Geometry of Type ${G}_{2}$

Last update: 21 March 2013

Geometry of Type ${G}_{2}$

Let $𝔤$ denote the Lie algebra of the group of type ${G}_{2}\text{.}$ Then $𝔤$ is generated by ${e}_{i}={e}_{{\alpha }_{i}},$ ${f}_{i}={f}_{{\alpha }_{i}},$ and ${h}_{i}={h}_{{\alpha }_{i}},$ for $i=1$ or 2, subject to the Serre relations (see [Hum1994] for one discussion of these relations). The relations are:

$[hi,hj]=0 (3.10) [xi,yi]= hi, [xi,yi]= 0 for i≠j (3.11) [hi,xj]= cijxj, [hi,yj]= -cijyj (3.12) ad(xi)-cij+1 (xj)=0,i≠j (3.13) ad(yi)-cij+1 (yj)=0,i≠j (3.14)$

where ${c}_{ij}=⟨{\alpha }_{j},{\alpha }_{i}^{\vee }⟩\text{.}$ Specifically, ${c}_{11}={c}_{22}=2,$ ${c}_{12}=-3,$ and ${c}_{21}=-1,$ so that the Cartan matrix for type ${G}_{2}$ is

$C=[cij]= [ 2-3 -12 ] .$

Then the smallest faithful representation of $𝔤$ is the 7-dimensional representation $\varphi ,$ given by:

$ϕ(e1)= [ 0100000 0000000 000-2000 0000100 0000000 0000001 0000000 ] , ϕ(e2)= [ 0000000 0010000 0000000 0000000 0000010 0000000 0000000 ] , ϕ(f1)= [ 0000000 1000000 0000000 00-10000 0002000 0000000 0000010 ] , ϕ(f2)= [ 0000000 0000000 0100000 0000000 0000000 0000100 0000000 ] , ϕ(h1)= [ 1000000 0-100000 0020000 0000000 0000-200 0000010 000000-1 ] ,and ϕ(h2)= [ 0000000 0100000 00-10000 0000000 0000100 00000-10 0000000 ] .$

For any positive root $\alpha ,$ the root space ${𝔤}_{\alpha }$ is 1-dimensional. We choose a distinguished basis vector ${e}_{\alpha }$ for each one: ${e}_{{\alpha }_{1}+{\alpha }_{2}}=\left[{e}_{1},{e}_{2}\right],$ ${e}_{2{\alpha }_{1}+{\alpha }_{2}}=\frac{1}{2}\left[{e}_{1},{e}_{{\alpha }_{1}+{\alpha }_{2}}\right],$ ${e}_{3{\alpha }_{1}+{\alpha }_{2}}=\frac{1}{3}\left[{e}_{1},{e}_{2{\alpha }_{1}+{\alpha }_{2}}\right],$ and ${e}_{3{\alpha }_{1}+2{\alpha }_{2}}=\left[{e}_{2},{e}_{3{\alpha }_{1}+{\alpha }_{2}}\right]\text{.}$ We also choose, for each positive root $\alpha ,$ basis vectors ${f}_{\alpha }$ for the root spaces ${𝔤}_{-\alpha }:$ ${f}_{{\alpha }_{1}+{\alpha }_{2}}=\left[{f}_{1},{f}_{2}\right],$ ${f}_{2{\alpha }_{1}+{\alpha }_{2}}=\frac{1}{2}\left[{f}_{1},{f}_{{\alpha }_{1}+{\alpha }_{2}}\right],$ ${f}_{3{\alpha }_{1}+{\alpha }_{2}}=\frac{1}{3}\left[{f}_{1},{f}_{2{\alpha }_{1}+{\alpha }_{2}}\right],$ and ${f}_{3{\alpha }_{1}+2{\alpha }_{2}}=\left[{f}_{2},{f}_{3{\alpha }_{1}+{\alpha }_{2}}\right]\text{.}$

Then

$ϕ(eα1+α2)= [ 0010000 0002000 0000000 0000010 000000-1 0000000 0000000 ] , ϕ(e2α1+α2)= [ 0002000 0000-100 00000-10 000000-1 0000000 0000000 0000000 ] , ϕ(e3α1+α2)= [ 0000-100 0000000 0000000 0000000 0000000 0000000 0000000 ] ,and ϕ(e3α1+2α2)= [ 0000010 0000001 0000000 0000000 0000000 0000000 0000000 ] .$

Also,

$ϕ(fα1+α2)= [ 0000000 0000000 -1000000 0-100000 0000000 000-2000 0000010 ] , ϕ(f2α1+α2)= [ 0000000 0000000 0000000 1000000 0-100000 00-10000 000-2000 ] , ϕ(f3α1+α2)= [ 0000000 0000000 0000000 0000000 1000000 0000000 00-10000 ] ,and ϕ(f3α1+2α2)= [ 0000000 0000000 0000000 0000000 0000000 1000000 0100000 ] .$

Let $G$ be the corresponding Lie group of type ${G}_{2},$ the group generated by all the ${x}_{\alpha }\left(c\right)=\text{exp} c{e}_{\alpha }$ for $\alpha \in R$ and $c\in ℂ\text{.}$ Define

$wi(c)=xαi (c)x-αi (-c-1)xαi (c),$

and

$Hαi(c)= wi(c) wi(1)-1.$

We use the exponentiated matrices to compute the following relations in $G\text{.}$ (These relations are described in [Ste1967].)

The first relations are commutator relations between unipotent elements.

$xα1(c) xα2(d) = xα2(d) xα1(c) xα1+α2(cd) x2α1+α2(-c2d) x3α1+α2(c3d) x3α1+2α2(-2c3d2) xα1(c) xα1+α2(d) = xα1+α2(d) xα1(c) x2α1+α2(2cd) x3α1+α2(-3c2d) x3α1+2α2(3cd2) xα1(c) xα1+α2(d) = x2α1+α2(d) xα1(c) x3α1+α2(3cd) xα1(c) x3α1+α2(d) = x3α1+α2(d) xα1(c) xα1(c) x3α1+2α2(d) = x3α1+2α2(d) xα1(c) xα2(c) xα1+α2(d) = xα1+α2(d) xα2(c) xα2(c) x2α1+α2(d) = x2α1+α2(d) xα2(c) xα2(c) x3α1+α2(d) = x3α1+α2(d) xα2(c) x3α1+2α2(cd) xα2(c) x3α1+2α2(d) = x3α1+2α2(d) xα2(c) xα1+α2(c) x2α1+α2(d) = x2α1+α2(d) xα1+α2(c) x3α1+2α2(-3cd) xα1+α2(c) x3α1+α2(c) = x3α1+α2(c) xα1+α2(c) xα1+α2(c) x3α1+2α2(c) = x3α1+2α2(c) xα1+α2(c) x2α1+α2(c) x3α1+α2(c) = x3α1+α2(c) x2α1+α2(c) x2α1+α2(c) x3α1+2α2(c) = x3α1+2α2(c) x2α1+α2(c) x3α1+α2(c) x3α1+2α2(c) = x3α1+2α2(c) x3α1+α2(c)$

The other needed relations describe how the Weyl group acts on the unipotent elements of $G\text{.}$

$w1(a) e1 (w1(a))-1 = -a-2f1 w1(a) e2 (w1(a))-1 = a3 e3α1+α2 w1(a) eα1+α2 (w1(a))-1 = -a e2α1+α2 w1(a) e2α1+α2 (w1(a))-1 = a-1 eα1+α2 w1(a) e3α1+α2 (w1(a))-1 = -a-3 eα2 w1(a) e3α1+2α2 (w1(a))-1 = e3α1+2α2 w2(b) e2 (w2(b))-1 = -b-2f2 w2(b) e1 (w2(b))-1 = -b eα1+α2 w2(b) eα1+α2 (w2(b))-1 = b-1eα1 w2(b) e2α1+α2 (w2(b))-1 = e2α1+α2 w2(b) e3α1+α2 (w2(b))-1 = b e3α1+2α2 w2(b) e3α1+2α2 (w2(b))-1 = -b-1 e3α1+α2$

Let ${G}_{ss}$ be the set of semisimple elements of $G$ and let $𝒩$ be the set of nilpotent elements of $𝔤\text{.}$ Then $G$ acts on the set

$Λ= { (s,n) ∣ s∈Gss,n∈ 𝒩qs } ,where 𝒩qs= { n∈𝒩 ∣ Ads(n)=q2n } ,$

by

$g·(s,n)= ( gsg-1, Adg(n) ) .$

Let $T$ be the group of weights of $\stackrel{\sim }{H},$ the affine Hecke algebra of type ${G}_{2}\text{.}$ Let ${t}_{z,w}\in T$ be given by

$t(Xα1) = z t(Xα2) = w.$

There is a bijection between $T$ and $D\subseteq G,$ given by

$ϕ:T ⟶ D (3.15) tz,w ⟼ stz,w= Hα1 (z3w2) Hα2 (z2w), (3.16)$

satisfying

$st·xα(c)= xα(t(Xα)c)$

for $\alpha \in R\text{.}$ Since ${W}_{0}=N\left(D\right)/D,$ ${W}_{0}$ acts on $D,$ and two diagonal elements $s$ and $s\prime$ are in the same $G\text{-orbit}$ if and only if they are in the same ${W}_{0}\text{-orbit.}$ The map $\varphi$ is ${W}_{0}\text{-equivariant,}$ giving a bijection

$θ: {W0-orbits on T} ⟷ {G-orbits on Gss} t ⟼ st.$

Define

$dα1(z) = stz,1 and dα2(z)= st1,z,$

for $z\in ℂ,$ so that

$dαi(z) ·eαi=z eαi and dαi(z) ·eαj= eαj,$

for $i,j=1,2$ and $i\ne j\text{.}$

Nilpotent Orbits and $\Lambda /G$

The relations in the previous subsection, along with 3.15, allow us to determine the $G\text{-orbits}$ on $\Lambda \text{.}$

Proposition 3.24. If ${q}^{2}$ is not a root of unity of order $\ell \le 6,$ then the following is a set of representatives of $\Lambda /G\text{.}$

$(st1,1,0) (st1,-1,0) (st11/3,-1,0) (st1,q2,0) (st1,q2,eα2) ( st1,q2, eα2+ e3α1+α2 ) ( st1,q2, eα1+α2 ) ( st1,q ,0 ) ( st1,q , e3α1+2α2 ) ( st1,-q , 0 ) ( st1,-q , e3α1+2α2 ) ( st1,z , 0 ) , for z≠±1 ,±q±1, q±2 ( stq2,1 , 0 ) ( stq2,1 , eα1 ) ( stq,1 , 0 ) ( stq,1 , e2α1+α2 ) ( st-q,1 , 0 ) ( st-q,1 , e2α1+α2 ) ( stq2/3,1 , 0 ) ( stq2/3,1 , e3α1+α2 ) ( stz,1 , 0 ) , for z≠q±2, ±q±1,± 11/3, q±2/3 ( stq2,-q-2 , 0 ) ( stq2,-q-2 , eα1 ) ( stq2,-q-2 , e3α1+2α2 ) ( stq2,-q-2 , eα1+ e3α1+2α2 ) ( st11/3,q2 , 0 ) ( st11/3,q2 , eα2 ) ( st11/3,q2 , e3α1+α2 ) ( st11/3,q2 , eα2+ e3α1+α2 ) ( stq2,q2 , 0 ) ( stq2,q2 , eα1 ) ( stq2,q2 , eα2 ) ( stq2,q2 , eα1+eα2 ) ( stq2,z , 0 ) , where P(t)={α1} ( stq2,z , eα1 ) , where P(t)={α1} ( stz,q2 , 0 ) , where P(t)={α2} ( stz,q2 , eα2 ) , where P(t)={α2} ( stz,w , 0 ) ,tz,w generic$

 Proof. Given an element $\left(s,n\right)\in \Lambda ,$ by 3.15, there is an element $\left(s\prime ,n\prime \right)$ in the $G\text{-orbit}$ of $\left(s,n\right)$ such that $s\prime$ is diagonal and that $s\prime ={s}_{t}$ for some ${t}_{z,w}$ such that ${t}_{z,w}$ is in the set of ${W}_{0}\text{-orbit}$ representatives listed in (2.6). Then by Lemma 3.9, it is sufficient to describe the ${C}_{G}\left({s}_{t}\right)\text{-orbits}$ in ${𝔤}_{q}^{s}$ for a set of representatives of possible central characters $t\text{.}$ Also, Theorem 3.6 shows that ${𝔤}_{q}^{s}$ is spanned by $\left\{{e}_{\alpha } \mid t\left({X}^{\alpha }\right)={q}^{2}\right\}\text{.}$ Case 1: $P\left(t\right)=\varnothing$ If $P\left(t\right)=\varnothing ,$ then ${𝔤}_{q}^{s}=\left\{0\right\},$ so that ${s}_{t}$ must be paired with 0. This includes ${t}_{1,1},$ ${t}_{1,-1},$ ${t}_{{1}^{1/3},1},$ ${t}_{1,z},$ ${t}_{z,1},$ and ${t}_{z,w}\text{.}$ Case 2: $\mid P\left(t\right)\mid =1$ If $\mid P\left(t\right)\mid =1,$ then ${𝔤}_{q}^{s}$ is 1-dimensional. Assume $\alpha \in P\left(t\right)\text{.}$ Since ${d}_{\alpha }\left({a}^{-1}\right)·a{e}_{\alpha }={e}_{\alpha },$ any non-zero element of ${𝔤}_{q}^{s}$ is in the same orbit as ${e}_{\alpha }\text{.}$ This case includes ${t}_{1,±q},$ ${t}_{{q}^{2},z},$ ${t}_{z,{q}^{2}},$ and ${t}_{±q,1}\text{.}$ Case 3: $Z\left(t\right)=\varnothing ,$ $\mid P\left(t\right)\mid =2$ If $Z\left(t\right)=\varnothing$ then ${C}_{G}\left(s\right)=D\text{.}$ and ${𝔤}_{q}^{s}$ is 2-dimensional, spanned by ${e}_{\alpha }$ and ${e}_{\beta }$ for some $\alpha$ and $\beta \text{.}$ However, the action of any element $D$ has both ${e}_{\alpha }$ and ${e}_{\beta }$ as eigenvectors. (Specifically, ${s}_{t}·{e}_{\alpha }=t\left({X}^{\alpha }\right){e}_{\alpha }\text{.)}$ Thus the non-zero ${C}_{G}\left(s\right)\text{-robits}$ in ${𝔤}_{q}^{s}$ are represented by ${e}_{\alpha },{e}_{\beta }$ and ${e}_{\alpha }+{e}_{\beta }\text{.}$ This case includes ${t}_{{q}^{2},{q}^{2}},$ ${t}_{{q}^{2},-{q}^{-2}},$ and ${t}_{{1}^{1/3},{q}^{2}}\text{.}$ Case 4: ${t}_{1,{q}^{2}}$ If $t={t}_{1,{q}^{2}},$ then ${𝔤}_{q}^{s}$ is spanned by ${e}_{{\alpha }_{2}},$ ${e}_{{\alpha }_{1}+{\alpha }_{2}},$ ${e}_{2{\alpha }_{1}+{\alpha }_{2}},$ and ${e}_{3{\alpha }_{1}+{\alpha }_{2}},$ while ${C}_{G}\left(s\right)$ contains ${x}_{{\alpha }_{1}}\left(c\right)$ and ${x}_{{\alpha }_{2}}\left(c\right)$ for $c\in {ℂ}^{×}\text{.}$ Let $n=a{e}_{{\alpha }_{2}}+b{e}_{{\alpha }_{1}+{\alpha }_{2}}+c{e}_{2{\alpha }_{1}+{\alpha }_{2}}+d{e}_{3{\alpha }_{1}+{\alpha }_{2}}\in {𝔤}_{q}^{s}\text{.}$ We claim that $n$ is in the ${C}_{G}\left(s\right)\text{-orbit}$ of either ${e}_{{\alpha }_{1}+{\alpha }_{2}},$ or some element in the span of ${e}_{{\alpha }_{2}}$ and ${e}_{3{\alpha }_{1}+{\alpha }_{2}}\text{.}$ If $a=0$ but $b\ne 0,$ then $x-α1(1)·n= (3b+3c+d)eα2+ (b+2c+d)eα1+α2 +(c+d)e2α1+α2 +de3α1+α2,$ so we can assume $a\ne 0\text{.}$ Then if $a\ne 0,$ then $xα1(-b/a)·n= aeα2+ (c-b2a) e2α1+α2+ ( d- 2b3a2- 3bca ) e3α1+α2.$ Then we may assume that $b=0\text{.}$ If $c=0,$ then $n$ is in the span of ${e}_{{\alpha }_{2}}$ and ${e}_{3{\alpha }_{1}+{\alpha }_{2}},$ so we assume $c\ne 0\text{.}$ Then $x-α1 ( az2+c -3cz-d- z3a ) xα1(z)·n = ( -a2z3d +2c3+ 3acdz+6 c2az2+ ad2 (3cz+d+z3a) 2 ) eα2 + ( acz2+adz- c2 3cz+dz3a ) eα1+α2+ (3cz+d+z3a) e3α1+α2.$ Hence if we choose $z$ to be a root of $ac{z}^{2}+adz-{c}^{2},$ then ${x}_{-{\alpha }_{1}}\left(\frac{a{z}^{2}+c}{-3cz-d-{z}^{3}a}\right){x}_{{\alpha }_{1}}\left(z\right)·n$ will be in the span of ${e}_{{\alpha }_{2}}$ and ${e}_{3{\alpha }_{1}+{\alpha }_{2}}\text{.}$ This calculation can fail only if both $ac{z}^{2}+adz-{c}^{2}$ and $3cz+d+{z}^{3}a$ are zero for the same value of $z\text{.}$ In this case, though, the GCD of these two polynomials will be non-trivial. However, $az3+3cz+d= (zc-dc2) (acz2+adz-c2) +4c3+ad2c2z.$ Hence $\text{GCD}\left(a{z}^{3}+3cz+d,-3cz-d-{z}^{3}a\right)=a$ exactly if $4{c}^{3}+a{d}^{2}\ne 0\text{.}$ Assume that this is the case, so that $n$ is conjugate to an element in the span of ${e}_{{\alpha }_{2}}$ and ${e}_{3{\alpha }_{1}+{\alpha }_{2}}\text{.}$ Then we note that $dα1 (a1/3d-1/3) dα2(a-1)· aeα2+d e3α1+α2= eα2+ e3α1+α2,$ while $w1·a e3α1+α2= aeα2$ and $dα2(a-1)· aeα2=eα2.$ Hence any element in the span of ${e}_{{\alpha }_{2}}$ and ${e}_{3{\alpha }_{1}+{\alpha }_{2}}$ is in the ${C}_{G}\left(s\right)\text{-orbit}$ of either ${e}_{{\alpha }_{2}}$ or ${e}_{{\alpha }_{2}}+{e}_{3{\alpha }_{1}+{\alpha }_{2}}\text{.}$ Finally, assume that $4{c}^{3}+a{d}^{2}=0\text{.}$ (Recall that we are assuming $c\ne 0,$ so that neither $a$ or $d$ is zero.) Then $a=\frac{-4{c}^{3}}{{d}^{2}}\text{.}$ But, $dα1 (-2c3d)· -4c3d2 eα2+c e2α1+α2+ de3α1+α2= -4c3d2 eα2+ 4c39d2 e2α1+α2- 8c327d2 e3α1+α2.$ But then $dα1 (3d2-4c3) x-α1(-1) xα1(1/3)· -4c3d2 eα2+ 4c39d2 e2α1+α2- 8c327d2 e3α1+α2= eα1+α2.$ Hence every non-zero $n\in {𝔤}_{q}^{s}$ is ${C}_{G}\left(s\right)\text{-conjugate}$ to either ${e}_{{\alpha }_{2}},$ ${e}_{{\alpha }_{1}+{\alpha }_{2}},$ or ${e}_{{\alpha }_{2}}+{e}_{3{\alpha }_{1}+{\alpha }_{2}}\text{.}$ These three elements cannot be conjugate to each other since in the basic representation of $𝔤,$ they have different Jordan forms. Case 5: ${t}_{{q}^{2},1}$ If $t={t}_{{q}^{2},1}$ then ${𝔤}_{q}^{s}$ is spanned by ${e}_{{\alpha }_{1}}$ and ${e}_{{\alpha }_{1}+{\alpha }_{2}}\text{.}$ Also, ${C}_{G}\left(s\right)$ contains ${x}_{{\alpha }_{2}}\left(c\right)$ and ${x}_{-{\alpha }_{2}}\left(c\right)$ for $c\in {ℂ}^{×},$ so in particular ${w}_{2}\in {C}_{G}\left(s\right)\text{.}$ If $ab\ne 0,$ then $x-α2(a/b)· ( aeα1+b eα1+α2 ) =beα1+α2.$ But $w2·beα1+α2 =beα1,$ and $dα1(b-1) ·beα1=eα1.$ Hence every non-zero element of ${𝔤}_{q}^{s}$ is ${C}_{G}\left(s\right)\text{-conjugate}$ to ${e}_{{\alpha }_{1}}\text{.}$ Case 6: ${t}_{{q}^{2/3},1}$ If $t={t}_{{q}^{2/3},1},$ then ${𝔤}_{q}^{s}$ is spanned by ${e}_{3{\alpha }_{1}+{\alpha }_{2}}$ and ${e}_{3{\alpha }_{1}+2{\alpha }_{2}}\text{.}$ Also, ${C}_{G}\left(s\right)$ contains ${x}_{{\alpha }_{2}}\left(c\right)$ and ${x}_{-{\alpha }_{2}}\left(c\right),$ so in particular ${w}_{2}\in {C}_{G}\left(s\right)\text{.}$ If $ab\ne 0,$ then $x-α2(-a/b)· ( ae3α1+α2+ be3α1+2α2 ) =be3α1+2α2.$ But $w2·b e3α1+2α2 =b e3α1+α2,$ and $dα2(b-1) ·b e3α1+α2= e3α1+α2.$ Hence every non-zero element of ${𝔤}_{q}^{s}$ is ${C}_{G}\left(s\right)\text{-conjugate}$ to ${e}_{3{\alpha }_{1}+{\alpha }_{2}}\text{.}$ $\square$

Proposition 3.25. If ${q}^{2}$ is a primitive sixth root of unity, then the following is a set of representatives of $\Lambda /G\text{.}$ (The notation ${1}^{1/3}$ is taken to mean a cube root of 1 besides ${q}^{4}\text{.)}$

$(st1,1,0) (st1,-1,0) (st11/3,-1,0) (st1,q2,0) (st1,q2,eα2) ( st1,q2, eα2+ e3α1+α2 ) ( st1,q2, eα1+α2 ) (st1,q,0) ( st1,q, e3α1+2α2 ) (st1,-q,0) ( st1,-q, e3α1+2α2 ) (st1,z,0), for z≠±1, ±q±1,q±2 (stq2,1,0) (stq2,1,eα1) (stq,1,0) ( stq,1, e2α1+α2 ) (st-q,1,0) ( st-q,1, e2α1+α2 ) (stq2/3,1,0) ( stq2/3,1, e3α1+α2 ) (stz,1,0) for z≠q±2, ±q±1,± 11/3, q±2/3 (stq2,-q-2,0) ( stq2,-q-2, eα1 ) ( stq2,-q-2, e3α1+2α2 ) ( stq2,-q-2, eα1+ e3α1+2α2 ) ( st11/3,q2 ,0 ) ( st11/3,q2 ,eα2 ) ( st11/3,q2 ,e3α1+α2 ) ( st11/3,q2 ,eα2+ e3α1+α2 ) ( stq2,q2,0 ) ( stq2,q2, eα1 ) ( stq2,q2, eα2 ) ( ss2s1s2tq2,q2 ,eα2 ) ( stq2,q2, eα1+eα2 ) ( ss2s1s2tq2,q2 ,e2α1+α2+ eα2 ) ( ss2s1s2s1tq2,q2 ,eα2+e3α1+α2 ) (stq2,z,0) , where P(t)= {α1} (stq2,z,eα1) , where P(t)= {α1} (stz,q2,0) , where P(t)= {α2} (stz,q2,eα2) , where P(t)= {α2} (stz,w,0), tz,w generic$

