## Type ${A}_{2}$

Last update: 9 March 2013

## Type ${A}_{2}$

The type ${A}_{2}$ root system is

$R= { ±α1, ±α2, ±(α1+α2) } ,$

where $⟨{\alpha }_{1},{\alpha }_{2}^{\vee }⟩=-1=⟨{\alpha }_{2},{a}_{1}^{\vee }⟩\text{.}$ Then $R$ is a root system as defined in (1.2.1), and the Weyl group ${W}_{0}\cong {S}_{3}\text{.}$ The simple roots are ${\alpha }_{1}$ and ${\alpha }_{2},$ and ${\alpha }_{1}+{\alpha }_{2}$ is the only other positive root.

$α1 α2 α1+α2 ω2 ω1 The type A2 root system$

The fundamental weights satisfy

$ω1=13 (2α1+α2), α1=2ω1- ω2, ω2=13 (2α2+α1), α2=2ω2- ω1.$

Let

$P=ℤ-span {ω1,ω2}$

be the weight lattice of $R\text{.}$

$Hα1 Hα2 ω1 ω2 0 The weight lattice P$

The affine Hecke algebra $\stackrel{\sim }{H}$ (see 1.2.2) is generated as a $ℂ\text{-algebra}$ by ${T}_{1},$ ${T}_{2},$ and $X=\left\{{X}^{\lambda } \mid \lambda \in P\right\},$ with relations

$XλXμ = Xλ+μ, for λ,μ∈P (2.4) T1T2T1 = T2T1T2 (2.5) Ti2 = (q-q-1)Ti +1, for i=1,2 (2.6) Xω1T1 = T1 Xω2-ω1+ (q-q-1) Xω1 (2.7) Xω1T2 = T2 Xω1-ω2+ (q-q-1) Xω2 (2.8) Xω1T2 = T2Xω1 (2.9) Xω2T1 = T1Xω2 (2.10)$

Let

$ℂ[X]= {Xλ ∣ λ∈P} ,$

a subalgebra of $\stackrel{\sim }{H},$ and let

$T=Homℂ-alg (ℂ[X],ℂ).$

Then ${W}_{0}$ acts on $X$ by

$s1·Xω1 = Xω2-ω1, s1·Xω2 = Xω2, s2·Xω1 = Xω1, and s2·Xω2 = Xω1-ω2,$

and ${W}_{0}$ acts on $T$ by

$(w·t) (Xλ)=t (Xw-1λ) .$

Let

$Q=ℤ-span(R)$

be the root lattice of $R\text{.}$ Let

$ℂ[Q]= {Xλ ∣ λ∈Q}$

and let

$TQ= Homℂ-alg (ℂ[Q],ℂ).$

Define

$tz,w: ℂ[Q] ⟶ ℂ by Xα1 ⟼ z Xα2 ⟼ w.$

We can also visualize ${T}_{Q}$ using the following picture. Here, the hyperplane ${H}_{\alpha }=\left\{t\in {T}_{Q} \mid t\left({X}^{\alpha }\right)=1\right\}$ is drawn as a solid line. The hyperplanes ${H}_{\alpha +\delta }=\left\{t\in {T}_{Q} \mid t\left({X}^{\alpha }\right)={q}^{2}\right\}$ are drawn in this picture as dashed lines. The weight ${t}_{{q}^{x},{q}^{y}}$ is the point $x$ units away from ${H}_{{\alpha }_{1}}$ and $y$ units away from ${H}_{{\alpha }_{2}}\text{.}$ The action of ${W}_{0}$ is also visible in this picture, as ${s}_{i}$ is given by reflection in the hyperplane ${H}_{{\alpha }_{i}}\text{.}$

$Hα1 Hα2 Hα1+α2 TQ$

For each ${t}_{z,w}\in {T}_{Q},$ there are 3 elements $t\in T$ with $t{\mid }_{Q}={t}_{z,w},$ determined by

