Zhichao Chen,Jiayi Cai,Lingchao Meng and Libin Li

School of Mathematical Sciences,Yangzhou University,Yangzhou 225002,China.

Abstract.The Z+-ring is an important invariant in the theory of tensor category.In this paper,by using matrix method,we describe all irreducible Z+-modules over a Z+-ring A,where A is a commutative ring with a Z+-basis{1,x,y,xy}and relations:

Key words:Non-negative integer,matrix representation,irreducible Z+-module,Z+-ring.

1 Introduction

Tensor categories are usually thought as counterparts of groups and rings in the world of categories.They are ubiquitous in noncommutative algebra and representation theory.The Z+-ring is an important invariant in the theory of tensor category.The terminology on Z+-rings comes from the paper by Lusztig[1]as well as[2,3].Such rings were also studied by Davydov in[4,5].The basic concepts and facts about Z+-rings can be found in[3,6].Examples of Z+-rings include the Green rings of Hopf algebras[7,8,9,10,11,12]and the Grothendieck rings of tensor categories[13,14,15,16,17].The terminology on Z+-modules(or Z+-representations)is from[2,3,6]which is different from the common representations over rings or algebras.A Z+-module M is called irreducible if any Z+-submodule of M is 0 or M.A Z+-module M is called indecomposable if it is not equivalent to a direct sum of two nonzero Z+-modules.A Z+-module over a based ring satisfying the rigid condition is called a based module in[3]or a NIM-representation(the non-negative integer matrix representation)in[18,19,20,21].It is quite difficult to classify all irreducible Z+-modules but the job is meaningful and interesting.

In[3],Ostrik proved that for a given Z+-ring of finite rank there exist only finite inequivalent irreducible Z+-modules.In the proofs,Ostrik indicated the rank of any irreducible Z+-module has an upper bound.However,for many situations,the upper bound seems quite large.For a given Z+-ring,it is difficult to determine whether the rank of an irreducible Z+-module is less than or equal to the rank of this ring.In this paper,by using matrix method,we describe all irreducible inequivalent Z+-modules over a Z+-ring A,where A is a commutative Z+-ring with a Z+-basis{1,x,y,xy}and relations:x2=1,y2=1+x+xy.In addition,we find the rank of each irreducible Z+-module is less than or equal to the rank of A.In fact,the problem is equivalent to studying the irreducible NIM solutions to a system of matrix equations:

We analyze all the situations when the order of the matrix is n,and we get the results completely.When the rank of Z+-modules n≥5,there does not exist irreducible Z+-modules and when the rank n≤4,there exists finite inequivalent irreducible Z+-modules,the number of which is respectively 1,3,3,2 when the rank runs from 1 to 4.

The rest of the paper is outlined as follows.In Section 2,we recall some relevant theorems and prove the preliminary propositions we will use in the next sections.In Section 3,we first exhibit the system of matrix equations,then we prove that there does not exist irreducible NIM solutions when order n≥5.In Section 4,we exhibit the concrete irreducible NIM solutions when order n≤4.More explicitly,the irreducible Z+-module over this Z+-ring A exists if and only if the rank of module is less than or equal to 4.The concrete irreducible NIM representations can be seen in table 4.1.Furthermore,in each case,we classify the irreducible Z+-modules under the equivalence in Table 4.2.

2 Preliminaries

2.1 Basic definitions and notation

2.1.1 The theory of matrix

We assume all matrices in this paper belong to Mn(N),where Mn(N)means the set consisting of n-order square matrices with only natural number elements.For any matrix A,we denote the element at the i-th row and j-th column of A by aij.If aij>0 for all i,j,then A is called positive.If aij≥0 for all i,j,then A is called non-negative.Besides,if A∈Mn(N),we call it non-negative integer matrix(NIM).

Definition 2.6.We call two Z+-modules M,M′over A with bases{mi}i∈Jand{m′j}j∈J′equivalent,if there exists a bijection φ:J→J′such that the induced Z-linear map?φ of abelian groups M,M′defined by?φ(mi)=m′φ(i)is an isomorphism of A-modules.In other words,we call two Z+-module M and M′of rank n equivalent if there exists an n×n permutation matrix P such that M′i=PMiP?1for all i∈I.

Definition 2.7.For a Z+-submodule N of a Z+-module M with a basis{ml}l∈J,we mean an abelian subgroup of M spaned by{ml}l∈J′,where J′is a subset of J such that N is an A-submodule.

