1. Introduction
The Heisenberg ferromagnet (HF) model is a fundamental integrable model that plays a vital role in condensed-matter physics and field theory, due to its connections with the nonlinear σ model and the confinement of quarks [1-3]. It has been shown that the HF model is geometrical and gauge equivalent to the nonlinear Schrödinger (NLS) equations [4-6]. Furthermore, Kundu [7] established the equivalence between a new HF model and the higher-order NLS equation. The extensions of the HF models in (1 + 1) and (2 + 1) dimensions have been widely investigated [8-12].
Supersymmetry is an elementary symmetry principle that unifies bosonic and fermionic degrees of infinite-dimensional dynamical systems. Supersymmetric integrable systems have attracted much attention from mathematical and physical viewpoints due to their applications in gravity theory [13], dark matter [14-16], the theory of strings and superstrings [17-19]. Great interest has been shown in generalizing various significant integrable models into the supersymmetric framework: for example, the supersymmetric NLS equation [20, 21] and the Heisenberg supermagnet (HS) model. The HS model is the superextension of the HF model [22]. With the help of gauge transformation, Makhankov and Pashaev [23] showed that the HS models under the square (S2 = S and S2 = 3S - 2I) and cubic constraints (S3 = S) are equivalent to related (super and fermionic) NLS equations. Moreover the higher-order and inhomogeneous HS models with square constraints in (1 + 1) and (2 + 1) dimensions have been investigated [24, 25]. Recently, a higher-order HS model with cubic constraint S3 = S in (1 + 1) dimensions was presented by developing an algebraic model, where the recursion operators of Lax representation are given by an algorithm and the supersymmetric NLS equation has been derived, in which the gauge transformation plays a fundamental role [26].
It is known that the generalized HS models with square constraints have been widely and deeply studied, but there are few related results in the generalized HS models with cubic constraints due to the inherent complexity, especially in the high-dimensional case. In this paper, our first objective is to construct the generalized inhomogeneous HS model with S3 = S in (1 + 1) dimensions and analyze the integrability and properties of the system. The second goal is to present (2 + 1)-dimensional deformations of HS models with S3 = S. Then based upon the gauge transformation, we aim to derive the gauge equivalent counterparts of these integrable systems. The main difficulty lies in how to construct the auxiliary matrix which has no commutation with the spin matrix.
The organization of this paper is as follows. In section 2 , we first recall some basic facts about the HS model and then construct the deformed inhomogeneous HS model with the cubic constraint. The corresponding gauge equivalent counterpart will be investigated. In section 3 , we establish the (2 + 1)-dimensional HS models with the cubic constraint and present their relations with the super NLS equation in (2 + 1) dimensions. Section 4 is devoted to a summary and discussion.
2. The inhomogeneous Heisenberg supermagnetic model in (1+1) dimensions
With the cubic constraint S3 = S, the HS model is the superextension of the HF model, which is given by
$\begin{eqnarray}{\rm{i}}{S}_{t}=[S,{S}_{xx}]-\frac{3}{2}{(S{S}_{x}{S}^{2})}_{x},\end{eqnarray}$
