Onthe McKay quivers and mCartan Matrices

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1、精品论文推荐On the McKay quivers and m-Cartan MatricesJin Yun GuoDepartment of Mathematics Xiangtan Universityemail: gjyAbstract In this paper, we introduce m-Cartan matrix and observe that some properties of the quadratic form associated to the Cartan matrix of an Euclidean diagram, such as it is positiv

2、e semi-definite, can be generalized to m-Cartan matrix of a McKay quiver. We also describe the McKay quiver for a finite abelian subgroup of a special linear group.Keywords: McKay quiver, $m$-Cartan matrix, linear group, finite abelian group1 IntroductionDynkin diagrams and Euclidean diagrams are ve

3、ry useful in mathematics, they appear in many classification problems. One way to describe them is to use Cartan matrices and relates them to positive definite and positive semidefinite quadratic forms. One of the applications of these diagrams is the represen- tation theory of finite dimensional he

4、reditary algebras. Recently, many peo- ple observed that this theory has some 2-dimensional feature and try to find some higher dimensional generalization9. In fact, Iyama and Yoshino find that McKay quivers are interesting in describing some higher dimensional Auslander- Reiten theory10. In this pa

5、per, we provide some further evidence that McKay quivers are the higher dimensional generalization of the Euclidean diagrams, by introducing m-Cartan matrices and proving that McKay quivers share some properties of the Euclidean diagrams. We show that the m-Cartan matrix of an m-dimensional McKay qu

6、iver defines a positive semidefinite quadratic form. We also describe the McKay quiver of an abelian group.In 1980, John McKay introduced McKay quiver for a finite subgroup of the general linear group. Let G GL(m, C) = GL(V ) be a finite subgroup, here V is an m-dimensional vector space over C. V is

7、 naturally a faithful representation of G. Let Si |i = 1, 2, . . . , n be a complete set of irreducible representations of1 This work is partly supported by Natural Science Foundation of China #10671061, SRFDP#2005050420042 Keywords:McKay quiver, m-Cartan matrix, linear group, finite abelian group10

8、G over C. For each Si , decompose the tensor product V Si as a direct sum ofirreducible representations, writeV Si = M ai,j Sj , i = 1, . . . , n,jhere ai,j Sj is a direct sum of ai,j copies of Sj . ai,j is finite since V is finite dimensional. The McKay quiver Q = Q(G) of G is defined as follow. Th

9、e vertex set Q0 is the set of the isomorphism classes of irreducible representations of G, and there are ai,j arrows from the vertex i to the vertex j. McKay observes that when G is a subgroup of SL(2, C), then Q = Q(G) is a double quiver of the Euclidean diagram of type A, D, E.McKay quiver has pla

10、yed an important role in many mathematician fields such as algebraic geometry, mathematics physics and representation theory (See, for example, 15). It also appears in representation theory of algebra, for ex- ample, in the study of the Auslander Reiten quiver of Cohen Macaulay modules 1, 2, preproj

11、ective algebras of tame hereditary algebras 3, 4, 6, 5 and quiver varieties 13, 14, etc. We find that it also plays a critical role in classification of selfinjective Koszul algebras of complexity 2 7.Given a diagram, one associates it with a matrix, its Cartan matrix. Us- ing the Cartan matrices we

12、 associate this diagram with a quadratic form. It is well known that the quadratic form is positive definite and respectively pos- itive semidefinite definite if and only if the diagram is a Dynkin diagram and respectively an Euclidean diagram. We know that in a Cartan matrix, we have2 as its diagon

13、al entries. One natural question is whether this 2 appears acci- dently? We observe that this 2 can be naturally explained as the dimension. In fact, the Cartan matrix in the case of SL(2, C) is just 2I minus the conjunction matrix of the McKay quiver. In the second section of this paper, we introdu

14、ce the corresponding m-Cartan matrix and try to generalize partially a property of the Euclidean diagram to a McKay quiver in general, it has a semidefinite quadratic form. This suggests that McKay quiver should be some kind of higher dimensional version of double Euclidean quiver.Though McKay quive

