span R N . This assumption is not necessary . That is, equalit y in inequalit y 2.1 already implies that the v ectors span the space [2]. The status of the equiangular line problem at this p oin t is summarized in the follow ing c hart [6, 2, 3] where N is the dimension of the Hilb ert space, M is the maximal n um b er of equiangular lines and these will o ccur at the ang le 1 /α . T able I: Maximal equiangular line sets N = 2 3 4 5 6 7 . . . 13 14 M = 3 6 6 10 16 2 8 . . . 28 2 8 − 30 α = 2 √ 5 3 3 3 3 . . . 3 5 N = 15 1 6 17 18 19 20 M = 36 ≥ 40 ≥ 48 ≥ 48 72 − 76 ∗ 9 2 − 96 ∗ α = 5 5 5 5 5 5 N = 21 22 2 3 . . . 41 42 43 M = 126 176 276 . . . 276 ≥ 2 76 34 4 α = 5 5 5 . . . 5 5 7 The ∗ in the c hart represen ts tw o cases whic h hav e b een report ed as solv ed in the literature but a ctually are still o p en. F o r recen t results o n the equiangular line problem see [2, 3]. Here w e will give concrete constructions f o r real equiangular line sets. The emphasis will b e on constructing building blo c ks whic h can b e put tog ether to build larger and larger equiangular line sets. 7 3 M = 3 v ecto r s in R 2 at angle 1 2 f 1 f 2 f 3 = 0 1 − √ 3 2 − 1 2 √ 3 2 − 1 2 4 M = 6 v ecto r s in R 3 at angle 1 √ 5 0 q 5 − √ 5 10 q 5+ √ 5 10 0 − q 5 − √ 5 10 q 5+ √ 5 10 q 5 − √ 5 10 q 5+ √ 5 10 0 − q 5 − √ 5 10 q 5+ √ 5 10 0 q 5+ √ 5 10 0 q 5 − √ 5 10 q 5+ √ 5 10 0 − q 5 − √ 5 10 8 5 Angle 1 / 3 (A) Building Blo c ks 5.1 M = 4 ve ctors in R 3 1 √ 3 1 1 1 − 1 1 1 1 − 1 1 1 1 − 1 5.2 M = 6 ve ctors in R 4 at angle 1 / 3 q 1 3 q 2 3 q 1 3 − q 2 3 q 1 3 q 2 3 q 1 3 − q 2 3 q 1 3 q 2 3 q 1 3 − q 2 3 5.3 M = 8 ve ctors in R 5 at angle 1 / 3 r 1 3 1 1 1 0 0 − 1 1 1 0 0 1 − 1 1 0 0 1 1 − 1 0 0 0 0 1 1 1 0 0 − 1 1 1 0 0 1 − 1 1 0 0 1 1 − 1 5.4 M = 8 ve ctors in R 7 at angle 1 / 3 r 1 10 1 1 1 1 √ 2 √ 2 √ 2 1 1 − 1 − 1 √ 2 √ 2 − √ 2 1 − 1 1 − 1 √ 2 − √ 2 − √ 2 1 − 1 − 1 1 − √ 2 √ 2 − √ 2 1 1 1 1 − √ 2 − √ 2 − √ 2 1 1 − 1 − 1 − √ 2 − √ 2 √ 2 1 − 1 1 − 1 − √ 2 √ 2 √ 2 1 − 1 − 1 1 √ 2 − √ 2 √ 2 9 (B) Sp ecific Dimensions 5.5 M = 6 ve ctors in R 4 at angle 1 / 3 1 0 0 0 − 1 3 1 3 q 3 2 1 3 √ 2 q 2 3 − 1 3 1 3 q 3 2 1 3 √ 2 − q 2 3 − 1 3 0 2 √ 2 3 0 − 1 3 − q 2 3 − √ 2 3 0 − 1 3 q 2 3 − √ 2 3 0 Remark 5.1. T he ab ove exam ple shows that an M elem ent e quiangular tight fr ame for R N ne e d not have the pr op erty that every N element subset is lin e arly indep endent? I n the ab ove example, we cle arly hav e four ve ctors which sit in R 3 . 