Translation Functors and the Shapovalov Determinant

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updated: 10 February 2015

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

Quantum Groups

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

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

•	${[m]}_{d} = \frac{q^{d m} - q^{- d m}}{q^{d} - q^{- d}}$
•	${[m]}_{d}! = {[m]}_{d} {[m - 1]}_{d} \dots {[2]}_{d} {[1]}_{d}$ for $m \geq 0$
•	${[\binom{m}{n}]}_{d} = \frac{{[m]}_{d}!}{{[n]}_{d}! {[m - n]}_{d}!}$ for $m \geq n > 0 .$

Let

𝔤

be a finite-dimensional complex semisimple Lie algebra, and let

A = {(a_{i j})}_{1 \leq i, j \leq n}

be the associated Cartan matrix. Choose

d_{i} \in ℤ_{> 0},

1 \leq i \leq k

minimal so that the matrix

{(d_{i} a_{i j})}_{1 \leq i, j \leq n}

is symmetric. To simplify notation, we will write

q_{i} = q^{d_{i}}

and

{[\cdot]}_{i} = {[\cdot]}_{d_{i}} .

There is some

k \in ℤ_{> 0}

so that

d_{i} = k \frac{⟨ α_{i}, α_{i} ⟩}{2} .

For

β \in R^{+},

define

d_{β} = k \frac{⟨ β, β ⟩}{2} .

Suppose

β = w α_{i}

for

w \in W

and

α_{i}

a simple root. Equation 1.2 implies

d_{β} = d_{i} .

The quantum group $U_{q} (𝔤)$ associated to $𝔤$ is the $ℚ (q) -algebra$ generated by the elements $E_{i},$ $F_{i},$ and $K_{i}^{\pm 1}$ $(1 \leq i \leq k)$ with relations $\begin{matrix} K_{i} K_{j} = K_{j} K_{i}, K_{i} K_{i}^{- 1} = 1 = K_{i}^{- 1} K_{i}; & (4.1) \end{matrix}$ $\begin{matrix} K_{i} E_{j} = q_{i}^{a_{i j}} E_{j} K_{i}, K_{i} F_{j} = q_{i}^{- a_{i j}} F_{j} K_{i}; & (4.2) \end{matrix}$ $\begin{matrix} E_{i} F_{j} - F_{j} E_{i} = δ_{i, j} \frac{K_{i} - K_{i}^{- 1}}{q_{i} - q_{i}^{- 1}}; & (4.3) \end{matrix}$ $\begin{matrix} \sum_{r = 0}^{1 - a_{i j}} {(- 1)}^{r} {[\binom{1 - a_{i j}}{r}]}_{i} E_{i}^{(1 - a_{i j}) - r} E_{j} E_{i}^{r} = 0, & (4.4) \end{matrix}$ $\begin{matrix} \sum_{r = 0}^{1 - a_{i j}} {(- 1)}^{r} {[\binom{1 - a_{i j}}{r}]}_{i} F_{i}^{(1 - a_{i j}) - r} F_{j} F_{i}^{r} = 0 for i \neq j . & (4.5) \end{matrix}$

The quantum group $U_{q} (𝔤)$ is a Hopf algebra with the following maps:

•	coproduct: $Δ (K_{i}) = K_{i} \otimes K_{i},$ $Δ (E_{i}) = E_{i} \otimes K_{i}^{- 1} + 1 \otimes E_{i},$ $Δ (F_{i}) = F_{i} \otimes 1 + K_{i} \otimes F_{i};$
•	counit: $ε (K_{i}) = 1,$ $ε (E_{i}) = 0 = ε (F_{i});$
•	antipode: $S (K_{i}) = K_{i}^{- 1},$ $S (E_{i}) = - E_{i} K_{i},$ $S (F_{i}) = - K_{i}^{- 1} F_{i} .$

(See ([CPr1994] 9.1.1) for a proof that these maps give a Hopf algebra structure.) We define the following

ℚ (q) -subalgebras

U_{q} (𝔤) :

•	$U_{q}^{+} (𝔤)$ is the subalgebra generated by ${E_{i} \| 1 \leq i \leq k} .$
•	$U_{q}^{0} (𝔤)$ is the subalgebra generated by ${K_{i}^{\pm 1} \| 1 \leq i \leq k} .$
•	$U_{q}^{-} (𝔤)$ is the subalgebra generated by ${F_{i} \| 1 \leq i \leq k} .$

([CPr1994] 9.1.3). $U_{q} (𝔤) ≅ U_{q}^{-} (𝔤) \otimes U_{q}^{0} (𝔤) \otimes U_{q}^{+} (𝔤) .$

There is a natural pairing between $U_{q}^{+} (𝔤)$ and $U_{q}^{-} (𝔤) .$ The following maps make use of this pairing. Define a $ℚ -anti-automorphism$ $τ : U_{q} (𝔤) \to U_{q} (𝔤)$ by $τ (E_{i}) = F_{i}; τ (F_{i}) = E_{i}; τ (K_{i}) = K_{i}^{- 1}; τ (q) = q^{- 1} .$ Define a $ℚ (q) -anti-automorphism$ $ϕ : U_{q} (𝔤) \to U_{q} (𝔤)$ by $ϕ (E_{i}) = K_{i}^{- 1} F_{i}; ϕ (F_{i}) = E_{i} K_{i}; ϕ (K_{i}) = K_{i} .$ Observe that $ϕ$ commutes with the coproduct map: $Δ \circ ϕ = (ϕ \otimes ϕ) \circ Δ .$ This property will be useful later when we consider tensor products of modules.

The Restricted Integral Form $U_{𝒜}^{res} (𝔤)$

Let $𝒜 = ℤ [q, q^{- 1}] .$ The restricted integral form $U_{𝒜}^{res} (𝔤)$ of $U_{q} (𝔤)$ is the $𝒜 -subalgebra$ of $U_{q} (𝔤)$ generated by $K_{i}^{\pm 1}; E_{i}^{(r)} = \frac{E_{i}^{r}}{{[r]}_{i}!}; F_{i}^{(r)} = \frac{F_{i}^{r}}{{[r]}_{i}!} for 1 \leq i \leq k, r > 0 .$ The algebra $U_{𝒜}^{res} (𝔤)$ inherits many properties from $U_{q} (𝔤),$ including the Hopf algebra structure and the triangular decomposition. The restricted integral form $U_{𝒜}^{res} (𝔤)$ is a Hopf algebra with the following maps

•	coproduct: $\begin{array}{rcl} Δ (K_{i}) & = & K_{i} \otimes K_{i} \\ Δ (E_{i}^{(r)}) & = & \sum_{j = 0}^{r} q_{i}^{j (r - j)} E_{i}^{(j)} \otimes K_{i}^{- j} E_{i}^{(r - j)} \\ Δ (F_{i}^{(r)}) & = & \sum_{j = 0}^{r} q_{i}^{- j (r - j)} F_{i}^{(j)} K_{i}^{r - j} \otimes F_{i}^{(r - j)} . \end{array}$
•	counit: $ε (K_{i}) = 1; ε (E_{i}^{(r)}) = 0 = ε (F_{i}^{(r)});$
•	antipode: $S (K_{i}) = K_{i}^{- 1}; S (E_{i}^{(r)}) = {(- 1)}^{𝔯} q^{r (r - 1)} E_{i}^{(r)} K_{i}^{r}; S (F_{i}^{(r)}) = {(- 1)}^{(r)} q^{- r (r - 1)} K_{i}^{- r} F_{i}^{(r)} .$

([Lus1990-4] 6.6). The algebra $U_{𝒜}^{res} (𝔤)$ decomposes as $U_{𝒜}^{res} (𝔤) ≃ U_{𝒜}^{res} {(𝔤)}^{+} \otimes U_{𝒜}^{res} {(𝔤)}^{0} \otimes U_{𝒜}^{res} {(𝔤)}^{-}$ where

•	$U_{𝒜}^{res} {(𝔤)}^{+} = U_{q}^{+} (𝔤) \cap U_{𝒜}^{res} (𝔤)$ is generated by ${E_{i}^{(r)} \| 1 \leq i \leq k, r > 0};$
•	${(U_{𝒜}^{res} (𝔤))}^{0} = U_{q}^{0} (𝔤) \cap U_{𝒜}^{res} (𝔤)$ is generated by ${K_{i}^{\pm 1}, [\binom{K_{i}; c}{r}] \| 1 \leq i \leq k, c \in ℤ, r > 0},$ with $[\binom{K_{i}; c}{r}] = \prod_{s = 1}^{r} \frac{K_{i} q_{i}^{c + 1 - s} - K_{i}^{- 1} q_{i}^{s - c - 1}}{q_{i}^{s} - q_{i}^{- s}};$
•	${(U_{𝒜}^{res} (𝔤))}^{-} = U_{q}^{-} (𝔤) \cap U_{𝒜}^{res} (𝔤)$ is generated by ${F_{i}^{(r)} \| 1 \leq i \leq k, r > 0} .$

A PBW Basis for $U_{𝒜}^{res} (𝔤)$

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

([Lus1990-4] 3.1). For each $1 \leq i \leq n,$ there is a unique $ℚ (q) -automorphism$ $T_{i} : U_{𝒜}^{res} (𝔤) \to U_{𝒜}^{res} (𝔤)$ defined by $\begin{array}{rcl} T_{i} (K_{j}) & = & K_{j} K_{i}^{- a_{i j}}; \\ T_{i} (E_{j}) & = & {\begin{cases} - F_{i} K_{i} & if i = j \\ \sum_{r = 0}^{- a_{i j}} {(- 1)}^{r - a_{i j}} q_{i}^{- r} E_{i}^{(- a_{i j} - r)} E_{j} E_{i}^{(r)} & if i \neq j \end{cases}; \\ T_{i} (F_{j}) & = & {\begin{cases} - K_{i}^{- 1} E_{i} & if i = j \\ \sum_{r = 0}^{- a_{i j}} {(- 1)}^{r - a_{i j}} q_{i}^{r} F_{i}^{(r)} F_{j} F_{i}^{(- a_{i j} - r)} & if i \neq j \end{cases} \end{array}$

Fix a reduced expression for the longest element of the Weyl group, $w_{0} = s_{i_{1}} \dots s_{i_{N}} .$ This gives an ordering of the positive roots: $β_{j} = s_{i_{1}} \dots s_{i_{j - 1}} α_{i_{j}}, 1 \leq j \leq N .$ Define $F_{β_{j}} \in {U_{𝒜}^{res} (𝔤)}^{-} \subseteq U_{q}^{-} (𝔤)$ to be $F_{β_{j}} = T_{i_{1}} \dots T_{i_{j - 1}} F_{j} .$ We similarly define $E_{β_{j}} .$ Finally, define $K_{β_{j}} = T_{i_{1}} \dots T_{i_{j - 1}} K_{j} = K_{i_{1}} \dots K_{i_{m}}$ where $β_{j} = α_{i_{1}} + \dots + α_{i_{m}} .$

([Lus1990-4], 6.7). The sets ${E_{β_{n}}^{(s_{n})} \dots E_{β_{1}}^{(s_{1})} | s_{i} \in ℤ_{\geq 0}} and {F_{β_{1}}^{(s_{1})} \dots F_{β_{N}}^{(s_{N})} | s_{i} \in ℤ_{\geq 0}}$ are bases for ${U_{𝒜}^{res} (𝔤)}^{+}$ and ${U_{𝒜}^{res} (𝔤)}^{-},$ respectively.

