Symmetric and alternating functions and their q-analogues

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

Last update: 26 September 2012

Symmetric and alternating functions and their q-analogues

Let 10 and ε0 be the elements of the finite Hecke algebra H which are determined by

102=10 and Ti10= q10, for all1≤i≤n, ε02=ε0 and Tiε0= (-q-1) ε0, for all1≤i≤n.

In terms of the basis {Tw∣w∈W} of H these elements have the explicit formulae

10= 1W0(q2) ∑w∈W qℓ(w)Tw, andε0= 1W0(q-2) ∑w∈W (-q)-ℓ(w) Tw, (2.1)

where W0(t)= ∑w∈W tℓ(w). (To define these elements one should adjoin the element W0(q2)-1 to 𝕂 or to H.) The elements 10 and ε0 are q-analogues of the elements in the group algebra of W given by

1=1∣W∣ ∑w∈Wwand ε=1∣W∣ ∑w∈W (-1)ℓ(w)w, (2.2)

and the vector spaces 10H∼10 and ε0H∼10 are q-analogues of the vector spaces (more precisely, free 𝕂=ℤ[q,q-1]-modules) of symmetric functions and alternating functions,

𝕂[P]W = { f∈𝕂[P]∣ wf=ffor allw∈W } =1𝕂[P], 𝒜 = { f∈𝕂[P]∣ wf=(-1)ℓ(w) ffor allw∈W } =ε𝕂[P], (2.3)

respectively, where the action of W on 𝕂[P] is as defined in 1.29.

For μ∈P let the orbit Wμ and the stabilizer Wμ of μ be defined by

Wμ= {wμ∣w∈W} andWμ= {w∈W∣wμ=μ} .

Then define

mμ= ∑γ∈Wμ xγ= ∣W∣ ∣Wμ∣ 1xμ, aμ=∑w∈W (-1)ℓ(w) wxμ=∣W∣ε xμ, Mμ=10 xμ10, Aμ=ε0xμ 10. (2.4)

Theorem 2.2 below shows that the elements in (2.4) which are indexed by elements of P+ and P++ form bases (over 𝕂) of 𝕂[P]W, 𝒜, 10H∼10, and ε0H∼10. This will be a consequence of the following straightening rules. The straightening law for the Mμ given in the following Proposition is a generalization of [Mac1995, III §2 Ex. 2].

For γ∈P let mγ, aγ, Mγ, and Aγ be as defined in (2.4). Let αi be a simple root and let μ∈P be such that d=⟨μ,αi∨⟩≥0. Then

msiμ=mμ, asiμ=-aμ, andAsiμ= -Aμ.

Letting t=q-2, Mμ=Msiμ if d=0, and if d>0 then

Msiμ = tMμ + ( ∑ j=1 ⌊d/2-1⌋ (t2-1) tj-1 Mμ-jαi ) + { (t-1)td/2-1 Mμ-(d/2)αi , ifdis even, 0 , ifdis odd.

Proof.

The first two equalities follow from the definitions of mλ and aμ and the fact that ℓ(si)=1.

Let μ∈P such that d=⟨μ,αi∨⟩≥0. Since xμ+xsiμ is in the center of the tiny little affine Hecke algebra generataed by Ti and the xγ, γ∈P,

Aμ+Asiμ = ε0 (xμ+xsiμ) 10=q-1ε0 (xμ+xsiμ) Ti10 = q-1ε0Ti (xμ+xsiμ) 10=-q-2 ε0 (xμ+xsiμ) 10 = -q-2 (Aμ+Asiμ) .

Thus Aμ+Asiμ =0 which establishes the third statement.

If d=0 then, by definition, Mμ= Msiμ. If d>0 then multiplying the fourth relation in (1.21) by 10 on both the left and the right (and then multiplying by q-1) gives

10 (xsiμ-xμ) 10=q-1 (q-q-1)10 ( xsiμ-xμ 1-x-αi ) 10.

