Canonical bases arising from quantum symmetric pairs

Abstract

We develop a general theory of canonical bases for quantum symmetric pairs $({\mathbf{U}}, {\mathbf{U}}^\imath )$ with parameters of arbitrary finite type. We construct new canonical bases for the finite-dimensional simple ${\mathbf{U}}$-modules and their tensor products regarded as ${\mathbf{U}}^\imath$-modules. We also construct a canonical basis for the modified form of the $\imath$quantum group ${\mathbf{U}}^\imath$. To that end, we establish several new structural results on quantum symmetric pairs, such as bilinear forms, braid group actions, integral forms, Levi subalgebras (of real rank one), and integrality of the intertwiners.

Appendix A. Integrality of the intertwiners of real rank one

The goal of this appendix is to provide a proof of Theorem 5.3(1) that the intertwiner $\Upsilon$ lies in (the completion of) the integral form ${}_\mathcal {A}{\mathbf{U}}^+$ for all quantum symmetric pairs of real rank one; see Table 1.

We shall first establish the integrality of $\Upsilon$ in type AIV and then $\hbox {AIII}_{11}$, which has the involution $\tau |_{\mathbb {I}_{\circ }} \ne 1$. These two types are easy and similar to the special case treated in [7] (denoted by ${\mathbf{U}}^\jmath$ therein). The integrality of $\Upsilon$ for type $\hbox {AI}_1$ was essentially known in [7, Lemma 4.6]. Then we will establish some general properties of $\Upsilon$ for the remaining types with $\tau |_{\mathbb {I}_{\circ }}=1$. Ultimately it requires a tedious type-by-type analysis to complete the proof for all types with $\tau |_{\mathbb {I}_{\circ }}=1$.

1.1 Type AIV of rank n

We recall the Satake diagram of type AIV from Table 1:

Proposition A.1

In type AIV, we have $\Upsilon _\mu \in {}_\mathcal {A}{\mathbf{U}}^+$ for any $\mu \in {{\mathbb {N}}}[\mathbb {I}]$.

Proof

Recall that $B_{1} = F_{1} + \varsigma _1 \texttt {T}_{w_\bullet } (E_{n}) {\widetilde{K}}^{-1}_1$ and $B_{n} = F_{n} + \varsigma _n \texttt {T}_{w_\bullet } (E_{1}) {\widetilde{K}}^{-1}_n$. Note that

$$\begin{aligned} F_{1} \texttt {T}_{w_\bullet } (E_{n}) {\widetilde{K}}^{-1}_1= & {} q_1^{-2}\texttt {T}_{w_\bullet } (E_{n}) {\widetilde{K}}^{-1}_1 F_1,\\ F_{n} \texttt {T}_{w_\bullet } (E_{1}) {\widetilde{K}}^{-1}_n= & {} q_n^{-2}\texttt {T}_{w_\bullet } (E_{1}) {\widetilde{K}}^{-1}_n F_n. \end{aligned}$$

Introduce the divided powers $B_i^{(a)} = B_i^a /[a]!$, for all $i \in \mathbb {I}$ and $a \in {{\mathbb {N}}}$. Then we have

$$\begin{aligned} B_1^{(a)}= & {} \sum _{s+t =a} q_1^{st}F^{(s)}_1 \big (\texttt {T}_{w_\bullet } (E_{n}) {\widetilde{K}}^{-1}_1\big )^{(t)},\nonumber \\ B_n^{(a)}= & {} \sum _{s+t =a} q_n^{st}F^{(s)}_n \big (\texttt {T}_{w_\bullet } (E_{1}) {\widetilde{K}}^{-1}_n\big )^{(t)}. \end{aligned}$$

(A.1)

So $B_i^{(a)}$ are $\psi _{\imath }$-invariant and integral (i.e. $B_i^{(a)} \in {}_\mathcal {A}{\mathbf{U}}$), for all $i \in \mathbb {I}$. Now for any $\lambda \in X^+$ and $x \in {}_\mathcal {A}L(\lambda )$, we can write

$$\begin{aligned} x =\sum c(a_1, \dots , a_k) B^{(a_1)}_{i_1} \dots B^{(a_k)}_{i_k} \eta _{\lambda }, \quad \text {for } c(a_1, \dots , a_k) \in \mathcal {A}. \end{aligned}$$

By Corollary 5.2 and using $\psi _{\imath }=\Upsilon \psi$ from (5.1), we have $\Upsilon (x) =\psi _{\imath }( \psi (x) ) \in {}_\mathcal {A}L(\lambda )$. Taking $x = \xi$ (the lowest weight vector) and $\lambda \gg 0$, we have $\Upsilon _\mu \in {}_\mathcal {A}{\mathbf{U}}^+$ for any $\mu$. $\square$

1.2 Type $\hbox {AIII}_{11}$

Recall the Satake diagram of type $\hbox {AIII}_{11}$ of real rank one from Table 1:

Note that the underline Dynkin diagram is not irreducible. We have $B_i = F_i + \varsigma _i E_{j} {\widetilde{K}}^{-1}_i$ for $1\le i \ne j \le 2$. Defining the divided powers $f_i^{(a)} = f_i^a /[a]!$ as usual, we have

$$\begin{aligned} B_i^{(a)} = \sum _{s+t =a} q_i^{st}F^{(s)}_i (E_{j} {\widetilde{K}}^{-1}_i)^{(t)} \in {}_\mathcal {A}{\mathbf{U}}. \end{aligned}$$

Now we are in a position to use (and choose to omit) the same argument as for Proposition A.1 to obtain the following.