These sets are also bases for $U_{q}^{+} (𝔤)$ and $U_{q}^{-} (𝔤),$ respectively. Note that $T_{i}$ commutes with the $ℚ -antiautomorphism$ $ϕ,$ and so we have $ϕ (E_{β}) = F_{β} .$

The Category $𝒪$

We begin by defining weights and weight spaces. A weight of $U_{q} (𝔤)$ is an element $Ω \in {Hom}_{ℚ (q)} (U_{q}^{0} (𝔤), ℚ (q)) .$ Since the set ${K_{i}^{\pm 1} | 1 \leq i \leq k}$ generates $U_{q}^{0} (𝔤)$ and $Ω (K_{i}^{- 1}) = {(Ω (K_{i}))}^{- 1},$ $Ω$ is determined by its action on each of the $K_{i} .$ Define $Ω_{i} = Ω (K_{i})$ and identify $Ω$ with the $k -tuple$ $(Ω_{1}, \dots, Ω_{k}) .$

We view $𝔥^{*}$ as sitting inside ${Hom}_{ℚ (q)} (U_{q}^{0} (𝔤), ℚ (q)) :$ for $ω \in 𝔥^{*},$ $ω (K_{i}) = q^{⟨ ω, α_{i}^{\lor} ⟩} .$ We will also be interested in certain "twistings" of the weights of $𝔤 .$ Let $σ = (σ_{1}, \dots, σ_{k}) \in {(\pm 1, \dots, \pm 1)}$ and $ω \in 𝔥^{*} .$ Define $ω^{σ} \in {Hom}_{ℚ (q)} (U_{q}^{0} (𝔤), ℚ (q))$ by $ω^{σ} (K_{i}) = σ_{i} q_{i}^{⟨ ω, α_{i}^{\lor} ⟩} = \pm q_{i}^{⟨ ω, α_{i}^{\lor} ⟩} .$

The dominance ordering on $𝔥^{*}$ induces a partial ordering on the weights of $U_{q} (𝔤) .$ For $Ω, Λ \in {Hom}_{ℚ (q)} (U_{q}^{0} (𝔤), ℚ (q)),$ $Ω \leq Λ if there is some ν \in Q_{+} such that \frac{Λ (K_{i})}{Ω (K_{i})} = q^{⟨ ν, α_{i}^{\lor} ⟩} for 1 \leq i \leq k .$ In this case we write $\frac{Ω}{Λ} = ν .$ For $ω, λ \in 𝔥^{*},$ this is the standard dominance ordering.

For a $U_{q} (𝔤)$ module $M,$ the $Ω -weight$ space of $M$ is given by $M^{Ω} = {v \in M | K_{i} v = Ω (K_{i}) v for 1 \leq i \leq k} .$

The Category $𝒪$ consists of $U_{q} (𝔤) -modules$ $M$ such that

•	$M$ is $U_{q} {(𝔤)}^{0} -diagonalizable:$ $M = ⨁_{Ω \in {Hom}_{ℚ (q)} (U_{q}^{0} (𝔤), ℚ (q))} M^{Ω};$
•	there are $Λ_{1}, \dots, Λ_{n} \in {Hom}_{ℚ (q)} (U_{q}^{0} (𝔤), ℚ (q))$ such that $M^{Ω} \neq 0$ only if $Ω \leq Λ_{i}$ for some $1 \leq i \leq n .$
•	$dim M^{Ω} < \infty$ for all $Ω \in {Hom}_{ℚ (q)} (U_{q}^{0} (𝔤), ℚ (q)) .$

A highest weight $U_{q} (𝔤) -module$ of highest weight $Ω$ is a $U_{q} (𝔤) -module$ $M$ with a vector $v^{+} \in M$ such that

•	$v^{+} \in M^{Ω};$
•	$M = U_{q} (𝔤) v^{+};$
•	$E_{i} v^{+} = 0$ for all $1 \leq i \leq k .$

The vector

v^{+}

is a highest weight vector. For such a module

M

the defining relations of

U_{q} (𝔤)

imply

M = ⨁_{Λ \leq Ω} M^{Λ} .

Note that all highest weight modules are in Category

𝒪 .

The Verma module of highest weight $Ω,$ is the $U_{q} (𝔤) -module$ $M_{q} (Ω) = U_{q} (𝔤) / I,$ where $I$ is the left ideal generated by $E_{i}$ and $K_{i} - Ω (K_{i})$ $(1 \leq i \leq k) .$ We can also define $M_{q} (Ω)$ as the induced module $M_{q} (Ω) = U_{q} (𝔤) \otimes_{U_{q}^{\geq 0} (𝔤)} ℚ {(q)}_{Ω},$ where $ℚ {(q)}_{Ω}$ is $ℚ (q)$ as a vector space; and, as a $U_{q}^{\geq 0} (𝔤) -module,$ $U_{q}^{0} (𝔤)$ acts by $Ω$ and $U_{q}^{+} (𝔤)$ acts by $0 .$ Using the same arguments as in Lemma 2.1.3, we can show that $M_{q} (Ω)$ has a maximal proper submodule, $J_{q} (Ω) .$ This implies that $M_{q} (Ω)$ has a unique simple quotient $L_{q} (Ω) = M_{q} (Ω) / J_{q} (Ω) .$

All of these objects can be similarly defined for $U_{𝒜}^{res} (𝔤) .$ The set of weights for $U_{𝒜}^{res} (𝔤)$ is ${Hom}_{𝒜} ({U_{𝒜}^{res} (𝔤)}^{0}, 𝒜) .$ We denote the $U_{𝒜}^{res} (𝔤) -Verma$ module of weight $Ω \in {Hom}_{𝒜} ({U_{𝒜}^{res} (𝔤)}^{0}, 𝒜)$ by $M_{res} (Ω) .$ In other words, $M_{res} (Ω) = U_{𝒜}^{res} (𝔤) / J,$ where $J$ is the ideal generated by $E_{i}^{(r)}, r > 0; K_{i} - Ω (K_{i}); [\frac{K_{i}; c}{r}] - \prod_{s = 1}^{r} \frac{Ω (K_{i}) q_{i}^{c + 1 - s} - Ω (K_{i}^{- 1}) q_{i}^{s - c - 1}}{q_{i}^{s} - q_{i}^{- s}} .$

From Theorem 4.2.2, we get the following.

Let $Ω, Λ \in {Hom}_{ℚ (q)} (U_{q}^{0} (𝔤), ℚ (q))$ so that $Ω \geq Λ .$ Suppose $\frac{Ω}{Λ} = ν \in Q_{+} .$ Define $F_{β_{j}}^{(t_{j})} = \frac{1}{[t_{j}]!} F_{β_{j}}^{t_{j}} .$ Then, the set ${F_{β_{1}}^{(t_{1})} \dots F_{β_{N}}^{(t_{N})} v^{+} | t_{1}, \dots, t_{N} \in ℤ_{\geq 0}, t_{1} β_{1} + \dots + t_{N} β_{N} = ν}$ is a basis for $M_{res} {(Ω)}^{Λ}$ and $M_{q} {(Ω)}^{Λ} .$

Embeddings of Verma Modules

Recall that for a weight $ω \in 𝔥^{*}$ of $𝔤$ and $σ = (σ_{1}, \dots, σ_{k}) \in {(\pm 1, \dots, \pm 1)},$ we can define a $U_{q} (𝔤) -weight$ $ω^{σ} \in {Hom}_{ℚ (q)} (U_{q}^{0} (𝔤), ℚ (q))$ by $ω^{σ} (K_{i}) = σ_{i} q_{i}^{⟨ ω, α_{i}^{\lor} ⟩} .$

Let $λ \in P,$ $σ \in {(\pm 1, \dots, \pm 1)},$ and $β \in R^{+} .$ If $s_{β} \circ λ \leq λ,$ then $M_{q} ({(s_{β} \circ λ)}^{σ}) \subseteq M_{q} (λ^{σ}) .$

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

Let $λ \in P$ and $σ \in {(\pm 1, \dots, \pm 1)} .$ If $s_{i} \circ λ \leq λ,$ $M_{q} ({(s_{i} \circ λ)}^{σ}) \subseteq M_{q} (λ^{σ}) .$ Moreover, if $v^{+}$ generates $M_{q} (λ^{σ}),$ then $F_{i}^{m} v^{+}$ is a highest weight vector of weight ${(s_{i} \circ λ)}^{σ},$ where $m = ⟨ λ + ρ, α_{i}^{\lor} ⟩ \geq 0 .$

Proof.

