Kazhdan-Lusztig polynomials

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

Last update: 6 November 2012

Bar invariant bases

Let (W,) be a poset such that for u,vW the interval [u,v] is finite. Let M be the free [q,q-1] modules with basis {TwwW},

M=[qq-1] -span {TwwW} ,

and let :MM be a -linear involution such that

q=q-1 and Tw=Tw+ v<wawv Tv,

where avw[qq-1]. Then

  1. There is a unique basis {Cw-wW} such that Cw=Cw andCw-= Tw+v<w Pvw-Tv withPvw- q-1[q-1] .
  2. There is a unique basis {Cw-wW} such that Cw=Cw andCw+= Tw+v<w Pvw+Tv withPvw- q[q] .

Proof.

(a) The pvw- are determined by induction:

Pww-=1and Puw-= k<0 fkqk

where

f=kfk qk=u<zw auz Pzw- ( =Puw-- Puw- ) .

(b) The Pvw+ are determined by induction:

Pww+=1and Puw+= k>0 fkqk

where

f=kfkqk =u<zw auzPzw+ ( =Puw+- Puw+ ) .

The dual module

M*=Hom[q,q-1] ( M, [q,q-1] )

is given a bar involution

:M* M*defined by φ,m= φ,m

If {TwwW} is the dual basis to {TwwW} then

Tw=v bvwTvwhere bvw= Tw, Tv = Tw, Tv = Tw,zv azvTz

so that B=At. If {CwwW} is the dual basis to {CwwW} then

Cw=Cwsince Cw,Cv = Cw, Cv = Cw, Cv =δvw,

and

Cw=PvwTv whereδvw Cw,Cv= uPuwTu, zPzvTz =uPuwPuv

so that

(Puw)= ((Puv)-1) t =(Pt)-1.

The affine Hecke algebra

The affine Hecke algebra H has [q,q-1] basis {TwwW},

H= [q,q-1]-span {TwwW}

with relations

Tw1Tw2= Tw1w2, if(w1w2) =(w1)+ (w2), TsiTw= (q-q-1)Tw+ Tsiw, if(siw)< (w) (0in).

The algebra H also has bases

{ XλTw wW,λP } and { TvXμ vW,μP } ,

where

Xλ=Ttλ, ifλP+, andXλ=Xμ (Xν)-1,

if λ=μ-ν with μ,νP+.

The bar involution on H is the -linear map :HH given by

q=q-1and Tw= Tw-1-1 forwW.

Define elements 10,ε0H by

102=10, and Tsi10=q 10,for 1in, ε02=ε0, and Tsiε0= q-1ε0, for1in,

and let

Aμ=ε0Xμ 10,forμ P.

  1. Xλ= Tw0Xw0λ Tw0-1, for λP,
  2. 10=10 and ε0=ε0.
  3. If z[P]W then z=z.
  4. q-(w0) Aλ+ρ =q-(w0) Aλ+ρ.

The τ-operators are given by

τi=Ti- q-q-1 1-X-αi .

Then

  1. Xλτi=τi Xsiλ,
  2. τi2= (q-q-1Xαi) (q-q-1X-αi) (1-Xαi) (1-X-αi)
  3. τiτjτi mijfactors = τjτiτj mijfactors

The shift operator is

Δ=αR+ ( qXα/2-q-1 X-α/2 ) .

Then

(Ti+q)Δ= (siΔ) (Ti-q)and Δ [P]W= { hH (ti+q-1h =0for1i n } ε0Eλ=Δ 10Eλ-ρ and Δf, Δg k =qNk f,g k+1 .

The ??-trace on H is the linear map τ:H given by

tr(h)=h1, or, more precisely,tr(Tw)= δw1.

Define an inner product H by

h1,h2= tr(h1h2),

so that

Tu,Tv = [Tu-1Tv] 1 and Tu,Tv =q(u)?? δuv-1.

