2.1 Cartan and root data

Throughout the present paper, we fix an indeterminate q, a Cartan datum $I = (I, \cdot )$ and a Y- and X-regular root datum $(Y, X, \langle , \rangle , \Pi ^\vee , \Pi )$ of type I. For each $i,j \in I$ and $n \in \mathbb {Z}_{\geq 0}$ , set

$$ \begin{align*} d_i &:= \frac{i \cdot i}{2}, \\ a_{i,j} &:= \frac{2i \cdot j}{i \cdot i}, \\ q_i &:= q^{d_i}, \\ [n]_i &:= \frac{q_i^n - q_i^{-n}}{q_i - q_i^{-1}}, \\ [n]_i! &:= [n]_i [n-1]_i \cdots [1]_i. \end{align*} $$

For each $i \in I$ , let $h_i \in \Pi ^\vee $ and $\alpha _i \in \Pi $ , denote the corresponding simple coroot and simple root, respectively; in particular, we have

$$\begin{align*}\langle h_i, \alpha_j \rangle = a_{i,j} \ \text{ for all } i,j \in I. \end{align*}$$

Let $\mathrm {Br}$ and W denote the braid group and Weyl group associated with I, respectively, and $s_i$ the simple reflection corresponding to $i \in I$ . They act on $\mathbb {Z}[I]$ , Y, and X by

$$\begin{align*}s_i j = j - a_{i,j} i, \quad s_i h = h - \langle h, \alpha_i \rangle h_i, \quad s_i \lambda = \lambda - \langle h_i, \lambda \rangle \alpha_i \end{align*}$$

for each $i,j \in I,\ h \in Y,\ \lambda \in X$ .

Let $Q := \sum _{i \in I} \mathbb {Z} \alpha _i \subseteq X$ denote the root lattice, and $Q^\pm := \pm \sum _{i \in I} \mathbb {Z}_{\geq 0} \alpha _i$ the positive and negative cones.

Let $\leq $ denote the dominance order on X, the partial ordering defined by declaring $\lambda \leq \mu $ to mean $\mu - \lambda \in Q^+$ .

Let $X^+$ denote the set of dominant weights:

$$\begin{align*}X^+ := \{ \lambda \in X \mid \langle h_i, \lambda \rangle \geq 0 \ \text{ for all } i \in I \}. \end{align*}$$

2.2 Quantum groups

Let $\mathbf {U}$ denote the quantum group associated with the root datum $(Y, X, \langle , \rangle , \Pi ^\vee , \Pi )$ . Namely, $\mathbf {U}$ is the unital associative $\mathbb {Q}(q)$ -algebra with generators

$$\begin{align*}\{ E_i,\ F_i,\ K_h \mid i \in I,\ h \in Y \} \end{align*}$$

subject to the following relations: for each $h,h_1,h_2 \in Y$ and $i,j \in I$ ,

$$ \begin{align*} &K_0 = 1, \\ &K_{h_1} K_{h_2} = K_{h_1 + h_2}, \\ &K_h E_i = q^{\langle h, \alpha_i \rangle} E_i K_h, \\ &K_h F_i = q^{\langle h, -\alpha_i \rangle} F_i K_h, \\ &E_i F_j - F_j E_i = \delta_{i,j} \frac{K_i - K_i^{-1}}{q_i - q_i^{-1}}, \\ &\sum_{r+s = 1-a_{i,j}} (-1)^s E_i^{(r)} E_j E_i^{(s)} = 0 \ \text{ if } i \ne j, \\ &\sum_{r+s = 1-a_{i,j}} (-1)^s F_i^{(r)} F_j F_i^{(s)} = 0 \ \text{ if } i \ne j, \end{align*} $$

where

$$\begin{align*}K_i := K_{d_i h_i}, \quad E_i^{(n)} := \frac{1}{[n]_i!} E_i^n, \quad F_i^{(n)} := \frac{1}{[n]_i!} F_i^n. \end{align*}$$

The braid group $\mathrm {Br}$ acts on $\mathbf {U}$ by $s_i \mapsto T^{\prime \prime }_{i,1}$ ([Reference Lusztig15, § 37.1]). For each $w \in W$ with reduced expression $w = s_{i_1} \cdots s_{i_l}$ , set

$$\begin{align*}T_w := T^{\prime\prime}_{i_1,1} \cdots T^{\prime\prime}_{i_l,1}. \end{align*}$$

Let $\mathbf {U}^-$ , $\mathbf {U}^0$ , and $\mathbf {U}^+$ denote the negative, Cartan and positive parts of $\mathbf {U}$ , respectively. We have a triangular decomposition:

(2.2.1) $$ \begin{align} \mathbf{U} = \mathbf{U}^- \mathbf{U}^0 \mathbf{U}^+ \simeq \mathbf{U}^- \otimes \mathbf{U}^0 \otimes \mathbf{U}^+. \end{align} $$

Let $\mathbf {B}(\pm \infty )$ denote the canonical bases of $\mathbf {U}^\mp $ , respectively. They are graded by $Q^\mp $ :

$$\begin{align*}\mathbf{B}(\pm \infty) = \bigsqcup_{\lambda \in Q^\mp} \mathbf{B}(\pm \infty)_\lambda, \end{align*}$$

where

$$\begin{align*}\mathbf{B}(\pm \infty)_\lambda := \mathbf{B}(\pm \infty) \cap \mathbf{U}^\mp_\lambda, \quad \mathbf{U}^\mp_\lambda := \{ u \in \mathbf{U}^\mp \mid K_h u K_{-h} = q^{\langle h, \lambda \rangle} u \ \text{ for all } h \in Y \}. \end{align*}$$

For each $b \in \mathbf {B}(\pm \infty )_\lambda $ , we set $\operatorname {wt}(b) := \lambda $ .

Let $\omega $ denote the Chevalley involution on $\mathbf {U}$ – that is, the algebra automorphism such that

$$\begin{align*}\omega(E_i) = F_i, \ \omega(K_h) = K_{-h} \ \text{ for all } i \in I,\ h \in Y. \end{align*}$$

Given a $\mathbf {U}$ -module V, let ${}^\omega V = \{ {}^\omega v \mid v \in V \}$ denote the $\mathbf {U}$ -module V twisted by $\omega $ :

$$\begin{align*}u \cdot {}^\omega v = {}^\omega(\omega(u) v) \ \text{ for each } u \in \mathbf{U},\ v \in V. \end{align*}$$

Let $\dot {\mathbf {U}}$ denote the modified form of $\mathbf {U}$ with idempotents $\{ 1_\lambda \mid \lambda \in X \}$ , and $\dot {\mathbf {B}}$ the canonical basis of $\dot {\mathbf {U}}$ .

2.3 Weight and integrable modules

A $\mathbf {U}$ -module V is said to be a weight module if it admits a linear space decomposition

$$\begin{align*}V = \bigoplus_{\lambda \in X} V_\lambda \end{align*}$$

such that

$$\begin{align*}V_\lambda = \{ v \in V \mid K_h v = q^{\langle h, \lambda \rangle} v \ \text{ for all } h \in Y \}. \end{align*}$$

For each $v \in V_\lambda \setminus \{ 0 \}$ , we set $\operatorname {wt}(v) := \lambda $ .

For each $\lambda \in X$ , let $M(\lambda )$ denote the Verma module of highest weight $\lambda $ with highest weight vector $m_\lambda $ .

For each $\lambda ,\mu \in X$ , set

$$\begin{align*}M(\lambda, \mu) := {}^\omega M(\lambda) \otimes M(\mu) \end{align*}$$

and

$$\begin{align*}m_{\lambda, \mu} := {}^\omega m_\lambda \otimes m_\mu \in M(\lambda, \mu). \end{align*}$$

The linear map

$$\begin{align*}\dot{\mathbf{U}} 1_{-\lambda + \mu} \rightarrow M(\lambda, \mu);\ u \mapsto u m_{\lambda, \mu} \end{align*}$$

is a $\mathbf {U}$ -module isomorphism ([Reference Lusztig15, 23.3.1 (c)]). The $\mathbf {U}$ -module $M(\lambda , \mu )$ (or rather $\dot {\mathbf {U}} 1_{-\lambda +\mu }$ ) is a universal weight module of weight $-\lambda + \mu $ in the following sense. For each weight $\mathbf {U}$ -module V and a weight vector $v \in V_{-\lambda +\mu }$ , there exists a unique $\mathbf {U}$ -module homomorphism $M(\lambda , \mu ) \rightarrow V$ which sends $m_{\lambda , \mu }$ to v.

A weight $\mathbf {U}$ -module V is said to be integrable if for each $\lambda \in X$ and $v \in V_\lambda $ , there exist $(a_i)_{i \in I}, (b_i)_{i \in I} \in \mathbb {Z}_{\geq 0}^I$ such that

$$\begin{align*}E_i^{(a_i+1)} v = F_i^{(b_i+1)} v = 0 \ \text{ for all } i \in I. \end{align*}$$

For each $\lambda \in X^+$ , let $V(\lambda )$ denote the integrable highest weight module of highest weight $\lambda $ . It is the maximal integrable quotient of the Verma module $M(\lambda )$ :

$$\begin{align*}V(\lambda) := M(\lambda)/\sum_{i \in I} \mathbf{U} F_i^{(\langle h_i, \lambda \rangle + 1)} m_\lambda. \end{align*}$$

Twisting it by the Chevalley involution $\omega $ , we obtain

(2.3.1) $$ \begin{align} {}^\omega V(\lambda) \simeq {}^\omega M(\lambda)/\sum_{i \in I} \mathbf{U} E_i^{(\langle h_i, \lambda \rangle + 1)} \cdot {}^\omega m_\lambda. \end{align} $$

Let $v_\lambda \in V(\lambda )$ denote the image of the highest weight vector $m_\lambda \in M(\lambda )$ . Let $\mathbf {B}(\lambda )$ denote the canonical basis of $V(\lambda )$ . The linear map

$$\begin{align*}\mathbf{U}^+ \rightarrow {}^\omega V(\lambda);\ u \mapsto u \cdot {}^\omega v_\lambda \end{align*}$$

gives rise to a bijection

$$\begin{align*}\mathbf{B}(-\infty)[-\lambda] := \{ b \in \mathbf{B}(-\infty) \mid b \cdot {}^\omega v_\lambda \neq 0 \} \rightarrow {}^\omega \mathbf{B}(\lambda) := \{ {}^\omega b \mid b \in \mathbf{B}(\lambda) \}. \end{align*}$$

Combining this fact with the presentation (2.3.1), we obtain

(2.3.2) $$ \begin{align} \sum_{b \in \mathbf{B}(-\infty) \setminus \mathbf{B}(-\infty)[-\lambda]} \mathbf{U} b \cdot {}^\omega m_\lambda = \sum_{i \in I} \mathbf{U} E_i^{(\langle h_i, \lambda \rangle + 1)} \cdot {}^\omega m_\lambda. \end{align} $$

For each $\lambda , \mu \in X^+$ , set

$$\begin{align*}V(\lambda, \mu) := {}^\omega V(\lambda) \otimes V(\mu) \end{align*}$$

and

$$\begin{align*}v_{\lambda, \mu} := {}^\omega v_\lambda \otimes v_\mu \in V(\lambda, \mu). \end{align*}$$

By [Reference Lusztig15, Proposition 23.3.6], we have

(2.3.3) $$ \begin{align} V(\lambda,\mu) \simeq M(\lambda,\mu)/(\sum_{i \in I} \mathbf{U} E_i^{(\langle h_i, \lambda \rangle + 1)} m_{\lambda,\mu} + \sum_{i \in I} \mathbf{U} F_i^{(\langle h_i, \mu \rangle + 1)} m_{\lambda,\mu}). \end{align} $$

The $\mathbf {U}$ -modules $V(\lambda , \mu )$ are universal integrable $\mathbf {U}$ -modules in the following sense. Let V be an integrable $\mathbf {U}$ -module and $v \in V$ a weight vector. Then, there exist $\lambda , \mu \in X^+$ and a $\mathbf {U}$ -module homomorphism $V(\lambda , \mu ) \rightarrow V$ which sends $v_{\lambda , \mu }$ to v ([Reference Lusztig15, Proposition 23.3.10]).

