Quantum coordinate ring in WZW model and affine vertex algebra extensions

Abstract

In this paper, we construct various simple vertex superalgebras which are extensions of affine vertex algebras, by using abelian cocycle twists of representation categories of quantum groups. This solves the Creutzig and Gaiotto conjectures (Creutzig and Gaiotto in Comm Math Phys 379:785–845, 2020, Conjecture 1.1 and 1.4) in the case of type ABC. If the twist is trivial, the resulting algebras correspond to chiral differential operators in the chiral case, and to WZW models in the non-chiral case.

Appendix

In Appendix, we will prove Conjecture 1 for the simple Lie algebras of type $B_n$ ($n \ge 1$). Here, we need to consider representations of quantum groups which are not type 1 (see Sect. 2.1 for type 1 representations). We will first review the non-type 1 representations of $U_q({\mathfrak {g}})$.

For $\lambda \in P$, let ${\mathbb {C}}\chi _\lambda$ be a one-dimensional representation of $U_q({\mathfrak {g}})$ defined by

$$\begin{aligned} E_\alpha \cdot \chi _\lambda =F_\alpha \cdot \chi _\lambda =0,\;\;\;\;\; K_\alpha \cdot \chi _\lambda = (-1)^{\langle \langle \alpha , \lambda \rangle \rangle }\chi _\lambda \text { for }\alpha \in \Pi . \end{aligned}$$

Then, for a type 1 module M, $M\otimes {\mathbb {C}}\chi _\lambda$ and ${\mathbb {C}}\chi _\lambda \otimes M$ are $U_q({\mathfrak {g}})$-modules, which are not of type 1. For example, for $L_q(\lambda )\otimes {\mathbb {C}}\chi _\lambda$, we have

$$\begin{aligned} K_\alpha \cdot v_\lambda \otimes \chi _\lambda = q^{\langle \langle \alpha ,\lambda \rangle \rangle }(-1)^{\langle \langle \alpha ,\lambda \rangle \rangle } v_\lambda \otimes \chi _\lambda \end{aligned}$$

for the highest weight vector $v_\lambda \in L_q(\lambda )$.

In order to define a braided tensor category structure on a non-type 1 representation category, the following lemma is very important.

Lemma A.1

For any type 1 module $M \in U_q({\mathfrak {g}}){\text {-mod}}$, define a linear map $h_M^\gamma :M\otimes {\mathbb {C}}\chi _\gamma \rightarrow {\mathbb {C}}\chi _\gamma \otimes M$ by

$$\begin{aligned} h_M^\gamma (m_\lambda \otimes \chi _\gamma )=\exp (\pi i (\gamma ,\lambda ))\chi _\gamma \otimes m_\lambda \end{aligned}$$

for any $\lambda \in P$ and $m_\lambda \in M_\lambda$. Then, $h_M^\gamma$ is a $U_q({\mathfrak {g}})$-module homomorphism. In particular, the family of the maps $\{h_M^{\gamma }\}_{M \in U_q({\mathfrak {g}}){\text {-mod}}}$ is a natural transformation of $- \otimes {\mathbb {C}}\chi _\gamma$ and ${\mathbb {C}}\chi _\gamma \otimes -$.

Proof

Let $\lambda \in P$ and $m_\lambda \in M_{\lambda }$ and $\alpha \in \Pi$. Since

$$\begin{aligned} E_\alpha \cdot (m_\lambda \otimes \chi _\gamma )&= \Delta (E_\alpha ) \cdot (m_\lambda \otimes \chi _\gamma ) \\&= (E_\alpha \otimes 1+K_\alpha \otimes E_\alpha )\cdot (m_\lambda \otimes \chi _\gamma )= (E_\alpha \cdot m_\lambda ) \otimes \chi _\gamma , \end{aligned}$$

and $E_\alpha \cdot m_\lambda \in M_{\lambda +\alpha }$ we have $h_{M}^\gamma (E_\alpha \cdot (m_\lambda \otimes \chi _\gamma ))= \exp (\pi i (\gamma ,\lambda +\alpha )) (\chi _\gamma \otimes E_\alpha \cdot m_\lambda ).$

