Schur–Weyl duality for partition algebras

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia

Last update: 9 April 2013

Schur–Weyl duality for partition algebras

Let n>0 and let V be a vector space with basis v1,,vn. Then the tensor product

Vk= VVVkfactors has basis { vi1vik 1i1,,ik n } .

For dAk and values i1,, ik,i1, ,ik {1,,n} define

(d) i1,,ik i1,,ik = { 1, ifir=is whenrand2 are in the same block ofd, 0, otherwise. (3.1)

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

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

With this notation, the formula

d ( vi1 vik ) =1i1,,ikn (d) i1,,ik i1,,ik vi1 vik (3.2)

defines actions

Φk:Ak End(Vk) andΦk+12 :Ak+12 End(Vk) (3.3)

of Ak and Ak+12 on Vk, where the second map Φk+12 comes from the fact that if dAk+12, then d acts on the subspace

Vk Vkvn= -span { vi1 vikvn 1i1,,ik n } V(k+1). (3.4)

In other words, the map Φk+12 is obtained from Φk+1 by restricting to the subspace Vkvn and identifying Vk with Vkvn.

The group GLn() acts on the vector spaces V and Vk by

gvi=j=1n gjivj,and g ( vi1 vi2 vik ) =gvi1g vi2g vik, (3.5)

for g=(gij)GLn(). View SnGLn() as the subgroup of permutation matrices and let

EndSn(Vk)= { bEnd(Vk) bσv=σbvfor allσ Snandv Vk } .

Theorem 3.6. Let n>0 and let {xddAk} be the basis of Ak(n) defined in (2.3). Then

  1. Φk:Ak (n)End(Vk) has imΦk= EndSn (Vk) andker Φk= -span { xddhas more than nblocks } ,and
  2. Φk+12:Ak+12 (n)End(Vk) has imΦk+12= EndSn-1 (Vk) andker Φk+12= -span { xddhas more than nblocks } .


(a) As a subgroup of GLn(), Sn acts on V via its permutation representation and Sn acts on Vk by

σ ( vi1vi2 vik ) =vσ(i1) vσ(i2) vσ(ik). (3.7)

Then bEndSn(Vk) if and only if σ-1bσ=b (as endomorphisms on Vk) for all σSn. Thus, using the notation of (3.1), bEndSn(Vk) if and only if

b i1,,ik i1,,ik = (σ-1bσ) i1,,ik i1,,ik = b σ(i1),,σ(ik) σ(i1),,σ(ik) ,for allσSn.

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 dAk. It follows from (2.3) and (3.1) that for all dAk,

(Φk(xd)) i1,,ik i1,,ik = { 1, ifir=is if and only ifrand sare in the same block ofd, 0, otherwise. (3.8)

Thus Φk(xd) has 1s in the matrix positions corresponding to d and 0s elsewhere, and so b is a linear combination of Φk(xd), dAk. Since xd,dAk, form a basis of Ak,imΦk=EndSn(Vk).

If d has more than n blocks, then by (3.8) the matrix entry (Φk(xd)) i1,,ik i1,,ik =0 for all indices i1,,ik,i1,,ik since we need a distinct ij{1,,n} for each block of d. Thus, xdkerΦk. If d has n blocks, then we can find an index set i1,,ik,i1,,ik with (Φk(xd)) i1,,ik i1,,ik =1 simply by choosing a distinct index from {1,,n} for each block of d. Thus, if d has n blocks then xdkerΦk, and so kerΦk= -span { xddhas more than nblocks } .

(b) The vector space VkvnV(k+1) is a submodule both for Ak+12CAk+1 and Sn-1Sn. If σSn-1, then σ ( vi1 vikvn ) = vσ(i1) vσ(ik)vn. Then as above bEndSn-1(Vk) if and only if

b i1,,ik,n i1,,ik,n = b σ(i1),,σ(ik),n σ(i1),,σ(ik),n ,for allσ Sn-1.

The Sn-1 orbits of the matrix coordinates of b correspond to set partitions dAk+12; that is, vertices ik+1 and i(k+1) must be in the same block of d. The same argument as in part (a) can be used to show that kerΦk+12 is the span of xd with dAk+12 having more than n blocks. We always choose the index n for the block containing k+1 and (k+1).

