## Type ${G}_{2}$

Last update: 17 November 2012

## Type ${G}_{2}$

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

$R= { ±α1, ±α2, ±(α1+α2), ±(2α1+α2), ±(3α1+α2), ±(3α1+2α2) } ,$

with $⟨{\alpha }_{1}{\alpha }_{2}^{\vee }⟩=-1$ and $⟨{\alpha }_{2},{\alpha }_{1}^{\vee }⟩=-3\text{.}$ Then $R$ is a root system as defined in (1.2.1), and the Weyl group is

$W0= ⟨ s1,s2 ∣ s12=s22=1, s1s2s1s2s1 s2=s2s1s2s1 s2s1 ⟩ ,$

isomorphic to the dihedral group of order 12. The simple roots are ${\alpha }_{1}$ and ${\alpha }_{2},$ and ${\alpha }_{1},$ ${\alpha }_{1}+{\alpha }_{2},$ $3{\alpha }_{1}+{\alpha }_{2}$ will be referred to as short roots, while ${\alpha }_{2},$ $2{\alpha }_{1}+{\alpha }_{2}$ and $3{\alpha }_{1}+2{\alpha }_{2}$ will be referred to as long roots.

$α1 α2 The typeG2root system$

The fundamental weights satisfy

$ω1=2α1+α2 α1=2ω1-ω2 ω2=3α1+2α2 α2=2ω2-3ω1.$

Let

$P=ℤ-span {ω1,ω2}.$

This is the same lattice spanned by ${\alpha }_{1}$ and ${\alpha }_{2}\text{.}$

$ω2 ω1 0 Hα1 Hα2$

The affine Hecke algebra $\stackrel{\sim }{H}$ is generated as a $ℂ\text{-algebra}$ by ${T}_{1},$ ${T}_{2},$ and $X=\left\{{X}^{\lambda }\phantom{\rule{0.2em}{0ex}}\mid \phantom{\rule{0.2em}{0ex}}\lambda \in P\right\},$ with relations

$XλXμ = {X}^{\lambda +\mu }, forλ,μ ∈P (2.21) T1T2 T1T2 T1T2 = T2T1 T2T1 T2T1 (2.22) Ti2 = (q-q-1) Ti+1, fori=1,2 (2.23) Xω1T1 = T1 Xs1·ω1 +(q-q-1) Xω1 (2.24) Xω2T2 = T2 Xs1·ω2 +(q-q-1) Xω2 (2.25) Xω1T2 = T2Xω1 (2.26) Xω2T1 = T1Xω2 (2.27)$

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

$s1·Xω1 = Xω2-ω1, s1·Xω2 = Xω2, s2·Xω1 = Xω1,and s2·Xω2 = X3ω1-ω2.$

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

$(w·t)(Xλ) =t(Xw-1λ) .$ $Hα1 H2α1+α2 H3α1+α2 Hα2 Hα1+α2 H3α1+2α2+δ H3α1+2α2 Hα1+α2+δ Hα2+δ H3α1+α2+δ H2α1+α2+δ Hα1+δ$

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 $P\left(t\right)=\left\{\alpha \in {R}^{+}\phantom{\rule{0.2em}{0ex}}\mid \phantom{\rule{0.2em}{0ex}}t\left({X}^{\alpha }\right)={q}^{±2}\right\}$ and $Z\left(t\right)=\left\{\alpha \in {R}^{+}\phantom{\rule{0.2em}{0ex}}\mid \phantom{\rule{0.2em}{0ex}}t\left({X}^{\alpha }\right)=1\right\}\text{.}$ For a generic weight $t,$ $P\left(t\right)$ and $Z\left(t\right)$ are empty, so we examine only the non-generic orbits.

Theorem 2.13. If ${q}^{2}$ is not a primitive $\ell \text{th}$ root of unity for $\ell \le 6$ and $Z\left(t\right)\cup P\left(t\right)\ne \varnothing ,$ then $t$ is in the same ${W}_{0}\text{-orbits}$ as one of the following weights.

$t1,1, t1,-1, t11/3,1, t1,q2, t1,±q, tq2,1, t±q,1, tq2/3,1, tq2,-q-2, t11/3,q2, tq2,q2, { t1,z∣ z∈ℂ×,z≠±1 ,q±2,± q±1 } , { tz,1∣ z∈ℂ×,z≠±1, 11/3,q±2 ,±q±1, q±2/3 } , { tq2,z ∣ z∈ℂ×, {1,q2,q-2} ∩ { z,q2z,q4z, q6z,q6z2 } =∅ } ,or { tz,q2 ∣ z∈ℂ×, {1,q2,q-2} ∩ { z,q2z,q2z2, q2z3,q4z3 } =∅ } .$

 Proof. In general, the third roots of unity in this theorem are assumed to be primitive, so that there are two different weights that we call ${t}_{{1}^{1/3},1}$ and ${t}_{{1}^{1/3},{q}^{2}}\text{.}$ Similarly, ${t}_{{q}^{2/3},1}$ typically refers to one of three different characters by corresponding to the three third roots of ${q}^{2}\text{.}$ Exceptions to this principle are noted as they arise. We refer to ${\alpha }_{1},$ ${\alpha }_{1}+{\alpha }_{2},$ and $2{\alpha }_{1}+{\alpha }_{2}$ as “short” roots. The other roots are referred to as “long” roots. Case 1: $\mid Z\left(t\right)\mid \ge 2$ If $Z\left(t\right)$ contains at least two roots, and one of them is short, then $Z\left(wt\right)$ contains ${\alpha }_{1}$ for some $w\in {W}_{0}\text{.}$ If $Z\left(wt\right)$ also contains any of ${\alpha }_{2},$ ${\alpha }_{1}+{\alpha }_{2},$ $2{\alpha }_{1}+{\alpha }_{2},$ or $3{\alpha }_{1}+{\alpha }_{2},$ then it contains both simple roots and thus $wt={t}_{1,1}\text{.}$ It is also possible that $Z\left(wt\right)=\left\{{\alpha }_{1},3{\alpha }_{1}+2{\alpha }_{2}\right\},$ in which case $wt\left({X}^{{\alpha }_{1}+{\alpha }_{2}}\right)=-1,$ and $wt\left({X}^{{\alpha }_{2}}\right)=-1,$ so that $wt={t}_{1,-1}\text{.}$ If $Z\left(t\right)$ contains no short roots, it contains two of ${\alpha }_{2},$ $3{\alpha }_{1}+{\alpha }_{2},$ and $3{\alpha }_{1}+2{\alpha }_{2}\text{.}$ But then it must also contain the third, and $Z\left(t\right)=\left\{{\alpha }_{2},3{\alpha }_{1}+{\alpha }_{2},3{\alpha }_{1}+2{\alpha }_{2}\right\}\text{.}$ In this case, $wt\left({X}^{{\alpha }_{2}}\right)=1,$ but $wt\left({X}^{{\alpha }_{1}}\right)$ is a third root of unity, so that $wt={t}_{{1}^{1/3},1}\text{.}$ Case 2: $\mid Z\left(t\right)\mid =1$ If $Z\left(t\right)$ has exactly one root, then there is some $w\in {W}_{0}$ with $Z\left(wt\right)=\left\{{\alpha }_{1}\right\}$ or $Z\left(wt\right)=\left\{{\alpha }_{2}\right\}\text{.}$ If $Z\left(wt\right)=\left\{{\alpha }_{1}\right\},$ then $P\left(t\right)$ either contains all of ${\alpha }_{2},$ ${\alpha }_{1}+{\alpha }_{2},$ $2{\alpha }_{1}+{\alpha }_{2},$ and $3{\alpha }_{1}+{\alpha }_{2},$ or it contains none of them. If it contains all of them, $wt\left({X}^{{\alpha }_{2}}\right)={q}^{±2},$ and $t$ is in the same ${W}_{0}\text{-orbit}$ as ${t}_{1,{q}^{2}}\text{.}$ If $3{\alpha }_{1}+2{\alpha }_{2}\in P\left(wt\right),$ then $wt\left({X}^{{\alpha }_{2}}\right)=±{q}^{±1},$ and $t$ is in the same orbit as ${t}_{1,±q}\text{.}$ Otherwise, $wt={t}_{1,z}$ for some $z$ besides $±{q}^{±1}$ and ${q}^{±2}\text{.}$ Also then, $z\ne ±1$ by assumption on $Z\left(t\right)\text{.}$ If $Z\left(wt\right)=\left\{{\alpha }_{2}\right\},$ then any two roots that differ by a multiple of ${\alpha }_{2}$ are either both in $P\left(wt\right)$ or both not in $P\left(wt\right)\text{.}$ By applying ${w}_{0}$ if necessary, we can assume that $wt\left({X}^{\alpha }\right)={q}^{2}$ for the $\alpha$ that are in $P\left(wt\right)\text{.}$ If ${\alpha }_{1}\in P\left(wt\right),$ then $wt\left({X}^{{\alpha }_{1}}\right)={q}^{2},$ and $wt={t}_{{q}^{2},1}\text{.}$ If $2{\alpha }_{1}+{\alpha }_{2}\in P\left(wt\right),$ then $wt={t}_{±q,1}\text{.}$ If $3{\alpha }_{1}+{\alpha }_{2}\in P\left(wt\right),$ then $wt\left({X}^{{\alpha }_{1}}\right)$ is a third root of ${q}^{2}$ and $wt={t}_{1,{q}^{2/3}}\text{.}$ Otherwise, $wt={t}_{z,1}$ for some $z$ so that none of $z,{z}^{2},{z}^{3}$ is equal to ${q}^{±2}$ or 1. That is, $z\ne ±1,$ ${1}^{1/3},$ ${q}^{±2},$ $±{q}^{±1},$ ${q}^{±2/3}\text{.}$ Case 3: $\mid Z\left(t\right)\mid =\varnothing$ If $Z\left(t\right)$ is empty but $P\left(t\right)$ contains a short root, then ${\alpha }_{1}\in P\left(wt\right)$ for some $w\in {W}_{0}\text{.}$ If $P\left(wt\right)$ contains another short root, then we can apply ${s}_{1}$ if necessary so that $P\left(t\right)$ contains ${\alpha }_{1}$ and ${\alpha }_{1}+{\alpha }_{2}\text{.}$ Then either $wt\left({X}^{{\alpha }_{1}}\right)=wt\left({X}^{{\alpha }_{1}+{\alpha }_{2}}\right)$ so that $wt\left({X}^{{\alpha }_{2}}\right)=1,$ or $wt\left({X}^{{\alpha }_{1}}\right)$ and $wt\left({X}^{{\alpha }_{1}+{\alpha }_{2}}\right)$ are ${q}^{2}$ and ${q}^{-2}$ in some order, so that $wt\left({X}^{2{\alpha }_{1}+{\alpha }_{2}}\right)=1\text{.}$ Thus $P\left(wt\right)$ contains at most one short root. If $P\left(wt\right)$ also contains a long root, then applying ${s}_{1}$ if necessary, we can assume $P\left(wt\right)$ contains either ${\alpha }_{2}$ or $3{\alpha }_{1}+2{\alpha }_{2}\text{.}$ If $P\left(wt\right)$ contains ${\alpha }_{1}$ and ${\alpha }_{2},$ then we can apply ${w}_{0}$ to assume that $wt\left({X}^{{\alpha }_{1}}\right)={q}^{2}\text{.}$ If $wt\left({X}^{{\alpha }_{2}}\right)={q}^{-2},$ then ${\alpha }_{1}+{\alpha }_{2}\in Z\left(wt\right)\text{.}$ Then $wt\left({X}^{{\alpha }_{2}}\right)={q}^{2}$ and $wt={t}_{{q}^{2},{q}^{2}}\text{.}$ If $P\left(wt\right)$ contains ${\alpha }_{1}$ and $3{\alpha }_{1}+2{\alpha }_{2},$ then since ${\alpha }_{1}$ is perpendicular to $3{\alpha }_{1}+2{\alpha }_{2},$ we can apply ${s}_{1}$ and/or ${s}_{3{\alpha }_{1}+2{\alpha }_{2}}$ to assume $wt\left({X}^{{\alpha }_{1}}\right)={q}^{2}=wt\left({X}^{3{\alpha }_{1}+2{\alpha }_{2}}\right)\text{.}$ Hence $wt\left({X}^{2{\alpha }_{2}}\right)={q}^{-4}$ and by assumption, $wt={t}_{{q}^{2},-{q}^{-2}}\text{.}$ If $P\left(wt\right)=\left\{{\alpha }_{1}\right\},$ then $wt={t}_{{q}^{2},z}$ does not take the value 1 or ${q}^{±2}$ on any other positive root. Then $\left\{1,{q}^{2},{q}^{-2}\right\}\cap \left\{z,{q}^{2}z,{q}^{4}z,{q}^{6}z,{q}^{6}{z}^{2}\right\}=\varnothing \text{.}$ If $P\left(t\right)$ contains no short roots, but at least two long roots, then $wt\left({X}^{{\alpha }_{2}}\right)={q}^{2}=wt\left({X}^{3{\alpha }_{1}+{\alpha }_{2}}\right)$ for some $w\in {W}_{0}\text{.}$ (If $wt\left({X}^{{\alpha }_{2}}\right)={q}^{-2},$ then $wt\left({X}^{3{\alpha }_{1}+2{\alpha }_{2}}\right)=1,$ a contradiction.) Hence $wt\left({X}^{{\alpha }_{1}}\right)$ is a primitive third root of unity and $wt={t}_{{1}^{1/3},{q}^{2}}\text{.}$ If $P\left(t\right)$ contains exactly one long root, then $wt={t}_{z,{q}^{2}}$ for some $z\in {ℂ}^{×}$ so that $wt$ does not take the value 1 or ${q}^{±2}$ on any other positive root. Thus $\left\{q,{q}^{2},{q}^{-2}\right\}\cap \left\{z,{q}^{2},{q}^{2},{z}^{2},{q}^{2},{z}^{3},{q}^{4},{z}^{3}\right\}=\varnothing \text{.}$ $\square$

