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Article

Module Algebra Structures of Non-Standard Quantum Group Xq(A1) on C[x,y,z]

School of Mathematics and Statistics, Henan University of Science and Technology, Luoyang 471023, China
Mathematics 2025, 13(14), 2227; https://doi.org/10.3390/math13142227
Submission received: 29 April 2025 / Revised: 5 July 2025 / Accepted: 7 July 2025 / Published: 8 July 2025

Abstract

In this paper, we investigate the module algebra structures of X q ( A 1 ) on the quantum polynomial algebra C q [ x , y , z ] . The automorphism group of C q [ x , y , z ] is denoted by Aut ( C q [ x , y , z ] ) , and we give a complete classification of X q ( A 1 ) -module algebra structures on C q [ x , y , z ] , where K 1 , K 2 Aut ( C q [ x , y , z ] ) .

1. Introduction

Hopf algebras are fundamental structures in algebra that combine the properties of algebras, coalgebras, and antipodes. Hopf algebras play a crucial role in various fields of mathematics and physics, such as invariant theory for knots and links, and representations closely connected to Lie theory, etc. The notion of Hopf algebra actions on algebras was introduced by Sweedler [1] in 1969. The Brauer groups of H-module and H-dimodule algebras were researched by Beattie [2]. A duality theorem for Hopf module algebras was studied by Blattner and Montgomery [3] in 1985. Moreover, the actions of Hopf algebras [4] and their generalizations (see, e.g., [5]) play an important role in quantum group theory [6,7] and in its various applications in physics [8]. Duplij and Sinel’shchikov used a general form of the automorphism of the quantum plane to give the notion of weight for U q ( s l 2 ) -actions and completely classified quantum group U q ( s l 2 ) -module algebra structures on the quantum plane [9,10], demonstrating that the results are much richer and consist of six nonisomorphic cases. Moreover, in [11] the authors used the method of weights [9,10] to classify actions in terms of action matrices, studied the module algebra structures of U q ( s l ( m + 1 ) ) on the coordinate algebra of the quantum vector spaces, and researched the module algebra structures of the quantum group U q ( s l 2 ) on C q [ x , y , z ] [11,12]. More relevant research can be found in [13,14].
The non-standard quantum groups were studied in Ge et al. [15], who obtained new solutions of Yang–Baxter equations and included the twisted extension quantum group structures related to these new solutions explicitly. In [16] one class of non-standard quantum deformation corresponding to simple Lie algebra s l n was given, which is denoted by X q ( A n 1 ) . For each vertex i ( i = 1 , , n 1 ) of the Dynkin diagram, the parameter q i is equal to q or q 1 ; if q i = q for all i, then X q ( A n 1 ) is just U q ( s l n ) . However, if q i q i + 1 for some 1 i n 1 , it has the relations E i 2 = F i 2 = 0 in X q ( A n 1 ) . Such a X q ( A n 1 ) is different to U q ( s l n ) . Jing et al. [17] derived a non-standard quantum group by employing the FRT constructive method and classified all finite dimensional irreducible representations of this non-standard quantum group. Cheng and Yang [18] considered the structures and representations of weak Hopf algebras w X q ( A 1 ) , which correspond to the non-standard quantum group X q ( A 1 ) . We [19] researched the representations of a class of small nonstandard quantum groups X ¯ q ( A 1 ) , over which the isomorphism classes of all indecomposable modules were classified and the decomposition formulas of the tensor product of arbitrary indecomposable modules and simple (or projective) modules were established. The projective class rings and Grothendieck rings of X ¯ q ( A 1 ) were also characterized. However, there are few research results on the module algebra of non-standard quantum groups. We [20] discussed the module algebra structures of the non-standard quantum group X q ( A 1 ) on the quantum plane. But the automorphism group of C q [ x , y , z ] is more complex, and the module algebra structures of the non-standard quantum group X q ( A 1 ) on the quantum polynomial algebra C q [ x , y , z ] are also more complex. Consequently, based on the research results of the module algebra of quantum groups, we consider the module algebra of the non-standard quantum group X q ( A 1 ) on the quantum polynomial algebra C q [ x , y , z ] . A complete list of X q ( A 1 ) -module algebra structures on C q [ x , y , z ] is produced and the isomorphism classes of these structures are described. Our research offer advantages through contributing to the classification of module algebra structures of X q ( A n ) on C q [ x , y , z ] for n 2 .
This paper is organized as follows. In Section 1, we introduce some necessary notations and concepts. In Section 2, for t = 0 , we discuss the module algebra structures of X q ( A 1 ) on the polynomial algebra C q [ x , y , z ] using the method of weights [9,10]. We study the concrete actions of X q ( A 1 ) on C q [ x , y , z ] and characterize all module algebra structures of X q ( A 1 ) on C q [ x , y , z ] . We present our classification in terms of a pair of symbolic matrices, which relies upon considering the low-dimensional (0-th and 1-st) homogeneous components of an action. In Section 3, we study the module algebra structures of X q ( A 1 ) on C q [ x , y , z ] for t 0 . In the same way as in Section 2, we study the concrete actions of X q ( A 1 ) on C q [ x , y , z ] and characterize all module algebra structures of X q ( A 1 ) on C q [ x , y , z ] .

2. Preliminaries

Throughout, we work over the complex field C unless otherwise stated. All algebras, Hopf algebras, and modules are defined over C ; all maps are C -linear.
Let ( H , m , η , Δ , ε , S ) be a Hopf algebra, where Δ and ε are the comultiplication and counit of H, respectively. Let A be a unital algebra with unit 1 . We will also use the Sweedler notation Δ ( h ) = i π h i h i [1].
Definition 1.
By a structure of H-module algebra on A, we mean a homomorphism π : H End C A such that the following hold:
  • For all h H , a , b A , π ( h ) ( a b ) = i π ( h i ) ( a ) · π ( h i ) ( b ) ;
  • For all h H , π ( h ) ( 1 ) = ε ( h ) 1 .
Let π 1 and π 2 be two H-module algebras on A; the structures π 1 , π 2 are said to be isomorphic if there exists an automorphism Ψ of the algebra A such that Ψ π 1 ( h ) Ψ 1 = π 2 ( h ) for all h H .
We assume that q C * = C { 0 } is not a root of the unit ( q n 1 for all non-zero integers n). A class of non-standard quantum algebra X q ( A 1 ) was studied by Jing etc. [17]. By definition the algebra X q ( A 1 ) is a unital associative C -algebra generated by E , F , K i , K i 1 ( i = 1 , 2 ) , and subject to the following relations:
K i K i 1 = K i 1 K i = 1 , K 1 K 2 = K 2 K 1 ,
K 1 E = q 1 E K 1 ,
K 1 F = q F K 1 ,
K 2 E = q 1 E K 2 ,
K 2 F = q F K 2 ,
E F F E = K 2 K 1 1 K 2 1 K 1 q q 1 ,
E 2 = F 2 = 0 .
The algebra X q ( A 1 ) is also a Hopf algebra, and the comultiplication Δ , counit ε , and antipode S are given as the following:
Δ ( K i ) = K i K i ,
Δ ( E ) = E 1 + K 2 K 1 1 E ,
Δ ( F ) = 1 F + F K 2 1 K 1 ,
ε ( K i ) = 1 , ε ( E ) = 0 , ε ( F ) = 0 , S ( K i ) = K i 1 , S ( E ) = K 1 K 2 1 E , S ( F ) = F K 2 K 1 1 .
The quantum polynomial algebra C q [ x , y , z ] is a unital algebra, generated by generators x , y , z and satisfying the following relations:
y x = q x y ,
z y = q y z ,
z x = q x z .
Denote by C q [ x , y , z ] s the s-th homogeneous component of C q [ x , y , z ] , which is a linear span of the monomials x m 1 y m 2 z m 3 with m 1 + m 2 + m 3 = s . For a given polynomial p C q [ x , y , z ] , denote by p s the s-th homogeneous component of p, that is the projection of p onto C q [ x , y , z ] s parallel to the direct sum of all other homogeneous components of C q [ x , y , z ] .
By [21,22,23,24], one has a description of automorphisms of the algebra C q [ x , y , z ] , as follows. Let Ψ be an automorphism of C q [ x , y , z ] ; then there exist nonzero constants α , β , γ C * and t C , such that
Ψ : x α x , y β y + t x z , z γ z .
For all such automorphisms form the automorphism group of C q [ x , y , z ] denoted by Aut ( C q [ x , y , z ] ) , one can get Aut ( C q [ x , y , z ] ) C C * 3 . In the following sections, we will explore the classificaion of X q ( A 1 ) -module algebra structures on C q [ x , y , z ] .

3. Classification of X q ( A 1 ) -Module Algebra Structures on C q [ x , y , z ] for T = 0

In this section, our aim is to describe the X q ( A 1 ) -module algebra structures on C q [ x , y , z ] for t = 0 , i.e., the automorphism of C q [ x , y , z ] , as follows:
Ψ ( x ) = α x , Ψ ( y ) = β y , Ψ ( z ) = γ z , ( α , β , γ C * ) ,
and Aut ( C q [ x , y , z ] ) C * 3 , where K 1 , K 2 Aut ( C q [ x , y , z ] ) .

