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 μ𝔥*, λ𝔥. 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 sαGL(𝔥*) such that, in GLn(𝔽), sα is conjugate to ( ξ 1 1 ) with ξ𝔽, ξ1. Then sαGL(𝔥) is conjugate to ( ξ-1 1 1 ) . Then 𝔥*=𝔥α αand𝔥= 𝔥αα where 𝔥α = (𝔥*)sα= {μ𝔥*|sαμ=μ} (1 eigenspace ofsα), α = (ξ-eigenspace ofsα), 𝔥α = 𝔥sα= { λ𝔥| sαλ=λ } =(1-eigenspace ofsα), α = (ξ-1-eigenspace ofsα) and 𝔥α = {μ𝔥*|μ,α=0}, 𝔥α = {λ𝔥|λ,α=0}. Choose α and α so that α,α=1-ξ=1-det(sα). Then sαμ=μ- μ,αα 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,ifμ𝔥α, sα-1λ = λ-λ,α α=λ-0,if λ𝔥α.

Weyl groups

Let 𝔥* be a -vector space. 𝔥*=-span {ω1,,ωn}, where ω1,,ωn is a -basis of 𝔥*. 𝔥* = 𝔥*= -span{ω1,,ωn}, 𝔥* = 𝔥*=-span {ω1,,ωn}, 𝔥* = 𝔥*=-span {ω1,,ωn}, 𝔥* = 𝔥*=-span {ω1,,ωn}. A Weyl group, or crystallographic reflection group, is a finite subgroup w0 of GL(𝔥*) generated by reflections.

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

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

Examples (Type GLn)

𝔥*=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)| 1i<jn } orR+= { εi-εj |1i<jn } and 𝔥εi-εj = (𝔥*)sij = { μ𝔥*| sijμ=μ } = { μ=μ1ε1++ μnεn| μ,εi-εj =0 } = { μ=μ1ε1++ μnεn| μi=μj } . The arrangement of hyperplanes 𝔥εi-εj in𝔥*, 1i<jn is the braid arrangement.

Remark Confn(n)= ( 𝔥*- (1i<jn𝔥εi-εj) ) has π1(Confn(n))= braid group.

(Type SL3) 𝔥*=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 𝔥*. W01-1 {chambers in𝔥*}. Let C be the closure of C. The dominant integral weights are P+=𝔥*C andP++= 𝔥*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|μ𝔥*} withXμXν= Xμ+ν. X is the same group as 𝔥*, except written multiplicatively. W0 acts on X by wXμ=Xwμ, forwW0,μ𝔥*. Two one-dimensional representations of W0 are W0 w 1 and W0 w det(w). The ring of symmetric functions is [X]W0= { f[X]| wf=ffor allwW0 } . The vector space of determinant symmetric functions is [x]det= { f[X]| wf=det(w)f, for allwW0 } .

(Type GL3) 𝔥*=span{ε1,ε2,ε3} andX={Xμ|μ𝔥*} where, for μ1,μ2,μ3, 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=γW0μ Xγ,forμP. Then mwμ=mμ, for wW0 and the orbit sums, or monomial symmetric functions, mλ=γW0λ Xγ,λP+ form a basis of [X]W0.

Let aμ=wW0 det(w-1)Xwμ, forμP. Then avμ = wW0det (w-1)Xwvμ = wW0det(v) det((wv)-1) Xwvμ = det(v)aμ, for μP, vW0.

If μ𝔥α so that sαμ=μ then aμ=asαμ= det(sα)aμ implies aμ=0, since det(sα)1. Thus aμ=wW0 det(w-1)Xμ, μP++, form a basis of [X]det.

Boson-Fermion correspondence

[X]W0 [X]det f aρf is well defined since, if f[X]W0 then w(aρf)= (waρ)(wf)= det(w)aρf, for wW0. In fact, this map is invertible!

Let g[X]det, g=μP gμXμ, with gμ. 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,αα = Xμ(1-X-μ,αα) = 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[X]det then g is divisible by αR+(1-X-α).

Claim: aρ = Xρ+lower stuff = XραR+ (1-X-α).

(Type GLn) Xi=sμXεi, where 𝔥*=span{ε1,,εn} and W0=Sn. Then aμ = wSn det(w-1) Xwμ = wSn det(w-1) Xw(1)μ1 Xw(n)μn = det(Xiμj). In this case C= { μ=μ1ε1++ μnεn| μi>μjfor 1i<jn } since 𝔥εi-εj={μ=μ1ε1++μnεu|μi=μj}. Hence P+ = { μ=μ1ε1++ μnεn| μi, μ1μ2 μn } , P++ = { μ=μ1ε1++ μnεn| μi, 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.

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