1.1 Conventions

We treat h as a formal variable and put $h^* = h$ when we consider $*$ -algebraic structures. We put

$$\begin{align*}q=e^h\quad\text{and}\quad\hbar = \frac{h}{\pi i},\\[-10pt] \end{align*}$$

the latter is mostly reserved for the KZ-equations. We denote the space of formal power series with coefficients in A by

and the space of Laurent series by

For

, we denote the smallest n such that $a^{(n)} \neq 0$ by $\operatorname {\mathrm {ord}}(a)$ .

We denote the h-adically completed tensor product of -modules by $\mathbin {\hat \otimes }$ . In particular, we have .

When $A = \mathbb {C}$ and has constant term $a^{(0)}>0$ , we take its nth root $b = a^{\frac {1}{n}}$ to be the unique solution of $b^n = a$ such that $b^{(0)}$ is positive. A similar convention is used for $\log $ .

1.2 Simple Lie groups

Throughout the entire paper, $\mathfrak {u}$ denotes a compact simple Lie algebra and $\mathfrak {g}$ denotes its complexification. The connected and simply connected Lie groups corresponding to $\mathfrak {g}$ and $\mathfrak {u}$ are denoted by G and U.

We denote by $(\cdot ,\cdot )_{\mathfrak {g}}$ the unique invariant symmetric bilinear form on $\mathfrak {g}$ such that, for any Cartan subalgebra $\mathfrak {h} < \mathfrak {g}$ , its dual form on $\mathfrak {h}^*$ has the property that $(\alpha ,\alpha )=2$ for every short root $\alpha $ . Let $t^{\mathfrak {u}}\in \mathfrak {u}^{\otimes 2}$ be the corresponding invariant tensor:

(1.1) $$ \begin{align} t^{\mathfrak{u}}=\sum_i X_i\otimes X^i,\\[-17pt]\nonumber \end{align} $$

where $(X_i)_i$ is a basis in $\mathfrak {g}$ and $(X^i)_i$ is the dual basis.

Recall that $(\cdot ,\cdot )_{\mathfrak {g}}$ is negative definite on $\mathfrak {u}$ . Therefore, if we define an antilinear involution $*$ on $\mathfrak {g}$ by letting $X^*=-X$ for $X\in \mathfrak {u}$ , then $(X,Y^*)_{\mathfrak {g}}$ becomes an $(\operatorname {\mathrm {Ad}} U)$ -invariant Hermitian scalar product on $\mathfrak {g}$ .

We denote the category of finite-dimensional algebraic representations of the linear algebraic group G (equivalently, finite-dimensional representations of $\mathfrak {g}$ ) by $\operatorname {\mathrm {Rep}} G$ . It is equivalent to the category of finite-dimensional unitary representations of U. We write $\pi \in \operatorname {\mathrm {Rep}} G$ to say that $\pi $ is a finite-dimensional representation of G, its underlying space is denoted by $V_{\pi }$ . We also fix a set $\operatorname {\mathrm {Irr}} G$ of representatives of the isomorphism classes of irreducible representations.

We will often have to extend the scalars to . Denote the category we get by . Thus, the objects of are the G-modules over that are isomorphic to the modules of the form for $\pi \in \operatorname {\mathrm {Rep}} G$ .

1.3 Multiplier algebras

For $n = 1, 2, \dots $ , we put

$$\begin{align*}\mathcal{U}(G^n) = \prod_{\substack{\pi_i \in \operatorname{\mathrm{Irr}} G,\\ i = 1, \dots, n}} \operatorname{\mathrm{End}}(V_{\pi_1}) \otimes \dots \otimes \operatorname{\mathrm{End}}(V_{\pi_n}).\\[-17pt] \end{align*}$$

We view G and $\mathfrak {g}$ as subsets of $\mathcal {U}(G) = \mathcal {U}(G^1)$ .

Since for every irreducible $\pi \in \operatorname {\mathrm {Rep}} G$ , there is a unique up to a scalar factor U-invariant Hermitian scalar product on $V_{\pi }$ , we have a canonical involution $*$ on $\mathcal {U}(G^n)$ . There is also a unique homomorphism

$$\begin{align*}\Delta\colon \mathcal{U}(G) \to \mathcal{U}(G^2)\\[-17pt] \end{align*}$$

characterized by the identities $(\pi _1 \otimes \pi _2)(\Delta (T))S=S\pi (T)$ for all intertwiners $S\colon V_{\pi }\to V_{\pi _1}\otimes V_{\pi _2}$ . Then $\Delta (g) = g \otimes g$ for $g \in G$ . This characterizes the elements of G among the nonzero elements of $\mathcal {U}(G)$ . Similarly, the identity $\Delta (X) = X \otimes 1 + 1 \otimes X$ for $X \in \mathfrak {g}$ characterizes $\mathfrak {g}$ inside $\mathcal {U}(G)$ .

Denote by $\mathcal {O}(G)$ the Hopf algebra of regular functions (matrix coefficients of finite-dimensional representations) on G. We occasionally write $\mathcal {O}(U)$ instead of $\mathcal {O}(G)$ when we think of it as a function algebra on U.

There is a nondegenerate pairing between $\mathcal {U}(G)$ and $\mathcal {O}(G)$ that allows us to identify $\mathcal {U}(G)$ with the dual space of $\mathcal {O}(G)$ . Concretely, if $\pi $ is irreducible, $T\in \operatorname {\mathrm {End}}(V_{\pi })$ , $v\in V_{\pi }$ , $\ell \in V_{\pi }^*$ , then for the matrix coefficient $a_{v,\ell }\in \mathcal {O}(G)$ , $a_{v,\ell }(g)=\ell (\pi (g)v)$ , we have

$$\begin{align*}\langle a_{v,\ell},T\rangle=\ell(T v),\\[-17pt] \end{align*}$$

and $\langle f,T\rangle =0$ for the matrix coefficients f of the irreducible representations $\pi '$ inequivalent to $\pi $ . Similarly, $\mathcal {U}(G^n)$ is the linear dual of $\mathcal {O}(G)^{\otimes n}$ . With respect to this duality, the bialgebra structures are related by

$$ \begin{align*} \langle f_1 \otimes f_2, \Delta(T) \rangle &= \langle f_1 f_2, T \rangle,& \langle \Delta(f), T_1 \otimes T_2 \rangle &= \langle f, T_1 T_2 \rangle\\[-17pt] \end{align*} $$

for $f_i \in \mathcal {O}(G)$ and $T_i \in \mathcal {U}(G)$ .

We can do the same constructions for any reductive linear algebraic group H over $\mathbb {C}$ . We then also define

for $0 \le n < \infty $ . In a more invariant form, $\mathcal {U}(H \times G^n)$ is the linear dual of $\mathcal {O}(H\times G^n)$ .

Assume in addition that H is an algebraic subgroup of G. Then the embedding $H\to G$ extends to an embedding of $\mathcal {U}(H^{n+1})$ into $\mathcal {U}(H\times G^n)$ . In particular, the comultiplication $\Delta \colon \mathcal {U}(H)\to \mathcal {U}(H^2)$ can be viewed as a homomorphism $\mathcal {U}(H)\to \mathcal {U}(H\times G)$ .

Note that in general H is not simply connected. In Lie algebraic terms, the category $\operatorname {\mathrm {Rep}} H$ consists of the finite-dimensional representations of $\mathfrak {h}$ that are subrepresentations of the finite-dimensional representations of $\mathfrak {g}$ restricted to $\mathfrak {h}$ .

1.4 Quasi-coactions and ribbon twist-braids

The notion of a quasi-bialgebra [Reference DrinfeldDri89b] has a straightforward adaptation to the setting of multiplier algebras, cf. [Reference Neshveyev and TusetNT11, Section 2]. We will be interested in multiplier quasi-bialgebras of the form . Thus, $\Delta _h$ is a nondegenerate homomorphism , meaning that the images of the idempotents $\Delta _h(\mathrm {id}_{V_{\pi }})$ ( $\pi \in \operatorname {\mathrm {Irr}} G$ ) in add up to $1$ , is a nondegenerate homomorphism, and is an invertible element (with $\Phi ^{(0)}=1$ ) satisfying the same identities as in [Reference DrinfeldDri89b, Section 1].

The assumption of nondegeneracy for the counit $\epsilon _h$ implies that it is determined by its restrictions to the blocks of . Since there are no nonzero ( -linear) homomorphisms for $\dim V>1$ and there is a unique such homomorphism for $\dim V=1$ , we conclude that $\epsilon _h$ coincides with the standard counit $\epsilon $ on . From now on, we will therefore omit $\epsilon _h$ from the notation for a multiplier quasi-bialgebra.

Given a reductive algebraic subgroup H of G, a quasi-coaction of on is given by a nondegenerate homomorphism and an associator satisfying $\Psi ^{(0)}=1$ ,

$$\begin{align*}(\mathrm{id}\otimes\epsilon)\alpha=\mathrm{id}, \end{align*}$$

(1.2)

the mixed pentagon equation

(1.3) $$ \begin{align} \Phi_{1,2,3} \Psi_{0, 12, 3} \Psi_{0, 1, 2} = \Psi_{0, 1, 23} \Psi_{01, 2, 3}, \end{align} $$

with $\Psi _{01, 2, 3} = (\alpha \otimes \mathrm {id})(\Psi )$ , $\Psi _{0, 12, 3} = (\mathrm {id}_{\mathcal {U}(H)} \otimes \Delta _h \otimes \mathrm {id})(\Psi )$ , etc., and the normalization conditions

$$\begin{align*}(\mathrm{id}\otimes\epsilon\otimes\mathrm{id})(\Psi) = (\mathrm{id}\otimes\mathrm{id}\otimes\epsilon)(\Psi)=1. \end{align*}$$

A multiplier quasi-bialgebra defines a tensor category , where the tensor product $\otimes _h$ on is defined using $\Delta _h$ and the associativity isomorphism is given by the action of $\Phi $ . A quasi-coaction as above defines then the structure of a right -module category on . Namely, the functor defining the module category structure is induced by $\alpha $ , while the associativity morphisms are defined by the action of $\Psi $ . See [Reference De Commer, Neshveyev, Tuset and YamashitaDCNTY19, Section 1] for more details, but note that in [Reference De Commer, Neshveyev, Tuset and YamashitaDCNTY19] we worked in the analytic setting, meaning that $q=e^h$ was a real number and $\Phi \in \mathcal {U}(G^3)$ , $\Psi \in \mathcal {U}(H\times G^2)$ .

Next, let be an R-matrix (with $\mathcal {R}^{(0)}=1$ ) for , that is, $\mathcal {R}\Delta _h(\cdot )=\Delta _h^{\mathrm {op}}(\cdot )\mathcal {R}$ and $\mathcal {R}$ satisfies the hexagon relations. Let $\beta $ be an automorphism of the quasi-triangular multiplier quasi-bialgebra . A ribbon $\beta $ -braid is given by an invertible element satisfying

(1.4)

(1.5) $$ \begin{align} (\alpha \otimes \mathrm{id})(\mathcal{E}) = \Psi^{-1} \mathcal{R}_{21} \Psi_{021} \mathcal{E}_{02} (\mathrm{id} \otimes\mathrm{id}\otimes \beta)(\Psi_{021}^{-1} \mathcal{R}_{12} \Psi), \end{align} $$

(1.6) $$ \begin{align} (\mathrm{id} \otimes \Delta_h)(\mathcal{E}) = \mathcal{R}_{21} \Psi_{021} \mathcal{E}_{02} (\mathrm{id} \otimes\mathrm{id}\otimes \beta)(\Psi_{021}^{-1} \mathcal{R}_{12} \Psi) \mathcal{E}_{01} (\mathrm{id} \otimes \beta \otimes \beta)(\Psi^{-1}). \end{align} $$

When $\beta $ is the identity map, we just say ‘ribbon braid’ instead of ‘ribbon $\mathrm {id}$ -braid’. We want to stress that, as opposed to $\Phi $ , $\Psi $ and $\mathcal {R}$ , we do not require $\mathcal {E}^{(0)}=1$ . A quadruple satisfying equations (1.4) and (1.5) is a version of a quasi-reflection algebra [Reference EnriquezEnr07]. In categorical terms, the action of $\mathcal {E}$ on $M\odot _{\alpha } N$ defines the structure of a ribbon $\beta $ -braided module category on . See again [Reference De Commer, Neshveyev, Tuset and YamashitaDCNTY19, Section 1] for more details.

