Representations of Hecke Algebras of Finite Groups with BN-Pairs of Classical Type

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

Last update: 8 June 2014

Notes and References

This is a typed copy of Peter Norbert Hoefsmit's thesis Representations of Hecke Algebras of Finite Groups with BN-Pairs of Classical Type published in August, 1974.

Chapter 2.Representations of the Generic Ring Corresponding to a Coxeter System of Classical Type

Partitions and Tableaux

Let $B$ be a subgroup of a finite group $G$ and let $k$ be a field of characteristic zero. Set $e = {| B |}^{- 1} \sum_{b \in B} b$ in the group algebra $k G .$ Then $e$ affords the $1 -representation$ $1_{B}$ of $B$ and the left $k G -module$ $k G e$ affords the induced representation $1_{B}^{G} .$

Definition (2.1.1) The Hecke algebra $H_{k} (G, B)$ is the subalgebra of $k G$ given by $e (k G) e .$

The Hecke algebra acts on $k G e$ by right multiplication and the action defines an isomorphism between $H_{k} (G, B)$ and the endomorphism algebra ${End}_{k G} (k G e) .$ The double coset sums $\sum_{x \in B g B} x,$ $g \in G,$ form a basis for $H_{k} (G, B)$ (see [Ste1967], Lemma 84).

In this thesis we will be concerned with finite groups $G$ with BN-pairs of subgroups $(B, N)$ satisfying the axioms of [Tit1969]. Then $H = B \cap N$ is a normal subgroup of $N$ and the Weyl group $W = N / H$ has a presentation with a set of distinguished involutionary generators $R$ and defining relations $\begin{matrix} (2.1.2) & \begin{matrix} r^{2} = 1, r \in R \\ {(r s \dots)}_{n_{r s}} = {(s r \dots)}_{n_{r s}}, r, s \in R, r \neq s \end{matrix} \end{matrix}$ where $n_{r s}$ is the order of $r s$ in $W$ and ${(x y \dots)}_{m}$ denotes a product of alternating $x's$ and $y's$ with $m$ factors. The pair $(W, R)$ is called a Coxeter system. The group $G$ is said to be of type $(W, R) .$

If $w \neq 1 \in W,$ we denote by $l (w)$ the least length $l$ of all expressions $\begin{matrix} (2.1.3) & w = r_{1} \dots r_{l}, r_{1}, \dots, r_{l} \in R . \end{matrix}$ (2.1.3) is called a reduced expression for $W$ if $l = l (w) .$

There is a bijection between the double cosets $B \ G / B$ and the elements $w \in W$ resulting in the Bruhat decomposition $G = ⋃_{w \in W} B w B .$ The structure of the Hecke algebra $H_{k} (G, B)$ of a finite group with a BN-pair with respect to a Borel subgroup $B$ was shown in [Iwa1964] and [Mat1969-2] to be as follows.

Theorem (2.1.4) $H_{k} (G, B)$ has $k -basis$ ${S_{w} : a \in W}$ where $S_{w} = {| B |}^{- 1} \sum_{x \in B w B} x$ with $S_{1}$ the identity element. Multiplication is determined by the formula $\begin{matrix} S_{w} S_{r} & = & S_{w r}, r \in R, l (w r) > l (w), \\ S_{w} S_{r} & = & q_{r} S_{w r} + (q_{r} - 1) S_{w}, r \in R, l (w r) < l (w) \end{matrix}$ where the ${q_{r}, r \in R}$ are the index parameters $\begin{matrix} (2.1.5) & q_{r} = | B : (B \cap r B r) | . \end{matrix}$ For any reduced expression $w = r_{1} \dots r_{l}$ for $W$ in $R,$ $w \neq 1$ $S_{w} = S_{r_{1}} \dots S_{r_{l}} .$ Thus $H_{k} (G, B)$ is generated by ${S_{r}, r \in R}$ and has defining relations $\begin{matrix} (2.1.6) & \begin{matrix} S_{r}^{2} = q_{r} S_{1} + (q_{r} - 1) S_{r}, r \in R \\ {(S_{r} S_{s} \dots)}_{n_{r s}} = {(S_{s} S_{r} \dots)}_{n_{r s}}, \end{matrix} \end{matrix}$ where $n_{r s}$ is as in (2.1.2).

Let $(W, R)$ be a Coxeter system and let ${μ_{r}, r \in R}$ be indeterminates over $k,$ chosen such that $μ_{r} = μ_{s}$ if and only if $r$ and $s$ are conjugate in $W .$ Let $D$ be the polynomial ring $D = k [μ_{r} : r \in R] .$ Then there exists an associative $D -algebra$ $𝒜$ with identity, free basis ${a_{w}, w \in W}$ over $D$ and multiplication determined by the formulas $\begin{matrix} (2.1.7) & \begin{matrix} a_{w} a_{r} & = & a_{w r}, r \in R, l (w r) > l (w), \\ a_{w} a_{r} & = & μ_{r} a_{w r} + (μ_{r} - 1) a_{w}, r \in R, l (w r) < l (w), \end{matrix} \end{matrix}$ (see [Bou1968], p. 55). The $D -algebra$ $𝒜$ is called the generic ring corresponding to the Coxeter system $(W, R) .$ Analogous to Theorem (2.1.4) the generic ring has a presentation with generators ${a_{r}, r \in R}$ and relations $\begin{matrix} (2.1.8) & \begin{matrix} a_{r}^{2} & = & μ_{r} a_{1} + (μ_{r} - 1) a_{r}, r \in R \\ {(a_{r} a_{s} \dots)}_{n_{r s}} & = & {(a_{s} a_{r} \dots)}_{n_{r s}}, r, s \in R, r \neq s \end{matrix} \end{matrix}$ with $n_{r s}$ is as in (2.1.2).

The Hecke algebra $H_{k} (G, B)$ can be compared with the group algebra $k W$ as follows. Let $L$ be any field of characteristic zero and $ϕ : D \to L$ a homomorphism. Consider $L$ as a $D -module$ by setting $d \cdot λ = ϕ (d) λ, d \in D, λ \in L .$ Then the specialized algebra $\begin{matrix} (2.1.9) & 𝒜_{ϕ, L} = L \otimes 𝒜 \end{matrix}$ is an algebra over $L$ with basis ${a_{w ϕ} = 1 \otimes a_{w}},$ generators ${a_{r ϕ}, r \in R}$ and defining relations obtained from (2.1.8) by applying $ϕ .$ Thus if $ϕ : D \to k$ is defined by $ϕ (μ_{r}) = q_{r},$ $r \in R,$ $q_{r}$ the index parameters (2.1.5), then $\begin{matrix} (2.1.10) & 𝒜_{ϕ, k} = H_{k} (G, B) \end{matrix}$ while if $ϕ_{0} : D \to k$ is defined by $ϕ_{0} (μ_{r}) = 1,$ for all $r \in R,$ then $\begin{matrix} (2.1.11) & 𝒜_{ϕ_{0}, k} = k W . \end{matrix}$

We say the Coxeter system $(W, R)$ is of classical type if $W$ is of type $A_{n},$ $B_{n},$ $n \geq 2,$ or $D_{n},$ $n \geq 4 .$ In this chapter we will determine the irreducible representations of the generic ring corresponding to a Coxeter system of classical type and by means of the appropriate specialized algebras the irreducible representations of the Hecke algebras $H_{k} (G, B)$ of groups with BN-pair of classical type.