 Proof. The proof differs from the proof of proposition 3.24 only for the central character ${t}_{{q}^{2},{2}^{2}}\text{.}$ Case 3a: ${t}_{{q}^{2},{q}^{2}}$ If $t={t}_{{q}^{2},{q}^{2}},$ then $P\left(t\right)=\left\{{\alpha }_{1},{\alpha }_{2},3{\alpha }_{1}+2{\alpha }_{2}\right\},$ and ${𝔤}_{q}^{s}$ is spanned by ${e}_{{\alpha }_{1}},$ ${e}_{{\alpha }_{2}},$ and ${e}_{-3{\alpha }_{1}-2{\alpha }_{2}}\text{.}$ However, if $n=a{e}_{{\alpha }_{1}}+b{e}_{{\alpha }_{2}}+c{e}_{-3{\alpha }_{1}-2{\alpha }_{2}},$ then $n$ is only nilpotent when at least one of $a,b,$ and $c$ is zero. However, the action of $D$ shows that there are 3 orbits of elements $a{e}_{{\alpha }_{1}}+b{e}_{{\alpha }_{2}}+c{e}_{-3{\alpha }_{1}-2{\alpha }_{2}}$ with exactly one of $a,b,$ and $c$ equal to zero: $dα1(a-1) dα2(b-1)· ( aeα1+ beα2 ) =eα1+eα2, dα1(q-1) dα2(a32c12) · ( aeα1+c e-3α1-2α2 ) =eα1+ e-3α1-2α2, and dα1(b23c13) dα2(b-1)· ( ce-3α1-2α2 +beα2 ) = e-3α1-2α2 +eα2.$ If exactly one of $a,b,$ and $c$ is non-zero, then $n=a{e}_{{\alpha }_{1}}+b{e}_{{\alpha }_{2}}+c{e}_{-3{\alpha }_{1}-2{\alpha }_{2}}$ is in the same ${C}_{G}\left(s\right)\text{-orbit}$ as ${e}_{{\alpha }_{1}},$ ${e}_{{\alpha }_{2}},$ or ${e}_{-3{\alpha }_{1}-2{\alpha }_{2}}\text{.}$ For ease of later computations, we choose an element in the $G\text{-orbit}$ of each pair $\left(s,n\right)$ so that $n$ is in the span of $\left\{{e}_{\alpha } \mid \alpha \in {R}^{+}\right\}\text{.}$ Specifically, ${s}_{2}{s}_{1}{s}_{2}·{e}_{-3{\alpha }_{1}-2{\alpha }_{2}}={e}_{{\alpha }_{2}},$ ${s}_{2}{s}_{1}{s}_{2}·\left({e}_{{\alpha }_{1}}+{e}_{-3{\alpha }_{1}-2{\alpha }_{2}}\right)={e}_{2{\alpha }_{1}+{\alpha }_{2}}+{e}_{{\alpha }_{2}},$ and ${s}_{2}{s}_{1}{s}_{2}{s}_{1}·\left({e}_{{\alpha }_{2}}+{e}_{-3{\alpha }_{1}-2{\alpha }_{2}}\right)={e}_{{\alpha }_{2}}+{e}_{3{\alpha }_{1}+{\alpha }_{2}}\text{.}$ $\square$

Proposition 3.26. If ${q}^{2}$ is a primitive fifth root of unity, then the following is a set of representatives of $\Lambda /G\text{.}$ (The notation ${q}^{2/3}$ is taken to mean a cube root of ${q}^{2}$ not equal to ${q}^{4}\text{.)}$

$( st1,1,0 ) ( st1,-1,0 ) ( st11/3,1 ,0 ) ( st1,q2, 0 ) ( st1,q2, eα2 ) ( st1,q2, eα2+ e3α1+α2 ) ( st1,q2, eα1+α2 ) ( st1,q,0 ) ( st1,q, e3α1+2α2 ) ( st1,-q,0 ) ( st1,-q, e3α1+2α2 ) ( st1,z,0 ) , for z≠ ±1,±q±1 ,q±2 ( stq2,1,0 ) ( stq2,1, eα1 ) ( stq2/3,1 ,0 ) ( stq2/3,1, e3α1+α2 ) ( stz,1,0 ) , for z≠ q±2,± q±1,± 11/3, q±2/3 ( stq2,-q-2,0 ) ( stq2,-q-2, eα1 ) ( stq2,-q-2, e3α1+2α2 ) ( stq2,-q-2, eα1+ e3α1+2α2 ) ( st11/3,q2 ,0 ) ( st11/3,q2 ,eα2 ) ( st11/3,q2 ,e3α1+α2 ) ( st11/3,q2 ,eα2+ e3α1+α2 ) ( st-q-4,1 , 0 ) ( st-q-4,1 , e2α1+α2 ) ( stq-4,1 , 0 ) ( stq-4,1 , e2α1+α2 ) ( stq-4,1 , e-3α1-α2 ) ( stq-4,1 , e2α1+α2+ e-3α1-α2 ) ( stq2,z,0 ) , where P(t)={α1} ( stq2,z, eα1 ) , where P(t)={α1} ( stz,q2,0 ) , where P(t)={α2} ( stz,q2, eα2 ) , where P(t)={α2} (stz,w,0) ,tz,w generic$

 Proof. The proof in this case differs from that of proposition 3.24 for the central characters ${t}_{{q}^{2},{q}^{2}}$ and ${t}_{{q}^{-4},1}\text{.}$ Note that ${t}_{{q}^{-4},1}={t}_{{q}^{6},1},$ and ${s}_{2}{s}_{1}{t}_{{q}^{6},1}={t}_{{q}^{2},{q}^{2}}\text{.}$ One notational note is also helpful - when ${q}^{2}$ is a primitive fifth root of unity, the central characters ${t}_{±q,1}$ are equal to ${t}_{±{q}^{-4},1}$ in some order. That order, however, depends on whether ${q}^{5}$ is 1 or -1. Calling the characters ${t}_{±{q}^{-4},1}$ allows us to treat the two cases at once. The weight ${t}_{-{q}^{-4},1}$ has $P\left(t\right)=\left\{2{\alpha }_{1}+{\alpha }_{2}\right\},$ so that the argument given in Case 2 of 3.24 classifies the orbits of pairs $\left({s}_{{t}_{-{q}^{-4},1}},n\right)\text{.}$ Case 3a: ${t}_{{q}^{-4},1}$ If $t={t}_{{q}^{-4},1}$ then ${𝔤}_{q}^{s}$ is spanned by ${e}_{2{\alpha }_{1}+{\alpha }_{2}},$ ${e}_{-3{\alpha }_{1}-{\alpha }_{2}},$ and ${e}_{-3{\alpha }_{1}-2{\alpha }_{2}}\text{.}$ Also, ${C}_{G}\left(s\right)$ contains ${x}_{±\left({\alpha }_{2}\right)}\left(c\right)$ for $c\in ℂ,$ as well as ${w}_{2}\left(1\right)\text{.}$ Let $n=a{e}_{2{\alpha }_{1}+{\alpha }_{2}}+b{e}_{-3{\alpha }_{1}-{\alpha }_{2}}+c{e}_{-3{\alpha }_{1}-2{\alpha }_{2}}\text{.}$ If $b=0,$ then $w2·n=a e2α1+α2-c e-3α1-α2,$ so we can assume $b\ne 0\text{.}$ Then $x-α2(-c/b)· ae2α1+α2+b e-3α1-α2+c e-3α1-2α2= ae2α1+α2+b e-3α1-α2.$ Hence every element of ${𝔤}_{q}^{s}$ is in the orbit of an element in the span of ${e}_{2{\alpha }_{1}+{\alpha }_{2}}$ and ${e}_{-3{\alpha }_{1}-{\alpha }_{2}}\text{.}$ However, ${x}_{±{\alpha }_{2}}\left(c\right)$ stabilizes ${e}_{2{\alpha }_{1}+{\alpha }_{2}},$ so that the orbits in ${𝔤}_{q}^{s}$ are represented by 0, ${e}_{2{\alpha }_{1}+{\alpha }_{2}},$ ${e}_{-3{\alpha }_{1}-{\alpha }_{2}},$ and ${e}_{2{\alpha }_{1}+{\alpha }_{2}}+{e}_{-3{\alpha }_{1}-{\alpha }_{2}}\text{.}$ $\square$

Proposition 3.27. If ${q}^{2}$ is a primitive fourth root of unity, then the following is a set of representatives of $\Lambda /G\text{.}$ (The notation ${q}^{2/3}$ is taken to mean a cube root of ${q}^{2}$ which is not equal to ${q}^{-2}\text{.)}$

$(st1,1,0) (st1,-1,0) (st11/3,1,0) (st1,q2,0) (st1,q2,eα2) (st1,q2,eα2+e3α1+α2) (st1,q2,eα1+α2) (st1,q,0) (st1,q,e3α1+2α2) (st1,-q,0) (st1,-q,e3α1+2α2) (st1,z,0), for z≠±1,±q±1,q±2 (st-q,1,0) (st-q,1,e2α1+α2) (stq,1,0) (stq,1,e2α1+α2) (stq2/3,1,0) (stq2/3,1,e3α1+α2) (stz,1,0), for z≠q±2,±q±1,±11/3,q±2/3 (st11/3,q2,0) (st11/3,q2,eα2) (st11/3,q2,e3α1+α2) (st11/3,q2,eα2+e3α1+α2) (stq2,q2,0) (stq2,q2,eα1) (stq2,q2,eα2) (stq2,q2,eα1+eα2) (stq2,q2,eα1+e3α1+2α2) (stq2,z,0), where P(t)={α1} (stq2,z,eα1), where P(t)={α1} (stz,q2,0), where P(t)={α2} (stz,q2,eα2), where P(t)={α2} (stz,w,0),tz,w generic$

 Proof. The proof differs from that of 3.24 for the central characters ${t}_{q,1},$ ${t}_{{q}^{2},-{q}^{-2}},$ and ${t}_{{q}^{2},{q}^{2}}\text{.}$ Since ${s}_{2}{s}_{1}{s}_{2}{s}_{1}{t}_{{q}^{2},1}={t}_{{q}^{2},{q}^{2}}$ and $-{q}^{-2}={q}^{2},$ all three of these central characters are in the same orbit. Case 5: ${t}_{{q}^{2},{q}^{2}}$ If $t={t}_{{q}^{2},{q}^{2}},$ then $P=\left\{{\alpha }_{1},{\alpha }_{2},2{\alpha }_{1}+{\alpha }_{2},3{\alpha }_{1}+2{\alpha }_{2}\right\},$ and ${𝔤}_{q}^{s}$ is spanned by ${e}_{{\alpha }_{1}},$ ${e}_{{\alpha }_{2}},$ ${e}_{-2{\alpha }_{1}-{\alpha }_{2}},$ and ${e}_{3{\alpha }_{1}+2{\alpha }_{2}}\text{.}$ Also, ${C}_{G}\left(s\right)$ contains ${x}_{±\left(3{\alpha }_{1}+{\alpha }_{2}\right)}\left(c\right)$ for $c\in ℂ,$ and ${s}_{1}{s}_{2}{s}_{1}={x}_{3{\alpha }_{1}+{\alpha }_{2}}\left(1\right){x}_{-3{\alpha }_{1}-{\alpha }_{2}}\left(-1\right){x}_{3{\alpha }_{1}+{\alpha }_{2}}\left(1\right)\text{.}$ Let $n=a{e}_{{\alpha }_{1}}+b{e}_{{\alpha }_{2}}+c{e}_{-2{\alpha }_{1}-{\alpha }_{2}}+d{e}_{3{\alpha }_{1}+2{\alpha }_{2}}\text{.}$ Assume $n$ is generic. Then $dα1(a-1) dα2(aab+cd) x3α1+α2 (adab+cd) x-3α1-α2 (ca)·n= eα1+eα2.$ The above calculation fails if either $a=0$ or if $ab+cd=0\text{.}$ Assume now that $a=0,$ but $ab+cd\ne 0\text{.}$ Then $c\ne 0$ and $d\ne 0,$ so that $x-3α1-α2(-b/d)· ( beα2+c e-2α1-α2+d e3α1+2α2 ) =ce-2α1-α2+ de3α1+2α2,$ and then $dα1(c-1) dα2(d-1) s1s2s1· ( ce-2α1-α2 +de3α1+2α2 ) =eα1+eα2.$ Next, assume $ab+cd=0$ and $a=0,$ so that $cd=0\text{.}$ If $a=c=d=0$ but $n\ne 0$ then $dα2(b-1) ·beα2=eα2.$ If $a=c=0$ but $d\ne 0$ then $dα1(d-1) s1s2s1 x-3α1-α2 (-b/d)· ( beα2+ de3α1+2α2 ) =eα2.$ If $a=d=0$ but $c\ne 0$ then $s1s2s1·b eα2+c e-2α1-α2 =be3α1+2α2 +ceα1,$ a case we will treat below. If $ab+cd=0$ and $a\ne 0,$ then $b=-cd/a\text{.}$ Then $x-3α1-α2 (cd)· ( aeα1+ -cdaeα2 +ce-2α1-α2 +de3α1+2α2 ) =aeα1+d e3α1+2α2.$ But $dα1(a-1) dα2(a-3d) ·aeα1+d e3α1+2α2= eα1+ e3α1+2α2.$ Hence every non-zero element of ${𝔤}_{q}^{s}$ is in the same orbit as either ${e}_{{\alpha }_{1}},$ ${e}_{{\alpha }_{2}},$ ${e}_{{\alpha }_{1}}+{e}_{{\alpha }_{2}},$ or ${e}_{{\alpha }_{1}}+{e}_{3{\alpha }_{1}+2{\alpha }_{2}}\text{.}$ $\square$

Proposition 3.28. If ${q}^{2}$ is a primitive third root of unity, then the following is a set of representatives of $\Lambda /G\text{.}$

$(st1,1,0) (st1,-1,0) (st1,q2,0) (st1,q2,eα2) (st1,q2,eα2+e3α1+α2) (st1,q2,eα1+α2) (st1,q2,e-3α1-2α2) (st1,q2,eα2+e-3α1-2α2) (st1,q2,eα1+α2+e-3α1-2α2) (st1,-q-2,0) (st1,-q-2,e3α1+2α2) (st1,z,0), for z≠±1,±q±1,q±2 (st-q-2,1,0) (st-q-2,1,e2α1+α2) (stq2,1,0) (stq2,1,eα1) (stq2/3,1,0) (stq2/3,1,e3α1+α2) (stz,1,0), for z≠q±2,±q±1,±11/3,q±2/3 (stq2,-q-2,0) (stq2,-q-2,eα1) (stq2,-q-2,e3α1+2α2) (stq2,-q-2,eα1+e3α1+2α2) (stq2,z,0), where P(t)={α1} (stq2,z,α1), where P(t)={α1} (stz,q2,0), where P(t)={α2} (stz,q2,eα2), where P(t)={α2} (stz,w,0),tz,w generic$

 Proof. The proof differs from that of 3.24 for the central characters ${t}_{{1}^{1/3},1},$ ${t}_{1,±q},$ ${t}_{±q,1},$ ${t}_{{1}^{1/3},{q}^{2}},$ ${t}_{{q}^{2},{q}^{2}},$ ${t}_{{q}^{2},1},$ and ${t}_{1,{q}^{2}}\text{.}$ 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, ${t}_{1,-q}={t}_{1,{q}^{4}}={w}_{0}{t}_{1,{q}^{2}},$ and ${t}_{-q,1}={t}_{{q}^{4},1}={w}_{0}{t}_{{q}^{2},1}\text{.}$ 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 ${t}_{1,{q}^{2}}\text{.}$ Note also that the central characters ${t}_{±q,1}$ are the same as ${t}_{±{q}^{-2},1}$ in some order, but the order depends on whether ${q}^{3}$ is 1 or $-1\text{.}$ We call these characters ${t}_{-{q}^{-2},1}$ and ${t}_{{q}^{2},1}={w}_{0}{t}_{{q}^{-2},1},$ so that we can treat both cases at once. Similarly, the central characters ${t}_{1,±q},$ are the same as ${t}_{q,±{q}^{-2}},$ but the order depends on whether ${q}^{3}$ is 1 or $-1\text{.}$ So we call these characters ${t}_{1,-{q}^{-2}}$ and ${t}_{1,{q}^{2}}\text{.}$ Thus it suffices to consider ${t}_{1,{q}^{2}}$ and ${t}_{{q}^{2},1}\text{.}$ Case 4: ${t}_{1,{q}^{2}}$ If $t={t}_{1,{q}^{2}},$ then ${𝔤}_{q}^{s}$ is spanned by ${e}_{{\alpha }_{2}},$ ${e}_{{\alpha }_{1}+{\alpha }_{2}},$ ${e}_{2{\alpha }_{1}+{\alpha }_{2}},$ ${e}_{3{\alpha }_{1}+{\alpha }_{2}},$ and ${e}_{-\left(3{\alpha }_{1}+2{\alpha }_{1}\right)}\text{.}$ However, the subspace $ℂ{e}_{-3{\alpha }_{1}-2{\alpha }_{2}}$ is fixed by the action of ${C}_{G}\left(s\right),$ and $𝔤qs= ⟨ eα2, eα1+α2, e2α1+α2, e3α1+α2 ⟩ ⊕ℂ e-3α1-2α2$ as ${C}_{G}\left(s\right)\text{-modules.}$ But the ${C}_{G}\left(s\right)$ orbits in the first summand are represented by 0, ${e}_{{\alpha }_{2}},$ ${e}_{{\alpha }_{1}+{\alpha }_{2}},$ and ${e}_{{\alpha }_{2}}+{e}_{3{\alpha }_{1}+{\alpha }_{2}},$ while the ${C}_{G}\left(s\right)\text{-orbits}$ in $ℂ{e}_{-3{\alpha }_{1}-2{\alpha }_{1}}$ are represented by 0 and ${e}_{-3{\alpha }_{1}-2{\alpha }_{1}}\text{.}$ Then the ${C}_{G}\left(s\right)\text{-orbits}$ in ${𝔤}_{q}^{s}$ are represented by 0, ${e}_{{\alpha }_{2}},$ ${e}_{{\alpha }_{1}+{\alpha }_{2}},$ ${e}_{{\alpha }_{2}}+{e}_{3{\alpha }_{1}+{\alpha }_{2}},$ ${e}_{-3{\alpha }_{1}-2{\alpha }_{2}},$ ${e}_{{\alpha }_{2}}+{e}_{-3{\alpha }_{1}-2{\alpha }_{2}},$ ${e}_{{\alpha }_{1}+{\alpha }_{2}}+{e}_{-3{\alpha }_{1}-2{\alpha }_{2}},$ and ${e}_{{\alpha }_{2}}+{e}_{3{\alpha }_{1}+{\alpha }_{2}}+c{e}_{-3{\alpha }_{1}-2{\alpha }_{2}},$ for $c\in {ℂ}^{×}\text{.}$ However, ${e}_{{\alpha }_{2}}+{e}_{3{\alpha }_{1}+{\alpha }_{2}}+c{e}_{-3{\alpha }_{1}-2{\alpha }_{2}}$ is not nilpotent unless $c=0\text{.}$ Case 5: ${t}_{{q}^{2},1}$ If $t={t}_{{q}^{2},1},$ then ${𝔤}_{q}^{{s}_{t}}$ is spanned by ${e}_{{\alpha }_{1}},$ ${e}_{{\alpha }_{1}+{\alpha }_{2}},$ and ${e}_{-2{\alpha }_{1}-{\alpha }_{2}}\text{.}$ ${C}_{G}\left(s\right)$ contains, in particular, ${x}_{3{\alpha }_{1}+{\alpha }_{2}}\left(1\right){x}_{-3{\alpha }_{1}-{\alpha }_{2}}\left(-1\right){x}_{3{\alpha }_{1}+{\alpha }_{2}}\left(1\right)={s}_{1}{s}_{2}{s}_{1}\text{.}$ Let $n=a{e}_{{\alpha }_{1}}+b{e}_{{\alpha }_{1}+{\alpha }_{2}}+c{e}_{-2{\alpha }_{1}-{\alpha }_{2}}\text{.}$ By Case 5 of Proposition 3.24, we may assume that $b=0\text{.}$ If $a\ne 0,$ then $dα1(a-1) x-3α1-α2 (-c/a)·aeα1 +ce-2α1-α2 =eα1.$ If $a=0$ and $c\ne 0,$ then $dα1(c-1) s1s2s1·c e-2α1-α2= eα1.$ If $a=c=0,$ then $n=0\text{.}$ So, every non-zero element of ${𝔤}_{q}^{s}$ is ${C}_{G}\left(s\right)\text{-conjugate}$ to ${e}_{{\alpha }_{1}}\text{.}$ $\square$

Proposition 3.29. If ${q}^{2}=-1,$ then the following is a set of representatives of $\Lambda /G\text{.}$ (The notation $-{1}^{1/3}$ is taken to mean a cube root of $-1$ besides $-1$ itself.)

