Examples of affine braid groups

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

Last update: 18 September 2012

Type $B_n$

The lattices are

$$P^{\vee }=\sum _{i=1}^n\mathbb{Z}{\epsilon }_i^{\vee },\text{and}{Q}^{\vee }=\{{\lambda }_1{\epsilon }_1^{\vee }+\dots +{\lambda }_n{\epsilon }_n^{\vee }\mid {\lambda }_1+\dots +{\lambda }_n=0\text{mod}2\}\text{.}$$

and

$$\begin{array}{ccc}{\alpha }_i& =& {\epsilon }_i-{\epsilon }_{i+1},\\ {\alpha }_n& =& {\epsilon }_n,\end{array}\begin{array}{ccc}{\alpha }_i^{\vee }& =& {\epsilon }_i^{\vee }-{\epsilon }_{i+1}^{\vee },\\ {\alpha }_n^{\vee }& =& 2{\epsilon }_n^{\vee },\end{array}\begin{array}{ccc}{\omega }_i^{\vee }& =& {\epsilon }_i^{\vee }+\dots +{\epsilon }_n^{\vee },\\ {\omega }_n^{\vee }& =& {\epsilon }_1^{\vee }+\dots +{\epsilon }_n^{\vee },\end{array}$$

The walls of the fundamental alcove are

$${𝔥}^{{\alpha }_0}={𝔥}^{-({\epsilon }_1+{\epsilon }_2)+\delta },{𝔥}^{{\alpha }_i}={𝔥}^{{\epsilon }_i-{\epsilon }_{i+1}},{𝔥}^{{\alpha }_n}={𝔥}^{{\epsilon }_n},$$

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

$$s_{{\epsilon }_1+{\epsilon }_2}=s_{{\epsilon }_1}{s}_{{\epsilon }_2}{s}_{{\epsilon }_1-{\epsilon }_2}=s_2{s}_3\dots s_{n-1}{s}_n{s}_{n-1}\dots s_2{s}_1{s}_2\dots s_{n-1}{s}_n{s}_{n-1}\dots s_2,$$

and

$${\omega }_1^{\vee }={\epsilon }_1^{\vee },w_1=s_{{\epsilon }_2}\dots s_{{\epsilon }_n},\text{and}{w}_0=s_{{\epsilon }_1}\dots s_{{\epsilon }_n},$$

so that

$$w_0{w}_1=s_{{\epsilon }_1}=s_1{s}_2\dots s_{n-1}{s}_n{s}_{n-1}\dots s_1$$

it follows (from formulas ???) that

$$Y^{{\epsilon }_1^{\vee }+{\epsilon }_2^{\vee }}=R_0{T}_{s_{{\epsilon }_1+{\epsilon }_2}}=R_0{T}_2\dots T_{n-1}{T}_n{T}_{n-1}\dots T_2{T}_1{T}_2\dots T_{n-1}{T}_n{T}_{n-1}\dots T_2,$$

and

$$Y^{{\epsilon }_1^{\vee }}=\gamma T_{s_{{\epsilon }_1}}=\gamma T_1{T}_2\dots T_{n-1}{T}_n{T}_{n-1}\dots T_2{T}_1,$$

and

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

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

$${\gamma }^2=1,\gamma R_0{\gamma }^{-1}=T_1,\text{and}\gamma T_i{\gamma }^{-1}=T_i,\text{for}i=2,\dots ,n\text{.}$$

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

$$\begin{array}{ccc}{Y}^{{\epsilon }_i^{\vee }}& =& T_{i-1}^{-1}{T}_{i-2}^{-1}\dots T_1^{-1}g{T}_1{T}_2\dots T_{n-1}{T}_n{T}_{n-1}\dots T_i\\ & =& g{T}_{i-1}^{-1}{T}_{i-2}^{-1}\dots T_2^{-1}{R}_0^{-1}{T}_1{T}_2\dots T_{n-1}{T}_n{T}_{n-1}\dots T_i\text{.}\end{array}$$