Note that $F_{i}^{m} v^{+} \in M_{q} {(λ^{σ})}^{{(s_{i} \circ λ)}^{σ}} .$ We must also check that $E_{j} F_{i}^{m} v^{+} = 0$ for all $1 \leq j \leq k .$ If $i \neq j,$ $E_{j} F_{i}^{m} v^{+} = F_{i}^{m} E_{j} v^{+} = 0 .$ If $i = j,$ we can use Equation 4.3 to show $\begin{array}{rcl} E_{i} F_{i}^{m} v^{+} & = & σ_{i} {[m]}_{i} {[⟨ λ, α_{i}^{\lor} ⟩ - m + 1]}_{i} F_{i}^{m - 1} v^{+} \\ = & σ_{i} {[m]}_{i} {[⟨ λ + ρ, α_{i}^{\lor} ⟩ - m]}_{i} F_{i}^{m - 1} v^{+} \\ = & 0 . \end{array}$

$□$

Let $λ \in P^{+},$ $σ \in {(\pm 1, \dots, \pm 1)},$ and $w \in W$ with reduced expression $w = s_{i_{n}} \dots s_{i_{1}} .$ Set $λ_{0} = λ, λ_{1} = s_{i_{1}} \circ λ_{0}, \dots, λ_{n} = s_{i_{n}} \circ λ_{n - 1} .$ Then, $M_{q} (λ_{n}^{σ}) \subseteq M_{q} (λ_{n - 1}^{σ}) \subseteq \dots \subseteq M_{q} (λ_{0}^{σ}) .$

Proof.

By definition, $λ_{j + 1} = s_{i_{j + 1}} \circ λ_{i} .$ Therefore, using the previous lemma, we need only check that $⟨ λ_{j} + ρ, α_{j + 1}^{\lor} ⟩ \geq 0$ to show that $M_{q} (λ_{j + 1}^{σ}) \subseteq M_{q} (λ_{j}^{σ}) .$ Note that $⟨ λ_{j} + ρ, α_{j + 1}^{\lor} ⟩ = ⟨ w_{j} (λ + ρ), α_{j + 1}^{\lor} ⟩ = ⟨ λ + ρ, w_{j}^{- 1} α_{j + 1}^{\lor} ⟩,$ where $w_{j} = s_{i_{j}} \dots s_{i_{1}} .$ Then $w_{j}^{- 1} α_{j + 1} \in R^{+} .$ (See [Dix1996] 11.1.8.) Because $λ \in P^{+},$ $⟨ λ + ρ, w_{j}^{- 1} α_{j + 1}^{\lor} ⟩ > 0 .$

$□$

Let $λ, μ \in P,$ and $σ \in {(\pm 1, \dots, \pm 1)} .$ Suppose $α_{i}$ is a simple root such that $M_{q} ({(s_{i} \circ μ)}^{σ}) \subseteq M_{q} (μ^{σ}) \subseteq M_{q} (λ^{σ}) .$ Then, $M_{q} ({(s_{i} \circ μ)}^{σ}) \subseteq M_{q} ({(s_{i} \circ λ)}^{σ}) .$

This gives the following picture, where arrows indicate inclusion: $\begin{matrix} M (λ^{σ}) \\ ↑ & M ({(s_{i} \circ λ)}^{σ}) \\ M (μ^{σ}) & ↑ \\ ↖ & M ({(s_{i} \circ μ)}^{σ}) \end{matrix}$

The picture indicates nothing about the relationship between $M_{q} (λ^{σ})$ and $M_{q} ({(s_{i} \circ λ)}^{σ}) .$ The proof will consider both possibilities: $M_{q} (λ^{σ}) \subseteq M_{q} ({(s_{i} \circ λ)}^{σ})$ and $M_{q} ({(s_{i} \circ λ)}^{σ}) . \subseteq M_{q} (λ^{σ}) .$

Proof.

Let $m = ⟨ μ + ρ, α_{i}^{\lor} ⟩ \geq 0$ and $n = ⟨ λ + ρ, α_{i}^{\lor} ⟩ .$ If $n \leq 0,$ then we have $M_{q} ({(s_{i} \circ μ)}^{σ}) \subseteq M_{q} (μ^{σ}) \subseteq M_{q} (λ^{σ}) \subseteq M_{q} ({(s_{i} \circ λ)}^{σ}),$ i.e. we have the following picture of embeddings. $\begin{matrix} \end{matrix}$ Then the lemma follows.

Suppose $n > 0 .$ Then $M_{q} ({(s_{i} \circ λ)}^{σ}) \subseteq M_{q} (λ^{σ}) .$ Let $v^{+}, {\tilde{v}}^{+} \in M_{q} (λ^{σ})$ be generators for $M_{q} (λ^{σ})$ and $M_{q} (μ^{σ})$ respectively. Then, $F_{i}^{m} {\tilde{v}}^{+}$ generates $M_{q} ({(s_{i} \circ μ)}^{σ}),$ and $F_{i}^{n} v^{+}$ generates $M_{q} ({(s_{i} \circ λ)}^{σ}) .$ Therefore, $M_{q} ({(s_{i} \circ μ)}^{σ}) \subseteq M_{q} ({(s_{i} \circ λ)}^{σ})$ if $F_{i}^{m} {\tilde{v}}^{+} \in M_{q} ({(s_{i} \circ λ)}^{σ}) .$

We first show that there is some $l \in ℤ_{\geq 0}$ such that $F_{i}^{l} {\tilde{v}}^{+} \in M_{q} ({(s_{i} \circ λ)}^{σ}) .$ In other words, we want to show that $F_{i}^{l} {\tilde{v}}^{+} = X F_{i}^{n} v^{+}$ for some $X \in U_{q} (𝔤) .$ We have ${\tilde{v}}^{+} = u v^{+}$ for some $u \in U_{q}^{-} (𝔤) .$ Since $u \in U_{q}^{-} (𝔤),$ $u$ is a $ℚ (q) -linear$ combination of elements of the form $F_{i_{1}} \dots F_{i_{j}}$ where $α_{i_{1}} + \dots + α_{i_{j}} = λ - μ .$ From the generating relations for $U_{q} (𝔤),$ we have $\begin{array}{rcl} F_{i}^{1 - a_{i j}} F_{j} & = & - \sum_{r = 1}^{1 - a_{i j}} {(- 1)}^{r} [\binom{1 - a_{i j}}{r}] F_{i}^{1 - a_{i j} - r} F_{j} F_{i}^{r} \\ = & (\sum_{r = 0}^{- a_{i j}} {(- 1)}^{r} {[\binom{1 - a_{i j}}{r + 1}]}_{i} F_{i}^{- a_{i j} - r} F_{j} F_{i}^{r}) F_{i} \end{array}$ for $i \neq j .$ Thus, $F_{i}^{t (1 - a_{i j})} F_{j} = X F_{i}^{t}$ for some $X \in U_{q}^{-} (𝔤) .$ Given the form of $u,$ this implies $F_{i}^{l} {\tilde{v}}^{+} \in M_{q} ({(s_{i} \circ λ)}^{σ})$ for some $l \in ℤ_{\geq 0} .$

We claim that this implies that $F_{i}^{m} {\tilde{v}}^{+} \in M_{q} ({(s_{i} \circ λ)}^{σ}) :$

•	If $l \leq m,$ then $F_{i}^{m} {\tilde{v}}^{+} = F_{i}^{m - l} F_{i}^{l} {\tilde{v}}^{+} \in M_{q} ({(s_{i} \circ λ)}^{σ}) .$
•	If $l > m,$ then $E_{i} F_{i}^{l} {\tilde{v}}^{+} = [l] [⟨ μ, α^{\lor} ⟩ - l + 1] F_{i}^{l - 1} {\tilde{v}}^{+} .$ Since $E_{i} F_{i}^{l} {\tilde{v}}^{+} \in M_{q} ({(s_{i} \circ λ)}^{σ}),$ $F_{i}^{m} {\tilde{v}}^{+} \in M_{q} ({(s_{i} \circ λ)}^{σ}) .$

In this case, we have the following picture:

\begin{matrix}  \end{matrix}

$□$

We now proceed with the proof of Proposition 4.4.1.

Proof of Proposition 4.4.1.

Let $μ = s_{β} \circ λ$ and $m = ⟨ λ + ρ, β^{\lor} ⟩ \geq 0 .$ If $m = 0,$ $μ = λ .$ Therefore, assume $m > 0 .$ Let $\tilde{μ} \in P^{+} \cap (W \circ μ)$ and choose $w \in W$ so that $μ = w \circ \tilde{μ} .$ Write $\tilde{λ} = w^{- 1} \circ λ .$ Fix a reduced expression $w = s_{i_{n}} \dots s_{i_{1}},$ and set $λ_{0} = \tilde{λ}, λ_{1} = s_{i_{1}} \circ λ_{0}, \dots, λ_{n} = s_{i_{n}} \circ λ_{n - 1} = λ;$ $μ_{0} = \tilde{μ}, μ_{1} = s_{i_{1}} \circ μ_{0}, \dots, μ_{n} = s_{i_{n}} \circ μ_{n - 1} = μ .$ Note that $\tilde{μ} = w^{- 1} \circ s_{β} \circ w \circ \tilde{λ} = s_{w^{- 1} β} \circ \tilde{λ} .$ Therefore $μ_{j} = s_{(s_{i_{j}} \dots s_{i_{1}} w^{- 1} β)} \circ λ_{j} .$ Since $s_{i}$ sends $α_{i}$ to $- α_{i}$ and permutes the other positive roots, $(s_{i_{j}} \dots s_{i_{1}} w^{- 1}) β = γ_{j} \in R^{+}$ or $(s_{i_{j}} \dots s_{i_{1}} w^{- 1}) β = - γ_{j} \in - R^{+} .$ Therefore, $μ_{j} = s_{γ_{j}} \circ λ_{j} = s_{- γ_{j}} \circ λ_{j}$ for some $γ_{j} \in R^{+} .$

Since $μ_{0} \in P^{+}$ and $λ_{0} \in W \circ μ_{0},$ we must have $μ_{0} - λ_{0} \in Q_{+} .$ On the other hand, $μ_{n} - λ_{n} = - m β .$ Let $j$ be minimal such that $μ_{j} - λ_{j} \in Q_{+}$ and $μ_{j + 1} - λ_{j + 1} \in - Q_{+} .$ Now, $μ_{j} - λ_{j} = s_{i_{j + 1}} \circ μ_{j + 1} - s_{i_{j + 1}} \circ λ_{j + 1} = s_{i_{j + 1}} (μ_{j + 1} - λ_{j + 1})$ and $μ_{j + 1} - λ_{j + 1} = s_{γ_{j + 1}} \circ λ_{j + 1} - λ_{j + 1} = - ⟨ λ_{j + 1} + ρ, γ_{j + 1} ⟩ γ_{j + 1} .$ This implies that $s_{i_{j + 1}} γ_{j + 1} \in R^{-}$ and so $γ_{j + 1} = α_{j + 1} .$