Similarly, since $E_\alpha \cdot (\chi _\gamma \otimes m_\lambda ) = (E_\alpha \otimes 1+K_\alpha \otimes E_\alpha )\cdot (\chi _\gamma \otimes m_\lambda )= (K_\alpha \cdot \chi _\gamma \otimes E_\alpha \cdot m_\lambda )$, and $K_\alpha \cdot \chi _\gamma = (-1)^{\langle \langle \gamma ,\alpha \rangle \rangle }\chi _\gamma$, we have

$$\begin{aligned} E_\alpha \cdot h_{M}^\gamma (m_\lambda \otimes \chi _\gamma )&= \exp (\pi i (\gamma ,\lambda )) E_\alpha \cdot (\chi _\gamma \otimes m_\lambda )\\&=\exp (\pi i (\gamma , \lambda +\alpha ))(\chi _\gamma \otimes E_\alpha \cdot m_\lambda ). \end{aligned}$$

Hence, $h_{M}^\gamma (E_\alpha \cdot (m_\lambda \otimes \chi _\gamma ))=E_\alpha \cdot h_{M}^\gamma (m_\lambda \otimes \chi _\gamma )$ for any $\alpha \in \Pi$. It is easy to check this for $F_\alpha$ and $K_\alpha$ and thus $h_M^\gamma$ is a $U_q({\mathfrak {g}})$-module homomorphism. The naturality is obvious. $\square$

It is easy to show that only for ${\mathfrak {g}}$ of type B one-dimensional representations satisfy the following important properties:

Lemma A.2

The following conditions are equivalent:

• 1. $\chi _{\lambda +\alpha }=\chi _\lambda$ for any $\alpha \in Q$ and $\lambda \in P$;

• 2. $\langle \langle \alpha ,\beta \rangle \rangle \in 2{\mathbb {Z}}$ for any $\alpha ,\beta \in Q$;

• 3. The simple Lie algebra ${\mathfrak {g}}$ is of type $A_1$ or of type $B_n$ ($n\ge 2$).

We will now proceed to the case of type B. According to [28], the root system of type $B_n$ can be written as

$$\begin{aligned} \{\pm e_i\pm e_j, e_i\}_{1 \le i,j \le n}, \end{aligned}$$

where $\{e_i\}_{i=1,2,\dots ,n}$ is the standard basis of ${\mathbb {R}}^n$, and the simple roots and the fundamental weights as

$$\begin{aligned} (\alpha _1,\alpha _2,\dots ,\alpha _{n-1},\alpha _n)&=(e_1-e_2,e_2-e_3,\dots , e_{n-1}-e_n, e_n),\\ (\lambda _1,\lambda _2,\dots ,\lambda _{n-1},\lambda _n)&=(e_1,e_1+e_2,\dots ,e_1+e_2+\dots +e_{n-1}, \frac{e_1+e_2+\dots +e_n}{2}). \end{aligned}$$

The weight lattice is spanned by $\{e_i, \lambda _n \}_{i=1,2,\dots ,n}$ and $P/Q\cong {\mathbb {Z}}_2$ is generated by $\lambda _n$. Note that by the normalization, $\langle \langle e_i, e_i \rangle \rangle =2$ for any $i=1,\dots ,n$ (see Lemma A.2) and $\langle \langle \lambda _n, \lambda _n \rangle \rangle =\frac{n}{2}$. Let us denote the generator of $U_q(\mathrm {so}({2n+1}))$, $E_{\alpha _i},F_{\alpha _i},K_{\alpha _i}$, by $E_i,F_i,K_i$ for short. We remark that among $\alpha _1,\alpha _2,\dots ,\alpha _n$, only $\alpha _n$ is a short root.

Proposition A.3

There exist Hopf algebra isomorphisms $\phi : U_q(\mathrm {so}(2n+1))\rightarrow U_{-q}(\mathrm {so}(2n+1))$ such that:

$$\begin{aligned} \phi (E_i)=E_i,\;\;\;\; \phi (F_i)&=F_{i},\;\;\;\; \phi (K_i)=K_i\;\;\;\;\;\; \text { for }i=1,\dots ,n-1 \end{aligned}$$

and

$$\begin{aligned} \phi (E_n)=- E_n,\;\;\;\; \phi (F_n)&=F_{n},\;\;\;\; \phi (K_n)=K_n. \end{aligned}$$