Remark. There are some redundancies in this list for specific values of $q\text{.}$ If ${q}^{2}$ is a primitive fifth root of unity, then $q$ and $-q$ are equal to ${q}^{-4}$ and $-{q}^{-4}$ in some order depending on whether ${q}^{5}=1$ or $-1\text{.}$ Then ${t}_{{q}^{2},{q}^{2}}$ is in the same orbit as ${t}_{{q}^{-4},1},$ which is equal to either ${t}_{q,1}$ or ${t}_{-q,1}\text{.}$

If ${q}^{2}$ is a primitive fourth root of unity, then one note is necessary on the weight ${t}_{{q}^{2/3},1}\text{.}$ Since ${q}^{-2}$ is a third root of unity, we take ${q}^{2/3}$ to mean a different third root of ${q}^{2}$ so that ${t}_{{q}^{2/3},1}$ and ${t}_{{q}^{2},1}$ are in different orbits. In addition, ${t}_{{q}^{2}-{q}^{-2}}={t}_{{q}^{2},{q}^{2}},$ which is in the same orbit as ${t}_{{q}^{2},1}\text{.}$

If ${q}^{2}$ is a primitive third root of unity, then ${1}^{1/3}={q}^{2},{q}^{-2},$ or 1. Then ${t}_{{1}^{1/3},1}$ is in the same orbit as ${t}_{{q}^{2},1}$ or ${t}_{1,1}\text{.}$ Also, ${t}_{{1}^{1/3},{q}^{2}}$ is in the same orbit as ${t}_{{q}^{2},{q}^{2}},$ which is in turn in the same orbit as ${t}_{1,{q}^{2}}\text{.}$ In addition $q$ and $-q$ are equal to ${q}^{-2}$ and $-{q}^{-2}$ in some order depending on whether ${q}^{3}$ is 1 or $-1\text{.}$ Then ${t}_{1,{q}^{2}}$ is in the same orbit as either ${t}_{1,q}$ or ${t}_{1,-q},$ and ${t}_{{q}^{2},1}$ is in the same orbit as either ${t}_{q,1}$ or ${t}_{-q,1}\text{.}$

If ${q}^{2}=-1,$ then ${t}_{{q}^{2},-{q}^{-2}}={t}_{{q}^{2},1}={t}_{-1,1},$ which is in the same orbit as ${t}_{{q}^{2},{q}^{2}},$ while ${t}_{1,{q}^{2}}={t}_{1,-1}\text{.}$ In fact, ${t}_{-1,1}={s}_{1}{s}_{2}{t}_{1,-1}\text{.}$ Also, since $q=-{q}^{-1},$ the weights ${t}_{1,±q}$ are in the same orbit as each other, as are the weights ${t}_{±1,q}\text{.}$ Finally, ${t}_{{1}^{1/3},{q}^{2}}$ is in the same orbit as ${t}_{{q}^{2/3},1}\text{.}$

Finally, if $q=-1,$ then ${t}_{1,1}={t}_{{q}^{2},1}={t}_{1,{q}^{2}}={t}_{{q}^{2},{q}^{2}}={t}_{1,-q}={t}_{-q,1}\text{.}$ Also, ${t}_{q,1}={t}_{-1,1},$ which is in the same orbit as ${t}_{1,-1}={t}_{1,q}={t}_{{q}^{2},-{q}^{-2}}\text{.}$ Finally, ${t}_{{1}^{1/3},1}={t}_{{q}^{2/3},1}={t}_{{1}^{1/3},{q}^{2}},$ while ${t}_{1,z}={t}_{{q}^{2},z}$ and ${t}_{z,1}={t}_{z,{q}^{2}}\text{.}$

### Analysis of the characters

Theorem 2.14. The 1-dimensional representations of $\stackrel{\sim }{H}$ are:

$Lq,q:H∼→ℂ Lq,-q-1:H∼→ℂ Lq,-q-1:H∼→ℂ L-q-1,-q-1:H∼→ℂ T1↦q T1↦q T1↦-q-1 T1↦-q-1 T2↦q T2↦-q-1 T2↦q T2↦-q-1 Xω1↦q6 Xω1↦q2 Xω1↦q-2 Xω1↦q-6 Xω2↦q10 Xω2↦q2 Xω2↦q-2 Xω2↦q-10$

 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 $Xω1T1=T1 Xω2-ω1+ (q-q-1) Xω1,and Xω2T2=T2 X3ω1-ω2+ (q-q-1) Xω2$ If ${T}_{1}v=qv,$ then the first equation forces $X2ω1-ω2v= q2v,$ but ${\alpha }_{1}=2{\omega }_{1}-{\omega }_{2},$ so this implies that ${X}^{{\alpha }_{1}}v={q}^{2}v\text{.}$ Similarly, if ${T}_{1}v=-{q}^{-1}v,$ then ${X}^{{\alpha }_{1}}v={q}^{-2}v\text{.}$ By the second relation, if ${T}_{2}v=qv,$ then ${X}^{{\alpha }_{2}}v={q}^{2}v$ and if ${T}_{2}v={q}^{-2}v,$ then ${X}^{{\alpha }_{2}}v={q}^{-2}v\text{.}$ $\square$

Remark: If ${q}^{2}=-1,$ then $q=-{q}^{-1}$ and all four 1-dimensional representations are isomorphic.

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

$M(t)= Indℂ[X]H∼ ℂt=H∼ ⊗ℂ[X] ℂt,$

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

$ℂt=span{vt} andXλvt =t(Xλ)vt.$

By (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 $t$ suffices to find all the irreducible $\stackrel{\sim }{H}\text{-modules.}$

Case 1: $P\left(t\right)=\varnothing \text{.}$

If $P\left(t\right)=\varnothing$ then by Kato’s criterion, Theorem 1.8, $M\left(t\right)$ is irreducible and thus 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 ${\tau }_{i}:\phantom{\rule{0.2em}{0ex}}M{\left(t\right)}_{wt}^{\text{gen}}\to M{\left(t\right)}_{{s}_{i}wt}^{\text{gen}}$ 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 $M{\left(t\right)}_{wt}^{\text{gen}}\text{.}$

$C C s2C s2s1C t1,1,q2≠1 t1,-1,q2≠±1$ $C s1C C s1C s1s2C s1s2s1C s1s2s1s2C s1s2s1s2s1C t13,1,q2 ≠1,q6≠1 tz,1,q2 ≠1,zgeneric$

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

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

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

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.