The maps ε12:End (Vk) End(Vk) and ε12:End (Vk) End(V(k-1))

If bEnd(Vk) let b i1,,ik i1,,ik be the coefficients in the expansion

b(vi1vik) =1i1,,ikn b i1,,ik i1,,ik vi1 vik. (3.9)

Define linear maps

ε12:End (Vk) End(Vk) andε12 :End(Vk) End(V(k-1)) by ε12 (b) i1,,ik i1,,ik = b i1,,ik i1,,ik δikik andε12 (b) i1,,i(k-1) i1,,ik-1 =j,=1n b i1,,i(k-1), i1,,ik-1,j . (3.10)

The composition of ε12 and ε12 is the map

ε1:End(Vk) End(V(k-1)) given byε1 (b) i1,,i(k-1) i1,,ik-1 =j=1n b i1,,i(k-1),j i1,,ik-1,j , (3.11)


Tr(b)=ε1k (b),forb End(Vk). (3.12)

The relation between the maps ε12,ε12 in (3.10) and the maps ε12,ε12, in Section 2 is given by

Φk-12 (ε12(b)) = ε12 (Φk(b)) V(k-1)vn, forbAk(n), Φk-1 (ε12(b)) = 1n ε12 (Φk(b)), forbAk-12(n), and Φk-1 (ε1(b)) = ε1 (Φk(b)), forbAk(n), (3.13)

where, on the right hand side of the middle equality b is viewed as an element of Ak via the natural inclusion Ak-12(n) Ak(n). Then

Tr(Φk(b)) =ε1k (Φk(b))= Φ0 (ε1k(b))= ε1k(b)= trk(b), (3.14)

and, by (3.4), if bAk-12(n) then

Tr(Φk-12(b)) =Tr ( Φk(b) V(k-1)vn ) =1nTr(Φk(b)) =1ntrk(b)= 1ntrk-12 (b). (3.15)

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

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

V IndSn-1Sn ResSn-1Sn (1n). (3.16)

More generally, for any Sn-module M,

IndSn-1Sn ResSn-1Sn (M) IndSn-1Sn ( ResSn-1Sn (M)1n-1 ) IndSn-1Sn ( ResSn-1Sn (M) ResSn-1Sn (1n) ) M IndSn-1Sn ResSn-1Sn (1n)MV, (3.17)

where the third isomorphism comes from the “tensor identity”,

IndSn-1Sn ( ResSn-1Sn (M)N ) M IndSn-1Sn N g(mn) gm(gn), (3.18)

for gSn,mM,nN, and the fact that IndSn-1Sn (W) = SnSn-1 W. By iterating (3.17) it follows that

( IndSn-1Sn ResSn-1Sn ) k (1)Vk and ResSn-1Sn ( IndSn-1Sn ResSn-1Sn ) k (1)Vk (3.19)

As Sn-modules, and Sn-1-modules, respectively.


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 A^(n) which encodes the decomposition of Vk,k0, by letting

vertices on levelk: A^k(n)= { λnk- λ>1 >0 } , vertices on levelk+12: A^k+12 (n)= { λn-1k- λ>1 0 } ,and an edgeλμ,if μA^k+12 (n)is obtained fromλ A^k(n) by removing a box, an edgeμλ,if λA^k+1 (n)is obtained fromμ A^k+12(n) by adding a box. (3.21)

For example, if n=5 then the first few levels of A^(n) are

k=0: k=0+12: k=1: k=1+12: k=2: k=2+12: k=3:

The following theorem is a consequence of Theorem 3.6 and the Centralizer Theorem, Theorem 5.4 (see also [GWa1998, Theorem 3.3.7]).

Theorem 3.22. Let n,k0. Let Snλ denote the irreducible Sn-module indexed by λ.

  1. As (Sn,Ak(n))-bimodules, Vk λA^k(n) SnλAkλ(n), where the vector spaces Akλ(n) are irreducible Ak(n)-modules and dim(Akλ(n))= ( number of paths from(n) A^0(n) toλ A^k(n) in the graphA^ (n) ) .
  2. As (Sn-1,Ak+12(n))-bimodules, Vk μA^k+12(n) Sn-1μ Ak+12μ(n), where the vector spaces Ak+12μ(n) are irreducible Ak+12(n)-modules and dim(Ak+12μ(n)) = ( number of paths from(n) A^0(n) toμ A^k+12 (n)in the graph A^(n) ) .

