Schur–Weyl duality for partition algebras

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 9 April 2013

Schur–Weyl duality for partition algebras

Let $n\in {\mathbb{Z}}_{>0}$ and let $V$ be a vector space with basis $v_1,\dots ,v_n\text{.}$ Then the tensor product

$$V^{\otimes k}=\underset{\underset{k\hspace{0.17em}\text{factors}}{⏟}}{V\otimes V\otimes \dots \otimes V}\hspace{0.17em}\text{has basis}\{v_{i_1}\otimes \dots \otimes v_{i_k}\mid 1\le i_1,\dots ,i_k\le n\}\text{.}$$

For $d\in A_k$ and values $i_1,\dots ,i_k,i_{1\prime },\dots ,i_{k\prime }\in \{1,\dots ,n\}$ define

$$\begin{array}{cc}{\left(d\right)}_{i_{1\prime },\dots ,i_{k\prime }}^{i_1,\dots ,i_k}=\{\begin{array}{cc}1,& \text{if}\hspace{0.17em}{i}_r=i_s\hspace{0.17em}\text{when}\hspace{0.17em}r\hspace{0.17em}\text{and}\hspace{0.17em}2\hspace{0.17em}\text{are in the same block of}\hspace{0.17em}d,\\ 0,& \text{otherwise.}\end{array}& \text{(3.1)}\end{array}$$

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

$$\begin{array}{c} i1 i2 i3 i4 i5 i6 i7 i8 i1′ i2′ i3′ i4′ i5′ i6′ i7′ i8′ \end{array}={\delta }_{i_1{i}_2}{\delta }_{i_1{i}_4}{\delta }_{i_1{i}_{2\prime }}{\delta }_{i_1{i}_{5\prime }}{\delta }_{i_5{i}_6}{\delta }_{i_5{i}_7}{\delta }_{i_5{i}_{3\prime }}{\delta }_{i_5{i}_{4\prime }}{\delta }_{i_5{i}_{6\prime }}{\delta }_{i_5{i}_{7\prime }}{\delta }_{i_8{i}_{8\prime }}\text{.}$$

With this notation, the formula

$$\begin{array}{cc}d(v_{i_1}\otimes \dots \otimes v_{i_k})=\sum _{1\le i_{1\prime },\dots ,i_{k\prime }\le n}{\left(d\right)}_{i_{1\prime },\dots ,i_{k\prime }}^{i_1,\dots ,i_k}{v}_{i_{1\prime }}\otimes \dots \otimes v_{i_{k\prime }}& \text{(3.2)}\end{array}$$

defines actions

$$\begin{array}{cc}{\Phi }_k:ℂA_k⟶\text{End}\left(V^{\otimes k}\right)\text{and}{\Phi }_{k+\frac{1}{2}}:ℂA_{k+\frac{1}{2}}⟶\text{End}\left(V^{\otimes k}\right)& \text{(3.3)}\end{array}$$

of $ℂA_k$ and $ℂA_{k+\frac{1}{2}}$ 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

$$\begin{array}{cc}{V}^{\otimes k}\cong V^{\otimes k}\otimes v_n=ℂ\text{-span}\{v_{i_1}\otimes \dots \otimes v_{i_k}\otimes v_n\mid 1\le i_1,\dots ,i_k\le n\}\subseteq V^{\otimes (k+1)}\text{.}& \text{(3.4)}\end{array}$$

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

$$\begin{array}{cc}g{v}_i=\sum _{j=1}^n{g}_{ji}{v}_j,\text{and}g(v_{i_1}\otimes v_{i_2}\otimes \dots \otimes v_{i_k})=g{v}_{i_1}\otimes g{v}_{i_2}\otimes \dots \otimes g{v}_{i_k},& \text{(3.5)}\end{array}$$

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

$${\text{End}}_{S_n}\left(V^{\otimes k}\right)=\{b\in \text{End}\left(V^{\otimes k}\right)\mid b\sigma v=\sigma bv\hspace{0.17em}\text{for all}\hspace{0.17em}\sigma \in S_n\hspace{0.17em}\text{and}\hspace{0.17em}v\in V^{\otimes k}\}\text{.}$$

Theorem 3.6. Let $n\in {\mathbb{Z}}_{>0}$ and let $\{x_d\mid d\in A_k\}$ be the basis of $ℂA_k\left(n\right)$ defined in (2.3). Then

1. ${\Phi }_k:ℂA_k\left(n\right)\to \text{End}\left(V^{\otimes k}\right)$ has $$\text{im}\hspace{0.17em}{\Phi }_k={\text{End}}_{S_n}\left(V^{\otimes k}\right)\text{and}\text{ker}\hspace{0.17em}{\Phi }_k=ℂ\text{-span}\{x_d\mid d\hspace{0.17em}\text{has more than}\hspace{0.17em}n\hspace{0.17em}\text{blocks}\},\text{and}$$
2. ${\Phi }_{k+\frac{1}{2}}:ℂA_{k+\frac{1}{2}}\left(n\right)\to \text{End}\left(V^{\otimes k}\right)$ has $$\text{im}\hspace{0.17em}{\Phi }_{k+\frac{1}{2}}={\text{End}}_{S_{n-1}}\left(V^{\otimes k}\right)\text{and}\text{ker}\hspace{0.17em}{\Phi }_{k+\frac{1}{2}}=ℂ\text{-span}\{x_d\mid d\hspace{0.17em}\text{has more than}\hspace{0.17em}n\hspace{0.17em}\text{blocks}\}\text{.}$$

