Partition Algebras

Tom Halverson† Arun Ram∗ Mathematics and Computer Science Department of Mathematics Macalester College University of Wisconsin–Madison Saint Paul, MN 55105 Madison, WI 53706 [email protected] [email protected]

June 24, 2003

0. Introduction

A centerpiece of representation theory is the Schur-Weyl duality, which says that,

(a) the general linear group GLn(C) and the symmetric group Sk both act on tensor space ⊗k ⊗···⊗ V = V V, with dim(V )=n, k factors

(b) these two actions commute and (c) each action generates the full centralizer of the other, so that

(d) as a (GLn(C),Sk)-bimodule, the tensor space has a multiplicity free decomposition, ⊗k ∼ ⊗ λ V = LGLn (λ) Sk , (0.1) λ

C λ where the LGLn (λ) are irreducible GLn( )-modules and the Sk are irreducible Sk-modules.

The decomposition in (0.1) essentially makes the study of the representations of GLn(C) and the study of representations of the symmetric group Sk two sides of the same coin. The group GLn(C) has interesting subgroups,

GLn(C) ⊇ On(C) ⊇ Sn ⊇ Sn−1, and corresponding centralizer algebras, CSk ⊆ CBk(n) ⊆ CAk(n) ⊆ CA 1 (n), k+ 2 which are combinatorially deﬁned in terms of the “multiplication of diagrams” (see Section 1) and which play exactly analogous “Schur-Weyl duality” roles with their corresponding subgroup

† Research supported in part by National Science Foundation Grant DMS-0100975. ∗ Research supported in part by the National Science Foundation (DMS-0097977) and the National Security Agency (MDA904-01-1-0032). 2 t. halverson and a. ram of GLn(C). The Brauer algebras CBk(n)were introduced in 1937 by R. Brauer [Bra] and the partition algebras CAk(n) arose in the 1990s in the work of V. Jones [Jo] and P. Martin [Ma1-3] as a generalization of the Temperley-Lieb algebra and the Potts model in statistical mechanics. Though the partition algebras CA 1 (n)have been used in the work of P. Martin and G. Rollet k+ 2 [MR], they have been studied minimally. Their existence and importance was pointed out to us by C. Grood. In this paper we show that if the algebras CA 1 (n) are given the same stature as the k+ 2 algebras Ak(n), then well-known methods from the theory of the Jones basic construction allow for easy analysis of the whole tower of algebras

CA0(n) ⊆ CA 1 (n) ⊆ CA1(n) ⊆ CA 1 (n) ⊆ ···, 2 1 2 all at once. 1 ∈ Z≥ Let 2 0.Inthis paper we prove: (a) A presentation by generators and relations for the algebras CA(n).

(b) CA(n) has C A(n) ∼ an ideal CI(n), with = CS, CI(n)

such that CI(n)isisomorphic to a Jones basic construction. Thus the structure of the ideal CI(n) can be analyzed with the general theory of the basic construction and the structure of the quotient CA(n)/(CI(n)) follows from the general theory of the representations of the symmetric group.

(c) The algebras CA(n) are in “Schur-Weyl duality” with the symmetric groups Sn and Sn−1 on V ⊗k. (d) The general theory of the basic construction provides a construction of “Specht modules” for the partition algebras, i.e. integral lattices in the (generically) irreducible CA(n)-modules.

(e) Except for a few special cases, the algebras CA(n) are semisimple if and only if ≤ (n+1)/2.

(f) There are “Murphy elements” Mi for the partition algebras that play exactly analogous roles to the classical Murphy elements for the group algebra of the symmetric group. In particular, the Mi commute with each other in CA(n), and when CA(n)issemisimple each irreducible CA(n)-module has a unique, up to constants, basis of simultaneous eigenvectors for the Mi. The primary new results in this paper are (a) and (f). There has been work towards a presentation theorem for the partition monoid by Fitzgerald and Leech [FL], and it is possible that by now they have proved a similar presentation theorem. The statement in (b) has appeared implicitly (though perhaps not explicitly) throughout the literature on the partition algebra and should not be considered new. However, we stress that it is the fact that we make this explicit which allows us to freely and efficiently exploit the powerful results from the theory of the Jones basic construction. We consider this an important step in the understanding of the structure of the partition algebras. The Schur-Weyl duality for the partition algebras CAk(n) appears in [Ma1] and [MR] and was one of the motivations for the introduction of these algebras in [Jo]. The Schur- Weyl duality for CA 1 (n) appears in [MW]. Most of the previous literature (for example [Ma3], k+ 2 [MW1-2], [DW]) on the partition algebras has studied the structure of the partition algebras using the “Specht” modules of (d). Our point here is that their existence follows from the general theory of the Jones basic construction. The statements about the semsimplicity of CA(n)have mostly, if not completely, appeared in the work of Martin and Saleur [Ma3], [MS]. The Murphy elements for the partition algebras are new. Their form was conjectured by Owens [Ow], who proved that the partition algebras 3 sum of the first k of them is a central element in CAk(n). Here we prove all of Owens’ theorems and conjectures (by a different technique than he was using). We have not taken the next natural step and provided formulas for the action of the generators of the partition algebra in the “seminormal” representations. We hope that someone will do this in the near future. Though this paper contains new results and new techniques in the study of partition algebras we have made a distinct effort to present this material in a “survey” style so that it may be accessible to nonexperts and to newcomers to the field. For this reason we have included, in separate sections, expositions, from scratch, of (a) the theory of the Jones basic construction (see also [GHJ, Ch. 2]), and (b) the theory of semisimple algebras, in particular, Maschke’s theorem, the Artin-Wedderburn theorem and the Tits deformation theorem. Our reworking of the theory of the basic construction removes many of the nondegeneracy assump- tions that appear in other papers where this theory is used. In fact, the basic construction theory is quite elementary and elegant, and it is a powerful tool for the study of algebras such as the partition algebra. The theory of semisimple algebras appears in many standard algebra and representation theory textbooks, but here the reader will find statements of the main theorems which are in exactly the correct form for our applications (generally difficult to find in the literature), and short slick proofs of all the results on semisimple algebras that we need for the study of the partition algebras. There are two sets of results on partition algebras that we have not had the space to treat in this paper: (a) The “Frobenius formula,” “Murnaghan-Nakayama” rule, and orthogonality rule for the irreducible characters given by Halverson [Ha] and Farina-Halverson [FH], and (b) The cellularity of the partition algebras proved by Xi [Xi] (see also Doran and Wales [DW]). The techniques in this paper apply, in exactly the same fashion, to the study of other diagram algebras; in particular, the planar partition algebras CPk(n), the Temperley-Lieb algebras CTk(n), and the Brauer algebras CBk(n). It was our original intent to include in this paper results (mostly known) for these algebras analogous to those which we have proved for the algebras CA(n), but the restrictions of time and space have prevented this. While perusing this paper, the reader should keep in mind that the techniques we have used do apply to these other algebras.

1. The Partition Monoid

For k ∈ Z>0, let A = set partitions of {1, 2,...,k,1, 2,...,k} , and k (1.1) A 1 = d ∈ Ak+1 (k +1)and (k +1) are in the same block . k+ 2

The propagating number of d ∈ Ak is the number of blocks in d that contain both an element pn(d)= . (1.2) of {1, 2,...,k} and an element of {1, 2,...,k}

For convenience, represent a set partition d ∈ Ak by a graph with k vertices in the top row, labeled 1,...,k left to right, and k vertices in the bottom row, labeled 1,...,k left to right, with vertex 4 t. halverson and a. ram

i and vertex j connected by a path if i and j are in the same block of the set partition d.For example,

12345678 •.... .•...... • ...•.. •... .•...... •.. •...... ...... ...... { } { } { } { } { } ...... represents 1, 2, 4, 2 , 5 , 3 , 5, 6, 7, 3 , 4 , 6 , 7 , 8, 8 , 1 , • ..•...... •... •...... •.. •..... •... •. 1 2 3 4 5 6 7 8

and has propagating number 3. The graph representing d is not unique. Deﬁne the composition d1 ◦ d2 of partition diagrams d1,d2 ∈ Ak to be the set partition d1 ◦ d2 ∈ Ak obtained by placing d1 above d2 and identifying the bottom dots of d1 with the top dots of d2, removing any connected components that live entirely in the middle row. For example,

•..... • ...•.... •... .•...... •.. • • •... •... .•.. •... •... .•...... if ..... and ...... then ...... d1 = ...... d2 = ...... • •... •... ..•... •.... • ..•.. • •...... • ..•...... •...... •...... •...

•.... • ..•..... •... •..... •.. • ...... •...... • ...... •..... •...... •...... •.. • ...... •. •. •. •. •. •. •...... ◦ ...... = ...... d1 d2 = ...... •. •..... •... ..•.. •..... •... ..•... • •..... • •.... •.... •.. ...•...... • •...... • ..•...... •...... •...... •...

•. •. •. . . . Diagram multiplication makes A into an associative monoid with identity, 1 = . . ··· . . The k •. •. •. propagating number satisﬁes

pn(d1 ◦ d2) ≤ min(pn(d1),pn(d2)). (1.3)

A set partition is planar [Jo] if it can be represented as a graph without edge crossings inside ∈ 1 Z of the rectangle formed by its vertices. For each k 2 >0, the following are submonoids of the partition monoid Ak:

Sk = {d ∈ Ak | pn(d)=k},Ik = {d ∈ Ak | pn(d)

Examples are

•.... •. ..•.... •... ..•... •...... •.... •.... •. ..•.... •... ..•... •...... •...... ∈ ...... ∈ 1 ...... I7, ...... I6+ , •. •. •...... •... ..•... •...... •.... •. •. •...... •... ..•... •...... •.... 2

•..... •...... •...... •...... •.... •...... •... •..... •...... •...... •..... •.... •...... •...... ∈ ...... ∈ 1 ...... P7, ...... P6+ , •...... •...... •... • ...... •... ..•... • •...... •...... •... • ...... •... ..•... ..•... 2

•...... •...... •.. •... .•.. •...... •.. •..... •...... •...... •...... •.. ...•.. •...... ∈ ...... ∈ ...... B7, ...... T7, •..... •...... •...... •...... •.. •...... •.. •...... •...... •.. •...... •...... •.. •.

•...... •. ..•... •...... •.. •.... .•...... ∈ ...... S7. •...... •. •...... •.. ..•... •...... •... partition algebras 5

∈ 1 Z For k 2 >0, there is an isomorphism of monoids

1−1 Pk ←→ T2k, (1.5) which is best illustrated by examples. For k =7wehave

•.... •...... •...... •.... •.. •... .•.. ◦. •.....◦.... ◦....•.....◦...... ◦....•...... ◦...... ◦....•.....◦.. ◦..•...◦.. ◦...•....◦.... .◦...•.. ◦.. ◦. ◦.... ◦..... ◦.... .◦... ◦.... .◦... ◦.. ◦.. ◦... ◦... ◦.... .◦... ◦...... ↔ ...... ↔ ...... , ...... •.... .•...... •... • ...•... •... • ◦. •....◦... .◦....•...... ◦... .◦.....•.....◦. ◦...•.◦.. ...◦...... •... ◦.. ◦.. •... ◦.. ◦...•.◦.. ◦. ◦... ◦... ◦... ◦... ..◦... ◦... ◦...... ◦... ◦... ◦... ◦... ◦... ◦...

1 and for k =6+2 we have

•.... •...... •...... •.... •.. •... .•.. ◦. •.....◦.... ◦....•.....◦...... ◦....•...... ◦...... ◦....•.....◦.. ◦..•...◦.. ◦..•....◦.... .◦...•... ◦. ◦.... ◦..... ◦.... .◦... ◦.... .◦... ◦.. ◦.. ◦.. ◦.. ◦.... .◦...... ↔ ...... ↔ ...... •.... .•...... •... • ...•... •... •. ◦. •....◦... .◦....•...... ◦... .◦.....•.....◦. ◦...•.◦.. ...◦...... •... ◦.. ◦.. •... ◦.. ◦.. •. ◦. ◦... ◦... ◦... ◦... ..◦... ◦... ◦...... ◦... ◦... ◦... ◦... ◦...

Let k ∈ Z>0.Bypermuting the vertices in the top row and in the bottom row each d ∈ Ak can be written as a product d = σ1tσ2, with σ1,σ2 ∈ Sk and t ∈ Pk, and so

...... •...... •...... •...... •.. •.... •.... •...... • ...... •...... •...... •...... •...... •... .•... .•... .•...... • •. • •...... •... •. • ...... •.. • • • ....•.. •. •. • •...... Ak = Sk Pk Sk. For example, ...... = ...... (1.6) • •...... •...... •...... •...... •...... •...... •.. • •...... •...... •...... •.. •...... •...... •...... •. •. •. •.... •...... •.. .•... .•...... •. •. •. •...... •... •.... •...... •..

For ∈ Z>0, deﬁne

the Bell number, B()=(the number of set partitions of {1, 2,...,}), 1 2 2 2 the Catalan number, C()= = − , (1.7) +1 +1 (2)!! = (2 − 1) · (2 − 3) ···5 · 3 · 1, and != · ( − 1) ···2 · 1, with generating functions (see [Sta, 1.24f, and 6.2]), √ z 1 − 1 − 4z B() = exp(ez − 1), C( − 1)z = , ! 2z ≥0 √ ≥0 (1.8) z 1 − 1 − 2z z 1 (2( − 1))!! = , ! = . ! z ! 1 − z ≥0 ≥0

Then

∈ 1 Z for k 2 >0, Card(Ak)=B(2k) and Card(Pk)=Card(T2k)=C(2k) , (1.9) for k ∈ Z>0, Card(Bk)=(2k)!!, and Card(Sk)=k!. 6 t. halverson and a. ram

Presentation of the Partition Monoid In this section, for convenience, we will write

d1d2 = d1 ◦ d2, for d1,d2 ∈ Ak

Let k ∈ Z>0.For1≤ i ≤ k − 1 and 1 ≤ j ≤ k, deﬁne

i i+1 j ...... •. •. •. •. •. •. •. •. • •. •...... 1 . ··· . . . . ··· . . ··· . . ··· . pi+ = ...... ,pj = . . . . , 2 •. •. •...... •.... •. •. •. •. • •. •. (1.10) i i+1 i i+1 •. •. •... •... •. •. •. •. •.. •.. •. •...... ··· . . ··· . . ··· ...... ··· ...... ei = ...... ,si = ...... •. •. •...... •... •. •. •. •. •...... •... •. •.

Note that ei = p 1 pipi+1p 1 . i+ 2 i+ 2 Theorem 1.11. (a) The monoid Tk is presented by generators e1,...,ek−1 and relations

2 | − | ei = ei,eiei±1ei = ei, and eiej = ejei, for i j > 1.

(b) The monoid Pk is presented by generators p 1 ,p1,p3 ,...,pk and relations 2 2

2 p = pi,pip ± 1 pi = pi, and pipj = pjpi, for |i − j| > 1/2. i i 2

2 | − | si =1,sisi+1si = si+1sisi+1, and sisj = sjsi, for i j > 1.

(d) The monoid Ak is presented by generators s1,...,sk−1 and p 1 ,p1,p3 ,...,pk and relations in 2 2 (b) and (c) and

sipipi+1 = pipi+1si = pipi+1,sip 1 = p 1 si = p 1 ,sipisi = pi+1, i+ 2 i+ 2 i+ 2 1 1 3 sisi+1p 1 si+1si = p 3 , and sipj = pjsi, for j = i − ,i,i+ ,i+1,i+ . i+ 2 i+ 2 2 2 2

Proof. Parts (a) and (c) are standard. See [GHJ, Prop. 2.8.1] and [Bou1, Ch. IV §1.3, Ex. 2], respectively. Part (b) is a consequence of (a) and the monoid isomorphism in (1.5).

