Higher Level q-Oscillator Representations for $U_q(C_n^{(1)}),U_q(C^{(2)}(n+1))$ and $U_q(B^{(1)}(0,n))$

Abstract

We introduce higher level q-oscillator representations for the quantum affine (super)algebras of type $C_n^{(1)},C^{(2)}(n+1)$ and $B^{(1)}(0,n)$. These representations are constructed by applying the fusion procedure to the level one q-oscillator representations which were obtained through the studies of the tetrahedron equation. We prove that these higher level q-oscillator representations are irreducible. For type $C_n^{(1)}$ and $C^{(2)}(n+1)$, we compute their characters explicitly in terms of Schur polynomials.

Appendices

Appendix A. Twistor

In this appendix, we prove Propositions 3.9 and 3.14. We first review the twistor introduced in [5] that relate quantum groups to quantum supergroups. Then we use it to relate the q-oscillator representation of $U_q(D_{n+1}^{(2)})$ in [15] to a representation of $U_q(C^{(2)}(n+1))$. An advantage to do so is that in the latter case we can take a classical limit $q\rightarrow 1$. We also obtain a representation of $U_q(B^{(1)}(0,n))$ from the q-oscillator representation of $U_q(A_{2n}^{(2)\dagger })$, where $A_{2n}^{(2)\dagger }$ is the same Dynkin diagram as $A_{2n}^{(2)}$ in [9] but the labeling of nodes are opposite.

1.1 The twistor of the covering quantum group

We review the covering quantum group and the twistor map introduced in [5]. Our notations for a Cartan datum is closer to Kac’s book [9]. Let I be the index set of the Dynkin diagram, $\{\alpha _i\}_{i\in I}$ the set of simple roots, $(a_{ij})_{i,j\in I}$ the Cartan matrix. The symmetric bilinear form $(\cdot ,\cdot )$ on the weight lattice is normalized so that it satisfies $d_i=(\alpha _i,\alpha _i)/2\in {\mathbb {Z}}$ for any $i\in I$. It is also assumed that $a_{ij}\in 2{\mathbb {Z}}$ if $d_i\equiv 1$ (mod 2) and $j\in I$. The parity function p(i) taking values in $\{0,1\}$ is consistent with $d_i$, namely, $p(i)\equiv d_i$ (mod 2). We set $q_i=q^{d_i},\pi _i=\pi ^{d_i}$.

Let $q,\pi$ be indeterminates and $\mathbf{i}=\sqrt{-1}$. For a ring R with 1, we set $R^\pi =R[\pi ]/(\pi ^2-1)$. The covering quantum group ${\mathbf {U}}$ associated to a Cartan datum is the ${\mathbb {Q}}^\pi (q,\mathbf{i})$-algebra with generators $E_i,F_i,K_i^{\pm 1},J_i^{\pm 1}$ for $i\in I$ subject to the following relations:

$$\begin{aligned}&J_iJ_j=J_jJ_i,\quad K_iK_j=K_jK_i,\quad J_iK_j=K_jJ_i,\\&K_iK_i^{-1}=K_i^{-1}K_i=J_iJ_i^{-1}=J_i^{-1}J_i=J_i^2=1,\\&J_iE_j=\pi ^{a_{ij}}E_jJ_i,\quad J_iF_j=\pi ^{-a_{ij}}F_jJ_i,\\&K_iE_j=q^{a_{ij}}E_jK_i,\quad K_iF_j=q^{-a_{ij}}F_jK_i,\\&E_iF_j-\pi ^{p(i)p(j)}F_jE_i=\delta _{ij}\frac{J_iK_i-K_i^{-1}}{\pi _iq_i-q_i^{-1}},\\&\sum _{l=0}^{1-a_{ij}}(-1)^l\pi ^{l(l-1)p(i)/2+lp(i)p(j)} {1-a_{ij}\brack l}_{q_i,\pi _i}E_i^{1-a_{ij}-l}E_jE_i^l=0\quad (i\ne j),\\&\sum _{l=0}^{1-a_{ij}}(-1)^l\pi ^{l(l-1)p(i)/2+lp(i)p(j)} {1-a_{ij}\brack l}_{q_i,\pi _i}F_i^{1-a_{ij}-l}F_jF_i^l=0\quad (i\ne j). \end{aligned}$$

Remark A.1

We changed the notations from [5]. We replaced v with q, ${\mathbf {t}}$ with $\mathbf{i}$, and $J_{d_ii}, K_{d_ii}, T_{d_ii}$ with $J_i, K_i, T_i$.

We extend $\mathbf{U}$ by introducing the generators $T_i,\Upsilon _i$ for $i\in I$. They commute with each other and with $J_i,K_i$. They also have the commutation relations with $E_i,F_i$ as

$$\begin{aligned} T_iE_j=\mathbf{i}^{d_ia_{ij}}E_jT_i,\quad T_iF_j=\mathbf{i}^{-d_ia_{ij}}F_jT_i,\quad \Upsilon _iE_j=\mathbf{i}^{\phi _{ij}}E_j\Upsilon _i,\quad \Upsilon _iF_j=\mathbf{i}^{-\phi _{ij}}F_j\Upsilon _i, \end{aligned}$$

where

$$\begin{aligned} \phi _{ij}={\left\{ \begin{array}{ll} d_ia_{ij}&{}\text {if }i>j,\\ d_i&{}\text {if }i=j,\\ -2p(i)p(j)&{}\text {if }i<j. \end{array}\right. } \end{aligned}$$

We denote this extended algebra by $\widehat{{\mathbf {U}}}$.

