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.

The Virasoro Algebra

The Virasoro algebra is the Lie algebra Vir=-span{z,dk|k} with bracket [,] given by [dk,z]=0, [dj,dk]= (j-k)dj+k+ δj,-k12 (j3-j)z.

([MPi1995], 1.9.4). The Virasoro algebra is the universal central extension of the Lie algebra W=-span{Dk|k} with relations [Dj,Dk]= (j-k)Dj+k.

See Section 1.3.5 for a definition of central extensions. The Lie algebra W is the Witt algebra. The Witt algebra can be identified with the Lie algebra of derivations on [t,t-1]. That is, the derivations {Dk=-tk+1ddt|k} are a basis for the derivations of [t,t-1], and they satisfy the relation DjDk-DkDj=(j-k)Dj+k.

The Virasoro algebra has a regular, finite Hermitian triangular decomposition Vir=Vir-𝔥Vir+ where Vir- = span{dn|n<0}; 𝔥 = span{d0,z}; Vir+ = span{dn|n>0}. The associated Hermitian anti-involution on Vir is ϕ:VirVir given by ϕ(dn)=d-n, ϕ(z)=z.

Let U(Vir) be the universal enveloping algebra of Vir. Proposition 2.0.5 shows that U(Vir) inherits a triangular decomposition from Vir: U(Vir)=U (Vir-)S (𝔥)U (Vir+), (3.1) where S(𝔥)=U(𝔥) is the symmetric algebra of 𝔥. Bases for U(Vir-) and U(Vir+) are {d-λ1d-λk|λi,λ1λk>0} and {d-λ1d-λk|λi,λ1λk>0}, respectively.

In order to write these bases more efficiently, we introduce the following notation. For a partition λ: λ1λk>0, the weight of λ is λ=λ1++λk. The number of parts of λ is denoted l(λ)=k. Define dλ = dλkdλ1 d-λ = d-λ1 d-λk. Then, the bases for U(Vir-) and U(Vir+) can be rewritten as {d-λ|λa partition} and {dλ|λa partition}, respectively.

Category 𝒪

Category 𝒪 was introduced in Section 2.1. We now make a few comments specific to the Virasoro algebra.

Recall 𝔥*=Hom(𝔥,). We can identify weights λ𝔥* with pairs in 2 by λ(λ(d0),λ(z)). Then, the partial ordering on 𝔥* is given by μ<λifμ(z)= λ(z)andλ(d0) -μ(d0)<0.

For λ𝔥*, the Verma module M(λ) is the induced module M(λ)=U(Vir) U(Vir+𝔥) λ. We write M(h,c) for M(λ), where (h,c) is the pair identified with λ. Similarly, we write J(h,c) for the unique maximal submodule of M(h,c) (Lemma 2.1.3) and L(h,c) for the unique irreducible quotient of M(h,c).

Using the PBW basis for U(Vir-), we see M(h,c)=n0 M(h,c)(h+n,c), where{d-λv+|λ=n} is a basis forM(h,c)(h+n,c). Then dimM(h,c)(h+n,c)=p(n), where p(n) is the number of partitions of n. Recall from Section 2.1.1 that the character of a module M records the dimensions of the weight spaces of M. For M(h,c), we have ch(M(h,c)) = n=1 p(n) e(h+n,c) (3.2) = e(h,c) j=1(1-qj) (3.3) where g=e(1,0).

Affine Lie Algebras

The Virasoro algebra has a close relationship with affine Lie algebras. In particular, it is possible to construct a representation of the Virasoro algebra on certain modules for affine Lie algebras. Before discussing this construction, we provide a brief introduction to affine Lie algebras.

Recall from Section 1.2.1 that a reductive Lie algebra 𝔤 is a direct sum of an abelian Lie algebra and simple Lie algebras. Let 𝔤 be a finite-dimensional reductive Lie algebra with nondegenerate bilinear form (,). (We assume this is the Killing form when 𝔤 is semisimple.) The affine Lie algebra 𝔤ˆ associated to 𝔤 is 𝔤ˆ=([t,t-1]𝔤) cd, with relations [tmx,tny]= tm+n[x,y]+m δm,-n(x,y)c, [c,tmx]=0, [c,d]=0, [d,tnx]=ntn x, for m,n and x,y𝔤. We define 𝔤ˆ=[𝔤ˆ,𝔤ˆ] =[t,t-1]𝔤+c. Finally, observe that 𝔤𝔤ˆ via the identification x1x. For m and x𝔤, we adopt the notation x(m)=tmx.

We extend the form (,) to a bilinear form on all of 𝔤ˆ by (x(n),y(m))= δn,-m(x,y), (x(n),c)=0, (x(n),d)=0, (c,d)=1, (c,d)=0, (d,d)=0. Then (,) is nondegenerate on 𝔤ˆ.

The amne Lie algebra 𝔤ˆ has a triangular decomposition with Cartan subalgebra 𝔥ˆ=𝔥dc. Then 𝔥ˆ* has a basis {α1,,αn,δ,ζ} where {αi} are the simple roots of 𝔤 and δ(𝔥)=0 δ(d)=1 δ(c)=0; ζ(𝔥)=0 ζ(d)=0 ζ(c)=1. The bilinear form , on 𝔥* extends to all of 𝔥ˆ* by δ,αi = 0=ζ,αi, δ,ζ = 1 δ,δ = 0=ζ,ζ.

Restricted Modules and the Casimir Element

A 𝔤ˆ module V is restricted if for all vV, x(n)v=0 for each x𝔤 for n sufficiently large. In particular, simple modules L(λ) (since they are highest weight modules) are restricted modules. The restricted completion U(𝔤ˆ)ˆ of U(𝔤ˆ) is the set of infinite sums i=1xi, xiU(𝔤ˆ), such that for any restricted module V and vV, xiv=0 for all but finitely many xi. Two sums are considered the same if they act the same on all restricted modules. (See [Kac1104219], 2.5 and 12.8 for more on these definitions.)

Let 𝔤 be a simple Lie algebra. Let {ui|1idim𝔤} be a basis for 𝔤, and let ui be a dual basis with respect to (,), so that (ui,uj)=δi,j. We define the Casimir element for 𝔤ˆ to be Ω=2(c+g)d+ iuiui+2 n=1i ui(-n)ui(n) where g=12θ+2ρ,θ is the dual Coxeter number of 𝔤. (Recall ρ=12βR+β.) Observe that ΩU(𝔤ˆ)ˆ.

([Kac1104219], Theorem 2.6 and Corollary 2.6).

(i) Let x𝔤ˆ. Then as operators on a restricted 𝔤ˆ-module, [Ω,x]=0.
(ii) For λ𝔥*, Ω acts on L(λ) by λ+2ρˆ,λ, where ρˆ=ρ+ζ.

The Virasoro Algebra and Affine Lie Algebras

We now construct an action of the Virasoro algebra on restricted 𝔤ˆ-modules. This construction, known as the Sugawara construction, follows [KRa1987].

Let 𝔤 be a simple or abelian Lie algebra and let 𝔤ˆ be the associated affine Lie algebra. Recall that the Virasoro algebra is the universal central extension of the Lie algebra of differential operators on [t,t-1]. Let Dk=tk+1 ddt. These operators have a natural action on 𝔤ˆ given by [Dk,x(n)] = tk+1ddt (tnx)=nx (n+k) [Dk,c] = 0. Note that [D0,x(n)]=nx(n)=[d,x(n)]; that is, the action of D0 on 𝔤ˆ coincides with the action of d. We would like to define an action of Vir on 𝔤ˆ-modules that is consistent with these relations, For k let Tk=12n 1idim𝔤: ui(-n)ui (n+k): U(𝔤ˆ)ˆ, where the normal ordering :·: is :x(-n)y(m): { x(-n)y(m) if-nm, y(m)x(-n) if-n>m. Note that T0=12Ω-2 (c+g)d. (3.4) As we will show, the operators TkU(𝔤ˆ)ˆ mimic the action of Dk on 𝔤ˆ.

For all k,m and x𝔤, [Tk,x(m)]=-(c+g)mx(m+k).