10 5.6 M = 8 ve ctors in R 5 at angle 1 / 3 1 √ 6 1 √ 6 1 √ 6 1 √ 6 1 √ 3 1 √ 6 1 √ 6 − 1 √ 6 − 1 √ 6 1 √ 3 1 √ 6 − 1 √ 6 1 √ 6 − 1 √ 6 1 √ 3 1 √ 6 − 1 √ 6 − 1 √ 6 1 √ 6 1 √ 3 1 √ 6 1 √ 6 1 √ 6 1 √ 6 − 1 √ 3 1 √ 6 1 √ 6 − 1 √ 6 − 1 √ 6 − 1 √ 3 1 √ 6 − 1 √ 6 1 √ 6 − 1 √ 6 − 1 √ 3 1 √ 6 − 1 √ 6 − 1 √ 6 1 √ 6 − 1 √ 3 5.7 M = 8 ve ctors in R 5 at angle 1 / 3 r 1 3 1 1 1 0 0 − 1 1 1 0 0 1 − 1 1 0 0 1 1 − 1 0 0 0 0 1 1 1 0 0 − 1 1 1 0 0 1 − 1 1 0 0 1 1 − 1 5.8 M = 10 v ectors in R 5 at angle 1 / 3 q 1 3 q 2 3 0 0 0 − q 1 3 q 2 3 0 0 0 q 1 3 0 q 2 3 0 0 − q 1 3 0 q 2 3 0 0 q 1 3 0 0 q 2 3 0 − q 1 3 0 0 q 2 3 0 0 1 3 q 3 2 1 3 q 3 2 1 3 q 3 2 q 1 2 0 1 3 q 3 2 − 1 3 q 3 2 1 3 q 3 2 − q 1 2 0 − 1 3 q 3 2 − 1 3 q 3 2 1 3 q 3 2 q 1 2 0 1 3 q 3 2 − 1 3 q 3 2 − 1 3 q 3 2 q 1 2 11 5.9 M = 12 v ectors in R 6 at angle 1 / 3 1 2 3 4 5 6 q 1 3 q 2 3 q 1 3 − q 2 3 q 1 3 q 2 3 q 1 3 − q 2 3 q 1 3 q 2 3 q 1 3 − q 2 3 q 1 3 q 2 3 q 1 3 − q 2 3 1 √ 6 1 √ 6 1 √ 6 1 √ 6 1 √ 3 1 √ 6 1 √ 6 − 1 √ 6 − 1 √ 6 1 √ 3 1 √ 6 − 1 √ 6 1 √ 6 − 1 √ 6 1 √ 3 1 √ 6 − 1 √ 6 − 1 √ 6 1 √ 6 1 √ 3 5.10 M = 12 v ectors in R 6 at angle 1 / 3 - another example W e use a building blo c k: 1 √ 3 1 1 1 − 1 1 1 1 − 1 1 1 1 − 1 Using this we ha v e (using + for 1 / √ 3 and - f or − 1 / √ 3): 1 2 3 4 5 6 1 + + + 2 − + + 3 + − + 4 + + − 5 + + + 6 − + + 7 + − + 8 + + − 9 + + + 10 − + + 11 + − + 12 + + − 12 5.11 M = 16 v ectors in R 6 at angle 1 / 3 1 √ 6 1 1 1 1 1 1 − 1 − 1 1 1 1 1 − 1 1 − 1 1 1 1 − 1 1 1 − 1 1 1 − 1 1 1 1 − 1 1 − 1 1 1 1 1 − 1 1 − 1 − 1 1 1 1 1 − 1 1 − 1 1 1 1 − 1 1 1 − 1 1 1 − 1 1 1 1 − 1 1 1 − 1 − 1 1 1 1 1 − 1 1 − 1 1 1 1 − 1 1 1 − 1 1 1 1 − 1 − 1 1 1 1 1 − 1 1 − 1 1 1 1 1 − 1 − 1 Pr o of. The ab o v e matrix has m utual inner pro ducts ± 1 / 3 since eac h ro w 2 ≤ m ≤ 16 has t w o − 1 ′ s . So their inner pro duct with row 1 ar e all 1 / 3. The other 15 row s come fro m putting − 1 ′ s in the 15 p ositions coming from 1 6 c ho ose 2. So an y t w o rows either hav e no − 1 ′ s in common and their inner pro duct is − 1 / 3 or they ha v e one − 1 in common and their inner pro duct is 1 / 3. Remark 5.2. This fam ily is p art of the M = 2 8 e quiangular lines in R 7 at angle 1 / 3 . 13 5.12 M = 16 v ectors in R 6 at angle 1 / 3 from bu ilding blo c ks r 1 3 1 1 1 0 0 0 − 1 1 1 0 0 0 1 − 1 1 0 0 0 1 1 − 1 0 0 0 0 0 1 1 1 0 0 0 − 1 1 1 0 0 0 1 − 1 1 0 0 0 1 1 − 1 0 1 0 0 0 1 1 − 1 0 0 0 1 1 1 0 0 0 − 1 1 1 0 0 0 1 − 1 0 1 0 1 0 1 0 − 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 − 1 14 5.13 M = 28 v ectors in R 7 at angle 1 / 3 . r 1 3 1 1 1 0 0 0 0 − 1 1 1 0 0 0 0 1 − 1 1 0 0 0 0 1 1 − 1 0 0 0 0 0 0 1 1 1 0 0 0 0 − 1 1 1 0 0 0 0 1 − 1 1 0 0 0 0 1 1 − 1 0 0 1 0 0 0 1 1 0 − 1 0 0 0 1 1 0 1 0 0 0 − 1 1 0 1 0 0 0 1 − 1 0 0 1 0 1 0 1 0 0 − 1 0 1 0 1 0 0 1 0 − 1 0 1 0 0 1 0 1 0 − 1 