## Translation Functors and the Shapovalov Determinant

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 $𝔤\text{.}$ The quantum group ${U}_{q}\left(𝔤\right)$ is a deformation of the universal enveloping algebra $U\left(𝔤\right)$ of $𝔤,$ and it possesses many properties that $𝔤$ does. In particular, we can use the Shapovalov determinant to study translation functors for ${U}_{q}\left(𝔤\right)\text{.}$ The background information on quantum groups presented in this chapter can be found in [CPr1994].

Let $q$ be an indeterminate. For integers $m,n,d\in ℤ,$ we define the following elements of $ℤ\left[q,{q}^{-1}\right]\text{:}$

 • ${\left[m\right]}_{d}=\frac{{q}^{dm}-{q}^{-dm}}{{q}^{d}-{q}^{-d}}$ • ${\left[m\right]}_{d}!={\left[m\right]}_{d}{\left[m-1\right]}_{d}\cdots {\left[2\right]}_{d}{\left[1\right]}_{d}$ for $m\ge 0$ • ${\left[\genfrac{}{}{0}{}{m}{n}\right]}_{d}=\frac{{\left[m\right]}_{d}!}{{\left[n\right]}_{d}!{\left[m-n\right]}_{d}!}$ for $m\ge n>0\text{.}$
Let $𝔤$ be a finite-dimensional complex semisimple Lie algebra, and let $A={\left({a}_{ij}\right)}_{1\le i,j\le n}$ be the associated Cartan matrix. Choose ${d}_{i}\in {ℤ}_{>0},$ $1\le i\le k$ minimal so that the matrix ${\left({d}_{i}{a}_{ij}\right)}_{1\le i,j\le n}$ is symmetric. To simplify notation, we will write ${q}_{i}={q}^{{d}_{i}}$ and ${\left[·\right]}_{i}={\left[·\right]}_{{d}_{i}}\text{.}$ There is some $k\in {ℤ}_{>0}$ so that ${d}_{i}=k\frac{⟨{\alpha }_{i},{\alpha }_{i}⟩}{2}\text{.}$ For $\beta \in {R}^{+},$ define ${d}_{\beta }=k\frac{⟨\beta ,\beta ⟩}{2}\text{.}$ Suppose $\beta =w{\alpha }_{i}$ for $w\in W$ and ${\alpha }_{i}$ a simple root. Equation 1.2 implies ${d}_{\beta }={d}_{i}\text{.}$

The quantum group ${U}_{q}\left(𝔤\right)$ associated to $𝔤$ is the $ℚ\left(q\right)\text{-algebra}$ generated by the elements ${E}_{i},$ ${F}_{i},$ and ${K}_{i}^{±1}$ $\left(1\le i\le k\right)$ with relations $KiKj=KjKi, KiKi-1=1=Ki-1Ki; (4.1)$ $KiEj=qiaij EjKi,KiFj =qi-aijFj Ki; (4.2)$ $EiFj-FjEi= δi,j Ki-Ki-1 qi-qi-1 ; (4.3)$ $∑r=01-aij (-1)r [1-aijr]i Ei(1-aij)-r EjEir=0, (4.4)$ $∑r=01-aij (-1)r [1-aijr]i Fi(1-aij)-r FjFir=0 for i≠j. (4.5)$

The quantum group ${U}_{q}\left(𝔤\right)$ is a Hopf algebra with the following maps:

 • coproduct: $\mathrm{\Delta }\left({K}_{i}\right)={K}_{i}\otimes {K}_{i},$ $\mathrm{\Delta }\left({E}_{i}\right)={E}_{i}\otimes {K}_{i}^{-1}+1\otimes {E}_{i},$ $\mathrm{\Delta }\left({F}_{i}\right)={F}_{i}\otimes 1+{K}_{i}\otimes {F}_{i}\text{;}$ • counit: $\epsilon \left({K}_{i}\right)=1,$ $\epsilon \left({E}_{i}\right)=0=\epsilon \left({F}_{i}\right)\text{;}$ • antipode: $S\left({K}_{i}\right)={K}_{i}^{-1},$ $S\left({E}_{i}\right)=-{E}_{i}{K}_{i},$ $S\left({F}_{i}\right)=-{K}_{i}^{-1}{F}_{i}\text{.}$
(See ([CPr1994] 9.1.1) for a proof that these maps give a Hopf algebra structure.) We define the following $ℚ\left(q\right)\text{-subalgebras}$ of ${U}_{q}\left(𝔤\right)\text{:}$
 • ${U}_{q}^{+}\left(𝔤\right)$ is the subalgebra generated by $\left\{{E}_{i} | 1\le i\le k\right\}\text{.}$ • ${U}_{q}^{0}\left(𝔤\right)$ is the subalgebra generated by $\left\{{K}_{i}^{±1} | 1\le i\le k\right\}\text{.}$ • ${U}_{q}^{-}\left(𝔤\right)$ is the subalgebra generated by $\left\{{F}_{i} | 1\le i\le k\right\}\text{.}$

([CPr1994] 9.1.3). ${U}_{q}\left(𝔤\right)\cong {U}_{q}^{-}\left(𝔤\right)\otimes {U}_{q}^{0}\left(𝔤\right)\otimes {U}_{q}^{+}\left(𝔤\right)\text{.}$

There is a natural pairing between ${U}_{q}^{+}\left(𝔤\right)$ and ${U}_{q}^{-}\left(𝔤\right)\text{.}$ The following maps make use of this pairing. Define a $ℚ\text{-anti-automorphism}$ $\tau :{U}_{q}\left(𝔤\right)\to {U}_{q}\left(𝔤\right)$ by $τ(Ei)=Fi; τ(Fi)=Ei; τ(Ki)=Ki-1; τ(q)=q-1.$ Define a $ℚ\left(q\right)\text{-anti-automorphism}$ $\varphi :{U}_{q}\left(𝔤\right)\to {U}_{q}\left(𝔤\right)$ by $ϕ(Ei)=Ki-1Fi; ϕ(Fi)=EiKi; ϕ(Ki)=Ki.$ Observe that $\varphi$ commutes with the coproduct map: $Δ∘ϕ= (ϕ⊗ϕ)∘Δ.$ This property will be useful later when we consider tensor products of modules.

### The Restricted Integral Form ${U}_{𝒜}^{\text{res}}\left(𝔤\right)$

Let $𝒜=ℤ\left[q,{q}^{-1}\right]\text{.}$ The restricted integral form ${U}_{𝒜}^{\text{res}}\left(𝔤\right)$ of ${U}_{q}\left(𝔤\right)$ is the $𝒜\text{-subalgebra}$ of ${U}_{q}\left(𝔤\right)$ generated by $Ki±1; Ei(r)=Eir[r]i!; Fi(r)=Fir[r]i! for 1≤i≤k,r>0.$ The algebra ${U}_{𝒜}^{\text{res}}\left(𝔤\right)$ inherits many properties from ${U}_{q}\left(𝔤\right),$ including the Hopf algebra structure and the triangular decomposition. The restricted integral form ${U}_{𝒜}^{\text{res}}\left(𝔤\right)$ is a Hopf algebra with the following maps

 • coproduct: $Δ(Ki) = Ki⊗Ki Δ(Ei(r)) = ∑j=0r qij(r-j) Ei(j)⊗Ki-j Ei(r-j) Δ(Fi(r)) = ∑j=0r qi-j(r-j) Fi(j) Kir-j⊗ Fi(r-j).$ • 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 ${U}_{𝒜}^{\text{res}}\left(𝔤\right)$ decomposes as $U𝒜res(𝔤)≃ U𝒜res(𝔤)+⊗ U𝒜res(𝔤)0⊗ U𝒜res(𝔤)-$ where

 • ${U}_{𝒜}^{\text{res}}{\left(𝔤\right)}^{+}={U}_{q}^{+}\left(𝔤\right)\cap {U}_{𝒜}^{\text{res}}\left(𝔤\right)$ is generated by ${Ei(r) | 1≤i≤k,r>0};$ • ${\left({U}_{𝒜}^{\text{res}}\left(𝔤\right)\right)}^{0}={U}_{q}^{0}\left(𝔤\right)\cap {U}_{𝒜}^{\text{res}}\left(𝔤\right)$ is generated by ${ Ki±1, [Ki;cr] | 1≤i≤k,c∈ℤ,r>0 } ,$ with $[Ki;cr]= ∏s=1r Kiqic+1-s-Ki-1qis-c-1 qis-qi-s ;$ • ${\left({U}_{𝒜}^{\text{res}}\left(𝔤\right)\right)}^{-}={U}_{q}^{-}\left(𝔤\right)\cap {U}_{𝒜}^{\text{res}}\left(𝔤\right)$ is generated by ${Fi(r) | 1≤i≤k,r>0}.$

### A PBW Basis for ${U}_{𝒜}^{\text{res}}\left(𝔤\right)$

In the following construction (due to [Lus1990-4]), a braid-group action on ${U}_{𝒜}^{\text{res}}\left(𝔤\right)$ is used to define the element of ${U}_{𝒜}^{\text{res}}{\left(𝔤\right)}^{±}$ corresponding to each positive root. These elements are then assembled into a PBW basis for ${{U}_{𝒜}^{\text{res}}\left(𝔤\right)}^{±}\text{.}$

([Lus1990-4] 3.1). For each $1\le i\le n,$ there is a unique $ℚ\left(q\right)\text{-automorphism}$ ${T}_{i}:{U}_{𝒜}^{\text{res}}\left(𝔤\right)\to {U}_{𝒜}^{\text{res}}\left(𝔤\right)$ defined by $Ti(Kj) = KjKi-aij; Ti(Ej) = { -FiKi if i=j ∑r=0-aij (-1)r-aij qi-r Ei(-aij-r) EjEi(r) if i≠j ; Ti(Fj) = { -Ki-1Ei if i=j ∑r=0-aij (-1)r-aij qirFi(r) FjFi(-aij-r) if i≠j$

Fix a reduced expression for the longest element of the Weyl group, ${w}_{0}={s}_{{i}_{1}}\cdots {s}_{{i}_{N}}\text{.}$ This gives an ordering of the positive roots: $βj=si1⋯ sij-1αij, 1≤j≤N.$ Define ${F}_{{\beta }_{j}}\in {{U}_{𝒜}^{\text{res}}\left(𝔤\right)}^{-}\subseteq {U}_{q}^{-}\left(𝔤\right)$ to be $Fβj=Ti1⋯ Tij-1Fj.$ We similarly define ${E}_{{\beta }_{j}}\text{.}$ Finally, define $Kβj=Ti1⋯ Tij-1Kj= Ki1⋯Kim$ where ${\beta }_{j}={\alpha }_{{i}_{1}}+\cdots +{\alpha }_{{i}_{m}}\text{.}$

([Lus1990-4], 6.7). The sets ${ Eβn(sn)⋯ Eβ1(s1) | si∈ℤ≥0 } and { Fβ1(s1)⋯ FβN(sN) | si∈ℤ≥0 }$ are bases for ${{U}_{𝒜}^{\text{res}}\left(𝔤\right)}^{+}$ and ${{U}_{𝒜}^{\text{res}}\left(𝔤\right)}^{-},$ respectively.