Theorem A.2

[5] There is a ${\mathbb {Q}}(\mathbf{i})$-algebra automorphism ${\widehat{\Psi }}$ on $\widehat{{\mathbf {U}}}$ such that

$$\begin{aligned}&E_i\mapsto \mathbf{i}^{-d_i}\Upsilon _i^{-1}T_iE_i,&\quad&F_i\mapsto F_i\Upsilon _i,&\quad&K_i\mapsto T_i^{-1}K_i,\\&J_i\mapsto T_i^2J_i,&\quad&T_i\mapsto T_i,&\quad&\Upsilon _i\mapsto \Upsilon _i,\\&q\mapsto \mathbf{i}^{-1}q,&\quad&\pi \mapsto -\pi . \end{aligned}$$

1.2 Image of the twistor ${\widehat{\Psi }}$

We apply the twistor ${\widehat{\Psi }}$ given in the previous subsection for the Cartan datum corresponding to $B_n$, namely, $I=\{1,2,\ldots ,n\}$ and the Cartan matrix is given by

$$\begin{aligned} (a_{ij})= \begin{pmatrix} 2&{}-1&{}\\ -1&{}2&{}-1\\ &{}-1&{} \\ &{}&{}&{}\ddots \\ &{}&{}&{}&{}&{}-1\\ &{}&{}&{}&{}-1&{}2&{}-1\\ &{}&{}&{}&{}&{}-2&{}2 \end{pmatrix} \end{aligned}$$

Through it, we are to regard the q-oscillator representation $\mathcal {W}=\bigoplus _{\mathbf{m}}{\mathbb {Q}}(q^{\frac{1}{2}})|\mathbf{m}\rangle$ of $U_q(B_n)$, the subalgebra of $U_q(D_{n+1}^{(2)})$ generated by $e_i,f_i,k_i$ for $i\in I\setminus \{0\}$, given in [15, Proposition 1] as a representation of $U_q(osp_{1|2n})$. Although we normalized the symmetric bilinear form on the weight lattice so that $(\alpha _i,\alpha _i)\in 2{\mathbb {Z}}$ for any $i\in I$ in the previous subsection, we renormalize it so that $(\alpha _n,\alpha _n)=1$ to adjust it to the notations in [15]. The generators $T_i,\Upsilon _i$ are represented on $\mathcal {W}$ as

$$\begin{aligned} T_i|\mathbf{m}\rangle ={\left\{ \begin{array}{ll} \mathbf{i}^{2(m_{i+1}-m_i)}|\mathbf{m}\rangle &{}(1\le i<n)\\ \mathbf{i}^{-2m_n}|\mathbf{m}\rangle &{}(i=n) \end{array}\right. },\quad \Upsilon _i|\mathbf{m}\rangle ={\left\{ \begin{array}{ll} \mathbf{i}^{-2m_i}|\mathbf{m}\rangle &{}(1\le i<n)\\ \mathbf{i}^{|\mathbf{m}|-2m_n}|\mathbf{m}\rangle &{}(i=n) \end{array}\right. }. \end{aligned}$$

Let $u_i$ ($i\in I$, $u=e,f,k$) be the generators of $U_q(B_n)$ ($\pi =1$) and ${\bar{u}}_i={\widehat{\Psi }}(u_i)$ be the image ($\pi =-1$) of the twistor ${\widehat{\Psi }}$. Then ${\bar{u}}_i$ satisfy the relations for $U_{{\bar{q}}}(osp_{1|2n})$ where ${\bar{q}}^{\frac{1}{2}}=\mathbf{i}^{-1}q^{\frac{1}{2}}$. On the space $\mathcal {W}$, they act as follows:

$$\begin{aligned} {\bar{e}}_i|\mathbf{m}\rangle&=\mathbf{i}^{2m_{i+1}}[m_i]|\mathbf{m}-\mathbf{e}_i+\mathbf{e}_{i+1}\rangle ,\\ {\bar{f}}_i|\mathbf{m}\rangle&=\mathbf{i}^{-2m_i}[m_{i+1}]|\mathbf{m}+\mathbf{e}_i-\mathbf{e}_{i+1}\rangle ,\\ {\bar{k}}_i|\mathbf{m}\rangle&=\mathbf{i}^{2m_i-2m_{i+1}}q^{-m_i+m_{i+1}}|\mathbf{m}\rangle ,\\ {\bar{e}}_n|\mathbf{m}\rangle&=\kappa \, \mathbf{i}^{1-|\mathbf{m}|}[m_n]|\mathbf{m}-\mathbf{e}_n\rangle ,\\ {\bar{f}}_n|\mathbf{m}\rangle&=\mathbf{i}^{|\mathbf{m}|-2m_n}|\mathbf{m}+\mathbf{e}_n\rangle ,\\ {\bar{k}}_n|\mathbf{m}\rangle&=\mathbf{i}^{2m_n+1}q^{-m_n-\frac{1}{2}}|\mathbf{m}\rangle , \end{aligned}$$

where $1\le i<n$, $\kappa =(q+1)/(q-1)$.

By introducing the actions of ${\bar{e}}_0,{\bar{f}}_0,{\bar{k}}_0$, we want to make $\mathcal {W}$ a representation of the quantum affine superalgebra associated to $C^{(2)}(n+1)$ or $B^{(1)}(0,n)$. For the former, we set

$$\begin{aligned} {\bar{e}}_0|\mathbf{m}\rangle&=x\,\mathbf{i}^{2m_1-|\mathbf{m}|}|\mathbf{m}+\mathbf{e}_1\rangle ,\\ {\bar{f}}_0|\mathbf{m}\rangle&=x^{-1}\kappa \,\mathbf{i}^{|\mathbf{m}|+1}[m_1]|\mathbf{m}-\mathbf{e}_1\rangle ,\\ {\bar{k}}_0|\mathbf{m}\rangle&=\mathbf{i}^{-2m_1-1}q^{m_1+\frac{1}{2}}|\mathbf{m}\rangle , \end{aligned}$$

and for the latter

$$\begin{aligned} {\bar{e}}_0|\mathbf{m}\rangle&=x(-1)^{|\mathbf{m}|}|\mathbf{m}+2\mathbf{e}_1\rangle ,\\ {\bar{f}}_0|\mathbf{m}\rangle&=x^{-1}(-1)^{|\mathbf{m}|}\frac{[m_1][m_1-1]}{[2]^2}|\mathbf{m}-2\mathbf{e}_1\rangle ,\\ {\bar{k}}_0|\mathbf{m}\rangle&=-q^{2m_1+1}|\mathbf{m}\rangle , \end{aligned}$$

where x is the so-called spectral parameter. We also note that the quantum parameter is still ${\bar{q}}^{\frac{1}{2}}=\mathbf{i}^{-1}q^{\frac{1}{2}}$.