where S is the superspin function and can be expressed as $\begin{eqnarray}\begin{array}{rcl}S & = & 2\displaystyle \sum _{a=1}^{3}{S}_{a}{E}_{a}+2\displaystyle \sum _{a=4}^{5}{C}_{a}{E}_{a}\\ & = & \left(\begin{array}{ccc}{S}_{3} & -{S}_{2}-{\rm{i}}{S}_{1} & \kappa (-{C}_{4}+{\rm{i}}{C}_{5})\\ {S}_{2}-{\rm{i}}{S}_{1} & -{S}_{3} & {C}_{4}+{\rm{i}}{C}_{5}\\ {C}_{4}+{\rm{i}}{C}_{5} & \kappa ({C}_{4}-{\rm{i}}{C}_{5}) & 0\end{array}\right),\end{array}\end{eqnarray}$
where κ = ± 1 and S1, S2, S3 and C4, C5 are the bosonic and fermionic variables, respectively. Here E1, E2, E3 and E4, E5 are the bosonic and fermionic generators of the superalgebra ospu(1, 1/1), respectively, and are given by $\begin{eqnarray}\begin{array}{rcl}{E}_{1} & = & \frac{1}{2}\left(\begin{array}{ccc}0 & -{\rm{i}} & 0\\ -{\rm{i}} & 0 & 0\\ 0 & 0 & 0\end{array}\right),\,{E}_{2}=\frac{1}{2}\left(\begin{array}{ccc}0 & -1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 0\end{array}\right),\\ {E}_{3} & = & \frac{1}{2}\left(\begin{array}{ccc}1 & 0 & 0\\ 0 & -1 & 0\\ 0 & 0 & 0\end{array}\right),\\ {E}_{4} & = & \frac{1}{2}\left(\begin{array}{ccc}0 & 0 & -\kappa \\ 0 & 0 & 1\\ 1 & \kappa & 0\end{array}\right),\,{E}_{5}=\frac{1}{2}\left(\begin{array}{ccc}0 & 0 & \kappa {\rm{i}}\\ 0 & 0 & {\rm{i}}\\ {\rm{i}} & -\kappa {\rm{i}} & 0\end{array}\right).\end{array}\end{eqnarray}$
This superalgebra is composed of even matrices A that satisfy str(A) = 0, AstG + GA = 0 and ΓA†Γ = A. Here $\begin{eqnarray}G=\left(\begin{array}{rcl}0 & -1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 1\end{array}\right),\,\,\,\,{\rm{\Gamma }}=\left(\begin{array}{rcl}1 & 0 & 0\\ 0 & -1 & 0\\ 0 & 0 & \kappa \end{array}\right),\end{eqnarray}$
str(A) is the supertrace of A, Ast is the supertranspose of A and A† is the Hermitian conjugate. The commutation relations of {Ei} satisfy $\begin{array}{l} {\left[E_{1}, E_{2}\right]=-\mathrm{i} E_{3},\left[E_{1}, E_{3}\right]=-\mathrm{i} E_{2},\left[E_{2}, E_{3}\right]=\mathrm{i} E_{1},} \\ {\left[E_{1}, E_{4}\right]=\frac{\kappa \mathrm{i}}{2} E_{4},\left[E_{1}, E_{5}\right]=-\frac{\kappa \mathrm{i}}{2} E_{5},\left[E_{2}, E_{4}\right]=\frac{\kappa \mathrm{i}}{2} E_{5},} \\ {\left[E_{2}, E_{5}\right]=\frac{\kappa \mathrm{i}}{2} E_{4},\left[E_{3}, E_{4}\right]=\frac{\mathrm{i}}{2} E_{5},\left[E_{3}, E_{5}\right]=-\frac{\mathrm{i}}{2} E_{4},} \\ \left\{E_{4}, E_{4}\right\}=E_{2}-\kappa E_{3},\left\{E_{4}, E_{5}\right\}=-E_{1}, \\ \left\{E_{5}, E_{5}\right\}=-E_{2}-\kappa E_{3} . \end{array}$
The superspin matrix S satisfies the constraint S3 = S, for S ∈ OSPU(1, 1/1)/U(1).The Lax pair of equation (2.1 ) [23, 27] is given by 2.1 ) is gauge equivalent to the following super NLS equation:
$\begin{eqnarray}U={\rm{i}}\lambda S,\quad V={\rm{i}}{\lambda }^{2}S+\lambda [S,{S}_{x}]-\frac{3}{2}S{S}_{x}{S}^{2}.\end{eqnarray}$
By using gauge transformation, the HS model ( $\begin{eqnarray}\begin{array}{l}{\rm{i}}{\varphi }_{t}+\frac{1}{2}{\varphi }_{xx}-(| \varphi {| }^{2}-\rho +2\kappa \bar{\psi }\psi )\varphi +2{\rm{i}}\psi {\psi }_{x}=0,\\ {\rm{i}}{\psi }_{t}+{\psi }_{xx}-\frac{1}{2}(| \varphi {| }^{2}-\rho )\psi +{\rm{i}}\kappa (\varphi {\bar{\psi }}_{x}+\frac{1}{2}{\varphi }_{x}\bar{\psi })=0,\end{array}\end{eqnarray}$
where φ(x, t) is a bosonic variable, ψ(x, t) is a fermionic variable and ρ is a quantity which does not depend on x. κ characterizes the interaction of the bosonic field φ and the fermionic field ψ; it shows repulsion with κ = 1 and attraction with κ = -1.Let us consider the generalized inhomogeneous HS model in (1+1) dimensions with the constraint S3 = S.