Proof

The assertion follows from an easy computation. The point is that $q_{\alpha _i}=q^{\frac{\langle \langle \alpha _i,\alpha _i\rangle \rangle }{2}}=q^2$ for any $i=1,2,\dots ,n-1$ since they are long roots and $\langle \langle \alpha ,\beta \rangle \rangle \in 2{\mathbb {Z}}$ for any $\alpha ,\beta \in Q$ (see Lemma A.2). In particular, $q_{\alpha } = (-q)_{\alpha }$ for long roots. The only non-trivial relation is

$$\begin{aligned} \sum _{r=0}^{1-a_{\alpha \beta }} (-1)^r \left( \begin{matrix} 1-a_{\alpha \beta } \\ r \end{matrix} \right) _{q_\alpha } E_\alpha ^r E_\beta E_\alpha ^{1-a_{\alpha \beta }-r} = 0, \end{aligned}$$

for $\alpha = \alpha _n$ and $\beta =\alpha _{n-1}$, which follows from $\left( \begin{matrix} 3 \\ 1 \end{matrix} \right) _{-q} = 3_{-q}=(-1)^{3+1}3_{q} = \left( \begin{matrix} 3 \\ 1 \end{matrix} \right) _{q}$. $\square$

Remark A.4

We note that the above proposition is also applicable to the case of $A_1 = B_1$, that is,

$$\begin{aligned} \phi : U_q(\mathrm {sl}_2)\rightarrow U_{-q}(\mathrm {sl}_2),\;\;\;\; \phi (E)=-E,\;\;\;\;\;\phi (F)=F,\;\;\;\;\;\; \phi (K)=K. \end{aligned}$$

It is noteworthy that the Hopf algebras $U_q(\mathrm {sl}_2)$ and $U_{q'}(\mathrm {sl}_2)$ are isomorphic if and only if $q'=\pm q^{\pm }$ (see [35, Proposition 6 in Section 3]).

Let M be a type 1 $U_{-q}(\mathrm {so}(2n+1))$-module and $\phi ^*M$ an $U_{q}(\mathrm {so}(2n+1))$-module defined by

$$\begin{aligned} a \cdot _\phi m = \phi (a) \cdot m \;\;\;\;\;\;\;\;\;\text { for any }a \in U_{q}(\mathrm {so}(2n+1)) \text { and }m\in M. \end{aligned}$$

Then, we have

$$\begin{aligned} K_i \cdot _\phi v = (-q)^{\langle \langle \alpha _i,\lambda \rangle \rangle }v =(-1)^{\langle \langle \alpha _i,\lambda \rangle \rangle } q^{\langle \langle \alpha _i,\lambda \rangle \rangle }v \text { for }v \in M_{\lambda }. \end{aligned}$$

Hence, $\phi ^* M$ is not necessarily a type 1 representation.

Based on this observation, we will define a type 2 module of $U_q(\mathrm {so}_{2n+1})$. We first observe that by Lemma A.2 the one dimensional representation ${\mathbb {C}}\chi _{\gamma }$ is only depends on $\gamma \in P/Q={\mathbb {Z}}/2{\mathbb {Z}}$. Denote $\chi _{\lambda _n}$ by $\chi$.

For each $\lambda \in P^+$, let $L_q^{{II}}(\lambda )$ be the unique irreducible highest module defined by

$$\begin{aligned} K_i v_\lambda = (-q)^{\langle \langle \alpha _i,\lambda \rangle \rangle } v_\lambda ,\;\;\;\; E_i v_\lambda = 0\;\;\;\;\text { for }i=1,\dots ,n. \end{aligned}$$

We say a $U_q(\mathrm {so}_{2n+1})$-module is of type 2 if it decomposes into a direct sum of $L_q^{{II}}(\lambda )$’s for $\lambda \in P^+$. Denote the category of type 2 (resp. of type 1) $U_q(\mathrm {so}_{2n+1})$-modules by $C^{{II}}$ (resp. $C^I$).

