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

Last update: 16 March 2013

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

Let

$J= [ 0010 0001 -1000 0-100 ] .$

Then we define the symplectic group

$SP4= { M∈GLn(ℂ) ∣ MTJM =J } .$

If we write $M=\left[\begin{array}{cc}A& B\\ C& D\end{array}\right],$ then the condition ${M}^{T}JM=J$ yields ${B}^{T}D-{D}^{T}B=0={A}^{T}C-{C}^{T}A,$ and ${A}^{T}D-{C}^{T}B=I={D}^{T}A-{B}^{T}C\text{.}$ Although it is not immediately obvious, $S{P}_{4}\subseteq S{L}_{4}\text{.}$

Let $G={\text{SP}}_{4}\left(ℂ\right)$ and $𝔤=𝔰{𝔭}_{4}\text{.}$ Each root space ${𝔤}_{\alpha }$ in $𝔤$ is 1-dimensional, and we choose distinguished vectors ${e}_{\alpha }$ in each one.

$eα1 = [ 0000 1000 000-1 0000 ] eα2 = [ 0010 0000 0000 0000 ] eα1+α2 = [ 0001 0010 0000 0000 ] e2α1+α2 = [ 0000 0001 0000 0000 ] e-α1 = [ 0100 0000 0000 00-10 ] e-α2 = [ 0000 0000 1000 0000 ] e-α1-α2 = [ 0000 0000 0100 1000 ] e-2α1-α2 = [ 0000 0000 0000 0100 ]$

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

$g·(s,n)= ( gsg-1, Adg(n) ) .$

For $\alpha \in R,$ let ${e}_{\alpha }$ be a non-zero element in the $\alpha$ root space of $𝔤\text{.}$ Let $D\subseteq G$ be the subgroup of diagonal matrices in $G\text{.}$ Then

$ϕ:T ⟶ D t ⟼ st= [ t(Xω2-ω1) 0 0 0 0 t(Xω1) 0 0 0 0 t(Xω1-ω2) 0 0 0 0 t(X-ω1) ] t(Xω1) = d2 t(Xω2) = d1d2 ⟵ [ d1000 0d200 00d1-10 000d2-1 ] ,$

is a bijection satisfying

$Adst·eα= t(Xα)eα$

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

$θ: {W0-orbits on T} ⟷ {G-orbits on Gss} t ⟼ st. (3.9)$

The weight ${t}_{z,w,i}$ satisfies $t\left({X}^{{\alpha }_{1}}\right)=z$ and $t\left({X}^{{\alpha }_{2}}\right)=w,$ for $i=1$ or 2, and $z,w\in {ℂ}^{×}\text{.}$ Then define

$dα1(z)= stz,1,1 and dα2(z) =st1,z,1,$

for $z\in ℂ,$ so that

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

for $i,j=1,2$ and $i\ne j\text{.}$ (We could have used ${t}_{z,1,2}$ and ${t}_{z,1,2}$ in our definitions or ${d}_{{\alpha }_{i}}\left(z\right),$ but all that matters is to have some elements of $D$ that satisfy 3.4.)

We begin by choosing a set of representatives for the orbits in $\Lambda /G$ for all values of $q\text{.}$ After this, we give the bijection between $\Lambda /G$ and the irreducible $\stackrel{\sim }{H}\text{-modules.}$

Theorem 3.19. If ${q}^{2}$ is not a primitive $\ell \text{th}$ root of unity with $\ell \le 4$ then the following is a set of representatives of the orbits in $\Lambda /G\text{.}$

$( ± [ 1000 0100 0010 0001 ] ,0 ) , ( [ -1000 0100 00-10 0001 ] ,0 ) , ( ± [ z1/2000 0z1/200 00z-1/20 000z-1/2 ] ,0 ) ,z≠1,q±2, ( ± [ q000 0q00 00q-10 000q-1 ] ,0 ) , ( ± [ q000 0q00 00q-10 000q-1 ] ,eα2 ) , ( ± [ q000 0q00 00q-10 000q-1 ] ,eα1+α2 ) , ( ± [ 1000 0q200 0010 000q-2 ] ,0 ) , ( ± [ 1000 0q200 0010 000q-2 ] ,eα1 ) , ( ± [ 1000 0q00 0010 000q-1 ] ,0 ) , ( ± [ 1000 0q00 0010 000q-1 ] ,e2α1+α2 ) , ( ± [ 1000 0-q00 0010 000-q-1 ] ,0 ) , ( ± [ 1000 0-q00 0010 000-q-1 ] ,e2α1+α2 ) , ( ± [ 1000 0z00 0010 000z ] ,0 ) ,z≠±1,± q±1, q±2, ( ± [ q000 0q300 00q-10 000q-3 ] ,0 ) , ( ± [ q000 0q300 00q-10 000q-3 ] ,eα1 ) , ( ± [ q000 0q300 00q-10 000q-3 ] ,eα2 ) , ( ± [ q000 0q300 00q-10 000q-3 ] ,eα1+ eα2 ) , ( ± [ z1/2000 0q2z1/200 00z-1/20 000q-2z-1/2 ] ,0 ) ,z≠1,q±2, q-4,q-6, ( ± [ z1/2000 0q2z1/200 00z-1/20 000q-2z-1/2 ] ,eα1 ) ,z≠1,q±2, q-4,q-6, ( [ -q000 0q00 00-q-10 000q-1 ] ,0 ) , ( [ -q000 0q00 00-q-10 000q-1 ] ,eα2 ) , ( [ -q000 0q00 00-q-10 000q-1 ] ,e2α1+α2 ) , ( [ -q000 0q00 00-q-10 000q-1 ] ,e2+ e2α1+α2 ) , ( ± [ q000 0zq00 00q-10 000z-1q-1 ] ,0 ) , z≠±1,q±2 ,q-4,-q-2, ( ± [ q000 0zq00 00q-10 000z-1q-1 ] ,eα2 ) , z≠±1,q±2 ,q-4,-q-2, ( ± [ w1/2000 0zw1/200 00w-1/20 000z-1w-1/2 ] ,0 ) ,tz,w generic, z≠-1,$

and

$( [ w1/2000 0-w1/200 00w-1/20 000-w-1/2 ] ,0 ) ,t-1,w generic.$

 Proof. Given an element $\left(s,n\right)\in \Lambda ,$ by 3.9, 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: ${t}_{1,1},{t}_{-1,1},{t}_{1,z},{t}_{z,1},{t}_{z,w}$ For the semisimples corresponding to ${t}_{1,1},{t}_{-1,1},{t}_{1,z},{t}_{z,1},{t}_{z,w},{𝔤}_{q}^{s}=0,$ so that 0 is the only nilpotent that can be paired with $s\text{.}$ Note that if $t{\mid }_{Q}$ takes the form ${t}_{-1,w},$ then $-{s}_{t}={s}_{{s}_{1}t},$ so that these characters are in the same ${W}_{0}\text{-orbit.}$ Hence only one pair $\left({s}_{t},0\right)$ is listed, and it listed separately from the other pairs involving ${t}_{z,1}$ or ${t}_{z,w}\text{.}$ Case 2: ${t}_{±q,1},{t}_{{q}^{2},z},{t}_{z,{q}^{2}}$ For the weights ${t}_{{q}^{2},1},{t}_{±q,1},{t}_{{q}^{2},z},{t}_{z,{q}^{2}},\mid P\left(t\right)\mid =1,$ so that ${𝔤}_{q}^{s}$ is 1-dimensional. Let $a{e}_{\alpha }\in {𝔤}_{q}^{{s}_{t}}$ with $a\in {ℂ}^{×}\text{.}$ Then $dα(a-1)· (st,aeα)= (st,eα).$ Then all the elements $\left(s,n\right)\in \Lambda$ with $s={s}_{t}$ are $G\text{-conjugate}$ to $\left({s}_{t},0\right)$ or $\left({s}_{t},{e}_{\alpha }\right)\text{.}$ Note that $±{s}_{{t}_{-1,{q}^{2}}}$ are in the same ${W}_{0}$ orbit, so the $±$ is deleted from these semisimples in the listing above. Case 3: ${s}_{t}$ for ${t}_{{q}^{2},1}$ If $s={s}_{{t}_{{q}^{2},1}},$ then ${𝔤}_{q}^{s}=ℂ{e}_{{\alpha }_{1}}+ℂ{e}_{{\alpha }_{1}+{\alpha }_{2}},$ and ${C}_{G}\left(s\right)$ is generated by $D$ and ${x}_{±{\alpha }_{2}}\left(c\right)$ for $c\in {ℂ}^{×}\text{.}$ Let $a{e}_{{\alpha }_{1}}+b{e}_{{\alpha }_{1}+{\alpha }_{2}}\in {𝔤}_{q}^{s}\text{.}$ If $a,b\in {ℂ}^{×}$ then $xα2(-b/a) dα1(a)· eα1=aeα1 +beα2,$ so that $a{e}_{{\alpha }_{1}}+b{e}_{{\alpha }_{2}}$ is ${C}_{G}\left(s\right)\text{-conjugate}$ to ${e}_{{\alpha }_{1}}\text{.}$ If $a=0$ then $xα1(b) x-α1 (-1/b)·b eα1+α2= eα1,$ and if $b=0$ then $dα1(a-1) ·aeα1=eα1.$ Then every element $\left(s,n\right)\in \Lambda$ with $s={s}_{t}$ is conjugate to $\left({s}_{t},0\right)$ or $\left({s}_{t},{e}_{{\alpha }_{1}}\right)\text{.}$ Case 4: ${t}_{{q}^{2},{q}^{2}}$ If $t={t}_{{q}^{2},{q}^{2}}$ then ${𝔤}_{q}^{{s}_{t}}=ℂ{e}_{{\alpha }_{1}}+ℂ{e}_{{\alpha }_{2}}\text{.}$ If $ab\ne 0$ then $dα1(a-1) dα2(b-1)· ( st,aeα1+ beα2 ) =(st,eα1+eα2), dα1(a-1) ·(st,aeα1) =(st,eα1), and dα2(b-1) ·(st,beα2) =(st,eα2).$ Hence every pair $\left({s}_{t},n\right)$ with $n\ne 0$ is conjugate to one of these three elements of $\Lambda ,$ but since $CG(s)= { ±dα1(c) dα2(d) ∣ c,d∈ ℂ× } ,$ none of these three elements are conjugate to each other. Thus $\left({s}_{t},0\right),\left({s}_{t},{e}_{{\alpha }_{1}}\right),\left({s}_{t},{e}_{{\alpha }_{2}}\right),$ and $\left({s}_{t},{e}_{{\alpha }_{1}}+{e}_{{\alpha }_{2}}\right)$ are the representatives of the orbits of elements of $\Lambda$ with $s={s}_{t}\text{.}$ Case 5: ${t}_{-1,{q}^{2}}$ If $t={t}_{-1,{q}^{2}}$ then ${𝔤}_{q}^{{s}_{t}}=ℂ{e}_{{\alpha }_{2}}+ℂ{e}_{2{\alpha }_{1}+{\alpha }_{2}}\text{.}$ If $ab\ne 0$ then $dα2(a-1) dα1(b-1a) · ( st,aeα2+b e2α1+α2 ) = ( st,eα2+ e2α1+α2 ) , dα2(a-1)· (st,aeα2)= (st,eα2), and dα2(b-1)· ( st,b e2α1+α2 ) = (st,e2α1+α2).$ Hence every pair $\left({s}_{t},n\right)$ with $n\ne 0$ is conjugate to one of these three elements of $\Lambda \text{.}$ Since ${C}_{G}\left({s}_{t}\right)$ consists of exactly the diagonal matrices, the ${C}_{G}\left({s}_{t}\right)\text{-orbit}$ of $n\in {𝔤}_{q}^{s}$ is $ℂn\text{.}$ Thus $\left({s}_{t},0\right),\left({s}_{t},{e}_{{\alpha }_{1}}\right),\left({s}_{t},{e}_{{\alpha }_{2}}\right),$ and $\left({s}_{t},{e}_{{\alpha }_{1}}+{e}_{{\alpha }_{2}}\right)$ are representatives of the orbits of elements of $\Lambda$ with $s={s}_{t}\text{.}$ Since $±{s}_{{t}_{-1,{q}^{2}}}$ are in the same orbit, the $±$ is deleted from the pairs involving this semisimple above. Case 6: ${t}_{1,{q}^{2}}$ If $t={t}_{1,{q}^{2}},$ then $𝔤qst=ℂeα2 +ℂeα1+α2+ ℂe2α1+α2.$ and $CG(st)= ⟨ xα1(c), ±dα1(a), ±dα2(b) ∣ a,b,c∈ ℂ× ⟩ .$ Let $f{e}_{{\alpha }_{2}}+g{e}_{{\alpha }_{1}+{\alpha }_{2}}+h{e}_{2{\alpha }_{1}+{\alpha }_{2}}\in {𝒩}_{q}^{s}$ with $f,g,h\in ℂ\text{.}$ If $gf=0$ then $x-α1(d)·h e2α1+α2=h d2eα2+hd eα1+α2+h e2α1+α2,$ so we assume $gf\ne 0\text{.}$ If $fh={g}^{2}$ then $xα1 (-ggd+f) x-α1(d)· ( feα2+g eα1+α2 +he2α1+α2 ) =(f+2gd+hd2) eα2.$ If $fh\ne {g}^{2}$ then $xα1 ( -h 2g2-fh ) x-α1 ( -g+g2-hf h ) · ( feα2+g eα1+α2+h e2α1+α2 ) =(g2-hf) eα1+α2.$ But $dα2 ( (f+2gd+hd2) -1 ) ·(f+2gd+hd2) eα2=eα2 and dα2 ( (g2-hf) -1 ) ·(g2-hf) eα1+α2= eα1+α2.$ Thus the elements $\left(s,n\right)$ in $\Lambda$ with $s={s}_{t}$ are represented by $\left({s}_{t},0\right),\left({s}_{t},{e}_{{\alpha }_{2}}\right),$ and $\left({s}_{t},{e}_{{\alpha }_{1}+{\alpha }_{2}}\right)\text{.}$ $\square$

Theorem 3.20. If ${q}^{2}$ is a primitive 4th root of unity then the following is a set of representatives of the orbits in $\Lambda /G\text{.}$