1.5 Twisting

We can transform a quasi-coaction of into a new one as follows. Suppose that we are given elements and such that $\mathcal {F}^{(0)}=1$ , $\mathcal {G}^{(0)}=1$ and

$$ \begin{align*} (\epsilon\otimes\mathrm{id})(\mathcal{F}) &= (\mathrm{id}\otimes\epsilon)(\mathcal{F})=1,& (\mathrm{id}\otimes\epsilon)(\mathcal{G}) &= 1. \end{align*} $$

Then the twisting of the quasi-coaction by $(\mathcal {F},\mathcal {G})$ is the quasi-coaction of the multiplier quasi-bialgebra , where

$$ \begin{align*} \Delta_{h,\mathcal{F}}&=\mathcal{F}\Delta_h(\cdot)\mathcal{F}^{-1}, & \Phi_{\mathcal{F}}&=(1\otimes \mathcal{F})(\mathrm{id}\otimes \Delta_h)(\mathcal{F})\Phi(\Delta_h\otimes\mathrm{id})(\mathcal{F}^{-1})(\mathcal{F}^{-1}\otimes 1),\\ \alpha_{\mathcal{G}}&=\mathcal{G}\alpha(\cdot)\mathcal{G}^{-1}, & \Psi_{\mathcal{F},\mathcal{G}}&=(1\otimes\mathcal{F})(\mathrm{id}\otimes \Delta_h)(\mathcal{G})\Psi (\alpha\otimes \mathrm{id})(\mathcal{G}^{-1})(\mathcal{G}^{-1}\otimes 1). \end{align*} $$

Twisting defines an equivalence relation on the quasi-coactions. In categorical terms, it means that we pass from

to the equivalent tensor category

, with the tensor product defined by $\Delta _{h,\mathcal {F}}$ , and, up to this equivalence, the

-module category

is equivalent to the

-module category

As the following result shows, twisting often allows one to push all the information on a quasi-coaction into the associators.

Next, given a quasi-coaction of , assume in addition we have an automorphism $\beta $ of . If $\mathcal {F}$ satisfies $(\beta \otimes \beta )(\mathcal {F})=\mathcal {F}$ , then $\beta $ remains an automorphism of . Assume also that is an R-matrix for that is fixed under $\beta $ . Then $\mathcal {R}_{\mathcal {F}}=\mathcal {F}_{21}\mathcal {R}\mathcal {F}^{-1}$ is an R-matrix for , again fixed by $\beta $ . Given a ribbon $\beta $ -braid $\mathcal {E}$ for the original quasi-coaction we get a ribbon $\beta $ -braid $\mathcal {E}_{\mathcal {G}}$ for the twisted quasi-coaction of defined by

(1.7) $$ \begin{align} \mathcal{E}_{\mathcal{G}}=\mathcal{G}\mathcal{E}(\mathrm{id}\otimes\beta)(\mathcal{G})^{-1}. \end{align} $$

The condition $(\beta \otimes \beta )(\mathcal {F})=\mathcal {F}$ can be relaxed; we will return to this in Section 5.1.

1.6 Symmetric pairs

Let $\mathfrak {k}$ be a proper Lie subalgebra of $\mathfrak {u}$ . We say that $\mathfrak {k}<\mathfrak {u}$ is a symmetric pair, or more precisely, an irreducible symmetric pair of type I, if there is a (necessarily unique) involutive automorphism $\sigma $ of $\mathfrak {u}$ such that $\mathfrak {k}=\mathfrak {u}^{\sigma }$ . Whenever convenient, we extend $\sigma $ to $\mathcal {U}(G)$ , in particular, to $\mathfrak {g}$ . Let $K=U^{\sigma }$ . The compact group K is connected by [Reference HelgasonHel01, Theorem VII.8.2]. Using the Cartan decomposition of G, we can also conclude that $G^{\sigma }$ is connected.

Given such a symmetric pair, put

$$\begin{align*}\mathfrak{m}=\{X\in\mathfrak{u}\mid \sigma(X)=-X\}, \end{align*}$$

which is the orthogonal complement of $\mathfrak {k}$ in $\mathfrak {u}$ with respect to the invariant inner product. We also write $\mathfrak {m}^{\mathbb {C}} = \mathfrak {m} \otimes _{\mathbb {R}} \mathbb {C}$ for its complexification.

We say that a symmetric pair $\mathfrak {k}<\mathfrak {u}$ is Hermitian, if $U/K$ is a Hermitian symmetric space. Such symmetric pairs are equivalently characterized by either of the following conditions; see [Reference BorelBor98, Proposition VI.1.3]:

• • The center $\mathfrak {z}(\mathfrak {k})$ is nontrivial (and $1$ -dimensional);

• • The space $\mathfrak {m}$ has a (unique up to a sign) $\mathfrak {k}$ -invariant complex structure.

The following closely related characterization will be crucial for us.

An irreducible symmetric pair of type II is an inclusion that is isomorphic to the diagonal inclusion of $\mathfrak {u}$ into $\mathfrak {u} \oplus \mathfrak {u}$ (with a simple compact Lie algebra $\mathfrak {u}$ ). This corresponds to the involution $\sigma (X,Y) = (Y,X)$ on $\mathfrak {u} \oplus \mathfrak {u}$ . For such a pair, we can put $\mathfrak {m} = \{(X, -X) \mid X \in \mathfrak {u}\}$ . Since both $\mathfrak {m}^{\mathfrak {u}}$ and $({\textstyle \bigwedge }^2 \mathfrak {m})^{\mathfrak {u}}$ are trivial, such pairs behave in many respects similarly to the non-Hermitian type I pairs. We will therefore mostly focus on the type I case and only make a few remarks on the type II case.

Back to type I symmetric pairs, in the Hermitian case, it is known that an invariant complex structure on $\mathfrak {m}$ is defined by an element of $\mathfrak {z}(\mathfrak {k})$ . The correct normalization is given by the following.

Lemma 1.4. Assuming that $\mathfrak {k}<\mathfrak {u}$ is a Hermitian symmetric pair, let $Z \in \mathfrak {z}(\mathfrak {k})$ be a vector such that $(Z,Z)_{\mathfrak {g}}=-1$ . Then on $\mathfrak {m}$ we have $(\operatorname {\mathrm {ad}} Z)^2=-a_{\sigma }^2\mathrm {id}$ , where

$$ \begin{align*} a_{\sigma} = \sqrt{\frac{2 h^{\vee} c}{\dim \mathfrak{m}}},\\[-15pt] \end{align*} $$

$c\in \{1, 2, 3\}$ is the ratio of the square lengths of long and short roots of $\mathfrak {g}$ and $h^{\vee }$ is the dual Coxeter number of $\mathfrak {g}$ .

Corollary 1.5. The $\mathfrak {k}$ -invariant complex structures on $\mathfrak {m}$ are given by $\pm \frac {1}{a_{\sigma }} (\operatorname {\mathrm {ad}} Z)|_{\mathfrak {m}}$ . For the involutive automorphism $\sigma $ such that $\mathfrak {k}=\mathfrak {u}^{\sigma }$ , we have

$$ \begin{align*} \sigma = \exp\left(\frac{\pi}{a_{\sigma}} \operatorname{\mathrm{ad}} Z\right).\\[-15pt] \end{align*} $$

In particular, we see that K is the stabilizer of Z in U with respect to the adjoint action. As the adjoint and coadjoint representations are equivalent, this leads to yet another known characterization of the Hermitian symmetric pairs: A symmetric pair $\mathfrak {k}<\mathfrak {u}$ is Hermitian if and only if the homogeneous U-space $U/K$ is isomorphic to a coadjoint orbit of U.

2.1 Co-Hochschild cohomology for multiplier algebras

The co-Hochschild cochains will play a central role in this paper. Let H be a reductive algebraic subgroup of G. Put $\tilde {B}_{G,H}^n = \mathcal {U}(H \times G^n)$ for $0 \le n < \infty $ , and define a differential $\tilde {B}_{G,H}^n\to \tilde {B}_{G,H}^{n+1}$ by

(2.1) $$ \begin{align} d_{\mathrm{cH}}(T) = T_{01,2,\dots,n+1} - T_{0,1 2, \dots, n+1} + \dots + (-1)^n T_{0,1,\dots,n(n+1)} + (-1)^{n+1} T_{0,1,\dots,n},\\[-15pt]\nonumber \end{align} $$

where $T_{0, \dots , j j + 1, \dots , n+1} = (\mathrm {id}_{\mathcal {U}(H \times G^{j-1})} \otimes \Delta \otimes \mathrm {id}_{\mathcal {U}(G^{n-j})})(T)$ and $T_{0,1,\dots ,n}=T\otimes 1$ . The group H acts diagonally by conjugation on $\mathcal {U}(H \times G^n)$ , the differential $d_{\mathrm {cH}}$ is equivariant with respect to this action. We put

$$\begin{align*}B_{G,H}^n=(\tilde{B}_{G,H}^n)^{H}. \end{align*}$$

Proof. The complex $\tilde {B}_{G,H}$ is the algebraic linear dual of $\tilde {B}_{G,H}' = (\mathcal {O}(H) \otimes \mathcal {O}(G)^{\otimes n})_{n=0}^{\infty }$ with the differential $d\colon \mathcal {O}(H) \otimes \mathcal {O}(G)^{\otimes n} \to \mathcal {O}(H) \otimes \mathcal {O}(G)^{\otimes n-1}$ given by

$$\begin{align*}d(f_0 \otimes f_1 \otimes \dots \otimes f_n) = \sum_{i=0}^{n-1} (-1)^i f_0 \otimes \dots \otimes f_i f_{i+1} \otimes \dots \otimes f_n + (-1)^n f_n(e) f_0 \otimes \dots \otimes f_{n-1}, \end{align*}$$

where $f_0 f_1$ is the product of $f_0$ and the restriction of $f_1$ to H. Thus, the cohomology of $\tilde {B}_{G,H}$ is the linear dual of the homology of $\tilde {B}_{G,H}'$ as an H-module.

The complex $\tilde {B}_{G,H}'$ is the standard complex computing the Hochschild homology

$$\begin{align*}\mathrm{HH}_*(\mathcal{O}(G), {}_{\mathrm{res}}\mathcal{O}(H)_{\epsilon}) = \operatorname{\mathrm{Tor}}^{\mathcal{O}(G) \otimes \mathcal{O}(G)}_*(\mathcal{O}(G), {}_{\mathrm{res}}\mathcal{O}(H)_{\epsilon}), \end{align*}$$

where the bimodule ${}_{\mathrm {res}}\mathcal {O}(H)_{\epsilon }$ has the underlying space $\mathcal {O}(H)$ with the bimodule structure $f.a.f' = f'(e) f a$ for $f, f' \in \mathcal {O}(G)$ and $a \in \mathcal {O}(H)$ . In other words, we are computing

$$\begin{align*}\operatorname{\mathrm{Tor}}^{\mathcal{O}(G\times G)}_*(\mathcal{O}(\Delta), \mathcal{O}(H\times\{e\})), \end{align*}$$

where $\Delta \subset G\times G$ is the diagonal. By [Reference Berthelot, Grothendieck, Ferrand, Jouanolou, Jussila, Kleiman and RaynaudBGI71, Proposition VII.2.5], this is the exterior algebra on $\operatorname {\mathrm {Tor}}^{\mathcal {O}(G\times G)}_1(\mathcal {O}(\Delta ), \mathcal {O}(H\times \{e\}))$ , and the latter is the conormal space of $H \subset G$ at the point e. Since this conormal space is the dual of $\mathfrak {g}/\mathfrak {h}$ , we obtain the assertion.

We will mainly need the following particular case.

2.2 Classification of associators and ribbon braids: non-Hermitian case

Assume the symmetric pair $\mathfrak {k}<\mathfrak {u}$ is non-Hermitian.

Consider a multiplier quasi-bialgebra such that $\Delta _h=\Delta $ modulo h. We claim that up to twisting by $(1,\mathcal {G})$ it has at most one quasi-coaction such that $\alpha =\Delta $ modulo h. Since by Lemma 1.1, we may assume that both $\Delta _h$ and $\alpha $ equal $\Delta $ , the following is an equivalent statement.

Proof. Suppose that $\Psi ^{(k)} = \Psi ^{\prime (k)}$ for $k < n$ . We claim that there is $T \in \mathcal {U}(G^{\sigma } \times G)^{\mathfrak {k}}$ such that $(\mathrm {id}\otimes \epsilon )(T)=0$ and $\Psi $ and $\mathcal {H}_{0,12} \Psi ' \mathcal {H}_{01,2}^{-1} \mathcal {H}_{0,1}^{-1}$ have the same terms up to (and including) order n for $\mathcal {H} = 1 - h^n T$ . The lemma is then proved by inductively applying this claim and taking the product of the elements $1 - h^n T$ we thus get. Note only that by $\mathfrak {k}$ -invariance the elements $T_{0,1}$ and $T_{01,2}$ obtained at different steps commute with each other.

Take the difference of identities (1.3) for $\Psi $ and $\Psi '$ and consider the terms of order n. Since $\Psi $ and $\Psi '$ have the same terms up to order $n-1$ , we get

$$\begin{align*}(\Psi^{(n)}-\Psi^{\prime (n)})_{0, 12, 3} + (\Psi^{(n)}-\Psi^{\prime (n)})_{0, 1, 2}=(\Psi^{(n)}-\Psi^{\prime (n)})_{0, 1, 23} + (\Psi^{(n)}-\Psi^{\prime (n)})_{01, 2, 3}.\\[-15pt] \end{align*}$$

Since $\Psi $ and $\Psi '$ are $\mathfrak {k}$ -invariant by equation (1.2), it follows that $\Psi ^{(n)} - \Psi ^{\prime (n)}$ is a cocycle in $B_{G, G^{\sigma }}^2$ . As we are in the non-Hermitian case, by Corollary 2.4 and Lemma 1.2, we have $\Psi ^{(n)}-\Psi ^{\prime (n)}=d_{\mathrm {cH}}(T)$ for some $T\in \mathcal {U}(G^{\sigma } \times G)^{\mathfrak {k}}$ . As $(\mathrm {id}\otimes \epsilon \otimes \mathrm {id})(\Psi ^{(n)}-\Psi ^{\prime (n)})=0$ , we have $(\mathrm {id}\otimes \epsilon )(T)=0$ . Thus, T satisfies our claim.

