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1 Some 2 2 Unitary Space-Time Codes from Sphere Packing Theory with Optimal Diversity Product of Code Size 6 Haiquan Wang Genyuan Wang Xiang-Gen Xia Abstract In this correspondence, we propose some new designs of 2 2 unitary space-time codes of sizes 6; 32; 48; 64 with best known diversity products (or product distances) by partially using sphere packing theory. In particular, we present an optimal 2 2 unitary space-time code of size 6 in the sense that it reaches the maximal possible diversity product for 2 2 unitary space-time codes of size 6. The construction and the optimality of the code of size 6 provide the precise value of the maximal diversity product of a 2 2 unitary space-time code of size 6. Keywords: Unitary space-time codes, dierential space-time modulation, optimal diversity product, packing theory. 1 Introduction Unitary space-time codes have been recently proposed in [6, 5] for dierential space-time modulation schemes and in [1, 2, 3, 4] for possibly other space-time modulation schemes. Unitary space-time codes in dierential encoding are useful not only when the channel information is not known at the receiver and non-coherent decoding is used but also when the channel information is known at the receiver and coherent decoding as a recursive trellis coding is used ointly with an error correction coding as a turbo type coding [19] where a super performance is achieved. There have been several unitary space-time code constructions in the literature: for example, group and optimal group constructions [6, 7, 5, 9]; orthogonal designs [8]; parametric codes [11]; Cayley transforms [10]; Lie groups [13, 16]; and Hamiltonian constellations or spherical codes using packing theory [9, 16]. It is known that the performance of a space-time code depends on its diversity product and having a good diversity product has become an important criterion in the design of a space-time code. In [11], some upper bounds on the diversity products of (unitary) space-time codes for a given size are presented. It is easy to reach the diversity product upper bound for 22 matrices of sizes below 4 and 22 unitary matrices of sizes 4 and 5 reaching the upper bound are also presented in [11] using the parametric forms of unitary matrices. In fact, 2 2 unitary matrices of sizes below 6 reaching the upper bound can be also constructed by The authors are with the Department of Electrical and Computer Engineering, University of Delaware, Newark, DE fhwang, gwang, xxiag@ece.udel.edu. This work was supported in part by the Air Force Oce of Scientic Research (AFOSR) under Grant No. F and the National Science Foundation under Grants CCR and CCR The material in this correspondence was presented in part at the IEEE International Symposium on Information Theory, Yokohama, Japan, June 29-July 4,

2 using the Hamiltonian constellations from the packing theory, i.e., the optimal sphere packing points. However, in [11] it is shown that the upper bound is not reachable when the 2 2 unitary code size is above 5 and a tight upper bound on the diversity products remains open. The optimal or best known sphere packing points of sizes above 5 do not provide optimal 2 2 unitary space-time codes with optimal diversity products. In this correspondence, we propose some 2 2 unitary space-time codes by partially using the optimal sphere packing points [20, 22]. We obtain a determinant relationship for dierence matrices between Hamiltonian and general 2 2 unitary constellations. We present some best-known designs for size L = 6; 32; 48; 64, and also show that the code with size 6 reaches the optimal diversity product. This paper is organized as follows. In Section 2, we present new best-known diversity product designs for size L = 6; 32; 48; 64. In Section 3, we show the optimality of the new code of size 6 presented in Section 2. Since the proof is heavily technical, we leave the most technical parts of the proof in Appendix. 2 Some 2 2 Unitary Codes with Best Known Diversity Products In this section, we present some new 2 2 unitary codes for sizes L = 6; 32; 48; 64 with best known diversity products. 2.1 Diversity Product Let G = fv 1 ; V 2 ; ; V L g be a 2 2 unitary space-time code of size L with V H l V l = I where H stands for the transpose and complex conugate. Dene and d L = max G (G) = min det(v l? V l 0): (1) V l ;V l 0 2G;l6=l 0 (G) = max G min det(v l? V l 0): (2) V l ;V l 0 2G;l6=l 0 Following the convention in the literature, the diversity product for a 2 2 code G is dened as follows: and the optimal diversity product for L-point constellation is dened as (G) = 1 2p (G); (3) (L) = max (G) = 1 G 2p dl : (4) We are interested in designing a code G with large or optimal diversity product. 2

3 Unitary Matrices The content of this subsection can be found in many literature, for example, [21, 16]. For the notational convenience for our later study, we briey introduce some concepts on 2 2 unitary matrices below. Let U(2) be the set of all 2 2 unitary matrices, i.e., U(2) = fa A is a 2 2 matrix with A H A = Ig: Between U(2) and the unit ball S 3 R 4, there exists a close relationship as follows. For any 2 2 matrix A with A H A = I, we have det(a) = 1 and thus there is a unique angle 2 [0; 2) such that det(a) = e. For any xed angle 2 [0; 2), let SU(2; ) = fa 2 U(2) det(a) = e g: (5) Thus, we have [ U(2) = SU(2; ): 2[0;2) We are particularly interested in the case of = 0 and denote the set SU(2; 0) by SU(2) for short, i.e., SU(2) = SU(2; 0). We now investigate the structure of SU(2). From some theory of unitary matrices (for example, see [21]), SU(2) can be isometrically embedded onto the 4-dimensional Euclidean real unit sphere. Let S 3 be the unit sphere of 4-dimensional real Euclidean space R 4, i.e., S 3 = fx 2 R 4 kxk = 1g; where k k denotes the conventional l 2 norm. Because det(a) = 1 and the unitariness for any element A in SU(2), it is not hard to see that there are two complex numbers, a = a 1 + a 2 and b = b 1 + b 2, such that a b A =?b a ; (6) where denotes the conugate, and a 1 ; a 2 ; b 1 ; b 2 are real numbers governed by the condition a a2 2 + b b2 2 = 1, i.e., a2 + b 2 = 1, [21]. From this expression, the following embedding from SU(2) onto S 3 can be obtained, also see for example [16]. Let i be the mapping i : A 7! i(a) from SU(2) into S 3 dened by i(a) = (a 1 ; a 2 ; b 1 ; b 2 ) = (Re(a); Im(a); Re(b); Im(b)); (7) where a 1 ; a 2 ; b 1 ; b 2 are the real numbers dened in (6) and Re and Im stand for the real and imaginary parts of a complex number, respectively. Clearly, the mapping i is one-to-one and onto. Furthermore, the following relationship holds: det(a? B) = ki(a)? i(b)k 2 : (8) 3

4 This equation also implies that all determinants of dierence matrices of two distinct 2 2 unitary matrices in SU(2) are positive. From (8), one can see that the problem to nd an optimal 2 2 space-time code in SU(2), i.e., it is restricted to have determinant 1, becomes to nd optimal packing points on the sphere S 3, which is called Hamiltonian constellations in [9]. Thus, as indicated in [9], if we denote D L as the maximal minimum-distance of L-point packing on S 3, then d L DL; 2 i.e., the square of the maximal minimum-distance of L-point packing on S 3 is a lower bound for d L. However, as we shall see later, the above Hamiltonian constellation may not be enough to have good codes and we need to consider the entire 2 2 unitary matrix space U(2). To do so, we need a determinant formula. 2.3 A Useful Determinant Formula Let us consider a relationship between SU(2) and U(2) or equivalently between SU(2) and SU(2; ) for any 2 [0; 2). For a xed, we dene a mapping J from SU(2; ) to SU(2) as follows: J (A) = e?=2 A; for A 2 SU(2; ): (9) Since det(j (A)) = e? det(a) = e? e = 1, this mapping is well-dened. Furthermore, it is not hard to see that it is one-to-one and onto. With this notation, one can see that any 2 2 unitary matrix A can be represented by A = e =2 J (A); for some 2 [0; 2): An important property from this mapping is that it also provides a determinant formula for a dierence matrix of two matrices selected from dierent sets SU(2; 1 ) and SU(2; 2 ), which is stated in the following proposition. Proposition 1 For any A 0 2 SU(2) and A 2 SU(2; ), we have Proof: Assume det(a? A 0 ) = det(a 0? J (A))? 4 sin 2 (=4): A 0 = a0 b 0?b 0 a 0 where a b 0 2 = 1 and a 2 + b 2 = 1. Then ; and J (A) = det(a 0? J (A)) = det( a0 b 0?b 0 a 0 a b?b a ;? a b?b a ) = 2? (a 0 a + b 0 b + a 0 a + b 0 b); 4