To obtain the representation for the quantum parameter q, we need to we switch $q^{\frac{1}{2}}$ to $\mathbf{i}q^{\frac{1}{2}}$ (${\bar{q}}^{\frac{1}{2}}$ to $q^{\frac{1}{2}}$). Also, the relations in Section A.1 and those in Section 2.3 are different. For the node i that is signified as $\bullet$ in the Dynkin diagram, there is a relation

$$\begin{aligned} e_if_i+f_ie_i=\frac{k_i-k_i^{-1}}{q^{\frac{1}{2}}-q^{-\frac{1}{2}}} \end{aligned}$$

in Section 2.3 rather than

$$\begin{aligned} e_if_i+f_ie_i=\frac{k_i-k_i^{-1}}{-q^{\frac{1}{2}}-q^{-\frac{1}{2}}} \end{aligned}$$

in Section A.1. The former relation is realized by deleting $\kappa$ from the action of ${\bar{e}}_i$ or ${\bar{f}}_i$ in the formulas of the q-oscillator representation above. By doing so, we obtain

$$\begin{aligned} {\bar{e}}_0|\mathbf{m}\rangle&= {\left\{ \begin{array}{ll} x\,\mathbf{i}^{2m_1-|\mathbf{m}|}|\mathbf{m}+\mathbf{e}_1\rangle &{}\text {for }U_q(C^{(2)}(n+1))\\ x(-1)^{|\mathbf{m}|}|\mathbf{m}+2\mathbf{e}_1\rangle &{}\text {for }U_q(B^{(1)}(1,n)) \end{array}\right. },\\ {\bar{f}}_0|\mathbf{m}\rangle&= {\left\{ \begin{array}{ll} x^{-1}\mathbf{i}^{|\mathbf{m}|+2m_1+1}[m_1]|\mathbf{m}-\mathbf{e}_1\rangle &{}\text {for }U_q(C^{(2)}(n+1))\\ x^{-1}(-1)^{|\mathbf{m}|+1}\frac{[m_1][m_1-1]}{[2]^2}|\mathbf{m}-2\mathbf{e}_1\rangle &{}\text {for }U_q(B^{(1)}(0,n)) \end{array}\right. },\\ {\bar{k}}_0|\mathbf{m}\rangle&= {\left\{ \begin{array}{ll} q^{m_1+\frac{1}{2}}|\mathbf{m}\rangle &{}\text {for }U_q(C^{(2)}(n+1))\\ q^{2m_1+1}|\mathbf{m}\rangle &{}\text {for }U_q(B^{(1)}(0,n)) \end{array}\right. },\\ {\bar{e}}_i|\mathbf{m}\rangle&=(-1)^{-m_i+m_{i+1}+1}[m_i]|\mathbf{m}-\mathbf{e}_i+\mathbf{e}_{i+1}\rangle ,\\ {\bar{f}}_i|\mathbf{m}\rangle&=(-1)^{-m_i+m_{i+1}+1}[m_{i+1}]|\mathbf{m}+\mathbf{e}_i-\mathbf{e}_{i+1}\rangle ,\\ {\bar{k}}_i|\mathbf{m}\rangle&=q^{-m_i+m_{i+1}}|\mathbf{m}\rangle ,\\ {\bar{e}}_n|\mathbf{m}\rangle&=\mathbf{i}^{1-|\mathbf{m}|+2m_n}[m_n]|\mathbf{m}-\mathbf{e}_n\rangle ,\\ {\bar{f}}_n|\mathbf{m}\rangle&=\mathbf{i}^{|\mathbf{m}|-2m_n}|\mathbf{m}+\mathbf{e}_n\rangle ,\\ {\bar{k}}_n|\mathbf{m}\rangle&=q^{-m_n-\frac{1}{2}}|\mathbf{m}\rangle , \end{aligned}$$

for $1\le i\le n-1$.

Finally, to obtain the actions of $U_q(C^{(2)}(n+1))$ and $U_q(B^{(1)}(0,n))$) in Propositions 3.9 and 3.14, respectively, we perform the basis change $|\mathbf{m}\rangle$ to

$$\begin{aligned} \mathbf{i}^{s(\mathbf{m})}q^{-|\mathbf{m}|/2}\prod _{j=1}^n[m_j]!|\mathbf{m}\rangle , \end{aligned}$$

where $s(\mathbf{m})=-|\mathbf{m}|(|\mathbf{m}|+1)/2-\sum _jm_j^2$. Next we apply the algebra automorphism sending $e_n\mapsto -e_n,f_n\mapsto -f_n$ and the other generators fixed. For $U_q(C^{(2)}(n+1))^\sigma$, we also apply $e_0\mapsto \sigma e_0$, $f_0\mapsto f_0\sigma$. Accordingly, the coproduct also changes. For $U_q(B^{(1)}(0,n))$, we alternatively apply $e_0\mapsto \mathbf{i}[2]e_0,f_0\mapsto \frac{1}{\mathbf{i}[2]}f_0$. This completes the proof.

Appendix B. Quantum R Matrix for $U_q(A_{2n}^{(2)\dagger })$

In this appendix, we consider the quantum R matrix for the q-oscillator representation of $U_q(A_{2n}^{(2)\dagger })$ where $A_{2n}^{(2)\dagger }$ is the Dynkin diagram whose nodes have the opposite labelings to $A^{(2)}_{2n}$. This will be used in Appendix C to derive the quantum R matrix for $U_q(B^{(1)}(0,n))$.

1.1 q-oscillator representation for $U_q(A^{(2)\dagger }_{2n})$

By $A^{(2)\dagger }_{2n}$ we denote the following Dynkin diagram.

Although we did not deal with the q-oscillator representation for $U_q(A^{(2)\dagger }_{2n})$ in [15], it is easy to guess from other cases given there. On the space $\mathcal {W}$, the actions are given as follows.

$$\begin{aligned} e_0|\mathbf{m}\rangle&=x|\mathbf{m}+2\mathbf{e}_1\rangle ,\\ f_0|\mathbf{m}\rangle&=x^{-1}\frac{[m_1][m_1-1]}{[2]^2}|\mathbf{m}-2\mathbf{e}_1\rangle ,\\ k_0|\mathbf{m}\rangle&=-q^{2m_1+1}|\mathbf{m}\rangle ,\\ e_i|\mathbf{m}\rangle&=[m_i]|\mathbf{m}-\mathbf{e}_i+\mathbf{e}_{i+1}\rangle ,\\ f_i|\mathbf{m}\rangle&=[m_{i+1}]|\mathbf{m}+\mathbf{e}_i-\mathbf{e}_{i+1}\rangle ,\\ k_i|\mathbf{m}\rangle&=q^{-2m_i+2m_{i+1}}|\mathbf{m}\rangle ,\\ e_n|\mathbf{m}\rangle&=\mathbf{i}\kappa [m_n]|\mathbf{m}-\mathbf{e}_n\rangle ,\\ f_n|\mathbf{m}\rangle&=|\mathbf{m}+\mathbf{e}_n\rangle ,\\ k_n|\mathbf{m}\rangle&=\mathbf{i}q^{-m_n-1/2}|\mathbf{m}\rangle , \end{aligned}$$

where $0<i<n$ and $\kappa =(q+1)/(q-1)$. Denote this representation map by $\pi _x$.