		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 $$\begin{array}{cc}\sigma (v_{i_1}\otimes v_{i_2}\otimes \dots \otimes v_{i_k})=v_{\sigma \left(i_1\right)}\otimes v_{\sigma \left(i_2\right)}\otimes \dots \otimes v_{\sigma \left(i_k\right)}\text{.}& \text{(3.7)}\end{array}$$ 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 $$b_{i_{1\prime },\dots ,i_{k\prime }}^{i_1,\dots ,i_k}={\left({\sigma }^{-1}b\sigma \right)}_{i_{1\prime },\dots ,i_{k\prime }}^{i_1,\dots ,i_k}=b_{\sigma \left(i_{1\prime }\right),\dots ,\sigma \left(i_{k\prime }\right)}^{\sigma \left(i_1\right),\dots ,\sigma \left(i_k\right)},\text{for all}\hspace{0.17em}\sigma \in S_n\text{.}$$ It follows that the matrix entries of $b$ are constant on the $S_n\text{-orbits}$ of its matrix coordinates. These orbits decompose $\{1,\dots ,k,1\prime ,\dots ,k\prime \}$ into subsets and thus correspond to set partitions $d\in A_k\text{.}$ It follows from (2.3) and (3.1) that for all $d\in A_k,$ $$\begin{array}{cc}{\left({\Phi }_k\left(x_d\right)\right)}_{i_{1\prime },\dots ,i_{k\prime }}^{i_1,\dots ,i_k}=\{\begin{array}{cc}1,& \text{if}\hspace{0.17em}{i}_r=i_s\hspace{0.17em}\text{if and only if}\hspace{0.17em}r\hspace{0.17em}\text{and}\hspace{0.17em}s\hspace{0.17em}\text{are in the same block of}\hspace{0.17em}d,\\ 0,& \text{otherwise.}\end{array}& \text{(3.8)}\end{array}$$ 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 $ℂA_k,\text{im}\hspace{0.17em}{\Phi }_k={\text{End}}_{S_n}\left(V^{\otimes k}\right)\text{.}$ 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 $i_j\in \{1,\dots ,n\}$ for each block of $d\text{.}$ Thus, $x_d\in \text{ker}\hspace{0.17em}{\Phi }_k\text{.}$ 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 $\{1,\dots ,n\}$ for each block of $d\text{.}$ Thus, if $d$ has $\le n$ blocks then $x_d\notin \text{ker}{\Phi }_k,$ and so $\text{ker}\hspace{0.17em}{\Phi }_k=ℂ\text{-span}\{x_d\mid d\hspace{0.17em}\text{has more than}\hspace{0.17em}n\hspace{0.17em}\text{blocks}\}\text{.}$ (b) The vector space $V^{\otimes k}\otimes v_n\subseteq V^{\otimes (k+1)}$ is a submodule both for $ℂA_{k+\frac{1}{2}}\subseteq C{A}_{k+1}$ and $ℂS_{n-1}\subseteq ℂS_n\text{.}$ If $\sigma \in S_{n-1},$ then $\sigma (v_{i_1}\otimes \dots \otimes v_{i_k}\otimes v_n)=v_{\sigma \left(i_1\right)}\otimes \dots \otimes v_{\sigma \left(i_k\right)}\otimes v_n\text{.}$ Then as above $b\in {\text{End}}_{S_{n-1}}\left(V^{\otimes k}\right)$ if and only if $$b_{i_{1\prime },\dots ,i_{k\prime },n}^{i_1,\dots ,i_k,n}=b_{\sigma \left(i_{1\prime }\right),\dots ,\sigma \left(i_{k\prime }\right),n}^{\sigma \left(i_1\right),\dots ,\sigma \left(i_k\right),n},\text{for all}\hspace{0.17em}\sigma \in S_{n-1}\text{.}$$ 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 $i_{(k+1)\prime }$ must be in the same block of $d\text{.}$ The same argument as in part (a) can be used to show that $\text{ker}\hspace{0.17em}{\Phi }_{k+\frac{1}{2}}$ 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 $(k+1)\prime \text{.}$ $\square$

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

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

$$\begin{array}{cc}b(v_{i_1}\otimes \dots \otimes v_{i_k})=\sum _{1\le i_{1\prime },\dots ,i_{k\prime }\le n}{b}_{i_{1\prime },\dots ,i_{k\prime }}^{i_1,\dots ,i_k}{v}_{i_{1\prime }}\otimes \dots \otimes v_{i_{k\prime }}\text{.}& \text{(3.9)}\end{array}$$

Define linear maps

$$\begin{array}{cc}\begin{array}{c}{\epsilon }_{\frac{1}{2}}:\text{End}\left(V^{\otimes k}\right)\to \text{End}\left(V^{\otimes k}\right)\text{and}{\epsilon }^{\frac{1}{2}}:\text{End}\left(V^{\otimes k}\right)\to \text{End}\left(V^{\otimes (k-1)}\right)\hspace{0.17em}\text{by}\\ {\epsilon }_{\frac{1}{2}}{\left(b\right)}_{i_{1\prime },\dots ,i_{k\prime }}^{i_1,\dots ,i_k}=b_{i_{1\prime },\dots ,i_{k\prime }}^{i_1,\dots ,i_k}{\delta }_{i_k{i}_{k\prime }}\hspace{0.17em}\text{and}\hspace{0.17em}{\epsilon }^{\frac{1}{2}}{\left(b\right)}_{i_{1\prime },\dots ,i_{(k-1)\prime }}^{i_1,\dots ,i_{k-1}}=\sum _{j,\ell =1}^n{b}_{i_{1\prime },\dots ,i_{(k-1)\prime },\ell }^{i_1,\dots ,i_{k-1},j}\text{.}\end{array}& \text{(3.10)}\end{array}$$

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

$$\begin{array}{cc}{\epsilon }_1:\text{End}\left(V^{\otimes k}\right)\to \text{End}\left(V^{\otimes (k-1)}\right)\hspace{0.17em}\text{given by}\hspace{0.17em}{\epsilon }_1{\left(b\right)}_{i_{1\prime },\dots ,i_{(k-1)\prime }}^{i_1,\dots ,i_{k-1}}=\sum _{j=1}^n{b}_{i_{1\prime },\dots ,i_{(k-1)\prime },j}^{i_1,\dots ,i_{k-1},j},& \text{(3.11)}\end{array}$$

and

$$\begin{array}{cc}\text{Tr}\left(b\right)={\epsilon }_1^k\left(b\right),\text{for}\hspace{0.17em}b\in \text{End}\left(V^{\otimes k}\right)\text{.}& \text{(3.12)}\end{array}$$

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

$$\begin{array}{cc}\begin{array}{cccc}{\Phi }_{k-\frac{1}{2}}\left({\epsilon }_{\frac{1}{2}}\left(b\right)\right)& =& {\epsilon }_{\frac{1}{2}}\left({\Phi }_k\left(b\right)\right){\mid }_{V^{\otimes (k-1)}\otimes v_n},& \text{for}\hspace{0.17em}b\in ℂA_k\left(n\right),\\ {\Phi }_{k-1}\left({\epsilon }^{\frac{1}{2}}\left(b\right)\right)& =& \frac{1}{n}{\epsilon }^{\frac{1}{2}}\left({\Phi }_k\left(b\right)\right),& \text{for}\hspace{0.17em}b\in ℂA_{k-\frac{1}{2}}\left(n\right),\text{and}\\ {\Phi }_{k-1}\left({\epsilon }_1\left(b\right)\right)& =& {\epsilon }_1\left({\Phi }_k\left(b\right)\right),& \text{for}\hspace{0.17em}b\in ℂA_k\left(n\right),\end{array}& \text{(3.13)}\end{array}$$

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

$$\begin{array}{cc}\text{Tr}\left({\Phi }_k\left(b\right)\right)={\epsilon }_1^k\left({\Phi }_k\left(b\right)\right)={\Phi }_0\left({\epsilon }_1^k\left(b\right)\right)={\epsilon }_1^k\left(b\right)={\text{tr}}_k\left(b\right),& \text{(3.14)}\end{array}$$

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

$$\begin{array}{cc}\text{Tr}\left({\Phi }_{k-\frac{1}{2}}\left(b\right)\right)=\text{Tr}\left({\Phi }_k\left(b\right){\mid }_{V^{\otimes (k-1)}\otimes v_n}\right)=\frac{1}{n}\text{Tr}\left({\Phi }_k\left(b\right)\right)=\frac{1}{n}{\text{tr}}_k\left(b\right)=\frac{1}{n}{\text{tr}}_{k-\frac{1}{2}}\left(b\right)\text{.}& \text{(3.15)}\end{array}$$

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 $V=ℂ\text{-span}\{v_1,\dots ,v_n\}$ be the permutation representation of $S_n$ given in (3.5). Then

$$\begin{array}{cc}V\cong {\text{Ind}}_{S_{n-1}}^{S_n}{\text{Res}}_{S_{n-1}}^{S_n}\left(1_n\right)\text{.}& \text{(3.16)}\end{array}$$