Let m,n. From Equations 1.3 and 1.4, and Proposition 1.2.3 we have i [ui(m),ui(n)] = i[ui,ui] (m+n)+κ(ui,ui) mδm,-nc = mδm,-ncdim𝔤; (3.5) i [[x,ui](m),ui(n)] = i[[x,ui],ui] (m+n) = 2gx(m+n); (3.6) and i[x,ui] (m)ui(n) = i,jκ ([x,ui],uj) uj(m)ui(n) = i,j-κ (ui,[x,uj]) uj(m)ui(n) = i,j-uj(m) κ([x,uj],ui) ui(n) = -juj(m) [x,uj](n). (3.7) We then have [x(m),Tk] = 12n i [x(m),ui(-n)ui(n+k)] Since[x(m),c]=0, Equation 3.5 implies that we can ignore the normal ordering. = 12n i [x(m),ui(-n)] ui(n+k)+ui (-n)[x(m),ui(n+k)] = 12n i[x,ui] (m-n)ui (n+k)+mδm,n κ(x,ui)c ui(n+k) +12n iui(-n) [x,ui] (m+n+k)+m δm,-(n+k) κ(x,ui)c ui(-n) = 12n<m-k2 i[x,ui] (m-n)ui(n+k) * + 12nm-k2 i[x,ui] (m-n)ui(n+k) ** + 12n<m+k2 iui(-n) [x,ui](m+n+k) + 12nm+k2 iui(-n) [x,ui](m+n+k) +mx(m+k)c = 12n i:[x,ui](m-n)ui(n+k): +n<m-k2 gx(m+k) We reorder*using Equation 3.6 and combine with**. + n i:ui(-n)[x,ui](m+n+k): -n<-m+k2 gx(m+k) We reorderusing Equation 3.6 and combine with. +mx(m+k)c = ni : [x,ui](-n) ui(m+n+k)+ ui(-n) [x,ui] (m+n+k) : We letnn+min the first sum from the previous line. Equation 3.7 implies this sum is0. +mx(m+k)c+ -m+k2n<m-k2 gx(m+k) = mx(m+k)(c+g).

For j,k, [Tj,Tk]=(c+g)(j-k)Tj+k+δj,-kj3-j12(dim𝔤)c(c+g).


[Tj,Tk] = 12n i[Tj,ui(-n)ui(n+k)] = 12ni Tj,ui(-n) ui(n+k)+ui (-n) [Tj,ui(n+k)] We can again ignore the normalordering from Equation 3.5 = 12ni nui(-n+j) ui(n+k) (c+g)-(n+k) ui(-n)ui (n+j+k)(c+g) From the previous lemma = 12(c+g)n in:ui(-n+j)ui(n+k): +12(c+g) n<j-k2 -n(-n+j) δj,-kdim (𝔤)c +12(c+g) ni -(n+k) :ui(-n)ui(n+j+k): +12(c+g) n<-j+k2 n(n+k)δj,-k dim(𝔤)c = 12(c+g)n i(j-k) :ui(-n)ui(n+j+k): +12c(c+g) 0n<jn (-n+j)dim(𝔤)c = (j-k)c(c+g) Tj+k+ j3-j12dim(𝔤) c(c+g)

Suppose that V is a 𝔤ˆ-module where c acts by a scalar M. We will call M the level of V.

Suppose that V is a restricted 𝔤ˆ-module with level M-g. Then dk1M+gTk defines an action of Vir on V with zMM+gdim𝔤.


This follows from the previous lemma.

Let 𝔤 be a reductive Lie algebra. Then 𝔤=i=1k𝔤i, where 𝔤i is simple or abelian. For each 1ik, suppose Vi is a restricted 𝔤ˆi-module, with level Mi. The above construction gives an action of Vir on Vi; denote the action of dk by dk𝔤i.

Note that 𝔤ˆ=i=1k𝔤ˆi. Therefore, the tensor product V1Vk is a 𝔤ˆ-module (where 𝔤ˆi acts on Vi). Define dk𝔤=i=1k dk𝔤i.

The map dkdk𝔤 defines a representation of Vir on V with zi=1k z𝔤i=i=1k MiMi+gi dim𝔤i.


Since 𝔤ˆi commutes with 𝔤ˆj, the operators dk𝔤i and dk𝔤j commute, and the result follows.

For the proof of Theorem 3.4.3, we will use a slight modification of the above construction. Let 𝔤 be a reductive Lie algebra and let 𝔭𝔤 be a reductive subalgebra of 𝔤. Then, for any restricted 𝔤-module, we can construct representations of Vir corresponding to both 𝔤 and 𝔭. Denote the Vir-operators corresponding to 𝔤 and 𝔭 by dk𝔤, z𝔤 and dk𝔭, z𝔭, respectively. Define an action of dk on restricted 𝔤-modules by dkdk𝔤-𝔭= dk𝔤-dk𝔭.

The operators dk form a representation of the Virasoro algebra with z acting by zz𝔤-𝔭=z𝔤-z𝔭. Moreover, the action of Vir commutes with the action of 𝔭ˆ.


Since [dk𝔤,x(n)]=nx(n+k)=[dk𝔭,tnx(n)] for x(n)𝔭ˆ, [dk𝔤-𝔭,𝔭ˆ]=0. This implies [dk𝔤-𝔭,dk𝔭]=0. Therefore [dj𝔤-𝔭,dk𝔤-𝔭] = [dj𝔤,dk𝔤]- [dj𝔭,dk𝔭] = (j-k)dj+k𝔤-𝔭 +δj,-kj3-j12 (z𝔤-z𝔭).

The Affine Lie Algebra sl2()ˆ

The proof of the determinant formula given in the next section relies specifically on the representions of the Virasoro algebra on sl2()ˆ-modules.

We fix the simple root α of sl2(). Then, 𝔥ˆ*=αδζ.

([KRa1987] 11.4 and 12.1). Let λ𝔥ˆ* such that λ=mζ+n2α, mn0. Set q=e-ζ. Then, ch(L(λ)) ch(L(ζ))= kI ψm,n,kch (L(ζ+λ-kα)) where I = { k|-12 (m+1-n)k n2 } ; ψm,n,k = (fm,n,k-fm,n,n+1-k) j=1(1-qj) ; fm,n,k = j q(m+2)(m+3)j2+(n+1+2k(m+2))j+k2 . Also, ch(L(λ)) ch(L(ζ)) =kI j0 Δm,n,kj ch(L(ζ+λ-kα-jδ)) (3.8) where Δm,n,kj0 are such that ψm,n,k=j0Δm,n,kjqj. The minimum value of j for which Δm,n,kj is nonzero is k2.

For dominant integral weights μ and γ of sl2()ˆ, the tensor product of L(μ)L(γ) is completely reducible ([Kac1104219], Corollary 10.7). (A weight μ𝔥ˆ* is dominant integral if μ,α,μ,δ,μ,ζ0.) Therefore, Equation 3.8 implies that as sl2()ˆ-modules L(ζ)L(λ) kIj0 L(ζ+λ-kα-jδ)Δm,n,kj kIjk2 L(ζ+λ-kα-jδ)Δm,n,kj.

We use the construction from the previous section, with 𝔭=sl2() and 𝔤=sl2()sl2(). (We embed sl2() in sl2()sl2() via the diagonal map: xxx.)

