Affine Braids, Markov Traces and the Category $𝒪$

Last update: 22 December 2013

The ${\stackrel{\sim }{ℬ}}_{k}$ Modules ${ℳ}^{\lambda /\mu }$ and ${ℒ}^{\lambda /\mu }$

Let $\lambda$ be integrally dominant and let $\mu \in {𝔥}^{*}\text{.}$ Define ${\stackrel{\sim }{ℬ}}_{k}$ modules $ℳλ/μ=Fλ (M(μ))and ℒλ/μ= Fλ(L(μ)). (4.1)$ The following lemma is the main tool for studying the structure of these ${\stackrel{\sim }{ℬ}}_{k}$ modules.

([Jan1980, Theorem 2.2], [Dix1996, Lemma 7.6.14]) Let $E$ be a finite dimensional ${U}_{h}𝔤$ module and let $\left\{{e}_{i}\right\}$ be a basis of $E$ consisting of weight vectors ordered so that $i if $\text{wt}\left({e}_{i}\right)<\text{wt}\left({e}_{j}\right)\text{.}$ Suppose $M$ is a ${U}_{h}𝔤$ module generated by a highest weight vector ${v}_{\mu }^{+}$ of weight $\mu \text{.}$ Set $Mi=∑j≥iUh 𝔫-(vμ+⊗ej).$ Then

 (a) $M\otimes E={M}_{1}\supseteq {M}_{2}\supseteq \cdots$ is a filtration of ${U}_{h}𝔤$ modules such that ${M}_{i}/{M}_{i+1}$ is $0$ or is a highest weight module of highest weight $\mu +\text{wt}\left({e}_{i}\right)\text{.}$ (b) If $M=M\left(\mu \right)$ then ${M}_{i}/{M}_{i+1}\cong M\left(\mu +\text{wt}\left({e}_{i}\right)\right)\text{.}$

The braid group ${ℬ}_{k}$ is the subgroup of ${\stackrel{\sim }{ℬ}}_{k}$ generated by ${T}_{1},\dots ,{T}_{k-1}\text{.}$ By restriction, both ${ℳ}^{\lambda /\mu }$ and ${V}^{\otimes k}=L\left(0\right)\otimes {V}^{\otimes k}$ are ${ℬ}_{k}$ modules.

There is a unique ${U}_{h}𝔤$ contravariant form ${⟨,⟩}_{M}$ on the Verma module $M\left(w\circ \mu \right)$ determined by ${⟨{v}_{w\circ \mu }^{+},{v}_{w\circ \mu }^{+}⟩}_{M}=1$ where ${v}_{w\circ \mu }^{+}$ is the generating highest weight vector of $M\left(w\circ \mu \right)\text{.}$ As in (2.15), this form together with a nondegenerate ${U}_{h}𝔤$ contravariant form ${⟨,⟩}_{V}$ on $V$ gives ${U}_{h}𝔤$ contravariant forms ${⟨,⟩}_{{V}^{\otimes k}}$ and $⟨,⟩$ on ${V}^{\otimes k}$ and $M\left(w\circ \mu \right)\otimes {V}^{\otimes k},$ respectively.

With these notations at hand we use Lemma 4.1 to prove the fundamental facts about the ${\stackrel{\sim }{ℬ}}_{k}$ modules ${ℳ}^{\lambda /\mu }$ and ${ℒ}^{\lambda /\mu }$ defined in (4.1).