2.4 Levi and parabolic subalgebras

In this subsection, we fix a subset $J \subseteq I$ , and set $K := I \setminus J$ . We often regard J as a Cartan datum and

$$\begin{align*}(Y, X, \langle , \rangle, \{ h_j \mid j \in J \}, \{ \alpha_j \mid j \in J \}) \end{align*}$$

as a root datum of type J.

Let $\mathbf {L} = \mathbf {L}_J$ denote the Levi subalgebra of the quantum group $\mathbf {U}$ associated with J – that is, the subalgebra generated by

$$\begin{align*}\{ E_j,\ F_j,\ K_h \mid j \in J,\ h \in Y \}. \end{align*}$$

The algebra $\mathbf {L}$ itself is the quantum group associated with the root datum of type J above. Let $\mathbf {L}^-$ and $\mathbf {L}^+$ denote the negative and the positive parts of $\mathbf {L}$ , respectively. Noting that the Cartan part of $\mathbf {L}$ is $\mathbf {U}^0$ , we have a triangular decomposition

(2.4.1) $$ \begin{align} \mathbf{L} = \mathbf{L}^- \mathbf{U}^0 \mathbf{L}^+ \simeq \mathbf{L}^- \otimes \mathbf{U}^0 \otimes \mathbf{L}^+. \end{align} $$

For each $\lambda \in X$ , let $M_{\mathbf {L}}(\lambda )$ denote the Verma module over $\mathbf {L}$ of highest weight $\lambda $ with highest weight vector $m_{\mathbf {L}, \lambda }$ . Similarly, for each $\lambda , \mu \in X$ , set $M_{\mathbf {L}}(\lambda , \mu ) := {}^\omega M_{\mathbf {L}}(\lambda ) \otimes M_{\mathbf {L}}(\mu )$ and $m_{\mathbf {L};\lambda ,\mu } := {}^\omega m_{\mathbf {L},\lambda } \otimes m_{\mathbf {L},\mu } \in M_{\mathbf {L}}(\lambda , \mu )$ . Then, we have $M_{\mathbf {L}}(\lambda , \mu ) \simeq \dot {\mathbf {L}} 1_{-\lambda + \mu }$ as $\mathbf {L}$ -modules, where $\dot {\mathbf {L}}$ denotes the modified form of $\mathbf {L}$ .

Let $\mathbf {P}^+ = \mathbf {P}^+_J$ and $\mathbf {P}^- = \mathbf {P}^-_J$ denote the parabolic and the opposite parabolic subalgebra of $\mathbf {U}$ associated with the subset J, respectively. Namely, $\mathbf {P}^+$ is the subalgebra of $\mathbf {U}$ generated by $\mathbf {L}$ and $\{ E_k \mid k \in K \}$ , and $\mathbf {P}^-$ generated by $\mathbf {L}$ and $\{ F_k \mid k \in K \}$ .

We have triangular decompositions:

(2.4.2) $$ \begin{align} &\mathbf{P}^+ = \mathbf{L}^- \mathbf{U}^0 \mathbf{U}^+ \simeq \mathbf{L}^- \otimes \mathbf{U}^0 \otimes \mathbf{U}^+; \end{align} $$

(2.4.3) $$ \begin{align} &\mathbf{P}^- = \mathbf{U}^- \mathbf{U}^0 \mathbf{L}^+ \simeq \mathbf{U}^- \otimes \mathbf{U}^0 \otimes \mathbf{L}^+; \end{align} $$

Although the algebras $\mathbf {P}^\pm $ are not quantum groups, we can construct their modified forms $\dot {\mathbf {P}}^\pm = \bigoplus _{\lambda \in X} \dot {\mathbf {P}}^\pm 1_\lambda $ in a canonical way. Just like the modified forms of quantum groups, the $\dot {\mathbf {P}}^\pm $ have natural $\mathbf {P}^\pm $ -bimodule structure. Furthermore, the $\mathbf {L}$ -bimodule structure on $\mathbf {P}^\pm $ gives rise to an $\dot {\mathbf {L}}$ -bimodule structure on $\dot {\mathbf {P}}^\pm $ .

Let $\mathbf {R}^\pm = \mathbf {R}^\pm _J$ denote the nilradical part of $\mathbf {P}^\pm $ . That is, $\mathbf {R}^+$ is the two-sided ideal of $\mathbf {U}^+$ generated by $\{ E_k \mid k \in K \}$ , and $\mathbf {R}^-$ generated by $\{ F_k \mid k \in K \}$ . Then, we have

(2.4.4) $$ \begin{align} \mathbf{U}^\pm = \mathbf{L}^\pm \oplus \mathbf{R}^\pm. \end{align} $$

Proposition 2.4.1. As a $\mathbb {Q}(q)$ -linear space, we have

(2.4.5) $$ \begin{align} \mathbf{P}^+ = \mathbf{L} \oplus \mathbf{L}^- \mathbf{U}^0 \mathbf{R}^+. \end{align} $$

Moreover, the subspace $\mathbf {L}^- \mathbf {U}^0 \mathbf {R}^+$ coincides with the two-sided ideal of $\mathbf {P}^+$ generated by $\{ E_k \mid k \in K \}$ .

Proof. By the decompositions (2.4.2), (2.4.4) and (2.4.1), we have

$$\begin{align*}\mathbf{P}^+ = \mathbf{L}^- \mathbf{U}^0 (\mathbf{L}^+ \oplus \mathbf{R}^+) = \mathbf{L} \oplus \mathbf{L}^- \mathbf{U}^0 \mathbf{R}^+. \end{align*}$$

This proves the first assertion.

To prove the second assertion, let us compute as

(2.4.6) $$ \begin{align} \mathbf{L}^- \mathbf{U}^0 \mathbf{R}^+ = \sum_{k \in K} \mathbf{L}^- \mathbf{U}^0 \mathbf{U}^+ E_k \mathbf{U}^+ = \sum_{k \in K} \mathbf{P}^+ E_k \mathbf{U}^+. \end{align} $$

For each $k \in K$ , we have

(2.4.7) $$ \begin{align} \mathbf{P}^+ E_k \mathbf{P}^+ = \mathbf{P}^+ E_k \mathbf{L}^- \mathbf{U}^0 \mathbf{U}^+ = \mathbf{P}^+ \mathbf{L}^- \mathbf{U}^0 E_k \mathbf{U}^+ = \mathbf{P}^+ E_k \mathbf{U}^+. \end{align} $$

Combining equations (2.4.6) and (2.4.7), we obtain

$$\begin{align*}\mathbf{L}^- \mathbf{U}^0 \mathbf{R}^+ = \sum_{k \in K} \mathbf{P}^+ E_k \mathbf{U}^+ = \sum_{k \in K} \mathbf{P}^+ E_k \mathbf{P}^+, \end{align*}$$

as desired. Thus, we complete the proof.

Lemma 2.4.2. Let $\lambda \in X$ . We denote by $1^{\mathbf {U}}_\lambda $ and $1^{\mathbf {P}^-}_\lambda $ the idempotents in $\dot {\mathbf {U}}$ and $\dot {\mathbf {P}}^-$ corresponding to $\lambda $ , respectively. Then, as $\mathbb {Q}(q)$ -linear spaces, we have

(2.4.8) $$ \begin{align} \mathbf{P}^- 1^{\mathbf{P}^-}_\lambda \simeq \mathbf{U} 1^{\mathbf{U}}_\lambda/\mathbf{U} \mathbf{R}^+ 1^{\mathbf{U}}_\lambda. \end{align} $$

Proof. By the decompositions (2.2.1) and (2.4.4), we have

$$\begin{align*}\mathbf{U} 1^{\mathbf{U}}_\lambda = \mathbf{U}^- \mathbf{U}^0 \mathbf{U}^+ 1^{\mathbf{U}}_\lambda = \mathbf{U}^- \mathbf{U}^0 (\mathbf{L}^+ \oplus \mathbf{R}^+) 1^{\mathbf{U}}_\lambda = (\mathbf{P}^- \oplus \mathbf{U} \mathbf{R}^+) 1^{\mathbf{U}}_\lambda. \end{align*}$$

Hence, we obtain a linear isomorphism

(2.4.9) $$ \begin{align} \mathbf{P}^- 1^{\mathbf{U}}_\lambda \simeq \mathbf{U} 1^{\mathbf{U}}_\lambda/\mathbf{U} \mathbf{R}^+ 1^{\mathbf{U}}_\lambda. \end{align} $$

Since the triangular decomposition (2.4.3) of $\mathbf {P}^-$ is consistent with the one (2.2.1) of $\mathbf {U}$ , both the linear spaces $\mathbf {P}^- 1^{\mathbf {U}}_\lambda $ and $\mathbf {P}^- 1^{\mathbf {P}^-}_\lambda $ are isomorphic to $\mathbf {U}^- \otimes \mathbf {L}^+$ in canonical ways. Thus, we complete the proof.

Proposition 2.4.4. Let V be a weight $\mathbf {L}$ -module. Then, as a $\mathbf {U}$ -module, we have

$$\begin{align*}\operatorname{Ind}_{\mathbf{P}^+}^{\mathbf{U}} V := \mathbf{U} \otimes_{\mathbf{P}^+} V = \dot{\mathbf{P}}^- \otimes_{\dot{\mathbf{L}}} V; \end{align*}$$

here, we regard V as a $\mathbf {P}^+$ -module via the projection $\mathbf {P}^+ \rightarrow \mathbf {L}$ with respect to the decomposition (2.4.5), and $\dot {\mathbf {P}}^- (= \bigoplus _{\lambda \in X} \mathbf {P}^- 1_\lambda )$ as a $\mathbf {U}$ -module via the linear isomorphism (2.4.8).