The Representations of $𝒜^{K} (B_{n})$

If a Coxeter system $(W, R)$ is of type $B_{n},$ $n \geq 2,$ $W (B_{n})$ is isomorphic to the hyperoctahedral group, the group of signed permutations on $n$ letters (see 2). Thus $W (B_{n})$ has a presentation with generators $R = {w_{1}, \dots, w_{n}}$ where $w_{i} = (i - 1, i),$ $i = 2, \dots, n,$ and $w_{1} = - (1),$ the first sign change and relations $\begin{matrix} w_{i}^{2} & = & 1, \\ w_{1} w_{2} w_{1} w_{2} & = & w_{2} w_{1} w_{2} w_{1}, \\ w_{i} w_{i + 1} w_{i} & = & w_{i + 1} w_{i} w_{i + 1}, i = 2, \dots, n - 1; \\ w_{i} w_{j} & = & w_{j} w_{i}, | i - j | > 1 \end{matrix}$ (see [Car1972]). Furthermore the set of generators $R$ is partitioned into 2 sets under conjugation; namely, $w_{i}$ is conjugate to $w_{j}$ for $i, j \geq 2$ while the negative one-cycle $w_{1}$ is not conjugate to any $w_{j},$ $j \geq 2 .$

For the Coxeter system $(W (B_{n}), R)$ taken as above, we take the generic ring $𝒜 (B_{n})$ to be defined over the polynomial ring $D = ℚ [x, y],$ $x, y$ indeterminates over $ℚ .$ It has a presentation with generators $a_{w_{i}} = a_{i},$ $a_{i} \in R,$ and relations

(B1)	$a_{1}^{2} = y 1 + (y - 1) a_{1},$
(B2)	$a_{i}^{2} = x 1 + (x - 1) a_{i}, i = 2, \dots, n;$
(B3)	$a_{1} a_{2} a_{1} a_{2} = a_{2} a_{1} a_{2} a_{1},$
(B4)	$a_{i} a_{j} = a_{j} a_{i}, \| i - j \| > 1 .$

We depart from the notations of 1 strictly for notational convenience, i.e., we switch from

ℚ (μ_{1}, μ_{2})

ℚ (x, y)

to avoid carrying around subscripts.

We now construct for each double partition $(μ) = (α, β)$ of $n,$ $n \geq 2,$ a $k -representation$ of $𝒜^{K} (B_{n}) = K \otimes 𝒜 (B_{n}),$ $K = ℚ (x, y) .$ The method involves the construction of $f^{μ} \times f^{μ}$ matrices over $k$ for each of the generators $a_{i}$ of $𝒜^{K} (B_{n})$ in a manner analogous to the construction of the matrices of the transpositions $(i - 1, i)$ for the outer tensor product representation $[α] \cdot [β]$ of $S_{n} .$

For any integer $k,$ let $Δ (k, y) = x^{k} y + 1$ Denote by $M (k, y)$ the $2 \times 2$ matrix $\begin{matrix} (2.2.1) & M (k, y) = \frac{1}{Δ (k, y)} (\begin{matrix} (x - 1) & Δ (k + 1, y) \\ x Δ (k - 1, y) & x^{k} y (x - 1) \end{matrix}) . \end{matrix}$ Then $trace M (k, y) = (x - 1),$ $det M (k, y) = - x,$ so the characteristic polynomial of $M (k, y)$ gives $\begin{matrix} (2.2.2) & M {(k, y)}^{2} = x I + (x - 1) M (k, y), \end{matrix}$ $I$ the $2 \times 2$ identity matrix.

For $k \geq 1,$ let $Δ (k, - 1) = \sum_{i = 0}^{k - 1} x^{i} .$ Denote by $M (k, - 1),$ $k \geq 2,$ the $2 \times 2$ matrix $\begin{matrix} (2.2.3) & M (k, - 1) = \frac{1}{Δ (k, - 1)} (\begin{matrix} - 1 & Δ (k + 1, - 1) \\ x Δ (k - 1, - 1) & x^{k} \end{matrix}) . \end{matrix}$ As $M (k, - 1)$ is obtained from $M (k, y)$ by setting $y = - 1,$ (2.2.1) shows $\begin{matrix} (2.2.4) & M {(k, - 1)}^{2} = x I + (x - 1) M (k, - 1) \end{matrix}$ Denote by $D (z, w)$ the $2 \times 2$ diagonal matrix $D (z, w) = (\begin{matrix} z & 0 \\ 0 & w \end{matrix}) .$ Then $\begin{matrix} (2.2.5) & D {(z, - 1)}^{2} = z I + (z - 1) D (z, - 1) . \end{matrix}$ In what follows, we employ the definitions and notations of (1) in regards to double partitions, standard tableaux, and axial distance.

Definition (2.2.6) Let $(μ) = (α, β)$ be a double partition of $n$ and let $T_{1}^{μ}, \dots, T_{f}^{μ},$ $f = f^{μ}$ be the ordering of the standard tableaux of shape $(μ)$ according to the last letter sequence. Construct $f \times f$ matrices $M^{μ} (i),$ $i = 1, \dots, n,$ over $K = ℚ (x, y)$ as follows:

(1)

Construct $M^{μ} (1)$ by placing

(i)	$y$ in the $p, p -th$ entry if the letter $1$ appears in $T_{p}^{α}$ of $T_{p}^{μ} = (T_{p}^{α}, T_{p}^{β}),$
(ii)	$- 1$ in the $p, p -th$ entry if the letter $1$ appears in $T_{p}^{β}$ of $T_{p}^{μ} = (T_{p}^{α}, T_{p}^{β}),$
(iii)	zeros elsewhere.

(2)

Construct $M^{μ} (i),$ $i = 2, \dots, n,$ by placing

(i)

$x$ in the $p, p -th$ entry if the letters $i - 1$ and $i$ appear in the same row of $T_{p}^{α}$ or $T_{p}^{μ} = (T_{p}^{α}, T_{p}^{β}),$

(ii)

$- 1$ in the $p, p -th$ entry if the letters $i - 1$ and $i$ appear in the same column of $T_{p}^{α}$ or $T_{p}^{β}$ of $T_{p}^{μ},$

(iii)

the matrix $M (k, - 1)$ in the $p, p -th,$ $p, q -th,,$ $q, p -th$ and $q, q -th$ entries corresponding to the tableaux $T_{p}^{μ}$ and $T_{q}^{μ}$ where

(a)	$p < q,$ $(i - 1, i) T_{p}^{μ} = T_{q}^{μ}$ and the letters $i - 1$ and $i$ appear either both in $T_{p}^{α}$ or $T_{p}^{β}$ of $T_{p}^{μ},$
(b)	$k$ is the axial distance from $i$ to $i - 1$ in $T_{p}^{μ},$

(iv)

the matrix $M (k, y)$ in the $p, p -th,$ $p, q -th,$ $q, p -th$ and $q, q -th$ entries corresponding to the tableaux $T_{p}^{μ}$ and $T_{q}^{μ}$ where

(a)	$p < q,$ $(i - 1, i) T_{p}^{μ} = T_{q}^{μ}$ and the letters $i - 1$ and $i$ appear in different tableaux of $T_{p}^{μ},$
(b)	$k$ is the axial distance from $i$ to $i - 1$ in $T_{p}^{μ},$

(v)

zeros elsewhere.

Let $V_{μ}$ denote the free $ℚ -module$ generated by $t_{1}, \dots, t_{f},$ $f = f^{μ}$ corresponding to the standard tableaux $T_{1}^{μ}, \dots, T_{f}^{μ}$ of shape $(μ)$ ordered according to the last letter sequence. For any field $L$ of characteristic zero set $V_{μ}^{L} = V_{μ} \otimes L .$ The corresponding basis elements $t_{i} \otimes 1$ of $V_{μ}^{L}$ will be denoted simply by $t_{i} .$ Set $K = ℤ (x, y) .$ Define linear operators $Z_{i}^{μ},$ $i = 1, \dots, n,$ on $V_{μ}^{K}$ such that the matrix of $Z_{i}^{μ}$ with respect to the basis ${t_{1}, \dots, t_{f}}$ of $V_{μ}^{K}$ is given by $M^{μ} (i) .$

Theorem (2.2.7) Let $K = ℤ (x, y)$ and let $𝒜^{K} (B_{n})$ denote the generic ring of the Coxeter system $(W (B_{n}), R)$ as before. Let $(μ)$ be a double partition of $n,$ $n \geq 2 .$ Then the $K -linear$ map $π^{μ} : 𝒜^{K} (B_{n}) \to End (V_{μ}^{K})$ defined by $π^{μ} (a_{i}) = Z_{i}^{μ}$ is a representation of $𝒜^{K} (B_{n}) .$

Proof.