(d) The right way to think of this is to realize that Ak is deﬁned as a presentation by the generators d ∈ Ak and the relations which specify the composition of diagrams. To prove the presentation in the statement of the theorem we need to establish that the generators and relations in each of these two presentations can be derived from each other. Thus it is suﬃcient to show that (1) The generators in (1.10) satisfy the relations in Theorem (1.11).

(2) Every set partition d ∈ Ak can be written as a product of the generators in (1.10).

(3) Any product d1 ◦ d2 can be computed using the relations in Theorem (1.11). (1) is established by a direct check using the deﬁnition of the multiplication of diagrams. (2) follows from (b) and (c) and the fact (1.6) that Ak = SkPkSk. The bulk of the work is in proving (3). partition algebras 7

Step 1. First note that the relations in (a–d) imply the following relations:

p 1 si−1p 1 = p 1 sisi−1p 1 = p 1 sisi−1p 1 si−1sisisi−1 (e1) i+ 2 i+ 2 i+ 2 i+ 2 i+ 2 i+ 2

= p − 1 p 1 sisi−1 = p − 1 p 1 si−1 = p 1 p − 1 si−1 = p 1 p − 1 . i 2 i+ 2 i 2 i+ 2 i+ 2 i 2 i+ 2 i 2

(e2) pisipi = sisipisipi = sipi+1pi = pi+1pi.

(f1) pip 1 pi+1 = pip 1 sipi+1 = pip 1 pisi = pisi. i+ 2 i+ 2 i+ 2

(f2) pi+1p 1 pi = pi+1sip 1 pi = sipip 1 pi = sipi. i+ 2 i+ 2 i+ 2

Step 2. Analyze how elements of Pk can be eﬃciently expressed in terms of the generators. Let t ∈ Pk. The blocks of t partition {1,...,k} into top blocks and partition {1 ,...,k } into bottom blocks.Int, some top blocks are connected to bottom blocks by an edge, but no top block is connected to two bottom blocks, for then by transitivity the two bottom blocks are actually a single block. Draw the diagram of t, such that if a top block connects to a bottom block, then it connects with a single edge joining the leftmost vertices in each block. The element t ∈ Pk can be decomposed in block form as

t =(p 1 ···p 1 )(pj ···pj )τ(p ···p )(p 1 ···p 1 )(1.12) i1+ 2 ir + 2 1 s 1 m r1+ 2 rn+ 2 with τ ∈ Sk, i1

•. •...... •...... •.... •. •...... •.... •...... •. •...... •...... •.... •. •...... •.... •. •. •...... •...... •.... •. •...... •.... •. •. •...... •...... •.... •. •...... •.... •. •. •. • • •. • • •...... •. •. • • •. • • •. •. •. • • •. • • •. •. •. • • •. • • •...... •. •...... • • •... • • •. •. •...... • • •... • • •. •. •...... •.. •.. •... ..•. •. •. •. •...... •.. •.. •... ..•. •. •...... t = ...... = ...... = ...... = ...... •...... •...... •.... • ....•.. •... • •. •. • • • ....•.. •... • •. •. •. •. •...... •.. •... •. •. •. •. •. •...... •.. •... •. •. •. • • • •. •. • •. •. • • • •. •. • •. •. • • • •. •. • •...... •...... •...... •.... • •. •. • •. •...... •...... •.... • •. •. • •. •. • • • •. •. • •. •...... •...... •.... •. •. •. •. •...... •...... •...... •.... •. •. •. •. •.

=(p 1 p 1 p 1 )(p3p4p6p7)τ(p2p3p4p7)(p 1 p 1 ), 2 2 3 2 6 2 1 2 2 2

=(p 1 p 1 p 1 )(p3p4p6p7)s2s3s5s4(p2p3p4p7)(p 1 p 1 ), 2 2 3 2 6 2 1 2 2 2

The dashed edges of τ are “non-propagating” edges, and they may be chosen so that they do not cross each other. The propagating edges of τ do not cross, since t is planar. Using the relations (f1) and (f2), the non-propagating edges of τ can be “removed”, leaving a planar diagram which is written in terms of the generators pi and p 1 .Inour example, this i+ 2 process will replace τ by p 1 p2p 1 p3p 1 p5p 1 p4, so that 2 2 3 2 5 2 4 2

•. •...... •...... •.... •...•...... •.... •...... •. •. • • •. • • •. (p 1 p 1 p 1 )(p3p4p6p7) 2 2 3 2 6 2 •...•...... •...... •.... •...... •.... •. •...... •. •...... •...... •.... •... •...... •.... •...... •. • • •. • •. •. •...... · 1 1 1 1 t = ...... = •. •. •. •...... •.... •. •. •. = p2 p2p3 p3p5 p5p4 p4 ...... 2 2 2 2 •...... •...... •.... • •. •. • •...... •. •. •. • •. •. •. •. •. • • • •. •. • •. . . . . · 1 1 . . . . (p2p3p4p7)(p p ). . . . . 1 2 2 2 . . . . •...... •...... •.... • •. •. • •. 8 t. halverson and a. ram

Step 3:Ift ∈ Pk and σ1 ∈ Sk which permutes the the top blocks of the planar diagram t, then there is a permutation σ2 of the bottom blocks of t such that σ1tσ2 is planar. Furthermore, this can be accomplished using the relations. For example, suppose T1 T2 ...... •. • • • • .•.. • ...... 1 1 1 1 1 1 1 1 t = . ... =(p1 p2 )(p2p3) (p6 )(p7) p4p5s5 (p2p3p4)(p1 p2 p3 ) (p6p7)(p5 p6 ) •....... •......... •...... .•....•........ .....•...... ....•.... 2 2 2 2 2 2 2 2 T1 T2 B1 B2 B1 B2

is a planar diagram with top blocks T1 and T2 connected respectively to bottom blocks B1 and B2

and T T •.... •.... •.. •.. .•... .•... .•... 2 1 ...... ...... •...... •...... •.. •.. ...•...... •...... •...... •...... •.... • • •...... •...... •...... •...... •...... •...... •.. •...... •...... •...... ...... σ1 = ...... so that σ1t = •. • • • • .•.. • = ...... = t , ...... •. •. •. •. •. •. •. . ... • • • • • • • . ... . ... •...... •...... •...... •....•...... •...... •.... B1 B2

then transposition of B1 and B2 can be accomplished with the permutation T T •...... •.... • • ...•...... •...... •.... 2 1 ...... ...... •.... •.... •.... •...... •.. ...•.. ...•...... •...... •.... • • •...... •...... •...... •...... •...... •...... •....•...... •...... •...... σ = ...... so that σ tσ = t σ = •.... •.... •... •... ..•.. ..•.. ..•.. = . ... 2 ...... 1 2 2 ...... •.... •.... •.... .•.. .•.. .•.. .•...... •...... •...... •.....•...... •...... •...... •...... ...... •.... •.... •.... .•... .•... .•... .•... B2 B1 is planar. It is possible to accomplish these products using the relations from the statement of the theorem. In our example, with σ1 = s2s1s3s2s4s3s5s2s4s6s1s3s5s2s4s3 and with σ2 = s4s5s6s3s4s5s2s3s4s1s2s3, −1 −1 σ1T1T2 p4p5 s5 B1B2σ2 =(σ1T1T2 p4p5 σ1 )(σ1s5σ2)(σ2 B1B2σ2)=T2T1 p3p4 s4 B2B1, where T T =(p 1 p2)(p 1 p 1 p6p7) and B B =(p2p3p 1 p 1 )(p5p6p7p 1 p 1 p 1 ). 2 1 1 2 5 2 6 2 2 1 1 2 2 2 4 2 5 2 6 2 Step 4: Let t, b ∈ Pk and let π ∈ Sk. Then tπb = txσ where x ∈ Pk and σ ∈ Sk, and this transformation can be acccomplished using the relations in (b), (c) and (d). Suppose T is a block of bottom dots of t containing more than one dot and which is connected, by edges of π,totwo top blocks B1 and B2 of b. Using Step 3 ﬁnd permutations γ1,γ2 ∈ Sk and σ1,σ2 ∈ Sk such that t = γ1tγ2 and b = σ1bσ2 are planar diagrams with T as the leftmost bottom block of t and B1 and B2 as the two leftmost top blocks of b. Then −1 −1 −1 −1 −1 −1 −1 −1 −1 −1 −1 −1 −1 −1 tπb = γ1 t γ2 πσ1 b σ2 = γ1 t (γ2 πσ1 )b σ2 = γ1 t (γ2 πσ1 )b σ2 = tπσ1 b σ2 , where b is a planar diagram with fewer top blocks than b has. This is best seen from the following picture, where tπb equals ...... −1. . γ . . 1 ...... − . . − . . . . 1. . 1...... T...... γ . . γ . . t ...... 1 . . 1 ...... − ...... 1...... . . ...... T...... γ ...... T...... T...... t ...... 2 ...... t ...... t ...... − − ...... − − ...... 1 1 ...... 1 1 ...... γ πσ ...... γ πσ . π ...... = π ...... = . .. .. 2 1 . = . .. .. 2 1 ...... − ...... 1...... . . . . . . σ ...... b . B1 B2 . 1 ...... b . . b ...... − . . − . . . . 1. . 1. . . . σ . . σ . . b .B1 B2 . 2 . . 2 ...... −1. . σ . . 2 ...... partition algebras 9

2 and the last equality uses the relations pi+ 1 = p 1 and fourth relation in (d) (multiple times). 2 i+ 2 −1 −1 −1 −1 −1 −1 Then tπb = γ1 t γ2 πσ1 b σ2 = tπ b σ2 , where π = πσ1 . By iteration of this process it is suﬃcient to assume that in proving Step 4 we are analyzing tπb where each bottom block of t connects to a single top block of b. Then, since π is a permutation, the bottom blocks of t must have the same sizes as the top blocks of b and π is the permutation that permutes the bottom blocks of t to the top blocks of b.Thus, by Step 1, there is σ ∈ Sk such that x = πbσ−1 is planar and tπb = t(πbσ−1)σ = txσ.

Completion of the proof:Ifd1,d2 ∈ Ak then use the decomposition Ak = SkPkSk to write d1 and d2 in the form

d1 = π1tπ2 and d2 = σ1bσ2, with t, b ∈ Pk,π1,π2,σ1,σ2 ∈ Sk, and use (b) and (c) to write these products in terms of the generators. Let π = π2σ1. Then Step 4 tell us that the relations give σ ∈ Sk and x ∈ Pk such that

d1d2 = π1tπ2σ1bσ2 = π1tπbσ2 = π1txσσ2,

Using Step 2 and that Ak = SkPkSk, this product can be identiﬁed with the product diagram d1d2.Thus, the relations are suﬃcient to compose any two elements of Ak.

2. Partition Algebras

∈ 1 Z ∈ C C C For k 2 >0 and n , the partition algebra Ak(n)isthe associative algebra over with basis Ak,

CAk(n)=Cspan-{d ∈ Ak}, and multiplication deﬁned by d1d2 = n (d1 ◦ d2), where, for d1,d2 ∈ Ak, d1 ◦ d2 is the product in the monoid Ak and is the number of blocks removed from the the middle row when constructing the composition d1 ◦ d2.For example,

. . . . •...... • ...... •...... •...... •...... •.. • ...... •...... • .....•...... •...... •...... •.. • ...... 2 ...... •. •. •. •. •. •. •...... = n ...... , (2.1) d d = ...... 1 2 • •...... •...... •.. •...... •...... •.. • •. • •. •. • •...... • •...... • ..•...... •...... •...... •... since two blocks are removed from the middle row. There are inclusions of algebras given by C → C CA − 1 → CAk Ak−1 A − 1 k 2 k 2 1 k 1 k and 1 k−1 1 k (2.2) •...... •...... •...... •...... •...... •...... •.. •...... •...... •...... •...... •...... •...... •.. •...... •...... •...... •...... •...... •.. •...... •...... •...... •...... •...... •.. •...... → . . . . → . . . . d . . d . . d . . d . . •...... •...... •...... •...... •...... •...... •.. •...... •...... •...... •...... •...... •...... •.. •...... •...... •...... •...... •...... •.. •...... •...... •...... •...... •...... •.. •. .

For d1,d2 ∈ Ak, deﬁne

d1 ≤ d2, if the set partition d2 is coarser than the set partition d1, 10 t. halverson and a. ram

i.e., i and j in the same block of d1 implies that i and j are in the same block of d2. Let {xd ∈ CAk|d ∈ Ak} be the basis of CAk uniquely deﬁned by the relation d = xd , for all d ∈ Ak. (2.3) d≤d

Under any linear extension of the partial order ≤ the transition matrix between the basis {d | d ∈ Ak} of CAk(n) and the basis {xd | d ∈ Ak} of CAk(n)isupper triangular with 1s on the diagonal and so the xd are well deﬁned.

1 The maps ε 1 : CAk → CA − 1 , ε 2 : CA − 1 → CAk−1 and trk: CAk → C 2 k 2 k 2

Let k ∈ Z>0. Deﬁne linear maps

1 ε 1 : CAk −→ CA − 1 ε 2 : CA − 1 −→ CAk−1 2 k 2 k 2 1 k 1 k k−1 ...... and 1 k 1 •. • • • • • •. •. • • • • • •.... •...... •...... •...... •...... •...... •...... •.... •...... •...... •...... •...... •...... •...... → ...... d . . d ...... → ...... d .... . d .... •...... •...... •...... •...... •...... •...... •.. •...... •...... •...... •...... •...... •...... •.... •...... •...... •...... •...... •...... •...... •..... •...... •...... •...... •...... •...... •...... so that ε 1 (d)isthe same as d except that the block containing k and the block containing k are 2 1 combined, and ε 2 (d) has the same blocks as d except with k and k removed. There is a factor of 1 n in ε 2 (d)ifthe removal of k and k reduces the number of blocks by 1. For example, ...... •...... • ...... •...... •.. • •...... • ...... •...... •.. •. 1 •...... • ...... •...... •.. •. •...... • ...... •...... •...... 2 ...... ε 1 ...... = ...... ,ε...... = ...... , 2 • •...... •.. •...... •.. • •...... •.. •...... •.. • •...... •.. •...... •.. • •...... •.. • and ...... •...... • ...... •...... •.. • •...... • ...... •...... •.. •. 1 •...... • ...... •...... •.. •. •...... • ...... •...... •...... 2 ...... ε 1 ...... = ...... ,ε...... = n ...... 2 • •...... •.. • • • •...... •.. • •. • •...... •.. • •. • •...... •.. •

1 ε1 1 The map ε 2 is the composition CA − 1 → CAk−→ CAk−1. The composition of ε 1 and ε 2 is the k 2 2 map ε1 : CAk −→ CAk−1

1 k 1 k−1 . (2.4) •...... •...... •...... •...... •...... •...... •.. •...... •...... •...... •...... •...... •...... → . . .. . d . . d .... •...... •...... •...... •...... •...... •...... •.. •...... •...... •...... •...... •...... •......

By drawing diagrams it is straightforward to check that, for k ∈ Z>0,

ε 1 (a1ba2)=a1ε 1 (b)a2, for a1,a2 ∈ A − 1 ,b∈ Ak 2 2 k 2 1 1 ε 2 (a1ba2)=a1ε 2 (b)a2, for a1,a2 ∈ Ak−1,b∈ A − 1 (2.5) k 2

ε1(a1ba2)=a1ε1(b)a2, for a1,a2 ∈ Ak−1,b∈ Ak. and p 1 bp 1 = ε 1 (b)p 1 = p 1 ε 1 (b), for b ∈ Ak k+ 2 k+ 2 2 k+ 2 k+ 2 2 1 1 pkbpk = ε 2 (b)pk = pkε 2 (b), for b ∈ A − 1 (2.6) k 2

ekbek = ε1(b)ek = ekε1(b), for b ∈ Ak.