Next, let us fix a $\sigma $ -invariant R-matrix for $(\Delta , \Phi )$ and look at compatible ribbon $\sigma $ -braids. Note that the left-hand side of equation (1.4) becomes $\mathcal {E} \Delta (T)$ in the present case.

The group $Z_U(K)$ is finite as remarked in Remark 1.3, so we have at most finitely many ribbon $\sigma $ -braids.

Proof. From equation (1.5), we get that $(\Delta \otimes \mathrm {id})(\mathcal {E}^{(0)})=\mathcal {E}_{02}^{(0)}$ . This implies that $\mathcal {E}^{(0)}=1\otimes g$ , with $g=(\epsilon \otimes \mathrm {id})(\mathcal {E}^{(0)})\in \mathcal {U}(G)$ . From equation (1.6) we then get $\Delta (g)=g\otimes g$ , hence $g\in G$ . But then equation (1.4) shows that $g\in Z_G(G^{\sigma })$ . Finally, as $\mathfrak {z}_{\mathfrak {g}}(\mathfrak {k})=\mathfrak {z}_{\mathfrak {u}}(\mathfrak {k})^{\mathbb {C}}=0$ , using the Cartan decomposition of G we see that $Z_G(K)=Z_U(K)$ , hence $g\in Z_U(K)$ .

Assume now that $\mathcal {E}'$ is another ribbon $\sigma $ -braid with ${\mathcal {E}'}{}^{(0)}=1\otimes g$ . We want to show that $\mathcal {E}'=\mathcal {E}$ . It will be convenient to first get rid of g. By multiplying both elements by $1\otimes g^{-1}$ on the right, we get new ribbon $\tilde {\sigma }$ -braids $\tilde {\mathcal {E}}$ and $\tilde {\mathcal {E}}'$ in with the order zero terms $1$ , where $\tilde {\sigma }=(\operatorname {\mathrm {Ad}} g)\circ \sigma $ .

We argue by induction on n that $\tilde {\mathcal {E}}^{(n)}=\tilde {\mathcal {E}}^{\prime (n)}$ . Suppose that we already know that $\tilde {\mathcal {E}}^{(k)} = \tilde {\mathcal {E}}^{\prime (k)}$ for $k < n$ . Comparing the terms of degree n in equations (1.5) and (1.6), we obtain

$$ \begin{align*} X_{01,2} &= X_{0,2},& X_{0,12} &= X_{0,2} + X_{0,1}\\[-15pt] \end{align*} $$

for $X = \tilde {\mathcal {E}}^{(n)} - \tilde {\mathcal {E}}^{\prime (n)}$ .

The first equality says that $X = 1 \otimes Y$ for $Y = (\epsilon \otimes \mathrm {id})(X) \in \mathcal {U}(G)$ . Then the second equality says $\Delta (Y) = Y_1 + Y_2$ , that is, Y is primitive, and we obtain $Y \in \mathfrak {g}$ .

Comparing the terms of degree n in equation (1.4), we see next that Y has to centralize $\mathfrak {k}$ . Hence $Y=0$ and $\tilde {\mathcal {E}}^{\prime (n)} = \tilde {\mathcal {E}}^{(n)}$ .

It remains to prove the last statement of the theorem. So assume $\mathcal {R}^{(1)}_{\phantom {1}}+\mathcal {R}^{(1)}_{21}\ne 0$ . Let us write $\tilde {B}_{G}$ , $B_{\mathfrak {g}}$ , etc., instead of $\tilde {B}_{G,\{e\}}$ , $B_{\mathfrak {g},0}$ .

By an analogue of [Reference DrinfeldDri89b, Proposition 3.1] for the multiplier algebras, by twisting $\Phi $ we may assume that $\Phi =1$ modulo $h^2$ . Such an analogue is proved in the same way as in [Reference DrinfeldDri89b] using that the embedding map $\tilde {B}_{\mathfrak {g}}\to \tilde {B}_{G}$ is a quasi-isomorphism by Corollary A.5.

Namely, consider the normalized skew-symmetrization map $\operatorname {\mathrm {Alt}}\colon \tilde {B}^3_{G}\to \tilde {B}^3_{G}$ . This map kills the coboundaries and transforms the cocycles of the form $X_1\otimes X_2\otimes X_3$ , $X_i\in \mathfrak {g}$ , into cohomologous ones by Remark A.4. But the classes of such cocycles span the entire space $\mathrm {H}^3(\tilde {B}_G)$ by the same remark and Corollary A.5. Therefore, if $T\in \tilde {B}^3_G$ is a cocycle killed by $\operatorname {\mathrm {Alt}}$ , then it is a coboundary. The hexagon relations imply that $\operatorname {\mathrm {Alt}}(\Phi ^{(1)})=0$ , so $\Phi ^{(1)}=d_{\mathrm {cH}}(T)$ for some $T\in \tilde {B}^2_G$ . We may assume that T is G- and $\sigma $ -invariant since $\Phi ^{(1)}$ has these invariance properties. Then the twisting by $\mathcal {F}=1-h T$ proves our claim.

Note that twisting does not change the element $\mathcal {R}^{(1)}_{\phantom {1}}+\mathcal {R}^{(1)}_{21}$ . The hexagon relations imply then that $\mathcal {R}^{(1)}\in \mathfrak {g}\otimes \mathfrak {g}$ (see the proof of [Reference DrinfeldDri89b, Proposition 3.1]), and since $\mathcal {R}$ commutes with the image of $\Delta $ , we get $\mathcal {R}^{(1)}\in (\mathfrak {g}\otimes \mathfrak {g})^{\mathfrak {g}}$ . Hence, $\mathcal {R}^{(1)}=-\lambda t^{\mathfrak {u}}$ for some $\lambda \ne 0$ , where $t^{\mathfrak {u}}$ is the normalized invariant $2$ -tensor defined by equation (1.1).

Next, identity (1.3) implies that $\Psi ^{(1)}$ is a $2$ -cocycle in $B_{G, G^{\sigma }}$ ; hence, by twisting we may assume that $\Psi =1$ modulo $h^2$ . We remind also that under twisting the ribbon twist-braids transform via formula (1.7).

Now, by looking at the first order terms in equation (1.5) for a $\tilde {\sigma }$ -braid $\tilde {\mathcal {E}}$ , with $\tilde {\mathcal {E}}^{(0)}=1$ , we get

$$\begin{align*}\tilde{\mathcal{E}}^{(1)}_{01,2}=-\lambda t^{\mathfrak{u}}_{2,1}+\tilde{\mathcal{E}}^{(1)}_{0,2}-(\mathrm{id}\otimes\mathrm{id}\otimes\tilde{\sigma})(\lambda t^{\mathfrak{u}}_{1,2}).\\[-10pt] \end{align*}$$

The tensor $t^{\mathfrak {u}}$ lies in $\mathfrak {k}\otimes \mathfrak {k}+\mathfrak {m}\otimes \mathfrak {m}$ . Denote the components of $t^{\mathfrak {u}}$ in $\mathfrak {k}\otimes \mathfrak {k}$ and $\mathfrak {m}\otimes \mathfrak {m}$ by $t^{\mathfrak {k}}$ and $t^{\mathfrak {m}}$ , resp. Then the above identity can be written as

$$\begin{align*}\tilde{\mathcal{E}}^{(1)}_{01,2}=-2\lambda t^{\mathfrak{k}}_{1,2}+\tilde{\mathcal{E}}^{(1)}_{0,2}+(\mathrm{id}\otimes\mathrm{id}\otimes\operatorname{\mathrm{Ad}} g)(\lambda t^{\mathfrak{m}}_{1,2})-\lambda t^{\mathfrak{m}}_{1,2}.\\[-10pt] \end{align*}$$

Applying $\epsilon $ to the $0$ th leg and letting $T=(\epsilon \otimes \mathrm {id})(\tilde {\mathcal {E}}^{(1)})$ , we obtain

(2.2) $$ \begin{align} \tilde{\mathcal{E}}^{(1)}=-2\lambda t^{\mathfrak{k}}+1\otimes T+(\mathrm{id}\otimes\operatorname{\mathrm{Ad}} g)(\lambda t^{\mathfrak{m}})-\lambda t^{\mathfrak{m}}.\\[-10pt]\nonumber \end{align} $$

But we must have $\tilde {\mathcal {E}}^{(1)}\in \mathcal {U}(G^{\sigma }\times G)$ . Since $t^{\mathfrak {m}}=\sum _j Y_j\otimes Y^j$ for a basis $(Y_j)_j$ in $\mathfrak {m}$ and the dual basis $(Y^j)_j$ , this is possible only when $\operatorname {\mathrm {Ad}} g$ acts trivially on $\mathfrak {m}$ . Hence, $g\in Z(U)$ .

2.3 Associators from cyclotomic KZ equations

We want to extend the results of the previous subsection to the Hermitian case. Since $\mathrm {H}^2(B_{G,G^{\sigma }})$ is now one-dimensional by Lemma 1.2, we should expect a one-parameter family of nonequivalent associators. In this subsection, we define a candidate for such a family arising from the cyclotomic KZ-equations.

Thus, assume $\mathfrak {k}<\mathfrak {u}$ is a Hermitian symmetric pair. We have an element $Z\in \mathfrak {z}(\mathfrak {k})$ , unique up to a sign, such that

$$\begin{align*}(Z,Z)_{\mathfrak{g}}=-a_{\sigma}^{-2}. \end{align*}$$

This normalization is equivalent to $(\operatorname {\mathrm {ad}} Z)^2=-1$ on $\mathfrak {m}$ by Lemma 1.4. We fix such Z for the rest of this section. The operator $\operatorname {\mathrm {ad}} Z$ has eigenvalues $\pm i$ on $\mathfrak {m}^{\mathbb {C}}$ . Denote by $\mathfrak {m}_{\pm }\subset \mathfrak {m}^{\mathbb {C}}$ the corresponding eigenspaces.

We remind that we denote the components of the normalized invariant $2$ -tensor $t^{\mathfrak {u}}$ in $\mathfrak {k}\otimes \mathfrak {k}$ and $\mathfrak {m}\otimes \mathfrak {m}$ by $t^{\mathfrak {k}}$ and $t^{\mathfrak {m}}$ , resp. The tensor $t^{\mathfrak {m}}$ lies in $\mathfrak {m}_+\otimes \mathfrak {m}_- + \mathfrak {m}_-\otimes \mathfrak {m}_+$ . We denote the components of $t^{\mathfrak {m}}$ in $\mathfrak {m}_{\pm }\otimes \mathfrak {m}_{\mp }$ by $t^{\mathfrak {m}_{\pm }}$ . We thus have

(2.3) $$ \begin{align} t^{\mathfrak{m}} &= t^{\mathfrak{m}_+}+t^{\mathfrak{m}_-},& (\operatorname{\mathrm{ad}} Z \otimes \mathrm{id})(t^{\mathfrak{m}_{\pm}}) &= \pm i t^{\mathfrak{m}_{\pm}},& (\mathrm{id}\otimes\operatorname{\mathrm{ad}} Z)(t^{\mathfrak{m}_{\pm}}) &= \mp i t^{\mathfrak{m}_{\pm}}. \end{align} $$

Given $s \in \mathbb {C}$ , consider the following elements of :

(2.4) $$ \begin{align} A_{-1} &= \hbar(t^{\mathfrak{k}}_{12} - t^{\mathfrak{m}}_{12}),& A_1 &= \hbar t^{\mathfrak{u}}_{12},& A_0 &= \hbar(2 t^{\mathfrak{k}}_{01} + C^{\mathfrak{k}}_1) + s Z_1, \end{align} $$

where $C^{\mathfrak {k}}$ is the Casimir element of $\mathfrak {k}$ , the image of $t^{\mathfrak {k}}$ under the product map $U(\mathfrak {k})\otimes U(\mathfrak {k})\to U(\mathfrak {k})$ . These lead to the shifted modified $2$ -cyclotomic KZ $_2$ -equation [Reference Enriquez and EtingofEE05, Reference De Commer, Neshveyev, Tuset and YamashitaDCNTY19]

(2.5) $$ \begin{align} G'(w) = \left( \frac{A_{-1}}{w + 1} + \frac{A_1}{w-1} + \frac{A_0}{w} \right) G(w). \end{align} $$

Remark 2.10. Consider a -valued character $\nu $ on $U(\mathfrak {k})$ such that $\nu (Z) = -(2 \hbar a_{\sigma }^2)^{-1}s$ . Then the slicing map $\varsigma _{\nu } = (\nu \otimes \mathrm {id})\Delta $ is an algebra homomorphism satisfying

$$\begin{align*}(\varsigma_{\nu} \otimes \mathrm{id})(2t^{\mathfrak{k}}) = 2 t^{\mathfrak{k}} + \hbar^{-1}(1 \otimes s Z) \end{align*}$$

and commuting with the right coaction $\Delta $ by $U(\mathfrak {g})$ . In particular, at least formally speaking, equation (2.5) is obtained from the case $s = 0$ by slicing. But since $\nu $ cannot be extended to , one should be careful with this construction.