Subtracting the same relation with μ replaced by μ-αi gives

10 (xsiμ-xμ) 10-10 ( xsiμ+αi- xμ-αi ) 10 = (1-q-2)10 ( xsiμ-xμ- xsiμ+αi+ xμ-αi 1-x-αi ) 10 = (1-q-2)10 ( -xsiμ+αi -xμ ) 10.

So

10xsiμ10 =q-210xμ 10-10 xμ-αi10+ q-210 xsiμ+αi10.

Inductively applying this relation yields the result. The first cases are

Msiμ= { Mμ , if⟨μ,αi∨⟩ =0, q-2Mμ , if⟨μ,αi∨⟩ =1, q-2Mμ+ (q-2-1) Mμ-αi , if⟨μ,αi∨⟩ =2, q-2Mμ+ (q-4-1) Mμ-αi , if⟨μ,αi∨⟩ =3, q-2Mμ+ (q-4-1) Mμ-αi +q-2 (q-2-1) Mμ-2αi , if⟨μ,αi∨⟩ =4.

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Proposition 2.1 implies that, for all μ∈P and w∈W,

mwμ=mμ, awμ= (-1)ℓ(w) aμ,and Awμ (-1)ℓ(w) Aμ. (2.5)

Let 𝕂=ℤ[q,q-1]. As free 𝕂-modules

𝕂[P]W has basis {mλ∣λ∈P+}, 10H∼10 has basis {Mλ∣λ∈P+}, 𝒜 has basis {aμ∣μ∈P++}, ε0H∼10 has basis {Aμ∣μ∈P++}.

Proof.

Since { xμTw∣ μ∈P,w∈W } form a basis of H∼ the elements Mμ=10xμ10 =q-ℓ(w)10 xμTw10, μ∈P, span 10H∼10. By Proposition 2.1, if μ is on the negative side of a hyperplane Hαi, that is, if ⟨μ,αi∨⟩<0, then Mμ can be rewritten as a linear combination of Mγ such that all terms have γ on the nonnegative side of Hαi. By repeatedly applying the relation in Proposition 2.1, Mμ can be rewritten as a linear combination of Mγ such that all terms have γ on the nonnegative side of Hα1,…, Hαn, that is, γ∈P+=P∩ C‾, where C‾= { x∈ℝn∣ ⟨x,αi∨⟩≥0 for all1≤i≤n } .

If λ∈P+, using the fourth relation in (1.21),

Mλ = 10xλ10= 1W0(q2) ∑w∈W qℓ(w)Tw xλ10= 1W0(q2) ∑γ,v,w qℓ(w) dv,γxγ Tv10 = 1W0(q2) ∑γ,v,w qℓ(w) dv,γxγ qℓ(v)10= 1W0(q2) ∑γdγxγ10,

where dv,γ and dγ are some polynomials in ℤ[q,(q-q-1)] such that dv,vλ=1 so that dw0λ=1. Furthermore dγ=0 unless γ is in the convex hull of the points in the orbit Wλ. Thus the coefficient of xw0λ in Mλ is W0(q2)-1 q2ℓ(w0) and the coefficient of xγTv can be nonzero only if γ≥w0λ. Thus the Mλ, λ∈P+, are linearly independent.

The proof for the cases of mμ, aμ and Aμ is easier, following directly from (2.5), the fact that C= { x∈ℝn∣ ⟨x,αi∨⟩>0 for all1≤i≤n } is a fundamental chamber for the action of W, and that if μ∈P+∖P++ then ⟨μ,αi∨⟩=0 and aμ=-asiμ=-aμ, in which case aμ=0 (similarly for Aμ).

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For λ∈P define the Schur function, or Weyl character, by

sλ=aλ+ρaρ, whereρ=12 ∑α∈R+α. (2.6)

The straightening law for aμ in (2.5) implies the following straightening law for the Schur functions. If μ∈P and w∈W then, by (2.5) and the definition of sμ,

(-1)ℓ(w) sμ= (-1)ℓ(w) aμ+ρ aρ = aw(μ+ρ)-ρ+ρ aρ =sw∘μ, (2.7)

where w∘μ=w(μ+ρ)-ρ.

