## Type ${C}_{2}$

Last update: 10 March 2013

## Type ${C}_{2}$

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

$R= { ±α1,±α2, ±(α1+α2) ,±(2α1+α2) } ,$

where $⟨{\alpha }_{1},{\alpha }_{2}^{\vee }⟩=-1$ and $⟨{\alpha }_{2},{\alpha }_{1}^{\vee }⟩=-2\text{.}$ Then $R$ is a root system as defined in 1.2.1, and the Weyl group is

$W0= ⟨ s1,s2 ∣ s12=s22=1 ,s1s2s1s2 =s2s1s2s1 ⟩ .$

The simple roots are ${\alpha }_{1}$ and ${\alpha }_{2},$ with additional positive roots ${\alpha }_{1}+{\alpha }_{2}$ and $2{\alpha }_{1}+{\alpha }_{2}\text{.}$ Then ${W}_{0}$ acts on $R$ by

$s1·α1=-α1, s2·α1=α1+α2, s1·α2=2α1+α2, and s2·α2=-α2.$ $α1 α1+α2 2α1+α2 α2 The type C2 root system$

The fundamental weights satisfy

$ω1=α1+12α2 α1=2ω1-ω2 ω2=α1+α2 α2=2ω2-2ω1.$

Let

$P=ℤ-span {ω1,ω2}$

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

$Hα1 Hα2 ω1 ω2 The weight lattice P.$

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

$XλXμ = Xλ+μ, for λ,μ∈P (2.11) T1T2T1T2 = T2T1T2T1 (2.12) Ti2 = (q-q-1)Ti +1, for i=1,2 (2.13) Xω1T1 = T1Xs1(w1) +(q-q-1)Xω1 (2.14) Xω2T2 = T2Xs2(ω2) +(q-q-1)Xω2 (2.15) Xω1T2 = T2Xω1 (2.16) Xω2T1 = T1Xω2 (2.17)$

Let

$ℂ[X]= {Xλ ∣ λ∈P},$

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

$T=Homℂ-alg (ℂ[X],ℂ).$

Then since ${\alpha }_{1}=2{\omega }_{1}-{\omega }_{2}$ and ${\alpha }_{2}=2{\omega }_{2}-2{\omega }_{1},$ ${W}_{0}$ acts on $X$ by

$s1·Xω1 = Xω2-ω1, s1·Xω2 = Xω2, s2·Xω1 = Xω1,and s2·Xω2 = X2ω1-ω2.$

Then ${W}_{0}$ 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

$tz,w: ℂ[Q] ⟶ ℂ by Xα1⟼z Xα2⟼w.$

Pictorially, a weight ${t}_{{q}^{x},{q}^{y}}\in {T}_{Q},$ for $x,y\in ℝ,$ is identified with with the point $x$ units from the hyperplane ${H}_{{\alpha }_{1}}$ and $y$ units from the hyperplane ${H}_{{\alpha }_{2}}\text{.}$ Then

$Hα= { t∈TQ ∣ t(Xα)=1 } ,$

and we define

$Hα±δ= { t∈TQ ∣ t(Xα)= q±2 } .$ $Hα1 Hα2 H2α1+α2 Hα1+α2 Hα1+α2+δ H2α1+α2+δ Hα1+δ Hα2+δ The torus TQ$

For all weights ${t}_{z,w}\in {T}_{Q},$ there are 2 elements $t\in T$ with $t{\mid }_{Q}={t}_{z,w},$ determined by

$t(Xω1)2= z2wand t(Xω2)= zw.$

We denote these two elements as ${t}_{z,w,1}$ and ${t}_{z,w,2}\text{.}$ Which particular weight ${t}_{z,w,i}$ is which is unimportant since we will always be examining them together. And in fact, most of the time, we will only refer to the restricted weight ${t}_{z,w},$ since the dimension of the modules with central character $t$ depends only on $t{\mid }_{Q}\text{.}$

One important remark, though, is that if $t\left({X}^{{\alpha }_{1}}\right)=-1,$ then the two weights $t$ with $t{\mid }_{Q}={t}_{-1,w}$ are in the same ${W}_{0}\text{-orbit}$ and represent the same central character. To see this, let $t{\mid }_{Q}={t}_{-1,w}\text{.}$ Then $t\left({X}^{{\omega }_{1}}\right)={w}^{1/2},$ and

$s1t(Xω1)= t(Xω2-ω1) =-w/t(Xω1)=- t(Xω1),$

while

$s1t(Xω2)= t(Xω2).$

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 2.6. 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:

$t1,1, t-1,1, t1,q2, tq2,1, t±q,1, tq2,q2, t-1,q2, { t1,z ∣ z≠1,q±2 } , { tz,1 ∣ z≠±1, q±2, ±q±1 } , { tq2,z ∣ z≠1,q±2, q-4,q-6 } , or { tz,q2 ∣ z≠±1,q±2, -q-2,q-4, ±q-1 } .$

