## On the weight space representations of the Brauer algebras

Last update: 10 September 2013

The following notes are not intended to be complete in any sense. They are merely facts that I don't wish to forget.

## Weight space representations

Let $n$ be a positive integer and let $I=\left\{-n,-\left(n-1\right),\dots ,-2,-1,0,1,2,\dots ,n-1,n\right\}\text{.}$ Let $\left\{{v}_{i} | i\in I\right\}$ be a set of independent noncommuting variables. Define $V$ to be the vector space over $ℂ$ with basis $\left\{{v}_{i} | i\in I\right\},$ and define $V⊗m=ℂ-span { vi1 vi2⋯ vim | ik∈I } ,$ so that the words (simple tensors) ${v}_{{i}_{1}}{v}_{{i}_{2}}\cdots {v}_{{i}_{m}}$ are a basis of ${V}^{\otimes m}\text{.}$

Let ${x}_{1},{x}_{2},\dots ,{x}_{n}$ be commuting, independent variables. Define ${x}_{0}=1$ and ${x}_{-i}={x}_{i}^{-1}$ for $i=1,2,\dots ,n,$ so that ${x}_{i}$ is defined for each $i\in I\text{.}$ Define the weight of each word ${v}_{{i}_{1}}\cdots {v}_{{i}_{m}}$ of ${V}^{\otimes m}$ to be $wt(vi1⋯vim) =xi1⋯xim.$ Note that the weight of a word is always of the form ${x}^{a}={x}_{1}^{{a}_{1}}{x}_{2}^{{a}_{2}}\cdots {x}_{n}^{{a}_{n}}$ where $a=\left({a}_{1},{a}_{2},\dots ,{a}_{n}\right)\in {ℤ}^{n}\text{.}$ For each sequence $a\in {ℤ}^{n}$ define $(V⊗m)a=ℂ -span { vi1⋯vim | wt (vi1⋯vim) =xa } .$

For $0\le k\le m-1,$ define an action of the generators ${G}_{k}$ and ${E}_{k}$ of ${B}_{m}\left(2n+1\right)$ on ${V}^{\otimes m}$ by $(vi1vi2⋯vim) Gk = vi1⋯vik-1 vik+1vik vik+2⋯ vim, (vi1vi2⋯vim) Ek = δik,-ik+1 ∑j∈Ivi1⋯ vik-1vj v-jvik+2 ⋯vim. (1.1)$ By writing out explicitly the action of a general $m\text{-diagram}$ one checks easily that the action defined in (1.1) extends to a well-defined action of ${B}_{m}\left(2n+1\right)$ on ${V}^{\otimes m}\text{.}$ Since the action of the Brauer algebra on ${V}^{\otimes m}$ does not change weights of the the words, ${\left({V}^{\otimes m}\right)}_{a}$ is always a ${B}_{m}\left(n\right)$ submodule of ${V}^{\otimes m}\text{.}$

Let ${H}_{n}$ denote the hyperoctahedral group of $n×n$ signed permutation matrices given in the usual way by generators and relations. Define an action of ${H}_{n}$ on the variables ${v}_{i},i\in I$ by $sivj= { v±(i+1), if j=±i, v±i, if j=±(i+1), vj, otherwise, for 1≤i≤n-1, andsnvj= { v∓n, if j=±n, vj, otherwise.$ and define an action of ${W}_{n}$ on ${V}^{\otimes m}$ by $w\left({v}_{{i}_{1}}\cdots {v}_{{i}_{m}}\right)={v}_{w\left({i}_{1}\right)}\cdots {v}_{w\left({i}_{m}\right)}\text{.}$ Define an action of ${H}_{n}$ on monomials ${x}_{{i}_{1}}\cdots {x}_{{i}_{m}}$ and on sequences $a=\left({a}_{1},\dots ,{a}_{n}\right)\in {ℤ}^{n}$ by requiring that for all words ${v}_{{i}_{1}}\cdots {v}_{{i}_{m}}$ and $w\in {H}_{n},$ $If wt (vi1⋯vim)= x1a1⋯xnan= xa, then wt (w(vi1⋯vim)) =w(xa)=xwa. (1.2)$

For each $a\in {ℤ}^{n}$ define $\lambda \left(a\right)$ to be the partition determined by rearranging the sequence $\left(|{a}_{1}|,|{a}_{2}|,\dots ,|{a}_{n}|\right)$ into decreasing order. Then $(V⊗m)a≃ (V⊗m)λ(a),$ as ${B}_{m}\left(2n+1\right)$ modules.