Deﬁne trk: CAk → C and tr − 1 : CA − 1 → C by the equations k 2 k 2

1 trk(b)=tr − 1 (ε 1 (b)), for b ∈ Ak, and tr − 1 (b)=trk−1(ε 2 (b)), for b ∈ A − 1 , (2.7) k 2 2 k 2 k 2 partition algebras 11

so that

k k−1 1 trk(b)=ε (b), for b ∈ Ak, and tr − 1 (b)=ε ε 2 (b), for b ∈ A − 1 . (2.8) 1 k 2 1 k 2

c Pictorially trk(d)=n where c is the number of connected components in the closure of the ...... diagram d, ...... •...... •...... •...... •...... •...... •...... •...... ∈ trk(d)= . d ...... , for d Ak. (2.9) •...... •...... •...... •...... •...... •...... •......

The ideal CIk(n) 1 ∈ Z≥ For k 2 0 deﬁne CIk(n)=C-span{d ∈ Ik}. (2.10) By (1.3), ∼ CIk(n)isanideal of CAk(n) and CAk(n)/CIk(n) = CSk, (2.11) since the set partitions with propagating number k are exactly the permutations in the symmetric ∈ Z group Sk (by convention S+ 1 = S for >0, see (2.2)). 2 ∼ View CIk(n)asanalgebra (without identity). Since CAk(n)/CIk = CSk and CSk is semisimple, Rad(CAk(n)) ⊆ CIk(n). Since CIk(n)/Rad(CAk(n)) is an ideal in CAk(n)/Rad(CAk(n)) the quotient CIk(n)/Rad(CAk(n)) is semisimple. Therefore Rad(CIk(n)) ⊆ Rad(CAk(n)). On the other hand, since Rad(CAk(n)) is an ideal of nilpotent elements in CAk(n), it is an ideal of nilpotent elements in CIk(n) and so Rad(CIk(n)) ⊇ Rad(CAk(n)). Thus

Rad(CAk(n)) = Rad(CIk(n)). (2.12)

Let k ∈ Z≥0.By(2.5) the maps

1 ε 1 : CAk −→ CA − 1 and ε 2 : CA − 1 −→ CAk−1 2 k 2 k 2

are (CA − 1 , CA − 1 )-bimodule and (CAk−1, CAk−1)-bimodule homomorphisms, respectively. The k 2 k 2 corresponding basic constructions are the algebras C ⊗ C C ⊗ C Ak(n) C Ak(n) and A − 1 (n) C − A − 1 (n)(2.13) Ak− 1 (n) k Ak 1(n) k 2 2 2 with products given by

1 (b1 ⊗ b2)(b3 ⊗ b4)=b1 ⊗ ε 1 (b2b3)b4, and (c1 ⊗ c2)(c3 ⊗ c4)=c1 ⊗ ε 2 (c2c3)c4, (2.14) 2

for b1,b2,b3,b4 ∈ CAk(n), and for c1,c2,c3,c4 ∈ CA − 1 (n). k 2 ∈ 1 Z Let k 2 >0. Then, by the relations in (2.6) and the fact that

every d ∈ Ik can be written as d = d1pkd2, with d1,d2 ∈ A − 1 ,(2.15) k 2 the maps C ⊗ C −→ C A − 1 (n) CA − (n) A − 1 (n) Ik(n) k 2 k 1 k 2 (2.16) b1 ⊗ b2 −→ b1pkb2 12 t. halverson and a. ram are algebra isomorphisms. Thus the ideal CIk(n)isalways isomorphic to a Jones basic construction.

Representations of the symmetric group

A partition λ is a collection of boxes in a corner. We shall conform to the conventions in [Mac] and assume that gravity goes up and to the left, i.e.,

Numbering the rows and columns in the same way as for matrices, let

λi = the number of boxes in row i of λ, λj = the number of boxes in column j of λ, and (2.17) |λ| = the total number of boxes in λ.

Any partition λ can be identiﬁed with the sequence λ =(λ1 ≥ λ2 ≥ ...) and the conjugate partition to λ is the partition λ =(λ1,λ2,...). The hook length of the box b of λ is

− − h(b)=(λi i)+(λj j)+1, if b is in position (i, j)ofλ.(2.18)

Write λ n if λ is a partition with n boxes. In the example above λ = (553311) and λ 18. See [Mac, §I.7] for details on the representation theory of the symmetric group. The irreducible C λ Sk-modules Sk are indexed by the elements of

ˆ { } λ k! Sk = λ n and dim(Sk )= . (2.19) h(b) b∈λ

For λ ∈ Sˆk, and µ ∈ Sˆk−1, Sk λ ∼ ν Sk µ ∼ ν Res (S ) S − and Ind (S ) S . (2.20) Sk−1 k = k 1 Sk−1 k−1 = k λ/ν= ν/µ= where the ﬁrst sum is over all partitions ν that are obtained from λ by removing a box, and the second sum is over all partitions ν which are obtained from µ by adding a box (this result follows, for example, from [Mac, §I.7 Ex. 22(d)]). The Young lattice is the graph Sˆ given by setting

vertices on level k: Sˆ = {partitions λ with k boxes}, and k (2.21) an edge λ → µ, λ ∈ Sˆk, µ ∈ Sˆk+1 if µ is obtained from λ by adding a box. partition algebras 13

It encodes the decompositions in (2.20). The ﬁrst few levels of Sˆ are given by

k =0: ∅ ...... k =1: ...... k=2: ...... k =3: ...... k =4:

For µ ∈ Sˆk deﬁne

(0) ∅ (k) ˆµ (0) (1) (k) T = , T = µ , and, for each , Sk = T =(T ,T ,...,T ) () () (+1) T ∈ Sˆ and T → T is an edge in Sˆ

ˆµ ∅∈ ˆ ∈ ˆ ˆ so that Sk is the set of paths from S0 to µ Sk in the graph S.Interms of the Young lattice

µ ˆµ dim(Sk )=Card(Sk ). (2.22)

§ µ This is a translation of the classical statement (see [Mac, I.7.6(ii)]) that dim(Sk )isthenumber of standard Young tableaux of shape λ (the correspondence is obtained by putting the entry in the box of λ which is added at the th step T (−1) → T () of the path).

Structure of the algebra CAk(n)

Build a graph Aˆ by setting

vertices on level k: Aˆk = {partitions µ | k −|µ|∈Z≥0}, 1 vertices on level k + : Aˆ 1 = Aˆk = {partitions µ | k −|µ|∈Z≥0}, 2 k+ 2 an edge λ → µ, λ ∈ Aˆk, µ ∈ Aˆ 1 if λ = µ or if µ is obtained from λ by removing a box, k+ 2 an edge µ → λ, µ ∈ Aˆ 1 , λ ∈ Aˆk+1,ifλ = µ or if λ is obtained from µ by adding a box. k+ 2 (2.23) 14 t. halverson and a. ram

The ﬁrst few levels of Aˆ are given by k =0: ∅ ...... 1 ∅ k =0+2 : ...... k =1: ∅ ...... 1 ∅ k =1+2 : ...... k=2: ∅ ...... 1 ∅ k =2+2 : ...... k =3: ∅

The following result is an immediate consequence of the Tits deformation theorem, Theorems 5.12 and 5.13 in this paper (see also [CR (68.17)]).

Theorem 2.24. (a) For all but a ﬁnite number of n ∈ C the algebra CAk(n) is semisimple. (b) If CAk(n) is semisimple then the irreducible CAk(n)-modules, µ ˆ { | −| |∈Z } Ak are indexed by elements of the set Ak = partitions µ k µ ≥0 , and µ ∅∈ ˆ ∈ ˆ ˆ dim(Ak )=(number of paths from A0 to µ Ak in the graph A). Let (0) (k) µ 1 − 1 T = ∅ , T = µ , and, for each , ˆ (0) ( 2 ) (k 2 ) (k) Ak = T =(T ,T ,...,T ,T ) () () (+ 1 ) T ∈ Aˆ and T → T 2 is an edge in Aˆ

ˆµ ∅∈ ˆ ∈ ˆ ˆ ∈ ˆ ∈ ˆ so that Ak is the set of paths from A0 to µ Ak in the graph A.Ifµ Sk then µ Ak and µ ∈ Aˆ 1 and, for notational convenience in the following theorem, k+ 2 (0) (1) (k) ∈ ˆµ identify P =(P ,P ,...,P ) Sk with the corresponding (0) (0) (1) (1) (k−1) (k−1) (k) ∈ ˆµ P =(P ,P ,P ,P ,...,P ,P ,P ) Ak , and (0) (0) (1) (1) (k−1) (k−1) (k) (k) ∈ ˆµ P =(P ,P ,P ,P ,...,P ,P ,P ,P ) A 1 . k+ 2

1 µ For ∈ Z≥0 and n ∈ C such that CA(n)issemisimple let χ , µ ∈ Aˆ,bethe irreducible 2 A(n) characters of CA(n). Let tr: CA(n) → C be the traces on CA(n) deﬁned in (2.8) and deﬁne µ ∈ ˆ constants tr (n), µ A,by µ µ tr = tr (n)χ . (2.25) A(n)

µ∈Aˆ partition algebras 15

Theorem 2.26. 1 ∈ C ∈ Z≥ (a) Let n and let k 2 0. Assume that

λ 1 ∈ ˆ ∈ Z≥ tr (n) =0, for all λ A, 2 0,

1 C ∈ Z≥ ≤ A(n) are semisimple for all 2 0, k. (2.27)

1 1 ∈ Z≥ ≤ − For each 2 0, k 2 , deﬁne

λ tr − 1 (n) λ 2 ε = for each edge µ → λ, µ ∈ Aˆ−1, λ ∈ Aˆ − 1 ,inthe graph Aˆ. µ µ 2 tr−1(n)

Inductively deﬁne elements in CA(n) by

µ 1 τ γ ∈ ˆ | |≤ − ∈ ˆµ ePQ = eP −T peTQ− , for µ A, µ 1, P, Q A , (2.28) τ γ εµεµ

− 1 − 1 − − 1 − 1 µ ( 2 ) ( 2 ) (0) ( 2 ) (0) ( 2 ) () ∈ ˆ where τ = P , γ = Q , R =(R ,...,R ) for R =(R ,...,R ,R ) A ˆµ λ and T is an element of A−1 (the element ePQ does not depend on the choice of T ). Then deﬁne λ − λ ∈ ˆ ∈ ˆλ µ ePQ =(1 z)sPQ, for λ S,P,Q S , where z = ePP (2.29) ∈ ˆ µ µ A P ∈Aˆ |µ|≤−1

{ λ | ∈ ˆ ∈ ˆλ} and sPQ λ S,P,Q S is any set of matrix units for the the group algebra of the symmetric group CS.Together, the elements in (2.28) and (2.29) form a set of matrix units in CA(n). 1 λ ∈ Z≥ ∈ Z ∈ ˆ (b) Let n 0 and let k 2 >0 be minimal such that trk (n)=0for some λ Ak. Then CA 1 (n) is not semisimple. k+ 2 1 ∈ Z≥ ∈ Z C C (c) Let n 0 and k 2 >0.If Ak(n) is not semisimple then Ak+j(n) is not semisimple for j ∈ Z>0.

µ Proof. (a) Assume that CA−1(n) and CA − 1 (n) are both semisimple and that tr − (n) =0 for 2 1 ∈ ˆ ∈ ˆ λ λ C all µ A−1.Ifλ A − 1 then ε =0if and only if tr 1 (n) =0,and, since the ideal I(n) 2 µ − C ⊗ 2 C is isomorphic to the Jones basic construction A − 1 (n) CA − (n) A − 1 (n)itthen follows from 2 1 2 C λ ∈ ˆ Theorem 4.28 that I(n)issemisimple if and only if tr − 1 (n) =0for all λ A− 1 .Thus, by 2 2 µ (2.12), if CA−1(n) and CA − 1 (n) are both semisimple and tr − (n) =0 for all µ ∈ Aˆ−1 then 2 1

C λ ∈ ˆ A(n)issemisimple if and only if tr − 1 (n) =0for all λ A− 1 .(2.30) 2 2

λ ∈ ˆ C By Theorem 4.28, when tr − 1 (n) =0for all λ A− 1 , the algebra I(n) has matrix units given 2 2 by the formulas in (2.28). The element z in (2.29) is the central idempotent in CA(n) such that CI(n)=zCA(n). Hence the complete set of elements in (2.28) and (2.29) form a set of matrix units for CA(n). This completes the proof of (a) and (b) follows from Theorem 4.28(b). 16 t. halverson and a. ram

Specht modules Let A be an algebra. An idempotent is a nonzero element p ∈ A such that p2 = p.A minimal idempotent is an idempotent p which cannot be written as a sum p = p1 + p2 with p1p2 = p2p1 =0.Ifp is an idempotent in A and pAp = Cp then p is a minimal idempotent of A 2 2 since, if p = p1 + p2 with p1 = p1, p2 = p2 and p1p2 = p2p1 =0then pp1p = kp for some constant p and so kp1 = kpp1 = pp1pp1 = p1 giving that either p1 =0ork =1,inwhich case p1 = pp1p = p. Let p be an idempotent in A. Then the map ∼ op (pAp) −→ EndA(Ap) , where φpbp(ap)=(ap)(pbp)=apbp, for ap ∈ Ap,(2.31) pbp → φpbp is a ring isomorphism. If p is a minimal idempotent of A and Ap is a semisimple A-module then Ap must be a simple A-module. To see this suppose that Ap is not simple so that there are A-submodules V1 and V2 of Ap such that Ap = V1 ⊕ V2. Let φ1,φ2 ∈ EndA(Ap)bethe A-invariant projections on V1 and V2.By(2.31) φ1 and φ2 are given by right multiplication by p1 = pp˜1p and p2 = pp˜2p, respectively, and it follows that p = p1 + p2, V1 = Ap1, V2 = Ap2, and Ap = Ap1 ⊕ Ap2. Then 2 2 2 p1 = φ1(p1)=φ1(p)=p1 and p1p2 = φ2(p1)=φ2(φ1(p)) = 0. Similarly p2 = p2 and p2p1 =0. Thus p is not a minimal idempotent. If p is an idempotent in A and Ap is a simple A-module then

op pAp = EndA(Ap) = C(p · 1 · p)=Cp, by (2.31) and Schur’s lemma (Theorem 5.3). The group algebra of the symmetric group Sk over the ring Z is

Sk,Z = ZSk and CSk = C ⊗Z Sk,Z , (2.32) where the tensor product is deﬁned via the inclusion Z → C. Let λ =(λ1,λ2,...,λ)beapartition of k. Deﬁne subgroups of Sk by

S = S ×···×S and S = S × ···×S , (2.33) λ λ1 λ λ λ1 λr where λ =(λ1,λ2,...,λr)isthe conjugate partition to λ, and let (w) 1λ = w and ελ = (−1) w. (2.34) ∈ ∈ w Sλ w Sλ

Let τ be the permutation in Sk that takes the row reading tableau of shape λ to the column reading tableau of shape λ.For example for λ = (553311),

12345 171115 17 678910 281216 18 τ =(2, 7, 8, 12, 9, 16, 14, 4, 15, 10, 18, 6)(3, 11)(5, 17), since τ · 11 12 13 = 3913 . 14 15 16 41014 17 5 18 6 partition algebras 17

The Specht module for Sk is the ZSk-module

− λ Z 1 Sk,Z =imΨSk =( Sk) pλ, where pλ = 1λτελ τ , and (2.35) Z where ΨSk is the Sk-module homomorphism given by ι π − Z −→ Z −→ Z 1 ΨSk :(Sk)1λ Sk ( Sk)τελ τ (2.36) −1 b 1λ −→ b 1λ −→ b 1λτελ τ

By induction and restriction rules for the representations of the symmetric groups, the CSk-modules −1 (CSk)1λ and (CSk)τελ τ have only one irreducible component in common and it follows (see [Mac,§I.7, Ex. 15]) that

λ C ⊗ λ C Sk = Z Sk,Z is the irreducible Sk-module indexed by λ,(2.37)

once one shows that ΨSk is not the zero map. ∈ 1 Z Z Let k 2 >0.For an indeterminate x, deﬁne the [x]-algebra by

Ak,Z = Z[x]-span{d ∈ Ak} (2.38) with multiplication given by replacing n with x in (2.1). For each n ∈ C,

evn: Z[x] → C CA (n)=C ⊗Z A Z, where the Z-module homomorphism (2.39) k [x] k, x → n

≤ ⊗ ⊗(k−|λ|) is used to deﬁne the tensor product. Let λ be a partition with k boxes. Let b pk denote the image of b ∈ A|λ|,Z under the map given by

A|λ|,Z −→ Ak,Z

•...... •.. •...... •.. • • ···• ...... −→ . . if k is an integer, and . b . . b . , •...... •.. •...... •.. • •···• k−|λ| A| | 1 Z −→ Ak,Z λ + 2 ,

...... •. •.... •. •.... • • ··· • • ...... 1 . .. −→ . .. if k − is an integer. . b ... . b ... , 2 •...... •.... •...... •...... ...•...... •...... ··· ...... •...... •... −| |− 1 k λ 2

1 ∈ Z Z For k 2 >0, deﬁne an Ak, -module homomorphism

⊗(k−|λ|) − ⊗(k−|λ|) ⊗ 1 ⊗ tλ = 1λ pk and sλ = τελ τ pk . (2.41) 18 t. halverson and a. ram

The Specht module for CAk(n)isthe Ak,Z-module λ ⊗(k−|λ|) Z Z ⊗ Ak,Z =imΨAk = image of Ak, eλ in Ak, /I|λ|,Z , where eλ = pλ pk . (2.42)

∈ 1 Z ≤ ∈ C Proposition 2.43. Let k 2 >0, and let λ be a partition with k boxes. If n such that CAk(n) is semisimple, then

λ C ⊗ λ C Ak (n)= Z[x] Ak,Z is the irreducible Ak(n)-module indexed by λ, where the tensor product is deﬁned via the Z-module homomorphism in (2.39).