5 By the denition of J in (5), we have A = e =2 J (A). Therefore, det(a 0? A) = det a0 b 0 e?b 0 a? =2 a e =2 b 0?e =2 b e =2 a = 1 + e? e =2 (a 0 a + b 0 b + a 0 a + b 0 b) = 1 + e? e =2 (2? det(a 0? J (A))) = e?=2 + e =2? (2? det(a 0? J (A))); which is the same as the one in the proposition. From this proposition, we immediately have the following corollary q.e.d. Corollary 1 For any A 1 2 SU(2; 1 ) and A 2 2 SU(2; 2 ), we have det(a 1? A 2 ) = det(j 1 (A 1 )? J 2 (A 2 ))? 4 sin 2 (( 1? 2 )=4): From the above proposition and corollary, one can see that the determinant absolute value of the dierence matrix of two 2 2 unitary matrices depends on the distance between their embeddings and their angle dierence. This motivates us to design a 2 2 unitary space-time code using two steps: one is to select good packing points on the sphere S 3 and the other is to select good angles. 2.4 Some New Codes with Best-Known Diversity Products With the help of the above determinant formulas, we can construct some 2 2 unitary codes with best-known diversity products Size L = 6 Let d =?5=2 + p 22. Select a 4-point packing on S 3 as follows: a 1 = (?a;?b; b;?b); a 2 = (?a; b; b; b); a 3 = (?a;?b;?b; b); a 4 = (?a; b;?b;?b); where a = p 1? 3d=8 and b = p (1? a 2 )=3. By mapping these points back to SU(2), we have the following four unitary matrices:?a? b b? b A 1 =?b? b?a + b?a? b?b + b A 3 = b + b?a + b ; A 2 = ; A 4 = For other two unitary matrices, we use angle. Let?a + b b + b?b + b?a? b?a + b?b? b b? b?a? b 1 = 2 arccos(d=2? a); and 2 = 2? 1 ; ; : and A 5 = e 1=2 I 2 SU(2; 1 ); A 6 =?e 2=2 I 2 SU(2; 2 ): 5

6 q It is easy to check that the diversity product of the code fa 1 ; A 2 ; ; A 6 g is 1?5=2 + p 22 0: q In next section, we shall prove that (6) = 1?5=2 + p 22, i.e., this code reaches the optimal diversity 2 product of any 2 2 unitary space-time codes of size L = Sizes L = 32; 48; 64 To construct 32, 48 or 64 unitary matrices with large diversity products, at rst, we rst construct four diamonds in S 3 as follows. Let t is a parameter, and a = r 1? 3 8 t2 ; r = p 1? a 2 ; b =? The four point coordinates of the rst diamond are p p t; r 1 = 3 t; = 2 3 : a 1 = (a; r; 0; 0); a 2 = (a; b; r 1 ; 0); a 3 = (a; b; r 1 cos(); r 1 sin()); a 4 = (a; b; r 1 cos(2); r 1 sin(2)): The ones of the second diamond are a 5 = (a;?r; 0; 0); a 6 = (a;?b;?r 1 ; 0); a 7 = (a;?b;?r 1 cos();?r 1 sin()); a 8 = (a;?b;?r 1 cos(2);?r 1 sin(2)): The ones of the third diamond are a 9 = (?a; r; 0; 0); a 10 = (?a; b; r 1 ; 0); a 11 = (?a; b; r 1 cos(); r 1 sin()); a 12 = (?a; b; r 1 cos(2); r 1 sin(2)): The ones of the forth diamond are a 13 = (?a;?r; 0; 0); a 14 = (?a;?b;?r 1 ; 0); a 15 = (?a;?b;?r 1 cos();?r 1 sin()); a 16 = (?a;?b;?r 1 cos(2);?r 1 sin(2)): Mapping these points back to SU(2) using the map i?1 given in (7), we obtain 16 matrices, denoted by Q, i.e., Q = i?1 (a ) for = 1; 2; ; 16. These matrices can be used to generate best-known diversity product unitary codes with L = 32, 48 and 64 as follows. For L = 32, let t = p 2, = arccos(3=4) and dene U i = Q i ; i = 1; 2; 3; 4; U i = Q 8+i ; i = 5; 6; 7; 8 U i = e (=4+=2) Q i?8 ; i = 9; 10; 11; 12; U i = e (=4+=2) Q i ; i = 13; 14; 15; 16; U i = e (=2) Q i?12 ; i = 17; 18; 19; 20; U i = e (=2) Q i?12 ; i = 21; 22; 23; 24; U i = e (3=4+=2) Q i?20 ; i = 25; 26; 27; 28; U i = e (3=4+=2) Q i?20 ; i = 29; 30; 31; 32: and put G 32 = fu 1 ; ; U 32 g, then the minimum determinant (G 32 ) = p 7?1, and the diversity product 2 p7?1 (G 32 ) is 2q 1 = 0:4536, which is best-known for size L = For L = 48, let t = p 2 and V i = Q i ; i = 1; 2; 3; 4; V i = Q 8+i ; i = 5; 6; 7; 8 V i = e (=6) Q i?4 ; i = 9; 10; 11; 12; V i = e (=6) Q i?4 ; i = 13; 14; 15; 16; V i = e (=3) Q i?16 ; i = 17; 18; 19; 20; V i = e (=6) Q i?8 ; i = 21; 22; 23; 24; V i = e (=2) Q i?20 ; i = 25; 26; 27; 28; V i = e (=2) Q i?20 ; i = 29; 30; 31; 32; V i = e (2=3) Q i?32 ; i = 33; 34; 35; 36; V i = e (2=3) Q i?24 ; i = 37; 38; 39; 40; V i = e (5=6) Q i?36 ; i = 41; 42; 43; 44; V i = e (5=6) Q i?36 ; i = 45; 46; 47; 48: 6

7 and dene G 48 = fv 1 ; ; V 48 g, then the minimum determinant (G 48 ) = p 3? 1, and the diversity product (G 48 ) is 1 2p p3? 1 = 0:4278, which is best-known for size L = 48. For L = 64, let t = p 1:3880, and dene W i = Q i ; i = 1; 2; 3; 4; W i = Q 8+i ; i = 5; 6; 7; 8 W i = e (=8) Q i?4 ; i = 9; 10; 11; 12; W i = e (=8) Q i?4 ; i = 13; 14; 15; 16; W i = e (=4) Q i?16 ; i = 17; 18; 19; 20; W i = e (=4) Q i?8 ; i = 21; 22; 23; 24; W i = e (3=8) Q i?20 ; i = 25; 26; 27; 28; W i = e (3=8) Q i?20 ; i = 29; 30; 31; 32; W i = e (=2) Q i?32 ; i = 33; 34; 35; 36; W i = e (=2) Q i?24 ; i = 37; 38; 39; 40; W i = e (5=8) Q i?36 ; i = 41; 42; 43; 44; W i = e (5=8) Q i?36 ; i = 45; 46; 47; 48; W i = e (3=4) Q i?48 ; i = 49; 50; 51; 52; W i = e (3=4) Q i?40 ; i = 53; 54; 55; 56; W i = e (7=8) Q i?52 ; i = 57; 58; 59; 60; W i = e (7=8) Q i?52 ; i = 61; 62; 63; 64: and dene G 64 = fw 1 ; ; W 64 g, then the minimum determinant (G 64 ) = 0:5406, and the diversity product (G 64 ) is 1 2p 0:5406 = 0:3676, which is best-known for size L = 64. The following table summarizes the above results and compares with some existing codes, where diversity sum means the minimum Euclidean distance between codeword matrices [11]. From Table 1, one can see that the optimal diversity sum, 0:7746, of the 2 by 2 unitary code of size 6 presented in [11] is slightly better than the one, 0:7400, of the 2 by 2 unitary code of size 6 with optimal diversity product presented in this paper. Fig. 1 shows the symbol error rates (SER) of these two codes of size 6 over a quasi static fading channel and one can see that the one with the optimal diversity product performs slightly better than the one with the optimal diversity sum at high SNR, which also conrms the argument between diversity product and diversity sum in [11]. Table 1: Diversity product and sum comparisons Hamiltonian Codes [9] Parametric Codes [11] New Codes Size Diversity product Diversity sum Diversity product Diversity sum Diversity product Diversity sum 6 0:7071 0:7071 0:7071 0:7746 (opt.) 0:7400 (opt.) 0: :4496 0:4496 0:4461 0:5621 0:4536 0: :3938 0:3938 0:3875 0:4278 0: :3609 0:3609 0:3535 0:4852 0:3676 0: Optimality of 2 2 Unitary Space-time Codes of Size L = 6. The main goal of this section is to prove the optimality of the code of size 6 presented in Section q Theorem 1 The maximal diversity product of a 22 unitary space-time code of size 6 is 1?5=2 + p 22, 2 i.e., q (6) = 1?5=2 + p 22: 2 7