$( ± [ 1000 0100 0010 0001 ] ,0 ) , ( [ -1000 0100 00-10 0001 ] ,0 ) , ( ± [ z1/2000 0z1/200 00z-1/20 000z-1/2 ] ,0 ) ,z≠1,q±2, ( ± [ q000 0q00 00q-10 000q-1 ] ,0 ) , ( ± [ q000 0q00 00q-10 000q-1 ] ,eα2 ) , ( ± [ q000 0q00 00q-10 000q-1 ] ,eα1+α2 ) , ( ± [ 1000 0q200 0010 000q-2 ] ,0 ) , ( ± [ 1000 0q200 0010 000q-2 ] ,eα1 ) , ( ± [ 1000 0q00 0010 000q-1 ] ,0 ) , ( ± [ 1000 0q00 0010 000q-1 ] ,e2α1+α2 ) , ( ± [ 1000 0-q00 0010 000-q-1 ] ,0 ) , ( ± [ 1000 0-q00 0010 000-q-1 ] ,e2α1+α2 ) , ( ± [ 1000 0z00 0010 000z ] , 0 ) , z≠±1,± q±1, q±2, ( [ q000 0q300 00q-10 000q-3 ] ,0 ) , ( [ q000 0q300 00q-10 000q-3 ] ,eα1 ) , ( [ q000 0q300 00q-10 000q-3 ] ,eα2 ) , ( [ q000 0q300 00q-10 000q-3 ] ,eα1+eα2 ) , ( [ q-3000 0q00 00q30 000q-1 ] ,eα2+ e2α1+α2 ) , ( [ q-3000 0q-100 00q30 000q ] ,eα1+eα2 ) , ( ± [ z1/2000 0q2z1/200 00z-1/20 000q-2z-1/2 ] ,0 ) ,z≠±1, q±2, q-4, q-6, ( ± [ z1/2000 0q2z1/200 00z-1/20 000q-2z-1/2 ] ,eα1 ) ,z≠±1, q±2, q-4, q-6, ( ± [ q000 0zq00 00q-10 000z-1q-1 ] ,0 ) ,z≠±1, q±2, q-4,- q-2, ( ± [ q000 0zq00 00q-10 000z-1q-1 ] ,eα2 ) ,z≠±1, q±2, q-4,- q-2, ( ± [ w1/2000 0zw1/200 00w-1/20 000z-1w-1/2 ] ,0 ) ,tz,w generic, z≠-1,$

and

$( ± [ w1/2000 0-w1/200 00w-1/20 000-w-1/2 ] ,0 ) ,tz,w generic.$

 Proof. The proof of Theorem 3.19 applies to this case as well, with changes only necessary in Case 4 and for the semisimple ${s}_{{t}_{-1,{q}^{2}}}\text{.}$ Since ${t}_{-1,{q}^{2}}$ is in the same ${W}_{0}\text{-orbit}$ as ${t}_{{q}^{2},{q}^{2}},{s}_{{t}_{-1,{q}^{2}}}$ is in the same $G\text{-orbit}$ as ${s}_{{t}_{{q}^{2},{q}^{2}}}$ and thus ${s}_{{t}_{-1,{q}^{2}}}$ is omitted from the list. Also, the two weights $t$ with $t{\mid }_{Q}={t}_{{q}^{2},{q}^{2}}$ are in the same ${W}_{0}\text{-orbit,}$ and thus the $±$ is omitted in the list of representatives. Case 4: ${t}_{{q}^{2},{q}^{2}}$ If $t={t}_{{q}^{2},{q}^{2}}$ then ${𝔤}_{q}^{{s}_{t}}=ℂ{e}_{{\alpha }_{1}}+ℂ{e}_{{\alpha }_{2}}+ℂ{e}_{-2{\alpha }_{1}-{\alpha }_{2}}\text{.}$ If $abc\ne 0$ then $a{e}_{{\alpha }_{1}}+b{e}_{{\alpha }_{2}}+c{e}_{-2{\alpha }_{1}-{\alpha }_{2}}$ is not nilpotent, since its minimal polynomial in the standard representation is ${x}^{4}+{a}^{2}bc\text{.}$ Then $dα1(a-1) dα2(a2b-1) · ( st,aeα1+b e-2α1-α2 ) = ( st,eα1+ eα2 ) , dα1 ((ab)-1/2) dα2(b-1)· ( st,a e-2α1-α2 +beα2 ) = ( st, e-2α1-α2 +eα2 ) , dα1(a-1) dα2(b-1)· ( st,aeα1 +beα2 ) = (st,eα1+eα2), dα2(a-1)· (st,ae-2α1-α2) =(st,e-2α1-α2) , and dα1(a-1)· (st,aeα1)= (st,eα1), and dα2(b-1)· (st,beα2)= (st,eα2).$ Hence every pair $\left({s}_{t},n\right)$ with $n\ne 0$ is conjugate to one of these six elements of $\Lambda ,$ but since $CG(s)= { ±dα1(c) dα2(d) ∣ c,d∈ℂ× } ,$ none of these elements are conjugate to each other. Thus the pairs $\left({s}_{t},0\right),\left({s}_{t},{e}_{{\alpha }_{1}}\right),\left({s}_{t},{e}_{{\alpha }_{2}}\right),\left({s}_{t},{e}_{-2{\alpha }_{1}-{\alpha }_{2}}\right),\left({s}_{t},{e}_{-2{\alpha }_{1}-{\alpha }_{2}}+{e}_{{\alpha }_{2}}\right),\left({s}_{t},{e}_{{\alpha }_{1}}+{e}_{{\alpha }_{2}}\right)$ and $\left({s}_{t},{e}_{{\alpha }_{1}}+{e}_{-2{\alpha }_{1}-{\alpha }_{2}}\right)$ are representatives of the orbits of elements of $\Lambda$ with $s={s}_{t}\text{.}$ However, later calculations are made easier by choosing nilpotent elements in ${𝒩}^{+}$ as representatives. Then we note: $w2w1· (st,e-2α1-α2) =(sw2w1t,eα2), w2w1· (st,e-2α1-α2+eα2) =(sw2w1t,eα2+e2α1+α2),$ and $w2w1w2· ( st,eα1+ e-2α1-α2 ) = ( sw1w2w1t, eα1+eα2 ) .$ These are the representatives listed above. $\square$

Theorem 3.21. If ${q}^{2}$ is a primitive third root of unity then the following is a set of representatives of the orbits in $\Lambda /G\text{.}$

$( ± [ 1000 0100 0010 0001 ] , 0 ) , ( [ -1000 0100 00-10 0001 ] , 0 ) , ( ± [ z1/2000 0z1/200 00z-1/20 000z-1/2 ] , 0 ) , z≠1,q±2, ( ± [ q000 0q00 00q-10 000q-1 ] , 0 ) , ( ± [ q000 0q00 00q-10 000q-1 ] , eα2 ) , ( ± [ q000 0q00 00q-10 000q-1 ] , eα1+α2 ) , ( ± [ 1000 0q200 0010 000q-2 ] , 0 ) , ( ± [ 1000 0q200 0010 000q-2 ] , eα1 ) , ( ± [ q2000 0100 00q-20 0001 ] , eα2 ) , ( ± [ q-2000 0100 00q20 0001 ] , eα1+eα2 ) , ( ± [ 1000 0q00 0010 000q-1 ] , 0 ) , ( ± [ 1000 0q00 0010 000q-1 ] , e2α1+α2 ) , ( ± [ 1000 0z00 0010 000z-1 ] , 0 ) , z≠±1, q±1, q±2, ( ± [ z1/2000 0q2z1/200 00z-1/20 000q-2z-1/2 ] , 0 ) , z≠1, q±2, q-4, q-6, ( ± [ z1/2000 0q2z1/200 00z-1/20 000q-2z-1/2 ] , eα1 ) , z≠1, q±2, q-4, q-6, ( [ -q000 0q00 00-q-10 000q-1 ] , 0 ) , ( [ -q000 0q00 00-q-10 000q-1 ] , eα2 ) , ( [ -q000 0q00 00-q-10 000q-1 ] , e2α1+α2 ) , ( [ -q000 0q00 00-q-10 000q-1 ] , eα2+ e2α1+α2 ) , ( ± [ q000 0zq00 00q-10 000z-1q-1 ] , 0 ) , z≠±1, q±2, q-4, -q-2, ( ± [ q000 0zq00 00q-10 000z-1q-1 ] , eα2 ) , z≠±1, q±2, q-4, -q-2, ( ± [ w1/2000 0zw1/200 00w-1/20 000z-1w-1/2 ] , 0 ) , tz,w generic, z≠-1, ( ± [ w1/2000 0-w1/200 00w-1/20 000-w-1/2 ] , 0 ) , tz,w generic.$

 Proof. The proof of Theorem 3.19 applies, with changes only necessary in cases 3 and 4, and omitting the characters ${t}_{-q,1}$ and ${t}_{{q}^{2},{q}^{2}}$ (Case 4), since these characters are in the same orbits as ${t}_{{q}^{2},1}$ and ${t}_{1,{q}^{2}},$ respectively. Case 3: ${s}_{t}$ for ${t}_{{q}^{2},1}$ If $s={s}_{{t}_{{q}^{2},1}}$ then ${𝔤}_{q}^{s}$ is spanned by ${e}_{{\alpha }_{1}},{e}_{{\alpha }_{1}+{\alpha }_{2}},$ and ${e}_{-2{\alpha }_{1}-{\alpha }_{2}},$ while ${C}_{G}\left(s\right)$ is generated by $D$ and ${x}_{±{\alpha }_{2}}\left(c\right)$ for $c\in {ℂ}^{×}\text{.}$ In particular, ${C}_{G}\left(s\right)$ contains ${w}_{2}\left(t\right)={x}_{{\alpha }_{2}}\left(t\right){x}_{-{\alpha }_{2}}\left(-{t}^{-1}\right){x}_{{\alpha }_{2}}\left(t\right),$ a representative of ${s}_{2}\in {W}_{0}$ in $N\left(T\right)/T\text{.}$ Let $n=w{e}_{{\alpha }_{1}}+x{e}_{{\alpha }_{1}+{\alpha }_{2}}+y{e}_{-2{\alpha }_{1}-{\alpha }_{2}}\in {𝔤}_{q}^{s},$ which is nilpotent. Since ${d}_{{\alpha }_{2}}\left(y\right)·\left(w{e}_{{\alpha }_{1}}+x{e}_{{\alpha }_{1}+{\alpha }_{2}}+y{e}_{-2{\alpha }_{1}-{\alpha }_{2}}\right)=w{e}_{{\alpha }_{1}}+x{e}_{{\alpha }_{1}+{\alpha }_{2}}+{e}_{-2{\alpha }_{1}-{\alpha }_{2}},$ we assume that $y=0$ or 1. If $y=0,$ then case 3 in the proof of Theorem 3.19 shows that either $n=0$ or $n$ is ${C}_{G}\left(s\right)\text{-conjugate}$ to ${e}_{{\alpha }_{1}}\text{.}$ Next assume $y=1\text{.}$ If $w,x\in {ℂ}^{×}$ then $dα2(w) dα1(w-1) xα2(x/w)· weα1+x eα1+α2+ e-2α2-α1= eα1+ e-2α1-α2.$ If $w=0$ then $xα1(x) x-α1(-1/x)· xeα1+α2+ e-2α1-α2= eα1+ e-2α1-α2,$ and if $x=0$ then $dα1(w-1)· weα1+ e-2α1-α2= eα1+ e-2α1-α2.$ Then every element $\left(s,n\right)\in \Lambda$ with $s={s}_{t}$ is conjugate to $\left({s}_{t},0\right),\left({s}_{t},{e}_{{\alpha }_{1}}\right),\left({s}_{t},{e}_{-2{\alpha }_{1}-{\alpha }_{2}}\right),$ or $\left({s}_{t},{e}_{{\alpha }_{1}}+{e}_{-2{\alpha }_{1}-{\alpha }_{2}}\right)\text{.}$ Later calculations are made easier if we choose representatives of $n$ in ${𝒩}^{+}\text{.}$ So we note: $w2w1· (st,e-2α1-α2) =(sw2w1t,eα2),$ and $w2w1w2· (st,eα1+e-2α1-α2)= (sw2w1w2t,eα1+eα2).$ These are the representatives listed above. $\square$

Theorem 3.22. If ${q}^{2}=-1$ then the following is a set of representatives of the orbits in $\Lambda /G\text{.}$

$( ± [ 1000 0100 0010 0001 ] , 0 ) , ( ± [ z1/2000 0z1/200 00z-1/20 000z-1/2 ] , 0 ) , z≠1,q±2 ( [ q000 0q00 00q-10 000q-1 ] , 0 ) , ( [ q000 0q00 00q-10 000q-1 ] , eα2 ) , ( [ q-1000 0q-100 00q0 000q ] , eα2 ) , ( [ q000 0q00 00q-10 000q-1 ] , eα1+α2 ) , ( [ q-1000 0q-100 00q0 000q ] , eα1+α2 ) , ( [ q000 0q-100 00q-10 000q ] , eα2+eα1 ) , ( [ q000 0q-100 00q-10 000q ] , e2α1+α2+ eα2 ) , ( [ q-1000 0q00 00q0 000q-1 ] , eα1+eα2 ) , ( [ 1000 0q200 0010 000q-2 ] , 0 ) , ( [ 1000 0q200 0010 000q-2 ] , eα1 ) , ( ± [ 1000 0q00 0010 000q-1 ] , 0 ) , ( ± [ 1000 0q00 0010 000q-1 ] , e2α1+α2 ) , ( ± [ 1000 0q-100 0010 000q ] , e2α1+α2 ) , ( ± [ 1000 0z00 0010 000z ] , 0 ) ,z≠1, q±1,q±2, ( [ z1/2000 0q2z1/200 00z-1/20 000q-2z-1/2 ] , 0 ) , z≠1, q±2, q-4, q-6, ( [ z1/2000 0q2z1/200 00z-1/20 000q-2z-1/2 ] , eα1 ) , z≠1, q±2, q-4, q-6, ( [ q2z1/2000 0z1/200 00q-2z-1/20 000z-1/2 ] , eα1 ) , z≠1, q±2, q-4, q-6, ( ± [ q000 0zq00 00q-10 000z-1q-1 ] , 0 ) , z≠±1, q±2, q-4, -q-2, ( ± [ q000 0zq00 00q-10 000z-1q-1 ] , eα2 ) , z≠±1, q±2, q-4, -q-2, ( ± [ q-1000 0zq00 00q0 000z-1q-1 ] , eα2 ) , z≠±1, q±2, q-4, -q-2, ( ± [ w1/2000 0zw1/200 00w-1/20 000z-1w-1/2 ] , 0 ) , tz,w generic, z≠-1,$