$(st1,1,0) (st1,q2,0) (st1,q2,eα2) (st1,q2,eα2+e3α1+α2) (st1,q2,eα1+α2) (st1,q2,eα2+e-α1-α2) (st1,q2,eα2+e-2α1-α2) (st11/3,1,0) (st1,q,0) (st1,q,e3α1+2α2) (st1,q,e-3α1-2α2) (st1,z,0), for z≠±1,±q,q2 (stq,1,0) (stq,1,e2α1+α2) (stq,1,e-2α1-α2) (st-11/3,1,0) (st-11/3,1,e3α1+α2) (st-11/3,1,e-3α1-α2) (st-11/3,1,e3α1+α2+e-3α1-2α2) (stz,1,0), for z≠q2,±q,-1,q2/3,11/3 (stq2,z,0), where P(t)={α1} (stq2,z,eα1), where P(t)={α1} (stq2,z,e-α1), where P(t)={α1} (stz,q2,0), where P(t)={α2} (stz,q2,eα2), where P(t)={α2} (stz,q2,e-α2), where P(t)={α2} (stz,w,0),tz,w generic$

 Proof. Since ${q}^{2}={q}^{-2},$ ${e}_{-\alpha }\in {𝔤}_{q}^{s}$ exactly when ${e}_{\alpha }\in {𝔤}_{q}^{s}\text{.}$ Note that ${t}_{{1}^{1/3},{q}^{2}}={t}_{{1}^{1/3},-1}$ is in the same orbit as ${t}_{-{1}^{1/3},1}\text{.}$ Case 1: $P\left(t\right)=\varnothing$ If $P\left(t\right)=\varnothing ,$ then ${𝔤}_{q}^{{s}_{t}}=0$ and 0 is the only orbit in ${𝔤}_{q}^{{s}_{t}}\text{.}$ This applies to ${t}_{1,1},$ ${t}_{{1}^{1/3},1},$ ${t}_{1,z},$ ${t}_{z,1},$ and ${t}_{z,w}\text{.}$ Case 2: $\mid P\left(t\right)\mid =1$ If $\mid P\left(t\right)\mid =1,$ then ${𝔤}_{q}^{{s}_{t}}$ is spanned by ${e}_{\alpha }$ and ${e}_{-\alpha },$ where $\alpha \in P\left(t\right)\text{.}$ However, $a{e}_{\alpha }+b{e}_{-\alpha }$ is nilpotent exactly when $ab=0\text{.}$ For the central characters ${t}_{{q}^{2},z}$ and ${t}_{z,{q}^{2}},$ $Z\left(t\right)=\varnothing ,$ and ${C}_{G}\left(s\right)=D\text{.}$ Then the nilpotent orbits in ${𝔤}_{q}^{{s}_{t}}$ are represented by ${e}_{\alpha }$ and ${e}_{-\alpha },$ since $dα(a-1)· aeα=eα$ and $dα(b)·b e-α=e-α.$ For the central characters ${t}_{1,q}$ and ${t}_{q,1},$ $Z\left(t\right)$ contains $\beta ,$ the positive root perpendicular to $\alpha \text{.}$ Thus ${x}_{±\beta }\left(c\right)$ fixes the subspace of ${𝔤}_{q}^{{s}_{t}}$ spanned by ${e}_{\alpha },$ and the subspace spanned by ${e}_{-\alpha }\text{.}$ Then again $dα(a-1)· aeα=eα$ and $dα(b)·b e-α=e-α$ and the nilpotent orbits are represented by ${e}_{\alpha }$ and ${e}_{-\alpha }\text{.}$ Case 3: ${t}_{1,{q}^{2}}$ If $t={t}_{1,{q}^{2}},$ then ${𝔤}_{q}^{{s}_{t}}$ is spanned by ${e}_{±{\alpha }_{2},}$ ${e}_{±\left({\alpha }_{1}+{\alpha }_{2}\right)},$ ${e}_{±\left(2{\alpha }_{1}+{\alpha }_{2}\right)},$ and ${e}_{±\left(3{\alpha }_{1}+{\alpha }_{2}\right)}\text{.}$ Also, ${C}_{G}\left(s\right)$ contains ${x}_{±{\alpha }_{1}}\left(c\right)$ and ${x}_{±\left(3{\alpha }_{1}+2{\alpha }_{2}\right)}\left(c\right)$ for $c\in ℂ\text{.}$ In particular, ${C}_{G}\left(s\right)$ contains ${w}_{1}$ and ${w}_{3{\alpha }_{1}+2{\alpha }_{2}}={w}_{2}{w}_{1}{w}_{2}{w}_{1}{w}_{2}\text{.}$ Let $n\in {𝔤}_{q}^{{s}_{t}}\text{.}$ Note that the action of ${x}_{±{\alpha }_{1}}$ on ${𝔤}_{q}^{{s}_{t}}$ fixes the subspace spanned by the $\left\{{e}_{\alpha } \mid \alpha \in P\left(t\right)\right\}$ and the subspace spanned by $\left\{{e}_{-\alpha } \mid \alpha \in P\left(t\right)\right\}\text{.}$ Then Case 4 of Proposition 3.24 shows that $n$ is conjugate to an element with positive part 0, ${e}_{{\alpha }_{2}},$ ${e}_{{\alpha }_{1}+{\alpha }_{2}},$ or ${e}_{{\alpha }_{2}}+{e}_{3{\alpha }_{1}+{\alpha }_{2}}\text{.}$ Case 3a: ${e}_{{\alpha }_{2}}$ First, assume $n={e}_{{\alpha }_{2}}+w{e}_{-{\alpha }_{2}}+x{e}_{-{\alpha }_{1}-{\alpha }_{2}}+y{e}_{-2{\alpha }_{1}-{\alpha }_{2}}+z{e}_{-3{\alpha }_{1}-{\alpha }_{2}}\text{.}$ Then if $n$ is nilpotent, $w=0\text{.}$ However, if $x\ne 0,$ $x-3α1-2α2 (-y2x) x-α1 ( 3y2-4xz 4ax ) · ( eα2+x e-α1-α2 +ye-2α1-α2 +ze-3α1-α2 ) =eα2+x e-α1-α2,$ and $dα1(x)·eα2 +xe-α1-α2= eα2+e-α1-α2.$ If $x=0,$ then $dα1(y) x-3α1-2α2 (za)·eα2+ ye-2α1-α2 +ze-3α1-α2 =eα2+e-2α1-α2.$ Hence $n$ is ${C}_{G}\left(s\right)\text{-conjugate}$ to either ${e}_{{\alpha }_{2}},$ ${e}_{{\alpha }_{2}}+{e}_{-{\alpha }_{1}-{\alpha }_{2}},$ or ${e}_{{\alpha }_{2}}+{e}_{-2{\alpha }_{1}-{\alpha }_{2}}\text{.}$ Case 3b: ${e}_{{\alpha }_{1}+{\alpha }_{2}}$ If $n={e}_{{\alpha }_{1}+{\alpha }_{2}}+w{e}_{-{\alpha }_{2}}+x{e}_{-{\alpha }_{1}-{\alpha }_{2}}+y{e}_{-2{\alpha }_{1}-{\alpha }_{2}}+z{e}_{-3{\alpha }_{1}-{\alpha }_{2}},$ then for $n$ to be nilpotent, $x$ must equal zero and either $w$ or $z$ is zero as well. In either case, $x-3α1-2α2 (y)· eα1+α2+w e-α2+y e-2α1-α2+ ze-3α1-α2 =we-α2+z e-3α1-α2.$ However, $dα1(w-1) dα2(w)· eα1+α2+ we-α2= eα1+α2+ e-α2,$ and $dα1(z-1) dα2(z)· eα1+α2+z e-3α1-α2= eα1+α2+ e-3α1-α2.$ Hence $n$ is ${C}_{G}\left(s\right)\text{-conjugate}$ to either ${e}_{{\alpha }_{1}+{\alpha }_{2}},$ ${e}_{{\alpha }_{1}+{\alpha }_{2}}+{e}_{-{\alpha }_{2}},$ or ${e}_{{\alpha }_{1}+{\alpha }_{2}}+{e}_{-3{\alpha }_{1}-{\alpha }_{2}}\text{.}$ Case 3c: ${e}_{{\alpha }_{2}}+{e}_{3{\alpha }_{1}+{\alpha }_{2}}$ If $n={e}_{{\alpha }_{2}}+{e}_{3{\alpha }_{1}+{\alpha }_{2}}+w{e}_{-{\alpha }_{2}}+x{e}_{-{\alpha }_{1}-{\alpha }_{2}}+y{e}_{-2{\alpha }_{1}-{\alpha }_{2}}+z{e}_{-3{\alpha }_{1}-{\alpha }_{2}},$ then the minimal polynomial of $n$ in the basic representation of $𝔤$ is ${X}^{7}+2\left(z-w\right){X}^{5}+{\left(z-w\right)}^{2}{X}^{3}+4\left({y}^{3}-{x}^{3}-xyz+wxy\right)X\text{.}$ Then for $n$ to be nilpotent, $z=w$ and ${x}^{3}={y}^{3}\text{.}$ By conjugating by ${d}_{{\alpha }_{1}}\left({1}^{1/3}\right)$ if necessary, we can assume that $y=x\text{.}$ If $w=x=0,$ then $n={e}_{{\alpha }_{2}}+{e}_{3{\alpha }_{1}+{\alpha }_{2}}\text{.}$ Assume $w\ne 0$ and $x\ne 0\text{.}$ If $w\ne x,$ then $xα1(1/2)w1 xα1(1) x3α1+2α2 ( 2wx+w2-3x24x ) w3α1+2α2 x3α1+2α2 (-1/(x-w))·n = 1/2(x-w) e2α1+α2- 8xx-w e-3α1-α2.$ But then, $dα1(-14x) dα2(w-x8x) w3α1+2α2· ( 1/2(x-w) e2α1+α2- 8xx-w e-3α1-α2 ) = eα2+ e-α1-α2.$ If $w=x\ne 0,$ then $xα1(1/2)w1 xα1(1) w3α1+2α2 x3α1+2α2 (-1/(4x))·n = (8x)eα2- (1/2) e-α1-α2.$ But then, $dα1(-4x) dα2(1/(8x))· ( (8x)eα2- (1/2) e-α1-α2 ) =eα2+ e-α1-α2.$ Finally, assume $w=0$ and $x\ne 0\text{.}$ Then, we compute that $xα1(1/2)w1 xα1(1) x3α1+2α2 (-3x/4) w3α1+2α2 x3α1+2α2 (-1/x) · ( eα2+e3α1+α2 +x ( eα1+α2+ e2α1+α2 ) ) = xα1(1/2)w1 xα1(1) x3α1+2α2 (-3x/4) w3α1+2α2 · ( eα2+eα1+α2 +e2α1+α2+ e3α1+α2+x e-α1-α2+x e-2α1-α2 ) = xα1(1/2)w1 xα1(1) x3α1+2α2 (-3x/4) · ( e-3α1-α2+ e-2α1-α2+ e-α1-α2+ e-α2-x e2α1+α2-x eα1+α2 ) = xα1(1/2)w1 xα1(1)· ( e-3α1-α2+ e-2α1-α2+ e-α1-α2+ e-α2 +( 3x/4) e3α1+α2+ (-x/4)e2α1+α2 +(-x/4) eα1+α2+ (3x/4)eα2 ) = xα1(1/2)w1· ( e-3α1-α2+ 2e-2α1-α2+ de-α1-α2+ 8e-α2+ (x/2)eα1+α2 +(3x/4)eα2 ) = xα1(1/2)· ( e-α2-2 e-α1-α2+d e-2α1-α2- 8e-3α1-α2- (x/2) e2α1+α2+ (3x/4) e3α1+α2 ) = -(x/2) e2α1+α2- 8e-3α1-α2$ But then, $dα1(x/4) dα2(-32/x3) · ( -(x/2) e2α1+α2- 8e-3α1-α2 ) = e2α1+α2+ e-3α1-α2.$ Thus we have seen that nilpotent elements in ${𝔤}_{q}^{s}$ are ${C}_{G}\left(s\right)\text{-conjugate}$ to either ${e}_{{\alpha }_{2}},$ ${e}_{{\alpha }_{2}}+{e}_{-{\alpha }_{1}-{\alpha }_{2}},$ ${e}_{{\alpha }_{2}}+{e}_{-2{\alpha }_{1}-{\alpha }_{2}},$ ${e}_{{\alpha }_{1}+{\alpha }_{2}},$ ${e}_{{\alpha }_{1}+{\alpha }_{2}}+{e}_{-{\alpha }_{2}},$ ${e}_{{\alpha }_{1}+{\alpha }_{2}}+{e}_{-3{\alpha }_{1}-{\alpha }_{2}},$ ${e}_{{\alpha }_{2}}+{e}_{3{\alpha }_{1}+{\alpha }_{2}},$ or ${e}_{2{\alpha }_{1}+{\alpha }_{2}}+{e}_{-3{\alpha }_{1}-{\alpha }_{2}}\text{.}$ However, we note that $w1 w3α1+α2· ( eα1+α2+ e-α2 ) =eα2+ e-α1-α2, w3α1+α2· ( eα1+α2+ e-3α1-α2 ) =eα2+ e-2α1-α2,$ and $w3α1+α2· ( e-3α1-α2+ e2α1+α2 ) =eα2+ e-α1-α2.$ Then a set of orbit representatives of nilpotent elements in ${𝔤}_{q}^{s}$ is ${e}_{{\alpha }_{2}},$ ${e}_{{\alpha }_{2}}+{e}_{-{\alpha }_{1}-{\alpha }_{2}},$ ${e}_{{\alpha }_{2}}+{e}_{-2{\alpha }_{1}-{\alpha }_{2}},$ ${e}_{{\alpha }_{1}+{\alpha }_{2}},$ and ${e}_{{\alpha }_{2}}+{e}_{3{\alpha }_{1}+{\alpha }_{2}}\text{.}$ Of these elements, the only two that are conjugate under the action of $G$ are ${e}_{{\alpha }_{2}}+{e}_{3{\alpha }_{1}+{\alpha }_{2}}$ and ${e}_{{\alpha }_{2}}+{e}_{-2{\alpha }_{1}-{\alpha }_{2}},$ since all the others have distinct Jordan forms in the basic representations of $𝔤\text{.}$ However, ${e}_{{\alpha }_{2}}+{e}_{3{\alpha }_{1}+{\alpha }_{2}}$ and ${e}_{{\alpha }_{2}}+{e}_{-2{\alpha }_{1}-{\alpha }_{2}}$ are not conjugate since the corresponding varieties ${ℬ}_{s,n}$ are nonisomorphic (see section 3.5.4). Thus these elements are in distinct ${C}_{G}\left(s\right)\text{-orbits.}$ Case 4: ${t}_{-{1}^{1/3},1}$ If $t={t}_{-{1}^{1/3},1},$ then $P\left(t\right)=\left\{3{\alpha }_{1}+{\alpha }_{2},3{\alpha }_{1}+2{\alpha }_{2}\right\}\text{.}$ Also, ${C}_{G}\left(s\right)$ contains ${x}_{±{\alpha }_{2}}\left(c\right)$ for $c\in ℂ,$ and thus contains ${w}_{2}\left(1\right)\text{.}$ Let $n=a{e}_{3{\alpha }_{1}+{\alpha }_{2}}+b{e}_{-3{\alpha }_{1}-{\alpha }_{2}}+c{e}_{3{\alpha }_{1}+2{\alpha }_{2}}+d{e}_{-3{\alpha }_{1}-2{\alpha }_{2}}\text{.}$ Then $n$ is nilpotent exactly when $ab-cd=0,$ so we assume $ab-cd=0\text{.}$ If $a\ne 0,$ then $xα2(-c/a)·n =ae3α1+α2+ de-3α1-2α2.$ If $c\ne 0,$ then $w2x-α2(-a/c) ·n=c e3α1+α2+b e-3α1-2α2,$ so if either $a$ or $c$ is non-zero, we can assume $n=a{e}_{3{\alpha }_{1}+{\alpha }_{2}}+d{e}_{-3{\alpha }_{1}-2{\alpha }_{2}},$ with $a\ne 0\text{.}$ If $d=0$ then $dα2(a-1)· ae3α1+α2= e3α1+α2.$ If $d\ne 0$ then $dα2(a3/2d1/2) dα1(a-1/3)· ae3α1+α2+d e-3α1-2α2= e3α1+α2+ e-3α1-2α2.$ Then assume $a=c=0\text{.}$ If $b\ne 0,$ then $dα2(b)x-α2 (-d/b)·b e-3α1-α2+d e-3α1-2α2= e-3α1-α2.$ If $b=0$ but $d\ne 0,$ then $dα2(d)w2· de-3α1-2α2 =e-3α1-α2.$ Hence in either case $n$ is ${C}_{G}\left(s\right)\text{-orbit}$ of ${e}_{-3{\alpha }_{1}-{\alpha }_{2}}\text{.}$ Thus the ${C}_{G}\left(s\right)\text{-orbits}$ in ${𝔤}_{q}^{{s}_{t}}$ are represented by 0, ${e}_{3{\alpha }_{1}+{\alpha }_{2}},$ ${e}_{-3{\alpha }_{1}-{\alpha }_{2}},$ and ${e}_{3{\alpha }_{1}+{\alpha }_{2}}+{e}_{-3{\alpha }_{1}-2{\alpha }_{2}}\text{.}$ $\square$

Proposition 3.30. If $q=-1,$ then the following is a set of representatives of $\Lambda /G\text{.}$

$(st1,1,0) (st1,1,eα1) (st1,1,eα2) (st1,1,eα1+e3α1+2α2) (st1,1,eα1+eα2) (st1,-1,0) (st1,-1,eα1) (st1,-1,eα1+e3α1+2α2) (st11/3,1,0) (st11/3,1eα2) (st11/3,1,eα2+e3α1+α2) (st1,z,0), for z≠±1,q,q2 (st1,z,eα1), for z≠±1,q,q2 (stz,1,0), for z≠q2,-1,q2/3 (stz,1,eα2), for z≠q2,-1,q2/3 (stz,w,0)$

 Proof. For each $t,$ ${𝔤}_{q}^{{s}_{t}}$ is a Lie subalgebra of $𝔤,$ since for $x,y\in {𝔤}_{q}^{{s}_{t}},$ ${s}_{t}·\left[x,y\right]=\left[x,y\right]$ and thus $\left[x,y\right]\in {𝔤}_{q}^{{s}_{t}}\text{.}$ Also, ${C}_{G}\left(s\right)$ contains $\text{exp}\left(x\right)$ for all $x\in {𝔤}_{q}^{{s}_{t}}\text{.}$ Also, since the Lie algebra ${𝔤}_{q}^{{s}_{t}}$ is generated for each $t$ by a root subsystem of the root system of type ${G}_{2},$ it is isomorphic to some rank two semisimple Lie algebra. Thus the nilpotent orbits in ${𝔤}_{q}^{{s}_{t}}$ are the same as those in the corresponding Lie algebra. (See [CMc1993] for a more thorough discussion of these nilpotent orbits.) If $t={t}_{1,1},$ the corresponding Lie algebra has type ${G}_{2},$ and the nilpotent orbits in ${𝔤}_{q}^{s}$ are represented by 0, ${e}_{{\alpha }_{1}},$ ${e}_{{\alpha }_{2}},$ ${e}_{{\alpha }_{1}}+{e}_{{\alpha }_{2}},$ and ${e}_{{\alpha }_{1}}+{e}_{3{\alpha }_{1}+2{\alpha }_{2}}\text{.}$ If $t={t}_{1,-1},$ then the corresponding Lie algebra has type ${A}_{1}×{A}_{1},$ and the nilpotent orbits in ${𝔤}_{q}^{s}$ are represented by 0, ${e}_{{\alpha }_{1}},$ ${e}_{3{\alpha }_{1}+2{\alpha }_{2}},$ and ${e}_{{\alpha }_{1}}+{e}_{3{\alpha }_{1}+2{\alpha }_{2}}\text{.}$ If $t={t}_{1/3,1},$ the corresponding Lie Algebra has type ${A}_{2},$ and the nilpotent orbits in ${𝔤}_{q}^{s}$ are represented by 0, ${e}_{{\alpha }_{2}},$ ${e}_{3{\alpha }_{1}+{\alpha }_{2}},$ and ${e}_{{\alpha }_{2}}+{e}_{3{\alpha }_{1}+{\alpha }_{2}}\text{.}$ If $t={t}_{1,z}$ or ${t}_{z,1},$ the corresponding Lie algebra has type ${A}_{1},$ and the nilpotent orbits in ${𝔤}_{q}^{s}$ are represented by 0 and either ${e}_{{\alpha }_{1}}$ or ${e}_{{\alpha }_{2}},$ respectively. If $t={t}_{z,w},$ ${𝔤}_{q}^{{s}_{t}}=0\text{.}$ $\square$

Cosets

We can also use the relations on $G$ to compute the action of elements of $G$ on cosets in $G/B\text{.}$ Specifically, we want to determine which cosets $gB$ are fixed by $\text{exp}\left(n\right)$ for a nilpotent element $n$ of $𝔤,$ since these are in bijection with the Borel subalgebras containing $n\text{.}$

Case: $\text{exp} c{e}_{{\alpha }_{1}}={x}_{{\alpha }_{1}}\left(c\right):$

The action of ${x}_{{\alpha }_{1}}\left(c\right)$ on cosets starting with ${x}_{{\alpha }_{1}}\left(d\right)$ is straightforward to compute:

$xα1(c)· B = B xα1(c)· xα1(d)w1 B = xα1(c+d)w1 B xα1(c)· xα1(d)w1 xα2(e)w2 B = xα1(c+d)w1 xα2(e)w2 B xα1(c)· xα1(d)w1 xα2(e)w2 xα1(f)w1 B = xα1(c+d)w1 xα2(e)w2 xα1(f)w1 B xα1(c)· xα1(d)w1 xα2(e)w2 xα1(f)w1 xα2(g)w2 B = xα1(c+d)w1 xα2(e)w2 xα1(f)w1 xα2(g)w2 B xα1(c)· xα1(d)w1 xα2(e)w2 xα1(f)w1 xα2(g)w2 xα1(h)w1 B = xα1(c+d)w1 xα2(e)w2 xα1(f)w1 xα2(g)w2 xα1(h)w1 B xα1(c)· xα1(d)w1 xα2(e)w2 xα1(f)w1 xα2(g)w2 xα1(h)w1 xα2(j)w2 B = xα1(c+d)w1 xα2(e)w2 xα1(f)w1 xα2(g)w2 xα1(h)w1 xα2(j)w2 B$

The action of ${x}_{{\alpha }_{1}}\left(c\right)$ on cosets starting with ${x}_{{\alpha }_{2}}\left(d\right)$ requires extensive use of the relations given above.

$xα1(c)· xα2(d)w2 B = xα2(d) xα1(c) xα1+α2(cd) x2α1+α2(-c2d) x3α1+α2(c3d) x3α1+2α2(-2c3d2) w2B = xα2(d)w2 xα1+α2(-c) xα1(cd) x2α1+α2 (-c2d) x3α1+2α2 (c3d) x3α1+α2 (2c3d2)B =xα2(d) w2B xα1(c)· xα2(d)w2 xα1(e)w1B = xα2(d)w2 xα1+α2(-c) xα1(cd) x2α1+α2 (-c2d) x3α1+2α2 (c3d) x3α1+α2 (2c3d2) xα1(e)w1B = xα2(d)w2 xα1+α2(-c) xα1(cd) x2α1+α2 (-c2d) xα1(e) x3α1+2α2 (c3d) x3α1+α2 (2c3d2)w1B = xα2(d)w2 xα1+α2(-c) xα1(e+cd) x2α1+α2 (-c2d) x3α1+2α2 (c3d) · x3α1+α2 (2c3d2+3c2de) w1B = xα2(d)w2 xα1(e+cd) xα1+α2(-c) x2α1+α2 (2(c2d+ce)) x3α1+α2 (-3c(e+cd)2) · x3α1+2α2 (-3c2(e+cd)) x2α1+α2 (-c2d) x3α1+2α2 (c3s) x3α1+α2 (2c3d2+3c2de) w1B = xα2(d)w2 xα1(e+cd) xα1+α2 (-c) x2α1+α2 (c2d+2c) x3α1+α2 (-3ce2-3c2de-c3d2) x3α1+2α2 (-3c2e+2c3d) w1B =xα2(d)w2 xα1(e+cd) w1x2α1+α2 (c) xα1+α2 (c2d+2c) xα2 (3ce2+3c2de+c3d2) x3α1+2α2 (-3c2e+2c3d) B =xα2(d)w2 xα1(e+cd)w1B xα1(c)· xα2(d)w2 xα1(e)w1 xα2(f)w2B = xα2(d)w2 xα1(e+cd)w1 x2α1+α2(c) xα1+α2 (c2d+2ce) xα2 (3ce2+3c2de+c3s2) x3α1+2α2 (-3c2e+2c3d) xα2(f)w2B = xα2(d)w2 xα1(e+cd)w1 x2α1+α2(c) xα1+α2 (c2d+2ce) xα2 (3ce2+3c2de+c3s2+f) x3α1+2α2 (-3c2e+2c3d) w2B = xα2(d)w2 xα1(e+cd)w1 xα2 (3ce2+3c2de+c3s2+f) x2α1+α2(c) xα1+α2 (c2d+2ce) x3α1+2α2 (-3c2e+2c3d) w2B = xα2(d)w2 xα1(e+cd)w1 xα2 (3ce2+3c2de+c3s2+f) w2 x2α1+α2(c) xα1 (c2d+2ce) x3α1+α2 (3c2e-2c3d) B = xα2(d)w2 xα1(e+cd)w1 xα2 (3ce2+3c2de+c3s2+f) w2B xα1(c)· xα2(d)w2 xα1(e)w1 xα2(f)w2 xα1(g)w1 B = xα2(d)w2 xα1(e+cd)w1 xα2(3ce2+3c2de+c3d2+f)w2 x2α1+α2(c) xα1(c2d+2ce) x3α1+α2(3c2e-2c3d) xα1(g)w1 B = xα2(d)w2 xα1(e+cd)w1 xα2(3ce2+3c2de+c3d2+f)w2 x2α1+α2(c) xα1(c2d+2ce+g) x3α1+α2(3c2e-2c3d) w1 B = xα2(d)w2 xα1(e+cd)w1 xα2(3ce2+3c2de+c3d2+f)w2 xα1(c2d+2ce+g) x2α1+α2(c) x3α1+α2 (-3(c3d+2c2e+cg)) x3α1+α2 (3c2e-2c3d) w1B = xα2(d)w2 xα1(e+cd)w1 xα2(3ce2+3c2de+c3d2+f)w2 xα1(c2d+2ce+g) x2α1+α2(c) x3α1+α2 (-3c2e-5c3d-3cg) w1B = xα2(d)w2 xα1(e+cd)w1 xα2(3ce2+3c2de+c3d2+f)w2 xα1(c2d+2ce+g) w1 xα1+α2(c) xα2(3c2e+5c3d+3cg) B = xα2(d)w2 xα1(e+cd)w1 xα2(3ce2+3c2de+c3d2+f)w2 xα1(c2d+2ce+g) w1B xα1(c)· xα2(d)w2 xα1(e)w1 xα2(f)w2 xα1(g)w1 xα2(h)w2 B = xα2(d)w2 xα1(e+cd)w1 xα2 (3ce23c2de+c3d2+f) w2xα1 (c2d+2ce+g)w1 xα1+α2(c) xα2(3c2e+5c3d+3cg) xα2(h) w2B = xα2(d)w2 xα1(e+cd)w1 xα2 (3ce23c2de+c3d2+f) w2xα1 (c2d+2ce+g)w1 xα1+α2(c) xα2(3c2e+5c3d+3cg+h) w2B = xα2(d)w2 xα1(e+cd)w1 xα2 (3ce23c2de+c3d2+f) w2xα1 (c2d+2ce+g)w1 xα2(3c2e+5c3d+3cg+h) xα1+α2(c) w2B = xα2(d)w2 xα1(e+cd)w1 xα2 (3ce23c2de+c3d2+f) w2xα1 (c2d+2ce+g)w1 xα2(3c2e+5c3d+3cg+h) w2xα1(c)B = xα2(d)w2 xα1(e+cd)w1 xα2 (3ce23c2de+c3d2+f) w2xα1 (c2d+2ce+g)w1 xα2(3c2e+5c3d+3cg+h) w2B$