Proof. Let r = |λ|. Since ∼ CAr(n)/CIr(n) = CSr and pλ is a minimal idempotent of CSr,itfollows from (4.20) that eλ, the image of eλ in (CAk(n))/(CIr(n)), is a minimal idempotent in (CAk(n))/(CIr(n)). Thus CAk(n) eλ is a simple (CAk(n))/(CIr(n))-module. CIr(n)

Since the projection CAk(n) → (CAk(n))/(CIr(n)) is surjective, any simple (CAk(n))/(CIr(n))- module is a simple CAk(n)-module.

3. Schur-Weyl Duality for Partition Algebras

Let n ∈ Z>0 and let V beavector space with basis v1,...,vn. Then the tensor product

⊗k ⊗ ⊗···⊗ { ⊗ ···⊗ | ≤ ≤ } V = V V V has basis vi1 vik 1 i1,...,ik n . k factors

For d ∈ Ak and values i1,...,ik,i1 ,...,ik ∈{1,...,n} deﬁne

i1,...,ik 1, if ir = is when r and s are in the same block of d, (d) = (3.1) i1 ,...,ik 0, otherwise.

i1,...,ik For example, viewing (d) as the diagram d with vertices labeled by the values i1,...,ik and i1 ,...,ik i1 ,...,ik ,wehave

i1 i2 i3 i4 i5 i6 i7 i8 •.... .•...... • ...•.. •... .•...... •.. •...... ...... = δi1i2 δi1i4 δi1i2 δi1i5 δi5i6 δi5i7 δi5i3 δi5i4 δi5i6 δi5i7 δi8i8 . • ..•...... •... •...... •.. •..... •... •. i1 i2 i3 i4 i5 i6 i7 i8

With this notation, the formula ⊗ ···⊗ i1,...,ik ⊗···⊗ d(vi vi )= (d) vi vi (3.2) 1 k i1 ,...,ik 1 k ≤ ≤ 1 i1 ,...ik n partition algebras 19 deﬁnes actions

⊗k ⊗k Φk : CAk −→ End(V ) and Φ 1 : CA 1 −→ End(V )(3.3) k+ 2 k+ 2

⊗k of CAk and CA 1 on V , where the second map Φ 1 comes from the fact that if d ∈ A 1 , k+ 2 k+ 2 k+ 2 then d acts on the subspace

⊗k ∼ ⊗k ⊗ C { ⊗ ···⊗ ⊗ | ≤ ≤ }⊆⊗(k+1) V = V vn = -span vi1 vik vn 1 i1,...,ik n V . (3.4)

⊗k In other words, the map Φ 1 is obtained from Φk+1 by restricting to the subspace V ⊗ vn and k+ 2 ⊗k ⊗k identifying V with V ⊗ vn. ⊗k The group GLn(C) acts on the vector spaces V and V by

n ⊗ ⊗···⊗ ⊗ ⊗ ···⊗ gvi = gjivj, and g(vi1 vi2 vik )=gvi1 gvi2 gvik , (3.5) j=1 for g =(gij) ∈ GLn(C). View Sn ⊆ GLn(C)asthe subgoup of permutation matrices and let ⊗k ∈ ⊗k | ∈ ∈ ⊗k EndSn (V )= b End(V ) bσv = σbv for all σ Sn and v V .

Theorem 3.6. Let n ∈ Z>0 and let {xd | d ∈ Ak} be the basis of CAk(n) deﬁned in (2.3). Then ⊗k (a) Φk : CAk(n) → End(V ) has

⊗k C { | } im Φk = EndSn (V ) and ker Φk = -span xd d has more than n blocks , and

⊗k (b) Φ 1 : CA 1 (n) → End(V ) has k+ 2 k+ 2

⊗k C { | } im Φ 1 = EndS − (V ) and ker Φ 1 = -span xd d has more than n blocks . k+ 2 n 1 k+ 2

Proof. (a) As a subgroup of GLn(C), Sn acts on V via its permutation representation and Sn acts on V ⊗k by ⊗ ⊗ ···⊗ ⊗ ⊗ ···⊗ σ(vi1 vi2 vik )=vσ(i1) vσ(i2) vσ(ik). (3.7) ∈ ⊗k −1 ⊗k ∈ Then b EndSn (V )ifand only if σ bσ = b (as endomorphisms on V ) for all σ Sn.Thus, ∈ ⊗k using the notation of (3.1), b EndSn (V )ifand only if

− i1,...,ik 1 i1,...,ik σ(i1),...,σ(ik) ∈ b =(σ bσ) = b , for all σ Sn. i1 ,...,ik i1 ,...,ik σ(i1 ),...,σ(ik )

It follows that the matrix entries of b are constant on the Sn-orbits of its matrix coordinates. These orbits decompose {1,...,k,1 ,...,k } into subsets and thus correspond to set partitions d ∈ Ak. It follows from (2.3) and (3.1) that for all d ∈ Ak,

i1,...,ik 1, if ir = is if and only if r and s are in the same block of d, (Φk(xd)) = (3.8) i1 ,...,ik 0, otherwise.

Thus Φk(xd) has 1s in the matrix positions corresponding to d and 0s elsewhere, and so b is a linear ∈ ∈ C ⊗k combination of Φk(xd),d Ak. Since xd,d Ak, form a basis of Ak,imΦk = EndSn (V ). 20 t. halverson and a. ram

i1,...,ik If d has more than n blocks, then by (3.8) the matrix entry (Φk(xd)) =0for all indices i1 ,...,ik i1,...,ik,i1 ,...,ik , since we need a distinct ij ∈{1,...,n} for each block of d.Thus, xd ∈ ker Φk. i ,...,i ≤ 1 k If d has n blocks, then we can ﬁnd an index set i1,...,ik,i1 ,...,ik with (Φk(xd)) =1 i1 ,...,ik simply by choosing a distinct index from {1,...,n} for each block of d.Thus, if d has ≤ n blocks then xd ∈ ker Φk, and so ker Φk = C-span{xd|d has more than n blocks}. ⊗k ⊗(k+1) (b) The vector space V ⊗ vn ⊆ V is a submodule both for CA 1 ⊆ CAk+1 and k+ 2 C − ⊆ C ∈ − ⊗ ···⊗ ⊗ ⊗ ···⊗ ⊗ Sn 1 Sn.Ifσ Sn 1, then σ(vi1 vik vn)=vσ(i1) vσ(ik) vn. Then as above ∈ ⊗k b EndSn−1 (V )ifand only if

i1,...,ik,n σ(i1),...,σ(ik),n ∈ b = b , for all σ Sn−1. i1 ,...,ik ,n σ(i1 ),...,σ(ik ),n

The Sn−1 orbits of the matrix coordinates of b correspond to set partitions d ∈ A 1 ; that is k+ 2 vertices ik+1 and i(k+1) must be in the same block of d. The same argument as part (a) can be used to show that ker Φ 1 is the span of xd with d ∈ A 1 having more than n blocks. We always k+ 2 k+ 2 choose the index n for the block containing k +1and (k +1).

⊗k ⊗k 1 ⊗k ⊗(k−1) The maps ε 1 : End(V ) → End(V ) and ε 2 : End(V ) → End(V ) 2 ∈ ⊗k i1,...,ik ∈ C If b End(V ) let b be the coeﬃcients in the expansion i1 ,...,ik ⊗ ···⊗ i1,...,ik ⊗ ···⊗ b(vi vi )= b vi vi . (3.9) 1 k i1 ,...,ik 1 k ≤ ≤ 1 i1 ,...ik n Deﬁne linear maps

⊗k ⊗k 1 ⊗k ⊗(k−1) ε 1 : End(V ) → End(V ) and ε 2 : End(V ) → End(V )by 2 n 1 i ,...,i i ,...,i i1,...,ik−1 i1,...,ik−1,j 1 k 1 k 2 ε 1 (b) = b δi i and ε (b) = b . (3.10) 2 i1 ,...,ik i1 ,...,ik k k i1 ,...,i(k−1) i1 ,...,i(k−1) , j,=1

1 The composition of ε 1 and ε 2 is the map 2 n ⊗k → ⊗(k−1) i1,...,ik−1 i1,...,ik−1,j ε1 : End(V ) End(V ) given by ε1(b) = b , (3.11) i1 ,...,i(k−1) i1 ,...,i(k−1) ,j j=1 and k ∈ ⊗k Tr(b)=ε1 (b), for b End(V ). (3.12) 1 1 The relation between the maps ε 2 , ε 1 in (3.10) and the maps ε 2 , ε 1 in Section 2 is given by 2 2 ∈ C Φ − 1 (ε 1 (b)) = ε 1 (Φk(b)) ⊗(k−1) , for b Ak(n), k 2 2 2 V ⊗vn 1 1 1 Φk−1(ε 2 (b)) = ε 2 (Φk(b)), for b ∈ CA − 1 (n), and (3.13) n k 2 Φk−1(ε1(b)) = ε1(Φk(b)), for b ∈ CAk(n), where, on the right hand side of the middle equality b is viewed as an element of CAk via the natural inclusion CA − 1 (n) ⊆ CAk(n). Then k 2

k k k Tr(Φk(b)) = ε1 (Φk(b))=Φ0(ε1 (b)) = ε1 (b)=trk(b), (3.14) partition algebras 21

and, by (3.4), if b ∈ CA − 1 (n) then k 2 1 1 1 Tr(Φ − 1 (b)) = Tr(Φk(b) ⊗(k−1) )= Tr(Φk(b)) = trk(b)= tr − 1 (b). (3.15) k 2 V ⊗vn n n n k 2

The representations (IndSn ResSn )k(1 ) and ResSn (IndSn ResSn )k(1 ) Sn−1 Sn−1 n Sn−1 Sn−1 Sn−1 n

(n) Let 1n = Sn be the trivial representation of Sn and let V = C-span{v1,...,vn} be the permutation representation of Sn given in (3.5). Then

∼ S S V Ind n Res n (1 ). (3.16) = Sn−1 Sn−1 n

More generally, for any Sn-module M,

S S ∼ S S Ind n Res n (M) Ind n (Res n (M) ⊗ 1 − ) Sn−1 Sn−1 = Sn−1 Sn−1 n 1 ∼ S S S Ind n (Res n (M) ⊗ Res n (1 )) (3.17) = Sn−1 Sn−1 Sn−1 n ∼ S S ∼ M ⊗ Ind n Res n (1 ) M ⊗ V, = Sn−1 Sn−1 n = where the third isomorphism comes from the “tensor identity,”

∼ IndSn (ResSn (M) ⊗ N) −→ M ⊗ IndSn N Sn−1 Sn−1 Sn−1 , (3.18) g ⊗ (m ⊗ n) → gm ⊗ (g ⊗ n)

Sn for g ∈ S , m ∈ M, n ∈ N, and the fact that Ind (W )=CS ⊗ − W .Byiterating (3.17) it n Sn−1 n Sn 1 follows that

S S ∼ ⊗ S S S ∼ ⊗ (Ind n Res n )k(1) V k and Res n (Ind n Res n )k(1) V k (3.19) Sn−1 Sn−1 = Sn−1 Sn−1 Sn−1 =

as Sn-modules and Sn−1-modules, respectively. If

λ =(λ1,λ2,...,λ) deﬁne λ>1 =(λ2,...,λ)(3.20) to be the same partition as λ except with the ﬁrst row removed. Build a graph Aˆ(n) which encodes ⊗k the decomposition of V , k ∈ Z≥0,byletting

vertices on level k: Aˆk(n)={λ n | k −|λ>1|∈Z≥0}, 1 ˆ 1 { − | −| |∈Z≥ } vertices on level k + 2 : Ak+ (n)= λ n 1 k λ>1 0 , and 2 (3.21) an edge λ → µ,ifµ ∈ Aˆ 1 (n)isobtained from λ ∈ Aˆk(n)byremoving a box, k+ 2 an edge µ → λ,ifλ ∈ Aˆk+1(n)isobtained from µ ∈ Aˆ 1 (n)byadding a box. k+ 2 22 t. halverson and a. ram

For example, if n =5then the ﬁrst few levels of Aˆ(n) are k =0: ...... 1 k =0+2 : ...... k =1: ...... 1 k =1+2 : ...... k=2: ...... 1 k =2+2 : ...... k =3:

The following theorem is a consequence of Theorem 3.5 and the Centralizer Theorem, Theorem 5.4, (see also [GW, Theorem 3.3.7]).

∈ Z λ Theorem 3.22. Let n, k ≥0. Let Sn denote the irreducible Sn-module indexed by λ. (a) As (CSn, CAk(n))-bimodules, ⊗k ∼ λ ⊗ λ V = Sn Ak (n),

λ∈Aˆk(n)

λ C where the vector spaces Ak (n) are irreducible Ak(n)-modules and λ ∈ ˆ ∈ ˆ ˆ dim(Ak (n))=(number of paths from (n) A0(n) to λ Ak(n) in the graph A(n)).

(b) As (CSn−1, CA 1 (n))-bimodules, k+ 2 ⊗k ∼ µ ⊗ µ V = Sn−1 A 1 (n), k+ 2 ∈ ˆ µ Ak+ 1 (n) 2 µ C where the vector spaces A 1 (n) are irreducible Ak+ 1 (n)-modules and k+ 2 2 µ ∈ ˆ ∈ ˆ ˆ dim(A 1 (n))=(number of paths from (n) A0(n) to µ Ak+ 1 (n) in the graph A(n)). k+ 2 2

Determination of the polynomials trµ(n) partition algebras 23

Let n ∈ Z>0.For a partition λ, let

λ>1 =(λ2,...,λ), if λ =(λ1,λ2,...,λ), i.e., remove the ﬁrst row of λ to get λ>1. Then, for n ≥ 2k, the maps Aˆ (n) ←→ Aˆ k k are bijections (3.23) λ −→ λ>1 which provide an isomorphism between levels 0 to n of the graphs Aˆ(n) and Aˆ.