The normalized monodromy of equation (2.5) from $w = 0$ to $w = 1$ is well-defined as long as the operator $\operatorname {\mathrm {ad}}(s Z)$ on $\mathcal {U}(G)$ does not have positive integers in its spectrum, cf. [Reference Neshveyev and TusetNT11, Proposition 3.1]. Since each matrix block $\operatorname {\mathrm {End}}(V_{\pi })$ in $\mathcal {U}(G)$ is generated by the image of $\mathfrak {g}$ , the eigenvalues of $\operatorname {\mathrm {ad}} Z$ are $i n$ for $n \in \mathbb {Z}$ by Lemma 1.4 and our choice of normalization. Therefore, $\Psi _{\mathrm {KZ},s}$ is well defined for all $s\not \in i\mathbb {Q}^{\times }$ . The element $\Psi _{\mathrm {KZ},s}$ together with the coproduct $\Delta \colon \mathcal {U}(G^{\sigma })\to \mathcal {U}(G^{\sigma }\times G)$ gives a quasi-coaction of on , where is Drinfeld’s KZ-associator for G.

In more detail, $\Psi _{\mathrm {KZ},s}$ is defined as follows. Under our restrictions on s, a standard argument (see, e.g., [Reference Neshveyev and TusetNT11, Proposition 3.1]) shows that there is a unique -valued solution $G_0$ of equation (2.5) on $(0,1)$ such that $G_0(w)w^{-A_0}$ extends to an analytic function in the unit disc with value $1$ at $w=0$ . Similarly, there is a unique solution $G_1$ of equation (2.5) such that $G_1(1-w)w^{-A_1}$ extends to an analytic function in the unit disc with value $1$ at $w=0$ . Then

$$\begin{align*}\Psi_{\mathrm{KZ},s}=G_1(w)^{-1}G_0(w) \end{align*}$$

for any $0<w<1$ . We can also write this as

(2.6) $$ \begin{align} \Psi_{\mathrm{KZ},s}=\lim_{w\to 1}(1-w)^{-A_1}G_0(w)=\lim_{w\to 1}(1-w)^{-A_1}G_0(w)w^{-A_0}. \end{align} $$

The case $s=0$ is special: In this case, it can be shown, using for example iterated integrals [Reference KasselKas95, Chapter XIX][Reference EnriquezEnr07], that $\Psi _{\mathrm {KZ},0}$ lives in the algebra rather than in its completion . Note also that this associator is well defined in the non-Hermitian case as well. We will denote it by $\Psi _{\mathrm {KZ}}$ .

Observe also that if $s\in \mathbb {R}$ , then $G_0$ is unitary; hence, $\Psi _{\mathrm {KZ},s}$ is unitary as well. Indeed, in this case $(G_0(w)^*)^{-1}$ has the defining properties of $G_0(w)$ , hence coincides with it.

Proposition 2.11 (cf. [Reference Enriquez and EtingofEE05, Proposition 4.7])

For every $s\not \in i\mathbb {Q}^{\times }$ , we have

$$\begin{align*}\Psi_{\mathrm{KZ},s}= 1 + \frac{h}{\pi i}\left((\log 2)t^{\mathfrak{u}}_{12}+\gamma t^{\mathfrak{m}}_{12}+ \psi\Big(\frac{1}{2}-\frac{is}{2}\Big)t^{\mathfrak{m}_+}_{12} + \psi\Big(\frac{1}{2}+\frac{is}{2}\Big)t^{\mathfrak{m}_-}_{12}\right)+ O(h^2), \end{align*}$$

where $\gamma $ is Euler’s constant and $\displaystyle \psi =\frac {\Gamma '}{\Gamma }$ is the digamma function.

Proof. If we restrict to a finite-dimensional block of $\mathcal {U}(G^{\sigma }\times G^2)$ , then $\operatorname {\mathrm {ad}}(s Z_1)$ has a finite number of eigenvalues there, so the corresponding component of $\Psi _{\mathrm {KZ},s}$ is well defined for all $s\not \in i N^{-1}\mathbb {Z}^{\times }$ for some N. As it is analytic in s in this domain, it therefore suffices to consider real s.

Put $H_0(w)=G_0(w)w^{-A_0}$ . Then $H_0$ satisfies the differential equation

$$\begin{align*}H^{\prime}_0(w)=\left( \frac{A_{-1}}{w + 1} + \frac{A_1}{w-1} \right)H_0(w)+\left[\frac{A_0}{w},H_0(w)\right] \end{align*}$$

and the initial condition $H_0(0)=1$ , and by equation (2.6) we have

$$\begin{align*}\Psi_{\mathrm{KZ},s}=\lim_{w\to 1}(1-w)^{-A_1}H_0(w). \end{align*}$$

Consider the expansion in h. For the order zero terms, we immediately get $\Psi ^{(0)}_{\mathrm {KZ},s}=H^{(0)}_0=1$ . Next, consider the order one terms. Let us write H for $\pi i H^{(1)}_0$ , so that $H_0=1+\hbar H+O(h^2)$ . Then

$$\begin{align*}H'(w)=\left( \frac{t^{\mathfrak{k}}_{12} - t^{\mathfrak{m}}_{12}}{w + 1} + \frac{t^{\mathfrak{u}}_{12}}{w-1} \right)+\left[\frac{s Z_1}{w},H(w)\right] \end{align*}$$

and $H(0)=0$ , while

$$\begin{align*}\pi i\Psi_{\mathrm{KZ},s}^{(1)}=\lim_{w\to1}(H(w)-\mathinner{\log(1-w)} t^{\mathfrak{u}}_{12}). \end{align*}$$

By equation (2.3), we have

$$ \begin{align*} H(w)&=\int^w_0\left(\frac{w}{u}\right)^{\operatorname{\mathrm{ad}} (s Z_1)}\left( \frac{t^{\mathfrak{k}}_{12} - t^{\mathfrak{m}}_{12}}{u + 1} + \frac{t^{\mathfrak{u}}_{12}}{u-1} \right) d u \\ &\ = \int^w_0 \left( t^{\mathfrak{k}}_{12} \Big(\frac1{u+1} + \frac1{u-1} \Big) + \Big( \left( \frac{w}{u} \right)^{i s} t^{\mathfrak{m}_+}_{12} + \left( \frac{w}{u} \right)^{-i s} t^{\mathfrak{m}_-}_{12} \Big) \Big( \frac{-1}{u+1} + \frac1{u-1} \Big) \right)d u. \end{align*} $$

Note that this integral is well defined for $0\le w<1$ as s is assumed to be real. We then get

$$\begin{align*}\pi i\Psi_{\mathrm{KZ},s}^{(1)}=b t^{\mathfrak{k}}_{12}+c(s)t^{\mathfrak{m}_+}_{12}+c(-s)t^{\mathfrak{m}_-}_{12}, \end{align*}$$

where

$$ \begin{align*} b&=\lim_{w\to 1}\left(\int^w_0\left( \frac{1}{u + 1} + \frac{1}{u-1} \right) d u-\log(1-w)\right)=\log 2,\\ c(s)&=\lim_{w\to 1}\left(\int^w_0\left(\frac{w}{u}\right)^{is}\left( \frac{1}{u - 1} - \frac{1}{u+1} \right) d u-\log(1-w)\right). \end{align*} $$

To compute $c(s)$ , we write $(u^2-1)^{-1}$ as a power series, integrate and get

$$\begin{align*}c(s)=\lim_{w\to 1}\bigg(-\sum^{\infty}_{n=0}\frac{w^{2n+1}}{n+\frac{1}{2}-\frac{is}{2}}-\log(1-w)\bigg). \end{align*}$$

Together with the Taylor expansion of $w^{-1}\log (1-w^2)$ and the standard formula

$$\begin{align*}\psi(z)+\gamma=\sum^{\infty}_{n=0}\left(\frac{1}{n+1}-\frac{1}{n+z}\right), \end{align*}$$

this gives

$$\begin{align*}c(s)=\psi\Big(\frac{1}{2}-\frac{is}{2}\Big)+\gamma+\lim_{w\to 1}(w^{-1}\log(1-w^2)-\log(1-w)) =\psi\Big(\frac{1}{2}-\frac{is}{2}\Big)+\gamma+\log2, \end{align*}$$

which completes the proof of the proposition.

Using the formula $\psi (1-z)-\psi (z)=\pi \cot (\pi z)$ , it will be convenient to rewrite the result as

(2.7) $$ \begin{align} \Psi_{\mathrm{KZ},s} &= 1 + \frac{h}{\pi i}\bigg((\log 2)t^{\mathfrak{u}}_{12}+\Big(\gamma+\frac{\psi\Big(\frac{1}{2}-\frac{is}{2}\Big)+ \psi\Big(\frac{1}{2}+\frac{is}{2}\Big)}{2}\Big)t^{\mathfrak{m}}_{12}\notag \\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad-\frac{\pi i}{2}\tanh\Big(\frac{\pi s}{2}\Big)\Big(t^{\mathfrak{m}_+}_{12}-t^{\mathfrak{m}_-}_{12}\Big)\bigg)+ O(h^2). \end{align} $$

Now, take . Replacing $s\not \in i\mathbb {Q}^{\times }$ by $s+\mu $ in equation (2.4), we can construct yet another associator, which we denote by $\Psi _{\mathrm {KZ},s;\mu }$ . If $s\in \mathbb {R}$ and , then $\Psi _{\mathrm {KZ},s;\mu }$ is unitary for the same reason as for $\Psi _{\mathrm {KZ},s}$ .

Remark 2.12. Similarly to Remark 2.10, $\Psi _{\mathrm {KZ},s;\mu }$ could be obtained from $\Psi _{\mathrm {KZ},s}$ by slicing by a character $\nu $ of $U(\mathfrak {k})$ satisfying $\nu (Z) = -(2\hbar a_{\sigma }^2)^{-1}\mu $ . Since such a character does not always extend to $\mathcal {U}(G^{\sigma })$ , to make sense of this we should have allowed in the construction of $\Psi _{\mathrm {KZ},s}$ arbitrary finite-dimensional representations of $\mathfrak {g}^{\sigma }$ instead of those in $\operatorname {\mathrm {Rep}} G^{\sigma }$ . Alternatively, with $s\not \in i\mathbb {R}$ fixed, both $\Psi _{\mathrm {KZ},s;\mu }$ and $\Psi _{\mathrm {KZ},s+z}$ for small z are specializations of an associator in constructed by treating $\mu $ as a second formal parameter. But this implies that $\Psi _{\mathrm {KZ},s;\mu }$ is obtained from the Taylor expansion of $\Psi _{\mathrm {KZ},s+z}$ at $z=0$ by simply taking $\mu $ as the argument:

(2.8) $$ \begin{align} \Psi_{\mathrm{KZ},s;\mu}=\sum^{\infty}_{k=0}\frac{\mu^k}{k!}\frac{d^k\Psi_{\mathrm{KZ},s}}{d s^k}=\sum_{n,k=0}^{\infty}\frac{h^n\mu^k}{k!}\frac{d^k\Psi^{(n)}_{\mathrm{KZ},s}}{d s^k}. \end{align} $$

This also works for $s\in i(\mathbb {R}\setminus \mathbb {Q}^{\times })$ if we consider only the components of $\Psi _{\mathrm {KZ},s;\mu }$ in finite-dimensional blocks of $\mathcal {U}(G^{\sigma }\times G^2)$ , which are well defined and analytic in a neighborhood of s.

Corollary 2.13. For all $s\not \in i\mathbb {Q}^{\times }$ and , we have

$$ \begin{align*} \Psi_{\mathrm{KZ},s;\mu + \nu} - \Psi_{\mathrm{KZ},s; \mu} &= h^{1+\operatorname{\mathrm{ord}}(\nu)}\nu^{(\operatorname{\mathrm{ord}}(\nu))}\bigg(\frac{1}{4\pi}\Big(\psi'\Big(\frac{1}{2}+\frac{is}{2}\Big)- \psi'\Big(\frac{1}{2}-\frac{is}{2}\Big)\Big)t^{\mathfrak{m}}_{12}\\ &\qquad\qquad\qquad\qquad\qquad -\frac{\pi}{4}\operatorname{\mathrm{sech}}^2\Big(\frac{\pi s}{2}\Big)\Big(t^{\mathfrak{m}_+}_{12}-t^{\mathfrak{m}_-}_{12}\Big)\bigg)+ O(h^{2 + \operatorname{\mathrm{ord}}(\nu)}). \end{align*} $$

Proof. By equation (2.8), we have

$$\begin{align*}\Psi_{\mathrm{KZ},s;\mu + \nu} - \Psi_{\mathrm{KZ},s; \mu} = h^{1+\operatorname{\mathrm{ord}}(\nu)}\nu^{(\operatorname{\mathrm{ord}}(\nu))}\frac{d\Psi^{(1)}_{\mathrm{KZ},s}}{d s}+ O(h^{2 + \operatorname{\mathrm{ord}}(\nu)}). \end{align*}$$

Hence, the result follows from equation (2.7).

2.4 Detecting co-Hochschild classes

To see that the associators $\Psi _{\mathrm {KZ},s;\mu }$ are not all equivalent, we need to see that a perturbation of the parameter $\mu $ gives rise to a nontrivial $2$ -cocycle in $B_{G,G^{\sigma }}$ . We can actually see that this is the case from results in Appendix A, but let us present a concrete cycle to detect this.