Case: $\text{exp} c{e}_{{\alpha }_{2}}={x}_{{\alpha }_{2}}\left(c\right):$

Cosets starting with ${x}_{{\alpha }_{2}}\left(d\right):$

$xα2(c)· xα2(d)w2 B = xα2(c+d)w2 B xα2(c)· xα2(d)w2 xα1(e)w1 B = xα2(c+d)w2 xα1(e)w1 B xα2(c)· xα2(d)w2 xα1(e)w1 xα2(f)w2 B = xα2(c+d)w2 xα1(e)w1 xα2(f)w2 B xα2(c)· xα2(d)w2 xα1(e)w1 xα2(f)w2 xα1(g)w1 B = xα2(c+d)w2 xα1(e)w1 xα2(f)w2 xα1(g)w1 B xα2(c)· xα2(d)w2 xα1(e)w1 xα2(f)w2 xα1(g)w1 xα2(h)w2 B = xα2(c+d)w2 xα1(e)w1 xα2(f)w2 xα1(g)w1 xα2(h)w2 B xα2(c)· xα2(d)w2 xα1(e)w1 xα2(f)w2 xα1(g)w1 xα2(h)w2 xα1(j)w1 B = xα2(c+d)w2 xα1(e)w1 xα2(f)w2 xα1(g)w1 xα2(h)w2 xα1(j)w1 B$

Cosets starting with ${x}_{{\alpha }_{1}}\left(d\right):$

$xα2(c)· xα1(d) w1B = xα1(d) xα2(c) x3α1+2α2 (2c2d3) x3α1+α2 (-cd3) x2α1+α2 (cd2) xα1+α2 (-cd) w1B = xα1(d)w1 x3α1+α2 (c) x3α1+2α2 (2c2d3) xα2(cd3) xα1+α2 (cd2) x2α1+α2 (cd)B = xα1(d)w1B xα2(c)· xα1(d)w1 w2B = xα1(d) xα2(c) x3α1+2α2 (2c2d3) x3α1+α2 (-cd3) x2α1+α2 (cd2) xα1+α2 (-cd)w1 xα2(e)w2B = xα1(d)w1 x3α1+α2 (c) x3α1+2α2 (2c2d3) xα2(cd3) xα1+α2 (cd2) x2α1+α2 (cd)xα2 (e)w2B = xα1(d)w1 x3α1+α2 (c) x3α1+2α2 (2c2d3) xα2(cd3+e) xα1+α2 (cd2) x2α1+α2 (cd)xα2 w2B = xα1(d)w1 xα2(cd3+e) x3α1+α2(c) x3α1+2α2 (-c2d3-ce) x3α1+2α2 (2c2d3) xα1+α2 (cd2) x2α1+α2 (cd)w2B = xα1(d)w1 xα2(cd3+e)w2 x3α1+2α2(c) x3α1+α2 (c-c2d3) xα1(cd2) x2α1+α2 (cd)B = xα1(d)w1 xα2(cd3+e)w2B xα2(c)· xα1(d)w1 xα2(e)w2 xα1(f)w1 B = xα1(d)w1 xα2(cd3+e) w2x3α1+2α2 (c) x3α1+α2 (ce-c2d3) xα1(cd2) x2α1+α2 (cd)xα1(f) w1B = xα1(d)w1 xα2(cd3+e) w2x3α1+2α2 (c) x3α1+α2 (ce-c2d3) xα1(cd2+g) x2α1+α2(cd) x3α1+α2 (-3cdf)w1B =xα1(d)w1 xα2(cd3+e) w2xα1 (cd2+f)w1 x3α1+2α2 (c)xα2 (c2d3-ce) xα1+α2 (cd)xα2 (3cdf)B =xα1(d)w1 xα2(cd3+e) w2xα1 (cd2+f)w1B xα2(c)· xα1(d)w1 xα2(e)w2 xα1(f)w1 xα2(g)w2 B xα1(d)w1 xα2(cd3+e) w2xα1 (cd2+f)w1 x3α1+2α2 (c)xα2 (c2d3-ce) xα1+α2 (cd)xα2 (3cdf)xα2 (g)w2B xα1(d)w1 xα2(cd3+e) w2xα1 (cd2+f)w1 xα2 (c2d3-ce+3cdf+g) x3α1+2α2 (cxα1+α2 (cd))w2B xα1(d)w1 xα2(cd3+e) w2xα1 (cd2+f)w1 xα2 (c2d3-ce+3cdf+g) w2 x3α1+α2 (-c)xα1 (cd)B xα1(d)w1 xα2(cd3+e) w2xα1 (cd2+f)w1 xα2 (c2d3-ce+3cdf+g) w2B xα2(c)· xα1(d)w1 xα2(e)w2 xα1(f)w1 xα2(g)w2 xα1(h)w1 B = xα1(d)w1 xα2(cd3+e) w2xα1 (cd2+f)w1 xα2 (c2d3-ce+3cdf+g) w2 x3α1+α2 (-c)xα1 (cd)xα1(h) w1B = xα1(d)w1 xα2(cd3+e) w2xα1 (cd2+f)w1 xα2 (c2d3-ce+3cdf+g) w2 xα1(cd+h) x3α1+α2 (-c)w1B = xα1(d)w1 xα2(cd3+e) w2xα1 (cd2+f)w1 xα2 (c2d3-ce+3cdf+g) w2 xα1(cd+h) w1xα2(c)B = xα1(d)w1 xα2(cd3+e) w2xα1 (cd2+f)w1 xα2 (c2d3-ce+3cdf+g) w2 xα1(cd+h) w1B$

Case: $\text{exp} c{e}_{{\alpha }_{1}+{\alpha }_{2}}={x}_{{\alpha }_{1}+{\alpha }_{2}}\left(c\right):$

Coset starting with ${x}_{{\alpha }_{1}}\left(d\right):$

$xα1+α2(c) ·xα1(d)w1B = xα1(d) xα1+α2(c) x3α1+2α2 (-3c2d) x3α1+α2 (3cd2) x2α1+α2 (-2cd)w1B =xα1(d)w1 x2α1+α2 (-c) x3α1+2α2 (-3c2d)xα2 (-3cd2) xα1+α2 (-2cd)B = xα1(d)w1B xα1+α2(c) ·xα1(d)w1 xα2(e)e2B =xα1(d)w1 x2α1+α2 (-c) x3α1+2α2 (-3c2d)xα2 (-3cd2) xα1+α2 (-2cd)xα2 (e)w2B =xα1(d)w1 x2α1+α2 (-c) x3α1+2α2 (-3c2d)xα2 (e-3cd2) xα1+α2 (-2cd)w2B = xα1(d)w1 xα2(e-3cd2) x2α1+α2 (-c) x3α1+2α2 (-3c2d) xα1+α2 (-2cd)w2B = xα1(d)w1 xα2(e-3cd2) w2x2α1+α2 (-c) x3α1+α2 (3c2d)xα1 (-2cd)B = xα1(d)w1 xα2(e-3cd2) w2B xα1+α2(c) ·xα1(d)w1 xα2(e)w2 xα1(f)w1B = xα1(d)w1 xα2(e-3cd2) w2x2α1+α2 (-c) x3α1+α2 (3c2d) xα1(-2cd) xα1(f)w1B = xα1(d)w1 xα2(e-3cd2) w2x2α1+α2 (-c) x3α1+α2 (3c2d) xα1(f-2cd) w1B = xα1(d)w1 xα2(e-3cd2) w2xα1 (f-2cd) x2α1+α2 (-c) ·x3α1+α2 (3c(f-2cd)) x3α1+α2 (3c2d)w1B = xα1(d)w1 xα2(e-3cd2) w2xα1 (f-2cd)w1 x2α1+α2 (-c)xα2 (3c2d-3cf)B = xα1(d)w1 xα2(e-3cd2) w2xα1 (f-2cd)w1B xα1+α2(c) ·xα1(d)w1 xα2(e)w2 xα1(f)w1 xα2(g)w2B = xα1(d)w1 xα2(e-3cd2)w2 xα1(f-2cd)w1 xα1+α2(-c) xα2(3c2d-3cf) xα2(g)w2B = xα1(d)w1 xα2(e-3cd2)w2 xα1(f-2cd)w1 xα2(3c2d-3cf+g) xα1+α2(-c)w2B = xα1(d)w1 xα2(e-3cd2)w2 xα1(f-2cd)w1 xα2(3c2d-3cf+g) w2xα1(-c)B = xα1(d)w1 xα2(e-3cd2)w2 xα1(f-2cd)w1 xα2(3c2d-3cf+g) w2B xα1+α2(c) ·xα1(d)w1 xα2(e)w2 xα1(f)w1 xα2(g)w2 xα1(h)w1B =xα1(d)w1 xα2(e-3cd2) w2xα1 (f-2cd)w1 xα2 (3c2d-3cf+g)w2 xα1(-c) xα1(h)w1B =xα1(d)w1 xα2(e-3cd2) w2xα1 (f-2cd)w1 xα2 (3c2d-3cf+g)w2 xα1(h-c)w1B xα1+α2(c) · xα1(d)w1 xα2(e)w2 xα1(f)w1 xα2(g)w2 xα1(h)w1 xα2(j)w2B =xα1(d)w1 xα2(e-3cd2) w1xα1 (f-2cd)w1 xα2 (3c2d-3cf+g)w2 xα1(h-c)w1 xα2(j)B$

Cosets starting with ${x}_{{\alpha }_{2}}\left(d\right):$

$xα1+α2(c) ·xα2(d)w2B = xα2(d)w2 xα1(c)B = xα2(d)w2B xα1+α2(c) · xα2(d)w2 xα1(e)w1 B = xα2(d)w2 xα1(c) xα1(e)w1 B = xα2(d)w2 xα1(c+e) w1B xα1+α2(c) · xα2(d)w2 xα1(e)w1 xα2(f)w2 B = xα2(d)w2 xα1(c+e) w1xα2(f) w2B xα1+α2(c) · xα2(d)w2 xα1(e)w1 xα2(f)w2 xα1(g)w1 B = xα2(d)w2 xα1(c+e)w1 xα2(f)w2 xα1(g)w1B xα1+α2(c) · xα2(d)w2 xα1(e)w1 xα2(f)w2 xα1(g)w1 xα2(h)w2B = xα2(d)w2 xα1(c+e)w1 xα2(f)w2 xα1(g)w1 xα2(h)w2B$

Case: $\text{exp} c{e}_{2{\alpha }_{1}+{\alpha }_{2}}={x}_{2{\alpha }_{1}+{\alpha }_{2}}\left(c\right):$

Cosets starting with ${x}_{{\alpha }_{1}}\left(d\right):$

$x2α1+α2(c) · xα1(d)w1 B = xα1(d) x2α1+α2(c) x3α1+α2(-3cd) w1 B = xα1(d)w1 x2α1+α2(c) xα2(3cd) B = xα1(d)w1 B x2α1+α2(c) · xα1(d)w1 xα2(e)w2 B = xα1(d)w1 xα1+α2(c) xα2(3cd+e) w2 B = xα1(d)w1 xα2(3cd+e)w2 xα1(c) B = xα1(d)w1 xα2(3cd+e)w2 B x2α1+α2(c) · xα1(d)w1 xα2(e)w2 xα1(f)w1 B = xα1(d)w1 xα2(3cd+e)w2 xα1(c+f)w1 B x2α1+α2(c) · xα1(d)w1 xα2(e)w2 xα1(f)w1 xα2(g)w2B = xα1(d)w1 xα2(3cd+e)w2 xα1(c+f)w1 xα2(g)w2B x2α1+α2(c) · xα1(d)w1 xα2(e)w2 xα1(f)w1 xα2(g)w2 xα1(h)w1B = xα1(d)w1 xα2(3cd+e)w2 xα1(c+f)w1 xα2(g)w2 xα1(h)w1B x2α1+α2(c) · xα1(d)w1 xα2(e)w2 xα1(f)w1 xα2(g)w2 xα1(h)w1 xα2(j)w2B = xα1(d)w1 xα2(3cd+e)w2 xα1(c+f)w1 xα2(g)w2 xα1(h)w1 xα2(j)w2B$

Cosets starting with ${x}_{{\alpha }_{2}}\left(d\right):$

$x2α1+α2(c) · xα2(d)w2 B = xα2(d)w2 x2α1+α2(c) B x2α1+α2(c) · xα2(d)w2 xα1(e)w1 B = xα2(d)w2 xα1(e)w1 B x2α1+α2(c) · xα2(d)w2 xα1(e)w1 xα2(f)w2 B = xα2(d)w2 x2α1+α2(c) xα1(e)w1 xα2(f)w2 B = xα2(d)w2 xα1(e)w1 xα2(3ce+f)w2 B x2α1+α2(c) · xα2(d)w2 xα1(e)w1 xα2(f)w2 xα1(g)w1 B = xα2(d)w2 x2α1+α2(c) xα1(e)w1 xα2(f)w2 xα1(g)w1 B = xα2(d)w2 xα1(e)w1 xα2(3ce+f)w2 xα1(c+g)w1 B x2α1+α2(c) · xα2(d)w2 xα1(e)w1 xα2(f)w2 xα1(g)w1 xα2(h)w2 B = xα2(d)w2 x2α1+α2(c) xα1(e)w1 xα2(f)w2 xα1(g)w1 xα2(h)w2 B = xα2(d)w2 xα1(e)w1 xα2(3ce+f)w2 xα1(c+g)w1 xα2(h)w2 B$

Case: $\text{exp} {e}_{3{\alpha }_{1}+{\alpha }_{2}}={x}_{3{\alpha }_{1}+{\alpha }_{2}}:$

Cosets starting with ${x}_{{\alpha }_{1}}\left(d\right):$

$x3α1+α2(c)· xα1(d)w1 B = xα1(d)w1 xα2(-c) B = xα1(d)w1 B x3α1+α2(c)· xα1(d)w1 xα2(e)w2 B = xα1(d)w1 xα2(-c) xα2(e)w2 B = xα1(d)w1 xα2(e-c)w2 B x3α1+α2(c)· xα1(d)w1 xα2(e)w2 xα1(f)w1 B = xα1(d)w1 xα2(-c) xα2(e)w2 xα1(f)w1 B = xα1(d)w1 xα2(e-c)w2 xα1(f)w1 B x3α1+α2(c)· xα1(d)w1 xα2(e)w2 xα1(f)w1 xα2(g)w2 B = xα1(d)w1 xα2(-c) xα2(e)w2 xα1(f)w1 xα2(g)w2 B = xα1(d)w1 xα2(e-c)w2 xα1(f)w1 xα2(g)w2 B x3α1+α2(c)· xα1(d)w1 xα2(e)w2 xα1(f)w1 xα2(g)w2 xα1(h)w1 B = xα1(d)w1 xα2(-c) xα2(e)w2 xα1(f)w1 xα2(g)w2 xα1(h)w1 B = xα1(d)w1 xα2(e-c)w2 xα1(f)w1 xα2(g)w2 xα1(h)w1 B x3α1+α2(c)· xα1(d)w1 xα2(e)w2 xα1(f)w1 xα2(g)w2 xα1(h)w1 xα2(j)w2 B = xα1(d)w1 xα2(-c) xα2(e)w2 xα1(f)w1 xα2(g)w2 xα1(h)w1 xα2(j)w2 B = xα1(d)w1 xα2(e-c)w2 xα1(f)w1 xα2(g)w2 xα1(h)w1 xα2(j)w2 B$

Cosets starting with ${x}_{{\alpha }_{2}}\left(d\right):$

$x3α1+α2(c) · xα2(d)w2 B = xα2(d) x3α1+α2(c) x3α1+2α2(-cd) w2 B = xα2(d)w2 x3α1+2α2(c) x3α1+α2(cd) B = xα2(d)w2 B x3α1+α2(c) · xα2(d)w2 xα1(e)w1 B = xα2(d)w2 x3α1+2α2(c) x3α1+α2(cd) xα1(e)w1 B = xα2(d)w2 xα1(e)w1 x3α1+2α2(c) xα2(-cd) B = xα2(d)w2 xα1(e)w1 B x3α1+α2(c) · xα2(d)w2 xα1(e)w1 xα2(f)w2 B = xα2(d)w2 xα1(e)w1 x3α1+2α2(c) xα2(-cd) xα2(f)w2 B = xα2(d)w2 xα1(e)w1 xα2(f-cd)w2 x3α1+α2(-c) B = xα2(d)w2 xα1(e)w1 xα2(f-cd)w2 B x3α1+α2(c) · xα2(d)w2 xα1(e)w1 xα2(f)w2 xα1(g)w1 B = xα2(d)w2 xα1(e)w1 xα2(f-cd)w2 x3α1+α2(-c) xα1(g)w1 B = xα2(d)w2 xα1(e)w1 xα2(f-cd)w2 xα1(g)w1 xα2(c) B = xα2(d)w2 xα1(e)w1 xα2(f-cd)w2 xα1(g)w1 B x3α1+α2(c) · xα2(d)w2 xα1(e)w1 xα2(f)w2 xα1(g)w1 xα2(h)w2 B = xα2(d)w2 xα1(e)w1 xα2(f-cd)w2 xα1(g)w1 xα2(c) xα2(h)w2 B = xα2(d)w2 xα1(e)w1 xα2(f-cd)w2 xα1(g)w1 xα2(c+h)w2 B$

Case: $\text{exp} c{e}_{3{\alpha }_{1}+2{\alpha }_{2}}={e}_{3{\alpha }_{1}+2{\alpha }_{2}}:$

Cosets starting with ${x}_{{\alpha }_{2}}\left(d\right):$

$x3α1+2α2(c) · xα2(d)w2 B = xα2(d)w2 x3α1+α2(-c) B = xα2(d)w2 B x3α1+2α2(c) · xα2(d)w2 xα1(e)w1 B = xα2(d)w2 x3α1+α2(-c) xα1(e)w1 B = xα2(d)w2 xα1(e)w1 B x3α1+2α2(c) · xα2(d)w2 xα1(e)w1 xα2(f)w2 B = xα2(d)w2 x3α1+α2(-c) xα1(e)w1 xα2(f)w2 B = xα2(d)w2 xα1(e)w1 xα2(f+c)w2 B x3α1+2α2(c) · xα2(d)w2 xα1(e)w1 xα2(f)w2 xα1(g)w1 B = xα2(d)w2 x3α1+α2(-c) xα1(e)w1 xα2(f)w2 xα1(g)w1 B = xα2(d)w2 xα1(e)w1 xα2(f+c)w2 xα1(g)w1 B x3α1+2α2(c) · xα2(d)w2 xα1(e)w1 xα2(f)w2 xα1(g)w1 xα2(h)w2 B = xα2(d)w2 x3α1+α2(-c) xα1(e)w1 xα2(f)w2 xα1(g)w1 xα2(h)w2 B = xα2(d)w2 xα1(e)w1 xα2(f+c)w2 xα1(g)w1 xα2(h)w2 B x3α1+2α2(c) · xα2(d)w2 xα1(e)w1 xα2(f)w2 xα1(g)w1 xα2(h)w2 xα1(j)w1 B = xα2(d)w2 x3α1+α2(-c) xα1(e)w1 xα2(f)w2 xα1(g)w1 xα2(h)w2 xα1(j)w1 B = xα2(d)w2 xα1(e)w1 xα2(f+c)w2 xα1(g)w1 xα2(h)w2 xα1(j)w1 B$

Cosets starting with ${x}_{{\alpha }_{1}}\left(d\right):$

$x3α1+2α2(c) · xα1(d)w1 B = xα1(d)w1 x3α1+2α2(c) B = xα1(d)w1 B x3α1+2α2(c) · xα1(d)w1 xα2(e)w2 B = xα1(d)w1 x3α1+2α2(c) xα2(e)w2 B = xα1(d)w1 xα2(e)w2 B x3α1+2α2(c) · xα1(d)w1 xα2(e)w2 xα1(f)w1 B = xα1(d)w1 x3α1+2α2(c) xα2(e)w2 xα1(f)w1 B = xα1(d)w1 xα2(e)w2 xα1(f)w1 B x3α1+2α2(c) · xα1(d)w1 xα2(e)w2 xα1(f)w1 xα2(g)w2 B = xα1(d)w1 x3α1+2α2(c) xα2(e)w2 xα1(f)w1 xα2(g)w2 B = xα1(d)w1 xα2(e)w2 xα1(f)w1 xα2(g+c)w2 B x3α1+2α2(c) · xα1(d)w1 xα2(e)w2 xα1(f)w1 xα2(g)w2 xα1(h)w1 B = xα1(d)w1 x3α1+2α2(c) xα2(e)w2 xα1(f)w1 xα2(g)w2 xα1(h)w1 B = xα1(d)w1 xα2(e)w2 xα1(f)w1 xα2(g+c)w2 xα1(h)w1 B$

The Varieties ${ℬ}_{s}$

The varieties ${ℬ}_{s}$ of Borel subgroups containing a fixed semisimple element $s$ can be determined using Theorem 3.8.

