2.1 Matched pairs of actions

Matched pairs of groups, enabling to define a Zappa–Szép product, have been known for a long time. On the other hand, the notion of a matched pair of Hopf algebras was introduced by Singer [Reference Singer37] in the graded case, and the current notion was introduced by Majid [Reference Majid25]. We consider matched pairs of actions on Hopf algebras, which will turn out to be a subclass of matched pairs of Hopf algebras [Reference Majid30, Definition 7.2.1]. Matched pairs of actions on groups have been already used, e.g., in the seminal work of Lu, Yan, and Zhu [Reference Lu, Yan and Zhu24], and then mutuated in several different contexts, such as for groupoids by Andruskiewitsch [Reference Andruskiewitsch3].

Definition 2.1 Let H be a bialgebra, $\bullet $ be its algebra product. A matched pair of actions $(H,\rightharpoonup ,\leftharpoonup )$ on H is the datum of a left action $\rightharpoonup $ and a right action $\leftharpoonup $ of H on itself, such that H is a left H-module coalgebra and a right H-module coalgebra with the respective actions, and the following conditions hold for all $a,b,c\in H$ :

(mp.1) $$ \begin{align} & a\rightharpoonup 1 = \epsilon(a)1,\text{ i.e.,}\ u_H\text{ is a morphism in }{}_H{\mathfrak{M}}; \end{align} $$

(mp.2) $$ \begin{align} &1\leftharpoonup a = \epsilon(a)1,\text{ i.e.,}\ u_H\text{ is a morphism in }{}{\mathfrak{M}}_H; \end{align} $$

(mp.3) $$ \begin{align} &\!\!\!\! a\rightharpoonup(b\bullet c)=(a_{1}\rightharpoonup b_{1})\bullet ((a_{2}\leftharpoonup b_{2})\rightharpoonup c); \end{align} $$

(mp.4) $$ \begin{align} &\!\!\!\!\! (a\bullet b)\leftharpoonup c=(a\leftharpoonup(b_{1}\rightharpoonup c_{1}))\bullet (b_{2}\leftharpoonup c_{2}); \end{align} $$

(⋆) $$ \begin{align} & a\bullet b = (a_1\rightharpoonup b_1)\bullet (a_2\leftharpoonup b_2).\end{align} $$

A morphism of matched pairs of actions between $(H,\rightharpoonup _H, \leftharpoonup _H)$ and $(K,\rightharpoonup _K, \leftharpoonup _K)$ is a morphism of bialgebras $H\to K$ that intertwines the two left actions and the two right actions, respectively, in the following sense:

$$\begin{align*}f(a\rightharpoonup_H b) = f(a)\rightharpoonup_K f(b),\quad f(a\leftharpoonup_H b) = f(a)\leftharpoonup_K f(b). \end{align*}$$

The category of matched pairs of actions (in ) will be denoted by .

A pair of actions $(\rightharpoonup ,\leftharpoonup )$ on a bialgebra H satisfying (mp.1)–(mp.4) allows one to construct the double cross product $H \bowtie H$ , as described in Majid [Reference Majid30], which is the vector space $H\otimes H$ endowed with the algebra structure given by the product

$$\begin{align*}(a\otimes h)(b\otimes g):=a\bullet (h_{1}\rightharpoonup b_{1})\otimes(h_{2}\leftharpoonup b_{2})\bullet g,\end{align*}$$

with unit $1_{H}\otimes 1_{H}$ , and the usual tensor product coalgebra structure. Then $H\bowtie H$ becomes a bialgebra if and only if the additional condition

(mp.5) $$ \begin{align} (a_1\rightharpoonup b_1)\otimes (a_2\leftharpoonup b_2) = (a_2\rightharpoonup b_2)\otimes (a_1\leftharpoonup b_1) \end{align} $$

holds; and in this case, $(H,H)$ is a matched pair of bialgebras in the sense of Majid [Reference Majid30, Definition 7.2.1]. Notice that (mp.5) is trivially satisfied in case H is cocommutative.

When H is a Hopf algebra, a matched pair of bialgebras $(H,H)$ will be usually called a matched pair on H.

The following result simply follows by an argument of naturality.

Lemma 2.2 Let $\mathcal {M}$ be a (strict) braided monoidal category with braiding $\sigma $ , let H be an object in $\mathcal {M}$ , and $m\colon H\otimes H\to H$ , any morphisms in $\mathcal {M}$ . Then, the following hold:

(br.1) $$ \begin{align} &\!\!\!\sigma_{H,H}(\mathrm{Id}\otimes u) = u\otimes \mathrm{Id};\qquad\qquad \end{align} $$

(br.2) $$ \begin{align} &\!\!\!\sigma_{H,H}(u\otimes \mathrm{Id}) = \mathrm{Id}\otimes u;\qquad\qquad \end{align} $$

(br.3) $$ \begin{align} &\sigma_{H,H} m_{23} =m_{12}(\sigma_{H,H})_{23}(\sigma_{H,H})_{12}; \end{align} $$

(br.4) $$ \begin{align} &\sigma_{H,H} m_{12} =m_{23}(\sigma_{H,H})_{12}(\sigma_{H,H})_{23}. \end{align} $$

Moreover, if m is a monomorphism in $\mathcal {M}$ and $\varsigma \colon H\otimes H\to H\otimes H$ is a morphism satisfying (br.1)–(br.4), then $\varsigma $ satisfies the hexagonal axioms, and hence $(H,\varsigma )$ is a braided object in $\mathcal {M}$ .

Lemma 2.3 (cf. Tambara [Reference Tambara40, Proposition 2.2])

Let H be a bialgebra. Suppose given a left action $\rightharpoonup :H\otimes H\to H$ and a right action $\leftharpoonup :H\otimes H\to H$ , such that H is a left H-module coalgebra and a right H-module coalgebra with the respective actions. Define $\sigma _{H,H}\colon H\otimes H\to H\otimes H$ by

$$\begin{align*}\sigma_{H,H}(a\otimes b):=(a_{1}\rightharpoonup b_{1})\otimes(a_{2}\leftharpoonup b_{2}). \end{align*}$$

Then, $\sigma _{H,H}$ satisfies (br.1)–(br.4) if and only if $\rightharpoonup $ and $\leftharpoonup $ satisfy (mp.1)–(mp.4).

Proof One has

$$\begin{align*}\sigma_{H,H}m_{12}(a\otimes b\otimes c)=\sigma_{H,H}(ab\otimes c)=(a_{1}b_{1}\rightharpoonup c_{1})\otimes(a_{2}b_{2}\leftharpoonup c_{2}), \end{align*}$$

while on the other hand

$$\begin{align*}\begin{aligned} &\hspace{-2em}m_{23}(\sigma_{H,H})_{12}(\sigma_{H,H})_{23}(a\otimes b\otimes c)\\ &=m_{23}(\sigma_{H,H})_{12}\big(a\otimes(b_{1}\rightharpoonup c_{1})\otimes(b_{2}\leftharpoonup c_{2})\big)\\&=m_{23}\big((a_{1}\rightharpoonup(b_{1}\rightharpoonup c_{1})_{1})\otimes(a_{2}\leftharpoonup(b_{1}\rightharpoonup c_{1})_{2})\otimes(b_{2}\leftharpoonup c_{2})\big)\\&=(a_{1}\rightharpoonup(b_{1}\rightharpoonup c_{1}))\otimes(a_{2}\leftharpoonup(b_{2}\rightharpoonup c_{2}))(b_{3}\leftharpoonup c_{3})\\&=(a_{1}b_{1}\rightharpoonup c_{1})\otimes(a_{2}\leftharpoonup(b_{2}\rightharpoonup c_{2}))(b_{3}\leftharpoonup c_{3}). \end{aligned} \end{align*}$$