We need to show the relations (B1 - B5) are satisfied with $Z_{i}^{μ}$ in place of $a_{i} .$ We argue by induction on $n .$ For $n = 2$ it is a case by case verification. The double partitions $((2), (0)),$ $((0), (2)),$ $((1^{2}), (0))$ and $((0), (1^{2}))$ are clearly seen to yield the well known one-dimensional representations of $𝒜^{K} (B_{2})$ ([CIK1971], 10). For the double partition $((1), (1))$ there are two standard tableaux, $\begin{matrix} \end{matrix} and \begin{matrix} \end{matrix} .$ From (2.2.6) $M^{((1), (1))} (1) = D (y, - 1)$ and $M^{((1), (1))} (2) = M (0, y) .$ Direct computation verifies the relation $M (0, y) D (y, - 1) M (0, y) D (y, - 1) = D (y, - 1) M (0, y) D (y, - 1) M (0, y) .$ Thus the relations (B1 - B3) are satisfied with $Z_{1}^{μ}$ and $Z_{2}^{μ}$ in place of $a_{1}$ and $a_{2}$ by the above computation, (2.2.2) and (2.2.5).

Now let $(μ) = (μ_{1}, \dots, μ_{s})$ be a double partition of $n .$ Deletion of the letter $n$ from a standard tableau automatically yields a standard tableau involving $n - 1$ letters. In fact deletion of $n$ from all standard tableaux having $n$ at the end of the $i -th$ column will yield all standard tableaux of shape $(μ_{1}, \dots, μ_{i} - 1, \dots, μ_{s}) .$ Denoting this partition by $(μ_{i} -)$ and using the fact that all standard tableaux with $n$ in the $i -th$ row precede all tableaux with $n$ in the $j -th$ row for $i > j$ when ordered according to the last letter sequence, we have $\begin{matrix} (2.2.8) & V_{μ}^{K} = V_{(μ_{s} -)}^{K} \oplus \dots \oplus V_{(μ_{1} -)}^{K} \end{matrix}$ and the corresponding matrix block form $M^{μ} (i) = M^{(μ_{s} -)} (i) ∔ \dots ∔ M^{(μ_{1} -)} (i), i < n$ as by (2.2.6), $M^{μ} (i)$ depends only on the letters $i - 1$ and $i .$ It is understood that $(μ_{i} -)$ is taken to equal zero if $n$ cannot appear in the $i -th$ row and the above summation, here and elsewhere, will be taken over those $(μ_{i} -)$ which are non-zero. By the induction hypothesis it therefore suffices to check the relations (B1 -B5) as they pertain to $Z_{n}^{μ} .$

The matrix $M^{μ} (n)$ from (2.2.6) is composed of the matrices $M (k, y)$ and $M (k, - 1)$ centered about the diagonal along with diagonal entries $x$ and $- 1$ . Thus the relation ${(Z_{n}^{μ})}^{2} = x I + (x - 1) Z_{n}^{μ}$ follows from (2.2.2), (2.2.4) and (2.2.5).

Let $V_{i, j}$ denote the subspace of $V_{μ}^{K}$ with basis $t_{1}^{i, j}, \dots, t_{s_{i, j}}^{i, j},$ corresponding to the standard tableaux of shape $(μ)$ with the letter $n$ appearing in the $i -th$ row and $n - 1$ appearing in the $j -th$ row, the ordering of the basis taken according to the last letter sequence. Then $V_{μ}^{K} = ⨁_{i, j} V_{i, j},$ the summation taken over all allowable $i, j$ such that $n$ appears in row $i$ and $n - 1$ appears in row $j,$ and this decomposition is consistent with the last letter sequence arrangement of the basis of $V_{μ}^{K} .$ Thus, whenever $n$ and $n - 1$ are in distinct rows and columns, we have $V_{i, j} ≅ V_{j, i}$ as $𝒜^{K} (B_{n - 2}) -modules$ for $n$ appearing in row $i,$ $n - 1$ appearing in row $j,$ where $W (B_{n - 2}) = ⟨ w_{1}, \dots, w_{n - 1} ⟩ .$

Suppose first that $n$ and $n - 1$ appear in distinct rows and columns, in the tableaux corresponding to $V_{i, j};$ $n$ in row $i,$ $n - 1$ in row $j .$ Then $n$ appears in row $j$ and $n - 1$ appears in row $i$ in the tableaux corresponding to $V_{j, i}$ and the map $t_{p}^{i, j} \mapsto t_{p}^{j, i},$ $p = 1, \dots, s_{i, j} = s_{j, i},$ gives an isomorphism $V_{i, j} ≅ V_{j, i}$ as $𝒜^{K} (B_{n - 2}) -modules,$ as the configuration of the first $n - 2$ letters in the tableau corresponding to $t_{p}^{i, j}$ is the same as the configuration of the first $n - 2$ letters in the tableau corresponding to $t_{p}^{j, i} .$ In particular the matrix of $Z_{k}^{μ},$ $k = 1, \dots, n - 2,$ on $V_{i, j} \oplus V_{j, i}$ is $S_{k} = (\begin{matrix} A_{k} \\ 0 & A_{k} \end{matrix})$ where $A_{k}$ is the matrix of $π^{μ} (a_{k})$ on $V_{i, j} .$ On the other hand, the matrix of $Z_{n}^{μ}$ on $K t_{p}^{i, j} \oplus K t_{p}^{j, i}$ is, by (2.2.6), $M (l, y)$ or $M (l, - 1),$ $l$ the axial distance from $n$ to $n - 1 .$ Thus the matrix of $Z_{n}^{μ}$ on $V_{i, j} \oplus V_{j, i}$ is $S_{n} = (\begin{matrix} a_{11} I & a_{12} I \\ a_{21} I & a_{22} I \end{matrix})$ where $(a_{i j}) = M (l, y)$ or $M (l, - 1),$ $l$ the $s_{i, j} \times s_{i, j}$ identity matrix. Then $S_{n} S_{k} = S_{k} S_{n}$ for $k = 1, \dots, n - 2$ and (B5) holds. The only other possibility is when $n$ and $n - 1$ appear in the same row or column of the tableaux corresponding to $V_{i, j} .$ But in this case the matrix of $Z_{n}^{μ}$ on $V_{i, j}$ is the scalar matrix $x I$ or $- I$ by (2.2.6) and thus commutes with $Z_{n}^{μ}$ on $V_{i, j},$ $k = 1, \dots, n - 2 .$ This proves (B5) for all cases.

To check the relation (B4), we consider the restriction of $Z_{n - 1}^{μ}$ and $Z_{n}^{μ}$ to subspaces with basis ${t_{i}}$ corresponding to all tableaux $T_{i}^{μ}$ having a fixed arrangement of the first $n - 3$ letters and all possible rearrangements of the letters $n - 2,$ $n - 1$ and $n .$ Let $V_{μ}^{k} = ⨁_{p} V_{p}$ denote the corresponding decomposition of $V_{μ}^{k},$ the ordering of the basis of each $V_{p}$ taken with respect to the last letter sequence. Then each $V_{p}$ is invariant under $Z_{n - 1}^{μ}$ and $Z_{n}^{μ}$ and it suffices to check (B4) for the various possible arrangements of the last 3 letters in a case by case basis.

In what follows, $M_{p} (i),$ $i = n$ or $n - 1,$ will denote the matrix of $Z_{i}^{μ}$ on $V_{p} .$

Case 1 — the letters $n - 2,$ $n - 1$ and $n$ in the same row or column. Then $V_{p}$ is one dimensional and $M_{p} (n) = M_{p} (n - 1) = x$ or $- 1$ by (2.2.6). Thus $M_{p} (n) M_{p} (n - 1) M_{p} (n) = M_{p} (n - 1) M_{p} (n) M_{p} (n - 1)$ and (B4) is satisfied.