Case 3: $Z\left(t\right),P\left(t\right)\ne \varnothing \text{.}$

For these central characters, rather than analyzing $M\left(t\right)$ directly, it is easier to construct several irreducible $\stackrel{\sim }{H}\text{-modules}$ and show that they include all the composition factors of $M\left(t\right)\text{.}$

Case 3a: ${t}_{1,{q}^{2}}$

Assume ${\alpha }_{1}\in Z\left(t\right)$ and ${\alpha }_{2}\in P\left(t\right)\text{.}$ Then $t={t}_{1,{q}^{±2}},$ but ${s}_{2}{s}_{1}{s}_{2}{s}_{1}{s}_{2}{t}_{1,{q}^{-2}}={t}_{1,{q}^{2}},$ so that analyzing $M\left({t}_{1,{q}^{2}}\right)$ is sufficient. Then let $t={t}_{1,{q}^{2}}\text{.}$

If ${q}^{2}=1,$ then $Z\left(t\right)=P\left(t\right)={R}^{+},$ and the irreducibles with this central character can be constructed using the results of (1.2.9). Specifically, there are four 1-dimensional modules and two 2-dimensional modules with central character $t\text{.}$

Otherwise, let ${w}_{0}={s}_{1}{s}_{2}{s}_{1}{s}_{2}{s}_{1}{s}_{2}$ and define

$H∼{2}=ℂ-span { T2Xλ,Xλ ∣λ∈P } ,$

the subalgebra of $\stackrel{\sim }{H}$ generated by ${T}_{2}$ and $ℂ\left[X\right]\text{.}$ Let $ℂ{v}_{t}$ and $ℂ{v}_{{w}_{0}t}$ be the 1-dimensional ${\stackrel{\sim }{H}}_{\left\{2\right\}}\text{-modules}$ spanned by ${v}_{t}$ and ${v}_{wt},$ respectively, and given by

$T2vt=qvt, Xλvt=t(Xλ)vt, T2vw0t= -q-1vw0t, and Xλvw0t= W0t(Xλ) vw0t.$

Then define

$M=ℂvt ⊗H∼{2} H∼and N=ℂvw0t ⊗H∼{2} H∼.$

Proposition 2.15. Assume ${q}^{2}\ne ±1\text{.}$ Let $M=ℂ{v}_{t}{\otimes }_{{\stackrel{\sim }{H}}_{\left\{2\right\}}}\stackrel{\sim }{H}$ and $N=ℂ{v}_{{w}_{0}t}{\otimes }_{{\stackrel{\sim }{H}}_{\left\{2\right\}}}\stackrel{\sim }{H},$ where $t={t}_{1,{q}^{2}}$.

1. ${M}_{{s}_{1}{s}_{2}{s}_{1}{s}_{2}t}$ is a 1-dimensional submodule of $M,$ and ${M}^{\prime },$ the image of the weight spaces ${M}_{{s}_{2}{s}_{1}{s}_{2}t}$ and ${M}_{{s}_{1}{s}_{2}t}$ in $M/{M}_{{s}_{1}{s}_{2}{s}_{1}{s}_{2}t},$ is a submodule of $M/{M}_{{s}_{1}{s}_{2}{s}_{1}{s}_{2}t}\text{.}$ The resulting quotient of $M$ is irreducible.
2. If ${q}^{2}$ is not a primitive third root of unity, then ${M}^{\prime }$ is irreducible.
3. If ${q}^{2}$ is a primitive third root of unity, then ${\left({M}^{\prime }\right)}_{{s}_{2}{s}_{1}{s}_{2}t}$ is a submodule of ${M}^{\prime }\text{.}$
4. ${N}_{{s}_{2}t}$ is a 1-dimensional submodule of $N,$ and ${N}^{\prime },$ the image of the weight spaces ${N}_{{s}_{1}{s}_{2}t}$ and ${N}_{{s}_{2}{s}_{1}{s}_{2}t}$ in $N/{N}_{{s}_{2}t},$ is a submodule of $N/{N}_{{s}_{2}t}\text{.}$ The resulting quotient of $N$ is irreducible.
5. If ${q}^{2}$ is not a primitive third root of unity, then ${N}^{\prime }$ is irreducible.
6. If ${q}^{2}$ is a primitive third root of unity, then ${\left({N}^{\prime }\right)}_{{s}_{1}{s}_{2}t}$ is a submodule of ${N}^{\prime }\text{.}$

 Proof. Assume ${q}^{2}\ne -1\text{.}$ Then $Z\left(t\right)=\left\{{\alpha }_{1}\right\}$ and $P\left(t\right)$ contains ${\alpha }_{2},$ ${\alpha }_{1}+{\alpha }_{2},$ $2{\alpha }_{1}+{\alpha }_{2},$ and $3{\alpha }_{1}+{\alpha }_{2}\text{.}$ If ${q}^{2}$ is a primitive third root of unity, then $P\left(t\right)$ also contains $3{\alpha }_{1}+2{\alpha }_{2}\text{.}$ Then $M$ has one 2-dimensional weight space ${M}_{t}^{\text{gen}}$ and four 1-dimensional weight spaces ${M}_{{s}_{2}t},$ ${M}_{{s}_{1}{s}_{2}t},$ ${M}_{{s}_{2}{s}_{1}{s}_{2}t},$ and ${M}_{{s}_{1}{s}_{2}{s}_{1}{s}_{2}t}\text{.}$ For $w\in \left\{{s}_{2},{s}_{1}{s}_{2},{s}_{2}{s}_{1}{s}_{2},{s}_{1}{s}_{2}{s}_{1}{s}_{2}\right\},$ let ${m}_{wt}$ be a non-zero vector in ${M}_{wt}\text{.}$ By a calculation as in (1.3), $mwt=TwT1 vt+∑w′ for $w\in \left\{{s}_{2},{s}_{1}{s}_{2},{s}_{2}{s}_{1}{s}_{2},{s}_{1}{s}_{2}{s}_{1}{s}_{2}\right\},$ where ${a}_{w,{w}^{\prime }}\in ℂ\text{.}$ Then if ${s}_{i}w>w,$ $τimwt≠0$ for $w\in \left\{{s}_{2},{s}_{1}{s}_{2},{s}_{2}{s}_{1}{s}_{2},{s}_{1}{s}_{2}{s}_{1}{s}_{2}\right\},$ since the term ${T}_{i}{T}_{w}{T}_{1}$ cannot be canceled by any other term in ${\tau }_{i}{m}_{wt}\text{.}$ Thus ${\tau }_{1}:\phantom{\rule{0.2em}{0ex}}{M}_{{s}_{1}{s}_{2}{s}_{1}{s}_{2}t}\to {M}_{{s}_{2}{s}_{1}{s}_{2}t}$ is the zero map since, by Theorem 1.4, ${\tau }_{1}^{2}:\phantom{\rule{0.2em}{0ex}}{M}_{{s}_{2}{s}_{1}{s}_{2}t}\to {M}_{{s}_{2}{s}_{1}{s}_{2}t}$ is the zero map. Hence ${M}_{{s}_{1}{s}_{2}{s}_{1}{s}_{2}t}$ is a submodule of $M\text{.}$ Let ${M}_{1}=M/{M}_{{s}_{1}{s}_{2}{s}_{1}{s}_{2}t}\text{.}$ Similarly, ${\tau }_{1}:\phantom{\rule{0.2em}{0ex}}{M}_{{s}_{1}{s}_{2}t}\to {M}_{{s}_{2}t}$ must be the zero map since, by Theorem 1.4, ${\tau }_{1}^{2}:\phantom{\rule{0.2em}{0ex}}{M}_{{s}_{2}t}\to {M}_{{s}_{2}t}$ is the zero map. Then ${M}^{\prime },$ the subspace spanned by $\stackrel{‾}{{m}_{{s}_{1}{s}_{2}t}}$ and $\stackrel{‾}{{m}_{{s}_{2}{s}_{1}{s}_{2}t}}$ in ${M}_{1},$ is a submodule of ${M}_{1}\text{.}$ Lemma 1.11 shows that ${M}_{2}={M}_{1}/{M}^{\prime }$ is irreducible. (b) If ${q}^{2}$ is not a primitive third root of unity, ${\tau }_{2}^{2}:\phantom{\rule{0.2em}{0ex}}{\left({M}^{\prime }\right)}_{{s}_{1}{s}_{2}t}\to {\left({M}^{\prime }\right)}_{{s}_{1}{s}_{2}t}$ is invertible, so that ${M}^{\prime }$ is irreducible. (c) If ${q}^{2}$ is a primitive third root of unity, then ${\tau }_{2}:\phantom{\rule{0.2em}{0ex}}{\left({M}_{1}^{\prime }\right)}_{{s}_{2}{s}_{1}{s}_{2}t}\to {\left({M}_{1}^{\prime }\right)}_{{s}_{1}{s}_{2}t}$ is the zero map and ${\left({M}_{1}^{\prime }\right)}_{{s}_{2}{s}_{1}{s}_{2}t}$ is a 1-dimensional submodule of ${M}_{1}^{\prime },$ and ${M}_{1}^{\prime }/{\left({M}_{1}^{\prime }\right)}_{{s}_{2}{s}_{1}{s}_{2}t}$ is 1-dimensional as well. (d)-(f) The same argument used in (a)-(c) applies, with each weight space ${M}_{wt}$ replaced by ${N}_{w{w}_{0}t}\text{.}$ $\square$

However, the composition factors of $M$ and $N$ are not distinct. If ${q}^{2}$ is not a primitive third root of unity, then ${M}^{\prime }$ and ${N}^{\prime }$ are irreducible 2-dimensional modules with the same weight space structure. Then Theorem 1.10 shows that ${M}^{\prime }\cong {N}^{\prime }\text{.}$ If ${q}^{2}$ is a primitive third root of unity, then note that two 1-dimensional modules $M$ and ${M}^{\prime }$ are isomorphic if and only if they have the same weight. Then ${M}_{1}^{\prime }$ and ${N}_{1}^{\prime }$ have the same composition factors. In any case, the 3-dimensional modules are different since their weight space structures are different.

Proposition 2.16. If ${q}^{2}\ne ±1$ then the composition factors of $M$ and $N$ are the only irreducibles with central character ${t}_{1,{q}^{2}}\text{.}$

 Proof. Counting multiplicities of weight spaces in $M\left(t\right)$ and the distinct composition factors of $M$ and $N$ shows that the remaining composition factor(s) of $M\left(t\right)$ must contain an ${s}_{1}{s}_{2}t$ weight space and an ${s}_{2}{s}_{1}{s}_{2}t$ weight space, each of dimension 1. If ${q}^{2}$ is not a primitive third root of unity then Theorem (1.8) shows that there must be one remaining composition factor with an ${s}_{1}{s}_{2}t$ weight space and an ${s}_{2}{s}_{1}{s}_{2}t$ weight space, and Theorem 1.10 shows that it is isomorphic to ${M}_{1}\text{.}$ If ${q}^{2}$ is a primitive third root of unity then ${\tau }_{2}^{2}:\phantom{\rule{0.2em}{0ex}}{M}_{{s}_{1}{s}_{2}t}\to {M}_{{s}_{2}{s}_{1}{s}_{2}}$ is not invertible. Hence there cannot be an irreducible module consisting of an ${s}_{1}{s}_{2}t$ weight space and an ${s}_{2}{s}_{1}{s}_{2}$ weight space, and the remaining composition factors of $M\left(t\right)$ are 1-dimensional. $\square$