If (br.4) holds true then, by applying $\epsilon \otimes \mathrm {Id}$ on both sides, we obtain $(ab)\leftharpoonup c=(a\leftharpoonup (b_{1}\rightharpoonup c_{1}))(b_{2}\leftharpoonup c_{2})$ , which is (mp.4). If conversely (mp.4) holds true, then $\sigma _{H,H}$ clearly satisfies (br.4). One analogously proves that (br.3) is equivalent to (mp.3). Observe that

$$\begin{align*}\sigma_{H,H}(1\otimes a)=(1\rightharpoonup a_{1})\otimes(1\leftharpoonup a_{2})=a_{1}\otimes(1\leftharpoonup a_{2}). \end{align*}$$

If (br.2) holds true, then

$$\begin{align*}a_{1}\otimes(1\leftharpoonup a_{2})=a\otimes1. \end{align*}$$

Hence, by applying $\epsilon \otimes \mathrm {Id}$ , we obtain $1\leftharpoonup a=\epsilon (a)1$ , which is (mp.2). Conversely, if (mp.2) holds true, then $\sigma _{H,H}$ clearly satisfies (br.2). The equivalence between (br.1) and (mp.1) is proven analogously.

Remark 2.4 Let H be a bialgebra, and suppose given

$$\begin{align*}\sigma_{H,H}(x\otimes y)=(x_{1}\rightharpoonup y_{1})\otimes(x_{2}\leftharpoonup y_{2}), \end{align*}$$

where $\rightharpoonup :H\otimes H\to H$ and $\leftharpoonup :H\otimes H\to H$ are such that $\epsilon (a\rightharpoonup b)=\epsilon (a)\epsilon (b)=\epsilon (a\leftharpoonup b)$ for all $a,b\in H$ . Then, $\rightharpoonup $ is retrieved as $(\mathrm {Id}\otimes \epsilon )\sigma _{H,H}$ , and $\leftharpoonup $ as $(\epsilon \otimes \mathrm {Id})\sigma _{H,H}$ (see [Reference Majid30, p. 300]).

Although there are examples of braidings that are obtained from a pair of actions satisfying (mp.1)–(mp.4) (see Section 5.2), it is not true that all braidings are obtained from a pair of maps as in Remark 2.4:

Counterexample 2.5 Consider, for instance, a coquasitriangular bialgebra $(H,\mathcal {R})$ (see Section 5.1) and the usual braiding (5.1) on ${\mathfrak {M}}^H$

$$\begin{align*}\sigma_{H,H}\colon H\otimes H\to H\otimes H,\quad a\otimes b\mapsto b_{1}\otimes a_{1}\mathcal{R}(a_{2}\otimes b_{2}). \end{align*}$$

If we suppose by contradiction that $\sigma _{H,H}(a\otimes b) = (a_1\rightharpoonup b_1)\otimes (a_2\leftharpoonup b_2)$ for some maps $\rightharpoonup ,\leftharpoonup $ , then Remark 2.4 yields

$$\begin{align*}a\rightharpoonup b=\mathcal{R}(a\otimes b_{2})b_{1},\qquad a\leftharpoonup b=a_{1}\mathcal{R}(a_{2}\otimes b), \end{align*}$$

but then

$$ \begin{align*} b_1\otimes a_1\mathcal{R}(a_2\otimes b_2)&=\sigma_{H,H}(a\otimes b)\\ &= (a_1\rightharpoonup b_1)\otimes (a_2\leftharpoonup b_2)\\ &= \mathcal{R}(a_1\otimes b_2)b_1\otimes a_2\mathcal{R}(a_3\otimes b_3), \end{align*} $$

and this is generally false. As an example, consider the coquasitriangular structure on the Sweedler’s Hopf algebra described in Section 6.1, with $a=b=g$ : one has $\mathcal {R}(g\otimes g) = -1$ , whence $b_1\otimes a_1\mathcal {R}(a_2\otimes b_2) = -g\otimes g$ and $\mathcal {R}(a_1\otimes b_2)b_1\otimes a_2\mathcal {R}(a_3\otimes b_3) = g\otimes g$ , which differ if .

Proof We compute

$$\begin{align*}1\rightharpoonup c=(\mathrm{Id}\otimes\epsilon)\sigma_{H,H}(1\otimes c)=(\mathrm{Id}\otimes\epsilon)(c\otimes 1)=c, \end{align*}$$

and also

$$ \begin{align*} ab\rightharpoonup c&=(\mathrm{Id}\otimes\epsilon)\sigma_{H,H}m_{12}(a\otimes b\otimes c)\\&=(\mathrm{Id}\otimes\epsilon)m_{23}(\sigma_{H,H})_{12}(\sigma_{H,H})_{23}(a\otimes b\otimes c)\\&=(\mathrm{Id}\otimes\epsilon\otimes\epsilon)(\sigma_{H,H})_{12}(\sigma_{H,H})_{23}(a\otimes b\otimes c)\\&=(\mathrm{Id}\otimes\epsilon)\sigma_{H,H}(\mathrm{Id}\otimes\mathrm{Id}\otimes\epsilon)(\mathrm{Id}\otimes\sigma_{H,H})(a\otimes b\otimes c)\\&=(\mathrm{Id}\otimes\epsilon)\sigma_{H,H}(a\otimes(b\rightharpoonup c))\\&=a\rightharpoonup(b\rightharpoonup c). \end{align*} $$

The proof is analogous for $\leftharpoonup $ .

In general, when $\sigma _{H,H}$ does not have the form $\sigma _{H,H}(x\otimes y)=(x_{1}\rightharpoonup y_{1})\otimes (x_{2}\leftharpoonup y_{2})$ , the actions $\rightharpoonup $ and $\leftharpoonup $ defined as in the previous lemma need not form a matched pair, as the following counterexample shows.

Counterexample 2.7 Let us consider again

$$\begin{align*}\sigma_{H,H}\colon H\otimes H\to H\otimes H,\quad a\otimes b\mapsto b_{1}\otimes a_{1}\mathcal{R}(a_{2}\otimes b_{2}) \end{align*}$$

and the left and right H-actions

$$\begin{align*}a\rightharpoonup b=\mathcal{R}(a\otimes b_{2})b_{1},\qquad a\leftharpoonup b=a_{1}\mathcal{R}(a_{2}\otimes b). \end{align*}$$

Then, we have $a\rightharpoonup bc=\mathcal {R}(a\otimes b_{2}c_{2})b_{1}c_{1}$ , while

$$ \begin{align*} (a_{1}\rightharpoonup b_{1})((a_{2}\leftharpoonup b_{2})\rightharpoonup c)&=\mathcal{R}(a_{1}\otimes b_{2})\mathcal{R}(a_{2}\otimes c_{2})\mathcal{R}(a_{3}\otimes b_{3})b_{1}c_{1}\\ &=\mathcal{R}(a\otimes b_{3}c_{2}b_{2})b_{1}c_{1}. \end{align*} $$

These are generally distinct (consider for instance the Sweedler’s Hopf algebra in Section 6.1, $a = b = c=g$ , ).

