The applications of a higher-dimensional Lie algebra and its decomposed subalgebras

Abstract

With the help of invertible linear transformations and the known Lie algebras, a higher-dimensional 6 × 6 matrix Lie algebra sμ(6) is constructed. It follows a type of new loop algebra is presented. By using a (2 + 1)-dimensional partial-differential equation hierarchy we obtain the integrable coupling of the (2 + 1)-dimensional KN integrable hierarchy, then its corresponding Hamiltonian structure is worked out by employing the quadratic-form identity. Furthermore, a higher-dimensional Lie algebra denoted by E, is given by decomposing the Lie algebra sμ(6), then a discrete lattice integrable coupling system is produced. A remarkable feature of the Lie algebras sμ(6) and E is used to directly construct integrable couplings.

1 Introduction

Integrable couplings were proposed in view of the Virasoro algebra and the soliton theory, a few ways to construct integrable couplings were presented by use of perturbations[1-3]. Tu Guizhang once proposed a simple and efficient method for generating integrable couplings and Hamiltonian of solition equations with infinite dimensions in Ref[4], Ma Wenxiu called the method as Tu scheme. By making use of Tu scheme, some well-known integrable hierarchies and corresponding Hamiltonian systems were worked out [5-18]. Its basis is the known algebra as follows:

$$\{\begin{array}{l}h=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right),e=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right),f=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right),\\ \left[h,e\right]=2e,\left[h\left(\text{m}\right),f\right]=-2f,\left[e,f\right]=h,\end{array}$$ (1)

Li[19] presented the Lax pair from the self-dual Yang-Mills equation[15]

$$\{\begin{array}{l}{\phi }_x=M\phi ,\\ {\phi }_t=P\left(\lambda \right){\phi }_y+N\phi .\end{array}$$ (2)

whose compatibility is the following (2 + 1)-dimensional partial-differential equation hierarchy

$$M_t-N_x+\left[M,N\right]-P\left(\lambda \right)M_y=0,$$ (3)

where P(λ) =α₀ +α₁λ + …, λ is a spectral parameter. If let P(λ) = 0, then ()reduce to the standard zero curvature equation

$$M_t-N_x+\left[M,N\right]=0$$ (4)

In this paper, with the help of invertible linear transformations and the known Lie algebras, we construct a higher-dimensional 6 × 6 matrix Lie algebra sμ(6), the integrable coupling of the (2 + 1)-dimensional KN integrable hierarchy is obtained under the framework of the (2 + 1)-dimensional partial-differential equation hierarchy, then we obtain its Hamiltonian structure by using the quadratic-form identity [9]. Finally, In Ref[19,20], a decomposed Lie algebra E from sμ(6) is presented whose two subalgebras are denoted by E₁ and E₂, which satisfy E =E₁ ⊕E₂, [E₁,E₂] ⊂E₂. We also require the closure between E₁ and E₂ under the matrix multiplication E₁E₂, E₂E₁ ⊂E₂. With the help of E, a discrete lattice Lax pair is given from which a discrete lattice integrable coupling system of an integrable hierarchy is obtained.

2 The application of a higher-dimensional Lie algebra

We construct a new higher-dimensional 6 × 6 matrix Lie algebra sμ(6) as follows[15]

$$\{\begin{array}{l}{h}_1=\left(\begin{array}{ccc}h& 0& 0\\ 0& h& 0\\ 0& 0& h\end{array}\right),h_2=\left(\begin{array}{ccc}e& 0& 0\\ 0& e& 0\\ 0& 0& e\end{array}\right),h_3=\left(\begin{array}{ccc}f& 0& 0\\ 0& f& 0\\ 0& 0& f\end{array}\right),\\ h_4=\left(\begin{array}{ccc}0& h& 0\\ 0& 0& h\\ 0& 0& 0\end{array}\right),h_5=\left(\begin{array}{ccc}0& e& 0\\ 0& 0& e\\ 0& 0& 0\end{array}\right),h_6=\left(\begin{array}{ccc}0& f& 0\\ 0& 0& f\\ 0& 0& 0\end{array}\right),\\ h_7=\left(\begin{array}{ccc}0& 0& h\\ 0& 0& 0\\ 0& 0& 0\end{array}\right),h_8=\left(\begin{array}{ccc}0& 0& e\\ 0& 0& 0\\ 0& 0& 0\end{array}\right),h_9=\left(\begin{array}{ccc}0& 0& f\\ 0& 0& 0\\ 0& 0& 0\end{array}\right),\\ \left[h_1,h_2\right]=2h_2,\left[h_1,h_3\right]=-2h_3,\left[h_2,h_3\right]=h_1,\left[h_1,h_4\right]=0,\left[h_1,h_5\right]=2h_5,\left[h_1,h_6\right]=-2h_6,\left[h_1,h_7\right]=0,\\ \left[h_1,h_8\right]=2h_8,\left[h_1,h_9\right]=-2h_9,\left[h_2,h_5\right]=0,\left[h_2,h_4\right]=-2h_5,\left[h_2,h_6\right]=h_4,\left[h_2,h_7\right]=-2h_8,\left[h_2,h_8\right]=0,\\ \left[h_2,h_9\right]=h_7,\left[h_3,h_4\right]=2h_6,\left[h_3,h_5\right]=-h_4,\left[h_3,h_6\right]=0,\left[h_3,h_7\right]=2h_9,\left[h_3,h_8\right]=-h_7,\left[h_3,h_9\right]=0,\\ \left[h_4,h_5\right]=2h_8,\left[h_4,h_6\right]=-2h_9,\left[h_4,h_7\right]=\left[h_4,h_8\right]=\left[h_4,h_9\right]=0,\left[h_5,h_6\right]=h_7,\\ \left[h_5,h_7\right]=\left[h_5,h_8\right]=\left[h_5,h_9\right]=\left[h_6,h_7\right]=\left[h_6,h_8\right]=\left[h_6,h_9\right]=\left[h_7,h_8\right]=\left[h_7,h_9\right]=\left[h_8,h_9\right]=0.\end{array}$$

