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# The kS3-Module Algebra Structures on M3(k)

by
Shiyin Zhao
1,*,
Yin Wang
2 and
Xiaojuan Chen
2
1
School of Literature and Science; Suqian College; Suqian 223800, China
2
School of Mathematical Science, Yangzhou University, Yangzhou 225002, China
*
Author to whom correspondence should be addressed.
Symmetry 2019, 11(2), 178; https://doi.org/10.3390/sym11020178
Submission received: 2 January 2019 / Revised: 27 January 2019 / Accepted: 1 February 2019 / Published: 2 February 2019

## Abstract

:
Let $S 3$ be the symmetric group on three elements. Let k be a field and $M 3 ( k )$ be the full matrix algebra of $3 × 3$-matrices over k. In this paper, the $k S 3$-module algebra structures on $M 3 ( k )$ are described, and classified up to isomorphism.
MSC:
16W30

## 1. Introduction

The theory of group actions and Hopf algebra actions on algebras is an important research topic in algebra, and many mathematicians have been working on the topic. In 1976, Beattie introduced the concept of Hopf algebra actions on algebras [1,2]. In 1985, Blattner and Montgomery studied a duality theorem for Hopf module algebras [3], which generalized the corresponding theorem of group actions. Then, many mathematicians were engaged in the theory of Hopf algebra actions. For example, Bergen and Cohen discussed the actions of commutative Hopf algebra in [4], and Cohen and Fishman studied general Hopf algebra actions in [5]. In 1993, Montgomery systematically summarized the achievement on Hopf algebra actions up to that point in reference [6]. For some other results about Hopf algebra actions on algebra, refer to [7,8,9,10,11,12,13]. It is well-known that a Yetter–Drinfeld module (resp., Yetter–Drinfeld module algebra) over a finite dimensional Hopf algebra H is the same as a module (resp., module algebra) over the Drinfeld double $D ( H )$ of H (see [6,11]). Chen and Zhang described the structures and classifications of 4-dimensional Yetter–Drinfeld module algebras over Sweedler 4-dimensional Hopf algebra $H 4$ in [14]. They also classified the Yetter–Drinfeld $H 4$-module algebra structures on the matrix algebra $M 2 ( k )$.
The aim of this article is to describe and classify all $k S 3$-module algebra structures on the matrix algebra $M 3 ( k )$, or the group actions of $S 3$ on $M 3 ( k )$, where $S 3$ is the symmetric group on three elements. In Section 1, we first recall some basic notions and results. Then, we discuss the weak similarity relation between the matrices in $M 3 ( k )$, and give the representative elements of equivalence classes with respect to the weak similarity for the two subsets of $G L 3 ( k )$ consisting of those matrices in $G L 3 ( k )$ that their square and cubic are scalar matrices, respectively. Finally, we describe the isomorphism classes of $k C 2$-module algebra structures and $k C 3$-module algebra structures on $M 3 ( k )$, respectively. In Section 2, we first discuss the weak similarity relation on the set $M n ( k ) × M n ( k )$, and describe the connection between the actions of $k S 3$ on the matrix algebra $M n ( k )$ and the weak similarity relation on $M n ( k ) × M n ( k )$. We show that the isomorphism classification of $k S 3$-module algebra structures on the matrix algebra $M n ( k )$ is equivalent to the equivalence classification on the set
${ ( X 1 , X 2 ) | X 1 , X 2 ∈ G L n ( k ) , X 1 2 ∈ k I n , X 2 3 ∈ k I n , ( X 1 X 2 ) 2 ∈ k I n }$
with respect to the weak similarity. Then, we give the representative elements of equivalence classes with respect to the weak similarity on the set for $n = 3$. Finally, we describe and classify all $k S 3$-module algebra structures on the matrix algebra $M 3 ( k )$.