Let $\lambda ,$ $\mu$ be integrally dominant weights and $w\in {W}^{\mu }\text{.}$

(a) As ${ℬ}_{k}$ modules, ${ℳ}^{\lambda /w\circ \mu }\cong {\left({V}^{\otimes k}\right)}_{\lambda -w\circ \mu }$
(b) ${ℳ}^{\lambda /w\circ \mu }\cong {ℳ}^{\lambda /y\circ \mu }$ if ${W}_{\lambda +\rho }w{W}_{\mu +\rho }={W}_{\lambda +\rho }y{W}_{\mu +\rho }\text{.}$
(c) Use the same notation $⟨,⟩$ for the ${U}_{h}𝔤$ contravariant form $⟨,⟩$ on $M\left(w\circ \mu \right)\otimes {V}^{\otimes k}$ and the ${\stackrel{\sim }{ℬ}}_{k}$ contravariant form on ${ℳ}^{\lambda /\left(w\circ \mu \right)}$ obtained by restriction of $⟨,⟩$ to the subspace ${\left(M\left(w\circ \mu \right)\otimes {V}^{\otimes k}\right)}_{\lambda }^{\left[\lambda \right]}\text{.}$ Then $ℒλ/w∘μ≅ ℳλ/(w∘μ) rad⟨,⟩ .$
(d) Assume $w\in {W}^{\mu }$ is maximal length in the coset $w{W}_{\mu +\rho }$ in ${W}^{\mu }\text{.}$ If ${ℒ}^{\lambda \left(w\circ \mu \right)}\ne 0$ then
 (1) $\lambda -w\circ \mu$ is a weight of ${V}^{\otimes k},$ (2) $w$ is maximal length in ${W}_{\lambda +\rho }w{W}_{\mu +\rho }\text{.}$
(e) If $\mu$ is a dominant integral weight then $ℒλ/μ≅ { v∈(V⊗k)λ-μ | Xi⟨μ+ρ,αi∨⟩ v=0, for all 1≤i≤n. } .$