It is easy to check that the superspin function S with the constraint S3 = S satisfies the relations
$\begin{eqnarray}\begin{array}{l}{S}^{2}{S}_{t}+S{S}_{t}S+{S}_{t}{S}^{2}={S}_{t},\\ {S}^{2}([S,{S}_{xx}]-\frac{3}{2}{(S{S}_{x}{S}^{2})}_{x})\\ \quad \quad +\,S([S,{S}_{xx}]-\frac{3}{2}{(S{S}_{x}{S}^{2})}_{x})S+([S,{S}_{xx}]\\ \quad \quad -\,\frac{3}{2}{(S{S}_{x}{S}^{2})}_{x}){S}^{2}\\ \quad \quad =\,[S,{S}_{xx}]-\frac{3}{2}{(S{S}_{x}{S}^{2})}_{x}.\end{array}\end{eqnarray}$
Based on the form in [25], suppose the inhomogeneous HS model with the cubic constraint is of the following form:
$\begin{eqnarray}{\rm{i}}{S}_{t}=f([S,{S}_{xx}]-\frac{3}{2}{(S{S}_{x}{S}^{2})}_{x})+g[S,{S}_{x}]+uS{S}_{x}{S}^{2}+w{S}_{x},\end{eqnarray}$
where f is a function of x and t, g, u and w need to be determined later.Then we derive
$\begin{eqnarray}\begin{array}{l}{S}^{2}[S,{S}_{x}]+S[S,{S}_{x}]S+[S,{S}_{x}]{S}^{2}=[S,{S}_{x}],\\ {S}^{2}(S{S}_{x}{S}^{2})+S(S{S}_{x}{S}^{2})S+(S{S}_{x}{S}^{2}){S}^{2}=S{S}_{x}{S}^{2},\\ {S}^{2}{S}_{x}+S{S}_{x}S+{S}_{x}{S}^{2}={S}_{x}.\end{array}\end{eqnarray}$
We assume the Lax pair of equation (2.9 ) has the following form:2.11 ) into the zero-curvature condition equation
$\begin{eqnarray}U={\rm{i}}\lambda S,V={\rm{i}}{\lambda }^{2}fS+\lambda f([S,{S}_{x}]-\frac{3}{2}S{S}_{x}{S}^{2})+v(S,{S}_{x}).\end{eqnarray}$
Substituting equation ( $\begin{eqnarray}{U}_{t}-{V}_{x}+[U,V]=0,\end{eqnarray}$
we obtain v(S, Sx) = -ihSx and the inhomogeneous HS model can be expressed as $\begin{eqnarray}\begin{array}{rcl}{\rm{i}}{S}_{t} & = & f([S,{S}_{xx}]-\frac{3}{2}{(S{S}_{x}{S}^{2})}_{x})+{f}_{x}[S,{S}_{x}]\\ & & -\,\frac{3}{2}{f}_{x}S{S}_{x}{S}^{2}-{\rm{i}}h{S}_{x},\end{array}\end{eqnarray}$
where h is a function of x and t, and λ is the spectral parameter satisfing $\begin{eqnarray}{\lambda }_{t}=2{\lambda }^{2}{f}_{x}-\lambda {h}_{x}.\end{eqnarray}$
Gauge transformations play a significant role in integrable systems. By studying the gauge equivalent equations of integrable systems, we can indirectly understand the structure and properties of integrable systems. To achieve a better understanding of the generalized inhomogeneous HS model (2.13 ), now we consider its gauge equivalent counterpart.
