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.

Lie Algebras with Triangular Decomposition

Lie algebras with triangular decomposition generalize important properties of the decomposition of semisimple Lie algebras given in Equation 1.1. Our study of the Virasoro algebra (Chapter 3) relies on its Hermitian triangular decomposition.

The material in this chapter through Section 2.4 can be found in [MPi1995].

Let 𝔤 be a Lie algebra over . Then 𝔤 has a triangular decomposition if

𝔤=𝔤<0𝔥𝔤>0, where 𝔤 is abelian and 𝔤>0, 𝔤<00;
𝔤>0=αQ+𝔤α where Q+𝔥*=Hom(𝔥,) is a free additive semigroup and 𝔤α= { x𝔤| [h,x]=α (h)xfor all h𝔥 } ;
there is a basis {αi|iI} of Q+ of linearly independent elements of 𝔥*, i.e. γQ+ can be written γ=iImiαi for mi0;
there is a linear anti-involution σ:𝔤𝔤 such that σ(𝔤>0)=𝔤<0 and σ|𝔥=id.
We also wish to consider Lie algebras with maps σ which are Hermitian (see section 1.3.1). Therefore, we say a Lie algebra 𝔤 has an Hermitian triangular decomposition if there is a Hermitian anti-involution σ:𝔤𝔤 and a Lie algebra 𝔤 over the real numbers such that
𝔤 is a Lie algebra with triangular decomposition with anti-involution σ|𝔤.
In general, the results in this chapter for Lie algebras with a triangular decomposition hold for Lie algebras with an Hermitian triangular decomposition. The exception is in defining a contravariant form (section 2.4). Therefore, we will only distinguish between the two types of Lie algebras in that section.

Suppose 𝔤 is a Lie algebra with a triangular decomposition. If x𝔤α for some αQ+, then σ(x)𝔤-α since [h,σ(a)]= [σ(h),σ(x)]= σ([x,h])= σ(-α(h)x)=- α(h)σ(x). Because σ(𝔤>0)=𝔤<0, we have 𝔤<0=αQ+𝔤-α.

The roots of 𝔤 are those 0α𝔥* such that 𝔤α0. An element 0x𝔤 is a root vector if x𝔤α for some root α. Then 𝔤<0 has a basis of root vectors. Fix an index set R for a basis of root vectors {xj|jR} of 𝔤<0 and a total order on R. Let βjQ+ so that xj𝔤-βj.

For a partition (t):t1tj>0, write x(t)=xtj xt1. Then Proposition 1.1.1 can be restated for 𝔤 as follows.

(See Proposition 1.1.1). Let 𝔤 be a Lie algebra with a triangular decomposition as described above. Then U(𝔤)U(𝔤<0)U(𝔥)U(𝔤>0), where

U(𝔥)S(𝔥), the symmetric algebra of 𝔥;
U(𝔤<0) has a basis {x(t)|(t):t1tj>0};
U(𝔤>0) has a basis {σ(x(t))|(t):t1tj>0}.

Let Q=-span{αi|iI}𝔥*. Define the height function ht:Q by ht(iImiαi) =iImi.

A triangular decomposition is regular if the weight spaces of 𝔤>0 are finite-dimensional. It is finite if the root lattice Q+ is finitely generated. All highest weight modules will be in Category 𝒪 only if 𝔤 has a regular triangular decomposition. The proof that Category 𝒪 decomposes into blocks will rely on 𝔤 having a finite triangular decomposition.

The Category 𝒪

Let 𝔤 be a Lie algebra with a regular triangular decomposition. Let M be a 𝔤-module and μ𝔥*. The μ-weight space of M is Mμ= {mM|hm=μ(h)mfor allh𝔥}. If Mμ0, then μ is a weight of M. The dominance order is the partial ordering on the set of weights 𝔥* given by λμifλ-μ Q+forλ,μ 𝔥*. The Category 𝒪 is the set of 𝔤-modules M such that

M is 𝔥-diagonalizable: M=μ𝔥*Mμ;
there exist λ1,,λn𝔥* such that Mμ0 only if μλi for some 1in;
for all μ𝔥*, dimMμ<.

([MPi1995] 2.6.2). Let M and N be modules in Category 𝒪. Then MN and MN are in Category 𝒪.

Within Category 𝒪, we pay special attention to highest weight modules. A 𝔤-module M is a highest weight module of highest weight λ𝔥* if there is vMλ such that v generates M, and 𝔤>0v=0. Such a vector is a highest weight vector. If M is a highest weight module of highest weight λ, then M=μλMμ.

Among highest weight modules, Verma modules are of particular interest. Let U(𝔤) be the universal enveloping algebra of 𝔤. The Verma module of weight λ𝔥* is the 𝔤-module M(λ)=U(𝔤)/I where I is the left U(𝔤)-ideal generated by 𝔤>0 and {h-λ(h)|h𝔥}. Equivalently, we can define the Verma module M(λ) as the induced module M(λ)=U(𝔤) U(𝔤>0𝔥) λ, where λ is the (𝔤>0𝔤)-module with action xC=0 and hC=λ(h)C for C, x𝔤>0, and h𝔥. We will write v+ for a choice of a highest weight vector of M(λ). (This choice is unique up to scalars.)

Let λ𝔥*. Then M(λ)𝒪, and M(λ)=μQ+M(λ)λ-μ. The weight space M(λ)λ-μ has a basis {x(t)v+|(t):t1tj>0,μ=βt1++βtj}.


As U(𝔤<0)-modules, M(λ)U(𝔤<0). Therefore, Proposition 2.0.5 implies that M(λ)=μλM(λ)μ and M(λ)μ has a basis {x(t)v+|(t):t1tj>0,λ-μ=βt1++βtj}.