Proof. Let

$$\begin{align*}\phi \colon \operatorname{Ind}_{\mathbf{P}^+}^{\mathbf{U}} V \rightarrow \dot{\mathbf{P}}^- \otimes_{\dot{\mathbf{L}}} V \text{ and } \psi \colon \dot{\mathbf{P}}^- \otimes_{\dot{\mathbf{L}}} V \rightarrow \operatorname{Ind}_{\mathbf{P}^+}^{\mathbf{U}} V \end{align*}$$

denote the linear maps given as follows. Let $\lambda \in X$ , $v \in V_\lambda $ , $\mu \in Q^+$ , $u^- \in \mathbf {U}^-$ , $u^0 \in \mathbf {U}^0$ , $u^+ \in \mathbf {U}^+_\mu $ , and $p \in \mathbf {P}^-$ . Then, we set

$$\begin{align*}\phi(u^-u^0u^+ \otimes v) := u^- 1_{\lambda+\mu} \otimes u^0u^+v \text{ and } \psi(p 1_\lambda \otimes v) := p \otimes v. \end{align*}$$

That these maps are well defined can be straightforwardly verified. Clearly, we have $\phi \circ \psi = \mathrm {id}$ , and see that $\psi $ is a $\mathbf {U}$ -module homomorphism. Thus, we complete the proof.

Corollary 2.4.5. Let $\lambda , \mu \in X$ . Then, as $\mathbf {U}$ -modules, we have

$$\begin{align*}\operatorname{Ind}_{\mathbf{P}^+}^{\mathbf{U}} M_{\mathbf{L}}(\lambda, \mu) \simeq \operatorname{Ind}_{\mathbf{P}^+}^{\mathbf{U}} \dot{\mathbf{L}} 1_{-\lambda + \mu} \simeq \dot{\mathbf{P}}^- \otimes_{\dot{\mathbf{L}}} \dot{\mathbf{L}} 1_{-\lambda + \mu} \simeq \dot{\mathbf{P}}^- 1_{-\lambda+\mu}. \end{align*}$$

3.1 Quantum symmetric pair

Let $(I, I_{\bullet }, \tau )$ be a generalized Satake diagram ([Reference Regelskis and Vlaar21, Definition 1]). From now on, we assume that the root datum $(Y, X, \langle , \rangle , \Pi ^\vee , \Pi )$ is a Satake datum of type $(I, I_{\bullet }, \tau )$ ([Reference Watanabe26, Definition 3.1.3]). In particular, $I_{\bullet }$ is a subdatum of I which is supposed to be of finite type, $\tau $ is an involutive automorphism on the Cartan datum I, and Y and X are supposed to be equipped with involutive automorphisms $\tau $ which are compatible with the one on I. Let $w_{\bullet } \in W$ denote the longest element of the Weyl group $W_{\bullet } \subseteq W$ associated with $I_{\bullet }$ . Set

$$ \begin{align*} \theta &:= -w_{\bullet} \tau \in \operatorname{Aut}(\mathbb{Z}[I]), \operatorname{Aut}(Y), \operatorname{Aut}(X); \\ Y^{\imath} &:= \{ h \in Y \mid \theta(h) = h \}; \\ X^{\imath} &:= X/\{ \lambda - \theta(\lambda) \mid \lambda \in X \}; \\ I_\circ &:= I \setminus I_{\bullet}. \end{align*} $$

Also, let

$$\begin{align*}\overline{\cdot} \colon X \rightarrow X^{\imath} \end{align*}$$

denote the quotient map, and

$$\begin{align*}\langle , \rangle \colon Y^{\imath} \times X^{\imath} \rightarrow \mathbb{Z} \end{align*}$$

the bilinear pairing induced from the one on $Y \times X$ .

For later use, let us recall one of the axioms for generalized Satake diagrams from ([Reference Regelskis and Vlaar21, (2.19)]):

(3.1.1) $$ \begin{align} \langle h_i, \theta(\alpha_i) \rangle \neq -1 \ \text{ for all } i \in I. \end{align} $$

Lemma 3.1.1. Let $k \in I_\circ $ . If $\tau (k) = k \neq w_{\bullet } k$ , then we have

$$\begin{align*}\langle h_k, w_{\bullet} \alpha_k - \alpha_k \rangle \leq -2. \end{align*}$$

Proof. Since $w_{\bullet } \alpha _k - \alpha _k \in \sum _{j \in I_{\bullet }} \mathbb {Z}_{\geq 0} \alpha _j$ , we have

$$\begin{align*}\langle h_k, w_{\bullet} \alpha_k - \alpha_k \rangle \in \mathbb{Z}_{\leq 0}. \end{align*}$$

Our assumption on k that $w_{\bullet } k \neq k$ implies that this value cannot be $0$ .

However, taking into account the assumption that $\tau (k) = k$ , we compute as follows:

$$\begin{align*}\langle h_k, w_{\bullet} \alpha_k - \alpha_k \rangle = \langle h_k, -\theta(\alpha_k) - \alpha_k \rangle = -2 - \langle h_k, \theta(\alpha_k) \rangle. \end{align*}$$

By the axiom (3.1.1), the rightmost-hand side cannot be $-1$ . Hence, the assertion follows.

Let $\mathbf {U}^{\imath }$ denote the $\imath $ quantum group associated with the Satake datum $(Y, X, \langle , \rangle , \Pi ^\vee , \Pi )$ and parameters ${\boldsymbol \varsigma } = (\varsigma _k)_{k \in I_\circ } \in (\mathbb {Q}(q)^\times )^{I_\circ }$ , ${\boldsymbol \kappa } = (\kappa _i)_{k \in I_\circ } \in \mathbb {Q}(q)^{I_\circ }$ . It is the subalgebra of the quantum group $\mathbf {U}$ generated by

$$\begin{align*}\{ E_j,\ F_j,\ B_k,\ K_h \mid j \in I_{\bullet},\ k \in I_\circ,\ h \in Y^{\imath} \}, \end{align*}$$

where

$$\begin{align*}B_k := F_k + \varsigma_k T_{w_{\bullet}}(E_{\tau(k)}) K_k^{-1} + \kappa_k K_k^{-1} \end{align*}$$

for each $k \in I_\circ $ . The parameters are supposed to satisfy the following for all $k \in I_\circ $ :

• • $\varsigma _k = \varsigma _{\tau (k)}$ if $k \cdot \theta (k) = 0$ .

• • $\kappa _k = 0$ unless $\tau (k) = k$ , $a_{k,j} = 0$ for all $j \in I_{\bullet }$ , and $a_{k,k'} \in 2\mathbb {Z}$ for all $k' \in I_\circ $ such that $\tau (k') = k'$ and $a_{k',j} = 0$ for all $j \in I_{\bullet }$ .

They are as general as [Reference Kolb11, Definition 5.6].

Let $\dot {\mathbf {U}}^{\imath }$ denote the modified form of $\mathbf {U}^{\imath }$ with idempotents $\{ 1_\zeta \mid \zeta \in X^{\imath } \}$ .

Let $\mathbf {L}$ , $\mathbf {P}^\pm $ , and $\mathbf {R}^\pm $ denote the Levi, parabolic subalgebras, and the nilradical parts associated with $I_{\bullet }$ , respectively.

Proposition 3.1.2 (cf. [Reference Bao and Wang2, Lemma 3.22])

Let $\lambda \in X$ and $\zeta := \bar {\lambda } \in X^{\imath }$ . As $\mathbf {U}^{\imath }$ -modules, we have

$$\begin{align*}\dot{\mathbf{U}}^{\imath} 1_\zeta \simeq \mathbf{U}^{\imath} 1_\lambda \simeq \dot{\mathbf{P}}^- 1_\lambda. \end{align*}$$

3.2 Weight modules

A $\mathbf {U}^{\imath }$ -module V is said to be a weight module ([Reference Watanabe26]) if it admits a linear space decomposition

$$\begin{align*}V = \bigoplus_{\zeta \in X^{\imath}} V_\zeta \end{align*}$$

such that

• • $V_\zeta = \{ v \in V \mid K_h v = q^{\langle h, \zeta \rangle } v \ \text { for all } h \in Y^{\imath } \}$ ;

• • $E_j V_\zeta \subseteq V_{\zeta + \overline {\alpha _j}}$ , $F_j V_\zeta \subseteq V_{\zeta - \overline {\alpha _j}}$ for all $j \in I_{\bullet }$ ;

• • $B_k V_\zeta \subseteq V_{\zeta - \overline {\alpha _k}}$ for all $k \in I_\circ $ ;

A weight $\mathbf {U}$ -module $V = \bigoplus _{\lambda \in X} V_\lambda $ has a canonical weight $\mathbf {U}^{\imath }$ -module structure:

(3.2.1) $$ \begin{align} V = \bigoplus_{\zeta \in X^{\imath}} V_\zeta, \ V_\zeta := \bigoplus_{\substack{\lambda \in X \\ \bar{\lambda} = \zeta}} V_\lambda. \end{align} $$

For each $\lambda , \mu \in X$ , set

$$\begin{align*}M^{\imath}(\lambda, \mu) := \operatorname{Ind}_{\mathbf{P}^+}^{\mathbf{U}} M_{\mathbf{L}}(-w_{\bullet} \lambda, \mu) \end{align*}$$

and

$$\begin{align*}m^{\imath}_{\lambda, \mu} := 1 \otimes m_{\mathbf{L}; -w_{\bullet} \lambda, \mu} \in M^{\imath}(\lambda, \mu). \end{align*}$$

Then, by Corollary 2.4.5, we have

(3.2.2) $$ \begin{align} M^{\imath}(\lambda, \mu) \simeq \operatorname{Ind}_{\mathbf{P}^+}^{\mathbf{U}} \dot{\mathbf{L}} 1_{w_{\bullet} \lambda + \mu} \simeq \dot{\mathbf{P}}^- 1_{w_{\bullet} \lambda + \mu} \end{align} $$

as $\mathbf {U}$ -modules. The $\mathbf {U}^{\imath }$ -module $M^{\imath }(\lambda , \mu )$ is a universal weight module of weight $\overline {w_{\bullet } \lambda + \mu }$ in the following sense.

Proof. By the isomorphism (3.2.2) and Proposition 3.1.2, we have

$$\begin{align*}M^{\imath}(\lambda,\mu) \simeq \dot{\mathbf{U}}^{\imath} 1_\zeta \end{align*}$$

as $\mathbf {U}^{\imath }$ -modules. Then, the assertion can be found in [Reference Watanabe26, after Definition 3.3.5].

3.3 Integrable modules

For each $\lambda , \mu \in X^+$ , set $V^{\imath }(\lambda , \mu )$ to be the quotient $\mathbf {U}$ -module of $M^{\imath }(\lambda , \mu )$ factored by the submodule generated by

(3.3.1) $$ \begin{align} \{ E_j^{(\langle h_j, -w_{\bullet} \lambda \rangle + 1)} m^{\imath}_{\lambda, \mu},\ F_j^{(\langle h_j, \mu \rangle + 1)} m^{\imath}_{\lambda, \mu},\ F_k^{(\langle h_k, w_{\bullet}\lambda + \mu \rangle + 1)} m^{\imath}_{\lambda,\mu} \mid j \in I_{\bullet},\ k \in I_\circ \}. \end{align} $$

Let $v^{\imath }_{\lambda ,\mu } \in V^{\imath }(\lambda ,\mu )$ denote the image of $m^{\imath }_{\lambda ,\mu } \in M^{\imath }(\lambda ,\mu )$ under the quotient map.