Recall that the $U_q(B_n)$-highest weight vectors $\{v_l\mid l\in {\mathbb {Z}}_{\ge 0}\}$ are calculated in [15, Proposition 4]. We take the coproduct (C.1) with $\pi =1$.

Lemma B.1

For $x,y\in {\mathbb {Q}}(q)$ we have

• (1) $(\pi _x\otimes \pi _y)\Delta (f_0f_1^{(2)}\cdots f_{n-1}^{(2)})v_l= -\frac{[l][l-1]}{[2]^2}(q^{2l-2}x^{-1}+q^{-1}y^{-1})v_{l-2}\quad (l\ge 2)$,

• (2) $(\pi _x\otimes \pi _y)\Delta (e_ne_{n-1}^{(2)}\cdots e_1^{(2)}e_0)v_0= \frac{\mathbf{i}\kappa [2]}{1-q}((y+qx)v_1-q(y+x)\Delta (f_n)v_0)$.

Define the quantum R matrix ${\check{R}}_{KO}(z,q)$ for $U_q(A^{(2)\dagger }_{2n})$ as in Proposition C.4. The existence of such ${\check{R}}_{KO}(z,q)$ is essentially given in [15, Theorem 13]. Namely, although $A^{(2)\dagger }_{2n}$ is not listed there, the corresponding gauge transformed quantum R matrix is $S^{2,1}(z)$ and the proof has been done as the cases (i),(iv) and (v). By using Lemma B.1, we have the following.

Proposition B.2

We have the following spectral decomposition

$$\begin{aligned} {\check{R}}_{KO}(z)=\sum _{l\in 2{\mathbb {Z}}_{+}} \prod _{j=1}^{l/2}\frac{z+q^{4j-1}}{1+q^{4j-1}z}P_l +\sum _{l\in 1+2{\mathbb {Z}}_{+}}\prod _{j=0}^{(l-1)/2}\frac{z+q^{4j+1}}{1+q^{4j+1}z}P_l, \end{aligned}$$

where $P_l$ is the projector on the subspace generated by the $U_q(B_n)$-highest weight vector $v_l\;(l\ge 0)$.

Appendix C. Quantum R Matrix for $U_q(C^{(2)}(n+1))$ and $U_q(B^{(1)}(0,n))$

In this appendix, we compare the quantum R matrix for the q-oscillator representation for $U_q(C^{(2)}(n+1))$ with the one for $U_q(D_{n+1}^{(2)})$ given in [15]. We also compare the quantum R matrix for $U_q(B^{(1)}(0,n))$ with the one for $U_q(A^{(2)\dagger }_{2n})$ in Appendix B. We keep the notations in Appendix A.

1.1 Gauge transformation

We take the following coproduct

$$\begin{aligned} \begin{aligned}&\Delta (k_i)=k_i\otimes k_i,\\&\Delta (e_i)=1\otimes e_i+e_i\otimes \sigma ^{\frac{1-\pi }{2} p(i)}k_i,\\&\Delta (f_i)=f_i\otimes \sigma ^{\frac{1-\pi }{2} p(i)}+k_i^{-1}\otimes f_i,\\ \end{aligned} \end{aligned}$$

(C.1)

for $i\in I$, where $\sigma$ satisfies (2.3). We also take the same coproduct (C.1) for $\overline{u}_i$. Let $\Gamma$ be an operator acting on $\mathcal {W}^{\otimes 2}$ by

$$\begin{aligned} \Gamma |\mathbf{m}\rangle \otimes |\mathbf{m}'\rangle =\mathbf{i}^{\sum _{k,l}\varphi _{kl}m_km'_l}|\mathbf{m}\rangle \otimes |\mathbf{m}'\rangle , \end{aligned}$$

(C.2)

for $\mathbf{m}=(m_1,\dots ,m_n)$ and $\mathbf{m}'=(m'_1,\dots ,m'_n)$. Here we have the constraint $\varphi _{kl}+\varphi _{lk}=0$. Then by [23] (see also [22]),

$$\begin{aligned} \Delta ^\Gamma (u)=\Gamma ^{-1}\Delta (u)\Gamma \end{aligned}$$

gives another coproduct of $U_q(B_n)$ acting on $\mathcal {W}^{\otimes 2}$. Take $\varphi _{kl}$ to be 1 for $k<l$. We also set

$$\begin{aligned} K|\mathbf{m}\rangle =\mathbf{i}^{c(\mathbf{m})}|\mathbf{m}\rangle , \end{aligned}$$

(C.3)

where

$$\begin{aligned} c(\mathbf{m})=-\frac{1}{2}\sum _km_k^2+\sum _k\left( k-n-\frac{1}{2}\right) m_k. \end{aligned}$$

Set

$$\begin{aligned} \gamma _i(\mathbf{m})&={\left\{ \begin{array}{ll} -|\mathbf{m}|+m_1&{}(i=0\text { and for }U_q(C^{(2)}(n+1)))\\ -2|\mathbf{m}|+2m_1&{}(i=0\text { and for }U_q(B^{(1)}(0,n)))\\ m_i+m_{i+1}&{}(0<i<n)\\ -|\mathbf{m}|+m_n&{}(i=n) \end{array}\right. },\\ \beta _i(\mathbf{m})&={\left\{ \begin{array}{ll} m_1+n&{}(i=0\text { and }U_q(C^{(2)}(n+1)))\\ 2m_1+2n+1&{}(i=0\text { and }U_q(B^{(1)}(0,n)))\\ -m_i+m_{i+1}&{}(0<i<n)\\ -m_n&{}(i=n) \end{array}\right. }. \end{aligned}$$

Let $\varvec{\alpha }_0=\mathbf{e}_1\text { for }U_q(C^{(2)}(n+1))),2\mathbf{e}_1\text { for } U_q(B^{(1)}(0,n)))$, $\varvec{\alpha }_i=-\mathbf{e}_i+\mathbf{e}_{i+1}$ $(0<i<n)$, and $\varvec{\alpha }_n=-\mathbf{e}_n$.