1 µ Proposition 3.24. For k ∈ Z≥0 and n ∈ C such that CAk(n) is semisimple, let χ , µ ∈ Aˆk, 2 Ak(n) be the irreducible characters of CA(n) and let trk: CAk(n) → C be the trace on CAk(n) deﬁned in (2.25). Use the notations for partitions in (2.17). For k>0 the coeﬃcients in the expansion |µ| µ µ µ 1 trk = tr (n)χ , are tr (n)= (n −|µ|−(µj − j)). Ak(n) h(b) ∈ j=1 µ∈Aˆk b µ

If n ∈ C is such that CA 1 (n) is semisimple then for k>0 the coeﬃcients in the expansion k+ 2 |µ| µ µ µ 1 · · − −| |− − trk+ 1 = tr 1 (n)χA (n), are tr 1 (n)= n (n 1 µ (µj j)). 2 2 k+ 1 2 h(b) ∈ ˆ 2 b∈µ j=1 µ Ak+ 1 2

Proof. Let λ be a partition with n boxes. Beginning with the vertical edge at the end of the ﬁrst row, label the boundary edges of λ sequentially with 0, 1, 2 ...,n. Then the vertical edge label for row i =(number of horizontal steps) + (number of vertical steps)

=(λ1 − λi)+(i − 1)=(λ1 − 1) − (λi − i), and the horizontal edge label for column j =(number of horizontal steps) + (number of vertical steps) − − − − =(λ1 j +1)+(λj 1)=(λ1 1)+(λj j)+1. Hence { } { − − − | ≤ ≤ }{ − − − | ≤ ≤ − } 1, 2,...,n = (λ1 1) (λj j)+1 1 j λ1 (λ1 1) (λi i) 2 i n λ1 +1

= {h(b) | b is in row 1 of λ}{(λ1 − 1) − (λi − i) | 2 ≤ i ≤ n − λ1 +1}. For example, if λ = (10, 7, 3, 3, 1) 24, then the boundary labels of λ and the hook numbers in the ﬁrst row of λ are 14 12 11 87653210 32 1 4 8 765 9 10 12 11 13 14 15 16 17 18 19 20 21 22 23 . 24 24 t. halverson and a. ram

Thus, since λ1 = n −|λ>1|,

| | λ>1 +1 λ n! 1 −| |− − − dim(Sn)= = (n λ>1 (λi (i 1))). (3.25) b∈λ h(b) h(b) b∈λ>1 i=2

Let n ∈ Z and let χλ denote the irreducible characters of the symmetric group S .By >0 Sn n taking the trace on both sides of the equality in Theorem 3.22, ⊗k λ λ ∈ C Tr(b, V )= χSn (1)χAk(n)(b)= dim(Sn)χAk(n)(b), for b Ak(n).

λ∈Aˆk(n) λ∈Aˆk(n)

Thus the equality in (3.25) and the bijection in (3.23) provide the expansion of trk for all n ∈ Z≥0 such that n ≥ 2k. The statement for all n ∈ C such that CAk(n)issemisimple is then a consequence of the fact that any polynomial is determined by its evaluations at an inﬁnite number of values of the parameter. The proof of the expansion of tr 1 is exactly analogous. k+ 2

µ µ | | | | Note that the polynomials tr (n) and tr 1 (n) (of degrees µ and µ +1,respectively) do not 2 depend on k.ByProposition 3.24,

{ µ | ∈ ˆ } { } roots of tr 1 (n) µ A 1 = 0 , 2 2 µ {roots of tr (n) | µ ∈ Aˆ1} = {1}, 1 (3.26) { µ | ∈ ˆ } { } roots of tr 1 (n) µ A1 1 = 0, 2 , and 1 2 2 µ 1 { | ∈ ˆ } { − } ∈ Z≥ ≥ roots of trk (n) µ Ak = 0, 1,...,2k 1 , for k 2 0, k 2.

µ µ For example, the ﬁrst few values of tr and tr 1 are 2

∅ ∅ tr (n)=1, tr 1 (n)=n − 2 − tr (n)=n 1, tr 1 (n)=n(n 2), 2 1 − 1 − − tr (n)= 2 n(n 3), tr 1 (n)= 2 n(n 1)(n 4), 2 1 − − 1 − − tr (n)= 2 (n 1)(n 2), tr 1 (n)= 2 n(n 2)(n 3), 2 1 − − 1 − − − tr (n)= 6 n(n 1)(n 5), tr 1 (n)= 6 n(n 1)(n 2)(n 6), 2 1 − − 1 − − − tr (n)= 6 n(n 2)(n 4), tr 1 (n)= 6 n(n 1)(n 3)(n 5), 2 1 − − − 1 − − − tr (n)= 6 (n 1)(n 2)(n 3), tr 1 (n)= 6 n(n 2)(n 3)(n 4), 2

1 ∈ Z≥ ∈ Z≥ Theorem 3.27. Let n 2 and k 2 0. Then

C ≤ n+1 Ak(n) is semisimple if and only if k 2 .

Proof. By Theorem 2.26(a) and the observation (3.26) it follows that CAk(n)issemisimple if n ≥ 2k − 1. partition algebras 25

Suppose n is even. Then Theorems 2.26(a) and 2.26(b) imply that

CA n 1 (n)issemisimple and CA n +1(n)isnot semisimple, 2 + 2 2

(n/2) (n/2) ∈ ˆ ∈ ˆ n n since (n/2) A n 1 and tr 1 (n)=0. Since (n/2) A (n), the A (n)-module A n (n) = + 2 +1 2 +1 +1 2 2 2 2 (n/2) 0. Since the path (∅,...,(n/2), (n/2), (n/2)) ∈ Aˆ n does not correspond to an element of 2 +1 (n/2) Aˆ n (n), 2 +1 (n/2) (n/2) Card(Aˆ n ) = Card(Aˆ n (n)). 2 +1 2 +1

Thus, Tits deformation theorem (Theorem 5.13) implies that CA n (n)iscannot be semisimple. 2 +1 C ≥ n 1 Now it follows from Theorem Theorem 2.26(c) that Ak(n)isnot semisimple for k 2 + 2 . If n is odd then Theorems 2.26(a) and 2.26(b) imply that

CA n 1 (n)issemisimple and CA n +1(n)isnot semisimple, 2 + 2 2

(n/2) n 1 since (n/2) ∈ Aˆ n 1 and tr (n)=0. Since ( − ) ∈ Aˆ n +1(n), the A n +1(n)-module 2 + 2 2 2 2 2 n − 1 n − 1 ( 2 2 ) n 1 n 1 n 1 ( 2 2 ) A n (n) =0 . Since the path (∅,...,( − ), ( + ), ( − )) ∈ Aˆ n does not correspond 2 +1 2 2 2 2 2 2 2 +1 n − 1 ( 2 2 ) to an element of Aˆ n (n), and since 2 +1

n − 1 n − 1 ( 2 2 ) ( 2 2 ) Card(Aˆ n ) = Card(Aˆ n (n)) 2 +1 2 +1 the Tits deformation theorem implies that CA n (n)isnot semisimple. Now it follows from 2 +1 C ≥ n 1 Theorem 2.26(c) that Ak(n)isnot semisimple for k 2 + 2 .

Murphy elements for CAk(n)

Let κn be the element of CSn given by κn = sm , (3.28) 1≤

c bS = {S ∪ S , {, } ∈S} and dI⊆S = {I,I , {, } ∈S}. (3.29)

For example, in A9,ifS = {2, 4, 5, 8} and I = {2, 4, 4 , 5, 8} then

•. •... •. .•.... •...... •. •. ..•... •. •. •... •. .•.... •...... •. •. ..•.. •...... ⊆ ...... bS = ...... and dI S = ...... •. •..... •. ...•...... •...... •. •...... •... •. •. •...... •. •...... •...... •. •...... •.. •.

For S ⊆{1, 2,...,k} deﬁne 1 c p = (−1)#({, }⊆I)+#({, }⊆I )d , (3.30) S 2 I I 26 t. halverson and a. ram

where the sum is over I ⊆ S ∪ S such that I = ∅, I = S ∪ S, I = {, } and I = {, }c.For S ⊆{1,...,k+1} such that k +1∈ S deﬁne 1 c p˜ = (−1)#({, }⊆I)+#({, }⊆I )d , (3.31) S 2 I I where the sum is over all I ⊆ S ∪ S such that {k +1, (k +1)}⊆I or {k +1, (k +1)}⊆Ic, I = S ∪ S, I = {k +1, (k +1)} and I = {k +1, (k +1)}c. Let Z1 =1and, for k ∈ Z>1, let k Z = + p + (n − k + |S|))(−1)|S|b . (3.32) k 2 S S S⊆{1,...,k} S⊆{1,...,k} |S|≥1 |S|≥2

View Zk ∈ CAk ⊆ CA 1 using the embedding in (2.2), and deﬁne Z 1 =1and k+ 2 2 |S| Z 1 = k + Zk + p˜S +(n − (k +1)+|S|)(−1) bS, (3.33) k+ 2 |S|≥2 k+1∈S

where the sum is over S ⊆{1,...,k+1} such that k +1∈ S and |S|≥2. Deﬁne

1 M 1 =1, and Mk = Zk − Z − 1 , for k ∈ Z>0.(3.34) 2 k 2 2

For example, the ﬁrst few Zk are • Z0 =1,Z1 =1,Z1 = , 2 •

p{1} •. •. • •. • •. •...... •.... •...... •...... 1 . . . − . − . . . Z1 = . . + . . . +n . . , and 2 •. •. • •. •...... •.... • •. •...... •....

Z1 p˜{1,2} b{1,2} •. •. • •. •. • • •. •. • •...... •.... •...... •.... •.. •.. •...... •.... •...... •...... − . − . − . − ...... Z2 = . . + . + . . . . . + ...... + +n . . , •. •. • •. •. • •...... •.... •...... •.... •. • • •. •... •... •...... •.... •...... •....

p{1} p{2} p{1,2} b{1,2}

and the ﬁrst few Mk are

...... • •. •. •. • •. • •. •. •. − . . . − . − . . . M0 =1,M1 =1,M1 = . ,M1 = . . . . + n . . , 2 • •. 1 2 •. •. •...... •.... • •. •...... •....

•. • •. • •...... •.... •...... •... •...... •...... M = . − . − . + ...... + , and 2 •. • •...... •.... •. • •...... •... •...... •....

•. •. •. • •. •. •... •. .•.. •. • •. •. •...... •.... •. •...... •.... •... •. .•...... 1 ...... − . . . − . . . M =2. . . + ...... + . . + . . + . . +(n 1) . . . +(n 1) ...... 2 2 •. •. •. •... •. .•... • •. •. •. •...... •.... •. • •. •. •...... •.... •... •. .•... • •...... •.... •...... •...... •.... •. • •. •...... •...... •.... •...... •...... •.... •...... •.... •. •...... •...... •.... •...... •... •...... + . + . + . . + . . − . − . − ...... − ...... •...... •...... •.... • •...... •.... •...... •...... •.... •. • •. •...... •.... •. •...... •...... •.... •...... •... •. •...... •...... •....

•...... •.... •. •. •...... •.... •...... •...... •... •...... •.... •. •...... •...... •...... + . . + . . + . . + ...... − n . . . •. •...... •.... •...... •.... •. •...... •.... •. •...... •...... •.. •...... •...... •....

Part (a) of the following theorem is well known. partition algebras 27

Theorem 3.35. ∈ Z C λ (a) For n ≥0, κn is a central element of Sn.Ifλ is a partition with n boxes and Sn is the irreducible Sn-module indexed by the partition λ, λ κn = c(b), as operators on Sn. b∈λ

⊗k (b) Let n, k ∈ Z≥0. Then, as operators on V , where dim(V )=n, − n − n − Zk = κn + kn, and Zk+ 1 = κn−1 +(k +1)n 1. 2 2 2

(c) Let n ∈ C, k ∈ Z≥0. Then Zk is a central element of CAk(n), and, if n ∈ C is such that CAk(n) is semisimple and λ n with |λ>1|≤k boxes then n Z = kn − + c(b), as operators on Aλ, k 2 k b∈λ

λ where A is the irreducible CAk(n)-module indexed by the partition λ.Furthermore, Z 1 k k+ 2 is a central element of CA 1 (n), and, if n is such that CA 1 (n) is semisimple and λ n is k+ 2 k+ 2 a partition with |λ>1|≤k boxes then n λ 1 − − Zk+ = kn + n 1 + c(b), as operators on Ak+ 1 , 2 2 2 b∈λ

λ C where A 1 is the irreducible Ak+ 1 (n)-module indexed by the partition λ. k+ 2 2

Proof. (a) The element κn is the class sum corresponding to the conjugacy class of transpositions C λ and thus κn is a central element of Sn. The constant by which κn acts on Sn is computed in [Mac, Ch. 1 §7 Ex. 7]. (c) The ﬁrst statement follows from parts (a) and (b) and Theorems 3.6 and 3.22 as follows. By C ∼ ⊗k ≥ ≥ Theorem 3.6, Ak(n) = EndSn (V )ifn 2k.Thus, by Theorem 3.22, if n 2k then Zk acts C λ on the irreducible Ak(n)-module Ak (n)bythe constant given in the statement. This means that Zk is a central element of CAk(n) for all n ≥ k.Thus, for n ≥ 2k, dZk = Zkd for all diagrams d ∈ Ak. Since the coeﬃcients in dZk (in terms of the basis of diagrams) are polynomials in n,it follows that dZk = Zkd for all n ∈ C. If n ∈ C is such that CA (n)issemisimple let χλ be the irreducible characters. Then k CAk(n) Z acts on Aλ(n)bythe constant χλ (Z )/ dim(Aλ(n)). If n ≥ k this is the constant in the k k CAk(n) k k statement, and therefore it is a polynomial in n, determined by its values for n ≥ 2k. The proof of the second statement is completely analogous using CA 1 , Sn−1, and the second k+ 2 statement in part (b).