## 2. The $kC 2$ and $kC 3$-Module Algebra Structures on $M 3 ( k )$

Throughout this paper, we work over a fixed field k. For a positive integer n, let $C n$ denote the cyclic group of order n, $S n$ the symmetric group on n elements. Let $M n ( k )$ be the full matrix algebra of $n × n$-matrices over k, and $I n$ the identity matrix in $M n ( k )$. Let $G L n ( k )$ be the multiplicative group of the invertible matrices in $M n ( k )$. Let $k * = k \ { 0 }$, the multiplicative group of nonzero elements in the field k, and $( k * ) 3 : = { α 3 | α ∈ k * }$. Then $( k * ) 3$ is a subgroup of $k *$, and one can form a quotient group $k * / ( k * ) 3$. For the basic notions and theory of Hopf algebras, the reader is directed to [6,15,16].
In this section we mainly investigate $k C 2$-module algebra structures and $k C 3$-module algebra structures on $M 3 ( k )$.
We remind the reader of the following useful fact from [6].
Proposition 1.
Let A be an algebra, G a group, and $A u t ( A )$ the automorphism group of A. Then, A is a left $k G$-module algebra if and only if there exists a group homomorphism map $ρ : G → A u t ( A )$, where the $k G$-action on A and the corresponding group homomorphism ρ are determined by
$g · a = ρ ( g ) ( a ) , g ∈ G , a ∈ A .$
Denote by $A ( ρ )$ the corresponding $k G$-module algebra A.
Let $C m = 〈 g 〉$ be a cyclic group of order m, and $A = M n ( k )$. Since A is a center simple algebra, all automorphisms of A are inner automorphisms. Hence, a group homomorphism $ρ : C m → A u t ( A )$ can be determined by an element $X ∈ G L n ( k )$ with $X m ∈ k I n$ as follows: $ρ ( g ) ( Y ) = X Y X − 1$ for any $Y ∈ M n ( k )$. We denote by $A ( X )$ the corresponding $k C m$-module algebra $A ( ρ )$.
Let $X , Y ∈ M n ( k )$. Recall from [17] that X and Y are weakly similar if $Z X Z − 1 = α Y$ for some $Z ∈ G L n ( k )$ and $α ∈ k *$. Obviously, two similar matrices are weakly similar, and weak similarity is an equivalence relation on the set $M n ( k )$. We have the following fact (see [17]).
Proposition 2.
Let $A = M n ( k )$. If $X 1 , X 2$ $∈ G L n ( k )$ with $X 1 m , X 2 m ∈ k I n$, then $A ( X 1 )$ is isomorphic to $A ( X 2 )$ as $k C m$-module algebras if and only if $X 1$ and $X 2$ are weakly similar.
Recall that two matrices in $M 3 ( k )$ are similar if and only if they have the same invariant factors. If $X ∈ M 3 ( k )$, then the set of invariant factors is one of the following: (1) $λ − α$, $λ − α$, $λ − α$ for some $α ∈ k$; (2) 1, $λ − α$, $( λ − α ) ( λ − β )$ for some $α , β ∈ k$; (3) 1, 1, $λ 3 − γ λ 2 − β λ − α$ for some $α , β , γ ∈ k$.
Lemma 1.
There are two equivalence classes on the set ${ X ∈ G L 3 ( k ) | X 2 ∈ k I 3 }$ with respect to the weak similarity, and their representatives are $I 3$ and $1 0 0 0 0 1 0 1 0$, respectively. In addition, if $c h a r ( k ) ≠ 2$, then we can choose $I 3$ and $1 0 0 0 1 0 0 0 − 1$ as their representative elements.
Proof.
Let $X ∈ G L 3 ( k )$ with $X 2 ∈ k I 3$. If the invariant factors of X are $λ − α$, $λ − α$, $λ − α$ for some $α ∈ k$, then X is similar to $α I 3$. Since $X ∈ G L 3 ( k )$, $α ≠ 0$, and hence X is weakly similar to $I 3$.
If the invariant factors of X are 1, $λ − α$, $( λ − α ) ( λ − β )$ for some $α$, $β ∈ k$, then X is similar to $α 0 0 0 0 − α β 0 1 α + β$. Since $X ∈ G L 3 ( k )$ and $X 2 ∈ k I 3$, we have
$0 ≠ α 0 0 0 0 − α β 0 1 α + β 2 = α 2 0 0 0 − α β − α β ( α + β ) 0 α + β α 2 + α β + β 2 ∈ k I 3 .$