Proposition A.2

In type $\hbox {AIII}_{11}$, we have $\Upsilon _\mu \in {}_\mathcal {A}{\mathbf{U}}^{+}$    for all $\mu$.

Remark A.3

The integrality of the standard divided powers $B_i^{(a)} \in {}_\mathcal {A}{\mathbf{U}}$ for $i \in \mathbb {I}_{\circ }$ in types AIV and $\hbox {AIII}_{11}$ distinguishes these two types from the remaining ones.

1.3 Type $\hbox {AI}_1$

Recall the Satake diagram of type $\hbox {AI}_{1}$ from Table 1:

Since there is only one element in $\mathbb {I}$, we shall drop the index 1 and write $B= F + q^{-1} E {K}^{-1} + \kappa {K}^{-1}$. This is the only real rank one case when $\kappa$ can be non-zero. Set $\Upsilon = \sum _{c \ge 0} \Upsilon _c$ where $\Upsilon _c= \gamma _c E^{(c)}$. Proposition A.11 below and its proof are adapted from [7, Lemma 4.6] (where $\kappa =1$).

Proposition A.4

In type $\hbox {AI}_1$, we have $\Upsilon _c \in {}_\mathcal {A}{\mathbf{U}}^+$ for all $c \ge 0$.

Proof

Recall from (3.6) we have $\overline{\kappa } =\kappa$. Equation (4.9) implies that

$$\begin{aligned} (F + q^{-1} E {K}^{-1} + \kappa {K}^{-1}) \Upsilon = \Upsilon ( F + q E {K} + \kappa {K}), \end{aligned}$$

which can be rewritten as

$$\begin{aligned} \gamma _{c+1} = -(q-q^{-1}) q^{-c} (q [c] \gamma _{c-1} + \kappa \gamma _c). \end{aligned}$$

It follows by induction on c that $\gamma _c \in \mathcal {A}$, since by definition we know $\gamma _0 =1$.

$\square$

1.4 Generalities when $\tau |_{\mathbb {I}_{\circ }} = 1$

In this subsection, we assume that the Satake diagrams are real rank one of types with $\tau |_{\mathbb {I}_{\circ }} =1$ and $\mathbb {I}_{\bullet }\ne \emptyset$ (i.e., of types $\hbox {AII}_3$, BII, CII, DII, FII); see Table 1. In all these cases, we have the parameters $\kappa _i=0$, for all $i\in \mathbb {I}_{\circ }$. Let

$$\begin{aligned} \mathbb {I}_{\circ }= \{ \mathbf i \}. \end{aligned}$$

Then following Theorem 4.8, we have

$$\begin{aligned} \Upsilon = \sum _{c \in {{\mathbb {N}}}} \Upsilon _{c}, \end{aligned}$$

(A.2)

where $\Upsilon _c = \Upsilon _{c(w_\bullet \mathbf{i} +\mathbf{i})}$ has weight $c(w_\bullet \mathbf{i} +\mathbf{i})$. Note that (4.9) implies that $\psi _{\imath }(B_\mathbf{i}) \Upsilon = \Upsilon \psi (B_\mathbf{i})$, that is,

$$\begin{aligned} \big (F_{\mathbf{i}} \!+ \!\varsigma _\mathbf{i} \texttt {T}_{w_\bullet }(E_{ \mathbf{i}}) {\widetilde{K}}^{-1}_\mathbf{i} \big ) \Upsilon = \Upsilon \big (F_{\mathbf{i}} \!+\! \varsigma ^{-1}_\mathbf{i} \psi (\texttt {T}_{w_\bullet }(E_{ \mathbf{i}})) {\widetilde{K}}_\mathbf{i}\big ), \, F_j \Upsilon \!=\! \Upsilon F_j (j \in \mathbb {I}_{\bullet }). \end{aligned}$$

Using [23, Proposition 3.16], we have (for $c \ge 1$ and $j \in \mathbb {I}_{\bullet }$)

$$\begin{aligned} r_\mathbf{i} (\Upsilon _c)&= -\big (q_\mathbf{i}-q_\mathbf{i}^{-1}\big ) \Upsilon _{c-1} \cdot \varsigma _\mathbf{i}^{-1} \cdot \psi (\texttt {T}_{w_\bullet }(E_{ \mathbf{i}})) \end{aligned}$$

(A.3)

$$\begin{aligned}&= -\big (q_\mathbf{i}-q_\mathbf{i}^{-1})(-q_\mathbf{i}\big )^{\langle \mathbf{i}, 2 \rho _{\bullet } \rangle } \cdot \varsigma _\mathbf{i}^{-1} \cdot \Upsilon _{c-1} \cdot \texttt {T}^{-1}_{w_\bullet }(E_{ \mathbf{i}}) , \nonumber \\ {}_\mathbf{i}r (\Upsilon _c)&= -\big (q_\mathbf{i}-q_\mathbf{i}^{-1}\big )\varsigma _\mathbf{i} \cdot q_\mathbf{i}^{\langle \mathbf{i}, w_\bullet \mathbf{i} \rangle } \cdot \texttt {T}_{w_\bullet }(E_{ \mathbf{i}}) \Upsilon _{c-1}, \end{aligned}$$