Lemma 3.3.2. Let $\lambda , \mu \in X^+$ . Then, there exist a finite number of dominant weights $\nu _1,\dots ,\nu _r \in X^+$ such that

$$\begin{align*}\mathbf{L} v^{\imath}_{\lambda, \mu} \simeq \bigoplus_{s=1}^r V_{\mathbf{L}}(\nu_s) \end{align*}$$

as $\mathbf {L}$ -modules and

$$\begin{align*}V^{\imath}(\lambda, \mu) \simeq \bigoplus_{s=1}^r V(\nu_s) \end{align*}$$

as $\mathbf {U}$ -modules, where $V_{\mathbf {L}}(\nu )$ denotes the integrable highest weight $\mathbf {L}$ -module of highest weight $\nu \in X_{\mathbf {L}}^+$ .

Proof. By the definition of $V^{\imath }(\lambda , \mu )$ and the presentation (2.3.3) of $V_{\mathbf {L}}(-w_{\bullet }\lambda , \mu )$ , there exists an $\mathbf {L}$ -module homomorphism

$$\begin{align*}V_{\mathbf{L}}(-w_{\bullet} \lambda, \mu) \rightarrow \mathbf{L} v^{\imath}_{\lambda, \mu} \end{align*}$$

which sends $v_{\mathbf {L};-w_{\bullet }\lambda , \mu }$ to $v^{\imath }_{\lambda , \mu }$ . This homomorphism is clearly surjective. Since the domain is finite-dimensional, so is the codomain. This implies that there exist $\nu _1,\dots ,\nu _r \in X_{\mathbf {L}}^+$ such that

$$\begin{align*}\mathbf{L} v^{\imath}_{\lambda, \mu} \simeq \bigoplus_{s=1}^r V_{\mathbf{L}}(\nu_s). \end{align*}$$

However, the $\mathbf {U}$ -module $V^{\imath }(\lambda , \mu )$ is integrable, and

(3.3.2) $$ \begin{align} \mathbf{U}^+ v^{\imath}_{\lambda, \mu} = \mathbf{L}^+ v^{\imath}_{\lambda, \mu} \end{align} $$

Since the right-hand side is finite-dimensional, the weights of $V^{\imath }(\lambda , \mu )$ are bounded above. Therefore, the $\mathbf {U}$ -module is a direct sum of submodules isomorphic to integrable highest weight modules. By equation (3.3.2), we see that the highest weight vectors in the $\mathbf {U}$ -module $V^{\imath }(\lambda , \mu )$ are exactly those in the $\mathbf {L}$ -module $\mathbf {L} v^{\imath }_{\lambda , \mu }$ . This implies that

$$\begin{align*}\nu_1,\dots,\nu_r \in X^+ \end{align*}$$

and

$$\begin{align*}V^{\imath}(\lambda, \mu) \simeq \bigoplus_{s=1}^r V(\nu_s). \end{align*}$$

Thus, we complete the proof.

Proposition 3.3.3. Let $L^{\imath }(\lambda ,\mu )$ denote the $\mathbf {U}$ -submodule of $V(\lambda ) \otimes V(\mu )$ generated by $v_{w_{\bullet } \lambda } \otimes v_\mu $ , where $v_{w_{\bullet } \lambda }$ denotes the unique element in $\mathbf {B}(\lambda )$ of weight $w_{\bullet } \lambda $ . Then, we have

$$\begin{align*}V^{\imath}(\lambda,\mu) \simeq L^{\imath}(\lambda,\mu) \end{align*}$$

as $\mathbf {U}$ -modules.

Proof. It is easily verified that $\mathbf {R}^+ (v_{w_{\bullet } \lambda } \otimes v_\mu ) = 0$ and

$$\begin{align*}E_j^{(\langle h_j, -w_{\bullet} \lambda \rangle + 1)} (v_{w_{\bullet} \lambda} \otimes v_\mu) = F_j^{(\langle h_j, \mu \rangle + 1)} (v_{w_{\bullet} \lambda} \otimes v_\mu) = F_k^{(\langle h_k, w_{\bullet} \lambda + \mu \rangle + 1)} (v_{w_{\bullet} \lambda} \otimes v_\mu) = 0 \end{align*}$$

for all $j \in I_{\bullet }$ , $k \in I_\circ $ . Hence, Lemma 3.3.1 implies that there exists a $\mathbf {U}$ -module homomorphism

$$\begin{align*}V^{\imath}(\lambda, \mu) \rightarrow L^{\imath}(\lambda,\mu) \end{align*}$$

which sends $v^{\imath }_{\lambda , \mu }$ to $v_{w_{\bullet } \lambda } \otimes v_\mu $ .

However, since $\mathbf {L}(v_{w_{\bullet } \lambda } \otimes v_\mu ) \simeq V_{\mathbf {L}}(-w_{\bullet } \lambda , \mu )$ , there exists an $\mathbf {L}$ -module homomorphism

$$\begin{align*}\mathbf{L}(v_{w_{\bullet} \lambda} \otimes v_\mu) \rightarrow \mathbf{L} v^{\imath}_{\lambda, \mu} \end{align*}$$

which sends $v_{w_{\bullet } \lambda } \otimes v_\mu $ to $v^{\imath }_{\lambda , \mu }$ .

Combining the two homomorphisms above, we see that $\mathbf {L} v^{\imath }_{\lambda , \mu } \simeq \mathbf {L}(v_{w_{\bullet } \lambda } \otimes v_\mu )$ . By Lemma 3.3.2, the $\mathbf {U}$ -module structure of $V^{\imath }(\lambda , \mu )$ is determined by the $\mathbf {L}$ -module structure of $\mathbf {L} v^{\imath }_{\lambda , \mu }$ . However, by [Reference Watanabe25, Proposition 3.4.3], the $\mathbf {U}$ -module structure of $L^{\imath }(\lambda ,\mu )$ is determined by the $\mathbf {L}$ -module structure of $\mathbf {L}(v_{w_{\bullet } \lambda } \otimes v_\mu )$ . Hence, the assertion follows.

As explained in Section 1, the $\mathbf {U}$ -modules $L^{\imath }(\lambda , \mu )$ in Proposition 3.3.3, which are isomorphic to $V^{\imath }(\lambda , \mu )$ , are counterparts of $V(\lambda ,\mu )$ in the theory of $\imath $ canonical bases [Reference Bao and Wang2], [Reference Bao and Wang3], and the latter can be used to define integrable $\mathbf {U}$ -modules. From this point of view, Definition 3.3.4 is quite natural.

4.1 Some elements $\mathfrak {B}_{k,\zeta }^{(n)}$

The aim of this subsection is to introduce a family $\{ \mathfrak {B}_{k,\zeta }^{(n)} \mid k \in I_\circ ,\ \zeta \in X^{\imath },\ n \in \mathbb {Z}_{\geq 0} \}$ of elements in $\dot {\mathbf {U}}^{\imath }$ , which will be used to express the $\mathbf {U}^{\imath }$ -modules $V^{\imath }(\lambda ,\mu )$ in terms of generators and relations.

4.1.1 Case $1$ : $\tau (k) = k = w_{\bullet } k$

First, let us consider the case where $\tau (k) = k = w_{\bullet } k$ . For each $\zeta \in X^{\imath }$ and $\lambda \in X$ such that $\bar {\lambda } = \zeta $ , the parity of integer $\langle h_k, \lambda \rangle $ is independent of the choice of $\lambda $ . We call it the parity of $\zeta $ at k. Set $p_k(\zeta ) \in \mathbb {Z}/2\mathbb {Z} = \{ \bar {0}, \bar {1} \}$ to be $\bar {0}$ if the parity of $\zeta $ at k is even, and to be $\bar {1}$ otherwise. Similarly, for each integer n, set $p(n)$ to be $\bar {0}$ if n is even, and to be $\bar {1}$ otherwise.

For each $\zeta \in X^{\imath }$ and $n \geq 0$ , define the element $\mathfrak {B}_{k, \zeta }^{(n)} \in \dot {\mathbf {U}}^{\imath } 1_\zeta $ inductively as follows:

• • $\mathfrak {B}_{k, \zeta }^{(0)} := 1_\zeta $ .

• • If $p_k(\zeta ) = \bar {0}$ , then $\mathfrak {B}_{k,\zeta }^{(1)} := (B_k - \kappa _k)1_\zeta $ .

• • If $p_k(\zeta ) = p(n)$ and $n \geq 1$ , then $\mathfrak {B}_{k,\zeta }^{(n)} := \frac {1}{[n]_k} B_k \mathfrak {B}_{k,\zeta }^{(n-1)}$ .

• • If $p_k(\zeta ) \neq p(n)$ and $n \geq 2$ , then

  $$\begin{align*}\mathfrak{B}_{k, \zeta}^{(n)} := \frac{1}{[n]_k[n-1]_k}(B_k^2 - (q_k^{n-1} + q_k^{-n+1})\kappa_k B_k + (\kappa_k^2 - [n-1]_k^2 q_k \varsigma_k)) \mathfrak{B}_{k, \zeta}^{(n-2)}. \end{align*}$$

Let us see where this definition comes from. Let $\mathbb {K}$ be an algebraic closure of the field $\mathbb {Q}(q)$ , and consider the following quadratic equations

(QE)_a,b $$\begin{align} x^2 - (q_k + q_k^{-1}) a x + a^2 - b = 0 \end{align} $$

with variable x and constants $a,b \in \mathbb {K}$ . Since this equation is symmetric with respect to x and a, it follows that if $c \in \mathbb {K}$ is a solution to $(\mathrm {QE})_{a,b}$ , then a is a solution to $(\mathrm {QE})_{c,b}$ .

Given scalars $a^{(0)}, b \in \mathbb {K}$ , define a family $(a^{(n)})_{n \in \mathbb {Z}} \in \mathbb {K}^{\mathbb {Z}}$ of scalars as follows: The $a^{(1)}$ and $a^{(-1)}$ are the solutions to the quadratic equation $(\mathrm {QE})_{a^{(0)},b}$ ; the ambiguity will not matter in the subsequent argument. Suppose that we have defined $a^{(m)}$ for $-n \leq m \leq n$ for some $n \geq 1$ . Then, set $a^{(\pm (n+1))} := (q_k + q_k^{-1}) a^{(\pm n)} - a^{(\pm (n-1))}$ .

By the construction above, it is clear that the $a^{(n+1)}$ and $a^{(n-1)}$ are the solutions to the quadratic equation $(\mathrm {QE})_{a^{(n)}, b}$ for each $n \in \mathbb {Z}$ . In particular, we have

$$ \begin{align*} &a^{(n+1)} + a^{(n-1)} = (q_k + q_k^{-1}) a^{(n)}, \\ &a^{(n+1)} a^{(n-1)} = (a^{(n)})^2 - b. \end{align*} $$

Then, one can straightforwardly verify that for each $n \geq 1$ , the $a^{(n)}$ and $a^{(-n)}$ are the solutions to the quadratic equation

$$\begin{align*}x^2 - (q_k^n + q_k^{-n}) a^{(0)} x + (a^{(0)})^2 - [n]_k^2 b = 0. \end{align*}$$

Now, let us consider the $(n+1)$ -dimensional irreducible representation $\mathbf {V}_n$ of $U_q(\mathfrak {sl}_2)$ , which we identify with the subalgebra of $\mathbf {U}$ generated by $E_k$ , $F_k$ , and $K_k^{\pm 1}$ . By a similar argument to [Reference Watanabe23, § 3.1], one can show that the $B_k \in \mathbf {U}^{\imath }$ acts on $\mathbf {V}_n$ semisimply with eigenvalues $\kappa ^{(n)}, \kappa ^{(n-2)}, \dots , \kappa ^{(-n)}$ , where we set $\kappa ^{(0)} := \kappa _k$ and $b := q_k \varsigma _k$ . Therefore, we see that

$$\begin{align*}\mathfrak{B}_{k,\zeta}^{(n+1)} = \frac{1}{[n]_k!}P_n(B_k) 1_\zeta \ \text{ for all } \zeta \in X^{\imath} \text{ with } p_k(\zeta) = p(n), \end{align*}$$

where $P_n(x)$ denotes the minimal polynomial of the action of $B_k$ on $\mathbf {V}_n$ .