More generally, for any $S_n\text{-module}$ $M,$

$$\begin{array}{cc}\begin{array}{ccc}{\text{Ind}}_{S_{n-1}}^{S_n}{\text{Res}}_{S_{n-1}}^{S_n}\left(M\right)& \cong & {\text{Ind}}_{S_{n-1}}^{S_n}({\text{Res}}_{S_{n-1}}^{S_n}\left(M\right)\otimes 1_{n-1})\\ & \cong & {\text{Ind}}_{S_{n-1}}^{S_n}({\text{Res}}_{S_{n-1}}^{S_n}\left(M\right)\otimes {\text{Res}}_{S_{n-1}}^{S_n}\left(1_n\right))\\ & \cong & M\otimes {\text{Ind}}_{S_{n-1}}^{S_n}{\text{Res}}_{S_{n-1}}^{S_n}\left(1_n\right)\cong M\otimes V,\end{array}& \text{(3.17)}\end{array}$$

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

$$\begin{array}{cc}\begin{array}{ccc}{\text{Ind}}_{S_{n-1}}^{S_n}({\text{Res}}_{S_{n-1}}^{S_n}\left(M\right)\otimes N)& \stackrel{\sim }{⟶}& M\otimes {\text{Ind}}_{S_{n-1}}^{S_n}N\\ g\otimes (m\otimes n)& ⟼& gm\otimes (g\otimes n),\end{array}& \text{(3.18)}\end{array}$$

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

$$\begin{array}{cc}{\left({\text{Ind}}_{S_{n-1}}^{S_n}{\text{Res}}_{S_{n-1}}^{S_n}\right)}^k\left(1\right)\cong V^{\otimes k}\text{and}{\text{Res}}_{S_{n-1}}^{S_n}{\left({\text{Ind}}_{S_{n-1}}^{S_n}{\text{Res}}_{S_{n-1}}^{S_n}\right)}^k\left(1\right)\cong V^{\otimes k}& \text{(3.19)}\end{array}$$

As $S_n\text{-modules,}$ and $S_{n-1}\text{-modules,}$ respectively.

If

$$\begin{array}{cc}\lambda =({\lambda }_1,{\lambda }_2,\dots ,{\lambda }_{\ell }),\text{define}{\lambda }_{>1}=({\lambda }_2,\dots ,{\lambda }_{\ell })& \text{(3.20)}\end{array}$$

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

$$\begin{array}{cc}\begin{array}{c}\text{vertices on level}\hspace{0.17em}k:{\hat{A}}_k\left(n\right)=\{\lambda ⊢n\mid k-\mid {\lambda }_{>1}\mid \in {\mathbb{Z}}_{>0}\},\\ \text{vertices on level}\hspace{0.17em}k+\frac{1}{2}:{\hat{A}}_{k+\frac{1}{2}}\left(n\right)=\{\lambda ⊢n-1\mid k-\mid {\lambda }_{>1}\mid \in {\mathbb{Z}}_{\ge 0}\},\text{and}\\ \text{an edge}\hspace{0.17em}\lambda \to \mu ,\hspace{0.17em}\text{if}\hspace{0.17em}\mu \in {\hat{A}}_{k+\frac{1}{2}}\left(n\right)\hspace{0.17em}\text{is obtained from}\hspace{0.17em}\lambda \in {\hat{A}}_k\left(n\right)\hspace{0.17em}\text{by removing a box,}\\ \text{an edge}\hspace{0.17em}\mu \to \lambda ,\hspace{0.17em}\text{if}\hspace{0.17em}\lambda \in {\hat{A}}_{k+1}\left(n\right)\hspace{0.17em}\text{is obtained from}\hspace{0.17em}\mu \in {\hat{A}}_{k+\frac{1}{2}}\left(n\right)\hspace{0.17em}\text{by adding a box.}\end{array}& \text{(3.21)}\end{array}$$

For example, if $n=5$ then the first few levels of $\hat{A}\left(n\right)$ 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 $n,k\in {\mathbb{Z}}_{\ge 0}\text{.}$ Let $S_n^{\lambda }$ denote the irreducible $S_n\text{-module}$ indexed by $\lambda \text{.}$

1. As $(ℂS_n,ℂA_k\left(n\right))\text{-bimodules,}$ $$V^{\otimes k}\cong \underset{\lambda \in {\hat{A}}_k\left(n\right)}{⨁}{S}_n^{\lambda }\otimes A_k^{\lambda }\left(n\right),$$ where the vector spaces $A_k^{\lambda }\left(n\right)$ are irreducible $ℂA_k\left(n\right)\text{-modules}$ and $$\text{dim}\left(A_k^{\lambda }\left(n\right)\right)=(\text{number of paths from}\hspace{0.17em}\left(n\right)\in {\hat{A}}_0\left(n\right)\hspace{0.17em}\text{to}\hspace{0.17em}\lambda \in {\hat{A}}_k\left(n\right)\hspace{0.17em}\text{in the graph}\hspace{0.17em}\hat{A}\left(n\right))\text{.}$$
2. As $(ℂS_{n-1},ℂA_{k+\frac{1}{2}}\left(n\right))\text{-bimodules,}$ $$V^{\otimes k}\cong \underset{\mu \in {\hat{A}}_{k+\frac{1}{2}}\left(n\right)}{⨁}{S}_{n-1}^{\mu }\otimes A_{k+\frac{1}{2}}^{\mu }\left(n\right),$$ where the vector spaces $A_{k+\frac{1}{2}}^{\mu }\left(n\right)$ are irreducible $ℂA_{k+\frac{1}{2}}\left(n\right)\text{-modules}$ and $$\text{dim}\left(A_{k+\frac{1}{2}}^{\mu }\left(n\right)\right)=(\text{number of paths from}\hspace{0.17em}\left(n\right)\in {\hat{A}}_0\left(n\right)\hspace{0.17em}\text{to}\hspace{0.17em}\mu \in {\hat{A}}_{k+\frac{1}{2}}\left(n\right)\hspace{0.17em}\text{in the graph}\hspace{0.17em}\hat{A}\left(n\right))\text{.}$$

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

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

$${\lambda }_{>1}=({\lambda }_2,\dots ,{\lambda }_{\ell }),\text{if}\lambda =({\lambda }_1,{\lambda }_2,\dots ,{\lambda }_{\ell }),$$

i.e., remove the first row of $\lambda$ to get ${\lambda }_{>1}\text{.}$ Then, for $n\ge 2k,$ the maps

$$\begin{array}{cc}\begin{array}{ccc}{\hat{A}}_k\left(n\right)& ⟷& {\hat{A}}_k\\ \lambda & ⟼& {\lambda }_{>1}\end{array}\text{are bijections}& \text{(3.23)}\end{array}$$

which provide an isomorphism between levels 0 to $n$ of the graphs $\hat{A}\left(n\right)$ and $\hat{A}\text{.}$

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

$${\text{tr}}_k=\sum _{\mu \in {\hat{A}}_k}{\text{tr}}^{\mu }\left(n\right){\chi }_{A_k\left(n\right)}^{\mu },\text{are}{\text{tr}}^{\mu }\left(n\right)=\left(\prod _{b\in \mu }\frac{1}{h\left(b\right)}\right)\prod _{j=1}^{\mid \mu \mid }(n-\mid \mu \mid -({\mu }_j-j))\text{.}$$