Let us take
$\begin{eqnarray}S(x,t)={g}^{-1}\left(x,t\right){\rm{\Sigma }}g\left(x,t\right),\end{eqnarray}$
where g(x, t) ∈ OSPU(1, 1/1), Σ = diag(1, -1, 0).Introducing the currents
$\begin{eqnarray}{J}_{1}={g}_{x}{g}^{-1},\,\,{J}_{0}={g}_{t}{g}^{-1},\end{eqnarray}$
J0 and J1 satisfy the following equation: $\begin{eqnarray}{\partial }_{t}{J}_{1}-{\partial }_{x}{J}_{0}+[{J}_{1},{J}_{0}]=0.\end{eqnarray}$
We set
$\begin{eqnarray}{J}_{1}={\rm{i}}\left(\begin{array}{ccc}0 & \bar{\varphi } & \kappa \bar{\psi }\\ -\varphi & 0 & -\psi \\ -\psi & -\kappa \bar{\psi } & 0\end{array}\right),\end{eqnarray}$
where φ is a boson variable and ψ is a fermionic variable. Here κ = ±1.From equations (2.15 ) and (2.16 ), we have 2.19 ), we have
$\begin{eqnarray}{\partial }_{\mu }S={g}^{-1}[{\rm{\Sigma }},{J}_{\mu }]g.\end{eqnarray}$
In terms of ( $\begin{eqnarray}{S}_{x}={\rm{i}}{g}^{-1}\left(\begin{array}{rcl}0 & 2\bar{\varphi } & \kappa \bar{\psi }\\ 2\varphi & 0 & \psi \\ \psi & -\kappa \bar{\psi } & 0\end{array}\right)g.\end{eqnarray}$
By means of the gauge transformation, U and V in equation (2.11 ) lead to $\hat{U}$ and $\hat{V}$, respectively:
$\begin{eqnarray}\begin{array}{rcl}\hat{U} & = & gU{g}^{-1}+{g}_{x}{g}^{-1}\\ & = & {\rm{i}}\lambda {\rm{\Sigma }}+{J}_{1},\\ \hat{V} & = & gV{g}^{-1}+{g}_{t}{g}^{-1}\\ & = & {\rm{i}}{\lambda }^{2}f{\rm{\Sigma }}+\lambda g(f[S,{S}_{x}]-\frac{3}{2}fS{S}_{x}{S}^{2}){g}^{-1}-{\rm{i}}\lambda h{\rm{\Sigma }}+{J}_{0}\\ & = & {\rm{i}}{\lambda }^{2}f{\rm{\Sigma }}+\lambda f{J}_{1}-{\rm{i}}\lambda h{\rm{\Sigma }}+{J}_{0}\\ & = & \lambda f\hat{U}-{\rm{i}}\lambda h{\rm{\Sigma }}+{J}_{0}.\end{array}\end{eqnarray}$
Substituting equations (2.18 ), (2.20 ) and (2.21 ) into
$\begin{eqnarray}{\hat{U}}_{t}-{\hat{V}}_{x}+[\hat{U},\hat{V}]=0,\end{eqnarray}$
we obtain $\begin{eqnarray}{\rm{i}}[{\rm{\Sigma }},{J}_{0}]={(f{J}_{1})}_{x}-{\rm{i}}h[{\rm{\Sigma }},{J}_{1}].\end{eqnarray}$
Denoting the elements of matrix J0 as ${({J}_{0})}_{ij},(i,j\,=1,2,3)$, we have 2.17 ), we get the diagonal elements of J0
$\begin{eqnarray}\begin{array}{l}\left(\begin{array}{ccc}0 & 2{({J}_{0})}_{12} & {({J}_{0})}_{13}\\ -2{({J}_{0})}_{21} & 0 & -{({J}_{0})}_{23}\\ -{({J}_{0})}_{31} & {({J}_{0})}_{32} & 0\end{array}\right)\\ \,=\,\left(\begin{array}{ccc}0 & {(f\bar{\varphi })}_{x}-2{\rm{i}}h\bar{\varphi } & \kappa {(f\bar{\psi })}_{x}-{\rm{i}}h\kappa \bar{\psi }\\ -{(f\varphi )}_{x}-2{\rm{i}}h\varphi & 0 & -{(f\psi )}_{x}-{\rm{i}}h\psi \\ -{(f\psi )}_{x}-{\rm{i}}h\psi & -\kappa {(f\bar{\psi })}_{x}+{\rm{i}}h\kappa \bar{\psi } & 0\end{array}\right).\end{array}\end{eqnarray}$
