Reflection groups and Braid groups

Reflection groups and Braid groups

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

Last updates: 15 November 2010

Groups

A group is a set G with a product G×GGg1,g2g1g2such that
  1. If g1,g2,g3G then g1g2g3=g1g2g3,
  2. There exists 1G such that if gG then 1g=g1=g,
  3. If gG then there exists g-1G such that gg-1=g-1g=1.

Example:

with product { , , , , , , , }
with product g1g2= g1 g2 so that = .

The symmetric group

Sn=graphs with n top vertices and n bottom vertices such thateach top vertex is connected to exactly one bottom vertex with product

with product g1g2= g1 g2

Examples:

S3= { , , , , , , , }
S2= { , } and S1= { }

Equivalently, Sn=n×n matrices with(a) exactly one nonzero entry in each row and each column(b) the nonzero entries are 1. with matrix multiplication 010100001010001100=001010100, or equivalently,

Complex numbers

=x+yix,ywith i2=-1,=ereiθr,θ[0,2π)0.

x θ 1 iy x+iy ereiθ eiθ

The cyclic group

/m=mth roots of unity=ggn=1=1,ξ,ξ2,,ξm-1withξ=e2πi/2m, and ξrξs=ξr+sandξ0=ξm=1. So

/5= 1 ξ4 ξ3 ξ2 ξ=e2πi/5

The groups Gm,1,n

Recall Sn=n×n matrices with(a) exactly one nonzero entry in each row and each column(b) the nonzero entries are 1. Then Gm,1,n=n×n matrices with(a) exactly one nonzero entry in each row and each column(b) the nonzero entries are in /m with product matrix multiplcation. So Sn=G1,1,n.

Example: If m=3 and n=3 /3=1,ξ,ξ2withξ=e2πi/3. G3,1,3=100010001,0ξ010000ξ2,ξ2000ξ2000ξ,00ξ2ξ200010, and the number of elements in G3,1,3 is G3,1,3=3!33=32133=162.

Homomorphisms and kernels

Let G and G be groups. A homomorphism from G to H is a function ϕ:GHgϕgsuch that
  1. If g1,g2G then ϕg1ϕg2=ϕg1g2,
  2. ϕ1=1,
  3. If gG then ϕg-1=ϕg-1.
The kernel of ϕ is kerϕ=gGϕg=1.

Example: /2=1,-1. A homomorphism is given by ϕ:Sn/2byϕg=-1# of crossings in g. So

ϕ ( ) =-15=-1.

The alternating group is An=kerϕ.

The groups Gm,l,n

Let l divide m. A homomorphism is ϕ:Gm,1,n/lgiven byϕg=non-zeroentriesgijm/l and Gm,l,n=kerϕ.

Example: If m=6,l=3 and n=5 then ϕ00ξ3000ξ40000000ξ10000000ξ20 =ξ3ξ4ξξ26/3=ξ102=ξ2=e2πi2/6=e2πi/3. So Gm,l,n=n×n matrices with(a) exactly one nonzero entry in each row and each column(b) the nonzero entries are in /m(c) non-zeroentriesgijm/l=1 with product matrix multiplication.

The dihedral group of order 2m is Gm,m,2.

Reflection groups

A reflection is a matrix with exactly one eigenvalue1.

Example: 1000ξ20001 is a reflection, and
if m=5 then g00ξ2010ξ300g-1=10101010-100ξ2010ξ30012012010120-12=-100010001, and so 00ξ2010ξ300is a reflection.

A reflection group is a group G of matrices generated by reflections.

Example: S3=100010001,010100001,, or

S3= { , , , , , , , }

has reflections

, ,

and every element of S3 is a product of reflections.

(Shephard-Todd) Except for 34 special cases, the Gm,l,n are all finite reflection groups.

Invariant rings

Let G be a group of matrices. Then Gacts onn=c1c2cnc1,c2,,cn. If gG and xi=ith00100 then gxi=g11g1ngn1gnn00100=g1ig2igni=g1ix1+g2ix2++gnixi. Then G acts on polynomials px1,,xn by gp1+p2=gp1+gp2andgp1p2=gp1gp2.

Example: 0ξ2ξ0x1=ξx2and0ξ2ξ0x2=ξ2x1, and 0ξ2ξ03x12x2+5x1x23=3ξx22ξ2x1+ξx2ξ2x13=3ξ4x1x22+5ξ7x13x2.

The invariant ring of G is x1,,xnG=px1,,xngp=p for all gG.

Example: If m=5 then 0ξ2ξ0x15x25=ξx25+ξ2x15=ξ5x25+ξ10x15=x15+x25.

(Chevalley, Shephard-Todd) G is a finite reflection group if and only if there exist polynomials p1,,pn such that x1,,xnG=p1,,pn.

Fundamental groups

Let X be a topological space with a fixed base point x0.

x0=

A path in X is a continuous map p:0,1withp0=x0. So

x0= [0,1]= and x2 is a path.

A loop in X is a path g:[0,1]X with g0=x0=g1

x0=

The fundamental group of X is π1X=loops in Xwith productg1g2t=g1t,if 0t12g2t,if 12t1.

Braid groups

The braid group on n strands is n=string diagrams with n top vertices and n bottom vertices andeach top vertex is connected to exactly one bottom vertex with product

with product g1g2= g1 g2

Example:

= ( ) ( )

Forgetting whetner crossings are over or under is a homomorphism ϕ:nSn.

Example:

ϕ =

Configuration space

The pure braid group is 𝒫n=kerϕ. Let n=c1,,cnc1,,cn, and let Hij=c1,,cnc1,,cnfor 1j<in, and let X=ni<jHij.

𝒫n=π1X.

Proof.

Let G be a reflection goup. For each reflection sα in G let Hα=c1,,cnnsαc1,,cn=c1,,cn. Let X=nreflectionsHα.

Problem: Describe and understand π1X.

References

[Bou] N. Bourbaki, Groupes et Algèbres de Lie, Masson, Paris, 1990.

[GW1] F. Goodman and H. Wenzl, The Temperly-Lieb algebra at roots of unity, Pacific J. Math. 161 (1993), 307-334. MR1242201 (95c:16020)

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