 Proof. Case 1: $\mid Z\left(t\right)\mid \ge 2\text{.}$ Case 1a: ${\alpha }_{1},{\alpha }_{1}+{\alpha }_{2}\in Z\left(t\right)\text{.}$ Then $t\left({X}^{{\alpha }_{1}}\right)=1=t\left({X}^{{\alpha }_{1}+{\alpha }_{2}}\right),$ so that $t\left({X}^{{\alpha }_{2}}\right)=1$ and thus $t={t}_{1,1}\text{.}$ Case 1b: Exactly one of ${\alpha }_{1},{\alpha }_{1}+{\alpha }_{2}$ is in $Z\left(t\right)\text{.}$ Choose an element $w$ with $wt\left({X}^{{\alpha }_{1}}\right)=1\text{.}$ Then ${\alpha }_{2}\in Z\left(wt\right)$ or $2{\alpha }_{1}+{\alpha }_{2}\in Z\left(wt\right)\text{.}$ Since $wt\left({X}^{2{\alpha }_{1}+{\alpha }_{2}}\right)=wt\left({X}^{{\alpha }_{1}}\right)wt\left({X}^{{\alpha }_{1}+{\alpha }_{2}}\right)=wt{\left({X}^{{\alpha }_{1}}\right)}^{2}wt\left({X}^{{\alpha }_{2}}\right),$ $wt\left({X}^{{\alpha }_{2}}\right)=wt\left({X}^{{\alpha }_{1}+{\alpha }_{2}}\right)=wt\left({X}^{2{\alpha }_{1}+{\alpha }_{2}}\right)=1,$ so that $t={t}_{1,1}\text{.}$ Case 1c: ${\alpha }_{1},{\alpha }_{1}+{\alpha }_{2}\notin Z\left(t\right)\text{.}$ Then $t\left({X}^{{\alpha }_{2}}\right)=1=t\left({X}^{2{\alpha }_{1}+{\alpha }_{2}}\right),$ so that $t\left({X}^{2{\alpha }_{1}}\right)=1,$ and $t\left({X}^{{\alpha }_{1}}\right)=±1\text{.}$ If $t\left({X}^{{\alpha }_{1}}\right)=1$ then $t={t}_{1,1}\text{.}$ By assumption, $t\left({X}^{{\alpha }_{1}}\right)=-1$ and $t={t}_{-1,1}\text{.}$ Case 2: $\mid Z\left(t\right)\mid =1\text{.}$ Case 2a: If ${\alpha }_{1}$ or ${\alpha }_{1}+{\alpha }_{2}\in Z\left(t\right),$ then choose a $w\in {W}_{0}$ with $wt\left({X}^{{\alpha }_{1}}\right)=1,$ so that $wt={t}_{1,z}$ for some $z\text{.}$ Then $wt\left({X}^{{\alpha }_{2}}\right)=wt\left({X}^{{\alpha }_{1}+{\alpha }_{2}}\right)=wt\left({X}^{2{\alpha }_{1}+{\alpha }_{2}}\right)\text{.}$ If $wt\left({X}^{{\alpha }_{2}}\right)={q}^{±2}$ then $t$ is in the same orbit as ${t}_{1,{q}^{2}}\text{.}$ Otherwise, $wt={t}_{1,z}$ where $z\ne 1,{q}^{±2}$ by assumption. Case 2b: If ${\alpha }_{1},{\alpha }_{1}+{\alpha }_{2}\ne Z\left(t\right),$ choose a $w\in {W}_{0}$ with $wt\left({X}^{{\alpha }_{2}}\right)=1\text{.}$ Then $t\left({X}^{{\alpha }_{1}}\right)=t\left({X}^{{\alpha }_{1}+{\alpha }_{2}}\right),$ so that if $t\left({X}^{{\alpha }_{1}}\right)={q}^{±2},$ then $t\left({X}^{2{\alpha }_{1}+{\alpha }_{2}}\right)={q}^{±4}$ and $P\left(t\right)=\left\{{\alpha }_{1},{\alpha }_{1}+{\alpha }_{2}\right\}\text{.}$ If $t\left({X}^{{\alpha }_{1}}\right)=±{q}^{±1},$ then $t\left({X}^{2{\alpha }_{1}+{\alpha }_{2}}\right)={q}^{±2},$ so that $P\left(t\right)=\left\{2{\alpha }_{1}+{\alpha }_{2}\right\}\text{.}$ Otherwise $wt={t}_{z,1}$ with $z\ne 1,±{q}^{±1},{q}^{±2}$ by assumption. Case 3: $Z\left(t\right)=\varnothing ,$ $P\left(t\right)\ne \varnothing$ Case 3a: ${\alpha }_{1},{\alpha }_{1}+{\alpha }_{2}\in P\left(t\right)\text{.}$ Choose $w\in {W}_{0}$ so that $wt\left({X}^{{\alpha }_{1}}\right)={q}^{2}\text{.}$ If $wt\left({X}^{{\alpha }_{1}+{\alpha }_{2}}\right)={q}^{2}$ then $wt\left({X}^{{\alpha }_{2}}\right)=1,$ contradicting the assumption on $Z\left(t\right)\text{.}$ If $wt\left({X}^{{\alpha }_{1}+{\alpha }_{2}}\right)={q}^{-2}$ then $wt\left({X}^{2{\alpha }_{1}+{\alpha }_{2}}\right)=1,$ contradicting the assumption on $Z\left(t\right)\text{.}$ Thus this case is impossible. Case 3b: $P\left(t\right)$ contains exactly one of ${\alpha }_{1}$ or ${\alpha }_{1}+{\alpha }_{2}\text{.}$ Choose $w\in {W}_{0}$ so that $wt\left({X}^{{\alpha }_{1}}\right)={q}^{2}\text{.}$ Then by the assumption on $Z\left(t\right),$ $wt\left({X}^{{\alpha }_{2}}\right)\ne {q}^{-2}$ and $wt\left({X}^{2{\alpha }_{1}+{\alpha }_{2}}\right)\ne {q}^{-2}\text{.}$ Thus if ${\alpha }_{2}\in P\left(wt\right)$ then $wt\left({X}^{{\alpha }_{2}}\right)={q}^{2}$ and $t={t}_{{q}^{2},{q}^{2}}\text{.}$ If $2{\alpha }_{1}+{\alpha }_{2}\in P\left(wt\right),$ then $wt\left({X}^{2{\alpha }_{1}+{\alpha }_{2}}\right)={q}^{-2}$ and $wt\left({X}^{{\alpha }_{2}}\right)={q}^{-6}\text{.}$ Then ${s}_{2}{s}_{1}{s}_{2}wt\left({X}^{{\alpha }_{1}}\right)={q}^{2}={s}_{2}{s}_{1}{s}_{2}wt\left({X}^{{\alpha }_{2}}\right),$ so that $t$ is in the same orbit as ${t}_{{q}^{2},{q}^{2}}\text{.}$ If neither ${\alpha }_{2}$ nor $2{\alpha }_{1}+{\alpha }_{2}$ is in $P\left(wt\right),$ then $wt={t}_{{q}^{2},z}$ for $z\ne 1,{q}^{±2},{q}^{-4},{q}^{-6}\text{.}$ Case 3c: Neither of ${\alpha }_{1},{\alpha }_{1}+{\alpha }_{2}$ is in $P\left(t\right)\text{.}$ Then at least one of ${\alpha }_{2}$ and $2{\alpha }_{1}+{\alpha }_{2}$ is in $P\left(t\right)\text{.}$ Then choose $w\in {W}_{0}$ so that $wt\left({X}^{{\alpha }_{2}}\right)={q}^{2}\text{.}$ If $2{\alpha }_{1}+{\alpha }_{2}\in P\left(t\right),$ then $wt\left({X}^{2{\alpha }_{1}}\right)=1$ or ${q}^{-4},$ so that $wt\left({X}^{{\alpha }_{1}}\right)=-1$ or $-{q}^{-2}$ by assumption. However, ${s}_{1}{s}_{2}{s}_{1}{t}_{-{q}^{-2},{q}^{2}}={t}_{-1,{q}^{2}},$ so both are in the same orbit as ${t}_{-1,{q}^{2}}\text{.}$ Otherwise, $wt={t}_{z,{q}^{2}}$ for some $z\text{.}$ By assumption on $P\left(t\right)$ and $Z\left(t\right),$ $z\ne ±1,{q}^{±2},-{q}^{-2},{q}^{-4},$ or $±{q}^{-1}\text{.}$ $Hα1 Hα1 Hα2 Hα1+δ Hα1 Hα1 Hα2 Hα1+δ Case 1: ∣Z(t)∣ ≥2, so that t Case 2: ∣Z(t)∣ =1, so that t lies on at least two hyperplanes Hα. lies on exactly one hyperplane Hα. Hα1 Hα1 Hα2 Hα1+δ Hα1 Hα1 Hα2 Hα1+δ Case 3: ∣Z(t)∣ =∅,α1∈P(t) , so that t Case 3: ∣Z(t)∣ =∅,α1∉P(t) , so that t lies on Hα1±δ , but not on any Hα. does not lie on Hα1±δ or any Hα.$ $\square$