Since 𝔤 has a regular triangular decomposition, dimM(λ)μ<. Therefore, M(λ)𝒪.

Since Verma modules are free U(𝔤<0)-modules, every highest weight module is the quotient of a Verma module. (If M is a highest weight module generated by vMλ, we can define a surjective homomorphism M(λ)M by v+v.) Therefore, the previous result implies that all highest weight modules are in 𝒪.

Let M𝒪 be a finitely-generated 𝔤-module. Then M has a maximal proper submodule. In particular, for λ𝔥*, M(λ) has a unique maximal proper submodule J(λ).


Since every element of M is a finite sum of weight vectors, M is generated by a finite sum of weight vectors. Let {m1,,mn} be a minimal generating set of weight vectors for M. Assume miMγi. Consider the set of submodules 𝒦={KM|mnK}. This set is nonempty (since {0}𝒦) and partially ordered by inclusion. We claim that this set has a maximal element. To prove this claim, let K1K2 be an increasing chain of submodules in 𝒦, and define K=iKi. Then K is a submodule of M.

Since dimMγn<, there is some k>0 so that Kiγn=Ki+1γn for all ik. Since mnKi for all i>0, this implies mnK. Therefore, K𝒦, and so every increasing chain in 𝒦 has a maximal element. Zorn's Lemma implies that 𝒦 has a maximal element K. By construction, any submodule M of M which properly contains K is not in 𝒦 and so must contain mn. Because we have chosen the generating set {m1,,mn} to be minimal, the submodule generated by mi (in) is contained in 𝒦, and so miKM for 1in-1. This implies M=M. Therefore, if K is a maximal proper submodule.

Any maximal proper submodule of M must not contain at least one of the generators of M. Since M(λ) is generated by one element, this implies M(λ) has a unique maximal proper submodule.

This result implies M(X) has a unique simple quotient L(λ)=M(λ)/ J(λ).

Up to isomorphism, the simple modules in Category 𝒪 are L(λ), λ𝔥*.


Let L𝒪 be a simple module. Since L𝒪, we can choose λ𝔥* so that Lλ0 and Lμ=0 for all μ>λ. Let 0vLλ. Then v is a highest weight vector, and so there is a map ϕ:M(λ)L given by ϕ:v+v. Since L is irreducible, ker(ϕ)=J(λ).

To show that the simple modules L(λ) are distinct, suppose λ,μ𝔥* with λμ. We may assume λμ. Therefore, L(μ)λ=0 and so L(μ)L(λ).

The following result is an analogue of the Krull-Schmidt theorem for groups (see [Isa1994], 10.9).

Let P𝒪 be finitely generated. Then P is a finite direct sum of indecomposable submodules which are also finitely generated.


Let {vi}i=1n be a generating set for P. Since P𝒪, every element of P can be written as a finite sum of weight vectors. Therefore, we assume viPνi for some νi𝔥*.

Suppose P can be written as a finite direct sum of nonzero submodules, P=P1Pm. Write vi=vi1++vim where vijPj. Then {vij}i=1n generates Pj. This implies that for each j there is at least one 1in such that vij0. Therefore, mi,jdim Pjνi=i dimPνi< since P𝒪. Therefore, we can continue to decompose P until the summands Pi are indecomposable; since m is bounded, the number of terms in this decomposition will be finite.


Let [𝔥*]={a1eλ1++aneλn|ai,λi𝔥*} be the group algebra of 𝔥*, and let [𝔥*]*={aλeλ|λ𝔥*,aλ} be the set of formal infinite sums. For M𝒪, the character of M, chM[𝔥*]*, is chM=λ𝔥* dimMλeλ. (2.1) For M,N𝒪, the product (chM)(chN) is well-defined and ch(MN)= (chM) (chN).

Special Filtrations in Category 𝒪

As we noted above, highest weight modules, and specifically Verma modules and irreducible highest weight modules, are special modules in Category 𝒪. Therefore, we would like to analyze other modules in Category 𝒪 by means of these modules. The following filtrations provide a way to do this. The concepts of local and Verma composition series lead us to define the decomposition numbers [M:L(λ)] and [M:M(λ)]. Highest weight series provide an important tool in proving the decomposition of Category 𝒪 into blocks (Theorem 2.3.12).

Local Composition Series

Let M𝒪. A composition series for M is a finite sequence of submodules M=MnMn-1M0=0 such that each factor Mi/Mi-1, 1in, is irreducible. Not all modules in Category 𝒪 have a composition series. Therefore, we use a more localized idea of a composition series. Let M𝒪 and λ𝔥*. A local composition series for M at λ is a finite sequence of submodules M=MnMn-1M0=0 and a subset I{1,,n} such that

for each iI, there is λiλ so that Mi/Mi-1L(λi);
for iI, (Mi/Mi-1)μ=0 for all μλ.

Example. Consider the Verma module M(0,1) for the Virasoro algebra. As we see in Chapter 3, this module has a unique chain of nested distinct submodules, which is infinite: M(0,1) M(1,1) M(n2,1). Therefore, M(0,1) does not have a composition series. However, M(0,1) does have a local composition series at λ for any λ𝔥*2. For example,

For λ=(h,c) with c1 or h0, the series M(0,1)0 is a local composition series at λ.
For λ=(h,1) with h>0, the series M(0,1)M(1,1)M(n2,1), where n0 so that n2h, is a local composition series at λ.

Let M𝒪 and λ𝔥*. Then M has a local composition series at λ.


We prove this claim by inducting on μλdimMμ. (This sum is always finite since dimMμ< and there are a finite number of weights between any two weights.)

If μλdimMμ=0, then M0 is a local composition series at λ. Suppose μλdimMμ=n, and assume the result holds for all modules M𝒪 with μλdim(M)μ<n. Choose μ𝔥* maximal so that Mμ0 and μλ. Let 0vMμ. By the maximality of μ, v must be a highest weight vector. Let V=U(𝔤)v and let U be the maximal submodule of V. Then V/UL(μ).