 Proof. (a) Let ${v}_{w\circ \mu }^{+}$ be the generating highest weight vector of $M\left(w\circ \mu \right)$ and, for $n\in {V}^{\otimes k}$ let $pr\left({v}_{w\circ \mu }^{+}\otimes n\right)$ be the image of ${v}_{w\circ \mu }^{+}\otimes n$ in $\left(M\otimes {V}^{\otimes k}\right)/{𝔫}^{-}\left(M\otimes {V}^{\otimes k}\right)\text{.}$ Then, since $\lambda$ is integrally dominant, Lemma 4.1 shows that $(V⊗k)λ-w∘μ ⟶ ℳλ/(w∘μ) n ⟼ pr(vw∘μ+⊗n) (4.2)$ is a vector space isomorphism. This is a ${ℬ}_{k}\text{-module}$ isomorphism since the ${ℬ}_{k}$ action on $M\left(w\circ \mu \right)\otimes {V}^{\otimes k}$ commutes with ${𝔫}^{-}$ and fixes ${v}_{w\circ \mu }^{+}\text{.}$ (b) By (a), $\lambda -w\circ \mu \in P$ and so ${W}^{\lambda }={W}^{\mu }\text{.}$ It is sufficient to show that ${ℳ}^{\lambda /w\circ \mu }\cong {ℳ}^{\lambda /\left({s}_{i}w\circ \mu \right)}$ for all simple reflections of ${W}^{\lambda }$ such that ${s}_{i}\in {W}_{\lambda +\rho }$ and ${s}_{i}w>w\text{.}$ Applying the exact functor ${F}_{\lambda }$ to the Verma module inclusion $M(siw∘μ)↪ M(w∘μ)gives ℳλ/siw∘μ ↪ℳλ/w∘μ,$ an inclusion of ${\stackrel{\sim }{ℬ}}_{k}\text{-modules.}$ Since ${s}_{i}\left(\lambda -w\circ \mu \right)={s}_{i}\left(\lambda +\rho \right)-{s}_{i}w\left(\mu +\rho \right)=\lambda +\rho -{s}_{i}w\left(\mu +\rho \right)=\lambda -\left({s}_{i}w\right)\circ \mu$ there is a (vector space) isomorphism of weight spaces $(V⊗k)λ-w∘μ ≅Vλ-siw∘μ⊗k.$ (This isomorphism can be realized by Lusztigâ€™s braid group action [CPr1994, §8.1-8.2], ${T}_{i}:{\left({V}^{\otimes k}\right)}_{\lambda -w\circ \mu }\to {\left({V}^{\otimes k}\right)}_{{s}_{i}\left(\lambda -w\circ \mu \right)}\text{).}$ Thus, by part (a), the ${\stackrel{\sim }{ℬ}}_{k}\text{-module}$ inclusion ${ℳ}^{\lambda /{s}_{i}w\circ \mu }↪{ℳ}^{\lambda /w\circ \mu }$ is an isomorphism. (c) Use the notations for the bilinear forms on $M\left(w\circ \mu \right)$ and ${V}^{\otimes k}$ as given in the paragraph before the statement of the proposition. Let $\left\{{b}_{i}\right\}$ be an orthonormal basis of ${V}^{\otimes k}$ with respect to ${⟨,⟩}_{{V}^{\otimes k}}\text{.}$ If $r\in \text{rad}{⟨,⟩}_{M}$ then $⟨r⊗b,s⊗b′⟩ =⟨r,s⟩M ⟨b,b′⟩V⊗k =0, for all s∈M(w∘μ) , b,b′∈V⊗k,$ and so $\left(\text{rad}{⟨,⟩}_{M}\right){V}^{\otimes k}\subseteq \text{rad}⟨,⟩\text{.}$ Conversely, if ${r}_{i}\in M\left(w\circ \mu \right)$ such that $\sum {r}_{i}\otimes {b}_{i}\in \text{rad}⟨,⟩$ then $0=⟨∑iri⊗bi,s⊗bj⟩ =∑i⟨ri,s⟩M δij=⟨ri,s⟩, for all s∈M(w∘μ).$ So ${r}_{i}\in \text{rad}{⟨,⟩}_{M}$ and thus $\text{rad}⟨,⟩\subseteq \text{rad}{⟨,⟩}_{M}\otimes {V}^{\otimes k}\text{.}$ By the ${U}_{h}𝔤$ contravariance of $⟨,⟩$ $⟨M(w∘μ)⊗V⊗k⟩λ[λ]⊥ ⟨M(w∘μ)⊗V⊗k⟩μ[μ]$ for integrally dominant weights $\lambda ,$ $\mu$ with $\lambda \ne \mu \text{.}$ Thus $rad⟨,⟩= (rad⟨,⟩M⊗V⊗k)λ[λ], (4.3)$ where $⟨,⟩$ is the restriction of the form on $M\left(w\circ \mu \right)\otimes {V}^{\otimes k}$ to ${\left(M\left(w\circ \mu \right)\otimes {V}^{\otimes k}\right)}_{\lambda }^{\left[\lambda \right]}\text{.