Consider the tensor

(2.9) $$ \begin{align} \Omega = [t^{\mathfrak{u}}_{12}, t^{\mathfrak{u}}_{13}] = \sum_{i, j} [X_i, X_j] \otimes X^i \otimes X^j\in \Big({\textstyle\bigwedge}^3 \mathfrak{u}\Big)^{\mathfrak{u}} \end{align} $$

with $(X_i)_i$ and $(X^i)_i$ as in equation (1.1). Every element $X\in \mathfrak {g}$ defines a function on G such that $g\mapsto (X,(\operatorname {\mathrm {Ad}} g)(Z))_{\mathfrak {g}}$ . This way $\Omega $ defines an element of $\mathcal {O}(G^{\sigma })\otimes \mathcal {O}(G)\otimes \mathcal {O}(G)$ , which by slightly abusing notation we continue to denote by $\Omega $ . Thus, for $(g,h,k)\in G^{\sigma }\times G\times G$ ,

$$\begin{align*}\Omega(g,h,k)=\Big([(\operatorname{\mathrm{Ad}} h)(Z),(\operatorname{\mathrm{Ad}} k)(Z)],(\operatorname{\mathrm{Ad}} g)(Z)\Big)_{\mathfrak{g}}=\Big([(\operatorname{\mathrm{Ad}} h)(Z),(\operatorname{\mathrm{Ad}} k)(Z)],Z\Big)_{\mathfrak{g}} \end{align*}$$

since $G^{\sigma }$ stabilizes Z. This is a $2$ -cycle in the complex $\tilde B^{\prime }_{G,G^{\sigma }}$ from the proof of Proposition 2.1, as

$$\begin{align*}\Omega(g,g,h)-\Omega(g,h,h)+\Omega(g,h,e)=0 \end{align*}$$

for all $(g,h)\in G^{\sigma }\times G$ . Hence, the map $\langle \Omega ,\cdot \rangle \colon \mathcal {U}(G^{\sigma }\times G^2)\to \mathbb {C}$ defined by pairing with $\Omega $ passes to $\mathrm {H}^2(B_{G,G^{\sigma }})$ . Explicitly, for $T\in \mathcal {U}(G^{\sigma }\times G^2)$ , we have

(2.10) $$ \begin{align} \langle \Omega,T \rangle=\epsilon(T_0)\Big([(\operatorname{\mathrm{ad}} T_1)(Z),(\operatorname{\mathrm{ad}} T_2)(Z)],Z\Big)_{\mathfrak{g}}, \end{align} $$

where $\operatorname {\mathrm {ad}}$ denotes the extension of the adjoint representation of $\mathfrak {g}$ to $\mathcal {U}(G)$ .

Proposition 2.14. The elements $t^{\mathfrak {k}}_{12}$ , $t^{\mathfrak {m}_{\pm }}_{12}$ are $2$ -cocycles in $B_{G,G^{\sigma }}$ . Furthermore, $t^{\mathfrak {k}}_{12}$ and $t^{\mathfrak {m}_{\phantom {-}}}_{12}=t^{\mathfrak {m}_+}_{12}+t^{\mathfrak {m}_-}_{12}$ are coboundaries, while

$$\begin{align*}\langle\Omega,t^{\mathfrak{m}_{\pm}}_{12}\rangle=\pm \frac{i}{2}\dim \mathfrak{m}. \end{align*}$$

In particular, $t^{\mathfrak {m}_+}_{12}$ and $-t^{\mathfrak {m}_-}_{12}$ represent the same nontrivial class in $\mathrm {H}^2(B_{G,G^{\sigma }})$ .

Proof. It is easy to check that $d_{\mathrm {cH}}(1\otimes X\otimes Y)=0$ for all $X,Y\in \mathfrak {g}$ . As $t^{\mathfrak {k}}_{12}$ and $t^{\mathfrak {m}_{\pm }}_{12}$ are $\mathfrak {k}$ -invariant, they are therefore $2$ -cocycles in $B_{G,G^{\sigma }}$ .

We have

$$\begin{align*}d_{\mathrm{cH}}(C^{\mathfrak{k}}_1) = C^{\mathfrak{k}}_2 - \Delta(C^{\mathfrak{k}})_{12} + C^{\mathfrak{k}}_1 = -2 t^{\mathfrak{k}}_{12}, \end{align*}$$

so $t^{\mathfrak {k}}_{12}$ is a coboundary. Similarly, $d_{\mathrm {cH}}(C^{\mathfrak {u}}_1) = -2 t^{\mathfrak {u}}_{12}$ so that $ t^{\mathfrak {u}}_{12}$ is also a coboundary, and hence $t^{\mathfrak {m}}_{12}=t^{\mathfrak {u}}_{12}-t^{\mathfrak {k}}_{12}$ is a coboundary as well.

Next, take a basis $(Y_j)_j$ in $\mathfrak {m}_+$ and the dual basis $(Y^j)_j$ in $\mathfrak {m}_-$ . Using that $\operatorname {\mathrm {ad}} Z$ acts by the scalar $\pm i$ on $\mathfrak {m}_{\pm }$ , we then compute:

$$ \begin{align*} \langle \Omega,t^{\mathfrak{m}_+}_{12}\rangle&=\sum_j \Big( [(\operatorname{\mathrm{ad}} Y_j)(Z), (\operatorname{\mathrm{ad}} Y^j)(Z)], Z \Big)_{\mathfrak{g}}=\sum_j ([Y_j, Y^j],Z)_{\mathfrak{g}} =\sum_j (Y^j,[Z,Y_j])_{\mathfrak{g}}\\ &=i\sum_j(Y^j,Y_j)_{\mathfrak{g}}=i\dim_{\mathbb{C}}\mathfrak{m}_+=\frac{i}{2}\dim_{\mathbb{R}} \mathfrak{m}. \end{align*} $$

The value $\langle \Omega ,t^{\mathfrak {m}_-}_{12}\rangle $ is obtained similarly, but it also follows from the above, as $t^{\mathfrak {m}_-}_{12}=t^{\mathfrak {m}_{\phantom {-}}}_{12}-t^{\mathfrak {m}_+}_{12}$ and $t^{\mathfrak {m}_{\phantom {-}}}_{12}$ is a coboundary.

Remark 2.15. Let us give a different perspective on the above pairing and its nontriviality.

We can view the tensor (2.9) also as a function on $(U/K)^3$ in the same way as above. Let us call this function $\omega $ . Then it is again easy to check that $\omega $ is a $2$ -cycle in the Hochschild chain complex $(C_n(A,A)=A^{\otimes (n+1)},b)$ for $A=\mathcal {O}(U)^{\mathfrak {k}}\subset C(U/K)$ . Under the Hochschild–Kostant–Rosenberg map, this cycle corresponds to the differential $2$ -form associated with the Kostant–Kirillov–Souriau bracket on the coadjoint orbit of $(\cdot ,Z)_{\mathfrak {g}}$ , which in turn defines a nonzero class in $\mathrm {H}^2(U/K;\mathbb {C})\cong \mathbb {C}$ .

We have a left $\mathcal {U}(G)$ -module structure on $\mathcal {O}(U)$ given by right translations: $T.a=a_{(0)}\langle a_{(1)},T\rangle $ . Given $T\in B^n_{G,G^{\sigma }}$ , we can then define an n-cocycle $D_T$ in the Hochschild cochain complex $(C^n(A,A)=\operatorname {\mathrm {Hom}}(A^{\otimes n},A),\delta )$ by

$$\begin{align*}D_T(a_1, \dots, a_n) = \epsilon(T_0)(T_1.a_1)\cdots (T_n.a_n). \end{align*}$$

The Hochschild cochains act on the chains by contractions: Given $D \in C^m(A, A)$ , we have

$$\begin{align*}i_D \colon C_n(A,A) \to C_{n-m}(A,A), \quad a_0\otimes \dots \otimes a_n \mapsto a_0 D(a_1, \dots, a_{m})\otimes a_{m+1}\otimes \dots \otimes a_n, \end{align*}$$

with the convention $i_D = 0$ if $n < m$ .

Now, if $T\in B^n_{G,G^{\sigma }}$ and $c\in C_n(A,A)$ is U-invariant (with respect to left translations), then $i_{D_T}c\in A$ is U-invariant, hence a scalar. It can be checked that if $b c=0$ , then this scalar depends only on the cohomology class of T. Taking $c=\omega $ , we recover pairing (2.10): $i_{D_T}\omega =\langle \Omega ,T\rangle $ .

2.5 Classification of associators: Hermitian case

We are now ready to establish, in the Hermitian case, a universality result for the associators $\Psi _{\mathrm {KZ},s;\mu }$ for generic quasi-coactions of such that $\alpha =\Delta $ modulo h. Similarly to Section 2.2, it suffices to consider the case $\alpha =\Delta $ .

Theorem 2.16. Let $\mathfrak {k} = \mathfrak {u}^{\sigma } <\mathfrak {u}$ be a Hermitian symmetric pair. Assume we are given a quasi-coaction of such that the number $\langle \Omega ,\Psi ^{(1)}\rangle $ defined by equation (2.10) is

$$\begin{align*}\text{neither}\quad \pm\frac{i}{2}\dim\mathfrak{m}, \quad \text{nor}\quad \frac{i(\zeta-1)}{2(\zeta+1)}\dim\mathfrak{m} \end{align*}$$

for a root of unity $\zeta \ne \pm 1$ . Then there exist $s \not \in i\mathbb {Q}^{\times }$ , and such that $(\mathrm {id}\otimes \epsilon )(\mathcal {H})=1$ and $\Psi _{\mathrm {KZ},s;\mu } =\mathcal {H}_{0, 12}\Psi \mathcal {H}_{01, 2}^{-1} \mathcal {H}_{0,1}^{-1}$ . Furthermore,

1. (i) The number s is unique up to adding $2i k$ ( $k\in \mathbb {Z}$ ), and once s is fixed, the element $\mu $ is uniquely determined;

2. (ii) We can choose $s\in \mathbb {R}$ if and only if $\langle \Omega ,\Psi ^{(1)}\rangle $ is a purely imaginary number in the interval

   $$\begin{align*}\left(-\frac{i}{2}\dim\mathfrak{m},\frac{i}{2}\dim\mathfrak{m}\right); \end{align*}$$

3. (iii) If $s\in \mathbb {R}$ and $\Psi $ is unitary, then and $\mathcal {H}$ can be chosen to be unitary.

We will later (Remark 2.20) slightly improve this result by showing that for any $\Psi $ the parameter $\mu $ is independent of the choice of s.

Proof. By Proposition 2.14 and our restrictions on $\Psi $ , we can choose $s \not \in i\mathbb {Q}^{\times }$ such that

(2.11) $$ \begin{align} -\frac{1}{2}\tanh\Big(\frac{\pi s}{2}\Big) \langle \Omega, t^{\mathfrak{m}_+}_{12}-t^{\mathfrak{m}_-}_{12}\rangle =-\frac{i}{2}\tanh\Big(\frac{\pi s}{2}\Big)\dim\mathfrak{m} = \langle \Omega,\Psi^{(1)}\rangle. \end{align} $$

We then start with $\mathcal {H}=1$ and $\mu =0$ and modify them by induction on n to have $\Psi _{\mathrm {KZ},s;\mu } =\mathcal {H}_{0, 12}\Psi \mathcal {H}_{01, 2}^{-1} \mathcal {H}_{0,1}^{-1}$ modulo $h^{n+1}$ .

Consider $n=1$ . By the proof of Theorem 2.6, $\Psi _{\mathrm {KZ},s;\mu }^{(1)}-\Psi ^{(1)}$ is a $2$ -cocycle in $B_{G,G^{\sigma }}$ . By Lemma 1.2 and Corollary 2.4, we have $\dim \mathrm {H}^2(B_{G,G^{\sigma }})=1$ . Hence, our choice of s, identity (2.7) and Proposition 2.14 imply that $\Psi _{\mathrm {KZ},s;\mu }^{(1)}-\Psi ^{(1)}$ is a coboundary so that $\Psi ^{(1)}-\Psi ^{(1)}_{\mathrm {KZ},s;\mu }=d_{\mathrm {cH}}(T)$ for some $T\in \mathcal {U}(G^{\sigma }\times G)^{\mathfrak {k}}$ . Letting $\mathcal {H}^{(1)}=T$ , we then get $\Psi _{\mathrm {KZ},s;\mu } = \mathcal {H}_{0, 12} \Psi \mathcal {H}_{01, 2}^{-1} \mathcal {H}_{0,1}^{-1}$ modulo $h^2$ .