3.1. Properties of X q ( A 1 ) -Module Algebras on C q [ x , y , z ]

By the definition of module algebra, it is easy to see that any action of X q ( A 1 ) on C q [ x , y , z ] is determined by the following 4 × 3 matrix with entries from C q [ x , y , z ] :
M = d e f i n i t i o n K 1 ( x ) K 1 ( y ) K 1 ( z ) K 2 ( x ) K 2 ( y ) K 2 ( z ) E ( x ) E ( y ) E ( z ) F ( x ) F ( y ) F ( z ) ,
which is called the full action matrix. Given a X q ( A 1 ) -module algebra structure on C q [ x , y , z ] , obviously the action of K 1 (or K 2 ) is determined by an automorphism of C q [ x , y , z ] ; in other words, the actions of K 1 and K 2 are determined by a matrix M K 1 K 2 as follows:
M K 1 K 2 = d e f i n i t i o n K 1 ( x ) K 1 ( y ) K 1 ( z ) K 2 ( x ) K 2 ( y ) K 2 ( z ) = α 1 ( x ) β 1 ( y ) γ 1 ( z ) α 2 ( x ) β 2 ( y ) γ 2 ( z ) ,
where α i , β i C * for i { 1 , 2 } . It is easy to see that every monomial x m 1 y m 2 z m 3 C q [ x , y , z ] is an eigenvector of K 1 (or K 2 ), and the associated eigenvalue α 1 m 1 β 1 m 2 γ 1 m 3 (or α 2 m 1 β 2 m 2 γ 2 m 3 ) is called the K 1 -weight (or K 2 -weight) of this monomial, which will be written as
w t K 1 ( x m 1 y m 2 z m 3 ) = α 1 m 1 β 1 m 2 γ 1 m 3 ,
w t K 2 ( x m 1 y m 2 z m 3 ) = α 2 m 1 β 2 m 2 γ 2 m 3 .
We will also need another matrix M E F as follows:
M E F = d e f i n i t i o n E ( x ) E ( y ) E ( z ) F ( x ) F ( y ) F ( z ) .
Obviously, all entries of M are weight vectors for K 1 and K 2 ; then
w t K i M = d e f i n i t i o n w t K i ( K 1 ( x ) ) w t K i ( K 1 ( y ) ) w t K i ( K 1 ( z ) ) w t K i ( K 2 ( x ) ) w t K i ( K 2 ( y ) ) w t K i ( K 2 ( z ) ) w t K i ( E ( x ) ) w t K i ( E ( y ) ) w t K i ( E ( z ) ) w t K i ( F ( x ) ) w t K i ( F ( y ) ) w t K i ( F ( z ) ) w t K i ( x ) w t K i ( y ) w t K i ( z ) w t K i ( x ) w t K i ( y ) w t K i ( z ) ( 1 ) i 1 q 1 w t K i ( x ) ( 1 ) i 1 q 1 w t K i ( y ) ( 1 ) i 1 q 1 w t K i ( z ) ( 1 ) i 1 q w t K i ( x ) ( 1 ) i 1 q w t K i ( y ) ( 1 ) i 1 q w t K i ( z ) = α i β i γ i α i β i γ i ( 1 ) i 1 q 1 α i ( 1 ) i 1 q 1 β i ( 1 ) i 1 q 1 γ i ( 1 ) i 1 q α i ( 1 ) i 1 q β i ( 1 ) i 1 q γ i ,
where the relation A = a s t B = b s t means that for every pair of indices s , t such that both a s t and b s t are nonzero, one has a s t = b s t .
We denote by M j the j-th homogeneous component of M, whose elements are just the j-th homogeneous components of the corresponding entries of M. Set
M 0 = 0 0 0 0 0 0 a 0 b 0 c 0 a 0 b 0 c 0 0 ,
where a 0 , b 0 , c 0 , a 0 , b 0 , c 0 C . Then, we obtain
w t K 1 M E F 0 q 1 α 1 q 1 β 1 q 1 γ 1 q α 1 q β 1 q γ 1 0 1 1 1 1 1 1 0 ,
w t K 2 M E F 0 q 1 α 2 q 1 β 2 q 1 γ 2 q α 2 q β 2 q γ 2 0 1 1 1 1 1 1 0 .
As q is not a root of the unit and relations (19) and (20), it means that each column of M E F should contain at least one 0.
Applying E and F to the relations (11)–(13) by using Equation (15), one has
E ( y ) x + β 1 1 β 2 y E ( x ) = q E ( x ) y + q α 1 1 α 2 x E ( y ) ,
E ( z ) y + γ 1 1 γ 2 z E ( y ) = q E ( y ) z + q β 1 1 β 2 y E ( z ) ,
E ( z ) x + γ 1 1 γ 2 z E ( x ) = q E ( x ) z + q α 1 1 α 2 x E ( z ) ,
y F ( x ) + α 2 1 α 1 F ( y ) x = q x F ( y ) + q β 2 1 β 1 F ( x ) y ,
z F ( y ) + β 2 1 β 1 F ( z ) y = q y F ( z ) + q γ 2 1 γ 1 F ( y ) z ,
z F ( x ) + α 2 1 α 1 F ( z ) x = q x F ( z ) + q γ 2 1 γ 1 F ( x ) z .
After projecting Equations (21)–(26) to C q [ x , y , z ] 1 , we obtain
b 0 1 q α 1 1 α 2 x + a 0 β 1 1 β 2 q y = 0 , c 0 1 q β 1 1 β 2 y + b 0 γ 1 1 γ 2 q z = 0 , c 0 1 q α 1 1 α 2 x + a 0 γ 1 1 γ 2 q z = 0 , a 0 1 q β 1 β 2 1 y + b 0 α 1 α 2 1 q x = 0 , b 0 1 q γ 1 γ 2 1 z + c 0 β 1 β 2 1 q y = 0 , a 0 1 q γ 1 γ 2 1 z + c 0 α 1 α 2 1 q x = 0 ,
which certainly implies
b 0 1 q α 1 1 α 2 = a 0 β 1 1 β 2 q = c 0 1 q β 1 1 β 2 = b 0 γ 1 1 γ 2 q = 0 , c 0 1 q α 1 1 α 2 = a 0 γ 1 1 γ 2 q = a 0 1 q β 1 β 2 1 = b 0 α 1 α 2 1 q = 0 , b 0 1 q γ 1 γ 2 1 = c 0 β 1 β 2 1 q = a 0 1 q γ 1 γ 2 1 = c 0 α 1 α 2 1 q = 0 .
We will determine the weight constants α i , β i and γ i ( i = 1 , 2 ) as follows:
a 0 0 α 1 = q , α 2 = q , β 1 1 β 2 = q , γ 1 1 γ 2 = q ; b 0 0 β 1 = q , β 2 = q , α 1 1 α 2 = q 1 , γ 1 1 γ 2 = q ; c 0 0 γ 1 = q , γ 2 = q , β 1 1 β 2 = q 1 , α 1 1 α 2 = q 1 ; a 0 0 α 1 = q 1 , α 2 = q 1 , β 1 β 2 1 = q 1 , γ 1 γ 2 1 = q 1 ; b 0 0 β 1 = q 1 , β 2 = q 1 , α 1 α 2 1 = q , γ 1 γ 2 1 = q 1 ; c 0 0 γ 1 = q 1 , γ 2 = q 1 , β 1 β 2 1 = q , α 1 α 2 1 = q .