Generic $q$

If $t={t}_{1,1},$ then ${ℬ}_{s}=ℬ\text{.}$

If $t={t}_{1,-1},$ then ${s}_{t}={H}_{{\alpha }_{1}}\left(-1\right){H}_{{\alpha }_{2}}\left(1\right)\text{.}$ Then ${ℬ}_{s}$ contains $B,$ ${x}_{{\alpha }_{1}}\left(c\right){w}_{1}B,$ ${x}_{{\alpha }_{1}}\left(c\right){w}_{1}{w}_{2}B,$ ${x}_{{\alpha }_{1}}\left(c\right){w}_{1}{w}_{2}{w}_{1}B,$ ${x}_{{\alpha }_{1}}\left(c\right){w}_{1}{w}_{2}{w}_{1}{x}_{{\alpha }_{2}}\left(d\right){w}_{2}B,$ ${x}_{{\alpha }_{1}}\left(c\right){w}_{1}{w}_{2}{w}_{1}{x}_{{\alpha }_{2}}\left(d\right){w}_{2}{w}_{1}B,$ ${x}_{{\alpha }_{1}}\left(c\right){w}_{1}{w}_{2}{w}_{1}{x}_{{\alpha }_{2}}\left(d\right){w}_{2}{w}_{1}{w}_{2}B,$ ${w}_{2}B,$ ${w}_{2}{w}_{1}B,$ ${w}_{2}{w}_{1}{x}_{{\alpha }_{2}}\left(c\right){w}_{2}B,$ ${w}_{2}{w}_{1}{x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}B,$ and ${w}_{2}{w}_{1}{x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}{w}_{2}B\text{.}$

If $t={t}_{{1}^{1/3},1},$ then ${s}_{t}={H}_{{\alpha }_{1}}\left({1}^{2/3}\right){H}_{{\alpha }_{2}}\left(1\right)\text{.}$ Then ${ℬ}_{s}$ contains $B,$ ${w}_{1}B,$ ${w}_{1}{x}_{{\alpha }_{2}}\left(c\right){w}_{2}B,$ ${w}_{1}{x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}B,$ ${w}_{1}{x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}{x}_{{\alpha }_{2}}\left(d\right){w}_{2}B,$ ${w}_{1}{x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}{x}_{{\alpha }_{2}}\left(d\right){w}_{2}{w}_{1}B,$ ${w}_{1}{x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}{x}_{{\alpha }_{2}}\left(d\right){w}_{2}{w}_{1}{x}_{{\alpha }_{2}}\left(e\right){w}_{2}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}{x}_{{\alpha }_{2}}\left(d\right){w}_{2}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}{x}_{{\alpha }_{2}}\left(d\right){w}_{2}{w}_{1}B,$ and ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}{x}_{{\alpha }_{2}}\left(d\right){w}_{2}{w}_{1}{x}_{{\alpha }_{2}}\left(e\right){w}_{2}B\text{.}$

If $t={t}_{1,{q}^{2}},$ then ${s}_{t}={H}_{{\alpha }_{1}}\left({q}^{2}\right){H}_{{\alpha }_{2}}\left({q}^{4}\right)\text{.}$ Then ${ℬ}_{s}$ contains $B,$ ${x}_{{\alpha }_{1}}\left(c\right){w}_{1}B,$ ${x}_{{\alpha }_{1}}\left(c\right){w}_{1}{w}_{2}B,$ ${x}_{{\alpha }_{1}}\left(c\right){w}_{1}{w}_{2}{w}_{1}B,$ ${x}_{{\alpha }_{1}}\left(c\right){w}_{1}{w}_{2}{w}_{1}{w}_{2}B,$ ${x}_{{\alpha }_{1}}\left(c\right){w}_{1}{w}_{2}{w}_{1}{w}_{2}{w}_{1}B,$ ${x}_{{\alpha }_{1}}\left(c\right){w}_{1}{w}_{2}{w}_{1}{w}_{2}{w}_{1}B,$ ${w}_{2}B,$ ${w}_{2}{w}_{1},$ ${w}_{2}{w}_{1}{w}_{2}B,$ ${w}_{2}{w}_{1}{w}_{2}{w}_{1}B,$ and ${w}_{2}{w}_{1}{w}_{2}{w}_{1}{w}_{2}B\text{.}$

If $t={t}_{1,±q},$ then ${s}_{t}={H}_{{\alpha }_{1}}\left(±1\right){H}_{{\alpha }_{2}}\left({q}^{2}\right)\text{.}$ Then ${ℬ}_{s}$ contains $B,$ ${x}_{{\alpha }_{1}}\left(c\right){w}_{1}B,$ ${x}_{{\alpha }_{1}}\left(c\right){w}_{1}{w}_{2}B,$ ${x}_{{\alpha }_{1}}\left(c\right){w}_{1}{w}_{2}{w}_{1}B,$ ${x}_{{\alpha }_{1}}\left(c\right){w}_{1}{w}_{2}{w}_{1}{w}_{2}B,$ ${x}_{{\alpha }_{1}}\left(c\right){w}_{1}{w}_{2}{w}_{1}{w}_{2}{w}_{1}B,$ ${x}_{{\alpha }_{1}}\left(c\right){w}_{1}{w}_{2}{w}_{1}{w}_{2}{w}_{1}{w}_{2}B,$ ${w}_{2}B,$ ${w}_{2}{w}_{1},$ ${w}_{2}{w}_{1}{w}_{2}B,$ ${w}_{2}{w}_{1}{w}_{2}{w}_{1}B,$ and ${w}_{2}{w}_{1}{w}_{2}{w}_{1}{w}_{2}B\text{.}$

If $t={t}_{1,z},$ then ${s}_{t}={H}_{{\alpha }_{1}}\left(z\right){H}_{{\alpha }_{2}}\left({z}^{2}\right)\text{.}$ Then ${ℬ}_{s}$ contains $B,$ ${x}_{{\alpha }_{1}}\left(c\right){w}_{1}B,$ ${x}_{{\alpha }_{1}}\left(c\right){w}_{1}{w}_{2}B,$ ${x}_{{\alpha }_{1}}\left(c\right){w}_{1}{w}_{2}{w}_{1}B,$ ${x}_{{\alpha }_{1}}\left(c\right){w}_{1}{w}_{2}{w}_{1}{w}_{2}B,$ ${x}_{{\alpha }_{1}}\left(c\right){w}_{1}{w}_{2}{w}_{1}{w}_{2}{w}_{1}B,$ ${x}_{{\alpha }_{1}}\left(c\right){w}_{1}{w}_{2}{w}_{1}{w}_{2}{w}_{1}{w}_{2}B,$ ${w}_{2}B,$ ${w}_{2}{w}_{1},$ ${w}_{2}{w}_{1}{w}_{2}B,$ ${w}_{2}{w}_{1}{w}_{2}{w}_{1}B,$ and ${w}_{2}{w}_{1}{w}_{2}{w}_{1}{w}_{2}B\text{.}$

If $t={t}_{{q}^{2},1},$ then ${s}_{t}={H}_{{\alpha }_{1}}\left({q}^{4}\right){H}_{{\alpha }_{2}}\left({q}^{6}\right)\text{.}$ Then ${ℬ}_{s}$ contains $B,$ ${w}_{1}B,$ ${w}_{1}{w}_{2}B,$ ${w}_{1}{w}_{2}{w}_{1}B,$ ${w}_{1}{w}_{2}{w}_{1}{w}_{2}B,$ ${w}_{1}{w}_{2}{w}_{1}{w}_{2}{w}_{1}B,$ ${w}_{1}{w}_{2}{w}_{1}{w}_{2}{w}_{1}{x}_{{\alpha }_{2}}\left(c\right){w}_{2}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}{w}_{2}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}{w}_{2}{w}_{1}B,$ and ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}{w}_{2}{w}_{1}{w}_{2}B\text{.}$

If $t={t}_{±q,1},$ then ${s}_{t}={H}_{{\alpha }_{1}}\left({q}^{2}\right){H}_{{\alpha }_{2}}\left(-{q}^{3}\right)\text{.}$ Then ${ℬ}_{s}$ contains $B,$ ${w}_{1}B,$ ${w}_{1}{w}_{2}B,$ ${w}_{1}{w}_{2}{w}_{1}B,$ ${w}_{1}{w}_{2}{w}_{1}{w}_{2}B,$ ${w}_{1}{w}_{2}{w}_{1}{w}_{2}{w}_{1}B,$ ${w}_{1}{w}_{2}{w}_{1}{w}_{2}{w}_{1}{x}_{{\alpha }_{2}}\left(c\right){w}_{2}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}{w}_{2}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}{w}_{2}{w}_{1}B,$ and ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}{w}_{2}{w}_{1}{w}_{2}B\text{.}$

If $t={t}_{{q}^{2/3},1},$ then ${s}_{t}={H}_{{\alpha }_{1}}\left({q}^{4/3}\right){H}_{{\alpha }_{2}}\left({q}^{2}\right)\text{.}$ Then ${ℬ}_{s}$ contains $B,$ ${w}_{1}B,$ ${w}_{1}{w}_{2}B,$ ${w}_{1}{w}_{2}{w}_{1}B,$ ${w}_{1}{w}_{2}{w}_{1}{w}_{2}B,$ ${w}_{1}{w}_{2}{w}_{1}{w}_{2}{w}_{1}B,$ ${w}_{1}{w}_{2}{w}_{1}{w}_{2}{w}_{1}{\alpha }_{2}\left(c\right){w}_{2}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}{w}_{2}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}{w}_{2}{w}_{1}B,$ and ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}{w}_{2}{w}_{1}{w}_{2}B\text{.}$

If $t={t}_{z,1},$ then ${s}_{t}={H}_{{\alpha }_{1}}\left({z}^{2}\right){H}_{{\alpha }_{2}}\left({z}^{3}\right)\text{.}$ Then ${ℬ}_{s}$ contains $B,$ ${w}_{1}B,$ ${w}_{1}{w}_{2}B,$ ${w}_{1}{w}_{2}{w}_{1}B,$ ${w}_{1}{w}_{2}{w}_{1}{w}_{2}B,$ ${w}_{1}{w}_{2}{w}_{1}{w}_{2}{w}_{1}B,$ ${w}_{1}{w}_{2}{w}_{1}{w}_{2}{w}_{1}{x}_{{\alpha }_{2}}\left(c\right){w}_{2}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}{w}_{2}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}{w}_{2}{w}_{1}B,$ and ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}{w}_{2}{w}_{1}{w}_{2}B\text{.}$

If $t={t}_{{q}^{2},-{q}^{-2}},$ ${t}_{{1}^{1/3},{q}^{2}},$ ${t}_{{q}^{2},{q}^{2}},$ ${t}_{{q}^{2},z},$ ${t}_{z,{q}^{2}},$ or ${t}_{z,w}$ for generic $z,w,$ then $Z\left(t\right)=\varnothing \text{.}$ In that case, ${ℬ}_{s}$ consists of $wB$ for $w\in {W}_{0}\text{.}$

For specific values of $q,$ we note that the sets ${ℬ}_{s}$ change only if $Z\left(t\right)$ differs from the generic case.

${q}^{12}=1$

None of the varieties ${ℬ}_{s}$ are different when ${q}^{2}$ is a primitive sixth root of unity.

${q}^{10}=1$

When ${q}^{2}$ is a primitive fifth root of unity, ${t}_{{q}^{2},{q}^{2}}$ and ${t}_{{q}^{-4},1}$ are in the same orbit. The central characters ${t}_{±q,1}$ are the same as ${t}_{±{q}^{-4},1}\text{.}$

For $s={s}_{{t}_{±{q}^{-4},1}},$ ${ℬ}_{s}$ contains $B,$ ${w}_{1}B,$ ${w}_{1}{w}_{2}B,$ ${w}_{1}{w}_{2}{w}_{1}B,$ ${w}_{1}{w}_{2}{w}_{1}{w}_{2}B,$ ${w}_{1}{w}_{2}{w}_{1}{w}_{2}{w}_{1}B,$ ${w}_{1}{w}_{2}{w}_{1}{w}_{2}{w}_{1}{x}_{{\alpha }_{2}}\left(c\right){w}_{2}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}{w}_{2}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}{w}_{2}{w}_{1}B,$ and ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}{w}_{2}{w}_{1}{w}_{2}B\text{.}$

For ease of calculation later, it is also useful to note that for $s={s}_{{t}_{{q}^{2},{q}^{2}}},$ ${ℬ}_{s}$ consists of $B,$ ${w}_{1}B,$ ${w}_{2}{w}_{1}B,$ ${w}_{1}{w}_{2}{w}_{1}B,$ ${w}_{2}{w}_{1}{x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}B,$ ${w}_{1}{w}_{2}{w}_{1}{x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}B,$ ${w}_{2}B,$ ${w}_{1}{w}_{2}B,$ ${w}_{2}{w}_{1}{x}_{{\alpha }_{2}}\left(c\right){w}_{2}B,$ ${w}_{1}{w}_{2}{w}_{1}{x}_{{\alpha }_{2}}\left(c\right){w}_{2}B,$ ${w}_{2}{w}_{1}{x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}{w}_{2}B,$ and ${w}_{2}{w}_{1}{x}_{{\alpha }_{1}}\left(c\right){w}_{2}{w}_{1}{w}_{2}{w}_{1}B\text{.}$

${q}^{8}=1$

If ${q}^{2}$ is a primitive fourth root of unity, ${t}_{{q}^{2},-{q}^{-2}}={t}_{{q}^{2},{q}^{2}},$ and both of these central characters are in the same orbits as ${t}_{{q}^{2},1}\text{.}$

For $s={s}_{{t}_{{q}^{2},{q}^{2}}},$ ${ℬ}_{s}$ contains $B,$ ${w}_{1}B,$ ${w}_{1}{x}_{{\alpha }_{2}}\left(c\right){w}_{2}B,$ ${w}_{1}{x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}B,$ ${w}_{1}{x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}{w}_{2}B,$ ${w}_{1}{x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}{w}_{2}{w}_{1}B,$ ${w}_{1}{x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}{w}_{2}{w}_{1}{w}_{2}B,$ ${w}_{2}B,$ ${w}_{2}{w}_{1}B,$ ${w}_{2}{w}_{1}{w}_{2}B,$ ${w}_{2}{w}_{1}{w}_{2}{w}_{1}B,$ and ${w}_{2}{w}_{1}{w}_{2}{w}_{1}{x}_{{\alpha }_{2}}\left(c\right){w}_{2}B\text{.}$ Geometrically, this is six copies of ${ℙ}^{1}\text{.}$

${q}^{6}=1$

If ${q}^{2}$ is a primitive third root of unity, then ${t}_{{1}^{1/3}}$ and ${t}_{-q,1}$ are in the same orbit as ${t}_{{q}^{2},1},$ while ${t}_{1,-q},$ ${t}_{{q}^{2},{q}^{2}},$ and ${t}_{{1}^{1/3},{q}^{2}}$ are all in the same orbit as ${t}_{{q}^{2},{q}^{2}}\text{.}$

The variety ${ℬ}_{{s}_{t}}$ changes only for $t={t}_{{q}^{2},1}={t}_{{1}^{1/3},1}\text{.}$ Then ${s}_{t}={H}_{{\alpha }_{1}}\left({1}^{2/3}\right){H}_{{\alpha }_{2}}\left(1\right)\text{.}$ Hence ${ℬ}_{s}$ contains $B,$ ${w}_{1}B,$ ${w}_{1}{x}_{{\alpha }_{2}}\left(c\right){w}_{2}B,$ ${w}_{1}{x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}B,$ ${w}_{1}{x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}{x}_{{\alpha }_{2}}\left(d\right){w}_{2}B,$ ${w}_{1}{x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}{x}_{{\alpha }_{2}}\left(d\right){w}_{2}{w}_{1}B,$ ${w}_{1}{x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}{x}_{{\alpha }_{2}}\left(d\right){w}_{2}{w}_{1}{x}_{{\alpha }_{2}}\left(e\right){w}_{2}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}{x}_{{\alpha }_{2}}\left(d\right){w}_{2}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}{x}_{{\alpha }_{2}}\left(d\right){w}_{2}{w}_{1}B,$ and ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}{x}_{{\alpha }_{2}}\left(d\right){w}_{2}{w}_{1}{x}_{{\alpha }_{2}}\left(e\right){w}_{2}B\text{.}$

${q}^{4}=1$

If ${q}^{2}=-1,$ then the central characters are represented by ${t}_{1,1},$ ${t}_{1,{q}^{2}}={t}_{1,-1},$ ${t}_{{1}^{1/3},1},$ ${t}_{1,q},$ ${t}_{1,z},$ ${t}_{{q}^{2},1},$ ${t}_{q,1},$ ${t}_{-{1}^{1/3},1},$ ${t}_{z,1},$ ${t}_{{q}^{2},z},$ ${t}_{z,{q}^{2}},$ and ${t}_{z,w}\text{.}$

The varieties ${ℬ}_{s}$ are the same as in the generic case for ${t}_{1,1},$ ${t}_{1,-1},$ ${t}_{{1}^{1/3},1},$ ${t}_{1,q},$ ${t}_{1,z},$ ${t}_{q,1},$ ${t}_{{q}^{2/3},1}={t}_{-{1}^{1/3},1}$ ${t}_{z,1},$ ${t}_{{q}^{2},z},$ ${t}_{z,{q}^{2}},$ and ${t}_{z,w}\text{.}$

If $t={t}_{{q}^{2},1},$ then ${s}_{t}={H}_{{\alpha }_{1}}\left(1\right){H}_{{\alpha }_{2}}\left({q}^{2}\right)\text{.}$ Then ${ℬ}_{s}$ contains $B,$ ${w}_{1}B,$ ${w}_{1}{w}_{2}B,$ ${w}_{1}{w}_{2}{x}_{{\alpha }_{1}}\left(c\right){w}_{1}B,$ ${w}_{1}{w}_{2}{x}_{{\alpha }_{1}}\left(c\right){w}_{1}{w}_{2}B,$ ${w}_{1}{w}_{2}{x}_{{\alpha }_{1}}\left(c\right){w}_{1}{w}_{2}{w}_{1}B,$ ${w}_{1}{w}_{2}{x}_{{\alpha }_{1}}\left(c\right){w}_{1}{w}_{2}{w}_{1}{x}_{{\alpha }_{2}}\left(d\right){w}_{2}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}{w}_{2}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}{w}_{2}{x}_{{\alpha }_{1}}\left(d\right){w}_{1}B,$ and ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}{w}_{2}{x}_{{\alpha }_{1}}\left(d\right){w}_{2}B\text{.}$

${q}^{2}=1$

When ${q}^{2}=1,$ the possible central characters are represented by ${t}_{1,1},$ ${t}_{1,-1},$ ${t}_{{1}^{1/3},1},$ ${t}_{1,z},$ ${t}_{z,1},$ and ${t}_{z,w}\text{.}$ The varieties ${ℬ}_{s}$ are the same as in the generic case for all these ${s}_{t}\text{.}$

The Varieties ${ℬ}_{s,n}$

We describe the varieties ${ℬ}_{s,n}$ for each representative of an orbit in $\Lambda /G\text{.}$ However, we note that ${ℬ}_{s,0}={ℬ}_{s},$ so we only describe ${ℬ}_{s,n}$ for $n\ne 0\text{.}$

Generic $q$

If $\left(s,n\right)=\left({s}_{{t}_{1,{q}^{2}}},{e}_{{\alpha }_{2}}\right),$ then ${ℬ}_{s,n}$ contains $B,$ ${x}_{{\alpha }_{1}}\left(c\right){w}_{1}B,$ ${w}_{1}{w}_{2}B,$ ${w}_{1}{w}_{2}{w}_{1}B,$ ${w}_{1}{w}_{2}{w}_{1}{w}_{2}B,$ and ${w}_{1}{w}_{2}{w}_{1}{w}_{2}{w}_{1}B.$ This is a ${ℙ}^{1}$ and four points.

If $\left(s,n\right)=\left({s}_{{t}_{1,{q}^{2}}},{e}_{{\alpha }_{2}}+{e}_{3{\alpha }_{1}+{\alpha }_{2}}\right),$ then we note that

$exp(eα2+e3α1+α2) =xα2(1) x3α1+α2(1) x3α1+2α2 (-1/2).$

Then ${ℬ}_{s,n}$ consists of $B,$ ${x}_{{\alpha }_{1}}\left(c\right){w}_{1}B$ and ${x}_{{\alpha }_{1}}\left({\zeta }^{i}\right){w}_{1}{w}_{2}B,$ where $\zeta$ is a primitve third root of unity and $i=0,1,$ or 2. Geometrically, this is ${ℙ}^{1}$ and three points.

The centralizer ${C}_{G}\left(s,n\right)$ is generated by ${H}_{{\alpha }_{1}}\left({\left(-1\right)}^{1/3}\right){H}_{{\alpha }_{2}}\left(-1\right){w}_{1},$ which has order 2, as well as ${H}_{{\alpha }_{1}}\left({1}^{1/3}\right),$ for some choice of a cube root of 1. Then ${C}_{G}\left(s,n\right)$ is isomorphic to ${S}_{3}\text{.}$ This group acts trivially on the ${ℙ}^{1}$ in ${ℬ}_{s,n},$ and permutes the three points via the natural action of ${S}_{3}\text{.}$ Then if ${H}_{•}$ is spanned by $\alpha ,\beta ,$ and $\gamma ,$ the span of $\alpha +\beta +\gamma$ is a copy of the trivial representation of ${S}_{3}$ in ${H}_{•}\left({ℬ}_{s,n}\right)\text{.}$ Thus ${H}_{•}{\left({ℬ}_{s,n}\right)}^{\text{triv}}$ is 3-dimensional, and as a $\stackrel{\sim }{H}\text{-module,}$ has a 2-dimensional generalized $t$ weight space, and a 1-dimensional ${s}_{2}t$ weight space. The remaining two dimensions of ${H}_{•}$ are $⟨\alpha ,\beta ,\gamma ⟩/\alpha +\beta +\gamma ,$ a copy of the 2-dimensional representation $\chi$ of ${S}_{3}\text{.}$ As a $\stackrel{\sim }{H}\text{-module,}$ it has a 2-dimensional ${s}_{2}t$ weight space.

If $\left(s,n\right)=\left({s}_{{t}_{1,{q}^{2}}},{e}_{{\alpha }_{1}+{\alpha }_{2}}\right)$ then ${ℬ}_{s,n}$ contains $B,$ ${w}_{2}B,$ ${x}_{{\alpha }_{1}}\left(c\right){w}_{1}B,$ ${w}_{1}{w}_{2}B,$ ${w}_{1}{w}_{2}{w}_{1}B,$ and ${w}_{1}{w}_{2}{w}_{1}{w}_{2}B\text{.}$ This is a ${ℙ}^{1}$ and 4 points.

$B,xα1(c)w1B w1w2B w1w2w1B w1w2w1w2B w1w2w1w2w1B B,xα1(c)w1B w2B,w1w2B w1w2w1B w1w2w1w2B (st1,q2,eα2) (st1,q2,eα1+α2) B,xα1(c)w1B xα1(ζi)w1w2B (st1,q2,eα2+e3α1+α2)$

If $\left(s,n\right)=\left({s}_{{t}_{1,±q}},{e}_{3{\alpha }_{1}+2{\alpha }_{2}}\right),$ then ${ℬ}_{s,n}$ consists of $B,$ ${x}_{{\alpha }_{1}}\left(c\right){w}_{1}B,$ ${x}_{{\alpha }_{1}}\left(c\right){w}_{1}{w}_{2}B,$ ${x}_{{\alpha }_{1}}\left(c\right){w}_{1}{w}_{2}{w}_{1}B,$ ${w}_{2}B,$ and ${w}_{2}{w}_{1}B\text{.}$ This is three ${ℙ}^{1}\text{s.}$

If $\left(s,n\right)=\left({s}_{{t}_{{q}^{2},1}},{e}_{{\alpha }_{1}}\right),$ then ${ℬ}_{s,n}$ contains $B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}B,$ ${w}_{2}{w}_{1}B,$ ${w}_{2}{w}_{1}{w}_{2}B,$ ${w}_{2}{w}_{1}{w}_{2}{w}_{1}B,$ and ${w}_{2}{w}_{1}{w}_{2}{w}_{1}{w}_{2}B\text{.}$ This is a ${ℙ}^{1}$ and four points.

If $\left(s,n\right)=\left({s}_{{t}_{±q,1}},{e}_{2{\alpha }_{1}+{\alpha }_{2}}\right),$ then ${ℬ}_{s,n}$ contains $B,$ ${w}_{1}B,$ ${w}_{1}{w}_{2}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}B,$ and ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}{w}_{2}B\text{.}$ This is three disjoint copies of ${ℙ}^{1}\text{.}$

$B,xα2(c)w2B w2w1B w2w1w2B w2w1w2w1B w2w1w2w1w2B B,xα1(c)w1B w2B,xα1(c)w1w2B w2w1B,xα1(c)w1w2w1B (stq2,1,eα1) (st1,±q,e3α1+α2) B,xα2(c)w2B w1B,xα2(c)w2w1B w1w2B,xα2(c)w2w1w2B (st1,±q,eα2+e3α1+α2)$

If $\left(s,n\right)=\left({s}_{{t}_{{q}^{2/3},1}},{e}_{3{\alpha }_{1}+{\alpha }_{2}}\right),$ then ${ℬ}_{s,n}$ consists of $B,$ ${w}_{1}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}B,$ ${w}_{2}{w}_{1}{w}_{2}B,$ and ${w}_{2}{w}_{1}{w}_{2}{w}_{1}B\text{.}$ This is two copies of ${ℙ}^{1}$ and two points.