and

$( ± [ w1/2000 0-w1/200 00w-1/20 000-w-1/2 ] , 0 ) , t-1,w generic.$

 Proof. The proof of Theorem 3.19 applies to this case as well, with essential changes only necessary in Cases 2, 3 and 4 and for the semisimple ${s}_{{t}_{{q}^{2},{q}^{2}}}\text{.}$ Case 1 now covers the central characters ${t}_{1,1},{t}_{1,z},{t}_{z,1},$ and ${t}_{z,w}\text{.}$ Since ${t}_{{q}^{2},{q}^{2}}$ is in the same ${W}_{0}\text{-orbit}$ as ${t}_{{q}^{2},1},$ ${s}_{{t}_{{q}^{2},{q}^{2}}}$ is in the same $G\text{-orbit}$ as ${s}_{{t}_{{q}^{2},1}}$ and thus ${s}_{{t}_{{q}^{2},{q}^{2}}}$ (Case 4) is omitted from the list. Case 2: ${s}_{t}$ for ${t}_{{q}^{2},z},{t}_{z,{q}^{2}},$ or ${t}_{±q,1}$ Since $P\left(t\right)=\left\{±\beta \right\},$ where $\beta$ is ${\alpha }_{1},{\alpha }_{2},$ or $2{\alpha }_{1}+{\alpha }_{2},$ ${𝔤}_{q}^{s}$ is spanned by ${e}_{±\beta }\text{.}$ However, $a{e}_{\beta }+b{e}_{-\beta }$ is nilpotent exactly when $ab=0\text{.}$ Since ${C}_{G}\left({s}_{t}\right)=D,$ and $dαi(c)· e±β=±c ⟨β,ωi⟩ e±β,$ every element of ${𝒩}_{s}^{q}$ is ${C}_{G}\left({s}_{t}\right)\text{-conjugate}$ to either ${e}_{\beta }$ or ${e}_{-\beta }\text{.}$ Case 3: ${s}_{t}$ for ${t}_{{q}^{2},1}$ If $s={s}_{{t}_{{q}^{2},1}}$ then ${𝔤}_{q}^{s}$ is spanned by ${e}_{±{\alpha }_{1}}$ and ${e}_{±\left({\alpha }_{1}+{\alpha }_{2}\right)\text{.}}$ And ${C}_{G}\left(s\right)$ is generated by $D,$ ${x}_{±{\alpha }_{2}}\left(c\right)$ and ${x}_{±\left(2{\alpha }_{1}+{\alpha }_{2}\right)}$ for $c\in {ℂ}^{×}\text{.}$ In particular, ${C}_{G}\left(s\right)$ contains representatives of ${s}_{2}$ and ${s}_{1}{s}_{2}{s}_{1},$ and thus of ${s}_{2}{s}_{1}{s}_{2}{s}_{1}$ as well. Let $n=f{e}_{{\alpha }_{1}}+g{e}_{{\alpha }_{1}+{\alpha }_{2}}+h{e}_{-{\alpha }_{1}}+j{e}_{-{\alpha }_{1}-{\alpha }_{2}}\in {𝔤}_{q}^{s}\text{.}$ Then $n$ is nilpotent exactly if $hf+gj=0,$ since its minimal polynomial in the standard representation of $𝔤$ is ${x}^{2}-\left(hf+gj\right)\text{.}$ By applying the argument in case 3 of Theorem 3.19, $n$ is ${C}_{G}\left(s\right)$ conjugate to an element of ${𝔤}_{q}^{s}$ with its positive part equal to either ${e}_{{\alpha }_{1}}$ or 0. Thus we may assume $g=0\text{.}$ Then if $n$ is nilpotent, $hf=0$ as well. If $g=h=0$ but $fj\ne 0,$ then $x-(2α1+α2) (1)x2α1+α2 (-fj-1)·f eα1+je-α1-α2 =-jeα1.$ If $f=g=0$ but $hj\ne 0,$ then $s2s1s2s1· he-α1+j e-α1-α2= heα1+j eα1+α2,$ which is in the same ${C}_{G}\left(s\right)\text{-orbit}$ as $h{e}_{{\alpha }_{1}}\text{.}$ If exactly one of $f,h,j$ is non-zero, then $n=a{e}_{\beta }$ for some $\beta \in \left\{±{\alpha }_{1},±\left({\alpha }_{1}+{\alpha }_{2}\right)\right\}\text{.}$ But the roots $±{\alpha }_{1}$ and $±\left({\alpha }_{1}+{\alpha }_{2}\right)$ are in the same orbit under the action of ${s}_{2}$ and ${s}_{1}{s}_{2}{s}_{1}\text{.}$ Hence, $n$ is conjugate to $a{e}_{{\alpha }_{1}},$ where $a$ is equal to the non-zero coefficient of $n\text{.}$ Finally, $dα1(f-1) ·feα1=eα1.$ Thus every element of ${𝒩}_{s}^{q}$ is in the same ${C}_{G}\left(s\right)\text{-orbit}$ as either ${e}_{{\alpha }_{1}}$ or 0. Case 4: ${s}_{t}$ for ${t}_{1,{q}^{2}}$ If $s={s}_{{t}_{1,{q}^{2}}}$ then ${𝔤}_{q}^{s}$ is spanned by ${e}_{±{\alpha }_{2}},{e}_{±\left({\alpha }_{1}+{\alpha }_{2}\right)},$ and ${e}_{±\left(2{\alpha }_{1}+{\alpha }_{2}\right)},$ while ${C}_{G}\left(s\right)$ is generated by $D$ and ${x}_{±{\alpha }_{1}}\left(c\right)\text{.}$ By the argument in case 4 of theorem 3.19, if $n\ne 0$ is in the span of ${e}_{{\alpha }_{2}},{e}_{{\alpha }_{1}+{\alpha }_{2}},$ and ${e}_{2{\alpha }_{1}+{\alpha }_{2}},$ then $n$ is ${C}_{G}\left(s\right)\text{-conjugate}$ to either ${e}_{{\alpha }_{2}}$ or ${e}_{{\alpha }_{1}+{\alpha }_{2}}\text{.}$ We also note that $CG(s)· ⟨ eα2, eα1+α2, e2α1+α2 ⟩ ⊆ ⟨ eα2, eα1+α2, e2α1+α2 ⟩ ,$ since ${C}_{G}\left(s\right)$ is the same as in the generic case. A similar argument using ${x}_{-{\alpha }_{1}}\left(c\right)$ in place of ${x}_{{\alpha }_{1}}\left(c\right)$ shows that if $n\ne 0$ is in the span of ${e}_{-{\alpha }_{2}},{e}_{-\left({\alpha }_{1}+{\alpha }_{2}\right)},$ and ${e}_{-\left(2{\alpha }_{1}+{\alpha }_{2}\right)},$ then $n$ is ${C}_{G}\left(s\right)\text{-conjugate}$ to either ${e}_{-{\alpha }_{2}}$ or ${e}_{-\left({\alpha }_{1}+{\alpha }_{2}\right)}\text{.}$ We now examine the general case. Let $n=f{e}_{{\alpha }_{2}}+g{e}_{{\alpha }_{1}+{\alpha }_{2}}+h{e}_{2{\alpha }_{1}+{\alpha }_{2}}+j{e}_{-{\alpha }_{2}}+k{e}_{-{\alpha }_{1}-{\alpha }_{2}}+l{e}_{-2{\alpha }_{1}-{\alpha }_{2}}\text{.}$ Assume that not all of $f,g,h$ are zero and not all of $j,k,l$ are zero, or else we are in one of the cases above. As noted above, $f{e}_{{\alpha }_{2}}+g{e}_{{\alpha }_{1}+{\alpha }_{2}}+h{e}_{2{\alpha }_{1}+{\alpha }_{2}}$ is ${C}_{G}\left(s\right)\text{-conjugate}$ to either ${e}_{{\alpha }_{2}}$ or ${e}_{{\alpha }_{1}+{\alpha }_{2}}\text{.}$ Let $x·\left(f{e}_{{\alpha }_{2}}+g{e}_{{\alpha }_{1}+{\alpha }_{2}}+h{e}_{2{\alpha }_{1}+{\alpha }_{2}}\right)={e}_{{\alpha }_{2}}$ or ${e}_{{\alpha }_{1}+{\alpha }_{2}},$ for some $x\in {C}_{G}\left(s\right)\text{.}$ Then because the positive and negative parts of ${𝔤}_{q}^{s}$ are fixed by the action of ${C}_{G}\left(s\right),$ $x·n=eα2+ j′e-α2+k′ e-α1-α2+l ′e-2α1-α2$ or $x·n=eα1+α2 j′e-α2+k′ e-α1-α2+l′ e-2α1-α2.$ Thus we can assume $f=1$ and $g=h=0,$ or $g=1$ and $f=h=0\text{.}$ Case 1: $f=1,g=h=0$ If $n={e}_{{\alpha }_{2}}+j{e}_{-{\alpha }_{2}}+k{e}_{-{\alpha }_{1}-{\alpha }_{2}}+l{e}_{-2{\alpha }_{1}-{\alpha }_{2}},$ then the minimal polynomial of $n$ in the defining representation of ${𝔰𝔭}_{4}$ is ${x}^{4}-j{x}^{2}\text{.}$ Hence if $n$ is nilpotent, $j=0\text{.}$ Then if $k\ne 0,$ $dα1(k) x-αl/(2k) · ( eα2+k e-α1-α2+l e-2α1-α2 ) =eα2+ e-α1-α2.$ If $k=0,$ then $dα1(l)· eα2+l e-2α1-α2= eα2+ e-2α1-α2.$ Case 2: $g=1,f=h=0$ If $n={e}_{{\alpha }_{1}+{\alpha }_{2}}+j{e}_{-{\alpha }_{2}}+k{e}_{-{\alpha }_{1}-{\alpha }_{2}}+l{e}_{-2{\alpha }_{1}-{\alpha }_{2}},$ then the minimal polynomial of $n$ in the defining representation of ${𝔰𝔭}_{4}$ is ${x}^{4}-2k+{k}^{2}-jl\text{.}$ Hence if $n$ is nilpotent, $k=0$ and either $j$ or $l$ is zero. If $j=0,$ then $w1· ( eα1+α2+l e-2α1-α2 ) =eα1+α2+ le-α2.$ Then $dα2(j) dα1(j-1)· ( eα1+α2+ je-α2 ) =eα1+α2+ e-α2.$ Thus, $n$ is in the ${C}_{G}\left(s\right)$ orbit of either $0,$ ${e}_{{\alpha }_{2}},$ ${e}_{{\alpha }_{1}+{\alpha }_{2}},$ ${e}_{-{\alpha }_{2}},$ ${e}_{-{\alpha }_{1}-{\alpha }_{2}},$ ${e}_{{\alpha }_{2}}+{e}_{-{\alpha }_{1}-{\alpha }_{2}},$ ${e}_{{\alpha }_{2}}+{e}_{-2{\alpha }_{1}-{\alpha }_{2}},$ or ${e}_{{\alpha }_{1}+{\alpha }_{2}}+{e}_{-{\alpha }_{2}}\text{.}$ Future computations will be easier if we choose orbit representatives $\left(s,n\right)$ with $n\in {𝒩}^{+}\text{.}$ Then we note: $w0· (st,e-α2) = ( sw0t, eα2 ) , w0· (st,e-α1-α2) = ( sw0t, eα1+α2 ) , w1w2w1· ( st,eα2+ e-α1-α2 ) = ( sw1w2w1t, eα2+eα1 ) , w1w2w1· ( st,eα2+ e-2α1-α2 ) = ( sw1w2W1t, eα2+e2α1+α2 ) ,$ and $w2· ( st,eα1+α2 +e-α2 ) = ( sw2t, eα1+eα2 ) .$ These are the orbit representatives listed above. $\square$

Theorem 3.23. If $q=-1$ then the following is a set of representatives of the orbits in $\Lambda /G\text{.}$

$( ± [ 1000 0100 0010 0001 ] , 0 ) , ( ± [ 1000 0100 0010 0001 ] , eα1 ) , ( ± [ 1000 0100 0010 0001 ] ,eα2 ) , ( ± [ 1000 0100 0010 0001 ] , eα1+eα2 ) , ( ± [ z1/2000 0z1/200 00z-1/20 000z-1/2 ] , 0 ) , z≠±1,q±2 ( ± [ z1/2000 0z1/200 00z-1/20 000z-1/2 ] , eα1 ) , z≠±1,q±2 ( [ 1000 0-100 0010 000-1 ] , 0 ) , ( [ 1000 0-100 0010 000-1 ] , e2α1+α2 ) , ( ± [ 1000 0-100 0010 000-1 ] , eα2+ e2α1+α2 ) , ( ± [ 1000 0z00 0010 000z ] , 0 ) , z≠±1, ( ± [ 1000 0z00 0010 000z ] , eα2 ) , z≠±1,$

