Translation Functors and the Shapovalov Determinant

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia

Last updated: 10 February 2015

This is an excerpt from the PhD thesis Translation Functors and the Shapovalov Determinant by Emilie Wiesner, University of Wisconsin-Madison, 2005.

Quantum Groups

We now consider quantum groups associated to semisimple Lie algebras 𝔤. The quantum group Uq(𝔤) is a deformation of the universal enveloping algebra U(𝔤) of 𝔤, and it possesses many properties that 𝔤 does. In particular, we can use the Shapovalov determinant to study translation functors for Uq(𝔤). The background information on quantum groups presented in this chapter can be found in [CPr1994].

Let q be an indeterminate. For integers m,n,d, we define the following elements of [q,q-1]:

[m]d!=[m]d[m-1]d[2]d[1]d for m0
[mn]d=[m]d![n]d![m-n]d! for mn>0.
Let 𝔤 be a finite-dimensional complex semisimple Lie algebra, and let A=(aij)1i,jn be the associated Cartan matrix. Choose di>0, 1ik minimal so that the matrix (diaij)1i,jn is symmetric. To simplify notation, we will write qi=qdi and [·]i=[·]di. There is some k>0 so that di=kαi,αi2. For βR+, define dβ=kβ,β2. Suppose β=wαi for wW and αi a simple root. Equation 1.2 implies dβ=di.

The quantum group Uq(𝔤) associated to 𝔤 is the (q)-algebra generated by the elements Ei, Fi, and Ki±1 (1ik) with relations KiKj=KjKi, KiKi-1=1=Ki-1Ki; (4.1) KiEj=qiaij EjKi,KiFj =qi-aijFj Ki; (4.2) EiFj-FjEi= δi,j Ki-Ki-1 qi-qi-1 ; (4.3) r=01-aij (-1)r [1-aijr]i Ei(1-aij)-r EjEir=0, (4.4) r=01-aij (-1)r [1-aijr]i Fi(1-aij)-r FjFir=0for ij. (4.5)

The quantum group Uq(𝔤) is a Hopf algebra with the following maps:

coproduct: Δ(Ki)=KiKi, Δ(Ei)=EiKi-1+1Ei, Δ(Fi)=Fi1+KiFi;
counit: ε(Ki)=1, ε(Ei)=0=ε(Fi);
antipode: S(Ki)=Ki-1, S(Ei)=-EiKi, S(Fi)=-Ki-1Fi.
(See ([CPr1994] 9.1.1) for a proof that these maps give a Hopf algebra structure.) We define the following (q)-subalgebras of Uq(𝔤):
Uq+(𝔤) is the subalgebra generated by {Ei|1ik}.
Uq0(𝔤) is the subalgebra generated by {Ki±1|1ik}.
Uq-(𝔤) is the subalgebra generated by {Fi|1ik}.

([CPr1994] 9.1.3). Uq(𝔤)Uq-(𝔤)Uq0(𝔤)Uq+(𝔤).

There is a natural pairing between Uq+(𝔤) and Uq-(𝔤). The following maps make use of this pairing. Define a -anti-automorphism τ:Uq(𝔤)Uq(𝔤) by τ(Ei)=Fi; τ(Fi)=Ei; τ(Ki)=Ki-1; τ(q)=q-1. Define a (q)-anti-automorphism ϕ:Uq(𝔤)Uq(𝔤) by ϕ(Ei)=Ki-1Fi; ϕ(Fi)=EiKi; ϕ(Ki)=Ki. Observe that ϕ commutes with the coproduct map: Δϕ= (ϕϕ)Δ. This property will be useful later when we consider tensor products of modules.

The Restricted Integral Form U𝒜res(𝔤)

Let 𝒜=[q,q-1]. The restricted integral form U𝒜res(𝔤) of Uq(𝔤) is the 𝒜-subalgebra of Uq(𝔤) generated by Ki±1; Ei(r)=Eir[r]i!; Fi(r)=Fir[r]i! for1ik,r>0. The algebra U𝒜res(𝔤) inherits many properties from Uq(𝔤), including the Hopf algebra structure and the triangular decomposition. The restricted integral form U𝒜res(𝔤) is a Hopf algebra with the following maps

coproduct: Δ(Ki) = KiKi Δ(Ei(r)) = j=0r qij(r-j) Ei(j)Ki-j Ei(r-j) Δ(Fi(r)) = j=0r qi-j(r-j) Fi(j) Kir-j Fi(r-j).
counit: ε(Ki)=1; ε(Ei(r))=0=ε(Fi(r));
antipode: S(Ki)=Ki-1; S(Ei(r))=(-1)𝔯 qr(r-1)Ei(r)Kir; S(Fi(r))=(-1)(r)q-r(r-1)Ki-rFi(r).

([Lus1990-4] 6.6). The algebra U𝒜res(𝔤) decomposes as U𝒜res(𝔤) U𝒜res(𝔤)+ U𝒜res(𝔤)0 U𝒜res(𝔤)- where

U𝒜res(𝔤)+=Uq+(𝔤)U𝒜res(𝔤) is generated by {Ei(r)|1ik,r>0};
(U𝒜res(𝔤))0=Uq0(𝔤)U𝒜res(𝔤) is generated by { Ki±1, [Ki;cr] |1ik,c,r>0 } , with [Ki;cr]= s=1r Kiqic+1-s-Ki-1qis-c-1 qis-qi-s ;
(U𝒜res(𝔤))-=Uq-(𝔤)U𝒜res(𝔤) is generated by {Fi(r)|1ik,r>0}.

