Affine Braids, Markov Traces and the Category 𝒪

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

Last update: 22 December 2013

The k Modules λ/μ and λ/μ

Let λ be integrally dominant and let μ𝔥*. Define k modules λ/μ=Fλ (M(μ))and λ/μ= Fλ(L(μ)). (4.1) The following lemma is the main tool for studying the structure of these k modules.

([Jan1980, Theorem 2.2], [Dix1996, Lemma 7.6.14]) Let E be a finite dimensional Uh𝔤 module and let {ei} be a basis of E consisting of weight vectors ordered so that i<j if wt(ei)<wt(ej). Suppose M is a Uh𝔤 module generated by a highest weight vector vμ+ of weight μ. Set Mi=jiUh 𝔫-(vμ+ej). Then

(a) ME=M1M2 is a filtration of Uh𝔤 modules such that Mi/Mi+1 is 0 or is a highest weight module of highest weight μ+wt(ei).
(b) If M=M(μ) then Mi/Mi+1M(μ+wt(ei)).

The braid group k is the subgroup of k generated by T1,,Tk-1. By restriction, both λ/μ and Vk=L(0)Vk are k modules.

There is a unique Uh𝔤 contravariant form ,M on the Verma module M(wμ) determined by vwμ+,vwμ+M=1 where vwμ+ is the generating highest weight vector of M(wμ). As in (2.15), this form together with a nondegenerate Uh𝔤 contravariant form ,V on V gives Uh𝔤 contravariant forms ,Vk and , on Vk and M(wμ)Vk, respectively.

With these notations at hand we use Lemma 4.1 to prove the fundamental facts about the k modules λ/μ and λ/μ defined in (4.1).

Let λ, μ be integrally dominant weights and wWμ.

(a) As k modules, λ/wμ(Vk)λ-wμ
(b) λ/wμλ/yμ if Wλ+ρwWμ+ρ= Wλ+ρyWμ+ρ.
(c) Use the same notation , for the Uh𝔤 contravariant form , on M(wμ)Vk and the k contravariant form on λ/(wμ) obtained by restriction of , to the subspace (M(wμ)Vk)λ[λ]. Then λ/wμ λ/(wμ) rad, .
(d) Assume wWμ is maximal length in the coset wWμ+ρ in Wμ. If λ(wμ)0 then
(1) λ-wμ is a weight of Vk,
(2) w is maximal length in Wλ+ρwWμ+ρ.
(e) If μ is a dominant integral weight then λ/μ { v(Vk)λ-μ |Xiμ+ρ,αi v=0,for all1in. } .

Proof.

(a) Let vwμ+ be the generating highest weight vector of M(wμ) and, for nVk let pr(vwμ+n) be the image of vwμ+n in (MVk)/𝔫-(MVk). Then, since λ is integrally dominant, Lemma 4.1 shows that (Vk)λ-wμ λ/(wμ) n pr(vwμ+n) (4.2) is a vector space isomorphism. This is a k-module isomorphism since the k action on M(wμ)Vk commutes with 𝔫- and fixes vwμ+.

(b) By (a), λ-wμP and so Wλ=Wμ. It is sufficient to show that λ/wμλ/(siwμ) for all simple reflections of Wλ such that siWλ+ρ and siw>w. Applying the exact functor Fλ to the Verma module inclusion M(siwμ) M(wμ)gives λ/siwμ λ/wμ, an inclusion of k-modules. Since si(λ-wμ)= si(λ+ρ)- siw(μ+ρ)= λ+ρ-siw(μ+ρ) =λ-(siw)μ there is a (vector space) isomorphism of weight spaces (Vk)λ-wμ Vλ-siwμk. (This isomorphism can be realized by Lusztig’s braid group action [CPr1994, §8.1-8.2], Ti:(Vk)λ-wμ(Vk)si(λ-wμ)). Thus, by part (a), the k-module inclusion λ/siwμλ/wμ is an isomorphism.