and

$( ± [ w1/2000 0zw1/200 00w-1/20 000z-1w-1/2 ] , 0 ) ,tz,w generic.$

 Proof. By (3.9) and the classification of central characters given in 2.4.1, the semisimple elements listed above are a set of representatives of the semisimple orbits in $G\text{.}$ Then it suffices to show that the nilpotent elements paired with each ${s}_{t}$ are a set of representatives of the ${C}_{G}\left(s\right)\text{-orbits}$ in ${𝒩}_{q}^{{s}_{t}}$ For $t\in T,$ $st=±dα1 (t(Xα1)) dα2 (t(Xα2)).$ The centralizer of ${s}_{t}$ in $G$ is generated by $\left\{{x}_{±\alpha }\left(c\right) \mid c\in {ℂ}^{×},t\left({X}^{\alpha }\right)=1\right\}$ and $D\text{.}$ Case 1: ${t}_{1,1}$ Since ${s}_{{t}_{1,1}}$ is central, ${C}_{G}\left({s}_{t}\right)=G$ and ${𝒩}_{s}^{q}=𝒩\text{.}$ Hence the nilpotent ${C}_{G}\left(s\right)\text{-orbits}$ in ${𝒩}_{s}^{q}$ are the nilpotent orbits in $𝔤\text{.}$ These are represented by 0, ${e}_{{\alpha }_{1}},$ ${e}_{{\alpha }_{2}},$ and ${e}_{{\alpha }_{1}}+{e}_{{\alpha }_{2}}\text{.}$ Case 2: ${t}_{1,z}$ and ${t}_{z,1}$ If $s={s}_{{t}_{1,z}}$ or ${s}_{{t}_{z,1}},$ then $P\left(t\right)=\left\{{e}_{{\alpha }_{i}}\right\},$ where $i=1$ or 2, respectively. In these cases, ${C}_{G}\left(s\right)$ is generated by $D$ and ${x}_{±{\alpha }_{i}}\left(c\right),$ for $c\in {ℂ}^{×}\text{.}$ In particular, ${C}_{G}\left(s\right)$ contains a representative of ${w}_{i}\in {W}_{0}\text{.}$ However, $a{e}_{{\alpha }_{i}}+b{e}_{-{\alpha }_{i}}$ is nilpotent exactly if $ab=0\text{.}$ If $a=0,$ then $wi·be-αi =beαi,$ so we assume that $n=a{e}_{{\alpha }_{i}}\text{.}$ Then $dαi(a-1) ·aeαi=eαi,$ so that every non-zero element of ${𝒩}_{q}^{s}$ is ${C}_{G}\left(s\right)\text{-conjugate}$ to ${e}_{{\alpha }_{i}}\text{.}$ Case 3: ${t}_{q,1}$ If $s={s}_{{t}_{q,1}},$ then $P\left(t\right)=\left\{{\alpha }_{1},2{\alpha }_{1}+{\alpha }_{2}\right\}$ and ${𝔤}_{q}^{s}$ is spanned by ${e}_{±\left(2{\alpha }_{1}+{\alpha }_{2}\right)}$ and ${e}_{±{\alpha }_{2}}\text{.}$ Also, ${C}_{G}\left(s\right)$ is generated by $D$ and ${x}_{±{\alpha }_{2}}\left(c\right)$ and ${x}_{±\left(2{\alpha }_{1}+{\alpha }_{2}\right)}\left(c\right)$ for $c\in {ℂ}^{×}\text{.}$ In particular, ${C}_{G}\left(s\right)$ contains a representative of ${w}_{2}$ and ${w}_{1}{w}_{2}{w}_{1}\text{.}$ However, $a{e}_{{\alpha }_{2}}+b{e}_{-{\alpha }_{2}}+c{e}_{2{\alpha }_{1}+{\alpha }_{2}}+d{e}_{-2{\alpha }_{1}-{\alpha }_{2}}$ is nilpotent exactly if $ab=0$ and $cd=0\text{.}$ However, $w2· ( be-α2+ ce2α1+α2+ de-2α1-α2 ) =beα2+c e2α1+α2+ de-2α1-α2,$ and $w1w2w1· ( aeα2+be-α2 +de-2α1-α2 ) =aeα2+be-α2 +de2α1+α2,$ so we can assume $n=a{e}_{{\alpha }_{2}}+c{e}_{2{\alpha }_{1}+{\alpha }_{2}}\text{.}$ Then if $ac\ne 0,$ $dα1(a/c) dα2(a-1)· ( aeα2+c e2α1+α2 ) =eα2+ e2α1+α2.$ If $a=0$ but $c\ne 0,$ then $dα2(c-1)· ce2α1+α2= e2α1+α2.$ if $c=0$ but $a\ne 0$ then $dα2(a-1)· aeα2=eα2.$ Hence $n$ is in the ${C}_{G}\left(s\right)\text{-orbit}$ of either 0, ${e}_{{\alpha }_{2}},$ or ${e}_{{\alpha }_{2}}+{e}_{2{\alpha }_{1}+{\alpha }_{2}}\text{.}$ Case 4: ${t}_{z,w}$ If $t={t}_{z,w},$ then $P\left(t\right)=\varnothing ,$ and ${𝒩}_{q}^{s}=0\text{.}$ $\square$

### Cosets

The irreducible $\stackrel{\sim }{H}\text{-modules}$ can also be constructed as the Borel-Moore homology of generalized Springer fibers. If $s$ is a semisimple element of $G$ and $n\in {𝔤}_{q}^{s},$ then ${ℬ}_{\left(s,n\right)}$ is the set of Borel subalgebras fixed by $s$ and containing $n\text{.}$ Identifying $G/B$ with $ℬ$ shows that ${ℬ}_{\left(s,n\right)}$ is the set of $B\text{-cosets}$ in $G$ fixed by the action of $s$ and $\text{exp}\left(n\right)\text{.}$ Thus, it is necessary to compute this action of $G$ on cosets in $G/B\text{.}$

The group $G$ is generated by:

$Hα(c)=wα (c)wα(-1) andxα(c) =exp(ceα), for α∈R,$

with some added relations (see [Ste1967]).

The first relations are commutator relations between the ${x}_{\alpha }\left(c\right)\text{.}$

$xα1(c) xα2(d) = xα2(d) xα1(c) xα1+α2(cd) x2α1+α2(-c2d) xα1(c) xα1+α2(d) = xα1+α2(d) xα1(c) x2α1+α2(2cd) xα1(c) x2α1+α2(d) = x2α1+α2(d) xα1(c) xα1(c) x-α2(d) = x-α2(d) xα1(c) xα1(c) x-α1-α2(d) = x-α1-α2(d) xα1(c) x-α2(-2cd) xα1(c) x-2α1-α2(d) = x-2α1-α2(d) xα1(c) x-α1-α2(-cd) x-α2(-c2d) x-α1(c) xα1(d) = x-α1(c) xα1(-c-1) xα1(d+c-1) = xα1(c-1) w1 xα1(d+c-1) xα2(c) xα1+α2(d) = xα1+α2(d) xα2(c) xα2(c) x2α1+α2(d) = x2α1+α2(d) xα2(c) xα2(c) x-α1-α2(d) = x-α1-α2(d) xα2(c) x-α1(cd) x-2α1-α2 (-d2c) xα2(c) x-2α1-α2(d) = x-2α1-α2(d) xα2(c) x-α2(c) xα2(d) = x-α2(c) xα2(-c-1) xα2(d+c-1) = xα2(c-1) w2 xα2(d+c-1)$

The next relations describe how the Weyl group interacts with the ${x}_{\alpha }\left(c\right)\text{.}$

$w1 xα(c) w1-1 = xs1(α)(-c) if α=± (α1+α2) or ±α1. w1xα(c) w1-1 =xs1(α)(c) if α=±α2 or ±(2α1+α2). w1 x±(α1+α2) (c)w1-1= x±(α1+α2) (c). w2xα(c) w2-1= xs2(α) (-c) if α= ±α1 or ±α2. w2xα(c) w2-1= xs2(α) (c) if α=± (α1+α2) or ±(2α1+α2).$

Explicitly, $G$ is generated by

$xα1(c)= [ 1000 c100 001-c 0001 ] xα2(c)= [ 10c0 0100 0010 0001 ] , xα1+α2(c)= [ 100c 01c0 0010 0001 ] x2α1+α2(c)= [ 1000 010c 0010 0001 ] , x-α1(c)= [ 1c00 0100 0010 00-c1 ] x-α2(c)= [ 1000 0100 c010 0001 ] , and xα1+α2(c)= [ 1000 0100 0c10 c001 ] x2α1+α2(c)= [ 1000 0100 0010 0c01 ] .$

### The Varieties ${ℬ}_{s}$

We use Theorem 3.8 to determine the varieties ${ℬ}_{s}$ for semisimple elements $s\in D\text{.}$ The following semisimple elements are defined only up to an element of $Z\left(G\right),$ but this does not affect ${ℬ}_{s}\text{.}$

First,

$st1,1= [ 1000 0100 0010 0001 ] =Hα1(1) Hα2(1)∈ ℤ(G),$

so that ${ℬ}_{{s}_{{t}_{1,1}}}=ℬ\text{.}$

Next,

$st-1,1= [ -1000 0100 00-10 0001 ] =st-1,1= Hα2(-1).$

Then ${ℬ}_{{s}_{{t}_{-1,1}}}$ consists of $B,$ ${w}_{1}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}B,$ ${w}_{1}{x}_{{\alpha }_{2}}\left(c\right){w}_{2}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}B,$ ${w}_{1}{x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}{x}_{{\alpha }_{2}}\left(d\right){w}_{2}B,$ and ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}{x}_{{\alpha }_{2}}\left(d\right){w}_{2}{w}_{1}B,$ for $c,d\in ℂ\text{.}$

If $z\ne 1,{q}^{±2}z\ne 1,{q}^{±2},$ then

$st1,z= [ z1/2000 0z1/200 00z-1/20 000z-1/2 ] =Hα1(z1/2) Hα2(z).$

Then ${ℬ}_{{s}_{{t}_{1,z}}}$ consists of $B,$ ${w}_{2}B,$ ${x}_{{\alpha }_{1}}\left(c\right){w}_{1}B,$ ${w}_{2}{w}_{1}B,$ ${x}_{{\alpha }_{1}}\left(c\right){w}_{1}{w}_{2}B,$ ${w}_{2}{w}_{1}{w}_{2}B,$ ${x}_{{\alpha }_{1}}\left(c\right){w}_{2}{w}_{1}{w}_{2}B,$ and ${x}_{{\alpha }_{1}}\left(c\right){w}_{1}{w}_{2}{w}_{1}{w}_{2}B,$ for $c\in ℂ\text{.}$ Geometrically, this consists of four disjoint copies of ${ℙ}^{1}\text{.}$

Next,

$st1,q2= [ q000 0q00 00q-10 000q-1 ] =Hα1(q) Hα2(q2).$

Then ${ℬ}_{{s}_{{t}_{1,{q}^{2}}}}$ contains $B,$ ${w}_{2}B$ ${x}_{{\alpha }_{1}}\left(c\right){w}_{1}B,$ ${w}_{2}{w}_{1}B,$ ${x}_{{\alpha }_{1}}\left(c\right){w}_{1}{w}_{2}B,$ ${w}_{2}{w}_{1}{w}_{2}B,$ ${x}_{{\alpha }_{1}}\left(c\right){w}_{1}{w}_{2}{w}_{1}B,$ and ${x}_{{\alpha }_{1}}\left(c\right){w}_{1}{w}_{2}{w}_{1}{w}_{2}B,$ for $c\in ℂ\text{.}$ (This changes for specific values of $q$ only if ${q}^{2}=1,$ in which case ${s}_{{t}_{1,{q}^{2}}}$ is central.) Geometrically, this consists of four disjoint copies of ${ℙ}^{1}$

Next,

$stq2,1= [ 1000 0q200 0010 000q-2 ] =Hα1(q2) Hα2(q2).$

Then ${ℬ}_{{s}_{{t}_{{q}^{2},1}}}$ contains $B,$ ${w}_{1}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}B,$ ${w}_{1}{w}_{2}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}B,$ ${w}_{1}{w}_{2}{w}_{1}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}{w}_{2}B,$ and ${w}_{1}{w}_{2}{w}_{1}{x}_{{\alpha }_{2}}\left(c\right){w}_{2}B,$ for $c\in ℂ\text{.}$ (This changes only if ${q}^{2}=±1,$ in which case ${s}_{t}$ was computed above.) Geometrically, this consists of four disjoint copies of ${ℙ}^{1}\text{.}$

Next,

$stq,1= [ 1000 0q00 0010 000q-1 ] =Hα1(q) Hα2(q).$

Then ${ℬ}_{{s}_{{t}_{q,1}}}$ consists of $B,$ ${w}_{1}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}B,$ ${w}_{1}{w}_{2}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}B,$ ${w}_{1}{w}_{2}{w}_{1}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}{w}_{2}B,$ and ${w}_{1}{w}_{2}{w}_{1}{x}_{{\alpha }_{2}}\left(c\right){w}_{2}B,$ for $c\in ℂ\text{.}$ (This changes only if $q=±1,$ in which case ${s}_{t}$ was computed above.) Geometrically, this consists of four disjoint copies of ${ℙ}^{1}\text{.}$

Next,

$st-q,1=± [ 1000 0-q00 0010 000-q-1 ] =Hα1(-q) Hα2(-q).$

Then ${ℬ}_{{s}_{{t}_{-q,1}}}$ contains $B,$ ${w}_{1}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}B,$ ${w}_{1}{w}_{2}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}B,$ ${w}_{1}{w}_{2}{w}_{1}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}{w}_{2}B,$ and ${w}_{1}{w}_{2}{w}_{1}{x}_{{\alpha }_{2}}\left(c\right){w}_{2}B,$ for $c\in ℂ\text{.}$ (This changes only if $q=±1,$ in which case ${s}_{t}$ was computed above.) Geometrically, this is four disjoint copies of ${ℙ}^{1}\text{.}$

If $z\ne 1,{q}^{±1},{q}^{±2},$ then

$stz,1= [ 1000 0z00 0010 000z ] =stz,1= Hα1(z) Hα2(z).$

Then ${ℬ}_{{s}_{{t}_{1,{q}^{2}}}}$ contains $B,$ ${w}_{1}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}B,$ ${w}_{1}{w}_{2}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}B,$ ${w}_{1}{w}_{2}{w}_{1}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}{w}_{2}B,$ and ${w}_{1}{w}_{2}{w}_{1}{x}_{{\alpha }_{2}}\left(c\right){w}_{2}B,$ for $c\in ℂ\text{.}$ Geometrically, this is four disjoint copies of ${ℙ}^{1}\text{.}$

Next,

$stq2,q2= [ q000 0q300 00q-10 000q-3 ] =Hα1(q4) Hα2(q3).$

Then ${ℬ}_{{s}_{{t}_{{q}^{2},{q}^{2}}}}$ consists of $wB$ for $w\in {W}_{0},$ eight disjoint points.

This changes if ${q}^{4}=1,$ ${q}^{6}=1,$ or ${q}^{2}=1\text{.}$ If ${q}^{2}=1,$ then ${s}_{t}$ is central. If ${q}^{4}=1,$ then ${s}_{t}$ is in the same orbit as ${s}_{{t}_{1,{q}^{2}}}$ above. If ${q}^{6}=1,$ then ${ℬ}_{{s}_{{t}_{{q}^{2},{q}^{2}}}}$ consists of $wB$ for $w\in {W}_{0},$ ${w}_{1}{x}_{{\alpha }_{2}}\left(c\right){w}_{2}B,$ ${w}_{1}{x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}B,$ ${w}_{1}{x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}{w}_{2}B,$ and ${w}_{2}{w}_{1}{x}_{{\alpha }_{2}}\left(c\right){w}_{2}B\text{.}$ Geometrically, this is four disjoint copies of ${ℙ}^{1}\text{.}$

If $z\ne 1,{q}^{±2},{q}^{-4},{q}^{-6},$ then

$stq2,z= [ z1/2000 0q2z1/200 00z-1/20 000q-2z-1/2 ] .$

Then ${ℬ}_{{s}_{{t}_{{q}^{2},z}}}$ consists of $wB$ for $w\in {W}_{0},$ which is eight disjoint points.

Next,

$st-1,q2= [ -q000 0q00 00-q-10 000q-1 ] =Hα1(q) Hα2(-q2).$

Then ${ℬ}_{{s}_{{t}_{-1,{q}^{2}}}}$ consists of $wB$ for $w\in {W}_{0},$ which is eight disjoint points.

If $z\ne ±1,{q}^{±2},{q}^{-4},-{q}^{-2},$ then

$stz,q2= [ q000 0zq00 00q-10 000z-1q-1 ] =Hα1(zq) Hα2(zq2).$

Then ${ℬ}_{{s}_{{t}_{z,{q}^{2}}}}$ consists of $wB$ for $w\in {W}_{0},$ which is eight disjoint points.