The generic degrees are dλ(q) given by

trλH^ dλ(q)χHλ.

The Kazhdan-Lusztig basis is defined by

{ hH h,h# 1+q-1 [q-1],h= h }

or by the usual bar invariance and triangularity conditions.

Kazhdan-Lusztig polynomials

The Iwahori-Hecke algebra is the algebra over [q] given by generators Tw, wW and relations

TsiTw = { Tsiw, ifsiw>w, qTsiw +(q-1)Tw, ifsiw<w.

The bar involution on H is the -algebra involution given by

q=q-1and Tw= Tw-1-1,

for wW. The Kazhdan-Lusztig basis of H is the basis {CwwW} given by

  1. Cw=Cw, and
  2. Cw=Tw+ vwpvw (q)Tv,where pvw(q) q[q].

Kazhdan-Lusztig polynomials

  1. Pww(q)=1,
  2. Pxw(q)=0, if xw,
  3. deg(Pxw(q)) 12 ((w)-(x)-1) , if xw.

Define

μ(x,w)= coefficient of the highest degree term in Pxw(q),

which is the term of degree 12 ( (w)- (x)-1 ) . Then, if sw<w

Pxw(q) = { Psx,w(q), ifsx>x, Psx,sw(q) +qPx,sw- sz<z q 12 ((w)-(z)) μ(z,sw) Px,z, ifsx<x.

The W-graph has

  1. Vertices: W
  2. Edges: xy if μ[x,y]= { μ(x,y), ifx<y, μ(y,x), ify<x,

Then

KL(s)xx = { -1, ifsx<x, 1, ifsx>x, KL(s)xy = { μ[x,y], ifsx<x, sy>yandx y, 0, otherwise,

Define a relation L by taking the closure of the relation

xLyif D(x) D(y)and xyis an edge.

and define

x=Lyif xLyand yLx.

The case of dihedral groups

In type A1,

H=span{1,T1} withT12= (q-1)T1+q.

So

T1=T1-1= (q-1-1)+ q-1T1,and C1=q-12 (1+T1),

since

q-12 (1+T1) =q12 (1+T1-1)= q12 ( 1+q-1T1+ (q-1-1) ) =q12q-1 (1+T1)= q-12 (1+T1).

In type A2, H=span { 1,T1,T2, T1T2, T2T1, T1T2T2 } and

C1 = q-12 (1+T1), C2 = q-12 (1+T2), C1C2 = q-1 ( 1+T1+T2+ T1T2=C12, ) C2C1 = q-1 ( 1+T1+T2+ T2T1=C21, ) C1C21 = q-32 ( T1T2T1+ T1T2+ (q-1)T1+q +T1+ T2T1+T1 +T2+1 ) , = q-32 ( T1T2T1+ T1T2+ T2T1+T1 +T2+1 ) +C1 ,

so that

C121=C1 C12-C1= q-32 ( T1T2T1+T1 T2+T2T1+T1 +T2+1 ) .

Note that C12= (q12+q-12) C1. Then, using that Ti=q12Ci-1, to produce the matrices for the regular representation in the KL-basis,

ρ(T1)= ( -1 0 0 0 0 0 q12 q q12 0 0 0 0 0 -1 0 0 0 0 0 0 -1 0 0 0 0 0 q12 q 0 0 0 q12 0 0 q ) andρ(T2)= ( -1 0 0 0 0 0 0 -1 0 0 0 0 0 q12 q 0 0 0 q12 0 0 q q12 0 0 0 0 0 -1 0 0 0 0 0 q12 q )

with rows and columns indexed by 1,C1,C21, C2,C12, C121.