Because q is not a root of the unit, q ± 1 . Therefore at least one of a 0 , b 0 , c 0 and a 0 , b 0 , c 0 is not zero. In summary, we have obtained the following results for the 0-th homogeneous component M E F 0 of M E F .
Lemma 1.
There are 7 cases for the 0-st homogeneous component M E F 0 of M E F , as follows:
a 0 0 0 0 0 0 0 α 1 = q , α 2 = q , β 1 1 β 2 = q , γ 1 1 γ 2 = q ;
0 b 0 0 0 0 0 0 β 1 = q , β 2 = q , α 1 1 α 2 = q 1 , γ 1 1 γ 2 = q ;
0 0 c 0 0 0 0 0 γ 1 = q , γ 2 = q , β 1 1 β 2 = q 1 , α 1 1 α 2 = q 1 ;
0 0 0 a 0 0 0 0 α 1 = q 1 , α 2 = q 1 , β 1 β 2 1 = q 1 , γ 1 γ 2 1 = q 1 ;
0 0 0 0 b 0 0 0 β 1 = q 1 , β 2 = q 1 , α 1 α 2 1 = q , γ 1 γ 2 1 = q 1 ;
0 0 0 0 0 c 0 0 , γ 1 = q 1 , γ 2 = q 1 , β 1 β 2 1 = q , α 1 α 2 1 = q ;
0 0 0 0 0 0 0 it does not determine the weight constants at all .
Next, for the 1-st homogeneous component, as q is not a root of the unit, one has
w t K 1 ( E ( x ) ) = q 1 α 1 = q 1 w t K 1 ( x ) w t K 1 ( x ) , w t K 2 ( E ( x ) ) = q 1 α 2 = q 1 w t K 2 ( x ) w t K 2 ( x ) ,
which implies
( E ( x ) ) 1 = a 1 y + a 2 z ,
for some a 1 , a 2 C , and in a similar way we have
M E F 1 = a 1 y + a 2 z b 1 x + b 2 z c 1 x + c 2 y a 1 y + a 2 z b 1 x + b 2 z c 1 x + c 2 y 1
where b 1 , b 2 , c 1 , c 2 , a 1 , a 2 , b 1 , b 2 , c 1 , c 2 C . Therefore
w t K i ( M E F ) 1 ( 1 ) i 1 q 1 α i ( 1 ) i 1 q 1 β i ( 1 ) i 1 q 1 γ i ( 1 ) i 1 q α i ( 1 ) i 1 q β i ( 1 ) i 1 q γ i β i or γ i α i or γ i α i or β i β i or γ i α i or γ i α i or β i ,
Now project (21)–(26) to C q [ x , y , z ] 2 to obtain
b 1 1 q α 1 1 α 2 x 2 + b 2 1 α 1 1 α 2 z x + a 1 β 1 1 β 2 q y 2 + a 2 β 1 1 β 2 q 2 y z = 0 , c 1 1 q 2 β 1 1 β 2 y x + c 2 1 q β 1 1 β 2 y 2 + b 2 γ 1 1 γ 2 q z 2 + b 1 q γ 1 1 γ 2 1 x z = 0 , c 1 1 q α 1 1 α 2 x 2 + q c 2 1 α 1 1 α 2 x y + q a 1 γ 1 1 γ 2 1 y z + a 2 γ 1 1 γ 2 q z 2 = 0 , a 1 1 q β 2 1 β 1 y 2 + a 2 1 q 2 β 2 1 β 1 y z + b 1 α 2 1 α 1 q x 2 + q b 2 α 2 1 α 1 1 x z = 0 , q b 1 1 γ 2 1 γ 1 x z + b 2 1 q γ 2 1 γ 1 z 2 + c 1 β 2 1 β 1 q 2 x y + c 2 β 2 1 β 1 q y 2 = 0 , q a 1 1 γ 2 1 γ 1 y z + a 2 1 q γ 2 1 γ 1 z 2 + c 1 α 2 1 α 1 q x 2 + q c 2 α 2 1 α 1 1 x y = 0 ,
which certainly implies
b 1 1 q α 1 1 α 2 = b 2 1 α 1 1 α 2 = a 1 β 1 1 β 2 q = a 2 β 1 1 β 2 q 2 = 0 , c 1 1 q 2 β 1 1 β 2 = c 2 1 q β 1 1 β 2 = b 2 γ 1 1 γ 2 q = b 1 q γ 1 1 γ 2 1 = 0 , c 1 1 q α 1 1 α 2 = q c 2 1 α 1 1 α 2 = q a 1 γ 1 1 γ 2 1 = a 2 q γ 1 1 γ 2 q = 0 , a 1 1 q β 1 β 2 1 = a 2 1 q 2 β 1 β 2 1 = b 1 α 1 α 2 1 q = q b 2 α 1 α 2 1 1 = 0 , q b 1 1 γ 1 γ 2 1 = b 2 1 q γ 1 γ 2 1 = c 1 β 1 β 2 1 q 2 = c 2 β 1 β 2 1 q = 0 , q a 1 1 γ 1 γ 2 1 = a 2 1 q γ 1 γ 2 1 = c 1 α 1 α 2 1 q = q c 2 α 1 α 2 1 1 = 0 .
As a consequence, we have
a 1 0 β 2 α 1 1 = q , γ 2 γ 1 1 = 1 , a 2 0 β 2 α 1 1 = q 2 , γ 2 γ 1 1 = q , b 1 0 α 2 α 1 1 = q 1 , γ 2 γ 1 1 = 1 , b 2 0 α 2 α 1 1 = 1 , γ 2 γ 1 1 = q , c 1 0 β 2 α 1 1 = q 2 α 2 α 1 1 = q 1 , c 2 0 β 2 α 1 1 = q 1 α 2 α 1 1 = 1 ,
a 1 0 β 2 α 1 1 = q , γ 2 γ 1 1 = 1 , a 2 0 β 2 α 1 1 = q 2 , γ 2 γ 1 1 = q , b 1 0 α 2 α 1 1 = q 1 , γ 2 γ 1 1 = 1 , b 2 0 α 2 α 1 1 = 1 , γ 2 γ 1 1 = q , c 1 0 β 2 α 1 1 = q 2 α 2 α 1 1 = q 1 , c 2 0 β 2 α 1 1 = q 1 α 2 α 1 1 = 1 .
From the above discussion, for the 1-st homogeneous component M E F 1 of M E F , we have the following lemma.
Lemma 2.
There are 13 cases for the 1-st homogeneous component M E F 1 of M E F , as follows:
a 1 y 0 0 0 0 0 1 β 1 = q 1 α 1 , β 2 = q 1 α 2 , β 1 1 β 2 = q , γ 1 1 γ 2 = 1 ;
a 2 z 0 0 0 0 0 1 γ 1 = q 1 α 1 , γ 2 = q 1 α 2 , β 1 1 β 2 = q 2 , γ 1 1 γ 2 = q ;
0 b 1 x 0 0 0 0 1 β 1 = q α 1 , β 2 = q α 2 , α 1 1 α 2 = q 1 , γ 1 1 γ 2 = 1 ;
0 b 2 z 0 0 0 0 1 γ 1 = q 1 β 1 , γ 2 = q 1 β 2 , α 1 1 α 2 = 1 , γ 1 1 γ 2 = q ;
0 0 c 1 x 0 0 0 1 α 1 = q 1 γ 1 , α 2 = q 1 γ 2 , α 1 1 α 2 = q 1 , β 1 1 β 2 = q 2 ;
0 0 c 2 y 0 0 0 1 β 1 = q 1 γ 1 , β 2 = q 1 γ 2 , α 1 1 α 2 = 1 , β 1 1 β 2 = q 1 ;
0 0 0 a 1 y 0 0 1 β 1 = q α 1 , β 2 = q α 2 , β 1 1 β 2 = q , γ 1 1 γ 2 = 1 ;
0 0 0 a 2 z 0 0 1 γ 1 = q α 1 , γ 2 = q α 2 , β 1 1 β 2 = q 2 , γ 1 1 γ 2 = q ;
0 0 0 0 b 1 x 0 1 β 1 = q 1 α 1 , β 2 = q 1 α 2 , α 1 1 α 2 = q 1 , γ 1 1 γ 2 = 1 ;
0 0 0 0 b 2 z 0 1 γ 1 = q β 1 , γ 2 = q β 2 , α 1 1 α 2 = 1 , γ 1 1 γ 2 = q ;
0 0 0 0 0 c 1 x 1 α 1 = q γ 1 , α 2 = q γ 2 , α 1 1 α 2 = q 1 , β 1 1 β 2 = q 2 ;
0 0 0 0 0 c 2 y 1 β 1 = q γ 1 , β 2 = q γ 2 , α 1 1 α 2 = 1 , β 1 1 β 2 = q 1 ;
0 0 0 0 0 0 1 it does not determine the weight constants at all .