Type $C_n$

The lattices are

$$P^{\vee }=\{{\lambda }_1{\epsilon }_1^{\vee }+\dots +{\lambda }_n{\epsilon }_n^{\vee }\mid \text{all}{\lambda }_i\in \frac{1}{2}{\mathbb{Z}}_{\ge 0}\text{or all}{\lambda }_i\in {\mathbb{Z}}_{\ge 0}\},Q^{\vee }=\sum _{i=1}^n\mathbb{Z}{\epsilon }_i^{\vee }\text{.}$$

and

$$\begin{array}{ccc}{\alpha }_i& =& {\epsilon }_i-{\epsilon }_{i+1},\\ {\alpha }_n& =& 2{\epsilon }_n,\end{array}\begin{array}{ccc}{\alpha }_i^{\vee }& =& {\epsilon }_i^{\vee }-{\epsilon }_{i+1}^{\vee },\\ {\alpha }_n^{\vee }& =& {\epsilon }_n^{\vee },\end{array}\begin{array}{ccc}{\omega }_i^{\vee }& =& {\epsilon }_1+\dots +{\epsilon }_i,\\ {\omega }_n^{\vee }& =& \frac{1}{2}({\epsilon }_1^{\vee }+\dots +{\epsilon }_n^{\vee }),\end{array}$$

The walls of the fundamental chamber are

$${𝔥}^{{\alpha }_0}={𝔥}^{-{\epsilon }_1+\delta },{𝔥}_{{\alpha }_i}={𝔥}^{{\epsilon }_i-{\epsilon }_{i+1}},{𝔥}^{{\alpha }_n}={𝔥}^{{\epsilon }_n}$$

for $i=1,\dots ,n-1\text{.}$ Then $ℬ$ is generated by $T_0,\dots ,T_n$ and $\Omega =\{1,\sigma \}$ with ${\sigma }^2=1$ and

$$T0 T1 T2 Tn-2 Tn-1 Tn \text{with}\sigma T_i{\sigma }^{-1}=T_{n-i},$$

for $i=0,\dots ,n\text{.}$ The abelian subgroup $Y=\{Y^{{\lambda }^{\vee }}\mid {\lambda }^{\vee }\in P^{\vee }\}$ is given by

$$Y^{{\epsilon }_1^{\vee }}=T_0{T}_1{T}_2\dots T_{n-1}{T}_n{T}_{n-1}\dots T_1,$$

and

$$Y^{\frac{1}{2}({\epsilon }_1^{\vee }+\dots +{\epsilon }_n^{\vee })}=g\left(T_n{T}_{n-1}\dots T_1\right)\left(T_n{T}_{n-1}\dots T_2\right)\left(T_n{T}_{n-1}\dots T_3\right)\dots \left(T_n{T}_{n-1}\right)T_n\text{.}$$

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

$$Y^{{\epsilon }_i^{\vee }}=T_{i-1}^{-1}{T}_{i-1}^{-2}\dots T_1^{-1}{T}_0{T}_1{T}_2\dots T_{n-1}{T}_n{T}_{n-1}\dots T_i\text{.}$$

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

$$s_{{\epsilon }_1}=s_1{s}_2\dots s_{n-1}{s}_n{s}_{n-1}\dots s_1,$$ $${\omega }_1^{\vee }=\frac{1}{2}({\epsilon }_1^{\vee }+\dots +{\epsilon }_n^{\vee })\text{and}{w}_0=s_{{\epsilon }_1}\dots s_{{\epsilon }_n}$$

and

$$w_1=s_{{\epsilon }_1-{\epsilon }_n}{s}_{{\epsilon }_2-{\epsilon }_{n-1}}{s}_{{\epsilon }_3-{\epsilon }_{n-2}}\dots ,$$

so that

$$w_0{w}_1=\left(s_n{s}_{n-1}\dots s_1\right)\left(s_n{s}_{n-1}\dots s_2\right)\left(s_n{s}_{n-1}\dots s_3\right)\left(s_n{s}_{n-1}\right)s_n\text{.}$$

A pictorial representation of $ℬ$ is

$$\begin{array}{cc}{T}_0=& \\ T_n=& & \text{and}\\ T_i=& i i+1 \\ \sigma =& \end{array}$$

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

$$\begin{array}{ccc}\sigma =& & \text{so that}\sigma T_0{\sigma }^{-1}=T_n,\text{and}\sigma g_i{\sigma }^{-1}=g_{n-i},\end{array}$$

produces the automorphism of the Dynkin diagram.

