## Schur–Weyl duality for partition algebras

Last update: 9 April 2013

## Schur–Weyl duality for partition algebras

Let $n\in {ℤ}_{>0}$ and let $V$ be a vector space with basis ${v}_{1},\dots ,{v}_{n}\text{.}$ Then the tensor product

$V⊗k= V⊗V⊗…⊗V⏟k factors has basis { vi1⊗…⊗vik ∣1≤i1,…,ik ≤n } .$

For $d\in {A}_{k}$ and values ${i}_{1},\dots ,{i}_{k},{i}_{1\prime },\dots ,{i}_{k\prime }\in \left\{1,\dots ,n\right\}$ define

$(d) i1′,…,ik′ i1,…,ik = { 1, if ir=is when r and 2 are in the same block of d, 0, otherwise. (3.1)$

For example, viewing ${\left(d\right)}_{{i}_{1\prime },\dots ,{i}_{k\prime }}^{{i}_{1},\dots ,{i}_{k}},$ as the diagram $d$ with vertices labeled by the values $i1,\dots ,{i}_{k}$ and ${i}_{1\prime },\dots ,{i}_{k\prime },$ 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

$d ( vi1⊗…⊗ vik ) =∑1≤i1′,…,ik′≤n (d) i1′,…,ik′ i1,…,ik vi1′⊗…⊗ vik′ (3.2)$

defines actions

$Φk:ℂAk⟶ End(V⊗k) andΦk+12 :ℂAk+12⟶ End(V⊗k) (3.3)$

of $ℂ{A}_{k}$ and $ℂ{A}_{k+\frac{1}{2}}$ on ${V}^{\otimes k},$ where the second map ${\Phi }_{k+\frac{1}{2}}$ comes from the fact that if $d\in {A}_{k+\frac{1}{2}},$ then $d$ acts on the subspace

$V⊗k≅ V⊗k⊗vn= ℂ-span { vi1⊗…⊗ vik⊗vn∣ 1≤i1,…,ik ≤n } ⊆V⊗(k+1). (3.4)$

In other words, the map ${\Phi }_{k+\frac{1}{2}}$ is obtained from ${\Phi }_{k+1}$ by restricting to the subspace ${V}^{\otimes k}\otimes {v}_{n}$ and identifying ${V}^{\otimes k}$ with ${V}^{\otimes k}\otimes {v}_{n}\text{.}$

The group ${GL}_{n}\left(ℂ\right)$ acts on the vector spaces $V$ and ${V}^{\otimes k}$ by

$gvi=∑j=1n gjivj,and g ( vi1⊗ vi2⊗…⊗ vik ) =gvi1⊗g vi2⊗…⊗g vik, (3.5)$

for $g=\left({g}_{ij}\right)\in {GL}_{n}\left(ℂ\right)\text{.}$ View ${S}_{n}\subseteq {GL}_{n}\left(ℂ\right)$ as the subgroup of permutation matrices and let

$EndSn(V⊗k)= { b∈End(V⊗k)∣ bσv=σbv for all σ ∈Sn and v∈ V⊗k } .$

Theorem 3.6. Let $n\in {ℤ}_{>0}$ and let $\left\{{x}_{d}\mid d\in {A}_{k}\right\}$ be the basis of $ℂ{A}_{k}\left(n\right)$ defined in (2.3). Then

1. ${\Phi }_{k}:ℂ{A}_{k}\left(n\right)\to \text{End}\left({V}^{\otimes k}\right)$ has $im Φk= EndSn (V⊗k) andker Φk=ℂ -span { xd∣d has more than n blocks } ,and$
2. ${\Phi }_{k+\frac{1}{2}}:ℂ{A}_{k+\frac{1}{2}}\left(n\right)\to \text{End}\left({V}^{\otimes k}\right)$ has $im Φk+12= EndSn-1 (V⊗k) andker Φk+12=ℂ -span { xd∣d has more than n blocks } .$