The following are straightforward consequences of the argument above.

Proposition 4.1.1. Let V be a weight $\mathbf {U}$ -module, $\lambda \in X$ , $v \in V_\lambda $ . Assume that $n := \langle h_k, \lambda \rangle \geq 0$ and $E_k \mathbf {L}^+ v = 0$ . Then, we have

$$\begin{align*}\mathfrak{B}_{k, \bar{\lambda}}^{(n+1)} v = F_k^{(n+1)} v. \end{align*}$$

Proposition 4.1.2. Let V be an integrable $\mathbf {U}$ -module, $\lambda \in X$ , and $v \in V_\lambda $ . Then, there exists $N \in \mathbb {Z}_{\geq 0}$ such that $\mathfrak {B}_{k, \bar {\lambda }}^{(n+1)} v = 0$ for all $n \geq N$ .

4.1.2 Case $2$ : $\tau (k) = k \neq w_{\bullet } k$

Next, let us consider the case where $\tau (k) = k \neq w_{\bullet } k$ . For each $\zeta \in X^{\imath }$ and $n \geq 0$ , set

$$\begin{align*}\mathfrak{B}_{k, \zeta}^{(n)} := \frac{1}{[n]_k!} B_k^n 1_\zeta. \end{align*}$$

Set $Y_k := B_k - F_k$ and $Z_k := F_k Y_k - q_k^{-2} Y_k F_k$ . Then, the following hold (cf. [Reference Bao and Wang3, § 5.2]):

• • $Z_k$ commutes with both $F_k$ and $Y_k$ .

• • $Z_k \in \mathbf {L}^+_{w_{\bullet }(\alpha _k) - \alpha _k}$ .

• • $Y_k \in \mathbf {L}^+ E_k \mathbf {L}^+$ .

Hence, for each $n \geq 0$ , we have

(4.1.1) $$ \begin{align} B_k^n = \sum_{f + 2z + y = n} a_{f,z,y} F_k^f Z_k^z Y_k^y \end{align} $$

for some $a_{f,z,y} \in \mathbb {Q}(q)$ .

Proposition 4.1.3. Let V be a weight $\mathbf {U}$ -module, $\lambda \in X$ , $v \in V_\lambda $ . Assume that $n := \langle h_k, \lambda \rangle \geq 0$ and $E_k \mathbf {L}^+ v = 0$ . Then, we have

$$\begin{align*}\mathfrak{B}_{k, \bar{\lambda}}^{(n+1)} v = \sum_{f+2z = n+1} a_{f,z} F_k^{(f)} Z_k^z v \end{align*}$$

for some $a_{f,z} \in \mathbb {Q}(q)$ such that $a_{n+1,0} = 1$ .

Proof. The assertion is immediate from equation (4.1.1).

Proposition 4.1.4. Let V be an integrable $\mathbf {U}$ -module, $\lambda \in X$ , and $v \in V_\lambda $ . Then, there exists $N \in \mathbb {Z}_{\geq 0}$ such that $\mathfrak {B}_{k, \bar {\lambda }}^{(n+1)} v = 0$ for all $n \geq N$ .

Proof. Since V is integrable, the element $Y_k$ , which is a scalar multiple of $T_{w_{\bullet }}(E_{\tau (k)}) K_k^{-1}$ , acts on V locally nilpotently. In particular, there exists $y_0 \geq 0$ such that

$$\begin{align*}Y_k^{y} v = 0 \ \text{ for all } y> y_0. \end{align*}$$

Recall that the algebra $\mathbf {L}^+$ is a quantum group of finite type. Hence, the $\mathbf {L}^+$ -submodule of the integrable module V generated by the finite set $\{ Y_k^y v \mid y \geq 0 \}$ is finite-dimensional. Hence, there exists $z_0 \geq 0$ such that

$$\begin{align*}Z_k^z Y_k^y v = 0 \ \text{ for all } y \geq 0,\ z> z_0. \end{align*}$$

Since the $F_k$ acts on V locally nilpotently, and the number of vectors of the form $Z_k^z Y_k^y v$ with $z,y \geq 0$ is finite, there exists $f_0 \geq 0$ such that

$$\begin{align*}F_k^f Z_k^z Y_k^y v = 0 \ \text{ for all } z,y \geq 0,\ f> f_0. \end{align*}$$

Now, set

$$\begin{align*}N := f_0 + 2z_0 + f_0. \end{align*}$$

By equation (4.1.1), we have

$$\begin{align*}B_k^{N+1} v = \sum_{f+2z+y=N+1} a_{f,z,y} F_k^f Z_k^z Y_k^y v. \end{align*}$$

All the summand in the right-hand side is $0$ by the definition of N. Hence, the assertion follows.

4.1.3 Case $3$ : $\tau (k) \neq k$

Finally, let us consider the case where $\tau (k) \neq k$ . For each $\zeta \in X^{\imath }$ and $n> 0$ , set

$$\begin{align*}\mathfrak{B}_{k, \zeta}^{(n)} := \frac{1}{[n]_k!} B_k^n 1_\zeta. \end{align*}$$

Set $Y_k := B_k - F_k$ . Then, the following hold (cf. [Reference Bao and Wang3, § 5.5]):

• • $F_k Y_k - q_k^{-2} Y_k F_k = 0$ .

• • $Y_k \in \mathbf {L}^+ E_{\tau (k)} \mathbf {L}^+$ .

Hence, for each $n \geq 0$ , we have

(4.1.2) $$ \begin{align} B_k^n = \sum_{f + y = n} a_{f,y} F_k^f Y_k^y \end{align} $$

for some $a_{f,y} \in \mathbb {Q}(q)$ .

Proposition 4.1.5. Let V be a weight $\mathbf {U}$ -module, $\lambda \in X$ , $v \in V_\lambda $ . Assume that $n := \langle h_k, \lambda \rangle \geq 0$ and $E_k \mathbf {L}^+ v = E_{\tau (k)} \mathbf {L}^+ v = 0$ . Then, we have

$$\begin{align*}\mathfrak{B}_{k, \bar{\lambda}}^{(n+1)} v = F_k^{(n+1)} v. \end{align*}$$

Proof. The assertion is immediate from equation (4.1.2).

Proposition 4.1.6. Let V be an integrable $\mathbf {U}$ -module, $\lambda \in X$ , and $v \in V_\lambda $ . Then, there exists $N \in \mathbb {Z}_{\geq 0}$ such that $\mathfrak {B}_{k, \bar {\lambda }}^{(n+1)} v = 0$ for all $n \geq N$ .

Proof. One can prove the assertion in the same way as Proposition 4.1.4.

4.1.4 Summary

Let us summarize in a unified manner what we have obtained in the preceding argument.

Proposition 4.1.7. Let V be a weight $\mathbf {U}$ -module, $\lambda \in X$ , $v \in V_\lambda $ , and $k \in I_\circ $ . Assume that $\langle h_k, \lambda \rangle \geq 0$ and $E_k \mathbf {L}^+ v = E_{\tau (k)} \mathbf {L}^+ v = 0$ . Let $\mathbf {U}_k^-$ denote the subalgebra of $\mathbf {U}$ generated by $F_k$ . Then, the following hold.

1. (1) $\mathfrak {B}_{k,\zeta }^{(\langle h_k, \lambda \rangle + 1)} v - F_k^{(\langle h_k, \lambda \rangle + 1)} v \in \sum _{b \in \mathbf {B}_{\mathbf {L}}(-\infty ) \setminus \{ 1 \}} \mathbf {U}^-_k F_k^{(\langle h_k, \lambda + \operatorname {wt}(b) \rangle + 1)} b v$ , where we understand that $F_k^{(n)} = 0$ if $n < 0$ .

2. (2) $E_k \mathfrak {B}_{k, \zeta }^{(\langle h_k, \lambda \rangle + 1)} v \in \sum _{b \in \mathbf {B}_{\mathbf {L}}(-\infty ) \setminus \{ 1 \}} \mathbf {U} F_k^{(\langle h_k, \lambda + \operatorname {wt}(b) \rangle + 1)} b v$ .

3. (3) $b' \mathfrak {B}_{k, \zeta }^{\langle h_k, \lambda \rangle + 1} v \in \sum _{\substack {b \in \mathbf {B}(-\infty ) \\ \operatorname {wt}(b) \geq \operatorname {wt}(b')}} \mathbf {U}^-_k F_k^{(\langle h_k, \lambda + \operatorname {wt}(b) \rangle + 1)} bv$ for all $b' \in \mathbf {B}_{\mathbf {L}}(-\infty )$ .

Proof. By Lemma 3.1.1, the first assertion follows from Propositions 4.1.1, 4.1.3 and 4.1.5. The other assertions are immediate from the first one.

Proposition 4.1.8. Let V be an integrable $\mathbf {U}$ -module, $\zeta \in X^{\imath }$ , and $v \in V_\zeta $ . Then, there exist $(a_j)_{j \in I_{\bullet }}, (b_j)_{j \in I_{\bullet }} \in \mathbb {Z}_{\geq 0}^{I_{\bullet }}$ , $(c_k)_{k \in I_\circ } \in \mathbb {Z}_{\geq 0}^{I_\circ }$ such that

$$\begin{align*}E_j^{(a_j+1)} v = F_j^{(b_j+1)} v = \mathfrak{B}_{k,\zeta}^{(c_k+1)} v = 0. \end{align*}$$

Proof. Since the weight vector v is a finite sum of weight vectors of the weight $\mathbf {U}$ -module V (see (3.2.1)), the assertion follows from Propositions 4.1.2, 4.1.4 and 4.1.6.