Type $D_{2n+1}$

Let

$$P^{\vee }=\{{\lambda }_1{\epsilon }_1^{\vee }+\dots +{\lambda }_n{\epsilon }_n^{\vee }\mid \text{all}{\lambda }_i\in \frac{1}{2}{\mathbb{Z}}_{\ge 0}\text{or all}{\lambda }_i\in {\mathbb{Z}}_{\ge 0}\},Q^{\vee }=\{{\lambda }_1{\epsilon }_1^{\vee }+\dots +{\lambda }_n{\epsilon }_n^{\vee }\mid {\lambda }_1+\dots +{\lambda }_n\in 2\mathbb{Z}\}\text{.}$$

and

$$\begin{array}{ccc}{\alpha }_i& =& {\epsilon }_i-{\epsilon }_{i+1},\\ {\alpha }_{n-1}& =& {\epsilon }_{n-1}-{\epsilon }_n,\\ {\alpha }_n& =& {\epsilon }_{n-1}+{\epsilon }_n,\end{array}\begin{array}{ccc}{\alpha }_i^{\vee }& =& {\epsilon }_i^{\vee }-{\epsilon }_{i+1}^{\vee },\\ {\alpha }_{n-1}^{\vee }& =& {\epsilon }_{n-1}-{\epsilon }_n^{\vee },\\ {\alpha }_n^{\vee }& =& {\epsilon }_{n-1}+{\epsilon }_n^{\vee },\end{array}\begin{array}{ccc}{\omega }_i^{\vee }& =& {\epsilon }_1+\dots +{\epsilon }_i,\\ {\omega }_{n-1}^{\vee }& =& \frac{1}{2}({\epsilon }_1^{\vee }+\dots +{\epsilon }_{n-1}^{\vee }-{\epsilon }_n^{\vee }),\\ {\omega }_n^{\vee }& =& \frac{1}{2}({\epsilon }_1^{\vee }+\dots +{\epsilon }_{n-1}^{\vee }+{\epsilon }_n^{\vee }),\end{array}$$

The walls of the fundamental chamber are

$${𝔥}^{{\alpha }_0}={𝔥}^{-({\epsilon }_1+{\epsilon }_2)+\delta },{𝔥}_{{\alpha }_i}={𝔥}^{{\epsilon }_i-{\epsilon }_{i+1}},{𝔥}^{{\alpha }_n}={𝔥}^{{\epsilon }_{n-1}+{\epsilon }_n}$$

for $i=1,\dots ,n-1\text{.}$ Then $ℬ$ is generated by $T_0,\dots ,T_n$ and $\Omega =\{1,g_1,g_{n-1},g_n\}$ with $h^4=1$ and

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

The abelian subgroup $Y=\{Y^{{\lambda }^{\vee }}\mid {\lambda }^{\vee }\in P^{\vee }\}$ is given by

$$\begin{array}{ccc}{Y}^{{\epsilon }_1^{\vee }}& =& g_1{T}_{w_0{w}_1},\\ Y^{\frac{1}{2}({\epsilon }_1^{\vee }+\dots +{\epsilon }_{n-1}^{\vee }-{\epsilon }_n^{\vee })}& =& g_{n-1}{T}_{w_0{w}_{n-1}},\\ Y^{\frac{1}{2}({\epsilon }_1^{\vee }+\dots +{\epsilon }_{n-1}^{\vee }+{\epsilon }_n^{\vee })}& =& g_n{T}_{w_0{w}_n},\end{array}$$

where

$$w_0=\{\begin{array}{ccc}{s}_{{\epsilon }_1}\dots s_{{\epsilon }_{n-1}},& & \text{if}n\text{is odd,}\\ s_{{\epsilon }_1}\dots s_{{\epsilon }_n},& & \text{if}n\text{is even,}\end{array}$$

since, if $n$ is even,

$$w_0({\epsilon }_i-{\epsilon }_j)=-{\epsilon }_i+{\epsilon }_j,\text{and}{w}_0({\epsilon }_i+{\epsilon }_j)=-{\epsilon }_i-{\epsilon }_j,$$

