Kazhdan-Lusztig polynomials

Kazhdan-Lusztig polynomials

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

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

Last updates: 10 June 2010

Bar invariant spaces

Let W be a poset such that for u,vW the interval ub is finite. Let M be the free q q -1 modules with basis Tw | wW , M= q q -1 -span Tw | wW , and let -:MM be a -linear involution such that q = q -1 and Tw = Tw + v<w a wv Tv , where a vw q q -1 . Then

  1. There is a unique basis Cw- | wW such that Cw = Cw and Cw- = Tw + v<w P vw - Tv with P vw - q -1 q -1
  2. There is a unique basis Cw- | wW such that Cw = Cw and Cw+ = Tw + v<w P vw + Tv with P vw + q q .

Proof.
  1. The p vw - are determined by induction: P ww - =1  and   P uw - = k <0 fk qk where f= k fk qk = u<vw a uz P zw - = P uw - - P uw - .
  2. The p vw + are determined by induction: P ww + =1  and   P uw + = k >0 fk qk where f= k fk qk = u<vw a uz P zw + = P uw + - P uw + .

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 Tw | wW is the dual basis to Tw | wW then Tw = v b vw Tv where b vw = Tw Tv = < Tw , Tv _ > = < Tw , zv a zv Tz > so that B= At . If Cw | wW is the dual basis to Cw | wW then Cw = Cw since Cw Cv = < Cw , Cv _ = < Cw , Cv > = δ vw , and Cw = P vw Tv where δ vw = Cw Cv = u P uw Tu z P zv Tz = u P uw P uv , so that P uw = `P uv -1 t = Pt -1 .

The affine Hecke algebras

The affine Hecke algebra H~ has q q -1 basis Tw | w W~ , with relations T w1 T w2 = T w1 w2 ,if  l w1 w2 =l w1 +l w2 , T si Tw = q- q -1 Tw + T si w ,if  l si w <l w , 0in . The algebra H~ also has bases Xλ Tw | wW,λP   and   Tv Xμ | vW,μP , where Xλ = T tλ ,  if  λ P+ ,  and   Xλ = Xμ Xν -1 , if λ=μ-ν with μ,ν P+ .

The bar involution on H~ is the -linear map -: H~ H~ given by q = q -1   and   Tw = T w -1 -1   for  w W~ .

Define elements 10 , ε0 H by 102 = 10 ,  and   T si 10 =q 10 ,  for  1in, ε02 = ε0 ,  and   T si ε0 = q -1 10 ,  for  1in, and let Aμ = ε0 Xμ 10 ,for  μP.

  1. Xλ = T w0 X w0 λ T w0 -1 , for λP,
  2. 10 = 10 and ε0 = ε0 .
  3. If z W W then z =z.
  4. q -l w0 A λ+ρ = q -l w0 A λ+ρ = q -l w0 A λ+ρ .

The τ -operators are given by τi = Ti - q- q -1 1- X - αi . Then

  1. Xλ τi = τi X si λ ,
  2. τi2 = q- q -1 X αi q- q -1 X - αi 1- X αi 1- X - αi
  3. τi τj τi m ij   factors = τj τi τj m ij   factors

The shift operator is Δ= α R+ q X α 2 - q -1 X - α 2 . Then Ti +q Δ= si Δ Ti -q   and  Δ P W = h H~ | ti + q -1 h =0  for  1in . Also, ε0 Eλ =Δ 10 E λ-ρ   and   Δf Δg k = q Nk fg k+1

The ????-trace on H~ is the linear map τ: H~ given by tr h =h | 1 ,or, more precisely,  tr Tw = δ w1 .

Define an inner product on H~ by h1 h2 =tr h1 h2 , so that Tu Tv = T u -1 Tv 1   and  , Tu Tv = q l u ? δ u v -1 . The generic degrees are dλ q given by tr= λ H~ 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 T si Tw = T si w , if si w>w, q T si w + q-1 Tw , if si w<w. The bar involution on H is the -algebra involution given by q = q -1 ,  and   Tw = T w -1 -1 , for wW. The Kazhdan-Lusztig basis of H is the basis Cw | wW given by

  1. Cw = Cw , and
  2. Cw = Tw + vw p vw q Tv ,where   p vw q q q .
The Kazhdan-Lusztig polynomials are
  1. P ww q =1,
  2. P xw q =0,  if  xw,
  3. deg P xw q 1 2 l w -l x -1 ,  if  xw.

Define μ xw =coefficient of the highest degree term in   P xw q , which is a term of degree 1 2 l w -l x -1 . Then, if sw<w P xw q = P sx,w q , if sx>x, P sx,w q +q P x,sw - sz<z q 12 l w -l z μ z sw P x,z , if sx<x.

The W -graph has

  1. Vertices: W
  2. Edges: xy if μ xy = μ xy , if x<y, μ yx , if y<x.
Then KL s xx = 1 if sx<x, -1 if sx>x, KL s xy = μ xy if sx>x,sy>y  and  xy, 0, otherwise. Define a relation L by taking the closure of the relation x L y  if   Dl x Dl y   and  xy  is an edge. Define x =L y  if  x L y  and  y L x.

The case of dihedral groups

In type A1 , H=span 1 T1 with T12 = q-1 T1 +q. So T1 = T1 -1 = q -1 -1 + q -1 T1 ,  and   C1 = q - 1 2 1+ T1 , since q -1/2 1+ T1 = q 1 2 1+ q -1 T1 + q -1 -1 = q 1 2 q -1 1+ T1 = q - 1 2 1+ T1 .