8 10 0 Two transmit and one receive antennas, 2 by 2 unitary code of size 6 Optimal diversity product code of size 6 in this paper Optimal diversity sum code of size 6 in [11] 10 1 Symbol error rate SNR(dB) at the receiver Figure 1: Symbol error rate comparison. This theorem implies that the code presented in Section has already reached the maximal diversity product. To prove this theorem, we need some preparations. First, we introduce the concept of dual. For any unitary matrix A = e =2 J (A), its dual is dened as e (2?)=2 (?J (A)) and denoted by ~ A. If = 0, then ~ A = e 2=2 (?J (A)) = A, i.e., the dual of A 2 SU(2) is itself. With the denition of a dual matrix we have the following corollary. Lemma 1 For any unitary matrices A 1 and A 2 with their duals ~ A1 and ~ A2, respectively, we have (i) det(a 1? B) = det( A1 ~? B); for any unitary matrix B 2 SU(2); (ii) det(a 1? A 2 ) = det( A1 ~? A2 ~ ): This lemma is a direct result of Proposition 1, we omit its proof. In what follows, for the notational convenience, we use ^A to denote J (A) for a matrix A 2 SU(2; ) by dropping the subscript without causing any confusion. Since there exists an embedding i from SU(2) onto S 3, we do not distinguish a matrix in SU(2) and a vector on S 3 and use the same notation A to express a matrix in SU(2) and a point on S 3. If A is treated as a point on S 3, it means its embedding, i.e., i(a) in (7). Lemma 2 Let A i 2 SU(2; i ), i = 1; 2; ; L, and fa 1 ; A 2 ; ; A L g be an optimal unitary space-time code of size L with the maximal diversity product d L 2. Then, det(a i? A ) = det( ^Ai? ^A )? 4 sin 2 (( i? )=4) d L ; if i? ; det(a i? A ) = 4 sin 2 (( i? )=4)? det( ^Ai? ^A ) d L ; if i? ; 8

9 where ^Al = J (A l ) is the proection of A l from SU(2; l ) to SU(2) as dened in Section 2.2. Its proof is in Appendix. Lemma 2 basically provides an expression of the absolute value of a dierence matrix determinant from the one of their proections to SU(2) and their angles for an optimal constellation. Lemma 3 Let fa 1 ; ; A L g be an unitary space-time code with the optimal diversity product d L > 2 and A 2 SU(2; ), = 1; 2; ; L 6. If 0 = 1 L < 2, then i+1? i < for i = 1; 2; ; L? 1, i.e., the dierence of two adacent angles is less than. Its proof is in Appendix. Lemma 4 Let fp 1 ; ; P L g be L points on the sphere S 3. Assume that kp i? P k 2 d for a constant d 2 and 1 i < L. Let P 0 be any a point on this sphere. Then, if L = 4, there exist s and t, 1 s; t L, such that kp 0? P s k 2 2? 2 p 1? 3d=8 and kp 0? P t k p 1? 3d=8; if L = 3, there exist s and t, 1 s; t L, such that kp 0? P s k 2 2? 2 p 1? d=3 and kp 0? P t k p 1? d=3; if L = 2, there exist s and t, 1 s; t L, such that kp 0? P s k 2 2? 2 p 1? d=4; and kp 0? P t k p 1? d=4: Its proof is in Appendix. Lemma 5 For any L points fp 1 ; ; P L g on the unit sphere S n in the n+1-dimensional real Euclidean space R n+1, we have X 1i<L Its proof can be found in, for example, [11]. kp i? P k 2 L 2 : Lemma 6 Let A i = e i=2 ^Ai 2 SU(2; i ), i = 1; 2; 3, be three unitary matrices. Assume If 3? 1, and for i 6=, det(a i? A ) d 6 p 22? 5=2, then, 3? 1 5=6: Its proof is in Appendix. 9

10 Lemma 7 Let 2 < d 2:5 and?1 < a; b 1? d=2. If arccos(d=2 + a) + arccos(d=2 + b) =2, then, 2 sin(arccos(a + d=2)=2)? b? cos(arccos(d=2 + b) + arccos(d=2 + a))? d=2 cos(arccos(?a)=2) d; (10) where 0 arccos(x). Its proof is in Appendix. Proposition 2 Let fa 1 ; A 2 ; ; A 6 g be an optimal constellation with A = e =2 ^A of the maximal diversity product d 6. Assume that 0 = and 6? 5, 6? 4. Then, d 6?5=2 + p 22. Its proof is in Appendix. Proposition 3 Let fa 1 ; A 2 ; ; A 6 g be an optimal constellation with A = e =2 ^A. Assume that 0 = and 6? 4, 6? 3. Then, d 6 <?5=2 + p 22. Its proof is in Appendix. Proposition 4 Let fa 1 ; A 2 ; ; A 6 g be an optimal constellation with A = e =2 ^A. Assume that 0 = and 6? 4, 6? 3. Then, d 6 <?5=2 + p 22. Its proof is in Appendix. Proposition 5 Let fa 1 ; A 2 ; ; A 6 g be an optimal constellation with A = e =2 ^A. Assume that 0 = and 6? 3, 6? 2. Then, d 6 <?5=2 + p 22. Its proof is in Appendix. Now we begin to prove Theorem 1. Proof of Theorem 1: Assume signal constellation G = fa 1 ; A 2 ; ; A 6 g is an optimal constellation with the maximal diversity product d 6 and A i 2 SU(2; i ), i = 1; 2; ; 6. By the construction in Section 2.4 (1), we have d 6?5=2 + p 22. We next need to show that d 6?5=2 + p 22. To do so, let us consider the dierent cases of the number of the zero angles of A i : p = ]fi i = 0g: Without loss of generality, we can assume that 1 p 6. In this proof and the proofs in Appendix, we always use 0 arccos(x). (i) p = 6. p = 6 means that all A i 2 SU(2), i.e., all six matrices A i are on the sphere S 3. In other words, there exist 6-point packing such that the minimal distance is greater than p 2, which contradicts with the packing result on S 3 (according to the result [20], the packing angle on S 3 is =2, that is, the maximal minimum distance is p 2). (ii) p = 5. 10

11 Assume 1 = = 5 = 0 and 6 > 0. Thus, A i = ^Ai ; i = 1; 2; ; 5. By Lemma 3, we have 6? 5, i.e., 6 : By Lemma 2, det( ^Ai? ^A6 ) d sin 2 ( 6 =4) > 2; i = 1; 2; ; 5: For 1 i 6= 5, from the condition, det( ^Ai? ^A ) > 2: Therefore, there exists six points f ^A1 ; ; ^A6 g on the sphere S 3 such that the minimum distance is greater than p 2, which contradicts with the packing result as in (i). (iii) p = 4. Assume 1 = = 4 = 0 and 0 < 5 6 < 2: By Lemma 3, we have 5 and 6? 5. If 6, then, as shown as (ii), f ^A1 ; ; ^A4 ; ^A5 ; ^A6 g consists of a 6-point packing on S 3 with minimum distance greater than p 2, which results in a contradiction. Therefore, we can assume that 5 6 and 6? 5. We next investigate the packing position of f ^A1 ; ; ^A4 ; ^A5 ;? ^A6 g on S 3. Denote ^Ai = (a i ; b i ; c i ; e i ); i = 1; ; 6. By a unitary transformation, we can assume ^A5 = I, i.e., a 5 = 1; b 5 = c 5 = e 5 = 0. We then convert this problem to a packing problem on the 3-dimensional unit sphere S 2 as follows. If a i 6= 1;?1, dene b i = 1 (b i ; c i ; e i ); i = 1; 2; 3; 4; 6; (11) r i q where r i = 1? a 2. Then i b i 2 S 2 and clearly, det( ^Ai? ^A ) = 2(1? a i a ) + r i r (kb i? b k 2? 2); (12) kb i? b k 2 = 2? 2(1? a ia )? det( ^Ai? ^A ) r i r : (13) Two remarks about this conversion are as follows. The mapping S 3 3 (a; b; c; e)! b = (b=r; c=r; e=r) 2 S 2 is not one-to-one. It is because, for dierent two points (a; b; c; e) and (?a; b; c; e), the images are the same. However, when we restrict a 0 or a 0, the mapping becomes one-to-one and onto. Another remark is that an image point b does not depend on a, when we restrict a to a 0 or a 0. To explain this, we use the polar coordination. For any point (a; b; c; e) 2 S 3, there exist three angles 1 ; 2 ; 3, such that a = sin( 1 ) and b = cos( 1 ) sin( 2 ); c = cos( 1 ) cos( 2 ) sin( 3 ); e = cos( 1 ) cos( 2 ) cos( 3 ). Hence b = (b=r; c=r; e=r) = (sin( 2 ); cos( 2 ) sin( 3 ); cos( 2 ) cos( 3 )), which is independent of 1, i.e., a. Therefore, when we restrict a to a 0 or a 0, the distance kb i? b k is independent of a i and a. For 1 i 4 and i = 6, because 5 and 6? 5, by Lemma 2, we have 2 < d 6 det(a 5? A i ) = det( ^A5? ^Ai )? 4 sin 2 ( 5 =4) = 2? 2a i? 4 sin 2 ( 5 =4): Therefore, a i < 0 for i = 1; 2; 3; 4; 6. Since 1 = 2 = 3 = 4 = 0, without loss of generality, we may assume that?1 a 1 a 2 a 3 a 4 < 0. If a 1 =?1; then b 1 = c 1 = e 1 = 0 and it is not hard to see that det( ^A4? ^A1 ) = 2 + 2a 4. It 11