Case 2 — the letters $n - 2,$ $n - 1$ and $n$ in two adjacent rows and two adjacent columns of the same diagram. Then $V_{p}$ is two-dimensional with basis elements corresponding to tableaux where the configuration of the last 3 letters is $(a) \begin{matrix} \end{matrix}, \begin{matrix} \end{matrix} or (b) \begin{matrix} \end{matrix}, \begin{matrix} \end{matrix}$ ordered according to last letter sequence. Then by (2.2.6), $M_{p} (n - 1) = M (2, 1)$ and $M_{p} (n) = D (x, - 1)$ in (a) and $M_{p} (n - 1) = D (x, - 1)$ and $M_{p} (n) = M (2, - 1)$ in (b). Thus (B4) is satisfied in both cases by direct verification of the relation $M (2, - 1) D (x, - 1) M (2, - 1) = D (x, - 1) M (2, - 1) D (x, - 1) .$

Case 3 — the letters $n - 2,$ $n - 1$ and $n$ in two rows and three columns or three rows and two columns. Then $V_{p}$ is three dimensional with basis elements corresponding to tableaux where the configuration of the last 3 letters is one of $\begin{matrix} (a) \begin{matrix} 1 \end{matrix} \begin{matrix} 2 \end{matrix} \begin{matrix} 3 \end{matrix} \\ (b) \begin{matrix} 1 \end{matrix} \begin{matrix} 2 \end{matrix} \begin{matrix} 3 \end{matrix} \\ (c) \begin{matrix} 1 \end{matrix} \begin{matrix} 2 \end{matrix} \begin{matrix} 3 \end{matrix} \\ (d) \begin{matrix} 1 \end{matrix} \begin{matrix} 2 \end{matrix} \begin{matrix} 3 \end{matrix} \end{matrix}$ ordered according to the last letter sequence. If we set $S_{1} = (\begin{matrix} c & \cdot & \cdot \\ \cdot & a_{11} & a_{12} \\ \cdot & a_{21} & a_{22} \end{matrix}), S_{2} = (\begin{matrix} b_{11} & b_{12} & \cdot \\ b_{21} & b_{22} & \cdot \\ \cdot & \cdot & c \end{matrix})$ we have, by (2.2.6), in case (a), $M_{p} (n - 1) = S_{1}$ and $M_{p} (n) = S_{2}$ where $(a_{i j}) = M (d_{1}, ϵ),$ $(b_{i j}) = M (d_{2}, ϵ),$ $ϵ = y$ or $- 1$ , and $c = x .$ Here $d_{1}$ is the axial distance form $n - 1$ to $n - 2$ in 2 and $d_{2}$ is the axial distance from $n$ to $n - 1$ in 1 so that $d_{2} = d_{1} + 1 .$ In case (b), $M_{p} (n - 1) = S_{2}$ and $M_{p} (n) = S_{1}$ with the same entries in $S_{i}$ as in case (a). The analysis of (c) and (d) is similar except that now $c = - 1$ in $S_{1}$ and $S_{2} .$ Thus in all cases (B4) is satisfied by

Lemma (2.2.9) Let $S_{1},$ $S_{2}$ be as above and let $(a_{i j}) = M (a, y),$ $(b_{i j}) = M (b, y) .$ Then for

(i)	$c = x$ and $b - 1 = a,$
(ii)	$c = - 1$ and $b + 1 = a,$

we have $S_{1} S_{2} S_{1} = S_{2} S_{1} S_{2} .$

Proof.

Observe that $S_{1} S_{2} S_{1} = S_{2} S_{1} S_{2}$ iff

(1)	$c b_{11} (c - b_{11}) = a_{11} b_{12} b_{21},$
(2)	$c a_{22} (c - a_{22}) = b_{22} a_{12} a_{21},$
(3)	$b_{i j} (a_{11} b_{22} + c (b_{11} - a_{11})) = 0, i \neq j,$
(4)	$a_{i j} (a_{11} b_{22} + c (b_{22} - a_{2})) = 0, i \neq j,$
(5)	$a_{11} b_{22} (b_{22} - a_{11}) = c (a_{12} a_{21} - b_{12} b_{21}) .$

For (1),

\begin{matrix} (2.2.10) & c b_{11} (c - b_{11}) - a_{11} b_{12} b_{21} = \frac{(x - 1)}{Δ {(b, y)}^{2} Δ (a, y)} (Δ (a, y) (c^{2} Δ (b, y) - c (x - 1)) - x Δ (b - 1, y) Δ (b + 1, y)) . \end{matrix}

Now for

c = x,

c^{2} Δ (b, y) - c (x - 1) = x Δ (b + 1, y)

and for

c = - 1,

c^{2} Δ (b, y) - c (x - 1) = x Δ (b - 1, y) .

Hence (2.2.10) equals zero for

c = x,

b + 1 = a

and for

c = - 1,

b - 1 = a .

The relation (2) is entirely similar.

For (3) and (4) we have $a_{11} b_{22} + c (b_{11} - a_{11}) = \frac{(x - 1)}{Δ (a, y) Δ (b, y)} (x^{b} y (x - 1) + c (Δ (a, y) - Δ (b, y)))$ and $a_{11} b_{22} + c (b_{22} - a_{22}) = \frac{(x - 1)}{Δ (a, y) Δ (b, y)} (x^{b} y (x - 1) + c (x^{a} y Δ (b, y) - x^{b} y Δ (a, y))) .$ But for $c = x,$ $b - 1 = a$ and $c = - 1,$ $b + 1 = a,$ $\begin{matrix} Δ (a, y) - Δ (b, y) & = & x^{a} y Δ (b, y) - x^{a} y Δ (a, y) \\ = & x^{b} y (1 - x) . \end{matrix}$ For (5), first note the useful factorization $\begin{matrix} (2.2.11) & \begin{matrix} (\frac{x Δ (a - 1, y) Δ (a + 1, y)}{Δ {(a, y)}^{2}}) - (\frac{x Δ (b - 1, z) Δ (b + 1, z)}{Δ {(b, z)}^{2}}) \\ = & \frac{{(x - 1)}^{2}}{Δ {(a, y)}^{2} Δ {(b, z)}^{2}} (x^{a} y {(x^{b} z + 1)}^{2} - x^{b} z {(x^{a} y + 1)}^{2}) \\ = & \frac{{(x - 1)}^{2} (x^{a + b} y z - 1) (x^{b} z - x^{a} y)}{Δ {(a, t)}^{2} Δ {(b, z)}^{2}} . \end{matrix} \end{matrix}$ Now $\begin{matrix} (2.2.12) & \frac{1}{c} a_{11} b_{22} (b_{22} - a_{11}) = \frac{{(x - 1)}^{2}}{Δ {(b, y)}^{2} Δ {(a, y)}^{2}} (\frac{x^{b + 1} y - x^{b} y}{c}) (x^{a + b} y^{2} - 1) . \end{matrix}$ But $\frac{1}{c} (x^{b + 1} y - x^{b} y) = x^{b} y - x^{a} y$ for $c = x,$ $b - 1 = a$ and for $c = - 1,$ $b + 1 = a .$ So for both cases, (2.2.12) equals $a_{12} a_{21} - b_{12} b_{21}$ using (2.2.11) with $z = y .$

$□$

Case 4 — the letters $n - 2,$ $n - 1$ and $n$ in three distinct rows and three distinct columns. Then $V_{p}$ is 6-dimensional with basis elements corresponding to tableaux where the configuration of the last 3 letters is $\begin{matrix} \begin{matrix} 1 \end{matrix} \begin{matrix} 2 \end{matrix} \begin{matrix} 3 \end{matrix} \\ \begin{matrix} 4 \end{matrix} \begin{matrix} 5 \end{matrix} \begin{matrix} 6 \end{matrix} \end{matrix}$ ordered according to the last letter sequence. Let $S_{1} = (\begin{matrix} a_{11} & a_{12} & \cdot & \cdot & \cdot & \cdot \\ \cdot & \cdot & b_{11} & b_{12} & \cdot & \cdot \\ \cdot & \cdot & b_{21} & b_{22} & \cdot & \cdot \\ \cdot & \cdot & \cdot & \cdot & c_{11} & c_{12} \\ \cdot & \cdot & \cdot & \cdot & c_{21} & c_{22} \end{matrix}), S_{2} = (\begin{matrix} c_{11} & \cdot & c_{12} & \cdot & \cdot & \cdot \\ \cdot & b_{11} & \cdot & \cdot & b_{12} & \cdot \\ c_{21} & \cdot & c_{22} & \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot & a_{11} & \cdot & a_{12} \\ \cdot & b_{21} & \cdot & \cdot & b_{22} & \cdot \\ \cdot & \cdot & \cdot & a_{21} & \cdot & a_{22} \end{matrix}) .$ Then if all rows are in the same diagram we have by (2.2.6), $M_{p} (n - 1) = S_{1}$ and $M_{p} (n) = S_{2}$ where $(a_{i j}) = M (d_{1}, - 1),$ $(b_{i j}) = M (d_{2}, - 1)$ and $(c_{i j}) = M (d_{3}, - 1) .$ Here $d_{1}$ is the axial distance from $n - 1$ to $n - 2$ in 1, $d_{2}$ is the axial distance from $n - 1$ to $n - 2$ in 3 and $d_{3}$ is the axial distance from $n - 1$ to $n - 2$ in 5 so that $d_{1} + d_{3} = d_{2}$ and all $d_{i} \geq 2 .$ If two rows are in one diagram and the third in the second diagram, we assume, without loss of generality, the lowest box to be in the second diagram. Superimposing the second diagram upon the first again does not alter the relation $d_{1} + d_{3} = d_{2}$ except that now only $d_{1} \geq 2 .$ In this case $M_{p} (n - 1) = S_{1}$ and $M_{p} (n) = S_{2},$ where now $(a_{i j}) = M (d_{1}, - 1),$ $(b_{i j}) = M (d_{2}, y)$ and $(c_{i j}) = M (d_{3}, y) .$ Thus for both cases (B4) is satisfied by