Hence, $β = − α ≠ 0$, and so X is similar to $α 0 0 0 0 α 2 0 1 0$. Furthermore, since $α 0 0 0 0 α 2 0 1 0$ and $α 0 0 0 0 α 0 α 0$ are similar, X is weakly similar to $1 0 0 0 0 1 0 1 0$. In addition, if $c h a r ( k ) ≠ 2$, then $1 0 0 0 0 1 0 1 0$ and $1 0 0 0 1 0 0 0 − 1$ are similar.
If the invariant factors of X are 1, 1, $λ 3 − γ λ 2 − β λ − α$ for some $α , β , γ ∈ k$, then X is similar to $0 0 α 1 0 β 0 1 γ$. Since $X ∈ G L 3 ( k )$ and $X 2 ∈ k I 3$, we have
$0 0 α 1 0 β 0 1 γ 2 = 0 α α γ 0 β α + β γ 1 γ β + γ 2 ∈ k I 3 .$
This is impossible.
Obviously, $I 3$ and $1 0 0 0 0 1 0 1 0$ are not weakly similar. □
Lemma 2.
If $c h a r ( k ) = 3$, then the representative elements of equivalence classes on the set ${ X ∈ G L 3 ( k ) | X 3 ∈ k I 3 }$ with respect to the weak similarity are $I 3$, $1 0 0 0 1 0 0 1 1$ and $0 0 α 1 0 0 0 1 0$, respectively, where $α ∈ k * / ( k * ) 3$.
Proof.
Let $X ∈ G L 3 ( k )$ with $X 3 ∈ k I 3$. If the invariant factors of X are $λ − α$, $λ − α$, $λ − α$ for some $α ∈ k$, then X is similar to $α I 3$. Since $X ∈ G L 3 ( k )$, we have $α ≠ 0$. Hence, X is weakly similar to $I 3$.
If the invariant factors of X are 1, $λ − α$, $( λ − α ) ( λ − β )$ for some $α , β ∈ k$, then X is similar to $α 0 0 0 0 − α β 0 1 α + β$. Since $X ∈ G L 3 ( k )$ and $X 3 ∈ k I 3$, we have
$0 ≠ α 0 0 0 0 − α β 0 1 α + β 3 = α 3 0 0 0 − α β ( α + β ) − α β ( α − β ) 2 0 ( α − β ) 2 ( α + β ) ( α 2 + β 2 ) ∈ k I 3 ,$
where we use $c h a r ( k ) = 3$. Hence, $α = β ≠ 0$, X is similar to $α 0 0 0 0 − α 2 0 1 − α$. Since $α 0 0 0 0 − α 2 0 1 − α$ is similar to $α 0 0 0 0 − α 0 α − α$, X is weakly similar to $1 0 0 0 0 − 1 0 1 − 1$. Furthermore, $1 0 0 0 0 − 1 0 1 − 1$ and $1 0 0 0 1 0 0 1 1$ are similar, hence X is weakly similar to $1 0 0 0 1 0 0 1 1$.
If the invariant factors of X are 1, 1, $λ 3 − γ λ 2 − β λ − α$ for some $α , β , γ ∈ k$, then X is similar to $0 0 α 1 0 β 0 1 γ$. Since $X ∈ G L 3 ( k )$ and $X 3 ∈ k I 3$, we have
$0 0 α 1 0 β 0 1 γ 3 = α α γ α β + α γ 2 β α + β γ α γ + β 2 + β γ 2 γ β + γ 2 α − β γ + γ 3 ∈ k I 3 .$
Hence $α ≠ 0$, $β = γ = 0$, and X is similar to $0 0 α 1 0 0 0 1 0$. In this case, the invariant factors of X are 1, 1, $λ 3 − α$. For any $α 1 , α 2 ∈ k *$, $α 1 0 0 α 2 1 0 0 0 1 0 = 0 0 α 1 α 2 α 1 0 0 0 α 1 0$ has invariant factors 1, 1, $λ 3 − α 1 3 α 2$. Hence $0 0 α 1 0 0 0 1 0$ and $0 0 α 2 1 0 0 0 1 0$ are weakly similar if and only if $α = α 2$ in the quotient group $k * / ( k * ) 3$, where we denote the image of $α$ under the natural homomorphism $k * → k * / ( k * ) 3$ still by $α$.
Finally, by a straightforward verification, one knows that $I 3$, $1 0 0 0 1 0 0 1 1$ and $0 0 α 1 0 0 0 1 0$ are not weakly similar to each other. □
Lemma 3.
If $c h a r ( k ) ≠ 3$, then the representative elements of equivalence classes on the set ${ X ∈ G L 3 ( k ) | X 3 ∈ k I 3 }$ with respect to the weak similarity can be described as follows:
(1) if k does not contain a primitive 3rd root of unity, then they are $I 3$ and $0 0 α 1 0 0 0 1 0$, respectively, where $α ∈ k * / ( k * ) 3$;
(2) if k contains a primitive 3rd root ω of unity, then they are $I 3 ,$ $ω 0 0 0 1 0 0 0 1 ,$ $ω 0 0 0 ω 0 0 0 1$, and $0 0 α 1 0 0 0 1 0$, respectively, where $α ∈ k * / ( k * ) 3$.
Proof.
Let $X ∈ G L 3 ( k )$ with $X 3 ∈ k I 3$. If the invariant factors of X are $λ − α$, $λ − α$, $λ − α$ for some $α ∈ k$, then X is similar to $α I 3$. Since $X ∈ G L 3 ( k )$, we have $α ≠ 0$. Hence, X is weakly similar to $I 3$.
If the invariant factors of X are 1, $λ − α$, $( λ − α ) ( λ − β )$ for some $α , β ∈ k$, then X is similar to $α 0 0 0 0 − α β 0 1 α + β$. Since $X ∈ G L 3 ( k )$ and $X 3 ∈ k I 3$, we have