The dot action of the Weyl group W on 𝔥ℝ* which is appearing here, w∘μ=t-ρw tρμ=(tρ-1) wtρμ, is ordinary action of W on 𝔥ℝ* except with the "center" shirted to -ρ. For the root system of type C2, in Example 1.1, the picture is

Hα1+α2 Hα1 Hα2 Hα1+2α2 C s1C s2C s1s2C s2s1C s1s2s1C s2s1s2C s1s2s1s2C ρ s1ρ s2ρ Hα1+α2 Hα1 Hα2 Hα1+2α2 C 0 s1∘0 s2∘0 -ρ the orbitWρ the orbitW∘0

The following proposition shows that the Weyl characters sλ are elements of 𝕂[P]W. The equaility in part (a) is the Weyl denominator formula, a generalization of the factorization of the Vandermonde determinant det(xin-j) =∏1≤i,j≤n (xi-xj). In the remainder of this section we shall abuse language and use the term "vector space" in place of "free 𝕂=ℤ [q,q-1] module".

Let P+, P++, 𝕂[P]W and 𝒜 be as in (1.7) and (2.4) and let ρ be as in (1.8).

  1. If f∈𝒜 then f is divisible by aρ and aρ=xρ ∏α∈R+ (1-x-α)
  2. The set {sλ∣λ∈P+} is a basis of 𝕂[P]W.
  3. The maps P+⟶P++ λ⟼λ+ρ and Φ:𝕂[P]W →𝒜 f⟼aρf sλ⟼aλ+ρ are a bijection and a vector space isomorphism, respectively.

Proof.

Since si takes αi to -αi and permutes the other elements of R+,

ρ- ⟨ρ,αi∨⟩ αi=siρ=ρ -αi,

and so

⟨ρ,αi∨⟩ =1,for all1≤ i≤n.

Thus the map P+→P++ given by λ↦λ+ρ is well defined and it is a bijection since it is invertible.

Let d=xρ ∏α∈R+ (1-x-α)= ∏α∈R+ (xα/2-x-α/2) .

Since si takes αi to -αi and permutes the other elements of R+, sid=-d for all 1≤i≤n and so wd=(-1)ℓ(w)d for all w∈W. Thus d is an element of 𝒜.

If α∈R+ and f=∑μ∈P cμxμ∈𝒜 then

∑μ∈Pcμxμ =f=-sαf= ∑μ∈P-cμ xsαμ,

and

f=∑⟨μ,α∨⟩≥0 cμ (xμ-xsαμ).

Since ( 1- x-⟨μ,α∨⟩α ) is divisible by (1-x-α) it follows that xμ-xsiμ=xμ ( 1- x-⟨μ,α∨⟩α ) is divisible by (1-x-α) and thus that f is divisible by (1-x-α) for all α∈R+. Since the elements (1-x-α) are relatively prime in the Laurent polynomial ring 𝕂[P] and xρ is a unit in 𝕂[P], f is divisible by d. Since both f and d are in 𝒜, the quotient f/d is an element of 𝕂[P]W.

The monomial xρ appears in aρ with the coefficient 1 and it is the unique term xμ in aρ with μ∈P+. Since d has highest term xρ with coefficient 1 and aρ is divisible by d and it follows that aρ/d=1. Thus aρ=d, the inverse of the map Φ in (c) is well defined, and Φ is an isomorphism.

Since {aλ+ρ∣λ∈P+} is a basis of 𝒜 and the map Φ is an isomorphism it follows that {sλ∣λ∈P+} is a 𝕂-basis of 𝕂[P]W.

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Acknowledgements

The research of A. Ram was partially supported by the National Science Foundation (DMS-0097977), the National Security Agency (MDA904-01-1-0032) and by EPSRC Grant GR K99015 at the Newton Institute for Mathematical Sciences. The research of K. Nelsen was partially supported by the National Science Foundation (DMS-0097977 and a VIGRE grant) and the National Security Agency (MDA904-01-1-0032).

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