(b) Let sii =1so that

n n 2κn + n = n +2 sij = sii + (sij + sji)= sij + sij = sij. 1≤i

Then   n n ⊗···⊗   ⊗ ···⊗ ⊗···⊗ (2κn + n)(vi1 vik )= sij (vi1 vik )= sijvi1 sijvik i,j=1 i,j=1 n − − ⊗···⊗ − − = (1 Eii Ejj + Eij + Eji)vi1 (1 Eii Ejj + Eij + Eji)vik i,j=1

⊗ ···⊗ and expanding this sum gives that (2κn + n)(vi1 vik )isequal to n

· δii S⊆{1,...,k} i ,...,i i,j=1 ∈Sc 1 k (3.36) #({, }⊆I)+#({, }⊆Ic) · − ⊗ ···⊗ ( 1) δii δij (vi1 vik ) I⊆S∪S ∈I ∈Ic where Sc ⊆{1,...,k} corresponds to the tensor positions where 1 is acting, and where I ⊆ S ∪ S corresponds to the tensor positions that must equal i and Ic corresponds to the tensor positions that must equal j. When |S| =0the set I is empty and the term corresponding to S in (3.36) is

n 2 ⊗···⊗ ⊗···⊗ δii (vi1 vik )=n (vi1 vik ). i,j=1 i1 ,...,ik ∈{1,...,k}

Assume |S|≥1 and separate the sum according to the cardinality of I. Note that the sum for I is equal to the sum for Ic, since the whole sum is symmetric in i and j. The sum of the terms in (3.36) which come from I = S ∪ S is equal to n |S| |S| − ⊗ ···⊗ − ⊗ ···⊗ n δii ( 1) δii (vi1 vik )=n( 1) bS(vi1 vik ). ∈ c ∈ ∪ i1 ,...,ik i=1 S S S

We get a similar contribution from the sum of the terms with I = ∅. If |S| > 1 then the sum of the terms in (3.36) which come from I = {, } is equal to   n |S|   − ⊗ ···⊗ δir ir ( 1) δiiδi i δir jδir j (vi1 vik ) ∈ c i1 ,...,ik i,j=1 r S r= − |S| ⊗···⊗ =( 1) bS−{}(vi1 vik ). and there is a corresponding contribution from I = {, }c. The remaining terms can be written as

n #({, }⊆I)+#({, }⊆Ic) − ⊗ δii ( 1) δii δij (vi1 vik ) ∈ c ⊆ ∪ ∈ ∈ c i1 ,...,ik i,j=1 S I S S I I ⊗ ···⊗ =2pS(vi1 vik ). partition algebras 29

Putting these cases together gives that 2κn + n acts on vi ⊗ ···⊗vi the same way that 1 k 2 1 2 2 n + 2n(−1) bS +2pS + 2n(−1) bS +2pS + (−1) 2bS−{} |S|=0 |S|=1 |S|=2 ∈S |S| |S| + 2n(−1) bS +2pS + (−1) 2bS−{} |S|>2 ∈S ⊗ ···⊗ | | ⊗ ···⊗ acts on vi1 vik . Note that bS =1if S =1. Hence 2κn + n acts on vi1 vik the same way that 2 |S| |S| n + (−2n +2pS)+ (2nbS +2+2pS)+ (−1) 2nbS +2pS + (−1) 2bS−{} |S|=1 |S|=2 |S|>2 ∈S k = n2 − 2nk +2 + 2p + 2(n − k + |S|)(−1)|S|b 2 S S |S|≥1 |S|≥2 ⊗ ···⊗ − 2 ⊗k acts on vi1 vik , and so Zk = κn +(n n +2nk)/2asoperators on V . This proves the ﬁrst statement. 0, if i = n or j = n, For the second statement, since (1 − δ )(1 − δ )= in jn 1, otherwise,   n−1 − ⊗ ···⊗ ⊗   ⊗···⊗ ⊗ (2κn−1 +(n 1))(vi1 vik vn)= sij (vi1 vik vn)   i,j=1 n  − −  ⊗ ···⊗ ⊗ = sij(1 δin)(1 δjn) (vi1 vik vn) i,j=1 n ⊗ ···⊗ ⊗ − − = sijvi1 sijvik (1 δin)(1 δjn)vn, i,j=1 n − − ⊗···⊗ − − = (1 Eii Ejj + Eij + Eji)vi1 (1 Eii Ejj + Eij + Eji)vik i,j=1

⊗ (1 − Eii − Ejj + EiiEjj)vn ⊗ ···⊗ ⊗ = sij (vi1 vik ) vn i,j n − − ⊗ ···⊗ − − ⊗ − − + (1 Eii Ejj + Eij + Eji)vi1 (1 Eii Ejj + Eij + Eji)vik ( Eii Ejj)vn i,j=1 n − − ⊗ ···⊗ − − ⊗ + (1 Eii Ejj + Eij + Eji)vi1 (1 Eii Ejj + Eij + Eji)vik EiiEjjvn i,j=1 ⊗ ···⊗ The ﬁrst sum is known to equal (2κn + n)(vi1 vik )bythe computation proving the ﬁrst statement, and the last sum has only one nonzero term, the term corresponding to i = j = n. Expanding the middle sum gives n

· δi,i S⊆{1,...,k+1} ∈ i1 ,...,ik i,j=1 S k+1∈S #({, }⊆I)+#({, }⊆Ic) · − ⊗ ···⊗ ( 1) δii δij (vi1 vik ) I ∈I ∈Ic 30 t. halverson and a. ram where the inner sum is over all I ⊆{1,...,k+1} such that {k+1, (k+1)}⊆I or {k+1, (k+1)}⊆ Ic.Asinpart (a) this sum is treated in four cases: (1) when |S| =0,(2) when I = S ∪ S or I = ∅, (3) when I = {, } or I = {, }c, and (4) the remaining cases. Since k +1∈ S, the ﬁrst case does not occur, and cases (2), (3) and (4) are as in part (a) giving |S| |S| −2n + (2nbS +2pS +2)+ 2n(−1) bS +2pS +2 (−1) bS−{} . |S|=1 |S|=2 |S|>2 ∈S k+1∈S k+1∈S k+1∈S

⊗ ···⊗ ⊗ ⊗ ⊗ ···⊗ ⊗ Combining this with the terms (2κn + n)(vi1 vik ) vn and 1 (vi1 vik vn) gives − ⊗ ···⊗ that 2κn−1 +(n 1) acts on vi1 vik as |S| (2κn + n)+1− 2n +2k + 2˜pS +2(n − (k +1)+|S|)(−1) bS. |S|≥2 k+1∈S

− ⊗ ···⊗ Thus κn−1 κn acts on vi1 vik as   1   n − (n − 1)+1− 2n +2k + 2˜p +2(n − (k +1)+|S|)(−1)|S|b  , 2 S S |S|≥2 k+1∈S

⊗k so, as operators on V ,wehave Z 1 = k+Zk+(κn−1−κn)−1+n−k = Zk+(κn−1−κn)+n−1. By k+ 2 n the ﬁrst statement in part (c) of this theorem we get Z 1 =(κn − +kn)+(κn−1 −κn)+n−1= k+ 2 2 n − − − κn 1 2 + kn + n 1.

1 ∈ Z≥ ∈ C Theorem 3.37. Let k 2 0 and let n . (a) The elements M 1 ,M1,...,M − 1 ,Mk, all commute with each other in CAk(n). 2 k 2

(b) Assume that CAk(n) is semisimple. Let µ ∈ Aˆk so that µ is a partition with ≤ k boxes, and let µ C Ak (n) be the irreducible Ak(n)-module indexed by µ. Then there is a unique, up to multiplication µ µ 1 µ { | ∈ ˆ } (0) ( 2 ) (k) ∈ ˆ by constants, basis vT T Ak of Ak (n) such that, for all T =(T ,T ,...,T ) Ak , and ∈ Z≥0 such that ≤ k, () (− 1 ) () (− 1 ) c(T /T 2 )vT , if T /T 2 = , MvT = () () (− 1 ) (n −|T |)vT , if T = T 2 , and () (+ 1 ) () (+ 1 ) n − c(T /T 2 ) vT , if T /T 2 = , M 1 vT = 1 + 2 () () (+ ) |T |vT , if T = T 2 , where λ/µ denotes the box where λ and µ diﬀer.

Proof. (a) View Z0,Z1 ,...,Zk ∈ CAk. Then Zk ∈ Z(CAk), so ZkZ = ZZk for all 0 ≤ ≤ k. 2 Since M = Z − Z − 1 ,wesee that the M commute with each other in CAk. 2 1 λ (b) The basis is deﬁned inductively. If k =0, 2 or 1, then dim(Ak (n)) = 1, so up to a constant C Ak(n) λ there is a unique choice for the basis. For k>1, we consider the restriction ResC (Ak (n)). Ak− 1 (n) 2 The branching rules for this restriction are multiplicity free, meaning that each CA − 1 (n)-irreducible k 2 partition algebras 31

λ that shows up in Ak (n)doessoexactly once. By induction, we can choose a basis for each λ CA − 1 (n)-irreducible, and the union of these bases forms a basis for A (n). For 1 = T (k− 1 ) and γ>1 = T 2 . Then by Theorem 3.35(c), Mk = Zk − Z − 1 acts on vT by the constant, k 2   n n c(b) − + kn −  c(b) − + kn − 1 = c(λ/γ)+1, 2 2 b∈λ b∈γ

− ∈ λ and Mk+ 1 = Zk+ 1 Zk acts on vT A 1 (n)bythe constant 2 2 k+ 2   n n  c(b) − + kn + n − 1 − c(b) − + kn = −c(λ/γ)+n − 1. 2 2 b∈γ b∈λ

The result now follows from (3.23) and the observation that (k) (k− 1 ) (k) (k+ 1 ) c(T /T 2 ) − 1, if T = T 2 + , c(λ/γ)= (k) (k) (k+ 1 ) n −|T |−1, if T = T 2 .

4. The Basic Construction

Let A ⊆ B be an inclusion of algebras. Then B ⊗C B is an (A, A)-bimodule where A acts on the left by left multiplication and on the right by right multiplication. Fix an (A, A)-bimodule homomorphism ε : B ⊗C B −→ A. (4.1)

The Jones basic construction is the algebra B ⊗A B with product given by

(b1 ⊗ b2)(b3 ⊗ b4)=b1 ⊗ ε(b2 ⊗ b3)b4, for b1,b2,b3,b4 ∈ B. (4.2)

More generally, let A be an algebra and let L bealeft A-module and R a right A-module. Let

ε : L ⊗C R −→ A, (4.3) be an (A, A)-bimodule homomorphism. The Jones basic construction is the algebra R ⊗A L with product given by

(r1 ⊗ 1)(r2 ⊗ 2)=r1 ⊗ ε(1 ⊗ r2)2, for r1,r2 ∈ R and 1,2 ∈ L. (4.4)

Let N = Rad(A) and let

A¯ = A/N, L¯ = L/NL, and R¯ = R/RN (4.5)

Deﬁne an (A,¯ A¯)-bimodule homomorphism

ε¯ : L¯ ⊗C R¯ −→ A¯ (4.6) ¯⊗ r¯ → ε( ⊗ r) 32 t. halverson and a. ram where ¯= + NL,¯r = r + RN, anda ¯ = a + N, for ∈ L, r ∈ R and a ∈ A. Then by basic tensor product relations [Bou2, §3.3 Cor. to Prop. 2 and §3.6 Cor. to Prop. 6], the surjective algebra homomorphism ⊗ −→ ¯ ⊗ ¯ π : R A L R A¯ L has ker(π)=R ⊗A NL. (4.7) r ⊗ → r¯ ⊗ ¯

The algebra A¯ is a split semisimple algebra (an algebra isomorphic to a direct sum of matrix algebras). Fix an algebra isomorphism ¯ −→∼ C A Mdµ ( ) µ∈Aˆ µ ← µ aPQ EPQ µ where EPQ is the matrix with 1 in the (P, Q)entry of the µth block and 0 in all other entries. Also, ﬁx isomorphisms ∼ −→µ ∼ ←−µ L¯ = A ⊗ Lµ and R¯ = Rµ ⊗ A (4.8) µ∈Aˆ µ∈Aˆ

−→µ ←−µ where A , µ ∈ Aˆ, are the simple left A¯-modules, A , µ ∈ Aˆ, are the simple right A¯-modules, and Lµ, Rµ, µ ∈ Aˆ are vector spaces. The practical eﬀect of this setup is that if Rˆµ is an index set for { µ | ∈ ˆµ} µ ˆµ { µ | ∈ ˆµ} µ ˆµ a basis rY Y R of R , L is an index set for a basis X X L of L , and A is an index set for bases −→ ←− {−→µ | ∈ ˆµ} µ {←−µ | ∈ ˆµ} µ a Q Q A of A and a P P A of A (4.9) such that λ −→µ −→µ ←−µ λ ←−µ aST a Q = δλµδTQ a S and a P aST = δλµδPS a T (4.10) then ¯ {−→µ ⊗ µ | ∈ ˆ ∈ ˆµ ∈ ˆµ} L has basis a P X µ A, P A ,X L and (4.11) ¯ { µ ⊗ µ | ∈ ˆ ∈ ˆµ ∈ ˆµ} R has basis rY a¯Q µ A, Q A ,Y R . With notations as in (4.9) and (4.11) the mapε ¯ : L¯ ⊗C R¯ → A¯ is determined by the constants µ ∈ C εXY given by −→µ ⊗ µ ⊗ µ ⊗ ←−µ µ µ ε( a Q X rY a P )=εXY aQP (4.12) µ and εXY does not depend on Q and P since −→λ ⊗ λ ⊗ µ ⊗ ←−µ λ −→λ ⊗ λ ⊗ µ ⊗ ←−µ µ ε( a S X rY a T )=ε(aSQ a Q X rY a P aPT) λ −→λ ⊗ λ ⊗ µ ⊗ ←−µ µ = aSQε( a Q X rY a P )aPT (4.13) µ µ µ µ µ µ = δλµaSQεXY aQP aPT = εXY aST .

For each µ ∈ Aˆ construct a matrix Eµ µ =(εXY )(4.14) µ µ µ µ µEµ µ and let D =(DST ) and C =(CZW )beinvertible matrices such that D C is a diagonal µ matrix with diagonal entries denoted εX , µEµ µ µ D C = diag(εX ). (4.15) partition algebras 33

µ µ Eµ µ In practice D and C are found by row reducing to its Smith normal form. The εP are the invariant factors of Eµ. ∈ ˆ ∈ ˆµ ∈ ˆµ ¯ ⊗ ¯ For µ A, X R ,Y L , deﬁne the following elements of R A¯ L, −→ ←− m¯ µ = rµ ⊗ a µ ⊗ a µ ⊗ µ , andn ¯µ = Cµ Dµ m¯ µ . (4.16) XY X P P Y XY Q1X YQ2 Q1Q2 Q1,Q2

Since λ ⊗ −→λ ⊗ ←−µ ⊗ µ λ ⊗ −→λ λ ⊗ ←−µ ⊗ µ (rS a W a Z T )=(rS a P aPW a Z T ) λ ⊗ −→λ ⊗ λ ←−µ ⊗ µ =(rS a P aPW a Z T ) (4.17) λ ⊗ −→λ ⊗ ←−λ ⊗ λ = δλµδWZ(rS a P a P T )

µ { µ | ∈ ˆ ∈ ˆµ ∈ ˆµ} ¯ ⊗ ¯ the elementm ¯ XY does not depend on P and m¯ XY µ A, X R ,Y L is a basis of R A¯ L. The following theorem is used by W.P. Brown in the study of the Brauer algebra. Part (a) is implicit in [Bro1,§2.2] and part (b) is proved in [Bro2].

⊗ → ¯⊗ ¯ { } ⊗ Theorem 4.18. Let π: R AL R A¯L be as in (4.7) and let ki beabasis of ker(π)=R ANL. Let µ ∈ ⊗ µ µ nYT R A L be such that π(nYT)=¯nYT, µ ∈ ¯ ⊗ ¯ where the elements n¯YT R A¯ L are as deﬁned in (4.16). { µ | ∈ ˆ ∈ ˆµ ∈ ˆµ} { µ | ∈ ˆ ∈ ˆµ ∈ ˆµ} (a) The sets m¯ XY µ A, X R ,Y L and n¯XY µ A, X R ,Y L are bases of ¯ ⊗ ¯ R A¯ L, which satisfy

λ µ µ µ λ µ µ µ m¯ ST m¯ QP = δλµεTQm¯ SP and n¯ST n¯QP = δλµδTQεT n¯SP .

(b) The radical of the algebra R ⊗A L is

⊗ C { µ | µ µ } Rad(R A L)= -span ki,nYT εY =0or εT =0

and the images of the elements

µ 1 µ µ µ eYT = µ nYT, for εY =0and εT =0, εT

are a set of matrix units in (R ⊗A L)/Rad(R ⊗A L).