}$ Thus $(M(w∘μ)⊗V⊗k)λ[λ] rad⟨,⟩ = (M(w∘μ)⊗V⊗k)λ[λ] (rad⟨,⟩M⊗V⊗k)λ[λ] ≅ (M(w∘μ)rad⟨,⟩M⊗V⊗k)λ[λ] = (L(w∘μ)⊗V⊗k)λ[λ] = ℒλ/(w∘μ),$ where the isomorphism is a consequence of the fact that, because $\lambda$ is an integrally dominant weight, the functor ${\left(·\otimes {V}^{\otimes k}\right)}_{\lambda }^{\left[\lambda \right]}$ is exact (3.8). (d) If $\lambda -w\circ \mu$ is not a weight of ${V}^{\otimes k}$ then, by part (a), ${ℳ}^{\lambda /w\circ \mu }=0\text{.}$ Since the functor ${F}_{\lambda }$ is exact and $L\left(w\circ \mu \right)$ is a quotient of $M\left(w\circ \mu \right),$ ${ℒ}^{\lambda /w\circ \mu }$ is a quotient of ${ℳ}^{\lambda /w\circ \mu }\text{.}$ Thus ${ℳ}^{\lambda /w\circ \mu }=0$ implies ${ℒ}^{\lambda /w\circ \mu }=0\text{.}$ Assume that $w\in {W}^{\mu }$ is not the longest element of the double coset ${W}_{\lambda +\rho }w{W}_{\mu +\rho }\in {W}^{\mu }\text{.}$ Then there is a positive root $\alpha >0$ such that ${s}_{\alpha }\in {W}_{\lambda +\rho }$ and ${s}_{\alpha }w>w\text{.}$ Since ${s}_{\alpha }w{W}_{\mu +\rho }\ne w{W}_{\mu +\rho }$ there is an inclusion of Verma modules $M\left({s}_{\alpha }w\circ \mu \right)\subseteq M\left(w\circ \mu \right)$ and ${F}_{\lambda }\left(L\left(w\circ \mu \right)\right)$ is a quotient of ${F}_{\lambda }\left(M\left(w\circ \mu \right)\right)/{F}_{\lambda }\left(M\left({s}_{\alpha }w\circ \mu \right)\right)\text{.}$ On the other hand, by part (b), $ℳλ/sαw∘μ≅ ℳλ/w∘μ,and so ℳλ/w∘μ ℳλ/sαw∘μ = Fλ(M(w∘μ)) Fλ(M(sαw∘μ)) =0.$ Thus ${F}_{\lambda }\left(L\left(w\circ \mu \right)\right)=0\text{.}$ (e) When $\mu$ is a dominant integral weight $ℒλ/μ≅ ( span-{pr(vμ+⊗n) | n∈V⊗k} span-{pr(Yi⟨μ+ρ,αi∨⟩vμ+⊗n) | n∈V⊗k} ) λ .$ see [Dix1996, 7.2.7]. Thus, by (c) and the vector space isomorphism (4.2) it follows that, as vector spaces, $ℒλ/μ≅ ( (span-{pr(vμ+⊗n) | n∈V⊗k})/ (span-{pr(Yi⟨μ+ρ,αi∨⟩vμ+⊗n) | n∈V⊗k}) ) λ .$ For any $k\ge 0,$ there is a nonzero constant $c$ such that $pr\left({Y}_{i}^{k+1}{v}_{\mu }^{+}\otimes n\right)=c·pr\left({Y}_{i}\left({Y}_{i}^{k}{v}_{\mu }^{+}\otimes n\right)-{Y}_{i}^{k}{v}_{\mu }^{+}\otimes {Y}_{i}n\right)=-c pr\left({Y}_{i}^{k}{v}_{\mu }^{+}\otimes {Y}_{i}n\right),$ and so, by induction, $pr\left({Y}_{i}^{k+1}{v}_{\mu }^{+}\otimes n\right)=\xi ·pr\left({v}_{\mu }^{+}\otimes {Y}_{i}^{k+1}n\right)$ for some constant $\xi \ne 0\text{.}$ Thus ${ℒ}^{\lambda /\mu }$ is isomorphic to the vector space $(V⊗k/(∑iYi⟨μ+ρ,αi∨⟩V⊗k))λ-μ≅ (∑iYi⟨μ+ρ,αi∨⟩V⊗k)λ-μ⊥.$ If $v\in {\left({Y}_{i}^{⟨\mu +\rho ,{\alpha }_{i}^{\vee }⟩}{V}^{\otimes k}\right)}^{\perp }$ then the ${U}_{h}𝔤$ contravariance of ${⟨,⟩}_{{V}^{\otimes k}}$ gives that $0=⟨Yi⟨μ+ρ,αi∨⟩n,b⟩V⊗k= ⟨v,Xi⟨μ+ρ,αi∨⟩b⟩V⊗k, for all n∈V⊗k.$ Thus, by the nondegeneracy of ${⟨,⟩}_{{V}^{\otimes k}},$ $ℒλ/μ≅ (∑iYi⟨μ+ρ,αi∨⟩V⊗k)λ-μ⊥= { b∈(V⊗k)λ-μ | Xi⟨μ+ρ,αi∨⟩b=0 } .$ $\square$

