## Examples of affine braid groups

Last update: 18 September 2012

## Type ${B}_{n}$

The lattices are

$P∨= ∑i=1nℤ εi∨,and Q∨= { λ1ε1∨+… +λnεn∨ ∣λ1+…+ λn=0mod2 } .$

and

$αi = εi-εi+1, αn = εn, αi∨ = εi∨- εi+1∨, αn∨ = 2εn∨, ωi∨ = εi∨+…+ εn∨, ωn∨ = ε1∨+…+ εn∨,$

The walls of the fundamental alcove are

$𝔥α0= 𝔥 -(ε1+ε2) +δ , 𝔥αi= 𝔥 εi-εi+1 , 𝔥αn= 𝔥εn,$

for $i=1,\dots ,n-1\text{.}$ Since

$sε1+ε2= sε1 sε2 sε1-ε2= s2s3… sn-1 sn sn-1…s2 s1s2… sn-1 sn sn-1…s2,$

and

$ω1∨=ε1∨, w1=sε2… sεn,and w0=sε1… sεn,$

so that

$w0w1=sε1= s1s2…sn-1 snsn-1…s1$

it follows (from formulas ???) that

$Y ε1∨+ε2∨ = R0 Tsε1+ε2 =R0T2…Tn-1 TnTn-1…T2 T1T2…Tn-1 TnTn-1…T2,$

and

$Yε1∨=γ Tsε1=γ T1T2…Tn-1 TnTn-1…T2 T1,$

and

$T1 R0 T2 T3 Tn-2 Tn-1 Tn$

with $\Omega =\left\{1,\gamma \right\}$ with

$γ2=1,γR0 γ-1=T1, andγTiγ-1 =Ti,fori=2 ,…,n.$

Then ${Y}^{{\epsilon }_{i}^{\vee }}={T}_{i-1}^{-1}{Y}^{{\epsilon }_{i-1}^{\vee }}{T}_{i-1}^{-1}$ and so

$Yεi∨ = Ti-1-1 Ti-2-1 …T1-1gT1 T2…Tn-1 TnTn-1…Ti = gTi-1-1 Ti-2-1 …T2-1 R0-1T1T2 …Tn-1 TnTn-1…Ti .$

## Type ${C}_{n}$

The lattices are

$P∨= { λ1ε1∨+…+ λnεn∨∣ allλi∈12 ℤ≥0or all λi∈ℤ≥0 } , Q∨= ∑i=1nℤ εi∨.$

and

$αi = εi-εi+1, αn = 2εn, αi∨ = εi∨- εi+1∨, αn∨ = εn∨, ωi∨ = ε1+…+εi, ωn∨ = 12 ( ε1∨+… +εn∨ ) ,$

The walls of the fundamental chamber are

$𝔥α0= 𝔥-ε1+δ, 𝔥αi= 𝔥εi-εi+1 ,𝔥αn= 𝔥εn$

for $i=1,\dots ,n-1\text{.}$ Then $ℬ$ is generated by ${T}_{0},\dots ,{T}_{n}$ and $\Omega =\left\{1,\sigma \right\}$ with ${\sigma }^{2}=1$ and

$T0 T1 T2 Tn-2 Tn-1 Tn with σTiσ-1= Tn-i,$

for $i=0,\dots ,n\text{.}$ The abelian subgroup $Y=\left\{{Y}^{{\lambda }^{\vee }}\phantom{\rule{0.2em}{0ex}}\mid \phantom{\rule{0.2em}{0ex}}{\lambda }^{\vee }\in {P}^{\vee }\right\}$ is given by

$Yε1∨=T0 T1T2…Tn-1 TnTn-1…T1,$

and

$Y 12 ( ε1∨+… +εn∨ ) =g ( TnTn-1…T1 ) ( TnTn-1…T2 ) ( TnTn-1…T3 ) … (TnTn-1) Tn.$