Proof. The ﬁrst statement in (a) follows from the equations in (4.17). If (C−1)µ and (D−1)µ are the inverses of the matrices Cµ and Dµ then (C−1)µ (D−1)µ n¯ = (C−1)µ Cµ m¯ Dµ (D−1)µ XS TY XY XS Q1X Q1Q2 YQ2 TY X,Y X,Y,Q 1,Q2 = δ δ m¯ µ =¯mµ , SQ1 Q2T Q1Q2 ST Q1,Q2 34 t. halverson and a. ram

µ µ and so the elementsm ¯ ST can be written as linear combinations of then ¯XY . This establishes the second statement in (a). By direct computation, using (4.10) and (4.12),

λ µ λ ⊗ −→λ ⊗ ←−λ ⊗ λ µ ⊗ −→µ ⊗ ←−µ ⊗ µ m¯ ST m¯ QP =(rS a W a W T )(rQ a Z a Z P ) λ ⊗ −→λ ⊗ ←−λ ⊗ λ ⊗ µ ⊗ −→µ ←−µ ⊗ µ = rS a W ε( a W T rQ a Z ) a Z P λ ⊗ −→λ ⊗ λ λ ←−λ ⊗ λ = δλµ(rS a W εTQa¯WZ a Z P ) λ λ ⊗ −→λ ⊗ ←−λ ⊗ λ λ λ = δλµεTQ(rS a W a W P )=δλµεTQm¯ SP , and n¯λ n¯µ = Cλ Dλ m¯ λ Cµ Dµ m¯ µ ST UV Q1S TQ2 Q1Q2 Q3U VQ4 Q3Q4 Q1,Q2,Q3,Q4 = δ Cλ Dλ εµ Cµ Dµ m¯ µ λµ Q1S TQ2 Q2Q3 Q3U VQ4 Q1Q4 Q1,Q2,Q3,Q4 = δ δ εµ Cµ Dµ m¯ µ = δ δ εµ n¯µ . λµ TU T Q1S VQ4 Q1Q4 λµ TU T SV Q1,Q4

i (b) Let N = Rad(A)asin(4.5). If r1 ⊗ n11,r2 ⊗ n22 ∈ R ⊗A NL with n1 ∈ N for some i ∈ Z>0 then

i+1 (r1 ⊗ n11)(r2 ⊗ n22)=r1 ⊗ ε(n11 ⊗ r2)n22 = r1 ⊗ n1ε(1 ⊗ r2)n22 ∈ R ⊗A N L.

Since N is a nilpotent ideal of A it follows that ker(π)=R ⊗A NL is a nilpotent ideal of R ⊗A L. So ker(π) ⊆ Rad(R ⊗A L). Let C { µ | µ µ } I = -span ki,nYT εY =0orεT =0 . ¯ ⊗ ¯ The multiplication rule for then ¯YT implies that π(I)isanideal of R A¯ L and thus, by the ¯ ⊗ ¯ ⊗ correspondence between ideals of R A¯ L and ideals of R A L which contain ker(π), I is an ideal of R ⊗A L. Ifn ¯µ , n¯µ , n¯µ ∈{n¯µ | εµ =0orεµ =0} then Y1T1 Y2T2 Y3T3 YT Y T

n¯µ n¯µ n¯µ = δ εµ n¯µ n¯µ = δ δ εµ εµ n¯µ =0, Y1T1 Y2T2 Y3T3 T1Y2 Y2 Y1T2 Y3T3 T1Y2 T2Y3 Y2 T2 Y1T3 since εµ =0orεµ =0.Thusany product nµ nµ nµ of three basis elements of I is in Y2 T2 Y1T1 Y2T2 Y3T3 ker(π). Since ker(π)isanilpotent ideal of R ⊗A L it follows that I is an ideal of R ⊗A L consisting of nilpotent elements. So I ⊆ Rad(R ⊗A L). Since

λ µ 1 1 λ µ 1 λ λ λ eYTeUV = λ µ nYTnUV = δλµδTU λ λ εT nYV = δλµδTUeYV mod I, εT εV εT εV

λ ⊗ the images of the elements eYT in (4.7) form a set of matrix units in the algebra (R A L)/I.Thus (R ⊗A L)/I is a split semisimple algebra and so I ⊇ Rad(R ⊗A L). partition algebras 35

Basic constructions for A ⊆ B

Let A ⊆ B be an inclusion of algebras. Let ε1 : B → A be an (A, A) bimodule homomorphism and use the (A, A)-bimodule homomorphism

ε : B ⊗C B −→ A (4.19) b1 ⊗ b2 −→ ε1(b1b2) and (4.2) to deﬁne the basic construction B ⊗A B. Theorem 4.28 below provides the structure of B ⊗A B in the case that both A and B are split semisimple. If p ∈ A and pAp = Cp then (p ⊗ 1)(B ⊗A B)(p ⊗ 1) = C · (p ⊗ 1), since, if b1,b2 ∈ B then (p ⊗ 1)(b1 ⊗ b2)(p ⊗ 1) = (p ⊗ ε1(b1)b2)(p ⊗ 1) = p ⊗ ε1(b1)ε1(b2p)=pε1(b1)ε1(b2)p ⊗ 1=ξp ⊗ 1, for some constant ξ ∈ C. If p is an idempotent of A and pAp = Cp then

2 pε1(1)p = ε1(p )=ε1(1 · p)=ε1(1)p giving that ε1(1) ∈ C.

2 If p ∈ A and pAp = Cp then (p ⊗ 1) = ε1(1)(p ⊗ 1) and so,

1 if ε (1) =0 , then (p ⊗ 1) is a minimal idempotent in B ⊗ B. (4.20) 1 ε(1) A

Assume A and B are split semisimple. Let Aˆ be an index set for the irreducible A-modules Aµ, Bˆ be an index set for the irreducible B-modules Bλ, and let µ µ Aˆ = { P →µ } be an index set for a basis of the simple A-module A , for each µ ∈ Aˆ (the composite P →µ is viewed as a single symbol). Deﬁne a two level graph

Γ with vertices on level A: Aˆ,vertices on level B: Bˆ, and λ → µ λ B λ (4.21) mµ edges µ λ if A appears with multiplicity mµ in ResA(B ).

For example the graph Γ for the symmetric group algebras A = CS3 and B = CS4 is

Aˆ : ...... Bˆ :

If λ ∈ Bˆ then

λ µ Bˆ = {P →µ → λ | µ ∈ Aˆ, P →µ ∈ Aˆ and µ → λ is an edge in Γ} (4.22) is an index set for a basis of the irreducible B-module Bλ. There are sets of matrix units in the algebras A and B,

{ | ∈ ˆ → → ∈ ˆµ} { | ∈ ˆ → → → → ∈ ˆλ} aP Q µ A, P µ, Q µ A and bP Q λ B, P µ λ, Q ν λ B , (4.23) µ µ ν λ 36 t. halverson and a. ram respectively, so that

aP QaST = δµν δQSaPT and bP QbST = δλσδQSδγτbPT, (4.24) µ ν µ µγ τν µ ν λ σ λ and such that, for all µ ∈ Aˆ, P, Q ∈ Aˆµ, µ λ aP Q = bP Q (4.25) µµ µ µ→λ λ where the sum is over all edges µ → λ in the graph Γ. Though is not necessary for the following it is conceptually helpful to let C = B ⊗A B, let Cˆ = Aˆ and extend the graph Γ to a graph Γˆ with three levels, so that the edges between level B and level C are the reﬂections of the of the edges between level A and level B. In other words,

Γˆ has vertices on level C: Cˆ, and (4.26) an edge λ → µ, λ ∈ Bˆ, µ ∈ Cˆ, for each edge µ → λ, µ ∈ Aˆ, λ ∈ Bˆ.

For each ν ∈ Cˆ deﬁne ∈ ˆ ∈ ˆ ∈ ˆ → ∈ ˆµ ν µ A, λ B, ν C, P µ A and Cˆ = P →µ → λ → ν , (4.27) µ → λ and λ → ν are edges in Γˆ so that Cˆν is the set of “paths to ν”inthe graph Γ.ˆ Continuing with our previous example, Γisˆ

Aˆ : ...... Bˆ : ...... Cˆ :

Theorem 4.28. Assume A and B are split semisimple, and let the notations and assumption be as in (4.21-4.25).

(a) The elements of B ⊗A B given by ⊗ bPT bT Q µγ γ ν λ σ

γ do not depend on the choice of T →γ ∈ Aˆ and form a basis of B ⊗A B. → λ ∈ C (b) For each edge µ λ in Γ deﬁne a constant εµ by λ ε1 bPP = εµ aPP (4.29) µµ µ λ partition algebras 37

λ → ∈ ˆµ Then εµ is independent of the choice of P µ A and ⊗ ⊗ σ ⊗ bPT bT Q bRX bXS = δγπδQRδντδσρεγ bPT bTS . µγ γ ν τπ π ξ π µ γ ξ λ σ ρ η γ η ⊗ ⊗ | λ σ Rad(B A B) has basis bPT bT Q εµ =0or εν =0 , µγ γ ν λ σ and the images of the elements 1 ⊗ λ σ eP Q = σ bPT bT Q , such that εµ =0and εν =0, µ ν εγ µγ γ ν λ σ λ σ γ

form a set of matrix units in (B ⊗A B)/Rad(B ⊗A B).

trA(ε1(b)) = trB(b), for all b ∈ B. (4.30)

µ ∈ ˆ λ ∈ ˆ Let χA, µ A, and χB, λ B,bethe irreducible characters of the algebras A and B, µ ∈ ˆ λ ∈ ˆ respectively. Deﬁne constants trA, µ A, and trB, λ B,bythe equations µ µ λ λ trA = trAχA and trB = trBχB, (4.31) µ∈Aˆ λ∈Bˆ

λ respectively. Then the constants εµ deﬁned in (4.29) satisfy

λ λ µ trB = εµ trA. ⊗ (d) In the algebra B A B, ⊗ ⊗ 1 1= bPP bPP µµ µµ P ↓ λ γ µ λ γ

(g) By left multiplication, the algebra B⊗AB is a left B-module. If Rad(B⊗AB) is a B-submodule of B ⊗A B and ι: B → (B ⊗A B)/Rad(B ⊗A B) is a left B-module homomorphism then ι bRS = eRS. τ β π→γ τ β π ππ γ

Proof. By (4.11) and (4.25), ∼ −→µ ∼ ←−ν B −→ A ⊗ Lµ B −→ Rν ⊗ A µ∈Aˆ ν∈Aˆ −→ −→ ⊗ µ and −→ ν ⊗ ←− (4.32) bP Q a P Q bP Q rP a Q µ ν µ µ ν µ ν µ ν ν λ λ λ λ 38 t. halverson and a. ram as left A-modules and as right A-modules, respectively. Identify the left and right hand sides of these isomorphisms. Then, by (4.17), the elements of C = B ⊗A B given by γ ⊗ ←− ⊗ −→ ⊗ γ ⊗ m¯ P Q = rP a T a T Q = bPT bT Q (4.33) µ ν µγ γ γ γ ν µγ γ ν λ σ λ σ λ σ γ

γ do not depend on T →γ ∈ Aˆ and form a basis of B ⊗A B. (b) By (4.12), the map ε: B ⊗C B → A is determined by the values µ ∈ C µ −→ ⊗ µ ⊗ µ ⊗ ←− εT Q given by εT QaPP = ε a P T rQ a P . (4.34) γ τ γ τ µ µ µγ τ µ µ λ σ λ σ λ σ µ µ

Since µ ⊗ ⊗ εT QaPP = ε bPT bQP = ε1 bPT bQP γ τ µ µγ τ µ µγ τ µ λ σ λ σ λ σ µ µ = δT Qε1 bPP = δT Qε1 bPPbPP = δT Qε PPaPP. γ τ µµ γ τ µµ µµ γ τ µµµ µ λσ λ λσ λ λ λσ λλ Eµ λ λ the matrix given by (4.14) is diagonal with entries εµ given by (4.15) and, by (4.17), εµ is µ independent of P →µ ∈ Aˆ .ByTheorem 4.18(a),

σ m¯ P Qm¯ RS = δγπεQRm¯ PS = δγπδQRεγ m¯ PS µ ν τ ξ ντ µ ξ ντ µ ξ λ σ ρη σ ρ λ η σρ λ η γ π γ γ γ in the algebra C. The rese of the statements in part (b) follow from Theorem 4.18(b). (c) Evaluating the equations in (4.31) and using (4.29) gives

λ λ λ µ trB =trB(bPP)=trA(ε1(bPP)) = εµtrA(aPP)=εµtrA, (4.35) µµ µµ µ λ λ

(d) Since 1= bPP in the algebra B, µµ P →µ→λ λ it follows from part (b) and (4.16) that ⊗ ⊗ ⊗ 1 1= bPP bQQ = δPQδµν bPP bQQ = m¯ PP µµ νν µµ νν µµ P →µ→λ Q→ν→γ P →µ→λ P λ γ Q→ν→γ λ γ ↓ λ γ µ µ λ γ giving part (d). ⊗ λ σ (e) By left multiplication, the algebra B A B is a left B-module. If εγ =0and εγ =0then 1 ⊗ 1 ⊗ bRSeP Q = σ bRS bPT bT Q = σ δSP bRT bT Q = δSPeRQ. τ β µ ν εγ τ β µγ γ ν εγ β µ τ γ γ ν β µ τν π λ σ π λ σ πλ λ σ πλ πσ γ γ partition algebras 39

Thus, if ι: B → (B ⊗A B)/Rad(B ⊗A B)isaleft B-module homomorphism then · ι bRS = ι bRS 1=bRS ePP = δSP eRP = eRS. τ β τ β τ β µµ β µ τ µ τ β P →µ→λ→γ P →µ→λ→γ π→γ π π π λλ πλ π λ ππ γ γ γ

5. Semisimple Algebras

Let R beaintegral domain and let AR be an algebra over R,sothat AR has an R-basis {b1,...,bd},

d { } k k ∈ AR = R-span b1,...,bd and bibj = rijbk, with rij R, k=1 making AR a ring with identity. Let F be the ﬁeld of fractions of R, let F¯ be the algebraic closure of F and set A = F¯ ⊗R AR = F¯-span{b1,...,bd}, with multiplication determined by the multiplication in AR. Then A is an algebra over F¯. A trace on A is a linear map t: A → F¯ such that

t(a1a2)=t(a2a1), for all a1,a2 ∈ A.

A trace t on A is nondegenerate if for each b ∈ A there is an a ∈ A such that t(ba) =0.

Lemma 5.1. Let A beafinite dimensional algebra over a field F, let t be a trace on A. Define a × → F ∈ symmetric bilinear form , : A A on A by a1,a2 = t(a1a2), for all a1,a2 A. Let B be a basis of A. Let G = b, b b,b∈B be the matrix of the form , with respect to B. The following are equivalent: (1) The trace t is nondegenerate. (2) det G =0 . (3) The dual basis B∗ to the basis B with respect to the form , exists.

Proof. (2) ⇔ (1): The trace t is degenerate if there is an element a ∈ A, a =0 ,such that t(ac)=0 for all c ∈ B.Ifab ∈ F¯ are such that a = abb, then 0 = a, c = abb, c b∈B b∈B for all c ∈ B.Soa exists if and only if the columns of G are linearly dependent, i.e. if and only if G is not invertible. (3) ⇔ (2): Let B∗ = {b∗} be the dual basis to {b} with respect to , and let P be the change of basis matrix from B to B∗. Then ∗ ∗ t d = Pdbb, and δbc = b, d = Pdcb, c =(GP )b,c. b∈B d∈B

So P t, the transpose of P ,isthe inverse of the matrix G.Sothe dual basis to B exists if and only if G is invertible, i.e. if and only if det G =0. 40 t. halverson and a. ram

Proposition 5.2. Let A be an algebra and let t beanondegenerate trace on A. Deﬁne a symmetric bilinear form , : A × A → F¯ on A by a1,a2 = t(a1a2), for all a1,a2 ∈ A. Let B beabasis of A and let B∗ be the dual basis to B with respect to , . (a) Let a ∈ A. Then [a]= bab∗ is an element of the center Z(A) of A b∈B and [a] does not depend on the choice of the basis B.