If ${q}^{2}=-1,$ then $\text{dim}\phantom{\rule{0.2em}{0ex}}\phantom{\rule{0.2em}{0ex}}{M}_{t}^{\text{gen}}=$ $\text{dim}\phantom{\rule{0.2em}{0ex}}\phantom{\rule{0.2em}{0ex}}{M}_{{s}_{2}t}^{\text{gen}}=$ $\text{dim}\phantom{\rule{0.2em}{0ex}}\phantom{\rule{0.2em}{0ex}}{M}_{{s}_{1}{s}_{2}t}^{\text{gen}}=2$ and $\text{dim}\phantom{\rule{0.2em}{0ex}}\phantom{\rule{0.2em}{0ex}}{N}_{t}^{\text{gen}}=$ $\text{dim}\phantom{\rule{0.2em}{0ex}}\phantom{\rule{0.2em}{0ex}}{N}_{{s}_{2}t}^{\text{gen}}=$ $\text{dim}\phantom{\rule{0.2em}{0ex}}{N}_{{s}_{1}{s}_{2}t}^{\text{gen}}=2\text{.}$

Proposition 2.17. Assume ${q}^{2}=-1$ and $t={t}_{1,{q}^{2}}\text{.}$ Let $M=\stackrel{\sim }{H}{\otimes }_{{\stackrel{\sim }{H}}_{\left\{2\right\}}}ℂ{v}_{t}$ and $N=\stackrel{\sim }{H}{\otimes }_{{\stackrel{\sim }{H}}_{\left\{1\right\}}}ℂ{v}_{{s}_{1}{s}_{2}t\text{.}}$

1. $M$ and $N$ each have 4 composition factors - two 1-dimensional modules and two 2-dimensional modules.
2. The composition factors of $M$ and $N$ are the only irreducible modules with central character $t\text{.}$

 Proof. By Proposition 1.10, there is a 2-dimensional module $P$ with $P={P}_{t}^{\text{gen}}\text{.}$ Let $v\in {P}_{t}$ be non-zero. The map $ℂvt ⟶ P vt ⟼ v$ is a ${\stackrel{\sim }{H}}_{\left\{2\right\}}\text{-module}$ homomorphism. Since $HomH∼(M,P)= HomH∼{2} (ℂvt,P),$ there is a non-zero map from $M$ to $P\text{.}$ Since $P$ is irreducible, this map is surjective and $P$ is a quotient of $M\text{.}$ The kernel of any map from $M$ to $P$ must be $M1= Ms2tgen⊕ Ms1s2tgen,$ which is then a submodule of $M\text{.}$ Then we note that $m={T}_{1}{T}_{2}{T}_{1}{T}_{2}{T}_{1}v-q{T}_{2}{T}_{1}{T}_{2}{T}_{1}v-{T}_{1}{T}_{2}{T}_{1}v+q{T}_{2}{T}_{1}v+{T}_{1}v-qv$ spans a 1-dimensional submodule of ${M}_{1}\text{.}$ Then let ${M}_{2}={M}_{1}/m,$ so that ${T}_{1}{T}_{2}{T}_{1}{T}_{2}{T}_{1}v=q{T}_{2}{T}_{1}{T}_{2}{T}_{1}v+{T}_{1}{T}_{2}{T}_{1}v-q{T}_{2}{T}_{1}v-{T}_{1}v+qv$ in ${M}_{2}\text{.}$ Then by Theorem 1.3, ${M}_{2}$ contains an element ${m}^{\prime }\in {M}_{{s}_{2}t}$ of the form $m′=T2T1vt +as1T1vt+ a1vt.$ Specifically, ${m}^{\prime }={T}_{2}{T}_{1}v-q{T}_{1}v-3v\text{.}$ Then $τ1(m′) = T1m′-qm′ = T1T2T1v-T1 v-qv- ( qT2T1v-q2 T1v-3qv ) = T1T2T1v-q T2T1v-2T1 v+2qv∈ Ms1s2t.$ Also, $T2·τ1(m′) = T2T1T2T1v -qT22T1v-2 T2T1v+2qT2v = T2T1T2T1v- q(q-q-1)T2 T1v-qT1v-2 T2T1v-2v = T2T1T2T1v- qT1v-2v.$ Then ${m}^{\prime },$ ${\tau }_{1}\left({m}^{\prime }\right)$ and ${T}_{2}·{\tau }_{1}\left({m}^{\prime }\right)$ are linearly independent and span ${M}_{2}\text{.}$ However, ${M}_{3}=⟨{\tau }_{1}\left({m}^{\prime }\right),{T}_{2}·{\tau }_{1}\left({m}^{\prime }\right)⟩$ is clearly closed under the action ${T}_{2}\text{.}$ Also, ${\tau }_{1}\left({m}^{\prime }\right)\in {M}_{{s}_{1}{s}_{2}t},$ so that $Xλ·T2τ1 (m′)=T2 Xs2λτ1 (m′)+ (q-q-1) Xλ-Xs2λ 1-X-α2 τ1(m′),$ which again lies in ${M}_{3}\text{.}$ Finally, $T1·τ1 (m′) = T12T2T1v-q T1T2T1v-2 T12v+2qT1v = (q-q-1)T1 T2T1v+T2T1 v-qT1T2T1v -2(q-q-1) T1v-2v+2qT1v = -q-1 ( T1T2T1v-q T2T1v-2T1v+ 2qv ) =-q-1τ1 (m′),$ and $T1·T2τ1 (m1) = T1T2T1T2 T1v-qT12v- 2T1v = qT2T1T2T1 v+T1T2T1v- qT2T1v-T1v +qv-q(q-q-1) T1v-qv-2T1v = qT2T1T2T1 v+T1T2T1v- qT2T1v-T1v = ( qT2T1T2T1 v-qT2T1v+T1 v-2qv ) + ( T1T2T1v-q T2T1v-2T1 v+2qv ) = q(τ1(m1)) +T2τ1(m1) .$ Thus ${M}_{3}$ is a submodule of ${M}_{2}\text{.}$ By Theorem 1.11, ${M}_{3}$ is irreducible, and ${M}_{2}/{M}_{3}$ is a 1-dimensional module which is isomorphic to the 1-dimensional module spanned by ${\tau }_{1}\left(m\right)\text{.}$ An analoguous argument proves the same result for $N\text{.}$ Let $Q$ be the 2-dimensional module with $Q={Q}_{{s}_{1}{s}_{2}t}^{\text{gen}}\text{.}$ Then there is a surjection from $N$ to $Q,$ and the kernel of this map, ${N}_{1},$ consists of the $t$ and ${s}_{2}t$ weight spaces of $N\text{.}$ Then $n={T}_{2}{T}_{1}{T}_{2}{T}_{1}{T}_{2}v-q{T}_{1}{T}_{2}{T}_{1}{T}_{2}v-{T}_{2}{T}_{1}{T}_{2}v+q{T}_{1}{T}_{2}v+{T}_{2}v-qv$ spans a 1-dimensional submodule of ${N}_{1}\text{.}$ Let ${N}_{2}={N}_{1}/n\text{.}$ Then ${N}_{{s}_{2}t}$ contains a non-zero element ${n}^{\prime },$ and ${n}^{\prime },$ ${\tau }_{2}\left({n}^{1}\right),$ and ${T}_{1}{\tau }_{2}\left({n}^{\prime }\right)$ are linearly independent and span ${N}_{2}\text{.}$ But ${\tau }_{2}\left({n}^{\prime }\right)$ and ${T}_{1}{\tau }_{2}\left({n}^{\prime }\right)$ span a submodule of ${N}_{2},$ which is irreducible by Theorem 1.11. (b) Let $ℂ{v}_{{s}_{2}t}$ be the one-dimensional ${\stackrel{\sim }{H}}_{\left\{1\right\}}\text{-module}$ with weight ${s}_{2}t,$ and define $L=\stackrel{\sim }{H}{\otimes }_{{\stackrel{\sim }{H}}_{\left\{1\right\}}}ℂ{v}_{{s}_{2}t}\text{.}$ We claim that the composition factors of $L$ are the same as those of $M\text{.}$ First, note that the one-dimensional $\stackrel{\sim }{H}\text{-module}$ ${L}_{q,q}$ restricted to ${\stackrel{\sim }{H}}_{\left\{1\right\}}$ is $ℂ{v}_{{s}_{2}t}\text{.}$ Then there is a ${\stackrel{\sim }{H}}_{\left\{1\right\}}\text{-module}$ map from $ℂ{v}_{{s}_{2}t}$ to ${L}_{q,q},$ and thus there is a map from $L$ to ${L}_{q,q}\text{.}$ Let ${L}_{1}$ be the kernel of this map. Then ${L}_{1}$ has a 1-dimensional ${s}_{2}t$ weight space, and 2-dimensional generalized $t$ and ${s}_{1}{s}_{2}t$ weight spaces. Also, ${L}_{1}$ contains $l={\tau }_{2}\left({v}_{{s}_{2}t}\right)={T}_{2}{v}_{{s}_{2}t}-q{v}_{{s}_{2}t},$ an element of the $t$ weight space of ${L}_{1}\text{.}$ Then we note that ${T}_{2}·\left({T}_{2}{v}_{{s}_{2}t}-q{v}_{{s}_{2}t}\right)=\left(q-{q}^{-1}\right){T}_{2}{v}_{{s}_{2}t}+{v}_{{s}_{2}t}-q{T}_{2}{v}_{{s}_{2}t}=q\left({T}_{2}{v}_{{s}_{2}t}-q{v}_{{s}_{2}t}\right),$ so that $l$ spans a 1-dimensional ${\stackrel{\sim }{H}}_{\left\{2\right\}}\text{-submodule}$ of ${L}_{1},$ with weight $t\text{.}$ Thus, there is a ${\stackrel{\sim }{H}}_{2}$ map from $ℂ{v}_{t}$ to ${L}_{1},$ and thus a $\stackrel{\sim }{H}$ map from $M$ to $L\text{.}$ This maps is surjective since $l,{T}_{2}l,{T}_{1}{T}_{2}l,{T}_{2}{T}_{1}{T}_{2}l,$ and ${T}_{1}{T}_{2}{T}_{1}{T}_{2}l$ are linearly independent and span ${L}_{1}\text{.}$ Then $L$ is a quotient of $M$ and its composition factors are composition factors of $M\text{.}$ Now, Let $P$ be any irreducible $\stackrel{\sim }{H}\text{-module}$ with central character ${t}_{1,{q}^{2}}\text{.}$ If $P$ is not a composition factor of $M$ or $N,$ then $P$ must be in the kernel of the (surjective) map from $M\left(t\right)$ to $M\text{.}$ Hence $P$ is at most 6-dimensional, and each of its generalized weight spaces is at most 2-dimensional. If $P={P}_{t}^{\text{gen}},$ then $P$ is 2-dimensional and must be the module described in Theorem 1.10. Otherwise, we note that ${P}_{{s}_{2}t}^{\text{gen}}\oplus {P}_{{s}_{1}{s}_{2}t}^{\text{gen}}$ is a ${\stackrel{\sim }{H}}_{\left\{1\right\}}\text{-submodule}$ of $P,$ since the action of ${\tau }_{1}$ fixes this subspace of $P\text{.}$ Thus ${P}_{{s}_{2}t}^{\text{gen}}\oplus {P}_{{s}_{1}{s}_{2}t}^{\text{gen}}$ contains an irreducible ${\stackrel{\sim }{H}}_{\left\{1\right\}}\text{-submodule.}$ This subspace must be either ${P}_{{s}_{1}{s}_{2}t}^{\text{gen}}$ or a 1-dimensional module with weight ${s}_{2}t\text{.}$ Hence $P$ is a quotient of either $L$ or $M$ and is isomorphic to a composition factor of $M\text{.}$ $\square$