We now have $μ_{j + 1} = s_{i_{j + 1}} \circ λ_{j + 1}$ with $μ_{j + 1} \leq λ_{j + 1} .$ Lemma 4.4.2 implies $M_{q} (μ_{j + 1}^{σ}) \subseteq M_{q} (λ_{j + 1}^{σ}) .$ From Lemma 4.4.3, we have $M_{q} (μ_{j + 2}^{σ}) \subseteq M_{q} (μ_{j + 1}^{σ}) .$ Since $M_{q} (μ_{j + 1}^{σ}) \subseteq M_{q} (λ_{j + 1}^{σ}),$ we can use the last lemma to conclude $M_{q} (μ_{j + 2}^{σ}) \subseteq M_{q} (λ_{j + 2}^{σ}) .$ Repeating this process, we get $M_{q} (μ^{σ}) \subseteq M_{q} (λ^{σ}) .$ The picture below describes the various embeddings used in the proof: $\begin{matrix} \end{matrix}$

$□$

Central characters

We now construct the central characters of $U_{q} (𝔤),$ following ([CPr1994] 9.2.B). Denote the center of $U_{q} (𝔤)$ by $Z_{q} (𝔤) = {X \in U_{q} (𝔤) | X Y = Y X for all Y \in U_{q} (𝔤)} .$ For $z \in Z_{q} (𝔤),$ we can write $z = \sum F_{β_{1}}^{(t_{1})} \dots F_{β_{N}}^{(t_{N})} K_{(s_{1}, \dots, s_{N})}^{(t_{1}, \dots, t_{N})} E_{β_{N}}^{(s_{N})} \dots E_{β_{1}}^{(s_{1})}$ where $K_{(s_{1}, \dots, s_{N})}^{(t_{1}, \dots, t_{N})} \in U_{q}^{0} (𝔤)$ and $s_{1}, \dots, s_{N}, t_{1}, \dots, t_{N} \in ℤ_{\geq 0}$ with $s_{1} β_{1} + \dots + s_{N} β_{N} = t_{1} β_{1} + \dots + t_{N} β_{N} .$ Define $h_{q} : Z_{q} (𝔤) \to U_{q}^{0} (𝔤)$ by $h_{q} (z) = K_{(0)}^{(0)} .$

The map $h_{q} : Z_{q} (𝔤) \to U_{q}^{0} (𝔤)$ is an algebra homomorphism.

Proof.

Let $z \in Z_{q} (𝔤)$ with $z = \sum F_{β_{1}}^{(t_{1})} \dots F_{β_{N}}^{(t_{N})} K_{(s_{1}, \dots, s_{N})}^{(t_{1}, \dots, t_{N})} E_{β_{N}}^{(s_{N})} \dots E_{β_{1}}^{(s_{1})},$ and let $v^{+}$ be a generator for the Verma module $M_{q} (Λ)$ $(Λ \in {Hom}_{ℚ (q)} (U_{q}^{0} (𝔤), ℚ (q))).$ Then, $z (v^{+}) = (K_{(0)}^{(0)}) v^{+} = Λ (h_{q} (z)) v^{+} .$ For $z, z' \in Z_{q} (𝔤),$ $Λ (h_{q} (z z')) v^{+} = z z' (v^{+}) = z Λ (h_{q} (z')) v^{+} = Λ (h_{q} (z)) Λ (h_{q} (z')) v^{+} .$ Therefore, $Λ (h_{q} (z z')) = Λ (h_{q} (z)) Λ (h_{q} (z'))$ for all $Λ,$ with implies $h_{q} (z z') = h_{q} (z) h_{q} (z') .$

$□$

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

For $Λ \in {Hom}_{ℚ (q)} (U_{q}^{0} (𝔤), ℚ (q)),$ we define the central character $χ_{Λ} : Z_{q} (𝔤) \to ℂ$ by $χ_{Λ} (z) = Λ (h_{q} (z)) .$ Let $M$ be a highest weight $U_{q} (𝔤) -module$ of highest weight $Λ,$ and let $v^{+}$ be a generator of $M .$ Then, $z v^{+} = χ_{Λ} (z) v^{+} .$ Since any $v \in M$ is of the form $v = x v^{+}$ for some $x \in U_{q} (𝔤),$ $z v = z x v^{+} = x z v^{+} = x z v^{+} = χ_{Λ} (z) x v^{+} = χ_{Λ} (z) v .$

Recall that the dot action of the Weyl group $W$ on $𝔥^{*}$ is $w \circ μ = w (μ + ρ) - ρ .$

([CPr1994], 9.1.8). Let $λ, μ \in P,$ integral weights of $𝔤 .$ Then $χ_{λ} = χ_{μ}$ if and only if $λ = w \circ μ$ for some $w \in W .$

The Shapovalov Determinant

Recall $τ : U_{q} (𝔤) \to U_{q} (𝔤)$ is the $ℚ -anti-automorphism$ given by $τ (E_{i}) = F_{i}; τ (F_{i}) = E_{i}; τ (K_{i}) = K_{i}^{- 1}; τ (q) = q^{- 1} .$ Using this map, we can define a Hermitian bilinear form $⟨, ⟩ : M_{q} (Ω) \times M_{q} (Ω) \to ℚ (q)$ by

•	$⟨ v^{+}, v^{+} ⟩ = 1,$ where $v^{+}$ is the generator of $M_{q} (Ω);$
•	$⟨ x u, w ⟩ = ⟨ u, τ (x) w ⟩$ for $x \in U_{q}$ and $u, w \in M_{q} (Ω) .$

Let $Ω, Λ \in {Hom}_{ℚ (q)} (U_{q}^{0} (𝔤), ℚ (q))$ with $Ω \geq Λ .$ Suppose $\frac{Ω}{Λ} = ν \in Q_{+} .$ We define $A {(Ω)}^{Λ} = {(⟨ F_{β_{1}}^{(t_{1})} \dots F_{β_{N}}^{(t_{N})} v^{+}, F_{β_{1}}^{(s_{1})} \dots F_{β_{N}}^{(s_{N})} v^{+} ⟩)}_{\underset{= s_{1} β_{1} + \dots + s_{N} β_{N}}{t_{1} β_{1} + \dots + t_{N} β_{N} = ν}}$ and $det (M_{q} {(Ω)}^{Λ}) = det A {(Ω)}^{Λ} .$ If $Ω, Λ \in {Hom}_{𝒜} (U_{q}^{0} (𝔤), 𝒜),$ we define $det M_{res} {(Ω)}^{Λ} = det (M_{q} {(Ω)}^{Λ}) .$

Example. Suppose $𝔤 = {s l}_{2} (ℂ) .$ Let $k \in ℤ_{\geq 0}$ and $Ω, Λ \in {Hom}_{ℚ (q)} (U_{q}^{0} ({s l}_{2} (ℂ)), ℚ (q))$ with $\frac{Ω}{Λ} = k α .$ Now consider $det (M_{q} {(Ω)}^{Λ}) .$

Equation 4.3 implies $E F^{j} v^{+} = [j] \frac{q^{- j + 1} Ω K - q^{j - 1} Ω {(K)}^{- 1}}{q - q^{- 1}} F^{j - 1} v^{+} .$ Therefore, $\begin{array}{rcl} det M_{q} {(Ω)}^{Λ} & = & ⟨ \frac{F^{k}}{[k]!} v^{+}, \frac{F^{k}}{[k]!} v^{+} ⟩ \\ = & ⟨ v^{+}, \frac{E^{k}}{[k]!} \frac{F^{k}}{[k]!} v^{+} ⟩ \\ = & \frac{1}{{([k]!)}^{2}} ⟨ v^{+}, \prod_{j = 1}^{k} [j] \frac{q^{j - 1} K - q^{- J + 1} K^{- 1}}{q - q^{- 1}} v^{+} ⟩ \\ = & \frac{1}{([k]!)} \prod_{j = 1}^{k} \frac{q^{- j + 1} Ω (K) - q^{j - 1} Ω (K^{- 1})}{q - q^{- 1}} \end{array}$ Theorem 4.5.2 gives a formula for $det M_{q} {(Ω)}^{Λ}$ for all $𝔤 .$

Recall that a weight $Ω$ is equivalent to a $k -tuple$ $(Ω_{1}, \dots, Ω_{k}) \in {(ℚ (q))}^{k} .$ We can view the determinant of $M_{q} {(Ω)}^{Λ}$ as living inside the Laurent polynomial ring $S = ℚ (q) [Ω_{i}^{\pm 1} | 1 \leq i \leq k] .$ 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 $D (Ω_{1}^{j_{1}} \dots Ω_{k}^{j_{k}}) = (j_{1}, \dots, j_{k}) .$ Also, define the following partial ordering on $ℤ^{k} :$ $(j_{1}, \dots, j_{k}) \leq (l_{1}, \dots, l_{k}) if j_{i} \leq l_{i} for all 1 \leq i \leq k .$

Let $Ω \geq Λ$ with $\frac{Ω}{Λ} = ν \in Q_{+} .$ The term with maximum degree in $det M_{q} {(Ω)}^{Λ}$ is $\prod_{β \in R^{+}} \prod_{r > 0} Ω_{β}^{P (ν - r β)} .$ Similarly, the term with minimal degree is $\prod_{β \in R^{+}} \prod_{r > 0} Ω_{β}^{- P (ν - r β)} .$

Proof.

