TitleIrreducible representations of the Author(s) Kosuda, Masashi Citation Osaka Journal of Mathematics. 43(2) Issue 2006-06 Date Text Version publisher URL http://hdl.handle.net/094/0396 DOI Rights Osaka University
Kosuda, M. Osaka J. Math. 43 (2006), 43 474 IRREDUCIBLE REPRESENTATIONS OF THE PARTY ALGEBRA Dedicated to Professor Noriaki Kawanaka on his sixtieth birthday MASASHI KOSUDA (Received December 8, 2004, revised June 2, 2005) Abstract In this paper we construct a complete set of representatives of the irreducible representations of the party algebra. which is the centralizer of the unitary reflection group (Ö ) in the endomorphism ring of the tensor space Î ÅÒ under the condition that Ò and Ö Ò.. Introduction Let be a group of linear transformations on a -dimensional vector space Î. Suppose that diagonally acts on the Ò-times tensor space Î ÅÒ. Then the question how the tensor space Î ÅÒ decomposes into irreducible representations of is a basic problem of the classical invariant theory. One way of studying this problem is to consider the centralizer algebra End ÅÒ Î. This approach was successfully done in cases = Ä (C) and Ç (C). These classical groups produced the centralizers CS Ò and Ò () (Brauer algebra [2, 3]) respectively, and the decompositions of the tensor representations of the original groups were obtained as well as the decompositions of their centralizers. In the 980s, the Õ-deformation of these centralizers were discovered and the various connections between the centralizers and other areas (such as knot theory, conformal field theory, etc.) were clarified [0, 4]. In the early 990s, Jones and Martin independently defined the partition algebra È Ò (É) as the generalization of the Temperley-Lieb algebra and the Potts model in statistical mechanics. This algebra corresponds to the case = S in the classical invariant theory above; if the parameter É of È Ò (É) is specialized to a positive integer, the partition algebra È Ò () surjectively mapped to the centralizer End ÅÒ Î where = S, and if further is large enough, (2 Ò É is sufficient), this map becomes injective [5]. In the paper [5, 9], they considered 2Ò, the set of all the set partitions of É Ò Ö Ö Ò, as a basis of È Ò (É) and defined the product among each element of. Further, they showed that 2Ò È Ò(É) is generated by the symmetric 2000 Mathematics Subject Classification. Primary 05E0; Secondary 20C30.
432 M. KOSUDA group S Ò acting on Î ÅÒ by tensor factors permutation and two special elements and 2. Inspired by the work of Jones, Tanabe considered the case is a unitary reflection group of type (Ö Ô ) where (Ö Ô ) is an index-ô subgroup of (Ö ), and (Ö ) is a group of monomial matrices whose non-zero entries are Ö-th roots of unity []. In the paper [2], Tanabe showed that End (ÖÔ) Î ÅÒ is generated by the symmetric group S Ò together with three further special operators, Ö and ÖÔ. (Note that the unitary reflection groups (Ö Ô ) include the symmetric group S = ( ). The operators and become 2 and respectively in È Ò (É), and the operator ÖÔ is not defined in case Ô =.) In this paper, we study further about the case = (Ö ) ( Ò, Ö Ò): we construct a complete set of irreducible representations, which corresponds to Hoefsmit-analogues of Young s seminormal representations of the symmetric group [4]. This paper is organized as follows. First we define the party algebra A Ò as an abstract algebra generated by the symmetric group S Ò and one of the above operators = = 2, which will turn out to be a subalgebra of the partition algebra É È Ò (É). In fact, a basis of A Ò has one to one correspondence with a subset of 2Ò called the set of seat-plans (see Section.). We showed in the previous paper [6] that any word of A Ò is reduced to one of the standard words of the generators under the defining relations and each standard word corresponds to one of the seat-plans (see Definition.). Similarly to the partition algebra, there exists a surjective homomorphism from A Ò to End ÅÒ Î. Moreover, if Ò and Ö Ò, this homomorphism becomes injective (Proposition.2 and.3). Next, we explicitly construct a complete set of representatives of the irreducible representations of the party algebra A Ò drawing the Bratteli diagram of the tower A 0 A A Ò and defining the tableaux on it. Finally we check that these representations are irreducible and nonequivalent each other. Comparing the square sum of the degrees of the irreducible representations with the number of the seat plans (standard words of the generators) we find that A Ò is semisimple... Definition of the party algebra. First, we define the party algebra A Ò. DEFINITION.. Let Z be the ring of rational integers. We put A 0 = A = Z. For an integer Ò, the party algebra A Ò is defined over Z by the following generators: and relations: 2 Ò 2 = ( Ò ) (È ) = ( Ò 2) (È 2)
IRREDUCIBLE REPRESENTATIONS OF THE PARTY ALGEBRA 433 Fig.. Generators of A Ò = ( Ò ) (È 3) = (È 4) = = (È 5) = (3 Ò ) (È 6) 2 2 = 2 2 (È 7) 2 3 2 2 3 2 = 2 3 2 2 3 2. (È 8) Putting = = ( 2 )( 3 2 ) ( 2 3 )( 2 ) we obtain another presentation of A Ò by s and s. ( = 2 3 Ò ), 2 = ( Ò ) (È ) = ( Ò 2) (È 2 ) = ( Ò ) (È 3 ) = ( Ò ) (È 4 ) = = ( Ò ) (È 5 ) = ( Ò ) (È 6 ) = ( Ò 2) (È 7 ) = ( Ò ) (È 8 ) For the new generators Ò, we give the diagrams figured in Fig.. In the following, to each word of the generators of A Ò, we give a diagram explanation.
434 M. KOSUDA Fig. 2. A seat-plan Let = Ò and Ê = Ö Ö 2 Ö Ò be two sets, each of which consists of Ò distinct elements. We further assume that Ê =. We decompose Ø Ê into subsets Ò (some of s might be empty) so that they satisfy Ò = = Ø Ê = if = Ò = Ê for = Ò We call such a partition into subsets a seat-plan of size Ò. Let È (Ò) be the set of partitions of an integer Ò. Then there exists a partition ¾ È (Ò) such that = ( Ò ) = ( 2 2 Ò 2). The number of seat-plans is Ò! 2 ()!! Ò! «! «2! «Ò! ¾È (Ò) where «= ; =. A seat-plan of size Ò is illustrated as in Fig. 2. Consider a rectangle with Ò marked points on the bottom and the same Ò on the top. The Ò marked points on the bottom are labeled by Ò from left to right. Similarly, the Ò marked points on the top is labeled by Ö Ö 2 Ö Ò from left to right. If Ø Ê is divided into nonempty Ñ subsets, then put Ñ shaded circles in the middle of the rectangle so that they have no intersections. Each of the circles corresponds to one of the non-empty s. Then we join the 2Ò marked points and the Ñ circles with 2Ò shaded bands so that the marked points labeled by the elements of are connected to the corresponding circle with bands. We associate generators Ò of A Ò to the following special seat-plans Ö Ö Ö Ö 2 Ö 2 Ò Ö Ò
IRREDUCIBLE REPRESENTATIONS OF THE PARTY ALGEBRA 435 Fig. 3. The product of seat-plans and Ö Ö Ö Ö 2 Ö 2 Ò Ö Ò respectively, which are illustrated in Fig.. Now we define the product Û Û 2 between two of rectangles Û Û 2 (each of which corresponds to a seat-plan) by placing Û on Û 2, gluing the corresponding boundaries and shrinking half along the vertical axis as in Fig. 3. We then have a new diagram possibly containing some closed loops. The product is the resulting diagram, with the closed loops removed. It is easy to define this product in terms of seat-plans (see for example Martin s paper [9]). The set of the seat-plans satisfies the relation (È ) (È 8 ). Moreover, in the paper [6], the author showed that there exist one to one correspondences between the set of seat-plans of size Ò and the set of standard words of A Ò and that using only the relations (È ) (È 8 ) any word of the generators becomes a standard word. This means that the linear combination of seat-plans is a surjective image of A Ò and it makes a finite dimensional algebra whose dimension is given by the expression (). The following proposition given by Tanabe [2] shows the relation between the party algebras and the centralizer algebras of the unitary reflection groups. Proposition.2 (Tanabe [2, Theorem 3.]). Let (Ö ) be the group of all the monomial matrices of size Ò whose non-zero entries are Ö-th roots of unity. Let Î be the C-vector space of dimension with the basis elements on which (Ö ) acts naturally. Let be the representation of the symmetric group S Ò on