Lemma C.1

The following formulas hold for $\mathbf{m}$, $\mathbf{m}'$, and $i\in I$:

• (1) $\Gamma ^{-1}(1\otimes e_i)\Gamma \, |\mathbf{m}\rangle \otimes |\mathbf{m}'\rangle =\mathbf{i}^{-\gamma _i(\mathbf{m})}|\mathbf{m}\rangle \otimes e_i|\mathbf{m}'\rangle$,

• (2) $\Gamma ^{-1}(e_i\otimes 1)\Gamma \,|\mathbf{m}\rangle \otimes |\mathbf{m}'\rangle =\mathbf{i}^{\gamma _i(\mathbf{m}')}e_i|\mathbf{m}\rangle \otimes |\mathbf{m}'\rangle$,

• (3) $\Gamma ^{-1}(1\otimes f_i)\Gamma \,|\mathbf{m}\rangle \otimes |\mathbf{m}'\rangle =\mathbf{i}^{\gamma _i(\mathbf{m}-\varvec{\alpha }_i)}|\mathbf{m}\rangle \otimes f_i|\mathbf{m}'\rangle$,

• (4) $\Gamma ^{-1}(f_i\otimes 1)\Gamma \, |\mathbf{m}\rangle \otimes |\mathbf{m}'\rangle =\mathbf{i}^{-\gamma _i(\mathbf{m}'-\varvec{\alpha }_i)} f_i|\mathbf{m}\rangle \otimes |\mathbf{m}'\rangle$.

Lemma C.2

The following formulas hold for $\mathbf{m}$ and $i\in I$:

• (1) $K^{-1}e_iK|\mathbf{m}\rangle =\mathbf{i}^{\beta _i(\mathbf{m})}e_i{|\mathbf{m}\rangle }$,

• (2) $K^{-1}f_iK|\mathbf{m}\rangle =\mathbf{i}^{-\beta _i(\mathbf{m}-\varvec{\alpha }_i)}f_i{|\mathbf{m}\rangle }$.

Proposition C.3

For $u_i$ $(i\in I$, $u=e,f,k)$, we have

$$\begin{aligned} \Delta ({\bar{u}}_i)|\mathbf{m}\rangle \otimes |\mathbf{m}'\rangle =\mathbf{i}^{\Lambda _i(\mathbf{m}+\mathbf{m}')} (K\otimes K)^{-1}\Delta ^\Gamma (u_i)(K\otimes K)|\mathbf{m}\rangle \otimes |\mathbf{m}'\rangle , \end{aligned}$$

on $\mathcal {W}^{\otimes 2}$. Here

$$\begin{aligned} \Lambda _i(\mathbf{m})={\left\{ \begin{array}{ll} m_i+m_{i+1}-(\delta _{i0}+\delta _{in})|\mathbf{m}|-n\delta _{i0}&{}(u=e)\\ m_i+m_{i+1}+(\delta _{i0}+\delta _{in})(|\mathbf{m}|+1)-2&{}(u=f)\\ 2m_i-2m_{i+1}&{}(u=k) \end{array}\right. }, \end{aligned}$$

except when $i=0$ and for $U_q(B^{(1)}(0,n))$, where

$$\begin{aligned} \Lambda _0(\mathbf{m})={\left\{ \begin{array}{ll} 2m_1-2|\mathbf{m}|-2n+1&{}(u=e)\\ 2m_1-2|\mathbf{m}|-2n+3&{}(u=f)\\ 0&{}(u=k) \end{array}\right. }. \end{aligned}$$

Here we should understand $m_0=m_{n+1}=0$.

Proof

It follows from Lemmas C.1 and C.2, and the following calculations. For instance, for $i=n$

$$\begin{aligned} \Delta ({\bar{e}}_n)|\mathbf{m}\rangle \otimes |\mathbf{m}'\rangle&= (1\otimes {\bar{e}}_n + {\bar{e}}_n\otimes \sigma {\bar{k}}_n)|\mathbf{m}\rangle \otimes |\mathbf{m}'\rangle \\&=\kappa (\mathbf{i}^{1-|\mathbf{m}'|}[m'_n] |\mathbf{m}\rangle \otimes |\mathbf{m}'-\mathbf{e}_n\rangle \\&\quad +(-1)^{|\mathbf{m}'|}\mathbf{i}^{2-|\mathbf{m}|+2m'_n}q^{-2m'_n-1}[m_n]|\mathbf{m}-\mathbf{e}_n\rangle \otimes |\mathbf{m}'\rangle ),\\ \Delta ^\Gamma (e_n)|\mathbf{m}\rangle \otimes |\mathbf{m}'\rangle&= (\Gamma ^{-1}(1\otimes e_n)\Gamma + \Gamma ^{-1}(e_n\otimes 1)\Gamma \cdot (1\otimes k_n))|\mathbf{m}\rangle \otimes |\mathbf{m}'\rangle \\&=\kappa (\mathbf{i}^{|\mathbf{m}|-m_n+1}[m'_n] |\mathbf{m}\rangle \otimes |\mathbf{m}'-\mathbf{e}_n\rangle \\&\quad +\mathbf{i}^{-|\mathbf{m}'|+m'_n+2}q^{-2m'_n-1}[m_n]|\mathbf{m}-\mathbf{e}_n\rangle \otimes |\mathbf{m}'\rangle ), \end{aligned}$$

and for $i\ne n$

$$\begin{aligned} \Delta ({\bar{e}}_i)|\mathbf{m}\rangle \otimes |\mathbf{m}'\rangle&= (1\otimes {\bar{e}}_i + {\bar{e}}_i\otimes {\bar{k}}_i)|\mathbf{m}\rangle \otimes |\mathbf{m}'\rangle \\&=\mathbf{i}^{2m'_{i+1}}[m'_i] |\mathbf{m}\rangle \otimes |\mathbf{m}'-\mathbf{e}_i+\mathbf{e}_{i+1}\rangle \\&\quad +\mathbf{i}^{2m_{i+1}+2m'_i-2m'_{i+1}}q^{-2m'_i+2m'_{i+1}}[m_i]|\mathbf{m}-\mathbf{e}_i+\mathbf{e}_{i+1}\rangle \otimes |\mathbf{m}'\rangle ,\\ \Delta ^\Gamma (e_i)|\mathbf{m}\rangle \otimes |\mathbf{m}'\rangle&= (\Gamma ^{-1}(1\otimes e_i)\Gamma + \Gamma ^{-1}(e_i\otimes 1)\Gamma \cdot (1\otimes k_i))|\mathbf{m}\rangle \otimes |\mathbf{m}'\rangle \\&=\mathbf{i}^{-m_i-m_{i+1}}[m'_i] |\mathbf{m}\rangle \otimes |\mathbf{m}'-\mathbf{e}_i+\mathbf{e}_{i+1}\rangle \\&\quad +\mathbf{i}^{m'_i+m'_{i+1}}q^{-2m'_i+2m'_{i+1}}[m_i]|\mathbf{m}-\mathbf{e}_i+\mathbf{e}_{i+1}\rangle \otimes |\mathbf{m}'\rangle . \end{aligned}$$