Let $M,N \in C^{{II}}$. Since $U_q({\mathrm {so}_{2n+1}})$ is a Hopf algebra, $M\otimes N$ is a $U_q({\mathrm {so}_{2n+1}})$-module and it is easy to show that $M\otimes N \in C^{{II}}$. Thus, $C^{{II}}$ is naturally a monoidal category. Let $\rho \in {\mathbb {C}}$ satisfy $\exp (\pi i \rho )=q$ and denote the braided tensor category $(U_q({\mathrm {so}_{2n+1}}),R(\rho )){\text {-mod}}$ by $C^I(\rho )$. In this section, we will prove Conjecture 1 for type B in three steps:

• 1. To give a braided tensor category structure on $C^{{II}}$, which depends on the choice of $\rho$. We denote it by $C^{{II}}(\rho +1)$;

• 2. To show that a Hopf algebra isomorphism $\phi :U_q({\mathrm {so}_{2n+1}})\rightarrow U_{-q}({\mathrm {so}_{2n+1}})$ induces an equivalence of braided tensor categories between $C^{{II}}(\rho +1)$ and $(U_{-q}({\mathrm {so}_{2n+1}}),R(\rho +1)){\text {-mod}}$;

• 3. To construct a functor $F:C^I(\rho )^{Q_{\mathrm {so}_{2n+1}}}$ and $C^{{II}}(\rho +1)$ which gives an equivalence of braided tensor categories.

We will first consider Step (1). For any type 2 module $M^{{II}}$, set for all $\lambda \in P$

$$\begin{aligned} M_{\lambda }^{{II}}= \{m\in M^{{II}}\mid K_i m= (-q)^{\langle \langle \lambda ,\alpha _i \rangle \rangle }m \text { for } i=1,\dots ,n \}. \end{aligned}$$

Then, we have

$$\begin{aligned} M^{{II}}= \bigoplus _{\lambda \in P} M_{\lambda }^{{II}}. \end{aligned}$$

In order to define the R-matrix for type 1 representations, we consider a linear map $f_\rho$ (see Sect. 2.2). Define for all type 2 $U_q({\mathrm {so}_{2n+1}})$-modules $M^{{II}}$ and $N^{{II}}$ a bijective linear map $f_\rho ^{{II}}:M^{{II}}\otimes N^{{II}}\rightarrow M^{{II}}\otimes N^{{II}}$ by

$$\begin{aligned} f_\rho (m\otimes n)=\exp \left( - \pi i (\rho +1) \langle \langle \lambda ,\mu \rangle \rangle \right) m\otimes n \text { for any }m\in M_\lambda ^{{II}}\text { and }n\in N_{\mu }^{{II}}\end{aligned}$$

and for all $\mu , \lambda \in P$. Then, a statement similar to Lemma 2.12 holds for type 2 modules by replacing $f_\rho$ with $f_\rho ^{{II}}$.

Lemma A.5

Let $u \in U_q({\mathrm {so}_{2n+1}})_{\mu }^-$ and $u' \in U_q({\mathrm {so}_{2n+1}})_{\mu }^+$ for $\mu \in Q$ with $\mu \ge 0$. For any type 2 $U_q({\mathrm {so}_{2n+1}})$-modules $M^{{II}}$ and $N^{{II}}$,

$$\begin{aligned} (f_{\rho }^{{II}})^{-1} \circ (u\otimes u') \circ f_{\rho }^{{II}}=uK_{\mu } \otimes K_{-\mu }u' \end{aligned}$$

as linear maps acting on $M^{{II}}\otimes N^{{II}}$.

Let us define a linear map $R(\rho )^{{II}}$ by

$$\begin{aligned} R(\rho )^{{II}}= (\Theta \circ f_\rho ^{{II}})^{-1}:M^{{II}}\otimes N^{{II}}\rightarrow M^{{II}}\otimes N^{{II}}. \end{aligned}$$

Then, by the above lemma, $R(\rho )^{{II}}$ satisfies the axiom of an R matrix (R1-R3 in Sect. 2.2) as an operator on $C^{{II}}$ (see for example [31, Section 3]). Denote by $C^{{II}}(\rho +1)$ the braided tensor category defined by $R(\rho )^{{II}}$.

Then, the following lemma follows from a similar argument in Sect. 2.3:

Lemma A.6

The Hopf algebra isomorphism $\phi :U_{q}(\mathrm {so}(2n+1)) \rightarrow U_{-q}(\mathrm {so}(2n+1))$ induces an equivalence between $C^{{II}}(\rho +1)$ and $(U_{-q}(\mathrm {so}(2n+1)),R(\rho +1)){\text {-mod}}$ as braided tensor categories.