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

$$\begin{array}{ccc}{w}_0({\epsilon }_i-{\epsilon }_n)=-{\epsilon }_i-{\epsilon }_n,& \text{and}& w_0({\epsilon }_i-{\epsilon }_j)=-{\epsilon }_i+{\epsilon }_j,\\ w_0({\epsilon }_i+{\epsilon }_n)=-{\epsilon }_i+{\epsilon }_n,& & w_0({\epsilon }_i+{\epsilon }_j)=-{\epsilon }_i-{\epsilon }_j,\end{array}$$

so that $w_0$ takes positive roots to negative roots. Then

$$w_1=\{\begin{array}{ccc}{s}_{{\epsilon }_2}\dots s_{{\epsilon }_n},& & \text{if}n\text{is odd,}\\ s_{{\epsilon }_2}\dots s_{{\epsilon }_{n-1}},& & \text{if}n\text{is even,}\end{array}$$ $$w_n=s_{{\epsilon }_1-{\epsilon }_n}{s}_{{\epsilon }_2-{\epsilon }_{n-1}}\dots \text{and}{w}_{n-1}=s_{{\epsilon }_1}{s}_{{\epsilon }_n}{w}_n\text{.}$$

If $n$ is even then $g_n^2=g_1$ and $g_n^3=g_{n-1}$ and $\Omega \cong \mathbb{Z}/4\mathbb{Z}\text{.}$ 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 $\Omega \cong \mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}\text{.}$ The lattices are

$$P^{\vee }\supseteq {𝔥}_{\mathbb{Z}}\supseteq Q^{\vee }\text{if}n\text{is odd},$$

and

$$\begin{array}{ccc}& Q^{\vee }+\mathbb{Z}{\omega }_n^{\vee }& \\ P^{\vee }& {𝔥}_{\mathbb{Z}}& Q^{\vee }\\ & Q^{\vee }+\mathbb{Z}{\omega }_{n-1}^{\vee }& \end{array}$$

with ${𝔥}_{\mathbb{Z}}=Q^{\vee }+\mathbb{Z}{\omega }_1^{\vee }\text{.}$

Then

$$w_0{w}_n=s_{{\epsilon }_1}{s}_{{\epsilon }_n}=s_1{s}_2\dots s_{n-2}{s}_{n-1}{s}_n{s}_{n-2}\dots s_1\text{.}$$

If $n$ is odd then

$$\begin{array}{ccc}{w}_0{w}_n& =& \left(s_{{\epsilon }_2}\dots s_{{\epsilon }_n}\right)\left(s_{{\epsilon }_1-{\epsilon }_n}{s}_{{\epsilon }_2-{\epsilon }_{n-2}}\dots \right)\\ & =& \left(s_n{s}_{n-2}\dots s_1\right)\left(s_{n-1}\dots s_2\right)\left(s_n{s}_{n-2}\dots s_3\right)\left(s_{n-1}\dots s_4\right)\dots \left(s_n{s}_{n-2}\right)s_{n-1}\end{array}$$

and

$$\begin{array}{ccc}{w}_0{w}_{n-1}& =& \left(s_{{\epsilon }_1}\dots s_{{\epsilon }_{n-1}}\right)\left(s_{{\epsilon }_1-{\epsilon }_n}{s}_{{\epsilon }_2-{\epsilon }_{n-2}}\dots \right)\\ & =& \left(s_{n-1}\dots s_1\right)\left(s_n{s}_{n-2}\dots s_2\right)\left(s_{n-1}\dots s_3\right)\left(s_n{s}_{n-2}\dots s_4\right)\dots \left(s_{n-1}{s}_{n-2}\right)s_n\text{.}\end{array}$$

If $n$ is even then

$$\begin{array}{ccc}{w}_0{w}_n& =& \left(s_{{\epsilon }_1}\dots s_{{\epsilon }_n}\right)\left(s_{{\epsilon }_1-{\epsilon }_n}{s}_{{\epsilon }_2-{\epsilon }_{n-2}}\dots \right)\\ & =& \left(s_n{s}_{n-2}\dots s_1\right)\left(s_{n-1}\dots s_2\right)\left(s_n{s}_{n-2}\dots s_3\right)\left(s_{n-1}\dots s_4\right)\dots \left(s_{n-1}{s}_{n-2}\right)s_n\end{array}$$

and

$$\begin{array}{ccc}{w}_0{w}_{n-1}& =& \left(s_{{\epsilon }_2}\dots s_{{\epsilon }_{n-2}}\right)\left(s_{{\epsilon }_1-{\epsilon }_n}{s}_{{\epsilon }_2-{\epsilon }_{n-2}}\dots \right)\\ & =& \left(s_{n-1}\dots s_1\right)\left(s_n{s}_{n-2}\dots s_2\right)\left(s_{n-1}\dots s_3\right)\left(s_n{s}_{n-2}\dots s_4\right)\dots \left(s_n{s}_{n-2}\right)s_{n-1}\text{.}\end{array}$$