 Proof. (a) As a subgroup of ${GL}_{n}\left(ℂ\right),$ ${S}_{n}$ acts on $V$ via its permutation representation and ${S}_{n}$ acts on ${V}^{\otimes k}$ by $σ ( vi1⊗vi2 ⊗…⊗vik ) =vσ(i1)⊗ vσ(i2)⊗… ⊗vσ(ik). (3.7)$ Then $b\in {\text{End}}_{{S}_{n}}\left({V}^{\otimes k}\right)$ if and only if ${\sigma }^{-1}b\sigma =b$ (as endomorphisms on ${V}^{\otimes k}\text{)}$ for all $\sigma \in {S}_{n}\text{.}$ Thus, using the notation of (3.1), $b\in {\text{End}}_{{S}_{n}}\left({V}^{\otimes k}\right)$ if and only if $b i1′,…,ik′ i1,…,ik = (σ-1bσ) i1′,…,ik′ i1,…,ik = b σ(i1′),…,σ(ik′) σ(i1),…,σ(ik) ,for all σ∈Sn.$ It follows that the matrix entries of $b$ are constant on the ${S}_{n}\text{-orbits}$ of its matrix coordinates. These orbits decompose $\left\{1,\dots ,k,1\prime ,\dots ,k\prime \right\}$ into subsets and thus correspond to set partitions $d\in {A}_{k}\text{.}$ It follows from (2.3) and (3.1) that for all $d\in {A}_{k},$ $(Φk(xd)) i1′,…,ik′ i1,…,ik = { 1, if ir=is if and only if r and s are in the same block of d, 0, otherwise. (3.8)$ Thus ${\Phi }_{k}\left({x}_{}\right)$ has 1s in the matrix positions corresponding to $d$ and 0s elsewhere, and so $b$ is a linear combination of ${\Phi }_{k}\left({x}_{d}\right),$ $d\in {A}_{k}\text{.}$ Since ${x}_{d},d\in {A}_{k},$ form a basis of $ℂ{A}_{k},\text{im} {\Phi }_{k}={\text{End}}_{{S}_{n}}\left({V}^{\otimes k}\right)\text{.}$ If $d$ has more than $n$ blocks, then by (3.8) the matrix entry ${\left({\Phi }_{k}\left({x}_{d}\right)\right)}_{{i}_{1\prime },\dots ,{i}_{k\prime }}^{{i}_{1},\dots ,{i}_{k}}=0$ for all indices ${i}_{1},\dots ,{i}_{k},{i}_{1\prime },\dots ,{i}_{k\prime }$ since we need a distinct ${i}_{j}\in \left\{1,\dots ,n\right\}$ for each block of $d\text{.}$ Thus, ${x}_{d}\in \text{ker} {\Phi }_{k}\text{.}$ If $d$ has $\le n$ blocks, then we can find an index set ${i}_{1},\dots ,{i}_{k},{i}_{1\prime },\dots ,{i}_{k\prime }$ with ${\left({\Phi }_{k}\left({x}_{d}\right)\right)}_{{i}_{1\prime },\dots ,{i}_{k\prime }}^{{i}_{1},\dots ,{i}_{k}}=1$ simply by choosing a distinct index from $\left\{1,\dots ,n\right\}$ for each block of $d\text{.}$ Thus, if $d$ has $\le n$ blocks then ${x}_{d}\notin \text{ker}{\Phi }_{k},$ and so $\text{ker} {\Phi }_{k}=ℂ\text{-span}\left\{{x}_{d}\mid d \text{has more than} n \text{blocks}\right\}\text{.}$ (b) The vector space ${V}^{\otimes k}\otimes {v}_{n}\subseteq {V}^{\otimes \left(k+1\right)}$ is a submodule both for $ℂ{A}_{k+\frac{1}{2}}\subseteq C{A}_{k+1}$ and $ℂ{S}_{n-1}\subseteq ℂ{S}_{n}\text{.}$ If $\sigma \in {S}_{n-1},$ then $\sigma \left({v}_{{i}_{1}}\otimes \dots \otimes {v}_{{i}_{k}}\otimes {v}_{n}\right)={v}_{\sigma \left({i}_{1}\right)}\otimes \dots \otimes {v}_{\sigma \left({i}_{k}\right)}\otimes {v}_{n}\text{.}$ Then as above $b\in {\text{End}}_{{S}_{n-1}}\left({V}^{\otimes k}\right)$ if and only if $b i1′,…,ik′,n i1,…,ik,n = b σ(i1′),…,σ(ik′),n σ(i1),…,σ(ik),n ,for all σ∈ Sn-1.$ The ${S}_{n-1}$ orbits of the matrix coordinates of $b$ correspond to set partitions $d\in {A}_{k+\frac{1}{2}};$ that is, vertices ${i}_{k+1}$ and ${i}_{\left(k+1\right)\prime }$ must be in the same block of $d\text{.}$ The same argument as in part (a) can be used to show that $\text{ker} {\Phi }_{k+\frac{1}{2}}$ is the span of ${x}_{d}$ with $d\in {A}_{k+\frac{1}{2}}$ having more than $n$ blocks. We always choose the index $n$ for the block containing $k+1$ and $\left(k+1\right)\prime \text{.}$ $\square$

The maps ${\epsilon }_{\frac{1}{2}}:\text{End}\left({V}^{\otimes k}\right)\to \text{End}\left({V}^{\otimes k}\right)$ and ${\epsilon }^{\frac{1}{2}}:\text{End}\left({V}^{\otimes k}\right)\to \text{End}\left({V}^{\otimes \left(k-1\right)}\right)$

If $b\in \text{End}\left({V}^{\otimes k}\right)$ let ${b}_{{i}_{1\prime },\dots ,{i}_{k\prime }}^{{i}_{1},\dots ,{i}_{k}}\in ℂ$ be the coefficients in the expansion

$b(vi1⊗…⊗vik) =∑1≤i1′,…,ik′≤n b i1′,…,ik′ i1,…,ik vi1′⊗… ⊗vik′. (3.9)$

Define linear maps

$ε12:End (V⊗k)→ End(V⊗k) andε12 :End(V⊗k)→ End(V⊗(k-1)) by ε12 (b) i1′,…,ik′ i1,…,ik = b i1′,…,ik′ i1,…,ik δikik′ and ε12 (b) i1′,…,i(k-1)′ i1,…,ik-1 =∑j,ℓ=1n b i1′,…,i(k-1)′,ℓ i1,…,ik-1,j . (3.10)$

The composition of ${\epsilon }_{\frac{1}{2}}$ and ${\epsilon }^{\frac{1}{2}}$ is the map

$ε1:End(V⊗k) →End(V⊗(k-1)) given by ε1 (b) i1′,…,i(k-1)′ i1,…,ik-1 =∑j=1n b i1′,…,i(k-1)′,j i1,…,ik-1,j , (3.11)$

and

$Tr(b)=ε1k (b),for b∈ End(V⊗k). (3.12)$

The relation between the maps ${\epsilon }^{\frac{1}{2}},{\epsilon }_{\frac{1}{2}}$ in (3.10) and the maps ${\epsilon }^{\frac{1}{2}},{\epsilon }_{\frac{1}{2}},$ in Section 2 is given by

$Φk-12 (ε12(b)) = ε12 (Φk(b)) ∣V⊗(k-1)⊗vn, for b∈ℂAk(n), Φk-1 (ε12(b)) = 1n ε12 (Φk(b)), for b∈ℂAk-12(n), and Φk-1 (ε1(b)) = ε1 (Φk(b)), for b∈ℂAk(n), (3.13)$

where, on the right hand side of the middle equality $b$ is viewed as an element of $ℂ{A}_{k}$ via the natural inclusion $ℂ{A}_{k-\frac{1}{2}}\left(n\right)\subseteq ℂ{A}_{k}\left(n\right)\text{.}$ Then