0 0 0 1 0 0 1 1 0 0 − 1 0 0 1 1 0 0 1 0 0 − 1 1 0 0 1 0 0 1 − 1 1 0 0 1 0 0 1 − 1 0 0 1 0 0 1 1 0 0 − 1 0 0 1 1 0 0 1 0 0 − 1 0 1 0 0 1 0 1 0 − 1 0 0 1 0 1 0 1 0 0 − 1 0 1 0 1 0 0 1 0 − 1 15 5.14 M = 28 v ectors in R 7 at angle 1 / 3 - simpler represen tation Let us try this without the zero es to see if it is clearer. Also, we add the dimension num b ers. r 1 3 1 2 3 4 5 6 7 1 1 1 -1 1 1 1 -1 1 1 1 - 1 1 1 1 -1 1 1 1 -1 1 1 1 - 1 1 1 1 -1 1 1 1 -1 1 1 1 -1 1 1 1 -1 1 1 1 - 1 1 1 1 -1 1 1 1 -1 1 1 1 -1 1 1 1 -1 1 1 1 -1 1 1 1 -1 1 1 1 -1 1 1 1 -1 1 1 1 -1 1 1 1 - 1 16 5.15 M = 28 v ectors in R 7 at angle 1 / 3 - Differen t format Y et another wa y to simplify this: First, we use our building blo c k for 4 vec tors in R 3 at angle 1 / 3 where + means 1 and − means − 1. B B 1 = r 1 3 + + + − + + + − + + + − No w, w e put these g roups of f o ur v ectors in to our c hart by just putting a dot where the columns g o. B B 1 4 B B 1 4 B B 1 4 B B 1 4 B B 1 4 B B 1 4 B B 1 4 1 2 3 4 5 6 7 • • • • • • • • • • • • • • • • • • • • • 17 5.16 M = 28 v ectors in R 7 at angle 1 / 3 - second example W e mak e a bulding blo ck b y embedding our 12 v ectors in R 6 in to R 7 . B B 9 = 1 √ 6 0 1 1 1 1 1 1 0 − 1 − 1 1 1 1 1 0 − 1 1 − 1 1 1 1 0 − 1 1 1 − 1 1 1 0 − 1 1 1 1 − 1 1 0 − 1 1 1 1 1 − 1 0 1 − 1 − 1 1 1 1 0 1 − 1 1 − 1 1 1 0 1 − 1 1 1 − 1 1 0 1 − 1 1 1 1 − 1 0 1 1 − 1 − 1 1 1 0 1 1 − 1 1 − 1 1 0 1 1 − 1 1 1 − 1 0 1 1 1 − 1 − 1 1 0 1 1 1 − 1 1 − 1 0 1 1 1 1 − 1 − 1 Under this building blo ck w e put our standard 12 ve ctors in R 7 at angle 1 / 3 1 2 3 4 5 6 7 q 1 3 q 2 3 q 1 3 − q 2 3 q 1 3 q 2 3 q 1 3 − q 2 3 q 1 3 q 2 3 q 1 3 − q 2 3 q 1 3 q 2 3 q 1 3 − q 2 3 q 1 3 q 2 3 q 1 3 − q 2 3 q 1 3 q 2 3 q 1 3 − q 2 3 18 (C) General Sets 6 M = 2( N − 1) v ector s in R N at angle 1 / 3 Prop osition 6.1. F or every N , R N has 2 ( N − 1) e quiangular li nes sp anning R N at angle 1 / 3 . This family is never a tight fr am e. Pr o of. W e will just write do wn the v ectors: q 1 3 q 2 3 0 0 · · · 0 0 q 1 3 − q 2 3 0 0 · · · 0 0 q 1 3 0 q 2 3 0 · · · 0 0 q 1 3 0 − q 2 3 0 · · · 0 0 . . . . . . . . . . . . · · · 0 0 q 1 3 0 0 0 · · · 0 q 2 3 q 1 3 0 0 0 · · · 0 − q 2 3 . 19 7 Angle 1 / 5 (A) Building Blo c ks 7.1 M = 4 ve ctors in R 4 at angle 1 / 5 q 2 5 q 1 5 q 1 5 q 1 5 q 2 5 − q 1 5 − q 1 5 q 1 5 q 2 5 − q 1 5 q 1 5 − q 1 5 q 2 5 q 1 5 − q 1 5 − q 1 5 7.2 M = 5 ve ctors in R 5 at angle 1 / 5 r 1 5 1 1 1 1 1 − 1 1 1 − 1 1 − 1 − 1 1 1 − 1 1 − 1 1 − 1 1 − 1 − 1 − 1 1 1 Remark 7.1. Note that the 4 × 4 submatrix in the upp er left c orner is an ortho gona l matrix. 