$\square$

For a quantum affine superalgebras such as $U_q(D^{(2)}_{n+1})$, $U_q(A^{(2)\dagger }_{2n})$, $U_q(C^{(2)}(n+1))$, and $U_q(B^{(1)}(0,n))$, a quantum R matrix R(z) is defined, if it exists, as an intertwiner satisfying

$$\begin{aligned} {\check{R}}(z)(\pi _x\otimes \pi _y)\Delta (u) =(\pi _y\otimes \pi _x)\Delta (u){\check{R}}(z), \end{aligned}$$

where ${\check{R}}(z)=PR(z)$, P is the transposition of the tensor components and $z=x/y$. We also note that the coproduct we use here is (C.1). For $U_q(D^{(2)}_{n+1})$ or $U_q(A^{(2)\dagger }_{2n})$, the existence of quantum R matrices are proved in [15] or Appendix B. We denote them by ${\check{R}}_{KO}(z)$. Let ${\check{R}}_{new}(z)$ be the quantum R matrices for the quantum groups $U_q(C^{(2)}(n+1))$ or $U_q(B^{(1)}(0,n))$. From Proposition C.3, we have

Proposition C.4

For generic $x, y\in {\mathbb {Q}}(q)$, ${\check{R}}_{new}(z)$ and ${\check{R}}_{KO}(z)$ have the following relation:

$$\begin{aligned} {\check{R}}_{new}(z,-q)=(K\otimes K)^{-1}\Gamma ^{-1}{\check{R}}_{KO}(z,q) \Gamma (K\otimes K). \end{aligned}$$

Appendix D. Irreducibility of $\mathcal {W}^{(s)}$

In this appendix, we prove Theorem 4.1. We adopt the arguments used in the finite-dimensional representations of the quantum affine algebras [12]. We assume that $X=C_n^{(1)}$ and $\mathcal {W}=\mathcal {W}_+$ since the proof for the other two cases are similar.

1.1 Normalized R matrix

Let us use the following notations.

• $\mathbf{k}$ : the base field, which is the algebraic closure of ${\mathbb {Q}}(q)$ in $\bigcup _n {\mathbb {C}}((q^{-1/n}))$,

• $\mathcal {W}[z] = \mathbf{k}[z^{\pm 1}]\otimes _\mathbf{k} \mathcal {W}(1)$ : the affinization of $\mathcal {W}(1)$, where z is a formal variable (see [14, Section 4.2]),

• $\mathcal {W}[z_1]\, {\widehat{\otimes }}\, \mathcal {W}[z_2]$, $\mathcal {W}[z_2]\, {\widetilde{\otimes }}\, \mathcal {W}[z_1]$ : the completions of $\mathcal {W}[z_1]{\otimes } \mathcal {W}[z_2]$ and $\mathcal {W}[z_2]{\otimes } \mathcal {W}[z_1]$, respectively, where $z_1, z_2$ are formal commuting variables (see [14, Section 7]),

• $\mathbf{k}\llbracket \frac{z_1}{z_2}\rrbracket$ : the ring of formal power series in $\frac{z_1}{z_2}$ over $\mathbf{k}$, where we regard $\mathbf{k}(\tfrac{z_1}{z_2}) \subset \mathbf{k}\llbracket \tfrac{z_1}{z_2}\rrbracket$ under the identification $\frac{1}{1-c(z_1/z_2)}=\sum _{k\ge 0}c^k(\frac{z_1}{z_2})^k\in \mathbf{k}\llbracket \tfrac{z_1}{z_2}\rrbracket$ for $c\in \mathbf{k}$.

Let $R^\mathrm{univ}$ be the universal R matrix for $U_q(C_n^{(1)})$. Then we have (cf. [14, (7.6)])

Lemma D.1

$\mathbf{k}(\frac{z_1}{z_2})\otimes _{\mathbf{k}[(\frac{z_1}{z_2})^{\pm 1}]}(\mathcal {W}[z_1]\otimes \mathcal {W}[z_2])$ is an irreducible module over $\mathbf{k}(\frac{z_1}{z_2})\otimes _{\mathbf{k}[(\frac{z_1}{z_2})^{\pm 1}]} U_q(C_n^{(1)})[z_1^{\pm 1},z_2^{\pm 1}]$.

Proof

See [15, Proposition 12]. $\quad \square$

Lemma D.2

The spaces $\mathcal {W}[z_1]\, {\widehat{\otimes }}\, \mathcal {W}[z_2]$ and $\mathcal {W}[z_2]\,{\widetilde{\otimes }}\, \mathcal {W}[z_1]$ are invariant under the action of $\mathbf{k}\llbracket \frac{z_1}{z_2}\rrbracket$.