Proposition 4.2d gives a necessary condition on $\lambda /w\circ \mu$ for the ${\stackrel{\sim }{ℬ}}_{k}\text{-module}$ ${ℒ}^{\lambda /w\circ \mu }$ to be nonzero. The following will be useful for analyzing the combinatorics of the examples in Section 6. If $P$ denotes the weight lattice $\lambda$ is an integrally dominant then the action of ${W}_{\lambda +\rho }$ on $\lambda -P$ by the dot action has fundamental domain $Cλ+ρ-= { μ∈λ-P | ⟨μ+ρ,α∨⟩ ∈ℤ≤0, for all α>0 such that ⟨λ+ρ,αi∨⟩=0 } .$ and the following are equivalent:

 (a) $\mu \in {C}_{\lambda +\rho }^{-},$ (b) $\mu ={w}^{\lambda }\circ \nu$ with $\nu$ integrally dominant and ${w}^{\lambda }\in {W}^{\lambda }$ longest in the coset ${W}_{\lambda +\rho }{w}^{\lambda }$ in ${W}^{\lambda },$ (c) $\mu ={w}_{\stackrel{\sim }{\mu }}^{\lambda }\circ \stackrel{\sim }{\mu }$ with ${w}_{\stackrel{\sim }{\mu }}^{\lambda }\in {W}^{\lambda }$ longest in the double coset ${W}_{\lambda +\rho }{w}_{\stackrel{\sim }{\mu }}^{\lambda }{W}_{\mu +\rho }$ in ${W}^{\lambda }\text{.}$

In the classical case, when $𝔤$ is type ${A}_{n}$ and $V=L\left({\omega }_{1}\right)$ is the $n+1$ dimensional fundamental representation the ${\stackrel{\sim }{ℬ}}_{k}\text{-module}$ ${ℒ}^{\lambda /w\circ \mu }$ is a simple ${\stackrel{\sim }{ℬ}}_{k}\text{-module}$ whenever it is nonzero (see [Suz1998]). As the following Proposition shows, this is a very special phenomenon.

Assume that $V=L\left(\nu \right)$ for a dominant integral weight $\nu \text{.}$ If the ${\stackrel{\sim }{ℬ}}_{k}\text{-module}$ ${ℒ}^{\lambda /\mu }$ is irreducible (or $0\text{)}$ for all $k,$ all dominant integral weights $\mu ,$ and all integrally dominant weights $\lambda$ then

 (a) $𝔤$ is type ${A}_{n},$ ${B}_{n},$ ${C}_{n}$ or ${G}_{2}$ and $V=L\left({\omega }_{1}\right),$ and (b) the action of the subgroup ${ℬ}_{k}$ of ${\stackrel{\sim }{ℬ}}_{k}$ generates ${\text{End}}_{{U}_{h}𝔤}\left({V}^{\otimes k}\right)\text{.}$

Proof.