3.2. The Structures of X q ( A 1 ) -Module Algebra on C q [ x , y , z ]

In this subsection, our aim is to describe the concrete X q ( A 1 ) -module algebra structures on C q [ x , y , z ] , where K 1 , K 2 Aut ( C q [ x , y , z ] ) C * 3 .
By Lemmas 1 and 2, and the fact that q is not a root of the unit, it follows that if both the 0-th homogeneous component and the 1-th homogeneous component of M E F are nonzero, it is easy to see that these series are empty. So, we need to consider following possibilities.
Lemma 3.
If the 0-th homogeneous component of M E F is zero and the 1-st homogeneous component of M E F is nonzero, then these series are empty.
Proof. 
Now, we show that the 0 0 0 0 0 0 0 , a 1 y 0 0 0 0 0 1 -series is empty. If we suppose the contrary, then it follows from
E F F E = K 2 K 1 1 K 2 1 K 1 q q 1
that within this series, one can have
K 2 K 1 1 K 2 1 K 1 q q 1 ( x ) = α 2 α 1 1 α 2 1 α 1 q q 1 x .
By a 1 0 , one can get β 1 = q 1 α 1 , β 2 = q 1 α 2 ,   β 2 β 1 1 = q , and γ 2 γ 1 1 = 1 ; hence α 2 α 1 1 = q and
K 2 K 1 1 K 2 1 K 1 q q 1 ( x ) = x .
On the other hand, projecting ( E F F E ) ( x ) to C q [ x , y , z ] we obtain
( E F F E ) ( x ) = E ( F ( x ) ) F ( E ( x ) ) = E ( 0 ) F ( a 1 y ) = 0 ,
However, 0 x . We have obtained contradictions and proved our claims.
In a similar way, one can prove that all other series with the 0-th homogeneous component of M E F are zero and those within the 1-th homogeneous component of M E F are nonzero and empty. □
Lemma 4.
If the 0-th homogeneous component of M E F is nonzero and the 1-st homogeneous component of M E F is zero, then these series are empty.
Proof. 
We only show that the a 0 0 0 0 0 0 0 , 0 0 0 0 0 0 1 -series is empty. in a similar way, one can prove that all other series are empty.
Considering this series, we obtain that
a 0 0 α 1 = q , α 2 = q , β 2 β 1 1 = q , γ 2 γ 1 1 = q .
and suppose that it is not empty. We set
K 1 ( x ) = q x , K 2 ( x ) = q x , K 1 ( y ) = β 1 y , K 2 ( y ) = β 2 y , K 1 ( z ) = γ 1 z , K 2 ( z ) = γ 2 z , E ( x ) = a 0 + m 1 + n 1 + l 1 2 ρ m 1 n 1 l 1 1 x m 1 y n 1 z l 1 for m 1 , n 1 , l 1 N , E ( y ) = m 2 + n 2 + l 2 2 ρ m 2 n 2 l 2 2 x m 2 y n 2 z l 2 for m 2 , n 2 , l 2 N , E ( z ) = m 3 + n 3 + l 3 2 ρ m 3 n 3 l 3 3 x m 3 y n 3 z l 3 for m 3 , n 3 , l 3 N , F ( x ) = m 4 + n 4 + l 4 2 ρ m 4 n 4 l 4 4 x m 4 y n 4 z l 4 for m 4 , n 4 , l 4 N , F ( y ) = m 5 + n 5 + l 5 2 ρ m 5 n 5 l 5 5 x m 5 y n 5 z l 5 for m 5 , n 5 , l 5 N , F ( z ) = m 6 + n 6 + l 6 2 ρ m 6 n 4 l 6 6 x m 6 y n 6 z l 6 for m 6 , n 6 , l 6 N ,
where β 1 , β 2 , γ 1 , γ 2 C * , and ρ m i n i l i i C , i = 1 , 2 , 3 , 4 , 5 , 6 . We have
( K 1 E q 1 E K 1 ) ( x ) = K 1 ( E ( x ) ) q 1 E ( K 1 ( x ) ) = K 1 ( a 0 + m 1 + n 1 + l 1 2 ρ m 1 n 1 l 1 1 x m 1 y n 1 z l 1 ) q 1 q E ( x ) = a 0 + m 1 + n 1 + l 1 2 ρ m 1 n 1 l 1 1 α 1 m 1 β 1 n 1 γ 1 l 1 x m 1 y n 1 z l 1 E ( x ) = m 1 + n 1 + l 1 2 ρ m 1 n 1 l 1 1 ( q m 1 β 1 n 1 γ 1 l 1 1 ) x m 1 y n 1 z l 1 = 0 ,
and then for all m 1 , n 1 , l 1 N with m 1 + n 1 + l 1 2 , one has ρ m 1 n 1 l 1 1 = 0 or q m 1 β 1 n 1 γ 1 l 1 = 1 .
And
( K 2 E + q 1 E K 2 ) ( x ) = K 2 ( E ( x ) ) + q 1 E ( K 2 ( x ) ) = K 2 ( a 0 + m 1 + n 1 + l 1 2 ρ m 1 n 1 l 1 1 x m 1 y n 1 z l 1 ) q 1 q E ( x ) = a 0 + m 1 + n 1 + l 1 2 ρ m 1 n 1 l 1 1 α 2 m 1 β 2 n 1 γ 2 l 1 x m 1 y n 1 z l 1 E ( x ) = m 1 + n 1 + l 1 2 ρ m 1 n 1 l 1 1 ( ( q ) m 1 β 2 n 1 γ 2 l 1 1 ) x m 1 y n 1 z l 1 = 0 ,
then for all m 1 , n 1 , l 1 N with m 1 + n 1 + l 1 2 , one has ρ m 1 n 1 l 1 1 = 0 or ( q ) m 1 β 2 n 1 γ 2 l 1 = 1 . If some ρ m 1 n 1 l 1 1 0 meet the conditions, i.e,
q m 1 β 1 n 1 γ 1 l 1 = 1 , ( q ) m 1 β 2 n 1 γ 2 l 1 = 1 ,
one can get ( 1 ) m 1 q ( n 1 + l 1 ) = 1 ; this contradicts with q not being unit root. Therefore, for all m 1 , n 1 , l 1 N with m 1 + n 1 + l 1 2 , we have E ( x ) = a 0 . By discussing E ( y ) , E ( z ) , F ( x ) , F ( y ) , and F ( z ) using methods similar to E ( x ) , we can obtain that
E ( y ) = 0 , E ( z ) = 0 ,
F ( x ) = 0 ρ 200 4 x 2 , F ( y ) = 0 ρ 110 5 x y , F ( z ) = 0 ρ 101 6 x z .
From E F F E = K 2 K 1 1 K 2 1 K 1 q q 1 , we have
K 2 K 1 1 K 2 1 K 1 q q 1 ( y ) = β 2 β 1 1 β 2 1 β 1 q q 1 y = q q 1 q q 1 y = y , K 2 K 1 1 K 2 1 K 1 q q 1 ( z ) = γ 2 γ 1 1 γ 2 1 γ 1 q q 1 y = q q 1 q q 1 z = z .
  • If F ( y ) = 0 , then ( E F F E ) ( y ) = E F ( y ) F E ( y ) = 0 y .
  • If F ( y ) = ρ 110 5 x y , then ( E F F E ) ( y ) = ρ 110 5 E ( x y ) = ρ 110 5 a 0 y = y ; hence ρ 110 5 = a 0 1 and F ( y ) = a 0 1 x y .
  • If F ( z ) = 0 , then ( E F F E ) ( z ) = E F ( z ) F E ( z ) = 0 z .
  • If F ( z ) = ρ 101 6 x z , then ( E F F E ) ( z ) = ρ 101 6 E ( x z ) = ρ 110 6 a 0 z = z ; hence ρ 101 6 = a 0 1 and F ( z ) = a 0 1 x z .
By E 2 = F 2 = 0 , one has F 2 ( y ) = F ( a 0 1 x y ) = a 0 1 ( x F ( y ) + q 1 F ( x ) y ) .
  • If F ( x ) = 0 , then F 2 ( y ) = a 0 2 x 2 y 0 ;
  • If F ( x ) = ρ 200 4 x 2 , then F 2 ( y ) = a 0 1 ( a 0 1 x 2 y + ρ 200 4 q 1 x 2 y ) = 0 ; hence ρ 200 4 = q a 0 1 and F ( x ) = q a 0 1 x 2 .
According to y x = q x y , then
F ( y x q x y ) = y F ( x ) q x F ( y ) + α 2 1 α 1 F ( y ) x β 2 1 β 1 q F ( x ) y = a 0 1 ( q 3 + q ) x 2 y 0 .
In summary, this series is empty. □
Next we turn to “nonempty” series; there is only one kind of “nonempty” series.
Theorem 1.
The 0 0 0 0 0 0 0 , 0 0 0 0 0 0 1 -series has X q ( A 1 ) −module algebra structures on C q [ x , y , z ] given by
K 1 ( x ) = λ 1 x , K 2 ( x ) = ± λ 1 x ,
K 1 ( y ) = λ 2 y , K 2 ( y ) = ± λ 2 y ,
K 1 ( z ) = λ 3 z , K 2 ( z ) = ± λ 3 z ,
E ( x ) = F ( x ) = E ( y ) = F ( y ) = E ( z ) = F ( z ) = 0 ,
where λ 1 , λ 2 , λ 3 C * ; they are pairwise nonisomorphic.
Proof. 
It is easy to check that (51)–(54) determine a well-defined X q ( A 1 ) -action consistent with the multiplication in X q ( A 1 ) and in C q [ x , y , z ] , as well as with comultiplication in X q ( A 1 ) . We prove that there are no other X q ( A 1 ) -actions here. Note that an application of (6) to x , y or z has zero projection to C q [ x , y , z ] 1 , i.e., ( E F F E ) ( x ) = ( E F F E ) ( y ) = ( E F F E ) ( z ) = 0 , because in this series E and F send any monomial to a sum of the monomials of higher degree. Therefore,
K 2 K 1 1 K 2 1 K 1 q q 1 ( x ) = α 2 α 1 1 α 1 α 2 1 q q 1 x = 0 , K 2 K 1 1 K 2 1 K 1 q q 1 ( y ) = β 2 β 1 1 β 1 β 2 1 q q 1 y = 0 , K 2 K 1 1 K 2 1 K 1 q q 1 ( z ) = γ 2 γ 1 1 γ 2 1 γ 1 q q 1 z = 0 ,
and hence,
α 2 α 1 1 α 2 1 α 2 = β 2 β 1 1 β 2 1 β 1 = γ 2 γ 1 1 γ 2 1 γ 2 = 0 ,
which leads to α 1 2 = α 2 2 , β 1 2 = β 2 2 and γ 1 2 = γ 2 2 . Let α 1 = λ 1 , β 1 = λ 2 , and γ 1 = λ 3 ; we have α 2 = ± λ 1 , β 2 = ± λ 2 , and γ 2 = ± λ 3 . To prove (54), note that if E ( x ) 0 or F ( y ) 0 , then they are a sum of the monomials with degrees are greater than 1. This is similar to the proof of Lemma 4; we find that this is impossible, because they can not satisfy conditions of X q ( A 1 ) -module algebra on C q [ x , y , z ] .
To see that the X q ( A 1 ) -module algebra structures are pairwise nonisomorphic, observe that all the automorphisms of C q [ x , y , z ] commute with the actions of K 1 and K 2 . □