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

and, by (3.4), if $b\in ℂ{A}_{k-\frac{1}{2}}\left(n\right)$ then

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

The representations ${\left({\text{Ind}}_{{S}_{n-1}}^{{S}_{n}}{\text{Res}}_{{S}_{n-1}}^{{S}_{n}}\right)}^{k}\left({1}_{n}\right)$ and ${\text{Res}}_{{S}_{n-1}}^{{S}_{n}}{\left({\text{Ind}}_{{S}_{n-1}}^{{S}_{n}}{\text{Res}}_{{S}_{n-1}}^{{S}_{n}}\right)}^{k}\left({1}_{n}\right)$

Let ${1}_{n}={S}_{n}^{\left(n\right)}$ be the trivial representation of ${S}_{n}$ and let $V=ℂ\text{-span}\left\{{v}_{1},\dots ,{v}_{n}\right\}$ be the permutation representation of ${S}_{n}$ given in (3.5). Then

$V≅ IndSn-1Sn ResSn-1Sn (1n). (3.16)$

More generally, for any ${S}_{n}\text{-module}$ $M,$

$IndSn-1Sn ResSn-1Sn (M) ≅ IndSn-1Sn ( ResSn-1Sn (M)⊗1n-1 ) ≅ IndSn-1Sn ( ResSn-1Sn (M)⊗ ResSn-1Sn (1n) ) ≅ M⊗ IndSn-1Sn ResSn-1Sn (1n)≅M⊗V, (3.17)$

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

$IndSn-1Sn ( ResSn-1Sn (M)⊗N ) ⟶∼ M⊗ IndSn-1Sn N g⊗(m⊗n) ⟼ gm⊗(g⊗n), (3.18)$

for $g\in {S}_{n},m\in M,n\in N,$ and the fact that ${\text{Ind}}_{{S}_{n-1}}^{{S}_{n}}\left(W\right)=ℂ{S}_{n}{\otimes }_{{S}_{n-1}}W\text{.}$ By iterating (3.17) it follows that

$( IndSn-1Sn ResSn-1Sn ) k (1)≅V⊗k and ResSn-1Sn ( IndSn-1Sn ResSn-1Sn ) k (1)≅V⊗k (3.19)$

As ${S}_{n}\text{-modules,}$ and ${S}_{n-1}\text{-modules,}$ respectively.

If

$λ= ( λ1,λ2, …,λℓ ) ,define λ>1= (λ2,…,λℓ) (3.20)$

to be the same partition as $\lambda$ except with the first row removed. Build a graph $\stackrel{^}{A}\left(n\right)$ which encodes the decomposition of ${V}^{\otimes k},k\in {ℤ}_{\ge 0},$ by letting

$vertices on level k: A^k(n)= { λ⊢n∣k- ∣λ>1∣∈ ℤ>0 } , vertices on level k+12: A^k+12 (n)= { λ⊢n-1∣k- ∣λ>1∣∈ ℤ≥0 } ,and an edge λ→μ, if μ∈A^k+12 (n) is obtained from λ∈ A^k(n) by removing a box, an edge μ→λ, if λ∈A^k+1 (n) is obtained from μ∈ A^k+12(n) by adding a box. (3.21)$

For example, if $n=5$ then the first few levels of $\stackrel{^}{A}\left(n\right)$ are

$k=0: k=0+\frac{1}{2}: k=1: k=1+\frac{1}{2}: k=2: k=2+\frac{1}{2}: k=3:$

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 $n,k\in {ℤ}_{\ge 0}\text{.}$ Let ${S}_{n}^{\lambda }$ denote the irreducible ${S}_{n}\text{-module}$ indexed by $\lambda \text{.}$

1. As $\left(ℂ{S}_{n},ℂ{A}_{k}\left(n\right)\right)\text{-bimodules,}$ $V⊗k≅ ⨁λ∈A^k(n) Snλ⊗Akλ(n),$ where the vector spaces ${A}_{k}^{\lambda }\left(n\right)$ are irreducible $ℂ{A}_{k}\left(n\right)\text{-modules}$ and $dim(Akλ(n))= ( number of paths from (n)∈ A^0(n) to λ∈ A^k(n) in the graph A^ (n) ) .$
2. As $\left(ℂ{S}_{n-1},ℂ{A}_{k+\frac{1}{2}}\left(n\right)\right)\text{-bimodules,}$ $V⊗k≅ ⨁μ∈A^k+12(n) Sn-1μ⊗ Ak+12μ(n),$ where the vector spaces ${A}_{k+\frac{1}{2}}^{\mu }\left(n\right)$ are irreducible $ℂ{A}_{k+\frac{1}{2}}\left(n\right)\text{-modules}$ and $dim(Ak+12μ(n)) = ( number of paths from (n)∈ A^0(n) to μ∈ A^k+12 (n) in the graph A^(n) ) .$

Determination of the polynomials ${\text{tr}}^{\mu }\left(n\right)$

Let $n\in {ℤ}_{>0}\text{.}$ For a partiton $\lambda ,$ let

$λ>1= (λ2,…,λℓ) ,ifλ= (λ1,λ2,…,λℓ) ,$

i.e., remove the first row of $\lambda$ to get ${\lambda }_{>1}\text{.}$ Then, for $n\ge 2k,$ 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 $\stackrel{^}{A}\left(n\right)$ and $\stackrel{^}{A}\text{.}$

Proposition 3.24. For $k\in \frac{1}{2}{ℤ}_{\ge 0}$ and $n\in ℂ$ such that $ℂ{A}_{k}\left(n\right)$ is semisimple, let ${\chi }_{{A}_{k}\left(n\right)}^{\mu },\mu \in {\stackrel{^}{A}}_{k},$ be the irreducible characters of $ℂ{A}_{\ell }\left(n\right)$ and let ${\text{tr}}_{k}:ℂ{A}_{k}\left(n\right)\to ℂ$ be the trace on $ℂ{A}_{k}\left(n\right)$ defined in (2.25). Use the notation for partitions in (2.17). For $k>0$ the coefficients in the expansion

$trk=∑μ∈A^k trμ(n) χAk(n)μ, are trμ(n)= ( ∏b∈μ 1h(b) ) ∏j=1∣μ∣ ( n-∣μ∣- (μj-j) ) .$