(a) If $\mu$ is large dominant integral weight (for example, we may take $\mu =n\rho ,$ $n\gg 0\text{)}$ then, as a ${U}_{h}𝔤\text{-module,}$ $L(μ)⊗V≅⨁b L(μ+wt(b)),$ where the sum is over a basis of $V$ consisting of weight vectors and $\text{wt}\left(b\right)$ is the weight of the vector $b\text{.}$ The group ${\stackrel{\sim }{ℬ}}_{1}$ is generated by the element ${X}^{{\epsilon }_{1}}$ which acts on a summand $L\left(\lambda \right)$ in $L\left(\mu \right)\otimes V$ by the constant ${q}^{⟨\lambda ,\lambda +2\rho ⟩-⟨\mu ,\mu +2\rho ⟩-⟨\nu ,\nu +2\rho ⟩}\text{.}$ Then ${F}_{\lambda }\left(L\left(\mu \right)\right)$ is the $L\left(\lambda \right)\text{-isotypic}$ component of $L\left(\mu \right)\otimes V$ and these are simple ${\stackrel{\sim }{ℬ}}_{1}$ modules (for the various $\lambda \text{)}$ 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 ${A}_{n},$ ${B}_{n},$ ${C}_{n},$ ${D}_{n},$ ${E}_{6},$ ${E}_{7}$ or ${G}_{2}$ and $V=L\left({\omega }_{1}\right),$ or (b) $𝔤$ is type ${A}_{n}$ and $V=L\left(k{\omega }_{1}\right)$ or $V=\left(k{\omega }_{n}\right)$ for some $k,$ or (c) $𝔤$ is type ${B}_{n}$ and $V=L\left({\omega }_{n}\right),$ or (d) $𝔤$ is type ${D}_{n}$ and $V=L\left({\omega }_{n-1}\right)$ or $V=L\left({\omega }_{n}\right)\text{.}$
Most of the weights of these representations lie in a single $W\text{-orbit.}$ If $\gamma$ and $\gamma \prime$ are two distinct weights of $V$ which are in the same $W\text{-orbit}$ then $⟨\gamma ,\gamma ⟩=⟨\gamma \prime ,\gamma \prime ⟩\text{.}$ If $\mu =n\rho$ with $n\gg 0$ then the condition that all the values in (4.4) be distinct forces that $2(n+1)⟨ρ,γ⟩ =2⟨μ+ρ,γ⟩ ≠ 2⟨μ+ρ,γ′⟩ =2(n+1)⟨ρ,γ′⟩.$ Writing $\gamma =\nu -\sum _{i}{c}_{i}{\alpha }_{i}$ and $\gamma \prime =\nu -\sum _{i}{c}_{i}^{\prime }{\alpha }_{i}$ with ${c}_{i},{c}_{i}^{\prime }\in {ℤ}_{>0}$ 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 $W\text{-orbit}$ of the highest weight are those listed in the statement of the proposition.

(b) Let ${𝒵}_{k}={\text{End}}_{{U}_{h}𝔤}\left({V}^{\otimes k}\right)\text{.}$ As a $\left({U}_{h}𝔤,{𝒵}_{k}\right)\text{-bimodule}$ $V⊗k≅⨁λ L(λ)⊗𝒵kλ,$ where ${𝒵}_{k}^{\lambda }$ is an irreducible ${𝒵}_{k}\text{-module}$ and the sum is over all dominant integral weights for which the irreducible ${U}_{h}𝔤\text{-module}$ $L\left(\lambda \right)$ appears in ${V}^{\otimes k}\text{.}$ By restriction ${𝒵}_{k}^{\lambda }$ is an ${ℬ}_{k}\text{-module}$ and this is the ${\stackrel{\sim }{ℬ}}_{k}\text{-module}$ ${F}_{\lambda }\left(L\left(0\right)\right)$ which, by assumption, is simple. Since $L\left(0\right)$ is the trivial module ${X}^{{\epsilon }_{1}}$ acts on ${F}_{\lambda }\left(L\left(0\right)\right)$ by the identity and so ${F}_{\lambda }\left(L\left(0\right)\right)$ is simple as a ${ℬ}_{k}\text{-module}$ Thus the simple ${𝒵}_{k}\text{-modules}$ in ${V}^{\otimes k}$ coincide exactly with the simple ${ℬ}_{k}\text{-modules}$ in ${V}^{\otimes k}$ and it follows that ${ℬ}_{k}$ generates ${𝒵}_{k}={\text{End}}_{{U}_{h}𝔤}\left({V}^{\otimes k}\right)\text{.}$