(A.4)

$$\begin{aligned} {}_j r (\Upsilon _c)&= r_j (\Upsilon _c) = 0. \end{aligned}$$

(A.5)

Recall the shorthand notation $w^{\bullet } = w_{\bullet } w_0.$ Let $\ell (w^{\bullet }) = k$. Note $w^{\bullet }$ is of the form

$$\begin{aligned} w^{\bullet } = s_{i_1} s_{i_2} \dots s_{i_{k-1}} s_{i_k}, \quad i_1 = i_k =\mathbf{i} \in \mathbb {I}_{\circ }. \end{aligned}$$

Introduce a shorthand notation $\texttt {T}_{i_1i_2 \dots i_{k-1}} =\texttt {T}_{i_1} \texttt {T}_{i_2} \dots \texttt {T}_{i_{k-1}}$. Applying Proposition 4.15, we have

$$\begin{aligned} \Upsilon _c = \sum \gamma _c(c_1, c_2, \dots , c_k) E^{(c_1)}_\mathbf{i} \texttt {T}_\mathbf{i} \big (E^{(c_2)}_{i_2}\big ) \cdots \Big ( \texttt {T}_{i_1i_2 \cdots i_{k-1}} \big (E^{(c_k)}_{\mathbf{i}}\big ) \Big ). \end{aligned}$$

(A.6)

We adopt the convention that $E^{(a)}_{j} = 0$ for any $j \in \mathbb {I}$, with $a < 0$, and $\gamma _{c}(c_1, c_2, \dots , c_k) = 0$ unless all $c_j \ge 0$. We shall write $\gamma _c = \gamma _c(c_1, c_2, \dots , c_k)$ when there is no need to specify each component.

The following lemma shall be used extensively in this section.

Lemma A.5

[11, Proposition 8.20] For any $w \in W$, if $w (i^{\prime }) = j^{\prime } \in X$ for $i, j \in \mathbb {I}$, then we have $\texttt {T}_w (E_i) = E_j$.

Lemma A.6

We have

$$\begin{aligned} \texttt {T}^{-1}_{w_\bullet } (E_{ \mathbf{i}}) = \texttt {T}_{i_1i_2 \cdots i_{k-1}} (E_{ \mathbf{i}}). \end{aligned}$$

Proof

We have ${w_\bullet } s_{i_1} s_{i_2} \cdots s_{i_{k-1}} ({\mathbf{i}'}) = -w_0 (\mathbf{i}') =\mathbf{i}'$. The lemma follows by Lemma A.5. $\square$

Lemma A.7

Let $c \ge 1$. If we have $\gamma _{c-1} (c_1, c_2, \dots , c_k) \in \mathcal {A}$ for all $(c_1, c_2, \dots , c_k)$, then $(1-q_\mathbf{i}^{-2})^{-c_1}\gamma _{c} (c_1, c_2, \dots , c_k) \!\in \!\mathcal {A}$ for all $(c_1, c_2, \dots , c_k)$ with $c_1 \ge 1$.

Proof

Using Lemma A.5 and (A.3), we have

$$\begin{aligned} r_\mathbf{i} (\Upsilon _c)&=\sum \gamma _c(c_1, c_2, \dots , c_k) q_\mathbf{i}^{c_1-1} q_\mathbf{i}^{\langle \mathbf{i}, c(w_\bullet \mathbf{i} + \mathbf{i}) - c_1 \mathbf{i} \rangle }E^{(c_1-1)}_\mathbf{i} \texttt {T}_\mathbf{i} \big (E^{(c_2)}_{i_2}\big ) \cdots \\&\quad \quad \times \Big ( \texttt {T}_{i_1i_2 \cdots i_{k-1}} \big (E^{(c_k)}_{\mathbf{i}}\big ) \Big ) \\&=-\big (q_\mathbf{i}-q_\mathbf{i}^{-1}\big )(-q_\mathbf{i})^{\langle \mathbf{i}, 2 \rho _{\bullet } \rangle } \cdot \varsigma _\mathbf{i}^{-1} \cdot \Upsilon _{c-1} \cdot \texttt {T}^{-1}_{w_\bullet }(E_{ \mathbf{i}}) \\&=-\big (q_\mathbf{i}-q_\mathbf{i}^{-1}\big )(-q_\mathbf{i})^{\langle \mathbf{i}, 2 \rho _{\bullet } \rangle } \varsigma _\mathbf{i}^{-1} \\&\quad \cdot \sum \gamma _{c-1}(c_1, c_2, \dots , c_k) [c_k+1]_\mathbf{i} E^{(c_1)}_\mathbf{i} \texttt {T}_\mathbf{i} (E^{(c_2)}_{i_2}) \cdots \Big ( \texttt {T}_{i_1i_2 \cdots i_{k-1}}\times \big (E^{(c_k+1)}_{\mathbf{i}}\big ) \Big ). \end{aligned}$$

