Characteristic Numbers of Matrix Lie Algebras

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1 Commun. Theor. Phys. (Beijing China) 49 (8) pp c Chinese Physical Society Vol. 49 No. 4 April 15 8 Characteristic Numbers of Matrix Lie Algebras ZHANG Yu-Feng 1 and FAN En-Gui 1 Mathematical School Liaoning Normal University Dalian 1169 China Institute of Mathematics Fudan University Shanghai 433 China (Received April 7 7) Abstract A notion of characteristic number of matrix Lie algebras is defined which is devoted to distinguishing various Lie algebras that are used to generate integrable couplings of soliton equations. That is the exact classification of the matrix Lie algebras by using computational formulas is given. Here the characteristic numbers also describe the relations between soliton solutions of the stationary zero curvature equations expressed by various Lie algebras. PACS numbers:.9.+p 5.45.Yv.3.Ik Key words: characteristic number Lie algebra integrable couplings 1 Introduction Since the Tu scheme was proposed by Tu Gui-Zhang [1] some interesting soliton hierarchies were obtained by the following Lie algebras A 11 = span{h e f} (1) where h = ( 1 ) ( 1 e = 1 ) ( f = ) 1 ; A 1 = span{ h e ± } () ( 1 ) ±1. ( 1 1 ) e± = 1 where h = 1 In fact some integrable Hamiltonian hierarchies were generated from the corresponding various loop algebras under the zero curvature equations such as the results in Refs. [] [13]. Recently Ma proposed a kind of classification of Lie algebras expressed by spectral matrices [14] from which Zhang et al. [15] constructed a few Lie algebras to use to generate higher-dimensional integrable couplings by using the quadratic-form identity. [16] But some of the Lie algebras may only produce integrable couplingsowever their Hamiltonian structure cannot be obtained by means of the quadratic-form identity. For example the following Lie algebras belong to this case [15] A 1 = span{p i } 5 (3) 1 1 p 1 = 1 p = 1 p 3 = 1 p 4 = p 5 = 1 [p 1 p ] = p [p 1 p 3 ] = p 3 [p p 3 ] = p 1 [p 1 p 4 ] = p 4 [p 1 p 5 ] = p 5 [p p 4 ] = [p p 5 ] = p 4 [p 3 p 4 ] = p 5 [p 3 p 5 ] = [p 4 5 ] = ; A = span{g i } 5 (4) g 1 = g = g 3 = g 4 = g 5 = 1 [g 1 g ] = g 3 [g 1 g 3 ] = g [g g 3 ] = g 1 [g 1 g 4 ] = 1 g 5 [g 1 g 5 ] = 1 g 4 [g g 4 ] = 1 g 4 [g 3 g 4 ] = 1 g 5 [g g 5 ] = 1 g 5 [g 3 g 5 ] = 1 g 4 [g 4 g 5 ] =. In this paper we distinguish Lie algebras for generating integrable couplings by defining a notion of characteristic numbers. Various Lie algebras correspond to various characteristic numbers specially the Lie algebras sharing characteristic numbers could produce Hamiltonian structure of integrable couplings by adopting the quadraticform identity. Characteristic Numbers of Lie Algebras When we use the Tu scheme to deduce soliton systems we usually first set up an isospectral problem by using Lie algebras (actually loop algebras) ϕ x = Uϕ ϕ t = V ϕ (5) The project supported by National Natural Science Foundation of China under Grant Nos and Shanghai Shuguang Project under Grant No. SG

