Classification of Calibrated Representations

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

Last update: 27 April 2013

Classification of Calibrated Representations

Structural results

We first examine some properties that hold for irreducible modules that are calibrated, i.e., can be decomposed into a direct sum of weight spaces (see (2.10)). This section follows closely similar results for affine Hecke algebras in [Ram1998].

Lemma 4.1. Let $M$ be an irreducible calibrated module. Then, for all $γ \in 𝔥_{ℂ}$ such that $M_{γ} \neq 0,$

$γ (α_{i}) \neq 0$ for all $1 \leq i \leq n,$
$dim (M_{γ}) = 1 .$

Proof.

The proof is by contradiction. Assume $γ (α_{i}) = 0 .$ Let $ℍ A_{1}$ be the subalgebra of $ℍ$ generated by $t_{s_{i}}$ and all $x \in 𝔥_{ℂ}^{*} .$ Then the two-dimensional $ℍ A_{1}$ principal series module $M (γ)$ is irreducible and there is an $ℍ A_{1} -module$ homomorphism given by

\begin{matrix} M (γ) & ⟶ & M \\ v_{γ} & ⟼ & m_{γ} \end{matrix}

where $m_{γ}$ is a nonzero element of $M_{γ} .$ Since $M (γ)$ is simple, this is an injection and thus, $M$ is not calibrated since $M (γ)$ is not calibrated. Thus $γ (α_{i}) \neq 0 .$
(b) The proof is by contradiction. Assume $γ \in 𝔥_{ℂ}$ is such that $dim (M_{γ}) > 1 .$ Let $m_{γ}$ be a nonzero element of $M_{γ} .$ Since $M$ is calibrated $τ_{i}$ acts on $m_{γ}$ as a linear combination of the action of $t_{s_{i}}$ and a multiple of the identity. Since $M$ is irreducible, it follows from Proposition 2.5(b) that the action of the $τ -operators$ must generate all of $M .$ Thus, since $dim (M_{γ}) > 1,$ there is a sequence of $τ -operators$ such that

n_{γ} = τ_{i_{1}} τ_{i_{2}} \dots τ_{i_{p}} m_{γ}

is a nonzero vector in $M_{γ}$ which is not a multiple of $m_{γ} .$

Assume that the sequence $τ_{i_{1}} τ_{i_{2}} \dots τ_{i_{p}}$ is chosen so that $p$ is minimal. Since the $τ -operators$ in this sequence are all well defined, the elements $s_{i_{k}} \dots s_{i_{p}} γ, 1 \leq k \leq p,$ in the orbit $W γ$ correspond (under the bijection in (2.7)) to a sequence of chambers in $𝔥_{ℝ}^{*}$ on the positive side of all $H_{α}, α \in Z (γ) .$ Each chamber in this sequence shares a face with the next chamber in the sequence. Since both $n_{γ}$ and $m_{γ}$ are in $M_{γ},$ this is a sequence which begins and ends at the chamber $C .$ Since the chambers are in bijection with the elements of $W,$ it follows that $s_{i_{1}} \dots s_{i_{p}} = 1$ in $W .$

This means that there is some $1 < k \leq p$ such that $s_{i_{1}} \dots s_{i_{k}}$ is not reduced and we can use the braid relations to rewrite this word as $s_{i_{1}^{'}} \dots s_{i_{k - 2}^{'}} s_{i_{k}} s_{i_{k}} .$ By Proposition 2.5(e) the $τ -operators$ also satisfy the braid relations and so

n_{γ} = τ_{i_{1}^{'}} τ_{i_{2}^{'}} \dots τ_{i_{k - 2}^{'}} τ_{i_{k}} τ_{i_{k}} \dots τ_{i_{p}} m_{γ} .