$Hα1 Hα2 Hα1+α2 H2α1+α2 t1,1 t1,q2 t1,z tq2,q2 tq,1 tq2,z tz,w tz,q2 tq2,1 t-q,1 Figure 5: Representatives of some possible central characters of H∼ -modules with generic q.$

Remark: For specific values of $q,$ there is a redundancy in the list of characters given above. Essentially, this is a result of the periodicity in ${T}_{Q}$ when $q$ is a root of unity.

If ${q}^{2}$ is a primitive fourth root of unity, then ${t}_{{q}^{2},{q}^{2}}={s}_{1}{s}_{2}{s}_{1}{t}_{-1,{q}^{2}}\text{.}$

$Hα1 Hα2 Hα1+α2 H2α1+α2 t1,1 t1,q2 t1,z tq2,q2 tq,1 tq2,z tz,w tz,q2 tq2,1 t-1,1 tz,1 Figure 6: Representatives of the possible central characters of modules over H∼ , with q a primitive eighth root of unity.$

If ${q}^{2}$ is a primitive third root of unity, ${t}_{{q}^{2},{q}^{2}}={s}_{2}{s}_{1}{s}_{2}{t}_{{q}^{2},1}\text{.}$ Also in this case, $q$ and $-q$ are equal to ${q}^{-2}$ and $-{q}^{-2}$ in some order, depending on whether ${q}^{3}=1$ or $-1\text{.}$ Then one of ${t}_{q,1}$ and ${t}_{-q,1}$ is equal to ${t}_{{q}^{-2},1}$ and is in the same orbit as ${t}_{{q}^{2},1}\text{.}$

$Hα1 Hα2 Hα1+α2 H2α1+α2 t1,1 t1,q2 t1,z tq2,q2 t±q,1 tq2,z tz,w tz,q2 tq2,1 t-1,1 tz,1 Figure 7: Representatives of the possible central characters of modules over H∼ , with q2 a primitive third root of unity.$

If ${q}^{2}=-1,$ then ${t}_{-1,1}={t}_{{q}^{2},1},$ and ${t}_{1,{q}^{2}}$ is in the same orbit as ${t}_{{q}^{2},{q}^{2}}={t}_{-1,{q}^{2}}\text{.}$ Also in this case, ${t}_{z,{q}^{2}}={z}^{-1}={s}_{2}{t}_{-z,-1}\text{.}$

$Hα1 Hα2 Hα1+α2 H2α1+α2 t1,1 tq2,1 t1,z tq,1 tq2,z tz,w tz,q2 t-1,1 tz,1 Figure 8: Representatives of the possible central characters of modules over H∼ , with q2=-1.$

Finally, if $q=-1,$ we have ${t}_{1,1}={t}_{{q}^{2},1}={t}_{1,{q}^{2}}={t}_{{q}^{2},{q}^{2}}={t}_{-q,1}\text{.}$ Also, ${t}_{-1,1}={t}_{-1,{q}^{2}},$ while ${t}_{{q}^{2},z}={t}_{1,z}$ and ${t}_{z,{q}^{2}}={t}_{z,1}\text{.}$

$Hα1 Hα2 Hα1+α2 H2α1+α2 t1,1 t1,z tz,w t-1,1 tz,1 Figure 9: Representatives of the possible central characters of modules over H∼ , with q2=1.$

### Analysis of the Characters

Theorem 2.7 The 1-dimensional representations of $\stackrel{\sim }{H}$ are

$Lq,q,1: H∼ ⟶ ℂ T1 ⟼ q T2 ⟼ q Xω1 ⟼ q3 Xω2 ⟼ q4 Lq,q,-1: H∼ ⟶ ℂ T1 ⟼ q T2 ⟼ q Xω1 ⟼ -q3 Xω2 ⟼ q4 Lq,-q-1,1: H∼ ⟶ ℂ T1 ⟼ q T2 ⟼ -q-1 Xω1 ⟼ q Xω2 ⟼ 1 Lq,-q-1,-1: H∼ ⟶ ℂ T1 ⟼ q T2 ⟼ -q-1 Xω1 ⟼ -q Xω2 ⟼ 1 L-q-1,q,1: H∼ ⟶ ℂ T1 ⟼ -q-1 T2 ⟼ q Xω1 ⟼ q-1 Xω2 ⟼ 1 L-q-1,q,-1: H∼ ⟶ ℂ T1 ⟼ -q-1 T2 ⟼ q Xω1 ⟼ -q-1 Xω2 ⟼ 1 L-q-1,-q-1,1: H∼ ⟶ ℂ T1 ⟼ -q-1 T2 ⟼ -q-1 Xω1 ⟼ q-3 Xω2 ⟼ q-4 L-q-1,-q-1,-1: H∼ ⟶ ℂ T1 ⟼ -q-1 T2 ⟼ -q-1 Xω1 ⟼ -q-3 Xω2 ⟼ q-4$