$t(Xω1)3= z2wandt (Xω2)=t (X-ω1) ·zw.$

The dimension of the modules with central character $t$ and the submodule structure of $M\left(t\right)$ depends only on $t{\mid }_{Q}\text{.}$ Thus we begin by examining the ${W}_{0}\text{-orbits}$ in ${T}_{Q}\text{.}$ The structure of the modules with weight $t$ depends virtually exclusively on $P\left(t\right)=\left\{\alpha \in {R}^{+} \mid t\left({X}^{\alpha }\right)={q}^{±2}\right\}$ and $Z\left(t\right)=\left\{\alpha \in {R}^{+} \mid t\left({X}^{\alpha }\right)=1\right\}\text{.}$ For a generic weight $t,$ $P\left(t\right)$ and $Z\left(t\right)$ are empty, so we examine only the non-generic orbits.

Proposition 2.3. If $t\in {T}_{Q},$ and $P\left(t\right)\cup Z\left(t\right)\ne \varnothing ,$ then $t$ is in the ${W}_{0}\text{-orbit}$ of one of the following weights:

$t1,1, t1,q2, tq2,1, tq2,q2, { t1,z ∣ z∈ℂ×z≠1, q±2 } , or { tq2,z ∣ z∈ℂ×z≠1, q±2,q-4 }$

 Proof. First, assume generic $q\text{.}$ Case 1: If $Z\left(t\right)$ contains two positive roots, then it must contain the third. This implies $t={t}_{1,1}\text{.}$ Case 2: If $Z\left(t\right)$ contains only one root, by applying an element of ${W}_{0},$ assume that it is ${\alpha }_{1}\text{.}$ Then $t\left({X}^{{\alpha }_{2}}\right)=t\left({X}^{{\alpha }_{1}+{\alpha }_{2}}\right),$ so either $P\left(t\right)=\varnothing$ or $P\left(t\right)=\left\{{\alpha }_{2},{\alpha }_{1}+{\alpha }_{2}\right\}\text{.}$ The first central character is ${t}_{1,z}$ for some $z\ne 1$ or ${q}^{±2}\text{.}$ (If $z=1$ or $z=±2,$ either $P\left(t\right)$ or $Z\left(t\right)$ would be larger.) For the second case, there are two potential choices for the orbit, arising from choosing $t\left({X}^{{\alpha }_{2}}\right)={q}^{2}$ or ${q}^{-2}\text{.}$ However, ${t}_{1,{q}^{-2}}$ is in the same orbit as ${t}_{{q}^{2},1}\text{.}$ Case 3: Now assume that $Z\left(t\right)=\varnothing \text{.}$ If $P\left(t\right)$ is not empty, assume that ${\alpha }_{1}\in P\left(t\right)$ and $t\left({X}^{{\alpha }_{1}}\right)={q}^{2}\text{.}$ Then $t\left({X}^{{\alpha }_{2}}\right)\ne {q}^{-2}$ by assumption on $Z\left(t\right)\text{.}$ Then it is possible that ${\alpha }_{2}\in P\left(t\right),$ in which case $t={t}_{{q}^{2},{q}^{2}}\text{.}$ If ${\alpha }_{1}+{\alpha }_{2}\in P\left(t\right)$ then $t\left({X}^{{\alpha }_{2}}\right)={q}^{-4}$ and $t={t}_{{q}^{2},{q}^{-4}}={s}_{2}{s}_{1}{t}_{{q}^{2}{q}^{2}}\text{.}$ Otherwise, $t={t}_{{q}^{2},z}$ for some $z\ne 1,{q}^{±2},{q}^{-4}\text{.}$ $\square$

Remark: There is some redundancy present in the list of central characters above for specific values of $q\text{.}$ If ${q}^{2}=-1,$ then ${t}_{1,{q}^{2}},$ ${t}_{{q}^{2},1},$ and ${t}_{{q}^{2},{q}^{2}}$ are all in the same ${W}_{0}\text{-orbit.}$ Also in this case, ${t}_{{q}^{2},z}={t}_{-1,z}={s}_{1}{t}_{-1,-z}\text{.}$ If ${q}^{2}=1,$ then ${t}_{1,1}={t}_{{q}^{2},1}={t}_{1,{q}^{2}}={t}_{{q}^{2},{q}^{2}},$ and ${t}_{1,z}={t}_{{q}^{2},z}\text{.}$ Also note that for every generic weight ${t}_{z,w},$ there are six weights in its ${W}_{0}\text{-orbits,}$ all of which are of course generic.