A PBW Basis for U𝒜res(𝔤)

In the following construction (due to [Lus1990-4]), a braid-group action on U𝒜res(𝔤) is used to define the element of U𝒜res(𝔤)± corresponding to each positive root. These elements are then assembled into a PBW basis for U𝒜res(𝔤)±.

([Lus1990-4] 3.1). For each 1in, there is a unique (q)-automorphism Ti:U𝒜res(𝔤)U𝒜res(𝔤) defined by Ti(Kj) = KjKi-aij; Ti(Ej) = { -FiKi ifi=j r=0-aij (-1)r-aij qi-r Ei(-aij-r) EjEi(r) ifij ; Ti(Fj) = { -Ki-1Ei ifi=j r=0-aij (-1)r-aij qirFi(r) FjFi(-aij-r) ifij

Fix a reduced expression for the longest element of the Weyl group, w0=si1siN. This gives an ordering of the positive roots: βj=si1 sij-1αij, 1jN. Define FβjU𝒜res(𝔤)-Uq-(𝔤) to be Fβj=Ti1 Tij-1Fj. We similarly define Eβj. Finally, define Kβj=Ti1 Tij-1Kj= Ki1Kim where βj=αi1++αim.

([Lus1990-4], 6.7). The sets { Eβn(sn) Eβ1(s1) |si0 } and { Fβ1(s1) FβN(sN) |si0 } are bases for U𝒜res(𝔤)+ and U𝒜res(𝔤)-, respectively.

These sets are also bases for Uq+(𝔤) and Uq-(𝔤), respectively. Note that Ti commutes with the -antiautomorphism ϕ, and so we have ϕ(Eβ)=Fβ.

The Category 𝒪

We begin by defining weights and weight spaces. A weight of Uq(𝔤) is an element ΩHom(q)(Uq0(𝔤),(q)). Since the set {Ki±1|1ik} generates Uq0(𝔤) and Ω(Ki-1)=(Ω(Ki))-1, Ω is determined by its action on each of the Ki. Define Ωi=Ω(Ki) and identify Ω with the k-tuple (Ω1,,Ωk).

We view 𝔥* as sitting inside Hom(q)(Uq0(𝔤),(q)): for ω𝔥*, ω(Ki)=qω,αi. We will also be interested in certain "twistings" of the weights of 𝔤. Let σ=(σ1,,σk){(±1,,±1)} and ω𝔥*. Define ωσHom(q)(Uq0(𝔤),(q)) by ωσ(Ki)=σi qiω,αi =±qiω,αi.

The dominance ordering on 𝔥* induces a partial ordering on the weights of Uq(𝔤). For Ω,ΛHom(q)(Uq0(𝔤),(q)), ΩΛif there is some νQ+such that Λ(Ki)Ω(Ki) =qν,αi for1ik. In this case we write ΩΛ=ν. For ω,λ𝔥*, this is the standard dominance ordering.

For a Uq(𝔤) module M, the Ω-weight space of M is given by MΩ= { vM|Kiv= Ω(Ki)v for1ik } .

The Category 𝒪 consists of Uq(𝔤)-modules M such that

M is Uq(𝔤)0-diagonalizable: M=ΩHom(q)(Uq0(𝔤),(q)) MΩ;
there are Λ1,,ΛnHom(q)(Uq0(𝔤),(q)) such that MΩ0 only if ΩΛi for some 1in.
dimMΩ< for all ΩHom(q)(Uq0(𝔤),(q)).

A highest weight Uq(𝔤)-module of highest weight Ω is a Uq(𝔤)-module M with a vector v+M such that

Eiv+=0 for all 1ik.
The vector v+ is a highest weight vector. For such a module M the defining relations of Uq(𝔤) imply M=ΛΩMΛ. Note that all highest weight modules are in Category 𝒪.

The Verma module of highest weight Ω, is the Uq(𝔤)-module Mq(Ω)=Uq (𝔤)/I, where I is the left ideal generated by Ei and Ki-Ω(Ki) (1ik). We can also define Mq(Ω) as the induced module Mq(Ω)= Uq(𝔤) Uq0(𝔤) (q)Ω, where (q)Ω is (q) as a vector space; and, as a Uq0(𝔤)-module, Uq0(𝔤) acts by Ω and Uq+(𝔤) acts by 0. Using the same arguments as in Lemma 2.1.3, we can show that Mq(Ω) has a maximal proper submodule, Jq(Ω). This implies that Mq(Ω) has a unique simple quotient Lq(Ω)= Mq(Ω)/ Jq(Ω).

All of these objects can be similarly defined for U𝒜res(𝔤). The set of weights for U𝒜res(𝔤) is Hom𝒜(U𝒜res(𝔤)0,𝒜). We denote the U𝒜res(𝔤)-Verma module of weight ΩHom𝒜(U𝒜res(𝔤)0,𝒜) by Mres(Ω). In other words, Mres(Ω)=U𝒜res(𝔤)/J, where J is the ideal generated by Ei(r),r>0; Ki-Ω(Ki); [Ki;cr]- s=1r Ω(Ki)qic+1-s-Ω(Ki-1)qis-c-1qis-qi-s.

From Theorem 4.2.2, we get the following.