In type A2 , H=span 1 T1 T2 T1 T 2 T2 T1 T1 T 2 T2 and C1 = q - 1 2 1+ T1 , C2 = q - 1 2 1+ T2 , C1 C2 = q -1 1+ T1 + T2 + T1 T2 = C12 , C2 C1 = q -1 1+ T1 + T2 + T2 T1 = C21 , C1 = q - 3 2 T1 T2 T1 + T1 T2 + q-1 T1 +q+ T1 + T2 T1 + T1 + T2 +1 = q - 3 2 T1 T2 T1 + T1 T2 + T2 T1 + T1 + T2 +1 + C1 , so that C 121 = C1 C12 - C1 = q - 3 2 T1 T2 T1 + T1 T2 + T2 T1 + T1 + T2 +1 . Note that C12 = q 1 2 + q - 1 2 C1 . Then, using that Ti = q 1 2 Ci -1, to produce the matrices for the regular representation in the KL -basis, ρ T1 = -1 0 0 0 0 0 q 1 2 q q 1 2 0 0 0 0 0 -1 0 0 0 0 0 0 -1 0 0 0 0 0 q 1 2 q 0 0 0 q 1 2 0 0 q and ρ T2 = -1 0 0 0 0 0 0 -1 0 0 0 0 0 q 1 2 q 0 0 0 q 1 2 0 0 q q 1 2 0 0 0 0 0 -1 0 0 0 0 0 q 1 2 q with rows and columns indexed by 1, C1 , C21 , C2 , C12 , C121 .

In type B2 , H=span 1 T1 T2 T1 T2 T2 T1 T1 T2 T1 T2 T1 T2 T1 T2 T1 T2 , and C1 C2 = C12 , C2 C1 = C21 , C1 C21 = C121 + C1 , C2 C12 = C212 + C2 , C2 C121 = C2121 + C21 , where C1 = q - 1 2 1+ T1 C2 = q - 1 2 1+ T2 C12 = q -1 1+ T1 + T2 + T1 T2 C21 = q -1 1+ T1 + T2 + T2 T1 C121 = q - 3 2 1+ T1 + T2 + T1 T2 + T2 T1 + T1 T2 T1 C212 = q - 3 2 1+ T1 + T2 + T1 T2 + T2 T1 + T2 T1 T2 C1212 = q -2 1+ T1 + T2 + T1 T2 + T2 T1 + T1 T2 T1 + T2 T1 T2 + T1 T2 T1 T2 . and the matrices of the regular representation in the KL-basis are ρ T1 = -1 0 0 0 0 0 0 0 q 1 2 q q 1 2 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 q 1 2 q 0 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 q 1 2 q q 1 2 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 q 1 2 q ρ T1 = -1 0 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 q 1 2 q q 1 2 0 0 0 0 0 0 0 -1 0 0 0 0 q 1 2 0 0 0 q q 1 2 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 q 1 2 q 0 0 0 0 q 1 2 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 - 1 2 l w vw Tv ,so that   p vw q =1,  for all  vw.

Proof.
Let Cw be defined by the formula in the statement of the theorem. If s1 w>w so that w= s2 s1 s2 s1 then C s1 = q -l w /2 q -1/2 vw Tv + v s1 w, s1 v<v Tv + q-1 v<w, s1 v<v Tv +q v<w, s2 v<v Tv = q -l w /2 q -1/2 v s1 w, s2 v<v Tv + v<w, s1 v<v Tv + v s1 w, s1 v<v Tv - v<w, s1 v<v Tv +q v s2 w Tv = C s1 v + q -l w /2 q 1 2 v s2 w Tv = C s1 w + C s2 w , and, if s1 w<w so that w= s1 s2 s1 s2 then let w' = s1 w and w" = s2 s1 w so that C s1 = C s1 C s1 w' = C s1 C s1 C w' - C s2 w' = C s1 C s1 C w' - C w" = q 12 + q - 12 C s1 C w' - C s1 C w" = q 1 2 + q - 1 2 C s1 C w' - q 1 2 + q - 1 2 C w" , by induction, = q 1 2 + q - 1 2 C s1 C w' - C w" = q 1 2 + q - 1 2 Cw . So, C s1 Cw = C s1 w + C s2 w , if s1 w>w,   ie  w= s2 s1 s2 s1 q 1 2 + q - 1 2 Cw , if s1 w<w,   ie  w= s1 s2 s1 s2 In the first case, l s2 w <l w and so, by induction, C s1 w = C s1 Cw - C s2 w is bar invariant.

From equation (???) T1 Cw = q 1 2 C s1 w - Cw + q 1 2 C s2 w , if s1 w>w,   ie  w= s2 s1 s2 s1 q Cw , if s1 w<w,   ie  w= s1 s2 s1 s2

For example, in the case I2 5 , C2 C1 = C21 , C1 C21 = C121 + C1 , C2 C121 = C2121 + C21 , C1 C2121 = C12121 + C121 , C1 C2 = C12 , C2 C12 = C212 + C2 , C1 C212 = C1212 + C12 , C2 C1212 = 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 q 1 2 q q 1 2 0 0 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 q 1 2 q q 1 2 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 q 1 2 q q 1 2 0 0 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 q 1 2 q 0 0 0 0 0 q 1 2 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 0 q 1 2 q q 1 2 0 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 q 1 2 q 0 0 0 0 0 q 0 0 0 0 q q 1 2 0 0 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 q 1 2 q q 1 2 0 0 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 q 1 2 q

References [PLACEHOLDER]

[Cu1] C. Curtis, Representations of Hecke Algebras, Asterisque 9, (1988), 13-60; arXiv:math/9909077v2, MR1828302 (2002e:20083)

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