### Nilpotent Elements

The next goal is to understand the action of $\text{exp}\left(c{e}_{\alpha }\right)$ on the elements of $G/B\text{.}$

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

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

$xα1(c)· xα1(d)w1B = xα1(c+d)w1B , xα1(c)· xα1(d)w1 xα2(e)w2 B = xα1(c+d)w1 xα2(e)w2 B, xα1(c)· xα1(d)w1 xα2(e)w2 xα1(f)w1 B = xα1(c+d)w1 xα2(e)w2 xα1(f)w1 B, xα1(c)· xα1(d)w1 xα2(e)w2 xα1(f)w1 xα2(g)w2 B = xα1(c+d)w1 xα2(e)w2 xα1(f)w1 xα2(g)w2 B,$

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

$xα1(c)· xα2(d)w2 B = xα2(d) xα1(c) xα1+α2(cd) x2α1+α2(-c2d)w2 B = xα2(d)w2 xα1+α2(-c) xα1(cd) x2α1+α2(-c2d) B = xα2(d)w2 B xα1(c)· xα2(d)w2 xα1(e)w1B = xα2(d) xα1(c) xα1+α2(cd) x2α1+α2(-c2d)w2 xα1(e)w1 B = xα2(d)w2 xα1+α2(-c) xα1(cd) x2α1+α2(-c2d) xα1(e)w1 B = xα2(d)w2 xα1+α2(-c) xα1(cd+e) x2α1+α2(-c2d) w1 B = xα2(d)w2 xα1(cd+e) xα1+α2(-c) x2α1+α2(c2d+2ce) w1B = xα2 (d) w2 xα1 (cd+e) w1 xα1+α2 (c) xα1 (c2d+2ce) B = xα2 (d) w2 xα1 (cd+e) w1 B xα1(c)· xα2(d)w2 xα1(e)w1 xα2(f)B = xα2 (d) w2 xα1 (cd+e) w1 xα1+α2 (c) xα1 (c2d+2ce) xα2 (f) w2 B = xα2 (d) w2 xα1 (cd+e) w1 xα2 (f) xα1+α2 (c) xα1 (c2d+2ce) · xα1+α2 (c2df+2cef) x2α1+α2 (-f(c2d+2ce)2) w2 B = xα2 (d) w2 xα1 (cd+e) w1 xα2 (f) w2 xα1 (c) xα1+α2 (-c2d-2ce) · xα1 (+c2df+2cef) x2α1+α2 (-f(+c2d+2ce)2) B = xα2(d)w2 xα1(cd+e) w1xα2(f) w2B$

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

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

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

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

$xα2(c)· xα1(d)w1B = xα1(d) xα2 (c) xα1+α2 (-cd) x2α1+α2 (c2d) w1 B = xα1(d) w1 xα1+α2 (c) x2α1+α2 (cd) xα2 (c2d) B = xα1(d) w1 B xα2(c)· xα1(d)w1 xα2(e)w2B = xα1(d)w1 x2α1+α2 (c) xα1+α2 (cd) xα2 (c2d) xα2 (e) w2 B = xα1(d)w1 xα2 (c2d+e) w2 x2α1+α2 (c) xα1 (cd) B = xα1(d)w1 xα2 (c2d+e) w2 B xα2(c)· xα1(d)w1 xα2(e)w2 xα1(f)w1B = xα1(d)w1 xα2(c2d+e)w2 x2α1+α2 (c) xα1 (cd) xα1 (d) w1 B = xα1(d)w1 xα2(c2d+e)w2 xα1(f+cd)w1 xα2(c) B = xα1(d)w1 xα2(c2d+e)w2 xα1(f+cd)w1 B$

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

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

$xα1+α2(c) ·xα1(d)w1B = xα1(d) xα1+α2 (c) x2α1+α2 (-2cd) w1 B = xα1(d) w1 xα1+α2 (-c) xα2 (-2cd) B = xα1(d) w1 B xα1+α2 (c)·xα1 (d)w1xα2 (e)w2B = xα1(d)w1 xα1+α2 (c) xα2 (-2cd) xα2 (e) w2 B = xα1(d)w1 xα1+α2 (c) xα2 (e-2cd) w2 B = xα1(d)w1 xα2 (e-2cd) w2 xα1(c) B = xα1(d)w1 xα2 (e-2cd) w2 B xα1+α2 (c)·xα1 (d)w1xα2 (e)w2xα1 (f)w1B = xα1(d)w1 xα2(e-2cd) w2xα1 (c) xα1 (f) w1B = xα1(d)w1 xα2(e-2cd) w2xα1 (c+f) w1B$

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

$xα1+α2(c) ·xα2(d)w2B = xα2(d) xα1+α2(c) w2B = xα2(d)w2 xα1(c)B = xα2(d)w2B xα1+α2(c) ·xα2(d)w2 xα1(e)w1B = xα2(d)w2xα1 (c)xα1(e) w1B = xα2(d)w2xα1 (c+e)w1B xα1+α2(c) ·xα2(d)w2 xα1(e)w1 xα2(f)w2B = xα2(d) w2xα1 (c)xα1(e) w1xα2(f) w2B = xα2(d) w2xα1 (c+e) w1xα2(f) w2B xα1+α2(c) ·xα2(d)w2 xα1(e)w1 xα2(f) w2xα1(g)w1 B = xα2(d)w2α1 (c)xα1(e) w1xα2(f) w2xα1(g) w1B = xα2(d)w2α1 (c+e) w1xα2(f) w2xα1(g) w1B$

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

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

$x2α1+α2 (c)·xα1(d) w1B = xα1(d) x2α1+α2 (c)w1B = xα1(d) w1xα2(c)B = xα1(d) w1B x2α1+α2(c) ·xα1(d)w1 xα2(e)w2B = xα1(d)w1xα2 (e)xα2(e) w2B = xα1(d)w1xα2 (e+c) w2B x2α1+α2(c)· xα1(d)w1 xα2(e)w2 xα1(f)w1B = xα1(d)w1 xα2 (c)xα2(e) w2xα1(f) w1B = xα1(d)w1 xα2 (e+c) w2xα1(f) w1B x2α1+α2(c)· xα1(d)w1 xα2(e)w2 xα1(f)w1 xα2(g)w2B = xα1(d)w1xα2 (c)xα2(e) w2xα1(f) w1xα2(g) w2B = xα1(d)w1xα2 (e+c) w2xα1(f) w1xα2(g) w2B$

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

$x2α1+α2(c) ·xα2(d)w2B = xα2(d)w2 x2α1+α2(c) B = xα2(d)w2 B x2α1+α2(c)· xα2(d)w2xα1 (e)w1B = xα2(d)w2 x2α1+α2 (c) xα1 (e) w1 B = xα2(d)w2 (e) w1 xα2 (c) B = xα2(d)w2 (e) w1 B x2α1+α2(c)· xα2(d)w2 xα1(e)w1 xα2(f)w2B = xα2(d)w2 x2α1+α2 (c) xα1 (e) w1 xα2 (f) w2B = xα2(d)w2 xα1(e)w1 xα2(c) xα2(f)w2B = xα2(d)w2 xα1(e)w1 xα2 (f+c)w2B$

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

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

$xα2(1) x2α1+α2(1) ·xα2(c)w2B = xα2(c+1)w2 x2α1+α2(1) B = xα2(c+1)w2B xα2(1) x2α1+α2(1)· xα2(c)w2 xα1(d)w1 B = xα2(c+1)w2 x2α1+α2(1) xα1(d)w1B = xα2(c+1)w2 xα1(d)w1 xα2(1)B = xα2(c+1)w2 xα1(d)w1B xα2(1) x2α1+α2(1) · xα2(c)w2 xα1(d)w1 xα2(e)w2B = xα2(c+1)w2 x2α1+α2(1) xα1(d)w1 xα2(e)w2B = xα2(c+1)w2 xα1(d)w1 xα2(e)w2 B = xα2(c+1)w2 xα1(d)w1 xα2(e+1)w2B xα2(1) x2α1+α2(1) · xα2(c)w2 xα1(d)w1 xα2(e)w2 xα1(f)w1B = xα2(c+1)w2 x2α1+α2(1) xα1(d)w1 xα2(e)w2 xα1(f)w1B = xα2(c+1)w2 xα1(d)w1 xα2(1) xα2(e)w2 xα1(f)w1B = xα2(c+1)w2 xα1(d)w1 xα2(e+1)w2 xα1(f)w1B$

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

$xα2(1) x2α1+α2(1) · xα1(c)w1 B = xα2(1)· xα1(c)w1 B = xα1(c)w1 B xα2(1) x2α1+α2(1) · xα1(c)w1 xα2(d)w2 B = xα2(1)· xα1(c)w1 xα2(d+1)w2 B = xα1(c)w1 xα2(c+d+1)w2 B xα2(1) x2α1+α2(1) · xα1(c)w1 xα2(d)w2 xα1(e)w1 B = xα2(1)· xα1(c)w1 xα2(d+1)w2 xα1(e)w1 B = xα1(c)w1 xα2(c+d+1)w2 xα1(c+f)w1 B$

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

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

$xα1+α2(1/2) xα2(1) xα1(1) x2α1+α2(1) · xα1(c)w1 B = xα1+α2(1/2) xα2(1) xα1(1) · xα1(c)w1 B = xα1+α2(1/2) xα2(1) · xα1(c+1)w1 B = xα1+α2(1/2) · xα1(c+1)w1 B = xα1(c+1)w1 B xα1+α2(1/2) xα2(1) xα1(1) x2α1+α2(1) · xα1(c)w1 xα2(d)w2 B = xα1+α2(1/2) xα2(1) xα1(1) · xα1(c)w1 xα2(d+1)w2 B = xα1+α2(1/2) xα2(1) · xα1(c+1)w1 xα2(d+1)w2 B = xα1+α2(1/2) · xα1(c+1)w1 xα2(c+d+2)w2 B = xα1(c+1)w1 xα2(d+1)w2 B xα1+α2(1/2) xα2(1) xα1(1) x2α1+α2(1) · xα1(c)w1 xα2(d)w2 xα1(e)w1 B = xα1+α2(1/2) xα2(1) xα1(1) · xα1(c)w1 xα2(d+1)w2 xα1(e)w1 B = xα1+α2(1/2) xα2(1) · xα1(c+1)w1 xα2(d+1)w2 xα1(e)w1 B = xα1+α2(1/2) · xα1(c+1)w1 xα2(d+c+2)w2 xα1(e+c+1)w1 B = xα1(c+1)w1 xα2(d+1)w2 xα1(e+c+3/2)w1 B xα1+α2(1/2) xα2(1) xα1(1) x2α1+α2(1) · xα1(c)w1 xα2(d)w2 xα1(e)w1 xα2(f)w2 B = xα1+α2(1/2) xα2(1) xα1(1) · xα1(c)w1 xα2(d+1)w2 xα1(e)w1 xα2(f)w2 B = xα1+α2(1/2) xα2(1) · xα1(c+1)w1 xα2(d+1)w2 xα1(e)w1 xα2(f)w2 B = xα1+α2(1/2) · xα1(c+1) xα2(1) xα1+α2(-cd) x2α1+α2(c2d)w1 xα2(d+1)w2 xα1(e)w1 xα2(f)w2 B = xα1+α2(1/2) · xα1(c+1)w1 xα2(1) x2α1+α2(1)w1 xα1+α2(cd) xα2(c2d) xα2(d+1)w2 xα1(e)w1 xα2(f)w2 B = xα1+α2(1/2) · xα1(c+1)w1 xα2(1) x2α1+α2(1)w1 xα1+α2(cd) xα2(c2d+d-1)w2 xα1(e)w1 xα2(f)w2 B = xα1+α2(1/2) · xα1(c+1)w1 xα2(c2d+d-1)w2 x2α1+α2(1) xα1(cd) xα1(e)w1 xα2(f)w2 B = xα1+α2(1/2) · xα1(c+1)w1 xα2(c2d+d-1)w2 xα1(e+cd) x2α1+α2(1)w1 xα2(f)w2 B = xα1+α2(1/2) · xα1(c+1)w1 xα2(c2d+d-1)w2 xα1(e+cd)w1 xα2(1) xα2(f)w2 B = xα1+α2(1/2) · xα1(c+1)w1 xα2(c2d+d-1)w2 xα1(e+cd)w1 xα2(f+1)w2 B = xα1(c+1)w1 xα2(c2d+d-c-2)w2 xα1(e+cd+1/2)w1 xα2(f+1)w2 B$

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

$xα1+α2(1/2) xα2(1) xα1(1) x2α1+α2(1) · xα2(c)w2B = xα1+α2(1/2) xα2(1) xα1(1) · xα2(c)w2B = xα1+α2(1/2) xα2(1) · xα2(c)w2B = xα1+α2(1/2) · xα2(c+1)w2B = xα2(c+1)w2B xα1+α2(1/2) xα2(1) xα1(1) x2α1+α2(1) · xα2(c)w2 xα1(d)w1 B = xα1+α2(1/2) xα2(1) xα1(1) · xα2(c)w2 xα1(d)w1 B = xα1+α2(1/2) xα2(1) · xα2(c)w2 xα1(c+d)w1 B = xα1+α2(1/2) · xα2(c+1)w2 xα1(c+d)w1 B = xα2(c+1)w2 xα1(c+d+1/2)w1 B xα1+α2(1/2) xα2(1) xα1(1) x2α1+α2(1) · xα2(c)w2 xα1(d)w1 xα2(e)w2B = xα1+α2(1/2) xα2(1) xα1(1) · xα2(c)w2 xα1(d)w1 xα2(e+1)w2B = xα1+α2(1/2) xα2(1) · xα2(c)w2 xα1(c+d)w1 xα2(e+1)w2B = xα1+α2(1/2) · xα2(c+1)w2 xα1(c+d)w1 xα2(e+1)w2B = xα2(c+1)w2 xα1(c+d+1/2)w1 xα2(e+1)w2B$

### The varieties ${ℬ}_{\left(s,n\right)}$

We examine the varieties ${ℬ}_{\left(s,n\right)},$ which are the cosets in $G/B$ fixed by both $s$ and $\text{exp}\left(n\right),$ for each pair $\left(s,n\right)$ listed in Theorems 3.19-3.23. The Kazhdan-Lusztig classification calls for an examination of the action of the simultaneous centralizer in $G$ of $s$ and $n$ on ${ℬ}_{s,n}\text{.}$ Note that $Z\left(G\right)\cong {ℤ}_{2}$ is always contained in this centralizer, but acts trivially on ${ℬ}_{s,n}\text{.}$ Thus in the discussion below, we let $C\left(s,n\right)$ be the quotient of the centralizer of $s$ and $n$ by the center.