2.2 Hopf braces

A skew brace, defined by Guarnieri and Vendramin [Reference Guarnieri and Vendramin18] generalizing Rump [Reference Rump35], is the datum of two group operations on the same set, satisfying a compatibility. Skew braces are ubiquitous in the theory of braidings and the YBE, and hence yearned for a Hopf-theoretic version. This was defined by Angiono, Galindo, and Vendramin [Reference Angiono, Galindo and Vendramin5]:

Definition 2.2 A Hopf brace $(H,\cdot ,\bullet ,1,\Delta , \epsilon ,S,T)$ is the datum of two Hopf algebra structures $(H,\cdot ,1,\Delta ,\epsilon ,S)$ and $(H,\bullet ,1',\Delta ,\epsilon ,T)$ on the same coalgebra $(H,\Delta ,\epsilon )$ , satisfying the following compatibility:

(hbc) $$ \begin{align} a\bullet(b\cdot c)=(a_{1}\bullet b)\cdot S(a_{2})\cdot(a_{3}\bullet c). \end{align} $$

Notice that the compatibility condition forces $1 = 1'$ (see [Reference Angiono, Galindo and Vendramin5, Remark 1.3]).

When no confusion arises, we shall simply indicate a Hopf brace by $(H,\cdot , \bullet )$ . We recall the following characterization.

Proposition 2.8 (Angiono, Galindo and Vendramin [Reference Angiono, Galindo and Vendramin5, Theorems 1.12, 3.3, and Corollary 2.4])

Given a Hopf algebra H, the following data are equivalent:

1. i. a Hopf brace structure $(H,\cdot ,\bullet )$ on H;

2. ii. a Hopf algebra A, an action $\rightharpoonup $ of A on H such that H is a left A-module algebra, and an isomorphism of coalgebras $\pi \colon A\to H$ which is a 1-cocycle of bialgebras; i.e., $\pi (ab) = \pi (a_1)(a_2\rightharpoonup \pi (b))$ .

In case H is cocommutative, the previous two are also equivalent to

1. iii. a matched pair on H, satisfying the additional condition (⋆).

Moreover, in the cocommutative case,

$$\begin{align*}\sigma_{H,H}\colon a\otimes b \mapsto (a_{1}\rightharpoonup b_{1})\otimes(a_{2}\leftharpoonup b_{2}) \end{align*}$$

is a coalgebra isomorphism and a solution to the braid equation on H.

The additional condition (⋆) is the braided commutativity of $\bullet $ with respect to the braiding operator $\sigma _{H,H}$ .

Remark 2.9 In the cocommutative case, the matched pair obtained from a Hopf brace $(H,\cdot ,\bullet )$ is given by $h\rightharpoonup k=S(h_{1})\cdot (h_{2}\bullet k)$ and $h\leftharpoonup k=T(h_{1}\rightharpoonup k_{1})\bullet h_{2}\bullet k_{2}$ . Then, the corresponding solution of the braid equation is

$$\begin{align*}\sigma_{H,H}(x\otimes y)=\big(S(x_{1})\cdot(x_{2}\bullet y_{1})\big)\otimes\big(T(S(x_{3})\cdot(x_{4}\bullet y_{2}))\bullet x_{5}\bullet y_{3}\big). \end{align*}$$

3.1 Yetter–Drinfeld braces

Let $(H,\bullet ,1,\Delta , \epsilon , T) $ be a Hopf algebra, $(H,\rightharpoonup ,\leftharpoonup )$ be a matched pair of actions on H. We begin by defining a second operation $\cdot $ and a map $S\colon H\to H$ , mimicking [Reference Angiono, Galindo and Vendramin5, Proposition 3.2]:

(3.1) $$ \begin{align} & a\cdot b:= a_1\bullet (T(a_2)\rightharpoonup b); \end{align} $$

(3.2) $$ \begin{align} & S(a):= a_1\rightharpoonup T(a_2).\qquad \end{align} $$

Notice that $\bullet $ is retrieved from $\cdot $ as

(3.3) $$ \begin{align} a\bullet b= a_{1}\bullet(T(a_{2})\rightharpoonup(a_{3}\rightharpoonup b))= a_{1}\cdot(a_{2}\rightharpoonup b). \end{align} $$

Recall that the linearity of $\Delta $ and $\epsilon $ with respect to $\rightharpoonup $ read as:

$$\begin{align*}\Delta(a\rightharpoonup b)=(a_{1}\rightharpoonup b_{1})\otimes(a_{2}\rightharpoonup b_{2}),\qquad \epsilon(a\rightharpoonup b)=\epsilon(a)\epsilon(b), \end{align*}$$

similarly for the linearity of $\Delta $ and $\epsilon $ with respect to $\leftharpoonup $ .

Remark 3.1 The condition (⋆) allows one to retrieve the right action from the left action, and vice versa. Indeed, by the linearity of $\Delta $ and $\epsilon $ with respect to $\rightharpoonup $ , (⋆) is equivalent to

(3.4) $$ \begin{align} &a\leftharpoonup b = T(a_1\rightharpoonup b_1)\bullet a_2\bullet b_2. \end{align} $$

Moreover, (3.4) is clearly equivalent to

(3.5) $$ \begin{align} &T(a\rightharpoonup b)=(a_{1}\leftharpoonup b_{1})\bullet T(b_{2})\bullet T(a_{2}). \end{align} $$

From (3.5) one can easily deduce

(3.6) $$ \begin{align} T(a_{1}\rightharpoonup T(a_{2}))=a_{1}\leftharpoonup T(a_{2}). \end{align} $$

Remark 3.2 Let us observe that $S^{2}$ is given on an element $a\in H$ by

$$ \begin{align*} S^{2}(a)&=S(a_{1}\rightharpoonup T(a_{2}))\\&=(a_{1}\rightharpoonup T(a_{4}))\rightharpoonup T(a_{2}\rightharpoonup T(a_{3}))\\&\kern-5pt\overset{({3.6})}{=}(a_{1}\rightharpoonup T(a_{4}))\rightharpoonup(a_{2}\leftharpoonup T(a_{3})). \end{align*} $$

Proof We compute

$$ \begin{align*} \Delta(a\cdot b)&=\Delta(a_{1}\bullet(T(a_{2})\rightharpoonup b))\\&=(a_{1}\bullet(T(a_{3})\rightharpoonup b)_{1})\otimes(a_{2}\bullet(T(a_{3})\rightharpoonup b)_{2}) \\&=(a_{1}\bullet(T(a_{4})\rightharpoonup b_{1}))\otimes(a_{2}\bullet(T(a_{3})\rightharpoonup b_{2}))\\&=(a_{1}\bullet(T(a_{3})\rightharpoonup b_{1}))\otimes(a_{2}\cdot b_{2}).\\[-35pt] \end{align*} $$

Although $\cdot $ is not a morphism of coalgebras in , this will be sorted out by setting ourselves in a suitable category.

An immediate computation shows the form of the twice iterated $\Delta $ , which is going to come in handy later:

(3.7) $$ \begin{align} \begin{aligned} (\Delta\otimes\mathrm{Id})\Delta\,(a\cdot b)&=\Delta(a_{1}\bullet(T(a_{3})\rightharpoonup b_{1}))\otimes(a_{2}\cdot b_{2})\\&= (a_{1}\bullet(T(a_{5})\rightharpoonup b_{1}))\otimes(a_{2}\bullet(T(a_{4})\rightharpoonup b_{2}))\otimes(a_{3}\cdot b_{3})\\&=(a_{1}\bullet(T(a_{6})\rightharpoonup b_{1}))\otimes(a_{2}\bullet(T(a_{5})\rightharpoonup b_{2}))\\ & \hspace{3.38cm}\otimes(a_{3}\bullet(T(a_{4})\rightharpoonup b_{3})). \end{aligned} \end{align} $$

Proposition 3.4 Let $(H,\bullet ,1,\Delta ,\epsilon ,T)$ be a Hopf algebra, $(H,\rightharpoonup ,\leftharpoonup )$ a matched pair of actions. Define $\cdot $ and S as in (3.1), (3.2). Then, $(H,\cdot ,1)$ becomes an algebra, $\epsilon $ is a morphism of algebras with respect to this structure, and S satisfies $a_{1}\cdot S(a_{2})=\epsilon (a)1=S(a_{1})\cdot a_{2}$ . Moreover, one has

(3.8) $$ \begin{align} a\rightharpoonup b=S(a_{1})\cdot(a_{2}\bullet b) \end{align} $$

and the two operations $\bullet $ and $\cdot $ satisfy the Hopf brace compatibility (hbc).