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

$$\begin{array}{ccc}{\text{tr}}_{k+\frac{1}{2}}& =& \sum _{\mu \in {\hat{A}}_{k+\frac{1}{2}}}{\text{tr}}_{\frac{1}{2}}^{\mu }\left(n\right){\chi }_{A_{k+\frac{1}{2}}\left(n\right)}^{\mu },\text{are}\\ {\text{tr}}_{\frac{1}{2}}^{\mu }\left(n\right)& =& \left(\prod _{b\in \mu }\frac{1}{h\left(b\right)}\right)·n·\prod _{j=1}^{\mid \mu \mid }(n-\mid \mu \mid -({\mu }_j-j))\text{.}\end{array}$$

		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 $$\begin{array}{ccc}\text{vertical edge label for row}\hspace{0.17em}i& =& \text{(number of horizontal steps)}+\text{(number of vertical steps)}\\ & =& ({\lambda }_1-{\lambda }_i)+(i-1)\\ & =& ({\lambda }_1-1)-({\lambda }_i-i),\text{and the}\end{array}$$ $$\begin{array}{ccc}\text{horizontal edge label for row}\hspace{0.17em}j& =& \text{(number of horizontal steps)}+\text{(number of vertical steps)}\\ & =& ({\lambda }_1-j+1)+({\lambda }_j^{\prime }-1)\\ & =& ({\lambda }_1-1)+({\lambda }_j^{\prime }-j)+1\text{.}\end{array}$$ Hence $$\begin{array}{ccc}\{1,2,\dots ,n\}& =& \{({\lambda }_1-1)-({\lambda }_j^{\prime }-j)+1\mid 1\le j\le {\lambda }_1\}\bigsqcup \{({\lambda }_1-1)-({\lambda }_i-i)\mid 2\le i\le n-{\lambda }_1+1\}\\ & =& \{h\left(b\right)\mid b\hspace{0.17em}\text{is in row 1 of}\hspace{0.17em}\lambda \}\bigsqcup \{({\lambda }_1-1)-({\lambda }_i-i)\mid 2\le i\le n-{\lambda }_1+1\}\end{array}$$ For example, if $\lambda =(10,7,3,3,1)⊢24,$ 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 ${\lambda }_1=n-\mid {\lambda }_{>1}\mid ,$ $$\begin{array}{cc}\text{dim}\left(S_n^{\lambda }\right)=\frac{n!}{\prod _{b\in \lambda }h\left(b\right)}=\left(\prod _{b\in {\lambda }_{>1}}\frac{1}{h\left(b\right)}\right)\prod _{i=2}^{\mid {\lambda }_{>1}\mid +1}(n-\mid {\lambda }_{>1}\mid -({\lambda }_i-(i-1)))\text{.}& \text{(3.25)}\end{array}$$ Let $n\in {\mathbb{Z}}_{>0}$ 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, $$\text{Tr}(b,V^{\otimes k})=\sum _{\lambda \in {\hat{A}}_k\left(n\right)}{\chi }_{S_n}^{\lambda }\left(1\right){\chi }_{A_k\left(n\right)}\left(b\right)=\sum _{\lambda \in {\hat{A}}_k\left(n\right)}\text{dim}\left(S_n^{\lambda }\right){\chi }_{A_k\left(n\right)}\left(b\right),\text{for}\hspace{0.17em}b\in ℂA_k\left(n\right)\text{.}$$ Thus the equality in (3.25) and the bijection in (3.23) provide the expansion of ${\text{tr}}_k$ for all $n\in {\mathbb{Z}}_{>0}$ such that $n\ge 2k\text{.}$ The statement for all $n\in ℂ$ such that $ℂA_k\left(n\right)$ 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,

$$\begin{array}{cc}\begin{array}{ccc}\{\text{roots of}\hspace{0.17em}{\text{tr}}_{\frac{1}{2}}^{\mu }\left(n\right)\mid \mu \in {\hat{A}}_{\frac{1}{2}}\}& =& \left\{0\right\},\\ \{\text{roots of}\hspace{0.17em}{\text{tr}}_1^{\mu }\left(n\right)\mid \mu \in {\hat{A}}_1\}& =& \left\{1\right\},\\ \{\text{roots of}\hspace{0.17em}{\text{tr}}_{1\frac{1}{2}}^{\mu }\left(n\right)\mid \mu \in {\hat{A}}_{1\frac{1}{2}}\}& =& \{0,2\},\text{and}\\ \{\text{roots of}\hspace{0.17em}{\text{tr}}_k^{\mu }\left(n\right)\mid \mu \in {\hat{A}}_k\}& =& \{0,1,\dots ,2k-1\},\text{for}\hspace{0.17em}k\in \frac{1}{2}{\mathbb{Z}}_{\ge 0},k\ge 2\text{.}\end{array}& \text{(3.26)}\end{array}$$

For example, the first few values of ${\text{tr}}^{\mu }$ and ${\text{tr}}_{\frac{1}{2}}^{\mu }$ are

$$\begin{array}{cc}{\text{tr}}^{\varnothing }\left(n\right)=1,& {\text{tr}}_{\frac{1}{2}}^{\varnothing }\left(n\right)=n,\\ {\text{tr}}^{ }\left(n\right)=n-1,& {\text{tr}}_{\frac{1}{2}}^{ }\left(n\right)=n(n-2),\\ {\text{tr}}^{ }\left(n\right)=\frac{1}{2}n(n-3),& {\text{tr}}_{\frac{1}{2}}^{ }\left(n\right)=\frac{1}{2}n(n-1)(n-4),\\ {\text{tr}}^{ }\left(n\right)=\frac{1}{2}(n-1)(n-2),& {\text{tr}}_{\frac{1}{2}}^{ }\left(n\right)=\frac{1}{2}n(n-2)(n-3),\\ {\text{tr}}^{ }\left(n\right)=\frac{1}{6}n(n-1)(n-5),& {\text{tr}}_{\frac{1}{2}}^{ }\left(n\right)=\frac{1}{6}n(n-1)(n-2)(n-6),\\ {\text{tr}}^{ }\left(n\right)=\frac{1}{6}n(n-2)(n-4),& {\text{tr}}_{\frac{1}{2}}^{ }\left(n\right)=\frac{1}{6}n(n-1)(n-3)(n-5),\\ {\text{tr}}^{ }\left(n\right)=\frac{1}{6}(n-1)(n-2)(n-3),& {\text{tr}}_{\frac{1}{2}}^{ }\left(n\right)=\frac{1}{6}n(n-2)(n-3)(n-4),\end{array}$$