12 implies d a 4, i.e., a 4 > 0, which contradicts with the result a 4 < 0 we derived before. Therefore, a 1 >?1: For 1 i 6= 4, from (12) and the fact that i = = 0, we have d 6 det(a i? A ) = det( ^Ai? ^A )? 4 sin 2 (( i? )=4) = 2? 2a i a + r i r (kb i? b k 2? 2): Because a i a > 0 and r i r > 0, we have kb i? b k 2? 2 > 0. Furthermore, it is easy to see that for a xed a i, the right hand side of the above inequality is increasing for a. Therefore, d 6 2? 2a i a + r i r (kb i? b k 2? 2) 2? 2a r 2 4(kb i? b k 2? 2); which implies that a 4? p 1? d 6 =kb i? b k 2. Because fb 1 ; ; b 4 g are on S 2, by the packing theory on S 2, there is at least one pair fb i ; b g such that kb i? b k 2 8=3. Hence, Since a 5 = 1;?1 a 6 < 0, and 0 6? 5, from Lemma 2 we have Therefore, a 4? p 1? 3d 6 =8: (14) d 6 det(a 6? A 5 ) = det( ^A6? ^A5 )? 4 sin 2 (( 6? 5 )=4) = 2? 2a 6? 4 sin 2 (( 6? 5 )=4) 2? 2(?1)? 4 sin 2 (( 6? 5 )=4): cos(( 6? 5 )=2) d 6 =2? 1: (15) Using the fact that d 6 det(a 5?A 4 ) = det( ^A5? ^A4 )?4 sin 2 ( 5 =4), and noting that det( ^A5? ^A4 ) = 2? 2a 4, we have d 6 2? 2a 4? 4 sin 2 ( 5 =4) p 1? 3d 6 =8? 4 sin 2 ( 5 =4) = 2 cos( 5 =2) + 2 p 1? 3d 6 =8; (16) where the second inequality is from (14). Inequality (16) implies cos( 5 =2) d 6 =2? p 1? 3d 6 =8: (17) We now replace A 5 ; A 6 by their duals ~ A5 ; ~ A6. From the denition, we have ~A 6 = e (2? 6)=2 (? ^A6 ); ~ A5 = e (2? 5)=2 (? ^A5 ): Furthermore, fa 1 ; ; A 4 ; ~ A6 ; ~ A5 g is also an optimal signal constellation by Lemma 1. normalization by multiplying? ^AH 6 from left to the constellation to get a new constellation: G 1 = f? ^AH 6 A 1 ;? ^AH 6 A 2 ;? ^AH 6 A 3 ;? ^AH 6 A 4 ;? ^AH 6 ~ A 6 ;? ^AH 6 ~ A 5 g = f? ^AH 6 A 1;? ^AH 6 A 2;? ^AH 6 A 3;? ^AH 6 A 4; e (2? 6)=2 I;? ^AH 6 ~ A 5 g: 12 We make a

13 Since? ^AH 6 is a unitary matrix, G 1 is also an optimal constellation. Furthermore, G 1 and fa 1 ; ; A 6 g have the same angle relationships. Therefore, inequality (17) corresponding to this new constellation G 1 also holds: cos((2? 6 )=2) d 6 =2? p 1? 3d 6 =8: (18) From (17) and (18), we have 5 2 arccos(d 6 =2? p 1? 3d 6 =8); 6 2? 2 arccos(d 6 =2? p 1? 3d 6 =8): Hence, 6? 5 2? 4 arccos(d 6 =2? p 1? 3d 6 =8): From (15), we know that 6? 5 2 arccos(d 6 =2? 1). Therefore, 2? 4 arccos(d 6 =2? p 1? 3d 6 =8) 2 arccos(d 6 =2? 1): Hence, p d 6?4(d 6 =2? 1? 3d 6 =8) 2 + 4; which implies the desired result d 6?5=2 + p 22: (iv) p 3. Assume 0 = Using the same argument as in the beginning of Case (iii) when p = 4 and Lemma 3, we can also assume that 6 and 2, 6? 5. We divide the proof into several cases according to the relationships among the angles. Case I 6 and 5. We divide this case into 4 subcases. Case I.1 6, 5 and 6? 4. This subcase is Proposition 2. Case I.2 6, 5 and 6? 4, 6? 3. This subcase is Proposition 3. Case I.3 6, 5 and 6? 3, 6? 2. By taking the rotation of angle? 5 to G, we obtain a new constellation G 0 = fa 0 1; A 0 2; ; A 0 6g; where A 0 = e? 5=2 A. For = 1; 2; 3; 4; A 0 = e? 5=2 e =2 ^A = e (2?( 5? ))=2 (? ^A5 ): For = 5, we have A 0 5 = e? 5=2 A 5 = ^A5. For = 6, we have A 0 6 = e? 5=2 e 6=2 ^A6 = e ( 6? 5 )=2 ^A6 : 13

14 Therefore, the relationship between G 0 and G is and f ^A0 1; ^A0 2; ^A0 3; ^A0 4; ^A0 5; ^A0 6g = f? ^A1 ;? ^A2 ;? ^A3 ;? ^A4 ; ^A5 ; ^A6 g; f 0 1; 0 2; 0 3; 0 4; 0 5; 0 6g = f2? 5 ; 2? ( 5? 2 ); 2? ( 5? 3 ); 2? ( 5? 4 ); 0; 6? 5 g: Clearly, the diversity product of G 0 is still d 6. We now consider the dual of G 0, denoted by ~ G 0, which has the same diversity product as G 0 Corollary 1: G0 ~ = f A ~ 0 1 ; A ~ 0 2 ; ; A ~ 0 6 g, where A ~ 0 = e00=2 ^~ A 0, 1 6. By the denition of a dual, the relationship between G ~ 0 and G 0 or G is and the corresponding angles are f ^~ A 0 1 ; ^~ A 0 2 ; ^~ A 0 3 ; ^~ A 0 4 ; ^~ A 0 5 ; ^~ A 0 6 g = f? ^A0 1;? ^A0 2;? ^A0 3;? ^A0 4; ^A0 5;? ^A0 6g = f ^A1 ; ^A2 ; ^A3 ; ^A4 ; ^A5 ;? ^A6 g; f 00 1; 00 2; 00 3; 00 4; 00 5; 00 6g = f2? 0 1; 2? 0 2; 2? 0 3; 2? 0 4; 0; 2? 0 6g = f 5 ; 5? 2 ; 5? 3 ; 5? 4 ; 0; 2? ( 6? 5 )g: It is easy to see that 6 00 and 00 for = 1; 2; 3; 4; 5. Furthermore, 00 6? 00 1 ; 00 6? 00 2 ; 00 6? 00 ; = 3; 4; 5: by Thus, if we rearrange ~ G0 into G 00 = f ~ A 0 5 ; ~ A 0 4 ; ~ A 0 3 ; ~ A 0 2 ; ~ A 0 1 ; ~ A 0 6 g; then the conditions on G 00 are exactly the same as the ones in Case I.2. By Proposition 3, we have proved this theorem in this subcase. Case I.4 6, 5 and 6? 2 Make a rotation angle? 2 to the constellation G as done in Case I.3 and we nd that the new constellation has the same conditions as in Case I.1. Therefore, by Proposition 2, we have proved this theorem in this subcase. Case II 6 ; 5 and 4. We divide this proof into 4 subcases. Case II.1 6 ; 5, 4 and 6? 4. In this case, we make a rotation to the constellation as follows. Let A 0 = e? 6=2 A for = 1; 2; ; 6. Then fa 0 1 ; A0 2 ; ; A0 6 g is also an optimal constellation. Since, for = 1; 2; 3; 4; 5, A0 = e? 6=2 A = e (2?( 6? ))=2 (? ^A ), we obtain ^A0 =? ^A and the angle 0 of A0 is 2? ( 6? ). For = 6, 6 0 = 0, i.e., A 0 6 belongs to SU(2) and A0 6 = ^A6. Furthermore, we have that 6 0 ; 0 1 ; 0 2 ; 0 3 ; 0 4 are all less than or 14