Denote J0 as $\begin{eqnarray}{J}_{0}=\left(\begin{array}{ccc}{({J}_{0})}_{11} & \frac{1}{2}{(f\bar{\varphi })}_{x}-{\rm{i}}h\bar{\varphi } & \kappa {(f\bar{\psi })}_{x}-{\rm{i}}h\kappa \bar{\psi }\\ \frac{1}{2}{(f\varphi )}_{x}+{\rm{i}}h\varphi & {({J}_{0})}_{22} & {(f\psi )}_{x}+{\rm{i}}h\psi \\ {(f\psi )}_{x}+{\rm{i}}h\psi & -\kappa {(f\bar{\psi })}_{x}+{\rm{i}}h\kappa \bar{\psi } & {({J}_{0})}_{33}\\ \end{array}\right).\end{eqnarray}$
Using ( $\begin{eqnarray}\begin{array}{rcl}{({J}_{0})}_{11} & = & \displaystyle \frac{{\rm{i}}}{2}(f| \varphi {| }^{2}-\rho +2\kappa f\bar{\psi }\psi \\ & & +\,{\int }_{-\infty }^{x}{f}_{x^{\prime} }(\bar{\varphi }\varphi +\bar{\psi }\psi )\,{\rm{d}}x^{\prime} ),\\ {({J}_{0})}_{22} & = & -\displaystyle \frac{{\rm{i}}}{2}(f| \varphi {| }^{2}-\rho +2\kappa f\bar{\psi }\psi \\ & & +\,{\int }_{-\infty }^{x}{f}_{x^{\prime} }(\bar{\varphi }\varphi +\bar{\psi }\psi )\,{\rm{d}}x^{\prime} ),{({J}_{0})}_{33}=0.\end{array}\end{eqnarray}$
Then the Lax representation of the linear problem is given by $\begin{eqnarray}\begin{array}{l}\hat{U}={\rm{i}}\left(\begin{array}{ccc}\lambda & \bar{\varphi } & \kappa \bar{\psi }\\ -\varphi & -\lambda & -\psi \\ -\psi & -\kappa \bar{\psi } & 0\end{array}\right),\\ \hat{V}=\left(\begin{array}{ccc}{\hat{V}}_{11} & \frac{1}{2}{(f\bar{\varphi })}_{x}+{\rm{i}}(\lambda f-h)\bar{\varphi } & \kappa {(f\bar{\psi })}_{x}+{\rm{i}}(\lambda f-h)\kappa \bar{\psi }\\ \frac{1}{2}{(f\varphi )}_{x}-{\rm{i}}(\lambda f-h)\varphi & {\hat{V}}_{22} & {(f\psi )}_{x}-{\rm{i}}(\lambda f-h)\psi \\ {(f\psi )}_{x}-{\rm{i}}(\lambda f-h)\psi & -\kappa {(f\bar{\psi })}_{x}-{\rm{i}}(\lambda f-h)\kappa \bar{\psi } & 0\\ \end{array}\right),\end{array}\end{eqnarray}$
where $\begin{eqnarray}\begin{array}{rcl}{\hat{V}}_{11} & = & {\rm{i}}[{\lambda }^{2}f-\lambda h+\displaystyle \frac{1}{2}(f| \varphi {| }^{2}-\rho +2\kappa f\bar{\psi }\psi \\ & & +\,{\int }_{-\infty }^{x}{f}_{x^{\prime} }(\bar{\varphi }\varphi +\bar{\psi }\psi )\,{\rm{d}}x^{\prime} )],\\ {\hat{V}}_{22} & = & -{\rm{i}}[{\lambda }^{2}f-\lambda h+\displaystyle \frac{1}{2}(f| \varphi {| }^{2}-\rho +2\kappa f\bar{\psi }\psi \\ & & +\,{\int }_{-\infty }^{x}{f}_{x^{\prime} }(\bar{\varphi }\varphi +\bar{\psi }\psi )\,{\rm{d}}x^{\prime} )],\end{array}\end{eqnarray}$
which leads to the inhomogeneous super NLS model $\begin{eqnarray}\begin{array}{l}{\rm{i}}{\varphi }_{t}+\displaystyle \frac{1}{2}{(f\varphi )}_{xx}+{\rm{i}}{(h\varphi )}_{x}-[f| \varphi {| }^{2}-\rho +2\kappa f\bar{\psi }\psi \\ \,+\,{\int }_{-\infty }^{x}{f}_{x^{\prime} }(\bar{\varphi }\varphi +\bar{\psi }\psi )\,{\rm{d}}x^{\prime} ]\varphi +2{\rm{i}}\psi {(f\psi )}_{x}=0,\\ {\rm{i}}{\psi }_{t}+{(f\psi )}_{xx}+{\rm{i}}{(h\psi )}_{x}-\displaystyle \frac{1}{2}[f| \varphi {| }^{2}-\rho \\ \,+\,{\int }_{-\infty }^{x}{f}_{x^{\prime} }(\bar{\varphi }\varphi +\bar{\psi }\psi )\,{\rm{d}}x^{\prime} ]\psi \\ \,+\,{\rm{i}}\kappa [\varphi {(f\bar{\psi })}_{x}+\displaystyle \frac{1}{2}{(f\varphi )}_{x}\bar{\psi }]=0.\end{array}\end{eqnarray}$