Since ${Y}^{{\epsilon }_{i}^{\vee }}={T}_{i-1}^{-1}{Y}^{{\epsilon }_{i-1}^{\vee }}{T}_{i-1}^{-1},$

$Yεi∨= Ti-1-1 Ti-1-2… T1-1T0T1 T2…Tn-1Tn Tn-1…Ti.$

The formulas (???) and (???) follow from

$sε1=s1s2 …sn-1sn sn-1…s1,$ $ω1∨=12 ( ε1∨+… +εn∨ ) andw0=sε1 …sεn$

and

$w1= sε1-εn sε2-εn-1 sε3-εn-2 …,$

so that

$w0w1= ( snsn-1…s1 ) ( snsn-1…s2 ) ( snsn-1…s3 ) (snsn-1)sn .$

A pictorial representation of $ℬ$ is

$T0= Tn= and Ti= i i+1 σ=$

It may be helpful to as $\stackrel{\sim }{ℬ}$ the full twist

$σ= so thatσT0 σ-1=Tn, andσgiσ-1 =gn-i,$

produces the automorphism of the Dynkin diagram.

## Type ${D}_{2n+1}$

Let

$P∨= { λ1ε1∨+… +λnεn∨ ∣allλi ∈12ℤ≥0 or allλi∈ ℤ≥0 } ,Q∨= { λ1ε1∨+… +λnεn∨ ∣λ1+…+ λn∈2ℤ } .$

and

$αi = εi-εi+1, αn-1 = εn-1-εn, αn = εn-1+εn, αi∨ = εi∨- εi+1∨, αn-1∨ = εn-1-εn∨, αn∨ = εn-1+εn∨, ωi∨ = ε1+…+εi, ωn-1∨ = 12 ( ε1∨+…+ εn-1∨- εn∨ ) , ωn∨ = 12 ( ε1∨+…+ εn-1∨+ εn∨ ) ,$

The walls of the fundamental chamber are

$𝔥α0= 𝔥 -(ε1+ε2) +δ ,𝔥αi= 𝔥 εi-εi+1 ,𝔥αn= 𝔥 εn-1+εn$

for $i=1,\dots ,n-1\text{.}$ Then $ℬ$ is generated by ${T}_{0},\dots ,{T}_{n}$ and $\Omega =\left\{1,{g}_{1},{g}_{n-1},{g}_{n}\right\}$ with ${h}^{4}=1$ and

$T1 R0 T2 T3 Tn-3 Tn-2 Tn-1 Tn$

The abelian subgroup $Y=\left\{{Y}^{{\lambda }^{\vee }}\phantom{\rule{0.2em}{0ex}}\mid \phantom{\rule{0.2em}{0ex}}{\lambda }^{\vee }\in {P}^{\vee }\right\}$ is given by

$Yε1∨ = g1Tw0w1, Y 12 ( ε1∨+…+ εn-1∨- εn∨ ) = gn-1 Tw0wn-1, Y 12 ( ε1∨+…+ εn-1∨+ εn∨ ) = gn Tw0wn,$

where

$w0= { sε1… sεn-1 , ifnis odd, sε1… sεn , ifnis even,$

since, if $n$ is even,

$w0(εi-εj) =-εi+εj, andw0 (εi+εj)= -εi-εj,$

for $1\le i and, if $n$ is odd

$w0 (εi-εn) =-εi-εn, and w0 (εi-εj) =-εi+εj, w0 (εi+εn) =-εi+εn, w0 (εi+εj)= -εi-εj,$

so that ${w}_{0}$ takes positive roots to negative roots. Then

$w1= { sε2… sεn , ifnis odd, sε2… sεn-1 , ifnis even,$ $wn= sε1-εn sε2-εn-1 …andwn-1= sε1sεn wn.$