These sets are also bases for ${U}_{q}^{+}\left(𝔤\right)$ and ${U}_{q}^{-}\left(𝔤\right),$ respectively. Note that ${T}_{i}$ commutes with the $ℚ\text{-antiautomorphism}$ $\varphi ,$ and so we have $ϕ(Eβ)=Fβ.$

### The Category $𝒪$

We begin by defining weights and weight spaces. A weight of ${U}_{q}\left(𝔤\right)$ is an element $\mathrm{\Omega }\in {\text{Hom}}_{ℚ\left(q\right)}\left({U}_{q}^{0}\left(𝔤\right),ℚ\left(q\right)\right)\text{.}$ Since the set $\left\{{K}_{i}^{±1} | 1\le i\le k\right\}$ generates ${U}_{q}^{0}\left(𝔤\right)$ and $\mathrm{\Omega }\left({K}_{i}^{-1}\right)={\left(\mathrm{\Omega }\left({K}_{i}\right)\right)}^{-1},$ $\mathrm{\Omega }$ is determined by its action on each of the ${K}_{i}\text{.}$ Define ${\mathrm{\Omega }}_{i}=\mathrm{\Omega }\left({K}_{i}\right)$ and identify $\mathrm{\Omega }$ with the $k\text{-tuple}$ $\left({\mathrm{\Omega }}_{1},\dots ,{\mathrm{\Omega }}_{k}\right)\text{.}$

We view ${𝔥}^{*}$ as sitting inside ${\text{Hom}}_{ℚ\left(q\right)}\left({U}_{q}^{0}\left(𝔤\right),ℚ\left(q\right)\right)\text{:}$ for $\omega \in {𝔥}^{*},$ $\omega \left({K}_{i}\right)={q}^{⟨\omega ,{\alpha }_{i}^{\vee }⟩}\text{.}$ We will also be interested in certain "twistings" of the weights of $𝔤\text{.}$ Let $\sigma =\left({\sigma }_{1},\dots ,{\sigma }_{k}\right)\in \left\{\left(±1,\dots ,±1\right)\right\}$ and $\omega \in {𝔥}^{*}\text{.}$ Define ${\omega }^{\sigma }\in {\text{Hom}}_{ℚ\left(q\right)}\left({U}_{q}^{0}\left(𝔤\right),ℚ\left(q\right)\right)$ by $ωσ(Ki)=σi qi⟨ω,αi∨⟩ =±qi⟨ω,αi∨⟩.$

The dominance ordering on ${𝔥}^{*}$ induces a partial ordering on the weights of ${U}_{q}\left(𝔤\right)\text{.}$ For $\mathrm{\Omega },\mathrm{\Lambda }\in {\text{Hom}}_{ℚ\left(q\right)}\left({U}_{q}^{0}\left(𝔤\right),ℚ\left(q\right)\right),$ $Ω≤Λif there is some ν∈Q+ such that Λ(Ki)Ω(Ki) =q⟨ν,αi∨⟩ for 1≤i≤k.$ In this case we write $\frac{\mathrm{\Omega }}{\mathrm{\Lambda }}=\nu \text{.}$ For $\omega ,\lambda \in {𝔥}^{*},$ this is the standard dominance ordering.

For a ${U}_{q}\left(𝔤\right)$ module $M,$ the $\mathrm{\Omega }\text{-weight}$ space of $M$ is given by $MΩ= { v∈M | Kiv= Ω(Ki)v for 1≤i≤k } .$

The Category $𝒪$ consists of ${U}_{q}\left(𝔤\right)\text{-modules}$ $M$ such that

 • $M$ is ${U}_{q}{\left(𝔤\right)}^{0}\text{-diagonalizable:}$ $M=⨁Ω∈Homℚ(q)(Uq0(𝔤),ℚ(q)) MΩ;$ • there are ${\mathrm{\Lambda }}_{1},\dots ,{\mathrm{\Lambda }}_{n}\in {\text{Hom}}_{ℚ\left(q\right)}\left({U}_{q}^{0}\left(𝔤\right),ℚ\left(q\right)\right)$ such that ${M}^{\mathrm{\Omega }}\ne 0$ only if $\mathrm{\Omega }\le {\mathrm{\Lambda }}_{i}$ for some $1\le i\le n\text{.}$ • $\text{dim} {M}^{\mathrm{\Omega }}<\infty$ for all $\mathrm{\Omega }\in {\text{Hom}}_{ℚ\left(q\right)}\left({U}_{q}^{0}\left(𝔤\right),ℚ\left(q\right)\right)\text{.}$

A highest weight ${U}_{q}\left(𝔤\right)\text{-module}$ of highest weight $\mathrm{\Omega }$ is a ${U}_{q}\left(𝔤\right)\text{-module}$ $M$ with a vector ${v}^{+}\in M$ such that

 • ${v}^{+}\in {M}^{\mathrm{\Omega }}\text{;}$ • $M={U}_{q}\left(𝔤\right){v}^{+}\text{;}$ • ${E}_{i}{v}^{+}=0$ for all $1\le i\le k\text{.}$
The vector ${v}^{+}$ is a highest weight vector. For such a module $M$ the defining relations of ${U}_{q}\left(𝔤\right)$ imply $M=\underset{\mathrm{\Lambda }\le \mathrm{\Omega }}{⨁}{M}^{\mathrm{\Lambda }}\text{.}$ Note that all highest weight modules are in Category $𝒪\text{.}$

The Verma module of highest weight $\mathrm{\Omega },$ is the ${U}_{q}\left(𝔤\right)\text{-module}$ $Mq(Ω)=Uq (𝔤)/I,$ where $I$ is the left ideal generated by ${E}_{i}$ and ${K}_{i}-\mathrm{\Omega }\left({K}_{i}\right)$ $\left(1\le i\le k\right)\text{.}$ We can also define ${M}_{q}\left(\mathrm{\Omega }\right)$ as the induced module $Mq(Ω)= Uq(𝔤) ⊗Uq≥0(𝔤) ℚ(q)Ω,$ where $ℚ{\left(q\right)}_{\mathrm{\Omega }}$ is $ℚ\left(q\right)$ as a vector space; and, as a ${U}_{q}^{\ge 0}\left(𝔤\right)\text{-module,}$ ${U}_{q}^{0}\left(𝔤\right)$ acts by $\mathrm{\Omega }$ and ${U}_{q}^{+}\left(𝔤\right)$ acts by $0\text{.}$ Using the same arguments as in Lemma 2.1.3, we can show that ${M}_{q}\left(\mathrm{\Omega }\right)$ has a maximal proper submodule, ${J}_{q}\left(\mathrm{\Omega }\right)\text{.}$ This implies that ${M}_{q}\left(\mathrm{\Omega }\right)$ has a unique simple quotient $Lq(Ω)= Mq(Ω)/ Jq(Ω).$

All of these objects can be similarly defined for ${U}_{𝒜}^{\text{res}}\left(𝔤\right)\text{.}$ The set of weights for ${U}_{𝒜}^{\text{res}}\left(𝔤\right)$ is ${\text{Hom}}_{𝒜}\left({{U}_{𝒜}^{\text{res}}\left(𝔤\right)}^{0},𝒜\right)\text{.}$ We denote the ${U}_{𝒜}^{\text{res}}\left(𝔤\right)\text{-Verma}$ module of weight $\mathrm{\Omega }\in {\text{Hom}}_{𝒜}\left({{U}_{𝒜}^{\text{res}}\left(𝔤\right)}^{0},𝒜\right)$ by ${M}_{\text{res}}\left(\mathrm{\Omega }\right)\text{.}$ In other words, ${M}_{\text{res}}\left(\mathrm{\Omega }\right)={U}_{𝒜}^{\text{res}}\left(𝔤\right)/J,$ 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 $\mathrm{\Omega },\mathrm{\Lambda }\in {\text{Hom}}_{ℚ\left(q\right)}\left({U}_{q}^{0}\left(𝔤\right),ℚ\left(q\right)\right)$ so that $\mathrm{\Omega }\ge \mathrm{\Lambda }\text{.}$ Suppose $\frac{\mathrm{\Omega }}{\mathrm{\Lambda }}=\nu \in {Q}_{+}\text{.}$ Define ${F}_{{\beta }_{j}}^{\left({t}_{j}\right)}=\frac{1}{\left[{t}_{j}\right]!}{F}_{{\beta }_{j}}^{{t}_{j}}\text{.}$ Then, the set $\left\{{F}_{{\beta }_{1}}^{\left({t}_{1}\right)}\cdots {F}_{{\beta }_{N}}^{\left({t}_{N}\right)}{v}^{+} | {t}_{1},\dots ,{t}_{N}\in {ℤ}_{\ge 0},{t}_{1}{\beta }_{1}+\cdots +{t}_{N}{\beta }_{N}=\nu \right\}$ is a basis for ${M}_{\text{res}}{\left(\mathrm{\Omega }\right)}^{\mathrm{\Lambda }}$ and ${M}_{q}{\left(\mathrm{\Omega }\right)}^{\mathrm{\Lambda }}\text{.}$

### Embeddings of Verma Modules

Recall that for a weight $\omega \in {𝔥}^{*}$ of $𝔤$ and $\sigma =\left({\sigma }_{1},\dots ,{\sigma }_{k}\right)\in \left\{\left(±1,\dots ,±1\right)\right\},$ we can define a ${U}_{q}\left(𝔤\right)\text{-weight}$ ${\omega }^{\sigma }\in {\text{Hom}}_{ℚ\left(q\right)}\left({U}_{q}^{0}\left(𝔤\right),ℚ\left(q\right)\right)$ by ${\omega }^{\sigma }\left({K}_{i}\right)={\sigma }_{i}{q}_{i}^{⟨\omega ,{\alpha }_{i}^{\vee }⟩}\text{.}$

Let $\lambda \in P,$ $\sigma \in \left\{\left(±1,\dots ,±1\right)\right\},$ and $\beta \in {R}^{+}\text{.}$ If ${s}_{\beta }\circ \lambda \le \lambda ,$ then ${M}_{q}\left({\left({s}_{\beta }\circ \lambda \right)}^{\sigma }\right)\subseteq {M}_{q}\left({\lambda }^{\sigma }\right)\text{.}$