It is helpful to draw a picture of the weights $\left\{{t}_{{q}^{x},{q}^{y}} \mid x,y\in ℝ\right\}$ for various values of $q\text{.}$ Solid lines in these pictures show sets of the form

$Hα= { t∈TQ ∣ t(Xα)=1 } ,$

for $\alpha \in {R}^{+},$ while dashed lines denote sets of the form

$Hα±δ= { t∈TQ ∣ t(Xα)= q±2 } ,$

for $\alpha \in {R}^{+}\text{.}$

$Hα1 Hα2 Hα1+α2 Hα1+α2+δ Hα1+δ Hα2+δ t1,1 t1,q2 t1,z tz,w tq2,1 tq2,q2 tq2,z Figure 1: Representatives of some central characters of modules over H∼, with general q. Hα1 Hα1+δ Hα1-δ Hα1 Hα2-δ Hα2+δ Hα2 Hα1+α2 Hα1+α2-δ Hα1+α2+δ t1,1 t1,q2 t1,z tz,w tq2,1 tq2,q2 tq2,z Figure 2: q2 a primitive third root of unity. Hα1 Hα1±δ Hα2±δ Hα2 Hα1+α2 Hα1+α2±δ t1,1 t1,q2 t1,z tz,w tq2,z Figure 3: q2-1. Hα1 Hα2 Hα1+α2 t1,1 tz,w tq2,z Figure 4: q2=1.$

### Analysis of the characters

Proposition 2.4. Fix $\epsilon ={e}^{2\pi i/3}\text{.}$ The 1-dimensional $\stackrel{\sim }{H}\text{-modules}$ are

$Lq2,q2: H∼ ⟶ ℂ T1⟼q T2⟼q Xω1⟼q2 Xω2⟼q2 Lεq2,ε2q2: H∼ ⟶ ℂ T1⟼q T2⟼q Xω1⟼εq2 Xω2⟼ε2q2 Lε2q2,εq2: H∼ ⟶ ℂ T1⟼q T2⟼q Xω1⟼ε2q2 Xω2⟼εq2 Lq-2,q-2: H∼ ⟶ ℂ T1⟼-q-1 T2⟼-q-1 Xω1⟼q-2 Xω2⟼q-2 Lεq-2,ε2q-2: H∼ ⟶ ℂ T1⟼-q-1 T2⟼-q-1 Xω1⟼εq-2 Xω2⟼ε2q-2 Lε2q-2,εq-2: H∼ ⟶ ℂ T1⟼-q-1 T2⟼-q-1 Xω1⟼ε2q-2 Xω2⟼εq-2$

 Proof. A straightforward check shows that the maps above respect the defining relations for $\stackrel{\sim }{H}$ (2.4) - (2.10), so that the maps are homomorphisms. Let $ℂv$ be any 1-dimensional $\stackrel{\sim }{H}\text{-module.}$ By (2.6) and (2.5), $Tiv=qvor Tiv=-q-1v andT1v= T2v.$ Case 1: ${T}_{1}v={T}_{2}v=qv\text{.}$ By (2.7) and (2.8), $X2ω1v=q2 Xω2v,and X2ω2v=q2 Xω1v.$ Thus $Xα1v= X2ω1-ω2v= q2v,and Xα2v= X2ω2-ω1= q2v,$ so that $X3ω1v= X2α1+α2v =q6v,and X3ω2v= Xα1+2α2 v=q6v.$ Hence, ${X}^{{\omega }_{1}}v={\epsilon }^{i}{q}^{2}v,$ where $i=0,1,$ or 2, and ${X}^{{\omega }_{2}}v={\epsilon }^{2i}{q}^{2}v\text{.}$ Case 2: ${T}_{1}v={T}_{2}v=-{q}^{-1}v\text{.}$ Then $X2ω1v=q-2 Xω2vand X2ω2v=q-2 Xω1v.$ This implies that $Xα1v= X2ω1-ω2v= q-2v,and Xα2v= X2ω2-ω1v =q-2v,$ so that $X3ω1v= X2α1+α2 v=q-6v, andX3ω2v= Xα1+2α2v= q-6v.$ Hence, ${X}^{{\omega }_{1}}v={\epsilon }^{i}{q}^{-2}v,$ where $i=0,1,$ or 2, and ${X}^{{\omega }_{2}}v={\epsilon }^{2i}{q}^{-2}v\text{.}$ $\square$