For the induction step, assume we have $\Psi _{\mathrm {KZ},s;\mu } =\mathcal {H}_{0, 12}\Psi \mathcal {H}_{01, 2}^{-1} \mathcal {H}_{0,1}^{-1}$ modulo $h^{n+1}$ for some $n\ge 1$ . Then, again by the proof of Theorem 2.6,

$$\begin{align*}\Psi_{\mathrm{KZ},s;\mu}^{(n+1)}-(\mathcal{H}_{0, 12}\Psi \mathcal{H}_{01, 2}^{-1} \mathcal{H}_{0,1}^{-1})^{(n+1)} \end{align*}$$

is a $2$ -cocycle in $B_{G,G^{\sigma }}$ . On the other hand, by Corollary 2.13, for any $a\in \mathbb {C}$ , we have

$$\begin{align*}\Psi_{\mathrm{KZ},s;\mu+h^n a}^{(n+1)}-\Psi_{\mathrm{KZ},s;\mu}^{(n+1)}=-a\frac{\pi}{4}\operatorname{\mathrm{sech}}^2\Big(\frac{\pi s}{2}\Big)\Big(t^{\mathfrak{m}_+}_{12}-t^{\mathfrak{m}_-}_{12}\Big)+b t^{\mathfrak{m}}_{12} \end{align*}$$

for some $b\in \mathbb {C}$ , and $\Psi _{\mathrm {KZ},s;\mu +h^n a}^{(k)}=\Psi _{\mathrm {KZ},s;\mu }^{(k)}$ for $k\le n$ . As $t^{\mathfrak {m}_+}_{12}-t^{\mathfrak {m}_-}_{12}$ represents a nontrivial cohomology class, the value of $\operatorname {\mathrm {sech}}$ is nonzero for our s and $t^{\mathfrak {m}}_{12}$ is cohomologically trivial, we see that with different choices of a the above difference can represent arbitrary classes in $\mathrm {H}^2(B_{G,G^{\sigma }})\cong \mathbb {C}$ . In particular, we can find $a\in \mathbb {C}$ such that

(2.12) $$ \begin{align} (\mathcal{H}_{0, 12}\Psi \mathcal{H}_{01, 2}^{-1} \mathcal{H}_{0,1}^{-1})^{(n+1)}-\Psi_{\mathrm{KZ},s;\mu+h^n a}^{(n+1)}=d_{\mathrm{cH}}(T) \end{align} $$

for some $T\in \mathcal {U}(G^{\sigma }\times G)^{\mathfrak {k}}$ . Replacing $\mathcal {H}$ by $(1+h^{n+1}T)\mathcal {H}$ and $\mu ^{(n)}$ by $\mu ^{(n)}+a$ , we then get $\Psi _{\mathrm {KZ},s;\mu } =\mathcal {H}_{0, 12}\Psi \mathcal {H}_{01, 2}^{-1} \mathcal {H}_{0,1}^{-1}$ modulo $h^{n+2}$ , proving the induction step.

As at the step n of our induction process we only modify $\mu ^{(n-1)}$ and $\mathcal {H}^{(k)}$ for $k\ge n$ , in the limit we get the required $\mu $ and $\mathcal {H}$ . It remains to prove (i)–(iii).

(iii): Assume $s\in \mathbb {R}$ and that $\Psi $ is unitary. In this case, we slightly modify the above inductive procedure to make sure that at every step we have unitarity of $\mathcal {H}$ and that .

Consider $n=1$ . We found $\mathcal {H}$ such that $\Psi _{\mathrm {KZ},s;\mu } = \mathcal {H}_{0, 12} \Psi \mathcal {H}_{01, 2}^{-1} \mathcal {H}_{0,1}^{-1}$ modulo $h^2$ . The unitarity of $\Psi _{\mathrm {KZ},s;\mu }$ and $\Psi $ implies then that the same identity holds for the unitary $\mathcal {H}(\mathcal {H}^*\mathcal {H})^{-1/2}$ instead of $\mathcal {H}$ , cf. the proof of [Reference Neshveyev and TusetNT11, Proposition 2.3].

For the induction step, we assume that we have $\Psi _{\mathrm {KZ},s;\mu } =\mathcal {H}_{0, 12}\Psi \mathcal {H}_{01, 2}^{-1} \mathcal {H}_{0,1}^{-1}$ modulo $h^{n+1}$ for some $n\ge 1$ , and unitary $\mathcal {H}$ . Then we take the unique $a\in \mathbb {C}$ such that

$$\begin{align*}(\mathcal{H}_{0, 12}\Psi \mathcal{H}_{01, 2}^{-1} \mathcal{H}_{0,1}^{-1})^{(n+1)}-\Psi_{\mathrm{KZ},s;\mu}^{(n+1)}+a\frac{\pi}{4}\operatorname{\mathrm{sech}}^2 \Big(\frac{\pi s}{2}\Big)\Big(t^{\mathfrak{m}_+}_{12}-t^{\mathfrak{m}_-}_{12}\Big) \end{align*}$$

is a coboundary. By taking adjoints and using that $(t^{\mathfrak {m}_+})^*=t^{\mathfrak {m}_-}$ , we also get that

$$\begin{align*}\Big((\mathcal{H}_{0, 12}\Psi \mathcal{H}_{01, 2}^{-1} \mathcal{H}_{0,1}^{-1})^{(n+1)}-\Psi_{\mathrm{KZ},s;\mu}^{(n+1)}\Big)^*-\bar a\frac{\pi}{4}\operatorname{\mathrm{sech}}^2\Big(\frac{\pi s}{2}\Big)\Big(t^{\mathfrak{m}_+}_{12}-t^{\mathfrak{m}_-}_{12}\Big) \end{align*}$$

is a coboundary. Hence, in order to conclude that $a\in \mathbb {R}$ it suffices to show that if we have two unitaries $\Psi _1$ and $\Psi _2$ in such that $\Psi _1=\Psi _2$ modulo $h^{n+1}$ , then the element $\Psi _1^{(n+1)}-\Psi _2^{(n+1)}$ is skew-adjoint. But this is clear from the identity

$$\begin{align*}(\Psi_1-\Psi_2)^*=-\Psi_1^*(\Psi_1-\Psi_2)\Psi_2^*. \end{align*}$$

Then we take T satisfying equation (2.12) and replace $\mu ^{(n)}$ by $\mu ^{(n)}+a$ and $\mathcal {H}$ by

$$\begin{align*}(1+h^{n+1}T)\Big((1+h^{n+1}T^*)(1+h^{n+1}T)\Big)^{-1/2}\mathcal{H}, \end{align*}$$

similarly to the step $n=1$ .

(i), (ii): If $\Psi _{\mathrm {KZ},s;\mu } =\mathcal {H}_{0, 12}\Psi \mathcal {H}_{01, 2}^{-1} \mathcal {H}_{0,1}^{-1}$ , then $\Psi ^{(1)}-\Psi ^{(1)}_{\mathrm {KZ},s}=d_{\mathrm {cH}}(\mathcal {H}^{(1)})$ . Hence, equation (2.11) is not only sufficient but also necessary for the existence of $\mathcal {H}$ and $\mu $ . Therefore, s is determined uniquely up to adding $2i k$ ( $k\in \mathbb {Z}$ ). This also makes (ii) obvious.

Next, assume $\Psi _{\mathrm {KZ},s;\mu } =\mathcal {H}_{0, 12}\Psi _{\mathrm {KZ},s;\mu '} \mathcal {H}_{01, 2}^{-1} \mathcal {H}_{0,1}^{-1}$ for some $s\not \in i\mathbb {Q}^{\times }$ , and . We have to show that $\mu =\mu '$ . Assume this is not the case.

Let $n\ge 1$ be the smallest order such that $\mu ^{(n)}\ne \mu ^{\prime (n)}$ . By Corollary 2.13, we have $\Psi _{\mathrm {KZ},s;\mu } = \Psi _{\mathrm {KZ},s;\mu '}$ modulo $h^{n+1}$ . We claim that we can modify $\mathcal {H}$ so that we still have $\Psi _{\mathrm {KZ},s;\mu } =\mathcal {H}_{0, 12}\Psi _{\mathrm {KZ},s;\mu '} \mathcal {H}_{01, 2}^{-1} \mathcal {H}_{0,1}^{-1}$ , but $\mathcal {H}=1$ modulo $h^{n+1}$ .

We will modify $\mathcal {H}$ by induction on $k\le n$ to get $\Psi _{\mathrm {KZ},s;\mu } =\mathcal {H}_{0, 12}\Psi _{\mathrm {KZ},s;\mu '} \mathcal {H}_{01, 2}^{-1} \mathcal {H}_{0,1}^{-1}$ and $\mathcal {H}=1$ modulo $h^{k+1}$ . Assume we have these two properties for some $k<n$ . Then

$$\begin{align*}d_{\mathrm{cH}}(\mathcal{H}^{(k+1)})=\Psi_{\mathrm{KZ},s;\mu'}^{(k+1)} -\Psi_{\mathrm{KZ},s;\mu}^{(k+1)}=0. \end{align*}$$

As $\mathrm {H}^1(B_{G,G^{\sigma }})=0$ by Lemma 1.2 and Corollary 2.4, there exists a central element $S\in \mathcal {U}(G^{\sigma })$ such that $\mathcal {H}^{(k+1)}=S_{01}-S_0$ . Putting $\mathcal {H}'=\exp (-h^{k+1}(S_{01}-S_0))$ , we see that $\Psi _{\mathrm {KZ},s;\mu }$ commutes with $\mathcal {H}^{\prime }_{0,12}$ , hence we have $\mathcal {H}^{\prime }_{0, 12}\Psi _{\mathrm {KZ},s;\mu } \mathcal {H}_{01, 2}^{\prime -1} \mathcal {H}_{0,1}^{\prime -1}=\Psi _{\mathrm {KZ},s;\mu }$ . It follows that by replacing $\mathcal {H}$ by $\mathcal {H}' \mathcal {H}$ we get $\Psi _{\mathrm {KZ},s;\mu } =\mathcal {H}_{0, 12}\Psi _{\mathrm {KZ},s;\mu '} \mathcal {H}_{01, 2}^{-1} \mathcal {H}_{0,1}^{-1}$ and $\mathcal {H}=1$ modulo $h^{k+2}$ . Thus, our claim is proved.

It follows now that $\Psi _{\mathrm {KZ},s;\mu '}^{(n+1)} -\Psi _{\mathrm {KZ},s;\mu }^{(n+1)}=d_{\mathrm {cH}}(\mathcal {H}^{(n+1)})$ . Since $t^{\mathfrak {m}_+}_{12}-t^{\mathfrak {m}_-}_{12}$ is not a coboundary, this contradicts Corollary 2.13. Hence, $\mu =\mu '$ .

2.6 Classification of ribbon braids

We complement our classification of associators by describing compatible ribbon twist-braids, both in the Hermitian and non-Hermitian cases.

In the following, we always take the universal R-matrix

for

.

Theorem 2.18. If $\mathfrak {k} = \mathfrak {u}^{\sigma } <\mathfrak {u}$ is a non-Hermitian symmetric pair, then the ribbon $\sigma $ -braids for the quasi-coaction of are the elements

$$\begin{align*}\mathcal{E}_{\mathrm{KZ}} g_1 = \exp(-h(2t^{\mathfrak{k}}_{01}+C^{\mathfrak{k}}_1))g_1, \end{align*}$$

where $g\in Z(U)$ .

A similar result holds in the Hermitian case, but the proof is more involved.

Theorem 2.19. If $\mathfrak {k} = \mathfrak {u}^{\sigma } < \mathfrak {u}$ is a Hermitian symmetric pair, $s\not \in i\mathbb {Q}^{\times }$ and , then the ribbon $\sigma $ -braids for the quasi-coaction of are the elements

$$\begin{align*}\mathcal{E}_{\mathrm{KZ},s;\mu} g_1 = \exp(-h(2t^{\mathfrak{k}}_{01}+C^{\mathfrak{k}}_1)-\pi i(s+\mu)Z_1)g_1, \end{align*}$$

where $g\in Z(U)$ .

Proof. The fact that the element $\mathcal {E} = \mathcal {E}_{\mathrm {KZ},s;\mu }=\exp (-h(2t^{\mathfrak {k}}_{01}+C^{\mathfrak {k}}_1)-\pi i(s+\mu )Z_1)$ is a ribbon $\sigma $ -braid, and hence that the elements $\mathcal {E}(1\otimes g)$ ( $g\in Z(U)$ ) are ribbon $\sigma $ -braids as well, follows again from the proof of [Reference De Commer, Neshveyev, Tuset and YamashitaDCNTY19, Theorem 3.8].