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

Let $\lambda \in P$ and $\sigma \in \left\{\left(±1,\dots ,±1\right)\right\}\text{.}$ If ${s}_{i}\circ \lambda \le \lambda ,$ ${M}_{q}\left({\left({s}_{i}\circ \lambda \right)}^{\sigma }\right)\subseteq {M}_{q}\left({\lambda }^{\sigma }\right)\text{.}$ Moreover, if ${v}^{+}$ generates ${M}_{q}\left({\lambda }^{\sigma }\right),$ then ${F}_{i}^{m}{v}^{+}$ is a highest weight vector of weight ${\left({s}_{i}\circ \lambda \right)}^{\sigma },$ where $m=⟨\lambda +\rho ,{\alpha }_{i}^{\vee }⟩\ge 0\text{.}$

 Proof. Note that ${F}_{i}^{m}{v}^{+}\in {M}_{q}{\left({\lambda }^{\sigma }\right)}^{{\left({s}_{i}\circ \lambda \right)}^{\sigma }}\text{.}$ We must also check that ${E}_{j}{F}_{i}^{m}{v}^{+}=0$ for all $1\le j\le k\text{.}$ If $i\ne j,$ ${E}_{j}{F}_{i}^{m}{v}^{+}={F}_{i}^{m}{E}_{j}{v}^{+}=0\text{.}$ If $i=j,$ we can use Equation 4.3 to show $EiFimv+ = σi[m]i [⟨λ,αi∨⟩-m+1]i Fim-1v+ = σi[m]i [⟨λ+ρ,αi∨⟩-m]i Fim-1v+ = 0.$ $\square$

Let $\lambda \in {P}^{+},$ $\sigma \in \left\{\left(±1,\dots ,±1\right)\right\},$ and $w\in W$ with reduced expression $w={s}_{{i}_{n}}\cdots {s}_{{i}_{1}}\text{.}$ Set $λ0=λ, λ1=si1∘λ0, ⋯, λn=sin∘λn-1.$ Then, $Mq(λnσ)⊆ Mq(λn-1σ)⊆ ⋯⊆ Mq(λ0σ).$

 Proof. By definition, ${\lambda }_{j+1}={s}_{{i}_{j+1}}\circ {\lambda }_{i}\text{.}$ Therefore, using the previous lemma, we need only check that $⟨{\lambda }_{j}+\rho ,{\alpha }_{j+1}^{\vee }⟩\ge 0$ to show that ${M}_{q}\left({\lambda }_{j+1}^{\sigma }\right)\subseteq {M}_{q}\left({\lambda }_{j}^{\sigma }\right)\text{.}$ Note that $⟨λj+ρ,αj+1∨⟩= ⟨wj(λ+ρ),αj+1∨⟩= ⟨λ+ρ,wj-1αj+1∨⟩,$ where ${w}_{j}={s}_{{i}_{j}}\cdots {s}_{{i}_{1}}\text{.}$ Then ${w}_{j}^{-1}{\alpha }_{j+1}\in {R}^{+}\text{.}$ (See [Dix1996] 11.1.8.) Because $\lambda \in {P}^{+},$ $⟨λ+ρ,wj-1αj+1∨⟩>0.$ $\square$

Let $\lambda ,\mu \in P,$ and $\sigma \in \left\{\left(±1,\dots ,±1\right)\right\}\text{.}$ Suppose ${\alpha }_{i}$ is a simple root such that ${M}_{q}\left({\left({s}_{i}\circ \mu \right)}^{\sigma }\right)\subseteq {M}_{q}\left({\mu }^{\sigma }\right)\subseteq {M}_{q}\left({\lambda }^{\sigma }\right)\text{.}$ Then, ${M}_{q}\left({\left({s}_{i}\circ \mu \right)}^{\sigma }\right)\subseteq {M}_{q}\left({\left({s}_{i}\circ \lambda \right)}^{\sigma }\right)\text{.}$

This gives the following picture, where arrows indicate inclusion: $M(λσ) ↑ M((si∘λ)σ) M(μσ) ↑ ↖ M((si∘μ)σ) A$

The picture indicates nothing about the relationship between ${M}_{q}\left({\lambda }^{\sigma }\right)$ and ${M}_{q}\left({\left({s}_{i}\circ \lambda \right)}^{\sigma }\right)\text{.}$ The proof will consider both possibilities: ${M}_{q}\left({\lambda }^{\sigma }\right)\subseteq {M}_{q}\left({\left({s}_{i}\circ \lambda \right)}^{\sigma }\right)$ and ${M}_{q}\left({\left({s}_{i}\circ \lambda \right)}^{\sigma }\right)\text{.}\subseteq {M}_{q}\left({\lambda }^{\sigma }\right)\text{.}$

Proof.

Let $m=⟨\mu +\rho ,{\alpha }_{i}^{\vee }⟩\ge 0$ and $n=⟨\lambda +\rho ,{\alpha }_{i}^{\vee }⟩\text{.}$ If $n\le 0,$ then we have $Mq((si∘μ)σ)⊆ Mq(μσ)⊆ Mq(λσ)⊆ Mq((si∘λ)σ),$ i.e. we have the following picture of embeddings. ${M}_{q}\left({\left({s}_{i}\circ \lambda \right)}^{\sigma }\right) {M}_{q}\left({\lambda }^{\sigma }\right) {M}_{q}\left({\mu }^{\sigma }\right) {M}_{q}\left({\left({s}_{i}\circ \mu \right)}^{\sigma }\right)$ Then the lemma follows.

Suppose $n>0\text{.}$ Then ${M}_{q}\left({\left({s}_{i}\circ \lambda \right)}^{\sigma }\right)\subseteq {M}_{q}\left({\lambda }^{\sigma }\right)\text{.}$ Let ${v}^{+},{\stackrel{\sim }{v}}^{+}\in {M}_{q}\left({\lambda }^{\sigma }\right)$ be generators for ${M}_{q}\left({\lambda }^{\sigma }\right)$ and ${M}_{q}\left({\mu }^{\sigma }\right)$ respectively. Then, ${F}_{i}^{m}{\stackrel{\sim }{v}}^{+}$ generates ${M}_{q}\left({\left({s}_{i}\circ \mu \right)}^{\sigma }\right),$ and ${F}_{i}^{n}{v}^{+}$ generates ${M}_{q}\left({\left({s}_{i}\circ \lambda \right)}^{\sigma }\right)\text{.}$ Therefore, ${M}_{q}\left({\left({s}_{i}\circ \mu \right)}^{\sigma }\right)\subseteq {M}_{q}\left({\left({s}_{i}\circ \lambda \right)}^{\sigma }\right)$ if ${F}_{i}^{m}{\stackrel{\sim }{v}}^{+}\in {M}_{q}\left({\left({s}_{i}\circ \lambda \right)}^{\sigma }\right)\text{.}$

We first show that there is some $l\in {ℤ}_{\ge 0}$ such that ${F}_{i}^{l}{\stackrel{\sim }{v}}^{+}\in {M}_{q}\left({\left({s}_{i}\circ \lambda \right)}^{\sigma }\right)\text{.}$ In other words, we want to show that ${F}_{i}^{l}{\stackrel{\sim }{v}}^{+}=X{F}_{i}^{n}{v}^{+}$ for some $X\in {U}_{q}\left(𝔤\right)\text{.}$ We have ${\stackrel{\sim }{v}}^{+}=u{v}^{+}$ for some $u\in {U}_{q}^{-}\left(𝔤\right)\text{.}$ Since $u\in {U}_{q}^{-}\left(𝔤\right),$ $u$ is a $ℚ\left(q\right)\text{-linear}$ combination of elements of the form ${F}_{{i}_{1}}\cdots {F}_{{i}_{j}}$ where ${\alpha }_{{i}_{1}}+\cdots +{\alpha }_{{i}_{j}}=\lambda -\mu \text{.}$ From the generating relations for ${U}_{q}\left(𝔤\right),$ 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 $i\ne j\text{.}$ Thus, ${F}_{i}^{t\left(1-{a}_{ij}\right)}{F}_{j}=X{F}_{i}^{t}$ for some $X\in {U}_{q}^{-}\left(𝔤\right)\text{.}$ Given the form of $u,$ this implies ${F}_{i}^{l}{\stackrel{\sim }{v}}^{+}\in {M}_{q}\left({\left({s}_{i}\circ \lambda \right)}^{\sigma }\right)$ for some $l\in {ℤ}_{\ge 0}\text{.}$

We claim that this implies that ${F}_{i}^{m}{\stackrel{\sim }{v}}^{+}\in {M}_{q}\left({\left({s}_{i}\circ \lambda \right)}^{\sigma }\right)\text{:}$

 • If $l\le m,$ then ${F}_{i}^{m}{\stackrel{\sim }{v}}^{+}={F}_{i}^{m-l}{F}_{i}^{l}{\stackrel{\sim }{v}}^{+}\in {M}_{q}\left({\left({s}_{i}\circ \lambda \right)}^{\sigma }\right)\text{.}$ • If $l>m,$ then ${E}_{i}{F}_{i}^{l}{\stackrel{\sim }{v}}^{+}=\left[l\right]\left[⟨\mu ,{\alpha }^{\vee }⟩-l+1\right]{F}_{i}^{l-1}{\stackrel{\sim }{v}}^{+}\text{.}$ Since ${E}_{i}{F}_{i}^{l}{\stackrel{\sim }{v}}^{+}\in {M}_{q}\left({\left({s}_{i}\circ \lambda \right)}^{\sigma }\right),$ ${F}_{i}^{m}{\stackrel{\sim }{v}}^{+}\in {M}_{q}\left({\left({s}_{i}\circ \lambda \right)}^{\sigma }\right)\text{.}$
In this case, we have the following picture: ${M}_{q}\left({\lambda }^{\sigma }\right) {M}_{q}\left({\left({s}_{i}\circ \lambda \right)}^{\sigma }\right) {M}_{q}\left({\mu }^{\sigma }\right) {M}_{q}\left({\left({s}_{i}\circ \mu \right)}^{\sigma }\right)$

$\square$

We now proceed with the proof of Proposition 4.4.1.

