## Weights and weight spaces

Last update: 3 March 2013

## Weights and weight spaces

A finite dimensional ${\stackrel{\sim }{H}}_{n}\text{-module}$ is calibrated if it has a basis of simultaneous eigenvectors for the ${x}_{i},1\le i\le n\text{.}$ In other words, $M$ is calibrated if it has a basis $\left\{{v}_{t}\right\}$ such that for all ${v}_{t}$ in the basis and all $1\le i\le n,$

$xivt=tivt, for some ti∈ℂ*.$

Weights. Let $X$ be the abelian group generated by the elements ${x}_{1},\dots ,{x}_{n}\in {\stackrel{\sim }{H}}_{n}$ and let

$T={group homomorphisms X→ℂ*} .$

The torus $T$ can be identified with ${\left({ℂ}^{}\right)}^{n}$ by identifying the element $t=\left({t}_{1},\dots ,{t}_{n}\right)\in {\left({ℂ}^{*}\right)}^{n}$ with the homomorphism given by $t\left({x}_{i}\right)={t}_{i},$ $1\le i\le n\text{.}$ The symmetric group ${S}_{n}$ acts on ${ℤ}^{n}$ by permuting coordinates and this action induces an action of ${S}_{n}$ on $T$ given by

$(wt)(xγ) =xw-1γ, for w∈Sn,γ ∈ℤn,$

with notation as in (1.2).

Weight spaces. Let $M$ be a finite dimensional ${\stackrel{\sim }{H}}_{n}\text{-module.}$ For each $t=\left({t}_{1},\dots ,{t}_{n}\right)\in T$ the $t\text{-weight}$ space of $M$ and the generalized $t\text{-weight}$ space are the subspaces

$Mt = { m∈M ∣ xim=tim for all 1≤i≤n } and Mtgen = { m∈M ∣ for each 1≤i≤n, (xi-ti)k m=0 for some k∈ ℤ>0 } ,$

respectively. From the definitions, ${M}_{t}\subseteq {M}_{t}^{\text{gen}}$ and $M$ is calibrated if and only if ${M}_{t}^{\text{gen}}={M}_{t}$ for all $t\in T\text{.}$ If ${M}_{t}^{\text{gen}}\ne 0$ then ${M}_{t}\ne 0\text{.}$ In general $M\ne {⨁}_{t\in T}{M}_{t},$ but we do have

$M=⨁t∈TMtgen.$

This is a decomposition of $M$ into Jordan blocks for the action of $ℂ\left[X\right]=ℂ\left[{x}_{1}^{±1},\dots ,{x}_{n}^{±1}\right]\text{.}$ The set of weights of $M$ is the set

$supp(M)= {t∈T ∣ Mtgen≠0} . (3.1)$

An element of ${M}_{t}$ is called a weight vector of weight $t\text{.}$

The $\tau$ operators. The maps ${\tau }_{i}:{M}_{t}^{\text{gen}}\to {M}_{{s}_{i}t}^{\text{gen}}$ defined below are local operators on $M$ in the sense that they act on each generalized weight space ${M}_{t}^{\text{gen}}$ of $M$ seperately. The operators ${\tau }_{i}$ is only defined on the generalized weight spaces ${M}_{t}^{\text{gen}}$ such that ${t}_{i}\ne {t}_{i+1}\text{.}$

Proposition 3.2. Let $t=\left({t}_{1},\dots ,{t}_{n}\right)\in T$ be such that ${t}_{i}\ne {t}_{i+1}$ and let $M$ be a finite dimensional ${\stackrel{\sim }{H}}_{n}\text{-module.}$ Define

$τi Mtgen ⟶ Msitgen m ⟼ ( Ti- (q-q-1)xi+1 xi+1-xi ) m.$
1. The map ${\tau }_{i}:{M}_{t}^{\text{gen}}⟶{M}_{{s}_{i}t}^{\text{gen}}$ is well defined.
2. As operators on ${M}_{t}^{\text{gen}}$ $xiτi = τixi+1, if j≠i,i+1, xi+1τi = τixi, if j≠i,i+1, xjτi = τixj, if j≠i,i+1, τiτi = (qxi+1-q-1xi) (qxi-q-1xi+1) (xi+1-xi) (xi-xi+1) , if 1≤i≤n-1, τiτj = τjτi, if ∣i-j∣>1, τiτi+1τi = τi+1τi τi+1, 1≤i≤n-1,$ whenever both sides are well defined.