Let V and W be restricted sl2()ˆ-modules. For xysl2()ˆsl2()ˆ and uwVW, define (xy)(vw)=xvw+vyw. Therefore, dk𝔤=dksl2() 1+1dksl2(). (3.10)

Let λ𝔥ˆ* such that λ=mζ+n2α, mn0. We consider the action of Vir on L(λ)L(ζ). Proposition 3.2.1 and Equation 3.4 imply that

d0sl2() acts on L(λ) by (λ+2ρˆ,λ)m+2-2d;
d0sl2() acts on L(ζ) by (ζ+2ρˆ,ζ)1+2-2d;
d0𝔭 acts on L(λ)L(ζ) by 1m+1+2Ω-2(d1+1d).
(The dual Coxeter number of sl2() is g=2.) Therefore, d0𝔤-𝔭 acts on L(λ)L(ζ) by d0𝔤-𝔭 = d0𝔤-d0𝔭= d0sl2() 1+1d0sl2() -d0𝔭 (3.11) = 12 ( λ+2ρˆ,λm+2+ ζ+2ρˆ,ζ1+2- 1m+1+2Ω ) = n(n+2)4(m+2)- 12(m+3)Ω (3.12) Also, z𝔤-𝔭= mm+23+ 11+23- m+1m+1+23= 1-6(m+2)(m+3). (3.13)

A Determinant Formula for M(λ)

Recall the Hermitian anti-involution ϕ:U(Vir)U(Vir) denned by ϕ(dk)=d-k, ϕ(z)=z. For (h,v)2, we use this to define an Hermitian form ,:M(h,c)×M(h,c) by

v+,v+=1, where v+ is a (fixed) generator of M(h,c);
xv,v=v,ϕ(x)v for xU(Vir), v,vM(h,c).
As we saw in the previous chapter, the form , has two important properties:
M(h,c)(h+n,c)M(h,c)(h+m,c) for mn (Lemma 2.4.1);
Rad,=J(h,c) (Lemma 2.4.2).
Therefore, the determinant det(M(h,c)(h+n,c))=det(d-λv+,d-λv+)λ=n=λ provides a tool to study J(h,c).

Example. Below, detM(h,c)(h+n,c) (for (h,c)2) is computed for n=1,2. detM(h,c)(h+1,c) = det(d-1v+,d-1v+) = v+,d1d-1v+ = v+,(d-1d1+2d0)v+ = v+,2hv+ = 2h. detM(h,c)(h+2,c) = det ( d-12v+,d-12v+ d-2v+,d-12v+ d-12v+,d-2v+ d-2v+,d-2v+ ) = det ( 8h2+4h 6h 6h 4h+c/2 ) = 2h(16h2+2(c-5)h+c) Theorem 3.4.3 gives a general formula for detM(h,c)(h+n,c).

The highest power of h in detM(h,c)(h+n,c) is r,s>01rsn p(n-rs), and the coefficient of this term is r,s>01rsn ((2r)ss!) p(n-rs)- p(n-r(s+1)) .


Consider the entries of A(h,c)(h+n,c)=(d-λv+,d-λ)λ=n=λ. Let λ and λ be partitions of n. Writing dλd-λ in terms of the decomposition of U(Vir) in Equation 3.1, we have that dλd-λ= ν,μpartitions d-νpν,μ (d0,z)d-μ, (3.14) where pν,μ(d0,z) is a polynomial in d0 and z. Then d-λv+,d-λv+=v+,dλd-λv+=p0,0(h,c).

Now consider p0,0(h,c) more closely. We can use the relations djdk = dkdj+(j-k) dj+kif j-k dkd-k = 2kd0+ k3-k12z to rearrange dλd-λ as in (3.14). These imply that, as a polynomial in h, the degree of (p0,0(h,c)) is less than or equal to l(λ),l(λ); and the degree of (p0,0(h,c))=l(λ) if and only if λ=λ. Therefore, for any given row of A(h,c)(h+n,c), the entry with the highest powers of h is the diagonal entry. Thus, the highest power of h in the determinant comes from the product of the diagonal entries in A(h,c)(h+n,c), The degree of this term is λ=n l(λ) = r,s>01rsn the number of partitions ofnwith at leastsparts of sizer A partition withtparts of sizerwill be counted fors=1,2,,t. = r,s>01rsn p(n-rs) By removingsparts of sizerwe obtain a partition ofn-rs. We now compute the coefficient of this term. A partition may be written as λ=(r1s1,,rjsj). Note that drd-rs = ( d-rdr+2rd0 +r3-r12z ) d-rs-1 = d-rdrd-rs-1 +d-rs-1 ( 2rd0+2r2(s-1) +r3-r12z ) = d-rsdr+ d-rs-1 ( 2rsd0+r2s(s-1) +(r3-r)s12z ) , and so drsd-rs=d-rsdrs+(2r)ss!d0s+ terms of lower degree in d0. Therefore, the coefficient of the highest power of h in a diagonal entry d-r1s1d-rjsjv+,d-r1s1d-rjsjv+ is (2rj)sj(sj)!, and the coefficient of the highest power of h in detM(h,c)(h+n,c) is λ=nλ=(r1s1,,rjsj) (2rj)sj(sj)!= r,s>0,1rsn ((2r)ss!) p(n-rs)- p(n-r(s+1)) .

Since the highest power of h in detM(h,c)(h+n,c) does not involve c, we fix c and think of detM(h,c)(h+n,c) as a polynomial in h.

Fix c. Let h0 and suppose (h-h0) divides detM(h,c)(h+k,c). Then (h-h0)p(n-k) divides detM(h,c)(h+n,c).


Suppose (h-h0) divides detM(h,c)(h+k,c). This implies Ak(h0,c) is degenerate. In other words, there is a vector (a1,,ap(k))T, ai and ai0 for at least one i, such that Ak(h0,c)(a1,,ap(k))T=0. Then. Ak(h,c) (a1ap(k))= (P1Pp(k)), where the Pi are polynomials in h which are divisible by (h-h0). Define v=i=1p(k)aid-λ(i)v+. We then have (h-h0) divides Pi=d-λ(i)v+,v.

Consider B={d-λ(i=1p(k)aid-λ(i))|λ=n-k}U(Vir-)(n,0). (Here we view U(Vir-) as a Vir-module under the adjoint action.) This set is linearly independent in U(Vir-) and can be extended to a set B of basis vectors for U(Vir-)(n,0). Let P be the matrix taking B to {d-λ|λ=n}. Then the entries of P are in and det(P)0.

Now, for d-λ(i=1p(k)aid-λ(i))B, d-λi=1p(k) aid-λ(i)v+ =d-λv. Also, (h-h0) divides d-λv,w for all wM(h,c). Then (h-h0)p(n-k) |det (Xiv+,Xjv+) Xi,XjB .

Finally, detM(h,c)(h+n,c) = det ( Pt (Xiv+,Xjv+) Xi,XjB P ) = det(Pt) det(P)det (Xiv+,Xjv+) Xi,XjB = det(P)2det (Xiv+,Xjv+) Xi,XjB . Since det(P)0, this implies (h-h0)p(n-k) divides detM(h,c)(h+n,c).

([KRa1987], [FFu1990]). For (h,c)2 and n0, detM(h,c)(h+n,c)= r,s>0,1rsn ((2r)ss!) p(n-rs)-p(n-r(s+1)) r,s>0 (h-hr,s) p(n-rs) , where hr,s(c)=148 ( (13-c)(r2+s2) +(c-1)(c-25) (r2-s2)-24rs-2 +2c ) .


Given lemmas 3.4.1 and 3.4.2, we only need to show that (h-hr,s(c)) divides detM(h,c)(h+rs,c). We will use the representation of Vir on restricted sl2()ˆ-modules to prove this. Recall (from Equation 3.9) that, for λ=mζ+n2α (mn>0), we can write the tensor product L(ζ)L(λ) of sl2()ˆ-modules as L(ζ)L(λ) = kIj0 L(ζ+λ-kα-jδ)Δm,n,kj, = kIjk2 L(ζ+λ-kα-jδ)Δm,n,kj. Let Um,n,kj be the space of highest weight vectors of weight ζ+λ-kα-jδ in L(ζ)L(λ). Then, dimUm,n,kj=Δm,n,kj and Um,n,k=j0Um,n,kj is the space of highest weight vectors of weight ζ+λ-kα for sl2()ˆ.

From Section 3.3, we know that L(λ)L(ζ) is a Vir-module. Since the action of sl2()ˆ and Vir commute (Proposition 3.3.5), we also have that Um,n,k is a Vir-module. Moreover, given Equations 3.11 and 3.13 and Proposition 3.2.1, it is clear that Um,n,kj is a weight space for the action of Vir such that

for all vUm,n,k, zv=(1-6(m+2)(m+3))v;
for vUm,n,kj, d0v = ( n(n+2)4(m+2)- 12(m+3)Ω ) v = ( n(n+2)4(m+2)+ j-(n-2k)(n-2k+2)4(m+3) ) v.
Define hr,sm = ((m+3)r-(m+2)s)2-1 4(m+2)(m+3) cm = 1-6(m+2)(m+3), where r = n+1, r = m-n+1, s = n+1-2k ifk0 s = m-n+2+2k, ifk<0. Note that (r,s)(m+2-r,m+3-s) switches these definitions.