(c) Use the notations for the bilinear forms on M(wμ) and Vk as given in the paragraph before the statement of the proposition. Let {bi} be an orthonormal basis of Vk with respect to ,Vk. If rrad,M then rb,sb =r,sM b,bVk =0,for allsM(wμ) ,b,bVk, and so (rad,M)Vkrad,. Conversely, if riM(wμ) such that ribirad, then 0=iribi,sbj =iri,sM δij=ri,s, for allsM(wμ). So rirad,M and thus rad,rad,MVk. By the Uh𝔤 contravariance of , M(wμ)Vkλ[λ] M(wμ)Vkμ[μ] for integrally dominant weights λ, μ with λμ. Thus rad,= (rad,MVk)λ[λ], (4.3) where , is the restriction of the form on M(wμ)Vk to (M(wμ)Vk)λ[λ]. Thus (M(wμ)Vk)λ[λ] rad, = (M(wμ)Vk)λ[λ] (rad,MVk)λ[λ] (M(wμ)rad,MVk)λ[λ] = (L(wμ)Vk)λ[λ] = λ/(wμ), where the isomorphism is a consequence of the fact that, because λ is an integrally dominant weight, the functor (·Vk)λ[λ] is exact (3.8).

(d) If λ-wμ is not a weight of Vk then, by part (a), λ/wμ=0. Since the functor Fλ is exact and L(wμ) is a quotient of M(wμ), λ/wμ is a quotient of λ/wμ. Thus λ/wμ=0 implies λ/wμ=0.

Assume that wWμ is not the longest element of the double coset Wλ+ρwWμ+ρWμ. Then there is a positive root α>0 such that sαWλ+ρ and sαw>w. Since sαwWμ+ρwWμ+ρ there is an inclusion of Verma modules M(sαwμ)M(wμ) and Fλ(L(wμ)) is a quotient of Fλ(M(wμ))/Fλ(M(sαwμ)). On the other hand, by part (b), λ/sαwμ λ/wμ,and so λ/wμ λ/sαwμ = Fλ(M(wμ)) Fλ(M(sαwμ)) =0. Thus Fλ(L(wμ))=0.

(e) When μ is a dominant integral weight λ/μ ( span-{pr(vμ+n)|nVk} span-{pr(Yiμ+ρ,αivμ+n)|nVk} ) λ . see [Dix1996, 7.2.7]. Thus, by (c) and the vector space isomorphism (4.2) it follows that, as vector spaces, λ/μ ( (span-{pr(vμ+n)|nVk})/ (span-{pr(Yiμ+ρ,αivμ+n)|nVk}) ) λ . For any k0, there is a nonzero constant c such that pr(Yik+1vμ+n)= c·pr(Yi(Yikvμ+n)-Yikvμ+Yin)= -cpr(Yikvμ+Yin), and so, by induction, pr(Yik+1vμ+n)= ξ·pr(vμ+Yik+1n) for some constant ξ0. Thus λ/μ is isomorphic to the vector space (Vk/(iYiμ+ρ,αiVk))λ-μ (iYiμ+ρ,αiVk)λ-μ. If v(Yiμ+ρ,αiVk) then the Uh𝔤 contravariance of ,Vk gives that 0=Yiμ+ρ,αin,bVk= v,Xiμ+ρ,αibVk, for allnVk. Thus, by the nondegeneracy of ,Vk, λ/μ (iYiμ+ρ,αiVk)λ-μ= { b(Vk)λ-μ |Xiμ+ρ,αib=0 } .

Proposition 4.2d gives a necessary condition on λ/wμ for the k-module λ/wμ to be nonzero. The following will be useful for analyzing the combinatorics of the examples in Section 6. If P denotes the weight lattice λ is an integrally dominant then the action of Wλ+ρ on λ-P by the dot action has fundamental domain Cλ+ρ-= { μλ-P| μ+ρ,α 0, for allα>0such that λ+ρ,αi=0 } . and the following are equivalent:

(a) μCλ+ρ-,
(b) μ=wλν with ν integrally dominant and wλWλ longest in the coset Wλ+ρwλ in Wλ,
(c) μ=wμλμ with wμλWλ longest in the double coset Wλ+ρwμλWμ+ρ in Wλ.

In the classical case, when 𝔤 is type An and V=L(ω1) is the n+1 dimensional fundamental representation the k-module λ/wμ is a simple k-module whenever it is nonzero (see [Suz1998]). As the following Proposition shows, this is a very special phenomenon.