15 equal to, and 5 0 is greater than or equal to. Therefore, fa0 1 ; A0 2 ; ; A0 6g satises the conditions in Case I. Thus we have proved this theorem in this subcase. Case II.2 6 ; 5, 4 and 6? 4, 6? 3. It is proved in Proposition 4. Case II.3 6 ; 5, 4 and 6? 3, 6? 2. It is proved in Proposition 5. Case II.4 6 ; 5, 4 and 6? 2. Let A 0?1 = e? 2=2 A for = 2; 3; 4; 5; 6, and A 0 6 = e(2? 2)=2 A 1. Note that ^A1 = A 1 since 1 = 0. Then, fa 0 1 ; A0 2 ; A0 3 ; A0 4 ; A0 5 ; A0 6g satises the conditions of Case I. Case III 6 ; 5 ; 4 and 3 : Under this assumption, we consider the dual constellation: 1 = 0 is xed, and 2 ; 3 are changed to 2? 2 ; 2? 3, which belong to [; 2], and 4 ; 5 : 6 are transferred to 2? 4 ; 2? 5 ; 2? 6, which belong to [0; ]. Therefore, through this duality, we change this subcase into Case II. Case IV 6 ; 5 ; 4 ; 3 and 2 : Also we consider its dual constellation and nd that this case can be converted to Case I. By summarizing all the above cases, this theorem is proved. 4 Conclusion q.e.d. In this correspondence, we have partially used sphere packing theory to construct 2 2 unitary spacetime codes. Although the optimal ones of sizes L below 6 can be constructed from the sphere packings on S 3, i.e., Hamiltonian constellations [9, 16] that reach the upper bound 1 2p 2L=(L? 1) of the maximal diversity products derived in [11]. This upper bound can not be reached when the sizes are above 5 as shown in [11]. The critical boundary on the sizes is size L = 6. In this correspondence, we have constructed 2 2 unitary space-time code of size 6 that has been shown q in this correspondence to have the optimal diversity product. The optimal diversity product d 6 = 1?5=2 + p 22 0:74 < 0: p 2L=(L? 1) when L = 6. Some constructions of 2 2 unitary space-time codes of sizes 32; 48; 64 of non-hamiltonian constellations with best known diversity products have been also presented by partially using sphere packing theory. To obtain these results, we have presented a determinant identity between the dierence matrices of two matrices in a Hamiltonian constellation and two matrices in non-hamiltonian constellations. Appendix In this Appendix, we always assume that 0 arccos(x). 15

16 Proof of Lemma 2 Condition d L 2 implies that for any i;, det(a i? A ) 2. Therefore, from Corrollary 1, we have det( ^Ai? ^A )? 4 sin 2 (( i? )=4) 2: If i?, then 0 4 sin 2 (( i? )=4) 2: Therefore, by noting that det( ^Ai? ^A ) 0 from (4) and Corollary 1, we obtain det(a i? A ) = det( ^Ai? ^A )? 4 sin 2 (( i? )=4): The second inequality can be similarly proved. q.e.d. Proof of Lemma 3 Assume that there is an index u such that u+1? u, we want to derive a contradiction. Since u+1? u and 0 u u+1 < 2, we have u u+1. Let us consider a new constellation f ^A1 ; ; ^Au ;? ^Au+1 ; ;? ^AL g SU(2). We want to show that this new constellation on S 3 has the minimum Euclidean distance p d L. For i < u, 0 i u, hence? i. By Lemma 2, det( ^Ai? ^A ) = det(a i? A ) + 4 sin 2 (( i? )=4) d L > 2: For i > u + 1, since u+1, we have i?. Therefore, by Lemma 2, we have det(? ^Ai? (? ^A )) = det( ^Ai? ^A ) = det(a i? A ) + 4 sin 2 (( i? )=4) d L > 2: For i u < u + 1, since? i u+1? u, by Lemma 2 we have det( ^A? ^Ai ) = 4 sin 2 ((? i )=4)? det(a? A i ): Note that det(? ^A? ^Ai ) = 4? det( ^A? ^Ai ). Thus, det(? ^A? ^Ai ) = 4? 4 sin 2 ((? i )=4) + det(a? A i ) det(a? A i ) d L : Therefore, using (4) we have shown that the minimum Euclidean distance of the points f ^A1 ; ; ^Au ;? ^Au+1 ;? ^AL g on S 3 is greater than p d L > p 2. This contradicts with the fact that, when L 6, the maximal minimum distance of L-point packing on the sphere S 3 is p 2 from the packing theory [20]. q.e.d. 16

17 Proof of Lemma 4 We only prove the case of L = 4 and the other cases can be similarly proved. p We rst prove that there exists t, 1 t 4, such that kp 0? P t k ? 3d=8. Since an orthogonal transformation does not change the distance between any two points, without loss of generality, we may assume P 0 = (1; 0; 0; 0). Let P i = (a i ; b i ; c i ; e i ). Then, kp? P 0 k 2 = 2? 2a. We p may assume that a i 6= 1;?1. In fact, if a i = 1, we let t = i, which is because kp t? P 0 k 2 = 0 < ? 3d=8. If a i =?1, then kp i? P 0 k 2 = 4. Let t 6= i and we have kp t? P 0 k 2 = 2? 2a t = 4? (2 + 2a t ) = 4? kp t? P i k 2 4? d p 1? 3d=8; where the third equality is from the assumption a i =?1, and the rst inequality is from the assumption kp t? P i k 2 d. We next derive a contradiction by assuming kp i? P 0 k 2 > p 1? 3d=8 for i = 1; 2; 3; 4. Since kp i? P 0 k 2 = 2? 2a i, we have On the other hand, since kp i? P k 2 d, we have a i <? p 1? 3d=8: (19) 2? 2a i a + (kb i? b k 2? 2)r i r d; where r i ; r ; b i ; b are the same as those described in (11)-(13) in the proof of Theorem 1. Since a i ; a 6= 1 or?1, we have r i ; r > 0. Therefore, we obtain kb i? b k d? 2 + 2a ia r i r ; i; = 1; 2; 3; 4; i 6= : (20) From (19), we know that a 0. It is not hard to check that, on interval (?1; 0], the right hand side of (20) is strictly decreasing for a i and a. Therefore, by using (19), we obtain kb i? b k d? 2 + 2a ia r i r d? 2 + 2(1? 3d=8) > 2 + 1? (1? 3d=8) = 8=3; i; = 1; 2; 3; 4; i 6= : (21) Clearly, (21) contradicts with the result of 4-point packing on S 2. t 2 f1; 2; 3; 4g such that kp 0? P t k p 1? 3d=8: This proves that there exists a We next prove that there exists an s 2 f1; 2; 3; 4g such that kp 0? P s k 2 2? 2 p 1? 3d=8: (22) 17