4.2 A Presentation of the $\mathbf {U}^{\imath }$ -module $V^{\imath }(\lambda , \mu )$

In this subsection, we fix $\lambda , \mu \in X^+$ , and set

$$ \begin{align*} &\nu := w_{\bullet} \lambda + \mu, \\ &\mathcal{E} := \sum_{j \in I_{\bullet}} \mathbf{U} E_j^{(\langle h_j, -w_{\bullet} \lambda \rangle + 1)} m^{\imath}_{\lambda, \mu}, \\ &\mathcal{E}' := \sum_{j \in I_{\bullet}} \mathbf{U}^{\imath} E_j^{(\langle h_j, -w_{\bullet} \lambda \rangle + 1)} m^{\imath}_{\lambda, \mu}, \\ &\mathcal{F} := \sum_{j \in I_{\bullet}} \mathbf{U} F_j^{(\langle h_j, \mu \rangle + 1)} m^{\imath}_{\lambda, \mu}, \\ &\mathcal{F}' := \sum_{j \in I_{\bullet}} \mathbf{U}^{\imath} F_j^{(\langle h_j, \mu \rangle + 1)} m^{\imath}_{\lambda, \mu}, \\ &\mathcal{F}_\circ := \sum_{k \in I_\circ} \mathbf{U} F_k^{(\langle h_k, w_{\bullet} \lambda + \mu \rangle + 1)} m^{\imath}_{\lambda, \mu}, \\ &\mathcal{B} := \sum_{k \in I_\circ} \sum_{b \in \mathbf{B}_{\mathbf{L}}(-\infty)[w_{\bullet} \lambda]} \mathbf{U} F_k^{(\langle h_k, \nu + \operatorname{wt}(b) \rangle + 1)} b m^{\imath}_{\lambda, \mu}, \\ &\mathcal{B}' := \sum_{k \in I_\circ} \sum_{b \in \mathbf{B}_{\mathbf{L}}(-\infty)[w_{\bullet} \lambda]} \mathbf{U}^{\imath} \mathfrak{B}_{k, \overline{\nu + \operatorname{wt}(b)}}^{(\langle h_k, \nu + \operatorname{wt}(b) \rangle + 1)} b m^{\imath}_{\lambda, \mu}. \end{align*} $$

The aim of this subsection is to show that

$$\begin{align*}\mathcal{E} + \mathcal{F} + \mathcal{F}_\circ = \mathcal{E} + \mathcal{F} + \mathcal{B} = \mathcal{E}' + \mathcal{F}' + \mathcal{B}'. \end{align*}$$

This provides us with a presentation of the $\mathbf {U}^{\imath }$ -module $V^{\imath }(\lambda , \mu )$ .

Lemma 4.2.2. We have

$$\begin{align*}\mathcal{E} = \mathcal{E}'. \end{align*}$$

Proof. Let $j \in I_{\bullet }$ and set $v := E_j^{(\langle h_j, -w_{\bullet } \lambda \rangle + 1)} m^{\imath }_{\lambda , \mu } \in M^{\imath }(\lambda , \mu )$ . Since $\mathbf {R}^+ E_j \subset \mathbf {R}^+$ , we have

$$\begin{align*}\mathbf{R}^+ v \subset \mathbf{R}^+ m^{\imath}_{\lambda, \mu} = 0. \end{align*}$$

Applying Lemma 4.2.1 to the $1$ -dimensional subspace spanned by v, we obtain

$$\begin{align*}\mathbf{U} v = \mathbf{U}^{\imath} v. \end{align*}$$

This implies the assertion.

Lemma 4.2.3. We have

$$\begin{align*}\mathcal{F} = \mathcal{F}'. \end{align*}$$

Proof. Let $j \in I_{\bullet }$ and set $v := F_j^{(\langle h_j, \mu \rangle + 1)} m^{\imath }_{\lambda , \mu } \in M^{\imath }(\lambda , \mu )$ . Let $x \in \mathbf {L}^+$ and $k \in I_\circ $ . By the triangular decomposition (2.4.1) of $\mathbf {L}$ , the vector $E_k x v$ is, up to scalar multiplies, of the form $E_k x^- x^+ m^{\imath }_{\lambda , \mu }$ with $x^- \in \mathbf {L}^-$ , $x^+ \in \mathbf {L}^+$ . Since $E_k$ commutes with the elements in $\mathbf {L}^-$ , we have

$$\begin{align*}E_k x^- x^+ m^{\imath}_{\lambda, \mu} = x^- E_k x^+ m^{\imath}_{\lambda, \mu} \in x^- \mathbf{R}^+ m^{\imath}_{\lambda, \mu} = 0. \end{align*}$$

The argument above implies that $\mathbf {R}^+ v = 0$ . Hence, the assertion follows by applying Lemma 4.2.1 as in the proof of Lemma 4.2.2.

Lemma 4.2.4. We have

$$\begin{align*}\mathcal{B} + \mathcal{E} = \mathcal{B}' + \mathcal{E}. \end{align*}$$

Proof. For each $k \in I_\circ $ and $b \in \mathbf {B}_{\mathbf {L}}(-\infty )$ , set

• • $\lambda _b := \nu + \operatorname {wt}(b)$ ,

• • $v_{k,b} := F_k^{(\langle h_k, \lambda _b \rangle + 1)} b m^{\imath }_{\lambda , \mu }$ ,

• • $v^{\prime }_{k,b} := \mathfrak {B}_{k, \overline {\lambda _b}}^{(\langle h_k, \lambda _b \rangle + 1)} b m^{\imath }_{\lambda , \mu }$ ,

• • $W_{> b} := \sum _{k \in I_\circ } \sum _{\substack {b' \in \mathbf {B}_{\mathbf {L}}(-\infty ) \\ \operatorname {wt}(b') > \operatorname {wt}(b)}} \mathbf {U} v_{k,b'} + \mathcal {E}$ ,

• • $W^{\prime }_{> b} := \sum _{k \in I_\circ } \sum _{\substack {b' \in \mathbf {B}_{\mathbf {L}}(-\infty ) \\ \operatorname {wt}(b') > \operatorname {wt}(b)}} \mathbf {U}^{\imath } v^{\prime }_{k,b'} + \mathcal {E}$ ,

• • $W_{\geq b} := \mathbb {Q}(q) v_{k,b} + W_{> b}$ ,

• • $W^{\prime }_{\geq b} := \mathbb {Q}(q) v^{\prime }_{k,b} + W^{\prime }_{> b}$ ,

Since $b m^{\imath }_{\lambda ,\mu } \in \mathcal {E}$ for all $b \in \mathbf {B}_{\mathbf {L}}(-\infty ) \setminus \mathbf {B}_{\mathbf {L}}(-\infty )[w_{\bullet }\lambda ]$ by equation (2.3.2), we have $v_{k,b},v^{\prime }_{k,b} \in \mathcal {E}$ for all $k \in I_\circ $ and such $b \in \mathbf {B}_{\mathbf {L}}(-\infty )$ . Hence, it holds that

$$\begin{align*}W_{>b} = \sum_{k \in I_\circ} \sum_{\substack{b' \in \mathbf{B}_{\mathbf{L}}(-\infty)[w_{\bullet}\lambda] \\ \operatorname{wt}(b') > \operatorname{wt}(b)}} \mathbf{U} v_{k,b'} + \mathcal{E}, \text{ and } W^{\prime}_{>b} = \sum_{k \in I_\circ} \sum_{\substack{b' \in \mathbf{B}_{\mathbf{L}}(-\infty)[w_{\bullet}\lambda] \\ \operatorname{wt}(b') > \operatorname{wt}(b)}} \mathbf{U}^{\imath} v^{\prime}_{k,b'} + \mathcal{E}. \end{align*}$$

In particular,

$$\begin{align*}\mathbf{U} W_{\geq 1} = \mathcal{B} + \mathcal{E} \text{ and } \mathbf{U}^{\imath} W^{\prime}_{\geq 1} = \mathcal{B}' + \mathcal{E}. \end{align*}$$

We shall show that

$$\begin{align*}\mathbf{U}^{\imath} W^{\prime}_{\geq b} = \mathbf{U} W_{\geq b} \end{align*}$$

for all $b \in \mathbf {B}_{\mathbf {L}}(-\infty )$ by descending induction on $\operatorname {wt}(b)$ . If this is the case, then we obtain

$$\begin{align*}\mathcal{B} + \mathcal{E} = \mathbf{U} W_{\geq 1} = \mathbf{U}^{\imath} W^{\prime}_{\geq 1} = \mathcal{B}' + \mathcal{E}, \end{align*}$$

as desired.

When $\operatorname {wt}(b) \gg 0$ , it holds that $b' \notin \mathbf {B}_{\mathbf {L}}(-\infty )[w_{\bullet } \lambda ]$ for all $b' \in \mathbf {B}_{\mathbf {L}}(-\infty )$ such that $\operatorname {wt}(b') \geq \operatorname {wt}(b)$ . These imply that $v_{k,b'}, v^{\prime }_{k,b'} \in \mathcal {E}$ and hence the subspaces $W_{\geq b}$ and $W^{\prime }_{\geq b}$ are equal to $\mathcal {E}$ . Hence, our claim follows in this case.

Assume that the claim holds for all $b' \in \mathbf {B}_{\mathbf {L}}(-\infty )$ with $\operatorname {wt}(b')> \operatorname {wt}(b)$ . Since

$$\begin{align*}W_{> b} = \sum_{\substack{b' \in \mathbf{B}_{\mathbf{L}}(-\infty) \\ \operatorname{wt}(b') > \operatorname{wt}(b)}} \mathbf{U} W_{\geq b'} \text{ and } W^{\prime}_{> b} = \sum_{\substack{b' \in \mathbf{B}_{\mathbf{L}}(-\infty) \\ \operatorname{wt}(b') > \operatorname{wt}(b)}} \mathbf{U}^{\imath} W^{\prime}_{\geq b'}, \end{align*}$$

these spaces are identical to each other.

Let us show that $\mathbf {U} \mathbf {R}^+ W^{\prime }_{\geq b} \subseteq W^{\prime }_{\geq b}$ . Since $W^{\prime }_{\geq b} = \mathbb {Q}(q) v^{\prime }_{k,b} \oplus W_{> b}$ and $\mathbf {U} W_{> b} = W_{> b}$ , we only need to show that $\mathbf {R}^+ v^{\prime }_{k,b} \in W_{> b}$ . To this end, we will prove that $E_{k'} b" v^{\prime }_{k,b} \in W_{> b}$ for all $k' \in I_\circ $ and $b" \in \mathbf {B}_{\mathbf {L}}(-\infty )$ . It is easily seen that the vector $E_{k'} b" v^{\prime }_{k,b}$ is equal to $0$ unless $k' = k$ . Hence, we may assume that $k' = k$ .

When $b" = 1$ , by Proposition 4.1.7 (2), we have

$$\begin{align*}E_k v^{\prime}_{k,b} \in \sum_{b_1 \in \mathbf{B}(-\infty) \setminus \{ 1 \}} \mathbf{U} F_k^{(\langle h_k, \nu + \operatorname{wt}(b) + \operatorname{wt}(b_1) \rangle + 1)} b_1 b m^{\imath}_{\lambda, \mu}. \end{align*}$$

Since $b_1 b$ is a linear combination of $b_2 \in \mathbf {B}(-\infty )$ with $\operatorname {wt}(b_2) = \operatorname {wt}(b) + \operatorname {wt}(b_1)$ , we see that

$$\begin{align*}F_k^{(\langle h_k, \nu + \operatorname{wt}(b) + \operatorname{wt}(b_1) \rangle + 1)} b_1 b m^{\imath}_{\lambda, \mu} \in \sum_{\operatorname{wt}(b_2)> \operatorname{wt}(b)} \mathbb{Q}(q) F_k^{(\langle h_k, \nu+\operatorname{wt}(b_2) \rangle + 1)} b_2 m^{\imath}_{\lambda,\mu}. \end{align*}$$

Therefore, we obtain

$$\begin{align*}E_k v^{\prime}_{k,b} \in W_{> b}, \end{align*}$$

as desired.