It is easy to verify that $s\mu (6)=\mathit{\text{span}}{\left\{h_i\right\}}_{i=1}^9$is a Lie algebra. Its resulting loop algebra sμ̃(6) with powers of λ being λ²ⁿ, λ²ⁿ⁺¹ is given by

$$\{\begin{array}{l}{h}_1(n)=h_1{\lambda }^{2n},h_2(n)=h_2{\lambda }^{2n+1},h_3(n)=h_3{\lambda }^{2n+1},h_4(n)=h_4{\lambda }^{2n},h_5(n)=h_5{\lambda }^{2n+1}\\ h_6(n)=h_6{\lambda }^{2n+1},h_7(n)=h_7{\lambda }^{2n},h_8(n)=h_8{\lambda }^{2n+1},h_9(n)=h_9{\lambda }^{2n+1},\end{array}$$ (5)

with a commutative operation defined as

$$\{\begin{array}{l}[h_1(m),h_2(n)]=2h_2(m+n),[h_1(m),h_{3(n)}]=-2h_3(m+n),[h_2(m),h_3(n)]=h_1(m+n+1),[h_1(m),h_4(n)]=0,\\ [h_1(m),h_5(n)]=2h_5(m+n),[h_{1(m)},h_6(n)]=-2h_6(m+n),[h_1(m),h_7(n)]=0,[h_1(m),h_8(n)]=2h_8(m+n),\\ [h_1(m),h_9(n)]=-2h_9(m+n),[h_2(m),h_5(n)]=0,[h_2(m),h_4(n)]=-2h_5(m+n),[h_2(m),h_6(n)]=h_4(m+n+1),\\ [h_2(m),h_7(n)]=-2h_8(m+n),[h_2(m),h_8(n)]=0,[h_2,h_9(n)]=h_7(m+n+1),[h_3(m),h_4(n)]=2h_6(m+n),\\ [h_3(m),h_5(n)]=-h_4(m+n+1),[h_3(m),h_6(n)]=0,[h_3(m),h_7(n)]=2h_9(m+n),[h_3(m),h_8(n)]=-h_7(m+n+1),\\ [h_3(m),h_9(n)]=0,[h_4(m),h_5(n)]=2h_8(m+n),[h_4(m),h_6(n)]=-2h_9(m+n),\\ [h_4(m),h_7(n)]=[h_4(m),h_8(n)]=[h_4(m),h_9(n)]=0,[h_5(m),h_6(n)]=h_7(m+n+1),\\ [h_5(m),h_7(n)]=[h_5(m),h_8(n)]=[h_5(m),h_9(n)]=[h_6(m),h_7(n)]=[h_6(m),h_8(n)]=[h_6(m),h_9]=0,\\ [h_7(m),h_8(n)]=[h_7(m),h_9(n)]=[h_8(m),h_9(n)]=0,m,n\in Z,n=0,\pm 1,\pm 2\dots \end{array}$$ (6)

Consider an isospectral problem

$${\phi }_x=M\phi ,M=h_1(1)+u_1{h}_2(0)+u_2{h}_3(0)+u_3{h}_5(0)+u_4{h}_6(0)+u_5{h}_8(0)+u_6{h}_9(0).$$ (7)

