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 𝔥=i=1n εi and𝔥*= i=1nεj with εi,εj =δij defining the bilinear map ,:𝔥×𝔥*. Let αn=2εn andαi= εi-εi+1, fori {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, fori,j {1,,n}, (b2) TiYεj =YεjTi, fori {1,2,,n-1}, j{1,2,,n} withji,i+1 ,and (b3) TnYεj= YεjTn, forj {1,2,,n-1}. (b4)
(c) Generators: T1,,Tn and Yλ, for λ𝔥,
Relations: T1 T2 Tn-2 Tn-1 Tn (c1) YλYσ= Yλ+σ andTi-1 Yλ= { Ysiλ Ti-1, if λ,αi =0, YsiλTi, if λ,αi =1, (c2)
for λ𝔥 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= T0T1Tn T1and Yεi+1= Ti-1Yεi Ti-1,fori {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, forλ=λ1ε1 ++λnεn.

Relations (b) from relations (a): The relations (b1) are the same as the relations (a1).

The picture PICTURE gives YεiYεj=YεjYεi, for i,j{1,,n}, proving relations (b2).

From the picture of Yεj TiYεj=YεjTi for j{1,,n} and i{1,,k-1} with ij-1,j. So TiYεj= YεjTi, forj{1,,n} ,i{1,,n-1} withji,i+1, proving relations (b3).

From the picture of Yεj, TnYεj= YεjTn, forj{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 = T1TnT1Ti =T1Tk Ti+1TiTi-1 TiTi-2T1 = T1TkTi+1 Ti-1TiTi-1 Ti-2T1 = T1Ti-2 Ti-1Ti Ti-1Ti+1 TkTi+1 TiTi-1 Ti-2T1 = T1Ti-2Ti Ti-1Ti Ti+1TkT1 =TiT1TkT1 =TiTsφ, which is also visible from the picture Tsφ= Tsε1= =T1TnT1.

If i{2,3,,n} then TiTsε1-1=Tsε1-1Ti and T0Ti= Yε1 Tsε1-1Ti =Yε1Ti Tsε1-1= TiYε1 Tsε1-1= TiT0 which proves the second set of relations in (a2).

Since Yε1=T0T1TnT1 and Yε2=T1-1Yε1T1-1=T1-1T0T1TnT2 we have Yε1 Yε2= T0T1Tn T2T0T1 TnT2=T0T1 T0T2Tn T2T1Tn T2. Then T1-1Yε1Yε2 = T1-1Yε1Yε2 = T1-1Yε1T1-1Yε1T1-1 = Yε2Yε1T1-1 = Yε1Yε2T1-1 gives T1Yε1Yε2 = T1T0T1T0T2 TnT2T1 TnT2 =Yε1 Yε2T1 = T0T1T0T2 TnT2Tsφ =T0T1T0Tsφ T2TnT2 = T0T1T0T1T2 TnT2T1 TnT2, since TsφTi=TiTsφ for i=2,,n. Multiplying on the right by (T2TnT2T1TnT2)-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, ifλi=λi+1, 1, ifλi+1= λi-1, and λ,αn ==λ,2εn =2λn= { 0, ifλn=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= YεjTi, forj {1,,n} withji,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= YεjTn, ifj{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 YεiTi-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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