In type B2, H=span { 1,T1,T2,T1 T2,T2T1,T1 T2T1,T2T1 T2,T1T2T1 T2 } , and

C1C2=C12, C2C1=C21, C1C21=C121 +C1,C2C12 =C212+C2, C2C121= C2121+C21.

where

C1 = q-12 (1+T1), C2 = q-12 (1+T2), C12 = q-1 ( 1+T1+T2+ T1T2, ) C21 = q-1 ( 1+T1+T2+ T2T1, ) C121 = q-32 ( 1+T1+T2+ T1T2+ T2T1+ T1T2T1 ) , C212 = q-32 ( 1+T1+T2+ T1T2+ T2T1+ T2T1T2 ) , C1212 = q-32 ( 1+T1+T2+ T1T2+ T2T1+ T1T2T1+ T2T1T2+ T2T1T2T1 ) .

and the matrices of the regular representation in the KL-basis are

ρ(T1) = ( -1 0 0 0 0 0 0 0 q12 q q12 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 q12 q 0 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 q12 q q12 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 q12 q ) ρ(T2) = ( -1 0 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 q12 q q12 0 0 0 0 0 0 0 -1 0 0 0 0 q12 0 0 0 q q12 0 0 0 0 0 0 0 -1 0 0 0 0 0 q 0 q12 q 0 0 0 0 q12 0 0 0 q )

with rows and columns indexed by 1,C1,C21, C121,C2, C12,C212 ,C1212.

Let W be the dihedral group of order 2m. Then

Cw=q-12(w) (vwTv), so thatpvw (q)=1,for all vw.

Proof.

Let Cw be defined by the formula in the statement of the Theorem. If s1w>w so that w=s2s1s2s1 then

Cs1Cw = q-(w)/2 q-12 ( vwTv+ vs1w s1v<v Tv+(q-1) v<w s1v<v Tv+q v<w s2v<v Tv ) = q-(w)/2 q-12 ( vs1w s2v<v Tv+ v<w s1v<v Tv+ vs1w s1v<v Tv- v<w s1v<v Tv+q vs2w Tv ) = Cs1v+ q-(w)/2 q1/2 ( vs2w Tv ) = Cs1w+ Cs2w,

and, if s1w<w so that w=s1s2s1s2 then let w=s1w and w=s2s1w so that

Cs1Cw = Cs1Cs1w =Cs1 ( Cs1Cw -Cs2w ) = Cs1 ( Cs1 Cw- Cw ) = ( q1/2+ q-1/2 ) Cs1Cw -Cs1 Cw = ( q1/2+ q-1/2 ) Cs1 Cw- ( q1/2+ q-1/2 ) Cw, by induction, = ( q1/2+ q-1/2 ) ( Cs1 Cw- Cw ) = ( q1/2+ q-1/2 ) Cw.

So,

Cs1Cw= { Cs1w+ Cs2w, ifs1w<w, i.e.w= s2s1s2 s1, ( q12+ q-12 ) Cw, ifs1w<w, i.e.w=s1 s2s1s2. (3.1)

In the first case, (s2w)<(w) and so, by induction, Cs1w= Cs1Cw- Cs2w is bar invariant.

From equation (???)

T1Cw= { qCw, ifs1w<w, i.e.w=s1s2 s1s2, q12Cs1w -Cw+q12 Cs2w, ifs1w>w, i.e.w=s2s1 s2s1.

For example, in the case I2(5),

C2C1=C21, C1C21= C121+C1, C2C121= C2121+C21, C1C2121= C12121+ C121, C1C2=C12, C2C12= C212+C2, C1C212= C1212+ C12, C2C1212= C21212+ C212,

and the matrices of the regular representation in the KL-basis are

ρ(T1) = ( -1 0 0 0 0 0 0 0 0 0 q12 q q12 0 0 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 q12 q q12 0 0 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 q12 q q12 0 0 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 q12 q 0 0 0 0 0 q12 0 0 0 0 q ) ρ(T2) = ( -1 0 0 0 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 q12 q q12 0 0 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 q12 q 0 0 0 0 0 q12 0 0 0 0 q q12 0 0 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 q12 q q12 0 0 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 q12 q )

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