According to Proposition 3.3.6, the minimum value of j for which Um,n,kj0 is j=k2. Therefore, as a Vir-module, Um,n,k has highest weight (hr,sm,cm). Since Δm,n,kj<, this shows that Um,n,k𝒪Vir. The character of Um,n,k is chUm,n,k = e(hr,sm,cm) jk2dim Um,n,kjqj-k2 = e(hr,sm,cm) q-k2ψm,n,k = e(hr,sm,cm) q-k2 i=1(1-qi) (fm,n,k-fm,n,n+1-k) = e(hr,sm,cm) 1 i=1(1-qi) × ( j q(m+2)(m+3)j2+((m+3)r-(m+2)s)- j q(m+2)(m+3)j2+((m+3)r+(m+2)s)+rs ) = e(hr,sm,cm) 1i=1(1-qi) =chM(hr,sm,cm) ( 1-qrs- q(m+2-r)(m+3-s)+ terms of degree>rs ) . (Here we are intentionally confusing q=e(1,0) (for the Virasoro algebra) and q=e-ζ (for sl2()ˆ).) Since the maximum weight for Um,n,k is (hr,sm,cm), L(hr,sm,cm)Um,n,k. Therefore, chL(cm,hr,sm)chUm,n,km. Let (r,s) be whichever of the pairs (r,s), (m+2-r,m+3-s) has minimum product. Note that hr,sm=hr,sm. The coefficient of qrs in chL(hr,sm,cm) is less than the coefficient of qrs in chM(hr,sm,cm), implying dimL(hr,sm,cm)(hr,sm+rs,cm)<dimM(hr,sm,cm)(hr,sm+rs,cm). Then, J(hr,sm,cm)(hr,sm+rs,cm)=Rad,(hr,sm+rs,cm)0. Since rsrs, J(hr,sm,cm)(hr,sm+rs,cm)0. We then have det(M(hr,sm,cm)(hr,sm+rs,cm))=0. Since hr,s(cm)=hr,sm, detM(h,c)(h+rs,c) vanishes at infinitely many points along the curve h=hr,s(c). Therefore, (h-hr,s(c)) divides detM(h,c)(h+rs,c).


Recall that we define an equivalence relation on the weights 𝔥* of Vir generated by the relation λμ if [M(λ):L(μ)]>0. The blocks of the Virasoro algebra are the equivalence classes of . Prom Theorem 2.3.6, we know that [M(λ):L(μ)]>0 if and only if M(μ)M(λ). We will use this alternative formulation of in order to describe the blocks of Vir.

Blocks and the Determinant Formula

For r,s, define 𝒞r,s(h,c)= { (h-148((13-c)(r2+s2)-24rs-2+2c))2 rs -1482(c-1) (c-25)(r2-s2)2 h-(r2-1)(1-c)24 r=s Viewing 𝒞r,s(h,c) as a polynomial in h, we have 𝒞r,s(h,c)= { (h-hr,s(c)) (h-hs,r(c)) rs, h-hr,r(c) r=s. Therefore, the determinant formula for detM(h,c)(h+n,c) can be rewritten as detM(h,c)(h+n,c)= r,s>0,1rsn ((2r)ss!) p(n-rs)-p(n-r(s+1)) (𝒞r,s(h,c))p(n-rs). (3.15) For (h,c)2, we know that detM(h,c)(h+n,c)=0 if and only if J(h,c)(h+n,c)0. Equation 3.15 implies that if r,s>0 are such that the product rs is minimal with 𝒞r,s(h,c)=0, then J(h,c)(h+rs,c)0 and J(h,c)(h+n,c)=0 for all n<rs. Therefore, any vector 0vJ(h,c)(h+rs,c) is a highest weight vector and so M(h+rs,c)M(h,c). Theorem 3.5.1 shows that for any r,s>0 such that 𝒞r,s(h,c)=0, M(h+rs,c)M(h,c) and that these embeddings produce a complete description of the submodule structure of M(h,c).

For fixed r,s, rs, the curves 𝒞r,s(h,c)=0 are hyperbolas. Below is the curve 𝒞1,2(h,c)=0. -20 c 20 40 -4 -2 h 4 2 Also note that

𝒞-r,s(h,c)=0 if and only if 𝒞r,s(h-rs,c)=0,

For fixed (h,c)2, 𝒞r,s(h,c) can be factored into terms linear in r and s: 𝒞r,s(h,c)= K(pr+qs+m) (pr+qs-m) (qr+ps+m) (qr+ps-m), where K,p,q,m such that pq+qp= c-136; 4pqh+(p+q)2 =m2;K= 16p2q2. Thus, for fixed (h,c), the solutions to the equation 𝒞r,s(h,c)=0 form two sets of parallel lines. The figure below illustrates the example 𝒞r,s(0,0)=0. -1.0 -0.5 0.5 1.0 -2 -1 1 2

To find all integer solutions to 𝒞r,s(h,c)=0, we only need to consider one line, say pr+qs+m=0. (If (r,s) is a point on any of the other lines, (-r,-s), (s,r) or (-s,-r) will lie on the line pr+qs+m=0.) We fix one of the lines and call it (h,c). Theorem 3.5.1 will show that the integer points (r,s) on this line encode the embeddings M(h,c)M(h,c)M(h,c).

Note that a line passes through 0, 1, or infinitely many integer points. (If the line passes through two integer points, it has rational slope and therefore passes through infinitely many integer points.) In other words, there are 0, 1, or infinitely many curves 𝒞r,s(h,c)=0, r,s, passing through a fixed point (h,c). Below we include a partial picture of curves 𝒞r,s(h,c)=0 for values of c near c=1. 0.2 0.4 0.6 0.8 1.0 c 1 2 3 4 5 h There are three points in the picture where multiple curves intersect: (h,c)=(0,1),(1,1),(4,1). As Theorem 3.5.1 shows, these weights belong to the block [(0,1)]={(m2,1)|m}.

The line (h,c) has nonzero slope. Thus, if it passes through infinitely many integer points (r,s) with rs>0 it must pass through finitely many points (r,s) with rs<0, and vice versa.

([FFu1990]). Suppose r,s>0 such that 𝒞r,s(h,c)=0. Then, M(h+rs,c)M(h,c). All embeddings of Verma modules arise in this way. Therefore, we have the following description of Verma module embeddings.
Fix a pair (h,c)2, and let (h,c) be one of the lines defined by this pair. Then the Verma module embeddings involving M(h,c) are described by one of the following four cases.

(i) Suppose (h,c) passes through no integer points. The Verma module M(h,c) is irreducible and does not embed in any other Verma modules. The block [(h,c)] is given by [(h,c)]={(h,c)}.
(ii) Suppose (h,c) passes through exactly one integer point (r,s).
(a) If rs>0, the embeddings for M(h,c) look like M(h,c) M(h+rs,c) where the arrow indicates inclusion.
(b) If rs<0, the embeddings for M(h,c) look like M(h+rs,c) M(h,c) The block [(h,c)] is given by [(h,c)]={(h,c),(h+rs,c)}.
(iii) Suppose (h,c) passes through infinitely many integer points and crosses an axis at an integer point. Label these points (ri,si) so that <r-2s-2<r-1s-1<0<r1s1<r2s2. (We exclude points (r,s) where r=0 or s=0; these correspond to the embedding M(h,c)=M(h+0,c)M(h,c).) (r-1,s-1) (r1,s1) (r2,s2) (r3,s3) (r4,s4) (r5,s5) (r6,s6) (r7,s7) (r8,s8) (r9,s9) The embeddings between the corresponding Verma modules take one of the following forms: M(h+r-1s-1,c) M(h,c) M(h+r1s1,c) slope((h,c))>0 M(h+r-1s-1,c) M(h,c) M(h+r1s1,c) slope((h,c))<0 The block [(h,c)] is given by [(h,c)]={(h,c),(h+risi,c)}.
(iv) Suppose (h,c) passes through infinitely many integer points and does not cross either axis at an integer point. Again label the integer points (ri,si) on (h,c) so that <r-2s-2<r-1s-1<0<r1s1<r2s2. Also consider the auxiliary line (h,c) with the same slope as (h,c) passing through the point (-r1,s1). Label the integer points on this line (rj,sj) as above. The embeddings between the corresponding Verma modules take one of the forms slope((h,c))>0 slope((h,c))<0 M(h+r1s1+r-3s-3,c) M(h+r1s1+r-4s-4,c) M(h+r-1s-1,c) M(h+r-2s-2,c) M(h+r1s1+r-1s-1,c) M(h,c)=M(h+r1s1+r-2s-2,c) M(h+r1s1,c) M(h+r2s2,c) M(h+r1s1+r1s1,c) M(h+r1s1+r2s2,c) M(h+r1s1+r-3s-3,c) M(h+r1s1+r-4s-4,c) M(h+r-1s-1,c) M(h+r-2s-2,c) M(h+r1s1+r-1s-1,c) M(h,c)=M(h+r1s1+r-2s-2,c) M(h+r1s1,c) M(h+r2s2,c) M(h+r1s1+r1s1,c) M(h+r1s1+r2s2,c) The block [(h,c)] is given by [(h,c)]={(h+risi,c),(h+r1s1+rjsj,c)}.

