## The affine braid group of type C

Last update: 10 September 2013

## The affine braid group of type C

Let $𝔥ℤ=∑i=1n ℤεi∨ and𝔥ℤ*= ∑i=1nℤεj with ⟨εi∨,εj⟩ =δij$ defining the bilinear map $⟨,⟩:{𝔥}_{ℤ}×{𝔥}_{ℤ}^{*}\to ℤ\text{.}$ Let $αn=2εn andαi= εi-εi+1, for i∈ {1,2,…,n-1}.$ The following are equivalent presentations of the affine braid group.

 (a) Generators: ${T}_{0},{T}_{1},\dots ,{T}_{n}$ Relations: $T0 T1 T2 Tn-2 Tn-1 Tn (a)$ (b) Generators: ${T}_{1},\dots ,{T}_{n}$ and ${Y}^{{\epsilon }_{1}^{\vee }},\dots ,{Y}^{{\epsilon }_{n}^{\vee }}$ Relations: $T1 T2 Tn-2 Tn-1 Tn (b1) Yεi∨ Yεj∨= Yεj∨ Yεi∨, for i,j ∈{1,…,n}, (b2) TiYεj∨ =Yεj∨Ti, for i∈ {1,2,…,n-1}, j∈{1,2,…,n} with j≠i,i+1 , and (b3) TnYεj∨= Yεj∨Tn, for j∈ {1,2,…,n-1}. (b4)$ (c) Generators: ${T}_{1},\dots ,{T}_{n}$ and ${Y}^{{\lambda }^{\vee }},$ for ${\lambda }^{\vee }\in {𝔥}_{ℤ},$ Relations: $T1 T2 Tn-2 Tn-1 Tn (c1) Yλ∨Yσ∨= Yλ∨+σ∨ andTi-1 Yλ∨= { Ysiλ∨ Ti-1, if ⟨λ∨,αi⟩ =0, Ysiλ∨Ti, if ⟨λ∨,αi⟩ =1, (c2)$
for ${\lambda }^{\vee }\in {𝔥}_{ℤ}$ and $i\in \left\{1,\dots ,n\right\}\text{.}$