 Proof. (a) Note that $\left(q-{q}^{-1}\right){x}_{i+1}/\left({x}_{i+1}-{x}_{i}\right)$ is not a well defined element of ${\stackrel{\sim }{H}}_{n}$ or $ℂ\left[{x}_{1}^{±1},\dots ,{x}_{n}^{±1}\right]$ since it is a power series and not a Laurent polynomial. Because of this we will be careful to view $\left(q-{q}^{-1}\right){x}_{i+1}/\left({x}_{i+1}-{x}_{i}\right)$ only as an operator on ${M}_{t}^{\text{gen}}\text{.}$ Let us describe this operator more precisely. The element ${x}_{i}{x}_{i+1}^{-1}$ acts on ${M}_{t}^{\text{gen}}$ by ${t}_{i}{t}_{i+1}^{-1}$ times a unipotent transformation. As an operator on ${M}_{t}^{\text{gen}},$ $\left(1-{x}_{i}{x}_{i+1}^{-1}\right)={x}_{i+1}/\left({x}_{i+1}-{x}_{i}\right)$ is invertible since it has determinant ${\left(1-{t}_{i}{t}_{i+1}^{-1}\right)}^{d}$ where $d=\text{dim}\left({M}_{t}^{\text{gen}}\right)\text{.}$ Since this determinant is nonzero $\left(q-{q}^{-1}\right){x}_{i+1}/\left({x}_{i+1}-{x}_{i}\right)=\left(q-{q}^{-1}\right){\left(1-{x}_{i}{x}_{i+1}^{-1}\right)}^{-1}$ is a well defined operator on ${M}_{t}^{\text{gen}}\text{.}$ Thus the definition of ${\tau }_{i}$ makes sense. The operator identities ${x}_{i}{\tau }_{i}={\tau }_{i}{x}_{i+1}{x}_{i+1}{\tau }_{i}={\tau }_{i}{x}_{i},$ and ${x}_{j}{\tau }_{j}={\tau }_{i}{x}_{j},$ if $i\ne i,i+1$ now follow easily from the definition of the ${\tau }_{i}$ and the identities in (1.4). These identities imply that ${\tau }_{i}$ maps ${M}_{t}^{\text{gen}}$ into ${M}_{{s}_{i}t}^{\text{gen}}\text{.}$ All of the operator identities in part (b) are proved by straightforward calculations of the same flavour as the calculation of ${\tau }_{i}{\tau }_{i}$ given below. We shall not give the details for the other cases. The only one which is really tedious is the calculation for the proof of ${\tau }_{i+1}{\tau }_{i}{\tau }_{i+1}={\tau }_{i}{\tau }_{i+1}{\tau }_{i}\text{.}$ For a more pleasant (but less elementary) proof of this identity see Proposition 2.7 in [Ram1998]. Since ${t}_{i}\ne {t}_{i+1}$ both ${\tau }_{i}:{M}_{t}^{\text{gen}}\to {M}_{{s}_{i}t}^{\text{gen}}$ and ${\tau }_{i}:{M}_{{s}_{i}t}^{\text{gen}}\to {M}_{t}^{\text{gen}}$ are well defined. Let $m\in {M}_{t}^{\text{gen}}\text{.}$ then $τiτim = ( Ti- (q-q-1)xi+1 xi+1-xi ) ( Ti- (q-q-1)xi+1 xi+1-xi ) m = ( Ti2-Ti (q-q-1)xi+1 xi+1-xi - (q-q-1)xi+1 xi+1-xi Ti+ (q-q-1)2 xi+12 (xi+1-xi)2 ) m = ( (q-q-1)Ti +1-Ti (q-q-1)xi+1 xi+1-xi -Ti (q-q-1)xi xi-xi+1 -(q-q-1)2 xi+1 xi+1-xi ( xi+1 xi+1-xi - xi xi-xi+1 ) + (q-q-1)2 xi+12 (xi+1-xi)2 ) m = ( (q-q-1)Ti +1-(q-q-1) Ti+(q-q-1)2 xixi+1 (xi+1-xi) (xi-xi+1) ) m = q2xixi+1 -2xixi+1+ q-2xixi+1 -xi2+2xi xi+1- xi+12 (xi+1-xi) (xi-xi+1) m = (qxi+1-q-1xi) (qxi-q-1xi+1) (xi+1-xi) (xi-xi+1) m.$ $\square$

Let $w\in {S}_{n}\text{.}$ Let $w={s}_{{i}_{1}}\dots {s}_{{i}_{p}}$ be a reduced word for $w$ and define

$τw= τi1… τip. (3.3)$

Since the $\tau \text{-operators}$ satisfy the braid relations the operator ${\tau }_{w}$ is independent of the choice of the reduced word for $w\text{.}$ The operator $\tau$ is a well defined operator on ${M}_{t}^{\text{gen}}$ if $t=\left({t}_{1},\dots ,{t}_{n}\right)$ is such that ${t}_{i}\ne {t}_{j}$ for all pairs $i such that $w\left(i\right)>w\left(j\right)\text{.}$ One may use the relations in (1.5) to rewrite ${\tau }_{w}$ in the form

$τw=∑u≤w Twauw (x1,…,xn)$

where ${a}_{uw}\left({x}_{1},\dots ,{x}_{n}\right)$ are rational functions in the variables ${x}_{1},\dots ,{x}_{n}\text{.}$ (The functions ${a}_{uw}\left({x}_{1},\dots ,{x}_{n}\right)$ are analogues of the Harish-Chandra $c\text{-function}\text{,}$ see [Mac1971, 4.1] and [Opd1995, Theorem 5.3].) If $t=\left({t}_{1},\dots ,{t}_{n}\right)$ is such that ${t}_{i}\ne {t}_{j}$ for all pairs $i such that $w\left(i\right)>w\left(j\right)$ then the expression