15、r are well known for SL(2, C). It is difficult to deter- mine the McKay quiver in general. In the last section of this paper, we describe the McKay quiver for the finite abelian subgroup of an arbitrary special linear group, using the skew group algebra construction we introduced early 8. Thencase o

16、f a cyclic group can be regarded as m-dimensional version of double Aquiver.2 The Quadratic Form associated to a McKay quiverQQQLet Q = (Q0 , Q1 ) be a quiver, where Q0 = 1, . . . , n is the vertex set and Q1 is the arrow set. Let MQ = (ai,j ) be the conjunction matrix of Q, that is, an n n matrix w

17、ith ai,j the number of arrows from i to j in Q. MQ is an integral matrix with nonnegative entries. Define the m-Cartan matrix of a quiver Q to be the n n matrix C = C (m) = mI MQ . When m = 2, this is exactly the Cartan matrix of the diagram of a double quiver. (Recall that a double quiver can be re

18、garded as the quiver obtained from a diagram by replacing each edge in the diagram with a pair of arrows pointing at the opposite directions). As for the Cartan matrix, we may defined a bilinear form BQ = B(m) and a quadratic form qQ = q(m) on Qn for Q. This bilinear form and the quadratic form are

19、called respectively the m-bilinear form and the m-quadratic form of the quiver x1 y1 Q. They are defined as follow: Let X = . , Y = . Qn , define . xn . ynBQ (X, Y ) = X t C Yand qQ (X ) = X t C X.Q2Here X t denote the transpose of X . For m 2, B(m) is usually not symmetric. Let C 0 = 1 (C + C t ),

20、then C 0 is symmetric and we have that qQ (X ) = X t C X =X t C 0X .Let G be a finite subgroup of GL(m, C). Let Q = Q(G) be the McKay quiver of G. Denote by MQ = (ai,j ) the conjunction matrix of Q. Let k be an algebraically closed field containing C. Consider mod kG, the category of finite generate

21、d left kG modules. Denote by k the trivial representation. Thus mk is a direct sum of m copies of the trivial representation k, and its dimension is the same as the natural representation V . For any two modules M, M 0, define an integral function on mod kG as in 12:(M, M 0) = dimk HomkG (M mk, M 0)

22、 dimk HomkG (M V, M 0).This is, in fact, a function on the isomorphism classes on mod kG. And we have the following lemma.精品论文推荐Lemma 2.1 1. (M, M 0) = BQ (dim M, dim M 0);2. (M, M ) 0;3. (M, M ) = 0 if M kG;Proof. Since dim(M M 0) = dim M +dim M 0, and we have dimk HomkG (Si , Sj ) =i,j . So, for k

23、G-modules M, M 0, if dim M = (d1 , . . . , dn ) and dim M 0 =(d0 , . . . , d0 ), we get that1 nndimk HomkG (M mk, M 0) = dimk (M HomkG (mdi Si , d0 Si ) = X mdi d0andiiii=1dimk HomkG (M V, M 0) = dimk (M HomkG (M di ai,j Sj , d0 Sj ) = X ai,j di d0 .jjj ii,jThis proves that (M, M 0) = BQ (dim M, dim

24、 M 0).The case m = 2 of the second and the third assertions are proven in 12. By the orthogonality of the characters of the simple representations, and by the tensor formula, we get that for any kG-modules M, M 0,1(M, M 0) =|G|( X tr (g, M )tr (g1 , M 0)(m tr (g, V ).gGIt follows from definition tha

25、t (M, M ) is integral, hence is real. Since the two factors are conjugate, we have that tr (g, M )tr (g1 , M ) 0. So the imaginary part of (m tr (g, V ) is zero. But since g SL(m, C), and it has finite order, each root of its characteristic polynomial of g is a root of the unit. This shows that the

26、real part of these m roots lie between 1 and 1. So we get (m tr (g, V ) 0 and hence (M, M ) 0. This proves the second assertion.i=1Note that dim kG = Pndimk Si dim Si = (dimk S1 , . . . , dimk Sn ). By thedefinition of McKay quiver, comparing the dimensions, we get m dimk S1 dimk S1 . = MG . . m dim