We have μλdim(M/V)μ<n and μλdimUμ<μλdimVμn. Therefore, by the inductive hypothesis, both M/V and U have local composition series at λ: M/V=VkV1 V0=0 U=UlU1 U0=0. For each 0ik, let VMiM be a submodule of M so that Mi/VVi. Then, M=MkM0= VU=Ul U0=0 is a local composition series for M at λ.

([MPi1995] 2.6.9). Let M𝒪 and λ,μ𝔥*. Suppose that M has a local composition series at λ, M=MkM0, and at μ, M=MlM0. Then there exist series M=MkM0, a refinement of {Mi}, and M=MlM0, a refinement of {Mj}, such that k=l and the two series have isomorphic factors.

For M𝒪 and μ𝔥*, define [M:L(μ)] to be the number of occurrences of L(μ) in a local composition series for M at μ.

Verma Composition Series

Let M𝒪. A Verma composition series for M is a finite sequence of submodules M=MnMn-1M0=0 such that each factor Mi/Mi-1, 1in, is a Verma module.

Let M𝒪. Then any two Verma composition series for M have isomorphic factors.


Let M = MnMn-1 M0=0 M = Mm Mm-1 M0=0 be Verma composition series for M with Mi/Mi-1M(λi), λi𝔥* and Mi/Mi-1M(μj), μj𝔥*. Then, i=1n chM(λi)= chM= j=1m chM(μj). (2.2) Choose λk maximal in {λ1,,λm}. This implies the coefficient of eλk in (2.2) is 1 and so μl=λi for some l. Therefore, ikchM(λi)=jlchM(μj). Induct on m to get the result.

Suppose M𝒪 has a Verma composition series. For μ𝔥*, define [M:M(μ)] to be the number of occurrences of M(μ) in a Verma composition series of M.

Highest Weight Series

Let M𝒪. Then a highest weight series for M is an increasing filtration of submodules 0=M0M1 such that

Mi/Mi-1 is a highest weight module.
Recall that a triangular decomposition for 𝔤 is finite if Q+ is finitely generated.

Let 𝔤 be a Lie algebra with regular finite triangular decomposition. Then all modules in Category 𝒪 have a highest weight series.


Let M𝒪. Choose λ1,,λn𝔥* so that Mμ0 only for μλi. We may assume that any weight μ of M is comparable to only one λi. (If μλi and μλj, then we can replace λi and λj by some λ𝔥* so that λ>λi,λj.) Let μ be a weight of M and suppose μλi. Then define d(μ)=ht(λi-μ).

Choose μ1𝔥* a weight of M so that d(μ1) is minimal. This implies that any nonzero vector in Mμ1 is a highest weight vector. Therefore, choose 0v1Mμ1 and let M1 be the 𝔤-submodule of M generated by v1.

Now choose a weight μ2 of M/M1 so that d(μ2) is minimal. Then we can choose 0v2(M/M1)μ2. Let M2M1 be the submodule of M so that M2/M1 is generated by v2. In this way we inductively define Mi. The following picture illustrates this process of choosing the weights μi: λi λj d(μ1) d(μ2) d(μ3) μ1 μ2 μ3 In this way, we construct a filtration 0=M0M1M2 such that each quotient is a highest weight module. Since Q is finitely generated, there are only a finite number of weights μ of M such that d(μ) has a given value. This, along with the fact that dimMμ<, show that the construction outlined above yields a filtration so that M=i=1Mi.


Let 𝔤 be a Lie algebra with a regular, finite triangular decomposition. Recall, for M𝒪 and λ𝔥*, [M:L(λ)] is the number of times L(λ) appears in a local composition series for M. Define a relation on 𝔥* by λμ if [M(λ):L(μ)]>0. Extend this to an equivalence relation on 𝔥*. The blocks of 𝔤 are the equivalence classes of 𝔥* under . For λ𝔥*, let [λ] represent the block containing λ. Denote the set of blocks of 𝔤 by [𝔥*].

For λ𝔥*, define 𝒪[λ] to be the full subcategory of 𝒪 consisting of modules M such that [M:L(μ)]>0 only if μλ. As we will see in Theorem 2.3.12, the category 𝒪 can be decomposed into these subcategories. This decomposition is the goal of this section, which follows Sections 2.10-2.12 of [MPi1995].

Projectives and BGG Reciprocity

In this section, we introduce a new class of 𝔤-modules: P(μ), the projective cover of the irreducible module L(μ) (μ𝔥*). We then establish a reciprocity between these projective modules and Verma modules on the one hand and Verma modules and irreducibles on the other.

The irreducible modules L(μ) do not in general have a projective cover in the full category 𝒪. Therefore, we fix λ𝔥* and restrict to the subcategory 𝒪(λ) which consists of modules M𝒪 such that Mμ0 only if μλ.

Rather than defining P(μ) directly, we construct a larger projective module P(μ) in 𝒪(λ). Let μλ. We will view U(𝔤>0) as a 𝔥𝔤>0-module with the following action: (h+x)·u (ad(h)+(μ,h)) u+xu for h𝔥, x𝔤>0, and uU(𝔤>0). Under this action, U(𝔤>0) has a weight-space decomposition U(𝔤>0)=νμU(𝔤>0)ν. Then the subspace νλU(𝔤>0)ν is a 𝔥𝔤>0-submodule of U(𝔤>0). Define the 𝔥𝔤>0-module W(μ)=Wλ(μ) =U(𝔤>0)/ νλ U(𝔤>0)ν. Then P(μ) is the induced 𝔤-module P(μ)= Pλ(μ) =U(𝔤) U(𝔥𝔤>0) W(μ).