Theorem 3.27. Let $n\in {\mathbb{Z}}_{\ge 2}$ and $k\in \frac{1}{2}{\mathbb{Z}}_{\ge 0}\text{.}$ Then $ℂA_k\left(n\right)$ 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 $ℂA_k\left(n\right)$ is semisimple if $n\ge 2k-1\text{.}$ Suppose $n$ is even. Then Theorems 2.26(a) and 2.26(b) imply that $$ℂA_{\frac{n}{2}+\frac{1}{2}}\left(n\right)\hspace{0.17em}\text{is semisimple}\text{and}ℂA_{\frac{n}{2}+1}\left(n\right)\hspace{0.17em}\text{is not semisimple,}$$ since $(n/2)\in {\hat{A}}_{\frac{n}{2}+\frac{1}{2}}$ and ${\text{tr}}_{\frac{1}{2}}^{(n/2)}\left(n\right)=0\text{.}$ Since $(n/2)\in {\hat{A}}_{\frac{n}{2}+1}\left(n\right),$ the $A_{\frac{n}{2}+1}\left(n\right)\text{-module}$ $A_{\frac{n}{2}+1}^{(n/2)}\left(n\right)\ne 0\text{.}$ Since the path $(\varnothing ,\dots ,(n/2),(n/2),(n/2))\in {\hat{A}}_{\frac{n}{2}+1}^{(n/2)}$ does not correspond to an element of ${\hat{A}}_{\frac{n}{2}+1}^{(n/2)}\left(n\right),$ $$\text{Card}\left({\hat{A}}_{\frac{n}{2}+1}^{(n/2)}\right)\ne \text{Card}\left({\hat{A}}_{\frac{n}{2}+1}^{(n/2)}\left(n\right)\right)\text{.}$$ Thus, Tits deformation theorem (Theorem 5.13) implies that $ℂA_{\frac{n}{2}+1}\left(n\right)$ cannot be semisimple. Now it follows from Theorem 2.26(c) that $ℂA_k\left(n\right)$ 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 $$ℂA_{\frac{n}{2}+\frac{1}{2}}\left(n\right)\hspace{0.17em}\text{is semisimple}\text{and}ℂA_{\frac{n}{2}+1}\left(n\right)\hspace{0.17em}\text{is not semisimple,}$$ since $(n/2)\in {\hat{A}}_{\frac{n}{2}+\frac{1}{2}}$ and ${\text{tr}}^{(n/2)}\left(n\right)=0\text{.}$ Since $(\frac{n}{2}-\frac{1}{2})\in {\hat{A}}_{\frac{n}{2}+1}\left(n\right),$ the $A_{\frac{n}{2}+1}\left(n\right)\text{-module}$ $A_{\frac{n}{2}+1}^{(\frac{n}{2}-\frac{1}{2})}\left(n\right)\ne 0\text{.}$ Since the path $(\varnothing ,\dots ,(\frac{n}{2}-\frac{1}{2}),(\frac{n}{2}+\frac{1}{2}),(\frac{n}{2}-\frac{1}{2}))\in {\hat{A}}_{\frac{n}{2}+1}^{(\frac{n}{2}-\frac{1}{2})}$ does not correspond to an element of ${\hat{A}}_{\frac{n}{2}+1}^{(\frac{n}{2}-\frac{1}{2})}\left(n\right),$ and since $$\text{Card}\left({\hat{A}}_{\frac{n}{2}+1}^{(\frac{n}{2}-\frac{1}{2})}\right)\ne \text{Card}\left({\hat{A}}_{\frac{n}{2}+1}^{(\frac{n}{2}-\frac{1}{2})}\left(n\right)\right),$$ the Tits deformation theorem implies that $ℂA_{\frac{n}{2}+1}\left(n\right)$ is not semisimple. Now it follows from Theorem 2.26(c) that $ℂA_k\left(n\right)$ is not semisimple for $k\ge \frac{n}{2}+\frac{1}{2}\text{.}$ $\square$

Murphy elements for $ℂA_k\left(n\right)$

Let ${\kappa }_n$ be the element of $ℂS_n$ given by

$$\begin{array}{cc}{\kappa }_n=\sum _{1\le \ell <m\le n}{s}_{\ell m},& \text{(3.28)}\end{array}$$

where $s_{\ell m}$ is the transposition in $S_n$ which switches $\ell$ and $m\text{.}$ Let $S\subseteq \{1,2,\dots ,k\}$ and let $I\subseteq S\cup S\prime \text{.}$ Define $b_S,d_I\in A_k$ by

$$\begin{array}{cc}{b}_S=\{S\cup S\prime ,{\{\ell ,\ell \prime \}}_{\ell \notin S}\}\text{and}{d}_{I\subseteq S}=\{I,I^c,{\{\ell ,\ell \prime \}}_{\ell \notin S}\}& \text{(3.29)}\end{array}$$

For example, in $A_9,$ if $S=\{2,4,5,8\}$ and $I=\{2,4,4\prime ,5,8\}$ then

$$\begin{array}{ccccc}{b}_S=& & \text{and}{d}_{I\subseteq S}=& & \text{.}\end{array}$$

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

$$\begin{array}{cc}{p}_s=\sum _I\frac{1}{2}{(-1)}^{\#(\{\ell ,\ell \prime \}\subseteq I)+\#(\{\ell ,\ell \prime \}\subseteq I^c)}{d}_I,& \text{(3.30)}\end{array}$$

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

$$\begin{array}{cc}{\stackrel{\sim }{p}}_s=\sum _I\frac{1}{2}{(-1)}^{\#(\{\ell ,\ell \prime \}\subseteq I)+\#(\{\ell ,\ell \prime \}\subseteq I^c)}{d}_I,& \text{(3.31)}\end{array}$$

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

Let $Z_1=1$ and, for $k\in Z_{>1},$ let

$$\begin{array}{cc}{Z}_k=\left(\genfrac{}{}{0ex}{}{k}{2}\right)+\sum _{\underset{\mid S\mid \ge 1}{S\subseteq \{1,\dots ,k\}}}{p}_S+\sum _{\underset{\mid S\mid \ge 2}{S\subseteq \{1,\dots ,k\}}}(n-k+\mid S\mid ){(-1)}^{\mid S\mid }{b}_S\text{.}& \text{(3.32)}\end{array}$$

View $Z_k\in ℂA_k\subseteq ℂA_{k+\frac{1}{2}}$ using the embedding in (2.2), and define $Z_{\frac{1}{2}}=1$ and

$$\begin{array}{cc}{Z}_{k+\frac{1}{2}}=k+Z_k+\sum _{\underset{k+1\in S}{\mid S\mid \ge 2}}{\stackrel{\sim }{p}}_S(n-(k+1)+\mid S\mid ){(-1)}^{\mid S\mid }{b}_S,& \text{(3.33)}\end{array}$$

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

$$\begin{array}{cc}{M}_{\frac{1}{2}}=1,\text{and}{M}_k=Z_k-Z_{k-\frac{1}{2}},\text{for}\hspace{0.17em}k\in \frac{1}{2}{\mathbb{Z}}_{>0}\text{.}& \text{(3.34)}\end{array}$$

For example, the first few $Z_k$ are

$$\begin{array}{c}{Z}_0=1,Z_{\frac{1}{2}}=1,Z_1=\underset{\underset{p_{\left\{1\right\}}}{⏟}}{\begin{array}{c} \end{array}},\\ Z_{1\frac{1}{2}}=\begin{array}{c} \end{array}+\underset{\underset{Z_1}{⏟}}{\begin{array}{c} \end{array}}-\underset{\underset{{\stackrel{\sim }{p}}_{\{1,2\}}}{⏟}}{\begin{array}{c} \end{array}-\begin{array}{c} \end{array}}+n\underset{\underset{b_{\{1,2\}}}{⏟}}{\begin{array}{c} \end{array}},\text{and}\\ Z_2=\begin{array}{c} \end{array}+\underset{\underset{p_{\left\{1\right\}}}{⏟}}{\begin{array}{c} \end{array}}+\underset{\underset{p_{\left\{2\right\}}}{⏟}}{\begin{array}{c} \end{array}}-\underset{\underset{p_{\{1,2\}}}{⏟}}{\begin{array}{c} \end{array}-\begin{array}{c} \end{array}-\begin{array}{c} \end{array}-\begin{array}{c} \end{array}+\begin{array}{c} \end{array}+\begin{array}{c} \end{array}}+n\underset{\underset{b_{\{1,2\}}}{⏟}}{\begin{array}{c} \end{array}},\end{array}$$