27、k Sn. dimk SnThat is, (mI MG )dim kG = 0, this implies (dim kG)t (mI MG )(dim kG) =0. This shows that (kG, kG) = 0, by the first assertion. We have the following theorem.Theorem 2.2 Let G be a finite subgroup of SL(m, C) and let qG be its quadratic form. Then1. qG is positive semidifinite with one d

28、imensional radical.2. dim kG is a positive generator for the radical of qG .Proof. This follows from Lemma 2.1, Theorem 4.3 and Lemma 4.5 of 11.精品论文推荐Remarks In the case of m = 2, it is classical that the condition (1) in Theorem 2.2 is sufficient and necessary for a double quiver to be McKay quiver

29、(double Euclidean quiver). We have no idea in which class of the quivers this condition will be sufficient and necessary for the McKay quiver in general. In the Theorem 3.1 and Theorem 3.2 in the next section, we have a kind of m-cyclic condition for the McKay quivers of finite abelian subgroups in

30、SL(2, C), we wonder if such condition could be suitable for definingm-tiple quiver. Using our computer program seeking symmetric m-Cartan matrix with positive semidefinite quadratic forms, we find 14 4 4 matrices and among them, only one is possible to be made into McKay quivers by suitable orientat

31、ion, according to our criteria developed in 8. The example of 3- McKay quivers of 4 vertices will be discussed at the end of this paper. We also find 115 5 5 symmetric m-Cartan matrices with positive semidefinite quadratic forms. But our algorithm seem not powerful enough to work with the matrices o

32、f higher order using current PC. We dont know if there is something similar for generalizing Dynkin dia- grams.3 The McKay Quiver of a Finite Abelian Sub- groupIn this section, we describe the McKay quiver of a finite abelian subgroup. We first consider the case of a cyclic subgroup.Theorem 3.1 Let

33、G = (g) be a finite cyclic subgroup of SL(m, C), and assume that |G| = n. Identify the vertex set Q0 of QG with Z/nZ = 1, . . . , n, the set of the residues modulo n. Then1. There are m nonnegative integers n1 , n2 , . . . , nm such that n1 + n2 + + nm 0 mod (n) and for each vertex i Q0 , there are

34、m arrows from i going to i + n1 , i + n2 , . . . , i + nm , (here i + nt is understood as an element in Z/nZ), respectively.The arrows starting from the same vertex are determined by a basis of V . The arrows determined by the same basic element are said to be of the same type2. Any m successive arr

35、ows of different types form a cycle in Q(G).Proof.nLet Si |i = 1, . . . , n be a complete set of representatives of the isomorphism classes of irreducible representations, then we have that dimk Si = 1 for all i since G is abelian. Now let i be the character of Si , then we may assume that i (g) = i

36、 , where n is a primitive nth root of the unit.t=1We may decompose V as V = LmTit Li ci Si . Since Si Sj Si+j , wesee that there are cl arrows from i to i + l for all i and l. Write it i provided that Tit Si . We see that in Q(G), there are m arrows starting at each vertex, and m arrows ending at ea

37、ch vertex. In fact, take a nonzero element xit from each Tit , then (xi1 , . . . , xim ) is a basis of V .Now consider the exterior algebra V of V over k. It is known that thequiver of V G is the McKay quiver of G, that is, there is an idempotent e such that e V Ge kQ(G)/I for some admissible ideal

38、I of kQ(G) 8. In fact, e = 1 since G is abelian and all the irreducible representationsare one dimensional. Let r(V G) be the radical of V G. We havet=0tthat V G = Lm k G as V G/r(V G) kG-modules, andtt k G r (V G)/rt+1(V G). Now we have a set of orthogonal prim-itive idempotents e1 , . . . , en suc

39、h that 1 = e1 + + en 0 V G kGt and Si kGei 0 V Gei . So we have that ei V Gej Lii Tit Sj .And ej+l xit ej represents an arrow from j to j + l provided that it l, this is the arrow of type t, and ej+l xit ej = 0 if it 6 i. Let nt = it , then we have that for each vertex i, xit defines an arrow from i