Proof We first observe that

$$ \begin{align*} &a\cdot1=a_{1}\bullet(T(a_{2})\rightharpoonup 1)=a_{1}\bullet\epsilon(T(a_{2}))1=a_{1}\epsilon(a_{2})=a,\\ & 1\cdot a=1\bullet(T(1)\rightharpoonup a)=a, \end{align*} $$

for all $a\in H$ . We now prove that $\cdot $ is associative. Given $a,b,c\in H$ we have:

$$ \begin{align*} a\cdot (b\cdot c)&= a\cdot \big(b_1\bullet (T(b_2)\rightharpoonup c)\big)\\&= a_1\bullet \Big( T(a_2)\rightharpoonup \big(b_1\bullet (T(b_2)\rightharpoonup c)\big) \Big)\\& \kern-8pt\overset{({\rm mp.3})}{=} a_1\bullet \Big( T(a_3)\rightharpoonup b_1 \Big)\bullet \Big( (T(a_2)\leftharpoonup b_2)\rightharpoonup (T(b_3)\rightharpoonup c)\Big)\\&\kern-5pt\overset{({3.4})}{=} a_1\bullet (T(a_4)\rightharpoonup b_1)\\&\hspace{0.9cm}\bullet \Big( T\big(T(a_3)\rightharpoonup b_2\big)\rightharpoonup \big( T(a_2)\rightharpoonup (b_3\rightharpoonup (T(b_4)\rightharpoonup c)) \big) \Big)\\&= a_1\bullet (T(a_4)\rightharpoonup b_1)\bullet \Big( T\big(T(a_3)\rightharpoonup b_2\big)\rightharpoonup \big( T(a_2)\rightharpoonup c \big) \Big)\\&= a_1\bullet (T(a_4)\rightharpoonup b_1) \bullet \Big(T\big(a_2\bullet (T(a_3)\rightharpoonup b_2)\big)\rightharpoonup c\Big)\\&= a_1\bullet (T(a_3)\rightharpoonup b_1)\bullet \Big(T(a_2\cdot b_2)\rightharpoonup c\Big)\\&= (a\cdot b)_1\bullet \Big(T\big((a\cdot b)_2\big)\rightharpoonup c\Big)\\&= (a\cdot b)\cdot c, \end{align*} $$

as desired. We then compute

$$ \begin{align*} \epsilon(a\cdot b)=\epsilon(a_{1}\bullet(T(a_{2})\rightharpoonup b))&=\epsilon(a_{1})\epsilon(T(a_{2})\rightharpoonup b)\\ &=\epsilon(T(a))\epsilon(b)=\epsilon(a)\epsilon(b). \end{align*} $$

Moreover, one has

$$ \begin{align*} a_{1}\cdot S(a_{2})&=a_{1}\bullet( T(a_{2})\rightharpoonup S(a_{3}))\\&=a_{1}\bullet(T(a_{2})\rightharpoonup(a_{3}\rightharpoonup T(a_{4})))\\&=a_{1}\bullet(T(a_{2})\bullet a_{3}\rightharpoonup T(a_{4}))\\&=a_{1}\bullet T(a_{2})\\&=\epsilon(a)1 \end{align*} $$

and

$$ \begin{align*} S(a_{1})\cdot a_{2}&=S(a_{1})_{1}\bullet(T(S(a_{1})_{2})\rightharpoonup a_{2})\\&=(a_{1_{1}}\rightharpoonup T(a_{2})_{1})\bullet(T(a_{1_{2}}\rightharpoonup T(a_{2})_{2})\rightharpoonup a_{3})\\&=(a_{1}\rightharpoonup T(a_{4}))\bullet(T(a_{2}\rightharpoonup T(a_{3}))\rightharpoonup a_{5})\\ &\kern-4pt\overset{({3.6})}{=}(a_{1}\rightharpoonup T(a_{4}))\bullet\Big(\big(a_{2}\leftharpoonup T(a_{3})\big)\rightharpoonup a_{5}\Big)\\&=(a_{1_{1}}\rightharpoonup T(a_{2})_{1})\bullet\Big(\big(a_{1_{2}}\leftharpoonup T(a_{2})_{2}\big)\rightharpoonup a_{3}\Big)\\&\kern-8pt\overset{({\rm mp.3})}{=}a_{1}\rightharpoonup(T(a_{2})\bullet a_{3})\\&=\epsilon(a)1. \end{align*} $$

As for (3.8), observe that

$$\begin{align*}a\rightharpoonup b=S(a_{1})\cdot a_{2}\cdot(a_{3}\rightharpoonup b)\overset{({3.3})}{=}S(a_{1})\cdot(a_{2}\bullet b). \end{align*}$$

We finally prove the Hopf brace compatibility (hbc):

$$ \begin{align*} a\bullet(b\cdot c)&=a\bullet b_{1}\bullet(T(b_{2})\rightharpoonup c)\\&=a_{1}\bullet b_{1}\bullet(T(b_{2})\bullet T(a_{2})\bullet a_{3}\rightharpoonup c)\\&=a_{1}\bullet b_{1}\bullet(T(a_{2}\bullet b_{2})\rightharpoonup(a_{3}\rightharpoonup c))\\&=(a_{1}\bullet b)\cdot(a_{2}\rightharpoonup c)\\&\kern-5pt\overset{({3.8})}{=}(a_{1}\bullet b)\cdot S(a_{2})\cdot(a_{3}\bullet c).\\[-33pt] \end{align*} $$

As promised, we now reinterpret this additional operation $\cdot $ in a suitable category, which is going to be the category of Yetter–Drinfeld modules over H.

Definition 3.1 Let $(H,\bullet ,1,\Delta , \epsilon , T)$ be a Hopf algebra. A (left-left) Yetter–Drinfeld module on H is the datum of a left H-module X with an action $\rightharpoonup \colon H\otimes X\to X$ , which is also a left H-comodule with a coaction $\rho \colon X\to H\otimes X$ , satisfying the following compatibility for all $x\in X$ , $a\in H$ :

(yd) $$ \begin{align} \rho(a\rightharpoonup x)=a_{1}\bullet x_{-1}\bullet T(a_{3})\otimes(a_{2}\rightharpoonup x_{0}). \end{align} $$

A morphism of Yetter–Drinfeld modules is a morphism of both left H-modules and left H-comodules. We denote by $^H_H\mathcal {YD}$ the category of Yetter–Drinfeld modules over H.