Lemma (2.2.13) Let $S_{1}$ and $S_{2}$ be as above and let $(a_{i j}) = M (s, w),$ $(b_{i j}) = M (p, y),$ $(c_{i j}) = M (t, z)$ with $s + t = p .$ Then for

(i)	$w = - 1,$ $y = z,$ $s \geq 2,$
(ii)	$z = - 1,$ $y = w,$ $t \geq 2,$
(iii)	$w = y = z = - 1,$ $s, t, p \geq 2,$

we have

S_{1} S_{2} S_{1} = S_{2} S_{1} S_{2} .

Proof.

First, either (i) or (ii) clearly imply (iii). Now $S_{1} S_{2} S_{1} = S_{2} S_{1} S_{2}$ iff

(1)	$a_{k l} (c_{i i} a_{i i} + b_{i i} (a_{j j} - c_{i i})) = 0,$ $k \neq l,$ $i \neq j;$
(2)	$c_{k l} (c_{i i} a_{i i} + b_{i i} (c_{j j} - a_{i i})) = 0,$ $k \neq l,$ $i \neq j;$
(3)	$b_{k l} (c_{i i} a_{i i} - c_{i i} b_{j j} - a_{j j} b_{i i}) = 0,$ $k \neq l,$ $i \neq j;$
(4)	$a_{i i} c_{i i} (a_{i i} - c_{i i}) = b_{i i} (c_{i j} c_{j i} - a_{i j} a_{j i}),$ $i \neq j;$
(5)	$a_{i i} b_{j j} (a_{i i} - b_{j j}) = c_{j i} (b_{i j} b_{j i} - a_{i j} a_{j i}),$ $i \neq j;$
(6)	$c_{i i} b_{j j} (c_{i i} - b_{j j}) = a_{j i} (b_{i j} b_{j i} - c_{i j} c_{j i}),$ $i \neq j;$

and as these relations are symmetric in the

(a_{i j})

and

(c_{i j}),

it suffices to prove the lemma for (i).

For (1), (2), and (3), we observe that $\begin{matrix} E_{1} & = & Δ (s, - 1) + x^{s} Δ (t, y) - Δ (p, y) = 0, \\ E_{2} & = & x^{t} y Δ (s, - 1) - Δ (t, y) + Δ (p, q) = 0 . \end{matrix}$ Also set $D = \frac{{(x - 1)}^{2}}{Δ (p, y) Δ (t, y) Δ (s, - 1)} .$ We then obtain $\begin{matrix} c_{11} a_{11} + b_{11} (a_{22} - c_{11}) = - D E_{1} & = & 0, \\ c_{22} a_{22} + b_{22} (a_{11} - c_{22}) = x^{p} y D E_{2} & = & 0, \\ c_{11} a_{11} + b_{11} (c_{22} - a_{11}) = D E_{2} & = & 0, \\ c_{22} a_{22} + b_{22} (c_{11} - a_{22}) = x^{p} y D E_{1} & = & 0, \\ c_{11} a_{22} - c_{11} c_{22} - a_{22} b_{11} = x^{s} D E_{2} & = & 0, \\ c_{22} a_{11} - c_{22} b_{11} - a_{11} b_{22} = - x^{t} y D E_{1} & = & 0 . \end{matrix}$

For (4) we have $\frac{1}{b_{11}} a_{11} c_{11} (a_{11} - c_{11}) = \frac{1}{b_{22}} a_{22} c_{22} (a_{22} - c_{22}) = \frac{{(x - 1)}^{2} (x^{s + t} y + 1) (x^{t} y + x^{s})}{Δ {(s, - 1)}^{2} Δ {(t, y)}^{2}} = c_{12} c_{21} - a_{12} a_{21}$ from (2.2.11), setting $a = t,$ $b = s,$ $z = - 1 .$

The relations (5) and (6) are handled in an entirely similar manner.

$□$

This completes the proof of the lemma and the proof of the theorem.

$□$

Theorem (2.2.14) Let $k$ be as before. The representations $π^{μ}$ of $𝒜^{K} (B_{n})$ are irreducible, pairwise inequivalent and are, up to isomorphism, a complete set of irreducible, inequivalent representations of $𝒜^{K} (B_{n}) .$ In particular $k$ is a splitting field for $𝒜^{K} (B_{n}) .$

Proof.

By induction in $n .$ For the representations of $𝒜^{K} (B_{2})$ it is a matter of direct computation to check irreducibility and inequivalence. Consideration of degrees shows a complete set of inequivalent representations is obtained. For $𝒜^{K} (B_{n})$ we employ the decomposition (2.2.8) afforded by the last letter sequence and the position of the letter $n$ in a standard tableau. Let $(μ)$ be a double partition of $n .$ The $𝒜^{K} (B_{n}) -module$ $V_{μ}^{K}$ is either irreducible or (2.2.8) is the decomposition of $V_{μ}^{K}$ into irreducible inequivalent $𝒜^{K} (B_{n})$ components, inequivalent because each of the double partitions $(μ_{i} -)$ of $n - 1$ is distinct. But for each pair $(μ_{i} -),$ $(μ_{j} -),$ $i \neq j,$ there exists a tableau $T_{p}^{μ}$ with $n$ in row $i,$ $n - 1$ in row $j$ and $(i - 1, i) T_{p}^{μ} = T_{q}^{μ}$ a tableau with $n$ in row $j$ and $n - 1$ in row $i .$ Thus the action of $π^{μ} (a_{n})$ does not decompose with respect to the $V^{(μ_{i} -)},$ $i = 1, \dots, s + r .$ Hence $V_{μ}^{K}$ is irreducible. Furthermore the double partition $(μ)$ is completely determined by the set of double partitions $(μ_{i},),$ $i = 1, \dots, s + r,$ of $n - 1 .$ Thus by the induction hypothesis $V_{μ}^{K} ≅ V_{μ'}^{K}$ as $𝒜^{K} (B_{n}) -modules$ implies $(μ) = (μ') .$ From ([You1929]), $\sum_{μ} {(f^{μ})}^{2} = 2^{n} n!,$ $(μ)$ a double partition of $n .$ Thus $𝒜^{K} (B_{n})$ is semisimple and as the $π^{μ}$ are defined over $k,$ $k$ is a splitting field for $𝒜^{K} (B_{n}) .$ This completes the proof.

$□$

It is clear that the above representations of the generic ring yield representations of a wide variety of specialized algebras of $𝒜^{K} (B_{n}) .$ Specifically, set $P (B_{n}) = x \prod_{i = 0}^{n - 1} (x^{i} + y) (x^{i} y + 1) (1 + \dots + x^{i}) \in D = ℚ [x, y] .$

Corollary (2.2.15) Let $L$ be any field of characteristic zero, $ϕ : D \to L$ a homomorphism such that $ϕ (P (B_{n})) \neq 0 .$ Let $(μ)$ be a double partition of $n$ and let $Z_{i ϕ}^{μ}$ denote the linear operator on $V_{μ}^{L}$ obtained by the substitution $x \mapsto ϕ (x),$ $y \mapsto ϕ (y)$ in the entry of $M^{μ} (i) .$ Then $Z_{i ϕ}^{μ}$ is well defined and the $L -linear$ map $π_{ϕ, L}^{μ} : 𝒜_{ϕ, L} (B_{n}) \to End (V_{μ}^{L})$ defined by $π_{ϕ, L}^{μ} (a_{i}) = Z_{i ϕ}^{μ}$ is a representation of $𝒜_{ϕ, L} (B n) .$ The representations ${π_{ϕ, L}^{μ}}$ are a complete set of irreducible inequivalent representations of $𝒜_{ϕ, L} (B_{n}) .$

Proof.