 Proof. A check of the defining relations for $\stackrel{\sim }{H}$ shows that these maps are homomorphisms. Let $ℂv$ be a 1-dimensional $\stackrel{\sim }{H}\text{-module}$ with weight $t\text{.}$ Then $\stackrel{\sim }{H}$ has relations $Xω1T1 = T1Xω2-ω1 +(q-q-1) Xω1, and (2.18) Xω2T2 = T2X2ω1-ω2 +(q-q-1) Xω2. (2.19)$ By (2.18), $X2ω1-ω2v= q2v if T1v= qvand X2ω2-ω2v= q-2v if T1 v=-q-1v,$ so that $Xα1v= { q2v, if T1v=qv q-2v, if T1v=- q-1v,$ since ${\alpha }_{1}=2{\omega }_{1}-{\omega }_{2}\text{.}$ Similarly, by (2.19), $Xα2v= { q2v, if T2v=qv q-2v, if T2v=- q-1v.$ Since $2{\omega }_{1}=2{\alpha }_{1}+{\alpha }_{2},$ $t(Xω1)2= t(X2α1+α2) and t(Xω2)= t(Xα1+α2) .$ This gives two choices of $t$ for each of the four possibilities for $t\left({X}^{{\alpha }_{1}}\right)$ and $t\left({X}^{{\alpha }_{2}}\right)\text{.}$ These are the eight representations given above. $\square$

Remark: If $q$ is a primitive fourth root of unity, then ${L}_{q,q,±1}\cong {L}_{q,-{q}^{-1},\mp 1}\cong {L}_{-{q}^{-1},q,±1}\cong {L}_{-{q}^{-1},-{q}^{-1},\mp 1}\text{.}$

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

$ℂt=span{vt} andXλvt= t(Xλ)vt.$

By Theorem 1.6, every irreducible $\stackrel{\sim }{H}$ module with central character $t$ is a composition factor of $M\left(t\right)\text{.}$ A local region at a weight $t$ is

$ℱ(t,J)= { w∈W0 ∣ R(w)∩Z(t) =∅,R(w)∩ P(t)=J } ,$

for $J\subseteq P\left(t\right),$ and we identify a local region ${ℱ}^{\left(t,J\right)}$ with the union of the chambers ${w}^{-1}C$ for $w\in {ℱ}^{\left(t,J\right)}\text{.}$

Case 1: $P\left(t\right)=\varnothing$

If $P\left(t\right)=\varnothing ,$ then by Kato’s criterion (Theorem 1.8), $M\left(t\right)$ is irreducible and is the only irreducible module with central character $t\text{.}$ 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 $M{\left(t\right)}_{wt}^{\text{gen}}\text{.}$

$M(t)=M(t)tgen M(t)tgen M(t)s2tgen M(t)s1s2tgen M(t)s2s1s2tgen M(t)tgen M(t)s1tgen M(t)s2s1tgen M(t)s1s2s1tgen t1,1,q2≠1 t1,z,q2≠1 tz,1,q2≠1 M(t)tgen M(t)s1tgen M(t)t M(t)s1t M(t)s2s1t M(t)s1s2s1t M(t)s2t M(t)s1s2t M(t)s2s1s2t M(t)s2s1s2s1t t-1,1,q≠±i tz,w$

Case 2: $Z\left(t\right)=\varnothing ,$ but $P\left(t\right)\ne \varnothing \text{.}$

If $Z\left(t\right)=\varnothing$ then $t$ is a regular central character. Then the irreducibles with central character $t$ are in bijection with the connected components of the calibration graph for $t,$ and can be constructed using Theorem 1.14. In particular, there is one irreducible $\stackrel{\sim }{H}\text{-module}$ ${L}^{\left(t,J\right)}$ for each $J\subseteq P\left(t\right)$ such that ${ℱ}^{\left(t,J\right)}\ne \varnothing ,$ and

$dimLwt(t,J)= { 1, if w∈ℱ(t,J), 0, if w∉ℱ(t,J).$

The following pictures show the local regions ${ℱ}^{\left(t,J\right)}$ for each $t$ with $Z\left(t\right)=\varnothing \text{.}$ There is one dot in the chamber ${w}^{-1}C$ for each $w\in {ℱ}^{\left(t,J\right)},$ and the dots corresponding to $w$ and ${s}_{i}w$ are connected exactly when $w$ and ${s}_{i}w$ are in the same local region.

$M(t)t M(t)s1t M(t)s2s1t M(t)s1s2s1t M(t)s2t M(t)s1s2t M(t)s2s1s2t M(t)s2s1s2s1t M(t)t M(t)s1t M(t)s2s1t M(t)s1s2s1t M(t)s2t M(t)s1s2t M(t)s2s1s2t M(t)s2s1s2s1t M(t)t M(t)s1t M(t)s2s1t M(t)s1s2s1t M(t)s2t M(t)s1s2t M(t)s2s1s2t M(t)s2s1s2s1t tq2,z,q2≠1 tz,q2,q2≠1 t-1,q2,q4≠1,q8≠1$

$M(t)t M(t)s1t M(t)s2s1t M(t)s1s2s1t M(t)s2t M(t)s1s2t M(t)s2s1s2t M(t)s2s1s2s1t M(t)t M(t)s1t M(t)s2s1t M(t)s1s2s1t M(t)s2t M(t)s1s2t M(t)s2s1s2t M(t)s2s1s2s1t tq2,q2, q generic tq2,q2, q a primitive eighth root of unity$

Case 3: Z(t),P(t) .

The only central characters not covered in cases 1 and 2 are those in the orbits of ${t}_{1,{q}^{2}},$ ${t}_{{q}^{2},1},$ and ${t}_{±q,1}\text{.}$ In these cases, rather than analyzing $M\left(t\right)$ directly, it is easier to construct several irreducible modules with central character $t$ and show that their composition factors exhaust the composition factors of $M\left(t\right)\text{.}$

$t{\mid }_{Q}={t}_{{q}^{2},1}\text{.}$

If ${q}^{2}=1,$ then the results of section 1.2.9 show that $\stackrel{\sim }{H}$ has five irreducible representations - four of them 1-dimensional, and one 2-dimensional. Specifically, an irreducible $\stackrel{\sim }{H}$ module is an irreducible ${W}_{0}\text{-module}$ (via the identification $ℂ\left[{W}_{0}\right]=H\text{)}$ on which ${X}^{\lambda }$ acts by the constant $t\left({X}^{\lambda }\right),$ and $M\left(t\right)$ is isomorphic to the regular representation of $ℂ\left[{W}_{0}\right]$ as a ${W}_{0}\text{-module.}$