If $n\in ℂ$ is such that $ℂ{A}_{k+\frac{1}{2}}\left(n\right)$ is semisimple then for $k>0$ the coefficients in the expansion

$trk+12 = ∑μ∈A^k+12 tr12μ(n) χAk+12(n)μ ,are tr12μ(n) = ( ∏b∈μ 1h(b) ) ·n· ∏j=1∣μ∣ ( n-∣μ∣- (μj-j) ) .$

 Proof. Let $\lambda$ be a partition with $n$ boxes. Beginning with the vertical edge at the end of the first row, label the boundary edges of $\lambda$ sequentially with $0,1,2,\dots ,n\text{.}$ Then the $vertical edge label for row i = (number of horizontal steps)+(number of vertical steps) = (λ1-λi)+ (i-1) = (λ1-1)- (λi-i), and the$ $horizontal edge label for row j = (number of horizontal steps)+(number of vertical steps) = (λ1-j+1)+ (λj′-1) = (λ1-1)+ (λj′-j)+1.$ 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 $\lambda =\left(10,7,3,3,1\right)⊢24,$ then the boundary labels of $\lambda$ and the hook numbers in the first row of $\lambda$ 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 ${\lambda }_{1}=n-\mid {\lambda }_{>1}\mid ,$ $dim(Snλ)= n!∏b∈λh(b) = ( ∏b∈λ>1 1h(b) ) ∏i=2∣λ>1∣+1 ( n-∣λ>1∣- (λi-(i-1)) ) . (3.25)$ Let $n\in {ℤ}_{>0}$ and let ${\chi }_{{S}_{n}}^{\lambda }$ denote the irreducible characters of the symmetric group ${S}_{n}\text{.}$ By taking the trace on both sides of the equality in Theorem 3.22, $Tr(b,V⊗k)= ∑λ∈A^k(n) χSnλ(1) χAk(n)(b) =∑λ∈A^k(n) dim(Snλ) χAk(n)(b), for b∈ℂAk(n).$ Thus the equality in (3.25) and the bijection in (3.23) provide the expansion of ${\text{tr}}_{k}$ for all $n\in {ℤ}_{>0}$ such that $n\ge 2k\text{.}$ The statement for all $n\in ℂ$ such that $ℂ{A}_{k}\left(n\right)$ 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 ${\text{tr}}_{k+\frac{1}{2}}$ is exactly analogous. $\square$

Note that the polynomials ${\text{tr}}^{\mu }\left(n\right)$ and ${\text{tr}}_{\frac{1}{2}}^{\mu }\left(n\right)$ (of degrees $\mid \mu \mid$ and $\mid \mu \mid +1,$ respectively) do not depend on $k\text{.}$ By Proposition 3.24,

${ roots of tr12μ (n)∣ μ∈A^12 } = {0}, { roots of tr1μ (n)∣ μ∈A^1 } = {1}, { roots of tr112μ (n)∣ μ∈A^112 } = {0,2},and { roots of trkμ (n)∣ μ∈A^k } = {0,1,…,2k-1}, for k∈12ℤ≥0 ,k≥2. (3.26)$

For example, the first few values of ${\text{tr}}^{\mu }$ and ${\text{tr}}_{\frac{1}{2}}^{\mu }$ 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 $n\in {ℤ}_{\ge 2}$ and $k\in \frac{1}{2}{ℤ}_{\ge 0}\text{.}$ Then $ℂ{A}_{k}\left(n\right)$ is semisimple if and only if $k\le \frac{n+1}{2}\text{.}$

 Proof. By Theorem 2.26(a) and the observation (3.26) it follows that $ℂ{A}_{k}\left(n\right)$ is semisimple if $n\ge 2k-1\text{.}$ Suppose $n$ is even. Then Theorems 2.26(a) and 2.26(b) imply that $ℂAn2+12(n) is semisimpleandℂ An2+1(n) is not semisimple,$ since $\left(n/2\right)\in {\stackrel{^}{A}}_{\frac{n}{2}+\frac{1}{2}}$ and ${\text{tr}}_{\frac{1}{2}}^{\left(n/2\right)}\left(n\right)=0\text{.}$ Since $\left(n/2\right)\in {\stackrel{^}{A}}_{\frac{n}{2}+1}\left(n\right),$ the ${A}_{\frac{n}{2}+1}\left(n\right)\text{-module}$ ${A}_{\frac{n}{2}+1}^{\left(n/2\right)}\left(n\right)\ne 0\text{.}$ Since the path $\left(\varnothing ,\dots ,\left(n/2\right),\left(n/2\right),\left(n/2\right)\right)\in {\stackrel{^}{A}}_{\frac{n}{2}+1}^{\left(n/2\right)}$ does not correspond to an element of ${\stackrel{^}{A}}_{\frac{n}{2}+1}^{\left(n/2\right)}\left(n\right),$ $Card(A^n2+1(n/2)) ≠ Card(A^n2+1(n/2)(n)).$ Thus, Tits deformation theorem (Theorem 5.13) implies that $ℂ{A}_{\frac{n}{2}+1}\left(n\right)$ cannot be semisimple. Now it follows from Theorem 2.26(c) that $ℂ{A}_{k}\left(n\right)$ is not semisimple for $k\ge \frac{n}{2}+\frac{1}{2}\text{.}$ If $n$ is odd then Theorems 2.26(a) and 2.26(b) imply that $ℂAn2+12(n) is semisimpleandℂ An2+1(n) is not semisimple,$ since $\left(n/2\right)\in {\stackrel{^}{A}}_{\frac{n}{2}+\frac{1}{2}}$ and ${\text{tr}}^{\left(n/2\right)}\left(n\right)=0\text{.}$ Since $\left(\frac{n}{2}-\frac{1}{2}\right)\in {\stackrel{^}{A}}_{\frac{n}{2}+1}\left(n\right),$ the ${A}_{\frac{n}{2}+1}\left(n\right)\text{-module}$ ${A}_{\frac{n}{2}+1}^{\left(\frac{n}{2}-\frac{1}{2}\right)}\left(n\right)\ne 0\text{.}$ Since the path $\left(\varnothing ,\dots ,\left(\frac{n}{2}-\frac{1}{2}\right),\left(\frac{n}{2}+\frac{1}{2}\right),\left(\frac{n}{2}-\frac{1}{2}\right)\right)\in {\stackrel{^}{A}}_{\frac{n}{2}+1}^{\left(\frac{n}{2}-\frac{1}{2}\right)}$ does not correspond to an element of ${\stackrel{^}{A}}_{\frac{n}{2}+1}^{\left(\frac{n}{2}-\frac{1}{2}\right)}\left(n\right),$ and since $Card(A^n2+1(n2-12))≠ Card(A^n2+1(n2-12)(n)),$ the Tits deformation theorem implies that $ℂ{A}_{\frac{n}{2}+1}\left(n\right)$ is not semisimple. Now it follows from Theorem 2.26(c) that $ℂ{A}_{k}\left(n\right)$ is not semisimple for $k\ge \frac{n}{2}+\frac{1}{2}\text{.}$ $\square$