Let {wi}i=1n be a basis of weight vectors for W(μ). (This set is finite since dimW(μ)νdimU(𝔤>0)ν< for all ν and W(μ)ν=0 unless μνλ.) As a U(𝔤<0)-module, P(μ) is (freely) generated by 1wi. Therefore, P(μ) is finitely generated as a 𝔤-module This implies P(μ)ν0 only if νλ and P(μ)ν<.

P(μ) is projective in 𝒪(λ).


Let M and N be modules in 𝒪(λ) such that P(μ) ϕ N ψ M 0. Note that elements of P(μ) are sums of elements of the form xy, where xU(𝔤) and yW(μ) is the image of yU(𝔤>0). Suppose ϕ(11)=m0M. Let n0N be such that ψ(n0)=m0. Define a map ϕ:P(μ)N by ϕ(xy)= xyn0. If this map is well-defined, it is clearly a 𝔤-module homomorphism such that ψϕ=ϕ. Therefore we check that ϕ is well-defined. Suppose that y=y. Then y=y+u for some uνλU(𝔤>0)ν. This implies xyn0=xyn0+ xun0. However, n0Nμ and so un0νλNν=0.

P(μ) has a Verma composition series and [P(μ):M(ν)] =dimHom𝔤 (P(μ),M(ν)) =dim(M(ν)μ).


Note that P(μ) is a free U(𝔤<0)-module.

Let {wi}i=1n be a basis of weight vectors for W(μ). Assume that wiW(μ)γi and γi>γj implies i>j. Let Mi be the submodule of P(μ) generated by 1wi. Then the following is a Verma composition series for P(μ): 0MnMn-1+ MnM1++ Mn=P (μ). Also, for μνλ, [P(μ):M(ν)] =dimW(μ)ν= dimM(ν)μ. Since P(μ) is projective, dimM(ν)μ= dimHom𝔤(P(μ),M(ν)).

Let P𝒪(λ) be a finitely generated indecomposable projective module. Then P has a unique maximal submodule.


Let μλ. From Lemma 2.1.5, we have P(μ)=1inPi, where each Pi is an indecomposable projective module. Moreover, each Pi is generated by one element, namely the projection of 11P(μ). Lemma 2.1.3 shows that Pi has a maximal proper submodule. Since Pi is generated by one element, this submodule is unique.

Now consider any finitely generated indecomposable projective module P. Let {v1,,vn} be a generating set for P so that viPνi. Then there is a surjective map ϕ:iP(νi)P defined by ϕ(vνi)=vi, where vνi=11P(νi). Since P is projective and indecomposable, it must be isomorphic to an indecomposable summand of iP(νi). The previous paragraph implies that P has a unique maximal submodule.

For each μλ the irreducible module L(μ) has a projective cover P(μ)𝒪(λ). The modules P(μ), μλ, are all of the indecomposable projectives in 𝒪.


Let μλ. Then L(μ) is a quotient of P(μ) and therefore must be a quotient of a projective indecomposable summand P of P(μ). Then P is a projective cover of L(μ), which we denote by P(μ).

The previous proposition shows that each finitely generated indecomposable module P has a unique irreducible quotient, and so PP(μ) for some μλ.

([RCW1982]). Let μ,ν𝔥*. Then, [M(μ):L(ν)]= [P(ν):M(μ)].


We begin by showing [M(μ):L(ν)]=dim(Hom𝔤(P(ν),M(μ))). According to Lemma 2.2.1, M(μ) has a local composition series at ν, M(μ)=M1Mn=0. Lemma 1.3.3 implies dim(Hom𝔤(P(ν),M(ν))) =k=1n-1 dim(Hom𝔤(P(ν),Mk/Mk+1)). Consider dim(Hom𝔤(P(ν),Mk/Mk+1)). By definition of a local composition series Mk/Mk+1L(γ) for some γν or (Mk/Mk+1)γ=0 for all γν. Now, dim(Hom𝔤(P(ν),L(γ)))= { 0 ifνγ ifν=γ since P(ν) has a unique maximal submodule L(ν). Suppose (Mk/Mk+1)γ=0 for all γν. Since L(ν) is the unique irreducible quotient of P(ν) and L(ν)ν0, any homomorphism ϕ:P(ν)Mk/Mk+1 must be trivial. Therefore, dimHom𝔤(P(ν),Mk,Mk+1)=0.

We now have dim(Hom𝔤(P(ν),M(μ)))= k=1ndim (Hom𝔤(P(ν),Mk/Mk+1)) =[M(μ):L(ν)].

Next we show dim(Hom𝔤(P(ν),M(μ)))=[P(ν),M(μ)]. Recall dim(Hom𝔤(P(ν),M(μ)))=[P(ν),M(μ)] and P(ν)=γmν(γ)P(γ) for some mν(γ)0. Therefore, γmν(γ) dim(Hom𝔤(P(γ),M(μ))) = dim(Hom𝔤(P(ν),M(μ))) = [P(ν):M(μ)] = γmν(γ) [P(γ):M(μ)] (2.3) Observe that dim(Hom𝔤(P(ν),L(ω))) =γmν(γ)dim (Hom𝔤(P(γ),L(ω))) =mν(ω). On the other hand, dim(Hom𝔤(P(ν),L(ω)))=dimL(ω)ν since P(ν) acts like a free module in 𝒪(λ). Therefore, mν(ν)=1 and mν(γ)=0 unless νγ. Using Equation 2.3, we can then induct on ht(λ-ν) to show [P(ν):M(μ)]=dim(Hom𝔤(P(ν):M(μ))).

Ext1(·,·) and a Decomposition of Category 𝒪

Ext1(·,·) provides a useful tool for studying blocks, particularly in the proof of the decomposition of Category 𝒪 by blocks (Theorem 2.3.12). For a definition of Ext1(·,·) see Section 1.3.4.