Let Ω,ΛHom(q)(Uq0(𝔤),(q)) so that ΩΛ. Suppose ΩΛ=νQ+. Define Fβj(tj)=1[tj]!Fβjtj. Then, the set {Fβ1(t1)FβN(tN)v+|t1,,tN0,t1β1++tNβN=ν} is a basis for Mres(Ω)Λ and Mq(Ω)Λ.

Embeddings of Verma Modules

Recall that for a weight ω𝔥* of 𝔤 and σ=(σ1,,σk){(±1,,±1)}, we can define a Uq(𝔤)-weight ωσHom(q)(Uq0(𝔤),(q)) by ωσ(Ki)=σiqiω,αi.

Let λP, σ{(±1,,±1)}, and βR+. If sβλλ, then Mq((sβλ)σ)Mq(λσ).

The proof of this proposition follows the proof of the same result for the classical case given in [Dix1996].

Let λP and σ{(±1,,±1)}. If siλλ, Mq((siλ)σ)Mq(λσ). Moreover, if v+ generates Mq(λσ), then Fimv+ is a highest weight vector of weight (siλ)σ, where m=λ+ρ,αi0.


Note that Fimv+Mq(λσ)(siλ)σ. We must also check that EjFimv+=0 for all 1jk. If ij, EjFimv+=FimEjv+=0. If i=j, we can use Equation 4.3 to show EiFimv+ = σi[m]i [λ,αi-m+1]i Fim-1v+ = σi[m]i [λ+ρ,αi-m]i Fim-1v+ = 0.

Let λP+, σ{(±1,,±1)}, and wW with reduced expression w=sinsi1. Set λ0=λ, λ1=si1λ0, , λn=sinλn-1. Then, Mq(λnσ) Mq(λn-1σ) Mq(λ0σ).


By definition, λj+1=sij+1λi. Therefore, using the previous lemma, we need only check that λj+ρ,αj+10 to show that Mq(λj+1σ)Mq(λjσ). Note that λj+ρ,αj+1= wj(λ+ρ),αj+1= λ+ρ,wj-1αj+1, where wj=sijsi1. Then wj-1αj+1R+. (See [Dix1996] 11.1.8.) Because λP+, λ+ρ,wj-1αj+1>0.

Let λ,μP, and σ{(±1,,±1)}. Suppose αi is a simple root such that Mq((siμ)σ)Mq(μσ)Mq(λσ). Then, Mq((siμ)σ)Mq((siλ)σ).

This gives the following picture, where arrows indicate inclusion: M(λσ) M((siλ)σ) M(μσ) M((siμ)σ) A

The picture indicates nothing about the relationship between Mq(λσ) and Mq((siλ)σ). The proof will consider both possibilities: Mq(λσ)Mq((siλ)σ) and Mq((siλ)σ).Mq(λσ).


Let m=μ+ρ,αi0 and n=λ+ρ,αi. If n0, then we have Mq((siμ)σ) Mq(μσ) Mq(λσ) Mq((siλ)σ), i.e. we have the following picture of embeddings. Mq((siλ)σ) Mq(λσ) Mq(μσ) Mq((siμ)σ) Then the lemma follows.

Suppose n>0. Then Mq((siλ)σ)Mq(λσ). Let v+,v+Mq(λσ) be generators for Mq(λσ) and Mq(μσ) respectively. Then, Fimv+ generates Mq((siμ)σ), and Finv+ generates Mq((siλ)σ). Therefore, Mq((siμ)σ)Mq((siλ)σ) if Fimv+Mq((siλ)σ).

We first show that there is some l0 such that Filv+Mq((siλ)σ). In other words, we want to show that Filv+=XFinv+ for some XUq(𝔤). We have v+=uv+ for some uUq-(𝔤). Since uUq-(𝔤), u is a (q)-linear combination of elements of the form Fi1Fij where αi1++αij=λ-μ. From the generating relations for Uq(𝔤), we have Fi1-aij Fj = -r=11-aij (-1)r[1-aijr] Fi1-aij-r FjFir = ( r=0-aij (-1)r[1-aijr+1]i Fi-aij-r FjFir ) Fi for ij. Thus, Fit(1-aij)Fj=XFit for some XUq-(𝔤). Given the form of u, this implies Filv+Mq((siλ)σ) for some l0.

We claim that this implies that Fimv+Mq((siλ)σ):

If lm, then Fimv+=Fim-lFilv+Mq((siλ)σ).
If l>m, then EiFilv+=[l][μ,α-l+1]Fil-1v+. Since EiFilv+Mq((siλ)σ), Fimv+Mq((siλ)σ).
In this case, we have the following picture: Mq(λσ) Mq((siλ)σ) Mq(μσ) Mq((siμ)σ)

We now proceed with the proof of Proposition 4.4.1.

Proof of Proposition 4.4.1.

Let μ=sβλ and m=λ+ρ,β0. If m=0, μ=λ. Therefore, assume m>0. Let μP+(Wμ) and choose wW so that μ=wμ. Write λ=w-1λ. Fix a reduced expression w=sinsi1, and set λ0=λ, λ1=si1λ0, , λn=sinλn-1=λ; μ0=μ, μ1=si1μ0, , μn=sinμn-1=μ. Note that μ=w-1sβwλ=sw-1βλ. Therefore μj=s(sijsi1w-1β)λj. Since si sends αi to -αi and permutes the other positive roots, (sijsi1w-1)β=γjR+ or (sijsi1w-1)β=-γj-R+. Therefore, μj=sγjλj=s-γjλj for some γjR+.

