The symmetric group and Brauer algebras

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

Department of Mathematics
University of Wisconsin, Madison
Madison, WI 53706 USA
ram@math.wisc.edu

Last updates: 20 June 2010

The symmetric group $ℂ S_{k}$ and the Brauer algebra

If $λ \in {\hat{GL}}_{n}$ then $L_{{𝔤 𝔩}_{n}} (λ) \otimes V ≅ \underset{μ / λ = □}{\otimes} L_{{𝔤 𝔩}_{n}} (μ) as {𝔤 𝔩}_{n} (ℂ) -modules,$ where the sum is over $μ \in {\hat{𝔤 𝔩}}_{n}$ that are obtained from $λ$ by adding a box. The Young lattice is the graph $\hat{S}$ given by setting $\begin{array}{rcl} vertices on level k : & {\hat{S}}_{k} = \{partitions λ with k boxes\}, & and \\ a labeled edge λ \overset{c (λ / μ)}{\to} μ, λ \in {\hat{S}}_{k}, μ \in {\hat{S}}_{k + 1} & if μ is obtained from λ by adding a box. \end{array}$ It encodes the decompositions in (????).

Define elements $m_{1}, \dots, m_{k} \in ℂ S_{k}$ by $m_{1} = 0, and m_{i} = \sum_{l = 1}^{i - 1} s_{l i}, for i > 1 .$ Then

$m_{i} m_{j} = m_{j} m_{i}, for 1 \leq i, j \leq n .$
The eigenvalues of the elements $m_{i}$ are given by the diagram $\hat{S}$
in the sense that if ${\hat{S}}_{k} is the set of vertices on level k and$ ${\hat{S}}_{k}^{λ} = \{paths p = (\emptyset \to p^{(1)} \to p^{(2)} \to \dots \to p^{(k)} = λ) to λ in \hat{S}\}, for λ \in {\hat{S}}_{k},$ then ${\hat{S}}_{k} is an index set for the simple ℂ S_{k} modules S_{k}^{λ} and$ $S_{k}^{λ} has basis \{v_{p} | p \in {\hat{S}}_{k}^{λ}\} with m_{i} v_{p} = c (p (i)) v_{p},$ where $p (i) = p^{(i)} / p^{(i - 1)}$ is the box added at step $i$ in $p$ and $c (b)$ denotes the content of the box $b .$
$κ = m_{k} + m_{k - 1} + \dots + m_{2}$ is a central element of $ℂ S_{k}$ and $κ acts on S_{k}^{λ} by the constant \sum_{b \in λ} c (b) .$

Proof.

The tensor product rule for ${GL}_{n}$ is $L_{{𝔤 𝔩}_{n}} (μ) \otimes V ≅ \underset{λ / μ = □}{\otimes} L_{{𝔤 𝔩}_{n}} (μ),$ where the sum is over all partitions $λ$ such that $l (λ) \leq n, λ \supseteq μ$ and $λ$ differs from $μ$ by a single box. Since the $S_{k}$ action and the ${GL}_{n}$ action commute on $V^{\otimes k}$ it follows that $as (U {𝔤 𝔩}_{n}, ℂ S_{k}) bimodules, V^{\otimes k} ≅ \underset{λ ⊢ k, l (λ) \leq n}{\otimes} L_{{𝔤 𝔩}_{n}} (λ) \otimes S_{k}^{λ},$ where $S_{k}^{λ}$ are some $S_{k}$ -modules. Comparing the $L_{{𝔤 𝔩}_{n}} (λ)$ components on each side of $\underset{λ}{\otimes} L_{{𝔤 𝔩}_{n}} (λ) \otimes S_{k}^{λ} ≅ V^{\otimes k} = V^{\otimes (k - 1)} \otimes V ≅ (\underset{μ}{\otimes} L_{{𝔤 𝔩}_{n}} (μ) \otimes S_{k - 1}^{μ}) \otimes V$ $≅ \underset{μ}{\otimes} \underset{λ / μ = □}{\otimes} L_{{𝔤 𝔩}_{n}} (λ) \otimes S_{k - 1}^{μ} ≅ \underset{λ}{\otimes} (L_{{𝔤 𝔩}_{n}} (λ) \otimes (\underset{λ / μ}{\otimes} S_{k - 1}^{μ}))$ gives $S_{k}^{λ} ≅ \underset{λ / μ = □}{\otimes} S_{k - 1}^{μ} .$

