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.

Basics and Background

Some Definitions for Lie Algebras

Let 𝔤 be a Lie algebra over a field 𝔽. We view 𝔤 as a 𝔤-module via the adjoint action ad:𝔤×𝔤𝔤 given by ad(x)y= [x,y]. Suppose 𝔤 is finite-dimensional. The Killing form κ(,):𝔤×𝔤𝔽 is given by κ(x,y)=Tr (ad(x)ad(y)), where Tr is the trace. The Killing form is invariant, that is, κ([x,y],z)=-κ(y,[x,z]).

Fix a Lie algebra 𝔤. Write 𝔤n=𝔤𝔤 (n times). The universal enveloping algebra of 𝔤 is the associative algebra U(𝔤)=n0 gn/ xy-yx-[x,y]|x,y𝔤. We can view U(𝔤) as a Lie algebra Lie(U(𝔤)) with bracket [u,v]=uv-vu for u,vU(𝔤). Then the defining relation of U(𝔤) implies that there is a Lie algebra homomorphism 𝔤Lie(U(𝔤)).

([Dix1996] 2.1.11). Let 𝔤 be a Lie algebra with basis {xi|iI}, where I is a totally ordered set. Then the following set is a basis for U(𝔤): { xi1 xin| i1in } .

Any Lie algebra 𝔤 has a Hopf algebra structure given by

coproduct: Δ(x)=x1+1x;
counit: ε(x)=0;
antipode: s(x)=-x.

Semisimple Lie Algebras

In this section we give a brief introduction to semisimple Lie algebras. More on semisimple Lie algebras can be found in [Dix1996] and [Hum1978].

A Lie algebra 𝔤 is simple if dim𝔤>1 and the only ideals of 𝔤 are 0 and 𝔤. A Lie algebra 𝔤 is semisimple if 𝔤 is the direct sum of simple Lie algebras.

Define 𝔤(1)=[𝔤,𝔤] and 𝔤(n)=[𝔤(n-1),𝔤(n-1)]. The radical of 𝔤, Rad𝔤, is the largest ideal of 𝔤 such that (Rad𝔤)(n)=0 for some n.

([Dix1996], 1.5.2 and 1.5.12). Let 𝔤 be a Lie algebra over . The following are equivalent:

𝔤 is semisimple;
the Killing form κ(,) on 𝔤 is nondegenerate;

Let 𝔤 be a finite-dimensional complex semisimple Lie algebra. Then 𝔤 contains a unique maximal abelian subalgebra 𝔥 (with respect to inclusion) such that the linear transformations ad(h), h𝔥, are simultaneously diagonalizable on 𝔤. We call 𝔥 the Cartan subalgebra of 𝔤.

Let 𝔥*=Hom(𝔥,). Define 𝔤α= {x𝔤|ad(h)x=α(h)x} and note that 𝔤=α𝔥*𝔤α. A nonzero element α𝔥* is a root of 𝔤 if 𝔤α0. The set of roots of 𝔤 is denoted R.

The Killing form κ(,) is nondegenerate on 𝔥. Therefore, we can identify an element h𝔥 with an element α𝔥* so that α(x)=κ(x,h) for all x𝔥. We use the Killing form to define a form ,:𝔥*×𝔥* on 𝔥*. For α,α𝔥*, α,α= κ(h,h), where α(x)=κ(x,h) and α(x)=κ(x,h) for all x𝔥.

We choose a basis of 𝔥* of simple roots α1,,αkR so that for each βR we can write β=ikiαi, where either ki0 for all i or ki0 for all i. Given a choice of simple roots, the positive roots are R+= { βR|β= ikiαi, ki0 } . The height function ht:R is given by ht(β)=iki ,ifα=i kiαi. The highest root is the unique element θR+ such that ht(θ)ht(β) for all βR. Define ρ=12βR+β. The dual Coxeter number g is g=12 θ,θ+2ρ.

Define αi=2αiαi,αi and let hi𝔥 so that αi(x)=κ(x,hi) for all x𝔥.