It should be pointed out that equation (
3. The Heisenberg supermagnet model in (2+1) dimensions
Now we consider a (2+1)-dimensional HS model with S3 = S,3.1 ) admits the following Lax representations:
$\begin{eqnarray}{\rm{i}}{S}_{t}=[S,{S}_{xy}]-\frac{3}{2}{(S{S}_{y}{S}^{2})}_{x}+E,\end{eqnarray}$
where the function E needs to be determined later. The linear problem of ( $\begin{eqnarray}{\phi }_{x}=F\phi ,\,\,\,\,{\phi }_{t}=G\phi +\lambda {\phi }_{y},\end{eqnarray}$
where $\phi ={({\phi }_{1},{\phi }_{2},{\phi }_{3})}^{{\rm{T}}},\,{\phi }_{j},\,j=1,2$ and Φ3 are the Grassman even and odd functions, respectively.The compatibility condition Φxt = Φtx of (3.2 ) leads to
$\begin{eqnarray}{F}_{t}-{G}_{x}+[F,G]-\lambda {F}_{y}=0.\end{eqnarray}$
Let us take
$\begin{eqnarray}F={\rm{i}}\lambda S.\end{eqnarray}$
Note that the superspin variable S under the constraint S3 = S satisfies the following relation: $\begin{eqnarray}[S,{S}^{2}u+SuS+u{S}^{2}-u]=[S,{S}^{2}u{S}^{2}+SuS]=0,\end{eqnarray}$
where u is an auxiliary matrix variable which depends on the superspin S and its derivative with respect to the x and y variables, respectively. Thus we may take G as $\begin{eqnarray}\begin{array}{rcl}G & = & \lambda [S,{S}_{y}]-\frac{3}{2}\lambda S{S}_{y}{S}^{2}\\ & & +\,\lambda ({S}^{2}u+SuS+u{S}^{2}-u+{S}^{2}u{S}^{2}+SuS).\end{array}\end{eqnarray}$
Substituting (3.4 ) and (3.6 ) into (3.3 ), we obtain 3.1 ) can be expressed by
$\begin{eqnarray}\begin{array}{l}{\rm{i}}{\lambda }^{2}:[S,[S,{S}_{y}]-\frac{3}{2}S{S}_{y}{S}^{2}]-{S}_{y}\\ \,=\,{S}^{2}{S}_{y}-2S{S}_{y}S+{S}_{y}{S}^{2}+3S{S}_{y}S-{S}_{y}=0,\\ \lambda :{\rm{i}}{S}_{t}-\{[S,{S}_{y}]-\frac{3}{2}S{S}_{y}{S}^{2}\\ \quad +\,({S}^{2}u+SuS+u{S}^{2}-u+{S}^{2}u{S}^{2}+SuS){\}}_{x}\\ \,=\,{\rm{i}}{S}_{t}-[{S}_{x},{S}_{y}]-[S,{S}_{xy}]+\frac{3}{2}{(S{S}_{y}{S}^{2})}_{x}\\ \quad -\,{({S}^{2}u+SuS+u{S}^{2}-u+{S}^{2}u{S}^{2}+SuS)}_{x}=0.\end{array}\end{eqnarray}$
Thus E in equation ( $\begin{eqnarray}E=[{S}_{x},{S}_{y}]+{({S}^{2}u+SuS+u{S}^{2}-u+{S}^{2}u{S}^{2}+SuS)}_{x},\end{eqnarray}$
and the parameter λ satisfies $\begin{eqnarray}{\lambda }_{t}=\lambda {\lambda }_{y}.\end{eqnarray}$
Under the constraint S3 = S, we note that St satisfies S2St + SStS + StS2 = St and [S, Sxy] satisfies S2[S, Sxy] + S[S, Sxy]S + [S, Sxy]S2 = [S, Sxy], thus the deformed function E should satisfy the transformation relation
$\begin{eqnarray}{S}^{2}E+SES+E{S}^{2}=E.\end{eqnarray}$
Noticing that SSyS2, [Sx, Sy] and ${({S}^{2}u{S}^{2}+SuS)}_{x}$ do not satisfy the same relation as (3.10 ), after a tedious but straightforward calculation, we obtain a solution
$\begin{eqnarray}\begin{array}{rcl}E & = & \displaystyle \frac{1}{4}({S}^{2}[{S}_{x},{S}_{y}]{S}^{2}+S[{S}_{x},{S}_{y}]S)\\ & & -\,\displaystyle \frac{3}{8}({S}^{2}{S}_{x}S{S}_{y}S-S{S}_{y}S{S}_{x}{S}^{2})+{({S}^{2}u{S}^{2}+SuS)}_{x},\end{array}\end{eqnarray}$