Denoting V = Σ_m≥0(a_mh₁(−m) + b_mh₂(−m) + c_mh₃(−m) + d_mh₄(−m) + e_mh₅(−m) + f_mh₆(−m) + g_mh₇(−m) + k_mh₈(−m) + l_mh₉(−m)), then the stationary linear equation

$$N_x=[M,N]$$ (8)

yields

$$\{\begin{array}{l}{a}_{mx}=u_1{c}_{m+1}-u_2{b}_{m+1},b_{mx}=2b_{m+1}-2u_1{a}_m\\ c_{mx}=-2c_{m+1}+2u_2{a}_m\\ d_{mx}=-u_2{e}_{m+1}+u_1{f}_{m+1}+u_3{c}_{m+1}-u_4{b}_{m+1},\\ e_{mx}=2e_{m+1}-2u_1{d}_m-2u_3{a}_m,\\ f_{mx}=-2f_{m+1}+2u_2{d}_m+2u_4{a}_m,\\ g_{mx}=u_1{l}_{m+1}-u_2{k}_{m+1}+u_3{f}_{m+1}-u_4{e}_{m+1}+u_5{c}_{m+1}-u_6{b}_{m+1},\\ k_{mx}=2k_{m+1}-2u_1{g}_m-2u_3{d}_m-2u_5{a}_m,\\ l_{mx}=-2l_{m+1}+2u_2{g}_m+2u_4{d}_m+2u_6{a}_m,\\ a_0=\alpha =\mathit{\text{const}},b_0=c_0=d_0=e_0=f_0=g_0=k_0=l_0=0,\\ a_1=-\frac{\alpha }{2}{u}_1{u}_2,b_1=\alpha u_1,c_1=\alpha u_2,d_1=0,e_1=\alpha u_3,f_1=\alpha u_4,g_1=0,k_1=\alpha u_5,l_1=\alpha u_6.\end{array}$$ (9)

Note

$$\{\begin{array}{l}{N}_{+}^{(n)}=\sum _{m=0}^n(a_m{h}_1(n-m)-b_m{h}_2(n-m)+c_m{h}_3(n-m)+d_m{h}_4(n-m)+e_m{h}_5(n-m)\\ +f_m{h}_6(n-m)+g_m{h}_7(n-m)+k_m{h}_8(n-m)+l_m{h}_9(n-m),\\ N_{-}^{(n)}={\lambda }^{n}N-N_{+}^{(n)},\end{array}$$

then Eq.(8) can be written as

$$-N_{+x}^{(n)}+[M,N_{+}^{(n)}]=N_{-x}^{(n)}-[M,N_{-}^{(n)}].$$ (10)

a direct calculation reads

$$\begin{array}{c}-N_{+x}^{(n)}+[M,N_{+}^{(n)}]=-a_{nx}{h}_1(0)-2b_{n+1}{h}_2(0)+2c_{n+1}{h}_3(0)-d_{nx}{h}_4(0)-2e_{n+1}{h}_5(0)\\ +2f_{n+1}{h}_6(0)-g_{nx}{h}_7(0)-2k_{n+1}{h}_8(0)+2l_{n+1}{h}_9(0).\end{array}$$

Taking $V^{(n)}=V_{+}^{(n)}+{\Delta }_n$, with Δn = −a_nh₁(0) − d_nh₄(0) − g_nh₇(0), then we obtain that

$$\begin{array}{c}-N_{+x}^{(n)}+[M,N_{+}^{(n)}]=(-2b_{n+1}+2u_1{a}_n)h_2(0)+(2c_{n+1}-2u_2{a}_n)h_3(0)-(2e_{m+1}-2u_1{d}_m-2u_3{a}_m)h_5(0)\\ -(-2f_{m+1}+2u_2{d}_m+2u_4{a}_m)h_6(0)-k_{nx}{h}_8(0)-l_{nx}{h}_9(0)\\ =-b_{nx}{h}_2(0)-c_{nx}{h}_3(0)-e_{nx}{h}_5(0)-f_{nx}{h}_6(0)-k_{nx}{h}_8(0)-l_{nx}{h}_9(0).\end{array}$$