$τw∣t= ∑u≤wTw auw (t1,…,tn) (3.4)$

is a well defined element of the Iwahori-Hecke algebra ${H}_{n}\text{.}$ If $w=uv$ with $\ell \left(w\right)=\ell \left(u\right)+\ell \left(v\right)$ then

$τw∣t= τu∣vt τv∣t. (3.5)$

The following result will be crucial to the proof of Theorem 5.5. This result is due to D. Barbasch and P. Diaconis [Dia1997] (in the $q=1$ case). The proof given below is a $q\text{-version}$ of a proof for the $q=1$ case given by S. Fomin [Fom1997].

Proposition 3.6. Let ${w}_{0}$ be the longest element of ${S}_{n},$

$w0= ( 12…n-1n nn-1…21 ) .$

Let $a\in {ℂ}^{*}$ and fix $t=\left(a,a{q}^{2},a{q}^{4},\dots ,a{q}^{2\left(n-1\right)}\right)\text{.}$ Then

$τw0∣t= ∑w∈SnTw (-q)ℓ(w0)-ℓ(w) ,$

where $\ell \left({w}_{0}\right)=\left(\genfrac{}{}{0}{}{n}{2}\right)\text{.}$

 Proof. Let $1\le k\le n\text{.}$ Then there is a $v\in {S}_{n}$ such that ${w}_{0}=v{s}_{k}$ and $\ell \left({w}_{0}\right)=\ell \left(v\right)+1\text{.}$ So $τw0=τvτk =τv ( Tk- (q-q-1)xk+1 xk+1-xk )$ and $τw0∣t= τv∣skt ( Tk- (q-q-1)tk+1 tk+1-tk ) =τv∣skt ( Tk- (q-q-1)q2tk (q2-1)tk ) =τv∣skt (Tk-q).$ Right multiplying by ${T}_{k}+{q}^{-1}$ and using the relation (1.3) gives $τw0∣t (Tk+q-1)= τv∣skt (Tk-q) (Tk+q-1) =0.$ The element $h={\sum }_{w\in {S}_{n}}{T}_{w}{\left(-q\right)}^{\ell \left({w}_{0}\right)-\ell \left(w\right)}$ is a multiple of the minimal central idempotent in ${H}_{n}$ corresponding to the representation $\varphi$ given by $\varphi \left({T}_{k}\right)=-{q}^{-1},$ for all $1\le k\le n\text{.}$ Up to multiplication by constants, it is the unique element in ${H}_{n}$ such that $h\left({T}_{k}+{q}^{-1}\right)=0$ for all $1\le k\le n\text{.}$ The lemma follows by noting that the coefficients of ${T}_{{w}_{0}}$ in $h$ and ${\tau }_{{w}_{0}}{\mid }_{t}$ are both 1. $\square$

The action of the $\tau \text{-operators}$ on weight vectors will be particularly important to the proofs of the results in later sections. Let us record the following facts.

Let $M$ be an ${\stackrel{\sim }{H}}_{n}\text{-module}$ and let ${m}_{t}$ be a weight vector in $M$ of weight $t=\left({t}_{1},\dots ,{t}_{n}\right)\text{.}$

$\begin{array}{cc}\text{(3.7a)}& \text{If} {t}_{i}\ne {t}_{i+1} \text{then}\\ & {\tau }_{i}{m}_{t}=\left({T}_{i}-\frac{\left(q-{q}^{-1}\right){x}_{i+1}}{{x}_{i}-{x}_{i+1}}\right){m}_{t}=\left({T}_{i}-\frac{\left(q-{q}^{-1}\right){t}_{i+1}}{{t}_{i}-{t}_{i+1}}\right){m}_{t}\\ & \text{is a weight vector of weight} {s}_{i}t\text{.}\\ \text{(3.7b)}& \text{By the second set of identities in Proposition 3.2 (b),} {\tau }_{i}{\tau }_{i}{m}_{t}=\left(q{t}_{i+1}-{q}^{-1}{t}_{i}\right)\left(q{t}_{i}-{q}^{-1}{t}_{i+1}\right){\left({t}_{i+1}-{t}_{i}\right)}^{-1}{\left({t}_{i}-{t}_{i+1}\right)}^{-1}{m}_{t}\text{. Thus}\\ & \text{If} {t}_{i}\ne {t}_{i+1} \text{and} {t}_{i}\ne {q}^{±2}{t}_{i+1} \text{then} {\tau }_{i}{m}_{t}\ne 0\text{.}\end{array}$

## Notes and References

This is an excerpt of the preprint entitled Skew shape representations are irreducible authored by Arun Ram in 1998.

Research supported in part by National Science Foundation grant DMS-9622985, and a Postdoctoral Fellowship at Mathematical Sciences Research Institute.