This completes Step (1) and Step (2). Finally, we will show the last step. For $S=I,{{II}}$ and $i \in P/Q={\mathbb {Z}}/2{\mathbb {Z}}$, let $C_i^{S}$ be a full subcategory of $C^S$ consisting of modules which is isomorphic to a direct sum of $L_q^S(\lambda )$’s for $\lambda \in i\lambda _n+Q$. This grading coincides with the $P/Q$-grading introduced in Sect. 2.4. We can define a (grading preserving) functor $F: C^I \rightarrow C^{{II}}$ by $F(M)=({\mathbb {C}}\chi ^0\otimes M_0) \oplus ({\mathbb {C}}\chi \otimes M_1)$ for any $M=M_0\oplus M_1 \in C^I=C_0^I\oplus C_1^I$, where ${\mathbb {C}}\chi ^0$ is the trivial representation. Then, F gives an equivalence of ${\mathbb {Z}}/2{\mathbb {Z}}$-graded abelian categories. For any $M \in C^I$, define a linear map $h_M:M\otimes {\mathbb {C}}\chi \rightarrow {\mathbb {C}}\chi \otimes M$ by

$$\begin{aligned} h_M(m_\lambda \otimes \chi )=\exp (\pi i (\lambda _n,\lambda ))\chi \otimes m_\lambda \end{aligned}$$

for any $\lambda \in P$ and $m_\lambda \in M_\lambda$. Then, $h_\bullet$ is a natural transformation by Lemma A.1.

Note that $C^{{II}}$ is a strict monoidal category since it is a full subcategory of the category of all $U_q({\mathrm {so}_{2n+1}})$-modules, which is clearly strict. Hence, we can identify $\chi ^0\otimes M=M$ for any $M \in C^{{II}}$. Define a $U_q({\mathrm {so}_{2n+1}})$-module isomorphism $\epsilon _2:{\mathbb {C}}\chi \otimes {\mathbb {C}}\chi \rightarrow {\mathbb {C}}$ by $\epsilon _2(\chi \otimes \chi )=1$.

Let $M_i \in C_i^I$ and $N_j \in C_j^I$ for $i,j=0,1$. Define a natural transformation $g_{M_i,N_j}:F(M_i) \otimes F(N_j) \rightarrow F(M_i\otimes N_j)$ by

$$\begin{aligned} g_{M_0,N_0}&: (\chi ^0 \otimes M_0)\otimes (\chi ^0\otimes N_0) = \chi ^0 \otimes M_0 \otimes N_0,\\ g_{M_1\otimes N_0}&: (\chi ^1 \otimes M_1)\otimes (\chi ^0\otimes N_0)= \chi ^{1}\otimes M_1 \otimes N_0,\\ g_{M_0\otimes N_1}&: (\chi ^0 \otimes M_0) \otimes (\chi ^1\otimes N_1) {\mathop {\longrightarrow }\limits ^{{\mathrm {id}}_{\chi ^0}\otimes h_{M_0} \otimes {\mathrm {id}}_{N_1}}} \chi ^{1}\otimes M_0\otimes N_1,\\ g_{M_1\otimes N_1}&: (\chi ^1 \otimes M_1)\otimes (\chi ^1\otimes N_1) {\mathop {\longrightarrow }\limits ^{{\mathrm {id}}_{\chi ^1}\otimes h_{M_1} \otimes {\mathrm {id}}_{N_1}}} \chi ^{2}\otimes M_1\otimes N_1 {\mathop {\rightarrow }\limits ^{\epsilon _2\otimes {\mathrm {id}}_{M_1\otimes N_1}}}\chi ^{0}\\&\quad \otimes M_1\otimes N_1. \end{aligned}$$

Then, we have:

Proposition A.7

The functor $F: C^I\rightarrow C^{{II}}$ together with the natural transformation $g_{M,N}:F(M)\otimes F(N) \rightarrow F(M\otimes N)$ and $\epsilon :{\varvec{1}} = F({\varvec{1}})$ give an equivalence of braided tensor categories from $(C^I)^{Q_{\mathrm {so}_{2n+1}}}$ to $C^{{II}}$.