Using the basis $\{v_{i_{1}} \otimes \dots \otimes v_{i_{k}} | 1 \leq i_{1}, \dots, i_{k} \leq n\}$ of $V^{\otimes k},$ the direct computation $\begin{array}{rcl} κ (v_{i_{1}} \otimes \dots \otimes v_{i_{n}}) & = & \sum_{l = 1}^{k} (v_{i_{1}} \otimes \dots \otimes \sum_{i, j = 1}^{n} E_{i j} E_{j l} v_{j l} \otimes \dots \otimes v_{i_{k}}) \\ + \sum_{1 \leq l < m \leq k} \sum_{i, j = 1}^{n} (v_{i_{1}} \otimes \dots \otimes E_{j i} v_{l} \otimes \dots \otimes E_{i j} v_{i_{m}} \otimes \dots \otimes v_{i_{k}} + v_{i_{1}} \otimes \dots \otimes E_{i j} v_{i_{l}} \otimes \dots \otimes E_{j i} v_{i_{m}} \otimes \dots \otimes v_{i_{k}}) \\ = & (k n + 2 \sum_{1 \leq l < m \leq k} s_{l m}) (v_{i_{1}} \otimes \dots \otimes v_{i k}) = (k n + 2 z_{k}) (v_{i_{1}} \otimes \dots \otimes v_{i_{k}}) \end{array}$

Shows that $κ = k n + 2 \sum_{1 \leq l < m \leq k} s_{l m}, as operators on V^{\otimes k} .$ Since $κ$ is a central element of $U {𝔤 𝔩}_{n}$ and $V^{\otimes k} ≅ \underset{λ ⊢ k, l (λ) \leq n}{\otimes} L (λ) \otimes S_{k}^{λ} as (U {𝔤 𝔩}_{n}, ℂ S_{k}) bimodules,$ it follows from (???) that $z_{k} = \sum_{1 \leq l < m \leq k} s_{l m} acts on S_{k}^{λ} by \sum_{b \in λ} c (b) .$ Thus, in (???), $m_{k} = \sum_{1 \leq l < k} s_{l k} = z_{k} - z_{k - 1} acts on S_{k - 1}^{μ} by the constant c (λ / μ) .$ Since the values $c (λ / μ)$ are distinct for the distinct summands in (???), $S_{k - 1}^{μ} = \{v \in S_{k}^{λ} | m_{k} v = c (λ / μ) v\},$ the $c (λ / μ),$ eigenspace of $m_{k}$ in $S_{k}^{λ} .$ Iterating the decomposition (???) gives $S_{k}^{λ} = \underset{p \in {\hat{S}}_{k}^{λ}}{\otimes} S_{1}^{□},$ and, since $S_{1}^{□}$ is one dimensional, this determines, up to constants, a unique $basis of S_{k}^{λ} \{v_{p} | p \in {\hat{S}}_{k}^{λ}\} such that m_{i} v_{p} = c (p (i)) v_{p} .$

In $ℂ S_{k},$ $s_{i} m_{i} s_{i} + s_{i} = m_{i + 1},$ and so $s_{i} m_{i} + 1 = m_{i + 1} s_{i} and s_{i} m_{j} = m_{j} s_{i}, for j \neq i, i + 1 .$ Write ${(s_{i})}_{p q}$ to denote the $(p, q)$ entry of the matrix determined by the action of $s_{i}$ on $S_{k}^{λ}$ with respect to the basis in (???). Then ${(s_{i})}_{p p} {(m_{i})}_{p p} + 1 = {(m_{i + 1})}_{p p} {(s_{i})}_{p p} giving {(s_{i})}_{p p} = \frac{1}{{(m_{i})}_{p p} - {(m_{i})}_{p p}} .$ $Then, COPY FROM NOTES ..... □$

As $(U {𝔤 𝔩}_{n}, ℂ S_{k})$ bimodules, $V^{\otimes k} ≅ \underset{λ ⊢ k, l (λ) \leq n}{\otimes} L_{{𝔤 𝔩}_{n}} (λ) \otimes S_{k}^{λ},$ where $S_{k}^{λ}$ are simple $S_{k}$ modules.