Case 3b: ${t}_{1,±q}$

Let ${t}^{\prime }\in T$ and assume ${\alpha }_{1}\in Z\left({t}^{\prime }\right)$ but ${\alpha }_{2}\notin P\left({t}^{\prime }\right),$ so that none of ${\alpha }_{1}+{\alpha }_{2},$ $2{\alpha }_{1}+{\alpha }_{2},$ or $3{\alpha }_{1}+{\alpha }_{2}$ are in $P\left({t}^{\prime }\right)\text{.}$ Since $P\left({t}^{\prime }\right)\ne \varnothing ,$ $3{\alpha }_{1}+2{\alpha }_{2}\in P\left({t}^{\prime }\right)$ so that ${t}^{\prime }\left({X}^{2{\alpha }_{2}}\right)={q}^{±2}$ and ${t}^{\prime }\left({X}^{{\alpha }_{2}}\right)=±{q}^{±1}\text{.}$ By applying ${w}_{0}$ if necessary, we may assume ${t}^{\prime }\left({X}^{{\alpha }_{2}}\right)=±q\text{.}$ If $q$ is a primitive third root of unity then $q={q}^{-2}$ and ${q}^{-1}={q}^{2},$ so that ${t}_{1,q}$ is the weight analyzed in case 3a. If $q$ is a primitive sixth root of unity then $-q={q}^{-2}$ and $-{q}^{-1}={q}^{2}$ so that ${t}_{1,-q}$ is the weight analyzed in Case 3a. Thus these cases are excluded from the following analysis, which then describes ${t}_{1,±q}$ if ${q}^{2}$ is not a primitive third root of unity, ${t}_{1,-q}$ if ${q}^{3}=1,$ and ${t}_{1,q}$ if ${q}^{3}=-1\text{.}$

If ${q}^{2}=1,$ then $Z\left({t}^{\prime }\right)=P\left({t}^{\prime }\right)=\left\{{\alpha }_{1},3{\alpha }_{1}+2{\alpha }_{2}\right\},$ and the irreducibles with central character $t$ can be constructed using the results of (1.2.9). Specifically, there are four 3-dimensional modules with central character ${t}^{\prime }\text{.}$

Otherwise, ${s}_{1}{s}_{2}{t}^{\prime }\left({X}^{{\alpha }_{1}}\right)={t}^{\prime }\left({X}^{-{\alpha }_{1}-{\alpha }_{2}}\right)=±{q}^{\mp 1}$ and ${s}_{1}{s}_{2}{t}^{\prime }\left({X}^{{\alpha }_{2}}\right)={t}^{\prime }\left({X}^{3{\alpha }_{1}+2{\alpha }_{2}}\right)={q}^{±2}\text{.}$ Then by theorem 1.5, $M\left({t}^{\prime }\right)$ and $M\left(t\right)$ have the same composition factors, where $t={s}_{1}{s}_{2}{t}^{\prime }\text{.}$ Also, $Z\left(t\right)=\left\{2{\alpha }_{1}+{\alpha }_{2}\right\}$ and $P\left(t\right)=\left\{{\alpha }_{2}\right\}\text{.}$ Let ${w}_{0}={s}_{1}{s}_{2}{s}_{1}{s}_{2}{s}_{1}{s}_{2}$ and define

$H∼{2}=ℂ-span { T2Xλ,Xλ ∣λ∈P } ,$

the subalgebra of $\stackrel{\sim }{H}$ generated by ${T}_{2}$ and $ℂ\left[X\right]\text{.}$ Let $ℂ{v}_{t}$ and $ℂ{v}_{{w}_{0}t}$ 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

$T2vt=qvt, Xλvt=t (Xλ)vt, T2vw0t=- q-1vw0t, and Xλvw0t= w0t(Xλ) vw0t.$

Then define

$M=ℂvt ⊗H∼{2} H∼and N=ℂvw0t ⊗H∼{2} H∼.$

Proposition 2.18. Assume ${q}^{2}\ne 1\text{.}$ Let ${t}^{\prime }={t}_{1,±q},$ and define $M$ and $N$ as above. Assume that it is not true that ${t}^{\prime }\left({X}^{{\alpha }_{2}}\right)={q}^{-2}\text{.}$ Then $M$ and $N$ are irreducible.

 Proof. Let $t={s}_{1}{s}_{2}t\text{.}$ Under the assumptions, $Z\left(t\right)=\left\{2{\alpha }_{1}+{\alpha }_{2}\right\}$ and $P\left(t\right)=\left\{{\alpha }_{2}\right\}\text{.}$ Then $\text{dim}\phantom{\rule{0.2em}{0ex}}{M}_{t}^{\text{gen}}=$ $\text{dim}\phantom{\rule{0.2em}{0ex}}{M}_{{s}_{1}t}^{\text{gen}}=$ $\text{dim}\phantom{\rule{0.2em}{0ex}}{M}_{{s}_{2}{s}_{1}t}^{\text{gen}}=2\text{.}$ By Lemma 1.11, $M$ has some composition factor ${M}^{\prime }$ with $\text{dim}\phantom{\rule{0.2em}{0ex}}{M}_{{s}_{2}{s}_{1}t}^{\text{gen}}=2,$ and by Theorem 1.9, ${M}^{\prime }=M$ and $M$ is irreducible. Similarly, Lemma 1.11 and Theorem 1.9 show that $N$ is irreducible. $\square$

Note that in this theorem, $t\left({X}^{{\alpha }_{2}}\right)={q}^{-2}$ only if $t\left({X}^{{\alpha }_{2}}\right)=q$ and ${q}^{3}=1$ or if $t\left({X}^{{\alpha }_{2}}\right)=-q$ and ${q}^{3}=-1\text{.}$ In these cases, $t$ is in the same orbit as ${t}_{1,{q}^{2}},$ which is described in case 3a. Otherwise, since $M$ and $N$ are each 6-dimensional, they must be the only composition factors of $M\left(t\right)\text{.}$

If $t$ is any weight such that $Z\left(t\right)$ contains ${\alpha }_{1},$ ${\alpha }_{1}+{\alpha }_{2},$ or $2{\alpha }_{1}+{\alpha }_{2},$ there exists $w\in {W}_{0}$ so that ${\alpha }_{1}\in Z\left(wt\right)\text{.}$ Then $t$ is in the orbit of one of the weights in case 3a or 3b. Then for the following cases, assume ${\alpha }_{1},$ ${\alpha }_{1}+{\alpha }_{2},$ $2{\alpha }_{1}+{\alpha }_{2}\notin Z\left(t\right)\text{.}$

Case 3c: ${t}_{{q}^{2},1}$

If ${\alpha }_{2}\in Z\left(t\right)$ and ${\alpha }_{1}\in P\left(t\right),$ then ${\alpha }_{1}+{\alpha }_{2}\in P\left(t\right)$ as well. If ${q}^{2}=-1$ then $t\left({X}^{2{\alpha }_{1}}+{\alpha }_{2}\right)=1,$ so that $t$ is in the orbit of one of the weights considered in cases 3a and 3b. If ${q}^{2}=1,$ then $t={t}_{1,1}$ which has also already been considered.

Let ${w}_{0}={s}_{1}{s}_{2}{s}_{1}{s}_{2}{s}_{1}{s}_{2}$ and define

$H∼{1}=ℂ-span { T1Xλ,Xλ ∣λ∈P } ,$

the subalgebra of $\stackrel{\sim }{H}$ generated by ${T}_{1}$ and $ℂ\left[X\right]\text{.}$ Let $ℂ{v}_{t}$ and $ℂ{v}_{{w}_{0}t}$ be the 1-dimensional ${\stackrel{\sim }{H}}_{\left\{1\right\}}\text{-modules}$ spanned by ${v}_{t}$ and ${v}_{wt},$ respectively, and given by

$T1vt=qvt, Xλvt=t (Xλ)vt, T1vw0t= -q-1vw0t, and Xλvw0t= w0t(Xλ) vw0t.$

Then define

$M=ℂvt ⊗H∼{1} H∼and N=ℂvw0t ⊗H∼{1} H∼.$

Proposition 2.19. If ${q}^{2}$ is a primitive third root of unity, then $M$ and $N$ are irreducible.

 Proof. If ${q}^{2}$ is a primitive third root of unity, then $Z\left(t\right)=\left\{{\alpha }_{2},3{\alpha }_{1}+{\alpha }_{2},3{\alpha }_{1}+2{\alpha }_{2}\right\}$ and $P\left(t\right)=\left\{{\alpha }_{1},{\alpha }_{1}+{\alpha }_{2},2{\alpha }_{1}+{\alpha }_{2}\right\}\text{.}$ Then $\text{dim}\phantom{\rule{0.2em}{0ex}}{M}_{t}^{\text{gen}}=4$ and $\text{dim}\phantom{\rule{0.2em}{0ex}}{M}_{{s}_{1}t}^{\text{gen}}=2\text{.}$ By Lemma 1.11, if ${M}^{\prime }\subseteq M$ is a submodule of $M,$ then $\text{dim}\phantom{\rule{0.2em}{0ex}}{\left({M}^{\prime }\right)}_{t}^{\text{gen}}\ge 2$ and $\text{dim}\phantom{\rule{0.2em}{0ex}}{\left({M}^{\prime }\right)}_{{s}_{i}t}^{\text{gen}}\ge 2\text{.}$ Then $\text{dim}\phantom{\rule{0.2em}{0ex}}{\left(M/{M}^{\prime }\right)}_{t}^{\text{gen}}\le 2,$ but $\text{dim}\phantom{\rule{0.2em}{0ex}}{\left(M/{M}^{\prime }\right)}_{{s}_{1}t}^{\text{gen}}0,$ so that Lemma 1.11 implies that ${\left(M/{M}^{\prime }\right)}_{t}=0\text{.}$ Thus ${M}^{\prime }=M$ and $M$ is irreducible. Lemma 1.11 similarly implies that $N$ is irreducible. $\square$