436 M. KOSUDA Fig. 4. 4 The Bratteli diagram for the sequence A Å C 4 =0 Î ÅÒ obtained by permuting the tensor product factors, i.e., for Ú Ú 2 Ú Ò ¾ Î and for Û ¾ S n, (Û) Ú Å Ú 2 Å Å Ú Ò := ÚÛ () Å Ú Û (2) Å Å Ú Û (Ò) Define further ( ) as follows: ( ) Ô Å Ô2 Å Å ÔÒ := Ô Å Ô2 Å Å ÔÒ if Ô = Ô 2 0 otherwise If Ö Ò, then End ÅÒ Î is generated by (S Ò ) and ( ) and defines a homomorphism from A Ò Å C to End ÅÒ Î. Proposition.3. Let be the map previously defined. If Ò, then is injective. Proof. Using Schur-Weyl reciprocity and counting the dimension, the proposition will be easily checked..2. Bratteli diagram of the party algebras. In this subsection, first we make a diagram Ò, which will turn out to be the Bratteli diagram of the sequence A Å C Ò =0. Then we define the sets of the tableaux on the diagram. Fig. 4 will help the reader to understand the recipe. Fix a positive integer Ò. Let «= [«() «(Ò)] be an Ò-tuple of Young diagrams. The -th coordinate of the tuple is referred to the
IRREDUCIBLE REPRESENTATIONS OF THE PARTY ALGEBRA 437 -th board. The height «of «is defined as the weight sum of the sizes of all the «()s. Namely, «is defined by Let «= Ò = «() Ä Ò () = «= [«() «(Ò)] «= be a set of Ò-tuples of height. For «¾ Ä Ò (), we set «(0) = Ò (the horizontal Young diagram of depth and of width Ò ) if necessary. Let ««or ««denote that «is obtained from «by removing one box from the Young diagram on the -th board and adding the box to the Young diagram on the ( )-st board for some (0 Ò ). The diagram Ò is defined as the Hasse diagram Ò of =0Ò Ä Ò () with respect to the order generated by s. Finally we define the sets of the tableaux on Ò. For «¾ Ä Ò (Ò), The set T(«) of tableaux of shape «is defined by T(«) = Ò È = «(0) «() «(Ò) «(0) = [ ] «(Ò) = ««() «() for 0 Ò Ó.3. Construction of the irreducible representation. Now we have defined the sets of tableaux on Ò, we define linear transformations of the tableaux. Let Q be the field of rational numbers and à 0 = Q( Ô Ô Ô 2 3 Ò) its extension. In the following, the linear transformations are defined over à 0. They will turn out to be a complete set of representatives of the irreducible representations of A Ò (à 0 ) = A Ò Å Ã 0. Similar methods are used for example in the references [, 3, 0, 3, 4]. Let V(«) = È ¾T(«)à 0 Ú È be a vector space over à 0 with the standard basis Ú È È ¾ T(«). For a generator of A Ò (à 0 ), we define a linear map «( ) on V(«) giving the matrix Å with respect to the basis Ú È È ¾ T(«). Namely, for a tableaux È = («(0) «() «(Ò) ) of T(«), define «( )(Ú È ) = ɾT(«) (Å ) ÉÈ Ú É. Let É = («(0) «() «(Ò) ). If there is an 0 ¾ Ò Ò such that «( 0) = «(0), then we put (Å ) ÉÈ = 0 In the following, we consider the case that «(0) = «(0) for 0 ¾ Ò Ò. CASE. First, we assume that «( ) and «() of the tableau È coincide with each other except on the -th and the ( )-st boards. In this case, «() is obtained from «( ) by moving a box in the Young diagram on the -th board to the Young
438 M. KOSUDA diagram on the ( )-st board and «() is obtained from «() by moving another box in the Young diagram on the -th board to the Young diagram on the ( )-st ( ) board. Denote the Young diagram on the -th board of «(resp. «(), «() ) by ( ) (resp. (), () ( ) ) and denote the Young diagram on the ( )-st board of «(resp. «(), «() ) by ( ) (resp. (), () ). Let or denote that is obtained from by removing one box. Recall that if then we can define the axial distance = ( ). Namely, if differs from in its Ö 0 -th row and 0 -th column only, and if differs from in its Ö -th row and -th column only, then = ( ) is defined by (Ö 0 ) if Ö 0 Ö (2) = ( ) = ( Ö ) ( 0 Ö 0 ) = (Ö 0 ) if Ö 0 Ö Here ( ) is the hook-length at ( ) in and for = ( ) the hook-length ( ) is defined by ( ) = ; Since ( ) () (), we can define the axial distance = ( () () ( ) ). Similarly, since ( ) () (), we can define the axial distance = ( ( ) () () ). If (resp. ), then there is a unique Young diagram = () (resp. = () ) which satisfies ( ) () (resp. ( ) () ). Let É É 2 É 3 be tableaux of shape «which are obtained from È by replacing ( () () ) on the -th and the ( )-st board of «() with ( () ), ( () ), ( ) respectively. For the basis elements given by the above tableaux, we define the linear map «( ) by the following matrix: where Å = «( ): (Ú È Ú É Ú É2 Ú É3 ) (Ú È Ú É Ú É2 Ú É3 )Å 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ½
IRREDUCIBLE REPRESENTATIONS OF THE PARTY ALGEBRA 439 If we put (3) = and = Õ then Å is written as follows: (4) Å = 2 Å 2 2 2 Even if = (resp. = ), we still adopt the matrix (4) since = 0 (resp. 2 = 0). CASE 2. Next, we consider the case that «() is obtained from «( ) by removing one box from the Young diagram on the -th board and adding the box to the Young diagram on the ( 2)-nd board. Let «be the Young diagrams on the -th, the ( )-st and the ( 2)-nd boards of «( ) respectively and «the Young diagrams on the corresponding three boards of «() respectively. Let Ö = () (resp. (Ö ) Ö = ()) be the set of all the Young diagrams which satisfy (resp. (Ö ) ) and let È È 2 È () (resp. É É 2 É ()) be all the tableaux which are obtained from È by replacing the Young diagrams on the -th, the ( )-st and the ( 2)-nd board of «() with «(resp. «(Ö ) ). For the basis elements given by the above tableaux, we define the linear map by the following matrix: (5) (Å ) ÈÖ È Ö = Ú ÙØ (Ö ) () 2 (Å ) ÈÖ É Ö = (Å ) ÉÖ È Ö = Ú ÙØ (Å ) ÉÖ É Ö = 0 (Ö ) () 2 Here () is the product of all the hook-lengths in : () = ( ) Putting (6) ()¾ = ()() ()() (Ö )
440 M. KOSUDA and combining the expression (2) and (3), we can write Å (5 ) (Å ) ÈÖ È Ö = (Ö ) (Å ) ÉÖ È Ö = (Å ) ÈÖ É Ö = (Å ) ÉÖ É Ö = 0 (Ö ) as follows: (Ö ) CASE 3. Finally, we consider the remaining cases. In these cases, if «() is obtained from «( ) by moving one box to the next board and if «() is obtained from «() by moving another box to the next board in a tableau È, then exchanging the -th step and the ( )-st step, we have another tableau É. For the basis elements given by the above tableaux, we define the linear map by the following matrix: (7) (Ú È Ú É ) (Ú È Ú É )Å = (Ú È Ú É ) 0 0.4. Main theorem. Now we have completed the preparation, we state the following main result. Theorem.4. Let «= [«() «(Ò)] be an Ò-tuple of Young diagrams in Ä Ò (Ò). Let Q be the field of rational numbers and à 0 = Q( Ô Ô Ô 2 3 Ò) its extension.. Define «as follows: «( )Ú È = «( )Ú È = ɾT(«) (Å ) ÉÈ Ú É Ú È if «(2) = [ ] 0 otherwise Then ( «V(«)) defines an absolutely irreducible representation of A Ò (à 0 ). 2. For ««¾ Ä Ò (Ò), the irreducible representations «and «of A Ò (à 0 ) are equivalent if and only if «= «. 3. Conversely, for any irreducible representation of A Ò (à 0 ), there exists an «¾ Ä Ò (Ò) such that and «are equivalent. In other words, ««¾ Ä Ò (Ò) make a complete set of the representatives of the irreducible representations of A Ò (à 0 ). Corollary.5. The party algebras A Ò (à 0 ) are absolutely semisimple, and the Bratteli diagram of the sequence A (à 0 ) =0Ò is given by the graph Ò.