See Section 2.5 for more on Jantzen filtrations.

([FFu1990]). Let (h,c)2 and classify (h,c) according to the cases given above. Then the Jantzen filtration of M(h,c) is given as follows:

(i), (iia) M(h,c)j=0 for all j>0.
(iib) M(h,c)1=M(h+rs,c) and M(h,c)j=0 for all j>1.
(iii) M(h,c)j=M(h+rjsj,c) and M(h,c)j=0 if there is no point (rj,sj) on the line (h,c). We have the following picture of the Jantzen filtration of M(h,c): M(h,c) M(h,c)1 M(h,c)2
(iv) Write n1,1 = r1s1 n1,2 = r2s2 n2,1 = r1s1+ r1s1 n2,2 = r1s1+ r2s2 n3,1 = r3s3 n3,2 = r4s4 Then M(h,c)j=M(h+nj,1,c)+M(h+nj,2,c), and M(h+nj,1,c)M(h+nj,2,c)=M(h,c)j+1. We have the following picture of the Jantzen filtration of M(h,c): M(h,c) M(h,c)1 M(h,c)2 M(h,c)3

Partial Proof of Theorems 3.5.1 and 3.5.2.

We give a proof of cases (i) and (ii) for both theorems simultaneously.

We note that if we set γ=(1,0), then for any λ𝔥*, ,λ+tγ will be nondegenerate. Therefore, Theorem 2.5.1 holds.

Case (i): Suppose (h,c) passes through no integer points. Then, detM(h,c)(h+n,c)0 for all n0 and so M(h,c) is irreducible.

Case (ii)a: Suppose (h,c) passes through one integer point (r,s) with rs>0. Since 𝒞r,s(h,c)=𝒞-r,-s(h,c), we can assume that r,s>0 and that r and s are the only positive integers such 𝒞r,s(h,c)=0. Therefore, M(h+rs,c)M(h,c). This means M(h+rs,c)J(h,c), and so dimJ(h,c)(h+n,c) dimM(h+rs,c)(h+n,c)= p(n-rs) for all x0. However, from Theorem 2.5.1, dimJ(h,c)(h+n,c) j>0dim M(h,c)j(h+n,c)= ord(detM(h+t,c)(h+t+n,c)) =p(n-rs). Therefore M(h+rs,c)=J(h,c)=M(h,c)1 and M(h,c)j=0 for j>1. Since 𝒞r,s(h+rs,c)=𝒞-r,s(h,c), there are no integers r,s>0 such that 𝒞r,s(h+rs,c)=0. This implies M(h+rs,c) is irreducible.

Case (ii)b: Suppose (h,c) passes through one integer point (r,s) with rs<0. Since 𝒞r,s(h,c)=𝒞-r,s(h+rs,c), the point (-r,s) is on the line (h,c) (if we choose the line (h,c) carefully out of the four possible lines.) Also, (-r,s) is the only integer point on (h-rs,c). Then (h-rs,c) falls into case (ii)a.

We do not provide a proof of cases (hi) and (iv). The proof of these cases can be found in [FFu1990], Part II, Section 1. However, we do make a few comments to show that these results are reasonable.

Case (iii): Suppose (h,c) passes through infinitely many integer points (ri,si) and crosses an axis at an integer point. We will assume the slope μ is positive and (h,c) passes through a the point (0,s0) for some s0>0. (If μ<0 or (h,c) crosses the axis at a different point, we can still make arguments similar to those below.) Write s0=kp+s0 where k0 and 0s0<p. We observe that (r1,s1)=(-(k+1)q,s0-p). (If s0=0, then there are two points on the line (h,c), (-(k+1)q,-p) and (q,(k+1)p), with the same product. In this case, there is not a unique choice for (r1,s1). We choose either point.)

We have M(h+r1s1,c)M(h,c), and (r1,-s1) is on the line (h+r1s1,c) (for a careful choice of this line). The line (h,c)=(h+r1s1,c) so passes through infinitely many integer points and crosses an axis at an integer point, so that we can use the same arguments as above. We see that (r1,s1)=(-(k+2)q,-s0), which implies M(h+r1s1+r1s1,c)M(h+r1s1,c)M(h,c). Since r1s1+r1s1=r2s2, we have M(h+r2s2,c)M(h+r1s1,c)M(h,c). We can continue this argument to get M(h,c)M(h+r1s1,c)M(h+r2s2,c)M(h+r3s3),

To show that M(h,c)M(h+r-1s-1,c)M(h+r-2s-2,c), we use the fact that 𝒞r,s(h,c)=𝒞-r,s(h+rs,c) and apply the above argument to the Verma modules M(h+r-is-i,c).

Therefore, the Verma module embeddings for M(h,c) are at least those indicated in Theorem 3.5.1.

Case (iv): We again have M(h+r1s1,c)M(h,c). Since (h,c) does not cross at axis at an integer point, detM(h+r1s1,c)(h+r2s2,c)0. Therefore, M(h+r1s1,c)(h+r2s2,c)M(h,c)j=0 for j>0. However, ordM(h+t,c)(h+t+r2s2,c)=p(r2s2-r1s1)+1=dimM(h+r1s1,c)(h+r2s2,c)+1. Using Theorem 2.5.1, we see that there must be some vector 0vM(h,c)1(h+r2s2,c) so that vM(h+r1s1,c). It remains to show that v is a highest weight vector.

Another Description of Blocks

We can use the line (h,c) to generate lines corresponding to the entire block [(h,c)] in the following way. If (r,s) is an integer point on the line (h,c) let (h,c) be the line with the same slope as (h,c) and passing through the point (-r,s). Then (h,c)=(h+rs,c) corresponds to the weight (h+rs,c)[(h,c)]. Using this approach we can construct a set of lines corresponding to the weights in a given block. In this section, we begin with a line and generate the weights in a given block.

Let (μ,a,b) be a line where μ is the slope of the line and (a,b) is a point on the line. Then (μ,a,b) determines a weight (h,c) by h=(aμ-b)2-(μ-1)24μ, c=13-6(μ+1μ). We will write [(μ,a,b)] for [(h,c)] if (μ,a,b) determines (h,c).

Blocks of size two are indexed by triples { (μ,a,b)|μ -withμ <1anda,b>0 } { (μ,a,a)| μ-,μ =1,a>0 } . The weights in a block of size two [(μ,a,b)] are indexed by triples {(μ,a,±b)}.
Infinite blocks with a maximal element are indexed by triples { (pq,a,b)| p,q>0, withgcd(p,q)=1,p<q If2q, then0a<q2,0b<p If2|q, then0a<q,0b<p2 } . Infinite blocks with a minimal element are indexed by triples { (-pq,-a,b)| p,q>0, withgcd(p,q)=1,p<q If2q, then0a<q2,0b<p If2|q, then0a<q,0b<p2 } . For a block [(±pq,±a,b)] with a0, the weights in the block are indexed by triples {(pq,a,±b+2kp)|k} or {(-pq,-a,±b+2kp)|k}. For a block [(±pq,0,b)], the weights in the block are indexed by triples {(pq,0,b),(pq,0,±b+2kp)|k>0} or {(-pq,0,b),(-pq,0,±b+2kp)|k>0}.