Principal Series Modules and Local Regions

Let $t\in T\text{.}$ The principal series module is

$M(t)= Ind ℂ[X] H∼ ℂt=H∼ ⊗ℂ[X] ℂt,$

where ${ℂ}_{t}$ is the one-dimensional $ℂ\left[X\right]\text{-module}$ given by

$ℂt=span{vt} andXλ vt=t(Xλ) vt.$

By (1.6), every irreducible $\stackrel{\sim }{H}$ module is a quotient of some principal series module $M\left(t\right)\text{.}$ Thus, finding all the composition factors of $M\left(t\right)$ for all central characters $t$ will find all the irreducible $\stackrel{\sim }{H}\text{-modules.}$

Assume for now that ${q}^{2}\ne 1\text{.}$

Case 1: $P\left(t\right)$ empty. By 1.8, if $P\left(t\right)=\left\{\alpha \in {R}^{+} \mid t\left({X}^{{\alpha }_{1}}\right)=1\right\}$ is empty, then $M\left(t\right)$ is irreducible and is the only irreducible module with central character $t\text{.}$ This case includes the central characters ${t}_{1,1},$ ${t}_{1,z},$ and ${t}_{z,w}$ for generic $z,w\text{.}$

Since $P\left(t\right)=\varnothing ,$ there is one local region

$ℱt,∅=W0 /Wt,$

the set of minimal length coset representatives of ${W}_{t}$ cosets in ${W}_{0},$ where ${W}_{t}$ is the stabilizer of $t$ in ${W}_{0}\text{.}$ If $w$ and ${s}_{i}w$ are both in ${ℱ}^{\left(t,\varnothing \right)}$ then ${\tau }_{i}:M{\left(t\right)}_{wt}^{\text{gen}}\to M{\left(t\right)}_{{s}_{i}wt}^{\text{gen}}$ is a bijection. The following pictures show ${ℱ}^{\left(t,\varnothing \right)}$ with one dot in the chamber ${w}^{-1}C$ for each basis element of $M{\left(t\right)}_{wt}^{\text{gen}}\text{.}$

These pictures also show the weight space structure of $M\left(t\right)\text{.}$ For $t={t}_{1,1},$ all of $M\left(t\right)$ is in the $t$ weight space, so all the dots lie in the same chamber. For $t={t}_{1,z},$ ${s}_{1}t=t,$ so that the weights of $M\left(t\right)$ are $t,$ ${s}_{2}t,$ and ${s}_{1}{s}_{2}t,$ and each weight space is 2-dimensional. The weight $t={t}_{z,w}$ is regular, so there are six different weights. In each case, all the dots lie in the same local region, a region bounded by solid and/or dashed lines.

$(M(t)) tgen =M(t) (M(t)) tgen (M(t)) s2tgen (M(t)) s1s2tgen (M(t))t (M(t))s2t (M(t))s1s2t (M(t))s1s2s1t (M(t))s2s1t (M(t))s1t t1,1,q2≠1 t1,z,q2≠1 tz,w$

Case 2: $Z\left(t\right)=\varnothing ,$ $P\left(t\right)\ne \varnothing \text{.}$ This case includes the central characters ${t}_{{q}^{2},{q}^{2}},$ and ${t}_{{q}^{2},z}\text{.}$ If $Z\left(t\right)$ is empty, then $M\left(t\right)$ is calibrated and the irreducible modules with central character $t$ are in one-to-one correspondence with the local regions and the components of the calibration graph. Each local region in the following pictures is a set of chambers between two dashed lines.