18 To do so, let us consider point?p 0. By the above result, there exists an s 2 f1; 2; 3; 4g such that k? P 0? P s k p 1? 3d=8: Since k? P 0? P s k 2 = 2 + 2a s = 4? (2? 2a s ) = 4? kp 0? P s k 2 ; we obtain 4? kp 0? P s k p 1? 3d=8; or kp 0? P s k 2 2? 2 p 1? 3d=8: q.e.d. Proof of Lemma 6 Let ^A = (a ; b ; c ; e ) for = 1; 2; 3. We want to convert these three 4-dimensional unit vectors equivalently into three 3-dimensional unit vectors by employing orthogonal transformations. We may rst assume ^A1 = I. By using a rotation R on S 3, we can assume ^A2 = (a 2 ; r 2 ; 0; 0), where R is a 3 3 orthogonal matrix, and r 2 = p 1? a 2 2. Similarly, using a rotation T on S 3, we can assume ^A3 = (a 3 ; r 3 ; c 3 ; 0), where T is a 2 2 orthogonal matrix. Thus, after normalizations, we may assume ^A1 ; ^A2 ; ^A3 of the following forms: 1 A ^A 1 = (1; 0; 0; 0); ^A2 = (a 2 ; r 2 ; 0; 0); ^A3 = (a 3 ; b 3 ; c 3 ; 0); which are equivalent to three 3-dimensional vectors on the 2-dimensional sphere S 2. Furthermore, we may assume 1 = 0. Because 3? 1 and 0 = 1 2 3, it is obvious that 2 ; 3 and 3? 2. Therefore, by Lemma 2 and the condition det(a i? A ) d 6, we have det( ^Ai? ^A ) d sin 2 (( i? )=4); 1 i < 3: (23) From (23), we have det( ^A1? ^A2 ) > 2, det( ^A1? ^A3 ) > 2, and det( ^A3? ^A2 ) > 2. This means that the points ^A2 ; ^A3 are on the dierent half sphere from the point ^A1. 18

19 Let O be the original point of coordinates and 12 be the angle \A 1 OA 2, 13 be the angle \A 1 OA 3. Then, 2? 2 cos( 12 ) = det( ^A1? ^A2 ) d sin 2 ( 2 =4); (24) 2? 2 cos( 13 ) = det( ^A1? ^A3 ) d sin 2 ( 3 =4); (25) Clearly, when ^A1 ; ^A2 ; ^A3 are on the same circle, the distance between ^A2 and ^A3 achieves the maximum. Therefore, det( ^A2? ^A3 ) 2? 2 cos(2? 12? 13 ) = 2? 2 cos( ): Applying (23) for i = 2 and = 3, we get that d sin 2 (( 3? 2 )=4) 2? 2 cos( ): That is d 6 =2 + cos( ) cos(( 3? 2 )=2): (26) From (24),(25), we have 12 arccos(cos( 2 =2)? d 6 =2) and 13 arccos(cos( 3 =2)? d 6 =2). But from (26), noticing that , we have arccos(d 6 =2? cos(( 3? 2 )=2)). Hence, + arccos(d 6 =2? cos(( 3? 2 )=2))? arccos(cos( 2 =2)? d 6 =2)? arccos(cos( 3 =2)? d 6 =2) 0: (27) We can check that the left hand side of (27), is decreasing for d 6. By condition d 6 p 22? 5=2, we then have + arccos(( p 22? 5=2)=2? cos(( 3? 2 )=2))? arccos(cos( 2 =2)? ( p 22? 5=2)=2)? arccos(cos( 3 =2)? ( p 22? 5=2)=2) 0: (28) Assume 3 > 5=6, we want to derive a contradiction. In fact, by investigating the left hand side of (28), we nd that it is decreasing for 3. Therefore, we have + arccos(( p 22? 5=2)=2? cos((5=6? 2 )=2))? arccos(cos( 2 =2)? ( p 22? 5=2)=2)? arccos(cos(5=12)? ( p 22? 5=2)=2) 0; (29) which is impossible since the maximum of the left hand side of (29) for 2 2 [0; ] is less than?0:02. Therefore, the lemma is proved. q.e.d Proof of Lemma 7 Let f(x) = 2 sin(arccos(a + d=2)=2)? x? cos(arccos(d=2 + x) + arccos(d=2 + a))? d=2 cos(arccos(?a)=2)? d: 19

20 Then, to prove the lemma, it is enough to prove that f(x) 0 for?1 < x 1? d=2. Obviously, f(x) is an innitely dierentiable function in the interval (?1; 1? d=2). Moreover, its derivative is f 0 (x) =? p 1? (d=2 + x)2? sin(arccos(d=2 + x) + arccos(d=2 + a)) p : 1? (d=2 + x)2 cos(arccos(?a)=2) Hence, equation f 0 (x) = 0 becomes sin(arccos(d=2 + x) + arccos(d=2 + a)) = sin(arccos(d=2 + x)): Since arccos(d=2 + x) + arccos(d=2 + a) =2 and arccos(d=2 + x) < =2, the unique solution x 0 of the equation f 0 (x) = 0 satises arccos(d=2 + x 0 ) + arccos(d=2 + a) =? arccos(d=2 + x 0 ): Hence, x 0 = sin(arccos(d=2 + a)=2)? d=2: The second derivative of f(x) at x 0 is f 00 (x 0 ) = 2 sin(arccos(d=2 + a)=2) cos 2 (arccos(d=2 + a)=2) cos(arccos(?a)=2) > 0: Thus, we have shown that, the equation f 0 (x) = 0 has unique solution x 0 in the interval (?1; 1? d=2) and f 00 (x 0 ) > 0. Therefore, f(x 0 ) is the minimum value of f(x) in this interval, that is, f(x) f(x 0 ) for x 2 (?1; 1? d=2). On the other hand, f(x 0 ) = (1? cos(arccos(?a)=2))(d? 2 sin(arccos(d=2 + a)=2)) cos(arccos(?a)=2) 0; where the inequality comes from the assumption d > 2 and a 1? d=2 < 0. This proves that, when b 2 (?1; 1? d=2), we have f(b) 0. If b = 1? d=2, then we have arccos(d=2 + b) = 0, which implies arccos(d=2 + b) + arccos(d=2 + a) = arccos(d=2 + a) =2 that contradicts with the condition. This shows that b 6= 1? d=2 in the lemma. Therefore, we have proved the lemma. q.e.d. Proof of Proposition 2 We denote G = fa 1 ; ; A 6 g and inherit the previous notations (a i ; b i ; c i ; e i ) for ^Ai ; i = 1; ; 6, and assume that ^A5 = I. Since for 1 4 and = 6, 5? ; from Lemma 2, 2 < d 6 det(a 5? A ) = 2? 2a? 4 sin 2 (( 5? )=4): (30) This implies a cos(( 5? )=2)? d 6 =2 1? d 6 =2 < 0 for = 1; 2; ; 6. We divide the proof of this proposition into two cases according to the number a 6 : one is a 6 =?1 and the other is a 6 >?1. Case 1 a 6 =?1: 20

21 Since a 6 =?1, we have ^A6 = (?1; 0; 0; 0). If there exists a 2 f1; 2; 3g such that a =?1, then b = c = e = 0 and det( ^A4? ^A ) = 2 + 2a 4. Thus, from Lemma 2 we have 2 < d 6 det(a 4? A ) = det( ^A4? ^A )? 4 sin 2 (( 4? )=4) = 2 cos(( 4? )=2) + 2a 4 ; which contradicts with the fact a 4 < 0. Hence, a >?1 for = 1; 2; 3: Similarly, we can prove a 4 >?1. Therefore, b 1 ; b 2 ; b 3 ; b 4 are well-dened in (11) and belong to S 2. From the result of [11], the optimal determinant for 5 unitary matrices is 12=5, so we have d 6 12=5 = 2:4. We next show that there exists at least one a for 2 f1; 2; 3; 4g such that a? p 1? 3d 6 =8. Otherwise, assume, for all 2 f1; 2; 3; 4g, a <? p 1? 3d 6 =8. Then, from the optimal constellation conditions, we have d 6 det( ^Ai? ^A )? 4 sin 2 (( i? )=4) det( ^Ai? ^A ): Using (13), we obtain kb i? b k d 6? 2 + 2a i a r i r > 2 + d 6? 2 + 2(? p 1? 3d 6 =8) 2 (1? (? p 1? 3d 6 =8) 2 ) = 8 3 : where the second inequality is from the assumption a i ; a <? p 1? 3d 6 =8 and the fact that the right hand side of the rst inequality above is decreasing for 1 < a i ; a < 0. Therefore, X 1i<4 kb i? b k 2 > = 16; which contradicts with the p result in Lemma 5. Thus, we have proven that there exists an a, 2 f1; 2; 3; 4g, such that a? 1? 3d 6 =8: Since a 6 =?1 and 6?, from Lemma 2 we have d 6 det(a 6? A ) = 4 sin 2 (( 6? )=4)? det( ^A6? ^A ) = 4 sin 2 (( 6? )=4)? (2 p + 2a )?2 cos(( 6? )=2) + 2 1? 3d 6 =8; which implies p cos(( 6? )=2) 1? 3d 6 =8? d 6 =2: (31) On the other hand, from det(a 6? A 5 ) d 6, i.e., d 6 det( ^A6? ^A5 )? 4 sin 2 (( 6? 5 )=4) = 2 + 2? 4 sin 2 (( 6? 5 )=4); 21

