## The Hyperoctahedral Group

Last update: 31 March 2012

## The Hyperoctahedral Group

The hyperoctahedral group $W{B}_{n}$ is the group of signed permutations of $1,2,...,n,$ i.e. bijections $w:\left\{-n,...,-2,-1,1,2,...,n\right\}\to \left\{-n,...,-2,-1,1,2,...,n\right\}$ such that $w\left(-i\right)=-w\left(i\right).$ There are multiple notations for signed permutations

1. Two line notations: where $w\left(i\right)$ in the second line is below $i$ in the first line, for $1\le i\le n.$ $w= 1 2 3 4 5 6 3 -1 5 -6 2 -4$
2. In cycle notation as permutations in the symmetric group ${S}_{2n}.$ $w= (1,3,5,2,-1,-3,-5,-2) (4,-6).$
3. Matrix notation: where the $\left(|w\left(i\right)|,i\right)$ entry is $1$ if $w\left(i\right)$ is positive and $-1$ if $w\left(i\right)$ is negative, and all other entries are $0.$ $w= 0 -1 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 -1 0 0 1 0 0 0 0 0 0 -1 0 0$
4. Diagram notation: where the ${i}^{\mathrm{th}}$ dot in the top row is connected to the $|w\left(i\right)|$th dot in the bottom row and the edge is labeled by $-1$ if $w\left(i\right)$ is negative. $w= -1 -1 -1$
The hyperoctahedral group is also called the Weyl group of type ${B}_{n}$ and the Weyl group of type ${C}_{n}$ and is the same as the group It is the group of $n×n$ matrices such that
1. there is exactly one nonzero entry in each row and each column,
2. each nonzero entry is $±1.$
The group $W{B}_{n}$ is isomorphic to the wreath product $\left(ℤ/2ℤ\right)\wr {S}_{n}$ and has order ${2}^{n}n!.$

The reflections in $W{B}_{n}$ are the elements $sεi = (i,-i) sεi-εj = (i,j) (-i,-j), sεi+εj = (i,-j) (-i,j).$ The simple reflections are $s1 = (1,-1) = -1 ⋯ and si = (i-1,i) = ⋯ ⋯$ for $2\le i\le n.$

The hyperoctahedral group $W{B}_{n}$ can be presented by generators ${s}_{1},{s}_{2},...,{s}_{n-1}$ and relations

 Proof. There are three things to show: The simple reflections in $W{B}_{n}$ satisfy the given relations. The simple reflections in $W{B}_{n}$ generate $W{B}_{n}.$ The group given by generators ${s}_{1},...,{s}_{n}$ and the relations in the statement has ${2}^{n}n!.$ The following pictures show that ${s}_{1},...,{s}_{n}\in W{B}_{n}$ satisfy the relations in the statement of the theorem. $-1 -1 = -1 = -1 = -1 -1 -1 = -1 -1 = -1 -1$ For each $1\le i\le n$ let $ti = (i,-i) = -1 ith ⋯ ⋯ = sisi-1⋯s3s2s1s2s3⋯si-1si.$ Every $w\in W{B}_{n}$ can be written as $w=\pi {t}_{{i}_{1}}\cdots {t}_{{i}_{k}}$ where $\pi \in {S}_{n}$ is the permutation given by $\pi \left(i\right)=|w\left(i\right)|$ and Pictorially $w = -1 -1 -1 = -1 -1 -1 = πt1t3t6$ Since ${s}_{2},...,{s}_{n}$ generate ${S}_{n}$ and ${t}_{k}={s}_{k}\cdots {s}_{2}{s}_{1}{s}_{2}\cdots {s}_{k}$ for $1\le k\le n,$ it follows that ${s}_{1},...,{s}_{n}$ generate $W{B}_{n}.$ Let ${G}_{n}$ be the free group generated by ${s}_{1},...,{s}_{n}$ modulo the relations in the statement of the theorem. We will show that every element $w\in {G}_{n}$ is either $w\in {G}_{n-1},$ $w={w}_{1}{s}_{n}{s}_{n-1}\cdots {s}_{k},$ with $w={w}_{1}{s}_{n}{s}_{n-1}\cdots {s}_{2}{s}_{1}{s}_{2}\cdots {s}_{k},$ with Let $w\in {G}_{n}$ and assume that First we will show that every element of ${G}_{n}$ can be written in the form Suppose $w={w}_{1}{s}_{n}{w}_{2}{s}_{n}{w}_{3},$ with ${w}_{1},{w}_{2},{w}_{3}\in {G}_{n-2}.$ Then, by the induction assumption, $w = w1snasn-1bsn-1sncsn-1dsn-1 = w1asn sn-1b sn-1c ⏟ snsn-1dsn-1 = w1asnxsn-1ysn-1snsn-1dsn-1 = w1axsnsn-1ysnsn-1sndsn-1 = w1axsnsn-1snysn-1sndsn-1 = w1axsn-1 ⏟ sn sn-1ysnd ⏟ snsn-1 = Asnβsn-1γsn-1snsn-1 = Aβsnsn-1γsnsn-1sn = Aβsnsn-1snγsn-1sn = Aβsn-1 ⏟ sn sn-1γsn-1 ⏟ sn = w′1sn w′2sn.$ It follows that if $w={w}_{1}{s}_{n}{w}_{2}{s}_{n}{w}_{3}{s}_{n}$ with ${w}_{1},{w}_{2},{w}_{3}\in {G}_{n-2}$ then $w={w\prime }_{1}{s}_{n}{w\prime }_{2}{s}_{n}{s}_{n}={w\prime }_{1}{s}_{n}{w\prime }_{2}.$ Then, by the induction assumption $w = w1sn w2sn,$ with ${w}_{2}\in {G}_{n-2},$ or ${w}_{2}=a{s}_{n-1}\cdots {s}_{k}$ and $a\in {G}_{n-2},$ or ${w}_{2}=a{s}_{n-1}\cdots {s}_{2}{s}_{1}{s}_{2}\cdots {s}_{k}$ and $a\in {G}_{n-2}.$ Case 1. $w = w1snw2sn = w1w2snsn = w1w2 ∈ Gn-1.$ Case 2. $w = w1sna sn-1⋯sksn = w1asn sn-1 snsn-2⋯sk = w1asn-1sn sn-1 sn-2 ⋯sk = w′1 snsn-1 ⋯sk,$ with ${w\prime }_{1}\in {G}_{n-1}.$ Case 3. with ${w\prime }_{1}\in {G}_{n-1}.$ So $\mathrm{Card}\left({G}_{n}\right)\le \mathrm{Card}\left({G}_{n-1}\right)\cdot 2n.$ So $\mathrm{Card}\left({G}_{n}\right)\le {2}^{n}n!.$ $\square$