IRREDUCIBLE REPRESENTATIONS OF THE PARTY ALGEBRA 44 Let «be an Ö-tuple of Young diagrams such that ««and T(«; «) = «(0) «() «(Ò ) «(Ò) ¾ T(«) «(Ò ) = «Let V(«; «) be a subspace of V(«) spanned by Ú É É ¾ T(«; «). Since A Ò (Ã 0 ) is isomorphic to the subalgebra A = Ò 2 of A Ò (Ã 0 ), considering the definition of «we find that the subspace V(«; «) is stable under the action of A. Further, applying the theorem above replacing Ò with Ò, we find that V(«; «) affords an irreducible representation of A Ò (Ã 0 ). In the proof of the theorem above, we will further obtain the following restriction rule. Corollary.6. For «¾ Ä Ò (Ò), the branching rule of the restriction of irreducible representation of A Ò (Ã 0 ) to the subalgebra A = A Ò (Ã 0 ) is given as follows: V(«) = Å «Ô «V(«Ô ; «) as A Ò (Ã 0 )-modules 2. Preliminary results for the axial distances and the hook-lengths To prove the main theorem, our main task is to show the well-definedness of the representations «. Since «( ) is defined by the matrix Å in the theorem and the entries of Å are written in terms of and (()), the task will be done by showing the various relations among them. In this section, we show miscellaneous relations among (()) defined by the expressions (2), (3) and (6). First we note that by the definition of and we immediately have 2 =. Using this we obtain the following relations among and by direct calculation: Lemma 2.. Let 0 be non-zero integers such that 0 =. Then we have the following.. 2 0 2 0 = 0, 2. 0 = 2 0. Let «Ö=(«) (resp. «(Ö ) Ö = («)) be the set of all the Young diagrams which satisfy ««(resp. «(Ö ) «). If and are a pair of Young diagrams such that, then we have and (Ö ). In the following, we assume that () = and () =. For = 2 (), we put ( ) = () ( ) = ( ). Further, let = Ò(Ò). If is a Young diagram, then there exists an index such that = ( ). More precisely, we have the following:
442 M. KOSUDA Lemma 2.2. Let Ö be axial distances defined by Ö =. Then we have Ö = 0 Ö = () = ( ) ( ) = 2 () In other words, there exists a bijection from the set = 2 () to the set Ö Ö = () Ö = 0 such that (( )) (( )) = ( ) ( ). Similarly, for Ö = 2 (), we put (Ö ) = () (Ö ) = (Ö ) and ( ) = ( ) ( Ò ). Then we have the following: Lemma 2.3. Let be axial distances defined by = ( ). Then we have Ò Ó Ò Ó ( ) ( ) = 0 = () = (Ö ) (Ö ) Ö = 2 () In other words, there exists a bijection from the set Ö Ö = 2 () to the set = () = 0 such that ( (Ö )) ( (Ö )) = (Ö ) (Ö ). We have also the following relations among and : Lemma 2.4. Let be Young diagrams such that and = ( ) their axial distance. If =, then there exists a Young diagram such that which differs from. Further in this case, we have the following: = Let be the set of all the Young diagrams of any size. Consider the vector space à 0 whose natural basis is indexed by the set [] ¾. Combining the result of the previous three lemmas, we have the following: (8) (9) Ö = = = ( ) ( ) = ( ) ( ) = (Ö ) (Ö ) = 0 Ö = () Ö µ ( ) ( ) = 2 () ( ) ( ) ( ) = (Ö ) (Ö ) (Ö ) = 0 = ()µ Ö = 2 () µ
IRREDUCIBLE REPRESENTATIONS OF THE PARTY ALGEBRA 443 Similarly, let à 0 ( ) be the vector space whose natural basis is indexed by the set = ( ) ¾ Then we have the following: (0) () Ö = = ( ) ( ) ( ) ( ) ( ) (Ö ) (Ö ) (Ö ) Ö = 0 Ö = () µ = 2 () = ( ) ( ) ( ) ( ) = = ( ) = ( ) ( ) (Ö ) (Ö ) ( ) (Ö ) (Ö ) = 0 = ()µ Ö = 2 ()µ Under the notation in Lemma 2.2, 2.3, in case is a Young diagram, we have Ö = = By Lemma 2.2, if Ö =, this is also equal to the following: (2) Ö = ( ) = ( ) = ( ) ( ) ( ) Similarly, in case ( ) is a Young diagram, we have = ( ) = ( ) ( ) By Lemma 2.3, if =, this is also equal to the following: (3) = (Ö ) = (Ö ) = (Ö ) (Ö ) (Ö ) On the other hand, in case () is a Young diagram, put (4) () = and Ö = If is a Young diagram, then we have Ö = = () Further if Ö =, then we have (5) Ö = ( ) = ( ) = () ( ) ( )
444 M. KOSUDA In case =, we have (6) () = = Similarly, in case () is a Young diagram, put () (7) () = and = ( ) If ( ) is a Young diagram, then we have = ( ) = ( ) () and (8) (9) Finally, we put (20) (2) = (Ö ) = (Ö ) = (Ö ) () () = = () = ( ) ( ) Ö Ö = (Ö ) (Ö ) Using the notation above we finally obtain the following relations among and (): 2. 3. 4. Lemma 2.5.. È () Ö= () Æ Ö = 0, È () Ö= () Æ Ö Ö Ö = È () Ö= () Æ =, È () = () Æ ( ) = 5. () 3 È () = 3 ()() (Ö = ) 0 (Ö = ) ()() ( = ) ( = ) Æ ( ) = () 3 È () Ö= 3 Ö Æ. Proof. The above relations are proved by specializing the parameter Õ to in the equations of Theorem 0. 0.2 in the paper [7]. 3. Well-definedness of the representations In this section, we show that «in the main theorem preserve the defining relations of the party algebra. First in Section 3. we check that «preserve the braid relation (È 2) in Definition.. Actually, this is the main part of the paper. Next in Section 3.2, we check that they also preserve the other relations.