2 846 ZHANG Yu-Feng and FAN En-Gui Vol. 49 whose compatibility manifests the zero curvature equation U t V x + [U V ] =. (6) In order to derive soliton integrable systems the resulting stationary zero curvature equation is first solved for V V x = [U V ] (7) which is a linear matrix equation with respect to V. Therefore if Z and V are two solutions to Eq. (7) then they should be linear-dependent. Hence we may assume where γ is a constant. Z = γv (8) Definition The γ satisfying Eq. (8) is called the characteristic number. In order to derive the trace identity and the quadraticform identity references [1] and [16] adopted Eq. (8) and took Z = [W V ] V λ [W V ] V λ (9) where the matrix W satisfies W x = U λ + [U W]. (1) After direct calculations we find that equation (9) with Eq. (1) satisfies the stationary zero curvature equation (7). Guo and Zhang in Ref. [17] deduced the computational formulae on the γ under the situation of the vector Lie algebras. In this paper we want to seek the characteristic numbers γ under the matrix Lie algebras for generating integrable couplings. From this we give a classification of the matrix Lie algebras in more detail than that given by Ma et al. [14] Next we consider the Lie algebra [115] sµ(4) 1 = span{t i } 6 (11) T 1 = 1 T = T 3 = T 1 4 = 1 1 T 5 = T 1 6 = [T 1 T ] = T [T 1 T 3 ] = T 3 [T T 3 ] = T 1 [T 1 T 4 ] = [T 1 T 5 ] = T 5 [T 1 T 6 ] = T 6 [T T 4 ] = T 5 [T T 5 ] = [T T 6 ] = T 4 [T 3 T 4 ] = T 6 [T 3 T 5 ] = T 4 [T 3 T 6 ] = [T 4 T 5 ] = [T 4 T 6 ] = [T 5 T 6 ] =. Set up a Lax pair appearing in Eq. (5) as follows from Eq. (11) 6 U = u i T i V = at 1 + bt + ct 3 + dt 4 + et 5 + ft 6. (1) Equation (7) gives a solution V by recurrence a x = u c u 3 b b x = u 1 b u a c x = u 1 c + u 3 a d x = u f u 3 e + u 5 c u 6 b e x = u 1 e u d + u 4 b u 5 a f x = u 1 f + u 3 d u 4 c + u 6 a. (13) From b x c + bc x = u ac + u 3 ab aa x = u ac + u 3 ab we have b x c+bc x +aa x =. Denote P 1 = a +bc then P 1x =. Since we have ce x + bf x = u 1 ce u cd u 5 ca u 1 fb + u 3 bd + u 6 ab c x e + b x f = u 1 ce + u 3 ae + u 1 bf u af (ce + bf) x + ad x = u cd + u 3 bd = a x d (ce + bf + ad) x =. Denote P = ce + bf + ad then P x =. Take W = 6 w it i from Eqs. (8) (1) we have a λ + γa = w c w 3 b b λ + γb = w 1 b w a c λ + γc = w 1 c + w 3 a d λ + γd = w f w 3 e + w 5 c w 6 b e λ + γe = w 1 e w d + w 4 b w 5 a f λ + γf = w 1 f + w 3 d w 4 c + w 6 a. (14) Since aa λ + cb λ + c λ b = γ(a + bc) then dp 1 dλ = γp 1 γ 11 γ = 1 λ ln P 1. (15) From Eq. (14) we see that (ce + bf + ad) λ = γ(ce + bf + ad) i.e. dp /dλ = γp thus γ 1 γ = 1 λ ln P (16) where P = ce + bf + da. Remark 1 Since the Lie algebra (11) may be decomposed into a direct sum of two subalgebras G 1 and G i.e. sµ(4) 1 = G 1 G [G 1 G ] G G 1 = A11 (17) then the characteristic numbers (15) and (16) correspond to the Lie algebras G 1 and G respectively. Again due to G 1 = A11 then the Lie algebras G 1 and A 11 should