Determination of the polynomials trμ(n)

Let n>0. For a partiton λ, let

λ>1= (λ2,,λ) ,ifλ= (λ1,λ2,,λ) ,

i.e., remove the first row of λ to get λ>1. Then, for n2k, the maps

A^k(n) A^k λ λ>1 are bijections (3.23)

which provide an isomorphism between levels 0 to n of the graphs A^(n) and A^.

Proposition 3.24. For k120 and n such that Ak(n) is semisimple, let χAk(n)μ,μA^k, be the irreducible characters of A(n) and let trk:Ak(n) be the trace on Ak(n) defined in (2.25). Use the notation for partitions in (2.17). For k>0 the coefficients in the expansion

trk=μA^k trμ(n) χAk(n)μ, are trμ(n)= ( bμ 1h(b) ) j=1μ ( n-μ- (μj-j) ) .

If n is such that Ak+12(n) is semisimple then for k>0 the coefficients in the expansion

trk+12 = μA^k+12 tr12μ(n) χAk+12(n)μ ,are tr12μ(n) = ( bμ 1h(b) ) ·n· j=1μ ( n-μ- (μj-j) ) .


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 rowi = (number of horizontal steps)+(number of vertical steps) = (λ1-λi)+ (i-1) = (λ1-1)- (λi-i), and the horizontal edge label for rowj = (number of horizontal steps)+(number of vertical steps) = (λ1-j+1)+ (λj-1) = (λ1-1)+ (λj-j)+1.


{1,2,,n} = { (λ1-1)- (λj-j) +11jλ1 } { (λ1-1)- (λi-i) 2in-λ1+1 } = { h(b)bis in row 1 of λ } { (λ1-1)- (λi-i) 2in-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 8 7 6 5 4 3 2 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

Thus, since λ1=n-λ>1,

dim(Snλ)= n!bλh(b) = ( bλ>1 1h(b) ) i=2λ>1+1 ( n-λ>1- (λi-(i-1)) ) . (3.25)

Let n>0 and let χSnλ denote the irreducible characters of the symmetric group Sn. By taking the trace on both sides of the equality in Theorem 3.22,

Tr(b,Vk)= λA^k(n) χSnλ(1) χAk(n)(b) =λA^k(n) dim(Snλ) χAk(n)(b), forbAk(n).

Thus the equality in (3.25) and the bijection in (3.23) provide the expansion of trk for all n>0 such that n2k. The statement for all n such that Ak(n) is semisimple 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 e expansion of trk+12 is exactly analogous.

Note that the polynomials trμ(n) and tr12μ(n) (of degrees μ and μ+1, respectively) do not depend on k. By Proposition 3.24,

{ roots of tr12μ (n) μA^12 } = {0}, { roots of tr1μ (n) μA^1 } = {1}, { roots of tr112μ (n) μA^112 } = {0,2},and { roots of trkμ (n) μA^k } = {0,1,,2k-1}, fork120 ,k2. (3.26)

For example, the first few values of trμ and tr12μ are

tr(n) = 1 , tr12(n) = n , tr (n) = n-1 , tr 12 (n) = n(n-2) , tr (n) = 12n(n-3) , tr 12 (n) = 12n(n-1) (n-4) , tr (n) = 12(n-1) (n-2) , tr 12 (n) = 12n(n-2) (n-3) , tr (n) = 16n(n-1) (n-5) , tr 12 (n) = 16n(n-1) (n-2)(n-6) , tr (n) = 16n(n-2) (n-4) , tr 12 (n) = 16n(n-1) (n-3)(n-5) , tr (n) = 16(n-1) (n-2)(n-3) , tr 12 (n) = 16n(n-2) (n-3)(n-4) ,

Theorem 3.27. Let n2 and k120. Then Ak(n) is semisimple if and only if kn+12.


By Theorem 2.26(a) and the observation (3.26) it follows that Ak(n) is semisimple if n2k-1.

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

An2+12(n) is semisimpleand An2+1(n) is not semisimple,

since (n/2)A^n2+12 and tr12(n/2)(n)=0. Since (n/2)A^n2+1(n), the An2+1(n)-module An2+1(n/2)(n)0. Since the path ( ,, (n/2), (n/2), (n/2) ) A^n2+1(n/2) does not correspond to an element of A^n2+1(n/2)(n),

Card(A^n2+1(n/2)) Card(A^n2+1(n/2)(n)).