Let $\left(\alpha ,\beta \right)$ be a pair of partitions sucht that the total number of boxes in $\alpha$ and $\beta$ is $n.$ A standard tableau of shape $\left(\alpha ,\beta \right)$ is a filling $T$ of the boxes of $\alpha$ and $\beta$ with $1,2,...,n$ such that, in each partition,

1. the rows of $T$ are increasing left to right,
2. the columns of $T$ are increasing top to bottom.
$T = 1 5 7 4 10 12 8 13 α 2 3 9 6 11 β$ The rows and columns of each partition are indexed as for matrices, The numbers $c\left(b\right)$ and $\mathrm{sgn}\left(b\right)$ are the content and the sign of the box $b,$ respectively. $0 1 2 -1 0 1 -2 -1 0 1 2 -1 0 Contents of boxes 1 1 1 1 1 1 1 1 -1 -1 -1 -1 -1 Signs of boxes$

1. The irreducible representations ${S}^{\left(\alpha ,\beta \right)}$ of the hyperoctahedral group $W{B}_{n}$ are indexed by pairs of partitions $\left(\alpha ,\beta \right)$ with $n$ boxes total.
2. $\mathrm{dim}{S}^{\left(\alpha ,\beta \right)}=#$ of standard tableaux of shape $\left(\alpha ,\beta \right).$
3. The irreducible $W{B}_{n}-$module ${S}^{\left(\alpha ,\beta \right)}$ is given by with basis $\left\{{v}_{T}\right\}$ and with $W{B}_{n}$ action given by $s1vT = sgn(T(1))vT, sivT = (si)TTvT + (1+(si)TT) vsiT, i=2,3,...,n,$ where
1. ${s}_{1}=\left(1,-1\right)$ and ${s}_{i}=\left(i,i-1\right)\left(-i,-\left(i-1\right)\right),$
2. $c\left(T\left(i\right)\right)$ is the content of the box containing $i$ in $T,$
3. $\mathrm{sgn}\left(T\left(i\right)\right)$ is the sign of the box containing $i$ in $T,$
4. ${s}_{i}T$ is the same as $T$ except that $i$ and $i-1$ are switched, and
5. ${v}_{{s}_{i}T}=0,$ if ${s}_{i}T$ is not a standard tableau.

 Proof. We must show that The ${S}^{\left(\alpha ,\beta \right)}$ are $W{B}_{n}$ modules, The ${S}^{\left(\alpha ,\beta \right)}$ are irreducible $W{B}_{n}$ modules, The ${S}^{\left(\alpha ,\beta \right)}$ are inequivalent $W{B}_{n}$ modules, These are all the simple $W{B}_{n}$ modules. The proofs are similar to the proofs of the analogous statements in the symmetric group case. $\square$

Define elements $1\le k\le n,$ in the group algebra of the hyperoctahedral group $W{B}_{n}$ by The action of these elements of the $W{B}_{n}-$module ${S}^{\left(\alpha ,\beta \right)}$ is given by $ykvT = sgn(T(k))vT and xkvT = c(T(k))vT,$ where $\mathrm{sgn}\left(T\left(k\right)\right)$ and $c\left(T\left(k\right)\right)$ are the sign and the content of the box containing $k$ in $T,$ respectively.

 Proof. By induction $ykvT = skyk-1skvT = skyk-1 ((sk)TTvT + (1+(sk)TT)vskT) = sk (sgn(T(k-1)) (sk)TT + sgn(T(k)) (1+(sk)TT)vskT) = sgn(T(k))sk ( skvT + (sgn(T(k-1)) -sgn(T(k))) ( 1+sgn(T(k)) sgn( T(k-1) )) c(T(k)) -c( T(k-1)) )vT = sgn(T(k))vT.$ $\square$

Note that $dim(S(α,β)) = ( n|α| ) fαfβ,$ where ${f}^{\alpha }$ is the number of standard tableaux of shape $\alpha .$ Thus ${f}^{\alpha }$ is the dimension of the irreducible module ${S}^{\alpha }$ for the symmetric group ${S}_{|\alpha |}.$ Then $∑α,β dim(S(α,β))2 = ∑α,β ( n|α| )2 (fα)2 (fβ)2 = ∑k=0n ( nk )2 ∑α⊢k (fα)2 ∑β⊢n-k (fβ)2 = ∑k=0n n! k!(n-k)! k!(n-k)! (nk) = ∑k=0n n!(nk) = 2nn!.$

## Notes and References

Where are these from?

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