Murphy elements for $ℂ{A}_{k}\left(n\right)$

Let ${\kappa }_{n}$ be the element of $ℂ{S}_{n}$ given by

$κn= ∑1≤ℓ

where ${s}_{\ell m}$ is the transposition in ${S}_{n}$ which switches $\ell$ and $m\text{.}$ Let $S\subseteq \left\{1,2,\dots ,k\right\}$ and let $I\subseteq S\cup S\prime \text{.}$ Define ${b}_{S},{d}_{I}\in {A}_{k}$ by

$bS= { S∪S′, {ℓ,ℓ′}ℓ∉S } anddI⊆S= { I,Ic, {ℓ,ℓ′}ℓ∉S } (3.29)$

For example, in ${A}_{9},$ if $S=\left\{2,4,5,8\right\}$ and $I=\left\{2,4,4\prime ,5,8\right\}$ then

$bS= anddI⊆S= .$

For $S\subseteq \left\{1,2,\dots ,k\right\}$ define

$ps=∑I12 (-1) #({ℓ,ℓ′}⊆I)+ #({ℓ,ℓ′}⊆Ic) dI, (3.30)$

where the sum is over $I\subseteq SUS\prime$ such that $I\ne \varnothing ,$ $I\ne S\cup S\prime ,$ $I\ne \left\{\ell ,\ell \prime \right\},$ and $I\ne {\left\{\ell ,\ell \prime \right\}}^{c}\text{.}$ For $S\subseteq \left\{1,\dots ,k+1\right\}$ such that $k+1\in S,$ define

$p∼s=∑I12 (-1) #({ℓ,ℓ′}⊆I)+ #({ℓ,ℓ′}⊆Ic) dI, (3.31)$

where the sum is over all $I\subseteq S\cup S\prime$ such that $\left\{k+1,\left(k+1\right)\prime \right\}\subseteq I$ or $\left\{k+1,\left(k+1\right)\prime \right\}\subseteq {I}^{C},$ $I\ne S\cup S\prime ,$ $I\ne \left\{k+1,\left(k+1\right)\prime \right\},$ and $I\ne {\left\{k+1,\left(k+1\right)\prime \right\}}^{c}\text{.}$

Let ${Z}_{1}=1$ and, for $k\in {Z}_{>1},$ let

$Zk=(k2)+ ∑S⊆{1,…,k}∣S∣≥1 pS+ ∑S⊆{1,…,k}∣S∣≥2 (n-k+∣S∣) (-1)∣S∣bS. (3.32)$

View ${Z}_{k}\in ℂ{A}_{k}\subseteq ℂ{A}_{k+\frac{1}{2}}$ using the embedding in (2.2), and define ${Z}_{\frac{1}{2}}=1$ and

$Zk+12=k+ Zk+ ∑∣S∣≥2k+1∈S p∼S (n-(k+1)+∣S∣) (-1)∣S∣bS, (3.33)$

where the sum is over $S\subseteq \left\{1,\dots ,k+1\right\}$ such that $k+1\in S$ and $\mid S\mid \ge 2\text{.}$ Define

$M12=1,and Mk=Zk- Zk-12,for k∈12ℤ>0. (3.34)$

For example, the first few ${Z}_{k}$ 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 ${M}_{k}$ 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 $n\in {ℤ}_{\ge 0},{\kappa }_{n}$ is a central element of $ℂ{S}_{n}\text{.}$ If $\lambda$ is a partition with $n$ boxes and ${S}_{n}^{\lambda }$ is the irreducible ${S}_{n}\text{-module}$ indexed by the partition $\lambda ,$ $κn=∑b∈λ c(n),as operators on Snλ.$
2. Let $n,k\in {ℤ}_{\ge 0}\text{.}$ Then, as operators on ${V}^{\otimes k},$ where $\text{dim}\left(V\right)=n,$ $Zk=κn- (n2)+kn, and Zk+12= κn-1- (n2)+ (k+1)n-1.$
3. Let $n\in C,k\in {ℤ}_{\ge 0}\text{.}$ Then ${Z}_{k}$ is a central element of $ℂ{A}_{k}\left(n\right),$ and, if $n\in ℂ$ is such that $ℂ{A}_{k}\left(n\right)$ is semisimple and $\lambda ⊢n$ with $\mid {\lambda }_{>1}\mid \ge k$ boxes, then $Zk=kn- (n2)+ ∑b∈λc(b), as operators on Akλ,$ where ${A}_{k}^{\lambda }$ is the irreducible $ℂ{A}_{k}\left(n\right)\text{-module}$ indexed by the partition $\lambda \text{.}$ Furthermore, ${Z}_{k+\frac{1}{2}}$ is a central element of $ℂ{A}_{k+\frac{1}{2}}\left(n\right),$ and, if $n$ is such that $ℂ{A}_{k+\frac{1}{2}}\left(n\right)$ is semisimple and $\lambda ⊢n$ is a partition with $\mid {\lambda }_{>1}\mid \le k$ boxes, then $Zk+12=kn+n- 1-(n2)+ ∑b∈λc(b), as operators on Ak+12λ,$ where ${A}_{k+\frac{1}{2}}^{\lambda }$ is the irreducible $ℂ{A}_{k+\frac{1}{2}}\left(n\right)\text{-module}$ indexed by the partition $\lambda \text{.}$