$0 ≠ α 0 0 0 0 − α β 0 1 α + β 3 = α 3 0 0 0 − α β ( α + β ) − α β ( α 2 + α β + β 2 ) 0 α 2 + α β + β 2 ( α + β ) ( α 2 + β 2 ) ∈ k I 3 .$
Thus, $α ≠ 0$ and $α 2 + α β + β 2 = 0$, which implies $( α − 1 β ) 2 + α − 1 β + 1 = 0$. If k does not contain a primitive 3rd root of unity, then for any $α , β ∈ k , ( α − 1 β ) 2 + α − 1 β + 1 ≠ 0$. In this case, there is no such X. If there exists a primitive 3rd root $ω$ of unity in k, then $β = ω α$ or $β = ω 2 α$. Hence, X is similar to $α 0 0 0 0 − ω α 2 0 1 − ω 2 α$ or $α 0 0 0 0 − ω 2 α 2 0 1 − ω α ,$ where $α ∈ k *$. One can easily check that $α 0 0 0 0 − ω α 2 0 1 − ω 2 α$ is weakly similar to $ω 0 0 0 1 0 0 0 1$, and that $α 0 0 0 0 − ω 2 α 2 0 1 − ω α$ is weakly similar to $ω 0 0 0 ω 0 0 0 1$. Moreover, $ω 0 0 0 1 0 0 0 1$ and $ω 0 0 0 ω 0 0 0 1$ are not weakly similar. Hence, X is weakly similar to $ω 0 0 0 1 0 0 0 1$ or $ω 0 0 0 ω 0 0 0 1$.
If the invariant factors of X are 1, 1, $λ 3 − γ λ 2 − β λ − α$ for some $α , β , γ ∈ k$, then X is similar to $0 0 α 1 0 β 0 1 γ$. Since $X ∈ G L 3 ( k )$ and $X 3 ∈ k I 3$, we have
$0 ≠ 0 0 α 1 0 β 0 1 γ 3 = α α γ α β + α γ 2 β α + β γ α γ + β 2 + β γ 2 γ β + γ 2 α + 2 β γ + γ 3 ∈ k I 3 .$
Hence, $α ≠ 0$ and $β = γ = 0$. Thus, X is similar to $0 0 α 1 0 0 0 1 0$, where $α ∈ k *$. For any $α , β ∈ k *$, by a straightforward verification, one can show that $0 0 α 1 0 0 0 1 0$ and $0 0 β 1 0 0 0 1 0$ are weakly similar if and only if $α β − 1 ∈ ( k * ) 3$.
Finally, it is easy to check that $0 0 α 1 0 0 0 1 0$ is not weakly similar to any one of $I 3$, $ω 0 0 0 1 0 0 0 1$, and $ω 0 0 0 ω 0 0 0 1$, where $α ∈ k * / ( k * ) 3$. □
Theorem 1.
Up to isomorphism, there are two $k C 2$-module algebra structures on $M 3 ( k )$ as follows:
(1) $g · X = X$;
(2) $g · x 11 x 12 x 13 x 21 x 22 x 23 x 31 x 32 x 33 =$ $x 11 x 13 x 12 x 31 x 33 x 32 x 21 x 23 x 22$,
where $X = ( x i , j ) ∈ M 3 ( k )$, g is the generator of cyclic group $C 2$. In addition, if $c h a r ( k ) ≠ 2$, then the $k C 2$-module algebra in (2) is isomorphic to the following one:
(3) $g · x 11 x 12 x 13 x 21 x 22 x 23 x 31 x 32 x 33 =$ $x 11 x 12 − x 13 x 21 x 22 − x 23 − x 31 − x 32 x 33$.
Proof.
It follows by Proposition 2 and Lemma 1. □
Theorem 2.
Up to isomorphism, any $k C 3$-module algebra structure on $M 3 ( k )$ is one of the following:
(a) if $c h a r ( k ) = 3$, then
(1) $g · X = X$;
(2) $g · x 11 x 12 x 13 x 21 x 22 x 23 x 31 x 32 x 33 =$ $x 11 x 12 − x 13 x 13 x 21 x 22 − x 23 x 23 x 21 + x 31 x 22 + x 32 − x 23 − x 33 x 23 + x 33$;
(3) $g · x 11 x 12 x 13 x 21 x 22 x 23 x 31 x 32 x 33 =$ $x 33 α x 31 α x 32 α − 1 x 13 x 11 x 12 α − 1 x 23 x 21 x 22$,
(b) if $c h a r ( k ) ≠ 3$ and k does not contain a primitive 3rd root of unity, then
(1) $g · X = X$;
(2) $g · x 11 x 12 x 13 x 21 x 22 x 23 x 31 x 32 x 33 =$ $x 33 α x 31 α x 32 α − 1 x 13 x 11 x 12 α − 1 x 23 x 21 x 22$,
(c) if $c h a r ( k ) ≠ 3$ and k contains a primitive 3rd root ω of unity, then
(1) $g · X = X$;
(2) $g · x 11 x 12 x 13 x 21 x 22 x 23 x 31 x 32 x 33 =$ $x 11 ω x 12 ω x 13 ω 2 x 21 x 22 x 23 ω 2 x 31 x 32 x 33$;
(3) $g · x 11 x 12 x 13 x 21 x 22 x 23 x 31 x 32 x 33 =$ $x 11 x 12 ω x 13 x 21 x 22 ω x 23 ω 2 x 31 ω 2 x 32 x 33$;
(4) $g · x 11 x 12 x 13 x 21 x 22 x 23 x 31 x 32 x 33 =$ $x 33 α x 31 α x 32 α − 1 x 13 x 11 x 12 α − 1 x 23 x 21 x 22$,
where $X = ( x i , j ) ∈ M 3 ( k )$, g is the generator of cyclic group $C 3$, and $α ∈ k * / ( k * ) 3$.
Proof.
It follows by Proposition 2, and Lemmas 2 and 3. □