$M(t)=M(t)t t1,1,q2=1$

Let

$w= { s1, if q2=-1, s1s2s1, if q2≠-1.$

Then let ${ℂ}_{{q}^{2},1}$ and ${ℂ}_{{q}^{-2},1}$ be the 1-dimensional ${\stackrel{\sim }{H}}_{\left\{1\right\}}\text{-modules}$ spanned by ${v}_{t}$ and ${v}_{wt},$ respectively, given by

$Xλvt=t (Xλ)vt, andT1vt=q vt,and Xλvwt= (wt)(Xλ) vwtand T1vwt=- q-1vwt.$

Then

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

are 4-dimensional $\stackrel{\sim }{H}\text{-modules.}$

Each dot in the chamber ${w}^{-1}C$ in the following picture represents a basis element of the $wt$ weight space of $M$ or $N\text{.}$ The dots that are connected by arcs represent basis vectors in the same module, $M$ or $N\text{.}$

$C s1C s1s2C s1s2s1C M N C s1C M N tq2,1,q2≠±1 tq2,1,q a primitive fourth root of unity$

Proposition 2.8. If ${q}^{2}=-1$ and $M=\stackrel{\sim }{H}{\otimes }_{{\stackrel{\sim }{H}}_{\left\{1\right\}}}{ℂ}_{{q}^{2},1}$ and $N=\stackrel{\sim }{H}{\otimes }_{{\stackrel{\sim }{H}}_{\left\{1\right\}}}{ℂ}_{{q}^{-2},1}$ then

1. $M$ is irreducible, and
2. The map $ϕ: N ⟶ M hvwt ⟼ hv, for h∈H∼$ is a $\stackrel{\sim }{H}\text{-module}$ isomorphism, where $v={T}_{1}{T}_{2}{v}_{t}-q{T}_{2}{v}_{t}-{v}_{t}\in M,$ and
3. Any irreducible $\stackrel{\sim }{H}\text{-module}$ $L$ with central character $t$ is isomorphic to $M\text{.}$

 Proof. (a) If $q$ is a primitive fourth root of unity, then M has weight spaces ${M}_{t}^{\text{gen}},$ and ${M}_{{s}_{1}t}^{\text{gen}},$ each of which is 2-dimensional. By Lemma 1.11, the only possible submodules of $M$ are the generalized weight spaces. However, Theorem 1.10 shows that neither of these weight spaces can be a submodule, and $M$ is irreducible. (b) Let $v={T}_{1}{T}_{2}{v}_{t}-q{T}_{2}{v}_{t}-{v}_{t}\text{.}$ Then we compute: $T1· ( T1T2vt- qT2vt-vt ) = (q-q-1) T1T2vt+ T2vt-qT1 T2vt-T1vt (2.20) = qT1T2vt+T2 vt-qvt = q ( T1T2vt-q T2vt-vt ) ,and Xω1· ( T1T2vt- qT2vt-vt ) = T1Xω2-ω1 T2vt+ (q-q-1) Xω1T2vt-q Xω1T2vt- Xω1t = T1Xω2T2 X-ω1vt- q-1T2Xω1 vt-Xω1vt = T1T2 X2ω1-ω2 X-ω1vt+ (q-q-1)T1 Xω2-ω1 -q-1T2Xω1 vt-Xω1vt = t(Xω1-ω2) T1T2vt+q t(Xω1)T2vt + ( (q2-1)t (Xω2-ω1) -t(Xω1) ) vt.$ But $t\left({X}^{{\omega }_{2}}\right)={q}^{2}=-1,$ and $( (q2-1) t(Xω2-ω1) -t(Xω1) ) =2t(X-ω1)- t(Xω1)= t(Xω1) (2t(X-2ω1)-1) =t(Xω1).$ Thus $Xω1·v=- t(Xω1)v=t (Xω2-ω1) v=s1t(Xω1) v.$ Similarly, $Xω2· ( T1T2vt-q T2vt-vt ) = (T1-q) (Xω2T2vt) -Xω2vt = (T1-q) ( T2Xα1+ (q-q-1) Xω2 ) vt+vt = -T1T2vt+ (q-q-1) T1Xω2vt -qT2Xα1vt -q(q-q-1) Xω2vt+vt = -T1T2vt+2 vt+qT2vt- 2vt+vt = - ( T1T2vt-q T2vt-vt ) = s1t(Xω2) v.$ This computation shows that $v$ spans a 1-dimensional ${\stackrel{\sim }{H}}_{\left\{1\right\}}\text{-submodule}$ of $M,$ given by $T1v=qv,and Xλv=s1t (Xλ)v.$ Then the ${\stackrel{\sim }{H}}_{\left\{1\right\}}\text{-module}$ map given by ${v}_{wt}↦v$ corresponds to $\varphi$ under the adjunction $HomH∼ ( H∼⊗H∼{1} ℂq-2,1,M ) =HomH∼{1} ( ℂq-2,1, M∣H∼{1} ) .$ Thus $\varphi$ is a $\stackrel{\sim }{H}\text{-module}$ map and since $M$ is irreducible, the map is surjective. Then since $M$ and $N$ have the same dimension, $\varphi$ is an isomorphism. (c) The weight spaces in the remaining composition factor(s) of $M\left(t\right)$ besides $M$ must be $t$ and ${s}_{1}t$ weight spaces, each of which appear with multiplicity 2. Let $L$ be an irreducible $\stackrel{\sim }{H}\text{-module}$ with central character $t\text{.}$ Then by Lemma 1.11, either $L$ is two dimensional, consisting of a single generalized weight space ${L}_{t}^{\text{gen}}$ or ${L}_{{s}_{1}t}^{\text{gen}},$ or else $L$ is 4-dimensional, with $\text{dim} {L}_{t}^{\text{gen}}=2=\text{dim} {L}_{{s}_{1}t}^{\text{gen}}\text{.}$ Then, viewing $L$ as a ${\stackrel{\sim }{H}}_{\left\{1\right\}}\text{-module,}$ it must have all 1-dimensional composition factors, and it must have a 1-dimensional submodule, either ${L}_{t}$ or ${L}_{{s}_{1}t}\text{.}$ First, assume ${L}_{t}$ is a submodule and $v\in {P}_{t},$ so that ${T}_{1}v=qv\text{.}$ Since $HomH∼ ( H∼⊗H∼{1} ℂq2,1,L ) =HomH∼{1} ( ℂq2,1, L∣H∼{1} ) ,$ and ${v}_{t}↦v$ gives a non-zero ${\stackrel{\sim }{H}}_{\left\{1\right\}}\text{-module}$ map from ${ℂ}_{{q}^{2},1}$ to $L,$ there is a non-zero $\stackrel{\sim }{H}\text{-module}$ map from $M$ to $L\text{.}$ Since $M$ and $L$ are irreducible, the map is an isomorphism. Alternatively, if ${L}_{{s}_{1}t}$ is a ${\stackrel{\sim }{H}}_{\left\{1\right\}}\text{-submodule}$ of $L,$ then choose $v\in {L}_{{s}_{1}t\text{.}}$ Then ${v}_{{s}_{1}t}↦v$ gives a non-zero ${\stackrel{\sim }{H}}_{\left\{1\right\}}\text{-module}$ map from ${ℂ}_{{q}^{-2},1}$ to $L,$ and $L$ is isomorphic to $N\text{.}$ $\square$