Since μ0P+ and λ0Wμ0, we must have μ0-λ0Q+. On the other hand, μn-λn=-mβ. Let j be minimal such that μj-λjQ+ and μj+1-λj+1-Q+. Now, μj-λj= sij+1 μj+1- sij+1 λj+1= sij+1 (μj+1-λj+1) and μj+1-λj+1= sγj+1 λj+1-λj+1 =-λj+1+ρ,γj+1 γj+1. This implies that sij+1γj+1R- and so γj+1=αj+1.

We now have μj+1=sij+1λj+1 with μj+1λj+1. Lemma 4.4.2 implies Mq(μj+1σ)Mq(λj+1σ). From Lemma 4.4.3, we have Mq(μj+2σ)Mq(μj+1σ). Since Mq(μj+1σ)Mq(λj+1σ), we can use the last lemma to conclude Mq(μj+2σ)Mq(λj+2σ). Repeating this process, we get Mq(μσ)Mq(λσ). The picture below describes the various embeddings used in the proof: Mq(μσ) Mq(λj+2σ) Mq(λj+1σ) Mq(μj+1σ) Mq(λσ) Mq(μj+2σ) Mq(μσ)

Central characters

We now construct the central characters of Uq(𝔤), following ([CPr1994] 9.2.B). Denote the center of Uq(𝔤) by Zq(𝔤)= { XUq(𝔤) |XY=YX for allYUq(𝔤) } . For zZq(𝔤), we can write z= Fβ1(t1) FβN(tN) K(s1,,sN)(t1,,tN) EβN(sN) Eβ1(s1) where K(s1,,sN)(t1,,tN)Uq0(𝔤) and s1,,sN,t1,,tN0 with s1β1++sNβN=t1β1++tNβN. Define hq:Zq(𝔤)Uq0(𝔤) by hq(z)= K(0)(0).

The map hq:Zq(𝔤)Uq0(𝔤) is an algebra homomorphism.


Let zZq(𝔤) with z=Fβ1(t1)FβN(tN)K(s1,,sN)(t1,,tN)EβN(sN)Eβ1(s1), and let v+ be a generator for the Verma module Mq(Λ) (ΛHom(q)(Uq0(𝔤),(q))). Then, z(v+)=(K(0)(0))v+=Λ(hq(z))v+. For z,zZq(𝔤), Λ(hq(zz))v+= zz(v+)= zΛ(hq(z))v+= Λ(hq(z))Λ(hq(z))v+. Therefore, Λ(hq(zz))=Λ(hq(z))Λ(hq(z)) for all Λ, with implies hq(zz)=hq(z)hq(z).

The map hq is the quantum analogue of the Harish-Chandra homomorphism for semisimple Lie algebras (see [Dix1996], Section 7.4).

For ΛHom(q)(Uq0(𝔤),(q)), we define the central character χΛ:Zq(𝔤) by χΛ(z)= Λ(hq(z)). Let M be a highest weight Uq(𝔤)-module of highest weight Λ, and let v+ be a generator of M. Then, zv+=χΛ(z)v+. Since any vM is of the form v=xv+ for some xUq(𝔤), zv= zxv+= xzv+= xzv+= χΛ(z)xv+= χΛ(z)v.

Recall that the dot action of the Weyl group W on 𝔥* is wμ=w(μ+ρ)-ρ.

([CPr1994], 9.1.8). Let λ,μP, integral weights of 𝔤. Then χλ=χμ if and only if λ=wμ for some wW.

The Shapovalov Determinant

Recall τ:Uq(𝔤)Uq(𝔤) is the -anti-automorphism given by τ(Ei)=Fi; τ(Fi)=Ei; τ(Ki)=Ki-1; τ(q)=q-1. Using this map, we can define a Hermitian bilinear form ,:Mq(Ω)×Mq(Ω)(q) by

v+,v+=1, where v+ is the generator of Mq(Ω);
xu,w=u,τ(x)w for xUq and u,wMq(Ω).

Let Ω,ΛHom(q)(Uq0(𝔤),(q)) with ΩΛ. Suppose ΩΛ=νQ+. We define A(Ω)Λ= (Fβ1(t1)FβN(tN)v+,Fβ1(s1)FβN(sN)v+) t1β1++tNβN=ν =s1β1++sNβN and det(Mq(Ω)Λ)= detA(Ω)Λ. If Ω,ΛHom𝒜(Uq0(𝔤),𝒜), we define detMres(Ω)Λ=det(Mq(Ω)Λ).

Example. Suppose 𝔤=sl2(). Let k0 and Ω,ΛHom(q)(Uq0(sl2()),(q)) with ΩΛ=kα. Now consider det(Mq(Ω)Λ).

Equation 4.3 implies EFjv+=[j]q-j+1ΩK-qj-1Ω(K)-1q-q-1Fj-1v+. Therefore, detMq(Ω)Λ = Fk[k]!v+, Fk[k]!v+ = v+, Ek[k]! Fk[k]!v+ = 1([k]!)2 v+,j=1k [j] qj-1K-q-J+1K-1q-q-1 v+ = 1([k]!) j=1k q-j+1Ω(K)-qj-1Ω(K-1)q-q-1 Theorem 4.5.2 gives a formula for detMq(Ω)Λ for all 𝔤.