## 3. The $kS 3$-Module Algebra Structures on $M 3 ( k )$

The aim of this section is to describe and classify $k S 3$-module algebra structures on $M 3 ( k )$. Let us describe $S 3$ by generators and relations as follows:
$S 3 = 〈 g 1 , g 2 | g 1 2 = 1 , g 2 3 = 1 , ( g 1 g 2 ) 2 = 1 〉 .$
Definition 1.
Let $X 1 , X 2 , Y 1 , Y 2 ∈ M n ( k )$. $( X 1 , X 2 )$ and $( Y 1 , Y 2 )$ are weakly similar if $X 1 = α X Y 1 X − 1 , X 2 = β X Y 2 X − 1$ for some $X ∈ G L n ( k )$ and $α , β ∈ k *$. Denoted by $( X 1 , X 2 ) ∼ ( Y 1 , Y 2 )$.
Remark 1.
A straightforward computation shows that ∼ is an equivalence relation on the set $M n ( k ) × M n ( k )$. Let
$Λ n = { ( X 1 , X 2 ) | X 1 , X 2 ∈ G L 3 ( k ) , X 1 2 ∈ k I 3 , X 2 3 ∈ k I 3 , ( X 1 X 2 ) 2 ∈ k I 3 } .$
Obviously, $Λ n$ is a disjoint union of some equivalence classes on $M n ( k ) × M n ( k )$ with respect to the weak similarity.
Lemma 4.
Let $A = M n ( k )$. Then, a group homomorphism $ρ : S 3 → A u t ( A )$ can be determined by an element $( X 1 , X 2 ) ∈ Λ n$. Denote the corresponding $k S 3$-module algebra $A ( ρ )$ by $A ( X 1 , X 2 )$.
Proof.
Let $ρ : S 3 → A u t ( A )$ be a group homomorphism. Then, both $ρ ( g 1 )$ and $ρ ( g 2 )$ are algebra automorphisms of A. Since A is a center simple algebra, all automorphisms of A are inner automorphisms. Hence, there exist two elements $X 1 , X 2 ∈ G L n ( k )$ such that $ρ ( g i ) ( Y ) = X i Y X i − 1$ for any $Y ∈ M n ( k )$, $i = 1 , 2$. Since $g 1 2 = 1$, $ρ ( g 1 2 ) = id$. Hence $ρ ( g 1 2 ) ( Y ) = ρ ( g 1 ) 2 ( Y ) = Y$ for any $Y ∈ M n ( k )$. That is, $X 1 2 Y = Y X 1 2$ for any $Y ∈ M n ( k )$. It follows that $X 1 2 ∈ k I n$.
Similarly, by $g 2 3 = 1$ and $( g 1 g 2 ) 2 = 1$, one gets that $X 2 3 ∈ k I n$ and $( X 1 X 2 ) 2 ∈ k I n$, respectively. □
Theorem 3.
If $( X 1 , X 2 ) , ( Y 1 , Y 2 ) ∈ Λ n$, then $A ( X 1 , X 2 )$ is isomorphic to $A ( Y 1 , Y 2 )$ as $k S 3$-module algebras if and only if $( X 1 , X 2 ) ∼ ( Y 1 , Y 2 )$.
Proof.
Let $( X 1 , X 2 ) , ( Y 1 , Y 2 ) ∈ Λ n$. Assume that $f : A ( X 1 , X 2 ) → A ( Y 1 , Y 2 )$ is a $k S 3$-module algebra isomorphism. As k-algebras, $A ( X 1 , X 2 ) = A ( Y 1 , Y 2 ) = M n ( k )$, which is a center simple algebra. It follows that f is an algebra automorphism of $M n ( k )$. Therefore, there is an $X ∈ G L n ( k )$ such that $f ( Y ) = X Y X − 1$ for all $Y ∈ A ( X 1 , X 2 )$. Since f is a $k S 3$-module homomorphism, we have $f ( g i · Y ) = g i · f ( Y )$ for any $Y ∈ A ( X 1 , X 2 )$, $i = 1 , 2$. However, $f ( g i · Y ) = f ( X i Y X i − 1 ) = X X i Y X i − 1 X − 1$ and $g i · f ( Y ) = Y i X Y X − 1 Y i − 1$, $i = 1 , 2$. Hence $X − 1 Y i − 1 X X i Y = Y X − 1 Y i − 1 X X i$ for any $Y ∈ M n ( k )$, which implies that $X − 1 Y i − 1 X X i = α i I n$ for some $α i ∈ k *$, $i = 1 , 2$. Thus, $X i = α i X − 1 Y i X$ for $i = 1 , 2$, and so $( X 1 , X 2 ) ∼ ( Y 1 , Y 2 )$.
Conversely, assume that $( X 1 , X 2 ) ∼ ( Y 1 , Y 2 )$. Then, there exist $X ∈ G L n ( k )$ and $α 1 , α 2 ∈ k *$ such that $X i = α i X − 1 Y i X$, $i = 1 , 2$. Let $f : A ( X 1 , X 2 ) → A ( Y 1 , Y 2 )$ be defined by $f ( Y ) = X Y X − 1$, $Y ∈ A ( X 1 , X 2 )$. Then, from the above proof, it is easy to see that f is a $k S 3$-module algebra isomorphism from $A ( X 1 , X 2 )$ to $A ( Y 1 , Y 2 )$. □
Theorem 4.
(1) Let $( X 1 , X 2 ) ∈ Λ n$ and $Y 2 ∈ G L n ( k )$ such that $X 2$ and $Y 2$ are weakly similar. Then $( X 1 , X 2 ) ∼ ( Y 1 , Y 2 )$ for some $Y 1 ∈ G L n ( k )$, and $( Y 1 , Y 2 ) ∈ Λ n$.
(2) Let $( X 1 , X 2 ) ∈ Λ n$ and $Y 1 ∈ G L n ( k )$ such that $X 1$ and $Y 1$ are weakly similar. Then $( X 1 , X 2 ) ∼ ( Y 1 , Y 2 )$ for some $Y 2 ∈ G L n ( k )$, and $( Y 1 , Y 2 ) ∈ Λ n$.
Proof.