Thus, the (2 + 1)-dimensional partial-differential equation hierarchy

$$M_t-N_x^{(n)}+[M,N^{(n)}]-M_y=0,$$ (11)

admits the following solution

$${\left(\begin{array}{c}{u}_1\\ u_2\\ u_3\\ u_4\\ u_5\\ u_6\end{array}\right)}_{t-y}=\left(\begin{array}{c}\begin{array}{c}{b}_{nx}\\ c_{nx}\\ e_{nx}\\ f_{nx}\\ k_{nx}\\ l_{nx}\end{array}\end{array}\right)=\left(\begin{array}{cccccc}0& 0& 0& 0& 0& \partial \\ 0& 0& 0& 0& \partial & 0\\ 0& 0& 0& \partial & 0& -\partial \\ 0& 0& \partial & 0& -\partial & 0\\ 0& \partial & 0& -\partial & 0& 0\\ \partial & 0& -\partial & 0& 0& 0\end{array}\right)\left(\begin{array}{c}{c}_n+f_n+l_n\\ b_n+e_n+k_n\\ c_n+f_n\\ b_n+e_n\\ c_n\\ b_n\end{array}\right)=J{W}_n,$$ (12)

where J is a Hamiltonian operator. According to the definition of integrable couplings, we conclude that the system (12) is a type of the (2+1)-dimensional integrable model of the KN hierarchy since taking u₃ = u₄ = u₅ = u₆ = 0, u₁ = q, u₂ = r, ∂_y = 0 the system (12) reduces to the KN hierarchy. In what follows, we look for the Hamiltonian structure of the system (12). We make use of the following quadratic-form identity. Let a, b ∈ sμ(6), s-order matrix R(b) is determined by[9]

$${[a,b]}^T=(a_1,a_2,\dots ,a_9)R(b)=a^{T}R(b),$$

and constant matrix F =(f_ij)_s×s, is determined by

$$F=F^T,R(b)F=-{(R(b)F)}^T.$$ (13)

We introduce quadratic-form identity functional

$$\left\{a,b\right\}=a^{T}Fb,a,b\in s\stackrel{\sim }{\mu }(6),$$

then, we obtain

$$\frac{\delta }{\delta u_i}\left\{V,U_{\lambda }\right\}={\lambda }^{-\gamma }\frac{\partial }{\partial \lambda }\left({\lambda }^{\gamma }\{V,\frac{\partial U}{\partial u_i}\}\right),1\le i\le l,lisa\mathit{\text{constant}}$$ (14)

where λ is a constant to be determined, we call (33) the quadratic-form identity[9]. we have

$${[a,b]}^T=a^T\left(\begin{array}{ccccccccc}0& 2b_2& -b_3& 0& 2b_5& -2b_6& 0& 2b_8& -2b_9\\ b_3& -2b_1& 0& b_6& -2b_4& 0& b_9& -2b_7& 0\\ -b_2& 0& 2b_1& -b_5& 0& 2b_4& -b_8& 0& 2b_7\\ 0& 0& 0& 0& 2b_2& -2b_3& 0& 2b_5& -2b_6\\ 0& 0& 0& b_3& -2b_1& 0& b_6& -2b_4& 0\\ 0& 0& 0& -b_1& 0& 2b_1& -b_5& 0& 2b_4\\ 0& 0& 0& 0& 0& 0& 0& 2b_2& -2b_3\\ 0& 0& 0& 0& 0& 0& b_3& -2b_1& 0\\ 0& 0& 0& 0& 0& 0& -b_2& 0& 2b_1\end{array}\right)=a^{T}R(b).$$

Solving the matrix equation F^T = F and R(b)F = −(R(b)F)^T, give rise to

$$F=\left(\begin{array}{ccccccccc}2& 0& 0& 2& 0& 0& 2& 0& 0\\ 0& 0& 1& 0& 0& 1& 0& 0& 1\\ 0& 1& 0& 0& 1& 0& 0& 1& 0\\ 2& 0& 0& 2& 0& 0& 0& 0& 0\\ 0& 0& 1& 0& 0& 1& 0& 0& 0\\ 0& 1& 0& 0& 1& 0& 0& 0& 0\\ 2& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 1& 0& 0& 0& 0& 0& 0\\ 0& 1& 0& 0& 0& 0& 0& 0& 0\end{array}\right).$$

In order to get the Hamiltonian structure, Let

$$\{\begin{array}{l}{\phi }_x=\stackrel{\sim }{M}\phi ,\stackrel{\sim }{M}={({\lambda }^2,u_1\lambda ,u_2\lambda ,0,u_3\lambda ,u_4\lambda ,0,u_5\lambda ,u_6\lambda )}^T\\ {\phi }_t={\stackrel{\sim }{N}}^n\phi +{\phi }_y,{\stackrel{\sim }{N}}^n=\sum _{m=0}^n{(a_m,\lambda b_m,\lambda c_m,d_m,\lambda e_m,\lambda f_m,g_m,\lambda k_m,\lambda l_m)}^T{\lambda }^{-2m}-{(a_n,0,0,d_n,0,0,g_n,0,0)}^T.\end{array}$$