and the first few $M_k$ are

$$\begin{array}{ccc}{M}_0& =& 1,M_{\frac{1}{2}}=1,M_1=\begin{array}{c} \end{array}-\begin{array}{c} \end{array},M_{1\frac{1}{2}}=\begin{array}{c} \end{array}-\begin{array}{c} \end{array}-\begin{array}{c} \end{array}+n\begin{array}{c} \end{array},\\ M_2& =& \begin{array}{c} \end{array}-\begin{array}{c} \end{array}-\begin{array}{c} \end{array}+\begin{array}{c} \end{array}+\begin{array}{c} \end{array},\text{and}\\ M_{2\frac{1}{2}}& =& 2\begin{array}{c} \end{array}+\begin{array}{c} \end{array}+\begin{array}{c} \end{array}+\begin{array}{c} \end{array}+\begin{array}{c} \end{array}+(n-1)\begin{array}{c} \end{array}+(n-1)\begin{array}{c} \end{array}\\ & & +\begin{array}{c} \end{array}+\begin{array}{c} \end{array}+\begin{array}{c} \end{array}+\begin{array}{c} \end{array}-\begin{array}{c} \end{array}-\begin{array}{c} \end{array}-\begin{array}{c} \end{array}-\begin{array}{c} \end{array}\\ & & +\begin{array}{c} \end{array}+\begin{array}{c} \end{array}+\begin{array}{c} \end{array}+\begin{array}{c} \end{array}-n\begin{array}{c} \end{array}\end{array}$$

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

Theorem 3.35.