For $λ \in {\hat{S}}_{k}$ and $μ \in {\hat{S}}_{k - 1},$ ${Res}_{S_{k - 1}}^{S_{k}} (S_{k}^{λ}) ≅ \underset{λ / ν = □}{\otimes} S_{k - 1}^{ν} and {Ind}_{S_{k - 1}}^{S_{k}} (S_{k - 1}^{μ}) ≅ \underset{μ / μ = □}{\otimes} S_{k}^{ν},$ where the first sum is over all partitions $ν$ that are obtained from $λ$ by removing a box, and the second sum is over all partitions $ν$ which are obtained from $μ$ by adding a box.

Let ${(ℂ S_{k})}_{p q}$ be the $(p, q)$ (simultaneous) eigenspace of $ℂ S_{k}$ with respect to the action of $m_{1}, \dots, m_{k}$ by left and right multiplication, ${(ℂ S_{k})}_{p q} = \{a \in ℂ S_{k} | for 1 \leq i, j \leq k, m_{i} a = c (p (i)) a and a m_{j} = c (q (j)) m_{j}\} .$ Then $\dim ({(ℂ S_{k})}_{p q}) = 1$ and there exist matrix units $e_{p q}^{λ}, λ \in {\hat{S}}_{k}, p, q \in {\hat{S}}_{k}^{λ}$ such that ${(ℂ S_{k})}_{p q} = ℂ e_{p q} and e_{p q}^{λ} e_{r s}^{μ} = δ_{λ μ} δ_{q r} e_{p s}^{λ} .$

Let $Φ : ℂ S_{k} \to End (V^{\otimes k})$ and $Ψ : U {𝔤 𝔩}_{n} \to End (V^{\otimes k})$ be the representations of $S_{k}$ and ${𝔤 𝔩}_{n}$ corresponding to their actions on $V^{\otimes k} .$ Then ${End}_{{GL}_{n} (ℂ)} (V^{\otimes k}) = Φ (ℂ S_{k}) and {End}_{ℂ S_{k}} (V^{\otimes k}) = Ψ (U {𝔤 𝔩}_{n}),$ and $\ker Φ = ⟨\sum_{w \in S_{n + 1}} \det (w) w⟩,$ the ideal of $ℂ S_{k}$ generated by the alternating sum of the permutations in the subgroup $S_{n}$ $(\ker Φ = 0 if n \leq k) .$

The tower $\hat{B}$

If $λ \in {\hat{O}}_{n}$ then $L_{O_{n}} (λ) \otimes V ≅ \underset{μ / λ = □ or λ / μ = □}{\otimes} L_{O_{n}} (μ), as O_{n} (ℂ) -modules,$ where the sum is over $μ \in {\hat{O}}_{n}$ that are obtained frm $λ$ by adding or removing a box. Build a graph $\hat{B} (n)$ which encodes the $O_{n} (ℂ)$ module decomposition of $V^{\otimes k}, k \in ℤ_{\geq 0},$ by setting $vertices on level k : {\hat{B}}_{k} (n) = \{λ \in {\hat{O}}_{n} | k - |λ| \in 2 ℤ_{\geq 0}\},$ and $an edge λ \to μ, if μ \in {\hat{B}}_{k + 1} (n) is obtained from λ \in {\hat{B}}_{k} (n) by adding or removing a box.$

Define elements $m_{1}, \dots, m_{k} \in ℂ B_{k} (n)$ by $m_{1} = 0, and m_{i} = \frac{k (n - 1)}{4} + \sum_{l = 1}^{i - 1} s_{l i} - e_{l i}, for i > 1 .$ Then