3 No. 4 Characteristic Numbers of Matrix Lie Algebras 847 have the same characteristic number. The characteristic number which labels the Lie algebra (11) is actually γ 1. Let us consider the Lie algebra [15] sµ(4) = span{f i } 6 (18) 1 f 1 = f = f 3 = f 4 = f 5 = 1 1 f 6 = 1 1 [f 1 f ] = f 3 [f 1 f 3 ] = f [f f 3 ] = f 1 [f 1 f 4 ] = [f 1 f 5 ] = f 6 [f 1 f 6 ] = f 5 [f f 4 ] = f 5 [f f 5 ] = f 4 [f f 6 ] = [f 3 f 4 ] = f 6 [f 3 f 5 ] = [f 3 f 6 ] = f 4 [f 4 f 5 ] = [f 4 f 6 ] = [f 5 f 6 ] =. A Lax pair is given by Eq. (18) 6 U = u i f i V = af 1 + bf + cf 3 + df 4 + ef 5 + hf 6 (19) Solving Eq. (7) yields a x = u c u 3 b b x = u 1 c u 3 a c x = u 1 b u a d x = u e u 3 h u 5 b + u 6 c e x = u 1 h u d + u 4 b u 6 a h x = u 1 e u 3 d + u 4 c u 5 a. () Denote P 3 = b a c P 4 = ce+ad bh from Eq. () we obtain P 3x = P 4x =. Let W = 6 w if i then in terms of Eqs. (8) (9) and (1) we find that a λ + γa = w c w 3 b b λ + γb = w 3 a + w 1 c c λ + γc = w 1 b w a d λ + γd = w e w 3 h w 5 b + w 6 c e λ + γe = w 1 h w d + w 4 b w 6 a h λ + γh = w 1 e w 3 d + w 4 c w 5 a. (1) A direct computation shows from Eq. (1) bb λ aa λ cc λ + γ(b a c ) = dp 3 dλ = γp 3 γ 1 γ = 1 λ ln P 3. Similarly we obtain dp 4 dλ = γp 4 γ = 1 λ ln P 4. Remark From Eqs. (11) and (18) we find that these two Lie algebras really are various. This fact is expressed by the explicit mathematical tools i.e. the characteristic numbers γ 1 and γ which are used to distinguish the matrix Lie algebras for producing integrable couplings. In what follows we again present two more complicated Lie algebras than Eqs. (11) and (18) which illustrate the roles of the characteristic numbers. Case 1 Consider the Lie algebra [1] sµ(6) 1 = span{ T i } T 1 = T = T3 = T 4 = T5 = T6 = 1 1 ()

4 848 ZHANG Yu-Feng and FAN En-Gui Vol T 7 = T8 = T9 = [ T 1 T ] = T [ T 1 T 3 ] = T 3 [ T T 3 ] = T 1 [ T 1 T 4 ] = [ T 1 T 5 ] = T 5 [ T 1 T 6 ] = T 6 [ T 1 T 7 ] = [ T 1 T 8 ] = T 8 [ T 1 T 9 ] = T 9 [ T T 5 ] = [ T T 4 ] = T 5 [ T T 6 ] = T 4 [ T T 7 ] = T 8 [ T T 8 ] = [ T T 9 ] = T 7 [ T 3 T 4 ] = T 6 [ T 3 T 5 ] = T 4 [ T 3 T 6 ] = [ T 3 T 7 ] = T 9 [ T 3 T 8 ] = T 7 [ T 3 T 9 ] = [ T 4 T 5 ] = T 8 [ T 4 T 6 ] = T 9 [ T 4 T 7 ] = [ T 4 T 8 ] = [ T 4 T 9 ] = [ T 5 T 6 ] = T 7 [ T 5 T 7 ] = [ T 5 T 8 ] = [ T 5 T 9 ] = [ T 6 T 7 ] = [ T 6 T 8 ] = [ T 6 T 9 ] = [ T 7 T 8 ] = [ T 7 T 9 ] = [ T 8 T 9 ] =. Take a Lax pair from Eq. () 9 U = u i Ti V = a T 1 + b T + c T 3 + d T 4 + e T 5 + f T 6 + g T 7 + h T 8 + w T 9. (3) Equation (7) admits a solution for V a x = u c u 3 b b x = u 1 b u a c x = u 1 c + u 3 a d x = u f u 3 e + u 5 c u 6 b e x = u 1 e u d + u 4 b u 5 a f x = u 1 f + u 3 d u 4 c + u 6 a g x = u w u 3 h + u 5 f u 6 e + u 8 c u 9 b h x = u 1 h u g u 5 d + u 4 e + u 7 b u 8 a w x = u 1 w + u 3 g u 4 f + u 6 d u 7 c + u 9 a (4) from which we have again P 1 = a + bc. Since ce x + bf x = u 1 ce u cd u 