Assume that V=L(ν) for a dominant integral weight ν. If the k-module λ/μ is irreducible (or 0) for all k, all dominant integral weights μ, and all integrally dominant weights λ then

(a) 𝔤 is type An, Bn, Cn or G2 and V=L(ω1), and
(b) the action of the subgroup k of k generates EndUh𝔤(Vk).

Proof.

(a) If μ is large dominant integral weight (for example, we may take μ=nρ, n0) then, as a Uh𝔤-module, L(μ)Vb L(μ+wt(b)), where the sum is over a basis of V consisting of weight vectors and wt(b) is the weight of the vector b. The group 1 is generated by the element Xε1 which acts on a summand L(λ) in L(μ)V by the constant qλ,λ+2ρ-μ,μ+2ρ-ν,ν+2ρ. Then Fλ(L(μ)) is the L(λ)-isotypic component of L(μ)V and these are simple 1 modules (for the various λ) only if all the values μ+wt(b),μ+wt(b)+2ρ -μ,μ+2ρ- ν,ν+2ρ=2 μ+ρ,wt(b) +wt(b),wt(b) -ν,ν+2ρ, (4.4) as b ranges over a weight basis of V, are distinct. It follows that all weight spaces of V must be one dimensional. This means that

(a) 𝔤 is type An, Bn, Cn, Dn, E6, E7 or G2 and V=L(ω1), or
(b) 𝔤 is type An and V=L(kω1) or V=(kωn) for some k, or
(c) 𝔤 is type Bn and V=L(ωn), or
(d) 𝔤 is type Dn and V=L(ωn-1) or V=L(ωn).
Most of the weights of these representations lie in a single W-orbit. If γ and γ are two distinct weights of V which are in the same W-orbit then γ,γ=γ,γ. If μ=nρ with n0 then the condition that all the values in (4.4) be distinct forces that 2(n+1)ρ,γ =2μ+ρ,γ 2μ+ρ,γ =2(n+1)ρ,γ. Writing γ=ν-iciαi and γ=ν-iciαi with ci,ci>0 the last equation becomes 2(n+1)·ici 2(n+1)·i ci. Finally, an easy case by case check verifies that the only choices of V in (a-d) above which satisfy this last condition for all weights in the W-orbit of the highest weight are those listed in the statement of the proposition.

(b) Let 𝒵k=EndUh𝔤(Vk). As a (Uh𝔤,𝒵k)-bimodule Vkλ L(λ)𝒵kλ, where 𝒵kλ is an irreducible 𝒵k-module and the sum is over all dominant integral weights for which the irreducible Uh𝔤-module L(λ) appears in Vk. By restriction 𝒵kλ is an k-module and this is the k-module Fλ(L(0)) which, by assumption, is simple. Since L(0) is the trivial module Xε1 acts on Fλ(L(0)) by the identity and so Fλ(L(0)) is simple as a k-module Thus the simple 𝒵k-modules in Vk coincide exactly with the simple k-modules in Vk and it follows that k generates 𝒵k=EndUh𝔤(Vk).

Jantzen filtrations for affine braid group representations

Applying the functor Fλ to the Jantzen filtration of M(μ) produces a filtration of λ/μ, λ/μ=Fλ (M(μ))= Fλ(M(μ)(0)) Fλ(M(μ)(1)) . (4.5) An argument of Suzuki [Suz1998, Thm. 4.3.5], shows that this filtration can be obtained directly from the k-contravariant form ,t on λ+tδ/μ+tδ= Fλ+tδ (M(μ+tδ))= (M(μ+tδ)Vk)λ+tδ[λ+tδ] which is the restriction of the Uh𝔤 contravariant form ,t on (M(μ+tδ)Vk), see (2.5) and (3.5)). To do this define λ+tδ/μ+tδ(j)= { m(λ+tδ)/μ+tδ |m,nt tj[t]for alln λ+tδ/μ+tδ } and (λ/μ)(j)= image ofλ+tδ/μ+tδ (j)in λ+tδ/μ+tδ [t][t]/ t[t] to obtain a filtration λ/μ= (λ/μ)(0) (λ/μ)(1) (4.6) such that the quotients (λ/μ)(j)/(λ/μ)(j+1) carry nondegenerate k contravariant forms. Since, for different λ, the subspaces (M(μ+tδ)Vk)λ+tδ[λ+tδ] are mutually orthogonal with respect to the Uh𝔤 contravariant form ,t on (M(μ+tδ)Vk), (M(μ+tδ)(j)Vk)λ+tδ[λ+tδ] (M(μ+tδ)Vk)λ+tδ[λ+tδ](j)= (λ+tδ)/(μ+tδ)(j). On the other hand, if u(M(μ+tδ)Vk)λ+tδ[λ+tδ](j) then write u=iaibi where aiM(μ+tδ) and bi is an orthonormal basis of Vk. Then, for all vM(μ+tδ), and all k, ak,vt=u,vbkttj[t] and so u(M(μ+tδ)(j)Vk)λ+tδ[λ+tδ]. So Fλ+tδ (M(μ+tδ))(j)= (λ+tδ)/(μ+tδ) (j) and the filtrations in (4.6) and (4.5) are identical.

