The affine braid group of type C

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

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 ⟨,⟩:𝔥ℤ×𝔥ℤ*→ℤ. 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: T0,T1,…,Tn
Relations: T0 T1 T2 Tn-2 Tn-1 Tn (a)
(b) Generators: T1,…,Tn and Yε1∨,…,Yεn∨
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: T1,…,Tn and Yλ∨, for λ∨∈𝔥ℤ,
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 λ∨∈𝔥ℤ and i∈{1,…,n}.

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): Ti=Ti, for i∈{1,2,…,n}, 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∈{1,…,n} then Ti=Ti 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 Yεi∨Yεj∨=Yεj∨Yεi∨, for i,j∈{1,…,n}, proving relations (b2).

From the picture of Yεj∨ TiYεj∨=Yεj∨Ti for j∈{1,…,n} and i∈{1,…,k-1} with i≠j-1,j. So TiYεj∨= Yεj∨Ti, for j∈{1,…,n} , i∈{1,…,n-1}  with j≠i,i+1, proving relations (b3).

From the picture of Yεj∨, 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∈{2,3,…,n} 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∈{2,3,…,n} then TiTsε1-1=Tsε1-1Ti and T0Ti= Yε1∨ Tsε1-1Ti =Yε1∨Ti Tsε1-1= TiYε1∨ Tsε1-1= TiT0 which proves the second set of relations in (a2).

Since Yε1∨=T0T1⋯Tn⋯T1 and Yε2∨=T1-1Yε1∨T1-1=T1-1T0T1⋯Tn⋯T2 we have Yε1∨ Yε2∨= T0T1⋯Tn ⋯T2T0T1⋯ Tn⋯T2=T0T1 T0T2⋯Tn⋯ T2T1⋯Tn ⋯T2. Then T1-1Yε1∨Yε2∨ = T1-1Yε1∨Yε2∨ = T1-1Yε1∨T1-1Yε1∨T1-1 = Yε2∨Yε1∨T1-1 = Yε1∨Yε2∨T1-1 gives T1Yε1∨Yε2∨ = T1T0T1T0T2 ⋯Tn⋯T2T1⋯ Tn⋯T2 =Yε1∨ Yε2∨T1 = T0T1T0T2⋯ Tn⋯T2Tsφ =T0T1T0Tsφ T2⋯Tn⋯T2 = T0T1T0T1T2⋯ Tn⋯T2T1⋯ Tn⋯T2, since TsφTi=TiTsφ for i=2,…,n. Multiplying on the right by (T2⋯Tn⋯T2T1⋯Tn⋯T2)-1 gives T1T0T1T0=T0T1T0T1. 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∈{1,…,n-1} and λ∨=λ1ε1∨+⋯+λnεn∨ 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∈{1,…,n-1}, 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).

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