 Proof. (a) The element ${\kappa }_{n}$ is the class sum corresponding to the conjugacy class of transpositions and thus ${\kappa }_{n}$ is a central element of $ℂ{S}_{n}\text{.}$ The constant by which ${\kappa }_{n}$ acts on ${S}_{n}^{\lambda }$ 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, $ℂ{A}_{k}\left(n\right)\cong {\text{End}}_{{S}_{n}}\left({V}^{\otimes k}\right)$ if $n\ge 2k\text{.}$ Thus, by Theorem 3.22, if $n\ge 2k$ then ${Z}_{k}$ acts on the irreducible $ℂ{A}_{k}\left(n\right)\text{-module}$ ${A}_{k}^{\lambda }\left(n\right)$ by the constant given in the statement. This means that ${Z}_{k}$ is a central element of $ℂ{A}_{k}\left(n\right)$ for all $n\ge k\text{.}$ Thus, for $n\ge 2k,$ $d{Z}_{k}={Z}_{k}d$ for all diagrams $d\in {A}_{k}\text{.}$ Since the coefficients in $d{Z}_{k}$ (in terms of the basis of diagrams) are polynomials in $n,$ it follows that $d{Z}_{k}={Z}_{k}d$ for all $n\in ℂ\text{.}$ If $n\in ℂ$ is such that $ℂ{A}_{k}\left(n\right)$ is semisimple let ${\chi }_{ℂ{A}_{k}\left(n\right)}^{\lambda }$ be the irreducible characters. Then ${Z}_{k}$ acts on ${A}_{k}^{\lambda }\left(n\right)$ by the constant ${\chi }_{ℂ{A}_{k}\left(n\right)}^{\lambda }\left({Z}_{k}\right)/\text{dim}\left({A}_{k}^{\lambda }\left(n\right)\right)\text{.}$ If $n\ge k$ this is the constant in the statement, and therefore it is a polynomial in $n,$ determined by its values for $n\ge 2k\text{.}$ The proof of the second statement is completely analogous using $ℂ{A}_{k+\frac{1}{2}},{S}_{n-1},$ and the second statement in part (b). (b) Let ${s}_{ii}=1$ so that $2κn+n=n+2 ∑1≤i Then $(2κnn) ( vi1⊗…⊗ vik ) = ( ∑i,j=1n sij ) ( vi1⊗…⊗ vik ) = ∑i,j=1n sijvi1 ⊗…⊗ sijvik = ∑i,j=1n ( 1-Eii- Ejj+ Eij+Eji ) vi1 ⊗…⊗ ( 1-Eii- Ejj+ Eij+Eji ) vik$ and expanding this sum gives that $\left(2{\kappa }_{n}+n\right)\left({v}_{{i}_{1}}\otimes \dots \otimes {v}_{{i}_{k}}\right)$ is equal to $∑S⊆{1,…,k} ∑i1′,…,ik′ ∑i,j=1n ( ∏ℓ∈Sc δiℓiℓ′ ) × ∑I⊆S∪S′ (-1) #({ℓ,ℓ′}⊆I)+ #({ℓ,ℓ′}⊆Ic) ( ∏ℓ∈I δiℓi ) × ( ∏ℓ∈Ic δiℓj ) ( vi1′ ⊗…⊗ vik′ ) (3.36)$ where ${S}^{c}\subseteq \left\{1,\dots ,k\right\}$ corresponds to the tensor positions where 1 is acting, and where $I\subseteq S\cup S\prime$ corresponds to the tensor positions that must equal $i$ and ${I}^{c}$ corresponds to the tensor positions that must equal $j\text{.}$ When $\mid S\mid =0$ the set $I$ is empty and the term corresponding to $S$ in (3.36) is $∑i,j=1n ∑i1′,…,ik′ ( ∏ℓ∈{1,…,k} δiℓiℓ′ ) ( vi1′ ⊗…⊗ vik′ ) =n2 ( vi1⊗…⊗vik ) .$ Assume $\mid S\mid \ge 1$ and separate the sum according to the cardinality of $I\text{.}$ Note that the sum for $I$ is equal to the sum for ${I}^{c},$ since the whole sum is symmetric in $i$ and $j\text{.}$ The sum of the terms in (3.36) which come from $I=S\cup {S}^{\prime }$ is equal to $∑i1′,…,ik′ n∑i=1n ( ∏ℓ∈Sc δiℓiℓ′ ) (-1)∣S∣ ( ∏ℓ∈S∪S′ δiℓi ) ( vi1′ ⊗…⊗ vik′ ) =n(-1)∣S∣ bS (vi1⊗…⊗vik).$ We get a similar contribution from the sum of the terms with $I=\varnothing \text{.}$ If $\mid S\mid >1$ then the sum of the terms in (3.36) which come from $I=\left\{\ell ,\ell \prime \right\}$ is equal to $∑i1′,…,ik′ ∑i,j=1n ( ∏r∈Sc δirir′ ) (-1)∣S∣ δiℓi δiℓ′i ( ∏r≠ℓ δir′j ) ( vi1′⊗… ⊗vik′ ) = (-1)∣S∣ bS-{ℓ} (vi1⊗…⊗vik).$ and there is a corresponding contribution from $I={\left\{\ell ,,\ell \prime \right\}}^{c}\text{.}$ The remaining terms can be written as $∑i1′,…,ik′ ∑i,j=1n ( ∏ℓ∈Sc δiℓiℓ′ ) ∑I⊆S∪S′ (-1) #({ℓ,ℓ′}⊆I)+ #({ℓ,ℓ′}⊆Ic) × ( ∏ℓ∈I δiℓi ) ( ∏ℓ∈Ic δiℓj ) (vi1′⊗vik′) =2pS (vi1⊗…⊗vik).