(1) Since $X 2$ and $Y 2$ are weakly similar, we have $X X 2 X − 1 = β Y 2$ for some $X ∈ G L n ( k )$ and $β ∈ k *$. Hence $Y 2 3 = β − 3 X X 2 3 X − 1 ∈ k I n$. Now let $Y 1 = X X 1 X − 1$. Then $Y 1 ∈ G L n ( k )$, $Y 1 2 = X X 1 2 X − 1 ∈ k I n$ and $( Y 1 Y 2 ) 2 = β − 2 X ( X 1 X 2 ) 2 X − 1 ∈ k I n$. Thus, $( Y 1 , Y 2 ) ∈ Λ n$. Clearly $( X 1 , X 2 ) ∼ ( Y 1 , Y 2 )$.
(2) It follows similarly. □
From now on, we fix $n = 3$ and denote $Λ 3$ by $Λ$ for simplicity. Then, by Proposition 1, Lemma 4, and Theorem 3, in order to classify the $k S 3$-module algebra structures on the matrix algebra $M 3 ( k )$, we only need to consider the equivalence classification of the set $Λ$ with respect to the weak similarity ∼.
Lemma 5.
Assume that $c h a r ( k ) = 3$. Let $X 1 ∈ G L 3 ( k )$ with $( X 1 , I 3 ) ∈ Λ$. Then $( X 1 , I 3 )$ are weakly similar to $( I 3 , I 3 )$ or $( 1 0 0 0 1 0 0 0 − 1 , I 3 )$. Moreover, $( I 3 , I 3 )$ and $( 1 0 0 0 1 0 0 0 − 1 , I 3 )$ are not weakly similar.
Proof.
It follows from Lemma 1. □
Lemma 6.
Assume that $c h a r ( k ) = 3$. Let $X 1 ∈ G L 3 ( k )$ with $( X 1 , 1 0 0 0 1 0 0 1 1 ) ∈ Λ$. Then $( X 1 , 1 0 0 0 1 0 0 1 1 )$ is weakly similar to $( 1 0 0 0 1 0 0 0 − 1 , 1 0 0 0 1 0 0 1 1 )$ or $( 1 0 0 0 − 1 0 0 0 1 , 1 0 0 0 1 0 0 1 1 )$. Moreover, $( 1 0 0 0 1 0 0 0 − 1 , 1 0 0 0 1 0 0 1 1 )$ and $( 1 0 0 0 − 1 0 0 0 1 , 1 0 0 0 1 0 0 1 1 )$ are not weakly similar.
Proof.
Let $X 1 = x 11 x 12 x 13 x 21 x 22 x 23 x 31 x 32 x 33$ and $X 2 = 1 0 0 0 1 0 0 1 1$. Then $X 1 2 = m I 3$ and $( X 1 X 2 ) 2 = n I 3$ for some $m , n ∈ k *$. We have $X 1 − 1 = m − 1 X 1$, $X 1 X 2 = n X 2 − 1 X 1 − 1 = n m − 1 X 2 − 1 X 1 = α X 2 − 1 X 1$, where $α = n m − 1 ∈ k *$. That is,
$x 11 x 12 + x 13 x 13 x 21 x 22 + x 23 x 23 x 31 x 32 + x 33 x 33 = α x 11 x 12 x 13 x 21 x 22 x 23 − x 21 + x 31 − x 22 + x 32 − x 23 + x 33 .$
Hence $α = 1$, $x 13 = x 23 = x 21 = 0$, $x 33 = − x 22$, and so $X 1 = x 11 x 12 0 0 x 22 0 x 31 x 32 − x 22$. From $X 1 2 ∈ k I 3$, one gets that $x 11 = x 22$ and $x 12 = 0$, or $x 11 = − x 22$ and $x 31 = 0$. Thus, $X 1 = x 11 0 0 0 x 11 0 x 31 x 32 − x 11$ or $X 1 = x 11 x 12 0 0 − x 11 0 0 x 32 x 11$, where $x 11 ≠ 0$.
If $X 1 = x 11 0 0 0 x 11 0 x 31 x 32 − x 11$, take $X = 1 0 0 0 1 0 − 1 2 x 11 − 1 x 31 − 1 2 x 11 − 1 x 32 1$. Then $X ∈ G L 3 ( k )$. By a straightforward computation, we have
$X X 1 = x 11 1 0 0 0 1 0 0 0 − 1 X , X X 2 = X 2 X .$
Hence $( X 1 , X 2 )$ is weakly similar to $( 1 0 0 0 1 0 0 0 − 1 , 1 0 0 0 1 0 0 1 1 )$ in this case.
If $X 1 = x 11 x 12 0 0 − x 11 0 0 x 32 x 11$, take $X = 1 1 2 x 11 − 1 x 12 0 0 1 0 0 1 2 x 11 − 1 x 32 1$. Then $X ∈ G L 3 ( k )$. By a straightforward computation, one knows that
$X X 1 = x 11 1 0 0 0 − 1 0 0 0 1 X , X X 2 = X 2 X .$
Hence $( X 1 , X 2 )$ is weakly similar to $( 1 0 0 0 − 1 0 0 0 1 , 1 0 0 0 1 0 0 1 1 )$ in this case.
Finally, by a straightforward verification, one can show that
$( 1 0 0 0 1 0 0 0 − 1 , 1 0 0 0 1 0 0 1 1 ) and ( 1 0 0 0 − 1 0 0 0 1 , 1 0 0 0 1 0 0 1 1 )$
are not weakly similar. □
Lemma 7.
Assume $c h a r ( k ) = 3$. Let $X 1 ∈ G L 3 ( k )$ with $( X 1 , 0 0 α 1 0 0 0 1 0 ) ∈ Λ$, where $α ∈ k *$. Then $( X 1 , 0 0 α 1 0 0 0 1 0 )$ and $( 1 0 0 0 0 1 0 1 0 , 0 0 1 1 0 0 0 1 0 )$ are weakly similar.
Proof.
Let $X 1 = x 11 x 12 x 13 x 21 x 22 x 23 x 31 x 32 x 33$ and $X 2 = 0 0 α 1 0 0 0 1 0$. Then, by the proof of Lemma 6, we have $X 1 X 2 = β X 2 − 1 X 1$ for some $β ∈ k *$. That is,
$x 12 x 13 x 11 α x 22 x 23 x 21 α x 32 x 33 x 31 α = β x 21 x 22 x 23 x 31 x 32 x 33 x 11 α − 1 x 12 α − 1 x 13 α − 1 .$
Hence $x 12 = β x 21 = α − 1 β 2 x 33 = α − 2 β 3 x 12$, $x 13 = β x 22 = β 2 x 31 = α − 2 β 3 x 13$ and $x 11 = α − 1 β x 23 = α − 1 β 2 x 32 = α − 2 β 3 x 11$. It follows that $α − 2 β 3 = 1$ and so $α = ( α β − 1 ) 3 ∈ ( k * ) 3$. Thus, by Lemma 2 and Theorem 4, we may assume that $α = 1$. Since $c h a r ( k ) = 3$, $X 1 2 ∈ k I 3$ and $| X 1 | ≠ 0$, we have $β = 1$, $X 1 = x 11 x 33 x 22 x 33 x 22 x 11 x 22 x 11 x 33$, $x 11 x 22 + x 22 x 33 + x 33 x 11 = 0$, $x 11 2 + x 22 2 + x 33 2 ≠ 0$ and $x 11 + x 22 + x 33 ≠ 0$.