If $n$ is even then ${g}_{n}^{2}={g}_{1}$ and ${g}_{n}^{3}={g}_{n-1}$ and $\Omega \cong ℤ/4ℤ\text{.}$ If $n$ is odd then ${g}_{n}^{2}=1,$ ${g}_{1}^{2}=1,$ ${g}_{n-1}^{2}=1$ and ${g}_{n}{g}_{1}={g}_{1}{g}_{n}={g}_{n-1}$ so that $\Omega \cong ℤ/2ℤ×ℤ/2ℤ\text{.}$ The lattices are

$P∨⊇𝔥ℤ⊇Q∨ ifnis odd,$

and

$Q∨+ℤωn∨ P∨ 𝔥ℤ Q∨ Q∨+ℤωn-1∨$

with ${𝔥}_{ℤ}={Q}^{\vee }+ℤ{\omega }_{1}^{\vee }\text{.}$

Then

$w0 wn = sε1 sεn = s1s2… sn-2 sn-1 sn sn-2 …s1.$

If $n$ is odd then

$w0wn = (sε2…sεn) ( sε1-εn sε2-εn-2 … ) = (snsn-2…s1) (sn-1…s2) (snsn-2…s3) (sn-1…s4) … (snsn-2) sn-1$

and

$w0wn-1 = (sε1…sεn-1) ( sε1-εn sε2-εn-2 … ) = (sn-1…s1) (snsn-2…s2) (sn-1…s3) (snsn-2…s4) … (sn-1sn-2) sn.$

If $n$ is even then

$w0wn = (sε1…sεn) ( sε1-εn sε2-εn-2 … ) = (snsn-2…s1) (sn-1…s2) (snsn-2…s3) (sn-1…s4) … (sn-1sn-2) sn$

and

$w0wn-1 = (sε2…sεn-2) ( sε1-εn sε2-εn-2 … ) = (sn-1…s1) (snsn-2…s2) (sn-1…s3) (snsn-2…s4) … (snsn-2) sn-1.$

Finally

$sε1+ε2= sε1sε2 sε1-ε2= snsn-2…s1 s2…sn-2sn sn-2…s2$

and ${T}_{0}={g}_{1}{T}_{1}{g}_{1}^{-1}={Y}^{{\epsilon }_{1}^{\vee }+{\epsilon }_{2}^{\vee }}{T}_{{s}_{{\epsilon }_{1}+{\epsilon }_{2}}}\text{.}$

## Type ${A}_{1}$

The Weyl group ${W}_{0}=⟨{s}_{1}\phantom{\rule{0.2em}{0ex}}\mid \phantom{\rule{0.2em}{0ex}}{s}_{1}^{2}=1⟩$ has order two and acts on the lattices

$𝔥ℤ=ℤω∨and 𝔥ℤ*=ℤω bys1ω∨=- ω∨ands1 ω=-ω, (5.1)$

and REARRANGE??

$α∨=2ω∨, α=2ω,and ⟨ω∨,α⟩=1. (5.2)$ $g∨s0∨s1∨s0∨ g∨s0∨s1∨ g∨s0∨ g∨ g∨s1∨ g∨s1∨s0∨ g∨s1∨s0∨s1∨ X-3ω X-ωs1 X-ω Xωs1 Xω X3ωs1 X3ω s1∨s0∨s1∨ s1∨s0∨ s1∨ 1 s0∨ s0∨s1∨ s0∨s1∨s0∨ X-αs1 X-α s1 1 Xαs1 Xα X2αs1 𝔥α∨+2d 𝔥α∨+d 𝔥α∨ 𝔥-α∨+d 𝔥-α∨+2d 𝔥-α∨+3d$

The double affine braid group is generated by ${q}^{\frac{1}{2}},g,{T}_{0},{T}_{1},{g}^{\vee },{T}_{0}^{\vee }$ with relations

$g2=1,T0= gT1g-1, (g∨)2=1, T0∨=g∨ T1(g∨)-1 , andT1g∨g= q-12g g∨T1-1. (5.3)$

Define ${Y}^{{\omega }^{\vee }}$ and ${X}^{\omega }$ by

$Yω∨=gT1and Xω=g∨T1-1 , (5.4)$

and note that

$T0T1= Y2ω∨, and(T0∨)-1 T1-1=X2ω. (5.5)$

(Duality).