Let $\mathcal {E}'$ be another ribbon $\sigma $ -braid. The same argument as in the proof of Theorem 2.8 shows that $\mathcal {E}^{\prime (0)}=1\otimes g$ for an element $g\in Z_G(G^{\sigma })$ . We now use the same strategy as in the proof of the last statement of that theorem to get more restrictions on g. Namely, we replace $\mathcal {E}'$ by $\mathcal {E}'(1\otimes g^{-1})$ to get a ribbon $\tilde {\sigma }$ -braid, where $\tilde {\sigma }=(\operatorname {\mathrm {Ad}} g)\circ \sigma $ , and then twist $\mathcal {E}'$ and $\Psi _{\mathrm {KZ},s;\mu }$ further by an element H to get rid of the terms $t^{\mathfrak {u}}$ and $t^{\mathfrak {m}}$ in $\Psi ^{(1)}_{\mathrm {KZ},s;\mu }$ ; see equation (2.7) and recall that by Corollary 2.13 identity (2.7) is still valid for $\Psi _{\mathrm {KZ},s;\mu }$ . Note for a future use that by the proof of Proposition 2.14 we can take $\mathcal {H}$ of the form

(2.13) $$ \begin{align} \mathcal{H}=1+h(a C^{\mathfrak{u}}_1+b C^{\mathfrak{k}}_1) \end{align} $$

for appropriate constants a and b. Thus, our new associator $\Psi = \mathcal {H}_{0,12}\Psi _{\mathrm {KZ},s;\mu }\mathcal {H}^{-1}_{01,2}\mathcal {H}^{-1}_{0,1}$ satisfies

(2.14) $$ \begin{align} \Psi = 1 -\frac{h}{2}\tanh\Big(\frac{\pi s}{2}\Big)(t^{\mathfrak{m}_+}_{12}-t^{\mathfrak{m}_-}_{12})+ O(h^2). \end{align} $$

Looking at the order one terms in equation (1.5) and applying $\epsilon $ to the $0$ th leg, instead of equation (2.2), we now get

$$ \begin{align*} \mathcal{E}^{\prime (1)}=-2t^{\mathfrak{k}} &+ 1\otimes T+(\mathrm{id}\otimes\operatorname{\mathrm{Ad}} g)(t^{\mathfrak{m}})-t^{\mathfrak{m}}\\ &\qquad +\tanh\Big(\frac{\pi s}{2}\Big)(t^{\mathfrak{m}_+}-t^{\mathfrak{m}_-})+\tanh\Big(\frac{\pi s}{2}\Big)(\mathrm{id}\otimes\operatorname{\mathrm{Ad}} g)(t^{\mathfrak{m}_+}-t^{\mathfrak{m}_-}), \end{align*} $$

where $T=(\epsilon \otimes \mathrm {id})(\mathcal {E}^{\prime (1)})$ and we used that $t^{\mathfrak {m}_{\pm }}_{21}=t^{\mathfrak {m}_{\mp }}_{12}$ . As $\mathcal {E}^{\prime (1)}\in \mathcal {U}(G^{\sigma }\times G)$ , this means that

(2.15) $$ \begin{align} \operatorname{\mathrm{Ad}} g-\mathrm{id}\mp\tanh\Big(\frac{\pi s}{2}\Big)\mathrm{id}\mp\tanh\Big(\frac{\pi s}{2}\Big)\operatorname{\mathrm{Ad}} g=0\quad\text{on}\quad \mathfrak{m}_{\pm}. \end{align} $$

Hence, $\operatorname {\mathrm {Ad}} g=e^{\pm \pi s}\mathrm {id}$ on $\mathfrak {m}_{\pm }$ , which implies that $g=\exp (-\pi is Z)g'$ for some $g'\in Z(G)=Z(U)$ .

Without loss of generality, we may assume that $g'=e$ , and then we want to prove that our original $\mathcal {E}'$ coincides with $\mathcal {E}$ . It is more convenient to modify $\mathcal {E}$ in the same way as $\mathcal {E}'$ , that is, by replacing it by

$$\begin{align*}\mathcal{H}\mathcal{E}(1\otimes g^{-1})(\mathrm{id}\otimes\tilde{\sigma})(\mathcal{H})^{-1}=\exp(-h(2t^{\mathfrak{k}}_{01}+C^{\mathfrak{k}}_1)-\pi i\mu Z_1),\\[-15pt] \end{align*}$$

where we used that $\mathcal {H}$ has the form (2.13).

Thus, our new setup is that we have two ribbon $\tilde {\sigma }$ -braids $\mathcal {E}'$ and $\mathcal {E}$ , with $\tilde {\sigma }=(\operatorname {\mathrm {Ad}} \exp (-\pi is Z))\circ \sigma $ , with respect to an associator $\Psi $ satisfying equation (2.14), $\mathcal {E}^{\prime (0)}=\mathcal {E}^{(0)}=1$ ,

(2.16) $$ \begin{align} \mathcal{E}^{(1)}=-(2t^{\mathfrak{k}}_{01}+C^{\mathfrak{k}}_1+\pi i\mu^{(1)} Z_1),\\[-15pt]\nonumber \end{align} $$

and the goal is to show that $\mathcal {E}'=\mathcal {E}$ .

We will prove by induction on n that $\mathcal {E}^{\prime (n)}=\mathcal {E}^{(n)}$ . Consider the case $n = 1$ . Put

$$\begin{align*}p=\tanh\Big(\frac{\pi s}{2}\Big)\quad\text{and}\quad t=t^{\mathfrak{m}_+}-t^{\mathfrak{m}_-}.\\[-15pt] \end{align*}$$

By the argument in the proof of Theorem 2.8, we have $\mathcal {E}^{\prime (1)} = \mathcal {E}^{(1)} + 1 \otimes Y$ for some $Y \in \mathfrak {z}_{\mathfrak {g}}(\mathfrak {k})=\mathbb {C} Z$ . Put also $T=\mathcal {E}^{\prime (2)}-\mathcal {E}^{(2)}$ .

Using equation (2.14), formula (1.5) for $\mathcal {E}$ , modulo $h^3$ and terms depending only on $\mathcal {R}_{\mathrm {KZ}}$ and $\Psi $ , becomes

$$\begin{align*}(1 + h \mathcal{E}^{(1)} + h^2 \mathcal{E}^{(2)})_{01,2} &= (1 + h(pt_{1,2} - t^{\mathfrak{u}}_{1,2}))(1 + h \mathcal{E}^{(1)} + h^2 \mathcal{E}^{(2)})_{0,2}\\ &\quad \times (\mathrm{id}\otimes\mathrm{id} \otimes\tilde{\sigma})(1 - h(pt_{1,2} + t^{\mathfrak{u}}_{1,2})),\\[-15pt] \end{align*}$$

where we again used that $t_{21}=-t_{12}$ . We have a similar formula for $\mathcal {E}'$ . Taking the difference and comparing the coefficients of $h^2$ , we obtain

$$\begin{align*}T_{01, 2} = T_{0,2} + (p t_{1,2} - t^{\mathfrak{u}}_{1,2}) Y_2 - Y_2 (\mathrm{id} \otimes\mathrm{id}\otimes \tilde{\sigma})(p t_{1,2} + t^{\mathfrak{u}}_{1,2}).\\[-15pt] \end{align*}$$

Using the identity

(2.17) $$ \begin{align} (p t - t^{\mathfrak{m}})- (\mathrm{id}\otimes \tilde{\sigma})(p t + t^{\mathfrak{m}})=0,\\[-15pt]\nonumber \end{align} $$

which is an equivalent form of equation (2.15), we can write this as

(2.18) $$ \begin{align} T_{01, 2} = T_{0,2} -2t^{\mathfrak{k}}_{1,2}Y_2+[p t_{1,2} - t^{\mathfrak{m}}_{1,2},Y_2].\\[-15pt]\nonumber \end{align} $$

Similarly, formula (1.6) for $\mathcal {E}$ , modulo $h^3$ and terms depending only on $\mathcal {R}_{\mathrm {KZ}}$ and $\Psi $ , becomes

$$ \begin{align*} &(1 + h \mathcal{E}^{(1)} + h^2 \mathcal{E}^{(2)})_{0,12} = \Big(1 + h \Big(\frac{p}{2} t_{1,2} - t^{\mathfrak{u}}_{1,2}\Big)\Big) (1 + h \mathcal{E}^{(1)} + h^2 \mathcal{E}^{(2)})_{0,2}\\ &\qquad\quad \times(\mathrm{id} \otimes\mathrm{id}\otimes \tilde{\sigma})(1 - h (p t_{1,2} + t^{\mathfrak{u}}_{1,2})) (1 + h \mathcal{E}^{(1)} + h^2 \mathcal{E}^{(2)})_{0,1}(\mathrm{id} \otimes\tilde{\sigma}\otimes \tilde{\sigma})\Big(1+ h\frac{p}{2} t_{1,2}\Big),\\[-15pt] \end{align*} $$

and we have a similar identity for $\mathcal {E}'$ . Taking the difference and comparing the coefficients of $h^2$ , we obtain

$$ \begin{align*} &T_{0,12} = T_{0,2} + T_{0,1} + \Big(\frac{p}{2} t_{1,2} - t^{\mathfrak{u}}_{1,2}\Big) (Y_1 + Y_2) - Y_2 (\mathrm{id}\otimes\mathrm{id} \otimes \tilde{\sigma})(p t_{1,2} + t^{\mathfrak{u}}_{1,2}) \\ &\quad - (\mathrm{id} \otimes \mathrm{id}\otimes \tilde{\sigma})(p t_{1,2} + t^{\mathfrak{u}}_{1,2}) Y_1 + (Y_1 + Y_2) (\mathrm{id}\otimes \tilde{\sigma} \otimes \tilde{\sigma})\Big(\frac{p}{2} t_{1,2}\Big) + Y_2 \mathcal{E}^{(1)}_{0,1} + \mathcal{E}^{(1)}_{0,2} Y_1 + Y_1 Y_2.\\[-15pt] \end{align*} $$

Using that $\tilde {\sigma }\otimes \tilde {\sigma }$ is the identity on $\mathfrak {m}_{\pm }\otimes \mathfrak {m}_{\mp }$ and that $\operatorname {\mathrm {ad}} Y_1=-\operatorname {\mathrm {ad}} Y_2$ on $\mathfrak {m}_{\pm }$ , we can write this as

$$ \begin{align*} T_{0,12} = T_{0,2} + T_{0,1} &+ (p t_{1,2} - t^{\mathfrak{u}}_{1,2}) (Y_1 + Y_2) - Y_2 (\mathrm{id}\otimes\mathrm{id} \otimes \tilde{\sigma})(p t_{1,2} + t^{\mathfrak{u}}_{1,2}) \\ &\qquad\qquad\quad - (\mathrm{id} \otimes \mathrm{id}\otimes \tilde{\sigma})(p t_{1,2} + t^{\mathfrak{u}}_{1,2}) Y_1 + Y_2 \mathcal{E}^{(1)}_{0,1} + \mathcal{E}^{(1)}_{0,2} Y_1 + Y_1 Y_2, \end{align*} $$

and then, using again equation (2.17), we get

(2.19) $$ \begin{align} T_{0,12} = T_{0,2} + T_{0,1} -2t^{\mathfrak{k}}_{1,2}(Y_1+Y_2) + [p t_{1,2} - t^{\mathfrak{m}}_{1,2},Y_2]+ Y_2 \mathcal{E}^{(1)}_{0,1} + \mathcal{E}^{(1)}_{0,2} Y_1 + Y_1 Y_2. \end{align} $$

Subtracting equation (2.19) from equation (2.18), we obtain

$$\begin{align*}d_{\mathrm{cH}}(T) =2 t^{\mathfrak{k}}_{12} Y_1 - \mathcal{E}^{(1)}_{01} Y_2 - \mathcal{E}^{(1)}_{02} Y_1 - Y_1 Y_2. \end{align*}$$

By equation (2.16), this means that

$$\begin{align*}d_{\mathrm{cH}}(T)=(2 t^{\mathfrak{k}}_{12}+2t^{\mathfrak{k}}_{02}+C^{\mathfrak{k}}_2+\pi i\mu^{(1)} Z_2)Y_1+ (2t^{\mathfrak{k}}_{01}+C^{\mathfrak{k}}_1+\pi i\mu^{(1)} Z_1)Y_2 - Y_1Y_2. \end{align*}$$

As $Y\in \mathbb {C} Z$ , we can also write this as

$$\begin{align*}d_{\mathrm{cH}}(T)=(2 t^{\mathfrak{k}}_{12}+2t^{\mathfrak{k}}_{02}+C^{\mathfrak{k}}_2)Y_1+ (2t^{\mathfrak{k}}_{01}+C^{\mathfrak{k}}_1+2\pi i\mu^{(1)} Z_1)Y_2-Y_1Y_2. \end{align*}$$

On the other hand, a straightforward computation shows that the right-hand side of the above identity is the coboundary of

$$\begin{align*}2t^{\mathfrak{k}}_{01}Y_0+C^{\mathfrak{k}}_1Y_0+C^{\mathfrak{k}}_0Y_1+2\pi i\mu^{(1)}Z_0Y_1-Y_0Y_1. \end{align*}$$

As $\mathrm {H}^1(B_{G,G^{\sigma }})=0$ by Lemma 1.2 and Corollary 2.4, it follows that there exists S in the center of $\mathcal {U}(G^{\sigma })$ such that

$$\begin{align*}T=S_{01}-S_0+2t^{\mathfrak{k}}_{0,1}Y_0+C^{\mathfrak{k}}_1Y_0+C^{\mathfrak{k}}_0Y_1+2\pi i\mu^{(1)}Z_0Y_1-Y_0Y_1. \end{align*}$$

The only consequence of the above identity that we need is that $T\in \mathcal {U}(G^{\sigma }\times G^{\sigma })$ . By looking at equation (2.18), we see that this implies

$$\begin{align*}[pt-t^{\mathfrak{m}},1\otimes Y]=(p-1)[t^{\mathfrak{m}_+},1\otimes Y]-(p+1)[t^{\mathfrak{m}_-},1\otimes Y]\in\mathcal{U}(G^{\sigma}\times G^{\sigma}). \end{align*}$$

As Y is a scalar multiple of Z and $\operatorname {\mathrm {ad}} Z$ acts by nonzero operators on $\mathfrak {m}_{\pm }$ , this is possible only if $Y=0$ . This shows that $\mathcal {E}^{\prime (1)} = \mathcal {E}^{(1)}$ .