For the weight ${t}_{{q}^{2},{q}^{2}}$ for generic $q,$ there are four local regions. If ${q}^{6}=1,$ each weight is in a local region by itself and thus there are six different representations in that case. For the weight ${t}_{{q}^{2},z},$ there are two local regions. These local regions are bounded by dashed lines, which also serve as barriers of sorts between the composition factors. This is a combinatorial reflection of the fact that if the $\tau$ operator between two weight spaces of $M\left(t\right)$ is not invertible, those weight spaces will be in different composition factors. Weight spaces with invertible $\tau$ operators between them are separated by dotted lines since they are in the same composition factor and local region.

$(M(t))t (M(t))s2t (M(t))s1s2t (M(t))s1s2s1t (M(t))s2s1t (M(t))s1t (M(t))t (M(t))s2t (M(t))s1s2t (M(t))s1s2s1t (M(t))s2s1t (M(t))s1t (M(t))t (M(t))s2t (M(t))s1s2t (M(t))s1s2s1t (M(t))s2s1t (M(t))s1t tq2,q2, q4≠1,q6≠1 tq2,q2,q2 a primitive third root of unity tq2,z,q2 ≠1$

Case 3: $P\left(t\right)\ne \varnothing ,$ $Z\left(t\right)\ne \varnothing \text{.}$ The only central characters with both $Z\left(t\right)$ and $P\left(t\right)$ nonempty are ${t}_{1,z}={t}_{{q}^{2},z}$ when ${q}^{2}=1,$ and ${t}_{{q}^{2},1}$ and ${t}_{1,{q}^{2}}$ in all cases. If ${q}^{4}=1,$ then ${t}_{{q}^{2},1}$ and ${t}_{1,{q}^{2}}$ are in the same orbit, and are in the same orbit as ${t}_{{q}^{2},{q}^{2}}\text{.}$ If ${q}^{2}=1,$ then ${t}_{{q}^{2},1}={t}_{1,{q}^{2}}={t}_{1,1}\text{.}$ In each of these pictures, there are three local regions, but the structure of $M\left(t\right)$ is slightly more complicated in this case.

The composition factors of $M\left(t\right)$ each have some weight $wt$ with $t\left({X}^{{\alpha }_{i}}\right)=1$ for some simple root ${\alpha }_{i}\text{.}$ Then 1.11 shows that when ${q}^{2}\ne ±1,$ this composition factor must be at least 3-dimensional. In this case, these composition factors are shown with the two dots in that weight space $M{\left(t\right)}_{wt}$ connected to each other, as well as to a dot in the next chamber, since the basis vectors correpsonding to these dots must lie in the same irreducible. When ${q}^{2}=-1,$ the two dimensional weight space $M{\left(t\right)}_{wt}$ makes up an entire composition factor.

$(M(t))t (M(t))s2t (M(t))s1s2t (M(t))t (M(t))s2s1t (M(t))s1t (M(t))t (M(t))s2t (M(t))s1s2t t1,q2, q4≠1 tq2,1, q4≠1 t1,q2, q2=-1$

To prove that these pictures do reflect the structure of $M\left(t\right),$ rather than analyzing $M\left(t\right)$ directly, it is easier to construct several irreducible modules with central character $t$ and show that they must account for all the composition factors of $M\left(t\right)\text{.}$ Let ${ℂ}_{{q}^{2},1}$ be the 1-dimensional ${\stackrel{\sim }{H}}_{\left\{1\right\}}\text{-module}$ spanned by ${v}_{t}$ and let ${ℂ}_{1,{q}^{-2}}$ be the 1-dimensional ${\stackrel{\sim }{H}}_{\left\{2\right\}}\text{-module}$ spanned by ${v}_{{s}_{2}{s}_{1}t},$ given by

$Xλvt=t (Xλ)vt, andT1vt= qvt,and Xλvs2s1t =(s2s1t) (Xλ) vs2s1t andT1 vs2s1t=- q-1vs2s1t .$

Then

$M=H∼⊗H∼{1} ℂq2,1and N=H∼⊗H∼{2} ℂ1,q-2$

are 3-dimensional $\stackrel{\sim }{H}\text{-modules}$ with central character ${t}_{{q}^{2},1}\text{.}$