Consider an entry entry in $A {(Ω)}^{Λ} :$ $\begin{matrix} ⟨ F_{β_{1}}^{(t_{1})} \dots F_{β_{N}}^{(t_{N})} v^{+}, F_{β_{1}}^{(s_{1})} \dots F_{β_{N}}^{(s_{N})} v^{+} ⟩ = ⟨ v^{+}, E_{β_{N}}^{(t_{N})} \dots E_{β_{1}}^{(t_{1})} F_{β_{1}}^{(s_{1})} \dots F_{β_{N}}^{(s_{N})} v^{+} ⟩ & (4.6) \end{matrix}$ where $t_{1} β_{1} + \dots + t_{n} β_{n} = ν = s_{1} β_{1} + \dots + s_{n} β_{n} .$ Then, both $F_{β_{1}}^{(t_{1})} \dots F_{β_{n}}^{(t_{n})}$ and $F_{β_{1}}^{(s_{1})} \dots F_{β_{n}}^{(s_{n})}$ can be written as linear combinations of $F_{i_{1}} \dots F_{i_{m}},$ where $ν = α_{i_{1}} + \dots + α_{i_{m}}$ for some choice of (not necessarily distinct) simple roots. Therefore, Equation 4.3 implies that the maximum possible degree term in (4.6) is $\begin{matrix} Ω_{i_{1}} \dots Ω_{i_{m}} = Ω_{β_{1}}^{t_{1}} \dots Ω_{β_{n}}^{t_{n}} = Ω_{β_{1}}^{s_{1}} \dots Ω_{β_{n}}^{s_{n}} & (4.7) \end{matrix}$ This means the term with maximum degree in $A {(Ω)}^{Λ}$ is $\begin{matrix} \prod_{(t) | t_{1} β_{1} + \dots + t_{n} β_{n} = ν} Ω_{β_{1}}^{t_{1}} \dots Ω_{β_{n}}^{t_{1}} . & (4.8) \end{matrix}$ The power of $Ω_{β_{j}}$ in (4.8) is $\sum_{(t) | t_{1} β_{1} + \dots + t_{n} β_{n} = ω - λ} t_{j} .$ We rewrite this as $\sum_{(t) | t_{1} β_{1} + \dots + t_{n} β_{n} = ν} t_{j} = \sum_{r > 0} ∣ {(t) | t_{j} \geq r} ∣ = \sum_{r > 0} P (ν - r β_{j}) .$ 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 $ν - r β_{j}$ maps uniquely to a partition with $t_{j} \geq r .$

We determine the minimum degree with an analogous argument.

$□$

([DKa1992]). Let $Ω \geq Λ \in {Hom}_{ℚ (q)} (U_{q}^{0} (𝔤), ℚ (q))$ with $\frac{Ω}{Λ} = ν \in Q^{+} .$ Then, $det (M_{q} {(Ω)}^{Λ}) = \prod_{β \in R^{+}} \prod_{r > 0} {(\frac{1}{{[r]}_{d_{β}}} [\frac{q^{d_{β} (⟨ ρ, β^{\lor} ⟩ - r)} Ω (K_{β}) - q^{- d_{β} (⟨ ρ, β^{\lor} ⟩ - r)} Ω (K_{β}^{- 1})}{q^{d_{β}} - q^{- d_{β}}}])}^{P (ν - r β)} .$

If we restrict the choice of weights to $ω, λ \in 𝔥^{*} \subseteq {Hom}_{ℚ (q)} (U_{q}^{0} (𝔤), ℚ (q)),$ this formula becomes $\begin{matrix} det (M_{q} {(ω)}^{λ}) = \prod_{β \in R^{+}} \prod_{r > 0} {(\frac{{[⟨ ω + ρ, β^{\lor} ⟩ - r]}_{d_{β}}}{{[r]}_{d_{β}}})}^{P (ω - λ - r β)} . & (4.9) \end{matrix}$

A Partial Proof.

Note that $\frac{q^{d_{β} (⟨ ρ, β^{\lor} ⟩ - r)} Ω (K_{β}) - q^{- d_{β} (⟨ ρ, β^{\lor} ⟩ - r)} Ω (K_{β}^{- 1})}{q^{d_{β}} - q^{- d_{β}}}$ can be written as $q^{d_{β} (⟨ ρ, β^{\lor} ⟩ - r)} Ω {(K_{β})}^{- 1} \frac{(Ω (K_{β}) - q^{- d_{β} (⟨ ρ, β^{\lor} ⟩ - r)}) (Ω (K_{β}) + q^{- d_{β} (⟨ ρ, β^{\lor} ⟩ - r)})}{q^{d_{β}} - q^{- d_{β}}} .$ We now show that ${(Ω (K_{β}) \pm q^{- d_{β} (⟨ ρ, β^{\lor} ⟩ - r)})}^{p (ν - r β)} | det M_{q} {(Ω)}^{Λ} .$ Suppose that $ω_{0} \in P$ is such that $⟨ ω_{0} + ρ, β^{\lor} ⟩ = r,$ and let $σ \in {(\pm 1, \dots, \pm 1)} .$ Then, $s_{β} \circ ω_{0} = ω_{0} - r β$ and so from Proposition 4.4.1, $M_{q} ({(ω_{0} - r β)}^{σ}) \subseteq M_{q} (ω_{0}^{σ}) .$ Therefore, $M_{q} {(ω_{0}^{σ})}^{{(ω_{0} - r β)}^{σ}} \cap Rad ⟨, ⟩ \neq 0 .$ This implies that $det (M_{q} {(ω_{0}^{σ})}^{{(ω_{0} - r β)}^{σ}}) = 0,$ i.e. $det (M_{q} {(Ω)}^{\tilde{Ω}}),$ where $\tilde{Ω} = Ω - r β,$ has a zero at $Ω_{β} = {(ω_{0}^{σ})}_{β} = \pm q^{⟨ ω_{0}, β^{\lor} ⟩} = \pm q^{r - ⟨ ρ, β^{\lor} ⟩} .$ As in the proof of Theorem 3.4.3, this means there is $v_{r, β} \in M_{q} {(Ω)}^{\tilde{Ω}}$ such that

•	$v_{r, β} = \sum_{(t) \| t_{1} β_{1} + \dots + t_{n} β_{n} = r β} a_{(t)} F_{β_{1}}^{(t_{1})} \dots F_{β_{n}}^{(t_{n})} v^{+}$ for some $a_{(t)} \in ℚ (q)$ with $a_{(t)} \neq 0$ for at least one $(t);$
•	$(Ω_{β} \pm q^{r - ⟨ ρ, β^{\lor} ⟩}) \| ⟨ v_{r, β}, w ⟩$ for all $w \in M_{q} (Ω) .$

Now, the set

\tilde{B} = {F_{β_{1}}^{(s_{1})} \dots F_{β_{N}}^{(s_{N})} \sum_{(t) | t_{1} β_{1} + \dots + t_{n} β_{n} = r β} a_{(t)} F_{β_{1}}^{(t_{1})} \dots F_{β_{n}}^{(t_{n})} | s_{1} β_{1} + \dots + s_{N} β_{N} = ν - r β}

is linearly independent in

U_{q}^{-} (𝔤)

and can be extended to a linearly independent set

B

which spans

span {F_{β_{1}}^{(s_{1})} \dots F_{β_{N}}^{(s_{N})} | s_{1} β_{1} + \dots + s_{N} β_{N} = ν} .

Let

P

be the matrix taking

{F_{β_{1}}^{(s_{1})} \dots F_{β_{N}}^{(s_{N})} | s_{1} β_{1} + \dots + s_{N} β_{N} = ν}

B .

Define $det (M_{q} {(Ω)}^{Λ}) = det {(⟨ X_{i} v^{+}, X_{j} v^{+} ⟩)}_{X_{i}, X_{j} \in B} .$ We have

•	$F_{β_{1}}^{(s_{1})} \dots F_{β_{N}}^{(s_{N})} \sum_{(t) \| t_{1} β_{1} + \dots + t_{n} β_{n} = r β} a_{(t)} F_{β_{1}}^{(t_{1})} \dots F_{β_{n}}^{(t_{n})} v^{+} = F_{β_{1}}^{(s_{1})} \dots F_{β_{N}}^{(s_{N})} v_{r, β};$
•	$(Ω_{β} \pm q^{r - ⟨ ρ, β^{\lor} ⟩})$ divides $⟨ F_{β_{1}}^{(s_{1})} \dots F_{β_{N}}^{(s_{N})} v_{r, β}, w ⟩ .$

Therefore,

{(Ω (K_{pt}) \pm q^{- d_{β} (⟨ ρ, β^{\lor} ⟩ - r)})}^{p (ν - r β)}

divides

{det}_{B} M_{q} {(Ω)}^{Λ} .

Finally,

det (M_{q} {(Ω)}^{Λ}) = det (P^{t}) {det}_{B} (M_{q} {(Ω)}^{Λ}) det (P),

which implies that

{(Ω (K_{β}) \pm q^{- d_{β} (⟨ ρ, β^{\lor} ⟩ - r)})}^{p (ν - r β)}

divides

{det}_{B} M_{q} {(Ω)}^{Λ} .

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

$□$

An Alternative Determinant Formula

Recall that DeConcini and Kac [DKa1992] use the $ℚ -antiautomorphism$ $τ : U_{q} \to U_{q}$ defined by $τ (E_{i}) = F_{i}, τ (F_{i}) = E_{i}, τ (K_{i}) = K_{i}^{- 1}, τ (q) = q^{- 1} .$ Instead, we will use the $ℚ (q) -antiautomorphism$ $ϕ$ given by $ϕ (E_{i}) = K_{i}^{- 1} F_{i}, ϕ (F_{i}) = E_{i} K_{i}, ϕ (K_{i}) = K_{i} .$ This defines a new contravariant form. In this section, we will write ${⟨, ⟩}_{τ}$ for the contravariant form induced by $τ$ and ${⟨, ⟩}_{ϕ}$ for the contravariant form induced by $ϕ .$ Define ${det}_{ϕ} (M_{q} {(ω)}^{λ}) = det {({⟨ F_{β_{1}}^{(t_{1})} \dots F_{β_{N}}^{(t_{N})} v^{+}, F_{β_{1}}^{(s_{1})} \dots F_{β_{N}}^{(s_{N})} v^{+} ⟩}_{ϕ})}_{\underset{= s_{1} β_{1} + \dots + s_{N} β_{N}}{t_{1} β_{1} + \dots + t_{N} β_{N} = ν}} .$

Let $ω, λ \in P$ with $λ \leq ω .$ Then, ${det}_{ϕ} (M_{q} {(ω)}^{λ}) = q^{P (ω - λ) k_{λ}^{ω}} C_{ω - λ} \prod_{β \in R^{+}} \prod_{r > 0} {(\frac{{[⟨ ω + ρ, β^{\lor} ⟩ - r]}_{d_{β}}}{{[r]}_{d_{β}}})}^{P (ω - λ - r β)},$ where $0 \neq C_{ω - λ} \in ℚ (q)$ and $k_{λ}^{ω} \in ℤ$ is given as follows. Write $ω - λ = α_{i_{1}} + \dots + α_{i_{m}} .$ Then $k_{λ}^{ω} = ⟨ ω, α_{i_{1}}^{\lor} + \dots + α_{i_{m}}^{\lor} ⟩ - (⟨ α_{i_{1}} + \dots + α_{i_{m}}, α_{i_{1}}^{\lor} ⟩ + ⟨ α_{i_{2}} + \dots + α_{i_{m}}, α_{i_{2}}^{\lor} ⟩ + \dots + ⟨ α_{i_{m}}, α_{i_{m}}^{\lor} ⟩) .$

Proof.

