Type C2

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

Last update: 10 March 2013

Type C2

The type C2 root system is

R= { ±α1,±α2, ±(α1+α2) ,±(2α1+α2) } ,

where α1,α2 =-1 and α2,α1 =-2. 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 α1 and α2, with additional positive roots α1+α2 and 2α1+α2. Then W0 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 typeC2root system

The fundamental weights satisfy

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

Let

P=-span {ω1,ω2}

be the weight lattice of R.

Hα1 Hα2 ω1 ω2 The weight latticeP.

The affine Hecke algebra H is generated as a -algebra by T1,T2, and X={XλλP}, with relations

XλXμ = Xλ+μ,for λ,μP (2.11) T1T2T1T2 = T2T1T2T1 (2.12) Ti2 = (q-q-1)Ti +1,fori=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

[X]= {XλλP},

a subalgebra of H, and let

T=Hom-alg ([X],).

Then since α1=2ω1-ω2 and α2=2ω2-2ω1, W0 acts on X by

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

Then W0 acts on T by

(w·t)(Xλ) =t(Xw-1λ).

Let

Q=-span(R)

be the root lattice of R. Let

[Q]= {XλλQ}

and let

TQ=Hom-alg ([Q],).

Define

tz,w: [Q] by Xα1z Xα2w.

Pictorially, a weight tqx,qyTQ, for x,y, is identified with with the point x units from the hyperplane Hα1 and y units from the hyperplane Hα2. Then

Hα= { tTQ t(Xα)=1 } ,

and we define

Hα±δ= { tTQ t(Xα)= q±2 } . Hα1 Hα2 H2α1+α2 Hα1+α2 Hα1+α2+δ H2α1+α2+δ Hα1+δ Hα2+δ The torusTQ

For all weights tz,wTQ, there are 2 elements tT with tQ=tz,w, determined by

t(Xω1)2= z2wand t(Xω2)= zw.

We denote these two elements as tz,w,1 and tz,w,2. Which particular weight tz,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 tz,w, since the dimension of the modules with central character t depends only on tQ.

One important remark, though, is that if t(Xα1)=-1, then the two weights t with tQ=t-1,w are in the same W0-orbit and represent the same central character. To see this, let tQ=t-1,w. Then t(Xω1)=w1/2, 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(t) depends only on tQ. Thus we begin by examining the W0-orbits in TQ. The structure of the modules with weight t depends virtually exclusively on P(t)= { αR+ t(Xα)= q±2 } and Z(t)= { αR+ t(Xα)=1 } . For a generic weight t, P(t) and Z(t) are empty, so we examine only the non-generic orbits.

Proposition 2.6. If tTQ, and P(t)Z(t), then t is in the W0-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 z1,q±2 } , { tz,1 z±1, q±2, ±q±1 } , { tq2,z z1,q±2, q-4,q-6 } ,or { tz,q2 z±1,q±2, -q-2,q-4, ±q-1 } .

Proof.

Case 1: Z(t)2.
Case 1a: α1,α1+α2Z(t). Then t(Xα1)=1= t(Xα1+α2), so that t(Xα2)=1 and thus t=t1,1.
Case 1b: Exactly one of α1,α1+α2 is in Z(t). Choose an element w with wt(Xα1)=1. Then α2Z(wt) or 2α1+α2Z(wt). Since wt(X2α1+α2) =wt(Xα1)w t(Xα1+α2) =wt(Xα1)2 wt(Xα2), wt(Xα2) =wt(Xα1+α2) =wt(X2α1+α2) =1, so that t=t1,1.
Case 1c: α1,α1+α2Z(t). Then t(Xα2) =1 =t(X2α1+α2), so that t(X2α1)=1, and t(Xα1)=±1. If t(Xα1)=1 then t=t1,1. By assumption, t(Xα1)=-1 and t=t-1,1.

Case 2: Z(t)=1.
Case 2a: If α1 or α1+α2Z(t), then choose a wW0 with wt(Xα1)=1, so that wt=t1,z for some z. Then wt(Xα2) =wt(Xα1+α2) =wt(X2α1+α2). If wt(Xα2)=q±2 then t is in the same orbit as t1,q2. Otherwise, wt=t1,z where z1,q±2 by assumption.
Case 2b: If α1,α1+α2Z(t), choose a wW0 with wt(Xα2)=1. Then t(Xα1)=t(Xα1+α2), so that if t(Xα1)=q±2, then t(X2α1+α2)=q±4 and P(t)={α1,α1+α2}. If t(Xα1)=±q±1, then t(X2α1+α2)=q±2, so that P(t)={2α1+α2}. Otherwise wt=tz,1 with z1,±q±1,q±2 by assumption.