By Proposition 2.5(c), the operator $τ_{i_{k}} τ_{i_{k}}$ above will act (on $τ_{i_{k + 1}} \dots τ_{i_{p}} m_{γ})$ by a constant $ξ \in ℂ$ and so

n_{γ} = ξ \cdot τ_{i_{1}^{'}} τ_{i_{2}^{'}} \dots τ_{i_{k - 2}^{'}} τ_{i_{k}} τ_{i_{k}} \dots τ_{i_{p}} m_{γ},

where the constant $ξ$ is nonzero since $n_{γ}$ is nonzero. But the expression

ξ^{- 1} n_{γ} = τ_{i_{1}^{'}} τ_{i_{2}^{'}} \dots τ_{i_{k - 2}^{'}} τ_{i_{k}} τ_{i_{k}} \dots τ_{i_{p}} m_{γ},

is shorter than the original expression of $n_{γ}$ and this contradicts the minimality of $p .$ It follows that $dim (M_{γ}) \leq 1 .$

$□$

Lemma 4.2. Let $M$ be an irreducible calibrated module. Suppose that $M_{γ}$ and $M_{s_{i} γ}$ are both nonzero. Then the map $τ_{i} : M_{γ} \to M_{s_{i} γ}$ is a bijection.

Proof.

By Lemma 4.1(b), $dim (M_{γ}) = dim (M_{s_{i} γ}) = 1,$ and thus it is sufficient to show that $τ_{i}$ is not the zero map. Let $v_{γ}$ be a nonzero vector in $M_{γ} .$ Since $M$ is irreducible, there must be a sequence of $τ -operators$ such that

v_{s_{i} γ} = τ_{i_{1}} \dots τ_{i_{p}} v_{γ}

is a nonzero element of $M_{s_{i} γ} .$ Let $p$ be minimal such that this is the case. Since $τ_{i} τ_{i_{1}} \dots τ_{i_{p}} v_{γ} \in M_{γ},$ it follows, as in the second paragraph of the proof of Lemma 4.1(b), that $s_{i} s_{i_{1}} \dots s_{i_{p}} = 1$ in $W .$ For notational convenience let $i_{0} = i .$ Let $0 \leq k < p$ be maximal such that $s_{i_{k}} s_{i_{k + 1}} \dots s_{i_{p}}$ is not reduced. If $k \neq 0,$ then we can use the braid relations to get

v_{s_{i} γ} = τ_{i_{1}} \dots τ_{i_{k}} τ_{i_{k}} τ_{i_{k + 2}^{'}} \dots τ_{i_{p}^{'}} v_{γ} .

Since $τ_{i_{k}} τ_{i_{k}}$ acts on $τ_{i_{k + 2}^{'}} \dots τ_{i_{p}^{'}} v_{γ}$ by a constant $ξ \in ℂ,$

v_{s_{i} γ} = ξ \cdot τ_{i_{1}} \dots τ_{i_{k - 1}} τ_{i_{k + 2}^{'}} \dots τ_{i_{p}^{'}} v_{γ},

and $ξ \neq 0$ since $v_{s_{i} γ}$ is not 0. But this contradicts the minimality of $p .$ Thus we must have that $k = 0, p = 1$ and

v_{s_{i} γ} = τ_{i} v_{γ}

Thus, since $v_{s_{i} γ} \neq 0,$ $τ_{i} \neq 0 .$

$□$

For simple roots $α_{i}$ and $α_{j}$ in $R,$ let $R_{i j}$ be the rank two root subsystem of $R$ generated by $α_{i}$ and $α_{j} .$ A weight $μ \in 𝔥_{ℂ}$ is skew if

for all simple roots $α_{i},$ $1 \leq i \leq n,$ $μ (α_{i}) \neq 0,$
for all pairs of simple roots $α_{i},$ $α_{j}$ such that ${α \in R_{i j} | μ (α) = 0} \neq \emptyset,$ the set ${α \in R_{i j} | μ (α) = \pm c_{α}}$ contains more than two elements.

Condition (a) says that $μ$ is regular for all rank 1 subsystems of $R$ generated by simple roots. Condition (b) is an “almost regular” condition on $μ$ with respect to rank 2 subsystems generated by simple roots. By the analysis in Section 3, the weights which appear in calibrated modules for graded Hecke algebras corresponding to rank two root systems are skew.