Generic $q:$

If $n=0,$ then ${ℬ}_{s,n}={ℬ}_{n},$ so we only specify ${ℬ}_{s,n}$ if $n\ne 0\text{.}$ When $n=0,$ the homology ${H}_{*}\left({ℬ}_{{s}_{t}},0\right)$ is the principal series module $M\left(t\right)\text{.}$

If

$(s,n)= ( ± [ q000 0q00 00q-10 000q-1 ] ,eα2 ) ,$

then ${ℬ}_{s,n}$ consists of $B,$ ${x}_{{\alpha }_{1}}\left(c\right){w}_{1}B,$ ${w}_{1}{w}_{2}B,$ and ${w}_{1}{w}_{2}{w}_{1}B\text{.}$ This is a single ${ℙ}^{1}$ and two points. The centralizer of $s$ is generated by $D,$ ${x}_{{\alpha }_{1}}\left(c\right),$ and ${x}_{-{\alpha }_{1}}\left(c\right),$ so that ${C}_{G}\left(s,n\right)$ is generated by $Z\left(G\right),$ $\left\{{d}_{{\alpha }_{1}}\left(z\right) \mid z\in {ℂ}^{×}\right\}$ and ${x}_{-{\alpha }_{1}}\left(c\right),$ and $C\left(s,n\right)$ is trivial.

If

$(s,n)= ( ± [ q000 0q00 00q-10 000q-1 ] , eα1+α2 ) ,$

then ${ℬ}_{s,n}$ consists of $B,$ ${x}_{{\alpha }_{1}}\left(c\right){w}_{1}B,$ ${w}_{2}B,$ and ${w}_{1}{w}_{2}B\text{.}$ This is a single ${ℙ}^{1}$ and two points. The centralizer of $s$ is generated by $D,$ ${x}_{{\alpha }_{1}}\left(c\right),$ and ${x}_{-{\alpha }_{1}}\left(c\right),$ so that ${C}_{G}\left(s,n\right)$ is generated by $Z\left(G\right),$ $\left\{{d}_{{\alpha }_{1}}\left(z\right){d}_{{\alpha }_{2}}\left({z}^{-1}\right) \mid z\in {ℂ}^{×}\right\}$ and ${d}_{{\alpha }_{1}}\left(-1\right){w}_{1}\text{.}$ Then $C\left(s,n\right)=\left\{id,{d}_{{\alpha }_{1}}\left(-1\right){w}_{1}\right\}\text{.}$

We examine how $C\left(s,n\right)$ acts on ${H}_{•}\left({ℬ}_{s,n}\right)\text{.}$

$dα1(-1)w1 B=w1B, dα1(-1)w1 · xα1(c)w1B = dα1(-1) x-α1(-c) w1w1B = x-α1(c) Hα1(-1)B = xα1(c-1) w1B, Hα1(-1)w1 ·w2B=w1w2B , and Hα1(-1)w1 ·w1w2B=w2B.$

This is a homeomorphism on ${ℙ}^{1},$ and switches the two points. ${H}_{0}\left({ℬ}_{s,n}\right)={ℤ}^{3}$ be generated by $\alpha ,$ β, and $\gamma ,$ which are the generators of ${H}_{0}\left({ℙ}^{1}\right),$ and of the points ${H}_{0}\left({w}_{2}B\right)$ and ${H}_{0}\left({w}_{1}{w}_{2}\right)B,$ respectively. Then the component ${H}_{•}{\left({ℬ}_{s,n}\right)}^{\chi }$ is 1-dimensional, generated by $\beta -\gamma ,$ where $\chi$ is the non-trivial representation of ${ℤ}_{2}\text{.}$ ${H}_{•}{\left({ℬ}_{s,n}\right)}^{\text{triv}}$ is 3-dimensional, generated by $\beta +\gamma$ and ${H}_{•}\left({ℙ}^{1}\right),$ where triv is the trivial representation of ${ℤ}_{2}\text{.}$

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

If

$(s,n)= ( ± [ 1000 0q200 0010 000q-2 ] ,eα1 ) ,$

then ${ℬ}_{s,n}$ consists of $B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}B,$ ${w}_{2}{w}_{1}B,$ and ${w}_{2}{w}_{1}{w}_{2}B,$ which is one copy of ${ℙ}^{1}$ and two points. The centralizer of $s$ is generated by $D,$ ${x}_{{\alpha }_{2}}\left(c\right),$ and ${x}_{-{\alpha }_{2}}\left(c\right),$ so that ${C}_{G}\left(s,n\right)$ is generated by $Z\left(G\right),$ $\left\{{d}_{{\alpha }_{2}}\left(z\right) \mid z\in {ℂ}^{×}\right\}$ and ${x}_{-{\alpha }_{2}}\left(c\right),$ and $C\left(s,n\right)$ is trivial.

$B,xα2(c)w2B w2w1B w2w1w2C (stq2,1,eα1)$

If

$(s,n)= ( ± [ 1000 0q00 0010 000q-1 ] , e2α1+α2 ) ,$

then ${ℬ}_{s,n}$ consists of $B,$ ${w}_{1}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}B,$ and ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}B,$ which is two disjoint copies of ${ℙ}^{1}\text{.}$ The centralizer of $s$ is generated by $Z\left(G\right),$ $D,$ ${x}_{{\alpha }_{2}}\left(c\right),$ and ${x}_{-{\alpha }_{2}}\left(c\right),$ so that ${C}_{G}\left(s,n\right)$ is generated by $Z\left(G\right),$ $\left\{{d}_{{\alpha }_{1}}\left(z\right){d}_{{\alpha }_{2}}\left({z}^{-2}\right) \mid z\in {ℂ}^{×}\right\},$ ${x}_{{\alpha }_{2}}\left(c\right),$ and ${x}_{-{\alpha }_{2}}\left(c\right),$ and $C\left(s,n\right)$ is trivial.

If

$(s,n)= ( ± [ 1000 0-q00 0010 000-q-1 ] , e2α1+α2 ) ,$

then ${ℬ}_{s,n}$ consists of $B,$ ${w}_{1}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}B,$ and ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}B,$ which is two disjoint copies of ${ℙ}^{1}\text{.}$ The centralizer of $s$ is generated by $D,$ ${x}_{{\alpha }_{2}}\left(c\right),$ and ${x}_{-{\alpha }_{2}}\left(c\right),$ so that ${C}_{G}\left(s,n\right)$ is generated by $Z\left(G\right),$ $\left\{{d}_{{\alpha }_{1}}\left(z\right){d}_{{\alpha }_{2}}\left({z}^{-2}\right) \mid z\in {ℂ}^{×}\right\}$ and ${x}_{{\alpha }_{2}}\left(c\right),$ and $C\left(s,n\right)$ is trivial.

$B,xα2(c)w2B w1B,xα2(c)w2w1B (st±q,1,e2α1+α2)$

If

$(s,n)= ( ± [ q000 0q300 00q-10 000q-3 ] , eα1 ) ,$

then ${ℬ}_{s,n}$ consists of four points - $B,$ ${w}_{2}B,$ ${w}_{2}{w}_{1}B,$ and ${w}_{2}{w}_{1}{w}_{2}B\text{.}$ The centralizer of $s$ is $D,$ so that ${C}_{G}\left(s,n\right)=\left\{±{d}_{{\alpha }_{2}}\left(z\right) \mid z\in {ℂ}^{×}\right\}$ and $C\left(s,n\right)$ is trivial.

If

$(s,n)= ( ± [ q000 0q300 00q-10 000q-3 ] , eα2 ) ,$

then ${ℬ}_{s,n}$ consists of four points - $B,$ ${w}_{1}B,$ ${w}_{1}{w}_{2}B,$ and ${w}_{1}{w}_{2}{w}_{1}B\text{.}$ The centralizer of $s$ is $D,$ so that ${C}_{G}\left(s,n\right)=\left\{±{d}_{{\alpha }_{1}}\left(z\right) \mid z\in {ℂ}^{×}\right\}$ and $C\left(s,n\right)$ is trivial.

If

$(s,n)= ( ± [ q000 0q300 00q-10 000q-3 ] , eα1+ eα2 ) ,$

then ${ℬ}_{s,n}=\left\{B\right\}\text{.}$ The centralizer of $s$ is $D,$ so that ${C}_{G}\left(s,n\right)=\left\{±{d}_{{\alpha }_{1}}\left({z}^{-1}\right){d}_{{\alpha }_{2}}\left(z\right) \mid z\in {ℂ}^{×}\right\}$ and $C\left(s,n\right)$ is trivial.

$B w2B w2w1B w2w1w2B B w1B w1w2B w1w2w1B (stq2,q2,eα1) (stq2,q2,eα2) B (stq2,q2,eα1+α2)$

If

$(s,n)= ( ± [ -q000 0q00 00-q-10 000q-1 ] , eα2 ) ,$

then ${ℬ}_{s,n}$ consists of four points - $B,$ ${w}_{1}B,$ ${w}_{1}{w}_{2}B,$ and ${w}_{1}{w}_{2}{w}_{1}B\text{.}$ The centralizer of $s$ is $D,$ so that ${C}_{G}\left(s,n\right)=\left\{±{d}_{{\alpha }_{1}}\left(z\right) \mid z\in {ℂ}^{×}\right\}$ and its component group is trivial.

If

$(s,n)= ( [ -q000 0q00 00-q-10 000q-1 ] , e2α1+α2 ) ,$

then ${ℬ}_{s,n}$ consists of four points - $B,$ ${w}_{1}B,$ ${w}_{2}B,$ and ${w}_{2}{w}_{1}B\text{.}$ The centralizer of $s$ is $D,$ so that ${C}_{G}\left(s,n\right)=\left\{±{d}_{{\alpha }_{1}}\left(z\right){d}_{{\alpha }_{2}}\left({z}^{-2}\right) \mid z\in {ℂ}^{×}\right\}$ and $C\left(s,n\right)$ is trivial.

If

$(s,n)= ( [ -q000 0q00 00-q-10 000q-1 ] , eα2+ e2α1+α2 ) ,$

then ${ℬ}_{n}$ consists of two points - $B$ and ${w}_{1}B\text{.}$ The centralizer of $s$ is $D,$ so that ${C}_{G}\left(s,n\right)=\left\{±1,±{d}_{{\alpha }_{1}}\left(-1\right)\right\}\cong {ℤ}_{2}^{2}\text{.}$ However, $C\left(s,n\right)={C}_{G}\left(s,n\right)$ acts trivially on ${ℬ}_{s,n},$ so ${\left({ℬ}_{s,n}\right)}^{\text{triv}}={ℬ}_{n}^{s}\text{.}$

$B w1B w1w2B w1w2w1B B w1B w2B w1w2B ( st-1,q2 ,eα2 ) ( st-1,q2, e2α1+α2 ) B w1B ( st-1,q2 ,eα2+ e2α1+α2 )$

For $z\ne 1,{q}^{±2},{q}^{-4},{q}^{-6},$ if

$(s,n)= ( ± [ z1/2000 0q2z1/200 00z-1/20 000q-2z-1/2 ] , eα1 ) ,$

then ${ℬ}_{s,n}$ consists of four points - $B,$ ${w}_{2}B,$ ${w}_{2}{w}_{1}B,$ and ${w}_{2}{w}_{1}{w}_{2}B\text{.}$ The centralizer of $s$ is $D,$ so that ${C}_{G}\left(s,n\right)=\left\{±{d}_{{\alpha }_{2}}\left(z\right) \mid z\in {ℂ}^{×}\right\}$ and $C\left(s,n\right)$ is trivial.

For $z\ne ±1,{q}^{±2},{q}^{-4},-{q}^{-2},$ if

$(s,n)= ( ± [ q000 0zq00 00q-10 000z-1q-1 ] , eα2 ) ,$

then ${ℬ}_{s,n}$ consists of four points - $B,$ ${w}_{1}B,$ ${w}_{1}{w}_{2}B,$ and ${w}_{1}{w}_{2}{w}_{1}B\text{.}$ The centralizer of $s$ is $D,$ so that ${C}_{G}\left(s,n\right)=\left\{±{d}_{{\alpha }_{1}}\left(z\right) \mid z\in {ℂ}^{×}\right\}$ and $C\left(s,n\right)$ is trivial.

$B w2B w2w1B w2w1w2B B w1B w1w2B w1w2w1B (stq2,z,eα1) (stz,q2,eα2)$

${q}^{8}=1 \left(\ell =4\right):$

The varieties ${ℬ}_{s,n}$ from the previous section change for the orbit of the semisimple element $\left[\begin{array}{cccc}q& 0& 0& 0\\ 0& {q}^{3}& 0& 0\\ 0& 0& {q}^{-1}& 0\\ 0& 0& 0& {q}^{-3}\end{array}\right],$ which now includes $\left[\begin{array}{cccc}q& 0& 0& 0\\ 0& -q& 0\\ 0& 0& {q}^{-1}& 0\\ 0& 0& 0& -{q}^{-1}\end{array}\right]\text{.}$ In this case ${𝔤}_{q}^{s}$ contains 6 non-zero orbits, 3 of which are not present for generic $q\text{.}$

Note that $\left({s}_{{t}_{{q}^{-4},{q}^{2}}},{e}_{{\alpha }_{2}}\right)={w}_{2}{w}_{1}·\left({s}_{{t}_{{q}^{2},{q}^{2}}},{e}_{-2{\alpha }_{1}-{\alpha }_{2}}\right)\text{.}$ If

$(s,n)= ( [ q-3000 0q00 00q30 000q-1 ] , eα2 ) ,$

then $s={H}_{{\alpha }_{2}}\left({q}^{-2}\right){H}_{{\alpha }_{1}}\left(q\right)\text{.}$ Then ${ℬ}_{s,n}$ consists of four points - $B,$ ${w}_{1}B,$ ${w}_{1}{w}_{2}B,$ and ${w}_{1}{w}_{2}{w}_{1}B\text{.}$ The centralizer of $s$ is $D,$ so that ${C}_{G}\left(s,n\right)$ is $\left\{±{d}_{{\alpha }_{1}}\left(z\right) \mid z\in {ℂ}^{×}\right\}$ and $C\left(s,n\right)$ is trivial.