IRREDUCIBLE REPRESENTATIONS OF THE PARTY ALGEBRA 445 3.. Preservation of the braid relation. Let È = («(0) «(Ò) ) be a tableau of shape «. As we defined in Section.3 and Section.4, the linear map «( ) is defined by the matrix Å and it is defined by the way how «() is obtained from «( ) (see the expressions (4), (5) and (7)). Suppose that «() (resp. «(), «(2) ) is obtained from «( ) (resp. «(), «() ) by moving a box on the 0 -th (resp. -th, -th) board of «( ) (resp. «(), «() ) to the ( 0 )-st (resp. ( )-st, ( )-st) board. Then to know the actions of and, it will be sufficient to know that how they alter the following matrix whose entries are Young diagrams: (22) «() () 2 = 0 0 Since column indices of the matrix above do not necessarily distinct depending on the differences among 0 and, we have to consider Table. In Table, if we replace ( 0 0 ) with the reverse ( 0 0 ), then we find Case 3. (resp. Case 3.2) turns Case 3.3 (resp. Case 3.4). This means -action on Ú È in Case 3.3 (resp. Case 3.4) Table. Classification by differences among 0, and Case 0 0 Case 2 0 = = Case 3. (Case 3.3) 0 = 0 Case 3.2 (Case 3.4) 0 0 = (Case 3.3) 0 0 = (Case 3.4) 0 0 = Case 3.5 0 0 = Case 3.6 0 = 0 Case 4. (Case 4.2) 0 = 0 = (Case 4.2) = 0 = 0 Case 4.3 = 0 = 0 Case 5. = 0 = Case 5.2 (Case 5.3) = 0 = (Case 5.3) 0 = = 0 Case 5.4 (Case 5.5) = 0 = (Case 5.5) 0 = = 0 Case 5.6 = 0 = Case 6. (Case 6.4) = = 0 Case 6.2 (Case 6.5) = 0 = Case 6.3 (Case 6.6) 0 = = (Case 6.4) 0 = = 0 = (Case 6.5) = 0 = = (Case 6.6) = 0 = = 0
446 M. KOSUDA = «( ) ( 0 ) () (2) (3) () (2) (3) 2 = «(2) ( 0 ) 0 -th board = «( ) ( 0 ) () (2) (3) () (2) (3) = «(2) ( 0 ) ( 0 )-st board Fig. 5. Case 2 Hasse diagrams of Young diagrams on 0 -th and ( 0 )-st board is obtained from -action on Ú È in Case 3. (resp. Case 3.2) by changing indices of tableaux and vice versa. Similarly, the conditions Case 4.2, 5.3, 5.5, 6.4, 6.5, and Case 6.6 are obtained from the conditions Case 4., 5.2, 5.4, 6., 6.2 and Case 6.3, respectively. Hence, we have only to consider Case, 2, 3., 3.2, 3.5, 3.6, 4., 4.3, 5., 5.2, 5.4, 5.6 and Case 6. 6.3. In the following, to simplify the notation, we write merely Ú È instead of «( )(Ú È ). Further, for a Young diagram, we use the notation (resp. ) to denote a Young diagram which is obtained from by removing (resp. adding) a box from (resp. to). CASE. First we consider Case, the most general case. In this case, column indices 0 0 and of the matrix (22) are all distinct each other. Since in this case, the actions of and are both presented by the matrix (7), they merely exchange the entries of the matrix (22). By direct calculation, we can check that Ú È = Ú È in this case. CASE 2. Next we consider Case 2. The assumption 0 = = means that on the way from «( ) to «(2) of È, three boxes of = «( ) ( 0 ) are removed and they are attached to = «( ) ( 0 ) one by one. Let Fig. 5 be Hasse diagrams which respectively describe how Young diagrams on the 0 -th and the ( 0 )-st boards of È would transform themselves (Some of the Young diagrams may be virtual ones. However, we do not have to care about that, since the associate coefficients would be zero). Let (Õ) Å ( ) (Ø)
IRREDUCIBLE REPRESENTATIONS OF THE PARTY ALGEBRA 447 be the vector which corresponds to a tableau obtained from È by replacing Young diagrams on the 0 -th and the ( 0 )-st boards of «() (resp. «() ) with and ( ) (resp. (Õ) and (Ø) ). For this vector, gives the following matrix; (ÕÖ) (ÕÖ) (ÕÖ) (ÕÖ) ( Ø) Å ( Ø) ( Ø) ( Ø) and gives the following matrix; (ÕÖ) (ÕÖ) (ÕÖ) (ÕÖ) Å ( Ø) ( Ø) ( Ø) ( Ø) Here (Õ Ö) = ( (Õ) ), ( Ø) = ( ( ) (Ø) ), (Õ Ö) = ( (Õ) ) and ( Ø) = ( ( ) (Ø) ). Hence, using a result of the paper [4], we can easily check that in this case Ú È = Ú È holds. CASE 3. We consider mainly Case 3. (Case 3.3). For the explanation below, see Table 2. Put («( ) ( 0 ) «( ) ( ) «( ) ( )) = («). If «( ) ( ) = «() ( ) in the tableau È, then and both give linear transformations defined by the matrix of type (7) and we can apply Case. Hence we have only to consider the case «( ) ( ) = «() ( ), namely, we assume the following: ) One box in «on the 0 -th board is removed and it is attached to on the -th board. 2) Then the attached box on the -th board is again removed and it is attached to on the ( )-st board. We keep our eyes on the -th and the ( )-st coordinates of Table 2. Let «(0 ) «( (23) ) «( ) «( ) «( ) ( 0 ) ( ) ( ) ( ) ( ) be a vector which corresponds to a tableau obtained from È by replacing Young diagrams on the 0 -th, -th, ( )-st, -th and ( )-st boards of the -th (resp. ( )-st) coordinate with «( 0 ) «( ) «( ) «( ) «( ) (resp. ( 0 ) ( ) ( ) ( ) ( )). By the definition of tableaux, we find that all the entries of the Table 2. Case 3. (Case 3.3), 0, = 0 coordinate board 0 «() (2) «() () (2) «() (2) 2 «() (2)
448 M. KOSUDA matrix (23) will be recovered from the entries of the second column. Hence instead of using the matrix (23) we merely write Let (Ö ) «( ) ( ) Ö = () be the set of Young diagrams such that and Ö = () the set of Young diagrams such that (Ö ). Let Ú È = () be the vector indexed by the tableau È. Then by definition (5 ) and (7) and using notation (2), we can check that both Ú È and Ú È are equal to the following: () Ö= () () Ö = (Ö ) () Ö (Ö ) As for Case 3.2 (3.4), 3.5, and Case 3.6, using Table 3, 4 and Table 5 respectively we can more easily check that Ú È = Ú È. Table 3. Case 3.2 (Case 3.4), 0, 0 = Table 4. Case 3.5 0, 2, 0 = coordinate board 0 0 coordinate board 0 0 «() (2) «() () (2) «() (2) 2 «() (2) «() (2) «() () (2) «() () (2) 2 «() (2) Table 5. Case 3.6 0,, = 0 coordinate board 0 «() (2) «() () (2) «() () (2) 2 «() (2)