 Proof of Proposition 4.4.1. Let $\mu ={s}_{\beta }\circ \lambda$ and $m=⟨\lambda +\rho ,{\beta }^{\vee }⟩\ge 0\text{.}$ If $m=0,$ $\mu =\lambda \text{.}$ Therefore, assume $m>0\text{.}$ Let $\stackrel{\sim }{\mu }\in {P}^{+}\cap \left(W\circ \mu \right)$ and choose $w\in W$ so that $\mu =w\circ \stackrel{\sim }{\mu }\text{.}$ Write $\stackrel{\sim }{\lambda }={w}^{-1}\circ \lambda \text{.}$ Fix a reduced expression $w={s}_{{i}_{n}}\cdots {s}_{{i}_{1}},$ and set $λ0=λ∼, λ1=si1∘λ0, …, λn=sin∘λn-1=λ;$ $μ0=μ∼, μ1=si1∘μ0, …, μn=sin∘μn-1=μ.$ Note that $\stackrel{\sim }{\mu }={w}^{-1}\circ {s}_{\beta }\circ w\circ \stackrel{\sim }{\lambda }={s}_{{w}^{-1}\beta }\circ \stackrel{\sim }{\lambda }\text{.}$ Therefore ${\mu }_{j}={s}_{\left({s}_{{i}_{j}}\cdots {s}_{{i}_{1}}{w}^{-1}\beta \right)}\circ {\lambda }_{j}\text{.}$ Since ${s}_{i}$ sends ${\alpha }_{i}$ to $-{\alpha }_{i}$ and permutes the other positive roots, $\left({s}_{{i}_{j}}\cdots {s}_{{i}_{1}}{w}^{-1}\right)\beta ={\gamma }_{j}\in {R}^{+}$ or $\left({s}_{{i}_{j}}\cdots {s}_{{i}_{1}}{w}^{-1}\right)\beta =-{\gamma }_{j}\in -{R}^{+}\text{.}$ Therefore, ${\mu }_{j}={s}_{{\gamma }_{j}}\circ {\lambda }_{j}={s}_{-{\gamma }_{j}}\circ {\lambda }_{j}$ for some ${\gamma }_{j}\in {R}^{+}\text{.}$ Since ${\mu }_{0}\in {P}^{+}$ and ${\lambda }_{0}\in W\circ {\mu }_{0},$ we must have ${\mu }_{0}-{\lambda }_{0}\in {Q}_{+}\text{.}$ On the other hand, ${\mu }_{n}-{\lambda }_{n}=-m\beta \text{.}$ Let $j$ be minimal such that ${\mu }_{j}-{\lambda }_{j}\in {Q}_{+}$ and ${\mu }_{j+1}-{\lambda }_{j+1}\in -{Q}_{+}\text{.}$ 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 ${s}_{{i}_{j+1}}{\gamma }_{j+1}\in {R}^{-}$ and so ${\gamma }_{j+1}={\alpha }_{j+1}\text{.}$ We now have ${\mu }_{j+1}={s}_{{i}_{j+1}}\circ {\lambda }_{j+1}$ with ${\mu }_{j+1}\le {\lambda }_{j+1}\text{.}$ Lemma 4.4.2 implies ${M}_{q}\left({\mu }_{j+1}^{\sigma }\right)\subseteq {M}_{q}\left({\lambda }_{j+1}^{\sigma }\right)\text{.}$ From Lemma 4.4.3, we have ${M}_{q}\left({\mu }_{j+2}^{\sigma }\right)\subseteq {M}_{q}\left({\mu }_{j+1}^{\sigma }\right)\text{.}$ Since ${M}_{q}\left({\mu }_{j+1}^{\sigma }\right)\subseteq {M}_{q}\left({\lambda }_{j+1}^{\sigma }\right),$ we can use the last lemma to conclude ${M}_{q}\left({\mu }_{j+2}^{\sigma }\right)\subseteq {M}_{q}\left({\lambda }_{j+2}^{\sigma }\right)\text{.}$ Repeating this process, we get ${M}_{q}\left({\mu }^{\sigma }\right)\subseteq {M}_{q}\left({\lambda }^{\sigma }\right)\text{.}$ The picture below describes the various embeddings used in the proof: ${M}_{q}\left({\stackrel{\sim }{\mu }}^{\sigma }\right) {M}_{q}\left({\lambda }_{j+2}^{\sigma }\right) {M}_{q}\left({\lambda }_{j+1}^{\sigma }\right) {M}_{q}\left({\mu }_{j+1}^{\sigma }\right) {M}_{q}\left({\lambda }^{\sigma }\right) {M}_{q}\left({\mu }_{j+2}^{\sigma }\right) {M}_{q}\left({\mu }^{\sigma }\right)$ $\square$

#### Central characters

We now construct the central characters of ${U}_{q}\left(𝔤\right),$ following ([CPr1994] 9.2.B). Denote the center of ${U}_{q}\left(𝔤\right)$ by $Zq(𝔤)= { X∈Uq(𝔤) | XY=YX for all Y∈Uq(𝔤) } .$ For $z\in {Z}_{q}\left(𝔤\right),$ we can write $z=∑ Fβ1(t1)⋯ FβN(tN) K(s1,…,sN)(t1,…,tN) EβN(sN)⋯ Eβ1(s1)$ where ${K}_{\left({s}_{1},\dots ,{s}_{N}\right)}^{\left({t}_{1},\dots ,{t}_{N}\right)}\in {U}_{q}^{0}\left(𝔤\right)$ and ${s}_{1},\dots ,{s}_{N},{t}_{1},\dots ,{t}_{N}\in {ℤ}_{\ge 0}$ with ${s}_{1}{\beta }_{1}+\cdots +{s}_{N}{\beta }_{N}={t}_{1}{\beta }_{1}+\cdots +{t}_{N}{\beta }_{N}\text{.}$ Define ${h}_{q}:{Z}_{q}\left(𝔤\right)\to {U}_{q}^{0}\left(𝔤\right)$ by $hq(z)= K(0)(0).$

The map ${h}_{q}:{Z}_{q}\left(𝔤\right)\to {U}_{q}^{0}\left(𝔤\right)$ is an algebra homomorphism.

 Proof. Let $z\in {Z}_{q}\left(𝔤\right)$ with $z=\sum {F}_{{\beta }_{1}}^{\left({t}_{1}\right)}\cdots {F}_{{\beta }_{N}}^{\left({t}_{N}\right)}{K}_{\left({s}_{1},\dots ,{s}_{N}\right)}^{\left({t}_{1},\dots ,{t}_{N}\right)}{E}_{{\beta }_{N}}^{\left({s}_{N}\right)}\cdots {E}_{{\beta }_{1}}^{\left({s}_{1}\right)},$ and let ${v}^{+}$ be a generator for the Verma module ${M}_{q}\left(\mathrm{\Lambda }\right)$ $\text{(}\mathrm{\Lambda }\in {\text{Hom}}_{ℚ\left(q\right)}\left({U}_{q}^{0}\left(𝔤\right),ℚ\left(q\right)\right)\text{).}$ Then, $z\left({v}^{+}\right)=\left({K}_{\left(0\right)}^{\left(0\right)}\right){v}^{+}=\mathrm{\Lambda }\left({h}_{q}\left(z\right)\right){v}^{+}\text{.}$ For $z,z\prime \in {Z}_{q}\left(𝔤\right),$ $Λ(hq(zz′))v+= zz′(v+)= zΛ(hq(z′))v+= Λ(hq(z))Λ(hq(z′))v+.$ Therefore, $\mathrm{\Lambda }\left({h}_{q}\left(zz\prime \right)\right)=\mathrm{\Lambda }\left({h}_{q}\left(z\right)\right)\mathrm{\Lambda }\left({h}_{q}\left(z\prime \right)\right)$ for all $\mathrm{\Lambda },$ with implies ${h}_{q}\left(zz\prime \right)={h}_{q}\left(z\right){h}_{q}\left(z\prime \right)\text{.}$ $\square$

The map ${h}_{q}$ is the quantum analogue of the Harish-Chandra homomorphism for semisimple Lie algebras (see [Dix1996], Section 7.4).

For $\mathrm{\Lambda }\in {\text{Hom}}_{ℚ\left(q\right)}\left({U}_{q}^{0}\left(𝔤\right),ℚ\left(q\right)\right),$ we define the central character ${\chi }_{\mathrm{\Lambda }}:{Z}_{q}\left(𝔤\right)\to ℂ$ by $χΛ(z)= Λ(hq(z)).$ Let $M$ be a highest weight ${U}_{q}\left(𝔤\right)\text{-module}$ of highest weight $\mathrm{\Lambda },$ and let ${v}^{+}$ be a generator of $M\text{.}$ Then, $z{v}^{+}={\chi }_{\mathrm{\Lambda }}\left(z\right){v}^{+}\text{.}$ Since any $v\in M$ is of the form $v=x{v}^{+}$ for some $x\in {U}_{q}\left(𝔤\right),$ $zv= zxv+= xzv+= xzv+= χΛ(z)xv+= χΛ(z)v.$

Recall that the dot action of the Weyl group $W$ on ${𝔥}^{*}$ is $w\circ \mu =w\left(\mu +\rho \right)-\rho \text{.}$

([CPr1994], 9.1.8). Let $\lambda ,\mu \in P,$ integral weights of $𝔤\text{.}$ Then ${\chi }_{\lambda }={\chi }_{\mu }$ if and only if $\lambda =w\circ \mu$ for some $w\in W\text{.}$

### The Shapovalov Determinant

Recall $\tau :{U}_{q}\left(𝔤\right)\to {U}_{q}\left(𝔤\right)$ is the $ℚ\text{-anti-automorphism}$ 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 $⟨,⟩:{M}_{q}\left(\mathrm{\Omega }\right)×{M}_{q}\left(\mathrm{\Omega }\right)\to ℚ\left(q\right)$ by

 • $⟨{v}^{+},{v}^{+}⟩=1,$ where ${v}^{+}$ is the generator of ${M}_{q}\left(\mathrm{\Omega }\right)\text{;}$ • $⟨xu,w⟩=⟨u,\tau \left(x\right)w⟩$ for $x\in {U}_{q}$ and $u,w\in {M}_{q}\left(\mathrm{\Omega }\right)\text{.}$

Let $\mathrm{\Omega },\mathrm{\Lambda }\in {\text{Hom}}_{ℚ\left(q\right)}\left({U}_{q}^{0}\left(𝔤\right),ℚ\left(q\right)\right)$ with $\mathrm{\Omega }\ge \mathrm{\Lambda }\text{.}$ Suppose $\frac{\mathrm{\Omega }}{\mathrm{\Lambda }}=\nu \in {Q}_{+}\text{.}$ We define $A(Ω)Λ= (⟨Fβ1(t1)⋯FβN(tN)v+,Fβ1(s1)⋯FβN(sN)v+⟩) t1β1+⋯+tNβN=ν =s1β1+⋯+sNβN$ and $det(Mq(Ω)Λ)= det A(Ω)Λ.$ If $\mathrm{\Omega },\mathrm{\Lambda }\in {\text{Hom}}_{𝒜}\left({U}_{q}^{0}\left(𝔤\right),𝒜\right),$ we define $\text{det} {M}_{\text{res}}{\left(\mathrm{\Omega }\right)}^{\mathrm{\Lambda }}=\text{det}\left({M}_{q}{\left(\mathrm{\Omega }\right)}^{\mathrm{\Lambda }}\right)\text{.}$

Example. Suppose $𝔤={sl}_{2}\left(ℂ\right)\text{.}$ Let $k\in {ℤ}_{\ge 0}$ and $\mathrm{\Omega },\mathrm{\Lambda }\in {\text{Hom}}_{ℚ\left(q\right)}\left({U}_{q}^{0}\left({sl}_{2}\left(ℂ\right)\right),ℚ\left(q\right)\right)$ with $\frac{\mathrm{\Omega }}{\mathrm{\Lambda }}=k\alpha \text{.}$ Now consider $\text{det}\left({M}_{q}{\left(\mathrm{\Omega }\right)}^{\mathrm{\Lambda }}\right)\text{.}$