4. Classification of X q ( A 1 ) -Module Algebra Structures on C q [ x , y , z ] for T 0

In this section, we suppose the automorphism Ψ of C q [ x , y , z ] as follows:
Ψ ( x ) = α x , Ψ ( y ) = β y + t x z , Ψ ( z ) = γ z , ( α , β , γ , t C * ) ,
and Aut ( C q [ x , y , z ] ) C C * 3 . One can have
Ψ 1 ( x ) = α 1 x , Ψ 1 ( y ) = β 1 y t α 1 β 1 γ 1 x z , Ψ 1 ( z ) = γ 1 z .
In the following, we will begin to discuss the X q ( A 1 ) -module algebra structures on C q [ x , y , z ] with t 0 , i.e., where K 1 , K 2 C C * 3 . In this section, our research method is similar to that in Section 3.

4.1. Properties of X q ( A 1 ) -Module Algebras on C q [ x , y , z ]

It is easy to see that any action of X q ( A 1 ) on C q [ x , y , z ] is determined by the following 4 × 3 matrix with entries from C q [ x , y , z ] :
M = K 1 ( x ) K 1 ( y ) K 1 ( z ) K 2 ( x ) K 2 ( y ) K 2 ( z ) E ( x ) E ( y ) E ( z ) F ( x ) F ( y ) F ( z ) .
Given a X q ( A 1 ) -module algebra structure on C q [ x , y , z ] , obviously the action of K 1 (or K 2 ) is determined by an automorphism of C q [ x , y , z ] ; in other words, the actions of K 1 and K 2 are determined by a matrix M K 1 K 2 as follows:
M K 1 K 2 = d e f i n i t i o n K 1 ( x ) K 1 ( y ) K 1 ( z ) K 2 ( x ) K 2 ( y ) K 2 ( z ) = α 1 ( x ) β 1 ( y ) + t 1 x z γ 1 ( z ) α 2 ( x ) β 2 ( y ) + t 2 x z γ 2 ( z ) ,
where α i , β i , γ i , t i C * for i { 1 , 2 } .
Lemma 5.
For all α i , β i , γ i , t i C * , i { 1 , 2 } , either β i = α i γ i or t i = β i α i γ i t , where t C * .
Proof. 
For all α i , β i , γ i , t i C * , i { 1 , 2 } , we have
K i ( y ) = β i y + t i x z and K i 1 ( y ) = β i 1 y t i α i 1 β i 1 γ i 1 x z
by (56). It is to easy check K i K i 1 ( y ) = y , and
K 1 K 2 ( y ) = K 1 ( β 2 y + t 2 x z ) = β 1 β 2 y + β 2 t 1 x z + t 2 α 1 γ 1 x z , K 2 K 1 ( y ) = K 2 ( β 1 y + t 1 x z ) = β 1 β 2 y + β 1 t 2 x z + t 1 α 2 γ 2 x z .
By the definition of module algebra and (1), we have t 1 ( β 2 α 2 γ 2 ) = t 2 ( β 1 α 1 γ 1 ) for t i C * , i = 1 , 2 ; hence, either β i = α i γ i or t 1 t 2 = β 1 α 1 γ 1 β 2 α 2 γ 2 , and we can write the latter as t i = β i α i γ i t , where t C * .
It is easy to see that every monomial x m 1 z m 3 C q [ x , y , z ] is an eigenvector of K 1 (or K 2 ), and the associated eigenvalue α 1 m 1 γ 1 m 3 (or α 2 m 1 γ 2 m 3 ) is called the K 1 -weight (or K 2 -weight) of this monomial, which will be written as
w t K 1 ( x m 1 z m 3 ) = α 1 m 1 γ 1 m 3 ,
w t K 2 ( x m 1 z m 3 ) = α 2 m 1 γ 2 m 3 .
We will also need another matrix M E F as follows:
M E F = d e f i n i t i o n E ( x ) E ( y ) E ( z ) F ( x ) F ( y ) F ( z ) .
Obviously, K 1 ( x ) , K 2 ( x ) , E ( x ) , F ( x ) and K 1 ( z ) , K 2 ( z ) , E ( z ) , F ( z ) are weight vectors for K 1 and K 2 ; then
w t K i M = d e f i n i t i o n w t K i ( K 1 ( x ) ) w t K i ( K 1 ( z ) ) w t K i ( K 2 ( x ) ) w t K i ( K 2 ( z ) ) w t K i ( E ( x ) ) w t K i ( E ( z ) ) w t K i ( F ( x ) ) w t K i ( F ( z ) ) w t K i ( x ) w t K i ( z ) w t K i ( x ) w t K i ( z ) ( 1 ) i 1 q 1 w t K i ( x ) ( 1 ) i 1 q 1 w t K i ( z ) ( 1 ) i 1 q w t K i ( x ) ( 1 ) i 1 q w t K i ( z ) = α i γ i α i γ i ( 1 ) i 1 q 1 α i ( 1 ) i 1 q 1 γ i ( 1 ) i 1 q α i ( 1 ) i 1 q γ i .
As in Section 3, we denote by M j the j-th homogeneous component of M. Obviously, if M j is nonzero, one can calculate the associated eigenvalues.
Set a 0 , b 0 , c 0 , a 0 , b 0 , c 0 C ; we obtain the 0-th homogeneous component of M as follows:
M 0 = 0 0 0 0 0 0 a 0 b 0 c 0 a 0 b 0 c 0 0 .
Then, we have
w t K i M E F 0 = ( 1 ) i 1 q 1 α i ( 1 ) i 1 q 1 γ i ( 1 ) i 1 q α i ( 1 ) i 1 q γ i 0 1 1 1 1 0 ,
According to the fact that q is not a root of the unit, and relation (60), a 0 and a 0 ( c 0 and c 0 ) should contain at least one 0.
Applying E and F to the relations (11)–(13) by using Equation (56), one has
E ( y ) x q E ( x ) y + K 2 K 1 1 ( y ) E ( x ) q K 2 K 1 1 ( x ) E ( y ) = 0 ,
E ( z ) y q E ( y ) z + K 2 K 1 1 ( z ) E ( y ) q K 2 K 1 1 ( y ) E ( z ) = 0 ,
E ( z ) x q E ( x ) z + K 2 K 1 1 ( z ) E ( x ) q K 2 K 1 1 ( x ) E ( z ) = 0 ,
y F ( x ) q x F ( y ) + F ( y ) K 2 1 K 1 ( x ) q F ( x ) K 2 1 K 1 ( y ) = 0 ,
z F ( y ) q y F ( z ) + F ( z ) K 2 1 K 1 ( y ) q F ( y ) K 2 1 K 1 ( z ) = 0 ,
z F ( x ) q x F ( z ) + F ( z ) K 2 1 K 1 ( x ) q F ( x ) K 2 1 K 1 ( z ) = 0 .
which certainly implies
b 0 1 q α 1 1 α 2 = a 0 β 1 1 β 2 q = c 0 1 q β 1 1 β 2 = b 0 γ 1 1 γ 2 q = 0 , c 0 1 q α 1 1 α 2 = a 0 γ 1 1 γ 2 q = a 0 1 q β 1 β 2 1 = b 0 α 1 α 2 1 q = 0 , b 0 1 q γ 1 γ 2 1 = c 0 β 1 β 2 1 q = a 0 1 q γ 1 γ 2 1 = c 0 α 1 α 2 1 q = 0 .
Because q is not a root of the unit, q ± 1 , and from the above discussion, for the 0-st homogeneous component M E F 0 of M E F , we have following lemma.
Lemma 6.
There are eight cases for the 0-st homogeneous component M E F 0 of M E F , as follows:
a 0 0 0 0 0 0 1 α 1 = q , α 2 = q , β 1 1 β 2 = q , γ 1 1 γ 2 = q ;
0 b 0 0 0 0 0 1 α 1 1 α 2 = q 1 , γ 1 1 γ 2 = q ;
0 0 c 0 0 0 0 1 γ 1 = q , γ 2 = q , β 1 1 β 2 = q 1 , α 1 1 α 2 = q 1 ;
0 0 0 a 0 0 0 1 α 1 = q 1 , α 2 = q 1 , β 1 β 2 1 = q 1 , γ 1 γ 2 1 = q 1 ;
0 0 0 0 b 0 0 1 α 1 α 2 1 = q , γ 1 γ 2 1 = q 1 ;
0 0 0 0 0 c 0 1 γ 1 = q 1 , γ 2 = q 1 β 1 β 2 1 = q , α 1 α 2 1 = q ;
0 b 0 0 0 b 0 0 1 α 1 1 α 2 = q 1 , γ 1 1 γ 2 = q ;
0 0 0 0 0 0 1 it does not determine the weight constants at all .
Next, for the 1-st homogeneous component, as q is not a root of the unit, one has
w t K 1 ( E ( x ) ) = q 1 α 1 = q 1 w t K 1 ( x ) w t K 1 ( x ) , w t K 2 ( E ( x ) ) = q 1 α 2 = q 1 w t K 2 ( x ) w t K 2 ( x ) ,