Recall from Section 2.3 that a pair $(γ, J)$ is a local region if the set

ℱ^{(γ, J)} = {w \in W | R (w) \cap Z (γ) = \emptyset and R (w) \cap P (γ) = J}

is nonempty. A local region $(γ, J)$ is skew if, for all $w \in ℱ^{(γ, J)},$ the weight $w γ$ is skew for all pairs $α_{i}, α_{j}$ of simple roots in $R .$

The following theorem specifies the weight space structure of an irreducible calibrated $ℍ -module.$

Theorem 4.3. If $M$ is an irreducible calibrated $ℍ -module$ with central character $γ \in 𝔥_{ℂ},$ then there is a unique skew local region $(γ, J)$ such that

dim (M_{w γ}) = {\begin{matrix} 1 & for all w \in ℱ^{(γ, J)}, \\ 0 & otherwise. \end{matrix}

Proof.

By Lemma 4.1 all nonzero generalized weight spaces of $M$ have dimension 1 and by Lemma 4.2 all $τ -operators$ between these weight spaces are bijections. This already guarantees that there is a unique local region $(γ, J)$ which satisfies the condition. It only remains to show that this local region is skew.

Let $ℍ_{i j}$ be the subalgebra of $ℍ$ generated by $t_{s_{i}}, t_{s_{j}}$ and $S (𝔥_{ℂ}^{*}) .$ Since $M$ is calibrated as an $ℍ -module$ it is calibrated as an $ℍ_{i j} -module$ and so all factors of a composition series of $M$ as an $ℍ_{i j} -module$ are calibrated. Thus, by the classification in Section 3, the weights of $M$ are skew. So $(γ, J)$ is a skew local region.

$□$

Construction

The following Proposition shows that the weight structure of calibrated representations as determined in Theorem 4.3 essentially forces the $ℍ -action$ on a weight basis.

Proposition 4.4. Let $M$ be a calibrated $ℍ -module$ and for all $γ \in 𝔥_{ℂ}$ such that $M_{γ} \neq 0,$ assume that

\begin{matrix} γ (α_{i}) \neq 0 for all 1 \leq i \leq n, & (A1) \end{matrix}

and

\begin{matrix} dim (M_{γ}) = 1 . & (A2) \end{matrix}

For each $b \in 𝔥_{ℂ}$ such that $M_{b} \neq 0$ let $v_{b}$ be a nonzero vector in $M_{b} .$ The vectors ${v_{b}}$ form a basis of $M .$ Let ${(t_{s_{i}})}_{c b} \in ℂ$ and $b (x) \in ℂ$ be given by

t_{s_{i}} v_{b} = \sum_{c} {(t_{s_{i}})}_{c b} v_{c} and x v_{b} = b (x) v_{b} for x \in 𝔥_{ℂ}^{*} .

Then

${(t_{s_{i}})}_{b b} = \frac{c_{α_{i}}}{b (α_{i})}$ for all $v_{b}$ in the basis,
if ${(t_{s_{i}})}_{c b} \neq 0,$ then $c = s_{i} b,$
${(t_{s_{i}})}_{b, s_{i} b} {(t_{s_{i}})}_{s_{i} b, b} = 1 - {(t_{s_{i}})}_{b b}^{2} = (1 + {(t_{s_{i}})}_{b b}) (1 + {(t_{s_{i}})}_{s_{i} b, s_{i} b}) .$

Proof.

The relation

x t_{s_{i}} - t_{s_{i}} s_{i} (x) = c_{α_{i}} \frac{x - s_{i} (x)}{α_{i}}

forces

\sum_{c} (c (x) {(t_{s_{i}})}_{c b} - {(t_{s_{i}})}_{c b} b (s_{i} x)) v_{c} = c_{α_{i}} \frac{b (x) - b (s_{i} x)}{b (α_{i})} v_{b} .

Comparing coefficients yields

c (x) {(t_{s_{i}})}_{c b} - {(t_{s_{i}})}_{c b} b (s_{i} x) = 0, if b \neq c,

and

b (x) {(t_{s_{i}})}_{b b} - {(t_{s_{i}})}_{b b} b (s_{i} x) = c_{α_{i}} \frac{b (x) - b (s_{i} x)}{b (α_{i})} .

These equations imply that

if {(t_{s_{i}})}_{c b} \neq 0, then b (s_{i} x) = c (x) for all x \in 𝔥_{ℂ}^{*},

and

{(t_{s_{i}})}_{b b} = \frac{c_{α_{i}}}{b (α_{i})} if b (α_{i}) \neq 0 and b (x) \neq b (s_{i} x) for some x \in 𝔥_{ℂ}^{*} .