Next, $\left({s}_{{t}_{{q}^{-2},{q}^{-2}}},{e}_{{\alpha }_{2}}+{e}_{2{\alpha }_{1}+{\alpha }_{2}}\right)={w}_{2}{w}_{1}·\left({s}_{{t}_{{q}^{2},{q}^{2}}},{e}_{{\alpha }_{2}}+{e}_{-2{\alpha }_{1}-{\alpha }_{2}}\right)\text{.}$

If

$(s,n)= ( [ q-3000 0q00 00q30 000q-1 ] , eα2+ e2α1+α2 ) ,$

then ${ℬ}_{s,n}$ consists of $B$ and ${w}_{1}B\text{.}$ The centralizer of $s$ is $D,$ so that ${C}_{G}\left(s,n\right)=\left\{1±,±{d}_{{\alpha }_{1}}\left(-1\right)\right\}$ and $C\left(s,n\right)$ is ${ℤ}_{2}\text{.}$ However, $C\left(s,n\right)$ acts trivially on ${ℬ}_{s,n},$ so that ${\left({ℬ}_{s,n}\right)}^{\text{triv}}={ℬ}_{s,n}\text{.}$

Finally, $\left({s}_{{t}_{{q}^{2},{q}^{2},1}},{e}_{{\alpha }_{1}}+{e}_{{\alpha }_{2}}\right)={w}_{2}{w}_{1}{w}_{2}·\left({s}_{{t}_{{q}^{2},{q}^{2},0}},{e}_{{\alpha }_{1}}+{e}_{-2{\alpha }_{1}-{\alpha }_{2}}\text{.}\right)$ If

$(s,n)= ( [ q-3000 0q-100 00q30 000q ] ,eα1+eα2 ) ,$

then $s={H}_{{\alpha }_{1}}\left({q}^{-1}\right){H}_{{\alpha }_{2}}\left({q}^{-4}\right)\text{.}$ Then ${ℬ}_{s,n}$ consists of $B\text{.}$ The centralizer of $s$ is $D$ so that $C\left(s,n\right)$ is trivial.

$w1w2w1B w1w2B B w1B B w1B ( stq2,q2, e-2α1-α2 ) ( stq2,q2, eα2+ e-2α1-α2 ) B ( stq2,q2, eα1+ e-2α1-α2 )$

${q}^{6}=1 \left(\ell =3\right):$

The varieties ${ℬ}_{s,n}$ change from the generic case for $s$ in the orbit of $\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& {q}^{2}& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& {q}^{-2}\end{array}\right],$ which is now in the same orbit as $±\left[\begin{array}{cccc}q& 0& 0& 0\\ 0& {q}^{3}& 0& 0\\ 0& 0& {q}^{-1}& 0\\ 0& 0& 0& {q}^{-3}\end{array}\right],$ depending on whether ${q}^{3}$ is 1 or $-1\text{.}$ The ${C}_{G}\left({s}_{{t}_{{q}^{2},1}}\right)\text{-orbits}$ in ${𝒩}_{q}^{{s}_{{t}_{{q}^{2},1}}}$ are represented by ${e}_{{\alpha }_{1}},$ ${e}_{-2{\alpha }_{1}-{\alpha }_{2}},$ and ${e}_{{\alpha }_{1}}+{e}_{-2{\alpha }_{1}-{\alpha }_{2}}\text{.}$

If

$(s,n)= ( ± [ 1000 0q200 0010 000q-2 ] ,eα1 ) ,$

then ${ℬ}_{s,n}$ consists of $B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}B,$ ${w}_{2}{w}_{1}B,$ and ${w}_{2}{w}_{1}{w}_{2}B,$ which is one copy of ${ℙ}^{1}$ and two points. The centralizer of $s$ is generated by ${x}_{±{\alpha }_{2}}\left(c\right)$ and $D,$ so that ${C}_{G}\left(s,n\right)$ is generated by $\left\{{x}_{-{\alpha }_{2}}\left(c\right) \mid c\in ℂ\right\}$ and $\left\{±{d}_{{\alpha }_{2}}\left(z\right) \mid z\in {ℂ}^{×}\right\}\text{.}$ Then $C\left(s,n\right)$ is trivial.

Note that $\left({s}_{{t}_{{q}^{-2},{q}^{-2}}},{e}_{{\alpha }_{2}}\right)={w}_{1}{w}_{2}·\left({s}_{{t}_{{q}^{2},1}},{e}_{-2{\alpha }_{1}-{\alpha }_{2}}\right)\text{.}$

If

$(s,n)= ( ± [ q2000 0100 00q-20 0001 ] ,eα2 ) ,$

then ${ℬ}_{s,n}$ consists of $B,$ ${w}_{1}B,$ ${w}_{1}{x}_{{\alpha }_{2}}\left(c\right){w}_{2}B,$ and ${w}_{1}{x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}B,$ which is two copies of ${ℙ}^{1}\text{.}$ The centralizer of $s$ is generated by ${x}_{±\left(2{\alpha }_{1}+{\alpha }_{2}\right)}\left(c\right)$ and $D,$ so that ${C}_{G}\left(s,n\right)$ is generated by ${x}_{±\left(2{\alpha }_{1}+{\alpha }_{2}\right)}\left(c\right)$ for $c\in ℂ$ and $\left\{±{d}_{{\alpha }_{1}}\left(z\right) \mid z\in {ℂ}^{×}\right\}\text{.}$ Then $C\left(s,n\right)$ is trivial.

Finally, $\left({s}_{{t}_{{q}^{2},{q}^{2}}},{e}_{{\alpha }_{1}}+{e}_{{\alpha }_{2}}\right)={w}_{2}{w}_{1}{w}_{2}·\left({s}_{{t}_{{q}^{2},1}},{e}_{{\alpha }_{1}}+{e}_{-2{\alpha }_{1}-{\alpha }_{2}}\right)\text{.}$

If

$(s,n)= ( ± [ q2000 0100 00q-20 0001 ] ,eα2+ eα2 ) ,$

then ${ℬ}_{s,n}$ consists of only $B\text{.}$ The centralizer of $s$ is generated by ${x}_{±\left(2{\alpha }_{1}+{\alpha }_{2}\right)}\left(c\right)$ and $D,$ so that ${C}_{G}\left(s,n\right)$ is generated by ${x}_{2{\alpha }_{1}+{\alpha }_{2}}\left(c\right)$ for $c\in ℂ,$ and $\left\{±{d}_{{\alpha }_{1}}\left(z\right){d}_{{\alpha }_{2}}\left({z}^{-1}\right) \mid z\in {ℂ}^{×}\right\}\text{.}$ Hence $C\left(s,n\right)$ is trivial.

$B,xα2(c)w2B w2w1B w2w1w2B w1B,w1xα2(c)w2B B,w1xα2(c)w2w1B (stq2,1,eα1) ( stq2,1, e-2α1-α2 ) B ( stq2,1, eα1+ e-2α1-α2 )$

${q}^{4}=1 \left(\ell =2\right):$

Again, we note only the changes from the generic case. Also note that ${ℬ}_{s,n}={ℬ}_{-s,n}={ℬ}_{{q}^{2}s,n}\text{.}$

${t}_{1,{q}^{2}}$

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

We can check which nilpotent elements satisfy condition 3.1 using the following procedure. First, we note that an element of ${𝔤}_{q}^{s}$ takes the form

$z=aeα2+b eα1+α2+c e2α1+α2+ de-α2+f e-α1-α2+g e-2α1-α2.$

Then for each $n,$ we find an ${𝔰𝔩}_{2}\text{-triple}$ $\left(n,y,h\right)\text{.}$ By computing the commutator $\left[y,z\right],$ we can determine necessary and sufficient conditions for $z$ to commute with $y\text{.}$ We then check, using the standard representation of $𝔤,$ whether an element $z$ satisfying those conditions is necessarily nilpotent. We outline this check below.

$n y Condition on Z𝔤(y) Z𝔤(y)∩𝔤qs⊆𝒩 eα2 e-α2 a=b=0 no e-α2 e-α2 d=f=0 no eα1+α2 e-α1-α2 a=b=c=0 yes e-α1-α2 eα1+α2 d=f=g=0 yes eα2+e-α1-α2 4e-α2+3eα1+α2 a=f=g=0=4b-3d yes e-α2+eα1+α2 4eα2+3e-α1-α2 b=c=d=0=4f-3a yes eα2+e-2α1-α2 e-α2+e2α1+α2 a=g=0=b-f no$

Specifically, ${e}_{2{\alpha }_{1}+{\alpha }_{2}}+{f}_{2{\alpha }_{1}+{\alpha }_{2}}$ commutes with ${e}_{±{\alpha }_{2}},$ and ${e}_{{\alpha }_{1}+{\alpha }_{2}}-{e}_{-{\alpha }_{1}-{\alpha }_{2}}$ commutes with ${e}_{-{\alpha }_{2}}+{e}_{2{\alpha }_{1}+{\alpha }_{2}},$ so that ${e}_{±{\alpha }_{2}}$ and ${e}_{{\alpha }_{2}}+{e}_{-2{\alpha }_{1}-{\alpha }_{2}}$ are not considered.

If

$(s,n)= ( [ q000 0q00 00q-10 000q-1 ] , eα1+α2 ) ,$

then ${ℬ}_{s,n}$ consists of $B,$ ${x}_{{\alpha }_{1}}\left(c\right){w}_{1}B,$ ${w}_{2}B,$ and ${w}_{1}{w}_{2}B,$ a ${ℙ}^{1}$ and two points. The centralizer of $s$ is generated by $D$ and ${x}_{±{\alpha }_{1}}\left(c\right),$ so that ${C}_{G}\left(s,n\right)$ is generated by $\left\{{d}_{{\alpha }_{1}}\left(z\right){d}_{{\alpha }_{2}}\left({z}^{-1}\right) \mid z\in {ℂ}^{×}\right\}$ and ${d}_{{\alpha }_{1}}\left(-1\right){w}_{1}\text{.}$ Then

$C(s,n)= { 1,dα1(-1) w1 } ≅ℤ2.$

If

$(s,n)=w0· ( st1,q2, e-α1-α2 ) = ( [ q-1000 0q-100 00q0 000q ] ,eα1+α2 ) ,$

then ${ℬ}_{s,n}$ consists of $B,$ ${x}_{{\alpha }_{1}}\left(c\right){w}_{1}B,$ ${w}_{2}B,$ and ${w}_{1}{w}_{2}B,$ a ${ℙ}^{1}$ and two points. The centralizer of $s$ is generated by $D$ and ${x}_{±{\alpha }_{1}}\left(c\right),$ so that ${C}_{G}\left(s,n\right)$ is generated by $\left\{{d}_{{\alpha }_{1}}\left(z\right){d}_{{\alpha }_{2}}\left({z}^{-1}\right) \mid c\in {ℂ}^{×}\right\}$ and ${d}_{{\alpha }_{1}}\left(-1\right){w}_{1}\text{.}$ Then $C\left(s,n\right)=\left\{1,{d}_{{\alpha }_{1}}\left(-1\right){w}_{1}\right\}\cong {ℤ}_{2}\text{.}$

In both of these cases, as in the case of generic $q,$ the action of $C\left(s,n\right)$ on ${ℬ}_{s,n}$ switches the points ${w}_{2}B$ and ${w}_{1}{w}_{2}B,$ and is a homeomorphism on ${ℙ}^{1}=\left\{{x}_{{\alpha }_{1}}\left(c\right){w}_{1}B \mid c\in ℂ\right\}\cup \left\{B\right\}\text{.}$ Thus the component of ${H}_{•}\left({ℬ}_{s,n}\right)$ corresponding to the sign representation is 1-dimensional, spanned by $\left[{w}_{2}B\right]+\left[{w}_{1}{w}_{2}B\right],$ while the component corresponding to the trivial representation is 3-dimensional.

If

$(s,n)= w1w2w1· ( st1,q2, eα2+ e-α1-α2 ) = ( [ q000 0q-100 00q-10 000q ] , eα2+eα1 ) ,$

then $ℬ=\left\{B\right\}\text{.}$ The centralizer of $s$ is generated by $D$ and ${x}_{±{\alpha }_{1}}\left(c\right),$ so that $C\left(s,n\right)$ is trivial.

If

$(s,n)= w2 · ( st1,q2, eα1+α2 +e-α2 ) = ( [ q-1000 0q00 00q0 000q-1 ] , eα1+eα2 ) ,$

then ${ℬ}_{s,n}=\left\{B\right\}\text{.}$ The centralizer of $s$ is generated by $D$ and ${x}_{{\alpha }_{1}}\left(c\right),$ so that $C\left(s,n\right)$ is trivial.

$B,xα1(c)w1B w2B,w1w2B B,xα1(c)w1B w2B,w1w2B (st1,q2,eα1+α2) (st1,q2,e-α1-α2) B B ( st1,q2, eα2+ e-α1-α2 ) ( st1,q2, eα1+α2 +e-α2 )$

Thus, the triples $\left({s}_{{s}_{2}t},{e}_{{\alpha }_{1}}+{e}_{{\alpha }_{2}},1\right)$ and $\left({s}_{{s}_{1}{s}_{2}{s}_{1}t},{e}_{{\alpha }_{1}}+{e}_{{\alpha }_{2}},1\right)$ must correspond to the 1-dimensional modules with weights ${s}_{2}t$ and ${s}_{1}{s}_{2}{s}_{1}t,$ respectively, since the homology of those varieties are each 1-dimensional. Then $\left({s}_{t},{e}_{{\alpha }_{1}+{\alpha }_{2}},1\right)$ and $\left({s}_{{w}_{0}t},{e}_{{\alpha }_{1}+{\alpha }_{2}},1\right)$ correspond to the 2-dimensional modules, which have weights $t$ and ${w}_{0}t\text{.}$ However, this leaves us with two triples not yet assigned to a module, $\left({s}_{t},{e}_{{\alpha }_{1}+{\alpha }_{2}},-1\right)$ and $\left({s}_{t},{e}_{{\alpha }_{1}+{\alpha }_{2}},-1\right),$ and every module has been accounted for. This is the first case where Grojnowskiâ€™s condition 3.2 seems to be missing some information about the representation $\chi$ in the triple $\left(s,n,\chi \right)\text{.}$ In particular, eliminating the triples $\left({s}_{t},{e}_{{\alpha }_{1}+{\alpha }_{2}},-1\right)$ and $\left({s}_{t},{e}_{{\alpha }_{1}+{\alpha }_{2}},-1\right)$ from our indexing set makes the correspondence a bijection.

${t}_{{q}^{2},1}$

If

$(s,n) = ( [ 1000 0q200 0010 000q-2 ] , eα1 ) ,$

then ${ℬ}_{s,n}$ consists of $B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}B,$ ${w}_{2}{w}_{1}B,$ and ${w}_{2}{w}_{1}{x}_{{\alpha }_{2}}\left(c\right){w}_{2}B,$ two disjoint copies of ${ℙ}^{1}\text{.}$ The centralizer of $s$ is generated by ${x}_{±{\alpha }_{2}}\left(c\right)$ and $D,$ so that ${C}_{G}\left(s,n\right)$ is generated by $\left\{{x}_{-{\alpha }_{2}}\left(c\right) \mid c\in ℂ\right\}$ and $\left\{±{d}_{{\alpha }_{2}}\left(z\right) \mid z\in {ℂ}^{×}\right\}\text{.}$ Then $C\left(s,n\right)$ is trivial.