Let $P$ be the matrix which takes ${⟨ F_{β_{1}}^{(t_{1})} \dots F_{β_{N}}^{(t_{N})} v^{+}, F_{β_{1}}^{(s_{1})} \dots F_{β_{N}}^{(s_{N})} v^{+} ⟩}_{τ}$ to ${⟨ F_{β_{1}}^{(t_{1})} \dots F_{β_{N}}^{(t_{N})} v^{+}, F_{β_{1}}^{(s_{1})} \dots F_{β_{N}}^{(s_{N})} v^{+} ⟩}_{ϕ} .$ Then, ${det}_{ϕ} (M_{q} {(ω)}^{λ}) = det (P) {det}_{τ} (M_{q} {(ω)}^{λ}) .$ Therefore, we need only find $det (P) .$ To do this, note that since the set ${F_{i_{1}} \dots F_{i_{m}} v^{+} | α_{i_{1}} + \dots + α_{i_{m}} = ω - λ}$ spans $M_{q} {(ω)}^{λ},$ there is a subset $B$ which is a basis for the space. Therefore, we can write a standard basis element $F_{β_{1}}^{(t_{1})} \dots F_{β_{N}}^{(t_{N})} v^{+}$ as a linear combination of elements of $B :$ $F_{β_{1}}^{(t_{1})} \dots F_{β_{N}}^{(t_{N})} v^{+} = \sum_{F_{i_{1}} \dots F_{i_{m}} v^{+} \in B} a_{i_{1}, \dots, i_{m}} F_{i_{1}} \dots F_{i_{m}} v^{+}$ where $a_{i_{1}, \dots, i_{m}} \in ℚ (q) .$ Then, $\begin{array}{rcl} {⟨ F_{β_{1}}^{(t_{1})} \dots F_{β_{N}}^{(t_{N})} v^{+}, F_{β_{1}}^{(s_{1})} \dots F_{β_{N}}^{(s_{N})} v^{+} ⟩}_{ϕ} & = & \sum a_{i_{1}, \dots, i_{m}} ⟨ v^{+}, E_{i_{m}} K_{i_{m}} \dots E_{i_{1}} K_{i_{1}} F_{β_{1}}^{s_{1}} \dots F_{β_{n}}^{s_{n}} v^{+} ⟩ \\ = & \sum a_{i_{1}, \dots, i_{m}} q^{k_{λ}^{ω}} ⟨ v^{+}, E_{i_{m}} \dots E_{i_{1}} F_{β_{1}}^{s_{1}} \dots F_{β_{n}}^{s_{n}} v^{+} ⟩ \end{array}$ where the sum is over $α_{i_{1}}, \dots, α_{i_{m}}$ such that $F_{i_{1}} \dots F_{i_{m}} \in B$ and $α_{i_{1}} + \dots + α_{i_{m}} = ω - λ;$ and where $\begin{array}{rcl} k_{λ}^{ω} & = & ⟨ ω - λ, α_{i_{1}} ⟩ + ⟨ ω - λ + α_{i_{1}}, α_{i_{2}} ⟩ + \dots + ⟨ ω - λ + (α_{i_{1}} + \dots + α_{i_{m}}), α_{i_{m}} ⟩ \\ = & ⟨ ω, α_{i_{1}} + \dots + α_{i_{m}} ⟩ - \sum_{j \geq k} ⟨ α_{i_{j}}, α_{i_{k}} ⟩ . \end{array}$ (Since $⟨ α_{i_{j}}, α_{i_{k}} ⟩ = ⟨ α_{i_{k}}, α_{i_{j}} ⟩,$ $k_{λ}^{ω}$ is independent of the order of the simple roots $α_{i_{1}}, \dots, α_{i_{m}} .)$

On the other hand, ${⟨ F_{β_{1}}^{(t_{1})} \dots F_{β_{N}}^{(t_{N})} v^{+}, F_{β_{1}}^{(s_{1})} \dots F_{β_{N}}^{(s_{N})} v^{+} ⟩}_{τ} = \sum_{α_{i_{1}} + \dots + α_{i_{m}} = ω - λ} \overline{a_{i_{1}, \dots, i_{m}}} ⟨ v^{+}, E_{i_{m}} \dots E_{i_{1}} F_{β_{1}}^{s_{1}} \dots F_{β_{n}}^{s_{n}} v^{+} ⟩,$ where $\overline{\cdot} : ℚ (q) \to ℚ (q)$ is the $ℚ -automorphism$ taking $q$ to $q^{- 1} .$

Therefore, $P = q^{k_{λ}^{ω}} A \overline{A^{- 1}},$ where $A$ is the matrix taking the basis $B$ to the standard basis ${F_{β_{1}}^{(t_{1})} \dots F_{β_{N}}^{(t_{N})} v^{+} | t_{1} β_{1} + \dots + t_{N} β_{N} = ω - λ}$ of $M_{q} {(ω)}^{λ} .$ Then, $det (P) = q^{P (ω - λ) k_{λ}^{ω}} det (A) det ({\overline{A}}^{- 1}) = q^{P (ω - λ) k_{λ}^{ω}} \frac{det (A)}{\overline{det (A)}} .$ Since $A$ is a matrix taking one basis to another, $det (A) \neq 0 .$ Therefore, ${det}_{ϕ} (M_{q} {(ω)}^{λ}) = q^{P (ω - λ) k_{λ}^{ω}} C_{ω - λ} {det}_{τ} (M_{q} {(ω)}^{λ}),$ where $C_{ω - λ} = det (A) / \overline{det (A)} .$

$□$

Translation Functors

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

Let $ω, ω_{0} \in P .$ Then $M_{q} (ω) \otimes L_{q} (ω_{0}) = ⨁_{λ \in P_{+}} {(M_{q} (ω) \otimes L_{q} (ω_{0}))}^{[λ]},$ where ${(M_{q} (ω) \otimes L_{q} (ω_{0}))}^{[λ]}$ has composition factors $L_{q} (μ)$ only for $μ \in [λ] .$

Proof.

Note that the module $M_{q} (ω) \otimes L_{q} (ω_{0})$ decomposes as a direct sum according to central characters. Theorem 4.4.6 implies that this decomposition corresponds to the decomposition by blocks stated in the proposition.

$□$

We again consider the translation $M_{q} (ω) ⟼ ⨁ {(M_{q} (ω) \otimes L_{q} (ω_{0}))}^{[λ]} .$ When possible, we will state our results in the more general setting of the restricted integral form.

For $μ \in Q_{+},$ define $n (μ) = dim (L_{res} {(ω_{0})}^{ω_{0} - μ}),$ and let ${w_{μ, i} | 1 \leq i \leq n (μ)}$ be a basis for $L_{res} {(ω_{0})}^{(ω_{0} - μ)} .$

For $λ \in Q_{+},$ the sets $ℬ_{1} = {F_{β_{1}}^{(t_{1})} \dots F_{β_{n}}^{(t_{n})} v^{+} \otimes w_{μ, i} | t_{1} β_{1} + \dots + t_{n} β_{n} = λ - μ, 1 \leq i \leq n (μ)}$ and $ℬ_{2} = {F_{β_{1}}^{(t_{1})} \dots F_{β_{n}}^{(t_{n})} (v^{+} \otimes w_{μ, i}) | t_{1} β_{1} + \dots + t_{n} β_{n} = λ - μ, 1 \leq i \leq n (μ)}$ are bases for ${(M_{res} (ω) \otimes L_{res} (ω_{0}))}^{(ω + ω_{0} - λ)} .$ Moreover, the transition matrix between the two bases has determinant one.

Proof.

The set $ℬ_{1}$ is clearly a basis for ${(M_{res} (ω) \otimes L_{res} (ω_{0}))}^{(ω + ω_{0} - λ)} .$ Since $ℬ_{2} \subseteq {(M_{res} (ω) \otimes L_{res} (ω_{0}))}^{(ω + ω_{0} - λ)},$ there is a matrix $A$ taking $ℬ_{1}$ to $ℬ_{2} .$

Let $F_{β_{1}}^{(t_{1})} \dots F_{β_{n}}^{(t_{n})} (v \otimes w_{μ, i}) \in ℬ_{2} .$

Since ${U_{𝒜}^{res} (𝔤)}^{-}$ is a Hopf algebra, $Δ (F_{β_{1}}^{(t_{1})} \dots F_{β_{n}}^{(t_{n})}) = F_{β_{1}}^{(t_{1})} \dots F_{β_{n}}^{(t_{n})} \otimes 1 + \sum a_{s_{1}, \dots, s_{N}}^{r_{1}, \dots, r_{N}} (F_{β_{1}}^{(s_{1})} \dots F_{β_{n}}^{(t_{N})} K_{s_{1}, \dots, s_{N}}^{r_{1}, \dots, r_{N}}) \otimes (F_{β_{1}}^{(r_{1})} \dots F_{β_{n}}^{(r_{n})}),$ where $a_{s_{1}, \dots, s_{N}}^{r_{1}, \dots, r_{N}} \in 𝒜,$ $K_{s_{1}, \dots, s_{N}}^{r_{1}, \dots, r_{N}} \in {U_{𝒜}^{res} (𝔤)}^{0},$ and the sum is over $s_{1}, \dots, s_{N}, r_{1}, \dots, r_{N} \in ℤ_{\geq 0}$ such that $r_{1} β_{1} + \dots + r_{N} β_{N} < t_{1} β_{1} + \dots + t_{N} β_{N}$ and $(r_{1} + s_{1}) β_{1} + \dots + (r_{n} + s_{N}) β_{N} = t_{1} β_{1} + \dots + t_{N} β_{N} .$ Therefore, $F_{β_{1}}^{(t_{1})} \dots F_{β_{n}}^{(t_{n})} (v^{+} \otimes w_{μ, i}) = F_{β_{1}}^{(t_{1})} \dots F_{β_{n}}^{(t_{n})} v \otimes w_{μ, i} + \sum_{\underset{1 \leq j \leq n (ν)}{ν > μ, (s) = ν - μ,}} a_{ν, j}^{(s)} F_{β_{1}}^{(s_{1})} \dots F_{β_{n}}^{(s_{n})} v^{+} \otimes w_{ν, j}$ for some $a_{ν, j}^{(s)} \in 𝒜 .$

Therefore, we can order vectors in such a way that the matrix $A'$ taking $F_{β_{1}}^{(t_{1})} \dots F_{β_{n}}^{(t_{n})} v^{+} \otimes w_{μ, i}$ to $F_{β_{1}}^{(t_{1})} \dots F_{β_{n}}^{(t_{n})} (v^{+} \otimes w_{μ, i})$ is upper-triangular with ones on the diagonal. Then, $A'$ is invertible with determinant one.