Case 3: Z(t)=, P(t)
Case 3a: α1,α1+α2P(t). Choose wW0 so that wt(Xα1)=q2. If wt(Xα1+α2)=q2 then wt(Xα2)=1, contradicting the assumption on Z(t). If wt(Xα1+α2)=q-2 then wt(X2α1+α2)=1, contradicting the assumption on Z(t). Thus this case is impossible.
Case 3b: P(t) contains exactly one of α1 or α1+α2. Choose wW0 so that wt(Xα1)=q2. Then by the assumption on Z(t), wt(Xα2)q-2 and wt(X2α1+α2)q-2. Thus if α2P(wt) then wt(Xα2)=q2 and t=tq2,q2. If 2α1+α2P(wt), then wt(X2α1+α2)=q-2 and wt(Xα2)=q-6. Then s2s1s2wt(Xα1) =q2 =s2s1s2wt(Xα2), so that t is in the same orbit as tq2,q2. If neither α2 nor 2α1+α2 is in P(wt), then wt=tq2,z for z1,q±2,q-4,q-6.
Case 3c: Neither of α1,α1+α2 is in P(t). Then at least one of α2 and 2α1+α2 is in P(t). Then choose wW0 so that wt(Xα2)=q2. If 2α1+α2P(t), then wt(X2α1)=1 or q-4, so that wt(Xα1)=-1 or -q-2 by assumption. However, s1s2s1 t-q-2,q2= t-1,q2, so both are in the same orbit as t-1,q2. Otherwise, wt=tz,q2 for some z. By assumption on P(t) and Z(t), z±1,q±2,-q-2,q-4, or ±q-1.

Hα1 Hα1 Hα2 Hα1+δ Hα1 Hα1 Hα2 Hα1+δ Case 1:Z(t) 2, so thatt Case 2:Z(t) =1, so thatt lies on at least two hyperplanesHα. lies on exactly one hyperplaneHα. Hα1 Hα1 Hα2 Hα1+δ Hα1 Hα1 Hα2 Hα1+δ Case 3:Z(t) =,α1P(t) , so thatt Case 3:Z(t) =,α1P(t) , so thatt lies onHα1±δ , but not on anyHα. does not lie onHα1±δ or anyHα.

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 ofH -modules with genericq.

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 TQ when q is a root of unity.

If q2 is a primitive fourth root of unity, then tq2,q2= s1s2s1 t-1,q2.

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 overH , withqa primitive eighth root of unity.

If q2 is a primitive third root of unity, tq2,q2= s2s1s2 tq2,1. Also in this case, q and -q are equal to q-2 and -q-2 in some order, depending on whether q3=1 or -1. Then one of tq,1 and t-q,1 is equal to tq-2,1 and is in the same orbit as tq2,1.

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 overH , withq2a primitive third root of unity.

If q2=-1, then t-1,1=tq2,1, and t1,q2 is in the same orbit as tq2,q2=t-1,q2. Also in this case, tz,q2 =z-1 =s2t-z,-1.

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 overH , withq2=-1.

Finally, if q=-1, we have t1,1 =tq2,1 =t1,q2 =tq2,q2 =t-q,1. Also, t-1,1=t-1,q2, while tq2,z=t1,z and tz,q2=tz,1.

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 overH , withq2=1.

Analysis of the Characters

Theorem 2.7 The 1-dimensional representations of H are

Lq,q,1: H T1 q T2 q Xω1 q3 Xω2 q4 Lq,q,-1: H T1 q T2 q Xω1 -q3 Xω2 q4 Lq,-q-1,1: H T1 q T2 -q-1 Xω1 q Xω2 1 Lq,-q-1,-1: H T1 q T2 -q-1 Xω1 -q Xω2 1 L-q-1,q,1: H T1 -q-1 T2 q Xω1 q-1 Xω2 1 L-q-1,q,-1: H T1 -q-1 T2 q Xω1 -q-1 Xω2 1 L-q-1,-q-1,1: H T1 -q-1 T2 -q-1 Xω1 q-3 Xω2 q-4 L-q-1,-q-1,-1: H T1 -q-1 T2 -q-1 Xω1 -q-3 Xω2 q-4

Proof.