Equation 4.3 implies $E{F}^{j}{v}^{+}=\left[j\right]\frac{{q}^{-j+1}\mathrm{\Omega }K-{q}^{j-1}\mathrm{\Omega }{\left(K\right)}^{-1}}{q-{q}^{-1}}{F}^{j-1}{v}^{+}\text{.}$ Therefore, $det Mq(Ω)Λ = ⟨ Fk[k]!v+, Fk[k]!v+ ⟩ = ⟨ v+, Ek[k]! Fk[k]!v+ ⟩ = 1([k]!)2 ⟨ v+,∏j=1k [j] qj-1K-q-J+1K-1q-q-1 v+ ⟩ = 1([k]!) ∏j=1k q-j+1Ω(K)-qj-1Ω(K-1)q-q-1$ Theorem 4.5.2 gives a formula for $\text{det} {M}_{q}{\left(\mathrm{\Omega }\right)}^{\mathrm{\Lambda }}$ for all $𝔤\text{.}$

Recall that a weight $\mathrm{\Omega }$ is equivalent to a $k\text{-tuple}$ $\left({\mathrm{\Omega }}_{1},\dots ,{\mathrm{\Omega }}_{k}\right)\in {\left(ℚ\left(q\right)\right)}^{k}\text{.}$ We can view the determinant of ${M}_{q}{\left(\mathrm{\Omega }\right)}^{\mathrm{\Lambda }}$ as living inside the Laurent polynomial ring $S=ℚ\left(q\right)\left[{\mathrm{\Omega }}_{i}^{±1} | 1\le i\le k\right]\text{.}$ To prove Theorem 4.5.2, we will bound the degree of the polynomial and then find its irreducible factors. The degree $D:S\to {ℤ}^{k}$ is given by $D\left({\mathrm{\Omega }}_{1}^{{j}_{1}}\cdots {\mathrm{\Omega }}_{k}^{{j}_{k}}\right)=\left({j}_{1},\dots ,{j}_{k}\right)\text{.}$ Also, define the following partial ordering on ${ℤ}^{k}\text{:}$ $(j1,…,jk)≤ (l1,…,lk) if ji≤li for all 1≤i≤k.$

Let $\mathrm{\Omega }\ge \mathrm{\Lambda }$ with $\frac{\mathrm{\Omega }}{\mathrm{\Lambda }}=\nu \in {Q}_{+}\text{.}$ The term with maximum degree in $\text{det} {M}_{q}{\left(\mathrm{\Omega }\right)}^{\mathrm{\Lambda }}$ is $∏β∈R+ ∏r>0 ΩβP(ν-rβ).$ Similarly, the term with minimal degree is $∏β∈R+ ∏r>0 Ωβ-P(ν-rβ).$

 Proof. Consider an entry entry in $A{\left(\mathrm{\Omega }\right)}^{\mathrm{\Lambda }}\text{:}$ $⟨ Fβ1(t1)⋯ FβN(tN)v+, Fβ1(s1)⋯ FβN(sN)v+ ⟩ = ⟨ v+, EβN(tN)⋯ Eβ1(t1) Fβ1(s1)⋯ FβN(sN)v+ ⟩ (4.6)$ where ${t}_{1}{\beta }_{1}+\cdots +{t}_{n}{\beta }_{n}=\nu ={s}_{1}{\beta }_{1}+\cdots +{s}_{n}{\beta }_{n}\text{.}$ Then, both ${F}_{{\beta }_{1}}^{\left({t}_{1}\right)}\cdots {F}_{{\beta }_{n}}^{\left({t}_{n}\right)}$ and ${F}_{{\beta }_{1}}^{\left({s}_{1}\right)}\cdots {F}_{{\beta }_{n}}^{\left({s}_{n}\right)}$ can be written as linear combinations of ${F}_{{i}_{1}}\cdots {F}_{{i}_{m}},$ where $\nu ={\alpha }_{{i}_{1}}+\cdots +{\alpha }_{{i}_{m}}$ for some choice of (not necessarily distinct) simple roots. Therefore, Equation 4.3 implies that the maximum possible degree term in (4.6) is $Ωi1⋯Ωim= Ωβ1t1⋯Ωβntn= Ωβ1s1⋯Ωβnsn (4.7)$ This means the term with maximum degree in $A{\left(\mathrm{\Omega }\right)}^{\mathrm{\Lambda }}$ is $∏(t)|t1β1+⋯+tnβn=ν Ωβ1t1⋯Ωβnt1. (4.8)$ The power of ${\mathrm{\Omega }}_{{\beta }_{j}}$ in (4.8) is $\sum _{\left(t\right)|{t}_{1}{\beta }_{1}+\cdots +{t}_{n}{\beta }_{n}=\omega -\lambda }{t}_{j}\text{.}$ We rewrite this as $∑(t)|t1β1+⋯+tnβn=ν tj=∑r>0 ∣{(t) | tj≥r}∣= ∑r>0P(ν-rβj).$ To understand the first equality, note that each partition in the second sum gets counted ${t}_{j}$ times. To understand the second equality, note that each partition of $\nu -r{\beta }_{j}$ maps uniquely to a partition with ${t}_{j}\ge r\text{.}$ We determine the minimum degree with an analogous argument. $\square$

([DKa1992]). Let $\mathrm{\Omega }\ge \mathrm{\Lambda }\in {\text{Hom}}_{ℚ\left(q\right)}\left({U}_{q}^{0}\left(𝔤\right),ℚ\left(q\right)\right)$ with $\frac{\mathrm{\Omega }}{\mathrm{\Lambda }}=\nu \in {Q}^{+}\text{.}$ Then, $det(Mq(Ω)Λ)= ∏β∈R+∏r>0 ( 1[r]dβ [qdβ(⟨ρ,β∨⟩-r)Ω(Kβ)-q-dβ(⟨ρ,β∨⟩-r)Ω(Kβ-1)qdβ-q-dβ] ) P(ν-rβ) .$

If we restrict the choice of weights to $\omega ,\lambda \in {𝔥}^{*}\subseteq {\text{Hom}}_{ℚ\left(q\right)}\left({U}_{q}^{0}\left(𝔤\right),ℚ\left(q\right)\right),$ this formula becomes $det(Mq(ω)λ)= ∏β∈R+ ∏r>0 ([⟨ω+ρ,β∨⟩-r]dβ[r]dβ) P(ω-λ-rβ) . (4.9)$

A Partial Proof.

Note that $\frac{{q}^{{d}_{\beta }\left(⟨\rho ,{\beta }^{\vee }⟩-r\right)}\mathrm{\Omega }\left({K}_{\beta }\right)-{q}^{-{d}_{\beta }\left(⟨\rho ,{\beta }^{\vee }⟩-r\right)}\mathrm{\Omega }\left({K}_{\beta }^{-1}\right)}{{q}^{{d}_{\beta }}-{q}^{-{d}_{\beta }}}$ 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 $(Ω(Kβ)±q-dβ(⟨ρ,β∨⟩-r))p(ν-rβ) |det Mq(Ω)Λ.$ Suppose that ${\omega }_{0}\in P$ is such that $⟨{\omega }_{0}+\rho ,{\beta }^{\vee }⟩=r,$ and let $\sigma \in \left\{\left(±1,\dots ,±1\right)\right\}\text{.}$ Then, ${s}_{\beta }\circ {\omega }_{0}={\omega }_{0}-r\beta$ and so from Proposition 4.4.1, ${M}_{q}\left({\left({\omega }_{0}-r\beta \right)}^{\sigma }\right)\subseteq {M}_{q}\left({\omega }_{0}^{\sigma }\right)\text{.}$ Therefore, ${M}_{q}{\left({\omega }_{0}^{\sigma }\right)}^{{\left({\omega }_{0}-r\beta \right)}^{\sigma }}\cap \text{Rad}⟨,⟩\ne 0\text{.}$ This implies that $\text{det}\left({M}_{q}{\left({\omega }_{0}^{\sigma }\right)}^{{\left({\omega }_{0}-r\beta \right)}^{\sigma }}\right)=0,$ i.e. $\text{det}\left({M}_{q}{\left(\mathrm{\Omega }\right)}^{\stackrel{\sim }{\mathrm{\Omega }}}\right),$ where $\stackrel{\sim }{\mathrm{\Omega }}=\mathrm{\Omega }-r\beta ,$ has a zero at $Ωβ=(ω0σ)β= ±q⟨ω0,β∨⟩= ±qr-⟨ρ,β∨⟩.$ As in the proof of Theorem 3.4.3, this means there is ${v}_{r,\beta }\in {M}_{q}{\left(\mathrm{\Omega }\right)}^{\stackrel{\sim }{\mathrm{\Omega }}}$ such that

 • ${v}_{r,\beta }=\sum _{\left(t\right)|{t}_{1}{\beta }_{1}+\cdots +{t}_{n}{\beta }_{n}=r\beta }{a}_{\left(t\right)}{F}_{{\beta }_{1}}^{\left({t}_{1}\right)}\cdots {F}_{{\beta }_{n}}^{\left({t}_{n}\right)}{v}^{+}$ for some ${a}_{\left(t\right)}\in ℚ\left(q\right)$ with ${a}_{\left(t\right)}\ne 0$ for at least one $\left(t\right)\text{;}$ • $\left({\mathrm{\Omega }}_{\beta }±{q}^{r-⟨\rho ,{\beta }^{\vee }⟩}\right)|⟨{v}_{r,\beta },w⟩$ for all $w\in {M}_{q}\left(\mathrm{\Omega }\right)\text{.}$
Now, the set $\stackrel{\sim }{B}=\left\{{F}_{{\beta }_{1}}^{\left({s}_{1}\right)}\cdots {F}_{{\beta }_{N}}^{\left({s}_{N}\right)}\sum _{\left(t\right)|{t}_{1}{\beta }_{1}+\cdots +{t}_{n}{\beta }_{n}=r\beta }{a}_{\left(t\right)}{F}_{{\beta }_{1}}^{\left({t}_{1}\right)}\cdots {F}_{{\beta }_{n}}^{\left({t}_{n}\right)} | {s}_{1}{\beta }_{1}+\cdots +{s}_{N}{\beta }_{N}=\nu -r\beta \right\}$ is linearly independent in ${U}_{q}^{-}\left(𝔤\right)$ and can be extended to a linearly independent set $B$ which spans $\text{span}\left\{{F}_{{\beta }_{1}}^{\left({s}_{1}\right)}\cdots {F}_{{\beta }_{N}}^{\left({s}_{N}\right)} | {s}_{1}{\beta }_{1}+\cdots +{s}_{N}{\beta }_{N}=\nu \right\}\text{.}$ Let $P$ be the matrix taking $\left\{{F}_{{\beta }_{1}}^{\left({s}_{1}\right)}\cdots {F}_{{\beta }_{N}}^{\left({s}_{N}\right)} | {s}_{1}{\beta }_{1}+\cdots +{s}_{N}{\beta }_{N}=\nu \right\}$ to $B\text{.}$