Suppose (μ,a,b)(h,c) and [(h,c)]>1. Then, (μ,a,b) must pass through at least one integer point. Therefore, we can restrict to triples (μ,a,b) with a,b.

We now consider what values of μ will determine real values for h and c. Note that c=13-6(μ+1μ) only if μ or μ with μ=1. Suppose μ- with μ=1. Then μ=A+Bi= with B0. It is straightforward to check that h=(aμ-b)2-(μ-1)24μ only if a2=b2.

Recall that (μ,a,b), (μ,-a,-b), (1μ,a,b), and (1μ,-a,-b) all determine to the same weight (h,c). Therefore, we restrict our attention to triples (μ,a,b) with μ so that 0<μ<1 and a>0, b0; or with μ so that μ=1, a>0, and b=±a.

We first consider blocks of size two. A pair (h,c) belonging to a block of size two lies on exactly one curve 𝒞r,s(h,c)=0, and so any line determining the pair (h,c) passes through exactly one integer point. Therefore, triples in the set {(μ,a,b)|μ-withμ<1anda>0,b}{(μ,a,±a)|μ-,μ=1,a>0} are in one-to-one correspondence with such pairs (h,c).
If (h,c) is the pair defined by (μ,a,b), then M(h+ab,c)M(h,c). This implies that (h+ab,c) corresponds to the line with slope μ passing through the point (-a,b). Therefore, any block of size two can be identified with a set {(μ,a,±b)}, with μ, a, and b as in the previous paragraph. Taking one triple from each of these pairs of triples, we see that the set { (μ,a,b)| μ-withμ <1anda,b>0 } { (μ,a,a)| μ-,μ =1,a>0 } indexes the blocks of size two.
Now we consider infinite blocks. Let (h,c) be a pair in an infinite block. We assume 0<μ1 and μ. (The arguments for 1μ<0 are the similar.) Write μ=pq such that p and q are relatively prime. Consider weights (h,c) which are maximal in their own block [(h,c)]. Since M(h,c) does not embed in any other Vermas, any line determined by (h,c) must pass through only integer points (a,b) such that ab>0. It is clear that the triples corresponding to maximal weights are contained in the set {(μ,a,b)|0a<q2,0b<p} (or {(μ,a,b)|0a<q,0b<p2} if q is even). However, (pq,a,b) and (pq,q-a,p-b) determine the same weight. Therefore, the set {(μ,a,b)|0a<q2,0b<p} (or {(μ,a,b)|0a<q,0b<p2}) contains exactly one triple corresponding to each such pair (h,c).
We can also describe the block [(h,c)]. Let (μ,a,b) (μ=pq0 with μ1 and a,b with 0a<q2 be a triple which determines (h,c). Then the integer points lying on (pq,a,b) are (a+kq,b+kp), k. This implies that M(h+(a+kq)(b+kp),c)M(h,c). Therefore, the line given by (μ,-a+kq,-(b+kp)) must determine (h+(a+kq)(b+kp),c). This may not produce all pairs (h,c) in the block (as in case (iv) of Theorem 3.5.1). Therefore, we also consider the pair (h+ab,c), which is determined by the triple (μ,a,-b). Using the same argument as above, we get the triples (μ,a,-b). Therefore, the set { (pq,a-kq,±b+kp) |k } { (pq,a,±b+2kp) |k } is in general a set of representatives for the elements of the block corresponding to (pq,a,b). If a=0, the triples (pq,0,b+2kp) and (pq,0,-b-2kp) correspond to distinct lines but still determine the same weight. In this case, the set {(pq,0,±b+2kp)|k0} forms set of representatives of the elements of the block.
Consider the example with μ=23. s=8 s=4 s=0 s=-4 (23,-2,1) (23,1,1) (23,1,-1) (23,4,-3) (-2,-1) (1,1) (4.3) The set of integer points {(a,b)2|0a<32,0b<2} indexes the infinite blocks with c=13-6(23+32)=0. The line (23,1,1) determines the weight (0,0). From the integer points (1,1), (-2,-1), and (4,3) on the line (23,1,1), we get the lines (1,-1), (23,-2,1), and (23,4,-3); these lines determine the weights (1,0), (2,0), and (12,0) respectively. In general, the set of points {(1,4k±1)|k} correspond to the block {((12k+2±3)2-124,0)|k}= {(j(3j±1)2,0)|j0}.

Define the group W=s0,s1|si2=1. We can define an action of W on the triples (μ,a,b) so that

(i) a block of size two [(μ,a,b)] is the orbit of the subgroup s0W;
(ii) a infinite block [(±pq,±a,b)], with a0, is the orbit of W;
(iii) a infinite block [(±pq,0,b)], b0, is the orbit of the subgroup s1,s0s1s0W;
(iv) a infinite block [(±pq,0,0)] is of the form {(s1s0)k(±pq,0,0)|k0}.


We define an action of W on triples (μ,a,b) as follows:

s0 is the reflection about 0: s0(μ,a,b)=(μ,a,-b);
for μ=pq, s1 is the reflection about p: s1(±pq,±a,b)=(±pq,±a,-(b-p)+p)=(±pq,±a,-b+2p).
Then (i) and (ii) follow from the previous proposition.

For (iii), note that s0s1s0 is the reflection about -p. Also, (pq,0,-b+2kp) and (pq,0,b-2kp) determine the same weight. Then s1,s0s1s0 generates [(±pq,0,b)], where we replace (pq,0,-b+2kp) with (pq,0,b-2kp) for k even.

Finally, we have that s1s0 is translation by 2p. Then, (iv) follows.

Translation Functors

We now consider M(h,c)L(h,c). From Theorem 2.3.12, we know that M(h,c) L(h,c)= [μ][𝔥*] (M(h,c)L(h,c))[μ] and (M(h,c)L(h,c))[μ]0 only if [μ]=[(h+h+k,c+c)] for some k0. Moreover, we know this submodule has a filtration by Verma modules. In this section, we use the contravariant form to better describe (M(h,c)L(h,c))[μ].

Recall ,:M(h,c)L(h,c)×M(h,c)L(h,c) is defined by vw,vw= v,vw,w where v,vM(h,c) and w,wL(h,c). This form on M(h,c)L(h,c) is contravariant.

Let (h,c),(h,c)2, and let {wk,j|1jdim(L(h,c)(h+k,c))} be a basis for L(h,c)(h+k,c). From Lemma 2.6.2, the following sets are bases for (M(h,c)L(h,c))(h+h+n,c+c): { d-λv+ wk,i| λ=n, 1idim (L(h,c)(h+k,c)) } ; (3.16) { d-λ (v+wk,i) | λ=n, 1idim (L(h,c)(h+k,c)) } . (3.17) We defined det(M(h,c)L(h,c))(h+h+k,c+c) det(d-λv+wm,j,d-λ*v+wm,j) where the entries in the matrix are indexed over partitions λ and λ* and positive integers m,m,j,j such that λ=k-m, λ*=k-m, 1jdimL(h,c)(h+m,c), and 1jdimL(h,c)(h+m,c). From Lemma 2.6.3, we have det(M(h,c)L(h,c))(h+h+k,c+c) = jk (detM(h,c)(h+k-j,c)) dimL(h,c)(h+j,c) × (detL(h,c)h+j,c) p(k-j) . (3.18)

For (h,c),(h,c)2 and k0, det(M(h,c)L(h,c))(h+h+k,c+c) is given by 0jk (detM(h+h+j,c+c)(h+h+k,c+c)) dimL(h,c)(h+j,c) (aj(h,c)(h,c)detL(h,c)(h+j,c))p(k-j), where aj(h,c)(h,c)= 1rsrsj (𝒞r,s(h,c)𝒞r,s(h+h+j-rs,c+c)) dimL(h,c)(h+j-rs,c) .