which indicates that the auxiliary matrix u satisfies the following constraint: $\begin{eqnarray}\begin{array}{l}{({S}^{2}u+SuS+u{S}^{2}-u)}_{x}\\ \,=\,-[{S}_{x},{S}_{y}]+\frac{1}{4}({S}^{2}[{S}_{x},{S}_{y}]{S}^{2}+S[{S}_{x},{S}_{y}]S)\\ \quad -\,\displaystyle \frac{3}{8}({S}^{2}{S}_{x}S{S}_{y}S-S{S}_{y}S{S}_{x}{S}^{2}).\end{array}\end{eqnarray}$
Thus a (2 + 1)-dimensional HS model with constraint S3 = S is given by $\begin{eqnarray}\begin{array}{rcl}{\rm{i}}{S}_{t} & = & [S,{S}_{xy}]-\frac{3}{2}{(S{S}_{y}{S}^{2})}_{x}\\ & & +\,\displaystyle \frac{1}{4}({S}^{2}[{S}_{x},{S}_{y}]{S}^{2}+S[{S}_{x},{S}_{y}]S)\\ & & -\,\displaystyle \frac{3}{8}({S}^{2}{S}_{x}S{S}_{y}S-S{S}_{y}S{S}_{x}{S}^{2})\\ & & +\,{({S}^{2}u{S}^{2}+SuS)}_{x}.\end{array}\end{eqnarray}$
When the reduction x = y is imposed, the equation (
Let us turn to construct the gauge equivalent counterpart of equation (3.13 ).
We diagonalize S such that
$\begin{eqnarray}S={g}^{-1}(t,x,y){\rm{\Sigma }}g(t,x,y),\end{eqnarray}$
where g(t, x, y) ∈ OSPU(1, 1/1) and diagonal matrix Σ = diag(1, - 1, 0).Introduce ${{\mathbb{J}}}_{\mu }\,(\mu =0,1,2)$, which denote the currents of t, x and y
$\begin{eqnarray}\begin{array}{l}{{\mathbb{J}}}_{0}=g{(t,x,y)}_{t}g{(t,x,y)}^{-1},\\ {{\mathbb{J}}}_{1}=g{(t,x,y)}_{x}g{(t,x,y)}^{-1},\\ {{\mathbb{J}}}_{2}=g{(t,x,y)}_{y}g{(t,x,y)}^{-1},\end{array}\end{eqnarray}$
and assume the form of ${{\mathbb{J}}}_{1}$ is $\begin{eqnarray}{{\mathbb{J}}}_{1}={\rm{i}}\left(\begin{array}{ccc}0 & \bar{\varphi } & \kappa \bar{\psi }\\ -\varphi & 0 & -\psi \\ -\psi & -\kappa \bar{\psi } & 0\end{array}\right),\end{eqnarray}$
where φ is a bosonic field and ψ is a fermionic field. Here κ = ±1.Based on (3.14 ) and (3.15 ), it follows that3.3 ) lead to $\hat{F}$ and $\hat{G}$, respectively:3.12 ), (3.15 ) and (3.17 ), it is easy to verify that 3.18 )-(3.20 ) into
$\begin{eqnarray}\begin{array}{l}{S}_{x}={g}^{-1}[{\rm{\Sigma }},{{\mathbb{J}}}_{1}]g,\\ {S}_{y}={g}^{-1}[{\rm{\Sigma }},{{\mathbb{J}}}_{2}]g.\end{array}\end{eqnarray}$
By means of the gauge transformation, F and G in ( $\begin{eqnarray}\begin{array}{rcl}\hat{F} & = & gF{g}^{-1}+{g}_{x}{g}^{-1}\\ & = & {\rm{i}}\lambda {\rm{\Sigma }}+{{\mathbb{J}}}_{1},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}\hat{G} & = & gG{g}^{-1}-\lambda {g}_{y}{g}^{-1}+{g}_{t}{g}^{-1}\\ & = & \lambda g([S,{S}_{y}]-\frac{3}{2}S{S}_{y}{S}^{2}\\ & & +\,({S}^{2}u+SuS+u{S}^{2}-u+{S}^{2}u{S}^{2}+SuS)){g}^{-1}\\ & & -\,\lambda {{\mathbb{J}}}_{2}+{{\mathbb{J}}}_{0}.\end{array}\end{eqnarray}$
Using ( $\begin{eqnarray}\begin{array}{rcl}{{\mathbb{J}}}_{2} & = & g([S,{S}_{y}]-\frac{3}{2}S{S}_{y}{S}^{2}\\ & & +\,({S}^{2}u+SuS+u{S}^{2}-u+{S}^{2}u{S}^{2}+SuS)){g}^{-1}.\end{array}\end{eqnarray}$