22 we have Similarly, from det(a 5? A ) d 6 and 5?, we have which implies cos(( 6? 5 )=2) d 6 =2? 1: (32) d 6 det(a 5? A ) = det( ^A5? ^A )? 4 sin 2 (( 5? )=4) = 2? 2a? 4 sin 2 (( 5? )=4); cos(( 5? )=2) d 6 =2 + a d 6 =2? p 1? 3d 6 =8: (33) Since (31), (32), and (33) have the same forms as the ones of (15), (17), and (18), we can use the same technique used in the proof of Case (iii) when p = 4 in the proof of Theorem 1 as follows. From (31) and (33), we have Hence, 6? 2 arccos( p 1? 3d 6 =8? d 6 =2) and 5? 2 arccos(d 6 =2? p 1? 3d 6 =8): 6? 5 2 arccos( p 1? 3d 6 =8? d 6 =2)? 2 arccos(d 6 =2? p 1? 3d 6 =8) = 2? 4 arccos(d 6 =2? p 1? 3d 6 =8): From (32), we have 6? 5 2 arccos(d 6 =2? 1). Therefore, which implies d 6?5=2 + p 22. Case 2 a 6 >?1. 2? 4 arccos(d 6 =2? p 1? 3d 6 =8) 2 arccos(d 6 =2? 1); The main idea of the following proof of this case is to construct a new constellation, G, that also has the diversity product d 6 and satises the conditions of Case 1. To do so, we rst take some rotations and duals of G to generate a constellation G 00. Using G 00 and G, we can obtain a desired G. We next divide the proof into three steps. The rst step is to diagonalize matrix ^A6 without altering ^A5 = I and other properties and to establish an equality on a 6. The second step is to construct a constellation G 00 through rotations and duals of G. The third one is to construct G. Step 1. Diagonalization of ^A6 and an equality on a 6 Since a 6 >?1, vector b 6 is a well-dened point on the sphere S 2. Then, there exists a real-valued rotation T on S 2 such that b 6 T = (1; 0; 0). Let 1 0 Q = 0 T Then, Q is an orthogonal matrix and (a 6 ; b 6 ; c 6 ; e 6 ) Q = (a 6 ; (b 6 ; c 6 ; e 6 ) T ) = (a 6 ; r 6 (b 6 =r 6 ; c 6 =r 6 ; e 6 =r 6 ) T ) : = (a 6 ; r 6 b 6 T ) = (a 6 ; r 6 ; 0; 0); 22

23 p and (1; 0; 0; 0) Q = (1; 0; 0; 0), where r 6 = 1? a 2 6. If we let (~a ; ~ b ; ~c ; ~e ) = (a ; b ; c ; e ) Q; 1 6; then, these points are on the unit sphere S 3. By using the mapping i?1 dened in Section 2, we obtain six 22 unitary matrices belonging to SU(2). Denote these matrices by ^~ A for 1 6: Then, ^~ A5 = I and a6 + r ^~A 6 = 6 0 ; 0 a 6? r 6 which is diagonal. Furthermore, since Q is an orthogonal matrix, we have det(^~ Ai? ^~ A ) = ki(^~ Ai )? i(^~ A )k 2 = ki( ^Ai )? i( ^A )k 2 = det( ^Ai? ^A ); 1 i < 6; where the rst equality is from (8), and the second equality is from the fact that Q is orthogonal, and the last equality is also from (8). Set ~ A = e =2 ^~ A for 1 6. Then, by Corollary 1, det( ~ Ai? ~ A ) = det(a i? A ); 1 i < 6: Thus, we have obtained a constellation that has diversity product d 6 but ^~ A5 = I and ^~ A6 is diagonal. Therefore, in the following proof, we assume the constellation G has the property: ^A5 = I and ^A6 has the above diagonal form. We now establish an equality on a 6. According to the relationships among the angles, the following inequality are clear by using the optimality conditions and Corollary 1: d 6 det(a 6? A ) = 4 sin 2 (( 6? )=4)? det( ^A6? ^A ); = 1; 2; 3; 4; (34) d 6 det(a 5? A ) = det( ^A5? ^A )? 4 sin 2 (( 5? )=4) = 2? 2a? 4 sin 2 (( 5? )=4) = 2 cos(( 5? )=2)? 2a ; = 1; 2; 3; 4; (35) d 6 det(a 6? A 5 ) = det( ^A6? ^A5 )? 4 sin 2 (( 6? 5 )=4) = 2? 2a 6? 4 sin 2 (( 6? 5 )=4) = 2 cos(( 6? 5 )=2)? 2a 6 : (36) Furthermore, we may assume the equality holds in (36), i.e., In fact, if d 6 = det(a 6? A 5 ) = det( ^A6? ^A5 )? 4 sin 2 (( 6? 5 )=4) = 2? 2a 6? 4 sin 2 (( 6? 5 )=4) = 2 cos(( 6? 5 )=2)? 2a 6 : (37) d 6 < det( ^A6? ^A5 )? 4 sin 2 (( 6? 5 )=4) = 2? 2a 6? 4 sin 2 (( 6? 5 )=4) = 2 cos(( 6? 5 )=2)? 2a 6 ; then, d 6 =2 + a 6 < cos(( 6? 5 )=2) 1: 23

24 Since d 6 > 2 and a 6 >?1, we have d 6 =2+a 6 > 0. Therefore, 0 arccos(d 6 =2+a 6 ) < =2 is well-dened. Let 0 6 = arccos(d 6 =2 + a 6 ). Then 0 0 6? 5 <, 0 6 < 2 (due to the assumption 5 ), and Therefore, 0 6 > 6. Let A 0 6 = e0 6 =2 ^A6. Clearly, cos(( 0 6? 5)=2) = d a 6 < cos(( 6? 5 )=2): det(a 0 6? A 5) = det( ^A6? ^A5 )? 4 sin 2 (( 0 6? 5)=4) =?2a cos(( 0 6? 5 )=2) =?2a 6 + 2(d 6 =2 + a 6 ) = d 6 On the other hand, for = 1; 2; 3; 4; 0 6? > 6?, and hence, det(a 0 6? A ) = 4 sin 2 (( 0 6? )=4)? det( ^A6? ^A ) > 4 sin 2 (( 6? )=4)? det( ^A6? ^A ) d 6 : Therefore, fa 1 ; ; A 5 ; A 0 6 g is also an optimal design with d 6 = det( ^A6? ^A5 )? 4 sin 2 (( 0 6? 5)=4) and ^A 0 6 = ^A6. Thus, in the following proof of this proposition, we assume (37) holds. From (37), we obtain Step 2. Rotations and duals of G. a 6 = cos(( 6? 5 )=2)? d 6 =2: (38) Let us rst make a rotation of angle? 4 to G to generate a new constellation. Denote this new constellation as G, i.e., we dene G = fa 1 ; A 2 ; A 3 ; A 4 ; A 5 ; A 6 g where A = e? 4=2 A. For = 1; 2; 3, A 2 SU(2; 2? ( 4? )), and for = 4; 5; 6, A 2 SU(2; (? 4 )). Therefore, f ^A 1; ^A 2; ^A 3; ^A 4; ^A 5; ^A 6g = f? ^A1 ;? ^A2 ;? ^A3 ; ^A4 ; ^A5 ; ^A6 g; and f 1; 2; 3; 4; 5; 6g = f2? 4 ; 2? ( 4? 2 ); 2? ( 4? 3 ); 0; 5? 4 ; 6? 4 g: Note that 4 = 0 and ^A 4 2 SU(2). We next consider the dual of G and denote this dual as ~ G, i.e., ~G = f ~ A 1; ~ A 2; ~ A 3; ~ A 4; ~ A 5; ~ A 6g; where A ~ is the dual of A. By the denition of dual, we have and their corresponding angles f ^~ A 1 ; ^~ A 2 ; ^~ A 3 ; ^~ A 4 ; ^~ A 5 ; ^~ A 6 g = f? ^A 1;? ^A 2;? ^A 3; ^A4 ;? ^A 5;? ^A 6g = f ^A1 ; ^A2 ; ^A3 ; ^A4 ;? ^A5 ;? ^A6 g f ~ 1; ~ 2; ~ 3; ~ 4; ~ 5; ~ 6g = f2? 1; 2? 2; 2? 3; 4; 2? 5; 2? 6g = f 4 ; 4? 2 ; 4? 3 ; 0; 2? ( 5? 4 ); 2? ( 6? 4 )g: 24