Define $\text{det}\left({M}_{q}{\left(\mathrm{\Omega }\right)}^{\mathrm{\Lambda }}\right)=\text{det}{\left(⟨{X}_{i}{v}^{+},{X}_{j}{v}^{+}⟩\right)}_{{X}_{i},{X}_{j}\in B}\text{.}$ We have

 • ${F}_{{\beta }_{1}}^{\left({s}_{1}\right)}\cdots {F}_{{\beta }_{N}}^{\left({s}_{N}\right)}\sum _{\left(t\right)|{t}_{1}{\beta }_{1}+\cdots +{t}_{n}{\beta }_{n}=r\beta }{a}_{\left(t\right)}{F}_{{\beta }_{1}}^{\left({t}_{1}\right)}\cdots {F}_{{\beta }_{n}}^{\left({t}_{n}\right)}{v}^{+}={F}_{{\beta }_{1}}^{\left({s}_{1}\right)}\cdots {F}_{{\beta }_{N}}^{\left({s}_{N}\right)}{v}_{r,\beta }\text{;}$ • $\left({\mathrm{\Omega }}_{\beta }±{q}^{r-⟨\rho ,{\beta }^{\vee }⟩}\right)$ divides $⟨{F}_{{\beta }_{1}}^{\left({s}_{1}\right)}\cdots {F}_{{\beta }_{N}}^{\left({s}_{N}\right)}{v}_{r,\beta },w⟩\text{.}$
Therefore, ${\left(\mathrm{\Omega }\left({K}_{\text{pt}}\right)±{q}^{-{d}_{\beta }\left(⟨\rho ,{\beta }^{\vee }⟩-r\right)}\right)}^{p\left(\nu -r\beta \right)}$ divides ${\text{det}}_{B}{M}_{q}{\left(\mathrm{\Omega }\right)}^{\mathrm{\Lambda }}\text{.}$ Finally, $\text{det}\left({M}_{q}{\left(\mathrm{\Omega }\right)}^{\mathrm{\Lambda }}\right)=\text{det}\left({P}^{t}\right){\text{det}}_{B}\left({M}_{q}{\left(\mathrm{\Omega }\right)}^{\mathrm{\Lambda }}\right)\text{det}\left(P\right),$ which implies that ${\left(\mathrm{\Omega }\left({K}_{\beta }\right)±{q}^{-{d}_{\beta }\left(⟨\rho ,{\beta }^{\vee }⟩-r\right)}\right)}^{p\left(\nu -r\beta \right)}$ divides ${\text{det}}_{B}{M}_{q}{\left(\mathrm{\Omega }\right)}^{\mathrm{\Lambda }}\text{.}$

It only remains to check the coefficient of the highest degree term in $\text{det} {M}_{q}{\left(\mathrm{\Omega }\right)}^{\mathrm{\Lambda }}\text{.}$ We omit this part of the proof, but the original argument can be found in [DKa1992], 1.9.

$\square$

#### An Alternative Determinant Formula

Recall that DeConcini and Kac [DKa1992] use the $ℚ\text{-antiautomorphism}$ $\tau :{U}_{q}\to {U}_{q}$ defined by $τ(Ei)=Fi, τ(Fi)=Ei, τ(Ki)=Ki-1, τ(q)=q-1.$ Instead, we will use the $ℚ\left(q\right)\text{-antiautomorphism}$ $\varphi$ given by $ϕ(Ei)=Ki-1Fi, ϕ(Fi)=EiKi, ϕ(Ki)=Ki.$ This defines a new contravariant form. In this section, we will write ${⟨,⟩}_{\tau }$ for the contravariant form induced by $\tau$ and ${⟨,⟩}_{\varphi }$ for the contravariant form induced by $\varphi \text{.}$ Define $detϕ(Mq(ω)λ)=det (⟨Fβ1(t1)⋯FβN(tN)v+,Fβ1(s1)⋯FβN(sN)v+⟩ϕ) t1β1+⋯+tNβN=ν =s1β1+⋯+sNβN .$

Let $\omega ,\lambda \in P$ with $\lambda \le \omega \text{.}$ Then, $detϕ(Mq(ω)λ)= qP(ω-λ)kλω Cω-λ ∏β∈R+ ∏r>0 ([⟨ω+ρ,β∨⟩-r]dβ[r]dβ)P(ω-λ-rβ),$ where $0\ne {C}_{\omega -\lambda }\in ℚ\left(q\right)$ and ${k}_{\lambda }^{\omega }\in ℤ$ is given as follows. Write $\omega -\lambda ={\alpha }_{{i}_{1}}+\cdots +{\alpha }_{{i}_{m}}\text{.}$ Then $kλω= ⟨ω,αi1∨+⋯+αim∨⟩- ( ⟨αi1+⋯+αim,αi1∨⟩+ ⟨αi2+⋯+αim,αi2∨⟩+⋯+ ⟨αim,αim∨⟩ ) .$

 Proof. Let $P$ be the matrix which takes ${⟨{F}_{{\beta }_{1}}^{\left({t}_{1}\right)}\cdots {F}_{{\beta }_{N}}^{\left({t}_{N}\right)}{v}^{+},{F}_{{\beta }_{1}}^{\left({s}_{1}\right)}\cdots {F}_{{\beta }_{N}}^{\left({s}_{N}\right)}{v}^{+}⟩}_{\tau }$ to ${⟨{F}_{{\beta }_{1}}^{\left({t}_{1}\right)}\cdots {F}_{{\beta }_{N}}^{\left({t}_{N}\right)}{v}^{+},{F}_{{\beta }_{1}}^{\left({s}_{1}\right)}\cdots {F}_{{\beta }_{N}}^{\left({s}_{N}\right)}{v}^{+}⟩}_{\varphi }\text{.}$ Then, ${\text{det}}_{\varphi }\left({M}_{q}{\left(\omega \right)}^{\lambda }\right)=\text{det}\left(P\right){\text{det}}_{\tau }\left({M}_{q}{\left(\omega \right)}^{\lambda }\right)\text{.}$ Therefore, we need only find $\text{det}\left(P\right)\text{.}$ To do this, note that since the set $\left\{{F}_{{i}_{1}}\cdots {F}_{{i}_{m}}{v}^{+} | {\alpha }_{{i}_{1}}+\cdots +{\alpha }_{{i}_{m}}=\omega -\lambda \right\}$ spans ${M}_{q}{\left(\omega \right)}^{\lambda },$ there is a subset $B$ which is a basis for the space. Therefore, we can write a standard basis element ${F}_{{\beta }_{1}}^{\left({t}_{1}\right)}\cdots {F}_{{\beta }_{N}}^{\left({t}_{N}\right)}{v}^{+}$ as a linear combination of elements of $B\text{:}$ $Fβ1(t1)⋯ FβN(tN)v+ =∑Fi1⋯Fimv+∈B ai1,…,imFi1 ⋯Fimv+$ where ${a}_{{i}_{1},\dots ,{i}_{m}}\in ℚ\left(q\right)\text{.}$ 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 ${\alpha }_{{i}_{1}},\dots ,{\alpha }_{{i}_{m}}$ such that ${F}_{{i}_{1}}\cdots {F}_{{i}_{m}}\in B$ and ${\alpha }_{{i}_{1}}+\cdots +{\alpha }_{{i}_{m}}=\omega -\lambda \text{;}$ and where $kλω = ⟨ω-λ,αi1⟩+ ⟨ω-λ+αi1,αi2⟩+⋯+ ⟨ω-λ+(αi1+⋯+αim),αim⟩ = ⟨ω,αi1+⋯+αim⟩- ∑j≥k ⟨αij,αik⟩.$ (Since $⟨{\alpha }_{{i}_{j}},{\alpha }_{{i}_{k}}⟩=⟨{\alpha }_{{i}_{k}},{\alpha }_{{i}_{j}}⟩,$ ${k}_{\lambda }^{\omega }$ is independent of the order of the simple roots ${\alpha }_{{i}_{1}},\dots ,{\alpha }_{{i}_{m}}\text{.)}$ On the other hand, $⟨Fβ1(t1)⋯FβN(tN)v+,Fβ1(s1)⋯FβN(sN)v+⟩τ= ∑αi1+⋯+αim=ω-λ ai1,…,im‾ ⟨v+,Eim⋯Ei1Fβ1s1⋯Fβnsnv+⟩,$ where $\stackrel{‾}{·}:ℚ\left(q\right)\to ℚ\left(q\right)$ is the $ℚ\text{-automorphism}$ taking $q$ to ${q}^{-1}\text{.}$ Therefore, $P={q}^{{k}_{\lambda }^{\omega }}A\stackrel{‾}{{A}^{-1}},$ where $A$ is the matrix taking the basis $B$ to the standard basis $\left\{{F}_{{\beta }_{1}}^{\left({t}_{1}\right)}\cdots {F}_{{\beta }_{N}}^{\left({t}_{N}\right)}{v}^{+} | {t}_{1}{\beta }_{1}+\cdots +{t}_{N}{\beta }_{N}=\omega -\lambda \right\}$ of ${M}_{q}{\left(\omega \right)}^{\lambda }\text{.}$ 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, $\text{det}\left(A\right)\ne 0\text{.}$ Therefore, $detϕ(Mq(ω)λ)= qP(ω-λ)kλω Cω-λdetτ (Mq(ω)λ),$ where ${C}_{\omega -\lambda }=\text{det}\left(A\right)/\stackrel{‾}{\text{det}\left(A\right)}\text{.}$ $\square$

### Translation Functors

For $\lambda \in P,$ define the block $\left[\lambda \right]$ by $\left[\lambda \right]=\left\{w\circ \lambda | w\in W\right\}\text{.}$

Let $\omega ,{\omega }_{0}\in P\text{.}$ Then $Mq(ω)⊗ Lq(ω0)= ⨁λ∈P+ (Mq(ω)⊗Lq(ω0))[λ],$ where ${\left({M}_{q}\left(\omega \right)\otimes {L}_{q}\left({\omega }_{0}\right)\right)}^{\left[\lambda \right]}$ has composition factors ${L}_{q}\left(\mu \right)$ only for $\mu \in \left[\lambda \right]\text{.}$

 Proof. Note that the module ${M}_{q}\left(\omega \right)\otimes {L}_{q}\left({\omega }_{0}\right)$ 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. $\square$

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 $\mu \in {Q}_{+},$ define $n\left(\mu \right)=\text{dim}\left({L}_{\text{res}}{\left({\omega }_{0}\right)}^{{\omega }_{0}-\mu }\right),$ and let $\left\{{w}_{\mu ,i} | 1\le i\le n\left(\mu \right)\right\}$ be a basis for ${L}_{\text{res}}{\left({\omega }_{0}\right)}^{\left({\omega }_{0}-\mu \right)}\text{.}$

For $\lambda \in {Q}_{+},$ the sets $ℬ1= { Fβ1(t1)⋯ Fβn(tn)v+ ⊗wμ,i | t1β1+⋯+tnβn =λ-μ,1≤i≤n(μ) }$ and $ℬ2= { Fβ1(t1)⋯ Fβn(tn) (v+⊗wμ,i) | t1β1+⋯+tnβn =λ-μ,1≤i≤n(μ) }$ are bases for ${\left({M}_{\text{res}}\left(\omega \right)\otimes {L}_{\text{res}}\left({\omega }_{0}\right)\right)}^{\left(\omega +{\omega }_{0}-\lambda \right)}\text{.}$ Moreover, the transition matrix between the two bases has determinant one.