If $ϕ (P (B_{n})) \neq 0,$ (2.2.1) and (2.2.3) show the matrices $M (k, y)$ and $M (k, - 1)$ are well defined under the substitution $x \mapsto ϕ (x),$ $y \mapsto ϕ (y)$ for $- n + 1 \leq k \leq n - 1 .$ It is clear from the definition that axial distance in a Young diagram corresponding to a double partition of $n$ cannot exceed $n - 1$ in absolute value. Thus by (2.2.6), $Z_{i}^{μ}$ is well defined for all $i .$ As $𝒜_{ϕ, L}$ has a presentation with generators ${a_{i ϕ}}$ and relations obtained from (B1 - B5) by applying $ϕ,$ the proofs of Theorem (2.2.7) and (2.2.14) carry over to this case.

$□$

Let $A$ be a separable algebra over a field $L$ and let $\overline{L}$ be an algebraic closure of $L .$ Define the numerical invariants of $A$ to be the set of integers ${n_{i}}$ such that $A^{\overline{L}}$ is isomorphic to a direct sum of total matrix algebras $A^{\overline{L}} = ⨁_{i} M_{n_{i}} (\overline{L}) .$ Thus for $ϕ$ defined as in Corollary (2.2.15) the algebras $𝒜^{K} (B_{n})$ and $𝒜_{ϕ, L}$ have the same numerical invariants. In particular Corollary (2.2.15) gives the well known result (see [BCu1972]) that for $G$ a finite group with BN-pair with Coxeter system $(W, R)$ of type $B_{n},$ $H_{ℚ} (G, B) ≅ ℚ W .$ Indeed in ([BCu1972]) this is shown to be the case for all Coxeter system with the possible exception of $(W, R)$ of type $E_{7} .$

Finally we remark that, for the specialization $ϕ_{0} : D \to ℚ$ defined by $ϕ_{0} (x) = ϕ_{0} (y) = 1,$ the representations ${π_{ϕ_{0}, ℚ}^{μ}}$ are the irreducible representations of $W (B_{n})$ given by Theorem (1.2.3).

The Representations of $𝒜^{K} (A_{n})$ and $𝒜^{K} (D_{n})$

We now obtain the representations of the generic ring of a Coxeter system of type $A_{n}$ and $D_{n} .$

If $(W, R)$ is a Coxeter system of type $A_{n - 1},$ $W (A_{n - 1})$ is isomorphic to the symmetric group $S_{n}$ and we take the set $R$ to be ${w_{2}, \dots, w_{n}}$ where $w_{i} = (i - 1, i),$ $i = 2, \dots, n .$ We take the generic ring $𝒜 (A_{n - 1})$ to be defined over the polynomial ring $D = ℚ [x] .$ It has a presentation with generators $a_{w_{i}} = a_{i},$ $i = 2, \dots, n,$ and relations (B2, B4, B5).

Set $K = ℚ (x) .$ The representations of $𝒜^{K} (A_{n - 1}) = 𝒜 (A_{n - 1}) \otimes_{D} K$ are readily obtained from the results of the previous section. As the matrices $M (k, - 1)$ are defined in $ℚ (x),$ (2.2.6) shows the matrices $M^{(α, (0))} (i),$ $i = 2, \dots, n,$ are defined in $ℚ (x)$ and $Z_{i}^{(α, (0))}$ can be regarded as a linear operator on $V_{(α, (0))}^{K} .$ Thus

Theorem (2.3.1) Let $α$ be a partition of $n,$ $n \geq 2$ and $K = ℚ (x) .$ The $k -linear$ map $π^{α} : 𝒜^{K} (A_{n - 1}) \to End (V_{(α, (0))}^{K})$ defined by $π^{α} (a_{i}) = Z_{i}^{(α, (0))},$ $i = 2, \dots, n,$ is a representation of $𝒜^{K} (A_{n - 1}) .$ The representations ${π^{α}},$ are a complete set of irreducible, inequivalent representations of $𝒜^{K} (A_{n - 1}) .$

Proof.

Theorem (2.2.7) shows the ${π^{α}}$ are representations of $𝒜^{K} (A_{n - 1}) .$ Irreducibility and inequivalence follows from Theorem (2.2.14) as the matrix of $Z_{1}^{(α, (0))}$ on $V_{(α, (0))}$ is the scalar matrix $y I .$ As (see [You1929]) $\sum_{α} {(f^{α})}^{2} = n!, α a partition of n,$ the ${π^{α}}$ are a complete set of inequivalent representations and are absolutely irreducible.

$□$

The representations of the specialized algebra are handled entirely analogous to Corollary (2.2.15). Set $P (A_{n}) = x \prod_{i = 1}^{n} (1 + \dots + x^{i}) .$ Then from the above and Corollary (2.2.15) we have

Corollary (2.3.2) Let $L$ be any field of characteristic zero, $ϕ : D = ℚ [X] \to L$ a homomorphism such that $ϕ (P (A_{n})) \neq 0 .$ Then for $(α)$ a partition of $n \geq 2,$ the linear operators $Z_{i ϕ}^{(α, (0))},$ $i = 2, \dots, n,$ are well defined and the $L -linear$ maps $π_{ϕ, L}^{α} : 𝒜_{ϕ, L} (A_{n - 1}) \to End (V_{(α, (0))}^{L})$ defined by $π_{ϕ, L}^{α} (a_{i}) = Z_{i ϕ}^{(α, (0))}$ is a representation of $𝒜_{ϕ, L} (A_{n - 1}) .$ The ${π_{ϕ, L}^{α}}$ are a complete set of irreducible, inequivalent representations of $𝒜_{ϕ, L} (A_{n - 1}) .$

Thus for $ϕ$ as above the algebras $𝒜^{K} (A_{n})$ and $𝒜_{ϕ, L} (A_{n})$ have the same numerical invariants. We remark that for the specialization $x \mapsto 1$ the definitions of the matrices $M (k, - 1)$ shows the semi-normal matrix representation of $S_{n}$ is obtained (see Theorem (1.2.1)).

If $(W, R)$ is a Coxeter system of type $D_{n},$ $n \geq 4,$ $W (D_{n})$ can be regarded as a subgroup of index 2 in $W (B_{n});$ $W (D_{n})$ acting on an orthonormal basis of $ℝ^{n}$ by means of permutations and even sign changes. A set of distinguished generators for $W (D_{n})$ can be obtained from the set ${w_{1}, \dots, w_{n}}$ of $W (B_{n})$ given in section (1) by setting ${\overline{w}}_{1} = w_{1} w_{2} w_{1}$ and taking the set $R$ to be ${{\overline{w}}_{1}, w_{2}, \dots, w_{n}} .$ (see [Car1972]).

Let $ϕ : ℚ [x, y] \to ℚ [x]$ be defined by $ϕ (y) = 1 .$ Then the specialized ring $𝒜_{ϕ, ℚ [x]} (B_{n})$ has basis ${a_{w ϕ}, w \in W (B_{n})}$ with relations obtained by applying $ϕ$ to (B1 - B5). In particular ${(a_{1 ϕ})}^{2} = 1 .$ Set ${\overline{a}}_{1 ϕ} = a_{w_{1} w_{2} w_{1} ϕ} .$ As $w_{1} w_{2} w_{1}$ is reduced in $(W (B_{n}), R)$ we have $a_{w_{1} w_{2} w_{1} ϕ} = a_{1 ϕ} a_{2 ϕ} a_{1 ϕ}$ by (2.1.8). Applying $ϕ$ to (B1 - B5) it is readily seen that

(B'1)	${\overline{a}}_{1 ϕ}^{2} = x 1 + (x - 1) {\overline{a}}_{1 ϕ},$
(B'2)	${\overline{a}}_{1 ϕ} a_{3 ϕ} {\overline{a}}_{1 ϕ} = a_{3 ϕ} {\overline{a}}_{1 ϕ} a_{3 ϕ},$
(B'3)	${\overline{a}}_{1 ϕ} a_{j ϕ} = a_{j ϕ} {\overline{a}}_{1 ϕ}, j \neq 1, 3 .$

As any reduced expression of

w \neq 1 \in W (D_{n})

in the generators

{{\overline{w}}_{1}, w_{2}, \dots, w_{n}}

is a reduced expression for

W

in the generators

{w_{1}, w_{2}, \dots, w_{n}}

W (B_{n}),

the relations (B'1 - B'3) show the subring of

𝒜_{ϕ, ℚ [x]} (B_{n})

generated by

{{\overline{a}}_{1 ϕ}, a_{2 ϕ}, \dots, a_{n ϕ}}

has free basis

{a_{w ϕ}, w \in W (D_{n})} .