Let λ and μ be integrally dominant weights and let w,yWμ be elements of maximal length in Wλ+ρwWμ+ρ and Wλ+ρyWμ+ρ respectively. Assume that the k modules λ/vμ, vWμ, are simple. Then multiplicities of λ/yμ in the filtration (4.5) are given by j0 [ (λ/wμ)(j) (λ/wμ)(j+1) :λ/(yμ) ] v12((y)-(w)-j) =Pwy(v). where Pwy(v) is the Kazhdan-Lusztig polynomial for the Weyl group Wμ.

Proof.

Since the functor Fλ is exact this result follows from the Beilinson-Bernstein theorem (2.6). The condition on y is necessary for the module λ/yμ to be nonzero.

The BGG resolution for affine braid groups

Let μ𝔥* be such that -(μ+ρ) is dominant and regular and let WJμ be a parabolic subgroup of the integral Weyl group Wμ. Let w0 be the longest element of WJμ and fix ν=w0μ. Following the method of [Che1987-2], applying the exact functor Fλ to the BGG resolution in (2.7) produces an exact sequence of k-modules 0𝒞N 𝒞1𝒞0 λ/ν0 (4.7) where 𝒞k=(w)=jλ/wν, and the sum is over all wWJμ of length j (in WJμ). Thus, in the Grothendieck group of the category 𝒪k of finite dimensional k-modules, [λ/ν]= wWJμ (-1)(w) [λ/(wν)] (4.8) where ν=w0μ and w0 is the longest element of WJμ. This identity is a generalization of the classical Jacobi-Trudi identity [Mac1995, I (5.4)] for expanding Schur functions in terms of homogeneous symmetric functions, sλ/ν=wSn (-1)(w) hλ+δ-w(ν+δ). (4.9)

Restriction of λ/μ to the braid group

The braid group k is the quotient of the affine braid group by the relation Xε1=1 and so the modules ν/0 are k-modules. The following proposition determines the structure of Fλ(L(μ)) as a k module when L(μ) is finite dimensional. This is a generalization of the Littlewood-Richardson rule.

Let P+ be the set of dominant integral weights. Define the tensor product multiplicities cμνλ, λ,μ,νP+, by the Uh𝔤-module decompositions L(μ)L(ν) λP+ L(λ)cμνλ. Then Reskk (λ/μ)= νP+ (ν/0)cμνλ.

Proof.

Let us abuse notation slightly and write sums instead of direct sums. Then, as a (Uh𝔤,k) bimodule L(μ)Vk= λL(λ) λ/μ, where λ/μ=Fλ(L(μ)). As a (Uh𝔤,k) bimodule L(μ)Vk= L(μ) (νL(ν)ν/0) =λ,νcμνλ L(λ)ν/0. Comparing coefficients of L(λ) in these two identities yields the formula in the statement.

Notes and references

This is a typed version of the paper Affine Braids, Markov Traces and the Category 𝒪 by Rosa Orellana and Arun Ram*.

*Research supported in part by National Science Foundation grant DMS-9971099, the National Security Agency and EPSRC grant GR K99015.

This paper is a slightly revised version of a preprint of 2001. We thank F. Goodman, A. Henderson and an anonymous referee for their very helpful comments on the original preprint.

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