Thus

t_{s_{i}} v_{b} = {(t_{s_{i}})}_{b b} v_{b} + {(t_{s_{i}})}_{s_{i} b, b} v_{s_{i} b} with {(t_{s_{i}})}_{b b} = \frac{c_{α_{i}}}{b (α_{i})} .

This completes the proof of (a) and (b). The relation $t_{s_{i}}^{2} = 1$ in $ℍ$ implies that

\begin{matrix} v_{b} & = & t_{s_{i}}^{2} v_{b} & = & [{(t_{s_{i}})}_{b b}^{2} + {(t_{s_{i}})}_{b, s_{i} b} {(t_{s_{i}})}_{s_{i} b, b}] v_{b} + [{(t_{s_{i}})}_{b b} + {(t_{s_{i}})}_{s_{i} b, s_{i} b}] {(t_{s_{i}})}_{s_{i} b, b} v_{s_{i} b} \\ = & [{(t_{s_{i}})}_{b b}^{2} + {(t_{s_{i}})}_{b, s_{i} b} {(t_{s_{i}})}_{s_{i} b, b}] v_{b}, \end{matrix}

since ${(t_{s_{i}})}_{b b} + {(t_{s_{i}})}_{s_{i} b, s_{i} b} = 0 .$ Thus

{(t_{s_{i}})}_{b, s_{i} b} {(t_{s_{i}})}_{s_{i} b, b} = 1 - {(t_{s_{i}})}_{b b}^{2} = (1 + (t_{s_{i}}) b b) (1 + (t_{s_{i}}) s_{i} b, s_{i} b) .

$□$

Theorem 4.5. Let $(γ, J)$ be skew and let $ℱ^{(γ, J)}$ index the chambers in the local region $(γ, J) .$ Define

H^{(γ, J)} = ℂ -span {v w | w \in ℱ^{(γ, J)}},

so that the symbols $v w$ are a labeled basis of the vector space $ℍ^{(γ, J)} .$ Then the following formulas make $ℍ^{(γ, J)}$ into an irreducible $ℍ -module.$ For each $w \in ℱ^{(γ, J)},$

x v w = (w γ) (x) v w for x \in 𝔥_{ℂ}^{*},

and

t_{s_{i}} v w = \frac{c_{α_{i}}}{w γ (α_{i})} v w + (1 + \frac{c_{α_{i}}}{w γ (α_{i})}) v_{s_{i} w} for 1 \leq i \leq n,

where we set $v_{s_{i} w} = 0$ if $s_{i} w \notin ℱ^{(γ, J)} .$

Proof.

Since $(γ, J)$ is skew, $(w γ) (α_{i}) \neq 0$ for all $w \in ℱ^{(γ, J)}$ and all simple roots $α_{i} .$ This implies that the coefficients in $t_{s_{i}} v w$ are well defined for all $i$ and $w \in ℱ^{(γ, J)} .$

By construction, the nonzero weight spaces of $ℍ^{(γ, J)}$ are ${(ℍ^{(γ, J)})}_{w γ}^{gen} = {(ℍ^{(γ, J)})}_{w γ}$ where $w \in ℱ^{(γ, J)} .$ Since $dim ({(ℍ^{(γ, J)})}_{u γ}) = 1$ for $u \in ℱ^{(γ, J)},$ any proper submodule $N$ of $ℍ^{(γ, J)}$ must have $N_{w γ} \neq 0$ and $N_{w' γ} = 0$ for some $w \neq w',$ with $w, w' \in ℱ^{(γ, J)} .$ This is a contradiction to Corollary 2.6. So $ℍ^{(γ, J)}$ is irreducible if it is an $ℍ -module.$

It remains to show that the defining relations for $ℍ$ are satisfied. Let $w \in ℱ^{(γ, J)} .$ Then

\begin{matrix} (s_{i} (x) t_{s_{i}} + c_{α_{i}} \frac{x - s_{i} x}{α_{i}}) v w & = & s_{i} x [\frac{c_{α_{i}}}{w γ (α_{i})} v w + (1 + \frac{c_{α_{i}}}{w γ (α_{i})}) v_{s_{i} w}] + c_{α_{i}} \frac{w γ (x) - w γ (s_{i} x)}{w γ (α_{i})} v w \\ = & \frac{c_{α_{i}}}{w γ (α_{i})} w γ (x) v w + (1 + \frac{c_{α_{i}}}{w γ (α_{i})}) s_{i} w γ (s_{i} x) v_{s_{i} x} \\ = & t_{s_{i}} x v w . \end{matrix}