Before giving the proof, we remark that the value $\exp (\pi i p \langle \langle \lambda _n,\lambda _n\rangle \rangle )=\exp (\frac{\pi i np}{2})$ is not well-defined for $p \in {\mathbb {Z}}/2{\mathbb {Z}}$, which is the source of the 3-cocycle $\alpha :({\mathbb {Z}}/2{\mathbb {Z}})^3\rightarrow {\mathbb {C}}^\times$. In fact, let $\iota :{\mathbb {Z}}/2{\mathbb {Z}}\rightarrow {\mathbb {Z}}$ be a map defined by sending $\{\bar{0},\bar{1}\}\mapsto \{0,1\}$. Then, we have:

Lemma A.8

The explicit form of the abelian cocycle $(\alpha _n,c_n) \in Z_{\text {ab}}^3({\mathbb {Z}}/2{\mathbb {Z}},{\mathbb {C}}^\times )$ such that $c(a,a)=Q_{{\mathrm {so}_{2n+1}}}(a)$ for $a \in {\mathbb {Z}}/2{\mathbb {Z}}=P/Q$ can be give by

$$\begin{aligned} \alpha _n(a,b,c)&={\left\{ \begin{array}{ll} (-1)^n &{} (a=b=c=1)\\ 1 &{} (\text {otherwise})\\ \end{array}\right. }\\&=\exp \left( \pi i a\frac{\iota (b)+\iota (c)-\iota (a+b)}{2}\right) ,\\ c_n(a,b)&={\left\{ \begin{array}{ll} i^n &{} (a=b=1) \\ 1 &{} (\text {otherwise}). \end{array}\right. } \end{aligned}$$

Proof of Proposition A.7

We will verify the conditions (LM1) and (LM2) in Sect. 1.2. Since both $C^I$ and $C^{{II}}$ are strict monoidal categories, the associative isomorphisms are trivial before twisting. Let $p_i \in {\mathbb {Z}}/2{\mathbb {Z}}$ and $M_i \in C_{p_i}^I$, $\beta _i \in P$, and $v_i \in (M_i)_{\beta _i}$ for $i=1,2,3$. Then, we have

$$\begin{aligned} g_{M_1\otimes M_2,M_3}&\circ (g_{M_1,M_2}\otimes {\mathrm {id}}_{M_3}) \left( (\chi ^{p_1} \otimes v_1 \otimes \chi ^{p_2} \otimes v_2)\otimes \chi ^{p_3} \otimes v_3\right) \\&=g_{M_1\otimes M_2,M_3} (\exp (p_2\pi i \langle \langle \lambda _n,\beta _1\rangle \rangle ) (\chi ^{p_1+p_2} \otimes v_1 \otimes \otimes v_2) \otimes \chi ^{p_3} \otimes v_3)\\&= \exp (\pi i (p_2\langle \langle \lambda _n,\beta _1\rangle \rangle +p_3\langle \langle \lambda _n,\beta _1+\beta _2 \rangle \rangle ) (\chi ^{p_1+p_2+p_3}\\&\quad \otimes (v_1 \otimes \otimes v_2) \otimes v_3 \end{aligned}$$

and

$$\begin{aligned} g_{M_1,M_2\otimes M_3}&\circ ({\mathrm {id}}_{M_1}\otimes g_{M_2,M_3}) ( \chi ^{p_1} \otimes v_1 \otimes (\chi ^{p_2} \otimes v_2 \otimes \chi ^{p_3} \otimes v_3))\\&=g_{M_1,M_2\otimes M_3}( \exp (\pi i p_3 \langle \langle \lambda _n,\beta _2 \rangle \rangle ) \chi ^{p_1} \otimes v_1 \otimes (\chi ^{p_2+p_3} \otimes v_2 \otimes v_3) )\\&= \exp (\pi i (p_3 \langle \langle \lambda _n,\beta _2 \rangle \rangle +\iota (p_2,p_3)\langle \langle \lambda _n,\beta _1\rangle \rangle ) \chi ^{p_1+p_2+p_3} \otimes v_1\\&\quad \otimes (v_2 \otimes v_3). \end{aligned}$$

Thus, in order to verify (LM1), it suffices to show that

$$\begin{aligned}&\alpha (p_1,p_2,p_3)\exp (\pi i (p_2\langle \langle \lambda _n,\beta _1\rangle \rangle +p_3\langle \langle \lambda _n,\beta _1+\beta _2 \rangle \rangle )\\&\quad =\exp (\pi i (p_3 \langle \langle \lambda _n,\beta _2 \rangle \rangle +\iota (p_2,p_3)\langle \langle \lambda _n,\beta _1\rangle \rangle ), \end{aligned}$$

which follows from Lemma A.8. (LM2) is obvious. Hence, the assertion holds. $\square$