7.3 M = 5 ve ctors in R 5 at angle 1 / 5 and is c irc u lan t 1 √ 5 − 1 1 1 1 1 1 − 1 1 1 1 1 1 − 1 1 1 1 1 1 − 1 1 1 1 1 1 − 1 7.4 M = 5 ve ctors in R 5 at angle 1 / 5 and is c irc u lan t 1 √ 5 1 1 1 − 1 1 1 1 1 1 − 1 − 1 1 1 1 1 1 − 1 1 1 1 1 1 − 1 1 1 20 7.5 M = 4 ve ctors in R 7 at angle 1 / 5 1 2 3 4 5 6 7 q 4 10 q 1 10 q 1 10 q 1 10 q 1 10 q 1 10 q 1 10 q 4 10 − q 1 10 − q 1 10 − q 1 10 − q 1 10 q 1 10 q 1 10 q 4 10 q 1 10 − q 1 10 q 1 10 − q 1 10 − q 1 10 − q 1 10 q 4 10 − q 1 10 − q 1 10 q 1 10 q 1 10 − q 1 10 − q 1 10 7.6 M = 8 ve ctors in R 7 at angle 1 / 5 r 1 10 1 1 1 1 √ 2 √ 2 √ 2 1 1 − 1 − 1 √ 2 − √ 2 √ 2 1 − 1 1 − 1 √ 2 − √ 2 − √ 2 1 − 1 − 1 1 − √ 2 − √ 2 √ 2 1 1 1 1 − √ 2 − √ 2 − √ 2 1 1 − 1 − 1 − √ 2 √ 2 − √ 2 1 − 1 1 − 1 − √ 2 √ 2 √ 2 1 − 1 − 1 1 √ 2 √ 2 − √ 2 7.7 M = 8 ve ctors in R 8 at angle 1 / 5 r 1 10 1 1 1 1 1 1 1 √ 3 1 1 − 1 − 1 − 1 − 1 1 √ 3 1 − 1 1 − 1 1 − 1 − 1 √ 3 1 − 1 − 1 1 − 1 1 − 1 √ 3 1 1 1 1 − 1 − 1 − 1 − √ 3 1 1 − 1 − 1 1 1 − 1 − √ 3 1 − 1 1 − 1 − 1 1 1 − √ 3 1 − 1 − 1 1 1 − 1 1 − √ 3 21 (B) Sp ecific Dimensions 7.8 M = 28 v ectors in R 14 at angle 1 / 5 This example is built up from building blo c ks. The first is: B B 2 = q 2 5 q 1 5 q 1 5 q 1 5 q 2 5 − q 1 5 − q 1 5 q 1 5 q 2 5 − q 1 5 q 1 5 − q 1 5 q 2 5 q 1 5 − q 1 5 − q 1 5 This mat r ix will b e spread out and the columns represen ted by ”bullets”. The second is: B B 3 = 1 2 3 4 5 6 7 q 4 10 q 1 10 q 1 10 q 1 10 q 1 10 q 1 10 q 1 10 q 4 10 − q 1 10 − q 1 10 − q 1 10 − q 1 10 q 1 10 q 1 10 q 4 10 q 1 10 q 1 10 − q 1 10 − q 1 10 − q 1 10 − q 1 10 q 4 10 − q 1 10 − q 1 10 q 1 10 q 1 10 − q 1 10 − q 1 10 F o r this mat rix we will just put the first row in. No w we use our dot tric k to piece these together. B B 3 \ 4 B B 2 \ 4 B B 2 \ 4 B B 2 \ 4 B B 2 \ 4 B B 2 \ 4 B B 2 \ 4 1 2 3 4 5 6 7 8 9 10 11 12 13 14 q 4 10 q 1 10 q 1 10 q 1 10 q 1 10 q 1 10 q 1 10 q 2 5 • • • q 2 5 • • • q 2 5 • • • q 2 5 • • • q 2 5 • • • q 2 5 • • • 22 7.9 M = 30 in R 15 at angle 1 / 5 W e use a Building Blo c k and put ”bullets” eac h place it o ccurs. B B 4 = 1 √ 5 − 1 1 1 1 1 1 − 1 1 1 1 1 1 − 1 1 1 1 1 1 − 1 1 1 1 1 1 − 1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 B B 4 \ 4 • • • • • B B 4 \ 4 • • • • • B B 4 \ 4 • • • • • B B 4 \ 4 • • • • • B B 4 \ 4 • • • • • B B 4 \ 4 • • • • • 23 7.10 M = 36 v ectors in R 15 at angle 1 / 5 First we rev erse one o f our Building Blo c ks: B B 5 = q 1 5 q 1 5 q 1 5 q 2 5 q 1 5 − q 1 5 − q 1 5 q 2 5 q 1 5 − q 1 5 q 1 5 − q 2 5 q 1 5 q 1 5 − q 1 5 − q 2 5 B B 6 = r 1 10 1 1 1 1 1 1 1 √ 3 1 1 − 1 − 1 − 1 − 1 1 √ 3 1 − 1 1 − 1 1 − 1 − 1 √ 3 1 − 1 − 1 1 − 1 1 − 1 √ 3 1 1 1 1 − 1 − 1 − 1 − √ 3 1 1 − 1 − 1 1 1 − 