1. The double affine braid group $\stackrel{\sim }{ℬ}$ is generated by ${T}_{0},{T}_{1},g,{X}^{\omega },$ and ${q}^{\frac{1}{2}},$ with relations

$g2=1, T0=gT1 g-1, gXω= q12 X-ωg, T1XωT1= X-ω,and T0X-ωT0 =q-1Xω. (5.6)$
2. The double affine braid group $\stackrel{\sim }{ℬ}$ is generated by ${T}_{0}^{\vee },{T}_{1},{g}^{\vee },{Y}^{{\omega }^{\vee }}$ and ${q}^{\frac{1}{2}}$ with relations

$(g∨)2=1, T0∨=g∨ T1 (g∨)-1, g∨Yω∨= q-12 Y-ω∨g∨ ,T1-1 Yω∨ T1-1,and (T0∨)-1 Y-ω∨ (T0∨)-1 =qYω∨. (5.7)$

 Proof. We prove that the presentation in (5.7) is equivalent to the presentation in (5.3). The proof that the presentation in (5.6) is equivalent to the presentation in (5.3) is similar. (5.3)$⇒$(5.7): Use (5.4) to define ${Y}^{\omega }$ in terms of $g$ and ${T}_{1}\text{.}$ The first and second relations in (5.7) are the third and fourth relations (5.3). The proof of the third, fourth and fifth relations in (5.7) are $g∨Yω∨= g∨gT1= q-12 T1-1gg∨= q-12 Y-ω∨g∨, T1-1Yω∨ T1-1=T1-1 gT1T1-1= T1-1g= Y-ω∨,$ and $(T0∨)-1 Y-ω∨ (T0∨)-1 =g∨T1-1 g∨Y-ω∨ g∨T1-1g∨ =q12g∨ T1-1 Yω∨T1-1 g∨=q12 g∨Y-ω∨ g∨=qYω∨,$ respectively. (5.7)$⇒$(5.3): Define $g={Y}^{{\omega }^{\vee }}{T}_{1}^{-1}$ and ${T}_{0}={Y}^{2{\omega }^{\vee }}{T}_{1}^{-1}\text{.}$ The third and fourth relations of (5.3) are exactly the first and second relations of (5.7). The proof of the first, second and fifth relations in (5.3) are $g2=Yω∨ T1-1Yω∨ T1-1= Yω∨ Y-ω∨=1, gT1g-1= Yω∨T1-1 T1T1 Y-ω∨= Yω∨T1 Y-ω∨= Yω∨T1 Y-ω∨T1 T1-1= Y2ω∨,$ and $T1g∨g=T1g∨ Yω∨T1-1 =q-12T1 Y-ω∨g∨ T1-1= q-12gg∨ T1-1,$ respectively. $\square$

The Haiman relations do not occur in Type ${A}_{1}$ since there are no short roots (or even elements of the lattice) $\alpha$ with $⟨\alpha ,{\phi }^{\vee }⟩$ equal to 0 or 1.

The double affine Hecke algebra $\stackrel{\sim }{H}$ is $ℂ\stackrel{\sim }{ℬ}$ with the additional relations

$Ti2= ( t1/2- t-1/2 ) Ti+1,for i=0,1, (5.8)$

Using (5.8), the relations in Proposition (4.1) give

$g∨Yω∨= q-1/2 Y-ω∨g∨, T1Yω∨= Y-ω∨T1+ ( t1/2- t-1/2 ) Yω∨- Y-ω∨ 1-Y-α∨ ,and T0∨Yω∨= q-1Y-ω∨ T0∨+ ( t1/2- t-1/2 ) ( Yω∨-q-1 Y-ω∨ 1-qYω∨ ) .$

With ${Y}^{{\alpha }_{0}}=q{Y}^{\alpha }$ and ${Y}^{{\alpha }_{1}}={Y}^{\alpha },$ then

