Representation Theory

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

Last updated: 2 October 2014

Lecture 7

Dual vector spaces

Let 𝔽 be a field, 𝔥* = span{ω1,…,ωn}  a vector space, 𝔥 = Hom(𝔥*,𝔽) the dual vector space. Write ⟨μ,λ∨⟩= μ(λ∨), for μ∈𝔥*, λ∨∈𝔥. Let G=GL(𝔥*). G acts on 𝔥*. Define an action of G on 𝔥 by ⟨μ,gλ∨⟩= ⟨g-1μ,λ∨⟩. Let ω1,…,ωn be a basis of 𝔥* and identify g with its matrix in GLn(𝔽).

Let α1∨,…,αn∨ be the dual basis in 𝔥. The matrix of the action of g on 𝔥 is g∨=(gt)-1.

Reflections

A reflection is sα∈GL(𝔥*) such that, in GLn(𝔽‾), sα is conjugate to ( ξ 1 ⋱ 1 ) with ξ∈𝔽‾, ξ≠1. Then sα∨∈GL(𝔥) is conjugate to ( ξ-1 1 ⋱ 1 ) . Then 𝔥*=𝔥α∨⊕ ℂαand𝔥= 𝔥α⊕ℂα∨ where 𝔥α∨ = (𝔥*)sα= {μ∈𝔥* | sαμ=μ} (1 eigenspace of sα), ℂα = (ξ-eigenspace of sα), 𝔥α = 𝔥sα∨= { λ∨∈𝔥 |  sα∨λ∨=λ∨ } =(1-eigenspace of sα), ℂα∨ = (ξ-1-eigenspace of sα∨) and 𝔥α∨ = {μ∈𝔥* | ⟨μ,α∨⟩=0}, 𝔥α = {λ∨∈𝔥 | ⟨λ∨,α⟩=0}. Choose α and α∨ so that ⟨α,α∨⟩=1-ξ=1-det(sα). Then sαμ=μ- ⟨μ,α∨⟩α andsα-1 λ∨=λ∨- ⟨λ∨,α⟩ α∨. Check: sαα = α-⟨α,α∨⟩ α=(1-⟨α,α∨⟩) α=ξα, sα-1α∨ = α∨-⟨α,α∨⟩ α∨=(1-⟨α,α∨⟩) α∨=ξα∨, as it should be. sαμ = μ-⟨μ,α∨⟩ α=μ-0,if μ∈𝔥α∨, sα-1λ∨ = λ∨-⟨λ∨,α⟩ α∨=λ∨-0,if  λ∨∈𝔥α.

Weyl groups

Let 𝔥ℤ* be a ℤ-vector space. 𝔥ℤ*=ℤ-span {ω1,…,ωn}, where ω1,…,ωn is a ℤ-basis of 𝔥ℤ*. 𝔥ℚ* = ℚ⊗ℤ𝔥ℤ*= ℚ-span{ω1,…,ωn}, 𝔥ℝ* = ℝ⊗ℤ𝔥ℤ*=ℝ-span {ω1,…,ωn}, 𝔥ℂ* = ℂ⊗ℤ𝔥ℤ*=ℂ-span {ω1,…,ωn}, 𝔥ℚ‾* = ℚ‾⊗ℤ𝔥ℤ*=ℚ‾-span {ω1,…,ωn}. A Weyl group, or crystallographic reflection group, is a finite subgroup w0 of GL(𝔥ℤ*) generated by reflections.

Let R+ be an index set for the reflections in W0 so that sα, α∈R+, are the reflections in W0.

WARNING: A Weyl group is really a pair (W0,𝔥ℤ*). W0 cannot exist without 𝔥ℤ*.

Examples (Type GLn)

𝔥ℤ*=span{ε1,…,εn} with W0=Sn acting by permuting ε1,…,εn. The reflections are sij= sεi∨-εj∨= 1 ⋯ i ⋯ j ⋯ n = i j ( 1 ) ⋱ 1 i 0 1 1 ⋱ 1 j 1 0 1 ⋱ 1 R+= { (ij) |  1≤i<j≤n } orR+= { εi∨-εj∨  | 1≤i<j≤n } and 𝔥εi∨-εj∨ = (𝔥*)sij = { μ∈𝔥ℝ* |  sijμ=μ } = { μ=μ1ε1+⋯+ μnεn |  ⟨μ,εi∨-εj∨⟩ =0 } = { μ=μ1ε1+⋯+ μnεn |  μi=μj } . The arrangement of hyperplanes 𝔥εi∨-εj∨ in𝔥ℂ*, 1≤i<j≤n is the braid arrangement.

Remark Confn(ℂn)= ( 𝔥ℂ*- (⋃1≤i<j≤n𝔥εi∨-εj∨) ) has π1(Confn(ℂn))= braid group.

(Type SL3) 𝔥ℤ*=span{ω1,ω2} andW0= ⟨s1,s2 | s12=s22=1,s1s2s1=s2s1s2⟩ C 𝔥α1∨ 𝔥α2∨ s1C s2C ω1 ω2 s1s2C s2s1C s1s2s1C=s2s1s2C where s1 is reflection in 𝔥α1∨ and s2 is reflection in 𝔥α2∨.

Let C be a fundamental chamber for the action of W0 on 𝔥ℝ*. W0⟷1-1 {chambers in 𝔥ℝ*}. Let C‾ be the closure of C. The dominant integral weights are P+=𝔥ℤ*∩C‾ andP++= 𝔥ℤ*∩C are the strictly dominant integral weights.

There is a bijection P+ ⟶ P++ λ ⟼ λ+ρ where ρ is the point of P++ closest to 0.