Thus, Tits deformation theorem (Theorem 5.13) implies that An2+1(n) cannot be semisimple. Now it follows from Theorem 2.26(c) that Ak(n) is not semisimple for kn2+12.

If n is odd then Theorems 2.26(a) and 2.26(b) imply that

An2+12(n) is semisimpleand An2+1(n) is not semisimple,

since (n/2)A^n2+12 and tr(n/2)(n)=0. Since (n2-12)A^n2+1(n), the An2+1(n)-module An2+1(n2-12)(n)0. Since the path ( ,, (n2-12), (n2+12), (n2-12) ) A^n2+1(n2-12) does not correspond to an element of A^n2+1(n2-12)(n), and since

Card(A^n2+1(n2-12)) Card(A^n2+1(n2-12)(n)),

the Tits deformation theorem implies that An2+1(n) is not semisimple. Now it follows from Theorem 2.26(c) that Ak(n) is not semisimple for kn2+12.

Murphy elements for Ak(n)

Let κn be the element of Sn given by

κn= 1<mn sm, (3.28)

where sm is the transposition in Sn which switches and m. Let S{1,2,,k} and let ISS. Define bS,dIAk by

bS= { SS, {,}S } anddIS= { I,Ic, {,}S } (3.29)

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

bS= anddIS= .

For S{1,2,,k} define

ps=I12 (-1) #({,}I)+ #({,}Ic) dI, (3.30)

where the sum is over ISUS such that I, ISS, I{,}, and I{,}c. For S{1,,k+1} such that k+1S, define

ps=I12 (-1) #({,}I)+ #({,}Ic) dI, (3.31)

where the sum is over all ISS such that {k+1,(k+1)}I or {k+1,(k+1)}IC, ISS, I{k+1,(k+1)}, and I{k+1,(k+1)}c.

Let Z1=1 and, for kZ>1, let

Zk=(k2)+ S{1,,k}S1 pS+ S{1,,k}S2 (n-k+S) (-1)SbS. (3.32)

View ZkAkAk+12 using the embedding in (2.2), and define Z12=1 and

Zk+12=k+ Zk+ S2k+1S pS (n-(k+1)+S) (-1)SbS, (3.33)

where the sum is over S{1,,k+1} such that k+1S and S2. Define

M12=1,and Mk=Zk- Zk-12,for k12>0. (3.34)

For example, the first few Zk are

Z0=1, Z12=1, Z1= p{1} , Z112= + Z1 - - p{1,2} +n b{1,2} ,and Z2= + p{1} + p{2} - - - - + + p{1,2} +n b{1,2} ,

and the first few Mk are

M0 = 1, M12=1, M1= - ,M112= - - +n , M2 = - - + + ,and M212 = 2 + + + + +(n-1) +(n-1) + + + + - - - - + + + + -n

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

Theorem 3.35.

  1. For n0,κn is a central element of Sn. If λ is a partition with n boxes and Snλ is the irreducible Sn-module indexed by the partition λ, κn=bλ c(n),as operators on Snλ.
  2. Let n,k0. Then, as operators on Vk, where dim(V)=n, Zk=κn- (n2)+kn, and Zk+12= κn-1- (n2)+ (k+1)n-1.
  3. Let nC,k0. Then Zk is a central element of Ak(n), and, if n is such that Ak(n) is semisimple and λn with λ>1k boxes, then Zk=kn- (n2)+ bλc(b), as operators on Akλ, where Akλ is the irreducible Ak(n)-module indexed by the partition λ. Furthermore, Zk+12 is a central element of Ak+12(n), and, if n is such that Ak+12(n) is semisimple and λn is a partition with λ>1k boxes, then Zk+12=kn+n- 1-(n2)+ bλc(b), as operators on Ak+12λ, where Ak+12λ is the irreducible Ak+12(n)-module indexed by the partition λ.


(a) The element κn is the class sum corresponding to the conjugacy class of transpositions and thus κn is a central element of Sn. The constant by which κn acts on Snλ is computed in [Mac1995, Chapter 1, Section 7, Example 7].
(c) The first statement follows from parts (a) and (b) and Theorems 3.6 and 3.22 as follows. By Theorem 3.6, Ak(n) EndSn(Vk) if n2k. Thus, by Theorem 3.22, if n2k then Zk acts on the irreducible Ak(n)-module Akλ(n) by the constant given in the statement. This means that Zk is a central element of Ak(n) for all nk. Thus, for n2k, dZk=Zkd for all diagrams dAk. Since the coefficients in dZk (in terms of the basis of diagrams) are polynomials in n, it follows that dZk=Zkd for all n.