A check of the defining relations for H shows that these maps are homomorphisms. Let v be a 1-dimensional H-module with weight t. Then 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= q2vifT1v= qvand X2ω2-ω2v= q-2vifT1 v=-q-1v,

so that

Xα1v= { q2v, ifT1v=qv q-2v, ifT1v=- q-1v,

since α1=2ω1-ω2.

Similarly, by (2.19),

Xα2v= { q2v, ifT2v=qv q-2v, ifT2v=- q-1v.

Since 2ω1=2α1+α2,

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(Xα1) and t(Xα2). These are the eight representations given above.

Remark: If q is a primitive fourth root of unity, then Lq,q,±1 Lq,-q-1,1 L-q-1,q,±1 L-q-1,-q-1,1.

Let tT. The principal series module is

M(t)= Ind [X] H t=H [X] t,

where t is the one-dimensional [X]-module given by

t=span{vt} andXλvt= t(Xλ)vt.

By Theorem 1.6, every irreducible H module with central character t is a composition factor of M(t). A local region at a weight t is

(t,J)= { wW0 R(w)Z(t) =,R(w) P(t)=J } ,

for JP(t), and we identify a local region (t,J) with the union of the chambers w-1C for w(t,J).

Case 1: P(t)=

If P(t)=, then by Kato’s criterion (Theorem 1.8), M(t) is irreducible and is the only irreducible module with central character t. Since P(t)=, there is one local region

t,=W0/Wt,

the set of minimal length coset representatives of Wt cosets in W0, where Wt is the stabilizer of t in W0. If w and siw are both in (t,) then τi: M(t)wtgen M(t)siwtgen is a bijection. The following pictures show (t,) with one dot in the chamber w-1C for each basis element of M(t)wtgen.

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,q21 t1,z,q21 tz,1,q21 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(t)=, but P(t).

If Z(t)= 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 H-module L(t,J) for each JP(t) such that (t,J), and

dimLwt(t,J)= { 1, ifw(t,J), 0, ifw(t,J).

The following pictures show the local regions (t,J) for each t with Z(t)=. There is one dot in the chamber w-1C for each w(t,J), and the dots corresponding to w and siw are connected exactly when w and siw 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,q21 tz,q2,q21 t-1,q2,q41,q81

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,qgeneric tq2,q2,qa 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 t1,q2, tq2,1, and t±q,1. In these cases, rather than analyzing M(t) 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(t).

tQ=tq2,1.

If q2=1, then the results of section 1.2.9 show that H has five irreducible representations - four of them 1-dimensional, and one 2-dimensional. Specifically, an irreducible H module is an irreducible W0-module (via the identification [W0]=H) on which Xλ acts by the constant t(Xλ), and M(t) is isomorphic to the regular representation of [W0] as a W0-module.

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

Let

w= { s1, ifq2=-1, s1s2s1, ifq2-1.

Then let q2,1 and q-2,1 be the 1-dimensional H{1}-modules spanned by vt and vwt, respectively, given by

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

Then

M=HH{1} q2,1and N=HH{1} q-2,1

are 4-dimensional H-modules.

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

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 q2=-1 and M=HH{1}q2,1 and N=HH{1}q-2,1 then

  1. M is irreducible, and
  2. The map ϕ: N M hvwt hv, forhH is a H-module isomorphism, where v=T1T2vt-q T2vt-vtM, and
  3. Any irreducible H-module L with central character t is isomorphic to M.

Proof.

(a) If q is a primitive fourth root of unity, then M has weight spaces Mtgen, and Ms1tgen, 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=T1T2vt- qT2vt-vt. 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(Xω2)=q2=-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 H{1}-submodule of M, given by

T1v=qv,and Xλv=s1t (Xλ)v.

Then the H{1}-module map given by vwtv corresponds to ϕ under the adjunction

HomH ( HH{1} q-2,1,M ) =HomH{1} ( q-2,1, MH{1} ) .