([Dix1996], 1.10.2). Let 𝔤 be a finite-dimensional semisimple Lie algebra. Then, 𝔤 is generated by elements ei𝔤αi, fi𝔤-αi, and hi𝔥, 1ik, with relations

(1) [hi,hj]=0;
(2) [hi,ej]=aijej, [hi,fj]=-aijfj;
(3) [ei,fj]=δi,jhi;
(4) [ei,[ei1-aijtimes,ej]]=0, [fi,[fi1-aijtimes,fj]]=0, if ij
where aij=αj,αi=αj(hi).

The generators h1,,hk are a basis for 𝔥. Let 𝔤>0 be the subalgebra of 𝔤 generated by e1,,ek and 𝔤<0 the subalgebra generated by f1,,fk. The previous proposition implies 𝔤=𝔤>0𝔥 𝔤<0. (1.1) The Cartan matrix associated to 𝔤 is A=(aij)1ik.

We define the following subsets of 𝔥*:

the set of integral weights of 𝔤: P={λ𝔥*|λ,α};
the set of dominant integral weights of 𝔤: P+={λ𝔥*|λ,α0};
the root lattice of 𝔤: Q=1ik αi;
the positive root lattice: Q+=1ik 0αi.
The dominance ordering on 𝔥* is the partial ordering on 𝔥* given by λμ if λ-μQ+.

Associated to 𝔤 is a Weyl group W. The Weyl group is the group generated by reflections si, 1ik, which act on 𝔥* by si(λ)=λ-λ,αiαi. It is straightforward to check that αi,slαj= slαi,αj (1.2) for 1i,j,lk. The dot action of W is given by siλ=si(λ+ρ)-ρ.

Recall 𝔤 is finite-dimensional. Suppose dim𝔤=n. Let {ui|1in} be a basis of 𝔤, and let {ui|1in} be a dual basis with respect to κ(,), i.e. κ(ui,uj)=δi,j. Define the Casimir element Ω0U(𝔤) by Ω0=i=1n uiui.

([Hum1978], 22.1-22.2). For x𝔤,

ad(Ω0)x=i=1n[ui,[ui,x]]=2gx, where g=12θ,θ+2ρ is the dual Coxeter number of 𝔤.

For any choice of dual bases {ui} and {ui}, we can write x𝔤 as x=i=1nκ (x,ui)ui. (1.3) Using this, it is easy to check that Ω0 is independent of the initial choice of basis {ui}. In particular, we can switch the roles of ui and ui to get i=1n [ui,ui]= i=1n uiui- i=1n uiui=0. (1.4)

Reductive Lie Algebras

A natural weakening of the definition of a semisimple Lie algebra is a reductive Lie algebra. A Lie algebra 𝔤 is reductive if it is the direct sum of an abelian Lie algebra and a semisimple Lie algebra. Note that the Killing form is not necessarily nondegenerate on 𝔤. However, it is still possible to define a nondegenerate form on 𝔤 ([Dix1996], 1.7.3).

Odds and Ends

Bilinear Forms

Let V and W be vector spaces. Then a bilinear form on V and W is a map ,:V×W such that for all v,vV, w,wW and α, αv+v,w = αv,w+ v,w v,αw+w = αv,w+ v,w. A Hermitian bilinear form is a map ,:V×W such that for all v,vV, w,wW and α, αv+v,w = αv,w +v,w, v,αw+w = αv,w+ v,w. Here, α is the complex conjugate of α.

Let q be an indeterminant. For vector spaces V and W over (q), we will also say a map ,:V×W(q) is a Hermitian bilinear form if for all v,vV, w,wW and α(q), αv+v,w = αv,w+ v,w; v,αw+w = αv,w+ v,w where ·:(q)(q) is the -automorphism sending q to q-1.


For an algebra A, a derivation on A is a linear map D:AA such that for any x,yA, D(xy)= D(x)y+ xD(y). The set of derivations 𝒟 on A naturally has a Lie algebra structure. That is, we can define a bracket [,]:𝒟×𝒟𝒟 by [D,D]=DD-DD.

A Lemma for Determinants

Suppose AMn() and BMm() with B=(bkl). Define C= ( Ab11 Ab1m Abm1 Abmm ) . Then, detAB= (detA)m (detB)n.


If b1i=0 for all 1im, then detC=0 and detB=0; then the lemma holds. Therefore, we may assume b1i0 for at least one 1im. Since a rearrangement of the columns of a matrix does not affect its determinant, we may assume b110.