Proposition 2.5. Let $M=\stackrel{\sim }{H}{\otimes }_{{\stackrel{\sim }{H}}_{\left\{1\right\}}}{ℂ}_{{q}^{2},1}$ and $N=\stackrel{\sim }{H}{\otimes }_{{\stackrel{\sim }{H}}_{\left\{2\right\}}}{ℂ}_{1,{q}^{-2}\text{.}}$

1. If ${q}^{4}\ne 1$ then $M$ and $N$ are irreducible.
2. If ${q}^{2}=-1$ then ${M}_{{s}_{1}t}$ is an irreducible submodule of $M$ and ${N}_{{s}_{1}t}$ is an irreducible submodule of $N\text{.}$ The quotients $N/{N}_{{s}_{1}t}$ and $M/{M}_{{s}_{1}t}$ are irreducible.

 Proof. (a) Assume ${q}^{4}\ne 1\text{.}$ If either $M$ or $N$ were reducible, it would have a 1-dimensional submodule or quotient, which cannot happen since the 1-dimensional modules have central character ${t}_{{q}^{2},{q}^{2}}\text{.}$ Thus both $M$ and $N$ are reducible. (b) If ${q}^{2}=-1,$ then note that the map $θ: H∼ ⟶ H∼ T1⟼T2 T2⟼T1 Xω1⟼Xω2 Xω2⟼Xω1$ is an automorphism of $\stackrel{\sim }{H}\text{.}$ If ${\varphi }_{M}:\stackrel{\sim }{H}\to {\text{GL}}_{3}$ and ${\varphi }_{N}:\stackrel{\sim }{H}\to {\text{GL}}_{3}$ are the maps describing the action of $\stackrel{\sim }{H}$ on $M$ and $N$ respectively, then ${\varphi }_{M}\circ \theta ={\varphi }_{N}$ and ${\varphi }_{N}\circ \theta ={\varphi }_{M}\text{.}$ Thus if one of $M$ and $N$ is irreducible, the other is as well. However, this would account for all the composition factors of $M\left(t\right),$ a contradiction since there is a 1-dimensional module with central character $t\text{.}$ Thus $M$ and $N$ must both be reducible, so $M$ has a 1-dimensional submodule or quotient. It must have weight ${s}_{1}t$ since that is the only weight of $M$ that supports a 1-dimensional module. Then the 2-dimensional composition factor is ${M}_{t}^{\text{gen}},$ by Lemma 1.11. Then since $HomH∼ (M,Mtgen)= HomH∼{1} ( ℂvt, Mtgen ) ≠0,$ ${M}_{{s}_{1}t}$ is a submodule of $M\text{.}$ Similarly, ${N}_{{s}_{1}t}$ is a submodule of $N$ since ${N}_{{s}_{2}{s}_{1}t}^{\text{gen}}$ must be a quotient of $N\text{.}$ $\square$

$(M(t))t (M(t))s2t (M(t))s1s2t (M(t))t (M(t))s2s1t (M(t))s1t t1,q2, q4≠1 tq2,1, q4≠1$

To construct the irreducibles with $t{\mid }_{Q}={t}_{1,{q}^{2}},$ composing with

$ϕ: H∼ ⟶ H∼ T1⟼T2 T2⟼T1 Xω1⟼Xω2 Xω2⟼Xω1,$

an automorphism of $\stackrel{\sim }{H},$ gives a bijection between representations with central character ${t}_{{q}^{2},1}$ and those with central character ${t}_{1,{q}^{2}}\text{.}$