Finally, we will prove that $F:C^I(\rho )^{Q_{\mathrm {so}_{2n+1}}}\rightarrow C^{{II}}(\rho +1)$ is a braided monoidal functor. Let $p_i \in {\mathbb {Z}}/2{\mathbb {Z}}$ and $M_i \in C_{p_i}^I$, $\beta _i \in P$, and $m_i \in (M_i)_{\beta _i}$ for $i=1,2$. It suffices to show that the following diagram commutes:

where $c_n(p_1,p_2)$ is given in Lemma A.8. Recall $\Theta =\sum _{\mu \ge 0}\Theta _\mu$ and $\Theta _\mu =\sum _{i=0}^{r(\mu )}v_i^\mu \otimes u_i^\mu \in U_\mu ^-{\hat{\otimes }}U_\mu ^+$ (see Sect. 2.2). Then, we have:

$$\begin{aligned} g_{M_1,M_2}&\circ (B_{F(M_1),F(M_2)}^{{II}})^{-1} (\chi ^{p_2} \otimes m_2)\otimes (\chi ^{p_1} \otimes m_1)\\&=g_{M_1,M_2} \circ \Theta \circ f_\rho ^{{II}}\circ P_{21} (\chi ^{p_2} \otimes v_2)\otimes (\chi ^{p_1} \otimes v_1)\\&= \exp (-\pi i (\rho +1)\langle \langle \beta _1,\beta _2 \rangle \rangle ) \sum _{\mu \ge 0}\sum _{i=0}^{r(\mu )} g_{M_1,M_2} (v_i^\mu \cdot (\chi ^{p_1} \otimes m_1))\\&\quad \otimes (u_i^\mu \cdot (\chi ^{p_2} \otimes m_2)). \end{aligned}$$

Since $\Delta (E_i)=E_i \otimes 1 + K_i\otimes E_i$ and $\Delta (F_i)=F_i\otimes K_i^{-1}+1\otimes F_i$, we have

$$\begin{aligned} (v_i^\mu \cdot (\chi ^{p_1} \otimes m_1)) \otimes (u_i^\mu \cdot (\chi ^{p_2} \otimes m_2))&=(\chi ^{p_1} \otimes v_i^\mu \cdot m_1) \otimes (K_\mu \cdot \chi ^{p_2} \otimes u_i^\mu \cdot m_2)\\&=(-1)^{p_2 \langle \langle \lambda _n,\mu \rangle \rangle } (\chi ^{p_1} \otimes v_i^\mu \cdot m_1)\\&\quad \otimes (\chi ^{p_2} \otimes u_i^\mu \cdot m_2). \end{aligned}$$

Hence, we have:

$$\begin{aligned}&g_{M_1,M_2}\circ (B_{F(M_1),F(M_2)}^{{II}})^{-1} (\chi ^{p_2} \otimes m_2) \otimes (\chi ^{p_1} \otimes m_1)\\&\quad = \exp (\pi i (\rho +1)\langle \langle \beta _1,\beta _2 \rangle \rangle ) \sum _{\mu \ge 0}\sum _{i=0}^{r(\mu )} \Bigl ( \exp (\pi i \iota (p_2) \langle \langle \lambda _n, \beta _1+\mu \rangle \rangle )\\&\qquad \times (-1)^{p_2\langle \langle \lambda _n,\mu \rangle \rangle } (\chi ^{p_1+p_2} \otimes v_i^\mu \cdot m_1 \otimes u_i^\mu \cdot m_2) \Bigr )\\&\quad = \exp (\pi i (\rho +1)\langle \langle \beta _1,\beta _2 \rangle \rangle ) \exp (\pi i \iota (p_2) \langle \langle \lambda _n, \beta _1\rangle \rangle ) \sum _{\mu \ge 0}\sum _{i=0}^{r(\mu )} (\chi ^{p_1+p_2}\\&\qquad \otimes v_i^\mu \cdot m_1 \otimes u_i^\mu \cdot m_2). \end{aligned}$$