IRREDUCIBLE REPRESENTATIONS OF THE PARTY ALGEBRA 449 CASE 4. We consider mainly Case 4. (Case 4.2). For the explanation below, see Table 6. Put («( ) ( 0 ) «( ) ( 0 )) = ( ). The assumption 0 = means that on the way from «( ) of È to «() of È, two boxes of = «( ) ( 0 ) are removed and they are attached to = «( ) ( 0 ) one by one. Hence we can put («() ( 0 ) «() ( 0 )) = ( ) and («() ( 0 ) «() ( 0 )) = ( ) using the Young diagrams such that and. Further, put («( ) ( ) «( ) ( )) = ( ). The assumption 0 = means that we can put («() ( ) «() ( )) = («() ( ) «() ( )) = ( ) and («(2) ( ) «(2) ( )) = ( ) using the Young diagrams such that and. Let «(0 ) «( (24) 0 ) «( ) «( ) ( 0 ) ( 0 ) ( ) ( ) be a vector which corresponds to a tableau obtained from È by replacing Young diagrams on the 0 -th, ( 0 )-st, -th and ( )-st boards of the -th (resp. ()-st) coordinate of «with «( 0 ) «( 0 ) «( ) «( ) (resp. ( 0 ) ( 0 ) ( ) ( )). By the definition of tableaux, we find that all the entries of the matrix (24) will be recovered from the first two columns. Hence instead of using the matrix (24) we merely write «(0 ) «( 0 ) ( 0 ) ( 0 ) Let Ú È = be the vector indexed by the tableau È. Put = ( ) and = ( ). Let (resp. ) be a (possibly virtual) Young diagram which satisfies (resp. ) and = (resp. = ). Then by definition (4) and (7), we can check that both Ú È and Ú È are equal to the following: Table 6. Case 4. (Case 4.2) 0 =, 0 = coordinate board 0 0 2
450 M. KOSUDA As for Case 4.3, using Table 7 we can check that Ú È = Ú È. CASE 5. Let and (Ö ) be the sets of Young diagrams defined in Case 3.. We also use the notation Ö Ö as in (2). Similarly, let ( ) and ( ) be the sets of Young diagrams such that ( ) and ( ) respectively. Further, let be the set of axial distances defined by = ( ) (). First we consider Case 5.. For the explanation below, see Table 8. Put «( ) ( 0 ) «( ) ( ) «( ) ( ) «( ) ( ) = («) As we saw in Case 3., under the assumption = 0, if «( ) ( ) = «() ( ), then we can attribute this case to one of the previous ones. Hence we may assume «( ) ( ) = «() ( ). The same things also hold for «() ( ) and «(2) ( ). Hence we may further assume that «() ( ) = «(2) ( ). In other words, we have only to consider a tableau È of the form presented by the data in Table 8. As we saw in the previous cases, we keep our eyes on the -th and the ( )-st coordinates of Table 8. Let «(0 ) «( (25) ) «( ) «( ) ( 0 ) ( ) ( ) ( ) be a vector which corresponds to a tableau obtained from È by replacing Young diagrams on the 0 -th, -th, -th and ( )-st board of the -th (resp. ( )-st) coordinate with «( 0 ) «( ) «( ) «( ) (resp. ( 0 ) ( ) ( ) ( )). By the definition of tableaux, we find that all the entries of the matrix (25) will be recovered from the entries of the second and the third columns. Hence instead of using Table 7. Case 4.3 = 0, = 0 coordinate board 0 0 2 Table 8. Case 5. = 0, = coordinate board 0 ««() «() 2 «
IRREDUCIBLE REPRESENTATIONS OF THE PARTY ALGEBRA 45 the matrix (25) we merely write «( ) «( ) ( ) ( ) Using these notation, we write Ú È = () () Then by definition (5 ) and (7) we find that Ú È ( ) () = Õ= () ( ) Ö= = () ( ) Ö = = () ( ) Ö = = (Õ) () () (Ö ) () (Ö ) () ( ) () ( ) () () ( ) ( ) () (Õ) ( ) Ö Ö is equal to the following: ( ) ( ) (Ö ) ( ) (Ö ) ( ) (Ö ) Here we used Lemma 2.5, 2 to obtain the first line of the equation above. In this equation, if we exchange and, exchange the first and the second columns of matrices, and exchange the first and the second rows of the matrices, then we have the same equation. This means -action and -action on Ú È coincide. Next, we consider Case 5.2 (Case 5.3). According to Table 9, we put () Ú È = () () Table 9. Case 5.2 (Case 5.3) = 0, = coordinate board 0 ««() «() () 2 «
452 M. KOSUDA Then by definition (5 ) and (7) we find that Ú È is equal to the following: () Ö= () () ( ) () = Ö = () ( ) (Ö ) () Ö (Ö ) ( ) (Ö ) Here we applied Lemma 2.5 2, 3 to obtain the first line of the equation above. By direct calculation, we find that Ú È coincides with this equation. As for Case 5.4 (5.5) and Case 5.6, using Table 0 and Table respectively, we can more easily check that Ú È = Ú È. CASE 6. Throughout Case 6 we adopt the notation introduced in Section 2. Further let be an axial distance defined by = ( ). CASE 6.. (Case 6.4.) = = 0. For the explanation below, see Table 2 and Table 3. Put «( ) ( 0 ) «( ) ( ) «( ) ( ) = («) The assumption = = 0 means the following: ) On the way from «( ) to «(), one box of «= «( ) ( 0 ) on the 0 -th board is removed and it is attached to = «( ) ( ) on the -th board. We put «() ( 0 ) «() ( ) «() ( ) = («() ) 2) Then on the way from «() to «(2), two boxes of () = «() ( ) on the -th board are moved to the next board. If the two boxes are both distinct from the box which is attached at the former step, then we can attribute this case to one of the previous cases. So we may assume that «() ( ) = (Case 6..) or «() ( ) = (Case 6..2). Here is a Young diagram which satisfies () and =. Hence we Table 0. Case 5.4 (Case 5.5) =, 0 = coordinate board 0 0 ««() «() () 2 «Table. Case 5.6 =, 0 = coordinate board 0 0 ««() «() 2 «
IRREDUCIBLE REPRESENTATIONS OF THE PARTY ALGEBRA 453 may put («) «() ( 0 ) «() ( ) «() ( ) = («) «(2) ( 0 ) «(2) ( ) «(2) ( ) = («) (Case 6..) (Case 6..2) Note that and there may exist a Young diagram such that and =. CASE 6... = = 0, «( ) = «(). For the explanation below, see Table 2. Let «(0 ) «( (26) ) «( ) ( 0 ) ( ) ( ) be a vector which corresponds to a tableau obtained from È by replacing Young diagrams on the 0 -th, -th and ( )-st boards of the -th (resp. ( )-st) coordinate with «( 0 ) «( ) «( ) (resp. ( 0 ) ( ) ( )). As we show below, in the process of -action or -action to Ú È, a set of entries «( ) ( ) of the matrix (26) contains or. Hence under the assumption that the matrix (26) denotes a tableau which appears in the process of -action or -action on Ú È, we merely write (27) «( ) «( ) if «( ) ( ) if «( ) ( ) ( ) ( ) For example, () «= () «and () = «() «We note that all the tableaux which appear in the following calculation are distinguished by the notation (27). Let Ú È = () Table 2. Case 6.. = = 0, «( ) = «() coordinate board 0 = ««() «2 «