$ Putting these cases together gives that $2{\kappa }_{n}+n$ acts on ${v}_{{i}_{1}}\otimes \dots \otimes {v}_{{i}_{k}}$ the same way that $∑∣S∣=0 n2 + ∑∣S∣=1 ( 2n(-1)1bS +2pS ) + ∑∣S∣=2 ( 2n(-1)2bS +2pS+∑ℓ∈S (-1)22 bS-{ℓ} ) + ∑∣S∣>2 ( 2n(-1)∣S∣ bS+2pS+ ∑ℓ∈S (-1)∣S∣ 2bS-{ℓ} )$ acts on ${v}_{{i}_{1}}\otimes \dots \otimes {v}_{{i}_{k}}\text{.}$ Note that ${b}_{S}=1$ if $\mid S\mid =1\text{.}$ Hence $2{\kappa }_{n}+n$ acts on ${v}_{{i}_{1}}\otimes \dots \otimes {v}_{{i}_{k}}$ the same way that $n2= ∑∣S∣=1 (-2n+2pS) + ∑∣S∣=2 (2nbS+2+2pS) + ∑∣S∣>2 ( (-1)∣S∣ 2nbS+2pS+ ∑ℓ∈S (-1)∣S∣2 bS-{ℓ} ) =n2-2nk+ 2(k2)+ ∑∣S∣≥1 2pS + ∑∣S∣≥2 2(n-k+∣S∣) (-1)∣S∣bS$ acts on ${v}_{{i}_{1}}\otimes \dots \otimes {v}_{{i}_{k}},$ and so ${Z}_{k}={\kappa }_{n}+\left(n-{n}^{2}+2nk\right)/2$ as operators on ${V}^{\otimes k}\text{.}$ This proves the first statement. For the second statement, since $\left(1-{\delta }_{in}\right)\left(1-{\delta }_{jn}\right)=\left\{\begin{array}{cc}0,& \text{if} i=n \text{or} j=n,\\ 1,& \text{otherwise,}\end{array}$ $(2κn-1+(n-1)) ( vi1⊗…⊗ vik⊗vn ) = ( ∑i,j=1n-1 sij ) ( vi1⊗…⊗ vik⊗vn ) = ( ∑i,j=1n sij(1-δin) (1-δjn) ) ( vi1⊗…⊗ vik⊗vn ) = ∑i,j=1n sijvi1⊗ …⊗sijvik ⊗(1-δin) (1-δjn)vn, = ∑i,j=1n ( 1-Eii- Ejj+Eij +Eji ) vi1⊗…⊗ ( 1-Eii-Ejj+ Eij+Eji ) vik ⊗ ( 1-Eii-Ejj +EiiEjj ) vn = (∑i,jsij) (vi1⊗…⊗vik) ⊗vn+ ∑i,j=1n ( 1-Eii-Ejj +Eij+Eji ) vi1 ⊗…⊗ ( 1-Eii-Ejj +Eij+Eji ) vik⊗ (-Eii-Ejj) vn +∑i,j=1n ( 1-Eii-Ejj +Eij+Eji ) vi1⊗…⊗ ( 1-Eii-Ejj+ Eij+Eji ) vik⊗Eii Ejjvn.$ The first sum is known to equal $\left(2{\kappa }_{n}+n\right)\left({v}_{{i}_{1}}\otimes \dots \otimes {v}_{{i}_{k}}\right)$ by the computation proving the first statement, and the last sum has only one nonzero term, the term corresponding to $i=j=n\text{.}$ Expanding the middle sum gives $∑S⊆{1,…,k+1}k+1∈S ∑i1′,…,ik′ ∑i,j=1n ( ∏ℓ∈S δiℓ,iℓ′ ) ×∑I (-1) #({ℓ,ℓ′}⊆I)+ #({ℓ,ℓ′}⊆Ic) ( ∏ℓ∈I δiℓi ) ( ∏ℓ∈Ic δiℓj ) ( vi1′⊗… ⊗vik′ )$ where the inner sum is over all $I\subseteq \left\{1,\dots ,k+1\right\}$ such that $\left\{k+1,\left(k+1\right)\prime \right\}\subseteq I$ or $\left\{k+1,\left(k+1\right)\prime \right\}\subseteq {I}^{c}\text{.}$ As in part (a) this sum is treated in four cases: (1) when $\mid S\mid =0,$ (2) when $I=S\cup S\prime$ or $I=\varnothing ,$ (3) when $I=\left\{\ell ,\ell \prime \right\}$ or $I={\left\{\ell ,\ell \prime \right\}}^{c},$ and (4) the remaining cases. Since $k+1\in S,$ the first case does not occur, and cases (2)–(4) are as in part (a) giving $∑∣S∣=1k+1∈S -2n + ∑∣S∣=2k+1∈S (2nbS+2pS+2) + ∑∣S∣>2k+1∈S ( 2n(-1)∣S∣ bS+2pS+2 ∑ℓ∈S (-1)∣S∣ bS-{ℓ} ) .$ Combining this with the terms $\left(2{\kappa }_{n}+n\right)\left({v}_{{i}_{1}}\otimes \dots \otimes {v}_{{i}_{k}}\right)\otimes {v}_{n}$ and $1\otimes \left({v}_{{i}_{1}}\otimes \dots \otimes {v}_{{i}_{k}}\otimes {v}_{n}\right)$ gives that $2{\kappa }_{n-1}+\left(n-1\right)$ acts on ${v}_{{i}_{1}}\otimes \dots \otimes {v}_{{i}_{k}}$ as $(2κn+n)+1- 2n+2k+ ∑∣S∣≥2k+1∈S 2p∼S + 2(n-(k+1)+∣S∣) (-1)∣S∣bS.$ Thus ${\kappa }_{n-1}-{\kappa }_{n}$ acts on ${v}_{{i}_{1}}\otimes \dots \otimes {v}_{{i}_{k}}$ as $12 ( n-(n-1)+1-2n+2k+ ∑∣S∣≥2k+1∈S 2p∼S+2 (n-(k+1)+∣S∣) (-1)∣S∣bS ) ,$ so, as operators on ${V}^{\otimes k},$ we have ${Z}_{k+\frac{1}{2}}=k+{Z}_{k}+\left({\kappa }_{n-1}-{\kappa }_{n}\right)-1+n-k={Z}_{k}+\left({\kappa }_{n-1}-{\kappa }_{n}\right)+n-1\text{.}$ By the first statement in part (c) of this theorem we get ${Z}_{k+\frac{1}{2}}=\left(\kappa -\left(\frac{n}{2}\right)+kn\right)+\left({\kappa }_{n-1}-{\kappa }_{n}\right)+n-1={\kappa }_{n-1}-\left(\frac{n}{2}\right)+kn+n-1\text{.}$ $\square$