$C s1C s1s2C s1s2s1C C s1C s1s2C s1s2s1C tq2,1,q2 a primitive third root of unity tq2,1,q generic$

Proposition 2.9.

1. If ${q}^{2}$ is a primitive third root of unity then ${M}_{{s}_{2}{s}_{1}t}$ is a submodule of $M$ isomorphic to ${L}_{-{q}^{-1},q,±1}$ and $M/{M}_{{s}_{2}{s}_{1}t}$ is irreducible. In addition, ${N}_{{s}_{1}t}$ is a submodule of $N$ isomorphic to ${L}_{q,-{q}^{-1}},$ and $N/{N}_{{s}_{1}t}$ is irreducible.
2. If ${q}^{2}$ is not $±1$ or a primitive third root of unity then $M$ and $N$ are irreducible and nonisomorphic.

 Proof. (a) Assume ${q}^{2}$ is a primitive third root of unity. Then the same computation used in (1.3) shows that ${M}_{t}^{\text{gen}}$ is spanned by ${v}_{t}$ and ${T}_{2}{v}_{t},$ while ${M}_{{s}_{1}t}$ is spanned by $vs1t=T1T2 vt+c1T2vt +c0vt$ for some ${c}_{0},{c}_{1}\in ℂ\text{.}$ Then $τ2:Ms1t ⟶Ms2s1t$ is non-zero, since the ${T}_{2}{T}_{1}{T}_{2}{v}_{t}$ term in ${\tau }_{2}\left({v}_{{s}_{1}t}\right)$ cannot be cancelled. But ${s}_{2}{s}_{1}t\left({X}^{{\alpha }_{2}}\right)={q}^{2}$ so that ${\tau }_{2}^{2}:{M}_{{s}_{2}t}\to {M}_{{s}_{2}t}$ is the zero map. Hence ${\tau }_{2}:{M}_{{s}_{2}{s}_{1}t}\to {M}_{{s}_{1}t}$ is the zero map, and ${M}_{{s}_{1}{s}_{2}t}$ is a submodule of $M\text{.}$ By Lemma 1.11, $M/{M}_{{s}_{1}{s}_{2}t}$ is irreducible. A parallel argument shows that ${N}_{{s}_{1}t}$ is a submodule of $N,$ with $N/{N}_{{s}_{1}t}$ irreducible. (b) If ${q}^{4}\ne 1$ and ${q}^{6}\ne 1,$ then $P\left(t\right)=\left\{{\alpha }_{1},{\alpha }_{1}+{\alpha }_{2}\right\}\text{.}$ Then Lemma 1.11 shows that the composition factor $M\prime$ of $M$ with ${\left(M\prime \right)}_{t}\ne 0$ has $\text{dim} {\left(M\prime \right)}_{t}^{\text{gen}}\ge 2$ and ${\left(M\prime \right)}_{{s}_{1}t}\ne 0\text{.}$ Then by Theorem 1.9, ${\left(M\prime \right)}_{{s}_{2}{s}_{1}t}\ne 0,$ so that $M\prime =M\text{.}$ Similarly, Lemma 1.11 and Theorem 1.9 show that $N$ is irreducible. Since they have different weight spaces, they are not isomorphic. $\square$

$t{\mid }_{Q}={t}_{1,{q}^{2}}$

Let ${ℂ}_{1,{q}^{2}}$ and ${ℂ}_{1,{q}^{-2}}$ be the 1-dimensional ${\stackrel{\sim }{H}}_{\left\{2\right\}}\text{-modules}$ spanned by ${v}_{t}$ and ${v}_{{w}_{0}t},$ respectively, and given by

$T2vt=qvt andXλvt=t (Xλ)vt, and T2vw0t=- q-1vw0t andXλ vw0t=w0t (Xλ)vw0t .$

Then

$M=H∼⊗H∼{2} ℂ1,q2and N=H∼⊗H∼{2} ℂ1,q-2$

are 4-dimensional $\stackrel{\sim }{H}\text{-modules.}$

Each dot in the chamber ${w}^{-1}C$ in the following picture represents a basis element of the $wt$ generalized weight space of $M$ or $N\text{.}$ The dots that are connected by arcs represent basis vectors in the same module, $M$ or $N\text{.}$

$C s2C s2s1C s2s1s2C M N t1,q2,q generic$

Proposition 2.10. Assume ${q}^{2}=-1$ and let $M=\stackrel{\sim }{H}{\otimes }_{{\stackrel{\sim }{H}}_{\left\{2\right\}}}{ℂ}_{1,{q}^{2}}$ and $N=\stackrel{\sim }{H}{\otimes }_{{\stackrel{\sim }{H}}_{\left\{2\right\}}}{ℂ}_{1,{q}^{-2}}\text{.}$ Then

1. ${M}_{{s}_{1}{s}_{2}t}$ is a submodule of $M,$ and the image of ${M}_{{s}_{2}t}$ is a submodule of $M/{M}_{{s}_{1}{s}_{2}t}\text{.}$ The resulting 2-dimensional quotient of $M$ is irreducible. Also, ${N}_{{s}_{2}t}$ is a submodule of $N$ and the image of ${N}_{{s}_{1}{s}_{2}t}$ in $N/{N}_{{s}_{2}t}$ is a submodule of $N/{N}_{{s}_{2}t}\text{.}$ The resulting 2-dimensional quotient of $N$ is irreducible, and
2. Any composition factor of $M\left(t\right)$ is a composition factor of either $M$ or $N\text{.}$