Therefore we have for weight reason that, for $c_1\ge 1$,

$$\begin{aligned}&\gamma _c(c_1, c_2, \dots , c_k) q_\mathbf{i}^{c_1-1} q_\mathbf{i}^{\langle \mathbf{i}, c(w_\bullet \mathbf{i} + \mathbf{i}) - c_1 \mathbf{i} \rangle } \\&\quad = -\big (q_\mathbf{i}-q_\mathbf{i}^{-1}\big )(-q_\mathbf{i})^{\langle \mathbf{i}, 2 \rho _{\bullet } \rangle } \cdot \varsigma _\mathbf{i}^{-1} \gamma _{c-1}(c_1-1, c_2, \dots , c_k-1) [c_k]_\mathbf{i}, \end{aligned}$$

that is,

$$\begin{aligned} \begin{aligned}&\gamma _c(c_1, c_2, \dots , c_k) \\&\quad = -\big (q_\mathbf{i}-q_\mathbf{i}^{-1}\big )q_\mathbf{i}^{- \langle \mathbf{i}, c(w_\bullet \mathbf{i} + \mathbf{i}) - c_1 \mathbf{i} \rangle } q_\mathbf{i}^{1-c_1}(-q_\mathbf{i})^{\langle \mathbf{i}, 2 \rho _{\bullet } \rangle } \varsigma _\mathbf{i}^{-1} \gamma _{c-1}\\&\qquad \quad \times \,(c_1-1, c_2, \dots , c_k-1) [c_k]_\mathbf{i}. \end{aligned} \end{aligned}$$

(A.7)

It follows by an induction on $c_1$ that $(1-q_\mathbf{i}^{-2})^{-c_1} \gamma _c(c_1, c_2, \dots , c_k) \in \mathcal {A}$ as long as $c_1 \ge 1$. $\square$

Remark A.8

We proved a stronger result than just $\gamma _c(c_1, c_2, \dots , c_k) \in \mathcal {A}$ under our assumption. The importance shall be clear later in this section.

Our strategy is to prove that $\Upsilon _c \in {}_\mathcal {A}{\mathbf{U}}^+$ by induction on c. The base case at $c=0$ is always true since we have $\Upsilon _0 = 1$. For the induction step we shall compute the precise actions of $r_j$ on $\Upsilon _c$ for $j \in \mathbb {I}$ case by case. Then thanks to Lemma A.7, it suffices to prove that

$$\begin{aligned}&\gamma _c(c_1, c_2, \dots , c_k) \in \mathcal {A}\text { for all } c_i, \nonumber \\&\quad \text { if } (1-q_i^{-2})^{-c_1} \quad \gamma _{c} (c_1, c_2, \dots , c_k) \in \mathcal {A}\text { when } c_1 \ge 1. \end{aligned}$$

(A.8)

This is what we shall do later in this section case by case.

To facilitate the case-by-case analysis below, let us introduce some shorthand notations. For a sequence $i_1 i_2 \dots i_k$ with $i_j \in \mathbb {I}$, we shall often use the shorthand notation

$$\begin{aligned} \texttt {T}_{i_1i_2\dots i_k} = \texttt {T}_{i_1} \texttt {T}_{i_2} \dots \texttt {T}_{i_k}. \end{aligned}$$

In concrete cases below (with labelings as in Table 1), the sequence $i_1 \dots i_k$ is naturally partitioned into increasing and decreasing subsequences, and we shall insert indices $i_l$ to indicate the local maxima/minima of the sequence. For example, $\texttt {T}_{i_1 \dots i_l \dots i_k}$ means the subsequences $i_1 \dots i_l$ and $i_l \dots i_k$ are monotone, and $\texttt {T}_{i_1 \dots i_l \dots i_m \dots i_k}$ means the subsequences $i_1 \dots i_l$, $i_l \dots i_m$, and $i_m \dots i_k$ are monotone, and so on. Here is a concrete example which occurs in Type CII below: the shorthand ${2 \dots n \dots 1 \dots n \dots k}$, for some $1\le k \le n$, means ${2 \ 3 \dots n-1 \ n\ n-1 \dots 2\ 1 2 \dots n-1\ n \ n-1 \dots k}$.

For $x, y \in {\mathbf{U}}^+$, we write

$$\begin{aligned}{}[x,y]_{q^{-1}_i} = xy - q_i^{-1}yx. \end{aligned}$$

Since the ring $\mathcal {A}$ is invariant under multiplication by $q^a$ for any $a \in {{\mathbb {Z}}}$, we shall often use the notation $q^*$ to indicate q-powers without computing the precise exponent when it is irrelevant.

1.5 Type AII of rank 3

In this subsection we assume the quantum symmetric pair $({\mathbf{U}}, {{\mathbf{U}}^{\imath }})$ is of type AII. We label the Satake diagram as follows:

Take the reduced expression $w^{\bullet } = s_2 s_1 s_3 s_2$. Then we have

$$\begin{aligned} \Upsilon _{c} = \sum \gamma (c_1, c_2 , c_3, c_4) E^{(c_1)}_2 \cdot \texttt {T}_2 \big (E^{(c_2)}_1\big ) \cdot \texttt {T}_2 \big (E^{(c_3)}_3\big ) \cdot \texttt {T}_{213} \big (E^{(c_4)}_2\big ), \end{aligned}$$

where $c = \frac{1}{2}(c_1 + c_2 + c_3 +c_4)$. (For weight reason $\Upsilon _c = 0$ if c is not an integer.)

We then compute the actions of $r_{1}$ and $r_3$ on those root vectors. Note that we have

$$\begin{aligned} \texttt {T}_2 (E_1)&= E_2 E_1 - q^{-1} E_1 E_2;\\ \texttt {T}_2 (E_3)&= E_2 E_3 - q^{-1} E_3 E_2;\\ \texttt {T}_{213} (E_2)&= (E_2E_1- q^{-1} E_1 E_2)E_3 - q^{-1} E_3 (E_2 E_1 - q^{-1}E_1 E_2). \end{aligned}$$

The following lemmas follow from straightforward computation.