Recall that a weight Ω is equivalent to a k-tuple (Ω1,,Ωk)((q))k. We can view the determinant of Mq(Ω)Λ as living inside the Laurent polynomial ring S=(q)[Ωi±1|1ik]. To prove Theorem 4.5.2, we will bound the degree of the polynomial and then find its irreducible factors. The degree D:Sk is given by D(Ω1j1Ωkjk)=(j1,,jk). Also, define the following partial ordering on k: (j1,,jk) (l1,,lk) ifjili for all1ik.

Let ΩΛ with ΩΛ=νQ+. The term with maximum degree in detMq(Ω)Λ is βR+ r>0 ΩβP(ν-rβ). Similarly, the term with minimal degree is βR+ r>0 Ωβ-P(ν-rβ).


Consider an entry entry in A(Ω)Λ: Fβ1(t1) FβN(tN)v+, Fβ1(s1) FβN(sN)v+ = v+, EβN(tN) Eβ1(t1) Fβ1(s1) FβN(sN)v+ (4.6) where t1β1++tnβn=ν=s1β1++snβn. Then, both Fβ1(t1)Fβn(tn) and Fβ1(s1)Fβn(sn) can be written as linear combinations of Fi1Fim, where ν=αi1++αim for some choice of (not necessarily distinct) simple roots. Therefore, Equation 4.3 implies that the maximum possible degree term in (4.6) is Ωi1Ωim= Ωβ1t1Ωβntn= Ωβ1s1Ωβnsn (4.7) This means the term with maximum degree in A(Ω)Λ is (t)|t1β1++tnβn=ν Ωβ1t1Ωβnt1. (4.8) The power of Ωβj in (4.8) is (t)|t1β1++tnβn=ω-λtj. We rewrite this as (t)|t1β1++tnβn=ν tj=r>0 {(t)|tjr}= r>0P(ν-rβj). To understand the first equality, note that each partition in the second sum gets counted tj times. To understand the second equality, note that each partition of ν-rβj maps uniquely to a partition with tjr.

We determine the minimum degree with an analogous argument.

([DKa1992]). Let ΩΛHom(q)(Uq0(𝔤),(q)) with ΩΛ=νQ+. Then, det(Mq(Ω)Λ)= βR+r>0 ( 1[r]dβ [qdβ(ρ,β-r)Ω(Kβ)-q-dβ(ρ,β-r)Ω(Kβ-1)qdβ-q-dβ] ) P(ν-rβ) .

If we restrict the choice of weights to ω,λ𝔥*Hom(q)(Uq0(𝔤),(q)), this formula becomes det(Mq(ω)λ)= βR+ r>0 ([ω+ρ,β-r]dβ[r]dβ) P(ω-λ-rβ) . (4.9)

A Partial Proof.

Note that qdβ(ρ,β-r)Ω(Kβ)-q-dβ(ρ,β-r)Ω(Kβ-1)qdβ-q-dβ can be written as qdβ(ρ,β-r) Ω(Kβ)-1 (Ω(Kβ)-q-dβ(ρ,β-r)) (Ω(Kβ)+q-dβ(ρ,β-r)) qdβ-q-dβ . We now show that (Ω(Kβ)±q-dβ(ρ,β-r))p(ν-rβ) |detMq(Ω)Λ. Suppose that ω0P is such that ω0+ρ,β=r, and let σ{(±1,,±1)}. Then, sβω0=ω0-rβ and so from Proposition 4.4.1, Mq((ω0-rβ)σ)Mq(ω0σ). Therefore, Mq(ω0σ)(ω0-rβ)σRad,0. This implies that det(Mq(ω0σ)(ω0-rβ)σ)=0, i.e. det(Mq(Ω)Ω), where Ω=Ω-rβ, has a zero at Ωβ=(ω0σ)β= ±qω0,β= ±qr-ρ,β. As in the proof of Theorem 3.4.3, this means there is vr,βMq(Ω)Ω such that

vr,β=(t)|t1β1++tnβn=rβa(t)Fβ1(t1)Fβn(tn)v+ for some a(t)(q) with a(t)0 for at least one (t);
(Ωβ±qr-ρ,β)|vr,β,w for all wMq(Ω).
Now, the set B={Fβ1(s1)FβN(sN)(t)|t1β1++tnβn=rβa(t)Fβ1(t1)Fβn(tn)|s1β1++sNβN=ν-rβ} is linearly independent in Uq-(𝔤) and can be extended to a linearly independent set B which spans span{Fβ1(s1)FβN(sN)|s1β1++sNβN=ν}. Let P be the matrix taking {Fβ1(s1)FβN(sN)|s1β1++sNβN=ν} to B.

Define det(Mq(Ω)Λ)=det(Xiv+,Xjv+)Xi,XjB. We have

(Ωβ±qr-ρ,β) divides Fβ1(s1)FβN(sN)vr,β,w.
Therefore, (Ω(Kpt)±q-dβ(ρ,β-r))p(ν-rβ) divides detBMq(Ω)Λ. Finally, det(Mq(Ω)Λ)=det(Pt)detB(Mq(Ω)Λ)det(P), which implies that (Ω(Kβ)±q-dβ(ρ,β-r))p(ν-rβ) divides detBMq(Ω)Λ.

It only remains to check the coefficient of the highest degree term in detMq(Ω)Λ. We omit this part of the proof, but the original argument can be found in [DKa1992], 1.9.