We prove the result by inducting on m. Now, detAB = det ( Ab11 Ab12 Ab1m 0 A(b22-b12b21b11) A(b2m-b1mb21b11) 0 A(bm2-b12bm1b11) A(bmm-b1mbm1b11) ) = (b11)n detA×det ( A(b22-b12b21b11) A(b2m-b1mb21b11) A(bm2-b12bm1b11) A(bmm-b1mbm1b11) ) = (b11)n detA (detA)m-1det ( (b22-b12b21b11) (b2m-b1mb21b11) (bm2-b12bm1b11) (bmm-b1mbm1b11) ) n By the induction hypothesis = (detA)m ( b11det ( (b22-b12b21b11) (b2m-b1mb21b11) (bm2-b12bm1b11) (bmm-b1mbm1b11) ) ) n = (detA)m (detB)n.

Projectives and Extensions

Let R be a ring and 𝒞 an abelian category of left R modules.

A module P𝒞 is projective in the category 𝒞 if whenever A,B𝒞 with a map α:PA and a surjective map β:BA, then there exists γ:PB such that βγ=α. In other words, we have the following diagram: P γ α B β A 0 For M𝒞, a projective cover is a projective module P𝒞 along with a surjective homomorphism ϕ:PM such that ϕ(P)M for any proper submodule P of P.

A projective cover P of M𝒞 is unique up to isomorphism.


Suppose that P and P are both projective covers of M. We then have P γ P M 0 P γ P M 0 By definition of a projective cover, both γ and γ must be surjective. Therefore, γγ=id|P and so γ is an isomorphism.

Suppose P,M,M𝒞 with P projective and MM. Then HomR(P,M/M)= HomR(P,M)/ HomR(P,M) and dimHomR(P,M/M)= dimHomR(P,M)- dimHomR(P,M).


Clearly, we have a map ·:HomR(P,M)HomR(P,M/M), where ϕ:PM is sent to ϕ:PMM/M. Then ϕ=0 implies that ϕ(P)M, and so ϕ is a map from P to M. Thus, ker(·)HomR(P,M).

For the other direction, note that a homomorphism from P to M/M can be lifted to a homomorphism from P to M since P is projective.

Let A,B𝒞. An extension of A by B is a short exact sequence 0BEA0, where E𝒪. Two extensions 0BEA0, 0BEA0, are equivalent if there exists γ:EE such that the following diagram commutes. 0 B E A 0 id γ id 0 B E A 0 Define Ext1(A,B) to be the set of equivalence classes of extensions of A by B.

([MPi1995], 1.12.4, 1.12.4'). Let B,B,B,A𝒞, and suppose 0BBB0 is exact. Then, 0 HomR(A,B) HomR(A,B) HomR(A,B)δ Ext1(A,B) Ext1(A,B) Ext1(A,B) is exact. The map δ:Hom(A,B)Ext1(A,B) is given by δ(f)=, where :0BEA0 is such that the following diagram commutes. 0 B B B 0 f 0 B E A 0 . Also, 0 HomR(B,A) HomR(B,A) HomR(B,A)δ Ext1(B,A) Ext1(B,A) Ext1(B,A) is exact. The map δ:Hom(B,A)Ext1(B,A) is given by δ(f)=, where :0AEB0 is such that the following diagram commutes. 0 B B B 0 f 0 A E B 0 .

Central Extensions

Let 𝔤 and 𝔠 be Lie algebras. A central extension of 𝔤 by 𝔠 is a short exact sequence of Lie algebras 0𝔠i 𝔤 π𝔤0 (1.5) so that 𝔠 is contained in the center of 𝔤. We will also say that 𝔤 is a central extension of 𝔤, suppressing the choice of maps i and π. A Lie algebra 𝔤 is perfect if 𝔤=[𝔤,𝔤]; a central extension (1.5) so that 𝔤 is perfect is a covering of 𝔤.

A covering (1.5) of 𝔤 is a universal central extension if, for any central extension 0𝔠i 𝔤π 𝔤0, the following diagram commutes: 0 𝔠 i 𝔤 π 𝔤 0 0 𝔠 i 𝔤 π 𝔤 0 . A Lie algebra 𝔤 has a universal central extension if and only if 𝔤 is perfect ([MPi1995] 1.9.2). A universal central extension will be unique up to an equivalence of extensions ([MPi1995], 1.9.1).

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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