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

Proposition 2.20. Assume ${q}^{2}\ne ±1$

1. If ${q}^{2}$ is a primitive fourth root of unity then ${M}_{{s}_{2}{s}_{1}{s}_{2}{s}_{1}t}$ is a 1-dimensional submodule of $M,$ and ${M}^{\prime },$ the image of the weight spaces ${M}_{{s}_{1}{s}_{2}{s}_{1}t}$ and ${M}_{{s}_{2}{s}_{1}t}$ in $M/{M}_{{s}_{1}{s}_{2}{s}_{1}t},$ is an irreducible submodule of $M/{M}_{{s}_{1}{s}_{2}{s}_{1}t}\text{.}$ The resulting quotient of $M$ is irreducible.
2. If ${q}^{2}$ is a primitive fourth root of unity then ${N}_{{s}_{1}t}$ is a 1-dimensional submodule of $N,$ and ${N}^{\prime },$ the image of the weight spaces ${N}_{{s}_{2}{s}_{1}t}$ and ${N}_{{s}_{1}{s}_{2}{s}_{1}t}$ in $N/{N}_{{s}_{1}t},$ is an irreducible submodule of $N/{N}_{{s}_{1}t}\text{.}$ The resulting quotient of $N$ is irreducible.
3. The composition factors of $M$ and $N$ are the only composition factors of $M\left(t\right)\text{.}$
4. If ${q}^{2}$ is not a primitive third or fourth root of unity then $M$ and $N$ are irreducible, and are the only irreducible modules with central character $t\text{.}$

 Proof. If ${q}^{2}$ is not $±1$ or a primitive third root of unity, $Z\left(t\right)=\left\{{\alpha }_{2}\right\},$ so that $M$ has one 2-dimensional weight space ${M}_{t}^{\text{gen}}$ and four 1-dimensional weight spaces ${M}_{{s}_{1}t},$ ${M}_{{s}_{2}{s}_{1}t},$ ${M}_{{s}_{1}{s}_{2}{s}_{1}t},$ and ${M}_{{s}_{2}{s}_{1}{s}_{2}{s}_{1}t}\text{.}$ (a) If ${q}^{2}$ is a primitive fourth root of unity, then $P\left(t\right)=\left\{{\alpha }_{1},{\alpha }_{1}+{\alpha }_{2},3{\alpha }_{1}+{\alpha }_{2},3{\alpha }_{1}+2{\alpha }_{2}\right\}\text{.}$ For $w\in \left\{{s}_{1},{s}_{2}{s}_{1},{s}_{1}{s}_{2}{s}_{1},{s}_{2}{s}_{1}{s}_{2}{s}_{1}\right\},$ let ${m}_{wt}$ be a non-zero vector in ${M}_{wt}\text{.}$ By a calculation as in (1.3), $mwt=TwT2 vt+∑w′ for $w\in \left\{{s}_{1},{s}_{2}{s}_{1},{s}_{1}{s}_{2}{s}_{1},{s}_{2}{s}_{1}{s}_{2}{s}_{1}\right\},$ where ${a}_{w,{w}^{\prime }}\in ℂ\text{.}$ Then if ${s}_{i}w>w,$ $τimwt≠0$ for $w\in \left\{{s}_{1},{s}_{2}{s}_{1},{s}_{1}{s}_{2}{s}_{1},{s}_{2}{s}_{1}{s}_{2}{s}_{1}\right\},$ since the term ${T}_{i}{T}_{w}{T}_{2}$ cannot be canceled by any other term in ${\tau }_{i}{m}_{wt}\text{.}$ Thus ${\tau }_{2}:\phantom{\rule{0.2em}{0ex}}{M}_{{s}_{2}{s}_{1}{s}_{2}{s}_{1}t}\to {M}_{{s}_{1}{s}_{2}{s}_{1}t}$ is the zero map since, by Theorem 1.4, ${\tau }_{2}^{2}:\phantom{\rule{0.2em}{0ex}}{M}_{{s}_{1}{s}_{2}{s}_{1}t}\to {M}_{{s}_{1}{s}_{2}{s}_{1}t}$ is the zero map. Hence ${M}_{{s}_{2}{s}_{1}{s}_{2}{s}_{1}t}$ is a submodule of $M\text{.}$ Let ${M}_{1}=M/{M}_{{s}_{2}{s}_{1}{s}_{2}{s}_{1}t}\text{.}$ Similarly, ${\tau }_{2}:\phantom{\rule{0.2em}{0ex}}{M}_{{s}_{2}{s}_{1}t}\to {M}_{{s}_{1}t}$ must be the zero map since, by Theorem 1.4, ${\tau }_{2}^{2}:\phantom{\rule{0.2em}{0ex}}{M}_{{s}_{1}t}\to {M}_{{s}_{1}t}$ is the zero map. Then ${M}_{1}^{\prime },$ the subspace spanned by $\stackrel{‾}{{m}_{{s}_{2}{s}_{1}t}}$ and $\stackrel{‾}{{m}_{{s}_{1}{s}_{2}{s}_{1}t}}$ in ${M}_{1},$ is a submodule of ${M}_{1}\text{.}$ Since ${\tau }_{1}^{2}:\phantom{\rule{0.2em}{0ex}}{\left({M}_{1}^{\prime }\right)}_{{s}_{2}{s}_{1}t}\to {\left({M}_{1}^{\prime }\right)}_{{s}_{2}{s}_{1}t}$ is invertible, ${M}_{1}^{\prime }$ is irreducible, and Lemma 1.11 shows that ${M}_{2}={M}_{1}/{M}_{1}^{\prime }$ is irreducible. (b) Replacing $t$ by ${w}_{0}t$ in this argument shows that $N$ also has three composition factors. The weight space ${N}_{{s}_{1}t}$ is a submodule of $N,$ and ${N}_{1}=N/{N}_{{s}_{1}t}$ has an irreducible 2-dimensional submodule ${N}_{1}^{\prime },$ consisting of the image of ${N}_{{s}_{2}{s}_{1}t}$ and ${N}_{{s}_{1}{s}_{2}{s}_{1}t}$ in ${N}_{1}\text{.}$ Lemma 1.11 shows that ${N}_{1}/{N}_{1}^{\prime }$ is irreducible. (c) The composition factors of $M$ and $N$ are not distinct, since ${M}_{1}^{\prime }$ and ${N}_{1}^{\prime }$ are irreducible 2-dimensional modules with the same weight spaces, and Theorem 1.10 shows that ${M}_{1}^{\prime }={N}_{1}^{\prime }\text{.}$ The 1-dimensional composition factors of $M$ and $N$ are not isomorphic since they have different weights, and the 3-dimensional modules are different since their weight space structures are different. Counting multiplicities of weight spaces in $M,$ $N,$ and $M\left(t\right)$ shows that the remaining composition factor(s) of $M\left(t\right)$ must contain an ${s}_{2}{s}_{1}t$ weight space and an ${s}_{1}{s}_{2}{s}_{1}t$ weight space, each of dimension 1. But Theorem (1.8) shows that there must be one remaining composition factor, and Theorem 1.10 shows that it is isomorphic to ${M}_{1}\text{.}$ Then the composition factors of $M$ and $N$ are all the composition factors of $M\left(t\right)\text{.}$ (d) If ${q}^{2}$ is not a primitive third or fourth root of unity then $P\left(t\right)=\left\{{\alpha }_{1},{\alpha }_{1}+{\alpha }_{2}\right\}\text{.}$ Then if ${M}^{\prime }$ is the composition factor of $M$ with ${\left({M}^{\prime }\right)}_{t}^{\text{gen}}\ne 0,$ then by Lemma 1.11, $\text{dim}\phantom{\rule{0.2em}{0ex}}{\left({M}^{\prime }\right)}_{t}^{\text{gen}}=2$ and $\text{dim}\phantom{\rule{0.2em}{0ex}}{\left({M}^{\prime }\right)}_{{s}_{1}t}=1\text{.}$ But then Theorem 1.9 implies that ${M}^{\prime }$ must have a nonzero weight space for the weights ${s}_{2}{s}_{1}t,$ ${s}_{1}{s}_{2}{s}_{1}t,$ and ${s}_{2}{s}_{1}{s}_{2}{s}_{1}t,$ so that ${M}^{\prime }=M$ and $M$ is irreducible. Similarly, Lemma 1.11 and Theorem 1.9 show that $N$ is irreducible. Since $M$ and $N$ are not isomorphic and are each 6-dimensional, they are the only composition factors of $M\left(t\right)\text{.}$ $\square$

Case 3d: ${t}_{±q,1}$

If ${\alpha }_{2}\in Z\left({t}^{\prime }\right)$ and $2{\alpha }_{1}+{\alpha }_{2}\in P\left({t}^{\prime }\right),$ then ${t}^{\prime }\left({X}^{2{\alpha }_{1}}\right)={q}^{±2}$ and ${t}^{\prime }\left({X}^{{\alpha }_{1}}\right)=±{q}^{±1}\text{.}$ By replacing ${t}^{\prime }$ by ${w}_{0}{t}^{\prime }$ if necessary, it suffices to assume that ${t}^{\prime }\left({X}^{{\alpha }_{1}}\right)=±q\text{.}$ If ${t}^{\prime }\left({X}^{{\alpha }_{1}}\right)={q}^{-2},$ then ${t}^{\prime }$ was analyzed in case 3c. This occurs when ${q}^{3}=1$ and ${t}^{\prime }={t}_{q,1},$ or when ${q}^{3}=-1$ and ${t}^{\prime }={t}_{-q,1}\text{.}$ Thus the following analysis will apply to ${t}_{q,1}$ except if ${q}^{3}=1,$ and ${t}_{-q,1}$ except for when ${q}^{3}=-1\text{.}$

Also, if ${t}^{\prime }\left({X}^{3{\alpha }_{1}}\right)={q}^{-2},$ then $P\left(t\right)$ also contains $3{\alpha }_{1}+{\alpha }_{2}$ and $3{\alpha }_{1}+2{\alpha }_{2}\text{.}$ This occurs when ${q}^{5}=1$ and ${t}^{\prime }\left({X}^{{\alpha }_{1}}\right)=q$ or when ${q}^{5}=-1$ and ${t}^{\prime }\left({X}^{{\alpha }_{1}}\right)=-q\text{.}$ Thus these cases will be treated separately below.