which implies
( E ( x ) ) 1 = a 1 z ,
for some a 1 C , and in a similar way we have
M E F 1 = a 1 z b 1 x + b 2 y + b 3 z c 1 x a 1 z b 1 x + b 2 y + b 3 z c 1 x 1
where b 1 , b 2 , b 3 , c 1 , a 1 , b 1 , b 2 , b 3 , c 1 C . In fact,
w t K i ( M E F ) 1 ( 1 ) i 1 q 1 α i ( 1 ) i 1 q 1 γ i ( 1 ) i 1 q α i ( 1 ) i 1 q γ i γ i α i γ i α i .
Now, we project (61)–(66) to C q [ x , y , z ] 2 ; one can obtain the following conclusion.
If a 0 0 , then α 1 = q , α 2 = q , β 1 1 β 2 = q , γ 1 1 γ 2 = q , and we have
( E ( y ) ) 1 x q ( E ( x ) ) 1 y + q y ( E ( x ) ) 1 + ( t 2 β 1 1 + q t 1 β 1 1 ) x z ( E ( x ) ) 0 + q x ( E ( y ) ) 1 = 0 ,
b 1 1 + q x 2 + 2 q b 2 x y + a 1 q 1 q y z + ( 2 q b 3 + a 0 t 2 β 1 1 + q a 0 t 1 β 1 1 ) x z = 0 ,
a 1 = 0 , b 1 = 0 , b 2 = 0 , b 3 = a 0 ( t 2 + q t 1 ) 2 q β 1 .
If a 0 = 0 , then
( E ( y ) ) 1 x q ( E ( x ) ) 1 y + β 1 1 β 2 y ( E ( x ) ) 1 q α 1 1 α 2 x ( E ( y ) ) 1 = 0 ,
b 1 1 q α 1 1 α 2 x 2 + b 2 1 α 1 1 α 2 y x + b 3 1 α 1 1 α 2 z x + a 1 β 1 1 β 2 q 2 y z = 0 ,
b 1 0 α 1 1 α 2 = q 1 , b 2 0 α 1 1 α 2 = 1 , b 3 0 α 1 1 α 2 = 1 , a 1 0 β 1 1 β 2 = q 2 .
If c 0 0 , then γ 1 = q , γ 2 = q , β 1 1 β 2 = q 1 , α 1 1 α 2 = q 1 , and we have
( E ( z ) ) 1 y q ( E ( y ) ) 1 z z ( E ( y ) ) 1 y ( E ( z ) ) 1 q ( t 2 β 1 1 + q 1 t 1 β 1 1 ) x z ( E ( z ) ) 0 = 0 ,
c 1 1 q x y 2 q b 2 y z b 3 1 + q z 2 ( 2 q b 1 + q c 0 t 2 β 1 1 + c 0 t 1 β 1 1 ) x z = 0 ,
b 2 = 0 , b 3 = 0 , c 1 = 0 , b 1 = c 0 ( t 1 + q t 2 ) 2 q β 1 .
If c 0 = 0 , then
( E ( z ) ) 1 y q ( E ( y ) ) 1 z + γ 1 1 γ 2 z ( E ( y ) ) 1 q β 1 1 β 2 y ( E ( z ) ) 1 = 0 ,
c 1 1 q 2 β 1 1 β 2 x y + q b 1 γ 1 1 γ 2 1 x z + q b 2 γ 1 1 γ 2 1 y z + b 3 γ 1 1 γ 2 q z 2 = 0 ,
c 1 0 β 1 1 β 2 = q 2 , b 1 0 γ 1 1 γ 2 = 1 , b 2 0 γ 1 1 γ 2 = 1 , b 3 0 γ 1 1 γ 2 = q .
( E ( z ) ) 1 x q ( E ( x ) ) 1 z + γ 1 1 γ 2 z ( E ( x ) ) 1 q α 1 1 α 2 x ( E ( z ) ) 1 = 0 ,
c 1 ( 1 q α 1 1 α 2 ) x 2 + a 1 ( γ 1 1 γ 2 q ) z 2 = 0 ,
c 1 0 α 1 1 α 2 = q 1 , a 1 0 γ 1 1 γ 2 = q .
If a 0 0 , then α 1 = q 1 , α 2 = q 1 , β 1 β 2 1 = q 1 , γ 1 γ 2 1 = q 1 , and we have
y ( F ( x ) ) 1 q x ( F ( y ) ) 1 ( F ( y ) ) 1 x ( F ( x ) ) 1 y q ( F ( x ) ) 0 ( t 1 α 2 1 γ 2 1 q 1 t 2 α 2 1 γ 2 1 ) x z = 0 ,
a 1 1 q y z b 1 1 + q x 2 2 q b 2 x y + ( 2 q b 3 + a 0 t 2 α 2 1 γ 2 1 q a 0 t 1 α 2 1 γ 2 1 ) x z = 0 ,
a 1 = 0 , b 1 = 0 , b 2 = 0 , b 3 = a 0 ( t 2 q t 1 ) 2 q α 2 γ 2 .
If a 0 = 0 , then
y ( F ( x ) ) 1 q x ( F ( y ) ) 1 + α 1 α 2 1 ( F ( y ) ) 1 x q β 1 β 2 1 ( F ( x ) ) 1 y = 0 ,
b 1 α 1 α 2 1 q x 2 + q b 2 α 1 α 2 1 1 x y + q b 3 α 1 α 2 1 1 x z + a 1 1 q 2 β 1 β 2 1 y z = 0 ,
a 1 0 β 1 β 2 1 = q 2 , b 1 0 α 1 α 2 1 = q , b 2 0 α 1 α 2 1 = 1 , b 3 0 α 1 α 2 1 = 1 .
If c 0 0 , then γ 1 = q 1 , γ 2 = q 1 , β 1 β 2 1 = q , α 1 α 2 1 = q , and we have
z ( F ( y ) ) 1 q y ( F ( z ) ) 1 + q ( F ( z ) ) 1 y + ( F ( z ) ) 0 ( t 1 α 2 1 γ 2 1 q t 2 α 2 1 γ 2 1 ) x z + q ( F ( y ) ) 1 z = 0 ,
2 q b 2 y z + b 3 1 + q z 2 + q c 1 ( 1 q ) x y + ( 2 q b 1 q c 0 t 2 α 2 1 γ 2 1 + c 0 t 1 α 2 1 γ 2 1 ) x z = 0 ,
b 2 = 0 , b 3 = 0 , c 1 = 0 , b 1 = c 0 ( q t 2 t 1 ) 2 q α 2 γ 2 .
If c 0 = 0 , then
z ( F ( y ) ) 1 q y ( F ( z ) ) 1 + β 1 β 2 1 ( F ( z ) ) 1 y q γ 1 γ 2 1 ( F ( y ) ) 1 z = 0 ,
q b 1 1 γ 1 γ 2 1 x z + q b 2 1 γ 1 γ 2 1 y z + b 3 1 q γ 1 γ 2 1 z 2 + c 1 β 1 β 2 1 q 2 x y = 0 ,
b 1 0 γ 1 γ 2 1 = 1 , b 2 0 γ 1 γ 2 1 = 1 , b 3 0 γ 1 γ 2 1 = q 1 , c 1 0 β 1 β 2 1 = q 2 .
z ( F ( x ) ) 1 x q x ( F ( z ) ) 1 + α 1 α 2 1 ( F ( z ) ) 1 x q γ 1 γ 2 1 ( F ( x ) ) 1 z = 0 ,
c 1 ( α 1 α 2 1 q ) x 2 + a 1 ( 1 q γ 1 γ 2 1 ) z 2 = 0 ,
a 1 0 γ 1 γ 2 1 = q 1 , c 1 0 α 1 α 2 1 = q .
From the above discussion, and as q is not a root of the unit, we can obtain the following lemma.
Lemma 7.
There are 18 cases for the 1-st homogeneous component M E F 1 of M E F , as follows:
a 1 y 0 0 0 0 0 1 β 1 = q 1 α 1 , β 2 = q 1 α 2 , β 1 1 β 2 = q , γ 1 1 γ 2 = 1 ;
a 2 z 0 0 0 0 0 1 γ 1 = q 1 α 1 , γ 2 = q 1 α 2 , β 1 1 β 2 = q 2 , γ 1 1 γ 2 = q ;
0 b 1 x 0 0 0 0 1 β 1 = q α 1 , β 2 = q α 2 , α 1 1 α 2 = q 1 , γ 1 1 γ 2 = 1 ;
0 b 2 z 0 0 0 0 1 γ 1 = q 1 β 1 , γ 2 = q 1 β 2 , α 1 1 α 2 = 1 , γ 1 1 γ 2 = q ;
0 0 c 1 x 0 0 0 1 α 1 = q 1 γ 1 , α 2 = q 1 γ 2 , α 1 1 α 2 = q 1 , β 1 1 β 2 = q 2 ;
0 0 c 2 y 0 0 0 1 β 1 = q 1 γ 1 , β 2 = q 1 γ 2 , α 1 1 α 2 = 1 , β 1 1 β 2 = q 1 ;
0 0 0 a 1 y 0 0 1 β 1 = q α 1 , β 2 = q α 2 , β 1 1 β 2 = q , γ 1 1 γ 2 = 1 ;
0 0 0 a 2 z 0 0 1 γ 1 = q α 1 , γ 2 = q α 2 , β 1 1 β 2 = q 2 , γ 1 1 γ 2 = q ;
0 0 0 0 b 1 x 0 1 β 1 = q 1 α 1 , β 2 = q 1 α 2 , α 1 1 α 2 = q 1 , γ 1 1 γ 2 = 1 ;
0 0 0 0 b 2 z 0 1 γ 1 = q β 1 , γ 2 = q β 2 , α 1 1 α 2 = 1 , γ 1 1 γ 2 = q ;
0 0 0 0 0 c 1 x 1 α 1 = q γ 1 , α 2 = q γ 2 , α 1 1 α 2 = q 1 , β 1 1 β 2 = q 2 ;
0 0 0 0 0 c 2 y 1 β 1 = q γ 1 , β 2 = q γ 2 , α 1 1 α 2 = 1 , β 1 1 β 2 = q 1 ;
0 0 0 0 0 0 1 i t d o e s n o t d e t e r m i n e t h e w e i g h t c o n s t a n t s a t a l l ;
If a 0 0 , then α 1 = q , α 2 = q , β 1 1 β 2 = q , γ 1 1 γ 2 = q , and we have
a 1 = 0 , b 1 = 0 , b 2 = 0 , b 3 = a 0 ( t 2 + q t 1 ) 2 q β 1 .
If c 0 0 , then γ 1 = q , γ 2 = q , β 1 1 β 2 = q 1 , α 1 1 α 2 = q 1 , and we have
b 2 = 0 , b 3 = 0 , c 1 = 0 , b 1 = c 0 ( t 1 + q t 2 ) 2 q β 1 .
If a 0 0 , then α 1 = q 1 , α 2 = q 1 , β 1 β 2 1 = q 1 , γ 1 γ 2 1 = q 1 , and we have
a 1 = 0 , b 1 = 0 , b 2 = 0 , b 3 = a 0 ( t 2 q t 1 ) 2 q α 2 γ 2 .
If c 0 0 , then γ 1 = q 1 , γ 2 = q 1 , β 1 β 2 1 = q , α 1 α 2 1 = q , and we have
b 2 = 0 , b 3 = 0 , c 1 = 0 , b 1 = c 0 ( q t 2 t 1 ) 2 q α 2 γ 2 .