As all the generators

{{\overline{w}}_{1}, w_{2}, \dots, w_{n}}

are conjugate in

W (D_{n}),

the subring generated by

{{\overline{a}}_{1 ϕ}, a_{2 ϕ}, \dots, a_{n ϕ}}

is isomorphic to the generic ring of a Coxeter system of type

D_{n},

n \geq 4 .

Denote this subring by

𝒜 (D_{n}) .

Thus the representations of $𝒜_{ϕ, L} (B_{n})$ given by Corollary (2.2.15) provide us with representations of $𝒜^{K} (D_{n}) .$ Young ([You1929]) showed the restrictions of the representations of $W (B_{n})$ to $W (D_{n})$ corresponding to a double partition $(α, β)$ of $n$ remain irreducible if $(α) \neq (β)$ and decomposes into two irreducible components when $(α) = (β) .$ We show that this holds true in a generic sense.

Recall that a standard tableau $T$ for the double partition $(α, β)$ of $n$ is an ordered pair $T = (T^{α}, T^{β}) .$ Then the tableau $T^{*} = (T^{β}, T^{α})$ is a standard tableau of shape $(β, α),$ called the conjugate tableau of $T .$ Moreover the map $T \mapsto T^{*}$ is a bijection from the standard tableaux of shape $(α, β)$ to the standard tableaux of shape $(β, α) .$ Take $(α) \neq (β) .$ If $T_{1}, \dots T_{p}, \dots, T_{q}, \dots T_{f},$ $f = f^{α, β},$ is the arrangement of the standard tableaux of shape $(α, β)$ according to the last letter sequence, order the tableaux of shape $(β, α)$ according to the scheme; $T_{q}^{*}$ precedes $T_{p}^{*}$ if $T_{p}$ precedes $T_{q}$ in the last letter sequence. Call this the conjugate ordering of the tableaux of shape $(β, α) .$

Let $I_{n}^{*}$ denote the $n \times n$ matrix $I_{n}^{*} = (\begin{matrix} 0 & \dots & 0 & 1 \\ 0 & \dots & 1 & 0 \\ ⋮ & ⋮ \\ 1 & 0 & \dots & 0 \end{matrix}) .$

Lemma (2.3.3) Let $M_{ϕ}^{α, β} (a)$ denote the matrix of $π_{ϕ}^{α, β} (a)$ with respect to the basis ${t_{i}}$ of $V_{α, β}^{K}$ ordered according to the last letter sequence, $a \in 𝒜^{K} (D_{n}) .$ Then $I_{f}^{*} M_{ϕ}^{α, β} (a) I_{f}^{*},$ $f = f^{α, β},$ is the matrix of $π_{ϕ}^{β, α} (a)$ with respect to the conjugate ordering of the basis ${t_{i}}$ of $V_{β, α}^{K} .$ Thus the restrictions of the representations $π_{ϕ}^{α, β}$ and $π_{ϕ}^{β, α}$ to $𝒜^{K} (D_{n})$ are equivalent.

Proof.

Let $T_{1}, \dots, T_{q}, \dots, T_{f}$ be the arrangement of the standard tableaux of shape $(α, β)$ according to the last letter sequence. For fixed $i,$ $i = 2, \dots, n,$ write $V_{α, β}^{K}$ as the direct sum $V_{α, β}^{K} = ⨁ V_{p, q}$ of $Z_{i ϕ}^{α, β}$ invariant subspaces where $V_{p, p}$ is taken to have basis ${t_{p}}$ if $i - 1$ and $i$ appear in the same row or column of $T_{p}$ and $V_{p, q}$ has basis ${t_{p}, t_{q}}$ where $(i - 1, i) T_{p} = T_{q},$ $p < q$ in the last letter sequence. Let $V_{β, α}^{K} = ⨁ V_{p, q}^{*}$ denote the corresponding decomposition of $V_{β, α}^{K},$ where $V_{p, q}^{*}$ has basis $ ${t_{p}^{*}, t_{q}^{*}}$ corresponding to $T_{p}^{*},$ $T_{q}^{*},$ and where the ordering of $t_{p}^{*},$ $t_{q}^{*}$ is taken with respect to the conjugate ordering. We need to show that if the matrix of $Z_{i ϕ}^{α, β}$ on $V_{p, q}$ is $M$ the matrix of $Z_{i ϕ}^{α, β}$ on $V_{p, q}^{*}$ is $I^{*} M I^{*} .$ This is a simple case by case verification.

1. If $i - 1$ and $i$ are in the same row or column of $T_{p}$ they are likewise in $T_{p}^{*}$ and the lemma is shown for this case.

2. If $i - 1$ and $i$ are in distinct rows and columns of the same tableau $T_{p}^{α}$ or $T_{p}^{β}$ of $T_{p} = (T_{p}^{α}, T_{p}^{β}),$ set $T_{q} = (i - 1, i) T_{p}$ and take $p \leq q .$ Then $T_{q}^{*} < T_{p}^{*}$ in the arrangement according to the last letter sequence while $T_{q}^{*} < T_{p}^{*}$ in the conjugate ordering. The axial distance, $k,$ from $i$ to $i - 1$ is the same both in $T_{p}$ and $T_{p}^{*} .$ Thus from (2.2.6) the matrix of $Z_{i ϕ}^{α, β}$ on $V_{p, q}$ is $M (k, - 1)$ while the matrix of $Z_{i ϕ}^{β, α}$ on $V_{p, q}^{*}$ is $I^{*} M (k, - 1) I^{*}$ as is required.

3. If $i - 1$ and $i$ are in distinct tableaux of $T_{p} = (T_{p}^{α}, T_{p}^{β}),$ set $T_{q} = (i - 1, i) T_{p}$ and take $p < q .$ Then $T_{p}^{*} < T_{q}^{*}$ in both the ordering according to the last letter sequence and the conjugate ordering. If $k$ is the axial distance from $i$ to $i - 1$ in $T_{p},$ $- k$ is the axial distance from $i$ to $i - 1$ in $T_{p}^{*} .$ Let $M (k, 1)$ denote the $2 \times 2$ matrix obtained from $M (k, y)$ under substitution $y = 1 .$ Then (2.2.6) shows the matrix of $Z_{i ϕ}^{α, β}$ on $V_{p, q}$ is $M (k, 1)$ while the matrix of $Z_{i ϕ}^{β, α}$ on $V_{p, q}^{*}$ is $M (- k, 1) .$ Direct computation verifies the relation $\begin{matrix} (2.3.4) & I^{*} M (k, 1) I^{*} = M (- k, 1) \end{matrix}$ as is required.

It remains to show the lemma for $M_{ϕ}^{α, β} ({\overline{a}}_{i ϕ}) .$ As $Z_{1 ϕ}^{α, β}$ acts on the basis ${t_{i}}$ of $V_{α, β}^{K}$ by scalar multiplication, the decomposition of $V_{α, β}^{K}$ into $Z_{2 ϕ}^{α, β}$ invariant subspaces as above is valid for $Z_{1 ϕ}^{α, β} Z_{2 ϕ}^{α, β} Z_{1 ϕ}^{α, β}$ as well. It is furthermore clear from (2.2.6) that the action of $Z_{1 ϕ}^{α, β} Z_{2 ϕ}^{α, β} Z_{1 ϕ}^{α, β}$ differs from that of $Z_{2 ϕ}^{α, β}$ only on the spaces $V_{p, q}$ where the letters $1$ and $2$ appear in distinct tableaux of $T_{p} = (T_{p}^{α}, T_{p}^{β}) .$ In this case the matrix of $Z_{i ϕ}^{β, α}$ on $V_{p, q}$ and on $V_{p, q}^{*}$ is $D (1, - 1) .$ Using (2.3.4), a simple matrix calculation completes the proof for this case. This completes the proof of the lemma.