Lemma A.9

We have

$$\begin{aligned} E_1 \cdot \texttt {T}_2(E_1) = q^{-1} \texttt {T}_2(E_1) \cdot E_1&\quad \text { and } \quad E_3\cdot \texttt {T}_2(E_3) = q^{-1} \texttt {T}_2(E_3)\cdot E_3,\\ E_1 \cdot \texttt {T}_1(E_2) = q \texttt {T}_1(E_2) \cdot E_1&\quad \text { and } \quad E_3\cdot \texttt {T}_3(E_2) = q \texttt {T}_3(E_2)\cdot E_3, \end{aligned}$$

Lemma A.10

We have

• (1) $r_1 (\texttt {T}_2 (E^{(c_2)}_1)) = (1-q^{-2}) E_2 \cdot \texttt {T}_2 (E^{(c_2-1)}_1)$;

• (2) $r_3 (\texttt {T}_2 (E^{(c_2)}_1)) = 0$;

• (3) $r_1 (\texttt {T}_2 (E^{(c_3)}_3)) = 0$;

• (4) $r_3 (\texttt {T}_2 (E^{(c_3)}_3)) = (1-q^{-2}) E_2 \cdot \texttt {T}_2 (E^{(c_3-1)}_1)$;

• (5) $r_1 (\texttt {T}_{213} (E^{(c_4)}_2)) = (1-q^{-2}) \texttt {T}_2 (E _3) \cdot \texttt {T}_{213} (E^{(c_4-1)}_2)$;

• (6) $r_3 (\texttt {T}_{213} (E^{(c_4)}_2)) = (1-q^{-2}) \texttt {T}_2 (E _1) \cdot \texttt {T}_{213} (E^{(c_4-1)}_2)$.

Proposition A.11

In type AII, we have $\Upsilon _c \in {}_\mathcal {A}{\mathbf{U}}^+$ for all $c \ge 0$.

Proof

It suffices to prove the statement (A.8) by the general discussion in Sect. A.4. Let us assume

$$\begin{aligned} (1-q_i^{-2})^{-c_1}\gamma _{c} (c_1, c_2, \dots , c_k) \in \mathcal {A}\text { when } c_1 \ge 1. \end{aligned}$$

Since $r_1 (\Upsilon _c) = r_3 (\Upsilon _c) = 0$ by (A.5), we have

$$\begin{aligned} 0&= \frac{1}{1-q^{-2}} r_1 (\Upsilon _c) \\&= \sum \gamma _c (c_1, c_2, c_3, c_4) [c_3+1] E^{(c_1)}_2 \cdot \texttt {T}_2 \big (E^{(c_2)}_1\big ) \\&\quad \cdot \texttt {T}_2 \big (E^{(c_3+1)}_3\big ) \cdot \texttt {T}_{213} (E^{(c_4-1)}_2) \\&\quad +\, \sum \gamma _c (c_1, c_2, c_3, c_4) [c_1+1] q^{c_4 -c_3}E^{(c_1+1)}_2 \cdot \texttt {T}_2\big (E^{(c_2-1)}_1\big )\\&\quad \cdot \texttt {T}_2 (E^{(c_3)}_3) \cdot \texttt {T}_{213} \big (E^{(c_4)}_2\big ). \end{aligned}$$

It follows that

$$\begin{aligned} \gamma _c (c_1, c_2+1, c_3+1, c_4) [c_1+1] q^{c_4 -c_3} \!= \!-\! \gamma _c (c_1\!+\!1, c_2, c_3, c_4+1) [c_3+1] . \end{aligned}$$

Therefore we have

$$\begin{aligned} \gamma _c (0, c_2+1, c_3+1, c_4) = -q^{c_3-c_4} \gamma _c (1, c_2, c_3, c_4+1) [c_3+1] \in \mathcal {A}. \end{aligned}$$

It follows that $\gamma _c (c_1, c_2, c_3, c_4) \in \mathcal {A}$ if $c_1, c_2$ are not all zero. On the other hand, we have $\gamma _c (0, 0, c_3, c_4) =0$ for weight reason. The proposition follows. $\square$

Corollary A.12

We have $c_1 = c_4$ and $c_2 =c_3$ whenever $\gamma _c(c_1, c_2, c_3 ,c_4) \ne 0$.

1.6 Type DII of rank n $\ge 4$

In this subsection we assume the quantum symmetric pair $({\mathbf{U}}, {{\mathbf{U}}^{\imath }})$ is of type DII. We label the Satake diagram is as follows:

We take the reduced expression $w^{\bullet } = s_1 \cdot s_2 \cdots s_{n-2} \cdot s_{n-1} \cdot s_n \cdot s_{n-2} \cdots s_1$. Therefore we can write $\Upsilon _c$ as

$$\begin{aligned} \Upsilon _c = \sum \gamma _c(c_1, \dots , c_{2n-2}) \, E^{(c_{1})}_{1} \cdot \texttt {T}_{1}\big (E^{(c_{2})}_{2}\big )\cdots \big (\texttt {T}_{1\cdots 2} \big (E^{(c_{2n-2})}_{1}\big )\big ). \end{aligned}$$

For weight reason, we must have $\gamma _c(c_1, \dots , c_{2n-2}) =0$ unless $\sum ^{2n-2}_{i=1} c_i = c$.