Similarly, we have

$$\begin{aligned}&c_n(p_1,p_2)F(B_{M_1,M_2}^I)^{-1} \circ g_{M_1,M_2} (\chi ^{p_2} \otimes m_2)\otimes (\chi ^{p_1} \otimes m_1)\\&\quad =c_n(p_1,p_2)\exp (\pi i \iota (p_1)\langle \langle \lambda _n, \beta _2 \rangle \rangle ) F(B_{M_1,M_2}^I)^{-1} (\chi ^{p_1+p_2} \otimes m_2 \otimes m_1)\\&\quad =c_n(p_1,p_2)\exp (\pi i \iota (p_1)\langle \langle \lambda _n, \beta _2 \rangle \rangle ) \chi ^{p_1+p_2} \otimes \left( \Theta \circ f_\rho \circ P_{21} (m_2 \otimes m_1)\right) \\&\quad =c_n(p_1,p_2)\exp (\pi i \iota (p_1)\langle \langle \lambda _n, \beta _2 \rangle \rangle ) \exp (-\pi i \rho (\beta _1,\beta _2)) \sum _{\mu \ge 0}\sum _{i=0}^{r(\mu )} \chi ^{p_1+p_2}\\&\qquad \otimes (v_i^\mu \cdot m_1 \otimes u_i^\mu m_2). \end{aligned}$$

Thus, the proof of the conjecture comes down to the following lemma:

Lemma A.9

If $(M_i)_{\beta _i} \ne 0$ for $i=1,2$, then

$$\begin{aligned} c_n(p_1,p_2)= \exp (\pi i (\langle \langle \beta _1,\beta _2 \rangle \rangle + \iota (p_2) \langle \langle \lambda _n, \beta _1\rangle \rangle - \iota (p_1)\langle \langle \lambda _n, \beta _2 \rangle \rangle ). \end{aligned}$$

Proof

Let $k:P\times P\rightarrow {\mathbb {C}}^\times$ be a map defined by $k(\beta _1,\beta _2)=\exp (\pi i (\langle \langle \beta _1,\beta _2 \rangle \rangle + \iota (\beta _2) \langle \langle \lambda _n, \beta _1\rangle \rangle - \iota (\beta _1)\langle \langle \lambda _n, \beta _2 \rangle \rangle )$, where $\iota :P\rightarrow P/Q= \{0,1\}$ is defined by the composition of the projection and the identification.

We claim that $k(\beta _1+\alpha ,\beta _2)=k(\beta _1,\beta _2+\alpha )=k(\beta _1,\beta _2)$ for any $\alpha \in Q$. The difference $k(\beta _1+\alpha ,\beta _2)k(\beta _1,\beta _2)^{-1}$ is equal to $\exp (\pi i (\langle \langle \beta _2,\alpha \rangle \rangle +\iota (\beta _2)\langle \langle \lambda _n,\alpha \rangle \rangle ))$. Thus, if $\beta _2 \in Q$ i.e., $\iota (\beta _2)=0$, then $k(\beta _1+\alpha ,\beta _2)k(\beta _1,\beta _2)^{-1}$ is equal to 1 by Lemma A.2. Similarly, if $\iota (\beta _2)=1$, then $k(\beta _1+\alpha ,\beta _2)k(\beta _1,\beta _2)^{-1} =\exp (\pi i (\langle \langle \beta _2,\alpha \rangle \rangle +\langle \langle \lambda _n,\alpha \rangle \rangle )) =\exp (\pi i (\langle \langle \lambda _n,\alpha \rangle \rangle +\langle \langle \lambda _n,\alpha \rangle \rangle ))=1,$ thus the claim is proved.

Since $k(0,0)=k(\lambda _n,0)=k(0,\lambda _n)=1$ and $k(\lambda _n,\lambda _n)=i^n$, the assertion follows from Lemma A.8. $\square$

Hence, we have:

Theorem A.10

The composition of F and $\phi ^*$ gives a braided monoidal equivalence between $(U_q({\mathrm {so}_{2n+1}}),R(\rho )){\text {-mod}}^{Q_{\mathrm {so}_{2n+1}}}$ and $(U_{-q}({\mathrm {so}_{2n+1}}),R(\rho )){\text {-mod}}$ for any $n \ge 1$.