Let ${w}_{0}={s}_{1}{s}_{2}{s}_{1}{s}_{2}{s}_{1}{s}_{2}$ and define

$H∼{1}=ℂ-span { T1Xλ,Xλ ∣λ∈P } ,$

the subalgebra of $\stackrel{\sim }{H}$ generated by ${T}_{1}$ and $ℂ\left[X\right]\text{.}$ Define $t={s}_{2}{s}_{1}{t}^{\prime }$ so that $t\left({X}^{{\alpha }_{1}}\right)={q}^{2}$ and $t\left({X}^{{\alpha }_{2}}\right)=±{q}^{-3}\text{.}$ Let $ℂ{v}_{t}$ and $ℂ{v}_{{w}_{0}t}$ be the 1-dimensional ${\stackrel{\sim }{H}}_{\left\{1\right\}}\text{-modules}$ spanned by ${v}_{t}$ and ${v}_{wt},$ respectively, and given by

$T1vt=qvt, Xλvt=t (Xλ)vt, T1vw0t= -q-1vw0t, and Xλvw0t= w0t(Xλ) vw0t.$

Then define

$M=ℂvt ⊗H∼{1} H∼and N=ℂvw0t ⊗H∼{1} H∼.$

Proposition 2.21. Let ${t}^{\prime }={t}_{±q,1}\text{.}$ Assume that ${t}^{\prime }\left({X}^{{\alpha }_{1}}\right)\ne {q}^{-2}$ and ${t}^{\prime }\left({X}^{3{\alpha }_{1}}\right)\ne {q}^{-2}\text{.}$ Then $M$ and $N$ are irreducible.

 Proof. Let $t={s}_{2}{s}_{1}{t}^{\prime }\text{.}$ Under the assumptions, $Z\left(t\right)=\left\{3{\alpha }_{1}+2{\alpha }_{2}\right\}$ and $P\left(t\right)=\left\{{\alpha }_{1}\right\}\text{.}$ Then $\text{dim}\phantom{\rule{0.2em}{0ex}}{M}_{t}^{\text{gen}}=\text{dim}\phantom{\rule{0.2em}{0ex}}{M}_{{s}_{2}t}^{\text{gen}}=\text{dim}\phantom{\rule{0.2em}{0ex}}{M}_{{s}_{1}{s}_{2}t}^{\text{gen}}=2\text{.}$ If ${M}^{\prime }$ is the composition factor of $M$ with ${\left({M}^{\prime }\right)}_{t}^{\text{gen}}\ne 0,$ then by Lemma 1.11, $\text{dim}\phantom{\rule{0.2em}{0ex}}{\left({M}^{\prime }\right)}_{t}^{\text{gen}}=2\text{.}$ But then Theorem 1.9 implies that $\text{dim}\phantom{\rule{0.2em}{0ex}}{\left({M}^{\prime }\right)}_{t}^{\text{gen}}=\text{dim}\phantom{\rule{0.2em}{0ex}}{\left({M}^{\prime }\right)}_{{s}_{2}t}^{\text{gen}}=\text{dim}\phantom{\rule{0.2em}{0ex}}{\left({M}^{\prime }\right)}_{{s}_{1}{s}_{2}t}^{\text{gen}}=2,$ so that ${M}^{\prime }=M$ and $M$ is irreducible. Similarly, Lemma 1.11 and Theorem 1.9 show that $N$ is irreducible. $\square$

Note that ${t}^{\prime }\left({X}^{{\alpha }_{1}}\right)={q}^{-2}$ exactly if ${q}^{3}=1$ or $-1$ and ${t}^{\prime }\left({X}^{{\alpha }_{1}}\right)=q$ or $-q,$ respectively. Also, ${t}^{\prime }\left({X}^{{\alpha }_{1}}\right)={q}^{-4}$ exactly if ${q}^{5}=1$ or $-1$ and ${t}^{\prime }\left({X}^{{\alpha }_{1}}\right)=q$ or $-q,$ respectively. In these cases $t$ falls into another case as noted above. Otherwise, since $M$ and $N$ are not isomorphic and are each 6-dimensional, they are the only composition factors of $M\left(t\right)\text{.}$

Proposition 2.22. If $t\left({X}^{{\alpha }_{2}}\right)={q}^{-3}$ and $q$ is a primitive fifth root of unity, or if $t\left({X}^{{\alpha }_{2}}\right)=-{q}^{-3}$ and $q$ is a primitive tenth root of unity, then

1. $M$ has a 5-dimensional irreducible submodule ${M}^{\prime }$ and
2. $N$ has a 5-dimensional irreducible submodule ${N}^{\prime }\text{.}$

 Proof. Given these assumptions, $Z\left(t\right)=\left\{3{\alpha }_{1}+2{\alpha }_{2}\right\}$ and $P\left(t\right)=\left\{{\alpha }_{1},{\alpha }_{2},3{\alpha }_{1}+{\alpha }_{2}\right\}\text{.}$ Then $\text{dim}\phantom{\rule{0.2em}{0ex}}{M}_{t}^{\text{gen}}=\text{dim}\phantom{\rule{0.2em}{0ex}}{M}_{{s}_{2}t}^{\text{gen}}=\text{dim}\phantom{\rule{0.2em}{0ex}}{M}_{{s}_{1}{s}_{2}t}^{\text{gen}}=2\text{.}$ Let ${L}_{q,q}=ℂv$ be the 1-dimensional $\stackrel{\sim }{H}\text{-module}$ given by $Tiv=qv, Xαi=q2v, fori=1,2.$ Since $HomH∼ ( H∼ ⊗H∼{1} ℂvt, Lq,q ) =HomH∼{1} ( ℂvt, Lq,q ∣H∼{1} )$ and $ϕ: ℂvt ⟶ Lq,q vt ⟼ v$ is a map of ${\stackrel{\sim }{H}}_{\left\{1\right\}}\text{-modules,}$ there is a nonzero map $\theta :\phantom{\rule{0.2em}{0ex}}M\to {L}_{q,q}\text{.}$ Then let ${M}_{1}$ be the kernel of $\theta ,$ which is 5-dimensional. If ${M}^{\prime }$ is the composition factor of ${M}_{1}$ with ${\left({M}^{\prime }\right)}_{t}^{\text{gen}}\ne 0,$ then by Lemma 1.11, $\text{dim}\phantom{\rule{0.2em}{0ex}}{\left({M}^{\prime }\right)}_{t}^{\text{gen}}=2\text{.}$ But then Theorem 1.9 implies that $\text{dim}\phantom{\rule{0.2em}{0ex}}{\left({M}^{\prime }\right)}_{t}^{\text{gen}}=\text{dim}\phantom{\rule{0.2em}{0ex}}{\left({M}^{\prime }\right)}_{{s}_{2}t}^{\text{gen}}=2,$ and Lemma 1.11 shows that ${M}_{{s}_{1}{s}_{2}t}^{\text{gen}}\ne 0\text{.}$ Thus ${M}^{\prime }={M}_{1}$ is irreducible. Similarly, there is a map $\rho :\phantom{\rule{0.2em}{0ex}}N\to {L}_{{q}^{-1},{q}^{-1}},$ where ${L}_{{q}^{-1},{q}^{-1}}=ℂv$ is given by $Ti=-q-1v, Xαi=q-2v ,fori=1,2.$ Then if ${N}_{1}$ is the kernel of $\rho ,$ Lemma 1.11 and Theorem 1.9 show that a composition factor ${N}^{\prime }$ of ${N}_{1}$ with ${N}_{{w}_{0}t}^{\prime }\ne 0$ is at least 5-dimensional, with $\text{dim}\phantom{\rule{0.2em}{0ex}}{N}_{{w}_{0}t}^{\text{gen}}=\text{dim}\phantom{\rule{0.2em}{0ex}}{N}_{{s}_{2}{w}_{0}t}^{\text{gen}}=2$ and ${N}_{{s}_{1}{s}_{2}{w}_{0}t}^{\text{gen}}\ne 0\text{.}$ Then ${N}^{\prime }={N}_{1}$ is irreducible. $\square$

Case 3e: ${t}_{{q}^{2/3},1}$

If ${\alpha }_{2}\in Z\left(t\right)$ and $3{\alpha }_{1}+{\alpha }_{2}\in P\left(t\right),$ then $3{\alpha }_{1}+2{\alpha }_{2}\in P\left(t\right)$ as well. If $t\left({X}^{3{\alpha }_{1}+{\alpha }_{2}}\right)={q}^{-2},$ then ${w}_{0}t\left({X}^{{\alpha }_{2}}\right)=1$ and ${w}_{0}t\left({X}^{3{\alpha }_{1}+{\alpha }_{2}}\right)={q}^{2},$ so by replacing $t$ with ${w}_{0}t$ if necessary, assume that $t{\left({X}^{{\alpha }_{1}}\right)}^{3}={q}^{2}\text{.}$ If ${\alpha }_{1}\in P\left(t\right),$ then this weight was analyzed in case 3c, and if $2{\alpha }_{1}+{\alpha }_{2}\in P\left(t\right),$ then this weight was analyzed in case 3d.

Then for the weights not already analyzed, we have $Z\left(t\right)=\left\{{\alpha }_{2}\right\}$ and $P\left(t\right)=\left\{3{\alpha }_{1}+{\alpha }_{2},3{\alpha }_{1}+2{\alpha }_{2}\right\}\text{.}$ Let ${w}_{0}={s}_{1}{s}_{2}{s}_{1}{s}_{2}{s}_{1}{s}_{2}$ and define

$H∼{1}=ℂ-span { T1Xλ,Xλ ∣λ∈P } ,$

the subalgebra of $\stackrel{\sim }{H}$ generated by ${T}_{1}$ and $ℂ\left[X\right]\text{.}$ Let $ℂ{v}_{t}$ and $ℂ{v}_{{w}_{0}t}$ be the 1-dimensional ${\stackrel{\sim }{H}}_{\left\{1\right\}}\text{-modules}$ spanned by ${v}_{t}$ and ${v}_{wt},$ respectively, and given by

$T1vt=qvt, Xλvt=t (Xλ)vt, T1vw0t= -q-1vw0t, and Xλvw0t= w0t(Xλ) vw0t.$

Then define

$M=ℂvt ⊗H∼{1} H∼and N=ℂvw0t ⊗H∼{1} H∼.$

Proposition 2.23. Assume $t={t}_{{q}^{2/3},1},$ where ${q}^{2/3}$ is a third root of ${q}^{2}$ not equal to ${q}^{±2}$ or $±{q}^{±1}\text{.}$ Then $M$ and $N$ are irreducible.