40、 to i + nt . It follows from 8 that ei xi (m) xi (1) ej = 0 for j 6 i mod (n) and ej xi (m) xi (1) ej = ej+n(1) +n(m) xi (m) ej+n(1) +n(m1) ej+n(1) xi (1) ej = 0, here is apermutation of 1, . . . , m. This implies that ej+n(1) +n(m) xi (m) ej+n(1) +n(m1) ej+n(1) xi (1) ej is a path from j to j in Q(

41、G) and we have that i + n1 + +nm i mod (n). So n1 + + nm 0 mod (n). This proves our theorem. The m-McKay quiver of the cyclic group can be regarded as an m-version ofnthe A. It can be visualized as a quiver with colored arrows in m colors, suchthat each path going through m arrows of different color

42、s comes back to the starting vertex.Now let G be a finite abelian subgroup of SL(m, C). We determine its McKay quiver. What we need to do is to find suitable index sets of the vertices and the arrows, the proof is similar to the above case. By the structural theorems=1of finite abelian groups, G Qr(

43、gs ) is a finite product of finite many cyclicgroups. Assume that gs is an element of order ns with ns |ns0 provided that s s0. Now let S be a irreducible representation of G. Let be the character of S. Since G is commutative and dimk S = 1, we have that (gg0) = (g)(g0) for any g, g0 G and (gi ) is

44、an ni th root of the unit. is determined by its values on the generators g1 , . . . , gr . Let i be a primitive ni th root of the unit.Let Sj1 ,.,jr be the irreducible representation of G with character j1 ,.,jr suchthat j ,.,jsjs 01r (g ) = s . We may take Q as the index set of the representative o

45、f the isomorphism classes of irreducible representations. So its elements are r-tuples j = (j1 , . . . , jr ) with jt Z/nt Z. Similar to the cyclic case, decomposeV as the direct sum of irreducible representationmtV = M Tit=1Ml=(l1 ,.,lr )Q0cl Sl .Then P cl = m. Let xit be a nonzero element in Tit ,

46、 then xi1 , . . . , xim form abasis of V . Write it j provide that Tit Sj . In the exterior algebra V G, we have that 1 = PjQ0 ej as a sum of orthogonal primitive idempotents andtSj kGej 0 V Gej for all j Q0 . Thus ej+l xi xj represents an arrow from j to j + l provided that it l, we call this arrow

47、 an arrow of type t, and ej+l xit xj = 0 if it 6 l. So there are arrows of m different types from each vertex i. We can describe the McKay quiver of G as follow.i=1Theorem 3.2 Let G r(gi ) be a finite abelian subgroup of SL(m, C), andassume that the order of gi is ni . The vertex set Q0 of the McKay

48、 quiver Q(G)of G is a set of r-tuples j = (j1 , . . . , jr ) with jt Z/nt Z.1. In Q(G) there are m r-tuples lt = (l(t) , . . . , l(t) ) for t = 1, . . . , m, such1 rthat l(t) Z/n Zand l(1) + + l(m) , for s = 1, . . . , r. For each vertexssssj Q0 , there are m arrows going from j to j + l1 , . . . ,

49、j + lm , respectively. These are arrows of m different types from the vertex j.2. Any m successive arrows of different types form a cycle in Q(G).Proof. It is proven in the same manner as Theorem 3.1 Example We have developed a program to find symmetric 3-Cartan matri- ces with semidefinite quadrati

50、c forms for the quivers without loop. Among the14 matrices we found with semidefinite quadratic forms, the one related to theMcKay quivers is 31 1 1 131 1 . 1 131 1 1 13We know that there are only two 4-dimensional abelian groups, one is cyclic group G = Z4 , and another is Klein 4-group G0 = Z2 Z2

51、, which is the direct sum of two copies of the cyclic group Z2 . Their McKay quivers are23(0,1) (1,0)1 4(0,0) (1,1)McKay quiver for Z4 McKay quiver for Z2 Z2respectively. We see that the above matrix is exactly the 3-Cartan matrix for the second quiver, and the 3-Cartan matrix for the first quiver i

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