 Proof. This is clear one only needs to check that the action of ${H}_{n}$ and of ${B}_{m}\left(2n+1\right)$ commute and that the action of ${B}_{m}\left(2n+1\right)$ preserves weights. $\square$

Let $\lambda =\left({\lambda }_{1},\dots ,{\lambda }_{r}\right),$ ${\lambda }_{1}\ge \cdots \ge {\lambda }_{r}>0$ be a partition of $m\text{.}$ We say that the subalgebra ${B}_{\lambda }\left(2n+1\right)={B}_{{\lambda }_{1}}\left(2n+1\right)\otimes \cdots \otimes {B}_{{\lambda }_{r}}\left(2n+1\right)$ is a "Young subalgebra" of the Brauer algebra ${B}_{m}\left(2n+1\right)\text{.}$ Given a representation $M$ of the Young subalgebra ${B}_{\lambda }\left(2n+1\right)$ the induced representation of ${B}_{m}\left(2n+1\right)$ is given by ${B}_{m}\left(2n+1\right){\otimes }_{{B}_{\lambda }\left(2n+1\right)}M\text{.}$

In general the weight space representation ${\left({V}^{\otimes m}\right)}_{\mu }$ is not isomorphic to an induced representation from a Young subalgebra of ${B}_{m}\left(2n+1\right)\text{.}$ In fact if $\mu$ is a partition of $m$ then ${\left({V}^{\otimes m}\right)}_{\mu }$ is never an induced representation from a Young subalgebra.

Proof.

Let us use the notation of symmetric functions. Let $s{b}_{\lambda }\left({x}_{1}^{±1},\dots ,{x}_{n}^{±1},1\right)$ denote the orthogonal Schur function which describes the character of the irreducible $SO\left(2n+1\right)$ module indexed by $\lambda$ and let $m{b}_{\lambda }={\sum }_{\gamma \in {H}_{n}\lambda }{x}^{\gamma }$ denote the monomial symmetric function corresponding to the partition $\lambda \text{.}$ Here ${H}_{n}$ is the hyperoctahedral group. Let the weight multiplicities in the irreducibles are given by coefficients ${K}_{\lambda \mu }$ such that $sbλ=∑μ Kλμmbμ.$ It follows from the fact that ${K}_{\lambda \mu }=0$ unless $\mu \le \lambda$ in dominance (for the root system of type B) that $Kλμ=0 if |λ|<|μ|. (1.5)$

Let ${\beta }_{\mu }$ denote the character of the Brauer algebra ${B}_{m}\left(2n+1\right)$ action on the weight space ${\left({V}^{\otimes m}\right)}_{\mu }\text{.}$ Then it is easy to see from the Schur-Weyl duality that $βμ=∑λ∈Bˆm χλKλμ$ where ${\stackrel{ˆ}{B}}_{m}$ is an index set for the irreducibles of ${B}_{m}\left(2n+1\right)$ and ${\chi }^{\lambda }$ is the irreducible character of the Brauer algebra ${B}_{m}\left(2n+1\right)\text{.}$ If $\mu$ is a partition of $m$ then (1.5) $βμ= ∑λ⊢m χλKλμ (1.6)$

Now let us use the Frobenius characteristic map. Since induction from Young subalgebras of the Brauer algebra corresponds to taking tensor products of $SO\left(2n+1\right)$ representations under the Frobenius characteristic map, the ${\beta }_{\mu }$ will be an induced representation from a Young subalgebra ${B}_{\lambda }\left(2n+1\right),$ $\lambda =\left({\lambda }_{1},\dots ,{\lambda }_{r}\right)$ if and only if there are symmetric functions ${H}_{1},{H}_{2},\dots ,{H}_{r}$ such that