Finally

$$s_{{\epsilon }_1+{\epsilon }_2}=s_{{\epsilon }_1}{s}_{{\epsilon }_2}{s}_{{\epsilon }_1-{\epsilon }_2}=s_n{s}_{n-2}\dots s_1{s}_2\dots s_{n-2}{s}_n{s}_{n-2}\dots s_2$$

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 $W_0=⟨s_1\mid s_1^2=1⟩$ has order two and acts on the lattices

$$\begin{array}{cc}{𝔥}_{\mathbb{Z}}=\mathbb{Z}{\omega }^{\vee }\text{and}{𝔥}_{\mathbb{Z}}^{*}=\mathbb{Z}\omega \text{by}{s}_1{\omega }^{\vee }=-{\omega }^{\vee }\text{and}{s}_1\omega =-\omega ,& \text{(5.1)}\end{array}$$

and REARRANGE??

$$\begin{array}{cc}{\alpha }^{\vee }=2{\omega }^{\vee },\alpha =2\omega ,\text{and}⟨{\omega }^{\vee },\alpha ⟩=1\text{.}& \text{(5.2)}\end{array}$$ $$g∨s0∨s1∨s0∨ g∨s0∨s1∨ g∨s0∨ g∨ g∨s1∨ g∨s1∨s0∨ g∨s1∨s0∨s1∨ X-3ω X-ωs1 X-ω Xωs1 Xω X3ωs1 X3ω s1∨s0∨s1∨ s1∨s0∨ s1∨ 1 s0∨ s0∨s1∨ s0∨s1∨s0∨ X-αs1 X-α s1 1 Xαs1 Xα X2αs1 𝔥α∨+2d 𝔥α∨+d 𝔥α∨ 𝔥-α∨+d 𝔥-α∨+2d 𝔥-α∨+3d$$

The double affine braid group is generated by $q^{\frac{1}{2}},g,T_0,T_1,g^{\vee },T_0^{\vee }$ with relations

$$\begin{array}{cc}\begin{array}{c}{g}^2=1,T_0=g{T}_1{g}^{-1},{\left(g^{\vee }\right)}^2=1,T_0^{\vee }=g^{\vee }{T}_1{\left(g^{\vee }\right)}^{-1},\\ \text{and}{T}_1{g}^{\vee }g=q^{-\frac{1}{2}}g{g}^{\vee }{T}_1^{-1}\text{.}\end{array}& \text{(5.3)}\end{array}$$

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

$$\begin{array}{cc}{Y}^{{\omega }^{\vee }}=g{T}_1\text{and}{X}^{\omega }=g^{\vee }{T}_1^{-1},& \text{(5.4)}\end{array}$$

and note that

$$\begin{array}{cc}{T}_0{T}_1=Y^{2{\omega }^{\vee }},\text{and}{\left(T_0^{\vee }\right)}^{-1}{T}_1^{-1}=X^{2\omega }\text{.}& \text{(5.5)}\end{array}$$

(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

   $$\begin{array}{cc}\begin{array}{c}{g}^2=1,T_0=g{T}_1{g}^{-1},\\ g{X}^{\omega }=q^{\frac{1}{2}}{X}^{-\omega }g,T_1{X}^{\omega }{T}_1=X^{-\omega },\text{and}{T}_0{X}^{-\omega }{T}_0=q^{-1}{X}^{\omega }\text{.}\end{array}& \text{(5.6)}\end{array}$$

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

   $$\begin{array}{cc}\begin{array}{c}{\left(g^{\vee }\right)}^2=1,T_0^{\vee }=g^{\vee }{T}_1{\left(g^{\vee }\right)}^{-1},\\ g^{\vee }{Y}^{{\omega }^{\vee }}=q^{-\frac{1}{2}}{Y}^{-{\omega }^{\vee }}{g}^{\vee },T_1^{-1}{Y}^{{\omega }^{\vee }}{T}_1^{-1},\text{and}{\left(T_0^{\vee }\right)}^{-1}{Y}^{-{\omega }^{\vee }}{\left(T_0^{\vee }\right)}^{-1}=q{Y}^{{\omega }^{\vee }}\text{.}\end{array}& \text{(5.7)}\end{array}$$

