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 8

Set up

𝔥* is a -vector space,

W0GL(𝔥*) a finite group generated by reflections,

sα, αR+, are the reflections in W0,

sαμ=μ-μ,αα,

C is a fundamental chamber for W0 acting on 𝔥*,

𝔥α1,,𝔥αn are the walls of C and their reflections,

s1,,sn are the simple reflections, W01-1 {chambers on𝔥*}

(Type SL3) 𝔥*=span{ω1,ω2} 1 𝔥α2 𝔥α1 s2 s1 s2s1 s1s2 s1s2s1

P+=𝔥*C and P++=𝔥*C are isomorphic semigroups P+ P++ λ λ+ρ

(Type GLn) 𝔥*=span{ε1,,εn} and W0=Sn, which has reflections sij= 1 i j n with 𝔥ij= {μ=(μ1,,μn)|μi=μj}. Then C= { μ=(μ1,,μn) n| μ1>μ2>>μn } has walls 𝔥αi= { μ=(μ1,,μn) |μi=μi+1 } andsi= 1 2 i i+1 n . Then P+ = { λ=(λ1,,λn) n|λ1λ2 λn } , P++ = { μ=(μ1,,μn) n|μ1> >μn } and P+ P++ λ λ+ρ where ρ=(n-1,n-2,,1,0).

[X] = span{Xμ|μ𝔥*} withXμXν= Xμ+ν, [X]W0 = { f[X]| wf=f, for allwW0 } , [X]det = { f[X]| wf=det(w)f, for all wW0 } . Then [X]W0 has basis mλ=γW0λ Xγ,λP+, [X]det has basis aλ+ρ=wW0 det(w)Xw(λ+ρ).

As [X]W0-modules Φ: [X]W0 [X]det f aρf.

Proof.

(a) Φ is a [X]W0-homomorphism.
If g[X]W0 then Φ(gf)= aρgf=gaρf= gΦ(f).
(b) Φ is well defined.
If wW0 then wΦ(f) = waρf = (waρ) (wf) = det(w)aρf = det(w)Φ(f).
(c) Φ is invertible.
Let g[X]det, g=μ𝔥*gμXμ.
Let sα be a reflection in W0, sαμ=μ-μ,α α,withμ,α. Then sαg=det(sα)g =-g, so that g=12(g-sαg).
So g = 12(1-sα)g = 12μ𝔥* gμ(Xμ-Xsαμ) = 12μ𝔥* gμXμ(1-X-μ,αα). Note, for example, 1-X-5α = (1-X-α) ( 1+Xα+ X2α+ X3α+ X4α ) , 1-X5α = X5α(1-X-α) ( 1+Xα+X2α+ X3α+X4α ) (-1). In any case, g is divisible by 1-X-α.
Since 1-X-α, αR+, are relatively prime, g is divisible by αR+(1-X-α).

Note: aρ = (αR+Xα/2) αR+(1-X-α) = XραR+ (1-X-α) since aρ=Xρ++XW0ρ andρ=12αR+α because

(a) si permutes R+-{αi} and siαi=-αi.
(b) W0 sends R+={α} to R-={-α|αR+}.

The Weyl character, or Schur function is sλ=aλ+ρaρ so that [X]W0 [X]det sλ aλ+ρ

Crystals

A path is p:[0,1]𝔥* (piecewise linear) with p(0)=0 and p(1)𝔥*.

A crystal is a set of paths B, closed under the action of the root operators ei,fi. 𝔥αi p fip 𝔥αi p f˜ip=0 and eifi=p, if fip0 and fieip=p, if eip0.

The character of B is char(B)=pB Xwt(p).

Favourite example

𝔥α2 𝔥α1 f2f1 f1f2 f2 f1 f1f2 Let B be a crystal and let pB. The i-string of p is fid--- fip-p- eip- ei2p-- eid+p where eid++1p=0 and fid-+1p=0. If h=eid+p then the paths in the i-string fiμ,αi h--fi2h- fi-h have weights μ-μ,αiαi, ,μ-2αi,μ-αi,μ with siμ=μ-μ,αiαi. Define an action of W0 on B by setting sip to be the opposite of p in its i-string. fiμ,αih- fiμ,αi-1h- - fi2h- fih-h Then wt(sip)=si wt(p). So char(B)=sichar(B) for i=1,,n. Since s1,,sn generate W0 it follows that char(B)[x]W0.

A highest weight path is pC-ρ C C-ρ A path p is highest weight, if and only if eip=0 for all i=1,,n.

Let B be a crystal. Then char(B)= pBpC-ρ swt(p).

Proof.

Define sμ=aμ+ρaρ, forμP. Then define a new action of W0 on P by wμ=w(μ+ρ)-ρ, forμ𝔥*, wW0. 0 -ρ s10 s20 s1s20 s2s10 s1s2s10 This is the dot action of W0. We have swμ = sw(μ+ρ)-ρ = aw(μ+ρ)-ρ+ρaρ =aw(μ+ρ)aρ = det(w)aμ+ρaρ =det(w)sμ. (*) Now let ε=wW0 det(w)w so that ε(Xμ)=aμ. then char(B) = 1aρchar(B) aρ = 1aρchar(B) ε(Xρ) = 1aρε(char(B)Xρ) = 1aρpB ε(Xwt(ρ)+ρ) = pBswt(p). The equation (*) can cause some cancellation in this sum.

Let pB such that p is not highest weight. Let i be minimal such that p leaves C-p by crossing 𝔥αi+δ. Define sip to be the element of the i-string of p such that wt(sip)= sip t-eit- ei2t-- fi2h- fih-h Then swt(sip)= det(si)swt(p) =-swt(p) and swt(sip)+ swt(p)=0. Note that sip leaves C-ρ at the same place that p leaves C-ρ. thus char(B)= pBpC-ρ swt(p).

Let pλ+ be a path, pλ+C-ρ with wt(pλ+)=λ. Let B(λ) be the crystal generated by pλ+. Then char(B(λ))=sλ.

Let B be a crystal. Let J{1,,n}. By ignoring the action of ei,fi for iJ, B is a (WJ,𝔥*)-crystal, where WJ=sj|jJ. Let CJ-ρJ be the region on the positive side of 𝔥αj+δj for jJ. Then char(B)= pBpCJ-ρJ swt(p)J.

Let λ,μP+. Let B(λ) and B(μ) be the irreducible crystals of highest weights λ and μ, respectively. Then B(λ)B(μ)= { pq|p B(λ),qB(μ) } is a crystal (pq is the concatenation of p and q). Then char(B(λ)B(μ))= qB(μ)pλ+qC-ρ swt(q)+λ.

Proof.

pqC-ρ only if p=pλ+, in which case swt(pλ+q)= sλ+wt(q).

Notes and References

These are a typed copy of Lecture 8 from a series of handwritten lecture notes for the class Representation Theory given on September 16, 2008.

page history