Lemma 3.5 Let $(H,\bullet ,1,\Delta ,\epsilon ,T)$ be a Hopf algebra. Suppose that $\rightharpoonup \colon H\otimes H\to H$ is a morphism of coalgebras, $\leftharpoonup \colon H\otimes H\to H$ is a linear map satisfying $\epsilon (a\leftharpoonup b)=\epsilon (a)\epsilon (b)$ , and (⋆) is satisfied. The following conditions are equivalent, and they are all equivalent to (mp.5):

(3.9) $$ \begin{align} &(a\leftharpoonup b)_1\otimes (a\leftharpoonup b)_2 = (a_1\leftharpoonup b_1)\otimes (a_2\leftharpoonup b_2),\qquad\qquad\qquad\nonumber\\ & \hspace{2em}\text{ i.e.,}\ \leftharpoonup\text{ is a morphism of coalgebras}\,; \end{align} $$

(3.10) $$ \begin{align} & \big((a_1\rightharpoonup b_1)\bullet (a_3\leftharpoonup b_3)\big)\otimes \big((a_2\rightharpoonup b_2)\bullet (a_4\leftharpoonup b_4)\big)\nonumber\\ & \hspace{2em}= \big((a_1\rightharpoonup b_1)\bullet (a_3\leftharpoonup b_3)_1\big)\otimes \big((a_2\rightharpoonup b_2)\bullet (a_3\leftharpoonup b_3)_2\big); \end{align} $$

(3.11) $$ \begin{align} & (a_2\leftharpoonup b_2)\otimes \big((a_1\rightharpoonup b_1)\bullet (a_3\leftharpoonup b_3)\big) = (a_1\leftharpoonup b_1)\otimes (a_2\bullet b_2); \end{align} $$

(3.12) $$ \begin{align} &\big((a_{1}\rightharpoonup b_{1})\bullet(a_{3}\leftharpoonup b_{3})\big)\otimes(a_{2}\rightharpoonup b_{2})=(a_{1}\bullet b_{1})\otimes(a_{2}\rightharpoonup b_{2}). \end{align} $$

In particular, having a matched pair of actions $(H,\rightharpoonup ,\leftharpoonup )$ all the previous equalities hold true.

Proof First, we notice that

$$ \begin{align*} &\hspace{-2em}\big((a_1\rightharpoonup b_1)\bullet (a_3\leftharpoonup b_3)_1\big)\otimes \big((a_2\rightharpoonup b_2)\bullet (a_3\leftharpoonup b_3)_2\big)\\ &= \big((a_1\rightharpoonup b_1)_1\bullet (a_2\leftharpoonup b_2)_1\big)\otimes \big((a_1\rightharpoonup b_1)_2\bullet (a_2\leftharpoonup b_2)_2\big)\\ &= \big((a_1\rightharpoonup b_1)\bullet (a_2\leftharpoonup b_2)\big)_1\otimes \big((a_1\rightharpoonup b_1)\bullet (a_2\leftharpoonup b_2)\big)_2\\ &\kern-3pt\overset{({\star})}{=} (a\bullet b)_1\otimes (a\bullet b)_2\\ &= (a_{1}\bullet b_{1})\otimes(a_2\bullet b_2). \end{align*} $$

Now, we check that (3.10) and (3.9) are equivalent:

$$ \begin{align*} ({3.10})\iff\;&\big(T(a_1\rightharpoonup b_1)\bullet (a_2\rightharpoonup b_2)\bullet (a_4\leftharpoonup b_4)\big)\otimes \big((a_3\rightharpoonup b_3)\bullet (a_5\leftharpoonup b_5)\big)\\ &= \big(T(a_1\rightharpoonup b_1)\bullet (a_2\rightharpoonup b_2)\bullet (a_4\leftharpoonup b_4)_1\big)\\ & \hspace{4.5cm}\otimes \big((a_3\rightharpoonup b_3)\bullet (a_4\leftharpoonup b_4)_2\big)\\ \iff\;& (a_2\leftharpoonup b_2)\otimes \big( (a_1\rightharpoonup b_1)\bullet (a_3\leftharpoonup b_3)\big) \\& = (a_2\leftharpoonup b_2)_1\otimes \big( (a_1\rightharpoonup b_1)\bullet (a_2\leftharpoonup b_2)_2\big)\\ \iff\;& (a_3\leftharpoonup b_3)\otimes \big(T(a_1\rightharpoonup b_1)\bullet (a_2\rightharpoonup b_2)\bullet (a_4\leftharpoonup b_4)\big)\\ & = (a_3\leftharpoonup b_3)_1\otimes \big( T(a_1\rightharpoonup b_1)\bullet (a_2\rightharpoonup b_2)\bullet (a_3\leftharpoonup b_3)_2\big)\\ \iff\;& (a_1\leftharpoonup b_1)\otimes (a_2\leftharpoonup b_2) = (a\leftharpoonup b)_1\otimes (a\leftharpoonup b)_2. \end{align*} $$

We prove that (3.11) and (3.12) are both equivalent to (3.10). One has

$$ \begin{align*} &\hspace{-2em}(a_2\leftharpoonup b_2)\otimes \big((a_1\rightharpoonup b_1)\bullet (a_3\leftharpoonup b_3)\big) = (a_1\leftharpoonup b_1)\otimes (a_2\bullet b_2)\\ &\Leftrightarrow\big((a_1\rightharpoonup b_1)\bullet(a_3\leftharpoonup b_3)\big)\otimes \big((a_2\rightharpoonup b_2)\bullet (a_4\leftharpoonup b_4)\big) \\ &\hspace{2em}= \big((a_1\rightharpoonup b_1)\bullet (a_2\leftharpoonup b_2)\big)\otimes (a_3\bullet b_3)\\ &\Leftrightarrow \big((a_1\rightharpoonup b_1)\bullet(a_3\leftharpoonup b_3)\big)\otimes \big((a_2\rightharpoonup b_2)\bullet (a_4\leftharpoonup b_4)\big) \\ &\hspace{2em}= (a_1\bullet b_1)\otimes (a_2\bullet b_2)\\ &\Leftrightarrow({3.10}) \end{align*} $$

and

$$ \begin{align*} ({3.10})&\iff&&\!\!\big((a_{1}\rightharpoonup b_{1})\bullet(a_{3}\leftharpoonup b_{3})\big)\otimes\big((a_{2}\rightharpoonup b_{2})\bullet (a_4\leftharpoonup b_4)\big)\\ &&&\hspace{2em}=(a_{1}\bullet b_{1})\otimes(a_2\bullet b_2)\\ &\iff &&\!\!\big((a_{1}\rightharpoonup b_{1})\bullet(a_{3}\leftharpoonup b_{3})\big)\otimes\big((a_{2}\rightharpoonup b_{2})\bullet (a_4\leftharpoonup b_4)\big)\\ &&& \hspace{2em}=(a_{1}\bullet b_{1})\otimes\big((a_{2}\rightharpoonup b_{2})\bullet (a_3\leftharpoonup b_3)\big)\\ &\iff &&\!\!\big((a_{1}\rightharpoonup b_{1})\bullet(a_{3}\leftharpoonup b_{3})\big)\otimes(a_{2}\rightharpoonup b_{2})=(a_{1}\bullet b_{1})\otimes(a_{2}\rightharpoonup b_{2}), \end{align*} $$

where we use the fact that (3.10) implies (3.9) and $\epsilon (a\leftharpoonup b)=\epsilon (a)\epsilon (b)$ in the latter equivalence. This completes the proof that (3.9)–(3.12) are equivalent. Finally, using the assumption (⋆), (mp.5) clearly implies (3.11); and, conversely, assuming (3.11) (and thus (3.9)) we immediately retrieve (mp.5).