 Proof. The set ${ℬ}_{1}$ is clearly a basis for ${\left({M}_{\text{res}}\left(\omega \right)\otimes {L}_{\text{res}}\left({\omega }_{0}\right)\right)}^{\left(\omega +{\omega }_{0}-\lambda \right)}\text{.}$ Since ${ℬ}_{2}\subseteq {\left({M}_{\text{res}}\left(\omega \right)\otimes {L}_{\text{res}}\left({\omega }_{0}\right)\right)}^{\left(\omega +{\omega }_{0}-\lambda \right)}\text{,}$ there is a matrix $A$ taking ${ℬ}_{1}$ to ${ℬ}_{2}\text{.}$ Let ${F}_{{\beta }_{1}}^{\left({t}_{1}\right)}\cdots {F}_{{\beta }_{n}}^{\left({t}_{n}\right)}\left(v\otimes {w}_{\mu ,i}\right)\in {ℬ}_{2}\text{.}$ Since ${{U}_{𝒜}^{\text{res}}\left(𝔤\right)}^{-}$ 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 ${a}_{{s}_{1},\dots ,{s}_{N}}^{{r}_{1},\dots ,{r}_{N}}\in 𝒜,$ ${K}_{{s}_{1},\dots ,{s}_{N}}^{{r}_{1},\dots ,{r}_{N}}\in {{U}_{𝒜}^{\text{res}}\left(𝔤\right)}^{0},$ and the sum is over ${s}_{1},\dots ,{s}_{N},{r}_{1},\dots ,{r}_{N}\in {ℤ}_{\ge 0}$ such that ${r}_{1}{\beta }_{1}+\cdots +{r}_{N}{\beta }_{N}<{t}_{1}{\beta }_{1}+\cdots +{t}_{N}{\beta }_{N}$ and $\left({r}_{1}+{s}_{1}\right){\beta }_{1}+\cdots +\left({r}_{n}+{s}_{N}\right){\beta }_{N}={t}_{1}{\beta }_{1}+\cdots +{t}_{N}{\beta }_{N}\text{.}$ Therefore, $Fβ1(t1)⋯Fβn(tn) (v+⊗wμ,i)= Fβ1(t1)⋯Fβn(tn) v⊗wμ,i+ ∑ν>μ,(s)=ν-μ,1≤j≤n(ν) aν,j(s) Fβ1(s1)⋯Fβn(sn) v+⊗wν,j$ for some ${a}_{\nu ,j}^{\left(s\right)}\in 𝒜\text{.}$ Therefore, we can order vectors in such a way that the matrix $A\prime$ taking ${F}_{{\beta }_{1}}^{\left({t}_{1}\right)}\cdots {F}_{{\beta }_{n}}^{\left({t}_{n}\right)}{v}^{+}\otimes {w}_{\mu ,i}$ to ${F}_{{\beta }_{1}}^{\left({t}_{1}\right)}\cdots {F}_{{\beta }_{n}}^{\left({t}_{n}\right)}\left({v}^{+}\otimes {w}_{\mu ,i}\right)$ is upper-triangular with ones on the diagonal. Then, $A\prime$ is invertible with determinant one. By possibly reordering the bases, we may assume $A=A\prime \text{.}$ Therefore, $A\prime$ has entries in $𝒜\text{.}$ Since $A\prime$ is upper-triangular with ones on the diagonal, ${\left(A\prime \right)}^{-1}$ will also have entries in $𝒜\text{.}$ This proves that ${ℬ}_{2}$ is an $𝒜\text{-basis}$ for ${\left({M}_{\text{res}}\left(\omega \right)\otimes {L}_{\text{res}}\left({\omega }_{0}\right)\right)}^{\left(\omega +{\omega }_{0}-\lambda \right)}\text{.}$ $\square$

Let $\omega ,{\omega }_{0}\in P$ and $\lambda ,\mu \in {Q}_{+}\text{.}$ Define $det Lres(ω0)ω0-μ= det(⟨wμ,i,wμ,j⟩)1≤i,j≤dim Lres(ω0)ω0-μ$ 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 $\mu ,\mu \prime \le \lambda ,$ $1\le i\le n\left(\mu \right),$ $1\le j\le n\left(\mu \prime \right),$ and ${s}_{1},\dots ,{s}_{N},{t}_{1},\dots ,{t}_{N}\in {ℤ}_{\ge 0}$ are such that ${s}_{1}{\beta }_{1}+\cdots +{s}_{N}{\beta }_{N}=\lambda -\mu$ and ${t}_{1}{\beta }_{1}+\cdots +{t}_{N}{\beta }_{N}=\lambda -\mu \prime \text{.}$

Let $\omega ,{\omega }_{0}\in P\text{.}$ For $\lambda \in {Q}_{+},$ $\text{det}\left({\left({M}_{\text{res}}\left(\omega \right)\otimes {L}_{\text{res}}\left({\omega }_{0}\right)\right)}^{\omega +{\omega }_{0}-\lambda }\right)$ is given by $∏μ∈Q+μ≤λ (det(Mres(ω)ω-(λ-μ)))n(μ) (det(Lres(ω0)ω0-μ))P(λ-μ). (4.10)$

 Proof. Note that ${M}_{\text{res}}{\left(\omega \right)}^{{\nu }_{1}}\otimes {L}_{\text{res}}{\left({\omega }_{0}\right)}^{{\mu }_{1}}$ is orthogonal (with respect to the contravariant form) to ${M}_{\text{res}}{\left(\omega \right)}^{{\nu }_{1}}\otimes {L}_{\text{res}}{\left({\omega }_{0}\right)}^{{\mu }_{1}}$ for ${\nu }_{1}\ne {\nu }_{2}\text{.}$ As in the proof of Lemma 2.6.3, $det((Mres(ω)⊗Lres(ω0))ω+ω0-λ) = ∏μ∈Q+μ≤λ det(Mres(ω)ω-(λ-μ)⊗Lres(ω0)ω0-μ) = ∏μ∈Q+μ≤λ (det(Mres(ω)ω-(λ-μ))n(μ)) (det(Lres(ω0)ω0-μ))P(λ-μ).$ $\square$

Using the basis ${ℬ}_{2}$ for ${\left({M}_{\text{res}}\left(\omega \right)\otimes {L}_{\text{res}}\left({\omega }_{0}\right)\right)}^{\omega +{\omega }_{0}-\lambda },$ we see that ${M}_{\text{res}}\left(\omega \right)\otimes {L}_{\text{res}}\left({\omega }_{0}\right)$ has a filtration by Verma modules ${M}_{\text{res}}\left(\omega +{\omega }_{0}-\mu \right)$ where the Verma modules ${M}_{\text{res}}\left(\omega +{\omega }_{0}-\mu \right)$ is generated by the image of ${v}^{+}\otimes {w}_{\mu ,i}\text{.}$ The following form of the determinant formula better reflects the filtration of ${M}_{q}\left(\omega \right)\otimes {L}_{q}\left({\omega }_{0}\right)$ by Verma modules.

Let $\omega ,{\omega }_{0}\in P$ and $\lambda \in {Q}_{+}\text{.}$ Then, $det((Mres(ω)⊗Lres(ω0))ω+ω0-λ) = ∏μ∈Q+μ≤λ det(Mres(ω+ω0-μ)ω+ω0-λ)n(μ) aμω0(ω) det(Lres(ω0)ω0-μ)P(λ-μ)$ where $aμω0(ω)= qRμ∏β∈R+,r>0 ([⟨ω+ρ,β∨⟩-r][⟨ω+ω0-μ+rβ+ρ,β∨⟩-r])n(μ-rβ)$ and ${R}_{\mu }$ is defined inductively by $Rμ=-∑ν<μ,n(ν)≠0 P(μ-ν) (⟨ω0-ν,αi1∨+⋯+αim∨⟩n(ν)+Rν),$ for $\mu -\nu ={\alpha }_{{i}_{1}}+\cdots +{\alpha }_{{i}_{m}}\text{.}$

 Proof. Note that ${R}_{\mu }$ is always an integer and $∑μ≤λRμ P(λ-μ) = -∑μ≤λ kω-λ+μω- kω+ω0-λω+ω0-μ P(λ-μ)n(μ) = -∑μ≤λ ⟨ω0-μ,αi1∨+⋯+αim∨⟩ P(λ-μ)n(μ).$ From Lemma 4.6.3, $\text{det}\left({\left({M}_{\text{res}}\left(\omega \right)\otimes {L}_{\text{res}}\left({\omega }_{0}\right)\right)}^{\omega +{\omega }_{0}-\lambda }\right)$ is given by $= ∏μ∈Q+μ≤λ det(Lres(ω0)ω0-μ)P(λ-μ) qn(μ)P(λ-μ)kω-(λ-μ)ω ∏μ∈Q+μ≤λ Cλ-μ∏β∈R+,r>0 (([⟨ω+ρ,β∨⟩-r][r])P(λ-μ-rβ))n(μ) = ∏μ∈Q+μ≤λ det(Lres(ω0)ω0-μ)P(λ-μ) ∏μ∈Q+μ≤λ (qP(λ-μ)(kω+ω0-λω+ω0-μ+(kω-λ+μω-kω+ω0-λω+ω0-μ)))n(μ) = ∏μ∈Q+μ≤λ ∏β∈R+,r>0 ([⟨ω+ρ,β∨⟩-r][⟨ω+ω0-μ+ρ,β∨⟩-r])P(λ-μ-rβ)n(μ)) ⏟* ∏μ∈Q+μ≤λ Cλ-μ∏β∈R+,r>0 ([⟨ω+ω0-μ+ρ,β∨⟩-r][r])P(λ-μ-rβ)n(μ)) = ∏μ∈Q+μ≤λ det(Lres(ω0)ω0-μ)P(λ-μ) ∏μ∈Q+μ≤λ q(kω-λ+μω-kω+ω0-λω+ω0-μ)P(λ-μ)n(μ) ∏μ∈Q+μ≤λ ∏β∈R+,r>0 ([⟨ω+ρ,β∨⟩-r][⟨ω+ω0-μ+rβ+ρ,β∨⟩-r])P(λ-μ)n(μ-rβ) ⏟ We send μ→μ-rβ in *. Since P(λ-μ)=0 for μ>λ and n(μ)=0 for μ<0, this does not affect the index of the product. ∏μ∈Q+μ≤λ det(Mres(ω+ω0-μ)ω+ω0-λ)n(μ) = ∏μ∈Q+μ≤λ (aμ det(Lres(ω0)ω0-μ))P(λ-μ) det(Mres(ω+ω0-μ)ω+ω0-λ)n(μ)$ $\square$