 Proof. Rewrite the relations (a) in the form $T1 T2 Tn-2 Tn-1 Tn (a1) T1T0T1T0= T0T1T0T1 andT0Ti= TiTi, for i∈{2,3,…,n}. (a2)$ Generators (b) in terms of generators (a): ${T}_{i}={T}_{i},$ for $i\in \left\{1,2,\dots ,n\right\},$ and $Yε1∨= T0T1⋯Tn ⋯T1and Yεi+1∨= Ti-1Yεi∨ Ti-1, for i∈ {1,2,…,n-1}.$ Pictorially, $Yεj∨= j$ Generators (a) in terms of generators (b): $T0=Yε1∨ T1-1⋯ Tn-1⋯ T1-1and Ti=Ti, for i∈{1,2,…,n}.$ Generators (c) in terms of generators (b): If $i\in \left\{1,\dots ,n\right\}$ then ${T}_{i}={T}_{i}$ and $Yλ∨= (Yε1∨)λ1 (Yε2∨)λ2 ⋯ (Yεn∨)λn, for λ∨=λ1ε1∨ +⋯+λnεn∨.$ Relations (b) from relations (a): The relations (b1) are the same as the relations (a1). The picture $PICTURE$ gives ${Y}^{{\epsilon }_{i}^{\vee }}{Y}^{{\epsilon }_{j}^{\vee }}={Y}^{{\epsilon }_{j}^{\vee }}{Y}^{{\epsilon }_{i}^{\vee }},$ for $i,j\in \left\{1,\dots ,n\right\},$ proving relations (b2). From the picture of ${Y}^{{\epsilon }_{j}^{\vee }}$ ${T}_{i}{Y}^{{\epsilon }_{j}^{\vee }}={Y}^{{\epsilon }_{j}^{\vee }}{T}_{i}$ for $j\in \left\{1,\dots ,n\right\}$ and $i\in \left\{1,\dots ,k-1\right\}$ with $i\ne j-1,j\text{.}$ So $TiYεj∨= Yεj∨Ti, for j∈{1,…,n} , i∈{1,…,n-1} with j≠i,i+1,$ proving relations (b3). From the picture of ${Y}^{{\epsilon }_{j}^{\vee }},$ $TnYεj∨= Yεj∨Tn, for j∈{1,…,n-1},$ proving relations (b4). Relations (a) from relations (b): The relations (a1) are the same as the relations (b1). Note that if $i\in \left\{2,3,\dots ,n\right\}$ then $TsφTi = T1⋯Tn⋯T1Ti =T1⋯Tk⋯ Ti+1TiTi-1 TiTi-2⋯T1 = T1⋯Tk⋯Ti+1 Ti-1TiTi-1 Ti-2⋯T1 = T1⋯Ti-2 Ti-1Ti Ti-1Ti+1⋯ Tk⋯Ti+1 TiTi-1 Ti-2⋯T1 = T1⋯Ti-2Ti Ti-1Ti Ti+1⋯Tk⋯T1 =TiT1⋯Tk⋯T1 =TiTsφ,$ which is also visible from the picture $Tsφ= Tsε1= =T1⋯Tn⋯T1.$ If $i\in \left\{2,3,\dots ,n\right\}$ then ${T}_{i}{T}_{{s}_{{\epsilon }_{1}}}^{-1}={T}_{{s}_{{\epsilon }_{1}}}^{-1}{T}_{i}$ and $T0Ti= Yε1∨ Tsε1-1Ti =Yε1∨Ti Tsε1-1= TiYε1∨ Tsε1-1= TiT0$ which proves the second set of relations in (a2). Since ${Y}^{{\epsilon }_{1}^{\vee }}={T}_{0}{T}_{1}\cdots {T}_{n}\cdots {T}_{1}$ and ${Y}^{{\epsilon }_{2}^{\vee }}={T}_{1}^{-1}{Y}^{{\epsilon }_{1}^{\vee }}{T}_{1}^{-1}={T}_{1}^{-1}{T}_{0}{T}_{1}\cdots {T}_{n}\cdots {T}_{2}$ we have $Yε1∨ Yε2∨= T0T1⋯Tn ⋯T2T0T1⋯ Tn⋯T2=T0T1 T0T2⋯Tn⋯ T2T1⋯Tn ⋯T2.$ Then ${T}_{1}^{-1}{Y}^{{\epsilon }_{1}^{\vee }}{Y}^{{\epsilon }_{2}^{\vee }}={T}_{1}^{-1}{Y}^{{\epsilon }_{1}^{\vee }}{Y}^{{\epsilon }_{2}^{\vee }}={T}_{1}^{-1}{Y}^{{\epsilon }_{1}^{\vee }}{T}_{1}^{-1}{Y}^{{\epsilon }_{1}^{\vee }}{T}_{1}^{-1}={Y}^{{\epsilon }_{2}^{\vee }}{Y}^{{\epsilon }_{1}^{\vee }}{T}_{1}^{-1}={Y}^{{\epsilon }_{1}^{\vee }}{Y}^{{\epsilon }_{2}^{\vee }}{T}_{1}^{-1}$ gives $T1Yε1∨Yε2∨ = T1T0T1T0T2 ⋯Tn⋯T2T1⋯ Tn⋯T2 =Yε1∨ Yε2∨T1 = T0T1T0T2⋯ Tn⋯T2Tsφ =T0T1T0Tsφ T2⋯Tn⋯T2 = T0T1T0T1T2⋯ Tn⋯T2T1⋯ Tn⋯T2,$ since ${T}_{{s}_{\phi }}{T}_{i}={T}_{i}{T}_{{s}_{\phi }}$ for $i=2,\dots ,n\text{.}$ Multiplying on the right by ${\left({T}_{2}\cdots {T}_{n}\cdots {T}_{2}{T}_{1}\cdots {T}_{n}\cdots {T}_{2}\right)}^{-1}$ gives ${T}_{1}{T}_{0}{T}_{1}{T}_{0}={T}_{0}{T}_{1}{T}_{0}{T}_{1}\text{.}$ This establishes the first relation in (a2). Relations (c) from relations (b): Relations (c1) are the same as relations (b1). The first set of relations in (c2) are equivalent to the relations in (b2). If $i\in \left\{1,\dots ,n-1\right\}$ and ${\lambda }^{\vee }={\lambda }_{1}{\epsilon }_{1}^{\vee }+\cdots +{\lambda }_{n}{\epsilon }_{n}^{\vee }$ then $⟨λ∨,αi⟩ =⟨λ∨,ei-εi+1⟩ =λi-λi+1= { 0, if λi=λi+1, 1, if λi+1= λi-1,$ and $⟨λ∨,αn⟩ ==⟨λ∨,2εn⟩ =2λn= { 0, if λn=0, 1, never.$ Thus the second set of relations in (c2) are equivalent to $Ti ( Yεi∨ Yεi+1∨ ) = ( Yεi∨ Yεi+1∨ ) λi Ti, (c3) TiYεj∨= Yεj∨Ti, for j∈ {1,…,n} with j≠i,i+1, (c4) Ti-1 (Yεi∨)λi (Yεi+1∨)λi-1 =(Yεi∨)λi-1 (Yεi+1∨)λi Ti, (c5)$ for $i\in \left\{1,\dots ,n-1\right\},$ and $TnYεj= Yεj∨Tn, if j∈{1,…,n-1}. (c6)$ Relations (c4) and (c6) are relations (b3) and (b4) and the computations $Ti-1Yεi∨ Yεi+1∨= Ti-1Yεi∨ Ti-1Yεi∨ Ti-1= Yεi+1∨ Yεi∨Ti-1,$ and $Ti-1 (Yεi∨)λi (Yεi+1∨)λi-1 = Ti-1 Yεi∨ (Yεi∨)λi-1 (Yεi+1∨)λi-1 = Ti-1 Yεi∨ Ti-1 Ti (Yεi∨)λi-1 (Yεi+1∨)λi-1 = Yεi+1∨ Ti (Yεi∨)λi-1 (Yεi+1∨)λi-1 = Yεi+1∨ (Yεi∨)λi-1 (Yεi+1∨)λi-1 Ti = (Yεi∨)λi-1 (Yεi+1∨)λi Ti,$ establish the relations in (c3) and (c5). Relations (b) from relations (c): The relations (b1) are the same as the relations (c1). The relations in (b2) are equivalent to the first set of relations in (c2). In view of the equivalence of the second set of relations in (c2) with the relations in (c3-6), the relations in (b2), (b3) and (b4) are a subset of the relations in (c2). $\square$