Thus ϕ is a H-module map and since M is irreducible, the map is surjective. Then since M and N have the same dimension, ϕ is an isomorphism.

(c) The weight spaces in the remaining composition factor(s) of M(t) besides M must be t and s1t weight spaces, each of which appear with multiplicity 2. Let L be an irreducible H-module with central character t. Then by Lemma 1.11, either L is two dimensional, consisting of a single generalized weight space Ltgen or Ls1tgen, or else L is 4-dimensional, with dimLtgen =2 =dimLs1tgen. Then, viewing L as a H{1}-module, it must have all 1-dimensional composition factors, and it must have a 1-dimensional submodule, either Lt or Ls1t.

First, assume Lt is a submodule and vPt, so that T1v=qv. Since

HomH ( HH{1} q2,1,L ) =HomH{1} ( q2,1, LH{1} ) ,

and vtv gives a non-zero H{1}-module map from q2,1 to L, there is a non-zero H-module map from M to L. Since M and L are irreducible, the map is an isomorphism. Alternatively, if Ls1t is a H{1}-submodule of L, then choose vLs1t. Then vs1tv gives a non-zero H{1}-module map from q-2,1 to L, and L is isomorphic to N.

C s1C s1s2C s1s2s1C C s1C s1s2C s1s2s1C tq2,1,q2 a primitive third root of unity tq2,1,qgeneric

Proposition 2.9.

  1. If q2 is a primitive third root of unity then Ms2s1t is a submodule of M isomorphic to L-q-1,q,±1 and M/Ms2s1t is irreducible. In addition, Ns1t is a submodule of N isomorphic to Lq,-q-1, and N/Ns1t is irreducible.
  2. If q2 is not ±1 or a primitive third root of unity then M and N are irreducible and nonisomorphic.

Proof.

(a) Assume q2 is a primitive third root of unity. Then the same computation used in (1.3) shows that Mtgen is spanned by vt and T2vt, while Ms1t is spanned by

vs1t=T1T2 vt+c1T2vt +c0vt

for some c0,c1. Then

τ2:Ms1t Ms2s1t

is non-zero, since the T2T1T2vt term in τ2(vs1t) cannot be cancelled. But s2s1t(Xα2)=q2 so that τ22:Ms2tMs2t is the zero map. Hence τ2:Ms2s1tMs1t is the zero map, and Ms1s2t is a submodule of M. By Lemma 1.11, M/Ms1s2t is irreducible. A parallel argument shows that Ns1t is a submodule of N, with N/Ns1t irreducible.

(b) If q41 and q61, then P(t)={α1,α1+α2}. Then Lemma 1.11 shows that the composition factor M of M with (M)t0 has dim(M)tgen2 and (M)s1t0. Then by Theorem 1.9, (M)s2s1t0, so that M=M. Similarly, Lemma 1.11 and Theorem 1.9 show that N is irreducible. Since they have different weight spaces, they are not isomorphic.

tQ=t1,q2

Let 1,q2 and 1,q-2 be the 1-dimensional H{2}-modules spanned by vt and vw0t, respectively, and given by

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

Then

M=HH{2} 1,q2and N=HH{2} 1,q-2

are 4-dimensional H-modules.

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

C s2C s2s1C s2s1s2C M N t1,q2,qgeneric

Proposition 2.10. Assume q2=-1 and let M=HH{2} 1,q2 and N=HH{2} 1,q-2. Then

  1. Ms1s2t is a submodule of M, and the image of Ms2t is a submodule of M/Ms1s2t. The resulting 2-dimensional quotient of M is irreducible. Also, Ns2t is a submodule of N and the image of Ns1s2t in N/Ns2t is a submodule of N/Ns2t. The resulting 2-dimensional quotient of N is irreducible, and
  2. Any composition factor of M(t) is a composition factor of either M or N.

Proof.

(a) If q2=-1, then M has weight spaces Mtgen, which is two dimensional, and Ms2t and Ms1s2t, both of which are 1-dimensional. Similarly, N has weight spaces Ns2s1s2tgen, which is 2-dimensional, along with Ns1s2t and Ns2t, both 1-dimensional. Since there are two 1-dimensional modules with central character t, with weights s2t and s1s2t, it cannot be true that both M and N are irreducible.