$B,xα2(c)w2B w1B,xα2(c)w2w1B w2w1w2B w2w1w2w1B (stq2/3,1,eα2)$

If $\left(s,n\right)=\left({s}_{{t}_{{q}^{2},-{q}^{-2}}},{e}_{{\alpha }_{1}}\right),$ then ${ℬ}_{s,n}$ consists of $B,$ ${w}_{2}B,$ ${w}_{2}{w}_{1}B,$ ${w}_{2}{w}_{1}{w}_{2}B,$ ${w}_{2}{w}_{1}{w}_{2}{w}_{1}B,$ and ${w}_{2}{w}_{1}{w}_{2}{w}_{1}{w}_{2}B\text{.}$

If $\left(s,n\right)=\left({s}_{{t}_{{q}^{2},-{q}^{-2}}},{e}_{3{\alpha }_{1}+2{\alpha }_{2}}\right),$ then ${ℬ}_{s,n}$ consists of $B,$ ${w}_{2}B,$ ${w}_{2}{w}_{1}B,$ ${w}_{1}B,$ ${w}_{1}{w}_{2}B,$ and ${w}_{1}{w}_{2}{w}_{1}B\text{.}$

If $\left(s,n\right)=\left({s}_{{t}_{{q}^{2},-{q}^{-2}}},{e}_{{\alpha }_{1}}+{e}_{3{\alpha }_{1}+2{\alpha }_{2}}\right),$ then $\text{exp}\left({e}_{{\alpha }_{1}}+{e}_{3{\alpha }_{1}+2{\alpha }_{2}}\right)={x}_{{\alpha }_{1}}\left(1\right){x}_{3{\alpha }_{1}+2{\alpha }_{2}}\left(1\right)\text{.}$ Then ${ℬ}_{s,n}$ consists of $B,$ ${w}_{2}B,$ and ${w}_{2}{w}_{1}B\text{.}$

$B w2B w2w1B w2w1w2B w2w1w2w1B w2w1w2w1w2B w1w2w1B w1w2B w1B B w2B w2w1B (stq2,-q-2,eα1) (stq2,-q-2,e3α1+2α2) B w2B w2w1B (stq2,-q-2,eα1+e3α1+2α2)$

If $\left(s,n\right)=\left({s}_{{t}_{{1}^{1/3},{q}^{2}}},{e}_{{\alpha }_{2}}\right),$ then ${ℬ}_{s,n}$ consists of $B,$ ${w}_{1}B,$ ${w}_{1}{w}_{2}B,$ ${w}_{1}{w}_{2}{w}_{1}B,$ ${w}_{1}{w}_{2}{w}_{1}{w}_{2}B,$ and ${w}_{1}{w}_{2}{w}_{1}{w}_{2}{w}_{1}B\text{.}$

If $\left(s,n\right)=\left({s}_{{t}_{{1}^{1/3},{q}^{2}}},{e}_{3{\alpha }_{1}+{\alpha }_{2}}\right),$ then ${ℬ}_{s,n}$ consists of $B,$ ${w}_{1}B,$ ${w}_{2}B,$ ${w}_{2}{w}_{1}B,$ ${w}_{2}{w}_{1}{w}_{2}B,$ and ${w}_{2}{w}_{1}{w}_{2}{w}_{1}B\text{.}$

If $\left(s,n\right)=\left({s}_{{t}_{{1}^{1/3},{q}^{2}}},{e}_{{\alpha }_{2}}+{e}_{3{\alpha }_{1}+{\alpha }_{2}}\right),$ then $\text{exp}\left(n\right)={x}_{{\alpha }_{2}}\left(1\right){x}_{3{\alpha }_{1}+{\alpha }_{2}}\left(1\right){x}_{3{\alpha }_{1}+2{\alpha }_{2}}\left(\frac{-1}{2}\right)\text{.}$ Then ${ℬ}_{s,n}$ consists of $B,$ and ${w}_{1}B\text{.}$

$w1w2w1w2w1B w1w2w1w2B w1w2w1B w1w2B w1B B w1B B w2B w2w1B w2w1w2B w2w1w2w1B (st11/3,q2,eα2) (st11/3,q2,e3α1+α2) w1B B ( st11/3,q2, eα2+ e3α1+α2 )$

If $\left(s,n\right)=\left({s}_{{t}_{{q}^{2},{q}^{2}}},{e}_{{\alpha }_{1}}\right),$ then ${ℬ}_{s,n}$ consists of $B,$ ${w}_{2}B,$ ${w}_{2}{w}_{1}B,$ ${w}_{2}{w}_{1}{w}_{2}B,$ ${w}_{2}{w}_{1}{w}_{2}{w}_{1}B,$ and ${w}_{2}{w}_{1}{w}_{2}{w}_{1}{w}_{2}B\text{.}$

If $\left(s,n\right)=\left({s}_{{t}_{{q}^{2},{q}^{2}}},{e}_{{\alpha }_{2}}\right),$ then ${ℬ}_{s,n}$ consists of $B,$ ${w}_{1}B,$ ${w}_{1}{w}_{2}B,$ ${w}_{1}{w}_{2}{w}_{1}B,$ ${w}_{1}{w}_{2}{w}_{1}{w}_{2}B,$ and ${w}_{1}{w}_{2}{w}_{1}{w}_{2}{w}_{1}B\text{.}$

If $\left(s,n\right)=\left({s}_{{t}_{{q}^{2},{q}^{2}}},{e}_{{\alpha }_{1}}+{e}_{{\alpha }_{2}}\right),$ then note that

$exp(eα1+eα2) =xα1(1) xα2(1) xα1+α2 (-1/2) x2α1+α2 (1/3) x3α1+α2 (-1/4) x3α1+2α2 (-1/10).$

Then ${ℬ}_{s,n}$ consists only of $B\text{.}$

$w1w2w1w2w1B w1w2w1w2B w1w2w1B w1w2B w1B B B w2B w2w1B w2w1w2B w2w1w2w1B w2w1w2w1w2B (stq2,q2,eα2) (stq2,q2,eα1) B (stq2,q2,eα1+eα2)$

If $\left(s,n\right)=\left({s}_{{t}_{{q}^{2},z}},{e}_{{\alpha }_{1}}\right),$ then ${ℬ}_{s,n}$ consists of $B,$ ${w}_{2}B,$ ${w}_{2}{w}_{1}B,$ ${w}_{2}{w}_{1}{w}_{2}B,$ ${w}_{2}{w}_{1}{w}_{2}{w}_{1}B,$ and ${w}_{2}{w}_{1}{w}_{2}{w}_{1}{w}_{2}B\text{.}$

If $\left(s,n\right)=\left({s}_{{t}_{z,{q}^{2}}},{e}_{{\alpha }_{2}}\right),$ then ${ℬ}_{s,n}$ consists of $B,$ ${w}_{1}B,$ ${w}_{1}{w}_{2}B,$ ${w}_{1}{w}_{2}{w}_{1}B,$ ${w}_{1}{w}_{2}{w}_{1}{w}_{2}B,$ and ${w}_{1}{w}_{2}{w}_{1}{w}_{2}{w}_{1}B\text{.}$

$w1w2w1w2w1B w1w2w1w2B w1w2w1B w1w2B w1B B B w2B w2w1B w2w1w2B w2w1w2w1B w2w1w2w1w2B (stz,q2,eα2) (stz,q2,eα1)$

The generic case will differ from a specialized $q$ only if $\mid Z\left(t\right)\mid$ increases so that ${ℬ}_{{s}_{t}}$ is larger for that value of $q,$ or if ${𝔤}_{q}^{s}$ is larger leading to new orbits that were not present before.

${q}^{12}=1$

The only orbits that differ from the generic case when ${q}^{12}=1$ are those involving the semisimples in the orbit of ${s}_{{t}_{{q}^{2},{q}^{2}}}\text{.}$ The nilpotent orbits in ${𝔤}_{q}^{s}$ are represented by ${e}_{{\alpha }_{1}},$ ${e}_{{\alpha }_{2}},$ ${e}_{{\alpha }_{1}}+{e}_{{\alpha }_{2}},$ ${e}_{-3{\alpha }_{1}-2{\alpha }_{2}},$ ${e}_{{\alpha }_{1}}+{e}_{-3{\alpha }_{1}-2{\alpha }_{2}},$ and ${e}_{{\alpha }_{2}}+{e}_{-3{\alpha }_{1}-2{\alpha }_{2}}\text{.}$ However, computations become easier if we choose these orbit representatives in $\Lambda :$

$w2w1w2· ( stq2,q2, e-3α1-2α2 ) = ( sw2w1w2tq2,q2 ,eα2 ) , w2w1w2· ( stq2,q2, eα1+ e-3α1-2α2 ) = ( sw2w1w2tq2,q2 ,eα2+ e2α1+α2 ) , and w2w1w2w1· ( stq2,q2, eα2+ e-3α1-2α2 ) = ( sw2w1w2w1tq2,q2 ,eα2+ e3α1+α2 ) .$

If $\left(s,n\right)=\left({s}_{{s}_{2}{s}_{1}{s}_{2}{t}_{{q}^{2},{q}^{2}}},{e}_{{\alpha }_{2}}\right),$ then ${ℬ}_{s,n}$ consists of $B,$ ${w}_{1}B,$ ${w}_{1}{w}_{2}B,$ ${w}_{1}{w}_{2}{w}_{1}B,$ ${w}_{1}{w}_{2}{w}_{1}{w}_{2}B,$ and ${w}_{1}{w}_{2}{w}_{1}{w}_{2}{w}_{1}B\text{.}$

If $\left(s,n\right)=\left({s}_{{s}_{2}{s}_{1}{s}_{2}{t}_{{q}^{2},{q}^{2}}},{e}_{{\alpha }_{2}}+{e}_{2{\alpha }_{1}+{\alpha }_{2}}\right),$ then ${ℬ}_{s,n}$ consists of $B,$ ${w}_{1}B,$ and ${w}_{1}{w}_{2}B\text{.}$

If $\left(s,n\right)=\left({s}_{{s}_{2}{s}_{1}{s}_{2}{s}_{1}{t}_{{q}^{2},{q}^{2}}},{e}_{{\alpha }_{2}}+{e}_{3{\alpha }_{1}+{\alpha }_{2}}\right),$ then ${ℬ}_{s,n}$ consists of $B$ ${w}_{1}B\text{.}$

$w1w2w1B w1w2w1w2B w1w2w1w2w1B B w1B w1w2B B w1B w1w2B ( stq2,q2, e-3α1-2α2 ) ,q12=1 ( stq2,q2, eα1+ e-3α1-2α2 ) ,q12=1 B w2B w2w1B w2w1w2B w2w1w2w1B w2w1w2w1w2B B w1B ( stq2,q2, eα1 ) ,q12=1 ( stq2,q2, eα2+ e-3α1-2α2 ) ,q12=1 w1w2w1w2w1B w1w2w1w2B w1w2w1B w1w2B w1B B B ( stq2,q2, eα2 ) ,q12=1 ( stq2,q2, eα1+eα2 ) ,q12=1$

${q}^{10}=1$

If ${q}^{10}=1,$ then ${t}_{{q}^{2},{q}^{2}}$ is in the same orbit as ${t}_{{q}^{-4},1},$ and $Z\left({t}_{{q}^{2},{q}^{2}}\right)$ now contains $3{\alpha }_{1}+2{\alpha }_{2}\text{.}$

If $\left(s,n\right)=\left({s}_{{t}_{-{q}^{-4},1}},{e}_{2{\alpha }_{1}+{\alpha }_{2}}\right),$ then ${ℬ}_{s,n}$ consists of $B,$ ${w}_{1}B,$ ${w}_{1}{w}_{2}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}B,$ and ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}{w}_{2}B\text{.}$

If $\left(s,n\right)=\left({s}_{{t}_{{q}^{-4},1}},{e}_{-3{\alpha }_{1}-{\alpha }_{2}}\right),$ then we choose a different orbit representative. Specifically, we use the pair $\left({s}_{{s}_{2}{s}_{1}{t}_{{q}^{-4},1}},{e}_{{\alpha }_{2}}\right)=\left({s}_{{t}_{{q}^{2},{q}^{2}}},{e}_{{\alpha }_{2}}\right)\text{.}$ Then ${ℬ}_{s,n}$ contains $B,$ ${w}_{1}B,$ ${w}_{1}{w}_{2}B,$ ${w}_{1}{w}_{2}{w}_{1}B,$ ${w}_{1}{w}_{2}{w}_{1}{x}_{{\alpha }_{2}}\left(c\right){w}_{2}B,$ and ${w}_{1}{w}_{2}{w}_{1}{x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}B\text{.}$ Geometrically, this is two copies of ${ℙ}^{1}$ and two points.

If $\left(s,n\right)=\left({s}_{{t}_{{q}^{-4},1}},{e}_{2{\alpha }_{1}+{\alpha }_{2}}+{e}_{-3{\alpha }_{1}-{\alpha }_{2}}\right),$ then we choose a different orbit representative. Specifically, we use $\left({s}_{{s}_{2}{s}_{1}{t}_{{q}^{-4},1}},{e}_{{\alpha }_{2}}+{e}_{{\alpha }_{1}}\right)=\left({s}_{{t}_{{q}^{2},{q}^{2}}},{e}_{{\alpha }_{1}}+{e}_{{\alpha }_{2}}\right)\text{.}$ Then note that $\text{exp}\left({e}_{{\alpha }_{1}}+{e}_{{\alpha }_{2}}\right)={x}_{{\alpha }_{1}}\left(1\right){x}_{{\alpha }_{2}}\left(1\right){x}_{{\alpha }_{1}+{\alpha }_{2}}\left(-1/2\right){x}_{2{\alpha }_{1}+{\alpha }_{2}}\left(1/3\right){x}_{3{\alpha }_{1}+{\alpha }_{2}}\left(-1/4\right){x}_{3{\alpha }_{1}+2{\alpha }_{2}}\left(-1/10\right)\text{.}$ However, ${ℬ}_{s,n}$ consists only of $B\text{.}$

$w1w2w1xα2(c)w2B w1w2w1B, w1w2w1xα2(c)w2w1B w1w2B, w1B B B,xα2(c)w2B w1B,xα2(c)w2w1B w2w1w2B w2w1w2w1B ( stq-4,1, e-3α1-α2 ) ,q10=1 ( stq-4,1, e2α1+α2 ) ,q10=1 B ( stq-4,1, e2α1+α2+ e-3α1-α2 ) ,q10=1$

${q}^{8}=1$

If $q$ is a primitive eighth root of unity, then ${t}_{{q}^{2},{q}^{2}}={t}_{{q}^{2},{q}^{-2}},$ and both of these central characters are in the same orbit as ${t}_{{q}^{2},1}\text{.}$

If $\left(s,n\right)=\left({s}_{{t}_{{q}^{2},{q}^{2}}},{e}_{{\alpha }_{1}}\right),$ then ${ℬ}_{s,n}$ consists of $B,$ ${w}_{2}B,$ ${w}_{2}{w}_{1}B,$ ${w}_{2}{w}_{1}{w}_{2}B,$ ${w}_{2}{w}_{1}{w}_{2}{w}_{1}B,$ and ${w}_{2}{w}_{1}{w}_{2}{w}_{1}{x}_{{\alpha }_{2}}\left(c\right){w}_{2}B\text{.}$

If $\left(s,n\right)=\left({s}_{{t}_{{q}^{2},{q}^{2}}},{e}_{{\alpha }_{2}}\right),$ then ${ℬ}_{s,n}$ consists of $B,$ ${w}_{1}B,$ ${w}_{1}{x}_{{\alpha }_{2}}\left(c\right){w}_{2}B,$ ${w}_{1}{x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}B,$ ${w}_{1}{w}_{2}{w}_{1}{w}_{2}B,$ and ${w}_{1}{w}_{2}{w}_{1}{w}_{2}{w}_{1}B\text{.}$

If $\left(s,n\right)=\left({s}_{{t}_{{q}^{2},{q}^{2}}},{e}_{{\alpha }_{1}}+{e}_{{\alpha }_{2}}\right),$ then ${ℬ}_{s,n}$ consists only of $B\text{.}$

If $\left(s,n\right)=\left({s}_{{t}_{{q}^{2},{q}^{2}}},{e}_{{\alpha }_{1}}+{e}_{3{\alpha }_{1}+2{\alpha }_{2}}\right),$ then note that $\text{exp}\left({e}_{{\alpha }_{1}}+{e}_{3{\alpha }_{1}+2{\alpha }_{2}}\right)={x}_{{\alpha }_{1}}\left(1\right){x}_{3{\alpha }_{1}+2{\alpha }_{2}}\left(1\right)\text{.}$ Then ${ℬ}_{s,n}$ consists of $B,$ ${w}_{2}B,$ and ${w}_{2}{w}_{1}B\text{.}$

$B w2B w2w1B w2w1w2B w2w1w2w1B, w2w1w2w1xα2(c)w2B w1B,w1xα2(c)w2B B,w1w2w1B w1w2w1w2B w1w2w1w2w1B (stq2,q2,eα1) ,q8=1 (stq2,q2,eα2) ,q8=1 B B w2B w2w1B (stq2,q2,eα1+eα2) ,q8=1 (stq2,q2,eα1+e3α1+2α2) ,q8=1$

${q}^{6}=1$

If ${q}^{2}$ is a primitive third root of unity, then ${t}_{{1}^{1/3}}$ and ${t}_{-q,1}$ are in the same orbit as ${t}_{{q}^{2},1},$ while ${t}_{1,-q},$ ${t}_{{q}^{2},{q}^{2}},$ and ${t}_{{1}^{1/3},{q}^{2}}$ are all in the same orbit as ${t}_{{q}^{2},{q}^{2}}\text{.}$

It suffices to consider ${t}_{1,{q}^{2}}$ and ${t}_{{q}^{2},1}\text{.}$ The other weights are in the same orbit as these or the varieties ${ℬ}_{s,n}$ are the same as in the generic case.

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

If $s={s}_{{t}_{{q}^{2},1}},$ then the ${C}_{G}\left(s\right)\text{-orbits}$ of nilpotents in ${𝔤}_{q}^{s}$ are represented by 0 and ${e}_{{\alpha }_{1}}\text{.}$

If $\left(s,n\right)=\left({s}_{{t}_{{q}^{2},1}},{e}_{{\alpha }_{1}}\right),$ then ${ℬ}_{s,n}$ contains $B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}B,$ ${w}_{2}{w}_{1}B,$ ${w}_{2}{w}_{1}{w}_{2}B,$ ${w}_{2}{w}_{1}{w}_{2}{w}_{1}B,$ and ${w}_{2}{w}_{1}{w}_{2}{w}_{1}{w}_{2}B\text{.}$

$e-3α1-2α2,q6=1$

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

In the case that $t={t}_{1,{q}^{2}},$ the correspondence between the combinatorial and geometric classifications is unclear. We will give the details of the geometric classification here and describe the issues that arise.

First, we note that if $s={s}_{{t}_{1,{q}^{2}}},$ then the nilpotent orbits in ${𝔤}_{q}^{s}$ are represented by 0, ${e}_{{\alpha }_{2}},$ ${e}_{{\alpha }_{2}}+{e}_{3{\alpha }_{1}+{\alpha }_{2}},$ ${e}_{{\alpha }_{1}+{\alpha }_{2}},$ ${e}_{-3{\alpha }_{1}-2{\alpha }_{2}},$ ${e}_{{\alpha }_{2}}+{e}_{-3{\alpha }_{1}-2{\alpha }_{2}},$ and ${e}_{{\alpha }_{1}+{\alpha }_{2}}+{e}_{-3{\alpha }_{1}-2{\alpha }_{2}}\text{.}$ However, since ${e}_{{\alpha }_{2}}+{e}_{3{\alpha }_{1}+{\alpha }_{2}}+{e}_{-3{\alpha }_{1}-2{\alpha }_{2}},$ an element of ${𝔤}_{q}^{s},$ is not nilpotent, the pair $\left(s,0\right)$ does not satisfy Grojnowski’s condition (3.1), and so is excluded from the indexing.

To check that the other nilpotent orbits do satisfy condition 3.1, we note that an element of ${𝔤}_{q}^{{s}_{{t}_{1,{q}^{2}}}}$ takes the form $g=a{e}_{{\alpha }_{2}}+b{e}_{{\alpha }_{1}+{\alpha }_{2}}+c{e}_{2{\alpha }_{1}+{\alpha }_{2}}+d{e}_{3{\alpha }_{1}+{\alpha }_{2}}+f{e}_{-3{\alpha }_{1}-2{\alpha }_{2}}\text{.}$ We check condition 3.1 for n by first finding a ${𝔰𝔩}_{2}\text{-triple}$ $\left(n,y,h\right)\text{.}$ By computing $\left[y,g\right]$ (using the basic representation of $𝔤\text{,)}$ we determine necessary and sufficient conditions on the coefficients of $g$ for $g$ to be in ${Z}_{𝔤}\left(y\right)\text{.}$ If all $g$ satisfying these conditions are again nilpotent, then condition 3.1 is satisfied. The following table summarizes this check.

$n y Condition on Z𝔤(y) Z𝔤(y)∩𝔤qs⊆𝒩 00noneno eα2 e-α2 a=b=0 yes eα1+α2 e-α1-α2 a=b=c=0 yes eα2+e3α1+α2 2e-α2-2e-3α1-α2 a=b=c=d=0 yes e-3α1-2α2 e3α1+2α2 f=0 yes eα2+e-3α1-2α2 2e-α2+2e3α1+2α2 a=b=f=0 yes eα1+α2+e-3α1-2α2 -6e-α1-α2+10e3α1+2α2 a=b=c=f=0 yes$

If $\left(s,n\right)=\left({s}_{{t}_{1,{q}^{2}}},{e}_{{\alpha }_{2}}\right),$ then ${ℬ}_{s,n}$ contains $B,$ ${x}_{{\alpha }_{1}}\left(c\right){w}_{1}B,$ ${w}_{1}{w}_{2}B,$ ${w}_{1}{w}_{2}{w}_{1}B,$ ${w}_{1}{w}_{2}{w}_{1}{w}_{2}B,$ and ${w}_{1}{w}_{2}{w}_{1}{w}_{2}{w}_{1}B\text{.}$ This is a ${ℙ}^{1}$ and four points. The corresponding module has weights $t,$ ${s}_{2}t,$ ${s}_{1}{s}_{2}t,$ ${s}_{2}{s}_{1}{s}_{2}t,$ ${s}_{1}{s}_{2}{s}_{1}{s}_{2}t,$ and ${s}_{2}{s}_{1}{s}_{2}{s}_{1}{s}_{2}t,$ and the $t$ weight space has dimension 2.

If $\left(s,n\right)=\left({s}_{{t}_{1,{q}^{2}}},{e}_{{\alpha }_{2}}+{e}_{3{\alpha }_{1}+{\alpha }_{2}}\right),$ then we note that

$exp(eα2+e3α1+α2) =xα2(1) x3α1+α2(1) x3α1+2α2(-1/2).$

Then ${ℬ}_{s,n}$ consists of $B,$ ${x}_{{\alpha }_{1}}\left(c\right){w}_{1}B$ and ${x}_{{\alpha }_{1}}\left({\zeta }^{i}\right){w}_{1}{w}_{2}B,$ where $\zeta$ is a primitve third root of unity and $i=0,1,$ or 2. Geometrically, this is ${ℙ}^{1}$ and three points.

The centralizer ${C}_{G}\left(s,n\right)$ is generated by ${H}_{{\alpha }_{1}}\left({\left(-1\right)}^{1/3}\right){H}_{{\alpha }_{2}}\left(-1\right){w}_{1},$ which has order 2, as well as ${H}_{{\alpha }_{1}}\left({1}^{1/3}\right),$ for some choice of a cube root of 1. Then ${C}_{G}\left(s,n\right)$ is isomorphic to ${S}_{3}\text{.}$ This group acts trivially on the ${ℙ}^{1}$ in ${ℬ}_{s,n},$ and permutes the three points via the natural action of ${S}_{3}\text{.}$ Then if ${H}_{•}$ is spanned by $\alpha ,\beta ,$ and $\gamma ,$ the span of $\alpha +\beta +\gamma$ is a copy of the trivial representation of ${S}_{3}$ in ${H}_{•}\left({ℬ}_{s,n}\right)\text{.}$ Thus ${H}_{•}{\left({ℬ}_{s,n}\right)}^{\text{triv}}$ is 3-dimensional, and as a $\stackrel{\sim }{H}\text{-module,}$ has a 2-dimensional generalized $t$ weight space, and a 1-dimensional ${s}_{2}t$ weight space. The remaining two dimensions of ${H}_{•}$ are $⟨\alpha ,\beta ,\gamma ⟩/\alpha +\beta +\gamma ,$ a copy of the 2-dimensional representation $\chi$ of ${S}_{3}\text{.}$ As a $\stackrel{\sim }{H}\text{-module,}$ it has a 2-dimensional ${s}_{2}t$ weight space.

If $\left(s,n\right)=\left({s}_{{t}_{1,{q}^{2}}},{e}_{{\alpha }_{1}+{\alpha }_{2}}\right),$ then ${ℬ}_{s,n}$ consists of $B,$ ${x}_{{\alpha }_{1}}\left(c\right){w}_{1}B,$ ${w}_{1}{w}_{2}B,$ ${w}_{1}{w}_{2}{w}_{1}B,$ ${w}_{1}{w}_{2}{w}_{1}{w}_{2}B,$ and ${w}_{2}B\text{.}$ This is a ${ℙ}^{1}$ and 4 points. The corresponding module has weights $t,$ ${s}_{2}t,$ ${s}_{1}{s}_{2}t,$ ${s}_{2}{s}_{1}{s}_{2}t,$ and ${s}_{1}{s}_{2}{s}_{1}{s}_{2}t,$ and the $t$ and ${s}_{2}t$ weight spaces have dimension 2.