(b) Let M and N be A-modules and let φ ∈ HomF¯(M,N) and deﬁne [φ]= bφb∗. b∈B

Then [φ] ∈ HomA(M,N) and [φ] does not depend on the choice of the basis B.

Proof. (a) Let c ∈ A. Then c[a]= cbab∗ = cb, d∗dab∗ = da d∗c, bb∗ = dad∗c =[a]c, b∈B b∈B d∈B d∈B b∈B d∈B since cb, d∗ = t(cbd∗)=t(d∗cb)=d∗c, b.So[a] ∈ Z(A). ∗ Let D be another basis of A and let D be the dual basis to D with respect to , . Let −1 P = Pdb be the transition matrix from D to B and let P be the inverse of P . Then ∗ −1 ˜∗ d = Pdbb and d = (P )˜bdb , b∈B ˜b∈B since ˜∗ −1 ˜∗ −1 d, d = Pdbb, (P )˜bd˜b = Pdb(P )˜bd˜δb˜b = δdd˜. b∈B ˜b∈B b,˜b∈B So ∗ −1 ˜∗ ˜∗ ∗ dad = Pdbba (P )˜bdb = bab δb˜b = bab . d∈D d∈D b∈B ˜b∈B b,˜b∈B b∈B So [a]does not depend on the choice of the basis B. The proof of part (b) is the same as the proof of part (a) except with a replaced by φ.

Let A be an algebra and let M be an A-module. Deﬁne

EndA(M)={T ∈ End(M) | Ta = aT for all a ∈ A}.

Theorem 5.3. (Schur’s Lemma) Let A be a ﬁnite dimensional algebra over an algebraically closed ﬁeld F¯. λ λ F¯ · (1) Let A be a simple A-module. Then EndA(A )= IdAλ . partition algebras 41

λ µ λ µ (2) If A and A are nonisomorphic simple A-modules then HomA(A ,A )=0.

Proof. Let T : Aλ → Aµ beanonzero A-module homomorphism. Since Aλ is simple, ker T =0and so T is injective. Since Aµ is simple, imT = Aµ and so T is surjective. So T is an isomorphism. Thus we may assume that T : Aλ → Aλ. Since F¯ is algebraically closed T has an eigenvector and a corresponding eigenvalue α ∈ F¯. λ λ Then T − α · Id ∈ HomA(A ,A ) and so T − α · Id is either 0 an isomorphism. However, since λ det(T −α·Id)=0,T −α·Id is not invertible. So T −α·Id =0.SoT = α·Id.SoEndA(A )=F¯·Id.

Theorem 5.4. (The Centralizer Theorem) Let A be a ﬁnite dimensional algebra over an algebraically closed ﬁeld F¯. Let M be a semisimple A-module and set Z = EndA(M). Suppose that ∼ ⊕ M = (Aλ) mλ , λ∈Mˆ

λ where Mˆ is an index set for the irreducible A-modules A which appear in M and the mλ are positive integers. ∼ F¯ (a) Z = Mmλ ( ). λ∈Mˆ (b) As an (A, Z)-bimodule ∼ M = Aλ ⊗ Zλ, λ∈Mˆ where the Zλ, λ ∈ Mˆ , are the simple Z-modules.

λ Proof. Index the components in the decomposition of M by dummy variables i so that we may write mλ ∼ λ ⊗ λ M = A i . λ∈Mˆ i=1 ∈ ˆ ≤ ≤ λ λ ⊗ → λ ⊗ For each λ M,1 i, j mλ let φij: A j A i be the A-module isomorphism given by

λ ⊗ λ ⊗ λ ∈ λ φij(m j )=m i , for m A .

By Schur’s Lemma,   ∼  λ ⊗ λ µ ⊗ µ EndA(M)=HomA(M,M) = HomA A j , A i λ j µ i

mλ ∼ λ ⊗ λ µ ⊗ µ ∼ F¯ λ = δλµHomA(A j ,A i ) = φij. λ,µ i,j λ i,j=1

Thus each element z ∈ EndA(M) can be written as

mλ λ λ λ ∈ F¯ z = zijφij, for some zij , λ∈Mˆ i,j=1 42 t. halverson and a. ram F¯ λ µ λ and identiﬁed with an element of λ Mmλ ( ). Since φijφkl = δλµδjkφil it follows that ∼ F¯ EndA(M) = Mmλ ( ). λ∈Mˆ µ { µ | ≤ ≤ } F¯ (b) As a vector space Z = span i 1 i mµ is isomorphic to the simple λ Mmλ ( ) module of column vectors of length mµ. The decomposition of M as A ⊗ Z modules follows since

⊗ λ ⊗ µ ⊗ µ ∈ µ ∈ (a φij)(m k )=δλµδjk(a i ), for all m A , a A,

If A is an algebra then Aop is the algebra A except with the opposite multiplication, i.e.

op { op | ∈ } op op op ∈ A = a a A with a1 a2 =(a2a1) , for all a1,a2 A. The left regular representation of A is the vector space A with A action given by left multiplication. Here A is serving both as an algebra and as an A-module. It is often useful to distinguish the two roles of A and use the notation A for the A-module, i.e. A is the vector space −→ A = {b | b ∈ A} with A-action ab = ab, for all a ∈ A, b ∈ A.

Proposition 5.5. Let A be an algebra and let A be the regular representation of A. Then ∼ op EndA(A) = A . More precisely,

EndA(A)={φb | b ∈ A}, where φb is given by φb(a)=ab, for all a ∈ A.

Proof. Let φ ∈ EndA(A) and let b ∈ A be such that φ(1) = b.Forall a ∈ A,

φ(a)=φ(a · 1) = aφ(1) = ab = ab,

∼ op and so φ = φb. Then EndA(A) = A since ◦ (φb1 φb2 )(a)=ab2b1 = φb2b1 (a), for all b1,b2 ∈ A and a ∈ A.

Theorem 5.6. Suppose that A is a ﬁnite dimensional algebra over an algebraically closed ﬁeld F such that the regular representation A of A is completely decomposable. Then A is isomorphic to a direct sum of matrix algebras, i.e. ∼ F¯ A = Mdλ ( ), λ∈Aˆ for some set Aˆ and some positive integers dλ, indexed by the elements of Aˆ.

Proof. If A is completely decomposable then, by Theorem 5.4, EndA(A)isisomorphic to a direct sum of matrix algebras. By Proposition ??, op ∼ F¯ A = Mdλ ( ), λ∈Aˆ partition algebras 43 for some set Aˆ and some positive integers dλ, indexed by the elements of Aˆ. The map F¯ op −→ F¯ λ∈Aˆ Mdλ ( ) λ∈Aˆ Mdλ ( ) a −→ at, where at is the transpose of the matrix a,isanalgebra isomorphism. So A is isomorphic to a direct sum of matrix algebras.

If A is an algebra then the trace tr of the regular representation is the trace on A given by

tr(a)=Tr(A(a)), for a ∈ A, where A(a)isthe linear transformation of A induced by the action of a on A by left multiplication. F¯ Proposition 5.7. Let A = λ∈Aˆ Mdλ ( ). Then the trace tr of the regular representation of A is nondegenerate.

Proof. As A-modules, the regular representation ∼ ⊕ A = (Aλ) dλ , λ∈Aˆ

λ where A is the irreducible A-module consisting of column vectors of length dλ.Fora ∈ A let Aλ(a)bethe linear transformation of Aλ induced by the action of a. Then the trace tr of the regular representation is given by χλ : A → F¯ tr = d χλ, where A λ a −→ Tr(Aλ(a)) , λ∈Aˆ

λ where χA are the irreducible characters of A. Since the dλ are all nonzero the trace tr is nondegenerate.

Theorem 5.8. (Maschke’s theorem) Let A be a ﬁnite dimensional algebra over a ﬁeld F such that the trace tr of the regular representation of A is nondegenerate. Then every representation of A is completely decomposable.

Proof. Let B be a basis of A and let B∗ be the dual basis of A with respect to the form , : A×A → F¯ deﬁned by a1,a2 = tr(a1a2), for all a1,a2 ∈ A. The dual basis B∗ exists because the trace tr is nondegenerate. Let M be an A-module. If M is irreducible then the result is vacuously true, so we may assume that M has a proper submodule N. Let p ∈ End(M)beaprojection onto N, i.e. pM = N and p2 = p. Let [p]= bpb∗, and e = bb∗. b∈B b∈B For all a ∈ A, ∗ ∗ tr(ea)= tr(bb a)= ab, b = ab b = tr(a), b∈B b∈B b∈B 44 t. halverson and a. ram

So tr((e − 1)a)=0,for all a ∈ A.Thus, since tr is nondegenerate, e =1. Let m ∈ M. Then pb∗m ∈ N for all b ∈ B, and so [p]m ∈ N.So[p]M ⊆ N. Let n ∈ N. Then pb∗n = b∗n for all b ∈ B, and so [p]n = en =1· n = n.So[p]M = N and [p]2 =[p], as elements of End(M). Note that [1 − p]=[1] − [p]=e − [p]=1− [p]. So

M =[p]M ⊕ (1 − [p])M = N ⊕ [1 − p]M, and, by Proposition ??, [1 − p]M is an A-module. So [1 − p]M is an A-submodule of M which is complementary to M.Byinduction on the dimension of M, N and [1 − p]M are completely decomposable, and therefore M is completely decomposable.

Together, Theorems 5.6, 5.8 and Proposition 5.7 yield the following theorem.

Theorem 5.9. (Artin-Wedderburn theorem) Let A beafinite dimensional algebra over an algebraically closed field F¯. Let {b1,...,bd} beabasis of A and let tr be the trace of the regular representation of A. The following are equivalent: (1) Every representation of A is completely decomposable. (2) The regular representation of A is completely decomposable. ∼ F¯ ˆ ∈ Z (3) A = λ∈Aˆ Mdλ ( ) for some finite index set A, and some dλ >0. (4) The trace of the regular representation of A is nondegenerate.

(5) det(tr(bibj)) =0 .

Remark. Let R be an integral domain, and let AR be an algebra over R with basis {b1,...,bd}. Then det(tr(bibj)) is an element of R and det(tr(bibj)) =0in F¯ if and only if det(tr(bibj)) =0in R.Inparticular, if R = C[x], then det(tr(bibj)) is a polynomial. Since a polynomial has only a ﬁnite number of roots, det(tr(bibj))(n)=0for only a ﬁnite number of values n ∈ C.

Theorem 5.10. (Tits deformation theorem) Let R be an integral domain, F, the ﬁeld of fractions of R, F¯ the algebraic closure of F, and R¯, the integral closure of R in F¯. Let AR be an R-algebra and let {b1,...,bd} be a basis of AR.Fora ∈ AR let A(a) denote the linear transformation of AR induced by left multiplication by a. Let t1,...,td be indeterminates and let

p(t1,...,td; x)=det(x · Id − (t1A(b1)+···tdA(bd))) ∈ R[t1,...,td][x], so that p is the characteristic polynomial of a “generic” element of AR. F¯ ⊗ (a) Let AF¯ = R AR.If ∼ F¯ AF¯ = Mdλ ( ), λ∈Aˆ then the factorization of p(t1,...,td,x) into irreducibles in F¯[t1,...,td,x] has the form λ dλ λ λ p = (p ) , with p ∈ R¯[t1,...,td,x] and dλ = deg(p ). λ∈Aˆ

λ If χ (t1,...,td) ∈ R¯[t1,...,td] is given by

λ dλ λ dλ−1 p (t1,...,td,x)=x − χ (t1,...,td)x + ···, partition algebras 45 then λ χ : AF¯ −→ F¯ A¯F ∈ ˆ λ λ A, α1b1 + ···+ αdbd −→ χ (α1,...,αd) , are the irreducible characters of AF¯. (b) Let K be a ﬁeld and let K¯ be the algebraic closure of K. Let γ: R → K be a ring homomorphism λ and let γ¯: R¯ → K¯ be the extension of γ. Let χ (t1,...,td) ∈ R¯[t1,...,td] be as in (a). If AK¯ = K¯ ⊗R AR is semisimple then

λ χ : AK¯ −→ K¯ ∼ K¯ AK¯ AK¯ = Mdλ ( ), and λ α1b1 + ···+ αdbd −→ (¯γχ )(α1,...,αd) , λ∈Aˆ for λ ∈ Aˆ, are the irreducible characters of AK¯ .

{ } Proof. First note that if b1,...,bd is another basis of AR and the change of basis matrix P =(Pij) is given by bi = Pijbj then the transformation ti = Pijtj, j j ∼ deﬁnes an isomorphism of polynomial rings R[t1,...,td] = R[t1,...,td]. Thus it follows that if the statements are true for one basis of AR (or AF¯) then they are true for every basis of AR (resp. AF¯). { µ ∈ ˆ ≤ ≤ } (a) Using the decomposition of AF¯ let eij,µ A, 1 i, j dλ be a basis of matrix units in AF¯ µ and let tij be corresponding variables. Then the decomposition of AF¯ induces a factorization µ λ dλ λ µ − µ λ p(tij,x)= (p ) , where p (tij; x)=det(x tijA (eij)). (5.11) λ∈Aˆ µ,i,j

λ µ The polynomial p (tij; x)isirreducible since specializing the variables gives

λ λ λ µ dλ − p (tj+1,j =1,t1,n = t, ti,j =0otherwise,x)=x t, (5.12)

λ which is irreducible in R¯[t; x]. This provides the factorization of p and establishes that deg(p )=dλ. By (5.11) − λ µ dλ − λ µ µ dλ 1 ··· p (tij; x)=x Tr(A ( tijeij))x + . µ,i,j which establishes the last statement. Any rootofp(t1,...,td,x)isanelement of R[t1,...,td]=R¯[t1,...,td]. So any root of λ λ p (t1,...,td,x)isanelement of R¯[t1,...,td] and therefore the coeﬃcients of p (t1,...,td,x) (sym- λ metric functions in the roots of p ) are elements of R¯[t1,...,td]. (b) Taking the image of the equation (5.11) give a factorization of γ(p), λ dλ γ(p)= γ(p ) , in K¯ [t1,...,td,x]. λ∈Aˆ

λ For the same reason as in (5.12) the factors γ(p ) are irreducible polynomials in K¯ [t1,...,td,x]. On the other hand, as in the proof of (a), the decomposition of AK¯ induces a factorization of γ(p)into irreducibles in K¯ [t1,...,td,x]. These two factorizations must coincide, whence the result. 46 t. halverson and a. ram

Applying the Tits deformation theorem to the case where R = C[x] (so that F = C(x)) gives the following theorem. The statement in (a) is a consequence of Theorem 5.6 and the remark which follows Theorem 5.9.

Theorem 5.13. Let CA(n) be a family of algebras defined by generators and relations such that the coefficients of the relations are polynomials in n. Assume that there is an α ∈ C such that CA(α) is semisimple. Let Aˆ be an index set for the irreducible CA(α)-modules Aλ(α). Then (a) CA(n) is semisimple for all but a finite number of n ∈ C. (b) If n ∈ C is such that CA(n) is semisimple then Aˆ is an index set for the simple CA(n)-modules Aλ(n) and dim(Aλ(n)) = dim(Aλ(α)) for each λ ∈ Aˆ. (c) Let x be an indeterminate and let {b1,...,bd} be a basis of C[x]A(x). Then there are poly- λ nomials χ (t1,...,td) ∈ C[t1,...,td,x], λ ∈ Aˆ, such that for every n ∈ C such that CA(n) is semisimple,

χλ : CA(n) −→ C A(n) ∈ ˆ λ λ A, α1b1 + ···+ αdbd −→ χ (α1,...,αd,n) ,

are the irreducible characters of CA(n).

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