Now let $X = 0 1 0 0 0 1 1 0 0$. Then $X ∈ G L 3 ( k )$. By a straightforward computation, we have
$X X 1 = x 22 x 11 x 33 x 11 x 33 x 22 x 33 x 22 x 11 X , X X 2 = X 2 X .$
Hence $( X 1 , X 2 )$ and $( x 22 x 11 x 33 x 11 x 33 x 22 x 33 x 22 x 11 , X 2 )$ are weakly similar. Similarly, one can show that $( x 22 x 11 x 33 x 11 x 33 x 22 x 33 x 22 x 11 , X 2 )$ and $( x 33 x 22 x 11 x 22 x 11 x 33 x 11 x 33 x 22 , X 2 )$ are weakly similar. Hence we may assume $x 11 ≠ 0$ since at least one of $x 11$, $x 22$, and $x 33$ is not equal to zero. Then, $X 1 = x 11 1 x 11 − 1 x 33 x 11 − 1 x 22 x 11 − 1 x 33 x 11 − 1 x 22 1 x 11 − 1 x 22 1 x 11 − 1 x 33$. By replacing $x 11 − 1 x 22$ and $x 11 − 1 x 33$ with $x 22$ and $x 33$, respectively, it follows that $( X 1 , X 2 )$ is weakly similar to $( 1 x 33 x 22 x 33 x 22 1 x 22 1 x 33 , X 2 )$ for some $x 22 , x 33 ∈ k$ with $x 22 + x 22 x 33 + x 33 = 0$ and $1 + x 22 + x 33 ≠ 0$. In this case, we have $x 22 ( 1 + x 33 ) = − x 33$. If $1 + x 33 = 0$, then $x 33 = − 1$, but on the other hand $x 33 = − x 22 ( 1 + x 33 ) = 0$, a contradiction. Hence $x 33 + 1 ≠ 0$, and so $x 22 = − x 33 ( 1 + x 33 ) − 1$. Thus, $( X 1 , X 2 )$ is weakly similar to $( 1 x 33 − x 33 ( 1 + x 33 ) − 1 x 33 − x 33 ( 1 + x 33 ) − 1 1 − x 33 ( 1 + x 33 ) − 1 1 x 33 , X 2 )$, where $x 33 ≠ − 1$, $x 33 2 + x 33 + 1 ≠ 0$ and $x 33 ≠ 1$.
Now let $Y = − 1 0 x 33 x 33 − 1 0 0 x 33 − 1$. Then $| Y | = x 33 3 − 1 = ( x 33 − 1 ) 3 ≠ 0$, and hence Y is an invertible matrix. Finally, by a direct computation, we have
$Y 1 x 33 − x 33 ( 1 + x 33 ) − 1 x 33 − x 33 ( 1 + x 33 ) − 1 1 − x 33 ( 1 + x 33 ) − 1 1 x 33 = ( x 33 2 + x 33 + 1 ) ( x 33 + 1 ) − 1 1 0 0 0 0 1 0 1 0 Y$
and $Y X 2 = X 2 Y$. Hence
$( 1 x 33 − x 33 ( 1 + x 33 ) − 1 x 33 − x 33 ( 1 + x 33 ) − 1 1 − x 33 ( 1 + x 33 ) − 1 1 x 33 , X 2 )$
is weakly similar to $( 1 0 0 0 0 1 0 1 0 , X 2 )$. Thus, $( X 1 , X 2 )$ is weakly similar to
$( 1 0 0 0 0 1 0 1 0 , 0 0 1 1 0 0 0 1 0 ) .$
□
Theorem 5.
Assume that $c h a r ( k ) = 3$. Then there are five equivalence classes on the set Λ with respect to the weak similarity ∼, and their representatives are the following:
$( I 3 , I 3 ) , ( 1 0 0 0 1 0 0 0 − 1 , I 3 ) , ( 1 0 0 0 1 0 0 0 − 1 , 1 0 0 0 1 0 0 1 1 ) ,$
$( 1 0 0 0 − 1 0 0 0 1 , 1 0 0 0 1 0 0 1 1 ) , ( 1 0 0 0 0 1 0 1 0 , 0 0 1 1 0 0 0 1 0 ) .$
Proof.
It follows by Lemma 2, Theorem 4, and Lemmas 5, 6, 7. □
Lemma 8.
Assume that $c h a r ( k ) ≠ 3$. Let $X 1 ∈ G L 3 ( k )$ with $( X 1 , I 3 ) ∈ Λ$. Then $( X 1 , I 3 )$ is weakly similar to $( I 3 , I 3 )$ or $( 1 0 0 0 0 1 0 1 0 , I 3 ) .$ Moreover, $( I 3 , I 3 )$ and $( 1 0 0 0 0 1 0 1 0 , I 3 )$ are not weakly similar.
Proof.
It follows by Lemma 1. □
Lemma 9.
Assume that $c h a r ( k ) ≠ 3$. Let $X 1 ∈ G L 3 ( k )$ and $X 2 = 0 0 α 1 0 0 0 1 0$, where $α ∈ k *$. If $( X 1 , X 2 ) ∈ Λ$. Then $( X 1 , X 2 )$ is weakly similar to
$( 1 0 0 0 0 1 0 1 0 , 0 0 1 1 0 0 0 1 0 ) .$
Proof.
It follows by Lemma 3 and an argument similar to the proofs of Lemmas 6 and 7. □
Lemma 10.
Assume that $c h a r ( k ) ≠ 3$ and that k contains a primitive 3rd root ω of unity. Then $( X 1 , ω 0 0 0 1 0 0 0 1 ) ∉ Λ$ and $( X 1 , ω 0 0 0 ω 0 0 0 1 ) ∉ Λ$ for any $X 1 ∈ G L 3 ( k )$.
Proof.
Suppose that there existed an $X 1 = x 11 x 12 x 13 x 21 x 22 x 23 x 31 x 32 x 33 ∈ G L 3 ( k )$ such that $( X 1 , ω 0 0 0 1 0 0 0 1 ) ∈ Λ$. Let $X 2 = ω 0 0 0 1 0 0 0 1$. Then, by the proof of Lemma 6, we have $X 1 X 2 = β X 2 − 1 X 1$ for some $β ∈ k *$. That is,