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We define the conjugate ordering of the standard tableaux $T = (T_{1}^{α}, T_{2}^{α})$ of shape $(α, α)$ as follows. Set $𝒯_{i} = {T = (T_{1}^{α}, T_{2}^{α}) : n appears in T_{i}^{α}}, i = 1, 2 .$ All standard tableaux belonging to $𝒯_{2}$ precede those belonging to $𝒯_{1}$ in the arrangement according to the last letter sequence. Rearrange the last $\frac{1}{2} f^{α, α}$ tableaux in the last letter sequence, i.e. those in $𝒯_{1},$ as follows; for $T_{1},$ $T_{2}$ in $𝒯_{1},$ $T_{a}$ precedes $T_{2}$ if $T_{2}^{*}$ precedes $T_{1}^{*}$ in the last letter sequence arrangement of the tableaux in $𝒯_{2} .$

Lemma (2.3.5) Let $M_{π}^{α, α} (a)$ denote the matrix of $π_{ϕ}^{α, α} (a)$ on $V_{α, α}^{K}$ with respect to the conjugate ordering of the basis ${t_{i}}$ of $V_{α, α}^{K},$ $a \in 𝒜^{K} (D_{n}) .$ Set $R_{f} = (\begin{matrix} - I_{\frac{1}{2} f} & I_{\frac{1}{2} f}^{*} \\ I_{\frac{1}{2} f}^{*} & I_{\frac{1}{2} f} \end{matrix}), f = f^{α, α}$ Then $\begin{matrix} (2.3.6) & R_{f} M_{ϕ}^{α, α} R_{f}^{- 1} = (\begin{matrix} M_{1} (a) & 0 \\ 0 & M_{2} (a) \end{matrix}) . \end{matrix}$

Proof.

If $(α, (α_{i} -))$ is a double partition of $n - 1$ contained in $(α, α)$ then so is $((α_{i} -), α),$ $(α_{i} -)$ as in the proof of Theorem (2.2.7). Thus we have the decomposition, in the conjugate ordering, $\begin{matrix} V_{α, α}^{K} = V_{(α, (α_{s} -))} \oplus \dots \oplus V_{(α, (α_{1} -))} \oplus V_{((α_{1} -), α)} \oplus \dots \oplus V_{((α_{s} -), α)} & (2.3.7) \end{matrix}$ of $V_{α, α}^{K}$ as $𝒜 (D_{n - 1}) -modules,$ $𝒜 (D_{n - 1})$ generated by ${{\overline{a}}_{1 ϕ}, a_{2 ϕ}, \dots, a_{(n - 1) ϕ}} .$ Thus Lemma (2.3.3) shows that for $a \in 𝒜 (D_{n - 1}),$ $M_{ϕ}^{α, α} (a)$ is of the form $\begin{matrix} (2.3.8) & (\begin{matrix} A & 0 \\ 0 & I^{*} A I^{*} \end{matrix}) \end{matrix}$ which is easily seen to commute with $R_{f} .$

Hence we need to show (2.3.6) only for $M_{ϕ}^{α, α} (a_{n ϕ}) .$ If the letters $n - 1$ and $n$ appear in the tableau $T_{2}^{α}$ of $T_{p} = (T_{1}^{α}, T_{2}^{α}) \in 𝒯_{2},$ the proof of lemma (2.3.3) shows the matrix of $Z_{n ϕ}^{α, α}$ on the subspaces with corresponding basis ${t_{p}, t_{p}^{*}}$ or ${t_{p}, t_{q}, t_{p}^{*}, t_{q}^{*}}$ if $(n - 1, n) T_{p} = T_{q},$ $p \neq q,$ is of the from (2.3.8) and the above reasoning applies. If the letters $n - 1$ and $n$ appear in distinct tableaux of $T_{p} = (T_{1}^{α}, T_{2}^{α}),$ with $T_{p} \in 𝒯_{2},$ then $(n - 1, n) T_{p} = T_{q} \in 𝒯_{1} .$ Thus $T_{q}^{*} \in 𝒯_{2}$ and we can choose $T_{p}$ such that $T_{p} < T_{q}^{*}$ in the last letter sequence arrangement of the tableaux belonging to $𝒯_{2} .$ Then $T_{p} < T_{q}^{*} < T_{q} < T_{p}^{*}$ is the arrangement of the tableaux according to the conjugate ordering. Taking the same ordering of the corresponding basis, the matrix of $Z_{n ϕ}^{α, α}$ on the subspace with basis ${t_{p}, t_{q}^{*}, t_{q}, t_{p}^{*}}$ is of the form $(\begin{matrix} a_{11} & \cdot & a_{12} & \cdot \\ \cdot & a_{22} & \cdot & a_{21} \\ a_{21} & \cdot & a_{22} & \cdot \\ \cdot & a_{12} & \cdot & a_{11} \end{matrix})$ where ${a_{i j}} = M (k, 1),$ $M (k, 1)$ defined as in lemma (2.3.3) and $k$ the axial distance from $i$ to $i - 1$ in $T_{p} .$ A simple matrix calculation shows $R_{f} A R_{f}^{- 1} = (\begin{matrix} A_{1} & 0 \\ 0 & A_{2} \end{matrix})$ This completes the proof.

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Let $V_{α, α}^{K} =_{1} V_{α, α}^{K} = \oplus_{2} V_{α, α}^{K} =$ where for basis elements $t_{p}$ corresponding to tableaux $T_{p} \in 𝒯_{2},$ $_{1} V_{α, α}^{K}$ has basis ${t_{p} + t_{p}^{*}}$ and $_{2} V_{α, α}^{K}$ has basis ${t_{p} - t_{p}^{*}} .$ By Lemma (2.3.5) the $k -linear$ maps $_{i} π_{ϕ}^{α, α} : 𝒜^{K} (D_{n}) \to End (V_{α, α}^{K})$ where the matrix of $_{i} π_{ϕ}^{α, α} (a_{w})$ with respect to the above basis is $M_{i} (a_{w}),$ $w \in W (D_{n}),$ are representations of $𝒜^{K} (D_{n}) .$

Theorem (2.3.9) For double partitions $(α, β),$ $(α, β),$ $| α | < | β |,$ and $(α, α)$ of $n \geq 4,$ the representations $π_{ϕ}^{α, β}$ and $_{i} π_{ϕ}^{α, α},$ $i = 1, 2,$ are a complete set of irreducible, inequivalent representations of $𝒜^{K} (D_{n}) .$

Proof.

By induction on $n .$ For $n = 4,$ it is a matter of direct verification. For $n > 4$ the induction assumption and the proof of Lemma (2.3.5) shows $dim {Hom}_{𝒜} (V_{α, α}^{K}) = 2,$ $𝒜 = 𝒜^{K} (D_{n}),$ ${Hom}_{𝒜} (V_{α, α}^{K})$ generated by $I_{f}$ and $I_{f}^{*},$ $f = f^{α, α} .$ Thus $_{1} π_{ϕ}^{α, α}$ and $_{2} π_{ϕ}^{α, α}$ are irreducible and inequivalent. The argument employed in Theorem (2.2.14) suffices for the irreducibility and inequivalence of the ${π_{ϕ}^{α, α}} .$ By (2.3.7) $\begin{matrix} (2.3.10) & _{i} V_{α, α}^{K} ≅ V_{((α_{s} -), α)} \oplus \dots \oplus V_{((α_{1} -), α)}, i = 1, 2, \end{matrix}$ as $𝒜^{K} (D_{n - 1}) -modules.$ Thus none of the $π_{ϕ}^{α, β}$ are equivalent to $_{i} π_{ϕ}^{α, α},$ $i = 1, 2 .$ Finally, consideration of degrees using the formula given in Theorem (2.2.14) shows a complete set of inequivalent representations is obtained, and the representations are absolutely irreducible. Thus $K = ℚ (x)$ is a splitting field for $𝒜^{K} (D_{n}) .$ This completes the proof.

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Representations of Hecke Algebras of Finite Groups with BN-Pairs of Classical Type

Notes and References

Chapter 2.Representations of the Generic Ring Corresponding to a Coxeter System of Classical Type

Partitions and Tableaux

The Representations of 𝒜K(Bn)

The Representations of 𝒜K(An) and 𝒜K(Dn)

The Representations of $𝒜^{K} (B_{n})$

The Representations of $𝒜^{K} (A_{n})$ and $𝒜^{K} (D_{n})$