 Proof. (a) If ${q}^{2}=-1,$ then $M$ has weight spaces ${M}_{t}^{\text{gen}},$ which is two dimensional, and ${M}_{{s}_{2}t}$ and ${M}_{{s}_{1}{s}_{2}t},$ both of which are 1-dimensional. Similarly, $N$ has weight spaces ${N}_{{s}_{2}{s}_{1}{s}_{2}t}^{\text{gen}},$ which is 2-dimensional, along with ${N}_{{s}_{1}{s}_{2}t}$ and ${N}_{{s}_{2}t},$ both 1-dimensional. Since there are two 1-dimensional modules with central character $t,$ with weights ${s}_{2}t$ and ${s}_{1}{s}_{2}t,$ it cannot be true that both $M$ and $N$ are irreducible. In $M,$ the weight space ${M}_{t}^{\text{gen}}$ is spanned by ${v}_{t}$ and ${T}_{1}{v}_{t},$ while ${M}_{{s}_{2}t}$ is spanned by ${T}_{2}{T}_{1}{v}_{t}-1{T}_{1}{v}_{t}-2{v}_{t}$ and ${M}_{{s}_{1}{s}_{2}t}$ is spanned by ${T}_{1}{T}_{2}{T}_{1}{v}_{t}+q{T}_{2}{T}_{1}{v}_{t}-{T}_{1}{v}_{t}+q{v}_{t}\text{.}$ Then ${M}_{{s}_{1}{s}_{2}t}$ is a submodule of $M,$ and the image of ${M}_{{s}_{2}t}$ is a submodule of $M/{M}_{{s}_{1}{s}_{2}t}\text{.}$ By Lemma 1.11, the resulting 2-dimensional quotient is irreducible. Similarly, ${N}_{{s}_{2}t}$ is spanned by ${T}_{1}{T}_{2}{T}_{1}{v}_{t}+q{T}_{2}{T}_{1}{v}_{t}-{T}_{1}{v}_{t}+q{v}_{t},$ and $N/{N}_{{s}_{2}t}$ has a 1-dimensional submodule spanned by the image of ${T}_{2}{T}_{1}{v}_{t}-q{T}_{1}{v}_{t}-2{v}_{t}\text{.}$ The resulting 2-dimensional quotient is irreducible. (b) There are at least two 2-dimensional modules and two 1-dimensional modules with central character ${t}_{1,{q}^{2}}\text{.}$ Counting dimensions of weight spaces, the remaining composition factor(s) of $M\left(t\right)$ must have weights ${s}_{2}t$ and ${s}_{1}{s}_{2}t\text{.}$ If there were only one composition factor $L$ left, it would contain both weight spaces which would each have dimension 1. But by Theorem 1.10, no irreducible module with this weight space structure exists. Hence there are two remaining composition factors of $M\left(t\right),$ each of which is 1-dimensional and isomorphic to one of those modules found in $M$ and $N$ above. $\square$

Proposition 2.11. Assume ${q}^{2}\ne ±1\text{.}$ Then ${M}_{{s}_{1}{s}_{2}t}$ is a submodule of $M$ and $M/{M}_{{s}_{1}{s}_{2}t}$ is irreducible. Similarly, ${N}_{{s}_{2}t}$ is a submodule of $N$ and $N/{N}_{{s}_{2}t}$ is irreducible.

 Proof. If ${q}^{4}\ne 1,$ then by the same reasoning as in Proposition 2.10, ${M}_{t}^{\text{gen}}$ is spanned by ${v}_{t}$ and ${T}_{1}{v}_{t}\text{.}$ And since ${M}_{{s}_{2}t}$ is spanned by ${T}_{2}{T}_{1}{v}_{t}+{f}_{{s}_{1}t}{T}_{1}{v}_{t}+{f}_{t}{v}_{t}$ for some ${f}_{{s}_{1}t}$ and ${f}_{t}\in ℂ\left[q,{q}^{-1}\right],$ ${\tau }_{1}:{M}_{{s}_{2}t}\to {M}_{{s}_{1}{s}_{2}t}$ is non-zero. But, since ${s}_{2}t\left({X}^{{\alpha }_{1}}\right)={q}^{2},$ ${\tau }_{1}:{M}_{{s}_{1}{s}_{2}t}\to {M}_{{s}_{2}t}$ must be the zero map. Hence, since ${\tau }_{2}:{M}_{{s}_{1}{s}_{2}t}\to {M}_{{s}_{2}{s}_{1}{s}_{2}t}$ is clearly 0, ${M}_{{s}_{1}{s}_{2}t}$ must be a submodule of $M\text{.}$ Similarly, ${N}_{{s}_{2}t}$ is a submodule of $N\text{.}$ Then Lemma 1.11 shows that the resulting 3-dimensional quotients of $M$ and $N$ are irreducible. $\square$

If ${q}^{2}\ne 1,$ the composition factors of $M$ and $N$ account for all 8 dimensions of $M\left(t\right)\text{.}$ The following pictures show the composition factors of $M\left(t\right)\text{.}$ Each dot in ${w}^{-1}C$ represents one basis element of $M{\left(t\right)}_{wt}^{\text{gen}},$ and the basis elements corresponding to connected dots are in the same composition factor.

$M(t)tgen M(t)s2t M(t)s1s2t M(t)s2s1s2tgen M(t)tgen M(t)s2t M(t)s1s2t M(t)s2s1s2tgen t1,q2,q a primitive fourth root of unity t1,q2,q generic$

$t{\mid }_{Q}={t}_{±q,1}\text{.}$

Let ${ℂ}_{±q,1}$ and ${ℂ}_{±{q}^{-1},1}$ be the 1-dimensional ${\stackrel{\sim }{H}}_{\left\{2\right\}}\text{-modules}$ spanned by ${v}_{{s}_{1}t}$ and ${v}_{{s}_{2}{s}_{1}t},$ respectively, and given by

$T2vs1t=q vs1tand Xλvs1t= s1t(Xλ) vs1t,and T2vs2s1t =-q-1 vs2s1t andXλ vs2s1t= s2s1t(Xλ) vs2s1t.$

Then

$M=H∼⊗H∼{2} ℂ±1,1andN= H∼⊗H∼{2} ℂ±q-1,1$

are 4-dimensional $\stackrel{\sim }{H}\text{-modules.}$

If $t{\mid }_{Q}={t}_{-q,1}$ and $q$ is a primitive sixth root of unity or if $t{\mid }_{Q}={t}_{q,1}$ and $q$ is a primitive third root of unity, then $t{\mid }_{Q}={t}_{{q}^{-2},1},$ which is in the same orbit as ${t}_{{q}^{2},1},$ and the irreducibles with central character $t$ have already been analyzed.