If Ext1(M(μ),L(λ))0, then [M(λ):L(μ)]>0.


We first claim λμ implies Ext1(M(μ),L(λ))=0. To prove this, assume λμ and let 0L(λ)N ϕM(μ) 0 be an extension of M(μ) by L(λ). Let vNμ so that ϕ(v)=v+M(μ). If λμ, then Nγ=0 for all γ>μ and so v is a highest weight vector in N. Therefore, we have a homomorphism ψ:M(μ)N given by ψ(v+)=v. Then ψϕ=idM(λ) and so the sequence splits. This implies Ext1(M(μ),L(λ))=0.

We now assume Ext1(M(μ),L(λ))0. Since λ>μ, M(μ)𝒪(λ). Using Theorem 2.3.6. we only need to show [P(μ),M(λ)]>0.

Since P(μ) has a Verma composition series (Proposition 2.3.3) and P(μ) is a direct summand of P(μ), P(μ) has a Verma composition series P(μ)=P0Pn=0 such that P(μ)/P1M(μ). We then have the following diagram P(μ) ϕ 0 L(λ) N ψ M(μ) 0. Since P(μ) is projective, there is a homomorphism ϕ:P(μ)N such that ψϕ=ϕ. Then, ϕ(P1)KerψL(λ). We must have either ϕ(P1)=0 or ϕ(P1)L(λ). If ϕ(P1)=0, ϕ induces a map ϕ:M(μ)N and the sequence splits. Therefore, ϕ(P1)L(λ). Applying ϕ to the Verma composition series for P(μ), we have ϕ(P1) L(λ)ϕ (Pn)=0. Therefore, there is some 1kn-1 such that ϕ(Pk)L(λ) and ϕ(Pk+1)=0. Then Pk/Pk+1M(λ).

Let M𝒪 be a highest weight module of highest weight λ. Suppose Ext1(M(μ),M)0. Then there is some ν𝔥* such that λνμ and [M(λ):L(ν)], [M(ν):L(μ)]>0. In particular, μ[λ].


We have λ>μ. Let N be the maximal submodule of M. Then we have a short exact sequence 0NML(λ)0. Proposition 1.3.4 implies that Ext1(M(μ),L(λ))0 or Ext1(M(μ),N)>0. If Ext1(M(μ),L(λ))0, the proposition holds. Therefore assume that Ext1(M(μ),N)0. Let N=N0N1 be a local composition series for N. Then, by Proposition 1.3.4, Ext1(M(μ),Ni/Ni+1)0 for some i.

Recall that Ni/Ni+1L(ν) for some λνμ or (Ni/Ni+1)ν=0 for all νμ. Suppose (Ni/Ni+1)ν=0 for all νμ. We claim this implies Ext1(M(μ),Ni/Ni+1)=0, a contradiction. Let 0Ni/Ni+1 NM(μ)0 be a short exact sequence. Choose vN so that vv+M(μ). Then v(N)μ. Since (Ni/Ni+1)ν=0 and M(μ)ν=0 for ν>μ, (N)ν=0. This implies that v is a highest weight vector. We then have a map M(μ)N given by v+v. Thus, the sequence splits, and so Ext1(M(μ),Ni/Ni+1)=0.

Therefore, Ni/Ni+1L(ν). By Lemma 2.3.7, [M(ν):L(μ)]>0. Since Ni/Ni+1 is a subquotient of M, and hence of M(λ), [M(λ):L(ν)]>0.

Recall that the blocks of 𝔤 are defined to be the equivalence classes of 𝔥* where λμ if [M(λ):L(μ)]>0. The following proposition provides alternative definitions of this equivalence relation.

([Nei1984]). Let λ,μ𝔥*. Then the following are equivalent:



Theorem 2.3.6 shows that the first two statements are equivalent. Also M(μ)M(λ) clearly implies [M(λ):L(μ)]>0. Therefore we only need to check that [M(λ):L(μ)]>0 implies M(μ)M(λ).

Suppose [M(λ):L(μ)]>0. We will show that M(μ)M(λ) by inducting on ht(λ-μ). If λ=μ, the result clearly holds. Now suppose λ>μ. Let J(λ) be the maximum proper submodule of M(λ). We may refine the series 0J(λ)M(λ) into a local composition series, which implies that [J(λ):L(μ)]>0. Let 0=J0J1 be a highest weight series for J(λ), where Ji/Ji-1 is a highest weight module of highest weight μi. There is some k such that (Jk)μ=J(λ)μ since the weight spaces of J(λ) are finite-dimensional. This means [J(λ)/Jk:L(μ)]=0 and so [Jk:L(μ)]>0. Therefore, 0=J0J1Jk can be refined to a local composition series with a factor of L(μ). Choose i minimal so that [Ji+1/Ji:L(μ)]>0. Then μi+1μ. Suppose μ=μi+1. Then we have a nontrivial map M(μ)J(λ)/Ji. We may assume i is minimal such that Hom𝔤(M(μ),J(λ)/Ji)0. The sequence 0Ji/Ji-1J(λ)/Ji-1J(λ)/Ji0 is exact. Therefore, Proposition 1.3.4 implies that Hom𝔤(M(μ),J(λ)/Ji-1) Hom𝔤(M(μ),J(λ)/Ji) Ext1(M(μ),Ji/Ji-1) is exact. The first term in this sequence is 0 and the second is nonzero. Therefore, the third term is nonzero: Ext1(M(μ),Ji/Ji-1)0. We use Lemma 2.3.8 to show there is some ν𝔥* such that λ>μiνμ and [M(μi):L(ν)]>0 and [M(ν):L(μ)]>0. Applying the inductive hypothesis, we have M(μ)M(ν)M(μi)M(λ).