Using Equation 3.18, we only need to show that jk (aj(h,c)(h,c)) p(k-j) = jk ( detM(h,c)(h+k-j,c) detM(h+h+j,c+c)(h+h+k,c+c) ) dimL(h,c)(h+j,c) . Note that 0jk ( detM(h,c)(h+k-j,c) detM(h+h+j,c+c)(h+h+k,c+c) ) dimL(h,c)(h+j,c) = 0jk1rs (𝒞r,s(h,c)𝒞r,s(h+h+j,c+c)) p(k-j-rs)dimL(h,c)(h+j,c) = j1rs (𝒞r,s(h,c)𝒞r,s(h+h+j,c+c)) p(k-j-rs)dimL(h,c)(h+j,c) We can let the product range over alljsince p(k-j-rs)=0forj>kanddimL(h,c)(h+j,c)=0forj<0. = j1rs (𝒞r,s(h,c)𝒞r,s(h+h+j-rs,c+c)) p(k-j)dimL(h,c)(h+j-rs,c) We shiftjj-rs = 0jk1rsrsj (𝒞r,s(h,c)𝒞r,s(h+h+j-rs,c+c)) p(k-j)dimL(h,c)(h+j-rs,c) .

Fix (h,c)2. Consider (h,c)2 such that (h+n,c)[(h,c)] for any n>0 (i.e. M(h,c) is irreducible). For each [μ][𝔥*] and n0, there is a projection map Prn[μ]: (M(h,c)L(h,c)) (h+h+n,c+c) ((M(h,c)L(h,c))[μ]) (h+h+n,c+c) given by Prn[μ](w)=w- [γ][μ] 1mw,vi[γ] vi[γ] where {v1[γ],,vm[γ]} and {v1[γ],,vm[γ]} are dual bases for ((M(h,c)L(h,c))[γ])(h+h+n,c+c).


Since (h+n,c)[(h,c)] for all n0, Equation 3.18 implies that the contravariant form is nondegenerate on (M(h,c)L(h,c))(h+h+n,c+c). From Proposition 2.7, distinct blocks are orthogonal with respect to the form. Therefore, the contravariant form is nondegenerate on each block.

Let {v1[γ],,vm[γ]} be a basis for ((M(h,c)L(h,c))[γ])(h+h+n,c+c). Since the contravariant form is nondegenerate on this space, there is a dual basis {v1[γ],,vm[γ]} for this space, i.e. vi[γ],vl[γ]=δi,l.

We define a map Prn[μ]: (M(h,c)L(h,c)) (h+h+n,c+c) (M(h,c)L(h,c)) (h+h+n,c+c) by Prn[μ](w)=w- [γ][μ] 1mw,vi[γ] vi[γ]. Note that Prn[μ](w),vi[γ]=0 whenever [γ][μ] since distinct blocks are orthogonal. Therefore, Prn[μ](w)((M(h,c)L(h,c))[μ])(h+h+n,c+c).

Also, for w((M(h,c)L(h,c))[μ])(h+h+n,c+c), Prn[μ](w)=w.

Fix (h,c)2 and n0. Suppose (h,c)2 is such that (h+j,c)[(h,c)] and (h+h+j,c+c)[(h+h+k,c+c)] for all j,kn with jk.
Then the submodule of M(h,c)L(h,c) generated by 0jn (M(h,c)L(h,c)) (h+h+j,c+c) is isomorphic to 0jn M(h+h+j,c+c)dimL(h,c)(h+j,c). For a suitable choice of generating highest weight vectors {vj,i+|1idim(L(h,c))(h+j,c)} of 0jnM(h+h+j,c+c)dimL(h,c)(h+j,c)M(h,c)L(h,c), this sum is orthogonal with respect to the contravariant form on M(h,c)L(h,c), and 1idim(L(h,c))(h+j,c) vj,i+,vj,i+= aj(h,c)(h,c) detL(h,c)(h+j,c).


Since [(h,c)][(h+j,c)] for all jn, the projection maps from the previous lemma are well-defined.

We have assumed [(h+h+j,c+c)][(h+h+k,c+c)] for j,kn and jk. Therefore, for μj=(h+h+j,c+c) with jn, the set {Prj[μj](v+wj,i)|1idimL(h,c)(h+j,c)} is a basis for ((M(h,c)L(h,c))[μj])(h+h+j,c+c), made up of highest weight vectors. Choose vectors {vj,i+} such that

the transition matrix from {Prj[μj](v+wj,i)} to {vj,i+} has determinant 1;
vj,i+,vj,k+=0 if ik.
Note that ivj,i+,vj,i+ =det ( Prj[μjj] (v+wj,i), Prj[μj] (v+wj,k) ) . Then det(d-λPrj[μj](v+wj,i),d-μPrj[μj](v+wj,k))=i(d-μvj,i+,d-λvj,i+) is (ivj,i+,vj,i+)p(n-j)× (det(h+h+j,c+c)(h+h+n,c+c)) dimL(h,c)(h+j,c) . Therefore, we only need to determine ivj,i+,vj,i+. We do this inductively. Suppose det(Prk[μk](v+wk,i),Prk[μk](v+wk,l)) =ak(h,c)(h,c) detL(h,c)(h+k,c) for k<j. Since distinct blocks are orthogonal, we have det(M(h,c)L(h,c))(h+h+j,c+c) is given by kjdet ( d-μPrj[γk] (v+d-λw+), d-μPrj[γk] (v+d-λw+) ) = det ( d-μPrj[μ] (v+d-λw+), d-μPrj[μ] (v+d-λw+) ) k<j ( (detM(h+h+k,c+c)(h+h+j,c+c)) dimL(h,c)(h+k,c) × (ak(h,c)(h,c)detL(h,c)(h+k,c)) p(k-j) ) . From Lemma 3.6.1, this implies det (Prj[μj](v+wj,i),Prj[μj](v+wj,l))= aj(h,c)(h,c) detL(h,c)(h+j,c).

For γ=(h+h+k,c+c), the set Bn[γ]= { d-μ(Prj[γ](v+d-λw+)) |(h+h+j,c+c) [γ],λ=j, μ=n-j } is a basis for ((M(h,c)L(h,c))[γ])(h+h+n,c+c). Define det((M(h,c)L(h,c))[γ])(h+h+n,c+c) =det(v,w)v,wBn[γ]

Let (h,c),(h,c)2, [γ][𝔥*], and n0. Suppose (h+k,c)[(h,c)] for all 0kn. Then, det((M(h,c)L(h,c))[γ])(h+h+n,c+c) is = (h+h+j,c+c)[γ] ( (detM(h+h+j,c+c)(h+h+n,c+c))dimL(h,c)(h+j,c) × (aj(h,c)(h,c)detL(h,c)(h+j,c))p(n-j) )


Let n0 and let K be any set of positive integers between 0 and n. Fix (h,c)2 and consider all (h,c)2 such that [(h+h+k,c+c)][(h+h+k,c+c)] for any k such that kK.

Let MK=kK (M(h,c)L(h,c))(h+h+k,c+c) We can construct projection maps PrjK:(M(h,c)L(h,c))(h+h+j,c+c)(MK)(h+h+n,c+c) analogous to those in Propostion 3.6.2. Applying these projection maps to the basis to {v+wij}, we can construct a basis {v1,,vm} for (MK)(h+h+n,c+c) which are linear combinations of the basis {d-λ(v+wij)|jn,λ=n-j} with coefficients which are rational functions of h and c.

Consider det(MK)(h+h+n,c+c) =det(vi,vj)1i,jm. This will be a rational function in h and c.

For most choices of (h,c), [(h+h+k,c+c)]={(h+h+k,c+c)} for each kK and so MKkKM (h+h+k,c+c)dimL(h,c)(h+k,c). Write γk=(h+h+k,c+c). Lemma 3.6.3 implies that for such choices of h and c, det(MK)(h+h+n,c+c) = kKdet ((M(h,c)L(h,c))[γk])(h+h+n,c+c) (3.19) = kK (ak(h,c)(h,c)L(h,c)(h+k,c))p(n-k) M(h+h+k,c+c)(h+h+n,c+c). (3.20) Since det(MK)(h+h+n,c+c) is a rational function of h and c, Equation 3.20 holds for all (h,c) where det(MK)(h+h+n,c+c) is defined. In particular, if [γ]{(h+h+j,c+c)|0jn}={(h+h+k,c+c)|kK}, then det((M(h,c)L(h,c))[γ])(h+h+n,c+c) is j[γ] (aj(h,c)(h,c)detL(h,c)(h+j,c))p(n-j) (detM(h+h+j,c+c)(h+h+n,c+c))dimL(h,c)(h+j,c).