4.2. The Structures of X q ( A 1 ) -Module Algebra on C q [ x , y , z ]

Through the previous discussion, we found that both the 0-st homogeneous component M E F 0 and the 1-st homogeneous component M E F 1 determine the eigenvalues of x and z. By Lemmas 6 and 7, and as q is not a root of the unit, it follows that there are 91 kinds of M E F 0 , M E F 1 , which are empty. Hence, we only discuss the following cases.
Lemma 8.
If the 0-th homogeneous component of M E F is zero and the 1-st homogeneous component of M E F is nonzero, then these series are empty.
Proof. 
The proof is similar to the proof of Lemma 3. □
Lemma 9.
If the 0-th homogeneous component of M E F is nonzero and the 1-st homogeneous component of M E F is zero, then these series are empty.
Proof. 
The proof is similar to the proof of Lemma 4. □
Lemma 10.
The a 0 0 0 0 0 0 0 , 0 a 0 ( t 2 + q t 1 ) 2 q β 1 z 0 0 0 0 1 -series is empty.
Proof. 
By (67), one has α 1 = q , α 2 = q , β 1 1 β 2 = q , γ 1 1 γ 2 = q , and
α 1 1 α 1 γ 1 1 γ 2 = q q = β 1 1 β 2 .
From Lemma 5, it can be concluded that t i = β i α i γ i t ( i = 1 , 2 ) , where t C * , and a 0 ( t 2 + q t 1 ) 2 q β 1 z = a 0 t z .
If we suppose this series is not empty, we have K 1 ( x ) = q x , K 2 ( x ) = q x and w t K i ( E ( x ) ) = ( 1 ) i 1 q 1 α i = 1 ; hence, E ( x ) = a 0 . Set
K 1 ( y ) = β 1 y + ( β 1 q γ 1 ) t x z , K 2 ( y ) = β 2 y + ( β 2 + q γ 2 ) t x z , K 1 ( z ) = γ 1 z , K 2 ( z ) = γ 2 z , E ( y ) = a 0 t z + m 2 + n 2 + l 2 2 ψ m 2 n 2 l 2 2 x m 2 y n 2 z l 2 for m 2 , n 2 , l 2 N , E ( z ) = m 3 + l 3 2 ψ m 3 l 3 3 x m 3 z l 3 for m 3 , l 3 N , F ( x ) = m 4 + l 4 2 ψ m 4 l 4 4 x m 4 z l 4 for m 4 , l 4 N , F ( y ) = m 5 + n 5 + l 5 2 ψ m 5 n 5 l 5 5 x m 5 y n 5 z l 5 for m 5 , n 5 , l 5 N , F ( z ) = m 6 + l 6 2 ψ m 6 l 6 6 x m 6 z l 6 for m 6 , l 6 N ,
where β 1 , β 2 , γ 1 , γ 2 , t C * , and ψ m 2 n 2 l 2 2 , ψ m 3 l 3 3 , ψ m 4 l 4 4 , ψ m 5 n 5 l 5 5 , ψ m 6 l 6 6 C . According to (61), one can obtain
m 2 + n 2 + l 2 2 ψ m 2 n 2 l 2 2 ( q n 2 + l 2 + q ) x m 2 + 1 y n 2 z l 2 = 0 ,
and then for all m 2 , n 2 , l 2 N with m 2 + n 2 + l 2 2 , one has ψ m 2 n 2 l 2 2 = 0 and E ( y ) = a 0 t z . Similarly, we can get E ( z ) = 0 . By (3) and (5), we have
( K 1 F q F K 1 ) ( x ) = K 1 ( F ( x ) ) q F ( K 1 ( x ) ) = K 1 ( m 4 + l 4 2 ψ m 4 l 4 4 x m 4 z l 4 ) q 2 F ( x ) = m 4 + l 4 2 ψ m 4 l 4 4 ( q m 4 γ 1 l 4 q 2 ) x m 4 z l 4 = 0 , ( K 2 F + q F K 2 ) ( x ) = K 2 ( F ( x ) ) + q F ( K 2 ( x ) ) = m 4 + l 4 2 ψ m 4 l 4 4 [ ( q ) m 4 γ 2 l 4 q 2 ] x m 4 z l 4 = 0 ,
and then for all m 4 , l 4 N with m 4 + l 4 2 , one has
ψ m 4 l 4 4 = 0 or ( q ) m 4 γ 2 l 4 = q 2 ,
and F ( x ) = 0 or F ( x ) = ψ 20 4 x 2 .
  • If F ( x ) = 0 , it is easy to get F ( y ) = F ( z ) = 0 ; then
    ( E F F E ) ( z ) = 0 K 2 K 1 1 K 2 1 K 1 q q 1 ( z ) = z ,
    This contradicts our hypothesis.
  • If F ( x ) = ψ 20 4 x 2 , by (66), one can get F ( z ) = ψ 20 4 ( q 2 1 ) 2 q x z and
    ( E F F E ) ( z ) = K 2 K 1 1 K 2 1 K 1 q q 1 ( z ) = z ,
    After calculation, we can conclude that F ( x ) = 2 q a 0 q 2 1 x 2 and F ( z ) = a 0 x z . However
    F 2 ( z ) = F ( a 0 x z ) = a 0 2 q 2 + 1 q 2 1 x 2 z 0 ,
    This contradicts our hypothesis.
In summary, this series is empty. □
Similar to Lemma 10, we can obtain
0 0 c 0 0 0 0 0 , 0 c 0 ( t 1 + q t 2 ) 2 q β 1 x 0 0 0 0 1 , 0 0 0 a 0 0 0 0 , 0 0 0 0 a 0 ( t 2 q t 1 ) 2 q α 2 γ 2 z 0 1 ,
0 0 0 C 0 0 0 0 , 0 0 z 0 0 c 0 ( q t 2 t 1 ) 2 q α 2 γ 2 z 0 1 , which are empty series.
Next we turn to “nonempty” series; there is only one “nonempty” series.
Theorem 2.
The 0 0 0 0 0 0 0 , 0 0 0 0 0 0 1 -series has two types of X q ( A 1 ) -module algebra structures on the C q [ x , y , z ] given by the following.
  • For all λ , μ , t C * , we have
    K ( x ) = λ x , K 2 ( x ) = ± λ x K ( y ) = λ μ y + t x z , K 2 ( x ) = ± ( λ μ y + t x z ) K ( z ) = μ z , K 2 ( z ) = ± μ z E ( x ) = E ( y ) = E ( z ) = 0 , F ( x ) = F ( y ) = F ( z ) = 0 ,
    which are pairwise nonisomorphic.
  • For all λ , σ , μ , t ˜ C * , we have
    K ( x ) = λ x , K 2 ( x ) = ± λ x K ( y ) = σ y + t ˜ x z , K 2 ( x ) = ± ( σ y + t ˜ x z ) K ( z ) = μ z , K 2 ( z ) = ± μ z E ( x ) = E ( y ) = E ( z ) = 0 , F ( x ) = F ( y ) = F ( z ) = 0 ,
    where t ˜ = ( λ μ σ ) t C * ; they are pairwise nonisomorphic.
Proof. 
The proof is similar to the proof of Theorem 1. □