$x 11 ω x 12 x 13 x 21 ω x 22 x 23 x 31 ω x 32 x 33 = β x 11 ω − 1 x 12 ω − 1 x 13 ω − 1 x 21 x 22 x 23 x 31 x 32 x 33 .$
Since $X 1$ is an invertible matrix, the third row of $X 1$ is not zero. Hence we have $β = ω$ or $β = 1$. If $β = ω$, then $x 11 = x 22 = x 23 = x 32 = x 33 = 0$, which implies that $X 1 = 0 x 12 x 13 x 21 0 0 x 31 0 0$ is not invertible, a contradiction. If $β = 1$, then $x 11 = x 12 = x 13 = 0$, which also implies that $X 1$ is not invertible, a contradiction. Thus, $( X 1 , ω 0 0 0 1 0 0 0 1 ) ∉ Λ$ for any $X 1 ∈ G L 3 ( k )$.
By a similar argument, one can show that $( X 1 , ω 0 0 0 ω 0 0 0 1 ) ∉ Λ$ for any $X 1 ∈ G L 3 ( k )$. □
Theorem 6.
Assume that $c h a r ( k ) ≠ 3$. Then, there are three equivalence classes on the set Λ with respect to the weak similarity ∼, and their representatives are $( I 3 , I 3 )$, $( 1 0 0 0 0 1 0 1 0 , I 3 )$ and $( 1 0 0 0 0 1 0 1 0 , 0 0 1 1 0 0 0 1 0 )$, respectively.
Proof.
It follows by Lemmas 3, 8, 9, 10 and Theorem 4. □
Theorem 7.
Assume that char $( k ) = 3$. Then, up to isomorphism, there are five $k S 3$-module algebra structures on $M 3 ( k )$, which are given respectively by
(1) $g 1 · X = X$, $g 2 · X = X$;
(2) $g 1 · x 11 x 12 x 13 x 21 x 22 x 23 x 31 x 32 x 33 = x 11 x 12 − x 13 x 21 x 22 − x 23 − x 31 − x 32 x 33$, $g 2 · X = X$;
(3) $g 1 · x 11 x 12 x 13 x 21 x 22 x 23 x 31 x 32 x 33 = x 11 x 12 − x 13 x 21 x 22 − x 23 − x 31 − x 32 x 33$,
$g 2 · x 11 x 12 x 13 x 21 x 22 x 23 x 31 x 32 x 33 = x 11 x 12 − x 13 x 13 x 21 x 22 − x 23 x 23 x 21 + x 31 x 22 − x 23 + x 32 − x 33 x 23 + x 33$;
(4) $g 1 · x 11 x 12 x 13 x 21 x 22 x 23 x 31 x 32 x 33 = x 11 − x 12 x 13 − x 21 x 22 − x 23 x 31 − x 32 x 33$,
$g 2 · x 11 x 12 x 13 x 21 x 22 x 23 x 31 x 32 x 33 = x 11 x 12 − x 13 x 13 x 21 x 22 − x 23 x 23 x 21 + x 31 x 22 − x 23 + x 32 − x 33 x 23 + x 33$;
(5) $g 1 · x 11 x 12 x 13 x 21 x 22 x 23 x 31 x 32 x 33 = x 11 x 13 x 12 x 31 x 33 x 32 x 21 x 23 x 22$,
$g 2 · x 11 x 12 x 13 x 21 x 22 x 23 x 31 x 32 x 33 = x 33 x 31 x 32 x 13 x 11 x 12 x 23 x 21 x 22$,
where $X = ( x i , j ) ∈ M 3 ( k )$.
Proof.
It follows by Proposition 1, Lemma 4, and Theorems 3 and 5. □
Theorem 8.
Assume that char $( k ) ≠ 3$. Then, up to isomorphism, there are three $k S 3$-module algebra structures on $M 3 ( k )$, which are given respectively by
(1) $g 1 · X = X$, $g 2 · X = X$;
(2) $g 1 · x 11 x 12 x 13 x 21 x 22 x 23 x 31 x 32 x 33 = x 11 x 13 x 12 x 31 x 33 x 32 x 21 x 23 x 22$, $g 2 · X = X$;
(3) $g 1 · x 11 x 12 x 13 x 21 x 22 x 23 x 31 x 32 x 33 = x 11 x 13 x 12 x 31 x 33 x 32 x 21 x 23 x 22$,
$g 2 · x 11 x 12 x 13 x 21 x 22 x 23 x 31 x 32 x 33 = x 33 x 31 x 32 x 13 x 11 x 12 x 23 x 21 x 22$,
where $X = ( x i , j ) ∈ M 3 ( k )$.
Proof.
It follows by Proposition 1, Lemma 4, and Theorems 3 and 6. □

## Author Contributions

All authors contributed equally.

## Funding

This research is supported by National Natural Science Foundation of China (Grant No. 11171291).

## Acknowledgments

The authors would like to express sincere thanks to the referees for careful reading and suggestions which helped us to improve the paper.

## Conflicts of Interest

The authors declare no conflict of interest.

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Zhao, S.; Wang, Y.; Chen, X. The kS3-Module Algebra Structures on M3(k). Symmetry 2019, 11, 178. https://doi.org/10.3390/sym11020178

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Zhao S, Wang Y, Chen X. The kS3-Module Algebra Structures on M3(k). Symmetry. 2019; 11(2):178. https://doi.org/10.3390/sym11020178

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Zhao, Shiyin, Yin Wang, and Xiaojuan Chen. 2019. "The kS3-Module Algebra Structures on M3(k)" Symmetry 11, no. 2: 178. https://doi.org/10.3390/sym11020178

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