We define a Jantzen-type filtration on M(λ)L(μ) in the following way. For an indeterminant t, we define the Vir-module M(h+t,c) as in Section 2.5. The map ε:[t] (t0) to a map ε:M(h+t,c) L(h,c) M(h,c)L(h,c). For each j0, define (M(h+t,c)L(h,c))j={vM(h+t,c)L(h,c)|tj|v,wfor allwM(h+t,c)L(h,c)} and (M(h,c)L(h,c))j= ε((M(h+t,c)L(h,c))j).

Let j0. Then (M(h,c)L(h,c))j =M(h,c)jL(h,c).


Let v(M(h+t,c)L(h,c))j. Since distinct weight spaces are orthogonal with respect to the contravariant form, we may assume v(M(h+t,c)L(h,c))(h+h+t+n,c+c) for some n0. For each jn, let {wj,i} be a basis for L(h,c) which is orthonormal with respect to the contravariant form. (Such a basis exists since the contravariant form is nondegenerate on L(h,c).) We may write v=j=0ni vj,iwj,i for some vj,iM(h+t,c). Then, for any vM(h+t,c), kn, and 1mdimL(h,c)(h+k,c), v,vwk,m = j=0ni vj,iwj,i,vwk,m = j=0ni vj,i,v wj,i,wk,m = vk,m,v. This implies tj|vk,m,w for all wM(h,c)L(h,c) and so vk,mM(h+t,c).

Let (h,c),(h,c)2 and [μ][𝔥*]. Then, for each n0 j>0dim ((M(h,c)L(h,c))j[μ])(h+h+n,c+c) is ord ( 0kn(h+h+k,c+c)[μ] (ak(h,c)(h+t,c))p(n-k) ) .


From the previous lemma and Theorem 3.5.2, we know that for each n0 j>0dim ((M(h,c)L(h,c))j)(h+h+n,c+c) is given by ord0kn (detM(h,c)(h+t+n-k,c)) dimL(h,c)(h+k,c) =ord0kn (ak(h,c)(h+t,c))p(n-k). (3.21) Therefore, to prove the result, we only need to show how these zeros are distributed.

From the previous lemma, we have (M(h,c)L(h,c))j=M(h,c)jL(h,c). Theorem 3.5.2 gives the structure of the Jantzen filtration for M(h,c). We will consider the cases of this result separately.

Case (i): In this case, M(h,c) is irreducible and so M(h,c)j=0 for all j. This corresponds to detM(h,c)(h+n,c)0 for all n0, implying ord(ak(h,c)(h+t,c))=0.

Cases (ii) and (iii): There are integer points (ri,si), 1ik for some k>0, on the line (h,c) such that

M(h,c)j=M(h+rjsj,c) for jk;
(M(h,c)j)(h+m,c)=0 for j>k and mn.
Then we have a correspondence between
distinct zeros in detM(h,c)(h+m,c), which will have the form 𝒞rj,sj(h,c);
j such that (M(h,c)j)(h+m,c)0.
Moreover, the multiplicity of the zero 𝒞rj,sj(h,c) in detM(h,c)(h+m,c) is p(m-rjsj)=dimM(h+rjsj,c)(h+m,c).

Now, if 1jk, we can describe the decomposition of M(h,c)jL(h,c)=M(h+rjsj,c)L(h,c) by blocks. In particular, by Proposition 2.6.1, we know that (M(h+rjsj,c)L(h,c))[μ] has a filtration by Verma modules 0=M0M1 such that

Mi/Mi-1M(h+rjsj+h+kj,i,c+c)dimL(h,c)(h+kj,i,c) for each kj,i0 such that (h+rjsj+h+kj,i,c+c)[μ].
This means that dim((M(h,c)jL(h,c))[μ])(h+h+n,c+c)= kj,ip(n-(rjsj+kj,i)) dimL(h,c)(h+kj,i,c), where we sum over {kj,i|(h+rjsj+h+kj,i,c+c)[μ]}. Then, jdim((M(h,c)jL(h,c))[μ])(h+h+n,c+c) is given by jkj,i|(h+rjsj+h+kj,i,c+c)[μ] p(n-(rjsj+kj,i)) dimL(h,c)(h+kj,i,c). (3.22) On the other hand, 0kn(h+h+k,c+c)[μ] (ak(h,c)(h+t,c))p(n-k) is ord ( 0kn(h+h+k,c+c)[μ] 1rsrsk (𝒞r,s(h+t,c)𝒞r,s(h+t+h+k-rs,c+c)) dimL(h,c)(h+k-rs,c)p(n-k) ) . (3.23) Given the correspondence stated earlier, we see that (3.23) is 0kn(h+h+k,c+c)[μ] j|rjsj<k dimL(h,c)(h+k-rjsj,c) p(n-k), which is equal to (3.22).

Case (iv): We have M(h,c)j=M (h+nj,1,c)+ M(h+nj,2,c) (3.24) where M(h+nj,1,c) M(h+nj,2,c)= M(h,c)j+1. (3.25) Consider nj0,i maximal so that nj0,in. Then, (M(h,c)j0L(h,c))(h+h+n,c+c) = (M(h+nj0,1,c)L(h,c))(h+h+n,c+c) = (M(h+nj0,2,c)L(h,c))(h+h+n,c+c). Again, we know the decomposition of each of these summands by blocks. The module (M(h+nj0,i,c)L(h,c))[μ] has a filtration by Verma modules where M(h+nj0,i+kj0,i,l,c+c) appears with multiplicity dimL(h,c)(h+kj0,i,l,c) for each kj0,i,l such that (h+nj0,i+kj0,i,l,c+c)[μ]. Therefore, dim((M(h,c)j0L(h,c))[μ])(h+h+n,c+c) is = dim((M(h+nj0,1,c)L(h,c))[μ])(h+h+n,c+c) + dim((M(h+nj0,2,c)L(h,c))[μ])(h+h+n,c+c) = kj0,i,ldim L(h,c)(h+kj0,i,l,c) p(n-(nj0,i+kj0,i,l)).

Using (3.24) and (3.25), we can similarly argue that dim((M(h,c)j0-1L(h,c))[μ])(h+h+n,c+c) is = dim((M(h+nj0-1,1,c)L(h,c))[μ])(h+h+n,c+c) +dim((M(h+nj0-1,2,c)L(h,c))[μ])(h+h+n,c+c) -dim((M(h+,c)j0L(h,c))[μ])(h+h+n,c+c) = kj0,i,l dimL(h,c)(h+kj0-1,i,l,c) p(n-(nj0-1,i+kj0-1,i,l)) -kj0-1,i,l dimL(h,c)(h+kj0,i,l,c) p(n-(nj0,i+kj0,i,l)). In general, dim((M(h,c)j0-mL(h,c))[μ])(h+h+n,c+c) is given by s=0m(-1)m-s kj0-s,i,l dimL(h,c)(h+kj0-s,i,l,c) p(n-(nj0-s,i+kj0+s,i,l)). Suppose that nj,in for jm and nj,i>n for j>m. (It may be the case that nj,in and nj,2>n. However, the same argument works with only minor modifications.) Then, j>0dim ((M(h,c)jL(h,c))[μ])(h+h+n,c+c) is s=0m-12 k2s+1,i,l dimL(h,c)(h+k2s+1,i,l,c) p(n-(n2s+1,i+k2s+1,i,l)) (3.26) Again, the distinct zeros in detM(h,c)(h+m,c) will be exactly of the form 𝒞rj,i,sj,i(h,c), where j=2s+1, 0sm-12, and rj,isj,i=nj,i. Moreover, the multiplicity of the zero 𝒞rj,i,sj,i(h,c) in detM(h,c)(h+m,c) is p(m-nj,i)=dimM(h+nj,i,c)(h+m,c). We then see that ord ( 0kn(h+h+k,c+c)[μ] (ak(h,c)(h+t,c))p(n-k) ) is 0kn(h+h+k,c+c)[μ] j,i|nj,i<k dimL(h,c)(h+k-nj,i,c) p(n-k), which is equal to (3.26).

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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