Substituting equations ( $\begin{eqnarray}{\hat{F}}_{t}-{\hat{G}}_{x}+[\hat{F},\hat{G}]-\lambda {\hat{F}}_{y}=0,\end{eqnarray}$
we obtain $\begin{eqnarray}{\rm{i}}[{\rm{\Sigma }},{{\mathbb{J}}}_{0}]={{\mathbb{J}}}_{1y},\end{eqnarray}$
$\begin{eqnarray}{{\mathbb{J}}}_{1t}-{{\mathbb{J}}}_{0x}+[{{\mathbb{J}}}_{1},{{\mathbb{J}}}_{0}]=0.\end{eqnarray}$
From (3.16 ) and (3.22 ), we get
$\begin{eqnarray}{{\mathbb{J}}}_{0}=\left(\begin{array}{lcc}\frac{{\rm{i}}}{2}{\partial }_{x}^{-1}{\partial }_{y}\{\varphi \bar{\varphi }+2\kappa \bar{\psi }\psi \} & \frac{\bar{{\varphi }_{y}}}{2} & \kappa \bar{{\psi }_{y}}\\ \frac{{\varphi }_{y}}{2} & -\frac{{\rm{i}}}{2}{\partial }_{x}^{-1}{\partial }_{y}\{\varphi \bar{\varphi }+2\kappa \bar{\psi }\psi \} & {\psi }_{y}\\ {\psi }_{y} & -\kappa \bar{{\psi }_{y}} & 0\end{array}\right).\end{eqnarray}$
Thus the (2 + 1)-dimensional super NLS equation is given by $\begin{eqnarray}\begin{array}{l}\displaystyle \frac{{\varphi }_{xy}}{2}+{\rm{i}}{\varphi }_{t}-\varphi {\partial }_{x}^{-1}{\partial }_{y}\{\varphi \bar{\varphi }+2\kappa \bar{\psi }\psi \}+2{\rm{i}}\psi {\psi }_{y}=0,\\ {\psi }_{xy}+{\rm{i}}{\psi }_{t}+{\rm{i}}\kappa (\varphi \bar{{\psi }_{y}}+\frac{1}{2}{\varphi }_{y}\bar{\psi })\\ \,-\,\displaystyle \frac{1}{2}{\partial }_{x}^{-1}{\partial }_{y}\{\varphi \bar{\varphi }+2\kappa \bar{\psi }\psi \}\psi =0.\end{array}\end{eqnarray}$
Equation (
4. Summary and discussion
In this paper, we construct two types of HS model with S3 = S, including the inhomogeneous HS model in (1 + 1) dimensions and the (2 + 1)-dimensional HS model. By virtue of gauge transformations, the corresponding gauge equivalent counterparts involving the (1 + 1)-dimensional inhomogeneous super NLS equation and the (2 + 1)-dimensional super NLS equation are presented. The method applied in this paper is used to construct many generalized higher-order HS models. The existing problem is that the form of the Lax representation is not easy to determine, and the physical significance of the related models still deserves further investigation.
As to the two HS models with S3 = S presented in this study, their applications should be of interest. It should be pointed out that the HS model has a core connection with the strong electron correlated Hubbard model [23]. The investigation of the high-temperature superconductivity phenomenon stimulated the interest of theoreticians in the Hubbard model. How do we investigate the relation between the two new HS models and the Hubbard model? Meanwhile, we will concentrate on other extended HS models with S3 = S in (2 + 1) dimensions, such as the Ishimori equation in the near future. In addition, a deeper exploration of the physical implications of these models presented in this paper, particularly in higher-dimensional contexts, is an interesting question worth discussing.