454 M. KOSUDA be the vector indexed by a tableau È. Then we have the following: Ú È = () () Ö= () Ö= () = () = () Ö =2 () Ö =2 () () () ( ) Ö () ( ) (Ö ) () (Ö ) () Ö (Ö ) () ( ) =2 ( ) ( ) (Ö ) (Ö ) (Ö ) (Ö ) (Ö ) (Ö ) () ( ) ( ) ( ) ( ) ( ) (Ö ) (Ö ) (Ö ) Here we applied Lemma 2.5 2, 3 to obtain the first line of the equation above. Applying Lemma 2.2 and 2.4 to the second line, the equation (3) and Lemma 2.3 to the third line from the bottom, and the equation (9) and Lemma 2.3 to the last line, we find that this equation is equal to the following: () = () Ö= () = () Ö =2 () ( ) () () ( ) ( ) (Ö ) () (Ö ) On the other hand, we have Ú È = () Ö= () = Ö ( ) 2 2 () (Ö ) () () ( ) =2 Ö ( ) () ( ) (Ö ) (Ö ) ( ) ( ) ( ) ( ) Ö (Ö ) (Ö ) ( ) ( )
IRREDUCIBLE REPRESENTATIONS OF THE PARTY ALGEBRA 455 () = () Ö =2 () = () = () ( ) (Ö ) () ( ) ( ) (Ö ) ( ) ( ) (Ö ) (Ö ) ( ) ( ) ( ) (Ö ) (Ö ) ( ) Now we apply the equation (0) and Lemma 2.2 to the first line, and multiply the last two lines of the equation above by () = if = 0 0 if = 0 Then this equation will be equal to the following: () () Ö Ö= () = () = () Ö =2 () () ( ) () ( ) (Ö ) () ( ) =2 ( ) () ( ) (Ö ) (Ö ) (Ö ) ( ) (Ö ) (Ö ) ( ) ( ) ( ) Applying Lemma 2. 2, to the last two lines of the equation above, we obtain the same equation as Ú È. CASE 6..2. = = 0, «( ) = «(). Table 3. Case 6..2 = = 0, «( ) = «() coordinate board 0 = ««() «2 «
456 M. KOSUDA In this case we assumed = «( ) ( ) and = «() ( ) are distinct. So we find = or = 0. By the same method shown in Case 6.., using Table 3 we can put (28) Ú È = Then we have (29) Ú È = Using the notation (7), we have () (30) Ú È = = ( ) ( ) () () = Hence using the notation in Lemma 2.2 and 2.3, we have (3) () Ú È = Ö= () = () Ö =2 ( ) Ö (Ö ) (Ö ) ( ) ( ) Ö ( ) (Ö ) (Ö ) ( ) (Ö ) (Ö ) Here we used the equation (0) and Lemma 2.2 (resp. the equations (), (8) and Lemma 2.3) to obtain the second (resp. bottom) line of the equation above. On the other hand, we have (32) Ú È = () () () () Æ Since = 0 in this case, using (8), (9) and = () ( ) we have Ú È = (33) () Ö= () Ö = (Ö ) (Ö ) (Ö ) (Ö )
IRREDUCIBLE REPRESENTATIONS OF THE PARTY ALGEBRA 457 Hence we have () Ú È = = () Ö= () Ö =2 ( ) Ö (Ö ) (Ö ) Ö (Ö ) (Ö ) ( ) (Ö ) (Ö ) ( ) Since =, using Lemma 2. we have =. Comparing the equation (3), in this case we find Ú È = Ú È. CASE 6.2. (Case 6.5.) = 0 =. For the explanation below, see Table 4 and Table 5. Put «( ) ( ) «( ) ( 0 ) «( ) ( 0 ) = («) The assumption = 0 = means the following. ) On the way from «( ) to «(), two boxes of = «( ) ( 0 ) on the 0 -th board are moved to the ( 0 )-st board. We put «() ( ) «() ( 0 ) «() ( 0 ) = «() 2) Then on the way from «() to «(2), one box of «= «() ( ) on the -th board is attached to «() ( 0 ) = () on the 0 -th board. We put «(2) ( ) «(2) ( 0 ) «(2) ( 0 ) = («) Similarly as in Case 6., we have only to consider the case «() ( 0 ) (). Namely, if we put = () ( Ò ), the following cases should be considered: «() ( ) «() ( 0 ) «() ( 0 ) = («) («) Case 6.2. Case 6.2.2 CASE 6.2.. = 0 =, «() ( 0 ) = «(2) ( 0 ). Table 4. Case 6.2. = 0 =, «() ( 0 ) = «(2) ( 0 ) coordinate board 0 = 0 «««() 2 «
458 M. KOSUDA Since = we have = and = 0. By the same method shown in Case 6.., using Table 4 we can put (34) Ú È = () Then we have (35) Ú È = () () () () Æ Since = 0, using the notation (5), (6) and = () ( ) we have (36) Ú È = () = ( ) ( ) Hence using the notation (4), we obtain () Ú È = Ö Ö= () =2 ( ) ( ) ( ) Ö ( ) ( ) ( ) ( ) ( ) On the other hand we have (37) Ú È = Hence we have (38) () Ú È = Ö= () =2 ( ) Ö Ö ( ) ( ) ( ) ( ) ( ) Here we used the equations (0), (5) and Lemma 2.2 to obtain the second line of the equation above. Since Ö = Ö, we have Ö Ö = Ö Ö by Lemma 2.. Hence in this case we obtain Ú È = Ú È.
IRREDUCIBLE REPRESENTATIONS OF THE PARTY ALGEBRA 459 CASE 6.2.2. = 0 =, «() ( 0 ) = «(2) ( 0 ). By the same method shown in Case 6.., using Table 5 we can put Ú È = () Then using the notation (5) and (6), we have Ú È = () By Lemma 2.4 we have () Ú È = Ö= = () ( ) () =2 () On the other hand we have () Ú È = Ö= () Ö= () =2 () =2 ( ) ( ) ( ) Æ = (). Using this, we find that ( Ö Ö ) () ( ) () () () ( ) () ( ) ( ) Ö ( ) ( ) Ö Ö ( ) ( ( ) ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) Table 5. Case 6.2.2 = 0 =, «() ( 0 ) = «(2) ( 0 ) coordinate board 0 = 0 «««() 2 «
460 M. KOSUDA Applying the equations (0), (2) (resp. the equations (5), (8) and Lemma 2.2) to the second (resp. bottom) line of the equation above, we obtain () Ú È = Ö= () =2 () Ö Ö Ö () ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) Since Ö = Ö and ( ) = ( ), using Lemma 2. 2, respectively we find Ú È = Ú È in this case. CASE 6.3. (Case 6.6.) 0 = =. For the explanation below, see Table 6, 7 and 8. Put «( ) ( ) «( ) ( 0 ) «( ) ( 0 ) = («) The assumption 0 = = means the following: ) On the way from «( ) to «(), one box of = «( ) ( 0 ) on the 0 -th board is removed and it is attached to = «( ) ( ) on the -th board. We put «() ( 0 ) =. 2) Then on the way from «() to «(), one box of «= «() ( ) on the -th board is removed and it is attached to «() ( 0 ) on the 0 -th board. We put «() ( ) = «. 3) Then on the way from «() to «(2), one box of «() ( 0 ) on the 0 -th board are moved to the next board. We put «(2) ( ) «(2) ( 0 ) «(2) ( 0 ) = («) As well as the Case 6., we find that the only following cases should be considered: «() ( 0 ) «() ( 0 ) = ( ) (Case 6.3.) ( ) (Case 6.3.2) ( ) (Case 6.3.3). Here and are Young diagrams such that,, = and =. Note also that in these cases. CASE 6.3.. 0 = =, «() ( 0 ) = «( 0 ), «(2) ( 0 ) = «() ( 0 ). By the same method shown in Case 6.., using Table 6 we can put (39) Ú È =
IRREDUCIBLE REPRESENTATIONS OF THE PARTY ALGEBRA 46 Using the calculation (37), and (38) in Case 6.2., in this case we have (40) Hence we have (4) () Ú È = () Ú È = Õ= () Õ = () =2 Ö= () =2 (Õ) (Õ ) ( ) ( ) Ö Ö ( ) ( ) ( ) () () ( Ö= ) Ö Ö (Õ) () () ( ) Ö Ö Õ Ö Ö= ( ) ( ) ( ) ( ) ( ) (Õ) (Õ ) (Õ ) The first line of the equation (4) should be 0 by Lemma 2.5 3. As for the second line, since Ö = Ö, we have Ö Ö = ( Ö Ö ) by Lemma 2.. Hence as well as the first line, using Lemma 2.5 3 we have () Ö= () ( ) Ö Ö Õ Ö = = () () ( Ö= )( Ö Õ Ö Ö Õ Ö ) () () if Õ = so (Õ ) = ( ) if () (Õ ) = 0 otherwise Table 6. Case 6.3. 0 = =, «() ( 0 ) = «( ) ( 0 ), «(2) ( 0 ) = «() ( 0 ). coordinate board 0 = 0 «««2 «