In terms of (13), it is easy to compute that

$$\begin{array}{lll}\left\{N,\frac{\partial M}{\partial u_1}\right\}={\lambda }^2(c+f+l),& \left\{N,\frac{\partial M}{\partial u_2}\right\}={\lambda }^2(b+e+k),& \left\{N,\frac{\partial M}{\partial u_3}\right\}={\lambda }^2(c+f),\\ \left\{N,\frac{\partial M}{\partial u_4}\right\}={\lambda }^2(b+e),& \left\{N,\frac{\partial M}{\partial u_5}\right\}={\lambda }^{2}c,& \left\{N,\frac{\partial M}{\partial u_6}\right\}={\lambda }^{2}b,\end{array}$$

$\left\{N,\frac{\partial M}{\partial \lambda }\right\}=2\lambda (2a+2d+2g)+u_1\lambda (c+f+l)+u_2\lambda (b+e+k)+u_3\lambda (c+f)+u_4\lambda (b+e)+u_5\lambda c+u_6\lambda b$ where $a=\text{∑}_{m\ge 0}{a}_m{\lambda }^{-2m},b=\text{∑}_{m\ge 0}{b}_m{\lambda }^{-2m}$, … Substituing the above formulae into the quadratic identity, comparison of the coefficient of λ⁻²ⁿ⁺¹ yields

$$\begin{array}{c}\frac{\delta }{\delta u}(4a_n+4d_n+4g_n+u_1(c_n+f_n+l_n)+u_2(b_n+e_n+k_n)+u_3(c_n+f_n)\\ +u_4(b_n+e_n)+u_5{c}_n+u_6{b}_n)=(\gamma -2n+2)\left(\begin{array}{c}{c}_n+f_n+l_n\\ b_n+e_n+k_n\\ c_n+f_n\\ b_n+e_n\\ c_n\\ b_n\end{array}\right).\end{array}$$

Taking n=1 gives γ = 0. Therefore,

$$\{\begin{array}{c}\frac{\delta H_n}{\delta u}=W_n,\\ H_n=\frac{4a_n+4d_n+4g_n+u_1(c_n+f_n+l_n)+u_2(b_n+e_n+k_n)+u_3(c_n+f_n)+u_4(b_n+e_n)+u_5{c}_n+u_6{b}_n}{2-2n}.\end{array}$$ (15)

Therefore, we obtain the Hamiltonian structure of the integrable coupling (12)

$$u_{t-y}={\left(\begin{array}{c}{u}_1\\ u_2\\ u_3\\ u_4\\ u_5\\ u_6\end{array}\right)}_{t-y}=J\frac{\delta H_n}{\delta u}$$ (16)

Remark 1

In the paper, we construct a higher-dimensional 6 × 6 matrix Lie algebra sμ(6), which can obtain the integrable hierarchy with more-potential functions and its Hamiltonian structure. The method also can be used to generalize other integrable hierarchies, such as TC hierarchy, BPT hierarchy, Burgers hierarchy, etc.

3 A decomposed Lie algebra E from the Lie algebra sμ(6)

Decompose h₁, h₄, h₇ in sμ(6) into the elements e₁, e₂, e₅, e₆, e₉, e₁₀,, we have

$$\begin{array}{l}{e}_1=\left(\begin{array}{ccc}{h}_1^{\ast }& 0& 0\\ 0& h_1^{\ast }& 0\\ 0& 0& h_1^{\ast }\end{array}\right),e_2=\left(\begin{array}{ccc}{h}_2^{\ast }& 0& 0\\ 0& h_2^{\ast }& 0\\ 0& 0& h_2^{\ast }\end{array}\right),e_3=\left(\begin{array}{ccc}e& 0& 0\\ 0& e& 0\\ 0& 0& e\end{array}\right),e_4=\left(\begin{array}{ccc}f& 0& 0\\ 0& f& 0\\ 0& 0& f\end{array}\right),\\ e_5=\left(\begin{array}{ccc}0& h_1^{\ast }& 0\\ 0& 0& h_1^{\ast }\\ 0& 0& 0\end{array}\right),e_6=\left(\begin{array}{ccc}0& h_2^{\ast }& 0\\ 0& 0& h_2^{\ast }\\ 0& 0& 0\end{array}\right)e_7=\left(\begin{array}{ccc}0& e& 0\\ 0& 0& e\\ 0& 0& 0\end{array}\right),e_8=\left(\begin{array}{ccc}0& f& 0\\ 0& 0& f\\ 0& 0& 0\end{array}\right),\\ e_9=\left(\begin{array}{ccc}0& 0& h_1^{\ast }\\ 0& 0& 0\\ 0& 0& 0\end{array}\right),e_{10}=\left(\begin{array}{ccc}0& 0& h_2^{\ast }\\ 0& 0& 0\\ 0& 0& 0\end{array}\right),e_{11}=\left(\begin{array}{ccc}0& 0& e\\ 0& 0& 0\\ 0& 0& 0\end{array}\right),e_{12}=\left(\begin{array}{ccc}0& 0& f\\ 0& 0& 0\\ 0& 0& 0\end{array}\right).\end{array}$$