Symmetric functions

Let X={Xμ | μ∈𝔥ℤ*} withXμXν= Xμ+ν. X is the same group as 𝔥ℤ*, except written multiplicatively. W0 acts on X by wXμ=Xwμ, for w∈W0,μ∈𝔥ℤ*. Two one-dimensional representations of W0 are W0 ⟶ ℂ w ⟼ 1 and W0 ⟶ ℂ w ⟼ det(w). The ring of symmetric functions is ℂ[X]W0= { f∈ℂ[X] |  wf=f for all w∈W0 } . The vector space of determinant symmetric functions is ℂ[x]det= { f∈ℂ[X] |  wf=det(w)f,  for all w∈W0 } .

(Type GL3) 𝔥ℤ*=span{ε1,ε2,ε3} andX={Xμ | μ∈𝔥ℤ*} where, for μ1,μ2,μ3∈ℤ, Xμ = Xμ1ε1+μ2ε2+μ3ε3 = (Xε1)μ1 (Xε2)μ2 (Xε3)μ3 = X1μ1 X2μ2 X3μ3,where  Xi=Xεi. Then m(1,1,-2)= X1X2X3-2+ X1X2-2X3+ X1-2X2X3 is symmetric, and a(1,1,-2)= X1X2X3-2- X2X1X3-2- X1-2X2X3- X1X2-2X3+ X2X3X1-2+ X3X1X2-2 is determinant symmetric since det(s1)=-1 and det(s2)=-1.

(Type SL3) C 𝔥α1∨ 𝔥α2∨ ω1 ω2 ρ s1ρ s2ρ mρ = Xρ+ Xs1ρ+ Xs2ρ+ Xs1s2ρ+ Xs2s1ρ+ Xs1s2s1ρ and m2ω1 = X2ω1+ X2ω2-2ω1+ X-2ω2 are symmetric and aρ = Xρ- Xs1ρ- Xs2ρ+ Xs1s2ρ+ Xs2s1ρ- Xs1s2s1ρ and a2ω1 = X2ω1- Xs12ω1- Xs22ω1+ Xs1s22ω1+ Xs2s12ω1- Xs1s2s12ω1 = X2ω1- X2ω1- X2ω2-2ω1+ X-2ω2+ X2ω2-2ω1- X-2ω2 are determinant symmetric.

Let mμ=∑γ∈W0μ Xγ,for μ∈P. Then mwμ=mμ, for w∈W0 and the orbit sums, or monomial symmetric functions, mλ=∑γ∈W0λ Xγ,λ∈P+ form a basis of ℂ[X]W0.

Let aμ=∑w∈W0 det(w-1)Xwμ, for μ∈P. Then avμ = ∑w∈W0det (w-1)Xwvμ = ∑w∈W0det(v) det((wv)-1) Xwvμ = det(v)aμ, for μ∈P, v∈W0.

If μ∈𝔥α∨ so that sαμ=μ then aμ=asαμ= det(sα)aμ implies aμ=0, since det(sα)≠1. Thus aμ=∑w∈W0 det(w-1)Xμ, μ∈P++, form a basis of ℂ[X]det.

Boson-Fermion correspondence

ℂ[X]W0 ⟶ ℂ[X]det f ⟼ aρf is well defined since, if f∈ℂ[X]W0 then w(aρf)= (waρ)(wf)= det(w)aρf, for w∈W0. In fact, this map is invertible!

Let g∈ℂ[X]det, g=∑μ∈P gμXμ, with gμ∈ℂ. Let sα be a reflection in W0. Then 12(g-sαg)= 12(g-det(sα)g)= 12(g+g)=g and Xμ-Xsαμ = Xμ-Xμ-⟨μ,α∨⟩α = Xμ(1-X-⟨μ,α∨⟩α) = Xμ(1-X-α) ( 1+ X-α+ X-2α+⋯+ X-(⟨μ,α∨⟩-1)α. ) Hence Xμ-Xsαμ 1-X-α =Xμ ( 1+X-α+ X-2α+⋯+ X-(⟨μ,α∨⟩-1)α ) and g=12(g-sαg) is divisible by 1-X-α. It follows that if g∈ℂ[X]det then g is divisible by ∏α∈R+(1-X-α).

Claim: aρ = Xρ+lower stuff = Xρ∏α∈R+ (1-X-α).

(Type GLn) Xi=sμXεi, where 𝔥ℤ*=span{ε1,…,εn} and W0=Sn. Then aμ = ∑w∈Sn det(w-1) Xwμ = ∑w∈Sn det(w-1) Xw(1)μ1⋯ Xw(n)μn = det(Xiμj). In this case C= { μ=μ1ε1+⋯+ μnεn |  μi>μj for  1≤i<j≤n } since 𝔥εi∨-εj∨={μ=μ1ε1+⋯+μnεu | μi=μj}. Hence P+ = { μ=μ1ε1+⋯+ μnεn |  μi∈ℤ, μ1≥μ2≥⋯≥ μn } , P++ = { μ=μ1ε1+⋯+ μnεn |  μi∈ℤ, μ1>μ2>⋯> μn } and P+ ⟶ P++ μ ⟼ μ+ρ where ρ=(n-1)ε1+(n-2)ε2+⋯+εn-1. Hence aρ=det(xin-j)= ( x1n-1 x1n-2 ⋯ x1 1 x2n-1 x2n-2 ⋯ x2 1 ⋮ xnn-1 xnn-2 ⋯ xn 1 ) =∏i<j(xi-xj).

Notes and References

These are a typed copy of Lecture 7 from a series of handwritten lecture notes for the class Representation Theory given on September 9, 2008.

page history