462 M. KOSUDA Hence we obtain Ú È = () =2 ( ) ( ) ( ) ( ) On the other hand, since we have Ú È = () using the calculation (34), (35) and (36) in Case 6.2. we have Ú È = () = ( ) ( ) ( ) ( ) Since () = by (6), we find that in this case Ú È = Ú È. CASE 6.3.2. 0 = =, «() ( 0 ) = «( ) ( 0 ), «(2) ( 0 ) = «() ( 0 ). By the same method shown in Case 6.., using Table 7 we can put Ú È = Since Ú È = () Table 7. Case 6.3.2 0 = =, «() ( 0 ) = «( ) ( 0 ), «(2) ( 0 ) = «() ( 0 ). coordinate board 0 = 0 «««2 «
IRREDUCIBLE REPRESENTATIONS OF THE PARTY ALGEBRA 463 using the calculation (28), (32) and (33) in Case 6..2 we have Ú È = () Ö= () Ö = (Ö ) (Ö ) (Ö ) (Ö ) On the other hand, using the calculation (29), (30) and (3) in Case 6..2, in this case we have () Ú È = Ö= () = () Ö =2 ( ) Ö (Ö ) (Ö ) Hence using the notation (20), we have () Ú È = Ö= () = () Ö =2 () ( ) (Ö ) Ö ( ) (Ö ) (Ö ) ( ) (Ö ) (Ö ) () ( = ( ) ) (Ö ) (Ö ) ( ) (Ö ) ( ) As for the second line of the equation above, since =, by Lemma 2. we have the following: = =
464 M. KOSUDA Hence using Lemma 2.5 4, we have () () = ( ) = = () () = ( ) () () if = so ( ) = ( ) if ( ) () = 0 otherwise. Thus we obtain () Ú È = Ö= () Ö =2 (Ö ) (Ö ) (Ö ) (Ö ) Since () =, in this case we obtain Ú È = Ú È. CASE 6.3.3. 0 = =, «() ( 0 ) = «( ) ( 0 ), «(2) ( 0 ) = «() ( 0 ). By the same method shown in Case 6.., using Table 8 we can put Ú È = Table 8. Case 6.3.3 0 = =, «() ( 0 ) = «( ) ( 0 ), «(2) ( 0 ) = «() ( 0 ). coordinate board 0 = 0 «««2 «
IRREDUCIBLE REPRESENTATIONS OF THE PARTY ALGEBRA 465 Replacing with in the calculation (39), (40) and (4) in Case 6.3., we have (42) () Ú È = Õ= ()2 () (Õ) () 3 Ö (Õ) Ö= () () () Õ =2 (Õ ) Ö= () Ö= ( ) ( ) (Õ) ( ) Ö Õ Ö (Õ ) ( ) (Õ ) Here we used Lemma 2.5 3 to obtain the first line of the equation above. On the other hand, replacing with in the calculation (29), (30) and (3) in Case 6..2, we have Hence we have () Ú È = (43) () Ú È = Ö= Ö= () = () ()2 () () =2 () Ö =2 = () = Ö ( ) (Ö ) 3 ( ) Ö ( ) (Ö ) (Ö ) ( ) (Ö ) (Ö ) () () ( ) = ( ) (Ö ) (Ö ) ( ) (Ö ) ( ) Now we see that the expressions (42) and (43) coincide. The first lines of them are obvious. As for the second lines of them, by Lemma 2.5 5 they coincide. We show
466 M. KOSUDA that the third line of the expression (42) and the last line of the expression (43) are equivalent. To prove this, we have only to show the following: (44) () () () Ö= () Ö Ö Ö = (Ö ) Since Ö (Ö ) = Ö Ö, by Lemma 2. we have Ö Ö Ö = Ö Ö (Ö ) Ö Ö (Ö ) = 2 Ö Ö Ö Ö (Ö ) Hence if we use Lemma 2.5 3, we can show that the equation (44) holds. Similarly, we can show that the last line of the expression (42) and the third line of the expression (43) are equal. Hence in this case, we obtain Ú Ô = Ú Ô. 3.2. Preservation of the other relations. In the following, we check that «preserve other relations (È ), (È 3) (È 8). By the definition of «, the relations (È 3), (È 4), (È 5), (È 6) are immediately checked. We focus on the relations (È ), (È 7) and (È 8). Proposition 3.. «preserves the relation (È ). Proof. Let Å be the matrix defined in Section. We have only to show that Å 2 = Á (the identity matrix). In case that Å is given by the matrix (4) or (7) this is easily checked. Consider the case that Å is given by the matrix (5). Since the matrix is symmetric, in order to show that Å 2 = Á, we find that the following equations must be checked: () Ö= () () Ö= () () () () Ö = (Ö ) = (Ö 0 ) (Ö 0 ) (Ö ) () () Ö= () (Ö 0 ) Ö = (Ö ) (Ö 0 ) (Ø ) (Ö ) () Ö= (Ö 0 ) (Ö ) () Ö= = (Ö 0 ) () = 0 (Ö ) () = 0 () (Ö 0 ) = 0 ()
IRREDUCIBLE REPRESENTATIONS OF THE PARTY ALGEBRA 467 Here Ö 0 Ö (), Ö 0 Ö () and Ö 0 = Ö, Ö 0 = Ö. Applying Lemma 2.5, we immediately obtain the equations above. Proposition 3.2. «preserves the relation (È 7). Proof. Consider the subalgebra 2 of A Ò (Ã 0 ). This algebra is isomorphic to the algebra A 3 (Ã 0 ). Hence, it is sufficient to prove that the proposition holds for A 3 (Ã 0 ). Put «= [«() «(2) «(3)]. Fig. 4 will help the reader to understand the following argument. If «() = 3, then none of the tableau of shape «has [ ] at its second coordinate Hence in this case we have «( ) = 0 and obviously the proposition holds. Next, consider the case «= [ ] and let É É 2 É 3 be all the tableaux of shape «defined by É = «(0) [ ] [ ] «É 2 = «(0) [ ] «É 3 = «(0) [ ] [ ] «Then the representation matrices of and 2 with respect to this basis become as follows: «( ) = 0 0 0 0 0 0 0 0 ½ «( 2 ) = 2 2 2 2 Ô Ô 0 2 2 Hence in this case, we can check that the proposition holds by direct calculation. Finally, consider the case «= [ ]. In this case, there exists only one tableau of shape «. The generators 2 and identically act on this tableau. Hence in this case the proposition holds. Ô 2 Ô 2 ½ Proposition 3.3. «preserves the relation (È 8). Proof. As we saw in the proof of the previous proposition, it is sufficient to prove the proposition holds for A 4 (Ã 0 ). Put «= [«() «(2) «(3) «(4)]. Again Fig. 4 will help the reader to understand the following argument. Similarly as in the proof of the previous proposition, we may assume that «() 4.
468 M. KOSUDA Consider the case «= [ ]. Let È È 6 be all the tableaux of shape «defined by È = «(0) «() [ ] [ ] «È 2 = È 3 = «(0) «() [ ] «(0) «() ««È 4 = «(0) «() [ ] [ ] «È 5 = «(0) «() [ ] «È 6 = («(0) «() [ ] [ ] «) Then the representation matrices of 2 and 3 with respect to this basis become as follows: «( ) = diag(0 0 0 0 0 ) «( ) = diag( ) «( 2 ) = «( 3 ) = 0 Ô 0 0 0 0 3 0 0 0 0 2 2 Ô 3 0 0 0 0 2 2 Ô 2 0 0 0 2 2 2 Ô 2 0 0 0 2 2 2 Ô Ô 2 2 0 0 0 0 Ô 2 2 Ô 2 6 ½ 0 0 0 3 3 3 Ô Ô 2 2 3 0 0 0 3 3 3 0 0 0 0 0 Ô Ô 6 3 0 0 0 0 3 3 0 0 0 0 0 0 0 0 0 0 ½ Using these matrices we can check that the relation (È 8) holds.