Lemma A.13

For $k \ne 1$, we have

$$\begin{aligned} { r_{k}} (\texttt {T}_{1 \cdots i} (E^{(a)}_{i+1}) ) = {\left\{ \begin{array}{ll} (1-q^{-2}) \, \texttt {T}_{1 \cdots (i-1)}(E_i)\cdot \big (\texttt {T}_{1 \cdots i} \big (E^{(a-1)}_{i+1}\big ) \big ), &{}\text {if } k =i+1;\\ 0, &{}\text {if } k \ne i+1. \end{array}\right. } \end{aligned}$$

Proof

Let us first assume that $i \le n-2$. The proof is divided into three cases:

• (1) If $k \ge i+2$, it is clear that ${ r_{k}} (\texttt {T}_{1 \cdots i} (E_{i+1}) ) = 0$.

• (2) For $k \le i$, we have

  $$\begin{aligned} {r_k } \big (\texttt {T}_{1 \dots (k-1)} \texttt {T}_{k} \texttt {T}_{(k+1) \dots i} (E_{i+1}) \big )&= \,{r_k } \big (\texttt {T}_{1 \dots (k-1)} [E_k, \texttt {T}_{(k+1) \dots i} (E_{i+1})]_{q^{-1}} \big )\\&= \,{r_k } \big ( [ \texttt {T}_{1 \dots (k-1)} (E_k), \texttt {T}_{(k+1) \dots i}(E_{i+1})]_{q^{-1}} \big )\\&= q^{-1} {r_k } \big ( \texttt {T}_{1 \dots (k-1)}(E_k) \big ) \cdot \texttt {T}_{(k+1) \dots i} (E_{i+1}) \\&\quad -\, q^{-1} \, \texttt {T}_{(k+1) \dots i} (E_{i+1})\\&\quad \cdot {r_k } \big (\texttt {T}_{1 \dots (k-1)} (E_k) \big ) = 0. \end{aligned}$$

• (3) For $k = i+1$, we have (for $i\le n-2$)

  $$\begin{aligned} \texttt {T}_{1\dots i} (E_{i+1}) = \texttt {T}_{1\dots (i-1)} ([E_i, E_{i+1}]_{q^{-1}}) = [\texttt {T}_{1\dots (i-1)}(E_i) , E_{i+1}]_{q^{-1}}. \end{aligned}$$

  It follows that

  $$\begin{aligned} { r_{i+1}} (\texttt {T}_{1\dots i} (E_{i+1}) ) = (1-q^{-2}) \texttt {T}_{1\dots (i-1)}(E_i). \end{aligned}$$

  Since $\texttt {T}_i(E_{i+1}) \cdot E_i = q^{-1} E_i \cdot \texttt {T}_{i} (E_{i+1})$, we have

  $$\begin{aligned} { r_{i+1}} \big (\texttt {T}_{1\dots i} (E^{(a)}_{i+1}) \big ) = (1-q^{-2}) \, \texttt {T}_{1\dots (i-1)}(E_i)\cdot \texttt {T}_{1\dots i} (E^{(a-1)}_{i+1}). \end{aligned}$$

The case $i = n-1$ is entirely similar to the case $i=n-2$, since $\texttt {T}_{n-1} (E_{n}) = E_n$. The lemma follows. $\square$

Lemma A.14

For $i \ne n, n-1$ and $k \ne 1$, we have

$$\begin{aligned}&{ r _{k}} \big (\texttt {T}_{1 \dots n \dots i} (E^{(a)}_{i-1}) \big )\\&\quad =\, {\left\{ \begin{array}{ll} (1 -q^{-2} )\texttt {T}_{1 \dots n \dots (i+1)} (E_{i}) \cdot \texttt {T}_{1 \dots n \dots i} (E^{(a-1)}_{i-1}), &{}\text { if } \,k = i \ne n-2;\\ (1 -q^{-2} )\texttt {T}_{1 \dots n} (E_{n-2}) \cdot \texttt {T}_{1 \dots n(n-2)} (E^{(a-1)}_{n-3}), &{}\text { if }\, k = i = n-2;\\ 0, &{}\text { if } k \ne i. \end{array}\right. } \end{aligned}$$

Proof

The computation is divided into six cases.

• (1) For $k \le i-2$, we have

  $$\begin{aligned} \texttt {T}_{1 \dots n \dots i} (E_{i-1}) = [\texttt {T}_{1 \dots (k-1)} (E_{k}), \texttt {T}_{(k+1) \dots n \dots i} (E_{i-1}) ]_{q^{-1}}. \end{aligned}$$

  Therefore we have

  $$\begin{aligned} r_k \Big ( \texttt {T}_{1 \dots n \dots i} (E_{i-1}) \Big )&=r_{k} \Big ( [\texttt {T}_{1 \dots (k-1)} (E_{k}), \texttt {T}_{(k+1) \dots n \dots i}(E_{i-1}) ]_{q^{-1}} \Big ) \\&= q^{-1} r_k (\texttt {T}_{1 \dots (k-1)} (E_{k})) \cdot \Big (\texttt {T}_{(k+1) \dots n \dots i}(E_{i-1})\Big ) \\&\qquad -\, q^{-1} \Big ( \texttt {T}_{(k+1) \dots n \dots i}(E_{i-1})\Big ) {\cdot } r_k (\texttt {T}_{1 \dots (k-1)} (E_{k})) \,{=}\, 0, \end{aligned}$$

  since $r_k (\texttt {T}_{1 \dots (k-1)} (E_{k}))$ and $\texttt {T}_{(k+1) \dots n \dots i}(E_{i-1})$ commute.