 Proof. Under the assumptions, $Z\left(t\right)=\left\{{\alpha }_{2}\right\}$ and $P\left(t\right)=\left\{3{\alpha }_{1}+{\alpha }_{2},3{\alpha }_{1}+2{\alpha }_{2}\right\},$ so that $\text{dim}\phantom{\rule{0.2em}{0ex}}{M}_{t}^{\text{gen}}=\text{dim}\phantom{\rule{0.2em}{0ex}}{M}_{{s}_{1}t}^{\text{gen}}=\text{dim}\phantom{\rule{0.2em}{0ex}}{M}_{{s}_{2}{s}_{1}t}^{\text{gen}}=2\text{.}$ If ${M}^{\prime }$ is the composition factor of $M$ with ${\left({M}^{\prime }\right)}_{t}^{\text{gen}}\ne 0,$ then by Lemma 1.11, $\text{dim}\phantom{\rule{0.2em}{0ex}}{\left({M}^{\prime }\right)}_{t}^{\text{gen}}=2\text{.}$ But then Theorem 1.9 implies that $\text{dim}\phantom{\rule{0.2em}{0ex}}{\left({M}^{\prime }\right)}_{t}^{\text{gen}}=\text{dim}\phantom{\rule{0.2em}{0ex}}{\left({M}^{\prime }\right)}_{{s}_{1}t}^{\text{gen}}=2,$ and Lemma 1.11 shows that ${M}_{{s}_{2}{s}_{1}t}^{\text{gen}}\ne 0\text{.}$ applying Theorem 1.9 again shows that ${\left(M\prime \right)}_{{s}_{1}{s}_{2}{s}_{1}t}\ne 0,$ so that $M\prime =M$ and $M$ is irreducible. Similarly, Lemma 1.11 and Theorem 1.9 show that ${N}^{\prime }$ is irreducible, so that $M$ and $N$ are the only irreducible modules with central character $t\text{.}$ $\square$

### Summary

We summarize the results of the previous theorems, including our choices of representatives for the various central characters, in the following tables. It should be noted that for any value of $q$ with ${q}^{2}$ not a root of unity of order 6 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 6 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 6 or less. When these hyperplanes coincide, the sets $P\left(t\right)$ and $Z\left(t\right)$ change for characters on those hyperplanes.

If ${q}^{10}=1,$ then note that the central characters ${t}_{±q,1}$ are replaced by ${t}_{±{q}^{-4},1}$ in some order.

If ${q}^{8}=1,$ then only the central characters ${t}_{{q}^{2},1},$ ${t}_{{q}^{2},-{q}^{-2}},$ and ${t}_{{q}^{2},{q}^{2}}$ change from the generic case. All three of these characters are now in the same orbit. Also, we assume that for the central character ${t}_{{q}^{2/3},1},$ we choose a cube root of ${q}^{2}$ besides ${q}^{-2}\text{.}$

If ${q}^{6}=1,$ the central characters ${t}_{±q,1}$ and ${t}_{1,±q}$ are replaced by ${t}_{±{q}^{-2},1}$ and ${t}_{1,±{q}^{-2}}$ in some order depending on whether ${q}^{3}$ is 1 or $-1\text{.}$ Then we note that ${t}_{{1}^{1/3},1}={t}_{{q}^{±2},1},$ and ${t}_{{q}^{-2},1}$ is in the same orbit as ${t}_{{q}^{2},1}\text{.}$ Also, ${t}_{1,{q}^{-2}}={w}_{0}{t}_{1,{q}^{2},}$ and ${t}_{{q}^{-2},1}={w}_{0}{t}_{{q}^{2},1}\text{.}$ Finally, ${t}_{{1}^{1/3},{q}^{2}}={t}_{{q}^{±2},{q}^{2}},$ but ${s}_{1}{t}_{{q}^{2},{q}^{2}}={t}_{{q}^{-2},{q}^{2}}={s}_{2}{t}_{1,{q}^{-2}}$ and so both are in the same orbit as ${t}_{1,{q}^{2}}\text{.}$

When ${q}^{2}=-1,$ a number of characters change from the general case. Now, ${t}_{1,-1}={t}_{1,{q}^{2}},$ which is in the same orbit as ${t}_{{q}^{2},-{q}^{-2}},$ ${t}_{{q}^{2},{q}^{2}}$ and ${t}_{{q}^{2},1}\text{.}$ Similarly, ${t}_{{1}^{1/3},{q}^{2}}$ is in the same orbit as ${t}_{{q}^{2/3},1}\text{.}$

When ${q}^{2}=1,$ $Z\left(t\right)=P\left(t\right)$ for all $t\in T\text{.}$

$t Z(t) P(t) Dim. of irreds. t1,1 R+ ∅ 12 t1,-1 {α1,3α1+2α2} ∅ 12 t11/3,1 {α2,3α1+α2,3α1+2α2} ∅ 12 t1,q2 {α1} R+\{α1,3α1+2α2} 1,1,2,3,3 t1,±q {α1} {3α1+2α2} 6,6 t1,z {α1} ∅ 12 tq2,1 {α2} {α1,α1+α2} 6,6 t±q,1 {α2} {2α1+α2} 6,6 tq2/3,1 {α2} {3α1+α2,3α1+2α2} 6,6 tz,1 {α2} ∅ 12 tq2,-q-2 ∅ {α1,3α1+2α2} 3,3,3,3 t11/3,q2 ∅ {α2,3α1+α2} 2,2,4,4 tq2,q2 ∅ {α1,α2} 1,1,5,5 tq2,z ∅ {α1} 6,6 tz,q2 ∅ {α2} 6,6 tz,w ∅ ∅ 12$

Table 13: Table of possible central characters in type ${G}_{2},$ when $q$ is generic.

$t Z(t) P(t) Dim. of irreds. t1,1 R+ ∅ 12 t1,-1 {α1,3α1+2α2} ∅ 12 t11/3,1 {α2,3α1+α2,3α1+2α2} ∅ 12 t1,q2 {α1} R+\{α1,3α1+2α2} 3,3,2,1,1 t1,±q {α1} {3α1+2α2} 6,6 t1,z {α1} ∅ 12 tq2,1 {α2} {α1,α1+α2} 6 t±q,1 {α2} {2α1+α2} 6,6 tq2/3,1 {α2} {3α1+α2,3α1+2α2} 6,6 tz,1 {α2} ∅ 12 tq2,-q-2 ∅ {α1,3α1+2α2} 3,3,3,3 t11/3,q2 ∅ {α2,3α1+α2} 4,4,2,2 tq2,q2 ∅ {α1,α2,3α1+2α2} 3,3,2,2,1,1 tq2,z ∅ {α1} 6,6 tz,q2 ∅ {α2} 6,6 tz,w ∅ ∅ 12$

Table 14: Table of possible central characters in type ${G}_{2},$ when ${q}^{12}=1\text{.}$

$t Z(t) P(t) Dim. of irreds. t1,1 R+ ∅ 12 t1,-1 {α1,3α1+2α2} ∅ 12 t11/3,1 {α2,3α1+α2,3α1+2α2} ∅ 12 t1,q2 {α1} R+\{α1,3α1+2α2} 3,3,2,1,1 t1,±q {α1} {3α1+2α2} 6,6 t1,z {α1} ∅ 12 tq2,1 {α2} {α1,α1+α2} 6 tq-4,1 {α2} {2α1+α2,3α1+α2,3α1+2α2} 5,5,1,1 t-q-4,1 {α2} {2α1+α2} 6,6 tq2/3,1 {α2} {3α1+α2,3α1+2α2} 6,6 tz,1 {α2} ∅ 12 tq2,-q-2 ∅ {α1,3α1+2α2} 3,3,3,3 t11/3,q2 ∅ {α2,3α1+α2} 4,4,2,2 tq2,z ∅ {α1} 6,6 tz,q2 ∅ {α2} 6,6 tz,w ∅ ∅ 12$

Table 15: Table of possible central characters in type ${G}_{2},$ when ${q}^{10}=1\text{.}$

$t Z(t) P(t) Dim. of irreds. t1,1 R+ ∅ 12 t1,-1 {α1,3α1+2α2} ∅ 12 t11/3,1 {α2,3α1+α2,3α1+2α2} ∅ 12 t1,q2 {α1} R+\{α1,3α1+2α2} 1,1,2,3,3 t1,±q {α1} {3α1+2α2} 6,6 tq2,1 {α2} {α1,α1+α2,3α1+α2,3α1+2α2} 3,3,2,1,1 t1,z {α1} ∅ 12 t±q,1 {α2} {2α1+α2} 6,6 tq2/3,1 {α2} {3α1+α2,3α1+2α2} 6,6 tz,1 {α2} ∅ 12 t11/3,q2 ∅ {α2,3α1+α2} 2,2,4,4 tq2,z ∅ {α1} 6,6 tz,q2 ∅ {α2} 6,6 tz,w ∅ ∅ 12$

Table 16: Table of possible central characters in type ${G}_{2},$ when ${q}^{8}=1\text{.}$

$t Z(t) P(t) Dim. of irreds. t1,1 R+ ∅ 12 t1,-1 {α1,3α1+2α2} ∅ 12 t11/3,1 {α1,3α1+2α2} ∅ 12 t1,q2 {α1} R+\{α1} 1,1,1,1,3,3 t1,-q-2 {α1} {3α1+2α2} 6,6 t1,z {α1} ∅ 12 tq2,1 {α1,3α1+α2,3α1+2α2} {α1,α1+α2,2α1+α2} 6,6 t-q-2,1 {α2} {2α1+α2} 6,6 tq2/3,1 {α2} {3α1+α2,3α1+2α2} 6,6 tz,1 {α2} ∅ 12 tq2,-q-2 ∅ {α1,3α1+2α2} 3,3,3,3 tq2,z ∅ {α1} 6,6 tz,q2 ∅ {α2} 6,6 tz,w ∅ ∅ 12$

Table 17: Table of possible central characters in type ${G}_{2},$ when ${q}^{2}=1\text{.}$

$t Z(t) P(t) Dim. of irreds. t1,1 R+ ∅ 12 t1,q2 {α1,3α1+2α2} R+\{α1,3α1+2α2} 2,2,1 t11/3,1 {α2,3α1+α2,3α1+2α2} ∅ 12 t1,q {α1} {3α1+2α2} 6,6 t1,z {α1} ∅ 12 tq,1 {α2} {2α1+α2} 6,6 tq2/3,1 {α2} {3α1+α2,3α1+2α2} 4,4,2 tz,1 {α2} ∅ 12 tq2,z ∅ {α1} 6,6 tz,q2 ∅ {α2} 6,6 tz,w ∅ ∅ 12$

Table 18: Table of possible central characters in type ${G}_{2},$ when ${q}^{4}=1\text{.}$

$t Z(t)=P(t) Dim. of irreds. t1,1 R+ 1,1,1,1,2,2 t1,-1 {α1,3α1+2α2} 3,3,3,3 t11/3,1 {α2,3α1+α2,3α1+2α2} 3,6,3 t1,z {α1} 6,6 tz,1 {α2} 6,6 tz,w ∅ 12$

Table 19: Table of possible central characters in type ${G}_{2},$ when $q=-1\text{.}$

## 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.