Where $h_1^{\ast }=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right),h_2^{\ast }=\left(\begin{array}{cc}0& 0\\ 0& 1\end{array}\right)$, it is easy to verify that E is a Lie algebra. Denote E₁ = span{e₁, e₂, e₃, e₄}, E₂ = span{e₅, e₆, e₇, e₈, e₉, e₁₀, e₁₁, e₁₂}, we also see that

$$E=E_1\oplus E_2,[E_1,E_2]\subset E_2.$$

Observe that E₁ and E₂ are closed under the following matrix multiplication

$$\begin{array}{l}{e}_1{e}_1=e_1,e_2{e}_2=e_2,e_5{e}_5=e_9,e_6{e}_6=e_{10},\\ e_1{e}_2=e_2{e}_1=e_3{e}_3=e_4{e}_4=e_7{e}_7=e_8{e}_8=e_9{e}_9=e_{10}{e}_{10}=0,\\ e_1{e}_{10}=e_{10}{e}_1=e_{11}{e}_1=e_1{e}_{11}=e_1{e}_{12}=e_{12}{e}_1=e_4{e}_8=e_8{e}_4=e_4{e}_{12}=e_{12}{e}_4=0,,\\ e_2{e}_5=e_5{e}_2=0,e_2{e}_6=e_6{e}_2=e_6,e_2{e}_9=e_9{e}_2=e_2{e}_{11}=e_{11}{e}_2=0,\\ \{\begin{array}{c}{e}_1{e}_3=e_3,\\ e_3{e}_1=0,\end{array}\{\begin{array}{c}{e}_1{e}_4=0,\\ e_4{e}_1=e_4,\end{array}\{\begin{array}{c}{e}_1{e}_7=e_7,\\ e_7{e}_1=0,\end{array}\{\begin{array}{c}{e}_1{e}_8=0,\\ e_8{e}_1=e_8,\end{array}\\ e_1{e}_5=e_5{e}_1=e_5,e_1{e}_6=e_6{e}_1=0,e_1{e}_9=e_9{e}_1=e_9,e_3{e}_7=e_7{e}_3=0,\\ \{\begin{array}{c}{e}_2{e}_3=0,\\ e_3{e}_2=e_3,\end{array}\{\begin{array}{c}{e}_2{e}_4=e_4,\\ e_4{e}_2=0,\end{array}\{\begin{array}{c}{e}_2{e}_7=0,\\ e_7{e}_2=e_7,\end{array}\{\begin{array}{c}{e}_2{e}_8=e_8,\\ e_8{e}_2=0,\end{array}\{\begin{array}{c}{e}_2{e}_{10}=0,\\ e_{10}{e}_2=e_{10},\end{array}\{\begin{array}{c}{e}_2{e}_{12}=e_{12},\\ e_{12}{e}_2=0,\end{array}\\ \{\begin{array}{c}{e}_3{e}_4=e_1,\\ e_4{e}_3=e_2,\end{array}\{\begin{array}{c}{e}_3{e}_5=0,\\ e_5{e}_3=e_7,\end{array}\{\begin{array}{c}{e}_3{e}_6=e_7,\\ e_6{e}_3=0,\end{array}\{\begin{array}{c}{e}_3{e}_8=e_5,\\ e_8{e}_3=e_6,\end{array}\{\begin{array}{c}{e}_3{e}_9=0,\\ e_9{e}_3=e_9,\end{array}\{\begin{array}{c}{e}_3{e}_{10}=e_{11},\\ e_{10}{e}_3=0,\end{array}\\ \{\begin{array}{c}{e}_4{e}_5=e_8,\\ e_5{e}_4=0,\end{array}\{\begin{array}{c}{e}_4{e}_6=0,\\ e_6{e}_4=e_8,\end{array}\{\begin{array}{c}{e}_4{e}_7=e_6,\\ e_7{e}_4=e_5,\end{array}\{\begin{array}{c}{e}_4{e}_9=e_{12},\\ e_9{e}_4=0,\end{array}\{\begin{array}{c}{e}_4{e}_{10}=0,\\ e_{10}{e}_4=e_{12},\end{array}\{\begin{array}{c}{e}_4{e}_{11}=e_{10},\\ e_{11}{e}_4=e_9,\end{array}\\ \{\begin{array}{c}{e}_5{e}_7=e_{11},\\ e_7{e}_5=0,\end{array}\{\begin{array}{c}{e}_5{e}_8=0,\\ e_8{e}_5=e_{12},\end{array}\{\begin{array}{c}{e}_6{e}_7=0,\\ e_7{e}_6=e_{11},\end{array}\{\begin{array}{c}{e}_6{e}_8=e_{12},\\ e_8{e}_6=0,\end{array}\{\begin{array}{c}{e}_7{e}_8=e_9,\\ e_8{e}_7=e_{10},\end{array}\{\begin{array}{c}{e}_3{e}_{12}=e_9,\\ e_{12}{e}_3=e_{10},\end{array}\\ e_5{e}_6=e_6{e}_5=e_5{e}_9=e_9{e}_5=e_6{e}_9=e_9{e}_6=e_6{e}_{10}=e_{10}{e}_6=e_7{e}_8=e_8{e}_7=0.\end{array}$$