		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)$\Rightarrow$(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 $$\begin{array}{c}{g}^{\vee }{Y}^{{\omega }^{\vee }}=g^{\vee }g{T}_1=q^{-\frac{1}{2}}{T}_1^{-1}g{g}^{\vee }=q^{-\frac{1}{2}}{Y}^{-{\omega }^{\vee }}{g}^{\vee },\\ T_1^{-1}{Y}^{{\omega }^{\vee }}{T}_1^{-1}=T_1^{-1}g{T}_1{T}_1^{-1}=T_1^{-1}g=Y^{-{\omega }^{\vee }},\end{array}$$ and $${\left(T_0^{\vee }\right)}^{-1}{Y}^{-{\omega }^{\vee }}{\left(T_0^{\vee }\right)}^{-1}=g^{\vee }{T}_1^{-1}{g}^{\vee }{Y}^{-{\omega }^{\vee }}{g}^{\vee }{T}_1^{-1}{g}^{\vee }=q^{\frac{1}{2}}{g}^{\vee }{T}_1^{-1}{Y}^{{\omega }^{\vee }}{T}_1^{-1}{g}^{\vee }=q^{\frac{1}{2}}{g}^{\vee }{Y}^{-{\omega }^{\vee }}{g}^{\vee }=q{Y}^{{\omega }^{\vee }},$$ respectively. (5.7)$\Rightarrow$(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 $$\begin{array}{c}{g}^2=Y^{{\omega }^{\vee }}{T}_1^{-1}{Y}^{{\omega }^{\vee }}{T}_1^{-1}=Y^{{\omega }^{\vee }}{Y}^{-{\omega }^{\vee }}=1,\\ g{T}_1{g}^{-1}=Y^{{\omega }^{\vee }}{T}_1^{-1}{T}_1{T}_1{Y}^{-{\omega }^{\vee }}=Y^{{\omega }^{\vee }}{T}_1{Y}^{-{\omega }^{\vee }}=Y^{{\omega }^{\vee }}{T}_1{Y}^{-{\omega }^{\vee }}{T}_1{T}_1^{-1}=Y^{2{\omega }^{\vee }},\end{array}$$ and $$T_1{g}^{\vee }g=T_1{g}^{\vee }{Y}^{{\omega }^{\vee }}{T}_1^{-1}=q^{-\frac{1}{2}}{T}_1{Y}^{-{\omega }^{\vee }}{g}^{\vee }{T}_1^{-1}=q^{-\frac{1}{2}}g{g}^{\vee }{T}_1^{-1},$$ 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

$$\begin{array}{cc}{T}_i^2=(t^{1/2}-t^{-1/2})T_i+1,\text{for}i=0,1,& \text{(5.8)}\end{array}$$

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

$$\begin{array}{c}{g}^{\vee }{Y}^{{\omega }^{\vee }}=q^{-1/2}{Y}^{-{\omega }^{\vee }}{g}^{\vee },T_1{Y}^{{\omega }^{\vee }}=Y^{-{\omega }^{\vee }}{T}_1+(t^{1/2}-t^{-1/2})\frac{Y^{{\omega }^{\vee }}-Y^{-{\omega }^{\vee }}}{1-Y^{-{\alpha }^{\vee }}},\text{and}\\ T_0^{\vee }{Y}^{{\omega }^{\vee }}=q^{-1}{Y}^{-{\omega }^{\vee }}{T}_0^{\vee }+(t^{1/2}-t^{-1/2})\left(\frac{Y^{{\omega }^{\vee }}-q^{-1}{Y}^{-{\omega }^{\vee }}}{1-q{Y}^{{\omega }^{\vee }}}\right)\text{.}\end{array}$$

With $Y^{{\alpha }_0}=q{Y}^{\alpha }$ and $Y^{{\alpha }_1}=Y^{\alpha },$ then

$${\tau }_g^{\vee }=g^{\vee },\text{and}{\tau }_i^{\vee }=T_i^{\vee }-(t^{1/2}-t^{-1/2})\left(\frac{1}{1-Y^{-{\alpha }_i^{\vee }}}\right),\text{for}i=0,1\text{.}$$