Let $w \in ℱ^{(γ, J)} .$ Then

\begin{matrix} t_{s_{i}}^{2} v w & = & t_{s_{i}} [\frac{c_{α_{i}}}{w γ (α_{i})} v w + (1 + \frac{c_{α_{i}}}{w γ (α_{i})}) v_{s_{i} w}] \\ = & \frac{c_{α_{i}}}{w γ (α_{i})} [\frac{c_{α_{i}}}{w γ (α_{i})} v w + (1 + \frac{c_{α_{i}}}{w γ (α_{i})}) v_{s_{i} w}] + (1 + \frac{c_{α_{i}}}{w γ (α_{i})}) [\frac{c_{α_{i}}}{s_{i} w γ (α_{i})} v_{s_{i} w} + (1 + \frac{c_{α_{i}}}{s_{i} w γ (α_{i})}) v w] \\ = & {(\frac{c_{α_{i}}}{w γ (α_{i})})}^{2} v w + (1 + \frac{c_{α_{i}}}{w γ (α_{i})}) (1 - \frac{c_{α_{i}}}{w γ (α_{i})}) v w + 0 \\ = & v w . \end{matrix}

Now let us check the braid relations. Write $t_{s_{i}} = τ_{i} + d_{i}$ where

τ_{i} v w = (1 + \frac{c_{α_{i}}}{(w g) (α_{i})}) v_{s_{i} w} and d_{i} v w = \frac{c_{α_{i}}}{(w g) (α_{i})} v w,

for $w \in ℱ^{(γ, J)} .$ Then di is a diagonal matrix and $τ_{i}$ is a pseudo-permutation matrix, in the sense that each row and each column contains at most one nonzero entry. For a sequence $j_{1}, \dots, j_{p}$ define a diagonal matrix $d_{i}^{j_{1}, \dots, j_{p}}$ by the relation

\begin{matrix} d_{i} τ_{j_{1}} \dots τ_{j_{p}} = τ_{j_{1}} \dots τ_{j_{p}} d_{i}^{j_{1}, \dots, j_{p}} . & (4.1) \end{matrix}

If $γ$ is generic, then, for all $w \in W,$

d_{i}^{j_{1}, \dots, j_{p}} v w = (\frac{c_{α_{i}}}{(s_{j_{p}} \dots s_{j_{1}} w γ) (α_{i})}) v w,

and all diagonal entries are nonzero, but, in general, some diagonal entries of $d_{i}^{j_{1}, \dots, j_{p}}$ may be 0. Use this method to expand the expression

\underset{\underset{m_{i j} factors}{⏟}}{t_{s_{i}} t_{s_{j}} t_{s_{i}} \dots} = \underset{\underset{m_{i j} factors}{⏟}}{(τ_{i} + d_{i}) (τ_{i} + d_{j}) (τ_{i} + d_{i}) \dots} = \sum_{z \in W} τ_{z} p_{z},

and move all the diagonal operators $d_{i}$ to the right of the $τ_{i}$ and obtain diagonal operators $p_{z} .$ The operators $τ_{w}$ are pseudo-permutation operators that may have some rows and columns without a nonzero entry. By replacing some diagonal entries of the $p_{z}$ operators by 0, we may "fix the $τ_{z} "$ and replace the $τ_{z}$ with operators $τ_{z}^{'}$ which have exactly one nonzero entry in each row and each column. This yields the expression

\begin{matrix} \underset{\underset{m_{i j} factors}{⏟}}{t_{s_{i}} t_{s_{j}} t_{s_{i}} \dots} = \sum_{z \in W} τ_{z}^{'} p_{z}^{'} . & (4.2) \end{matrix}