1 − √ 3 1 − 1 1 − 1 − 1 1 1 − √ 3 1 − 1 − 1 1 1 − 1 1 − √ 3 1 2 3 4 5 6 7 8 9 10 11 12 1 3 1 4 15 B B 5 4 • • • q 2 5 B B 5 4 • • • q 2 5 B B 5 4 • • • q 2 5 B B 5 4 • • • q 2 5 B B 5 4 • • • q 2 3 B B 5 4 • • • q 2 5 B B 5 4 • • • q 2 5 B B 6 8 q 1 10 q 1 10 q 1 10 q 1 10 q 1 10 q 1 10 q 1 10 q 3 10 24 7.11 M = 40 v ectors in R 16 at angle 1 / 5 F o r this we need a new building blo c k. B B 9 = q 1 5 − q 1 10 − q 1 10 − q 1 10 q 1 10 q 2 10 q 1 5 q 1 10 q 1 10 q 1 10 − q 1 10 q 2 10 − q 1 5 − q 1 10 q 1 10 − q 1 10 − q 1 10 q 2 10 − q 1 5 q 1 10 − q 1 10 q 1 10 q 1 10 q 2 10 In the next chart, the first sev en rows are building blo c k 5 and contain four ve ctors eac h. Ro w 8 is Building blo c k 6 and has eight v ectors. And Row 9 is Building blo c k 9 and has four ve ctors. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 • • • q 2 5 • • • q 2 5 • • • q 2 5 • • • q 2 5 • • • q 2 3 • • • q 2 5 • • • q 2 5 q 1 10 q 1 10 q 1 10 q 1 10 q 1 10 q 1 10 q 1 10 q 3 10 q 1 5 ± q 1 10 ± q 1 10 ± q 1 10 ± q 1 10 ± q 2 5 On t he next page we will giv e a hin t ab out ho w to pro v e this is an equiangular line set. 25 The first 36 ve ctors represen t 36 equiangular lines in R 15 and this is easily c hec k ed by sigh t. Also, the last fo ur vec tors ha v e q 1 5 in the first p osition and this hits exactly the groups in row s 1,3 and 5 in eactly one p osition - the first p osition and this yields 1 / 5. The last g r oup hits the elemen ts in ro ws 2,4,6, and 7 in exactly one pla ce and pro duces r 2 5 · r 1 10 = 1 5 . So w e only need to che c k how the la st four v ectors in teract with the eight just b efor e them. But these vec tors all hav e exactly four co ordinates in common (9,1 1 ,13, and 14) and they are resp ectiv ely: + + + + + − − + − − − − − + + − + + − − + − + − − − + + − + − + and − − − + + + + − − + − − + − + + No w, a visual c hec k sho ws that the inner pro ducts of the row s in the second group with the rows in the first group will yield ± " r 1 10 · r 1 10 + r 1 10 · r 1 10 # = ± 1 5 . 26 7.12 M = 60 v ectors in R 19 at angle 1 / 5 W e use a Building Blo c k and put ”bullets” eac h place it o ccurs. B B 4 = 1 √ 5 − 1 1 1 1 1 1 − 1 1 1 1 1 1 − 1 1 1 1 1 1 − 1 1 1 1 1 1 − 1 1 2 3 4 5 6 7 8 9 10 11 1 2 13 14 15 16 17 18 19 • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 27 7.13 M = 45 v ectors in R 21 at angle 1 / 5 W e use t w o building blo c ks. The first is a 4 × 4 orthogonal matrix with an extra column - and it is represen ted in our matrix b y ”bullets”. B B 7 = r 1 5 1 1 1 1 1 1 1 1 − 1 − 1 1 1 − 1 1 − 1 1 1 − 1 − 1 1 The second building blo ck is a 5 × 5 