$τg∨=g∨,and τi∨=Ti∨- ( t1/2- t-1/2 ) ( 1 1-Y-αi∨ ) ,fori=0,1.$

To illustrate Theorem 2.2, note that ${X}^{-2\omega }{s}_{1}^{\vee }{s}_{0}^{\vee }$ is a reduced word and

$τ1∨τ0∨ = ( T1∨+ t-1/2(1-t) 1-Y-α1∨ ) τ0∨ = T1∨T0∨+ T1∨ t-1/2(1-t) 1-Y-α0∨ +(T0∨)-1 t-1/2(1-t) 1-Y-s0α1∨ + ( t-1/2(1-t) 1-Y-s0α1∨ ) ( t-1/2(1-t) Y-α0∨ 1-Y-α0∨ ) = X-2ω + T1∨ t-1/2(1-t) 1-Y-α0∨ +X2ω T1∨ t-1/2(1-t) 1-Y-s0α1∨ + ( t-1/2(1-t) 1-Y-s0α1∨ ) ( t-1/2(1-t) Y-α0∨ 1-Y-α0∨ )$

The corresponding paths in $ℬ\left(1,\stackrel{\to }{-2\omega }\right)=B\left(\stackrel{\to }{-2\omega }\right)$ are

$X-2ω T1∨ t-1/2(1-t) 1-Y-α0∨ X2ω T1∨ t-1/2(1-t) 1-Y-s0α1∨ t-1/2(1-t) 1-Y-s0α1∨ t-1/2(1-t) Y-α0∨ 1-Y-α0∨$

The polynomial representations is defined by

$g1=1,T0 1=t121, andT11= t121.$

In this case

$ρc=12cα andW0= { X-ℓw∣ ℓ∈ℤ≥0 } ∪ { Xℓωs1∨ ∣ℓ∈ℤ>0 } , (5.9)$

is the set of minimal length coset representation of ${W}^{\vee }/{W}_{0}\text{.}$

Applying the expansion of ${\tau }_{1}^{\vee }{\tau }_{0}^{\vee }$ to $1$ and using

$Y-α0∨1=q Yα∨1=qqc 1=qt1,and Y-s0α1∨1= Yα∨+2d1= q2Yα∨1= q2t,$

gives

$E-2ω = τ1∨τ0∨1 = X-2ω+t1/2 t-1/2(1-t) 1-qt +X2ωt1/2 t-1/2(1-t) 1-q2t + ( t-1/2(1-t) 1-q2t ) ( t-1/2(1-t)qt 1-qt ) = X-2ω+ 1-t1-qt +X2ω 1-t1-q2t+ (1-t1-q2t) ((1-t)q1-qt) .$

Since ${1}_{0}={T}_{1}^{\vee }+{t}^{-1/2}$ the symmetric Macdonald polynomial ${P}_{2\omega }={1}_{0}{E}_{2\omega }={1}_{0}{\tau }_{0}^{\vee }1$ is

$P2ω = 10E2ω= (T1∨+t-1/2) τ0∨1 = ( T1∨T0∨+ T1∨ t-1/2(1-t) 1-Y-α0∨ +t-1/2 (T0∨)-1+ t-1/2 t-1/2(1-t) Y-α0∨ 1-Y-α0∨ ) 1 = ( X-2ω+ t1/2 t-1/2(1-t) 1-qt +t-1/2 X2ωT1∨+ t-1/2 t-1/2(1-t) qt 1-qt ) 1 = ( X-2ω+ t1/2 1-t 1-qt +X2ω+ t-1/2 (1-t) q 1-qt ) 1 = ( X2ω+ X-2ω+ t1/2+ (1+q) 1-t 1-qt ) 1.$