If μμi+1, we can use the induction hypothesis to conclude M(μ)M(μi+1) and M(μi+1)M(λ).

Recall that, for a block [λ][𝔥*], 𝒪[λ] is the full subcategory of 𝒪 consisting of modules M such that L(μ) is a local composition factor of M only if μ[λ].

Let [λ],[μ][𝔥*] with [λ][μ]. Suppose E𝒪[λ] and F𝒪[μ] are highest weight modules. Then, Ext1(E,F)=0.


From Proposition 2.3.8, if Ext1(M(μ),E)0, there exists ν𝔥*, λν>μ such that [M(λ):L(ν)]>0 and [M(ν):L(μ)]>0. This implies λνμ, a contradiction. Therefore, Ext1(M(μ),E)=0.

Since F is a highest weight module of highest weight μ, there is a submodule NM(μ) such that 0NM(μ)F0 is exact. Proposition 1.3.4 implies Hom𝔤(N,E) Ext1(F,E) Ext1(M(μ),E)=0. If Hom𝔤(N,E)=0, then Ext1(F,E)=0.

Therefore, we only need to check that Hom𝔤(N,E)=0. Let ϕHom𝔤(N,E). Since N𝒪[μ], ϕ(N)𝒪[μ]. On the other hand, ϕ(N)𝒪[λ]. This implies ϕ(N)=0.

([MPi1995], 2.12.3). Let [λ],[μ][𝔥*] with [λ][μ]. Suppose E𝒪[λ] and F𝒪[μ], with [λ][μ], have a filtration by highest weight modules of finite length. Then, Ext1(E,F)=0.

([MPi1995]). Let 𝔤 be a Lie algebra with a regular, finite triangular decomposition. Then, the category 𝒪 decomposes into subcategories: 𝒪=[λ][𝔥*] 𝒪[λ]. This means that for M𝒪, M=[λ][𝔥*] M[λ], where M[λ]𝒪[λ].


Let M𝒪. Then M has a filtration 0=M0M1M2 such that Mi/Mi-1 is a highest weight module.

For [λ][𝔥*], we construct M[λ] by constructing a highest weight series 0M0[λ] M1[λ] such that Mi = [λ][𝔥*] Mi[λ]; Mi[λ]/ Mi-1[λ] = { Mi/Mi-1 ifMi/Mi-1 𝒪[λ], 0 otherwise. We do this inductively. Suppose we have constructed these highest weight series for j<n, and suppose Mn/Mn-1𝒪[λ]. For [μ][λ], let Mn[μ]=Mn-1[μ]. To define Mn[λ], we use the short exact sequence 0Mn-1/ Mn-1[λ] Mn/Mn-1[λ] Mn/Mn-10. (2.4) Since Mn-1=[μ][𝔥*]Mn-1[μ], Mn-1/Mn-1[λ][μ][λ]Mn-1[μ][μ][λ]𝒪[μ]. Since Mn/Mn-1𝒪[λ], Lemma 2.3.11 implies Ext1(Mn/Mn-1,Mn-1/Mλn-1)=0, and so the sequence in (2.4) splits: Mn/Mn-1[λ] Mn-1/Mn-1[λ] Mn/Mn-1. Therefore there exists a module Mλn, Mλn-1Mλn, so that Mλn/ Mλn-1 Mn/Mn-1 Mn/Mλn-1 = Mλn/ Mλn-1 Mn-1/ Mλn-1. To finish the induction step, we claim Mn=[μ][𝔥*] Mn[μ]=Mn[λ] [μ][λ] Mn-1[μ]. Clearly, Mn=Mn[λ]+[μ][λ]Mn-1[μ]. Also, Mn[λ]Mn-1=Mn-1[λ], which shows Mn[λ] [μ][λ] Mn-1[μ]=0.

The Contravariant Form and the Shapovalov Determinant

Suppose 𝔤 has a triangular decomposition. Proposition 2.0.5 implies the following decomposition, U(𝔤)= U(𝔥) (𝔤<0U(𝔤)+U(𝔤)𝔤>0). Let τ:U(𝔤)U(𝔥) be the projection map. Recall that σ:U(𝔤)U(𝔤) is the anti-involution such that σ(𝔤>0)=𝔤<0 and σ|𝔥=id. We can then define a bilinear form ,:U(𝔤)×U(𝔤)S(𝔥) on U(𝔤). For x,yU(𝔤), define x,y= τ(σ(x)y). This induces a bilinear form ,:M(λ)×M(λ) on M(λ). Fix a highest weight vector v+M(λ) of M(λ). For x,yU(𝔤), define xv+,yv+= λ(τ(σ(x)y)).

If 𝔤 has an Hermitian triangular decomposition, we make the following modifications to the definition of the contravariant form. In this case there is a real Lie algebra 𝔤 such that 𝔤=𝔤, and σ is an Hermitian anti-involution such that σ|𝔥=id. For λ𝔥*, define λ𝔥* by λ(ch)=cλ(h) for c and h𝔥. Then the Hermitian form ,:M(λ)×M(λ) is defined by xv+,yv+=λ(τ(σ(x)y)), where x,yU(𝔤) and v+ is a highest weight vector of M(λ).

For the remainder of this chapter we assume

𝔤 has a regular, finite triangular decomposition; OR
𝔤 has a regular, finite Hermitian triangular decomposition, and we restrict all choices of weights to λ𝔥* such that λ=λ for some λ𝔥*.

For μ,ν𝔥* with μν, M(λ)μM(λ)ν with respect to ,.


Choose h𝔥 so that μ(h)ν(h). For 0vM(λ)μ and 0wM(λ)ν, we have μ(h)v,w= hv,w= v,hw= ν(h)v,w v,w =0.