Remark 2

For computing convenience, the multiplication relations among e₁₂ and the other elements in E are not presented here.

It is easy to see that E₁E₂, E₂E₁ ⊂ E₂, which is key idea to generate discrete integrable couplings. In addition, the spectral matrices) are of the form

$$\overline{N}=\left(\begin{array}{ccc}U& {\overline{\overline{U}}}_1& {\overline{\overline{U}}}_2\\ O& U& {\overline{\overline{U}}}_1\\ O& O& U\end{array}\right)$$ (17)

where U,U̿₁U̿₂ are 2 × 2 matrices, respectively, O stands for a 2 × 2 zero matrix. According to variuos Jordan blocks, a general classification of the Lie algebras is followed. More detailed cases will be discussed in another paper.

Consider an isospectral problem

$$E{U}_n=U_n\psi ,U_n=\left(\begin{array}{cccccc}0& -\frac{1}{r_n}& 0& u_{1n}& 0& u_{4n}\\ r_n& \lambda +s_n& u_{2n}& v_n& u_{5n}& u_{3n}\\ 0& 0& 0& -\frac{1}{r_n}& 0& u_{1n}\\ 0& 0& r_n& \lambda +s_n& u_{2n}& v_n\\ 0& 0& 0& 0& 0& -\frac{1}{r_n}\\ 0& 0& 0& 0& r_n& \lambda +s_n\end{array}\right)$$ (18)

$$=\lambda e_2+s_n{e}_2-\frac{1}{r_n}{e}_3+r_n{e}_4++v_n{e}_6+u_{1n}{e}_7+u_{2n}{e}_8+u_{3n}{e}_{10}+u_{4n}{e}_{11}+u_{5n}{e}_{12}.$$

Note Γ = Σ_i≥0(a_i(e₁ − e₂) + b_ie₃ + c_ie₄ + f_ie₅ − f_ie₆ + g_ie₇ + h_ie₈ + k_ie₉ − k_ie₁₀ + l_ie₁₁ + w_ie₁₂)λ⁻ⁱ ≡ a(e₁ − e₂) + be₃ + ce₄ + fe₅ − fe₆ + ge₇ + he₈ + ke₉ − ke₁₀ + le₁₁ + we₁₂,

solving the discrete stationary zero curvature equation

$$(E\text{Γ})U-U\text{Γ}=0$$ (19)