If $γ$ is generic, then the diagonal entries ${(p_{z}^{'})}_{w w}$ of $p_{z}^{'}$ are nonzero and have the form ${(p_{z}^{'})}_{w w} = w γ (P_{z}^{'}),$ $w \in W,$ where $P_{z}^{'}$ is a rational function in the $α_{i} .$ A similar expansion gives

\begin{matrix} \underset{\underset{m_{i j} factors}{⏟}}{t_{s_{i}} t_{s_{j}} t_{s_{i}} \dots} = \sum_{z \in W} τ_{z}^{'} q_{z}^{'}, & (4.3) \end{matrix}

where the $q_{z}^{'}$ are diagonal operators which, for generic $γ,$ have diagonal entries ${(q_{z}^{'})}_{w w} = w γ (Q_{z}^{'}),$ where $Q_{z}^{'}$ is a rational function of the $α_{i} .$ As in the proof of Proposition 2.5(e), $γ (P_{z}^{'}) = γ (Q_{z}^{'})$ for all generic $γ,$ and so it follows that $P_{z}^{'} = Q_{z}^{'}$ as rational functions.

When $γ$ is not generic, the operators $p_{z}^{'}$ and $q_{z}^{'}$ may have some diagonal entries equal to zero. From the classification of representations of rank two graded Hecke algebras we know that there exists a calibrated representation of $ℍ_{i j}$ when $(γ, J)$ is skew. This representation has a unique, up to constant multiples, basis of simultaneous eigenvectors for the action of $λ \in 𝔥_{ℂ}^{*},$ and Proposition 4.4 shows that the action on this basis is forced except for the values of the off diagonal elements of the $t_{s_{i}} .$ These values depend on the normalization of the basis. Because we know that this representation exists we know that there are choices of the nonzero entries in the $τ_{z}^{'}$ such that (4.2) and (4.3) are equal. If a diagonal entry ${(p_{z}^{'})}_{w w}$ of $p_{z}^{'}$ is nonzero, then it is equal to $(w γ) (P_{z}^{'})$ and ${(p_{z}^{'})}_{w w} = (w γ) (P_{z}^{'}) = (w γ) (Q_{z}^{'}) = {(q_{z}^{'})}_{w w},$ since (as shown above) $P_{z}^{'} = Q_{z}^{'} .$ Thus it follows that nonzero contributions from the terms $τ_{z}^{'} p_{z}^{'}$ and $τ_{z}^{'} q_{z}^{'}$ are equal and that $t_{s_{i}} t_{s_{j}} t_{s_{i}} \dots v w$ is equal to $t_{s_{j}} t_{s_{i}} t_{s_{j}} \dots v w .$

$□$

Remark 4.6. The action of $ℍ$ on a weight basis of $ℍ^{(γ, J)}$ is forced up to the freedom in Proposition 4.4(c). Our choice ${(t_{s_{i}})}_{s_{i} b, b} = 1 + {(t_{s_{i}})}_{b b}$ in Theorem 4.5 and the alternative choice ${(t_{s_{i}})}_{s_{i} b, b} = 1 + {(t_{s_{i}})}_{s_{i} b, s_{i} b}$ yield isomorphic modules. The change of basis $v_{b}^{'} = \frac{1}{(1 + {(t_{s_{i}})}_{b b})} v_{b}$ provides the isomorphism.

We summarize the results of this section with the following corollary of Theorem 4.3 and the construction in Theorem 4.5.

Theorem 4.7. Let $M$ be an irreducible calibrated $ℍ -module.$ Let $γ \in 𝔥_{ℂ}$ be (a fixed choice of) the central character of $M$ and let $J = R (w) \cap P (γ)$ for any $w \in W$ such that $M_{w γ} \neq 0 .$ Then $(γ, J)$ is skew and $M ≃ ℍ^{(γ, J)},$ where $ℍ^{(γ, J)}$ is the module defined in Theorem 4.5.

Notes and References

This is an excerpt of a paper entitled Representations of graded Hecke algebras, written by Cathy Kriloff (Department of Mathematics, Idaho State University, Pocatello, Idaho 83209-8085) and Arun Ram.

Research of the first author supported in part by an NSF-AWM Mentoring Travel Grant. Research of the second author supported in part by National Security Agency grant MDA904-01-1-0032 and EPSRC Grant GR K99015.

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