Definition 2.8.We call a Z+-module M irreducible if any Z+-submodule of M is 0 or M(In other words,the Z-span of any proper subset of the basis of M is not an A-submodule).

The irreducible Z+-module can also be defined as follows:

Definition 2.9.A Z+-module M is called reducible if Miis simultaneously reducible for any i∈I.M is called irreducible when all Micannot be simultaneously put into triangular block form and it is called irreducible NIM-representation.

2.2 Auxiliary results

To prove our results we will need the following statements.

Theorem 2.1.The necessary and sufficient condition for the non-negative matrix A with order n to be irreducible is(E+A)n?1>0.

Proof.A proof of the theorem above can be found in the book of Gantmacher[22,Chapter XIII,Section 1].

Proof.A proof of the theorem above can be found in the book of Gantmacher[22,Chapter XIII,Section 2].

2.3 Preliminary propositions

Proposition 2.1.If A,B∈Mn(N)and satisfy AB=E.Then A and B are both permutation matrices.

where ai1b1j≥0,ai2b2j≥0,...,aijbjj≥2bjj,...,ainbnj≥0.In this case,we can get bjj=0.Similarly,considering the elements δi1,δi2···δin,we can get bj1,bj2···bjn=0.But under this circumstance,there exists a line where all elements are 0 in B which contradicts with the fact that the determinant of B is not 0.So,the assumption does not holds,and any element of A is 0 or 1.According to AB=E,we can get BA=E.By using the same method,we can prove any element of B is 0 or 1 as well.Now we will prove that there is only one 1 in per row and per column of A and B.

Since A and B are both reversible, there exists at least one 1 in per row and per column.We give the following discussion to prove that there exists at most one 1 in per row and per column.We might as well assume ai1,ai2are both 1 and the remaining elements of the i-th row are 0.Considering the element δiiof E,δii=ai1b1i+ai2b2i+···+ainbni=1 and we have b1i=0 or b2i=0.We might as well assume that b1i=0,then b2i=1.Considering that the elements of E,δi1,δi2,···,δii?1,δii+1,···,δinare all equal to 0,we have that b11=0,b12=0,···,b1i?1=0,b1i+1=0,···,b1n=0.Combined with b1i=0 we know there exists a zero line in B which contradicts with the fact that the determinant of B is not equal to 0.Therefore,we have the conclusion that there exists at most one 1 in per row and per column of A and the same conclusion is true on matrix B.Hence,there is only one 1 in per row and per column of A and B.Then A and B are both permutation matrices.

Corollary 2.3.If A∈Mn(N)and satisfies Ak=E,k≥1,then A must be a permutation matrix.

Proof.Let B=Ak?1and we have AB=E according to Ak=E,then by proposition 2.1 we get A must be a permutation matrix.

3 The irreducible NIM representations when n≥5

In this section,we will prove there does not exist irreducible Z+-module M when the rank of M satisfies n≥5 over the ring A,where A is a commutative ring with a Z+-basis{1,x,y,xy}and relations:x2=1,y2=1+x+xy.This question is equivalent to proving that there does not exist irreducible NIM solutions to the following equations:

where A and B ∈ Mn(N)are NIM solutions to(?).Now in the following,we let b=E+A+B+AB.Then it is easy to see that b∈Mn(N).

In this case,A and B must be reducible solutions which contradicts with the assumption.Therefore,we claim b is irreducible.

Theorem 3.2.If A and B are irreducible NIM solutions,then b>0.

Proof.We already know that b is an irreducible non-negative matrix by theorem 3.1,so we can get(E+b)n?1>0 by theorem 2.1.According to the binomial expansion and the system of matrix equations(?),we have

Now we claim b=E+A+B+AB>0.Otherwise,if there exists one element bijof matrix b such that bij=0,due to the fact that A,B and AB are all non-negative integer matrices,the elements of the i-th row and the j-th column for each matrix E,A,B and AB are all 0,which contradicts with the fact that m1E+m2A+m3B+m4AB>0.Thereby we can get b=E+A+B+AB>0.

Theorem 3.3.If b>0,then b is irreducible and tr(b)=6.

Proof.Firstly,the irreducibility of b is from Theorem 2.1.According to the system of matrix equations(?),we have b2=6b.Assume λ is is the eigenvalue of b.Then we have λ2=6λ,λ=6 or 0.Then we claim there must exist one eigenvalue which equals to 6.Otherwise the trace of b satisfies tr(b)=0,and it contradicts with b>0,tr(b)>0.Considering that b is irreducible and b>0,by Theorem 2.2 we get 6 is the only non-zero single root of b.Therefore,we can get tr(b)=6.