Proof We already know that $(H,\rightharpoonup )$ is a left H-module and $(H,\mathrm {Ad}_{L})$ is a left H-comodule. Since $\leftharpoonup $ is a morphism of coalgebras, the equivalent conditions of Lemma 3.5 hold true. We check the compatibility of Yetter–Drinfeld modules (yd):

$$ \begin{align*} \mathrm{Ad}_L(a\rightharpoonup b)& =\Big((a_{1}\rightharpoonup b_{1})\bullet T(a_{3}\rightharpoonup b_{3})\Big)\otimes(a_{2}\rightharpoonup b_{2})\\&\kern-5pt\overset{({3.5})}{=}(a_{1}\rightharpoonup b_{1})\bullet(a_{3}\leftharpoonup b_{3})\bullet T(b_{4})\bullet T(a_{4})\otimes(a_{2}\rightharpoonup b_{2})\\&\kern-6pt\overset{({3.12})}{=}a_{1}\bullet b_{1}\bullet T(b_{3})\bullet T(a_{3})\otimes(a_{2}\rightharpoonup b_{2})\\&=a_{1}\bullet b_{-1}\bullet T(a_{3})\otimes(a_{2}\rightharpoonup b_{0}).\\[-35pt] \end{align*} $$

Remark 3.8 Notice that the braiding $\sigma ^{\mathcal {YD}}_{H,H}$ coincides with the flip map $\tau $ in case the adjoint coaction $\mathrm {Ad}_{L}$ is trivial, i.e., $a_{1}\bullet T(a_{3})\otimes a_{2}=1\otimes a$ , which clearly happens in the cocommutative case. Conversely, if $a_{1}\bullet T(a_{3})\otimes a_{2}=1\otimes a$ , then

$$\begin{align*}a_{1}\bullet T(a_{3})\otimes a_{2}=1\otimes a=1\otimes a_{1}\epsilon(a_{2})=a_{2}\bullet T(a_{3})\otimes a_{1} \end{align*}$$

and so $a_{1}\otimes a_{2}=a_{2}\otimes a_{1}$ , i.e., H is cocommutative. Furthermore, let us also observe that $\sigma ^{\mathcal {YD}}_{H,H}$ coincides with $\tau $ also in case $\rightharpoonup $ is the trivial action.

We merge the previous results in the following theorem.

Proof The compatibility (hbc) is satisfied by Proposition 3.4.

We first check that $\cdot $ is a morphism of coalgebras in $^{H}_{H}\mathcal {YD}$ . Using Lemma 3.3 we obtain:

$$\begin{align*}\begin{aligned} \Delta(a\cdot b)&=(a_{1}\bullet(T(a_{3})\rightharpoonup b_{1}))\otimes(a_{2}\cdot b_{2})\\&\kern-5pt\overset{({3.3})}{=}(a_{1}\cdot(a_{2}\rightharpoonup(T(a_{4})\rightharpoonup b_{1})))\otimes(a_{3}\cdot b_{2})\\&=a_{1}\cdot(a_{2}\bullet T(a_{4})\rightharpoonup b_{1})\otimes(a_{3}\cdot b_{2}), \end{aligned} \end{align*}$$

i.e., $\Delta m_{\cdot }=(m_{\cdot }\otimes m_{\cdot })(\mathrm {Id}\otimes \sigma ^{\mathcal {YD}}_{H,H}\otimes \mathrm {Id})(\Delta \otimes \Delta )$ . By Proposition 3.4, we also know that $\epsilon (a\cdot b)=\epsilon (a)\epsilon (b)$ . Together, they are the compatibility conditions of a bialgebra in $^{H}_{H}\mathcal {YD}$ .

Our next step is proving that $(H,\Delta , \epsilon )$ is a coalgebra in $^H_H\mathcal {YD}$ . We know that $\Delta $ and $\epsilon $ are left H-linear with respect to $\rightharpoonup $ . We thereby prove that they are also left H-colinear. We compute

$$ \begin{align*} (\mathrm{Id}\otimes\Delta)\mathrm{Ad}_{L}(a)&=(\mathrm{Id}\otimes\Delta)(a_{1}\bullet T(a_{3})\otimes a_{2})\\&=a_{1}\bullet T(a_{4})\otimes a_{2}\otimes a_{3}\\&=(a_{1}\bullet T(a_{3})\bullet a_{4}\bullet T(a_{6}))\otimes a_{2}\otimes a_{5}\\&=(a_{1_{1}}\bullet T(a_{1_{3}}))\bullet(a_{2_{1}}\bullet T(a_{2_{3}}))\otimes a_{1_{2}}\otimes a_{2_{2}} \end{align*} $$

and

This concludes the proof that $(H,\Delta ,\epsilon )$ is a coalgebra in $^{H}_{H}\mathcal {YD}$ .

Clearly, the unit is left H-linear and left H-colinear, so in order to conclude that $(H,\cdot ,1)$ is in $\mathrm {Mon}(^{H}_{H}\mathcal {YD})$ we only need to verify that $\cdot $ is left H-linear and left H-colinear.

First we show that $\cdot $ is left H-linear with respect to $\rightharpoonup $ . Therefore, we compute

$$ \begin{align*} a\rightharpoonup(b\cdot c)&\overset{({3.1})}{=}a\rightharpoonup(b_{1}\bullet(T(b_{2})\rightharpoonup c)) \\&\kern-4pt\overset{({\rm mp.3})}{=}(a_{1}\rightharpoonup b_{1})\bullet((a_{2}\leftharpoonup b_{2})\rightharpoonup(T(b_{3})\rightharpoonup c))\\&\kern-1pt\overset{({3.3})}{=}(a_{1}\rightharpoonup b_{1})\cdot((a_{2}\rightharpoonup b_{2})\rightharpoonup((a_{3}\leftharpoonup b_{3})\rightharpoonup(T(b_{4})\rightharpoonup c))\\&\kern4pt=(a_{1}\rightharpoonup b_{1})\cdot(((a_{2}\rightharpoonup b_{2})\bullet(a_{3}\leftharpoonup b_{3}))\rightharpoonup(T(b_{4})\rightharpoonup c))\\&\kern2pt\overset{({\star})}{=}(a_{1}\rightharpoonup b_{1})\cdot(a_{2}\bullet b_{2}\rightharpoonup(T(b_{3})\rightharpoonup c))\\&\kern4pt=(a_{1}\rightharpoonup b_{1})\cdot(a_{2}\bullet b_{2}\bullet T(b_{3})\rightharpoonup c))\\&\kern4pt=(a_{1}\rightharpoonup b)\cdot(a_{2}\rightharpoonup c). \end{align*} $$

We now check that $\cdot $ is left H-colinear with respect to $\mathrm {Ad}_L$ . We already know that (3.7) holds, i.e.,

$$\begin{align*}(\Delta\otimes\mathrm{Id})\Delta(a\cdot b)=(a_{1}\bullet(T(a_{6})\rightharpoonup b_{1}))\otimes(a_{2}\bullet(T(a_{5})\rightharpoonup b_{2}))\otimes(a_{3}\bullet(T(b_{4})\rightharpoonup b_{3})). \end{align*}$$

The request that $\cdot $ is colinear with respect to the adjoint coaction $\mathrm {Ad}_L$ reads as follows:

(3.13) $$ \begin{align} (a\cdot b)_{1}\bullet T((a\cdot b)_{3})\otimes(a\cdot b)_{2}=a_{1}\bullet T(a_{3})\bullet b_{1}\bullet T(b_{3})\otimes(a_{2}\cdot b_{2}). \end{align} $$