An Alternative Determinant Formula

Recall that DeConcini and Kac [DKa1992] use the -antiautomorphism τ:UqUq defined by τ(Ei)=Fi, τ(Fi)=Ei, τ(Ki)=Ki-1, τ(q)=q-1. Instead, we will use the (q)-antiautomorphism ϕ given by ϕ(Ei)=Ki-1Fi, ϕ(Fi)=EiKi, ϕ(Ki)=Ki. This defines a new contravariant form. In this section, we will write ,τ for the contravariant form induced by τ and ,ϕ for the contravariant form induced by ϕ. Define detϕ(Mq(ω)λ)=det (Fβ1(t1)FβN(tN)v+,Fβ1(s1)FβN(sN)v+ϕ) t1β1++tNβN=ν =s1β1++sNβN .

Let ω,λP with λω. Then, detϕ(Mq(ω)λ)= qP(ω-λ)kλω Cω-λ βR+ r>0 ([ω+ρ,β-r]dβ[r]dβ)P(ω-λ-rβ), where 0Cω-λ(q) and kλω is given as follows. Write ω-λ=αi1++αim. Then kλω= ω,αi1++αim- ( αi1++αim,αi1+ αi2++αim,αi2++ αim,αim ) .


Let P be the matrix which takes Fβ1(t1)FβN(tN)v+,Fβ1(s1)FβN(sN)v+τ to Fβ1(t1)FβN(tN)v+,Fβ1(s1)FβN(sN)v+ϕ. Then, detϕ(Mq(ω)λ)=det(P)detτ(Mq(ω)λ). Therefore, we need only find det(P). To do this, note that since the set {Fi1Fimv+|αi1++αim=ω-λ} spans Mq(ω)λ, there is a subset B which is a basis for the space. Therefore, we can write a standard basis element Fβ1(t1)FβN(tN)v+ as a linear combination of elements of B: Fβ1(t1) FβN(tN)v+ =Fi1Fimv+B ai1,,imFi1 Fimv+ where ai1,,im(q). Then, Fβ1(t1)FβN(tN)v+,Fβ1(s1)FβN(sN)v+ϕ = ai1,,im v+,EimKim Ei1Ki1 Fβ1s1 Fβnsnv+ = ai1,,im qkλω v+,Eim Ei1 Fβ1s1 Fβnsnv+ where the sum is over αi1,,αim such that Fi1FimB and αi1++αim=ω-λ; and where kλω = ω-λ,αi1+ ω-λ+αi1,αi2++ ω-λ+(αi1++αim),αim = ω,αi1++αim- jk αij,αik. (Since αij,αik=αik,αij, kλω is independent of the order of the simple roots αi1,,αim.)

On the other hand, Fβ1(t1)FβN(tN)v+,Fβ1(s1)FβN(sN)v+τ= αi1++αim=ω-λ ai1,,im v+,EimEi1Fβ1s1Fβnsnv+, where ·:(q)(q) is the -automorphism taking q to q-1.

Therefore, P=qkλωAA-1, where A is the matrix taking the basis B to the standard basis {Fβ1(t1)FβN(tN)v+|t1β1++tNβN=ω-λ} of Mq(ω)λ. Then, det(P)= qP(ω-λ)kλω det(A)det(A-1)= qP(ω-λ)kλω det(A)det(A). Since A is a matrix taking one basis to another, det(A)0. Therefore, detϕ(Mq(ω)λ)= qP(ω-λ)kλω Cω-λdetτ (Mq(ω)λ), where Cω-λ=det(A)/det(A).

Translation Functors

For λP, define the block [λ] by [λ]={wλ|wW}.

Let ω,ω0P. Then Mq(ω) Lq(ω0)= λP+ (Mq(ω)Lq(ω0))[λ], where (Mq(ω)Lq(ω0))[λ] has composition factors Lq(μ) only for μ[λ].


Note that the module Mq(ω)Lq(ω0) decomposes as a direct sum according to central characters. Theorem 4.4.6 implies that this decomposition corresponds to the decomposition by blocks stated in the proposition.

We again consider the translation Mq(ω) (Mq(ω)Lq(ω0))[λ]. When possible, we will state our results in the more general setting of the restricted integral form.

For μQ+, define n(μ)=dim(Lres(ω0)ω0-μ), and let {wμ,i|1in(μ)} be a basis for Lres(ω0)(ω0-μ).

For λQ+, the sets 1= { Fβ1(t1) Fβn(tn)v+ wμ,i| t1β1++tnβn =λ-μ,1in(μ) } and 2= { Fβ1(t1) Fβn(tn) (v+wμ,i) | t1β1++tnβn =λ-μ,1in(μ) } are bases for (Mres(ω)Lres(ω0))(ω+ω0-λ). Moreover, the transition matrix between the two bases has determinant one.


The set 1 is clearly a basis for (Mres(ω)Lres(ω0))(ω+ω0-λ). Since 2(Mres(ω)Lres(ω0))(ω+ω0-λ), there is a matrix A taking 1 to 2.

Let Fβ1(t1)Fβn(tn)(vwμ,i)2.

Since U𝒜res(𝔤)- is a Hopf algebra, Δ(Fβ1(t1)Fβn(tn))= Fβ1(t1)Fβn(tn)1+ as1,,sNr1,,rN ( Fβ1(s1)Fβn(tN) Ks1,,sNr1,,rN ) (Fβ1(r1)Fβn(rn)), where as1,,sNr1,,rN𝒜, Ks1,,sNr1,,rNU𝒜res(𝔤)0, and the sum is over s1,,sN,r1,,rN0 such that r1β1++rNβN<t1β1++tNβN and (r1+s1)β1++(rn+sN)βN=t1β1++tNβN. Therefore, Fβ1(t1)Fβn(tn) (v+wμ,i)= Fβ1(t1)Fβn(tn) vwμ,i+ ν>μ,(s)=ν-μ,1jn(ν) aν,j(s) Fβ1(s1)Fβn(sn) v+wν,j for some aν,j(s)𝒜.