$\square$

Jantzen filtrations for affine braid group representations

Applying the functor ${F}_{\lambda }$ to the Jantzen filtration of $M\left(\mu \right)$ produces a filtration of ${ℳ}^{\lambda /\mu },$ $ℳλ/μ=Fλ (M(μ))= Fλ(M(μ)(0)) ⊇Fλ(M(μ)(1)) ⊇⋯. (4.5)$ An argument of Suzuki [Suz1998, Thm. 4.3.5], shows that this filtration can be obtained directly from the ${\stackrel{\sim }{ℬ}}_{k}\text{-contravariant}$ form ${⟨,⟩}_{t}$ on $ℳλ+tδ/μ+tδ= Fλ+tδ (M(μ+tδ))= (M(μ+tδ)⊗V⊗k)λ+tδ[λ+tδ]$ which is the restriction of the ${U}_{h}𝔤$ contravariant form ${⟨,⟩}_{t}$ on $\left(M\left(\mu +t\delta \right)\otimes {V}^{\otimes k}\right),$ see (2.5) and (3.5)). To do this define $ℳλ+tδ/μ+tδ(j)= { m∈ℳ(λ+tδ)/μ+tδ | ⟨m,n⟩t ∈tjℂ[t] for all n∈ ℳλ+tδ/μ+tδ }$ and $(ℳλ/μ)(j)= image of ℳλ+tδ/μ+tδ (j) in ℳλ+tδ/μ+tδ ⊗ℂ[t]ℂ[t]/ tℂ[t]$ to obtain a filtration $ℳλ/μ= (ℳλ/μ)(0)⊇ (ℳλ/μ)(1)⊇ ⋯ (4.6)$ such that the quotients ${\left({ℳ}^{\lambda /\mu }\right)}^{\left(j\right)}/{\left({ℳ}^{\lambda /\mu }\right)}^{\left(j+1\right)}$ carry nondegenerate ${\stackrel{\sim }{ℬ}}_{k}$ contravariant forms. Since, for different $\lambda ,$ the subspaces ${\left(M\left(\mu +t\delta \right)\otimes {V}^{\otimes k}\right)}_{\lambda +t\delta }^{\left[\lambda +t\delta \right]}$ are mutually orthogonal with respect to the ${U}_{h}𝔤$ contravariant form ${⟨,⟩}_{t}$ on $\left(M\left(\mu +t\delta \right)\otimes {V}^{\otimes k}\right),$ $(M(μ+tδ)(j)⊗V⊗k)λ+tδ[λ+tδ] ⊆(M(μ+tδ)⊗V⊗k)λ+tδ[λ+tδ](j)= ℳ(λ+tδ)/(μ+tδ)(j).$ On the other hand, if $u\in {\left(M\left(\mu +t\delta \right)\otimes {V}^{\otimes k}\right)}_{\lambda +t\delta }^{\left[\lambda +t\delta \right]}\left(j\right)$ then write $u=\sum _{i}{a}_{i}\otimes {b}_{i}$ where ${a}_{i}\in M\left(\mu +t\delta \right)$ and ${b}_{i}$ is an orthonormal basis of ${V}^{\otimes k}\text{.}$ Then, for all $v\in M\left(\mu +t\delta \right),$ and all $k,$ ${⟨{a}_{k},v⟩}_{t}={⟨u,v\otimes {b}_{k}⟩}_{t}\in {t}^{j}ℂ\left[t\right]$ and so $u\in {\left(M\left(\mu +t\delta \right)\left(j\right)\otimes {V}^{\otimes k}\right)}_{\lambda +t\delta }^{\left[\lambda +t\delta \right]}\text{.}$ So $Fλ+tδ (M(μ+tδ))(j)= ℳ(λ+tδ)/(μ+tδ) (j)$ and the filtrations in (4.6) and (4.5) are identical.

Let $\lambda$ and $\mu$ be integrally dominant weights and let $w,y\in {W}^{\mu }$ be elements of maximal length in ${W}_{\lambda +\rho }w{W}_{\mu +\rho }$ and ${W}_{\lambda +\rho }y{W}_{\mu +\rho }$ respectively. Assume that the ${\stackrel{\sim }{ℬ}}_{k}$ modules ${ℒ}^{\lambda /v\circ \mu },$ $v\in {W}^{\mu },$ are simple. Then multiplicities of ${ℒ}^{\lambda /y\circ \mu }$ in the filtration (4.5) are given by $∑j≥0 [ (ℳλ/w∘μ)(j) (ℳλ/w∘μ)(j+1) :ℒλ/(y∘μ) ] v12(ℓ(y)-ℓ(w)-j) =Pwy(v).$ where ${P}_{wy}\left(v\right)$ is the Kazhdan-Lusztig polynomial for the Weyl group ${W}^{\mu }\text{.}$