We now rephrase the condition (3.13). One has

$$ \begin{align*} &(a\cdot b)_{1}\bullet T((a\cdot b)_{3})\otimes(a\cdot b)_{2}\\ &=\bigg(a_1\bullet (T(a_6)\rightharpoonup b_1)\bullet T\Big(a_3\bullet (T(a_4)\rightharpoonup b_3) \Big)\bigg)\otimes \Big(a_2\bullet (T(a_5)\rightharpoonup b_2)\Big)\\ &= \bigg(a_1\bullet (T(a_6)\rightharpoonup b_1)\bullet T\Big(T(a_4)\rightharpoonup b_3 \Big)\bullet T(a_3)\bigg)\otimes \Big(a_2\bullet (T(a_5)\rightharpoonup b_2)\Big)\\ &= \bigg(a_1\bullet (T(a_6)\rightharpoonup b_1)\bullet T\Big(T(a_4)\rightharpoonup b_3 \Big)\bullet T(a_3)\bullet b_4\bullet T(b_5)\bigg)\\ & \hspace{5.8cm}\otimes \Big(a_2\bullet (T(a_5)\rightharpoonup b_2)\Big)\\ &\kern-5pt\overset{({3.4})}{=} \Big(a_1\bullet (T(a_5)\rightharpoonup b_1)\bullet (T(a_3)\leftharpoonup b_3)\bullet T(b_4)\Big)\otimes\Big(a_2\bullet (T(a_4)\rightharpoonup b_2)\Big). \end{align*} $$

Therefore, (3.13) becomes

(3.14) $$ \begin{align} &\Big(a_1\bullet (T(a_5)\rightharpoonup b_1)\bullet (T(a_3)\leftharpoonup b_3)\bullet T(b_4)\Big)\otimes\Big( a_2\bullet (T(a_4)\rightharpoonup b_2)\Big)\nonumber\\ &\hspace{2em}= \Big(a_1\bullet T(a_4)\bullet b_1\bullet T(b_3)\Big)\otimes \Big( a_2\bullet (T(a_3)\rightharpoonup b_2) \Big). \end{align} $$

But now, using the antipode, we obtain

$$ \begin{align*} ({3.14})\iff\;&\Big((T(a_4)\rightharpoonup b_1)\bullet (T(a_2)\leftharpoonup b_3)\Big)\otimes\big( a_1\bullet (T(a_3)\rightharpoonup b_2)\Big)\\ &= \big(T(a_3)\bullet b_1\big)\otimes \Big( a_1\bullet (T(a_2)\rightharpoonup b_2) \Big)\\\iff\;&\Big((T(a_3)\rightharpoonup b_1)\bullet (T(a_1)\leftharpoonup b_3)\Big)\otimes\Big( T(a_2)\rightharpoonup b_2\Big)\\ &\hspace{2em}= \big(T(a_2)\bullet b_1\big)\otimes \big( T(a_1)\rightharpoonup b_2 \big). \end{align*} $$

Thus, we obtain that (3.13) is equivalent to

(3.15) $$ \begin{align} &(T(a_3)\rightharpoonup b_1)\bullet (T(a_1)\leftharpoonup b_3)\otimes (T(a_2)\rightharpoonup b_2)\nonumber\\ &\hspace{2em} = \big(T(a_2)\bullet b_1\big)\otimes \big( T(a_1)\rightharpoonup b_2\big). \end{align} $$

The latter is implied by (3.12), which holds true since $\leftharpoonup $ is a morphism of coalgebras (and is in fact equivalent to (3.12) when T is bijective). This concludes the proof that $(H,\cdot , 1)$ is an algebra in $^H_H\mathcal {YD}$ , and hence $(H,\cdot , 1,\Delta , \epsilon )$ is a bialgebra in $^H_H\mathcal {YD}$ for what already observed.

We already know from Proposition 3.4 that S satisfies $S(a_1)\cdot a_2 = \epsilon (a)1_{H}= a_1\cdot S(a_2)$ , thus we only need to prove that S is a morphism of Yetter–Drinfeld modules. This follows from Proposition 3.9.

In Remark 5.6, we shall see that this is indeed a generalization of Majid’s transmutation [Reference Majid28].

Corollary 3.13 Let $(H,\bullet ,1,\Delta ,\epsilon ,T)$ be a Hopf algebra and $(H,\rightharpoonup ,\leftharpoonup )$ a matched pair of actions. Define $\cdot $ as in (3.1), and S as in (3.2). Then, the following is a bialgebra structure on $H\otimes H$ :

$$ \begin{align*} &(a\otimes h) \cdot_{\#} (a'\otimes h') := a\cdot S(h_{1})\cdot (h_{2}\bullet a')\otimes h_3\bullet h',\\ &\Delta_{\#}(a\otimes h):= a_1\otimes a_2\bullet T(a_4)\bullet h_1\otimes a_3\otimes h_2,\\ &S_{\#}(a\otimes h):=S\big(T(T(a_{3})\bullet h_{3})\big)\cdot\big(T(T(a_{4})\bullet h_{2})\bullet T(a_{2})\big)\otimes T(a_{1}\bullet T(a_{5})\bullet h_{1}).\end{align*} $$

Proof Let $H^\bullet :=(H,\bullet )$ and denote the two multiplicative structures. By Theorem 3.10, we have that $(H,\cdot ,1,\Delta ,\epsilon )$ is in $\mathrm {Bimon}({}^{H}_{H}\mathcal {YD})$ , thus the bosonization is a bialgebra, where $(a\otimes h)\cdot _{\#}(a'\otimes h'):=a\cdot (h_{1}\rightharpoonup a')\otimes h_{2}\bullet h'$ and $\Delta _{\#}(a\otimes h):=a_{1}\otimes a_{2_{-1}}\bullet h_{1}\otimes a_{2_{0}}\otimes h_{2}$ denote the smash product and the smash coproduct, respectively, see e.g., Heckenberger and Schneider [Reference Heckenberger and Schneider19, Proposition 3.8.4]. Moreover, since $H^{\bullet }$ and are both Hopf algebras, we obtain an antipode on given by $S_{\#}(a\otimes h):=\big (T(a_{-1}\bullet h_{2})\rightharpoonup S(a_{0})\big )\otimes T(a_{-2}\bullet h_{1})$ ; see e.g., [Reference Heckenberger and Schneider19, Theorem 3.8.10]. Thus, we compute

$$ \begin{align*} S_{\#}(a\otimes h)&=\big(T(a_{-1}\bullet h_{2})\rightharpoonup S(a_{0})\big)\otimes T(a_{-2}\bullet h_{1})\\&=\big(T(a_{2}\bullet T(a_{4})\bullet h_{2})\rightharpoonup S(a_{3})\big)\otimes T(a_{1}\bullet T(a_{5})\bullet h_{1})\\&=\big(T(T(a_{5})\bullet h_2)\bullet T(a_{2})\rightharpoonup(a_{3}\rightharpoonup T(a_{4}))\big)\otimes T(a_{1}\bullet T(a_{6})\bullet h_{1})\\&=\big(T(T(a_{3})\bullet h_{2})\rightharpoonup T(a_{2})\big)\otimes T(a_{1}\bullet T(a_{4})\bullet h_{1})\\&=S\big(T(T(a_{3})\bullet h_{3})\big)\cdot\big(T(T(a_{4})\bullet h_{2})\bullet T(a_{2})\big)\otimes T(a_{1}\bullet T(a_{5})\bullet h_{1}). \end{align*} $$

Theorem 3.10 suggests us the following definition.