1. For $n\in {\mathbb{Z}}_{\ge 0},{\kappa }_n$ is a central element of $ℂS_n\text{.}$ If $\lambda$ is a partition with $n$ boxes and $S_n^{\lambda }$ is the irreducible $S_n\text{-module}$ indexed by the partition $\lambda ,$ $${\kappa }_n=\sum _{b\in \lambda }c\left(n\right),\text{as operators on}\hspace{0.17em}{S}_n^{\lambda }\text{.}$$
2. Let $n,k\in {\mathbb{Z}}_{\ge 0}\text{.}$ Then, as operators on $V^{\otimes k},$ where $\text{dim}\left(V\right)=n,$ $$Z_k={\kappa }_n-\left(\genfrac{}{}{0ex}{}{n}{2}\right)+kn,\text{and}{Z}_{k+\frac{1}{2}}={\kappa }_{n-1}-\left(\genfrac{}{}{0ex}{}{n}{2}\right)+(k+1)n-1\text{.}$$
3. Let $n\in C,k\in {\mathbb{Z}}_{\ge 0}\text{.}$ Then $Z_k$ is a central element of $ℂA_k\left(n\right),$ and, if $n\in ℂ$ is such that $ℂA_k\left(n\right)$ is semisimple and $\lambda ⊢n$ with $\mid {\lambda }_{>1}\mid \ge k$ boxes, then $$Z_k=kn-\left(\genfrac{}{}{0ex}{}{n}{2}\right)+\sum _{b\in \lambda }c\left(b\right),\text{as operators on}\hspace{0.17em}{A}_k^{\lambda },$$ where $A_k^{\lambda }$ is the irreducible $ℂA_k\left(n\right)\text{-module}$ indexed by the partition $\lambda \text{.}$ Furthermore, $Z_{k+\frac{1}{2}}$ is a central element of $ℂA_{k+\frac{1}{2}}\left(n\right),$ and, if $n$ is such that $ℂA_{k+\frac{1}{2}}\left(n\right)$ is semisimple and $\lambda ⊢n$ is a partition with $\mid {\lambda }_{>1}\mid \le k$ boxes, then $$Z_{k+\frac{1}{2}}=kn+n-1-\left(\genfrac{}{}{0ex}{}{n}{2}\right)+\sum _{b\in \lambda }c\left(b\right),\text{as operators on}\hspace{0.17em}{A}_{k+\frac{1}{2}}^{\lambda },$$ where $A_{k+\frac{1}{2}}^{\lambda }$ is the irreducible $ℂA_{k+\frac{1}{2}}\left(n\right)\text{-module}$ 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 $ℂS_n\text{.}$ 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, $ℂA_k\left(n\right)\cong {\text{End}}_{S_n}\left(V^{\otimes k}\right)$ if $n\ge 2k\text{.}$ Thus, by Theorem 3.22, if $n\ge 2k$ then $Z_k$ acts on the irreducible $ℂA_k\left(n\right)\text{-module}$ $A_k^{\lambda }\left(n\right)$ by the constant given in the statement. This means that $Z_k$ is a central element of $ℂA_k\left(n\right)$ 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 $ℂA_k\left(n\right)$ is semisimple let ${\chi }_{ℂA_k\left(n\right)}^{\lambda }$ be the irreducible characters. Then $Z_k$ acts on $A_k^{\lambda }\left(n\right)$ by the constant ${\chi }_{ℂA_k\left(n\right)}^{\lambda }\left(Z_k\right)/\text{dim}\left(A_k^{\lambda }\left(n\right)\right)\text{.}$ 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 $ℂA_{k+\frac{1}{2}},S_{n-1},$ and the second statement in part (b). (b) Let $s_{ii}=1$ so that $$2{\kappa }_n+n=n+2\sum _{1\le i<j\le n}{s}_{ij}=\sum _{i=1}^n{s}_{ii}+\sum _{1\le i<j\le n}(s_{ij}+s_{ji})=\sum _{i=j}{s}_{ij}+\sum _{i\ne j}{s}_{ij}=\sum _{i,j=1}^n{s}_{ij}\text{.}$$ Then $$\begin{array}{ccc}\left(2{\kappa }_{n}n\right)(v_{i_1}\otimes \dots \otimes v_{i_k})& =& \left(\sum _{i,j=1}^n{s}_{ij}\right)(v_{i_1}\otimes \dots \otimes v_{i_k})\\ & =& \sum _{i,j=1}^n{s}_{ij}{v}_{i_1}\otimes \dots \otimes s_{ij}{v}_{i_k}\\ & =& \sum _{i,j=1}^n(1-E_{ii}-E_{jj}+E_{ij}+E_{ji})v_{i_1}\\ & & \otimes \dots \otimes (1-E_{ii}-E_{jj}+E_{ij}+E_{ji})v_{i_k}\end{array}$$ and expanding this sum gives that $(2{\kappa }_n+n)(v_{i_1}\otimes \dots \otimes v_{i_k})$ is equal to $$\begin{array}{cc}\begin{array}{c}\sum _{S\subseteq \{1,\dots ,k\}}\sum _{i_{1\prime },\dots ,i_{k\prime }}\sum _{i,j=1}^n\left(\prod _{\ell \in S^c}{\delta }_{i_{\ell }{i}_{\ell \prime }}\right)\\ \times \sum _{I\subseteq S\cup S\prime }{(-1)}^{\#(\{\ell ,\ell \prime \}\subseteq I)+\#(\{\ell ,\ell \prime \}\subseteq I^c)}\left(\prod _{\ell \in I}{\delta }_{i_{\ell }i}\right)\\ \times \left(\prod _{\ell \in I^c}{\delta }_{i_{\ell }j}\right)(v_{i_{1\prime }}\otimes \dots \otimes v_{i_{k\prime }})\end{array}& \text{(3.36)}\end{array}$$ where $S^c\subseteq \{1,\dots ,k\}$ 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 $$\sum _{i,j=1}^n\sum _{i_{1\prime },\dots ,i_{k\prime }}\left(\prod _{\ell \in \{1,\dots ,k\}}{\delta }_{i_{\ell }{i}_{\ell \prime }}\right)(v_{i_{1\prime }}\otimes \dots \otimes v_{i_{k\prime }})=n^2(v_{i_1}\otimes \dots \otimes v_{i_k})\text{.}$$ 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 $$\sum _{i_{1\prime },\dots ,i_{k\prime }}n\sum _{i=1}^n\left(\prod _{\ell \in S^c}{\delta }_{i_{\ell }{i}_{\ell \prime }}\right){(-1)}^{\mid S\mid }\left(\prod _{\ell \in S\cup S\prime }{\delta }_{i_{\ell }i}\right)(v_{i_{1\prime }}\otimes \dots \otimes v_{i_{k\prime }})=n{(-1)}^{\mid S\mid }{b}_S(v_{i_1}\otimes \dots \otimes v_{i_k})\text{.}$$ 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 $I=\{\ell ,\ell \prime \}$ is equal to $$\sum _{i_{1\prime },\dots ,i_{k\prime }}\sum _{i,j=1}^n\left(\prod _{r\in S^c}{\delta }_{i_r{i}_{r\prime }}\right){(-1)}^{\mid S\mid }{\delta }_{i_{\ell }i}{\delta }_{i_{\ell \prime }i}\left(\prod _{r\ne \ell }{\delta }_{i_{r\prime }j}\right)(v_{i_{1\prime }}\otimes \dots \otimes v_{i_{k\prime }})={(-1)}^{\mid S\mid }{b}_{S-\left\{\ell \right\}}(v_{i_1}\otimes \dots \otimes v_{i_k})\text{.}$$ and there is a corresponding contribution from $I={\{\ell ,,\ell \prime \}}^c\text{.}$ The remaining terms can be written as $$\begin{array}{c}\sum _{i_{1\prime },\dots ,i_{k\prime }}\sum _{i,j=1}^n\left(\prod _{\ell \in S^c}{\delta }_{i_{\ell }{i}_{\ell \prime }}\right)\sum _{I\subseteq S\cup S\prime }{(-1)}^{\#(\{\ell ,\ell \prime \}\subseteq I)+\#(\{\ell ,\ell \prime \}\subseteq I^c)}\\ \times \left(\prod _{\ell \in I}{\delta }_{i_{\ell }i}\right)\left(\prod _{\ell \in I^c}{\delta }_{i_{\ell }j}\right)(v_{i_{1\prime }}\otimes v_{i_{k\prime }})=2p_S(v_{i_1}\otimes \dots \otimes v_{i_k})\text{.}\end{array}$$ 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 $$\sum _{\mid S\mid =0}{n}^2+\sum _{\mid S\mid =1}(2n{(-1)}^1{b}_S+2p_S)+\sum _{\mid S\mid =2}(2n{(-1)}^2{b}_S+2p_S+\sum _{\ell \in S}{(-1)}^{2}2b_{S-\left\{\ell \right\}})+\sum _{\mid S\mid >2}(2n{(-1)}^{\mid S\mid }{b}_S+2p_S+\sum _{\ell \in S}{(-1)}^{\mid S\mid }2b_{S-\left\{\ell \right\}})$$ 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 $$n^2=\sum _{\mid S\mid =1}(-2n+2p_S)+\sum _{\mid S\mid =2}(2n{b}_S+2+2p_S)+\sum _{\mid S\mid >2}({(-1)}^{\mid S\mid }2n{b}_S+2p_S+\sum _{\ell \in S}{(-1)}^{\mid S\mid }2b_{S-\left\{\ell \right\}})=n^2-2nk+2\left(\genfrac{}{}{0ex}{}{k}{2}\right)+\sum _{\mid S\mid \ge 1}2p_S+\sum _{\mid S\mid \ge 2}2(n-k+\mid S\mid ){(-1)}^{\mid S\mid }{b}_S$$ acts on $v_{i_1}\otimes \dots \otimes v_{i_k},$ and so $Z_k={\kappa }_n+(n-n^2+2nk)/2$ as operators on $V^{\otimes k}\text{.}$ This proves the first statement. For the second statement, since $(1-{\delta }_{in})(1-{\delta }_{jn})=\{\begin{array}{cc}0,& \text{if}\hspace{0.17em}i=n\hspace{0.17em}\text{or}\hspace{0.17em}j=n,\\ 1,& \text{otherwise,}\end{array}$ $$\begin{array}{ccc}\multicolumn{3}{c}{(2{\kappa }_{n-1}+(n-1))(v_{i_1}\otimes \dots \otimes v_{i_k}\otimes v_n)}\\ & =& \left(\sum _{i,j=1}^{n-1}{s}_{ij}\right)(v_{i_1}\otimes \dots \otimes v_{i_k}\otimes v_n)\\ & =& \left(\sum _{i,j=1}^n{s}_{ij}(1-{\delta }_{in})(1-{\delta }_{jn})\right)(v_{i_1}\otimes \dots \otimes v_{i_k}\otimes v_n)\\ & =& \sum _{i,j=1}^n{s}_{ij}{v}_{i_1}\otimes \dots \otimes s_{ij}{v}_{i_k}\otimes (1-{\delta }_{in})(1-{\delta }_{jn})v_n,\\ & =& \sum _{i,j=1}^n(1-E_{ii}-E_{jj}+E_{ij}+E_{ji})v_{i_1}\otimes \dots \otimes (1-E_{ii}-E_{jj}+E_{ij}+E_{ji})v_{i_k}\\ & & \otimes (1-E_{ii}-E_{jj}+E_{ii}{E}_{jj})v_n\\ & =& \left(\sum _{i,j}{s}_{ij}\right)(v_{i_1}\otimes \dots \otimes v_{i_k})\otimes v_n+\sum _{i,j=1}^n(1-E_{ii}-E_{jj}+E_{ij}+E_{ji})v_{i_1}\\ & & \otimes \dots \otimes (1-E_{ii}-E_{jj}+E_{ij}+E_{ji})v_{i_k}\otimes (-E_{ii}-E_{jj})v_n\\ & & +\sum _{i,j=1}^n(1-E_{ii}-E_{jj}+E_{ij}+E_{ji})v_{i_1}\otimes \dots \otimes (1-E_{ii}-E_{jj}+E_{ij}+E_{ji})v_{i_k}\otimes E_{ii}{E}_{jj}{v}_n\text{.}\end{array}$$ The first sum is known to equal $(2{\kappa }_n+n)(v_{i_1}\otimes \dots \otimes v_{i_k})$ 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 $$\sum _{\underset{k+1\in S}{S\subseteq \{1,\dots ,k+1\}}}\sum _{i_{1\prime },\dots ,i_{k\prime }}\sum _{i,j=1}^n\left(\prod _{\ell \in S}{\delta }_{i_{\ell },i_{\ell \prime }}\right)\times \sum _I{(-1)}^{\#(\{\ell ,\ell \prime \}\subseteq I)+\#(\{\ell ,\ell \prime \}\subseteq I^c)}\left(\prod _{\ell \in I}{\delta }_{i_{\ell }i}\right)\left(\prod _{\ell \in I^c}{\delta }_{i_{\ell }j}\right)(v_{i_{1\prime }}\otimes \dots \otimes v_{i_{k\prime }})$$ where the inner sum is over all $I\subseteq \{1,\dots ,k+1\}$ such that $\{k+1,(k+1)\prime \}\subseteq I$ or $\{k+1,(k+1)\prime \}\subseteq I^c\text{.}$ 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 $I=\{\ell ,\ell \prime \}$ or $I={\{\ell ,\ell \prime \}}^c,$ 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 $$\sum _{\underset{k+1\in S}{\mid S\mid =1}}-2n+\sum _{\underset{k+1\in S}{\mid S\mid =2}}(2n{b}_S+2p_S+2)+\sum _{\underset{k+1\in S}{\mid S\mid >2}}(2n{(-1)}^{\mid S\mid }{b}_S+2p_S+2\sum _{\ell \in S}{(-1)}^{\mid S\mid }{b}_{S-\left\{\ell \right\}})\text{.}$$ Combining this with the terms $(2{\kappa }_n+n)(v_{i_1}\otimes \dots \otimes v_{i_k})\otimes v_n$ and $1\otimes (v_{i_1}\otimes \dots \otimes v_{i_k}\otimes v_n)$ gives that $2{\kappa }_{n-1}+(n-1)$ acts on $v_{i_1}\otimes \dots \otimes v_{i_k}$ as $$(2{\kappa }_n+n)+1-2n+2k+\sum _{\underset{k+1\in S}{\mid S\mid \ge 2}}2{\stackrel{\sim }{p}}_S+2(n-(k+1)+\mid S\mid ){(-1)}^{\mid S\mid }{b}_S\text{.}$$ Thus ${\kappa }_{n-1}-{\kappa }_n$ acts on $v_{i_1}\otimes \dots \otimes v_{i_k}$ as $$\frac{1}{2}(n-(n-1)+1-2n+2k+\sum _{\underset{k+1\in S}{\mid S\mid \ge 2}}2{\stackrel{\sim }{p}}_S+2(n-(k+1)+\mid S\mid ){(-1)}^{\mid S\mid }{b}_S),$$ so, as operators on $V^{\otimes k},$ we have $Z_{k+\frac{1}{2}}=k+Z_k+({\kappa }_{n-1}-{\kappa }_n)-1+n-k=Z_k+({\kappa }_{n-1}-{\kappa }_n)+n-1\text{.}$ By the first statement in part (c) of this theorem we get $Z_{k+\frac{1}{2}}=(\kappa -\left(\frac{n}{2}\right)+kn)+({\kappa }_{n-1}-{\kappa }_n)+n-1={\kappa }_{n-1}-\left(\frac{n}{2}\right)+kn+n-1\text{.}$ $\square$