IRREDUCIBLE REPRESENTATIONS OF THE PARTY ALGEBRA 469 Similarly, for the cases «=, [ ],, [ ], [ ], we can concretely obtain the representation matrices and the relation (È 8) will be checked by the direct calculation. 4. Proof of the main theorem This section is devoted to prove the main theorem. In order to know whether two representations of A Ò (Ã 0 ) are equivalent or not, it is useful to check that how they split into irreducible ones as A Ò (Ã 0 )-modules. So we consider the following set. Let be an Ò-tuple of Young diagrams of height Ò. Then the restriction Ò of to A Ò (Ã 0 ) is defined by Ò = Ô Ô Lemma 4.. Let «= [«() «(Ò)] and = [ () (Ò)] be two Ò-tuples of Young diagrams whose heights are both Ò. For these «, define two integers 0 and by 0 = max «() = = max () = Assume that Ò 3. If «=, then the following statement holds: ) «Ò = Ò, or else 2) 0 =, «() = () = for 0 and «( 0 ) ( 0 ) = Ò Ó. Proof. Assuming that «= and «Ò = Ò, we show that the second statement holds. Without loss of generality, we may further assume that 0, or else = 0 and ( 0 ) «( 0 ). Since if 0 then ( 0 ) = and ( 0 ) «( 0 ), we may always assume that 0 and ( 0 ) «( 0 ). First we show that ( 0 ) = «( 0 ) (accordingly we have = 0 ). Assume that ( 0 ) «( 0 ). If there exists a (0 0 ) such that «() =, then there exists an Ò-tuple of Young diagrams «¾ «Ò such that «= «() «( ) «() «( 0 ) where =, «( ) «( ) and «() «(). The assumption ( 0 ) «( 0 ) makes us unable to obtain from «. This contradicts the assumption «Ò = Ò. Hence if ( 0 ) «( 0 ), then «must be of the form However, this requires that (45) «Ò = «= [ «( 0 ) ] Ò Ó «( 0 ) (Ô) «(0 ) (Ô) «( 0 )
470 M. KOSUDA Since 0, must be written as one of the following forms: ) = 0 and = [ ( 0 ) ], 2) 3 0 and = «( 0 ) (Ô) 3) 0 = 2 and = «( 0 ) (Ô), 4) 0 = 2 and = «( 0 ) (Ô). The first one contradicts the assumption ( 0 ) «( 0 ). In the second case, there exists an Ò-tuple of Young diagrams which is not contained in the set (45). In the remaining cases, in order that Ò = «Ò holds, «( 0 ) (Ô) must be the empty partition and this contradicts the assumption Ò 3. Hence we obtain «( 0 ) = ( 0 ). Accordingly, = 0 follows. Next we show that «( 0 ) ( 0 ) = Ò Ó. Since «Ò = Ò requires «(0 ) (Ô) «(0 ) (Ô) «( 0 ) = ( 0 ) (Õ) (0 ) (Õ) ( 0 ), we find that «( 0 ) = ( 0 ) or «( 0 ) = ( 0 ). If «( 0 ) = ( 0 ), then «Ò = Ò requires «( 0 ) = ( 0 ) and inductively we obtain «() = () forò 0. This contradicts the assumption «=. Hence we obtain «( 0 ) ( 0 ) = Finally, we show that «() = () = for 0. Note that since Ò 3, we can assume that 2 0. If there exists an «() = for 0, then there exists an Ò-tuple of Young diagrams «¾ «Ò such that «= [«() «( ) «() «( 0 ) ] However since «( 0 ) = ( 0 ) and «( 0 ) = ( 0 ), we can not obtain from «. This contradicts the assumption «Ò = Ò. The same argument also holds for. Hence we have «() = () = for 0. Thus we have proved the lemma. Lemma 4.2. Let «= [«() «(Ò)] be an Ò-tuple of Young diagrams of height Ò. For an arbitrary distinct pair «0 ««Ò, there exists an Ò-tuple of Young diagrams of height Ò 2 such that «0 and «. Proof. Assume that «0 (resp. «) is obtained from «by moving a box on the 0 -th (resp. -th) board. Namely, «0 and «are written as follows: «0 = «() «( 0 2) «( 0 ) (Ô) «( 0) (Ô ) «( 0 ) «(Ò) «= «() «( 2) «( ) (Õ) «( ) (Õ ) «( ) «(Ò) Here 0 = 0 and =. Since without loss of generality we may assume that 0, the following cases should be considered. CASE. 0 2. In this case, «() «( ) (Õ) «( ) (Õ ) «( 0 ) (Ô) «( 0) (Ô ) «(Ò) = Ó.
IRREDUCIBLE REPRESENTATIONS OF THE PARTY ALGEBRA 47 satisfies the required condition. CASE 2. = 0. In this case, «() «( ) (Õ) «( ) «( 0 ) (Ô ) «(Ò) = satisfies the required condition. CASE 3. = 0. In this case, satisfies the required condition. Here «( 0 ) = «( 0 ) = = «() «( 0 ) «( 0 ) «(Ò) «( 0 ) (Ô) «( 0 ) (Õ) if «( 0 ) (Ô) = «( 0 ) (Õ) one of the Young diagrams such that «( 0 ) «(Ô) ( 0 ) if «( 0 ) (Ô) = «( 0 ) (Õ) «( 0 ) (Ô ) «( 0) (Õ ) if «( 0 ) (Ô ) = «( 0) (Õ ) one of the Young diagrams such that «( 0 ) «( 0 ) (Ô ) if «( 0 ) (Ô ) = «( 0) (Õ ) Proof of Theorem.4. Since we have shown that the representations «are well-defined, we have only to show that they are also absolutely irreducible and mutually non-equivalent. We do this by induction on Ò. If Ò = 0, then the result is obvious. So is the case Ò =. If Ò = 2, we can easily check that the (three) representations are mutually non-isomorphic. Since they are all one-dimensional and dim A Ò (Ã 0 ) = 3, we find that they make a complete set of absolutely irreducible representations. Assume that Ò 3 and the theorem holds for Ò. Let A = Ò 2 be the subalgebra of A Ò (Ã 0 ). This algebra is isomorphic to the algebra A Ò (Ã 0 ). Consider the restriction of the representation «of A Ò (Ã 0 ) to the subalgebra A. Suppose that «Ò = ««. We divide the set T(«) of the standard tableaux of shape «into subsets T «T «. Here T «Ô is the subset of T(«) whose (Ò )-st coordinate is «Ô Ô. We define subspaces V «of V(«) corresponding to these subsets T «Ô. Namely, = V «Ô Ã 0 Ú È È ¾T(«Ô) Then the definition of the action of Ò 2 and implies that V «Ô is stable under the action of A and induction hypothesis shows that V «Ô gives an absolutely irreducible representation of A Ò (Ã 0 ) (hence A ) and that if Ô = Õ then V «Ô is not isomorphic to V «Õ as AÒ (Ã 0 )-modules.
472 M. KOSUDA Let Ï be a non-zero subspace of V(«) Å Ã 0 as A Ò Ã 0 -modules, where Ã0 denotes the algebraic closure of the field à 0 and A Ò Ã 0 = AÒ Å Ã 0. If we consider Ï as an A Åà 0 -module, then it contains some irreducible component V «Ô of A Åà 0. Let «Õ (Õ = Ô) be another Ò-tuple of Young diagrams contained in «Ò. Then by Lemma 4.2, there exists an Ò-tuple of Young diagrams contained in both «Ô Ò 2 and «Õ Ò 2. Let È be a tableau of shape «whose (Ò 2)-nd and (Ò )-st coordinates are and «Ô respectively. We can obtain another tableau É of shape «from È by replacing the (Ò )-st coordinate of È with «Õ. Now we claim that there exists a projection È É of A Ò (à 0 ) such that È É Ú È = Ú É. By induction assumption, A is an absolutely semisimple algebra with the minimal central idempotents Þ «, labeled by Ò-tuples of Young diagrams of height (Ò ). According to the classification in the proof of Lemma 4.2, consider the action of Ò ¾ A Ò (à 0 ) one by one. CASE. 0 2. In this case, we have Ò Ú È = Ú É and the claim is proved. CASE 2. = 0. In this case we have Þ «Ô Ò Ú È = «( ) (Õ ) «( ) «( ) (Ô) («( )) 2 «( ) (Õ ) «( ) (Ô) Since the coefficient of Ú É is not equal to zero, the claim is proved. CASE 3. = 0. Ú É In this case, the following four cases are considered. In each case, Þ «Ô Ò sends Ú È non-zero scalar multiple of Ú É, so the claim is proved. In the following, we put = «( 0 ) «( 0 ) (Ô) «( ) 0 and = «( 0 ) «( 0 ) (Ô ) «( 0). CASE 3.. «( 0 ) (Ô) = «( 0 ) (Õ) and «( 0) (Ô ) = «( 0) (Õ ). In this case we have Þ «Ô Ò Ú È = Ú É CASE 3.2. «( 0 ) (Ô) = «( 0 ) (Õ) and «( 0) (Ô ) = «( 0) (Õ ). In this case we have Þ «Ô Ò Ú È = Ú É