5. Conclusions

We investigated the module algebra structures of X q ( A 1 ) on the quantum polynomial algebra C q [ x , y , z ] . Our main contributions are as follows.
  • For t = 0 , the classification of X q ( A 1 ) -module algebra structures were given on C q [ x , y , z ] , as follows:
    M E F 0 , M E F 1 X q ( A 1 ) -module algebra structures
    0 0 0 0 0 0 0 , 0 0 0 0 0 0 1 K 1 ( x ) = λ 1 x , K 2 ( x ) = ± λ 1 x , K 1 ( y ) = λ 2 y , K 2 ( y ) = ± λ 2 y , K 1 ( z ) = λ 3 z , K 2 ( z ) = ± λ 3 z , E ( x ) = F ( x ) = E ( y ) = F ( y ) = E ( z ) = F ( z ) = 0 .
  • For t 0 , the classification of X q ( A 1 ) -module algebra structures were given on C q [ x , y , z ] , as follows:
      M E F 0 , M E F 1 X q ( A 1 ) -module algebra structures
    0 0 0 0 0 0 0 , 0 0 0 0 0 0 1 K ( x ) = λ x , K 2 ( x ) = ± λ x K ( y ) = λ μ y + t x z , K 2 ( x ) = ± ( λ μ y + t x z ) K ( z ) = μ z , K 2 ( z ) = ± μ z E ( x ) = E ( y ) = E ( z ) = 0 , F ( x ) = F ( y ) = F ( z ) = 0 .
    0 0 0 0 0 0 0 , 0 0 0 0 0 0 1 K ( x ) = λ x , K 2 ( x ) = ± λ x K ( y ) = σ y + t ˜ x z , K 2 ( x ) = ± ( σ y + t ˜ x z ) K ( z ) = μ z , K 2 ( z ) = ± μ z E ( x ) = E ( y ) = E ( z ) = 0 , F ( x ) = F ( y ) = F ( z ) = 0 .
    This research makes some preparations on the classification of module algebra structures of X q ( A n ) on C q [ x , y , z ] for n 2 .

Funding

This research was funded by National Natural Science Foundation of China (Grant No. 12201187) and Natural Science Foundation of Henan Province (Grant No. 222300420156).

Data Availability Statement

Data from this study are available upon request from the corresponding author.

Conflicts of Interest

The authors declare no conflicts of interest.

References

  1. Sweedler, M.E. Hopf Algebras; W. A. Benjamin, Inc.: New York, NY, USA, 1969; pp. 56–73. [Google Scholar]
  2. Beattie, M. A direct sum for the Brauer group of H-module algebras. J. Algebra 1976, 43, 686–693. [Google Scholar]
  3. Blattner, R.J.; Montgomery, S. A duality theorem for Hopf module algebras. J. Algebra 1985, 95, 153–172. [Google Scholar]
  4. Montgomery, S. Hopf Algebras and Their Actions on Rings; The Conference Board of the Mathematical Sciences: Washington, DC, USA, 1993; pp. 74–93. [Google Scholar]
  5. Drabant, B.; Daele, A.V.; Zhang, Y. Actions of multiplier Hopf algebras. Commun. Algebra 1999, 27, 4117–4172. [Google Scholar]
  6. Kassel, C. Quantum Groups; Springer: New York, NY, USA, 1995; pp. 146–157. [Google Scholar]
  7. Klimyk, A.; Schmüdgen, K. Quantum Groups and Their Representations; Springer: Berlin/Heidelberg, Germany, 1997; pp. 68–87. [Google Scholar]
  8. Castellani, L.; Wess, J. Quantum Groups and Their Applications in Physics; IOS Press: Amsterdam, The Netherlands, 1996; pp. 56–89. [Google Scholar]
  9. Duplij, S.; Sinel’shchikov, S. Classification of Uq(sl2)-module algebra structures on the quantum plane. J. Math. Phys. Anal. Geom. 2010, 6, 406–430+436,439. [Google Scholar]
  10. Duplij, S.; Sinel’shchikov, S. On Uq(sl2)-actions on the quantum plane. Acta Polytech. 2010, 50, 25–29. [Google Scholar]
  11. Duplij, S.; Hong, Y.; Li, F. Uq(sl(m+1))-module algebra structures on the coordinate algebra of a quantum vector space. J. Lie Theory 2015, 25, 327–361. [Google Scholar]
  12. Zhu, M. Classification of Quantum group An-Module Algebra Structures on Quantum Polynomial Algebras. Ph.D. Thesis, Yangzhou University, Yangzhou, China, 2012. [Google Scholar]
  13. Chan, K.; Wang, C.W.Y.; Zhang, J. Hopf actions on filtered regular algebras. J. Algebra 2014, 397, 68–90. [Google Scholar]
  14. Hu, N. Quantum divided power algebra, q-derivatives, and some new quantum groups. J. Algebra 2000, 232, 507–540. [Google Scholar]
  15. Ge, M.; Liu, G.; Xue, K. New solutions of Yang-Baxter equations: Birman-Wenzl algebra and quantum group structures. J. Phys. A Math. Gen. 1991, 24, 2679–2690. [Google Scholar]
  16. Aghamohammadi, A.; Karimipour, V.; Rouhani, S. The multiparametric non-standard deformation of An-1. J. Phys. A Math. Gen. 1993, 26, L75–L82. [Google Scholar]
  17. Jing, N.; Ge, M.; Wu, Y. A New Quantum Group Associated with a Non-standard Braid Group Representation. Lett. Math. Phys. 1991, 21, 193–203. [Google Scholar]
  18. Cheng, C.; Yang, S. Weak Hopf algebras corresponding to non-standard quantum groups. Bull. Korean Math. Soc. 2017, 54, 463–484. [Google Scholar]
  19. Su, D.; Yang, S. Representations of the small nonstandard quantum groups X ¯ q(A1). Comm. Algebra 2019, 47, 5039–5062. [Google Scholar]
  20. Su, D.; Gao, F.; Gao, Z. Module algebra structures of nonstandard quantum group Xq(A1) on the quantum plane. Electron. Res. Arch. 2025, 33, 3543–3560. [Google Scholar]
  21. Alev, J.; Chamarie, M. Dérivations et automorphismes de quelques algèbres quantiques. Comm. Algebra 1992, 20, 1787–1802. [Google Scholar]
  22. Artamonov, V.A. Quantum polynomial algebras. J. Math. Sci. 1997, 87, 3441–3462. [Google Scholar]
  23. Artamonov, V.A. Actions of pointlike Hopf algebras on quantum polynomials. Russian Math. Surveys 2000, 55, 1137–1138. [Google Scholar]
  24. Artamonov, V.A.; Wisbauer, R. Homological properties of quantum polynomials. Algebr. Represent. Theory 2001, 4, 219–247. [Google Scholar]
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Su, D. Module Algebra Structures of Non-Standard Quantum Group Xq(A1) on C[x,y,z]. Mathematics 2025, 13, 2227. https://doi.org/10.3390/math13142227

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Su D. Module Algebra Structures of Non-Standard Quantum Group Xq(A1) on C[x,y,z]. Mathematics. 2025; 13(14):2227. https://doi.org/10.3390/math13142227

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Su, Dong. 2025. "Module Algebra Structures of Non-Standard Quantum Group Xq(A1) on C[x,y,z]" Mathematics 13, no. 14: 2227. https://doi.org/10.3390/math13142227

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Su, D. (2025). Module Algebra Structures of Non-Standard Quantum Group Xq(A1) on C[x,y,z]. Mathematics, 13(14), 2227. https://doi.org/10.3390/math13142227

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