To illustrate Theorem 2.2, note that $X^{-2\omega }{s}_1^{\vee }{s}_0^{\vee }$ is a reduced word and

$$\begin{array}{ccc}{\tau }_1^{\vee }{\tau }_0^{\vee }& =& (T_1^{\vee }+\frac{t^{-1/2}(1-t)}{1-Y^{-{\alpha }_1^{\vee }}}){\tau }_0^{\vee }\\ & =& T_1^{\vee }{T}_0^{\vee }+T_1^{\vee }\frac{t^{-1/2}(1-t)}{1-Y^{-{\alpha }_0^{\vee }}}+{\left(T_0^{\vee }\right)}^{-1}\frac{t^{-1/2}(1-t)}{1-Y^{-s_0{\alpha }_1^{\vee }}}+\left(\frac{t^{-1/2}(1-t)}{1-Y^{-s_0{\alpha }_1^{\vee }}}\right)\left(\frac{t^{-1/2}(1-t)Y^{-{\alpha }_0^{\vee }}}{1-Y^{-{\alpha }_0^{\vee }}}\right)\\ & =& X^{-2\omega }+T_1^{\vee }\frac{t^{-1/2}(1-t)}{1-Y^{-{\alpha }_0^{\vee }}}+X^{2\omega }{T}_1^{\vee }\frac{t^{-1/2}(1-t)}{1-Y^{-s_0{\alpha }_1^{\vee }}}+\left(\frac{t^{-1/2}(1-t)}{1-Y^{-s_0{\alpha }_1^{\vee }}}\right)\left(\frac{t^{-1/2}(1-t)Y^{-{\alpha }_0^{\vee }}}{1-Y^{-{\alpha }_0^{\vee }}}\right)\end{array}$$

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

$$\begin{array}{cccc}& & & \\ X^{-2\omega }& T_1^{\vee }\frac{t^{-1/2}(1-t)}{1-Y^{-{\alpha }_0^{\vee }}}& X^{2\omega }{T}_1^{\vee }\frac{t^{-1/2}(1-t)}{1-Y^{-s_0{\alpha }_1^{\vee }}}& \frac{t^{-1/2}(1-t)}{1-Y^{-s_0{\alpha }_1^{\vee }}}\frac{t^{-1/2}(1-t)Y^{-{\alpha }_0^{\vee }}}{1-Y^{-{\alpha }_0^{\vee }}}\end{array}$$

The polynomial representations is defined by

$$g1=1,T_{0}1=t^{\frac{1}{2}}1,\text{and}{T}_{1}1=t^{\frac{1}{2}}1\text{.}$$

In this case

$$\begin{array}{cc}{\rho }_c=\frac{1}{2}c\alpha \text{and}{W}^0=\{X^{-\ell w}\mid \ell \in {\mathbb{Z}}_{\ge 0}\}\cup \{X^{\ell \omega }{s}_1^{\vee }\mid \ell \in {\mathbb{Z}}_{>0}\},& \text{(5.9)}\end{array}$$

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

$$Y^{-{\alpha }_0^{\vee }}1=q{Y}^{{\alpha }^{\vee }}1=q{q}^{c}1=qt1,\text{and}{Y}^{-s_0{\alpha }_1^{\vee }}1=Y^{{\alpha }^{\vee }+2d}1=q^2{Y}^{{\alpha }^{\vee }}1=q^{2}t,$$

gives

$$\begin{array}{ccc}{E}_{-2\omega }& =& {\tau }_1^{\vee }{\tau }_0^{\vee }1\\ & =& X^{-2\omega }+t^{1/2}\frac{t^{-1/2}(1-t)}{1-qt}+X^{2\omega }{t}^{1/2}\frac{t^{-1/2}(1-t)}{1-q^{2}t}+\left(\frac{t^{-1/2}(1-t)}{1-q^{2}t}\right)\left(\frac{t^{-1/2}(1-t)qt}{1-qt}\right)\\ & =& X^{-2\omega }+\frac{1-t}{1-qt}+X^{2\omega }\frac{1-t}{1-q^{2}t}+\left(\frac{1-t}{1-q^{2}t}\right)\left(\frac{(1-t)q}{1-qt}\right)\text{.}\end{array}$$