25 Notice that in ~ G, ~ 4 = 0, i.e., ~ A 4 2 SU(2). To have the right order of angles, we rearrange the order of ~ G as follows. Let If we write A 0 = =2 e0 ^A0 2 SU(2; 0 ), then and A 0 1 = ~ A 4 ; A0 2 = ~ A 3 ; A0 3 = ~ A 2 ; A0 4 = ~ A 1 ; A0 5 = ~ A 6 ; A0 6 = ~ A 5 ; f ^A0 1 ; ^A0 2 ; ^A0 3 ; ^A0 4 ; ^A0 5 ; ^A0 6 g = f ^~ A 4 ; ^~ A 3 ; ^~ A 2 ; ^~ A 1 ; ^~ A 6 ; ^~ A 5 g f 0 1 ; 0 2 ; 0 3 ; 0 4 ; 0 5 ; 0 6 g = f~ 4 ; ~ 3 ; ~ 2 ; ~ 1 ; ~ 6 ; ~ 5 g = f ^A4 ; ^A3 ; ^A2 ; ^A1 ;? ^A6 ;? ^A5 g; = f0; 4? 3 ; 4? 2 ; 4 ; 2? ( 6? 4 ); 2? ( 5? 4 )g: It is clear that and 0 = < 2; 0 6 ; 0 5 ; 0 6? 0 4 : This means that the conditions on f 0 1 ; 0 2 ; ; 0 6 g are the same as those on f 1; 2 ; ; 6 g. Therefore, constellation fa 0 g has the same properties as fa g does if they have the same normalizations on their proections f ^A0 g as fa g do, namely ^A0 5 = I and ^A0 6 is diagonal. It is assumed that ^A5 = I. We now want to convert ^A 0 5 to I. Because ^A0 5 =? ^A6, we multiply? ^AH 6 to fa 0 1 ; ; A0 6g from the left and the resultant constellation is denoted by fa 00 1 ; ; A00 6 g = G 00. If we let A 00 =2 ^A00, then we have and = e00 f ^A00 1; ^A00 2; ^A00 3; ^A00 4; ^A00 5; ^A00 6g = f? ^AH ^A 0 6 1;? ^AH ^A 0 6 2;? ^AH ^A 0 6 3;? ^AH ^A 0 6 4;? ^AH ^A 0 6 5;? ^AH ^A 0 6 6g f 00 1; 00 2; 00 3; 00 4; 00 5; 00 6g = f 0 1; 0 2; 0 3; 0 4; 0 5; 0 6g = f? ^AH 6 ^A 4 ;? ^AH 6 ^A 3 ;? ^AH 6 ^A 2 ;? ^AH 6 ^A 1 ; I; ^AH 6 g (39) = f0; 4? 3 ; 4? 2 ; 4 ; 2? ( 6? 4 ); 2? ( 5? 4 )g; (40) from which one can see that ^A00 5 = I and ^A00 6 is diagonal since ^AH 6 is diagonal. Therefore, constellation G 00 has the same properties as G does. Additionally, from (39), we have a 00 5 = 1 and a00 6 = a 6. Since G 00 has the same angle relationships as those of G, inequalities (34)-(36) are also true if a is replaced by a 00, and is replaced by 00. Furthermore, since 6? 5 = 6 00? 00 5 and a00 6 = a 6, equalities (37) and (38) hold for G 00. We next establish some relationships between a and a 00 for = 1; 2; 3; 4. 25

26 Let = arccos(?a 6 ). Then 0 < =2 because a 6 < 0. From the form of ^A6 in Step 1, we have? cos + sin 0 ^A 6 = : 0? cos? sin Going back to (39), we obtain, for = 1; 2; 3; 4; a 00 + b00 c 00 + d00?c 00 + d00 a 00? b00 cos + sin 0 = 0 cos? sin a4?+1 + b 4?+1 c 4?+1 + e 4?+1 :?c 4?+1 + e 4?+1 a 4?+1? b 4?+1 In other words, 0 a 00 B B@ b 00 c 00 d 00 1 C CA = 0 cos? sin 0 0 sin cos cos? sin 0 0 sin cos 10 C A We need more relationships between coecients a and a 00 (35), we have 5? 2 arccos(d 6 =2 + a ): By (40), we obtain 00 6? 00 4?+1 = 2? ( 5? ). Hence Therefore, a 4?+1 b 4?+1 c 4?+1 e 4? ? 00 4?+1 2? 2 arccos(d 6=2 + a ): 1 C A ; = 1; 2; 3; 4: (41) for = 1; 2; 3; 4. For = 1; 2; 3; 4, from 00 6? ? 2 arccos(d 6=2 + a )? ( 00 5? 00 4?+1): (42) Since (38) holds also for G 00, the left hand side of (42) is equal to For the right hand side of (42), since 2 arccos(d 6 =2 + a 00 6) = 2 arccos(d 6 =2 + a 6 ): d 6 det(a 00 5? A00 4?+1) = det( ^A00 5? ^A00 4?+1)? 4 sin 2 (( 00 5? 00 4?+1)=4) = 2? 2a 00 4?+1? 4 sin 2 (( 00 5? 00 4?+1)=4); i.e., 00 5? 00 4?+1 2 arccos(d 6 =2 + a 00 4?+1): Hence, (42) can be changed into arccos(d 6 =2 + a 00 4?+1)? arccos(d 6 =2 + a )? arccos(d 6 =2 + a 6 ): (43) 26

27 Or equivalently, a 00 4?+1? cos (arccos(d 6 =2 + a ) + arccos(d 6 =2 + a 6 ))? d 6 =2 < 0; (44) where the second inequality is from d 6 > 2. Properties (41), (43), and (44) are important for the following proof, which provides us some relationships between a and a 00 4? through a 6 and d 6. ^A 5 and Step 3. Construction of a new constellation G that satises Case 1. Let = =2. Let = I and ^A 6 = 4 =2 for = 1; 2; 3; 4; and let 5 = ( )=2, 6 = ( )=2. Let =?I. Dene, for = 1; 2; 3; 4, 0 a b c d 1 C A = 0 cos? sin 0 0 sin cos cos? sin 0 0 sin cos ^A = i?1 ((a ; b 10 C A a b c e 1 C A : (45) ; c ; d )); (46) where i is the isomorphic mapping dened in Section 2. From (45), we know that ^A, = 1; 2; 3; 4, are on S 3. Thus, we can view ^A as unitary matrices in SU(2). Furthermore, from (45), we have det( ^A i? ^A) = det( ^Ai? ^A ): G is dened as follows: A 6 = e 6 =2 (?I); A 5 = e 5 =2 I; A = e =2 ^A ; = 1; 2; 3; 4: (47) We next show that the diversity product of this new constellation G is d 6, i.e., we show det(a i? A ) d 6 for 1 i < 6. From the denition of G, we have For 1 i < 4, from (45) and i 6 = (2? )=2; 5 = (2? )=2; (48) det(a i 6? ) + ( 5? ) = 2; = 1; 2; 3; 4: (49) ( = = 4 =2, we have? A ) = det( ^A i? ^A ) = det( ^Ai? ^A ) where the rst inequality is from the conditions of G. For det(a d sin 2 (( i? )=4) d 6 ; (50) 6? A 5 ), we have det(a 6? A 5 ) = det(?i? I)? 4 sin 2 (( 6? 5 )=4) = cos(( 6? 5 )=2) = cos(( 6? 5 )=2) = cos(arccos(d 6 =2 + a 6 )) = 2 + 2a 6 + d 6 d 6 ; (51) 27

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