 Proof. Since the functor ${F}_{\lambda }$ is exact this result follows from the Beilinson-Bernstein theorem (2.6). The condition on $y$ is necessary for the module ${ℒ}^{\lambda /y\circ \mu }$ to be nonzero. $\square$

The BGG resolution for affine braid groups

Let $\mu \in {𝔥}^{*}$ be such that $-\left(\mu +\rho \right)$ is dominant and regular and let ${W}_{J}^{\mu }$ be a parabolic subgroup of the integral Weyl group ${W}^{\mu }\text{.}$ Let ${w}_{0}$ be the longest element of ${W}_{J}^{\mu }$ and fix $\nu ={w}_{0}\circ \mu \text{.}$ Following the method of [Che1987-2], applying the exact functor ${F}_{\lambda }$ to the BGG resolution in (2.7) produces an exact sequence of ${\stackrel{\sim }{ℬ}}_{k}\text{-modules}$ $0⟶𝒞N⟶⋯⟶ 𝒞1⟶𝒞0⟶ ℒλ/ν⟶0 (4.7)$ where ${𝒞}_{k}=\underset{\ell \left(w\right)=j}{⨁}{ℳ}^{\lambda /w\circ \nu },$ and the sum is over all $w\in {W}_{J}^{\mu }$ of length $j$ (in ${W}_{J}^{\mu }\text{).}$ Thus, in the Grothendieck group of the category ${\stackrel{\sim }{𝒪}}_{k}$ of finite dimensional ${\stackrel{\sim }{ℬ}}_{k}\text{-modules,}$ $[ℒλ/ν]= ∑w∈WJμ (-1)ℓ(w) [ℳλ/(w∘ν)] (4.8)$ where $\nu ={w}_{0}\circ \mu$ and ${w}_{0}$ is the longest element of ${W}_{J}^{\mu }\text{.}$ 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, $sλ/ν=∑w∈Sn (-1)ℓ(w) hλ+δ-w(ν+δ). (4.9)$

Restriction of ${ℒ}^{\lambda /\mu }$ to the braid group

The braid group ${ℬ}_{k}$ is the quotient of the affine braid group by the relation ${X}^{{\epsilon }_{1}}=1$ and so the modules ${ℒ}^{\nu /0}$ are ${ℬ}_{k}\text{-modules.}$ The following proposition determines the structure of ${F}_{\lambda }\left(L\left(\mu \right)\right)$ as a ${ℬ}_{k}$ module when $L\left(\mu \right)$ 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 ${c}_{\mu \nu }^{\lambda },$ $\lambda ,\mu ,\nu \in {P}^{+},$ by the ${U}_{h}𝔤\text{-module}$ decompositions $L(μ)⊗L(ν)≅ ⨁λ∈P+ L(λ)⊕cμνλ.$ Then $Resℬkℬ∼k (ℒλ/μ)= ⨁ν∈P+ (ℒν/0)⊕cμνλ.$

 Proof. Let us abuse notation slightly and write sums instead of direct sums. Then, as a $\left({U}_{h}𝔤,{ℬ}_{k}\right)$ bimodule $L(μ)⊗V⊗k= ∑λL(λ)⊗ ℒλ/μ,$ where ${ℒ}^{\lambda /\mu }={F}_{\lambda }\left(L\left(\mu \right)\right)\text{.}$ As a $\left({U}_{h}𝔤,{ℬ}_{k}\right)$ bimodule $L(μ)⊗V⊗k= L(μ)⊗ (∑νL(ν)⊗ℒν/0) =∑λ,νcμνλ L(λ)⊗ℒν/0.$ Comparing coefficients of $L\left(\lambda \right)$ in these two identities yields the formula in the statement. $\square$

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.