Theorem 3.37. Let $k\in \frac{1}{2}{\mathbb{Z}}_{\ge 0}$ 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 $ℂA_k\left(n\right)\text{.}$
2. Assume that $ℂA_k\left(n\right)$ is semisimple. Let $\mu \in {\hat{A}}_k$ so that $\mu$ is a partition with $\le k$ boxes, and let $A_k^{\mu }\left(n\right)$ be the irreducible $ℂA_k\left(n\right)\text{-module}$ indexed by $\mu \text{.}$ Then there is a unique, up to multiplication by constants, basis $\{v_T\mid T\in {\hat{A}}_k^{\mu }\}$ of $A_k^{\mu }\left(n\right)$ such that, for all $T=(T^{\left(0\right)},T^{\left(\frac{1}{2}\right)},\dots ,T^{\left(k\right)})\in {\hat{A}}_k^{\mu },$ and $\ell \in {\mathbb{Z}}_{\ge 0}$ such that $\ell \le k,$ $$M_{\ell }{v}_T=\{\begin{array}{cc}c(T^{\left(\ell \right)}/T^{(\ell -\frac{1}{2})})v_T,& \text{if}\hspace{0.17em}{T}^{\left(\ell \right)}/T^{(\ell -\frac{1}{2})}=▫,\\ (n-\mid T^{\left(\ell \right)}\mid )v_T,& \text{if}\hspace{0.17em}{T}^{\left(\ell \right)}=T^{(\ell -\frac{1}{2})},\end{array}$$ and $$M_{\ell +\frac{1}{2}}{v}_T=\{\begin{array}{cc}(n-c(T^{\left(\ell \right)}/T^{(\ell +\frac{1}{2})}))v_T,& \text{if}\hspace{0.17em}{T}^{\left(\ell \right)}/T^{(\ell +\frac{1}{2})}=▫,\\ \mid T^{\left(\ell \right)}\mid v_T,& \text{if}\hspace{0.17em}{T}^{\left(\ell \right)}=T^{(\ell +\frac{1}{2})},\end{array}$$ where $\lambda /\mu$ denotes the box where $\lambda$ and $\mu$ differ.

		Proof.
	(a) View $Z_0,Z_{\frac{1}{2}},\dots ,Z_k\in ℂA_k\text{.}$ Then $Z_k\in Z\left(ℂA_k\right),$ 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 $ℂA_k\text{.}$ (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 ${\text{Res}}_{ℂA_{k-\frac{1}{2}}\left(n\right)}^{ℂA_k\left(n\right)}\left(A_k^{\lambda }\left(n\right)\right)\text{.}$ The branching rules for this restriction are multiplicity free, meaning that each $ℂA_{k-\frac{1}{2}}\left(n\right)\text{-irreducible}$ that shows up in $A_k^{\lambda }\left(n\right)$ does so exactly once. By induction, we can choose a basis for each $ℂA_{k-\frac{1}{2}}\left(n\right)\text{-irreducible,}$ and the union of these bases forms a basis for $A_k^{\lambda }\left(n\right)\text{.}$ For $\ell <k,$ $M_{\ell }\in ℂA_{k-\frac{1}{2}}\left(n\right),$ 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 $\gamma ⊢(n-1)$ such that ${\lambda }_{>1}=T^{\left(k\right)}$ and ${\gamma }_{>1}=T^{(k-\frac{1}{2})}\text{.}$ Then by Theorem 3.35(c), $M_k=Z_k-Z_{k-\frac{1}{2}}$ acts on $v_T$ by the constant $$(\sum _{b\in \lambda }c\left(b\right)-\left(\genfrac{}{}{0ex}{}{n}{2}\right)+kn)-(\sum _{b\in \gamma }c\left(b\right)-\left(\genfrac{}{}{0ex}{}{n}{2}\right)+kn-1)=c(\lambda /\gamma )+1,$$ 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 $$(\sum _{b\in \gamma }c\left(b\right)-\left(\genfrac{}{}{0ex}{}{n}{2}\right)+kn+n-1)-(\sum _{b\in \lambda }c\left(b\right)-\left(\genfrac{}{}{0ex}{}{n}{2}\right)+kn)=-c(\lambda /\gamma )+n-1\text{.}$$ The result now follows from (3.23) and the observation that $$c(\lambda /\gamma )=\{\begin{array}{cc}c(T^{\left(k\right)}/T^{(k-\frac{1}{2})})-1,& \text{if}\hspace{0.17em}{T}^{\left(k\right)}=T^{(k+\frac{1}{2})}+▫,\\ n-\mid T^{\left(k\right)}\mid -1,& \text{if}\hspace{0.17em}{T}^{\left(k\right)}=T^{(k+\frac{1}{2})}\text{.}\end{array}$$ $\square$

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).

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