$B,xα2(c)w2B w2w1B,w2w1xα2(c)w2B (stq2,1,eα1)$

${t}_{q,1}$

If

$(s,n) = ( ± [ 1000 0q00 0010 000q-1 ] , e2α1+α2 ) ,$

then ${ℬ}_{s,n}$ consists of $B,$ ${w}_{1}B,$ ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}B,$ and ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}B,$ two disjoint copies of ${ℙ}^{1}\text{.}$ The centralizer of $s$ is generated by ${x}_{±{\alpha }_{2}}\left(c\right)$ and $D,$ so that ${C}_{G}\left(s,n\right)$ is generated by $Z\left(G\right),$ $\left\{{x}_{-{\alpha }_{2}}\left(c\right) \mid c\in ℂ\right\}$ and ${d}_{{\alpha }_{1}}\left(-1\right)\text{.}$ The group $C\left(s,n\right)$ is isomorphic to ${ℤ}_{2},$ generated by ${d}_{{\alpha }_{1}}\left(-1\right),$ but this group acts trivially on ${ℬ}_{s,n}\text{.}$

If

$(s,n) = w0 · ( stq,1, e-2α1-α2 ) = ( ± [ 1000 0q-100 0010 000q ] , e2α1+α2 ) ,$

then ${ℬ}_{s,n}$ consists of $B,$ ${w}_{1}B,$ ${x}_{{\alpha }_{2}}\left(c\right)B,$ and ${x}_{{\alpha }_{2}}\left(c\right){w}_{2}{w}_{1}B,$ two disjoint copies of ${ℙ}^{1}\text{.}$ The centralizer of $s$ is generated by ${x}_{±{\alpha }_{2}}\left(c\right)$ and $D,$ so that ${C}_{G}\left(s,n\right)$ is generated by $Z\left(G\right),$ $\left\{{x}_{-{\alpha }_{2}}\left(c\right) \mid c\in ℂ\right\}$ and ${d}_{{\alpha }_{1}}\left(-1\right)\text{.}$ The group $C\left(s,n\right)$ is isomorphic to ${ℤ}_{2},$ generated by ${d}_{{\alpha }_{1}}\left(-1\right),$ but this group acts trivially on ${ℬ}_{s,n}\text{.}$

$B,xα2(c)w2B w1B,w1xα2(c)w2B B,xα2(c)w2B w1B,xα2(c)w2w1B ( st±q,1, e-2α1-α2 ) ( st±q,1, e2α1+α2 )$

If $z\ne 1,{q}^{±2},{q}^{-4},{q}^{-6}$ and

$(s,n) = ( ± [ z1/2000 0q2z1/200 00z-1/20 000q-2z-1/2 ] , eα1 ) ,$

then ${ℬ}_{s,n}$ consists of four points - $B,$ ${w}_{2}B,$ ${w}_{2}{w}_{1}B,$ and ${w}_{2}{w}_{1}{w}_{2}B\text{.}$ The centralizer of $s$ is $D$ so that ${C}_{G}\left(s,n\right)$ is $\left\{±{d}_{{\alpha }_{2}}\left(z\right) \mid z\in {ℂ}^{×}\right\}\text{.}$ Then $C\left(s,n\right)$ is trivial.

If $z\ne 1,{q}^{±2},{q}^{-4},{q}^{-6}$ and

$(s,n) = ( ± [ q2z1/2000 0z1/200 00q-2z-1/20 000z-1/2 ] , eα1 ) ,$

then ${ℬ}_{s,n}$ consists of four points - $B,$ ${w}_{2}B,$ ${w}_{2}{w}_{1}B,$ and ${w}_{2}{w}_{1}{w}_{2}B\text{.}$ The centralizer of $s$ is $D$ so that ${C}_{G}\left(s,n\right)$ is $\left\{±{d}_{{\alpha }_{2}}\left(z\right) \mid z\in {ℂ}^{×}\right\}\text{.}$ Then $C\left(s,n\right)$ is trivial.

$B w2B w2w1B w2w1w2B B w2B w2w1B w2w1w2B (stq2,z,eα1) (stq2,z,e-α1)$

If $z\ne ±1,{q}^{±2},{q}^{-4},-{q}^{-2}$ and

$(s,n) = ( ± [ q000 0zq00 00q-10 000z-1q-1 ] , eα2 ) ,$

then ${ℬ}_{s,n}$ consists of four points - $B,$ ${w}_{1}B,$ ${w}_{1}{w}_{2}B,$ and ${w}_{1}{w}_{2}{w}_{1}B\text{.}$ The centralizer of $s$ is $D$ so that ${C}_{G}\left(s,n\right)$ is $\left\{±{d}_{{\alpha }_{1}}\left(z\right) \mid z\in {ℂ}^{×}\right\}\text{.}$ Then $C\left(s,n\right)$ is trivial.

If $z\ne ±1,{q}^{±2},{q}^{-4},-{q}^{-2}$ and

$(s,n) = ( ± [ q-1000 0zq00 00q-10 000z-1q-1 ] , eα2 ) ,$

then ${ℬ}_{s,n}$ consists of four points - $B,$ ${w}_{1}B,$ ${w}_{1}{w}_{2}B,$ and ${w}_{1}{W}_{2}{w}_{1}B\text{.}$ The centralizer of $s$ is $D$ so that ${C}_{G}\left(s,n\right)$ is $\left\{±{d}_{{\alpha }_{1}}\left(z\right) \mid z\in {ℂ}^{×}\right\}\text{.}$ Then $C\left(s,n\right)$ is trivial.

$B w1B w1w2B w1w2w1B B w1B w1w2B w1w2w1B (stz,q2,eα2) (stz,q2,e-α2)$

${q}^{2}=1 \left(\ell =1\right):$

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 summarize the bijections between irreducible representations of $\stackrel{\sim }{H}$ and orbits in

${ (s,n) ∣ s∈ Gss,n∈𝔤qs }$

paired with representations of $C{\left(s,n\right)}^{\circ }$ appearing in ${H}_{•}\left({ℬ}_{s,n}\right)\text{.}$

Generic $q$

$Central Character Dimension Indexing Weights t1,1 8 (st,0,1) W0t t-1,1 8 (st,0,1) W0t t1,z 8 (st,0,1) W0t t1,q2 1 (st,eα1+α2,-1) s2t t1,q2 1 (st,eα2,1) w0t t1,q2 3 (st,eα1+α2,1) t,s2t t1,q2 3 (st,0,1) w0t,s2w0t tq2,1 4 (st,0,1) s1t,s2s1t,s1s2s1t tq2,1 4 (st,eα1,1) t,s1t,s2s1t tq,1 4 (st,0,1) s2s1t,s1s2s1t tq,1 4 (st,e2α1+α2,1) t,s1t t-q,1 4 (st,0,1) s2s1t,s1s2s1t t-q,1 4 (st,e2α1+α2,1) t,s1t tz,1 8 (st,0,1) W0t tq2,q2 1 (st,0,1) W0t tq2,q2 1 (st,eα1+eα2,1) t tq2,q2 3 (st,eα1,1) s2t,s1s2t,s2s1s2t tq2,q2 3 (st,eα2,1) s1t,s2s1t,s1s2s1t tq2,z 4 (st,0,1) s1t,s2s1t,s1s2s1t,w0t tq2,z 4 (st,eα1,1) t,s2t,s1s2t,s2s1s2t t-1,q2 2 (st,0,1) s2s1s2t,w0t t-1,q2 2 (st,eα2,1) s2t,s1s2t t-1,q2 2 (st,e2α1+α2,1) s2s1t,s1s2s1t t-1,q2 2 (st,eα2+e2α1+α2,1) t,s1t tz,q2 4 (st,0,1) s2t,s1s2t,s2s1s2t,w0t tz,q2 4 (st,eα2,1) t,s1t,s2s1t,s1s2s1t tz,w 8 (st,0,1) W0t Table 27: Geometric Indexing in Type C2, with generic q.$

${q}^{8}=1$

$Central Character Dimension Indexing Weights t1,1 8 (st,0,1) W0t t-1,1 8 (st,0,1) W0t t1,z 8 (st,0,1) W0t t1,q2 1 (st,eα1+α2,-1) s2t t1,q2 1 (st,eα2,1) w0t t1,q2 3 (st,eα1+α2,1) t,s2t t1,q2 3 (st,0,1) s1s2t,s2s1s2t tq2,1 4 (st,0,1) s1t,s2s1t,s1s2s1t,w0t tq2,1 4 (st,eα1,1) t,s2t,s1s2t,s2s1s2t tq,1 4 (st,0,1) s2s1t,s1s2s1t tq,1 4 (st,e2α1+α2,1) t,s1t t-q,1 4 (st,0,1) s2s1t,s1s2s1t t-q,1 4 (st,e2α1+α2,1) t,s1t tz,1 8 (st,0,1) W0t tq2,q2 1 (ss1s2t,eα2,1) W0t tq2,q2 2 (ss1s2t,eα2+e2α1+α2,1) s2s1t,s1s2s1t tq2,q2 1 (ss2s1s2t,eα1+eα2,1) s2s1s2t tq2,q2 1 (st,eα1+eα2,1) t tq2,q2 2 (st,eα1,1) s2t,s1s2t tq2,q2 1 (st,eα2,1) s1t tq2,z 4 (st,0,1) s1t,s2s1t,s1s2s1t,w0t tq2,z 4 (st,eα1,1) t,s2t,s1s2t,s2s1s2t tz,q2 4 (st,0,1) s2t,s1s2t,s2s1s2t,w0t tz,q2 4 (st,eα2,1) t,s1t,s2s1t,s1s2s1t tz,w 8 (st,0,1) W0t Table 28: Geometric Indexing in Type C2, with q8=1.$

Note that $\left({s}_{{t}_{{q}^{2},{q}^{2}}},0\right)$ is omitted from this bijection since ${Z}_{𝔤}\left(0\right)\cap {𝔤}_{q}^{s}$ is not contained in $𝒩,$ and thus condition 3.1 omits it.

${q}^{6}=1$

$Central Character Dimension Indexing Weights t1,1 8 (st,0,1) W0t t-1,1 8 (st,0,1) W0t t1,z 8 (st,0,1) W0t t1,q2 1 (st,eα1+α2,-1) s2t t1,q2 1 (st,eα2,1) s1s2t t1,q2 3 (st,eα1+α2,1) t,s2t t1,q2 3 (st,0,1) s1s2t,s2s1s2t tq2,1 3 (st,0,1) s2s1t,s1s2s1t tq2,1 3 (st,eα1,1) t,s1t tq2,1 1 (ss2s1t,eα2,1) s1t tq2,1 1 (ss2s1t,eα1+eα2,1) s2s1t tq,1 4 (st,0,1) s2s1t,s1s2s1t tq,1 4 (st,e2α1+α2,1) t,s1t tz,1 8 (st,0,1) W0t tq2,z 4 (st,0,1) s1t,s2s1t,s1s2s1t,s2s1s2s1t tq2,z 4 (st,eα1,1) t,s2t,s1s2t,s2s1s2t t-1,q2 2 (st,0,1) s2s1s2t,s1s2s1s2t t-1,q2 2 (st,eα2,1) s2s1t,s1s2s1t t-1,q2 2 (st,e2α1+α2,1) s2t,s1s2t t-1,q2 2 (st,eα2+e2α1+α2,1) t,s1t tz,q2 4 (st,0,1) s2t,s1s2t,s2s1s2t,s1s2s1s2t tz,q2 4 (st,eα2,1) t,s1t,s2s1t,s1s2s1t tz,w 8 (st,0,1) W0t Table 29: Geometric Indexing in Type C2, with q6=1.$

${q}^{4}=1$

$Central Character Dimension Indexing Weights t1,1 8 (st,0,1) W0t t-1,1 4 (st,eα1,1) t,s1t t1,z 8 (st,0,1) W0t t1,q2 1 (st,e-α2+eα1+α2,1) s2t t1,q2 1 (st,eα2+e-α1+α2,1) s1s2t t1,q2 2 (st,eα1+α2,1) t t1,q2 ? (st,eα1+α2,1) ? t1,q2 2 ({s}_{{w}_{0}t},{e}_{{\alpha }_{1}+{\alpha }_{2}},1) s2s1s2t t1,q2 ? (sw0t,eα1+α2,-1) ? tq,1 4 (st,e2α1+α2,1) t,s1t tq,1 4 (st,e-2α1-α2,1) s2s1t,s1s2s1t tz,1 8 (st,0,1) W0t tq2,z 4 (st,eα1,1) t,s2t,s1s2t,s2s1s2t tq2,z 4 (st,e-α1,1) s1t,s2s1t,s1s2s1t,s2s1s2s1t tz,q2 4 (st,eα2,1) t,s1t,s2s1t,s1s2s1t tz,q2 4 (st,e-α2,1) s2t,s1s2t,s2s1s2t,s1s2s1s2t tz,w 8 (st,0,1) W0t Table 30: Geometric Indexing in Type C2, with q2=-1.$

For the central character ${t}_{1,{q}^{2}}$ when ${q}^{4}=1,$ we note that condition 3.1 omits the nilpotent ${C}_{G}\left({s}_{t}\right)\text{-orbits}$ of 0, ${e}_{±{\alpha }_{2}},$ and ${e}_{{\alpha }_{2}}+{e}_{-2{\alpha }_{1}-{\alpha }_{2}}$ from conisderation.

${q}^{2}=1$

$Central Character Dimension Indexing Weights t1,1 1 (st,0,1) sign t1,1 2 (st,eα1,1) t1,1 1 (st,eα1,-1) t1,1 1 (st,eα2,1) t1,1 1 (st,eα1+eα2,1) triv t1,z 4 (st,0,1) sign ⊗H∼{1}H∼ t1,z 4 (st,eα1,1) triv ⊗H∼{1}H∼ tq,1 4 (st,0,1) sign ⊗H∼{1}H∼ tq,1 4 (st,e2α1+α2,1) triv ⊗H∼{1}H∼ tz,1 4 (st,0,1) sign ⊗H∼{1}H∼ tz,1 4 (st,eα2,1) triv ⊗H∼{1}H∼ tz,w 8 (st,0,1) triv ⊗ℂ[X]H∼ Table 31: Geometric Indexing in Type C2, with q=-1.$

## 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.