If n is such that Ak(n) is semisimple let χAk(n)λ be the irreducible characters. Then Zk acts on Akλ(n) by the constant χAk(n)λ(Zk) /dim(Akλ(n)). If nk this is the constant in the statement, and therefore it is a polynomial in n, determined by its values for n2k.

The proof of the second statement is completely analogous using Ak+12,Sn-1, and the second statement in part (b).
(b) Let sii=1 so that

2κn+n=n+2 1i<jn sij=i=1n sii+ 1i<jn (sij+sji)= i=jsij+ ijsij= i,j=1n sij.


(2κnn) ( vi1 vik ) = ( i,j=1n sij ) ( vi1 vik ) = i,j=1n sijvi1 sijvik = i,j=1n ( 1-Eii- Ejj+ Eij+Eji ) vi1 ( 1-Eii- Ejj+ Eij+Eji ) vik

and expanding this sum gives that (2κn+n) (vi1vik) is equal to

S{1,,k} i1,,ik i,j=1n ( Sc δii ) × ISS (-1) #({,}I)+ #({,}Ic) ( I δii ) × ( Ic δij ) ( vi1 vik ) (3.36)

where Sc{1,,k} corresponds to the tensor positions where 1 is acting, and where ISS corresponds to the tensor positions that must equal i and Ic corresponds to the tensor positions that must equal j.

When S=0 the set I is empty and the term corresponding to S in (3.36) is

i,j=1n i1,,ik ( {1,,k} δii ) ( vi1 vik ) =n2 ( vi1vik ) .

Assume S1 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=SS is equal to

i1,,ik ni=1n ( Sc δii ) (-1)S ( SS δii ) ( vi1 vik ) =n(-1)S bS (vi1vik).

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

i1,,ik i,j=1n ( rSc δirir ) (-1)S δii δii ( r δirj ) ( vi1 vik ) = (-1)S bS-{} (vi1vik).

and there is a corresponding contribution from I={,,}c. The remaining terms can be written as

i1,,ik i,j=1n ( Sc δii ) ISS (-1) #({,}I)+ #({,}Ic) × ( I δii ) ( Ic δij ) (vi1vik) =2pS (vi1vik).

Putting these cases together gives that 2κn+n acts on vi1vik the same way that

S=0 n2 + S=1 ( 2n(-1)1bS +2pS ) + S=2 ( 2n(-1)2bS +2pS+S (-1)22 bS-{} ) + S>2 ( 2n(-1)S bS+2pS+ S (-1)S 2bS-{} )

acts on vi1vik. Note that bS=1 if S=1. Hence 2κn+n acts on vi1vik the same way that

n2= S=1 (-2n+2pS) + S=2 (2nbS+2+2pS) + S>2 ( (-1)S 2nbS+2pS+ S (-1)S2 bS-{} ) =n2-2nk+ 2(k2)+ S1 2pS + S2 2(n-k+S) (-1)SbS

acts on vi1vik, and so Zk=κn+(n-n2+2nk)/2 as operators on Vk. This proves the first statement.

For the second statement, since (1-δin) (1-δjn) = { 0, ifi=n orj=n, 1, otherwise,

(2κn-1+(n-1)) ( vi1 vikvn ) = ( i,j=1n-1 sij ) ( vi1 vikvn ) = ( i,j=1n sij(1-δin) (1-δjn) ) ( vi1 vikvn ) = i,j=1n sijvi1 sijvik (1-δin) (1-δjn)vn, = i,j=1n ( 1-Eii- Ejj+Eij +Eji ) vi1 ( 1-Eii-Ejj+ Eij+Eji ) vik ( 1-Eii-Ejj +EiiEjj ) vn = (i,jsij) (vi1vik) vn+ i,j=1n ( 1-Eii-Ejj +Eij+Eji ) vi1 ( 1-Eii-Ejj +Eij+Eji ) vik (-Eii-Ejj) vn +i,j=1n ( 1-Eii-Ejj +Eij+Eji ) vi1 ( 1-Eii-Ejj+ Eij+Eji ) vikEii Ejjvn.