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

$$\begin{array}{ccc}{P}_{2\omega }& =& 1_0{E}_{2\omega }=(T_1^{\vee }+t^{-1/2}){\tau }_0^{\vee }1\\ & =& (T_1^{\vee }{T}_0^{\vee }+T_1^{\vee }\frac{t^{-1/2}(1-t)}{1-Y^{-{\alpha }_0^{\vee }}}+t^{-1/2}{\left(T_0^{\vee }\right)}^{-1}+t^{-1/2}\frac{t^{-1/2}(1-t)Y^{-{\alpha }_0^{\vee }}}{1-Y^{-{\alpha }_0^{\vee }}})1\\ & =& (X^{-2\omega }+t^{1/2}\frac{t^{-1/2}(1-t)}{1-qt}+t^{-1/2}{X}^{2\omega }{T}_1^{\vee }+t^{-1/2}\frac{t^{-1/2}(1-t)qt}{1-qt})1\\ & =& (X^{-2\omega }+t^{1/2}\frac{1-t}{1-qt}+X^{2\omega }+t^{-1/2}\frac{(1-t)q}{1-qt})1=(X^{2\omega }+X^{-2\omega }+t^{1/2}+(1+q)\frac{1-t}{1-qt})1\text{.}\end{array}$$

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

$$\begin{array}{cccc}& & & \\ X^{-2\omega }& \frac{1-t}{1-tq}& X^{2\omega }& q\frac{(1-t)}{1-tq}\end{array}$$

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 $\tau T_1{\tau }^{-1}=s_0,$

$$\begin{array}{c}{Y}^{{\epsilon }_1^{\vee }}=\tau T_1,Y^{{\epsilon }_2^{\vee }}=\tau S_0,Y^{{\epsilon }_1^{\vee }-{\epsilon }_2^{\vee }}=S_0{T}_1,\\ X^{{\epsilon }_1}={\tau }^{\vee }{T}_1^{-1},X^{{\epsilon }_2}={\tau }^{\vee }{\left(S_0^{\vee }\right)}^{-1},X^{{\epsilon }_1-{\epsilon }_1}={\left(S_0^{\vee }\right)}^{-1}{T}_1^{-1}\text{.}\end{array}$$

Further ${ℬ}_{G{L}_2}\subseteq {ℬ}_2$ by

$$\tau =T_0{T}_1{T}_2,X^{{\epsilon }_1+{\epsilon }_2}={\tau }^2={\left(T_0{T}_1{T}_2\right)}^2,\text{and}{S}_0=\tau T_1{\tau }^{-1}=T_0{T}_2^{-1}{T}_1{T}_2{T}_0^{-1}\text{.}$$

To check these we note that

$$Y^{{\epsilon }_1^{\vee }}=T_0{T}_1{T}_2{T}_1\text{and}{Y}^{{\epsilon }_2^{\vee }}=T_1^{-1}{T}_0{T}_1{T}_2$$

so that

$$\tau T_1{\tau }^{-1}=T_0{T}_1{T}_2{T}_1{T}_2^{-1}{T}_1^{-1}{T}_0^{-1}=T_0{T}_2^{-1}{T}_1{T}_2{T}_1{T}_1^{-1}{T}_0^{-1}=T_0{T}_2^{-1}{T}_1{T}_2{T}_0^{-1}\text{.}$$

which should be $T_0$ in $G{L}_2\text{.}$

Type $B_2$

$$\begin{array}{cc} R0 T2 T1 & \text{with}{ℬ}_{S{O}_5}\subseteq {ℬ}_2\text{via}{R}_0=T_0{T}_1{T}_0\text{.}\end{array}$$

Further,

$$\text{with}\gamma =T_0,{ℬ}_{{\text{Spin}}_5}\text{is the quotient of}{ℬ}_{{Sp}_4}\text{by}{\gamma }^2=1\text{.}$$

The conversions come from the formulas

$$Y^{{\epsilon }_1^{\vee }+{\epsilon }_2^{\vee }}=R_0{T}_2{T}_1{T}_2,Y^{{\epsilon }_1^{\vee }}=\gamma T_1{T}_2{T}_1,$$

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 $A_1\times A_1$

There is an embedding of $ℬ\subseteq {ℬ}_2$ by

$$a_2=T_2{T}_1{T}_2,R_0=T_0{T}_1{T}_0{S}_0=T_2{T}_0{T}_1\text{.}$$ $$X2ε1 Xε1-ε2 Xε1+ε2 R0 S0 1 s1 a2 s1a2 X-2ε1 X-ε1+ε2 X-ε1-ε2 +- +- + - + - + - + - 𝔥-(ε1∨-ε2∨)+2d 𝔥-(ε1∨-ε2∨)+d 𝔥ε1∨-ε2∨ 𝔥ε1∨+ε2∨ 𝔥-(ε1∨-ε2∨)+2d 𝔥-(ε1∨-ε2∨)+d$$

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.

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