The corresponding paths in $𝒫\left(\stackrel{\to }{2\omega }\right)$ are

$X-2ω 1-t1-tq X2ω q (1-t)1-tq$

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

$S0s1S0 X2ε1 Xε1-ε2 Xε1 Xε1+ε2 S0s1 S0 τs1 τ X-ε2 Xε2 τ-1s1 τ-1 1 s1 τS0 τS0s1 X-ε1 X-ε1+ε2 X2ε2-ε1 τ-1S0 τ-1S0s1 s1S0 s1S0s1 -+ -+ -+ +- +- + - + - + - + - 𝔥-(ε1∨-ε2∨)+2d 𝔥-(ε1∨-ε2∨)+d 𝔥ε1∨-ε2∨ 𝔥-ε1∨+d 𝔥-ε1∨+d 𝔥ε1∨ 𝔥-ε2∨+d 𝔥-ε2∨+2d 𝔥ε2∨$

From $G{L}_{n}$ we get $\tau {T}_{1}{\tau }^{-1}={s}_{0},$

$Yε1∨=τT1, Yε2∨=τS0, Y ε1∨-ε2∨ =S0T1, Xε1=τ∨ T1-1, Xε2=τ∨ (S0∨)-1, Xε1-ε1 =(S0∨)-1 T1-1.$

Further ${ℬ}_{G{L}_{2}}\subseteq {ℬ}_{2}$ by

$τ=T0T1T2, Xε1+ε2= τ2= (T0T1T2)2 ,andS0=τT1 τ-1=T0T2-1 T1T2T0-1.$

To check these we note that

$Yε1∨=T0T1 T2T1and Yε2∨=T1-1 T0T1T2$

so that

$τT1τ-1=T0 T1T2T1T2-1 T1-1T0-1= T0T2-1T1 T2T1T1-1 T0-1=T0 T2-1T1T2 T0-1.$

which should be ${T}_{0}$ in $G{L}_{2}\text{.}$

## Type ${B}_{2}$

$R0 T2 T1 withℬSO5 ⊆ℬ2viaR0 =T0T1T0.$

Further,

$withγ=T0, ℬSpin5is the quotient of ℬSp4by γ2=1.$

The conversions come from the formulas

$Yε1∨+ε2∨ =R0T2T1T2, Yε1∨=γT1 T2T1,$

THIS $\gamma$ WANTS TO BE ${T}_{0}$ in ${ℬ}_{{\text{Spin}}_{5}}\text{.}$

$X2ε1 Xε1-ε2 Xε1+ε2 R0 1 s1 s2 s2s1s2 s2s1 s1s2 s1s2s1 s1s2s1s2 X-2ε1 X-ε1+ε2 X-ε1-ε2 -+ -+ -+ +- + - + - + - + - + - + - + - 𝔥-(ε1∨-ε2∨)+2d 𝔥-(ε1∨-ε2∨)+d 𝔥ε1∨-ε2∨ 𝔥-ε1∨+d 𝔥-ε1∨+d 𝔥ε1∨ 𝔥-ε2∨+d 𝔥-ε2∨+2d 𝔥ε2∨ 𝔥ε1∨+ε2∨ 𝔥-(ε1∨-ε2∨)+2d 𝔥-(ε1∨-ε2∨)+d$

## Type ${A}_{1}×{A}_{1}$

There is an embedding of $ℬ\subseteq {ℬ}_{2}$ by

$a2=T2T1T2, R0=T0T1 T0S0=T2 T0T1.$ $X2ε1 Xε1-ε2 Xε1+ε2 R0 S0 1 s1 a2 s1a2 X-2ε1 X-ε1+ε2 X-ε1-ε2 +- +- + - + - + - + - 𝔥-(ε1∨-ε2∨)+2d 𝔥-(ε1∨-ε2∨)+d 𝔥ε1∨-ε2∨ 𝔥ε1∨+ε2∨ 𝔥-(ε1∨-ε2∨)+2d 𝔥-(ε1∨-ε2∨)+d$

## Notes and References

This page is taken from a paper entitled Double affine braid groups and Hecke algebras of classical type by Arun Ram, March 5, 2009.