]> Reflections

Reflection groups

Arun Ram
Department of Mathematics
University of Wisconsin
Madison, WI 53706 USA
ram@math.wisc.edu

Reflections

Let 𝔽 be a field. A reflection is an element of G L n ( 𝔽 ) such that rank ⁡ ( 1 – s ) = 1 . In other words, if s is diagonalisable then

s is conjugate to ξ 1 ⋱ 1


in G L n ( 𝔽 ) . Let 𝔥 be an n-dimensional 𝔽 -vector space and view s as an element of G L ( 𝔥 ) . Then s is a reflection if

codim ⁡ ( 𝔥 s ) = 1 , where 𝔥 s = { λ ∨ ∈ 𝔥 ∣ s λ ∨ = λ ∨ }


is the reflecting hyperplane. The dual space of 𝔥 is 𝔥 * = Hom ( 𝔥 , 𝔽 ) , the vector space of linear maps μ : 𝔥 → 𝔽 .

Let s act on 𝔥 * by ⟨ s μ , λ ∨ ⟩ = ⟨ μ , s – 1 λ ⟩ , where ⟨ μ , λ ∨ ⟩ = μ ( λ ∨ ) , for μ ∈ 𝔥 * , λ ∨ ∈ 𝔥 .

Choose α ∈ 𝔥 * and α ∨ ∈ 𝔥 such that

s μ = μ – ⟨ μ, α ∨ ⟩ α and s – 1 λ ∨ = λ ∨ – ⟨ λ ∨ ,α ⟩ α ∨

(*)


for all μ ∈ h * , λ ∨ ∈ h . To do this suppose that the action of s – 1 on 𝔥 has matrix s – 1 = ξ – 1 1 ⋱ 1 with respect to β 1 ∨ , … , β n ∨ , a basis of 𝔥 . Then the matrix of the action of s on 𝔥 * is s = ξ – 1 1 ⋱ 1 with respect to β 1 , … , β n , the dual basis in 𝔥 * to the basis β 1 ∨ , … , β n ∨ of 𝔥 . Then α ∨ = β 1 ∨ and α = β 1 satisfies (*) and

s α = α – ⟨ α , α ∨ ⟩ α = ( 1 – ⟨ α , α ∨ ⟩ ) α = ξ – 1 α , s – 1 α ∨ = α ∨ – ⟨ α ∨ , α ⟩ α ∨ = ( 1 – ⟨ α , α ∨ ⟩ ) α ∨ = ξ – 1 α ∨ ,


and

⟨ α , α ∨ ⟩ = 1 – ξ – 1 = 1 – det 𝔥 * ( s ) .


Hence, if α ~ ∈ 𝔥 * , α ~ ∨ ∈ 𝔥 is another choice in (*) then there is a nonzero constant γ ∈ 𝔽 ¯ such that α ~ = γ α and α ~ ∨ = γ – 1 α ∨ .

The reflecting hyperplanes for s are

( 𝔥 * ) s = { μ ∈ 𝔥 * ∣ ⟨ μ , α ∨ ⟩ = 0 } and 𝔥 s = { λ ∨ ∈ 𝔥 ∣ ⟨ λ ∨ ,α ⟩ = 0 }


Reflection Groups

Let 𝔬 be an integral domain and let

𝔽 = { a b ∣ a , b ∈ 𝔬 , b ≠ 0 }


be the field of fractions of 𝔬 .

An 𝔬 -lattice is a free 𝔬 -module, i.e an 𝔬 -module with a basis, i.e a vector space over 𝔬 .

If P ∨ is an 𝔬 -lattice and w 1 ∨ , … , w n ∨ is a basis

P ∨ = 𝔬 -span { w 1 ∨ , … , w n ∨ } and 𝔥 = 𝔽 ⊗ 𝔬 P ∨ = 𝔽 -span { w 1 ∨ , … , w n ∨ }


is a vector space over 𝔽 . Use the basis to identify

G L ( P ∨ ) ⊆ G L ( 𝔥 ) with G L n ( 𝔬 ) ⊆ G L n ( 𝔽 ) .


A lattice is a free ℤ -module.

A reflection group is a pair ( W , P ∨ ) where P ∨ is an 𝔬 -lattice and W is a subgroup of G L ( P ∨ ) generated by reflections.

A crystallographic reflection-group is a ℤ -reflection group.

Examples

  1. P ∨ = ℤ n = ∑ i = 1 n ℤ e i ∨ = { λ ∨ = λ 1 e 1 ∨ + ⋯ + λ n e n ∨ ∣ λ 1 , … , λ n ∈ ℤ }


    with W = S n the symmetric group acting on P ∨ by permuting the e 1 ∨ , … , e n ∨ . With respect to the basis e 1 ∨ , … , e n ∨ , the reflections in W are

    S i j =   i j 1 0 ··· 0 0 0 1 ⋮ ⋮ 1 0 0 0 ··· 0 1 for  1 ≤ i < j ≤ n


    Each s i j is conjugate to

    – 1 1 ⋱ 1 in G L ( ℚ ¯ )


  2. Let

    𝔬 = ℤ [ ξ ] where ξ = ⅇ 2 π ⅈ / r


    is a primitive rth of unity. Let

    P ∨ = ∑ i = 1 n ℤ [ ξ ] e i ∨


    and, with respect to the basis e 1 ∨ , … , e i ∨ ,

    W = n × n matrices with (a) exactly one nonzero entry in each row and each coloumn (b) nonzero entries are in ℤ / r ℤ


    where ℤ / r ℤ = { 1 , ξ , ξ 2 , … , ξ r – 1 } . The reflections in W are

    S i j =   i j 1 0 ··· 0 0 0 1 ⋮ ⋮ 1 0 0 0 ··· 0 1 for  1 ≤ i < j ≤ n ,


    t i l = 1 ⋱ 1 ξ l 1 ⋱ 1 1 ≤ i ≤ n , 0 < l < r


    Let

    t μ = t 1 μ 1 … t n μ n = ξ μ 1 ⋱ ⋱ ξ μ n


    then

    W = { t μ w ∣ μ t ( ℤ / r ℤ ) n , w ∈ S n } ,


    where ℤ / r ℤ = { 0 , 1 , … , r – 1 } .

    The symmetric group S n is the case ξ = 1 = ⅇ 2 π ⅈ / 1 .

    The hyperoctahedral group W B n is the case ξ = – 1 = ⅇ 2 π ⅈ / 2 .