matrix - and it is represen ted by ”*”. B B 8 = 1 √ 5 1 1 1 1 1 − 1 − 1 − 1 1 1 − 1 1 1 − 1 1 − 1 − 1 1 1 − 1 1 − 1 1 − 1 1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 2 1 • • • • • • • • • • • • • • • • • • • • • • • • • ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ 28 7.14 M = 45 v ectors in R 21 at angle 1 / 5 W e use the a Building Blo c k and put ”bullets” each place it app ears. B B 4 = 1 √ 5 − 1 1 1 1 1 1 − 1 1 1 1 1 1 − 1 1 1 1 1 1 − 1 1 1 1 1 1 − 1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 2 1 • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 29 (C) General Sets 7.15 N − 1 ve ctors in R N at angle 1 / 5 1 q 1 5 q 2 5 q 2 5 · · · 2 q 1 5 q 2 5 − q 2 5 · · · 3 q 1 5 q 2 5 q 2 5 · · · 4 q 1 5 q 2 5 − q 2 5 · · · 5 q 1 5 q 2 5 q 2 5 · · · 6 q 1 5 q 2 5 − q 2 5 · · · . . . . . . . . . . . . . . . . . . . . . . . . 7.16 N − 1 ve ctors in R N at angle 1 / 5 1 q 1 5 q 1 5 q 3 5 · · · 2 q 1 5 q 1 5 − q 3 5 · · · 3 q 1 5 q 1 5 q 3 5 · · · 4 q 1 5 q 1 5 − q 3 5 · · · 5 q 1 5 q 1 5 q 3 5 · · · 6 q 1 5 q 1 5 − q 3 5 · · · . . . . . . . . . . . . . . . . . . . . . . . . 30 7.17 M = 4 N v ectors in R 3 N +1 at angle 1 / 5 W e use our building blo c k B B 5 = q 1 5 q 1 5 q 1 5 q 2 5 q 1 5 − q 1 5 − q 1 5 q 2 5 q 1 5 − q 1 5 q 1 5 − q 2 5 q 1 5 q 1 5 − q 1 5 − q 2 5 W e use ”bullets” to indicate where these columns go. • • • • · · · • • • • · · · • • • • · · · . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . · · · 31 8 Angle 1 / 7 (A) Building Blo c ks 8.1 M = 7 ve ctors in R 7 at angle 1 / 7 1 √ 7 1 1 1 1 1 1 1 − 1 − 1 − 1 1 1 1 1 1 1 − 1 − 1 − 1 1 1 1 − 1 1 1 − 1 − 1 1 1 1 − 1 1 1 − 1 − 1 − 1 1 1 1 1 − 1 1 − 1 1 − 1 1 − 1 − 1 1 8.2 M = 8 ve ctors in R 7 at angle 1 / 7 r 1 10 1 1 1 1 √ 2 √ 2 √ 2 1 1 − 1 − 1 √ 2 − √ 2 √ 2 1 − 1 1 − 1 √ 2 − √ 2 − √ 2 1 − 1 − 1 1 − √ 2 − √ 2 √ 2 1 1 1 1 − √ 2 − √ 2 − √ 2 1 1 − 1 − 1 − √ 2 √ 2 − √ 2 1 − 1 1 − 1 − √ 2 √ 2 √ 2 1 − 1 − 1 1 √ 2 √ 2 − √ 2 32 9 M = 2 N v ec tors in R N 9.1 M = 6 ve ctors in R 3 at angle 1 / √ 5 0 q 5 − √ 5 10 q 5+ √ 5 10 0 − q 5 − √ 5 10 q 5+ √ 5 10 q 5 − √ 5 10 q 5+ √ 5 10 0 − q 5 − √ 5 10 q 5+ √ 5 10 0 q 5+ √ 5 10 0 q 5 − √ 5 10 q 5+ √ 5 10 0 − q 5 − √ 5 10 9.2 M = 10 v ectors in R 5 at angle 1 / 3 q 1 3 q 2 3 0 0 0 − q 1 3 q 2 3 0 0 0 q 1 3 0 q 2 3 0 0 − q 1 3 0 q 2 3 0 0 q 1 3 0 0 q 2 3 0 − q 1 3 0 0 q 2 3 0 0 1 3 q 3 2 1 3 q 3 2 1 3 q 3 2 q 1 2 0 1 3 q 3 2 − 1 3 q 3 2 1 3 q 3 2 − q 1 2 0 − 1 3 q 3 2 − 1 3 q 3 2 1 3 q 3 2 q 1 2 0 1 3 q 3 2 − 1 3 q 3 2 − 1 3 q 3 2 q 1 2 33 9.3 M = 12 v ectors in R 6 at angle 1 / 3 1 2 3 4 5 6 q 1 3 q 2 3 q 1 3 − q 2 3 q 1 3 q 2 3 q 1 3 − q 2 3 q 1 3 q 2 3 q 1 