Therefore, we can order vectors in such a way that the matrix A taking Fβ1(t1)Fβn(tn)v+wμ,i to Fβ1(t1)Fβn(tn)(v+wμ,i) is upper-triangular with ones on the diagonal. Then, A is invertible with determinant one.

By possibly reordering the bases, we may assume A=A. Therefore, A has entries in 𝒜. Since A is upper-triangular with ones on the diagonal, (A)-1 will also have entries in 𝒜. This proves that 2 is an 𝒜-basis for (Mres(ω)Lres(ω0))(ω+ω0-λ).

Let ω,ω0P and λ,μQ+. Define detLres(ω0)ω0-μ= det(wμ,i,wμ,j)1i,jdimLres(ω0)ω0-μ and det((Mres(ω)Lres(ω0))ω+ω0-λ)= det(Fβ1(t1)FβN(tN)v+wμ,i,Fβ1(s1)FβN(sN)v+wμ,j), where μ,μλ, 1in(μ), 1jn(μ), and s1,,sN,t1,,tN0 are such that s1β1++sNβN=λ-μ and t1β1++tNβN=λ-μ.

Let ω,ω0P. For λQ+, det((Mres(ω)Lres(ω0))ω+ω0-λ) is given by μQ+μλ (det(Mres(ω)ω-(λ-μ)))n(μ) (det(Lres(ω0)ω0-μ))P(λ-μ). (4.10)


Note that Mres(ω)ν1Lres(ω0)μ1 is orthogonal (with respect to the contravariant form) to Mres(ω)ν1Lres(ω0)μ1 for ν1ν2. As in the proof of Lemma 2.6.3, det((Mres(ω)Lres(ω0))ω+ω0-λ) = μQ+μλ det(Mres(ω)ω-(λ-μ)Lres(ω0)ω0-μ) = μQ+μλ (det(Mres(ω)ω-(λ-μ))n(μ)) (det(Lres(ω0)ω0-μ))P(λ-μ).

Using the basis 2 for (Mres(ω)Lres(ω0))ω+ω0-λ, we see that Mres(ω)Lres(ω0) has a filtration by Verma modules Mres(ω+ω0-μ) where the Verma modules Mres(ω+ω0-μ) is generated by the image of v+wμ,i. The following form of the determinant formula better reflects the filtration of Mq(ω)Lq(ω0) by Verma modules.

Let ω,ω0P and λQ+. Then, det((Mres(ω)Lres(ω0))ω+ω0-λ) = μQ+μλ det(Mres(ω+ω0-μ)ω+ω0-λ)n(μ) aμω0(ω) det(Lres(ω0)ω0-μ)P(λ-μ) where aμω0(ω)= qRμβR+,r>0 ([ω+ρ,β-r][ω+ω0-μ+rβ+ρ,β-r])n(μ-rβ) and Rμ is defined inductively by Rμ=-ν<μ,n(ν)0 P(μ-ν) (ω0-ν,αi1++αimn(ν)+Rν), for μ-ν=αi1++αim.


Note that Rμ is always an integer and μλRμ P(λ-μ) = -μλ kω-λ+μω- kω+ω0-λω+ω0-μ P(λ-μ)n(μ) = -μλ ω0-μ,αi1++αim P(λ-μ)n(μ).

From Lemma 4.6.3, det((Mres(ω)Lres(ω0))ω+ω0-λ) is given by = μQ+μλ det(Lres(ω0)ω0-μ)P(λ-μ) qn(μ)P(λ-μ)kω-(λ-μ)ω μQ+μλ Cλ-μβR+,r>0 (([ω+ρ,β-r][r])P(λ-μ-rβ))n(μ) = μQ+μλ det(Lres(ω0)ω0-μ)P(λ-μ) μQ+μλ (qP(λ-μ)(kω+ω0-λω+ω0-μ+(kω-λ+μω-kω+ω0-λω+ω0-μ)))n(μ) = μQ+μλ βR+,r>0 ([ω+ρ,β-r][ω+ω0-μ+ρ,β-r])P(λ-μ-rβ)n(μ)) * μQ+μλ Cλ-μβR+,r>0 ([ω+ω0-μ+ρ,β-r][r])P(λ-μ-rβ)n(μ)) = μQ+μλ det(Lres(ω0)ω0-μ)P(λ-μ) μQ+μλ q(kω-λ+μω-kω+ω0-λω+ω0-μ)P(λ-μ)n(μ) μQ+μλ βR+,r>0 ([ω+ρ,β-r][ω+ω0-μ+rβ+ρ,β-r])P(λ-μ)n(μ-rβ) We sendμμ-rβin*. SinceP(λ-μ)=0forμ>λand n(μ)=0forμ<0, this does not affect the index of the product. μQ+μλ det(Mres(ω+ω0-μ)ω+ω0-λ)n(μ) = μQ+μλ (aμdet(Lres(ω0)ω0-μ))P(λ-μ) det(Mres(ω+ω0-μ)ω+ω0-λ)n(μ)

(Mq(ω)Lq(ω0))[λ](Mq(ω)Lq(ω0))[μ] for [μ][λ].