The radical of , is Rad,= { mM(λ)| m,m=0 for allmM(λ) } .

Let λ𝔥*. Then Rad,=J(λ), the unique maximal submodule of M(λ).


We first show that Rad, is a submodule of M(λ). Let x𝔤 and mRad,. Then for any mM(λ), xm,m=m,σ(x)m=0 since σ(x)mM(λ). Therefore, xmRad,. Clearly, linear combinations of elements in Rad, are in Rad,. Thus, Rad, is a submodule of M(λ).

Since v+Rad,, Rad,J(λ). We now show that J(λ)Rad,. Let vJ(λ). Then there is no xU(𝔤) such that xv=cv+ for some 0c. Therefore, for any xU(𝔤), xv+,v=v+,σ(x)v=0. Since any vector in M(λ) is of the form xv+ for some xU(𝔤), this implies vRad,.

Let M be any highest weight module of highest weight λ. Since MM(λ)/M for some submodule MJ(λ), the previous lemma implies that , can be extended to M.

The weight space M(λ)λ-μ has a basis {x(t)v+|(t):t1tj>0,βt1++βtj=μ}. (Recall the βtiQ+ are such that xti𝔤-βti.) We can associate to ,|M(λ)λ-μ a matrix A(λ)λ-μ defined by A(λ)λ-μ= (x(s)v+,x(t)v+) βs1++βsk=μ =βt1++βtj . According to Lemmas 2.4.1 and 2.4.2, the matrix A(λ)λ-μ is degenerate if and only if J(λ)λ-μ0. Therefore we define detM(λ)λ-μ= detA(λ)λ-μ. Then detM(λ)λ-μ=0 if and only if J(λ)λ-μ0.

Example. Consider sl2(), with the standard basis {h,e,f} where [h,e]=2e, [h,f]=-2f, [e,f]=h. The Cartan subalgebra is 𝔥=h and let α𝔥* so that α(h)=1. The anti-automorphism σ:sl2()sl2() is given by σ(h)=h, σ(e)=f, σ(f)=e. Let λ𝔥* and n0. Then M(λ)λ-nα has a basis fnv+. It is straightforward to check that efj=fje+jfj-1(h-j+1), and so detM(λ)μ = det(fnv+,fnv+) = v+,enfnv+ = j=1n j(λ(h)-j+1). More generally, Shapovalov [Sha1972] computed the determinant detM(λ)λ-μ for any finite-dimensional semisimple Lie algebra 𝔤: detM(λ)λ-μ= βR+,r>0 (ω+ρ,β-rr) p(μ-rβ) , where p(γ) is the number of ways that γ can be written as a sum of positive roots. As we will see, Kac also found an explicit formula for the determinant M(h,c)(h+n,c) for the Virasoro algebra (Theorem 3.4.3).

Jantzen Filtrations

Let t be an indeterminant. Define 𝔤[t]=[t]𝔤, with the Lie bracket on 𝔤[t] given by [p(t)x,q(t)y]= p(t)q(t)[x,y]. For λ,γ𝔥*, the Verma module M(λ+tγ) is given by M(λ+tγ)= U(𝔤[t])/I where I is the left U(𝔤[t])-ideal generated by 𝔤>0 and {h-(λ+tγ)(h)|h𝔥}. Let ε:[t] be the -linear map given by t0. This can be extended to surjective homomorphisms ε:𝔤[t]𝔤 and ε:M(λ+tδ)M(λ).

We can extend the Shapovalov form to a form ,(λ+tγ):M(λ+tδ)×M(λ+tδ)[t]. Define M(λ+tγ)j= { vM(λ+tγ) |tjdivides v,w(λ+tγ) for allwM(λ+tγ) } . Then M(λ+tγ) is a 𝔤[t]-submodule of M(λ+tγ). Define M(λ)j=ε (M(λ+tγ)j) The Jantzen filtration of M(λ) is M(λ)=M (λ)0 M(λ)1 M(λ)2. For x[t], define the order of x by ord(x)=k if tk|x and tk+1x.

(See [MPi1995], Theorem 2.9.4) Suppose γ𝔥* is chosen so that ,(λ+tγ) is nondegenerate. Then,

for μλ, ord(detM(λ+tγ)(μ+tγ)) =j=1dim M(λ)jμ;
M(λ)1=J(λ), the maximal proper submodule of M(λ);

Translation Functors

Let 𝔤 be a Lie algebra with a regular, finite triangular decomposition. Recall that Category 𝒪 is closed under tensor products. Theorem 2.3.12 then implies that M(λ)L(μ) decomposes by blocks. In this section, we consider the translation M(λ) (M(λ)L(μ))[ν].

Fix λ,μ𝔥*. For [ν][𝔥*], consider the set [ν] { γ𝔥*| λ+μ-γ Q+,L (μ)γ-λ 0 } . If this set is nonempty, enumerate the elements ν1,ν2, so that νiνj for i>j. The following proposition is a corollary of Theorem 2.3.12.

Let λ,μ𝔥*. Then, M(λ)L(μ)= [ν][𝔥*] (M(λ)L(μ))[ν]. The module (M(λ)L(μ))[ν]𝒪[ν] has a filtration 0=M0M1 such that (M(λ)L(μ))[ν]=Mi and Mi/Mi-1M(νi)dimL(μ)νi-λ.


Let v+ be a generator of M(λ). Enumerate the weights of L(μ) by ν1,ν2, so that νiνj for i>j. For each weight νi of L(μ), let {wνi,j|1jdimL(μ)νi} be a basis for L(μ)νi. Define Mνi,j to be the submodule of M(λ)L(μ) generated by v+wνi,j. Then 0 Mν1,1 Mνi,2 Mν2,1 gives a filtration by Verma modules. We then apply the proof of Theorem 2.3.12 to get the decomposition.