In M, the weight space Mtgen is spanned by vt and T1vt, while Ms2t is spanned by T2T1vt-1 T1vt-2vt and Ms1s2t is spanned by T1T2T1vt+ qT2T1vt- T1vt+qvt. Then Ms1s2t is a submodule of M, and the image of Ms2t is a submodule of M/Ms1s2t. By Lemma 1.11, the resulting 2-dimensional quotient is irreducible. Similarly, Ns2t is spanned by T1T2T1vt+q T2T1vt-T1 vt+qvt, and N/Ns2t has a 1-dimensional submodule spanned by the image of T2T1vt-qT1 vt-2vt. The resulting 2-dimensional quotient is irreducible.

(b) There are at least two 2-dimensional modules and two 1-dimensional modules with central character t1,q2. Counting dimensions of weight spaces, the remaining composition factor(s) of M(t) must have weights s2t and s1s2t. 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(t), each of which is 1-dimensional and isomorphic to one of those modules found in M and N above.

Proposition 2.11. Assume q2±1. Then Ms1s2t is a submodule of M and M/Ms1s2t is irreducible. Similarly, Ns2t is a submodule of N and N/Ns2t is irreducible.

Proof.

If q41, then by the same reasoning as in Proposition 2.10, Mtgen is spanned by vt and T1vt. And since Ms2t is spanned by T2T1vt+ fs1tT1 vt+ftvt for some fs1t and ft[q,q-1], τ1:Ms2tMs1s2t is non-zero. But, since s2t(Xα1)=q2, τ1:Ms1s2tMs2t must be the zero map. Hence, since τ2:Ms1s2tMs2s1s2t is clearly 0, Ms1s2t must be a submodule of M. Similarly, Ns2t is a submodule of N. Then Lemma 1.11 shows that the resulting 3-dimensional quotients of M and N are irreducible.

If q21, the composition factors of M and N account for all 8 dimensions of M(t). The following pictures show the composition factors of M(t). Each dot in w-1C represents one basis element of M(t)wtgen, 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,qgeneric

tQ=t±q,1.

Let ±q,1 and ±q-1,1 be the 1-dimensional H{2}-modules spanned by vs1t and vs2s1t, respectively, and given by

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

Then

M=HH{2} ±1,1andN= HH{2} ±q-1,1

are 4-dimensional H-modules.

If tQ=t-q,1 and q is a primitive sixth root of unity or if tQ=tq,1 and q is a primitive third root of unity, then tQ=tq-2,1, which is in the same orbit as tq2,1, and the irreducibles with central character t have already been analyzed.

M N t±q,1q21, (excludingt-q,1when qis a primitive sixth root of unity, and tq,1whenq is a primitive third root of unity.)

Proposition 2.12. Let M=HH{2} ±q,1 and N=HH{2} ±q-1,1. Unless tQ=t-q,1 and q is a primitive sixth root of unity or tQ=tq,1 and q is a primitive third root of unity, M and N are irreducible.

Proof.

Then if q21, 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 tq,1 when q is a primitive third root of unity. These characters are in the same orbit as t1,q2, which has already been analyzed, and is thus excluded from the following analysis. With this assumption, M has weights t and s1t, with dimMtgen=2= dimMs1tgen. Since P(t)={2α1+α2}, τ1:MtgenMs1tgen is invertible, so that the dimensions of Mtgen and Ms1tgen must be the same in any H-module. Then by Lemma 1.11, M is irreducible. Similar reasoning shows that N is irreducible.

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

Summary

The following tables summarize the classification. It should be noted that for any value of q with q2 not a root of unity of order 4 or less, the representation theory of H can be described in terms of q only. If q2 is a primitive root of unity of order 4 or less, then the representation theory of 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(t) and Z(t) for all possible central characters t can be described solely in terms of q. In the local region pictures, this is reflected in the fact that the hyperplanes Hα and Hα±δ are distinct unless q2 is a root of unity of order 4 or less. When these hyperplanes coincide, the sets P(t) and Z(t) change for characters on those hyperplanes.

tQ 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 typeC2 , with general.

tQ 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 typeC2 , withq=-1.

tQ 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 typeC2 , withqa primitive fourth root of unity

tQ 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 typeC2 , withqa primitive sixth root of unity.

tQ 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 typeC2 , withqa 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.

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