We note that ${w}_{2}{w}_{1}{w}_{2}·\left({s}_{{t}_{1,{q}^{2}}},{e}_{-3{\alpha }_{1}-2{\alpha }_{2}}\right)=\left({s}_{{t}_{{q}^{2},{q}^{2}}},{e}_{{\alpha }_{2}}\right)\text{.}$ Then if $\left(s,n\right)=\left({s}_{{t}_{{q}^{2},{q}^{2}}},{e}_{{\alpha }_{2}}\right),$ then ${ℬ}_{s,n}$ consists of $B,$ ${w}_{1}B,$ ${w}_{1}{w}_{2}B,$ ${w}_{1}{w}_{2}{w}_{1}B,$ ${w}_{1}{w}_{2}{w}_{1}{x}_{{\alpha }_{2}}\left(c\right){w}_{2}B,$ and ${w}_{1}{w}_{2}{w}_{1}{x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}B\text{.}$ This is two copies of ${ℙ}^{1}$ and two points. If $t={t}_{1,{q}^{2}},$ then the corresponding module has weights ${s}_{2}{s}_{1}{s}_{2}t,$ ${s}_{1}{s}_{2}{s}_{1}{s}_{2}t,$ and ${s}_{2}{s}_{1}{s}_{2}{s}_{1}{s}_{2}t,$ and each weight space has dimension 2.

If $\left(s,n\right)=\left({s}_{{t}_{{q}^{2},{q}^{2}}},{e}_{{\alpha }_{1}}+{e}_{{\alpha }_{2}}\right)={w}_{2}{w}_{1}{w}_{2}{w}_{1}·\left({s}_{{t}_{1,{q}^{2}}},{e}_{{\alpha }_{1}+{\alpha }_{2}}+{e}_{-3{\alpha }_{1}-2{\alpha }_{2}}\right),$ then we note that $\text{exp}\left({e}_{{\alpha }_{2}}+{e}_{{\alpha }_{1}}\right)={x}_{{\alpha }_{1}}\left(1\right){x}_{{\alpha }_{2}}\left(1\right){x}_{{\alpha }_{1}+{\alpha }_{2}}\left(-1/2\right){x}_{2{\alpha }_{1}+{\alpha }_{2}}\left(1/3\right){x}_{3{\alpha }_{1}+{\alpha }_{2}}\left(-1/4\right){x}_{3{\alpha }_{1}+2{\alpha }_{2}}\left(-1/10\right)\text{.}$ Then ${ℬ}_{s,n}$ consists of the single point $B\text{.}$ The module has a single weight space ${M}_{{s}_{2}{s}_{1}{s}_{2}}t$

If $\left(s,n\right)=\left({s}_{{t}_{{q}^{2},{q}^{2}}},{e}_{{\alpha }_{2}}+{e}_{3{\alpha }_{1}+{\alpha }_{2}}\right)={w}_{2}{w}_{1}{w}_{2}{w}_{1}·\left({s}_{{t}_{1,{q}^{2}}},{e}_{{\alpha }_{2}}+{e}_{-3{\alpha }_{1}-2{\alpha }_{2}}\right),$ then we note that $\text{exp}\left({e}_{{\alpha }_{2}}+{e}_{3{\alpha }_{1}+2{\alpha }_{2}}\right)={x}_{{\alpha }_{2}}\left(1\right){x}_{3{\alpha }_{1}+{\alpha }_{2}}\left(1\right){x}_{3{\alpha }_{1}+2{\alpha }_{2}}\left(-1/2\right)\text{.}$ Then ${ℬ}_{s,n}$ consists of the points $B$ and ${w}_{1}B\text{.}$ The corresponding module has weights ${s}_{2}{s}_{1}{s}_{2}t$ and ${s}_{1}{s}_{2}{s}_{1}{s}_{2}t\text{.}$

These pictures show the weight space structure of ${H}_{•}\left({ℬ}_{s,n}\right)\text{.}$ The nilpotent orbit of $n$ must index a module which is a composition factors of ${H}_{•}\left({ℬ}_{s,n}\right),$ which is what allows us to determine the correspondence.

$triv sign triv sign triv sign eα2,q6=1 eα1+α2,q6=1 eα2+e3α1+α2,q6=1 sign sign sign eα2+e-3α1-2α2,q6=1 eα1+α2+e-3α1-2α2,q6=1 e-3α1-2α2,q6=1$

The nilpotent orbit of ${e}_{{\alpha }_{1}+{\alpha }_{2}}+{e}_{-3{\alpha }_{1}-2{\alpha }_{2}}$ must correspond to the 1-dimensional module with weight ${s}_{2}{s}_{1}{s}_{2}t,$ since ${H}_{•}\left({ℬ}_{s,n}\right)$ is only 1-dimensional and must be precisely that module. The composition factors of ${H}_{•}\left({ℬ}_{s,{e}_{{\alpha }_{2}}+{e}_{-3{\alpha }_{1}-2{\alpha }_{2}}}\right)$ are 1-dimensional modules with weights ${s}_{2}{s}_{1}{s}_{2}t$ and ${s}_{1}{s}_{2}{s}_{1}{s}_{2}t,$ so that ${e}_{{\alpha }_{2}}+{e}_{-3{\alpha }_{1}-2{\alpha }_{2}}$ must correspond to the 1-dimensional module with weight ${s}_{1}{s}_{2}{s}_{1}{s}_{2}t\text{.}$ Similarly, ${e}_{-3{\alpha }_{1}-2{\alpha }_{2}}$ must correspond to the 3-dimensional module with weights ${w}_{0}t$ and ${s}_{2}{w}_{0}t\text{.}$

For $n={e}_{{\alpha }_{2}}+{e}_{3{\alpha }_{1}+{\alpha }_{2}},$ $C\left(s,n\right)\cong {S}_{3}$ as noted above. The module ${H}_{•}\left({ℬ}_{s,n}\right)$ contains the trivial representation of ${S}_{3}$ and the 2-dimensional representation $\chi$ of ${S}_{3}\text{.}$ Thus ${H}_{•}\left({ℬ}_{s,n}\right)$ contributes two modules, indexed by $\left({s}_{t},n,\text{triv}\right)$ and $\left({s}_{t},n,\chi \right)\text{.}$ The module corresponding to $\left({s}_{t},n,\chi \right)$ is 1-dimensional with weight ${s}_{2}t,$ while the module corresponding to $\left({s}_{t},n,\text{triv}\right)$ is 3-dimensional with weights $t$ and ${s}_{1}t\text{.}$

This leaves only one simple module left - a 1-dimensional module with weight ${s}_{2}{s}_{1}t\text{.}$ However, there are two nilpotent orbits remaining - ${e}_{{\alpha }_{2}}$ and ${e}_{{\alpha }_{1}+{\alpha }_{2}}\text{.}$ Moreover, ${H}_{•}\left({ℬ}_{s,n}\right)$ has this module as a composition factor if $n$ is either ${e}_{{\alpha }_{2}}$ or ${e}_{{\alpha }_{1}+{\alpha }_{2}}\text{.}$ In short, even after considering condition 3.1, the indexing set for the $\stackrel{\sim }{H}\text{-modules}$ is too large. If we were to remove either the triple $\left(s,{e}_{{\alpha }_{2}}+{e}_{3{\alpha }_{1}+{\alpha }_{2}},\chi \right)$ or one of $\left(s,{e}_{{\alpha }_{2}},1\right)$ or $\left(s,{e}_{{\alpha }_{1}+{\alpha }_{2}},1\right),$ from the indexing set, the correspondence would be clear. However, Theorem 3.2 as stated says to include all of those triples in the indexing.

It seems (although this is merely a conjecture) that the problem lies with the representations of $C\left(s,n\right)$ which are part of the indexing. Grojnowski’s combinatorial statement of the indexing set for the $\stackrel{\sim }{H}\text{-modules}$ (Theorem 3.2 above) does not seem to mention the representation $\chi$ of $C\left(s,n\right)\text{.}$ Thus the indexing set must include all possible $\left(s,n,\chi \right)$ or exclude all possible $\left(s,n,\chi \right)$ for a fixed $s$ and $n\text{.}$ Again, conjecturally, it seems that $\left(s,{e}_{{\alpha }_{2}}+{e}_{3{\alpha }_{1}+{\alpha }_{2}}\right)$ is the correct triple to exclude from the indexing, partly because doing so would solve a similar issue in the ${q}^{2}=-1$ case to follow. More importantly, the geometric statement of Grojnowski’s indexing([Gro1994-2], Theorem 1) seems to depend on the element $\chi$ in the triple $\left(s,n,\chi \right),$ so it seems strange that $\chi$ does not appear in the combinatorial statement.

${q}^{4}=1$

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

For the central character ${t}_{1,{q}^{2}},$ Grojnowski’s condition (3.1) rules out several nilpotent orbits from the indexing. The pair $\left({s}_{{t}_{1,{q}^{2}}},0\right)$ is excluded since ${𝔤}_{q}^{s}$ is not contained in $𝒩\text{.}$ The orbits of ${e}_{{\alpha }_{2}}$ and ${e}_{{\alpha }_{1}+{\alpha }_{2}}$ are excluded since ${e}_{3{\alpha }_{1}+{\alpha }_{2}}+{e}_{-3{\alpha }_{1}-{\alpha }_{2}},$ a non-nilpotent element of ${𝔤}_{q}^{s},$ commutes with ${f}_{{\alpha }_{2}}$ and ${f}_{{\alpha }_{1}+{\alpha }_{2}},$ respectively.

To check that the other nilpotent orbits do satisfy condition 3.1, we note that an element of ${𝔤}_{q}^{{s}_{{t}_{1,{q}^{2}}}}$ takes the form $x=a{e}_{{\alpha }_{2}}+b{e}_{{\alpha }_{1}+{\alpha }_{2}}+c{e}_{2{\alpha }_{1}+{\alpha }_{2}}+d{e}_{3{\alpha }_{1}+{\alpha }_{2}}+f{e}_{-{\alpha }_{2}}+g{e}_{-{\alpha }_{1}-{\alpha }_{2}}+h{e}_{-2{\alpha }_{1}-{\alpha }_{2}}+j{e}_{-3{\alpha }_{1}-{\alpha }_{2}}\text{.}$ We check condition 3.1 for $n$ by first finding a ${𝔰𝔩}_{2}\text{-triple}$ $\left(n,y,h\right)\text{.}$ By computing $\left[y,g\right]$ (using the basic representation of $𝔤\text{,)}$ we determine necessary and sufficient conditions on the coefficients of $g$ for $g$ to be in ${Z}_{𝔤}\left(y\right)\text{.}$ If all $g$ satisfying these conditions are again nilpotent, then condition 3.1 is satisfied. The following table summarizes this check.

$n y Condition on Z𝔤(y) Z𝔤(y)∩𝔤qs⊆𝒩 00noneno eα2 e-α2 a=b=0 no eα1+α2 e-α1-α2 a=b=c=h=0 no eα2+e3α1+α2 2e-α2-2e-3α1-α2 a=b=c=d=0=f+j yes eα2+e-α1-2α2 10e-α2-6eα1+α2 a=c=g=h=j=0=3e+5b yes eα2+e-2α1-α2 e-α2+e2α1+α2 a=b=g=h=j=0 yes$

If $\left(s,n\right)=\left({s}_{{t}_{1,{q}^{2}}},{e}_{{\alpha }_{2}}+{e}_{3{\alpha }_{1}+{\alpha }_{2}}\right),$ then we note that

$exp(eα2+e3α1+α2) =eα2(1) e3α1+α2(1) e3α1+2α2(-1/2).$

Then ${ℬ}_{s,n}$ consists of $B,$ ${x}_{{\alpha }_{1}}\left(c\right){w}_{1}B,$ and ${x}_{{\alpha }_{1}}\left(\zeta \right){w}_{1}{w}_{2}B,$ where ${\zeta }^{3}=1\text{.}$ This is a ${ℙ}^{1}$ and three points.

The centralizer ${C}_{G}\left(s,n\right)$ is generated by ${x}_{3{\alpha }_{1}+2{\alpha }_{2}}\left(c\right),$ ${H}_{{\alpha }_{1}}\left({\left(-1\right)}^{1/3}\right){H}_{{\alpha }_{2}}\left(-1\right){w}_{1},$ which has order 2, and ${H}_{{\alpha }_{1}}\left({1}^{1/3}\right),$ for some choice of a cube root of 1. The elements ${x}_{3{\alpha }_{1}+2{\alpha }_{2}}\left(c\right)$ are central in ${C}_{G}\left(s,n\right),$ so that the component group of ${C}_{G}\left(s,n\right)$ is isomorphic to ${S}_{3}\text{.}$ This group acts trivially on the ${ℙ}^{1}$ in ${ℬ}_{s,n},$ and permutes the three points via the natural action of ${S}_{3}\text{.}$ Then if ${H}_{•}$ is spanned by $\alpha ,\beta ,$ and $\gamma ,$ the span of $\alpha +\beta +\gamma$ is a copy of the trivial representation of ${S}_{3}$ in ${H}_{•}\left({ℬ}_{s,n}\right)\text{.}$ Thus ${H}_{•}{\left({ℬ}_{s,n}\right)}^{\text{triv}}$ is 3-dimensional, and as a $\stackrel{\sim }{H}\text{-module,}$ has a 2-dimensional generalized $t$ weight space, and a 1-dimensional ${s}_{2}t$ weight space. The remaining two dimensions of ${H}_{•}$ are $⟨\alpha ,\beta ,\gamma ⟩/\alpha +\beta +\gamma ,$ a copy of the 2-dimensional representation $\chi$ of ${S}_{3}\text{.}$ As a $\stackrel{\sim }{H}\text{-module,}$ it has a 2-dimensional ${s}_{2}t$ weight space.

If $\left(s,n\right)=\left({s}_{{t}_{1,{q}^{2}}},{e}_{{\alpha }_{2}}+{e}_{-{\alpha }_{1}-{\alpha }_{2}}\right),$ then we note that ${w}_{1}{w}_{2}{w}_{1}{w}_{2}{w}_{1}·\left({s}_{{t}_{1,{q}^{2}}},{e}_{{\alpha }_{2}}+{e}_{-{\alpha }_{1}-{\alpha }_{2}}\right)=\left({s}_{{t}_{{q}^{2},{q}^{2}}},{e}_{{\alpha }_{1}}+{e}_{{\alpha }_{2}}\right),$ which we choose as our orbit representative. Then ${ℬ}_{s,n}$ consists only of the point $B\text{.}$ The corresponding module has weight ${s}_{2}t\text{.}$

If $\left(s,n\right)=\left({s}_{{t}_{1,{q}^{2}}},{e}_{{\alpha }_{2}}+{e}_{-2{\alpha }_{1}-{\alpha }_{2}}\right),$ then we note that ${w}_{1}{w}_{2}{w}_{1}·\left({s}_{{t}_{1,{q}^{2}}},{e}_{{\alpha }_{2}}+{e}_{-2{\alpha }_{1}-{\alpha }_{2}}\right)=\left({s}_{{t}_{{q}^{2},1}},{e}_{{\alpha }_{1}}+{e}_{3{\alpha }_{1}+2{\alpha }_{2}}\right),$ which we choose as our orbit representative. Then note that $\text{exp}\left({e}_{{\alpha }_{1}}+{e}_{3{\alpha }_{1}+2{\alpha }_{2}}\right)={x}_{{\alpha }_{1}}\left(1\right){x}_{3{\alpha }_{1}+2{\alpha }_{2}}\left(1\right)\text{.}$ Then ${ℬ}_{s,n}$ consists of $B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}B,$ and ${w}_{2}{w}_{1}B\text{.}$ The resulting module has a 2-dimensional ${s}_{1}{s}_{2}t$ weight space and a 1-dimensional ${s}_{2}t$ weight space.

$B,xα1(c)w1B xα2(ζi)w1w2B B ( st1,q2, eα2+ e3α1+α2 ) ,q2=-1 ( st1,q2, eα2+ e-α1-α2 ) ,q2=-1 w1w2B B,xα2(c)w2B ( st1,q2, eα2+ e-2α1-α2 ) ,q2=-1$

As in the case ${t}_{1,{q}^{2}}$ when ${q}^{6}=-1$ it is unclear exactly how the combinatorial and geometric classifications match up in this case. The nilpotent ${e}_{{\alpha }_{2}}+{e}_{-{\alpha }_{1}-{\alpha }_{2}}$ must correspond to the 1-dimensional module with weight ${s}_{2}t$ since ${H}_{•}\left({ℬ}_{s,{e}_{{\alpha }_{2}}+{e}_{-{\alpha }_{1}-{\alpha }_{2}}}\right)$ is 1-dimensional. The 2-dimensional module with weight ${s}_{1}{s}_{2}t$ only appears in ${ℬ}_{s,n}$ if $n={e}_{{\alpha }_{2}}+{e}_{-2{\alpha }_{1}-{\alpha }_{2}},$ so it must correspond to ${e}_{{\alpha }_{2}}+{e}_{-2{\alpha }_{1}-{\alpha }_{2}}\text{.}$ But, ${H}_{•}\left({ℬ}_{s,{e}_{{\alpha }_{2}}}\right)$ contains two representations of $C\left(s,n\right)\cong {S}_{3}$ - the trivial and 2-dimensional representations.

As in the ${q}^{6}=1$ case, the indexing set for these modules is too large. However, eliminating either the triple $\left(s,{e}_{{\alpha }_{2}},\chi \right)$ or $\left(s,{e}_{{\alpha }_{2}}+{e}_{-{\alpha }_{1}-{\alpha }_{2}},1\right)$ would make the correspondence clear.

Other Cases:

We note that for any $t$ with $P\left(t\right)=\left\{\alpha \right\},$ ${𝔤}_{q}^{s}$ is spanned by ${e}_{\alpha }$ and ${e}_{-\alpha },$ and the nilpotent orbits in ${𝔤}_{q}^{s}$ are represented by 0, ${e}_{\alpha },$ and ${e}_{-\alpha }\text{.}$ However, ${e}_{\alpha }+{e}_{-\alpha }\in {Z}_{𝔤}\left(0\right),$ but ${e}_{\alpha }+{e}_{-\alpha }$ is not nilpotent. Hence, Grojnowski’s condition (3.1) shows that the pair $\left({s}_{t},0\right)$ should be excluded from the indexing.

If $\left(s,n\right)=\left({s}_{{t}_{1,q}},{e}_{3{\alpha }_{1}+2{\alpha }_{2}}\right),$ then ${ℬ}_{s,n}$ consists of $B,$ ${x}_{{\alpha }_{1}}\left(c\right){w}_{1}B,$ ${x}_{{\alpha }_{1}}\left(c\right){w}_{1}{w}_{2}B,$ ${x}_{{\alpha }_{1}}\left(c\right){w}_{1}{w}_{2}{w}_{1}B,$ ${w}_{2}B,$ and ${w}_{2}{w}_{1}B\text{.}$

Next, note that $\left({s}_{{t}_{1,q}},{e}_{-3{\alpha }_{1}-2{\alpha }_{2}}\right)={w}_{0}·\left({s}_{{t}_{1,{q}^{-1}}},{e}_{3{\alpha }_{1}+2{\alpha }_{2}}\right)\text{.}$ If $\left(s,n\right)=\left({s}_{{t}_{1,{q}^{-1}}},{e}_{3{\alpha }_{1}+2{\alpha }_{2}}\right),$ then ${ℬ}_{s,n}$ consists of $B,$ ${x}_{{\alpha }_{1}}\left(c\right){w}_{1}B,$ ${x}_{{\alpha }_{1}}\left(c\right){w}_{1}{w}_{2}B,$ ${x}_{{\alpha }_{1}}\left(c\right){w}_{1}{w}_{2}{w}_{1}B,$ ${w}_{2}B,$ and ${w}_{2}{w}_{1}B\text{.}$

$B,xα1(c)w1B w2B,xα1(c)w1w2B w2w1B,xα1(c)w1w2w1B w2w1B, xα1(c)w1w2w1B w2B, xα1(c)w1w2B B,xα1(c)w1B (st1,q,e3α1+2α2) ,q2=-1 (st1,q,e-3α1-2α2) ,q2=-1$

If $\left(s,n\right)=\left({s}_{{t}_{q,1}},{e}_{2{\alpha }_{1}+{\alpha }_{2}}\right),$ then ${ℬ}_{s,n}$ consists of $B,$ ${w}_{1}B,$ ${w}_{1}{w}_{2}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}B,$ and ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}{w}_{2}B\text{.}$

Next, note that $\left({s}_{{t}_{q,1}},{e}_{-2{\alpha }_{1}-{\alpha }_{2}}\right)={w}_{0}·\left({s}_{{t}_{{q}^{-1},1}},{e}_{2{\alpha }_{1}+{\alpha }_{2}}\right)\text{.}$ Then if $\left(s,n\right)=\left({s}_{{t}_{{q}^{-1},1}},{e}_{2{\alpha }_{1}+{\alpha }_{2}}\right),$ then ${ℬ}_{s,n}$ consists of $B,$ ${w}_{1}B,$ ${w}_{1}{w}_{2}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}B,$ and ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}{w}_{2}B\text{.}$

$B,xα2(c)w2B w1B, xα2(c)w2w1B w1w2B, xα2(c)w2w1w2B w1w2B, xα2(c)w2w1w2B w1B, xα2(c)w2w1B B,xα2(c)w2B (stq,1,e2α1+α2) ,q2=-1 (stq,1,e-2α1-α2) ,q2=-1$

If $\left(s,n\right)=\left({s}_{{t}_{-{1}^{1/3},1}},{e}_{3{\alpha }_{1}+{\alpha }_{2}}\right),$ then ${ℬ}_{s,n}$ consists of $B,$ ${w}_{1}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}B,$ ${w}_{2}{w}_{1}{w}_{2}B,$ and ${w}_{2}{w}_{1}{w}_{2}{w}_{1}B\text{.}$

Next, note that $\left({s}_{{t}_{-{1}^{1/3},1}},{e}_{-3{\alpha }_{1}-{\alpha }_{2}}\right)={w}_{0}·\left({s}_{{t}_{-{1}^{1/3},1}},{e}_{3{\alpha }_{1}+{\alpha }_{2}}\right)\text{.}$ Then for the pair $\left(s,n\right)=\left({s}_{{t}_{-{1}^{1/3},1}},{e}_{3{\alpha }_{1}+{\alpha }_{2}}\right),$ ${ℬ}_{s,n}$ consists of $B,$ ${w}_{1}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}B,$ ${w}_{2}{w}_{1}{w}_{2}B,$ and ${w}_{2}{w}_{1}{w}_{2}{w}_{1}B\text{.}$

Then, ${w}_{2}{w}_{1}{w}_{2}·\left({s}_{{t}_{-{1}^{1/3},1}},{e}_{3{\alpha }_{1}+{\alpha }_{2}}+{e}_{-3{\alpha }_{1}-2{\alpha }_{2}}\right)=\left({s}_{{t}_{{1}^{1/3},-1}},{e}_{{\alpha }_{2}}+{e}_{3{\alpha }_{1}+{\alpha }_{2}}\right),$ which we use as our orbit representative. Then $\text{exp}\left({e}_{{\alpha }_{2}}+{e}_{3{\alpha }_{1}+{\alpha }_{2}}\right)={x}_{{\alpha }_{2}}\left(1\right){x}_{3{\alpha }_{1}+{\alpha }_{2}}\left(1\right){x}_{3{\alpha }_{1}+2{\alpha }_{2}}\left(-1/2\right)\text{.}$ Then ${ℬ}_{s,n}$ consists of $B$ and ${w}_{1}B\text{.}$

$B,xα2(c)w2B w1B, xα2(c)w2w1B w2w1w2B w2w1w2w1B w2w1w2w1B w2w1w2B w1B, xα2(c)w2w1B B,xα2(c)w2B (st-11/3,1,e3α1+α2) ,q2=-1 (st-11/3,1,e-3α1-α2) ,q2=-1$

If $\left(s,n\right)=\left({s}_{{t}_{{q}^{2},z}},{e}_{{\alpha }_{1}}\right),$ then ${ℬ}_{s,n}$ consists of $B,$ ${w}_{2}B,$ ${w}_{2}{w}_{1}B,$ ${w}_{2}{w}_{1}{w}_{2}B,$ ${w}_{2}{w}_{1}{w}_{2}{w}_{1}B,$ and ${w}_{2}{w}_{1}{w}_{2}{w}_{1}{w}_{2}B\text{.}$