Recall that 𝔤 is a Hopf algebra with coproduct Δ(x)=x1+1x. This allows us to define a contravariant form on the tensor product M(λ)L(μ) using the contravariant forms on M(λ) and L(μ). For ν,νM(λ) and w,wL(μ), define ,:M(λ)L(μ)×M(λ)L(μ) by vw,v,w= v,v w,w. This form will be contravariant: x(vw),vw = xvw+ vxw, vw = xvw,vw+ vxw,vw = xv,v w,w+ v,v xw,w = v,σ(x)v w,w+ v,v w,σ(x)w = vw,σ(x)vw+vσ(x)w = vw,σ(x)(vw) for any x𝔤.

Let λ,μ𝔥*. For νQ+, let {wν,i|1idimL(μ)μ-ν} be a basis for L(μ)μ-ν. For γQ+, the following sets are bases for (M(λ)L(μ))λ+μ-γ: 1= { x(t)v+ wν,i| 1idimL (μ)μ-ν, (t):t1 tj>0,βt1 ++βtj= γ-ν } ; (2.5) 2= { x(t) (v+wν,i) | 1idimL (μ)μ-ν, (t):t1 tj>0,βt1 ++βtj= γ-ν } . (2.6) Moreover, the transition matrix between these two bases has determinant 1.


We have x(t) (v+wν,i)= x(t)v+ wν,i+ ζ>ν1jdimL(μ)μ-ζ(s)|βs1+βsk=γ-ζ x(s)v+ wζ,j. By appropriately ordering the elements of (2.5) and (2.6), we see that the transition matrix taking (2.6) to (2.5) is upper triangular with ones on the diagonal. Therefore, this matrix is invertible with determinant 1.

Define detL(μ)μ-ν =det (wνi,wνj) 1i,jdimL(μ)μ-ν . Note that detL(μ)μ-ν0. Also, define det(M(λ)L(μ))λ+μ-γ= det (x(s)wν,i,x(t)v+wζ,j) x(s)v+ wν,i,x(t) v+wζ,j 1 .

Let λ,μ𝔥* and γQ+. Then det(M(λ)L(μ))λ+μ-γ= ν+ζ=γ (detM(λ)λ-ν) dimL(μ)μ-ζ (detL(μ)μ-ζ)p(ν).


Enumerate the weights between 0 and γ: ζ1,,ζn.

We have shown that M(λ)λ-νM(λ)λ-ν for νν and L(μ)μ-ζL(μ)μ-ζ for ζζ. Therefore, the matrix (x(s)v+wν,i,x(t)v+wζ,j) has the form ( Bζ1 0 0 0 0 0 0 Bζn ) , where Bζi is the matrix (A(λ)λ-(γ-ζi)wζi,j,wζi,k) 1j,kdimL(μ)μ-ζi . Using Lemma 1.3.1, we have det(Bζi)=(det(A(λ)λ-(γ-ζi)))dimL(μ)μ-ζi(detL(μ)μ-ζi)p(ν).

Suppose [ν],[γ][𝔥*] with [ν][γ]. Then, (M(λ)L(μ))[ν](M(λ)L(μ))[γ] respect to the contravariant form.


First we write down a basis for (M(λ)L(μ))[ν]. Enumerate the set [ν] { γ𝔥*|λ+ μ-γQ+,L (μ)γ-λ0 } . so that νiνj for i>j. { νiQ+| λ+μ-νi[ν] L(μ)μ-νi 0 } . so that νiνj for i>j. Given the decomposition in Proposition 2.6.1, we can choose a set {vνi,j|1jdimL(μ)μ-νi}(M(λ,μ)[ν])λ+μ-νi so that for each weight space (M(λ)L(μ))λ+μ-δ, the set νi<δ { x(s)vνi,j |(s):s1 sk>0,βs1+ +βsk=δ-νi ,1jdimL(μ)μ-νi } is a basis for ((M(λ)L(μ))[ν])λ+μ-δ.

We will show ((M(λ)L(μ))[ν]) λ+μ-δ ((M(λ)L(μ))[γ]) λ+μ-δ (2.7) by inducting on max{ht(δ-νi)|νi[ν]}.

To prove the base case, we need to show that (2.7) holds for all δ=νi such that νiνj for any j. If ((M(λ)L(μ))[γ])λ+μ-νi=0, the result is obvious. Therefore, assume ((M(λ)L(μ))[γ])λ+μ-νi0. Since λ+μ-νi[γ], ((M(λ)L(μ))[γ])λ+μ-νi has a basis {x(s)vγk,l|γk>νi,(s):s1sl>0,βs1++βsl=γk-νi}. Since νiνj for any j, ((M(λ)L(μ))[ν])λ+μ-νi has a basis {vνi,j}. We then have x(s)vγk,l,vνi,j= vγk,l,σ(x(s))vνi,j =0 since σ(x(s))vνi,j((M(λ)L(μ))[ν])λ+μ-(νi-(βs1++βsk))=0.

For general δQ+, ((M(λ)L(μ))[ν])λ+μ-δ has a basis { x(s)vνi,j |νi<δ,(s) :s1sk>0, βs1++βsk =δ-νi } S1 {vνi,j|νi=δ}S2.

For vνi,jS2, we have x(s)vγk,l,vνi,j=vγk,l,σ(x(s)vνi,j). Since σ(x(s))vνi,j((M(λ)L(μ))[ν])λ+μ-(νi-(βs1++βsk)) we can use the induction hypothesis to conclude x(s)vγk,l,vνi,j=0.

For x(s)vνi,jS1 and v((M(λ)L(μ))[γ])λ+μ-δ, x(s)vνi,j,v= vνi,j,σ(x(s))v=0 by the induction hypothesis.

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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