${\left({M}_{q}\left(\omega \right)\otimes {L}_{q}\left({\omega }_{0}\right)\right)}^{\left[\lambda \right]}\perp {\left({M}_{q}\left(\omega \right)\otimes {L}_{q}\left({\omega }_{0}\right)\right)}^{\left[\mu \right]}$ for $\left[\mu \right]\ne \left[\lambda \right]\text{.}$

 Proof. The arguments from the proof of Proposition 2.6.4 can be applied here. $\square$

Let $\omega ,{\omega }_{0}\in P$ and let $\lambda \in {Q}_{+}\text{.}$ Suppose $\omega$ is such that there is no $\mu \in {Q}_{+}$ with $0<\mu \le \lambda$ and $\omega -\mu \in \left[\omega \right]\text{.}$ Then for each block $\gamma ,$ we can define projection maps $Prλ[γ]: Mq(ω)⊗Lq(ω0)⟶ ((Mq(ω)⊗Lq(ω0))[γ])ω+ω0-λ$ given by $Prλ[γ](v)=v- ∑[ν]≠[γ] ∑1m ⟨v,vi[ν]⟩ v‾i[ν]$ where $\left\{{v}_{1}^{\left[\nu \right]},\dots ,{v}_{m}^{\left[\nu \right]}\right\}$ and $\left\{{\stackrel{‾}{v}}_{1}^{\left[\nu \right]},\dots ,{\stackrel{‾}{v}}_{m}^{\left[\nu \right]}\right\}$ are dual bases for ${\left({\left({M}_{q}\left(\omega \right)\otimes {L}_{q}\left({\omega }_{0}\right)\right)}^{\left[\nu \right]}\right)}^{\omega +{\omega }_{0}-\lambda }\text{.}$

 Proof. Since there is no $\mu \in {Q}_{+}$ with $0<\mu \le \lambda$ and $\omega -\mu \in \left[\omega \right],$ $\text{det} {M}_{q}{\left(\omega \right)}^{\omega -\lambda }\ne 0\text{.}$ Lemma 4.6.3 then implies the contravariant form is nondegenerate on ${\left({M}_{q}\left(\omega \right)\otimes {L}_{q}\left({\omega }_{0}\right)\right)}^{\omega +{\omega }_{0}-\lambda }\text{.}$ We know from Lemma 4.6.5 that distinct blocks are orthogonal. This means that the contravariant form is nondegenerate on ${\left({\left({M}_{q}\left(\omega \right)\otimes {L}_{q}\left({\omega }_{0}\right)\right)}^{\left[\gamma \right]}\right)}^{\omega +{\omega }_{0}-\lambda }\text{.}$ Let $\left\{{v}_{1}^{\left[\nu \right]},\dots ,{v}_{m}^{\left[\nu \right]}\right\}$ be a basis for ${\left({\left({M}_{q}\left(\omega \right)\otimes {L}_{q}\left({\omega }_{0}\right)\right)}^{\left[\nu \right]}\right)}^{\omega +{\omega }_{0}-\lambda }\text{.}$ Since the contravariant form is nondegenerate on this space, there is a basis $\left\{{\stackrel{‾}{v}}_{1}^{\left[\nu \right]},\dots ,{\stackrel{‾}{v}}_{m}^{\left[\nu \right]}\right\}$ of ${\left({\left({M}_{q}\left(\omega \right)\otimes {L}_{q}\left({\omega }_{0}\right)\right)}^{\left[\nu \right]}\right)}^{\omega +{\omega }_{0}-\lambda }$ such that $⟨{v}_{i}^{\left[\nu \right]},{\stackrel{‾}{v}}_{l}^{\left[\nu \right]}⟩={\delta }_{i,l}\text{.}$ We define a map $Prλ[γ]: (Mq(ω)⊗Lq(ω))ω+ω0-λ⟶ ((Mq(ω)⊗Lq(ω))[γ])ω+ω0-λ$ by $Prλ[γ](v)= v-∑[ν]≠[γ] ∑1m⟨v,vi[ν]⟩ v‾i[ν].$ Note that $⟨{\text{Pr}}_{\lambda }^{\left[\gamma \right]}\left(v\right),{v}_{i}^{\left[\nu \right]}⟩=0$ whenever $\left[\nu \right]\ne \left[\gamma \right]\text{.}$ Therefore, $Prλ[γ](v)∈ ((Mq(ω)⊗Lq(ω))[γ])ω+ω0-λ.$ Also, for $v\in {\left({\left({M}_{q}\left(\omega \right)\otimes {L}_{q}\left(\omega \right)\right)}^{\left[\gamma \right]}\right)}^{\omega +{\omega }_{0}-\lambda },$ ${\text{Pr}}_{\lambda }^{\left[\gamma \right]}\left(v\right)=v\text{.}$ $\square$

Note: We can inductively construct a basis for each block of ${M}_{q}\left(\omega \right)\otimes {L}_{q}\left({\omega }_{0}\right)$ using a basis for ${M}_{\text{res}}\left(\omega \right)\otimes {L}_{\text{res}}\left({\omega }_{0}\right)\text{.}$ Elements of this basis will be $ℚ\left(q\right)\text{-linear}$ combinations of the basis for ${M}_{\text{res}}\left(\omega \right)\otimes {L}_{\text{res}}\left({\omega }_{0}\right)\text{.}$ Therefore, we have a basis for each block of ${M}_{q}\left(\omega \right)\otimes {L}_{q}\left({\omega }_{0}\right)$ such that a nonzero multiple of this basis is contained in ${M}_{\text{res}}\left(\omega \right)\otimes {L}_{\text{res}}\left({\omega }_{0}\right)\text{.}$

Fix $\omega ,{\omega }_{0}\in P$ and $\lambda \in {Q}_{+}\text{.}$ Suppose $\omega$ is such that there is no $w\in W$ with $\omega \ge w\circ \omega \ge \omega -\lambda$ or $\omega +{\omega }_{0}\ge w\circ \left(\omega +{\omega }_{0}\right)\ge \omega +{\omega }_{0}-\lambda \text{.}$

Then the submodule of ${M}_{q}\left(\omega \right)\otimes {L}_{q}\left({\omega }_{0}\right)$ generated by $⨁μ∈Q+,μ≤λ (Mq(ω)⊗Lq(ω0))ω+ω0-μ$ is isomorphic to $⨁μ∈Q+,μ≤λ Mq(ω+ω0-μ)⊕dim Lq(ω0)ω0-μ.$ For a suitable choice of generating highest weight vectors $\left\{{v}_{\mu ,i}^{+} | 1\le i\le \text{dim}{\left({L}_{q}\left({\omega }_{0}\right)\right)}^{{\omega }_{0}-\mu }\right\},$ this sum is orthogonal with respect to the contravariant form on ${M}_{q}\left(\omega \right)\otimes {L}_{q}\left({\omega }_{0}\right),$ and $∏1≤i≤dim(L(ω0))ω0-μ ⟨vμ,i,vμ,i⟩= aμω0(ω) det L(ω0)ω0-μ$

Proof.

Since there is no $w\in W$ with $\omega \ge w\circ \omega \ge \omega -\lambda ,$ the projection maps from the previous lemma are well-defined.

We have assumed there is no $w\in W$ such that $\omega +{\omega }_{0}\ge w\circ \left(\omega +{\omega }_{0}\right)\ge \omega +{\omega }_{0}-\lambda \text{.}$ Therefore, there is only one weight in $\left[\omega +{\omega }_{0}-\mu \right]$ between $\omega +{\omega }_{0}$ and $\omega +{\omega }_{0}-\lambda ,$ namely $\omega +{\omega }_{0}-\mu \text{.}$ This implies ${\text{Pr}}_{\mu }^{\left[\mu \right]}\left({v}^{+}\otimes {w}_{\mu ,i}\right)$ is a highest weight vector. Choose vectors $\left\{{v}_{\mu ,i}^{+}\right\}$ such that

 • the transition matrix from $\left\{{\text{Pr}}_{\mu }^{\left[\mu \right]}\left({v}^{+}\otimes {w}_{\mu ,i}\right)\right\}$ to $\left\{{v}_{\mu ,i}^{+}\right\}$ has determinant 1; • $⟨{v}_{\mu ,i}^{+},{v}_{\mu ,k}^{+}⟩=0$ if $i\ne k\text{.}$
Therefore, we only need to determine $( ∏i⟨vμ,i+,vμ,i+⟩ )$ 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 $\nu \in {P}_{+}$ with $\nu <\mu \text{.}$

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 $= ∏i ( ⟨ Fβ1(t1)⋯ Fβn(tn) vν,i+, Fβ1(s1)⋯ Fβn(sn) vν,i+ ⟩ ) (t),(s) = (∏i⟨vν,i+,vν,i+⟩)p(μ-ν) (det Mq(ω+ω0-ν)ω+ω0-μ)n(ν).$ Since distinct blocks are orthogonal, we have $\text{det}{\left({M}_{q}\left(\omega \right)\otimes {L}_{q}\left({\omega }_{0}\right)\right)}^{\omega +{\omega }_{0}-\mu }$ is $∏ν≤μdet ( ⟨ Fβ1(t1)⋯ Fβn(tn) Prν[ν] (v+⊗wν,i), Fβ1(s1)⋯ Fβn(sn) Prν[ν] (v+⊗wν,k) ⟩ ) = ∏ν<μ (det Mq(ω+ω0-ν)ω+ω0-μ)n(ν) ×(aνω0(ω)det Lq(ω0)ω0-ν) det(⟨Prμ[μ](v+⊗wμ,i),Prμ[μ](v+⊗wμ,l)⟩).$ This implies $\text{det}\left(⟨{\text{Pr}}_{\mu }^{\left[\mu \right]}\left({v}^{+}\otimes {w}_{\mu ,i}\right),{\text{Pr}}_{\mu }^{\left[\mu \right]}\left({v}^{+}\otimes {w}_{\mu ,l}\right)⟩\right)={a}_{\mu }^{{\omega }_{0}}\left(\omega \right)\text{det} {L}_{q}{\left({\omega }_{0}\right)}^{{\omega }_{0}-\mu }\text{.}$

$\square$

## 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.