Theorem 3.37. Let $k\in \frac{1}{2}{ℤ}_{\ge 0}$ and let $n\in ℂ\text{.}$

1. The elements ${M}_{\frac{1}{2}},{M}_{1},\dots ,{M}_{k-\frac{1}{2}},{M}_{k},$ all commute with each other in $ℂ{A}_{k}\left(n\right)\text{.}$
2. Assume that $ℂ{A}_{k}\left(n\right)$ is semisimple. Let $\mu \in {\stackrel{^}{A}}_{k}$ so that $\mu$ is a partition with $\le k$ boxes, and let ${A}_{k}^{\mu }\left(n\right)$ be the irreducible $ℂ{A}_{k}\left(n\right)\text{-module}$ indexed by $\mu \text{.}$ Then there is a unique, up to multiplication by constants, basis $\left\{{v}_{T}\mid T\in {\stackrel{^}{A}}_{k}^{\mu }\right\}$ of ${A}_{k}^{\mu }\left(n\right)$ such that, for all $T=\left({T}^{\left(0\right)},{T}^{\left(\frac{1}{2}\right)},\dots ,{T}^{\left(k\right)}\right)\in {\stackrel{^}{A}}_{k}^{\mu },$ and $\ell \in {ℤ}_{\ge 0}$ such that $\ell \le k,$ $MℓvT= { c(T(ℓ)/T(ℓ-12)) vT, if T(ℓ)/ T(ℓ-12) =▫, (n-∣T(ℓ)∣) vT, if T(ℓ) =T(ℓ-12),$ and $Mℓ+12vT= { (n-c(T(ℓ)/T(ℓ+12))) vT, if T(ℓ)/ T(ℓ+12) =▫, ∣T(ℓ)∣ vT, if T(ℓ) =T(ℓ+12),$ where $\lambda /\mu$ denotes the box where $\lambda$ and $\mu$ differ.

 Proof. (a) View ${Z}_{0},{Z}_{\frac{1}{2}},\dots ,{Z}_{k}\in ℂ{A}_{k}\text{.}$ Then ${Z}_{k}\in Z\left(ℂ{A}_{k}\right),$ so ${Z}_{k}{Z}_{\ell }={Z}_{\ell }{Z}_{k}$ for all $0\le \ell \le k\text{.}$ Since ${M}_{\ell }={Z}_{\ell }-{Z}_{\ell -\frac{1}{2}},$ we see that the ${M}_{\ell }$ commute with each other in $ℂ{A}_{k}\text{.}$ (b) The basis is defined inductively. If $k=0,\frac{1}{2}$ or 1, then $\text{dim}\left({A}_{k}^{\lambda }\left(n\right)\right)=1,$ so up to a constant there is a unique choice for the basis. For $k>1,$ we consider the restriction ${\text{Res}}_{ℂ{A}_{k-\frac{1}{2}}\left(n\right)}^{ℂ{A}_{k}\left(n\right)}\left({A}_{k}^{\lambda }\left(n\right)\right)\text{.}$ The branching rules for this restriction are multiplicity free, meaning that each $ℂ{A}_{k-\frac{1}{2}}\left(n\right)\text{-irreducible}$ that shows up in ${A}_{k}^{\lambda }\left(n\right)$ does so exactly once. By induction, we can choose a basis for each $ℂ{A}_{k-\frac{1}{2}}\left(n\right)\text{-irreducible,}$ and the union of these bases forms a basis for ${A}_{k}^{\lambda }\left(n\right)\text{.}$ For $\ell ${M}_{\ell }\in ℂ{A}_{k-\frac{1}{2}}\left(n\right),$ so ${M}_{\ell }$ acts on this basis as in the statement of the theorem. It remains only to check the statement for ${M}_{k}\text{.}$ Let $k$ be an integer, and let $\lambda ⊢n$ and $\gamma ⊢\left(n-1\right)$ such that ${\lambda }_{>1}={T}^{\left(k\right)}$ and ${\gamma }_{>1}={T}^{\left(k-\frac{1}{2}\right)}\text{.}$ Then by Theorem 3.35(c), ${M}_{k}={Z}_{k}-{Z}_{k-\frac{1}{2}}$ acts on ${v}_{T}$ by the constant $( ∑b∈λc(b)- (n2)+kn ) - ( ∑b∈γc(b)- (n2)+kn-1 ) =c(λ/γ)+1,$ and ${M}_{k+\frac{1}{2}}={Z}_{k+\frac{1}{2}}-{Z}_{k}$ acts on ${v}_{T}\in {A}_{k+\frac{1}{2}}^{\lambda }\left(n\right)$ by the constant $( ∑b∈γc(b)- (n2)+kn+n-1 ) - ( ∑b∈λc(b)- (n2)+kn ) =-c(λ/γ)+n-1.$ The result now follows from (3.23) and the observation that $c(λ/γ)= { c ( T(k)/ T(k-12) ) -1, if T(k)= T(k+12)+ ▫, n-∣T(k)∣-1, if T(k)= T(k+12).$ $\square$

## 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).