Proof

Let us consider $\mathcal {W}[z_2]\,{\widetilde{\otimes }}\,\mathcal {W}[z_1]$. The proof for $\mathcal {W}[z_1]\, {\widehat{\otimes }}\, \mathcal {W}[z_2]$ is the same. It suffices to check that $\mathcal {W}[z_2]\,{\widetilde{\otimes }}\,\mathcal {W}[z_1]$ is invariant under multiplication by $\mathbf{k}\llbracket \frac{z_1}{z_2}\rrbracket$. Let Q be the root lattice for $C_n^{(1)}$ and $Q_+$ the set of non-negative integral linear combinations of simple roots. By definition of $\mathcal {W}[z_2]\, {\widetilde{\otimes }}\, \mathcal {W}[z_1]$, we have

$$\begin{aligned} \mathcal {W}[z_2]\, {\widetilde{\otimes }}\, \mathcal {W}[z_1]=\sum _{(\lambda ,\mu )}F_{(\lambda ,\mu )}(\mathcal {W}[z_2]\otimes \mathcal {W}[z_1]), \end{aligned}$$

where $F_{(\lambda ,\mu )}(\mathcal {W}[z_2]\otimes \mathcal {W}[z_1])=\prod _{\beta \in Q_+} \mathcal {W}[z_2]_{\lambda -\beta }\times \mathcal {W}[z_1]_{\mu +\beta }$. Here we understand the weights $\lambda -\beta$ and $\mu +\beta$ of $\mathcal {W}[z_i]$ ($i=1,2$) as elements in the affine weight lattice, say P in [14, Section 2.1]. We have

$$\begin{aligned} (\tfrac{z_1}{z_2})^k F_{(\lambda ,\mu )}(\mathcal {W}[z_2]\otimes \mathcal {W}[z_1]) \subset F_{(\lambda ,\mu )}(\mathcal {W}[z_2]\otimes \mathcal {W}[z_1]), \end{aligned}$$

for $k\in {\mathbb {Z}}_{\ge 0}$, since for a given $\mu +\beta$ ($\beta \in Q_+$), there exist only finitely many $k\in {\mathbb {Z}}_{\ge 0}$ and $\beta '\in Q_+$ such that $\mu +\beta =\mu + \beta '+k\delta$, where $\delta$ is the null root of $C_n^{(1)}$. This proves the lemma. $\quad \square$

By Lemma D.2, we may regard

$$\begin{aligned} \begin{aligned}&\mathbf{k}(\tfrac{z_1}{z_2})\otimes _{\mathbf{k}[(\frac{z_1}{z_2})^{\pm 1}]}(\mathcal {W}[z_1]\otimes \mathcal {W}[z_2]) \ \subset \ \mathbf{k}\llbracket \tfrac{z_1}{z_2}\rrbracket \otimes _{\mathbf{k}[(\frac{z_1}{z_2})^{\pm 1}]} \left( \mathcal {W}[z_1]\otimes \mathcal {W}[z_2]\right) \ \subset \ \mathcal {W}[z_1]\ {\widehat{\otimes }}\ \mathcal {W}[z_2],\\&\mathbf{k}(\tfrac{z_1}{z_2})\otimes _{\mathbf{k}[(\frac{z_1}{z_2})^{\pm 1}]}(\mathcal {W}[z_2]\otimes \mathcal {W}[z_1]) \ \subset \ \mathbf{k}\llbracket \tfrac{z_1}{z_2}\rrbracket \otimes _{\mathbf{k}[(\frac{z_1}{z_2})^{\pm 1}]} \left( \mathcal {W}[z_2]\otimes \mathcal {W}[z_1]\right) \ \subset \ \mathcal {W}[z_2]\ {\widetilde{\otimes }}\ \mathcal {W}[z_1]. \end{aligned} \end{aligned}$$

Note that the weight of $z_i^k\otimes |\mathbf{m}\rangle$ is in $-\tfrac{1}{2}\varpi _n +{\mathbb {Z}}_{\ge 0}\delta$ if and only if $|\mathbf{m}\rangle =|\mathbf{0}\rangle$. Hence one can check without difficulty that

$$\begin{aligned} R^\mathrm{univ} (|\mathbf{0}\rangle \otimes |\mathbf{0}\rangle ) = a(\tfrac{z_1}{z_2})\otimes ( |\mathbf{0}\rangle \otimes |\mathbf{0}\rangle ), \end{aligned}$$

for some non-zero $a(\frac{z_1}{z_2})\in \mathbf{k}\llbracket \frac{z_1}{z_2}\rrbracket$, which is invertible. Now, we define the normalized R matrix by

$$\begin{aligned} R^\mathrm{norm}_{z_1,z_2}= a(\tfrac{z_1}{z_2})^{-1}R^\mathrm{univ}. \end{aligned}$$

Lemma D.3

The normalized R matrix $R^\mathrm{norm}_{z_1,z_2}$ gives a $\mathbf{k}(\frac{z_1}{z_2})\otimes _{\mathbf{k}[(\frac{z_1}{z_2})^{\pm 1}]} U_q(C_n^{(1)})[z_1^{\pm 1},z_2^{\pm 1}]$-linear map

where $R^\mathrm{norm}_{z_1,z_2}(|\mathbf{0}\rangle \otimes |\mathbf{0}\rangle )=|\mathbf{0}\rangle \otimes |\mathbf{0}\rangle$. Moreover, $R^\mathrm{norm}_{z_1,z_2}$ is a unique such map and hence

$$\begin{aligned} {\check{R}}_+(z_1,z_2)=R^\mathrm{norm}_{z_1,z_2}, \end{aligned}$$

where ${\check{R}}_+(z_1,z_2)$ is the quantum R matrix in (4.1).

Proof

The well-definedness and uniqueness of $R^\mathrm{norm}_{z_1,z_2}$ follows from Lemma D.1 and $R^\mathrm{norm}_{z_1,z_2}(|\mathbf{0}\rangle \otimes |\mathbf{0}\rangle )=|\mathbf{0}\rangle \otimes |\mathbf{0}\rangle$. In particular, we have $R^\mathrm{norm}_{z_1,z_2}={\check{R}}_+(z_1,z_2)$. $\quad \square$

1.2 Irreducibility of $\mathcal {W}^{(2)}$

Let us prove first that $\mathcal {W}^{(2)}$ is irreducible. Recall that $\mathcal {W}^{(2)}$ is the image of

which is well-defined by (4.1) and Lemma D.3. Let

$$\begin{aligned} \mathbf{K}=\mathbf{k}(\tfrac{z_1}{z_3})\otimes \mathbf{k}(\tfrac{z_2}{z_3}),\quad \mathbf{D}=\mathbf{k}[(\tfrac{z_1}{z_3})^{\pm 1},(\tfrac{z_2}{z_3})^{\pm 1}]. \end{aligned}$$