The arguments from the proof of Proposition 2.6.4 can be applied here.

Let ω,ω0P and let λQ+. Suppose ω is such that there is no μQ+ with 0<μλ and ω-μ[ω]. Then for each block γ, we can define projection maps Prλ[γ]: Mq(ω)Lq(ω0) ((Mq(ω)Lq(ω0))[γ])ω+ω0-λ given by Prλ[γ](v)=v- [ν][γ] 1m v,vi[ν] vi[ν] where {v1[ν],,vm[ν]} and {v1[ν],,vm[ν]} are dual bases for ((Mq(ω)Lq(ω0))[ν])ω+ω0-λ.


Since there is no μQ+ with 0<μλ and ω-μ[ω], detMq(ω)ω-λ0. Lemma 4.6.3 then implies the contravariant form is nondegenerate on (Mq(ω)Lq(ω0))ω+ω0-λ. We know from Lemma 4.6.5 that distinct blocks are orthogonal. This means that the contravariant form is nondegenerate on ((Mq(ω)Lq(ω0))[γ])ω+ω0-λ.

Let {v1[ν],,vm[ν]} be a basis for ((Mq(ω)Lq(ω0))[ν])ω+ω0-λ. Since the contravariant form is nondegenerate on this space, there is a basis {v1[ν],,vm[ν]} of ((Mq(ω)Lq(ω0))[ν])ω+ω0-λ such that vi[ν],vl[ν]=δi,l.

We define a map Prλ[γ]: (Mq(ω)Lq(ω))ω+ω0-λ ((Mq(ω)Lq(ω))[γ])ω+ω0-λ by Prλ[γ](v)= v-[ν][γ] 1mv,vi[ν] vi[ν]. Note that Prλ[γ](v),vi[ν]=0 whenever [ν][γ]. Therefore, Prλ[γ](v) ((Mq(ω)Lq(ω))[γ])ω+ω0-λ.

Also, for v((Mq(ω)Lq(ω))[γ])ω+ω0-λ, Prλ[γ](v)=v.

Note: We can inductively construct a basis for each block of Mq(ω)Lq(ω0) using a basis for Mres(ω)Lres(ω0). Elements of this basis will be (q)-linear combinations of the basis for Mres(ω)Lres(ω0). Therefore, we have a basis for each block of Mq(ω)Lq(ω0) such that a nonzero multiple of this basis is contained in Mres(ω)Lres(ω0).

Fix ω,ω0P and λQ+. Suppose ω is such that there is no wW with ωwωω-λ or ω+ω0w(ω+ω0)ω+ω0-λ.

Then the submodule of Mq(ω)Lq(ω0) generated by μQ+,μλ (Mq(ω)Lq(ω0))ω+ω0-μ is isomorphic to μQ+,μλ Mq(ω+ω0-μ)dimLq(ω0)ω0-μ. For a suitable choice of generating highest weight vectors {vμ,i+|1idim(Lq(ω0))ω0-μ}, this sum is orthogonal with respect to the contravariant form on Mq(ω)Lq(ω0), and 1idim(L(ω0))ω0-μ vμ,i,vμ,i= aμω0(ω)det L(ω0)ω0-μ


Since there is no wW with ωwωω-λ, the projection maps from the previous lemma are well-defined.

We have assumed there is no wW such that ω+ω0w(ω+ω0)ω+ω0-λ. Therefore, there is only one weight in [ω+ω0-μ] between ω+ω0 and ω+ω0-λ, namely ω+ω0-μ. This implies Prμ[μ](v+wμ,i) is a highest weight vector. Choose vectors {vμ,i+} such that

the transition matrix from {Prμ[μ](v+wμ,i)} to {vμ,i+} has determinant 1;
vμ,i+,vμ,k+=0 if ik.
Therefore, we only need to determine ( ivμ,i+,vμ,i+ ) We do this inductively. Assume vν,i+,vν,l+= det(Prν[ν](v+wν,i)mPrν[ν](v+wν,l)) =aνω0(ω)det Lq(ω0)ω0-ν for νP+ with ν<μ.

Note that det ( Fβ1(t1) Fβn(tn) Prν[ν] (v+wν,i), Fβ1(s1) Fβn(sn) Prν[ν] (v+wν,k) ) (t),s i,k is given by = i ( Fβ1(t1) Fβn(tn) vν,i+, Fβ1(s1) Fβn(sn) vν,i+ ) (t),(s) = (ivν,i+,vν,i+)p(μ-ν) (detMq(ω+ω0-ν)ω+ω0-μ)n(ν). Since distinct blocks are orthogonal, we have det(Mq(ω)Lq(ω0))ω+ω0-μ is νμdet ( Fβ1(t1) Fβn(tn) Prν[ν] (v+wν,i), Fβ1(s1) Fβn(sn) Prν[ν] (v+wν,k) ) = ν<μ (detMq(ω+ω0-ν)ω+ω0-μ)n(ν) ×(aνω0(ω)detLq(ω0)ω0-ν) det(Prμ[μ](v+wμ,i),Prμ[μ](v+wμ,l)). This implies det(Prμ[μ](v+wμ,i),Prμ[μ](v+wμ,l))=aμω0(ω)detLq(ω0)ω0-μ.

Notes and References

This is an excerpt from the PhD thesis Translation Functors and the Shapovalov Determinant by Emilie Wiesner, University of Wisconsin-Madison, 2005.

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