When $b" \neq 1$ , by Proposition 4.1.7 (3), we have

$$\begin{align*}E_k b" v^{\prime}_{k,b} \in \sum_{\operatorname{wt}(b') \geq \operatorname{wt}(b")} \mathbf{U} F_k^{(\langle h_k, \nu + \operatorname{wt}(b) + \operatorname{wt}(b') \rangle + 1)} b' b m^{\imath}_{\lambda, \mu} = W_{> b}. \end{align*}$$

This completes the proof of the claim that $\mathbf {U} \mathbf {R}^+ W^{\prime }_{\geq b} \subseteq W^{\prime }_{\geq b}$ .

Let us apply Lemma 4.2.1 to obtain

$$\begin{align*}\mathbf{U} W^{\prime}_{\geq b} = \mathbf{U}^{\imath} W^{\prime}_{\geq b}. \end{align*}$$

By Proposition 4.1.7 (1), we have

$$\begin{align*}v^{\prime}_{k,b} - v_{k,b} \in W_{> b}. \end{align*}$$

This shows that

$$\begin{align*}W^{\prime}_{\geq b} = W_{\geq b}. \end{align*}$$

Thus, we complete the proof.

Lemma 4.2.5. We have

$$\begin{align*}\mathcal{F}_\circ + \mathcal{E} = \mathcal{B} + \mathcal{E}. \end{align*}$$

Proof. Clearly, we have $\mathcal {F}_\circ \subseteq \mathcal {B}$ . Hence, we only need to show that $F_k^{(\langle h_k, \nu + \operatorname {wt}(b) \rangle + 1)} b m^{\imath }_{\lambda , \mu } \in \mathcal {F}_\circ $ for all $k \in I_\circ $ and $b \in \mathbf {B}_{\mathbf {L}}(-\infty )$ . Since $b m^{\imath }_{\lambda , \mu } \in \mathcal {E}$ if $b \notin \mathbf {B}_{\mathbf {L}}(-\infty )[w_{\bullet } \lambda ]$ , we may assume that $b \in \mathbf {B}_{\mathbf {L}}(-\infty )[w_{\bullet } \lambda ]$ . Noting that $\operatorname {wt}(b) \in \sum _{j \in I_{\bullet }} \mathbb {Z}_{\geq 0} \alpha _j$ and $(\lambda - w_{\bullet } \lambda ) - \operatorname {wt}(b) \in \sum _{j \in I_{\bullet }} \mathbb {Z}_{\geq 0} \alpha _j$ , we see that

$$\begin{align*}\langle h_k, \operatorname{wt}(b) \rangle \leq 0 \text{ and } \langle h_k, w_{\bullet} \lambda + \mu + \operatorname{wt}(b) \rangle \geq \langle h_k, \lambda + \mu \rangle \geq 0. \end{align*}$$

Therefore, the vector $F_k^{(\langle h_k, \nu + \operatorname {wt}(b) \rangle + 1)} b m^{\imath }_{\lambda , \mu }$ is a scalar multiple of the vector

$$\begin{align*}E_k^{-\langle h_k, \operatorname{wt}(b) \rangle} b F_k^{(\langle h_k, w_{\bullet} \lambda + \mu \rangle + 1)} m^{\imath}_{\lambda, \mu}. \end{align*}$$

This implies our claim, and the proof is completed.

Theorem 4.2.6. We have

$$\begin{align*}\mathcal{E} + \mathcal{F} + \mathcal{F}_\circ = \mathcal{E}' + \mathcal{F}' + \mathcal{B}'. \end{align*}$$

In other words, the $\mathbf {U}^{\imath }$ -module $V^{\imath }(\lambda , \mu )$ coincides with the quotient $\mathbf {U}^{\imath }$ -module of $M^{\imath }(\lambda , \mu )$ factored by the submodule generated by

$$\begin{align*}\{ E_j^{(\langle h_j, -w_{\bullet} \lambda \rangle + 1)} m^{\imath}_{\lambda, \mu},\ F_j^{(\langle h_j, \mu \rangle + 1)} m^{\imath}_{\lambda, \mu} \mid j \in I_{\bullet} \} \end{align*}$$

and

$$\begin{align*}\{ \mathfrak{B}_{k, \overline{w_{\bullet} \lambda + \mu + \mathrm{wt}(b)}}^{(\langle h_k, w_{\bullet} \lambda + \mu + \operatorname{wt}(b) \rangle + 1)} b m^{\imath}_{\lambda, \mu} \mid k \in I_\circ,\ b \in \mathbf{B}_{\mathbf{L}}(-\infty) \}. \end{align*}$$

Corollary 4.2.7. Let V be an integrable $\mathbf {U}^{\imath }$ -module. Then, for each $\zeta \in X^{\imath }$ and $v \in V_\zeta $ , there exist $(a_j)_{j \in I_{\bullet }}, (b_j)_{j \in I_{\bullet }} \in \mathbb {Z}_{\geq 0}^{I_{\bullet }}$ and $(c_k)_{k \in I_\circ } \in \mathbb {Z}_{\geq 0}^{I_\circ }$ such that

$$\begin{align*}E_j^{(a_j+1)} v = F_j^{(b_j+1)} v = 0 \ \text{ for all } j \in I_{\bullet} \end{align*}$$

and

$$\begin{align*}\mathfrak{B}_{k,\zeta}^{(c_k+1)} v = 0 \ \text{ for all } k \in I_\circ. \end{align*}$$

4.3 Some sufficient conditions

In this subsection, we give some sufficient conditions for a weight $\mathbf {U}^{\imath }$ -module to be integrable.

Proposition 4.3.1. Let V be a weight $\mathbf {U}^{\imath }$ -module generated by a weight vector v of weight $\zeta $ . Suppose that there exist $(a_j)_{j \in I_{\bullet }}, (b_j)_{j \in I_{\bullet }} \in \mathbb {Z}_{\geq 0}^{I_{\bullet }}$ and $(c_k)_{k \in I_\circ } \in \mathbb {Z}_{\geq 0}^{I_\circ }$ such that

$$\begin{align*}E_j^{(a_j+1)} v = F_j^{(b_j+1)} v = 0 \ \text{ for all } j \in I_{\bullet} \end{align*}$$

and

$$\begin{align*}\mathfrak{B}_{k,\zeta}^{(c_k+1)} v = 0 \ \text{ for all } k \in I_\circ. \end{align*}$$

Then, V is integrable.

Proof. The assumption on v implies that the $\mathbf {L}$ -module $\mathbf {L} v$ is integrable. Since $\mathbf {L}$ is a quantum group of finite type, every cyclic integrable module is finite-dimensional. Therefore, the set

$$\begin{align*}B_V := \{ b \in \mathbf{B}_{\mathbf{L}}(-\infty) \mid bv \neq 0 \} \end{align*}$$

is finite.

Let $\lambda ,\mu \in X^+$ be such that $\overline {w_{\bullet } \lambda + \mu } = \zeta $ . By replacing $\lambda $ and $\mu $ by $\lambda + \tau (\nu )$ and $\mu + \nu $ , respectively, for a suitable $\nu \in X^+$ , we may assume that

$$ \begin{align*} &\langle h_j, -w_{\bullet} \lambda \rangle \geq a_j, \\ &\langle h_j, \mu \rangle \geq b_j, \\ &\langle h_k, w_{\bullet} \lambda + \mu + \operatorname{wt}(b) \rangle \geq c_k \end{align*} $$

for all $j \in I_{\bullet }$ , $k \in I_\circ $ , and $b \in B_V$ . These imply that

$$ \begin{align*} &E_j^{(\langle h_j, -w_{\bullet} \lambda \rangle + 1)} v \in \mathbf{U}^{\imath} E_j^{(a_j+1)}v = 0, \\ &F_j^{(\langle h_j, \mu \rangle + 1)} v \in \mathbf{U}^{\imath} F_j^{(b_j+1)}v = 0, \\ &\mathfrak{B}_{k, \overline{w_{\bullet}\lambda + \mu} + \mathrm{wt}^{\imath}(b)}^{(\langle h_k, w_{\bullet}\lambda + \mu + \operatorname{wt}(b) \rangle + 1)} bv = b \mathfrak{B}_{k, \overline{w_{\bullet}\lambda + \mu}}^{(\langle h_k, w_{\bullet}\lambda + \mu + \operatorname{wt}(b) \rangle + 1)} v \in \mathbf{U}^{\imath} \mathfrak{B}_{k,\zeta}^{(c_k+1)} v = 0 \end{align*} $$

for all $j \in I_{\bullet }$ , $k \in I_\circ $ , and $b \in B_V$ . Hence, the assertion follows from Theorem 4.2.6.

Proof. Let V be an integrable $\mathbf {U}$ -module. It is a weight $\mathbf {U}^{\imath }$ -module. Let $\zeta \in X^{\imath }$ and $v \in V_\zeta $ . Let us write

$$\begin{align*}v = \sum_{r=1}^s v_r \end{align*}$$

for some $s \in \mathbb {Z}_{> 0}$ , $v_r \in V_{\lambda _r}$ , and $\lambda _r \in X$ with $\overline {\lambda _r} = \zeta $ .

Since V is an integrable $\mathbf {U}$ -module, there exist $a_j,b_j \in \mathbb {Z}_{\geq 0}$ for $j \in I_{\bullet }$ such that

$$\begin{align*}E_j^{(a_j+1)} v_r = F_j^{(b_j+1)} v_r = 0 \end{align*}$$

for all $r = 1,\dots ,s$ . Hence, we obtain

$$\begin{align*}E_j^{(a_j+1)} v = F_j^{(a_j+1)} v = 0 \ \text{ for all } j \in I_{\bullet}. \end{align*}$$

Also, for each $k \in I_\circ $ , by Proposition 4.1.8, there exists $c_k \geq 0$ such that

$$\begin{align*}\mathfrak{B}_{k,\zeta}^{(c_k+1)} b v = 0 \ \text{ for all } b \in \mathbf{B}_{\mathbf{L}}(-\infty). \end{align*}$$

This completes the proof.

5.1 Simple integrable modules

Let $\lambda , \mu \in X^+$ and consider the $\mathbf {U}^{\imath }$ -module $V^{\imath }(\lambda , \mu )$ . It is equipped with a nondegenerate symmetric bilinear form $(,) = (,)_{\lambda , \mu }$ such that $(v^{\imath }_{\lambda , \mu }, v^{\imath }_{\lambda , \mu }) = 1$ and

$$\begin{align*}(x u, v) = (u, \rho(x) v) \ \text{ for all } x \in \mathbf{U}^{\imath},\ u,v \in V^{\imath}(\lambda, \mu) \end{align*}$$

(cf. [Reference Bao and Wang2, § 6.6]). Since $V^{\imath }(\lambda , \mu )$ is finite-dimensional, the existence of such a bilinear form ensures that it is semisimple. In fact, if W is a submodule of $V^{\imath }(\lambda ,\mu )$ , and $W^\perp $ denotes the orthogonal complement of W with respect to $(,)_{\lambda ,\mu }$ , then $W^\perp $ is also a submodule of $V^{\imath }(\lambda ,\mu )$ since $\rho $ preserves $\mathbf {U}^{\imath }$ . With some index set $\Gamma $ , let

$$\begin{align*}\{ V(\gamma) \mid \gamma \in \Gamma \} \end{align*}$$

denote a complete set of representatives for the isomorphism classes of simple $\mathbf {U}^{\imath }$ -modules appearing as a submodule of $V^{\imath }(\lambda , \mu )$ for some $\lambda , \mu \in X^+$ . Since the $\mathbf {U}^{\imath }$ -module $V^{\imath }(\lambda ,\mu )$ is integrable by Proposition 4.3.2, so is each $V(\gamma )$ .