The first sum is known to equal (2κn+n) (vi1vik) by the 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

S{1,,k+1}k+1S i1,,ik i,j=1n ( S δi,i ) ×I (-1) #({,}I)+ #({,}Ic) ( I δii ) ( Ic δij ) ( vi1 vik )

where the inner sum is over all I{1,,k+1} such that {k+1,(k+1)}I or {k+1,(k+1)}Ic. As in part (a) this sum is treated in four cases: (1) when S=0, (2) when I=SS or I=, (3) when I={,} or I={,}c, and (4) the remaining cases. Since k+1S, the first case does not occur, and cases (2)–(4) are as in part (a) giving

S=1k+1S -2n + S=2k+1S (2nbS+2pS+2) + S>2k+1S ( 2n(-1)S bS+2pS+2 S (-1)S bS-{} ) .

Combining this with the terms (2κn+n) (vi1vik) vn and 1(vi1vikvn) gives that 2κn-1+(n-1) acts on vi1vik as

(2κn+n)+1- 2n+2k+ S2k+1S 2pS + 2(n-(k+1)+S) (-1)SbS.

Thus κn-1-κn acts on vi1vik as

12 ( n-(n-1)+1-2n+2k+ S2k+1S 2pS+2 (n-(k+1)+S) (-1)SbS ) ,

so, as operators on Vk, we have Zk+12=k+ Zk+(κn-1-κn) -1+n-k=Zk+ (κn-1-κn)+n- 1. By the first statement in part (c) of this theorem we get Zk+12= ( κ-(n2)+kn ) + (κn-1-κn) +n-1=κn-1- (n2)+kn+n- 1.

Theorem 3.37. Let k120 and let n.

  1. The elements M12,M1,,Mk-12,Mk, all commute with each other in Ak(n).
  2. Assume that Ak(n) is semisimple. Let μA^k so that μ is a partition with k boxes, and let Akμ(n) be the irreducible Ak(n)-module indexed by μ. Then there is a unique, up to multiplication by constants, basis {vTTA^kμ} of Akμ(n) such that, for all T= ( T(0), T(12),, T(k) ) A^kμ, and 0 such that k, MvT= { c(T()/T(-12)) vT, if T()/ T(-12) =, (n-T()) vT, ifT() =T(-12), and M+12vT= { (n-c(T()/T(+12))) vT, if T()/ T(+12) =, T() vT, ifT() =T(+12), where λ/μ denotes the box where λ and μ differ.


(a) View Z0,Z12,,ZkAk. Then ZkZ(Ak), so ZkZ=ZZk for all 0k. Since M=Z-Z-12, we see that the M commute with each other in Ak.

(b) The basis is defined inductively. If k=0,12 or 1, then dim(Akλ(n))=1, so up to a constant there is a unique choice for the basis. For k>1, we consider the restriction Res Ak-12(n) Ak(n) (Akλ(n)). The branching rules for this restriction are multiplicity free, meaning that each Ak-12(n)-irreducible that shows up in Akλ(n) does so exactly once. By induction, we can choose a basis for each Ak-12(n)-irreducible, and the union of these bases forms a basis for Akλ(n). For <k, MAk-12(n), so M acts on this basis as in the statement of the theorem. It remains only to check the statement for Mk. Let k be an integer, and let λn and γ(n-1) such that λ>1=T(k) and γ>1=T(k-12). Then by Theorem 3.35(c), Mk=Zk-Zk-12 acts on vT by the constant

( bλc(b)- (n2)+kn ) - ( bγc(b)- (n2)+kn-1 ) =c(λ/γ)+1,

and Mk+12=Zk+12-Zk acts on vTAk+12λ(n) by the constant

( bγc(b)- (n2)+kn+n-1 ) - ( bλc(b)- (n2)+kn ) =-c(λ/γ)+n-1.

The result now follows from (3.23) and the observation that

c(λ/γ)= { c ( T(k)/ T(k-12) ) -1, ifT(k)= T(k+12)+ , n-T(k)-1, ifT(k)= T(k+12).

Notes and References

This is an excerpt of a paper entitled Partition algebras, written by Tom Halverson (Mathematics and Computer Science, Macalester College, Saint Paul, MN 55105, United States) and Arun Ram.

This research was supported in part by National Science Foundation Grant DMS-0100975. This research was also supported in part by the National Science Foundation (DMS-0097977) and the National Security Agency (MDA904-01-1-0032).

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