3 − q 2 3 q 1 3 q 2 3 q 1 3 − q 2 3 1 √ 6 1 √ 6 1 √ 6 1 √ 6 1 √ 3 1 √ 6 1 √ 6 − 1 √ 6 − 1 √ 6 1 √ 3 1 √ 6 − 1 √ 6 1 √ 6 − 1 √ 6 1 √ 3 1 √ 6 − 1 √ 6 − 1 √ 6 1 √ 6 1 √ 3 9.4 M = 28 v ectors in R 14 at angle 1 / 3 First, w e use our building blo ck fo r 4 v ectors in R 3 at angle 1 / 3 where + means 1 and − means − 1. B B 1 = r 1 3 + + + − + + + − + + + − No w, w e put these g roups of f o ur v ectors in to our c hart by just putting a dot where the columns g o. B B 1 4 B B 1 4 B B 1 4 B B 1 4 B B 1 4 B B 1 4 B B 1 4 1 2 3 4 5 6 7 • • • • • • • • • • • • • • • • • • • • • 34 9.5 M = 28 v ectors in R 14 at angle 1 / 5 This example is built up from building blo c ks. The first is: B B 2 = q 2 5 q 1 5 q 1 5 q 1 5 q 2 5 − q 1 5 − q 1 5 q 1 5 q 2 5 − q 1 5 q 1 5 − q 1 5 q 2 5 q 1 5 − q 1 5 − q 1 5 This mat r ix will b e spread out and the columns represen ted by ”bullets”. The second is: B B 3 = 1 2 3 4 5 6 7 q 4 10 q 1 10 q 1 10 q 1 10 q 1 10 q 1 10 q 1 10 q 4 10 − q 1 10 − q 1 10 − q 1 10 − q 1 10 q 1 10 q 1 10 q 4 10 q 1 10 − q 1 10 q 1 10 − q 1 10 − q 1 10 − q 1 10 q 4 10 − q 1 10 − q 1 10 q 1 10 q 1 10 − q 1 10 − q 1 10 F o r this mat rix we will just put the first row in. No w we use our dot tric k to piece these together. B B 3 \ 4 B B 2 \ 4 B B 2 \ 4 B B 2 \ 4 B B 2 \ 4 B B 2 \ 4 B B 2 \ 4 1 2 3 4 5 6 7 8 9 10 11 12 13 14 q 4 10 q 1 10 q 1 10 q 1 10 q 1 10 q 1 10 q 1 10 q 2 5 • • • q 2 5 • • • q 2 5 • • q 2 5 • • • q 2 5 • • • q 2 5 • • • 35 10 M = N + 1 v ector s in R N Prop osition 10.1. F or every natur al numb er N , the ( N + 1) × ( N +1) matrix b elow r epr esents N + 1 unit norm ve ctors { f m } N +1 m =1 in R N +1 giving an e quiangular tight fr am e for an N - dimensional sp ac e and s atisfying: (1) k f m k = 1 . (2) F or any n 6 = m , h f m , f n i = − 1 N . Pr o of. F or an y m we ha v e k f m k 2 = 1 N ( N + 1 [ N 2 + N ] = 1 . Also, h f m , f n i = − 2 N + N − 1 N ( N + 1) = − ( N + 1) N ( N + 1) = − 1 N . 36 11 Unitary Matrice s 11.1 Circulan t self-adjoin t Matrix − b a a · · · a a − b a · · · a . . . . . . . . . · · · . . . a a a · · · a a a a · · · − b 11.2 Circulan t self-adjoin t Unitary Matrix If we rotate the ro ws of the matrix w e can get matrices for ev ery 2 N -dimensional Hilb ert space yielding a circulan t, self-adjoin t unitary matrix. a a . . . a − b a . . . a a a . . . a a − b . . . a . . . . . . . . . . . . . . . . . . . . . . . . a a . . . a a a . . . − b − b a . . . a a a . . . a a − b . . . a a a . . . a . . . . . . . . . . . . . . . . . . . . . . . . a a . . . − b a a . . . a Pr o of. W e ha v e for a ny tw o ro ws n 6 = m of this matrix: h f n , f m i = (2 N − 2) a 2 − 2 ab = a [(2( N − 1) a − 2 b ] . So this is zero if b = ( N − 1) a . 