$m_{i} m_{j} = m_{j} m_{i}$ for $1 \leq i, j \leq n .$
The eigenvalues of the elements $m_{i}$ are given by the diagram

in the sense that if ${\hat{B}}_{k} is the set of vertices on level k$ and ${\hat{B}}_{k}^{λ} = \{paths p = (\emptyset \to p^{(1)} \to p^{(2)} \to \dots \to p^{(k)}) to λ in \hat{B}\}, for λ \in {\hat{B}}_{k},$ then ${\hat{B}}_{k} is an index set for the simple ℂ B_{k} modules B_{k}^{λ},$ and $B_{k}^{λ} has a basis \{v_{p} | p \in {\hat{B}}_{k}^{λ}\} with m_{i} v_{p} = c (p (i)) v_{p},$ where ??????? $\{\begin{array}{rcl} c (p^{(i)} / p^{(i - 1)}) + \frac{n - 1}{2}, & if & p^{(i)} / p^{(i - 1)} = □, \\ - c (p^{(i - 1)} / p^{(i)}) - \frac{n - 1}{2}, & if & p^{(i - 1)} / p^{(i)} = □, \end{array}$
$κ = m_{k} + m_{k - 1} + \dots + m_{2} is a central element of ℂ B_{k} (n)$ and $κ acts on B_{k}^{λ} by the constant \frac{n - 1}{2} + \sum_{b \in λ} c (b)$

		Proof.
	Let $κ$ be the Casimir element of ${𝔰 𝔬}_{n}$ as in (???). Then $\begin{array}{rcl} κ (v_{i_{1}} \otimes \dots \otimes v_{i_{k}}) & = & - \frac{1}{4} (\sum_{l = 1}^{k} v_{i_{1}} \otimes \dots \otimes \sum_{i, j = 1}^{n} {(E_{i j} - E_{j i})}^{2} v_{i_{l}} \otimes \dots \otimes v_{i_{k}}) \\ - \frac{1}{4} 2 \sum_{1 \leq l < m \leq k} \sum_{i, j = 1}^{n} v_{i_{1}} \otimes \dots \otimes (E_{i j} - E_{j i}) v_{i_{l}} \otimes \dots \otimes (E_{i j} - E_{j i}) v_{i_{m}} \otimes \dots \otimes v_{i_{k}} \\ = & (- \frac{1}{4} (\sum_{l = 1}^{k} 1 - n - n + 1) - \frac{1}{4} 2.2 \sum_{1 \leq l < m \leq k} (e_{l m} - s_{l m})) (v_{i_{1}} \otimes \dots \otimes v_{i_{k}}) \end{array}$ since ${(E_{i j} - E_{j i})}^{2} = E_{j i}^{2} - E_{i j} E_{j i} - E_{j i} E_{i j} + E_{j i}^{2} and E_{j i}^{2} = 0$ unless $i = j = i_{l} .$ Thus, as operators on $V^{\otimes k},$ $κ = - \frac{1}{4} k (2 - 2 n) + \sum_{1 \leq l < m \leq k} (s_{l m} - e_{l m}) = \frac{k (n - 1)}{2} + \sum_{1 \leq l < m \leq k} s_{l m} - e_{l m} .$ Since $κ$ is a central element of $U {𝔤 𝔩}_{n}$ and $V^{\otimes k} ≅ \underset{λ ⊢ k, l (λ) \leq n}{\otimes} L (λ) \otimes B_{k}^{λ} as (U {𝔰 𝔬}_{n}, ℂ B_{k} (n)) bimodules,$ it follows from (???) that $\frac{k (n - 1)}{2} + \sum_{1 \leq l < m \leq k} s_{l m} - e_{l m} acts on B_{k}^{λ} by (n - 1) \|λ\| + \sum_{b \in λ} c (b) .$ The last statement follows since $m_{1} + \dots + m_{k} = \frac{k (n - 1)}{2} + \sum_{1 \leq l < m \leq k} s_{l m} - e_{l m},$ for every $k \in ℤ_{> 0} . □$

References [PLACEHOLDER]

[BG] A. Braverman and D. Gaitsgory, Crystals via the affine Grassmanian, Duke Math. J. 107 no. 3, (2001), 561-575; arXiv:math/9909077v2, MR1828302 (2002e:20083)

page history