 (1.7a) ${H}_{j}={\sum }_{\nu \in {\stackrel{ˆ}{B}}_{{\lambda }_{j}}}{c}_{j\nu }s{b}_{\nu }$ for some nonnegative integers ${c}_{j\nu },$ and 1.7b) ${H}_{1}{H}_{2}\cdots {H}_{r}={\sum }_{\lambda \in {\stackrel{ˆ}{B}}_{m}}s{b}_{\lambda }{K}_{\lambda \mu }\text{.}$
Furthermore, if we also require that $\mu ⊢m$ then (1.5) forces
 1.7c) ${H}_{1}{H}_{2}\cdots {H}_{r}={\sum }_{\lambda ⊢m}s{b}_{\lambda }{K}_{\lambda \mu }\text{.}$
Now we note three facts: If $\nu ⊢s$ and $\pi ⊢p$ then the decomposition of $s{b}_{\nu }s{b}_{\pi }$
 1.8a) Contains a nonzero term $s{b}_{\gamma }$ for some $\gamma ⊢s+t$ (1.8b) Contains a nonzero term $s{b}_{\gamma }$ for some partition $\gamma ⊢s+t-2\text{.}$ 1.8c) Does not contain a nonzero term $s{b}_{\gamma }$ for any partition $|\gamma |>s+t\text{.}$
Both of these facts can be proved easily by looking at the decomposition rule of Black-King-Wybourne [BKW1983] given in Theorem 5.3 of [Sun1990].

Let $\mu ⊢m$ and assume that symmetric functions ${H}_{1},\dots ,{H}_{r}$ exist satisfying (1.7abc). Then (1.8c) combined with (1.7c) implies that ${H}_{j}={\sum }_{\nu \in {\stackrel{ˆ}{B}}_{{\lambda }_{j}}}{c}_{j\nu }s{b}_{\nu }$ where ${c}_{j\nu }>0$ for at least one $\nu ⊢{\lambda }_{j}\text{.}$ Then (1.8b) implies that ${H}_{1}\cdots {H}_{r}$ contains a nonzero term $s{b}_{\gamma }$ for some partition $\gamma ⊢s+t-2\text{.}$ This is a contradiction to (1.7c). Thus ${\beta }_{\mu }$ is not equal to a character induced from a Young subalgebra.

$\square$

For each partition $\lambda \in {\stackrel{ˆ}{B}}_{m},$ the dimension of the $\lambda$ weight space is $dim((V⊗m)λ)= ∑s0+2(s1+⋯+sn)=m-|λ| ( m s0,λ1+s1, s1,λ2+s2, s2,…,λ+sn ,sn )$

 Proof. A basis vector $w={v}_{{i}_{1}}\cdots {v}_{{i}_{m}}$ in ${V}^{\otimes m}$ is of weight $\lambda =\left({\lambda }_{1},\dots ,{\lambda }_{n}\right)$ if there is a sequence of positive integers ${s}_{0},{s}_{1},\dots ,{s}_{n}$ such that $w$ contains ${s}_{0}$ ${v}_{0}\text{'s,}$ ${\lambda }_{1}+{s}_{1}$ ${v}_{1}\text{'s,}$ ${s}_{1}$ ${v}_{-1}\text{'s,}$ and so on. The multinomial coefficient is just the number of ways of choosing the positions of these letters. $\square$

In type $C$ the dimension of $V$ is $2n$ and there are no vectors ${v}_{0}$ of weight $0$ in $V\text{.}$ The dimension of the zero weight space in type $C$ is given by $dim((V⊗m)0)= { 0, if m is odd, (2r)!r!2 ∑s1+⋯+sn=r (rs1,s2,…,sn)2 , if m=2r.$

 Proof. This is clear from (1.9) and the fact that $(2rs1,s2,…,sn) = (2r)! s1!s1!⋯ sn!sn! =(2r)!r!2 (rs1,s2,…,sn)2 .$ $\square$

## Notes and References

This is a copy of the paper On the weight space representations of the Brauer algebras by Arun Ram, Department of Mathematics, University of Wisconsin, Madison, WI 53706, January 20, 1994. This paper was supported in part by a National Science Foundation postdoctoral fellowship.