Then, note that ${w}_{0}·\left({s}_{{t}_{{q}^{2},z}},{e}_{-{\alpha }_{1}}\right)-\left({s}_{{t}_{{q}^{2},{z}^{-1}}},{e}_{{\alpha }_{1}}\right),$ so we use this latter pair as our orbit representative. Then if $\left(s,n\right)=\left({s}_{{t}_{{q}^{2},{z}^{-1}}},{e}_{{\alpha }_{1}}\right),$ then ${ℬ}_{s,n}$ consists of $B,$ ${w}_{2}B,$ ${w}_{2}{w}_{1}B,$ ${w}_{2}{w}_{1}{w}_{2}B,$ ${w}_{2}{w}_{1}{w}_{2}{w}_{1}B,$ and ${w}_{2}{w}_{1}{w}_{2}{w}_{1}{w}_{2}B\text{.}$

$B w2B w2w1B w2w1w2B w1w2w2w1B w2w1w2w1w2B B w2B w2w1B w2w1w2B w2w1w2w1B w2w1w2w1w2B (stq2,z,e-α1) ,q2=-1 (stq2,z,eα1) ,q2=-1$

If $\left(s,n\right)=\left({s}_{{t}_{z,{q}^{2}}},{e}_{{\alpha }_{1}}\right),$ then ${ℬ}_{s,n}$ consists of $B,$ ${w}_{2}B,$ ${w}_{2}{w}_{1}B,$ ${w}_{2}{w}_{1}{w}_{2}B,$ ${w}_{2}{w}_{1}{w}_{2}{w}_{1}B,$ and ${w}_{2}{w}_{1}{w}_{2}{w}_{1}{w}_{2}B\text{.}$

Then, note that ${w}_{0}·\left({s}_{{t}_{z,{q}^{2}}},{e}_{-{\alpha }_{1}}\right)=\left({s}_{{t}_{{z}^{-1},{q}^{2}}},{e}_{{\alpha }_{1}}\right),$ so we use this latter pair as our orbit representative. Then if $\left(s,n\right)=\left({s}_{{t}_{{z}^{-1},{q}^{2}}},{e}_{{\alpha }_{1}}\right),$ then ${ℬ}_{s,n}$ consists of $B,$ ${w}_{2}B,$ ${w}_{2}{w}_{1}B,$ ${w}_{2}{w}_{1}{w}_{2}B,$ ${w}_{2}{w}_{1}{w}_{2}{w}_{1}B,$ and ${w}_{2}{w}_{1}{w}_{2}{w}_{1}{w}_{2}B\text{.}$

$w1w2w1w2w1B w1w2w1w2B w1w2w1B w1w2B w1B B w1w2w1w2w1B w1w2w1w2B w1w2w1B w1w2B w1B B (stz,q2,eα2) ,q2=-1 (stz,q2,e-α2) ,q2=-1$

${q}^{2}=1$

When ${q}^{2}=1,$ $Z\left(t\right)=P\left(t\right)$ for any $t,$ and ${C}_{G}\left(s\right)$ is the Lie group generated by $D$ and $\left\{{x}_{\alpha }\left(c\right) \mid \alpha \in {𝔤}_{q}^{s},c\in ℂ\right\}\text{.}$ In fact, if ${e}_{\alpha }$ and ${e}_{\beta }$ are elements of ${𝔤}_{q}^{s},$ then ${e}_{\alpha +\beta }\in {𝔤}_{q}^{s}$ as well. Then ${𝔤}_{q}^{s}$ is a Lie subalgebra of $𝔤$ and ${C}_{G}\left(s\right)$ is its associated Lie Group. Then the ${C}_{G}\left(s\right)$ orbits of ${𝒩}_{q}^{s}$ are exactly the (adjoint) nilpotent orbits of ${𝔤}_{q}^{s}\text{.}$ In addition, the set of Borel subgroups of ${C}_{G}\left(s\right)$ is precisely ${ℬ}_{s},$ and the Weyl group of ${C}_{G}\left(s\right)$ is ${W}_{t},$ the stabilizer of $t$ in ${W}_{0}\text{.}$

Then, the Springer correspondence gives a bijection between irreducible representations of ${W}_{t}$ and ${C}_{G}\left({s}_{t}\right)\text{-orbits}$ of pairs $\left(n,\chi \right),$ where $n$ is a nilpotent element of $𝔤$ and $\chi$ is a simple representation of the component group of ${C}_{G}\left({s}_{t},n\right)$ that appears in $H\left({ℬ}_{s,n}\right)\text{.}$ But these are exactly the $G\text{-orbits}$ of triples $\left({s}_{t},n,\chi \right)$ where $\chi$ is a simple representation of $C\left(s,n\right)$ that appears in ${H}_{•}\left({ℬ}_{s,n}\right)\text{.}$ Then the orbits of such triples are in bijection with the irreducible representations of ${W}_{t}\text{.}$ In turn, the results of section 1.2.9 show that the irreducible representations of ${W}_{t}$ are in bijection with the irreducible representations of $\stackrel{\sim }{H}$ with central character $t\text{.}$ Thus, if ${q}^{2}=1,$ using the Springer correspondences for all the potential groups ${W}_{t}$ gives a geometric indexing of the irreducible representations of $\stackrel{\sim }{H}\text{.}$

Bijections

We give explicitly the bijections between irreducible representations of $\stackrel{\sim }{H}$ and orbits in $\left\{\left(s,n\right) \mid s\in {G}^{ss},n\in {𝔤}_{q}^{s}\right\}$ paired with representations of $C{\left(s,n\right)}^{\circ }$ appearing in ${H}_{•}\left({ℬ}_{s,n}\right)\text{.}$

If ${q}^{12}=1,$ note that the pair $\left({s}_{{t}_{{q}^{2},{q}^{2}}},0\right)$ does not satisfy the condition to be included in the indexing set, since ${e}_{{\alpha }_{1}}+{e}_{{\alpha }_{2}}+{e}_{-3{\alpha }_{1}-2{\alpha }_{2}}$ is not nilpotent.

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.

If ${q}^{6}=1,$ 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 ${t}_{1,{q}^{2}}\text{.}$ For the central character ${t}_{1,{q}^{2}},$ the correspondence is unclear. We give our best guess at what the correspondence should be, which involves eliminating the only triple with a non-trivial representation $\chi$ of $C\left(s,n\right)\text{.}$

When ${q}^{4}=1,$ then we note that the correspondence for the central character ${t}_{1,{q}^{2}}$ is unclear. We have given here our best guess at what the correspondence should be, which involves eliminating the only triple with a non-trivial representation of $C\left(s,n\right)\text{.}$

In the table for the case ${q}^{2}=1,$ we use the notation of Carter ([Car1985], p. 412) to denote representations of ${W}_{0}\text{.}$

$(s,n,χ) Dimension weights ( st1,1,0,1 ) 12 Wt ( st1,-1,0,1 ) 12 Wt ( st11/3,1 ,0,1 ) 12 Wt ( st1,q2,0,1 ) 3 s2w0t,w0t ( st1,q2, eα2,1 ) 1 s2w0t ( st1,q2, eα2+ e3α1+α2, 1 ) 3 t,s2t ( st1,q2, eα2+ e3α1+α2, χ ) 1 s2t ( st1,q2, eα1+α2,1 ) 2 s1s2t,s2s1s2t ( st1,±q,0,1 ) 6 s2s1s2t,… s2s1s2s1s2t ( st1,±q, e3α1+2α2,1 ) 6 t,s2t,s1s2t ( st1,z,0,1 ) 12 Wt ( stq2,1,0,1 ) 6 s1t,s2s1t ,…w0t ( stq2,1,eα1,1 ) 6 t,s1t,s2s1t, s1s2s1t,s2 s1s2s1t ( st±q,1,0,1 ) 6 s1s2s1t,… s1s2s1s2 s1t ( st±q,1, e2α1+α2,1 ) 6 t,s1t,s2s1t ( stq2/3,1 ,0,1 ) 6 s2s1t,s1s2 s1t,…s2w0t ( stq2/3,1 e3α1+α2,1 ) 6 t,s1t,s2s1t ,s1s2s1t ( stz,1,0,1 ) 12 Wt ( stq2,-q-2 ,0,1 ) 3 s2s1s2s1t, s2w0t, w0t ( stq2,-q-2, eα1,1 ) 3 s2s1s2t, s1s2s1s2t, s1w0t ( stq2,-q-2, e3α1+2α2,1 ) 3 s1t, s2s1t, s1s2s1t ( stq2,-q-2, eα1+ e3α1+2α2,1 ) 3 t,s2t,s1s2t ( st11/3,q2 ,0,1 ) 2 w0t,s1w0t ( st11/3,q2, eα2,1 ) 4 s1s2t,s1s2 s1t,s1s2s1 s2t,s2w0t ( st11/3,q2, e3α1+α2,1 ) 4 s2t,s2s1t, s2s1s2t, s2s1s2s1t ( st11/3,q2, eα2+ e3α1+α2,1 ) 2 t,s1t ( stq2,q2, 0,1 ) 1 w0t ( stq2,q2, eα1,1 ) 5 s2t,…s1w0t ( stq2,q2, eα2,1 ) 5 s1t,…s2w0t ( stq2,q2, eα1+eα2,1 ) 1 t ( stq2,z,0,1 ) 6 s1t,s2s1t ,…w0t ( stq2,z, eα1,1 ) 6 t,s2t,… s1w0t ( stz,q2,0,1 ) 6 s2t,s1s2t ,…w0t ( stz,q2, eα2,1 ) 6 t,s1t,… s2w0t ( stz,w,0,1 ) 12 Wt Table 32: Geometric Indexing in Type G2, with generic q. (s,n,χ) Dimension weights ( st1,1,0,1 ) 12 Wt ( st1,-1,0,1 ) 12 Wt ( st11/3,1 ,0,1 ) 12 Wt ( st1,q2,0,1 ) 3 s2w0t,w0t ( st1,q2, eα2,1 ) 1 s2w0t ( st1,q2, eα2+ e3α1+α2, 1 ) 3 t,s2t ( st1,q2, eα2+ e3α1+α2, χ ) 1 s2t ( st1,q2, eα1+α2,1 ) 2 s1s2t,s2s1s2t ( st1,±q,0,1 ) 6 s2s1s2t, s1s2s1s2t, s2s1s2s1s2t ( st1,±q, e3α1+2α2,1 ) 6 t,s2t,s1s2t ( st1,z,0,1 ) 12 Wt ( stq2,1,0,1 ) 6 s1t,s2s1t ,…w0t ( stq2,1,eα1,1 ) 6 t,s1t,…,s2 s1s2s1t ( st±q,1,0,1 ) 6 s1s2s1t, s2s1s2s1t, s1s2s1s2s1t ( st±q,1, e2α1+α2,1 ) 6 t,s1t,s2s1t ( stq2/3,1 ,0,1 ) 6 s2s1t, s1s2s1t, s2s1s2s1t, s1s2s1s2s1t ( stq2/3,1 e3α1+α2,1 ) 6 t,s1t,s2s1t ,s1s2s1t ( stz,1,0,1 ) 12 Wt ( stq2,-q-2 ,0,1 ) 3 s2s1s2s1t, s2w0t, w0t ( stq2,-q-2, eα1,1 ) 3 s2s1s2t, s1s2s1s2t, s1w0t ( stq2,-q-2, e3α1+2α2,1 ) 3 s1t, s2s1t, s1s2s1t ( stq2,-q-2, eα1+ e3α1+2α2,1 ) 3 t,s2t,s1s2t ( st11/3,q2 ,0,1 ) 2 w0t,s1w0t ( st11/3,q2, eα2,1 ) 4 s1s2t,s1s2 s1t,s1s2s1 s2t,s2w0t ( st11/3,q2, e3α1+α2,1 ) 4 s2t,s2s1t, s2s1s2t, s2s1s2s1t ( st11/3,q2, eα2+ e3α1+α2,1 ) 2 t,s1t ( stq2,q2, eα1,1 ) 2 s2t,s1s2t ( stq2,q2, eα2,1 ) 3 s1t,s2s1t,s1s2s1t ( stq2,q2, eα1+eα2,1 ) 1 t ( ss2s1s2tq2,q2, eα2,1 ) 1 w0t ( ss2s1s2tq2,q2, eα2+ e2α1+α2,1 ) 3 s2s1s2t,…s1w0t ( ss2s1s2s1tq2,q2, e3α1+α2 +eα2,1 ) 2 s2s1s2s1t,s2w0t ( stq2,z,0,1 ) 6 s1t,s2s1t ,…w0t ( stq2,z, eα1,1 ) 6 t,s2t,… s2s1s2s1s2t ( stz,q2,0,1 ) 6 s2t,s1s2t ,…w0t ( stz,q2, eα2,1 ) 6 t,s1t, s2s1t,… s1s2s1s2s1t ( stz,w,0,1 ) 12 Wt Table 33: Geometric Indexing in Type G2, with q12=1. (s,n,χ) Dimension weights ( st1,1,0,1 ) 12 Wt ( st1,-1,0,1 ) 12 Wt ( st11/3,1 ,0,1 ) 12 Wt ( st1,q2,0,1 ) 3 s2w0t,w0t ( st1,q2, eα2,1 ) 1 s2w0t ( st1,q2, eα2+ e3α1+α2, 1 ) 3 t,s2t ( st1,q2, eα2+ e3α1+α2, χ ) 1 s2t ( st1,q2, eα1+α2,1 ) 2 s1s2t,s2s1s2t ( st1,±q,0,1 ) 6 s2s1s2t, s1s2s1s2t, s2s1s2s1s2t ( st1,±q, e3α1+2α2,1 ) 6 t,s2t,s1s2t ( st1,z,0,1 ) 12 Wt ( stq2,1,0,1 ) 6 s1t, s2s1t, … s1s2s1s2s1t ( stq2,1,eα1,1 ) 6 t,s1t,…,s2 s1s2s1t ( stq-4,1, 0, 1 ) 1 s1s2s1t ( stq-4,1, e2α1+α2, 1 ) 5 t, s1t, s2s1t ( stq-4,1, e-3α1-α2, 1 ) 5 s1s2s1t, s2s1s2s1t, s1s2s1s2s1t ( stq-4,1, e2α1+α2+ e-3α1-α2, 1 ) 1 s2s1t ( st-q-4,1, 0, 1 ) 6 s1s2s1t, s2s1s2s1t, s1s2s1s2s1t ( st-q-4,1, e2α1+α2, 1 ) 6 t, s1t, s2s1t ( stq2/3,1 ,0,1 ) 6 s2s1t, s1s2s1t, s2s1s2s1t, s1s2s1s2s1t ( stq2/3,1 e3α1+α2,1 ) 6 t,s1t,s2s1t ,s1s2s1t ( stz,1,0,1 ) 12 Wt ( stq2,-q-2 ,0,1 ) 3 s2s1s2s1t, s2w0t, w0t ( stq2,-q-2, eα1,1 ) 3 s2s1s2t, s1s2s1s2t, s1w0t ( stq2,-q-2, e3α1+2α2,1 ) 3 s1t, s2s1t, s1s2s1t ( stq2,-q-2, eα1+ e3α1+2α2,1 ) 3 t,s2t,s1s2t ( st11/3,q2 ,0,1 ) 2 w0t,s1w0t ( st11/3,q2, eα2,1 ) 4 s1s2t,s1s2 s1t,s1s2s1 s2t,s2w0t ( st11/3,q2, e3α1+α2,1 ) 4 s2t,s2s1t, s2s1s2t, s2s1s2s1t ( st11/3,q2, eα2+ e3α1+α2,1 ) 2 t,s1t ( stq2,z,0,1 ) 6 s1t,s2s1t ,…w0t ( stq2,z, eα1,1 ) 6 t, s2t, s1s2t, … s2s1s2s1s2t ( stz,q2,0,1 ) 6 s2t,s1s2t ,…w0t ( stz,q2, eα2,1 ) 6 t,s1t, s2s1t,… s1s2s1s2s1t ( stz,w,0,1 ) 12 Wt Table 34: Geometric Indexing in Type G2, with q10=1. (s,n,χ) Dimension weights ( st1,1,0,1 ) 12 Wt ( st1,-1,0,1 ) 12 Wt ( st11/3,1 ,0,1 ) 12 Wt ( st1,q2,0,1 ) 3 s2w0t,w0t ( st1,q2, eα2,1 ) 1 s2w0t ( st1,q2, eα2+ e3α1+α2, 1 ) 3 t,s2t ( st1,q2, eα2+ e3α1+α2, χ ) 1 s2t ( st1,q2, eα1+α2,1 ) 2 s1s2t,s2s1s2t ( st1,±q,0,1 ) 6 s2s1s2t, s1s2s1s2t, s2s1s2s1s2t ( st1,±q, e3α1+2α2,1 ) 6 t,s2t,s1s2t ( st1,z,0,1 ) 12 Wt ( st±q,1, 0,1 ) 6 s1s2s1t, s2s1s2s1t, s1s2s1s2s1t ( st±q,1, e2α1+α2,1 ) 6 t,s1t,s2s1t ( stq2/3,1 ,0,1 ) 6 s2s1t, s1s2s1t, s2s1s2s1t, s1s2s1s2s1t ( stq2/3,1 e3α1+α2,1 ) 6 t,s1t,s2s1t ,s1s2s1t ( stz,1,0,1 ) 12 Wt ( st11/3,q2 ,0,1 ) 2 w0t,s1w0t ( st11/3,q2, eα2,1 ) 4 s1s2t,s1s2 s1t,s1s2s1 s2t,s2w0t ( st11/3,q2, e3α1+α2,1 ) 4 s2t,s2s1t, s2s1s2t, s2s1s2s1t ( st11/3,q2, eα2+ e3α1+α2,1 ) 2 t,s1t ( stq2,q2, 0,1 ) 1 w0t ( stq2,q2, eα1,1 ) 3 s2s1s2t, s1s2s1s2t ( stq2,q2, eα2,1 ) 3 t, s1t ( stq2,q2, eα1+eα2,1 ) 1 t ( stq2,q2, eα1+ e3α1+2α2,1 ) 2 s2t, s1s2t ( stq2,z,0,1 ) 6 s1t,s2s1t ,…w0t ( stq2,z, eα1,1 ) 6 t, s2t, s1s2t, … s2s1s2s1s2t ( stz,q2,0,1 ) 6 s2t,s1s2t ,…w0t ( stz,q2, eα2,1 ) 6 t,s1t, s2s1t,… s1s2s1s2s1t ( stz,w,0,1 ) 12 Wt Table 35: Geometric Indexing in Type G2, with q8=1. (s,n,χ) Dimension weights ( st1,1,0,1 ) 12 Wt ( st1,-1,0,1 ) 12 Wt ( st1,q2, eα2,1 ) 1 s1s2t ( st1,q2, eα2+ e3α1+α2, 1 ) 3 t,s2t ( st1,q2, eα2+ e3α1+α2, χ ) ? ? ( st1,q2, eα1+α2,1 ) 1 s2t ( st1,q2, e-3α1-2α2,1 ) 3 s1s2s1s2t, s2s1s2s1s2t ( st1,q2, eα2- e-3α1-2α2,1 ) 1 s1s2s1s2t ( st1,q2, eα1+α2+ e-3α1-2α2,1 ) 1 s2s1s2t ( st1,-q-2,0,1 ) 6 s2s1s2t, s1s2s1s2t, s2s1s2s1s2t ( st1,-q-2, e3α1+2α2,1 ) 6 t,s2t,s1s2t ( st1,z,0,1 ) 12 Wt ( stq2,1, 0,1 ) 6 t, t, s1t, s1t, s1t, s1t ( stq2,1, eα1,1 ) 6 t, t, t, t, s1t, s1t ( st-q-2,1, 0, 1 ) 6 s1s2s1t, s2s1s2s1t, s1s2s1s2s1t ( st-q-2,1, e2α1+α2,1 ) 6 t,s1t,s2s1t ( stq2/3,1 ,0,1 ) 6 s2s1t, s1s2s1t, s2s1s2s1t, s1s2s1s2s1t ( stq2/3,1 e3α1+α2,1 ) 6 t,s1t,s2s1t ,s1s2s1t ( stz,1,0,1 ) 12 Wt ( stq2,-q-2 ,0,1 ) 3 s2s1s2s1t, s2w0t, w0t ( stq2,-q-2, eα1,1 ) 3 s2s1s2t, s1s2s1s2t, s1w0t ( stq2,-q-2, e3α1+2α2,1 ) 3 s1t, s2s1t, s1s2s1t ( stq2,-q-2, eα1+ e3α1+2α2,1 ) 3 t,s2t,s1s2t ( stq2,z,0,1 ) 6 s1t,s2s1t ,…w0t ( stq2,z, eα1,1 ) 6 t, s2t, s1s2t, … s2s1s2s1s2t ( stz,q2,0,1 ) 6 s2t,s1s2t ,…w0t ( stz,q2, eα2,1 ) 6 t,s1t, s2s1t,… s1s2s1s2s1t ( stz,w,0,1 ) 12 Wt Table 36: Geometric Indexing in Type G2, with q6=1. (s,n,χ) Dimension weights ( st1,1,0,1 ) 12 Wt ( st1,q2, eα2+ e3α1+α2, 1 ) 2 t ( st1,q2, eα2+ e3α1+α2, χ ) ? ? ( st1,q2, eα2+ e-α1-α2, 1 ) 1 s2t ( st1,q2, eα2+ e-2α1-α2, 1 ) 2 s1s2t ( st11/3,1, 0, 1 ) 12 Wt ( st1,q, e3α1+2α2, 1 ) 6 t, s1t, … s1s2s1s2s1t ( st1,q, e-3α1-2α2, 1 ) 6 s2t, s1s2t, … w0t ( st1,z,0,1 ) 12 Wt ( stq,1, e2α1+α2, 1 ) 6 t, s1t, s2s1t, … s1s2s1s2s1t ( stq,1, e-2α1-α2, 1 ) 6 s2t, s1s2t, … w0t ( stq2/3,1, e3α1+α2, 1 ) 4 t, s1t ( stq2/3,1 e-3α1-α2, 1 ) 4 w0t, s1w0t ( stq2/3,1 e3α1+α2+ e-3α1-2α2, 1 ) 2 s2s1t, s1s2s1t ( stz,1,0,1 ) 12 Wt ( stq2,z, eα1,1 ) 6 t, s2t, s1s2t, … s2s1s2s1s2t ( stq2,z, e-α1,1 ) 6 s1t,s2s1t ,…w0t ( stz,q2, eα2,1 ) 6 t,s1t, s2s1t,… s1s2s1s2s1t ( stz,q2, e-α2,1 ) 6 s2t,s1s2t ,…w0t ( stz,w,0,1 ) 12 Wt Table 37: Geometric Indexing in Type G2, with q4=1. (s,n,χ) Dimension Wt representation ( st1,1,0,1 ) 1 sign ( st1,1 , eα1 , 1 ) 2 ϕ2,2 ( st1,1 , eα2 , 1 ) 1 ϕ1,3′′ ( st1,1 , eα1+eα2 , 1 ) 1 triv ( st1,1 , eα1+ e3α1+2α2 , 1 ) 2 ϕ2,1 ( st1,1 , eα1+ e3α1+2α2 , χ ) 2 ϕ1,3′ ( st1,-1 , 0 , 1 ) 3 sign⊗sign ( st1,-1 , eα1 , 1 ) 3 sign⊗triv ( st1,-1 , e3α1+2α2 , 1 ) 3 triv⊗sign ( st1,-1 , eα1+ e3α1+2α2 , 1 ) 3 triv⊗triv ( st11/3,1 , 0 , 1 ) 3 sign ( st11/3,1 , eα2 , 1 ) 6 χ ( st11/3,1 , eα2+ e3α1+α2 , 1 ) 3 triv ( st1,z , 0 , 1 ) 6 sign ( st1,z , eα1 , 1 ) 6 triv ( stz,1 , 0 , 1 ) 6 sign ( stz,1 , eα2 , 1 ) 6 triv ( stz,w , 0 , 1 ) 12 triv Table 38: Geometric Indexing in Type G2, with q=-1.$

Notes and References

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