5 ac u 1 bf + u 3 bd + u 6 ab ec x + b x f = u 1 ce + u 3 ae + u 1 bf u af hence Again since (ce + bf) x = u cd u 5 ac + u 3 bd + u 6 ab + u 3 ae u af. da x + ad x = u cd u 3 bd u af u 3 ae + u 5 ac u 6 ab we see that (ce + bf + ad) x = again derive P x = which is consistent to that of the previous Lie algebra (11). From Eq. (4) therefore Since ch x + bw x = u 1 ch u cg u 5 cd + u 4 ce u 8 ac u 1 wb + u 3 gb u 4 bf + u 6 bd + u 9 ab c x h + b x w = u 1 ch + u 3 ah + u 1 bw u aw (ch + bw) x = u cg u 5 cd + u 4 ce u 8 ac + u 3 gb u 4 bf + u 6 bd + u 9 ab + u 3 ah u aw. a x g = u gc u 3 bg ag x = u aw u 3 ah + u 5 af u 6 ae + u 8 ac u 9 ab we have (ch + bw + ag) x = u 5 af u 5 cd + u 6 bd u 6 ae + u 4 ce u 4 bf. While Thus ef x + e x f = u 3 de u 4 ce + u 6 ae u df + u 4 bf u 5 af. (ch + bw + ag + ef) x = u 3 de u df u 5 cd + u 6 bd = dd x. Denote P 5 = ch + bw + ag + ef + d then P 5x =. From Eqs. (8)(9) and (1) we have a λ + γa = w c w 3 b b λ + γb = w 1 b w a c λ + γc = w 1 c + w 3 a d λ + γd = w f w 3 e + w 5 c w 6 b e λ + γe = w 1 e w d + w 4 b w 5 a f λ + γf = w 1 f + w 3 d w 4 c + w 6 a g λ + γg = w w w 3 h + w 5 f w 6 e + w 8 c w 9 b

5 No. 4 Characteristic Numbers of Matrix Lie Algebras 849 from which we obtain h λ + γh = w 1 h w g w 5 d + w 4 e + w 7 b w 8 a w λ + γw = w 1 w + w 3 g w 4 f + w 6 d w 7 c + w 9 a (5) γ 31 = γ 11 γ 3 = γ 1 γ 33 = 1 λ ln P 5. Remark 3 Due to the Lie algebra () being the expanding one of Eq. (11) we not only get γ 11 γ 1 but also obtain γ 33. The characteristic number γ 33 describes the feature of the Lie algebra (). However the following matrix Lie algebra cannot share the property. Case Consider the Lie algebra [15] sµ(6) = span{h i } 9 (6) h 1 = = = h 4 = = = h 7 = 1 8 = 1 9 = 1 [h 1 ] = h 3 [h 1 3 ] = h [h 3 ] = h 1 [h 1 4 ] = [h 1 5 ] = h 6 [h 1 6 ] = h 5 [h 1 7 ] = h 8 [h 1 8 ] = h 7 [h 1 9 ] = [h 4 ] = h 5 [h 5 ] = h 4 [h 6 ] = [h 7 ] = [h 8 ] = h 9 [h 9 ] = h 8 [h 3 4 ] = h 6 [h 3 5 ] = [h 3 6 ] = h 4 [h 3 7 ] = h 9 [h 3 8 ] = [h 3 9 ] = h 7 [h 4 5 ] = h 7 [h 4 6 ] = h 8 [h 4 7 ] = [h 4 8 ] = [h 4 9 ] = [h 5 6 ] = h 9 [h 5 7 ] = [h 5 8 ] = [h 5 9 ] = [h 6 7 ] = [h 6 8 ] = [h 6 9 ] = [h 7 8 ] = [h 7 9 ] = [h 8 9 ] =. Take a Lax pair from Eq. (6) 9 U = u i h i V = ah 1 + bh + ch 3 + dh 4 + eh 5 + fh 6 + gh 7 + jh 8 + kh 9. (7) Solving Eq. (7) one obtains a x = u c u 3 b b x = u 1 c u 3 a c x = u 1 b u a d x = u e + u 3 f u 5 b u 6 c e x = u 1 f u d + u 4 b + u 6 a f x = u 1 e + u 3 d u 4 c + u 5 a g x = u 1 j u 3 k + u 4 e u 5 d u 8 a + u 9 c j x = u 1 g u k u 4 f + u 6 d u 7 a + u 9 b k x = u j u 3 g + u 5 f u 6 e + u 7 c u 8 b (8) from which we have bb x cc x = aa x (b a c ) x = i.e. P 3x =. Since ad x + ce x = u ae + u 3 af u 5 ab u 1 fc u cd + u 4 bc a x d + c x e = u cd u 3 bd + u 1 be u ae one infers (ad + ce) x = u 3 af u 5ab u 1 fc + u 4 bc u 3 bd + u 1 be. Again due to b x f + f x b = u 1 cf u 3 af u 1 be + u 3 bd u 4 bc + u 5 ab one infers (bf + ad + ce) x =.