By possibly reordering the bases, we may assume $A = A' .$ Therefore, $A'$ has entries in $𝒜 .$ Since $A'$ is upper-triangular with ones on the diagonal, ${(A')}^{- 1}$ will also have entries in $𝒜 .$ This proves that $ℬ_{2}$ is an $𝒜 -basis$ for ${(M_{res} (ω) \otimes L_{res} (ω_{0}))}^{(ω + ω_{0} - λ)} .$

$□$

Let $ω, ω_{0} \in P$ and $λ, μ \in Q_{+} .$ Define $det L_{res} {(ω_{0})}^{ω_{0} - μ} = det {(⟨ w_{μ, i}, w_{μ, j} ⟩)}_{1 \leq i, j \leq dim L_{res} {(ω_{0})}^{ω_{0} - μ}}$ and $det ({(M_{res} (ω) \otimes L_{res} (ω_{0}))}^{ω + ω_{0} - λ}) = det (⟨ F_{β_{1}}^{(t_{1})} \dots F_{β_{N}}^{(t_{N})} v^{+} \otimes w_{μ, i}, F_{β_{1}}^{(s_{1})} \dots F_{β_{N}}^{(s_{N})} v^{+} \otimes w_{μ', j} ⟩),$ where $μ, μ' \leq λ,$ $1 \leq i \leq n (μ),$ $1 \leq j \leq n (μ'),$ and $s_{1}, \dots, s_{N}, t_{1}, \dots, t_{N} \in ℤ_{\geq 0}$ are such that $s_{1} β_{1} + \dots + s_{N} β_{N} = λ - μ$ and $t_{1} β_{1} + \dots + t_{N} β_{N} = λ - μ' .$

Let $ω, ω_{0} \in P .$ For $λ \in Q_{+},$ $det ({(M_{res} (ω) \otimes L_{res} (ω_{0}))}^{ω + ω_{0} - λ})$ is given by $\begin{matrix} \prod_{\underset{μ \leq λ}{μ \in Q_{+}}} {(det (M_{res} {(ω)}^{ω - (λ - μ)}))}^{n (μ)} {(det (L_{res} {(ω_{0})}^{ω_{0} - μ}))}^{P (λ - μ)} . & (4.10) \end{matrix}$

Proof.

Note that $M_{res} {(ω)}^{ν_{1}} \otimes L_{res} {(ω_{0})}^{μ_{1}}$ is orthogonal (with respect to the contravariant form) to $M_{res} {(ω)}^{ν_{1}} \otimes L_{res} {(ω_{0})}^{μ_{1}}$ for $ν_{1} \neq ν_{2} .$ As in the proof of Lemma 2.6.3, $\begin{array}{rcl} det ({(M_{res} (ω) \otimes L_{res} (ω_{0}))}^{ω + ω_{0} - λ}) & = & \prod_{\underset{μ \leq λ}{μ \in Q_{+}}} det (M_{res} {(ω)}^{ω - (λ - μ)} \otimes L_{res} {(ω_{0})}^{ω_{0} - μ}) \\ = & \prod_{\underset{μ \leq λ}{μ \in Q_{+}}} (det {(M_{res} {(ω)}^{ω - (λ - μ)})}^{n (μ)}) {(det (L_{res} {(ω_{0})}^{ω_{0} - μ}))}^{P (λ - μ)} . \end{array}$

$□$

Using the basis $ℬ_{2}$ for ${(M_{res} (ω) \otimes L_{res} (ω_{0}))}^{ω + ω_{0} - λ},$ we see that $M_{res} (ω) \otimes L_{res} (ω_{0})$ has a filtration by Verma modules $M_{res} (ω + ω_{0} - μ)$ where the Verma modules $M_{res} (ω + ω_{0} - μ)$ is generated by the image of $v^{+} \otimes w_{μ, i} .$ The following form of the determinant formula better reflects the filtration of $M_{q} (ω) \otimes L_{q} (ω_{0})$ by Verma modules.

Let $ω, ω_{0} \in P$ and $λ \in Q_{+} .$ Then, $\begin{array}{rcl} det ({(M_{res} (ω) \otimes L_{res} (ω_{0}))}^{ω + ω_{0} - λ}) & = & \prod_{\underset{μ \leq λ}{μ \in Q_{+}}} det {(M_{res} {(ω + ω_{0} - μ)}^{ω + ω_{0} - λ})}^{n (μ)} \\ a_{μ}^{ω_{0}} (ω) det {(L_{res} {(ω_{0})}^{ω_{0} - μ})}^{P (λ - μ)} \end{array}$ where $a_{μ}^{ω_{0}} (ω) = q^{R_{μ}} \prod_{β \in R^{+}, r > 0} {(\frac{[⟨ ω + ρ, β^{\lor} ⟩ - r]}{[⟨ ω + ω_{0} - μ + r β + ρ, β^{\lor} ⟩ - r]})}^{n (μ - r β)}$ and $R_{μ}$ is defined inductively by $R_{μ} = - \sum_{ν < μ, n (ν) \neq 0} P (μ - ν) (⟨ ω_{0} - ν, α_{i_{1}}^{\lor} + \dots + α_{i_{m}}^{\lor} ⟩ n (ν) + R_{ν}),$ for $μ - ν = α_{i_{1}} + \dots + α_{i_{m}} .$

Proof.

Note that $R_{μ}$ is always an integer and $\begin{array}{rcl} \sum_{μ \leq λ} R_{μ} P (λ - μ) & = & - \sum_{μ \leq λ} k_{ω - λ + μ}^{ω} - k_{ω + ω_{0} - λ}^{ω + ω_{0} - μ} P (λ - μ) n (μ) \\ = & - \sum_{μ \leq λ} ⟨ ω_{0} - μ, α_{i_{1}}^{\lor} + \dots + α_{i_{m}}^{\lor} ⟩ P (λ - μ) n (μ) . \end{array}$

From Lemma 4.6.3, $det ({(M_{res} (ω) \otimes L_{res} (ω_{0}))}^{ω + ω_{0} - λ})$ is given by $\begin{array}{cl} = & \prod_{\underset{μ \leq λ}{μ \in Q_{+}}} det {(L_{res} {(ω_{0})}^{ω_{0} - μ})}^{P (λ - μ)} q^{n (μ) P (λ - μ) k_{ω - (λ - μ)}^{ω}} \\ \prod_{\underset{μ \leq λ}{μ \in Q_{+}}} C_{λ - μ} \prod_{β \in R^{+}, r > 0} {({(\frac{[⟨ ω + ρ, β^{\lor} ⟩ - r]}{[r]})}^{P (λ - μ - r β)})}^{n (μ)} \\ = & \prod_{\underset{μ \leq λ}{μ \in Q_{+}}} det {(L_{res} {(ω_{0})}^{ω_{0} - μ})}^{P (λ - μ)} \\ \prod_{\underset{μ \leq λ}{μ \in Q_{+}}} {(q^{P (λ - μ) (k_{ω + ω_{0} - λ}^{ω + ω_{0} - μ} + (k_{ω - λ + μ}^{ω} - k_{ω + ω_{0} - λ}^{ω + ω_{0} - μ}))})}^{n (μ)} \\ = & \underset{\underset{*}{⏟}}{\prod_{\underset{μ \leq λ}{μ \in Q_{+}}} \prod_{β \in R^{+}, r > 0} {(\frac{[⟨ ω + ρ, β^{\lor} ⟩ - r]}{[⟨ ω + ω_{0} - μ + ρ, β^{\lor} ⟩ - r]})}^{P (λ - μ - r β) n (μ)})} \\ \prod_{\underset{μ \leq λ}{μ \in Q_{+}}} C_{λ - μ} \prod_{β \in R^{+}, r > 0} {(\frac{[⟨ ω + ω_{0} - μ + ρ, β^{\lor} ⟩ - r]}{[r]})}^{P (λ - μ - r β) n (μ)}) \\ = & \prod_{\underset{μ \leq λ}{μ \in Q_{+}}} det {(L_{res} {(ω_{0})}^{ω_{0} - μ})}^{P (λ - μ)} \\ \prod_{\underset{μ \leq λ}{μ \in Q_{+}}} q^{(k_{ω - λ + μ}^{ω} - k_{ω + ω_{0} - λ}^{ω + ω_{0} - μ}) P (λ - μ) n (μ)} \\ \underset{\underset{\underset{n (μ) = 0 for μ < 0, this does not affect the index of the product.}{We send μ \to μ - r β in * . Since P (λ - μ) = 0 for μ > λ and}}{⏟}}{\prod_{\underset{μ \leq λ}{μ \in Q_{+}}} \prod_{β \in R^{+}, r > 0} {(\frac{[⟨ ω + ρ, β^{\lor} ⟩ - r]}{[⟨ ω + ω_{0} - μ + r β + ρ, β^{\lor} ⟩ - r]})}^{P (λ - μ) n (μ - r β)}} \\ \prod_{\underset{μ \leq λ}{μ \in Q_{+}}} det {(M_{res} {(ω + ω_{0} - μ)}^{ω + ω_{0} - λ})}^{n (μ)} \\ = & \prod_{\underset{μ \leq λ}{μ \in Q_{+}}} {(a_{μ} det (L_{res} {(ω_{0})}^{ω_{0} - μ}))}^{P (λ - μ)} det {(M_{res} {(ω + ω_{0} - μ)}^{ω + ω_{0} - λ})}^{n (μ)} \end{array}$

$□$

${(M_{q} (ω) \otimes L_{q} (ω_{0}))}^{[λ]} ⊥ {(M_{q} (ω) \otimes L_{q} (ω_{0}))}^{[μ]}$ for $[μ] \neq [λ] .$

Proof.