$(M(t))t (M(t))s2t (M(t))s1s2t t1,q2,q2= -1$

If ${q}^{2}=1,$ then the results of section 1.2.9 suffice to classify the representations of $\stackrel{\sim }{H}$ with central characters ${t}_{1,1}$ and ${t}_{1,z}$ for $z\ne {q}^{±2}\text{.}$ Specifically, if $t{\mid }_{Q}={t}_{1,1},$ then ${W}_{t}={W}_{0}$ and a $\stackrel{\sim }{H}\text{-module}$ is merely a ${W}_{0}\text{-module}$ (via the isomorphism $H\cong ℂ\left[{W}_{0}\right]\text{)}$ on which ${X}^{\lambda }\in ℂ\left[X\right]$ acts by the scalar $t\left({X}^{\lambda }\right)\text{.}$ In fact, $M\left(t\right)$ considered as a ${W}_{0}$ module is the regular ${W}_{0}$ module. If $t{\mid }_{Q}={t}_{1,z}$ for $z\ne 1,$ then ${W}_{t}=\left\{1,{s}_{1}\right\}\text{.}$ Since ${W}_{t}$ has two 1-dimensional irreducible representations, $\stackrel{\sim }{H}$ has two irreducible 3-dimensional representations obtained by inducing up from ${\stackrel{\sim }{H}}_{\left\{1\right\}}\text{.}$

$(M(t))t=M(t) (M(t)) t (M(t)) s2t (M(t)) s1s2t (M(t))t (M(t))s2t (M(t))s1s2t (M(t))s2s1s2t (M(t))s2s1t (M(t))s1t t1,1,q2=1 t1,z,q2=1 tz,w,q2=1$

Note that in all cases, the local region picture is a picture of a small neighborhood around the point corresponding to $t$ in the picture of ${T}_{Q}\text{.}$ The necessary information to understand the composition factors of $M\left(t\right)$ is contained in the local region picture around $t\text{.}$

The following tables summarize the classification. It should be noted that for any value of $q$ with ${q}^{2}\ne ±1$ and ${q}^{2}$ not a primitive third root of unity, the representation theory of $\stackrel{\sim }{H}$ can be described in terms of $q$ only. If ${q}^{2}$ is a primitive root of unity of order 3 or less, then the representation theory of $\stackrel{\sim }{H}$ does not fit that same description. This fact can be seen through a number of different lenses. It is a reflection of the fact that the sets $P\left(t\right)$ and $Z\left(t\right)$ for all possible central characters $t$ can be described solely in terms of $q\text{.}$ In the local region pictures, this is reflected in the fact that the hyperplanes ${H}_{\alpha }$ and ${H}_{\alpha ±\delta }$ are distinct unless ${q}^{2}$ is a root of unity of order 3 or less. When these hyperplanes coincide, the sets $P\left(t\right)$ and $Z\left(t\right)$ change for characters on those hyperplanes.

$t∣Q Z(t) P(t) Dimensions of Irreds. t1,1 R+ ∅ 6 t1,z {α1} ∅ 6 t1,q2 {α1} {α2,α1+α2} 3,3 tq2,1 {α2} {α1,α1+α2} 3,3 tq2,q2 ∅ {α1,α2} 1,1,2,2 tq2,z ∅ {α1} 3,3 tz,w ∅ ∅ 6 Table 4: Table of possible central characters in Type A2, with q generic.$

$t∣Q Z(t) P(t) Dimensions of Irreds. t1,1 R+ ∅ 6 t1,z {α1} ∅ 6 t1,q2 {α1} {α2,α1+α2} 3,3 tq2,1 {α2} {α1,α1+α2} 3,3 tq2,q2 ∅ R+ 1,1,1,1,1,1 tq2,z ∅ {α1} 3,3 tz,w ∅ ∅ 6 Table 5: Table of possible central characters in Type A2, with q6=1.$

$t∣Q Z(t) P(t) Dimensions of Irreds. t1,1 R+ ∅ 6 t1,z {α1} ∅ 6 t1,q2 {α1} {α2,α1+α2} 1,2,2 tq2,z ∅ {α1} 3,3 tz,w ∅ ∅ 6 Table 6: Table of possible central characters in Type A2, with q2=-1.$

$t∣Q Z(t) P(t) Dimensions of Irreds. t1,1 R+ R+ 1,1,2 t1,z {α1} {α1} 3,3 tz,w ∅ ∅ 6 Table 7: Table of possible central characters in Type A2, with q=-1.$

## Notes and References

This is an excerpt from Matt Davis' Ph.D Thesis entitled Representations of Rank Two Affine Hecke Algebras at Roots of Unity, University of Wisconsin, 2010.