11.3 Another example of circulan t matrices a − b a · · · a a a − b · · · a . . . . . . . . . · · · . . . a a a · · · − b − b a a · · · a 37 12 Multiple angl e s 12.1 M = 2 N v ectors at t w o angles: 1 / √ 5 , 0 Prop osition 12.1. F or e very N , R N has 2 N lines at thr e e angles: ± 1 / √ 5 , 0 . These lines sp an R N . Pr o of. W e let x = s 5 − √ 5 10 y = s 5 + √ 5 10 . W e will just write do wn the ve ctors: f 1 f 2 f 3 f 4 f 5 f 6 . . . f 2 N − 3 f 2 N − 2 f 2 N − 1 f 2 N . = x y 0 0 · · · 0 0 0 − x y 0 0 · · · 0 0 0 0 x y 0 · · · 0 0 0 0 − x y 0 · · · 0 0 0 0 0 x y · · · 0 0 0 0 0 − x y · · · 0 0 0 . . . . . . . . . . . . . . . . . . . . . . . . 0 0 0 0 · · · 0 x y 0 0 0 0 · · · 0 − x y y 0 0 0 · · · 0 0 x y 0 0 0 · · · 0 0 − x It is ob vious that t hese lines span R N . Corollary 12.2. F or eve ry N , the 2 N unit ve ctors at angles 1 / √ 5 , 0 c a n b e divide d i nto two sets of c ir culant ve ctors. Pr o of. The tw o sets are: f 1 f 2 f 3 . . . f N = x y 0 0 · · · 0 0 0 x y 0 · · · 0 0 0 0 x y · · · 0 0 . . . . . . . . . . . . · · · 0 0 0 0 0 0 · · · x y y 0 0 0 · · · 0 x f 1 f 2 f 3 . . . f N = − x y 0 0 · · · 0 0 0 − x y 0 · · · 0 0 0 0 − x y · · · 0 0 . . . . . . . . . . . . · · · 0 0 0 0 0 0 · · · − x y y 0 0 0 · · · 0 − x 38 References [1] J. J. Benedetto and J. K olesar, Ge ometric pr op e rties of Gr assma nnian fr ame s for R 2 and R 3 , EURASIP J. Applied Signal Pro cessing, (2006 ), 17 pages. [2] P .G . Casazza, D. Redmond and J.C. T remain, R e al e quiangular tight fr am es , in prepa- ration. [3] P .G . Casazza, D. Redmond and J.C. T remain, R e al e quiangular fr ames , Pro ceedings of CISS - Princeton Univers ity (20 08). [4] J. Haan tjes, Equilater al p oint-sets in el l iptic two- and thr e e-dime nsional sp ac es , Nieuw Arc h. Wisk. 22 (1948) 355-362 . [5] R. B. Holmes and V. I. P aulsen, Optimal fr ames for er asur es , Linear Alg. a nd Appli- cations 377 (204) 31-51 . [6] P . Lemmens and J. Seidel, Equiangular lines , Journal of Algebra 24 (1973) 494- 512. [7] J. H. v an Lin t and J. J. Seidel, Equiangular p oint sets in el lip tic ge ometry , Pro c. Nederl. Ak ad. W etensc h. Series A 69 (1966 ) 335-348. [8] T. Str o hmer and R. W. Heath, Gr assmannian fr ames w ith applic ations to c o ding and c ommunic ation , Appl. Comp. Harmonic Anal. 14 No. 3 (2 003) 2 5 7-275. [9] L. R. W elc h, L ower b o unds on the maxim um cr oss-c o rr elation of s ignals , IEEE T rans. Inform. Theory 20 (1974 ) 3 97-399. 39
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