Consider the following maps

By the hexagon property of $R^\mathrm{univ}$, we have

$$\begin{aligned} \begin{aligned} (R^\mathrm{norm}_{z_1,z_3}\otimes \mathrm{id})\circ (\mathrm{id}\otimes R^\mathrm{norm}_{z_2,z_3})&=(a(\tfrac{z_1}{z_3})^{-1}R^\mathrm{univ}_{1,3}\otimes \mathrm{id})\circ (a(\tfrac{z_2}{z_3})^{-1} \mathrm{id}\otimes R^\mathrm{univ}_{2,3})\\&= a(\tfrac{z_1}{z_3})^{-1}a(\tfrac{z_2}{z_3})^{-1}(R^\mathrm{univ}_{1,3}\otimes \mathrm{id})\circ ( \mathrm{id}\otimes R^\mathrm{univ}_{2,3})\\&= a(\tfrac{z_1}{z_3})^{-1}a(\tfrac{z_2}{z_3})^{-1}R^\mathrm{univ}_{12,3}, \end{aligned} \end{aligned}$$

(D.1)

where the second equality follows from the fact that the action of $z_i$ commutes with those of $e_j$ and $f_j$, and $R^\mathrm{univ}_{12,3}$ in the last equality is understood as the map

Put

$$\begin{aligned} R = a(\tfrac{z_1}{z_3})^{-1}a(\tfrac{z_2}{z_3})^{-1}R^\mathrm{univ}_{12,3}=(R^\mathrm{norm}_{z_1,z_3}\otimes \mathrm{id})\circ (\mathrm{id}\otimes R^\mathrm{norm}_{z_2,z_3}). \end{aligned}$$

By (4.1), we have a well-defined non-zero $U_q(C_n^{(1)})$-linear map $\mathbf{r}:=R\vert _{z_1=z_2^{-1}=z_3=q}$:

By the hexagon property (D.1), we obtain the following.

Lemma D.4

Let $S\subset \mathcal {W}(q)\otimes \mathcal {W}(q^{-1})$ be a $U_q(C_n^{(1)})$-submodule. Then we have

$$\begin{aligned} \mathbf{r}(S \otimes \mathcal {W}(q^{-1}))\subset \mathcal {W}(q^{-1})\otimes S, \end{aligned}$$

and the following diagram commutes:

$\square$

Proposition D.5

$\mathcal {W}^{(2)}$ is irreducible.

Proof

Note that $\mathcal {W}(q^{\pm 1})$ is irreducible, and $\mathbf{r}= (\mathbf{r}_{1,3}\otimes \mathrm{id})\otimes (\mathrm{id}\otimes \mathbf{r}_{2,3})$, where $\mathbf{r}_{2,3}=R^\mathrm{norm}_{z_2,z_3}\vert _{z_2^{-1}=z_3=q}$ and $\mathbf{r}_{1,3}=R^\mathrm{norm}_{z_1,z_3}\vert _{z_1=z_3=q}=\mathrm{id}$. Hence, we may apply the same arguments as in [12, Theorem 3.12] to show that if S is a non-zero submodule of $\mathcal {W}(q)\otimes \mathcal {W}(q^{-1})$, then S includes $\mathrm{Im}(R^\mathrm{norm}_{q^{-1},q})$. This implies that $\mathcal {W}^{(2)}$ is irreducible. $\quad \square$

1.3 Proof of Theorem 4.1

Fix $s\ge 2$. Let $z_1,\dots ,z_s$ be formal commuting variables. Consider

$$\begin{aligned} \mathcal {W}[z_1]\otimes \dots \otimes \mathcal {W}[z_s]. \end{aligned}$$

For $a\in {\mathbb {Q}}(q)^\times$, let $x_i=aq^{2i-1-s}$ and $\mathcal {W}_i = \mathcal {W}(x_i)$ for $1\le i\le s$. We have a map $\mathbf{r}_{i,j}:= R^\mathrm{norm}_{x_i,x_j}: \mathcal {W}_i\otimes \mathcal {W}_j \longrightarrow \mathcal {W}_j\otimes \mathcal {W}_i$ for $1\le i<j\le s$ and

which is the composition of $\mathbf{r}_{i,j}$ associated to a reduced expression of $w_0\in \mathfrak {S}_s$.

Let us prove that $\mathcal {W}^{(s)}=\mathrm{Im}(\mathbf{r})$ is irreducible. Use induction on s. It is true for $s=2$ by Proposition D.5. Suppose that $s\ge 3$. Let $\mathbf{r}={\mathbf {r}}_s$ be the map in the statement and let ${\mathbf {r}}_{s-1}$ be the map corresponding to the first $s-1$ factors. We have the following commutative diagram:

where $\hat{{\mathbf {r}}}_i= \mathrm{id}^{\otimes s-i-1}\otimes {\mathbf {r}}_{i,s} \otimes \mathrm{id}^{i-1}$ for $1\le i\le s-1$. Note that ${\mathbf {r}}_s \ne 0$ since $\mathbf{r}(|\mathbf{0}\rangle ^{\otimes s})=|\mathbf{0}\rangle ^{\otimes s}$. Thus ${\mathbf {r}}_{s-1}$ has a nonzero image $\mathcal {W}^{(s-1)}$, which is irreducible by the induction hypothesis. Applying the hexagon property repeatedly, we have the following commutative diagram:

Here the map ${\mathbf {r}}_{\mathcal {W}^{(s-1)},\mathcal {W}_s}$ is given by

$$\begin{aligned} {\mathbf {r}}_{\mathcal {W}^{(s-1)},\mathcal {W}_s}= c R^\mathrm{univ}_{1\dots s-1,s}\vert _{z_i=aq^{2i-1-s}}, \end{aligned}$$

where c is an element in $\bigotimes _{i<j}\mathbf{k}(\frac{z_i}{z_j})$, and $R^\mathrm{univ}_{1\dots s-1,s}$ is the universal R matrix

Thus the image of ${\mathbf {r}}_s$ is equal to that of ${\mathbf {r}}_{\mathcal {W}^{(s-1)},\mathcal {W}_s}$.

Now, let S be a non-zero submodule of $\mathcal {W}_s\otimes \mathcal {W}^{(s-1)}$. Put $\mathbf{r}'=\mathbf{r}_{s,s}\circ {\mathbf {r}}_{\mathcal {W}^{(s-1)},\mathcal {W}_s}$. Then as in Lemma D.4, we can check the following commutative diagram:

Again by the same arguments as in [12, Theorem 3.12], we conclude that $\mathcal {W}^{(s)}\subset S$, which implies that $\mathcal {W}^{(s)}$ is irreducible. This completes the proof.

Remark D.6

The universal R matrix for $C^{(2)}(n+1)$ and $B^{(1)}(0,n)$ can be found in [7].