$M N t±q,1 q2≠1, (excluding t-q,1 when q is a primitive sixth root of unity, and tq,1 when q is a primitive third root of unity.)$

Proposition 2.12. Let $M=\stackrel{\sim }{H}{\otimes }_{{\stackrel{\sim }{H}}_{\left\{2\right\}}}{ℂ}_{±q,1}$ and $N=\stackrel{\sim }{H}{\otimes }_{{\stackrel{\sim }{H}}_{\left\{2\right\}}}{ℂ}_{±{q}^{-1},1}\text{.}$ Unless $t{\mid }_{Q}={t}_{-q,1}$ and $q$ is a primitive sixth root of unity or $t{\mid }_{Q}={t}_{q,1}$ and $q$ is a primitive third root of unity, $M$ and $N$ are irreducible.

 Proof. Then if ${q}^{2}\ne 1,$ both $M$ and $N$ are irreducible, except in the cases that $t={t}_{-q,1}$ when $q$ is a primitive sixth root of unity and ${t}_{q,1}$ when $q$ is a primitive third root of unity. These characters are in the same orbit as ${t}_{1,{q}^{2}},$ which has already been analyzed, and is thus excluded from the following analysis. With this assumption, $M$ has weights $t$ and ${s}_{1}t,$ with $\text{dim} {M}_{t}^{\text{gen}}=2=\text{dim} {M}_{{s}_{1}t}^{\text{gen}}\text{.}$ Since $P\left(t\right)=\left\{2{\alpha }_{1}+{\alpha }_{2}\right\},$ ${\tau }_{1}:{M}_{t}^{\text{gen}}\to {M}_{{s}_{1}t}^{\text{gen}}$ is invertible, so that the dimensions of ${M}_{t}^{\text{gen}}$ and ${M}_{{s}_{1}t}^{\text{gen}}$ must be the same in any $\stackrel{\sim }{H}\text{-module.}$ Then by Lemma 1.11, $M$ is irreducible. Similar reasoning shows that $N$ is irreducible. $\square$

Since they have different weight spaces and are not isomorphic, $M$ and $N$ are the only two irreducibles with central character $t\text{.}$

### Summary

The following tables summarize the classification. It should be noted that for any value of $q$ with ${q}^{2}$ not a root of unity of order 4 or less, 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 4 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 4 or less. When these hyperplanes coincide, the sets $P\left(t\right)$ and $Z\left(t\right)$ change for characters on those hyperplanes.

$t∣Q Z(t) P(t) Dim. of Irreds. t1,1 R+ ∅ 8 t-1,1 {α2,2α1+α2} ∅ 8 t1,z {α1} ∅ 8 t1,q2 {α1} R+\{α1} 1,1,3,3 tq2,1 {α2} {α1,α1+α2} 4,4 tq,1 {α2} {2α1+α2} 4,4 t-q,1 {α2} {2α1+α2} 4,4 tz,1 {α2} ∅ 8 tq2,q2 ∅ {α1,α2} 1,1,3,3 tq2,z ∅ {α1} 4,4 t-1,q2 ∅ {α2,2α1+α2} 2,2,2,2 tz,q2 ∅ {α2} 4,4 tz,w ∅ ∅ 8 Table 8: Table of possible central characters in type C2 , with general ℓ.$

$t∣Q Z(t) P(t) Dim. of Irreds. t1,1 R+ R+ 1,1,1,1,2 t-1,1 {α2,2α1+α2} {α2,2α1+α2} 2,2,2,2 t1,z {α1} {α1} 4,4 tz,1 {α2} {α2} 4,4 tz,w ∅ ∅ 8 Table 9: Table of possible central characters in type C2 , with q=-1.$

$t∣Q Z(t) P(t) Dim. of Irreds. t1,1 R+ ∅ 8 tq2,1 {α2,2α1+α2} {α1,α1+α2} 4 t1,z {α1} ∅ 8 t1,q2 {α1} {α2,α1+α2,2α1+α2} 1,1,2,2 tq,1 {α2} {2α1+α2} 4,4 tz,1 {α2} ∅ 8 tq2,z ∅ {α1} 4,4 tz,q2 ∅ {α2} 4,4 tz,w ∅ ∅ 8 Table 10: Table of possible central characters in type C2 , with q a primitive fourth root of unity$

$t∣Q Z(t) P(t) Dim. of Irreds. t1,1 R+ ∅ 8 t-1,1 {α2,2α1+α2} ∅ 8 t1,z {α1} ∅ 8 t1,q2 {α1} {α2,α1+α2,2α1+α2} 1,1,3,3 tq2,1 {α2} {α1,α1+α2,2α1+α2} 1,1,3,3 tq,1 {α2} {2α1+α2} 4,4 tz,1 {α2} ∅ 8 tq2,z ∅ {α1} 4,4 t-1,q2 ∅ {α2,2α1+α2} 2,2,2,2 tz,q2 ∅ {α2} 4,4 tz,w ∅ ∅ 8 Table 11: Table of possible central characters in type C2 , with q a primitive sixth root of unity.$

$t∣Q Z(t) P(t) Dim. of Irreds. t1,1 R+ ∅ 8 t-1,1 {α2,2α1+α2} ∅ 8 t1,z {α1} ∅ 8 t1,q2 {α1} {α2,α1+α2,2α1+α2} 1,1,3,3 tq2,1 {α2} {α1,α1+α2} 4,4 tq,1 {α2} {2α1+α2} 4,4 t-q,1 {α2} {2α1+α2} 4,4 tz,1 {α2} ∅ 8 tq2,q2 ∅ {α1,α2,2α1+α2} 1,1,1,1,2,2 tq2,z ∅ {α1} 4,4 tz,q2 ∅ {α2} 4,4 tz,w ∅ ∅ 8 Table 12: Table of possible central characters in type C2 , with q a primitive eighth root of unity.$

## 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.