6 85 ZHANG Yu-Feng and FAN En-Gui Vol. 49 Denote P 6 = bf + ad + ce then P 6x =. It is easy to find that P 6 is different form P 4. Since again from Eq. (8) cj x +ak x = u 1 gc u kc u 4 fc+u 6 dc+u 9 bc+u aj u 3 ag+u 5 af u 6 ae u 8 ab c x j +a x k = u 1 bj u aj +u ck u 3 bk we find that (cj + ak) x = u 1 gc u 4 fc + u 6 cd + u 9 bc u 3 ag + u 5 af u 6 ae u 8 ab + u 1 bj u 3 bk. Since b x g g x b = u 1 gc+u 3 ag u 1 jb+u 3 bk u 4 be+u 5 bd+u 8 ab u 9 bc one infers (cj+ak bg) x = u 4 fc+u 6 cd+u 5 af u 6 ae u 4 be+u 5 bd. From ff x + ee x + (cj + ak bg) x = u 6 dc + u 5 bd u 3 df u de = (u e + u 3 f u 5 b u 6 c)d = d x d we have (cj + ak bg 1 f + 1 (d + e )) x =. Denote P 7 = 1 (d + e f ) + ak + cj bg then P 7x =. Obviously P 7 is different from P 5. Assume W = 9 w ih i in terms of Eqs. (8) (9) and (1) we see that V λ + γv = [W V ] which admits a λ + γa = w c w 3 b b λ + γb = w 1 c w 3 a c λ + γc = w 1 b w a d λ + γd = w e + w 3 f w 5 b w 6 c e λ + γe = w 1 f w d + w 4 b + w 6 a f λ + γf = w 1 e + w 3 d w 4 c + w 5 a g λ + γg = w 1 j w 3 k + w 4 e w 5 d w 8 a + w 9 c j λ + γj = w 1 g w k w 4 f + w 6 d w 7 a + w 9 b k λ + γk = w j w 3 g + w 5 f w 6 e + w 7 c w 8 b (9) which generates that γ 41 = γ 1 γ 4 = (1/λ) ln P 6 γ 43 = (1/λ) ln P 7. From this we conclude that the Lie algebra (6) is not the completely extended form of the Lie algebra (18). From the above analyses we see that the characteristic numbers may describe differences among Lie algebras in more detail than that presented in Ref. [14]. They also give rise to the explicit relations of the soliton solutions of the stationary zero curvature equations under various Lie algebras. Remark 4 By similar computation to the Lie algebras (3) and (4) we cannot eliminate the potential functions in Eq. (7). Thus we cannot also obtain the corresponding characteristic numbers. Can we make a conclusion that the characteristic numbers of such Lie algebras do not exist? This is an open problem to be studied further. Again we find that the Hamiltonian structure of integrable couplings obtained by Eqs. (3) and (4) cannot be derived from the quadratic-form identity. However the Hamiltonian structure of integrable couplings obtained by sµ(4) i sµ(6) i (i = 1 ) may be worked out by the quadratic-form identity. This observing result is worthy of strict study logically in the future. References [1] Tu Gui-Zhang J. Math. Phys. 3 (1989) 33. [] Ma Wen-Xiu Chin. J. Contemp. Math. 13 (199) 79. [3] Hu Xing-Biao J. Phys. A 3 (1997) 619. [4] Ma Wen-Xiu and B. Fuchssteiner Chaos Solitons and Fractals 7 (1996) 17. [5] Ma Wen-Xiu Methods and Applications of Analysis 7 () 1. [6] Fu-Kui Guo and Yu-Feng Zhang J. Math. Phys. 44 (3) [7] Zhang Yu-Feng Chaos Solitons and Fractals 1 (4) 35. [8] Zhang Yu-Feng Tam Honwah and Zhao Jing Commun. Theor. Phys. (Beijing China) 45 (6) 411. [9] Zhang Yu-Feng and Guo Xiu-Rong Chaos Solitons and Fractals 7 (6) 555. [1] Zhang Yu-Feng and Fan En-Gui Phys. Lett. A 348 (6) 18. [11] Fan En-Gui and Zhang Yu-Feng Chaos Solitons and Fractals 8 (6) 966; Fan En-Gui et al. Phys. Lett. A 36 (6) 99. [1] Zhang Yu-Feng and Guo Fu-Kui Commun. Theor. Phys. (Beijing China) 46 (6) 81. [13] Zhang Yu-Feng and Wang Yan Phys. Lett. A 36 (6) 9. [14] Ma Wen-Xiu Xu Xi-Xiang and Zhang Yu-Feng Phys. Lett. A 351 (6) 15. [15] Zhang Yu-Feng Tam Honwah and Guo Fu-Kui Commun. in Nonlinear Science and Numerical Simulation 13 (8) 68. [16] Guo Fu-Kui and Zhang Yu-Feng J. Phys. A 38 (5) [17] Guo Fu-Kui and Zhang Yu-Feng Chaos Solitons and Fractals (7) doi:1.116/j.chaos

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