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

$□$

Let $ω, ω_{0} \in P$ and let $λ \in Q_{+} .$ Suppose $ω$ is such that there is no $μ \in Q_{+}$ with $0 < μ \leq λ$ and $ω - μ \in [ω] .$ Then for each block $γ,$ we can define projection maps ${Pr}_{λ}^{[γ]} : M_{q} (ω) \otimes L_{q} (ω_{0}) ⟶ {({(M_{q} (ω) \otimes L_{q} (ω_{0}))}^{[γ]})}^{ω + ω_{0} - λ}$ given by ${Pr}_{λ}^{[γ]} (v) = v - \sum_{[ν] \neq [γ]} \sum_{1}^{m} ⟨ v, v_{i}^{[ν]} ⟩ {\overline{v}}_{i}^{[ν]}$ where ${v_{1}^{[ν]}, \dots, v_{m}^{[ν]}}$ and ${{\overline{v}}_{1}^{[ν]}, \dots, {\overline{v}}_{m}^{[ν]}}$ are dual bases for ${({(M_{q} (ω) \otimes L_{q} (ω_{0}))}^{[ν]})}^{ω + ω_{0} - λ} .$

Proof.

Since there is no $μ \in Q_{+}$ with $0 < μ \leq λ$ and $ω - μ \in [ω],$ $det M_{q} {(ω)}^{ω - λ} \neq 0 .$ Lemma 4.6.3 then implies the contravariant form is nondegenerate on ${(M_{q} (ω) \otimes L_{q} (ω_{0}))}^{ω + ω_{0} - λ} .$ We know from Lemma 4.6.5 that distinct blocks are orthogonal. This means that the contravariant form is nondegenerate on ${({(M_{q} (ω) \otimes L_{q} (ω_{0}))}^{[γ]})}^{ω + ω_{0} - λ} .$

Let ${v_{1}^{[ν]}, \dots, v_{m}^{[ν]}}$ be a basis for ${({(M_{q} (ω) \otimes L_{q} (ω_{0}))}^{[ν]})}^{ω + ω_{0} - λ} .$ Since the contravariant form is nondegenerate on this space, there is a basis ${{\overline{v}}_{1}^{[ν]}, \dots, {\overline{v}}_{m}^{[ν]}}$ of ${({(M_{q} (ω) \otimes L_{q} (ω_{0}))}^{[ν]})}^{ω + ω_{0} - λ}$ such that $⟨ v_{i}^{[ν]}, {\overline{v}}_{l}^{[ν]} ⟩ = δ_{i, l} .$

We define a map ${Pr}_{λ}^{[γ]} : {(M_{q} (ω) \otimes L_{q} (ω))}^{ω + ω_{0} - λ} ⟶ {({(M_{q} (ω) \otimes L_{q} (ω))}^{[γ]})}^{ω + ω_{0} - λ}$ by ${Pr}_{λ}^{[γ]} (v) = v - \sum_{[ν] \neq [γ]} \sum_{1}^{m} ⟨ v, v_{i}^{[ν]} ⟩ {\overline{v}}_{i}^{[ν]} .$ Note that $⟨ {Pr}_{λ}^{[γ]} (v), v_{i}^{[ν]} ⟩ = 0$ whenever $[ν] \neq [γ] .$ Therefore, ${Pr}_{λ}^{[γ]} (v) \in {({(M_{q} (ω) \otimes L_{q} (ω))}^{[γ]})}^{ω + ω_{0} - λ} .$

Also, for $v \in {({(M_{q} (ω) \otimes L_{q} (ω))}^{[γ]})}^{ω + ω_{0} - λ},$ ${Pr}_{λ}^{[γ]} (v) = v .$

$□$

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

Fix $ω, ω_{0} \in P$ and $λ \in Q_{+} .$ Suppose $ω$ is such that there is no $w \in W$ with $ω \geq w \circ ω \geq ω - λ$ or $ω + ω_{0} \geq w \circ (ω + ω_{0}) \geq ω + ω_{0} - λ .$

Then the submodule of $M_{q} (ω) \otimes L_{q} (ω_{0})$ generated by $⨁_{μ \in Q_{+}, μ \leq λ} {(M_{q} (ω) \otimes L_{q} (ω_{0}))}^{ω + ω_{0} - μ}$ is isomorphic to $⨁_{μ \in Q_{+}, μ \leq λ} M_{q} {(ω + ω_{0} - μ)}^{\oplus dim L_{q} {(ω_{0})}^{ω_{0} - μ}} .$ For a suitable choice of generating highest weight vectors ${v_{μ, i}^{+} | 1 \leq i \leq dim {(L_{q} (ω_{0}))}^{ω_{0} - μ}},$ this sum is orthogonal with respect to the contravariant form on $M_{q} (ω) \otimes L_{q} (ω_{0}),$ and $\prod_{1 \leq i \leq dim {(L (ω_{0}))}^{ω_{0} - μ}} ⟨ v_{μ, i}, v_{μ, i} ⟩ = a_{μ}^{ω_{0}} (ω) det L {(ω_{0})}^{ω_{0} - μ}$

Proof.

Since there is no $w \in W$ with $ω \geq w \circ ω \geq ω - λ,$ the projection maps from the previous lemma are well-defined.

We have assumed there is no $w \in W$ such that $ω + ω_{0} \geq w \circ (ω + ω_{0}) \geq ω + ω_{0} - λ .$ Therefore, there is only one weight in $[ω + ω_{0} - μ]$ between $ω + ω_{0}$ and $ω + ω_{0} - λ,$ namely $ω + ω_{0} - μ .$ This implies ${Pr}_{μ}^{[μ]} (v^{+} \otimes w_{μ, i})$ is a highest weight vector. Choose vectors ${v_{μ, i}^{+}}$ such that

•	the transition matrix from ${{Pr}_{μ}^{[μ]} (v^{+} \otimes w_{μ, i})}$ to ${v_{μ, i}^{+}}$ has determinant 1;
•	$⟨ v_{μ, i}^{+}, v_{μ, k}^{+} ⟩ = 0$ if $i \neq k .$

Therefore, we only need to determine

(\prod_{i} ⟨ v_{μ, i}^{+}, v_{μ, i}^{+} ⟩)

We do this inductively. Assume

⟨ v_{ν, i}^{+}, v_{ν, l}^{+} ⟩ = det (⟨ {Pr}_{ν}^{[ν]} (v^{+} \otimes w_{ν, i}) m {Pr}_{ν}^{[ν]} (v^{+} \otimes w_{ν, l}) ⟩) = a_{ν}^{ω_{0}} (ω) det L_{q} {(ω_{0})}^{ω_{0} - ν}

for

ν \in P_{+}

with

ν < μ .

Note that $det {(⟨ F_{β_{1}}^{(t_{1})} \dots F_{β_{n}}^{(t_{n})} {Pr}_{ν}^{[ν]} (v^{+} \otimes w_{ν, i}), F_{β_{1}}^{(s_{1})} \dots F_{β_{n}}^{(s_{n})} {Pr}_{ν}^{[ν]} (v^{+} \otimes w_{ν, k}) ⟩)}_{\underset{i, k}{(t), s}}$ is given by $\begin{array}{cl} = & \prod_{i} {(⟨ F_{β_{1}}^{(t_{1})} \dots F_{β_{n}}^{(t_{n})} v_{ν, i}^{+}, F_{β_{1}}^{(s_{1})} \dots F_{β_{n}}^{(s_{n})} v_{ν, i}^{+} ⟩)}_{(t), (s)} \\ = & {(\prod_{i} ⟨ v_{ν, i}^{+}, v_{ν, i}^{+} ⟩)}^{p (μ - ν)} {(det M_{q} {(ω + ω_{0} - ν)}^{ω + ω_{0} - μ})}^{n (ν)} . \end{array}$ Since distinct blocks are orthogonal, we have $det {(M_{q} (ω) \otimes L_{q} (ω_{0}))}^{ω + ω_{0} - μ}$ is $\begin{array}{cl} \prod_{ν \leq μ} det (⟨ F_{β_{1}}^{(t_{1})} \dots F_{β_{n}}^{(t_{n})} {Pr}_{ν}^{[ν]} (v^{+} \otimes w_{ν, i}), F_{β_{1}}^{(s_{1})} \dots F_{β_{n}}^{(s_{n})} {Pr}_{ν}^{[ν]} (v^{+} \otimes w_{ν, k}) ⟩) \\ = & \prod_{ν < μ} {(det M_{q} {(ω + ω_{0} - ν)}^{ω + ω_{0} - μ})}^{n (ν)} \\ \times (a_{ν}^{ω_{0}} (ω) det L_{q} {(ω_{0})}^{ω_{0} - ν}) det (⟨ {Pr}_{μ}^{[μ]} (v^{+} \otimes w_{μ, i}), {Pr}_{μ}^{[μ]} (v^{+} \otimes w_{μ, l}) ⟩) . \end{array}$ This implies $det (⟨ {Pr}_{μ}^{[μ]} (v^{+} \otimes w_{μ, i}), {Pr}_{μ}^{[μ]} (v^{+} \otimes w_{μ, l}) ⟩) = a_{μ}^{ω_{0}} (ω) det L_{q} {(ω_{0})}^{ω_{0} - μ} .$

$□$

Notes and References

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

page history

Translation Functors and the Shapovalov Determinant

Quantum Groups

The Restricted Integral Form U𝒜res(𝔤)

A PBW Basis for U𝒜res(𝔤)

The Category 𝒪

Embeddings of Verma Modules

Central characters

The Shapovalov Determinant

An Alternative Determinant Formula

Translation Functors

Notes and References

The Restricted Integral Form $U_{𝒜}^{res} (𝔤)$

A PBW Basis for $U_{𝒜}^{res} (𝔤)$

The Category $𝒪$