Let V be a simple integrable $\mathbf {U}^{\imath }$ -module, and $v \in V$ a nonzero weight vector. Then, there exist $\lambda , \mu \in X^+$ and a $\mathbf {U}^{\imath }$ -module homomorphism $V^{\imath }(\lambda , \mu ) \rightarrow V$ which sends $v^{\imath }_{\lambda , \mu }$ to v. Since V is simple, this homomorphism must be surjective. The semisimplicity of $V^{\imath }(\lambda , \mu )$ implies that it has a simple submodule isomorphic to V. Therefore, V is isomorphic to $V(\gamma )$ for some $\gamma \in \Gamma $ .

The argument above shows that the set $\{ V(\gamma ) \mid \gamma \in \Gamma \}$ is a complete set of representatives for the isomorphism classes of simple integrable $\mathbf {U}^{\imath }$ -modules.

Lemma 5.1.1. Let $\gamma \in \Gamma $ . Then, almost all the elements in $\dot {\mathbf {B}}^{\imath }$ acts on the $\mathbf {U}^{\imath }$ -module $V(\gamma )$ as $0$ :

$$\begin{align*}\sharp \{ b \in \dot{\mathbf{B}}^{\imath} \mid b V(\gamma) \neq 0 \} < \infty. \end{align*}$$

Proof. Let $B(\gamma ) = \{ v^\gamma _i \mid i \in I_\gamma \}$ be a basis of $V(\gamma )$ consisting of weight vectors. Since $V(\gamma )$ is finite-dimensional, the index set $I_\gamma $ is finite. By the integrability of $V(\gamma )$ , for each $i \in I_\gamma $ , we may take $\lambda _i,\mu _i \in X^+$ such that there exists a $\mathbf {U}^{\imath }$ -module homomorphism $V^{\imath }(\lambda _i,\mu _i) \rightarrow V(\gamma )$ which sends $v^{\imath }_{\lambda _i,\mu _i}$ to $v^\gamma _i$ .

Since the $\dot {\mathbf {B}}^{\imath }$ possesses the stability property, we have

$$\begin{align*}\sharp \{ b \in \dot{\mathbf{B}}^{\imath} \mid b v^{\imath}_{\lambda,\mu} \neq 0 \} = \dim V^{\imath}(\lambda,\mu) \ \text{ for all } \lambda,\mu \in X^+. \end{align*}$$

Hence, the number of elements $b \in \dot {\mathbf {B}}^{\imath }$ such that $b v^{\imath }_{\lambda _i,\mu _i} \neq 0$ for some $i \in I_\gamma $ is finite.

Now, suppose that $b \in \dot {\mathbf {B}}^{\imath }$ acts on $V(\gamma )$ as nonzero. Then, there exists $i \in I_\gamma $ such that $b v^{\imath }_{\lambda _i,\mu _i} \neq 0$ . By the argument above, the number of such elements is finite.

5.2 Space of matrix coefficients

Given a linear space V, let $V^{*}$ denote the linear dual space $\operatorname {Hom}_{\mathbb {Q}(q)}(V, \mathbb {Q}(q))$ , and $\langle \ \mid \ \rangle \colon V \times V^{*} \rightarrow \mathbb {Q}(q)$ the canonical pairing.

Set $\dot {\mathbf {U}}^{\imath *} := (\dot {\mathbf {U}}^{\imath })^{*}$ . It is equipped with the following $\mathbf {U}^{\imath }$ -bimodule structure:

$$\begin{align*}\langle u \mid x \phi y \rangle = \langle y u x \mid \phi \rangle \ \text{ for all } x,y \in \mathbf{U}^{\imath},\ u \in \dot{\mathbf{U}}^{\imath},\ \phi \in \dot{\mathbf{U}}^{\imath *}. \end{align*}$$

For each $\gamma \in \Gamma $ , $v \in V(\gamma )$ and $\phi \in V(\gamma )^{*}$ , let $c^\gamma _{v, \phi } \in \dot {\mathbf {U}}^{\imath *}$ denote the corresponding matrix coefficient:

$$\begin{align*}\langle x \mid c^\gamma_{v, \phi} \rangle = \langle xv \mid \phi \rangle \ \text{ for all } x \in \dot{\mathbf{U}}^{\imath}. \end{align*}$$

The matrix coefficients $\{ c^\gamma _{v,\phi } \mid v \in V(\gamma ),\ \phi \in V(\gamma )^{*} \}$ form a simple $\mathbf {U}^{\imath }$ -bimodule isomorphic to $V(\gamma ) \otimes V(\gamma )^{*}$ ; the matrix coefficient $c^\gamma _{v, \phi }$ corresponds to the vector $v \otimes \phi $ .

Let $\mathbf {O}^{\imath }$ denote the subspace of $\dot {\mathbf {U}}^{\imath *}$ consisting of $\phi \in \dot {\mathbf {U}}^{\imath *}$ such that the $\mathbf {U}^{\imath }$ -submodule $\mathbf {U}^{\imath } \phi $ and the right $\mathbf {U}^{\imath }$ -submodule $\phi \mathbf {U}^{\imath }$ are integrable.

The following two statements can be proved by a similar way to classical results (e.g., see [Reference Procesi20, 6.2.6]).

Proposition 5.2.3. The $\mathbf {O}^{\imath }$ is a $\mathbf {U}^{\imath }$ -bisubmodule of $\dot {\mathbf {U}}^{\imath *}$ . Moreover, the assignment

$$\begin{align*}V(\gamma) \otimes V(\gamma)^{*} \rightarrow \dot{\mathbf{U}}^{\imath *};\ v \otimes \phi \mapsto c^\gamma_{v,\phi} \end{align*}$$

gives rise to a Peter-Weyl type decomposition:

$$\begin{align*}\bigoplus_{\gamma \in \Gamma} V(\gamma) \otimes V(\gamma)^{*} \simeq \mathbf{O}^{\imath} \ (\text{as } \mathbf{U}^{\imath}\text{-bimodules}). \end{align*}$$

Let $\dot {\mathbf {B}}^{\imath *} := \{ \delta _b \mid b \in \dot {\mathbf {B}}^{\imath } \}$ denote the dual basis to $\dot {\mathbf {B}}^{\imath }$ :

$$\begin{align*}\langle b_1 \mid \delta_{b_2} \rangle = \delta_{b_1, b_2} \ \text{ for all } b_1,b_2 \in \dot{\mathbf{B}}^{\imath}. \end{align*}$$

Proof. First, we shall show that $\dot {\mathbf {B}}^{\imath *} \subset \mathbf {O}^{\imath }$ .

Let $b \in \dot {\mathbf {B}}^{\imath }$ . Take $\lambda , \mu \in X^+$ in a way such that $b v^{\imath }_{\lambda , \mu } \neq 0$ . Set

$$\begin{align*}B := \{ b' \in \dot{\mathbf{B}}^{\imath} \mid b' v^{\imath}_{\lambda, \mu} \neq 0 \}. \end{align*}$$

Note that the set B is finite. In fact, since the $\mathbf {U}^{\imath }$ -module $V^{\imath }(\lambda , \mu )$ is integrable, there exist $(a_j)_{j \in I_{\bullet }}, (b_j)_{j \in I_{\bullet }} \in \mathbb {Z}_{\geq 0}^{I_{\bullet }}$ , $(c_k)_{k \in I_\circ } \in \mathbb {Z}_{\geq 0}^{I_\circ }$ such that

$$\begin{align*}E_j^{(a_j+1)} b' v^{\imath}_{\lambda,\mu} = F_j^{(b_j+1)} b' v^{\imath}_{\lambda,\mu} = \mathfrak{B}_{k, \zeta}^{(c_k+1)} b' v^{\imath}_{\lambda,\mu} \ \text{ for all } j \in I_{\bullet},\ k \in I_\circ,\ b' \in B, \end{align*}$$

where $\zeta \in X^{\imath }$ denotes the weight of $b' v^{\imath }_{\lambda , \mu }$ .

We shall show that $\delta _b x = 0$ for all

$$\begin{align*}x \in \{ E_j^{(a_j+1)},\ F_j^{(b_j+1)},\ \mathfrak{B}_{k, \zeta}^{(c_k+1)} \mid j \in I_{\bullet},\ k \in I_\circ,\ \zeta \in X^{\imath} \}. \end{align*}$$

Let $b_1 \in \dot {\mathbf {B}}^{\imath }$ and write

(5.2.1) $$ \begin{align} x b_1 = \sum_{b_2 \in \dot{\mathbf{B}}^{\imath}} c_{b_2} b_2 \end{align} $$

with $c_{b_2} \in \mathbb {Z}[q,q^{-1}]$ . Then, we have

$$\begin{align*}\langle b_1 \mid \delta_b x \rangle = \langle x b_1 \mid \delta_b \rangle = c_b. \end{align*}$$

We want to show that $c_b = 0$ . By equation (5.2.1), we have

$$\begin{align*}x b_1 v^{\imath}_{\lambda, \mu} = \sum_{b_2 \in B} c_{b_2} b_2 v^{\imath}_{\lambda, \mu}. \end{align*}$$

The left-hand side equals $0$ by the definition of x. Since the elements $\{ b_2 v^{\imath }_{\lambda ,\mu } \mid b_2 \in B \}$ form a basis of $V^{\imath }(\lambda ,\mu )$ , we must have $c_{b_2} = 0$ for all $b_2 \in B$ . In particular, we obtain $c_b = 0$ .

Next, we shall prove that the $\dot {\mathbf {B}}^{\imath *}$ spans $\mathbf {O}^{\imath }$ . Let $\{ v^\gamma _i \mid i \in I_\gamma \}$ be a linear basis of $V(\gamma )$ and $\{ \phi ^\gamma _i \mid i \in I_\gamma \}$ its dual basis. By the Peter-Weyl type decomposition 5.2.3, $\mathbf {O}^{\imath }$ has the following linear basis:

$$\begin{align*}\{ c^\gamma_{v^\gamma_i, \phi^\gamma_j} \mid i,j \in I_\gamma \}. \end{align*}$$

We can formally express these basis elements as

$$\begin{align*}c^\gamma_{v^\gamma_i, \phi^\gamma_j} = \sum_{b \in \dot{\mathbf{B}}^{\imath}} c^\gamma_{v^\gamma_i, \phi^\gamma_j}(b) \delta_b. \end{align*}$$

Since $c^\gamma _{v^\gamma _i,\phi ^\gamma _j}(b) = 0$ for almost all $b \in \dot {\mathbf {B}}^{\imath }$ (Lemma 5.1.1), the right-hand side is a finite sum. This proves the assertion.