• (2) For $k=i-1$, we consider

  $$\begin{aligned} \begin{aligned} \texttt {T}_{1 \dots n \dots i} (E_{i-1})&=\texttt {T}_{1 \dots n \dots (i+1)} ([E_{i},E_{i-1}]_{q^{-1}})\\&= [\texttt {T}_{1 \dots n \dots (i+1)} (E_{i}) , \texttt {T}_{1 \dots i} (E_{i-1}) ]_{q^{-1}}\\&= [\texttt {T}_{1 \dots n \dots (i+1)} (E_{i}) , E_i]_{q^{-1}}. \end{aligned} \end{aligned}$$

  (A.9)

  Then since $i-1 \le i+1 -2$, by Case  (1) we have

  $$\begin{aligned} {r_{i-1}}&\big (\texttt {T}_{1 \dots n \dots i} (E_{i-1}) \big ) \\&= q^{-1} {r_{i-1}} (\texttt {T}_{1 \dots n \dots (i+1)} (E_{i})) \cdot E_{i} - q^{-1} E_{i} \cdot {r_{i-1}} (\texttt {T}_{1 \dots n \dots (i+1)} (E_{i})) \\&= 0. \end{aligned}$$

• (3) For $k=i$, following (A.9) we have

  $$\begin{aligned} {r_{i}} (\texttt {T}_{1 \dots n \dots i} (E_{i-1}))&= \texttt {T}_{1 \dots n \dots (i+1)} (E_{i}) - q^{-2} \texttt {T}_{1 \dots n \dots (i+1)} (E_{i}) \\&= (1-q^{-2}) \texttt {T}_{1 \dots n \dots (i+1)} (E_{i}). \end{aligned}$$

  More generally we have

  $$\begin{aligned} {r_{i}} \big (\texttt {T}_{1 \dots n \dots i} (E^{(a)}_{i-1}) \big )&=(1-q^{-2}) \texttt {T}_{1 \dots n \dots (i+1)} (E_{i}) \cdot \texttt {T}_{1 \dots n \dots i}(E^{(a-1)}_{i-1}). \end{aligned}$$

• (4) For $n-3 \ge k \ge i+1$, we consider

  $$\begin{aligned} \texttt {T}_{1 \dots n \dots i} (E_{i-1})&= \texttt {T}_{1 \dots n \dots (k+1)} \big ( [E_{k}, \texttt {T}_{(k-1) \dots i}(E_{i-1}) ]_{q^{-1}} \big )\\&= [\texttt {T}_{1 \dots n \dots (k+1)} (E_{k}), \texttt {T}_{k \dots (i+1)} (E_{i})]_{q^{-1}}. \end{aligned}$$

  Note that ${ r_k} (\texttt {T}_{k \cdots (i+1)}(E_{i})) = 0$ unless $k=i$. Therefore by Case (2) we have

  $$\begin{aligned} r_{k} (\texttt {T}_{1 \dots n \dots i} (E_{i-1})) =0. \end{aligned}$$

• (5) For $k =n-2$, we consider that

  $$\begin{aligned} \texttt {T}_{1 \dots n \dots i} (E_{i-1})&= \texttt {T}_{1 \dots n } \big ( [E_{k}, \texttt {T}_{(k-1) \dots i}(E_{i-1}) ]_{q^{-1}} \big ) \\&= [\texttt {T}_{1 \dots n } (E_{k}), \texttt {T}_{k \dots (i+1)} (E_{i})]_{q^{-1}}. \end{aligned}$$

  Note that ${ r_k} \big (\texttt {T}_{k \dots (i+1)}(E_{i}) \big ) = 0$ unless $k=i$. Therefore by Case  (2) we have

  $$\begin{aligned} r_{k} (\texttt {T}_{1 \dots n \dots i} (E_{i-1})) =0. \end{aligned}$$

• (6) For $k=n-1$ (the case $k=n$ is similar), we have (true for $i = n-2$ as well)

  $$\begin{aligned} \texttt {T}_{1 \cdots n \dots i}(E_{i-1})&= \texttt {T}_{1 \dots (n-1)} \Big ( [E_{n}, \texttt {T}_{(n-2) \dots i} (E_{i-1}) ]_{q^{-1}} \Big )\\&= [ \texttt {T}_{1 \dots (n-1)} (E_{n}), \texttt {T}_{(n-1) \dots (i+1)} (E_{i})]_{q^{-1}}. \end{aligned}$$

  Note that since by Lemma A.13

  $$\begin{aligned} r_{n-1} \Big (\texttt {T}_{1 \dots (n-1)} (E_{n}) \Big ) = r_{n-1} \Big ( \texttt {T}_{(n-1) \dots (i+1)}(E_{i}) \Big ) = 0, \end{aligned}$$

  we have

  $$\begin{aligned} r_{n-1} (\texttt {T}_{1 \dots n \dots i} (E_{i-1})) =0. \end{aligned}$$

This completes the proof. $\square$

Remark A.15

The computation for (1)–(4) in the proof of Lemma A.14 is essentially a type A computation, and will appear very often for the other cases as well.