The induction step is similar. Assuming that $\mathcal {E}^{\prime (k)} = \mathcal {E}^{(k)}$ for $k<n$ for some $n\ge 2$ , we have $\mathcal {E}^{\prime (n)} = \mathcal {E}^{(n)} + 1 \otimes Y$ for an element $Y \in \mathfrak {z}_{\mathfrak {g}}(\mathfrak {k})=\mathbb {C} Z$ . Then, with $T=\mathcal {E}^{\prime (n+1)}-\mathcal {E}^{(n+1)}$ , comparing the coefficients of $h^{n+1}$ in equation (1.5), we get the same identity (2.18). If we do the same for equation (1.6), the only difference from equation (2.19) is that we do not get the term $Y_1Y_2$ at the end. But this has almost no effect on the rest of the argument; we just have to remove the terms $Y_1Y_2$ and $Y_0Y_1$ in the subsequent identities. Thus, we get $Y=0$ .

Remark 2.20. Theorem 2.19 implies that in the Hermitian case the ribbon twist-braids contain complete information about the associators. Namely, assume that $\Psi _{\mathrm {KZ},s';\mu '} =\mathcal {H}_{0, 12}\Psi _{\mathrm {KZ},s;\mu } \mathcal {H}_{01, 2}^{-1} \mathcal {H}_{0,1}^{-1}$ for some . By equation (1.7) and Theorem 2.19, it follows that

$$\begin{align*}\mathcal{H} \exp(-h(2t^{\mathfrak{k}}_{01}+C^{\mathfrak{k}}_1)-\pi i(s+\mu)Z_1) (\mathrm{id}\otimes\sigma)(\mathcal{H})^{-1} =\exp(-h(2t^{\mathfrak{k}}_{01}+C^{\mathfrak{k}}_1)-\pi i(s'+\mu')Z_1)g_1 \end{align*}$$

for some $g\in Z(U)$ . Since $(\epsilon \otimes \mathrm {id})(\mathcal {H})$ is a $\mathfrak {k}$ -invariant element of $\mathcal {U}(G)$ and $\sigma $ is an inner automorphism defined by an element of K, we have $\sigma ((\epsilon \otimes \mathrm {id})(\mathcal {H}) )=(\epsilon \otimes \mathrm {id})(\mathcal {H})$ , and then by applying $\epsilon $ to the $0$ th leg we get

$$\begin{align*}\exp(-h C^{\mathfrak{k}}-\pi i(s+\mu)Z)=\exp(-h C^{\mathfrak{k}}-\pi i(s'+\mu')Z)g. \end{align*}$$

This implies that $s=s'+2ik$ for some $k\in \mathbb {Z}$ , and $\mu =\mu '$ .

3.1 Root vectors and Poisson structure

Let us quickly review a standard Poisson–Lie group structure on U making $U^{\nu }$ a Poisson–Lie subgroup (for Hermitian symmetric pairs).

Let $\mathfrak {t}$ be a Cartan subalgebra of $\mathfrak {u}$ containing $\mathfrak {z}(\mathfrak {u}^{\nu })$ . Then $\mathfrak {t}$ is contained in $\mathfrak {u}^{\nu }$ , and its complexification $\mathfrak {h}$ is a Cartan subalgebra of $\mathfrak {g}$ .

Recall that a root $\alpha $ is called compact if $\mathfrak {g}_{\alpha }\subset \mathfrak {g}^{\nu }$ , and noncompact otherwise. As $\mathfrak {g}^{\nu }$ is the centralizer of $\mathfrak {z}(\mathfrak {u}^{\nu })$ , a root is compact if and only if it vanishes on $\mathfrak {z}(\mathfrak {u}^{\nu })$ .

As in Section 2.3, we fix $Z_{\nu }\in \mathfrak {z}(\mathfrak {u}^{\nu })$ such that $(Z_{\nu },Z_{\nu })_{\mathfrak {g}}=-a_{\nu }^{-2}$ . Let us fix an ordered basis in $i\mathfrak {t}$ , with $-i Z_{\nu }$ being the first element of the basis and consider the corresponding lexicographic order on the roots. Then, in this order, any noncompact positive root is bigger than any compact root. Furthermore, the noncompact positive roots are totally positive in the sense of [Reference Harish-ChandraHC55], meaning that if $\gamma $ is a noncompact positive root, $\alpha _1, \dots , \alpha _k$ are compact roots and $m_1,\dots , m_k$ are integers such that $\gamma ' = \gamma + \sum _{i=1}^km_i \alpha _i$ is a root, then $\gamma '$ is positive.

We denote by $\Phi $ the set of all roots and by $\Phi ^+$ that of positive roots. We further denote by $\Phi _{\mathrm {nc}}^+$ (resp. $\Phi _{\mathrm {nc}}^-$ ) the set of positive (resp. negative) noncompact roots. Let $\Pi = \{\alpha _i\}_{i\in I}$ be the set of simple positive roots. Recall that we denote by $\mathfrak {m}_{\nu \pm }\subset \mathfrak {m}_{\nu }^{\mathbb {C}}$ the eigenspaces of $\operatorname {\mathrm {ad}} Z_{\nu }$ corresponding to the eigenvalues $\pm i$ . It is clear from our choice of the ordering that

$$ \begin{align*} \mathfrak{m}_{\nu+}&=\bigoplus_{\alpha \in \Phi_{\mathrm{nc}}^+}\mathfrak{g}_{\alpha},& \mathfrak{m}_{\nu-}&=\bigoplus_{\alpha \in \Phi_{\mathrm{nc}}^-}\mathfrak{g}_{\alpha}.\\[-15pt] \end{align*} $$

Since $2i$ is not an eigenvalue of $\operatorname {\mathrm {ad}} Z_{\nu }$ , we have $[\mathfrak {m}_{\nu +},\mathfrak {m}_{\nu +}]=0$ . It follows that $\mathfrak {p}=\mathfrak {g}^{\nu }+\mathfrak {m}_{\nu +}$ is a parabolic subalgebra of $\mathfrak {g}$ . As $\mathfrak {g}^{\nu }$ acts irreducibly on $\mathfrak {g}/\mathfrak {p}\cong \mathfrak {m}_{\nu -}$ , this parabolic subalgebra is maximal. Hence, it corresponds to a maximal proper subset of $\Pi $ . This set must consist of compact roots, and hence its complement consists of one noncompact root. We denote this unique noncompact simple positive root by $\alpha _o$ . It corresponds to the black vertex in a standard Vogan diagram of $\nu $ .

For every root $\alpha $ , let $H_{\alpha }\in \mathfrak {h}$ be the element dual to the coroot $\alpha ^{\vee } = \frac {2}{(\alpha ,\alpha )}\alpha $ so that we have

$$\begin{align*}\alpha(H_{\beta}) = (\alpha,\beta^{\vee})\quad (\alpha,\beta \in \Phi),\\[-15pt] \end{align*}$$

where $(\cdot ,\cdot )$ is the scalar product dual to the restriction of $(\cdot ,\cdot )_{\mathfrak {g}}$ to $\mathfrak {h}$ . For positive roots $\alpha $ , we choose root vectors $X_{\alpha }\in \mathfrak {g}_{\alpha }$ such that the antilinear involution for $\mathfrak {u}$ satisfies

$$\begin{align*}(X_{\alpha},X_{\alpha}^*)_{\mathfrak{g}} = \frac{2}{(\alpha,\alpha)}.\\[-15pt] \end{align*}$$

Then $[X_{\alpha },X^*_{\alpha }]=H_{\alpha }$ , and we put $X_{-\alpha }=X^*_{\alpha }$ .

Let

$$ \begin{align*} Y_{\alpha} &= \frac{i}{2}(X_{\alpha}+ X_{-\alpha}) \in \mathfrak{u},& Z_{\alpha} &= \frac{1}{2}( X_{\alpha}-X_{-\alpha}) \in \mathfrak{u} \quad (\alpha \in \Phi^+), \end{align*} $$

and consider the antisymmetric tensor

$$ \begin{align*} r = \sum_{\alpha>0} (\alpha,\alpha) (Y_{\alpha} \otimes Z_{\alpha} - Z_{\alpha} \otimes Y_{\alpha}) \in \mathfrak{u}^{\otimes 2}. \end{align*} $$

Note that we can also write

(3.1) $$ \begin{align} r = i \sum_{\alpha>0} \frac{(\alpha,\alpha)}{2} (X_{-\alpha} \otimes X_{\alpha} - X_{\alpha} \otimes X_{-\alpha}) \in \mathfrak{g}^{\otimes 2}. \end{align} $$

Then $i r \pm t^{\mathfrak {u}}$ (with $t^{\mathfrak {u}}$ defined by equation (1.1)) satisfies the classical Yang–Baxter equation, and $\mathfrak {u}$ becomes a Lie bialgebra with the cobracket

$$\begin{align*}\delta_r(X) = [r, \Delta(X)]. \end{align*}$$

Thus, U becomes a Poisson–Lie group with the Sklyanin Poisson bracket

$$\begin{align*}\{f_1,f_2\}_{\mathrm{Sk}} = m (r^{(l,l)} - r^{(r,r)})(f_1 \otimes f_2), \end{align*}$$

where m is the product map, and for $X \in \mathfrak {u}$ and $f \in C^{\infty }(U)$ we put

$$ \begin{align*} (X^{(l)} f)(g) &= \frac{d}{d t} f(e^{t X} g) |_{t = 0},& (X^{(r)} f)(g) &= \frac{d}{d t} f(g e^{t X}) |_{t = 0}. \end{align*} $$

Note that if we as usual view $\mathfrak {u}$ and U as sitting inside $\mathcal {U}(G)$ , then we can write the Sklyanin bracket as

$$\begin{align*}\{f_1,f_2\}_{\mathrm{Sk}}(g) = \langle f_1 \otimes f_2, [r, g\otimes g] \rangle\quad (g\in U,\ f_1, f_2 \in \mathcal{O}(U)). \end{align*}$$

The Lie algebra $\mathfrak {u}^{\nu }$ is the intersection of $\mathfrak {u}$ with a parabolic subalgebra of $\mathfrak {g}$ , namely, $\mathfrak {u}^{\nu }=\mathfrak {u}\cap \mathfrak {p}$ . It is well known that this implies that $U^{\nu }$ is a Poisson–Lie subgroup of U.

3.2 Satake form

We will have to conjugate $U^{\nu }$ in order to go beyond the Poisson–Lie subgroups, making it closer to the coisotropic subgroup associated with symmetric pairs [Reference Foth and LuFL04]. In order to do so, let us review the Satake form of involutions.

For $X\subset I$ , we denote by $\Pi _X$ the subset $\{\alpha _j \mid j\in X\}\subset \Pi $ . Assume $\theta $ is a nontrivial involution on $\mathfrak {u}$ such that its extension to $\mathfrak {g}$ leaves $\mathfrak {h}$ globally invariant. We write $\Theta \in \operatorname {\mathrm {End}}(\mathfrak {h}^*)$ for the endomorphism dual to $\theta |_{\mathfrak {h}}$ .

The above set X is then uniquely determined, representing the black vertices in the corresponding Satake diagram. Then there exist unimodular $w_{\alpha } \in \mathbb {C}$ such that

$$ \begin{align*} \theta(X_{\alpha}) = w_{\alpha} X_{\Theta(\alpha)} \quad (\alpha\in \Phi). \end{align*} $$

Moreover, there exists a unique Dynkin diagram involution $\tau _{\theta }$ such that

(3.2) $$ \begin{align} \Theta(\alpha) = -w_X\tau_{\theta}(\alpha)\quad (\alpha \in \Phi), \end{align} $$

with $w_X$ the longest element of the Weyl group associated to X. This involution leaves the set X globally invariant. See [Reference Kac and WangKW92] and [Reference KolbKol14, Appendix A] for details.

Put

$$\begin{align*}I_{\mathrm{ns}} = \{i \in I\setminus X\mid \tau_{\theta}(i) = i \textrm{ and }(\alpha_i,\alpha_j)=0 \textrm{ for all } j \in X\}, \end{align*}$$

which corresponds to the $\tau _{\theta }$ -stable white vertices not connected to any black vertices in the Satake diagram. We then put

$$\begin{align*}I_{\mathcal{S}}=\{i\in I_{\mathrm{ns}}\mid ({\alpha}^{\vee}_j,\alpha_i) \in 2\mathbb{Z}\textrm{ for all }j\in I_{\mathrm{ns}}\}. \end{align*}$$

We also put

$$\begin{align*}I_{\mathcal{C}} = \{i \in I \setminus X \mid \tau_{\theta}(i)\neq i\textrm{ and }(\alpha_i,\Theta(\alpha_i)) = 0\}. \end{align*}$$

It can be shown that this is the set of white vertices not fixed by $\tau _{\theta }$ such that, if $\bar \alpha _i$ is the corresponding restricted root, then $2 \bar \alpha _i$ is not a restricted root; see [Reference ArakiAra62][Reference HelgasonHel01, p. 530].

We will use the following definition from [Reference De Commer, Neshveyev, Tuset and YamashitaDCNTY19].

See [Reference De Commer, Neshveyev, Tuset and YamashitaDCNTY19, Appendix C] for a concrete classification of the Hermitian symmetric pairs into these types. Recall that the restricted root system is always of type $\mathrm {C}$ or $\mathrm {BC}$ in the Hermitian case [Reference MooreMoo64, Theorem 2]. By a case-by-case analysis (see, e.g., [Reference Onishchik and VinbergOV90, Reference Chapter, table 9][Reference HelgasonHel01, Chapter X, Table VI][Reference KnappKna02, Appendix C]), one sees that we are in the S-type case exactly when the restricted root system is of type $\mathrm {C}$ , while we are in the C-type case when the restricted root system is of type $\mathrm {BC}$ .