yields

$$\{\begin{array}{l}{r}_n{b}_m^{(1)}-\frac{1}{r_n}{c}_m=0,\\ -\frac{1}{r_n}(a_m^{(1)}+a_m)+b_{m+1}^{(1)}+s_n{b}_m^{(1)}=0,\\ r_n(a_m^{(1)}+a_m)+c_{m+1}+s_n{c}_m=0,\\ \frac{1}{r_n}{c}_m^{(1)}+r_n{b}_m+s_n(a_m^{(1)}-a_m)+(a_{m+1}^{(1)}-a_{m+1})=0,\\ b_m^{(1)}{u}_{2n}+r_n{g}_m^{(1)}+\frac{1}{r_n}{h}_m-c_m{u}_{1n}=0,\\ c_m^{(1)}{u}_{1n}-a_m^{(1)}{v}_n-\frac{1}{r_n}{h}_m^{(1)}-s_n{f}_m^{(1)}-f_{m+1}^{(1)}-r_n{k}_m^{(1)}-r_n{g}_m+f_{m+1}+s_n{f}_m-b_m{u}_{2n}+a_m{v}_n=0,\\ a_m^{(1)}{u}_{1n}+v_n{b}_m^{(1)}+g_{m+1}^{(1)}+s_n{g}_m^{(1)}-\frac{1}{r_n}(f_m^{(1)}+f_m)+a_m{u}_{1n}=0,\\ a_m^{(1)}{u}_{2n}+r_n{f}_m^{(1)}+f_m{r}_n+h_{m+1}+s_n{h}_m+v_n{c}_m+a_m{u}_{2n}=0,\\ u_{5n}{b}_m^{(1)}+u_{2n}{g}_m^{(1)}+r_n{l}_m^{(1)}+\frac{1}{r_n}{w}_m-u_{1n}{h}_m-u_{4n}{c}_m=0,\\ u_{4n}{c}_m^{(1)}-a_m^{(1)}{u}_{3n}+h_m^{(1)}{u}_{1n}-f_m^{(1)}{v}_n-\frac{1}{r_n}{w}_m^{(1)}-k_{m+1}^{(1)}-s_n{k}_m^{(1)}+s_n{k}_m+k_{m+1}-r_n{l}_m-\\ u_{2n}{g}_m+u_{3n}{a}_m-u_{5n}{b}_m+v_n{f}_m=0,\\ u_{4n}{a}_m^{(1)}+b_m^{(1)}{u}_{3n}+k_m^{(1)}{u}_{1n}+g_m^{(1)}{v}_n-\frac{1}{r_n}{k}_m^{(1)}+u_{1n}{f}_m+l_{m+1}^{(1)}+s_n{l}_m^{(1)}-\frac{1}{r_n}{k}_m+u_{4n}{a}_m=0,\\ u_{5n}{a}_m^{(1)}+u_{2n}{f}_m^{(1)}+k_m^{(1)}{r}_n+k_m{r}_n+w_{m+1}+s_n{w}_m+f_m{u}_{2n}+v_n{h}_m+a_m{u}_{5n}+c_m{u}_{3n}=0.\end{array}$$ (20)

Throught some deduction, it is easy to see that the second and seventh and eleventh equation in (20) can be derived from the others. Hence, the three equations are superfluous.

Noting ${\Gamma }_{+}^{(m)}={\sum }_{i=0}^m(a_i(e_1-e_2)+b_i{e}_3+c_i{e}_4+f_i{e}_5-f_i{e}_6-g_i{e}_7+h_i{e}_8+k_i{e}_9-k_i{e}_{10}+l_i{e}_{11}+w_i{e}_{12}){\lambda }^{m-i}={\lambda }^m\Gamma -{\Gamma }_{-}^{(m)}$, then Eq.(20) can be written as

$$(E{\Gamma }_{+}^{(m)})U_n-U_n{\Gamma }_{+}^{(m)}=-(E{\Gamma }_{-}^{(m)})U_n+U_n{\Gamma }_{-}^{(m)}.$$ (21)

A direct calculation reads

$$(E{\Gamma }_{+}^{(m)})U_n-U_n{\Gamma }_{+}^{(m)}=-b_{m+1}^{(1)}{e}_3+c_{m+1}{e}_4+(a_{m+1}^{(1)}-a_{m+1})e_2+(f_{m+1}^{(1)}-f_{m+1})e_6-g_{m+1}^{(1)}{e}_7+h_{m+1}{e}_8+(k_{m+1}^{(1)}-k_{m+1})e_{10}-l_{m+1}^{(1)}{e}_{11}+w_{m+1}{e}_{12}$$

Take ${\Gamma }^{(m)}={\Gamma }_{+}^{(m)}$, the discrete zero curvature equation

$${(U_n)}_{t_m}=(E{\Gamma }^{(m)})U_n-U_n{\Gamma }^{(m)}$$ (22)

generates the discrete integrable hierarchy

$$\{\begin{array}{l}{(r_n)}_{t_m}=c_{m+1,}\\ {(s_n)}_{t_m}=a_{m+1}^{(1)}-a_{m+1,}\\ {(v_n)}_{t_m}=-f_{m+1}^{(1)}-f_{m+1},\\ {(u_{1n})}_{t_m}=-g_{m+1}^{(1)},\\ {(u_{2n})}_{t_m}=h_{m+1}\\ {(u_{3n})}_{t_m}=k_{m+1}^{(1)}-k_{m+1},\\ {(u_{4n})}_{t_m}=-l_{m+1}^{(1)},\\ {(u_{5n})}_{t_m}=w_{m+1}.\end{array}$$ (23)

The first two equations in (23) were presented in Ref.[21]. Hence, the discrete lattice hierarchy (23) is the discrete integrable couplings of the first two equations.

Remark 3

The above example indicates a way to generate discrete integrable couplings by the Lie algebra, but there is an open problem: Could we generalize the continuous-type quadratic-form identity in [9] into the discrete-type one so that the Hamiltonian structures of the discrete integrable couplings could be obtained? which is worth studying in the future.