Theorem 3.4.Let A and B in Mn(N)when n≥7 be solutions to the system of matrix equations(?).Then A and B must be reducible solutions.

Proof.Assume there exist irreducible NIM solutions A and B.We have known b=E+A+B+AB and by Theorem 3.2 we get b>0.Therefore tr(b)≥tr(E)=n≥7 which contradicts with the fact tr(b)=6 by Theorem 3.3.In a word,the assumption is incorrect,so the NIM solutions are always reducible.

Theorem 3.5.If n=5,there is no irreducible NIM solution to the system of matrix equations(?).

Proof.We assume there exist irreducible NIM solutions A and B,then by Theorem 3.3 we know that tr(b)=6.According to b>0,there are several cases about the values of the diagonal elements of b.We might as well let the diagonal elements of b respectively be b11=2,bii=1,(i=2,3,4,5)and other cases are similar.We notice A is a permutation matrix according to A2=E by corollary 2.1,and the order of A is odd,so there is at least one 1 in the diagonal elements of A and a11=1.Besides,since tr(b)=tr(E+A+B+AB),tr(A)≥1 and tr(E)=5,we have tr(AB)=0,tr(B)=0,and tr(A)=1.Considering A is NIM and satisfies A2=E,we get A is one of the following three matrices:

Notation:We only discuss one situation when A is of the form(1)above,because other situations about the choice of A are similar.

According to tr(AB)=0 and tr(B)=0,we get B must be the following matrix:

Since AB=BA,we know that a12=a13,a14=a15,a21=a31,a24=a35,a25=a34,a41=a51,a42=a53,a52=a43.We have the system of diagonal element equations as follows by B2=A+E+AB:

so we get 2a14a41=1,which contradicts with the fact that a14and a41are non-negative integers.It is obvious that the rest situations are similar to the above and we can get contradictions as well.Therefore,the assumption does not hold,that is to say when n=5 if the solutions A and B of the matrix equations exist,they must be reducible.In other words,there doesn’t exist irreducible Z+-modules when the rank of the module is 5.

Theorem 3.6.If n=6,there is no NIM irreducible solution to the system of matrix equations(?).

Proof.We assume that there exists irreducible NIM solutions A and B.By Theorem 3.3,we get tr(b)=6.Since b>0,we getwhere bii=1,(i=1,2,3,4,5,6).It’s easy to know that all of the diagonal elements of A,AB and B equal to 0.Then according to corollary 2.1,we get A is one of the following fifteen matrices:

Notation:We only discuss the situation when A is of the form(1)above because other situations about the choice of A are similar.

According to tr(AB)=0,tr(B)=0,we get B must be the following:

Since we have AB=BA,we get a13=a24,a14=a23,a15=a26,a16=a25,a31=a42,a32=a41,a35=a46,a36=a45,a51=a62,a52=a61,a53=a64,a54=a63.According to B2=A+E+AB we get the system of diagonal element equations as follows:

Hence we get 2(a13a31+a14a32)=1,which contradicts with the fact that a13,a31,a14,a32are non-negative integers.Therefore,the solution does not exist.The rest situations are the same as above and we can get contradictions as well.Therefore,the assumption doesn’t hold,that is to say if NIM solutions A and B of the matrix equations(?)exist,they must be reducible when the order is 6.In other words,there doesn’t exist irreducible Z+-modules when the rank of the module is 6.

4 The irreducible NIM representations when n≤4

We have proved that there do not exist irreducible NIM solutions when order n≥5.In this section,we will calculate all the irreducible NIM solutions to matrix equations(?)when n≤4.

?When n=1,there is only one irreducible NIM solution:

By the same analyzing and calculating method as the case when n=2,we can easily calculate all the irreducible NIM solutions to(?).Therefore,the calculating process is omitted and the results are shown in Table 1.

Table 1:

According to the discussion above,we calculate all the irreducible NIM solutions to the matrix equations(?)when n≤4 and get the following theorem.

Theorem 4.1.All the irreducible NIM solutions to(?)are as follows:

Now,according to Table 1,we classify all the irreducible Z+-modules under the equivalence by Definition 2.6,and we have the following theorem.

Table 2:

Theorem 4.2.All the inequivalent irreducible NIM solutions to(?)are as follows:

Acknowledgments

This work is supported in part by the National Natural Science Foundation of China(Grant No.11871063)and Science And Innovation Fund For College Students(Grant No.201811117037Z).