Lemma A.16

For $k \ne 1$, we have

$$\begin{aligned} { r _{k}} (\texttt {T}_{1\cdots n}(E^{(a)}_{n-2}))&= {\left\{ \begin{array}{ll} (1 -q^{-2} ) \texttt {T}_{1\dots (n-1)}(E_{n}) \cdot \texttt {T}_{1\dots n}(E^{(a-1)}_{n-2}), &{}\text {if } k = n-1;\\ (1 -q^{-2} ) \texttt {T}_{1 \dots (n-2) n}(E_{n-1}) \cdot \texttt {T}_{1\dots n}(E^{(a-1)}_{n-2}), &{}\text {if } k = n;\\ 0, &{}\text {otherwise. } \end{array}\right. } \end{aligned}$$

Proof

Note that

$$\begin{aligned}&\texttt {T}_{1\dots n} (E_{n-2})= \texttt {T}_{1\dots (n-3)} ( q^{-2} E_{n} E_{n-1} E_{n-2} -q^{-1}E_{n} E_{n-2} E_{n-1}\\&\quad - \,q^{-1} E_{n-1} E_{n-2} E_{n} + E_{n-2} E_{n-1} E_{n}). \end{aligned}$$

Therefore we have $r_k ( \texttt {T}_{1\cdots n} (E_{n-2})) = 0$ for $k < n-2$, thanks to Lemma A.13. It is easy to see $r_{n-2} ( \texttt {T}_{1\cdots n} (E_{n-2})) = 0$ as well by a direct computation using Lemma A.13.

On the other hand, we have

$$\begin{aligned} { r_{n-1}} (\texttt {T}_{1\cdots n} (E_{n-2}))&= q^{-3} E_{n} \cdot \texttt {T}_{1\cdots (n-3)} (E_{n-2}) -q^{-1}E_{n} \cdot \texttt {T}_{1\cdots (n-3)} (E_{n-2}) \\&\quad -\, q^{-2} \texttt {T}_{1\cdots (n-3)} (E_{n-2}) \cdot E_{n} + \texttt {T}_{1\cdots (n-3)}(E_{n-2}) \cdot E_{n} \\&= -q^{-2} \texttt {T}_{1\cdots (n-2)}(E_{n}) + \texttt {T}_{1\cdots (n-2)} (E_{n})\\&= (1-q^{-2}) \texttt {T}_{1\cdots (n-1)} (E_{n}). \end{aligned}$$

Now since $E_n \cdot \texttt {T}_n (E_{n-2}) = q \texttt {T}_n (E_{n-2}) \cdot E_n$, we have

$$\begin{aligned} { r _{n-1}} ( \texttt {T}_{1\cdots n} (E^{(a)}_{n-2})) = (1 -q^{-2} ) \texttt {T}_{1\cdots (n-1)} (E_{n}) \cdot \texttt {T}_{1\cdots n} (E^{(a-1)}_{n-2}). \end{aligned}$$

The computation of ${ r_{n}} ( \texttt {T}_{1\cdots n} (E^{(a)}_{n-2}))$ is entirely similar. The lemma follows. $\square$

Table 4 Satake diagrams of irreducible symmetric pairs

Proposition A.17

For quantum symmetric pairs of type DII of rank n $\ge 4$, we have $\Upsilon _c \in {_\mathcal {A}{\mathbf{U}}^+}$ for all $c \ge 0$.

Proof

Recall by the general discussion in Sect. A.4 it suffices to prove the following statement (which implies (A.8)):

$$\begin{aligned} \gamma _{c}(c_1,\dots ,c_{2n-2}) \in \mathcal {A}\text { for all } c_i, \text { if }\gamma _{c}(c_1,\dots ,c_{2n-2}) \in \mathcal {A}\text { when } c_1 >0. \end{aligned}$$

We compare the coefficient of the following terms in the identity $r_{k}(\Upsilon _c)=0$:

$$\begin{aligned} E^{(c_{1})}_{1} \cdot \texttt {T}_{1}(E^{(c_{2})}_{2})\cdots (\texttt {T}_{1\cdots 2} (E^{(c_{2n-2})}_{1})) \text { with } c_{k-1}=1, c_j=0 \text { for } j<k-1. \end{aligned}$$

We obtain that

$$\begin{aligned}&(1-q^{-2}) \gamma _c (0,\dots , c_{k-1}=0, c_k,\dots ) \in (1-q^{-2})\\&\quad \sum \gamma _c (\dots , c_{k-1}=1, c_k-1,\dots )\cdot \mathcal {A}. \end{aligned}$$

Therefore thanks to Lemma A.7 for the base case, we have inductively:

$$\begin{aligned} \gamma _c (c_1,\dots ) \in \mathcal {A}, \text { if } c_{k} >0. \end{aligned}$$

(A.10)

The proposition follows. $\square$

1.7 Other types

The proof of part (1) of Theorem 5.3 for QSP of type BII of rank $n\ge 2$, type CII of rank $n\ge 3$, and type FII follows from entirely similar computation as type DII of rank $n \ge 4$. The precise details can be found in the (longer) appendix of the arXiv Version  1 of this paper, and shall be omitted here.