Definition 3.3 A Yetter–Drinfeld brace (or $\mathcal {YD}$ -brace) $(H, \cdot , \bullet ,1,\Delta ,\epsilon ,S,T)$ is the datum of a Hopf algebra $H^\bullet = (H,\bullet , 1,\Delta , \epsilon , T)$ , a second operation $\cdot $ on H, and a linear map $S\colon H\to H$ such that:

1. i. $(H,\cdot ,1,\Delta ,\epsilon ,S)$ is in $\mathrm {Hopf}({}^{H^\bullet }_{H^\bullet }\mathcal {YD})$ with the action $\rightharpoonup $ defined by $a\rightharpoonup b := S(a_1)\cdot (a_2\bullet b)$ , and the coaction given by $\mathrm {Ad}_L$ ;

2. ii. if we define $a\leftharpoonup b:= T(a_1\rightharpoonup b_1)\bullet a_2\bullet b_2$ , the two maps $\rightharpoonup , \leftharpoonup $ satisfy (mp.5);

3. iii. the two operations $\bullet $ and $\cdot $ satisfy the Hopf brace compatibility (hbc).

Given two Yetter–Drinfeld braces H and K, a morphism of Yetter–Drinfeld braces is a map $f\colon H\to K$ that is a morphism in between the respective Hopf algebras in , and satisfies

$$\begin{align*}f(a\cdot_H b) = f(a)\cdot_K f(b).\end{align*}$$

In particular, this implies $S_{K}f=fS_{H}$ , $f\rightharpoonup _{H}\;=\;\rightharpoonup _{K}\!(f\otimes f)$ and $(\mathrm {Ad}_{L})_Kf=(f\otimes f)(\mathrm {Ad}_{L})_H$ . We denote the category of Yetter–Drinfeld braces by .

Remark 3.14 Let $f\colon H\to K$ be a morphism of Yetter–Drinfeld braces. Since $f\colon H^{\bullet }\to K^{\bullet }$ is a morphism of bialgebras, one can regard as a left $H^{\bullet }$ -module through f, so that the condition $f(a\rightharpoonup _{H} b)=f(a)\rightharpoonup _{K}f(b)$ means that f is a morphism in $_{H^{\bullet }}\mathfrak {M}$ ; and as a left $K^{\bullet }$ -comodule through f, so that the condition $(\mathrm {Ad}_{L})_{K}f=(f\otimes f)(\mathrm {Ad}_{L})_{H}$ means that f is a morphism in $^{K^{\bullet }}\mathfrak {M}$ . One can consider the category $^{K^{\bullet }}_{H^{\bullet }}\mathcal {YD}$ of relative Yetter–Drinfeld modules; see [Reference Caenepeel, Militaru and Zhu8, Section 4.4, p. 183]. Clearly and are in $^{K^{\bullet }}_{H^{\bullet }}\mathcal {YD}$ . Indeed, given $a,b\in H$ , we compute:

$$ \begin{align*}(a_{1}\rightharpoonup_{H} b)^K_{-1}\bullet_{K} f(a_{2})\otimes(a_{1}\rightharpoonup_{H} b)^{K}_{0}&=f((a_{1}\rightharpoonup_{H}b)_{-1}\bullet_{H}a_{2})\otimes(a_{1}\rightharpoonup_{H}b)_{0}\\ &=f(a_{1}\bullet_{H} b_{-1})\otimes(a_{2}\rightharpoonup_{H}b_{0})\\ &=f(a_{1})\bullet_{K} b_{-1}^{K}\otimes(a_{2}\rightharpoonup_{H} b_{0}^K), \end{align*} $$

where $a_{-1}\otimes a_0 = (\mathrm {Ad}_L)_H(a)$ is the usual adjoint $H^\bullet $ -coaction on H, while $a^K_{-1}\otimes a^K_0= (f\otimes \mathrm {Id})(\mathrm {Ad}_L)_H(a)$ denotes the $K^\bullet $ -coaction on induced by $\mathrm {Ad}_L$ through f. The Yetter–Drinfeld compatibility for is deduced analogously. Hence, we obtain that f is a morphism in $^{K^{\bullet }}_{H^{\bullet }}\mathcal {YD}$ . Moreover, one clearly has that and are in $\mathrm {Mon}(^{K^{\bullet }}_{H^{\bullet }}\mathcal {YD})$ and $\mathrm {Comon}(^{K^{\bullet }}_{H^{\bullet }}\mathcal {YD})$ ; thus the condition $f(a\cdot _{H}b)=f(a)\cdot _{K}f(b)$ means that is a morphism in $\mathrm {Mon}(^{K^{\bullet }}_{H^{\bullet }}\mathcal {YD})$ . Notice that $f $ is automatically a morphism in $\mathrm {Comon}(^{K^{\bullet }}_{H^{\bullet }}\mathcal {YD})$ .

From Theorem 3.10, we obtain:

Remark 3.16 Define $\leftharpoonup $ as in Definition 3.3 ii. Since $\rightharpoonup $ is a morphism of coalgebras, it is clear that $\rightharpoonup , \leftharpoonup $ satisfy (⋆):

$$ \begin{align*} (a_1\rightharpoonup b_1)\bullet (a_2\leftharpoonup b_2)&=(a_1\rightharpoonup b_1)\bullet T(a_{2}\rightharpoonup b_{2})\bullet a_{3}\bullet b_{3}\\&=\epsilon(a_{1}\rightharpoonup b_{1})a_{2}\bullet b_{2}=a\bullet b. \end{align*} $$

Observe that $\epsilon (a\leftharpoonup b) = \epsilon (a)\epsilon (b)$ , thus we are in the hypotheses of Lemma 3.5. Consequently, Definition 3.3 ii. is equivalent to the request that $\leftharpoonup $ is a morphism of coalgebras.

Clearly, in a Yetter–Drinfeld brace, the map $\rightharpoonup $ defined as in Definition 3.3 i. is a left $H^{\bullet }$ -action, and $(H,\cdot ,1)$ is a left $H^{\bullet }$ -module algebra. In fact, this is true under weaker assumptions:

Lemma 3.18 Let $H^\bullet =(H,\bullet ,1,\Delta ,\epsilon )$ be a bialgebra and $(H,\cdot ,1,\Delta ,\epsilon )$ be an algebra and a coalgebra, endowed with a map $S\colon H\to H$ satisfying $S(a_{1})\cdot a_{2}=\epsilon (a)1=a_{1}\cdot S(a_{2})$ . Suppose that the Hopf brace compatibility (hbc) between $\bullet $ and $\cdot $ is satisfied. Then

$$\begin{align*}a\rightharpoonup b:=S(a_{1})\cdot(a_{2}\bullet b) \end{align*}$$

is a left $H^{\bullet }$ -action on H. Moreover, $(H,\cdot ,1)$ is a left $H^{\bullet }$ -module algebra and, if $\epsilon $ is a morphism of algebras with respect to $\cdot $ , then $\epsilon $ is left $H^{\bullet }$ -linear with respect to $\rightharpoonup $ .

Proof First we compute $1\rightharpoonup b=S(1)\cdot (1\bullet b)=b$ . Moreover, we have

$$ \begin{align*} a\rightharpoonup(b\rightharpoonup c)&=a\rightharpoonup(S(b_{1})\cdot(b_{2}\bullet c))\\&=S(a_{1})\cdot\Big(a_{2}\bullet\big(S(b_{1})\cdot(b_{2}\bullet c)\big)\Big)\\&=S(a_{1})\cdot(a_{2}\bullet S(b_{1}))\cdot S(a_{3})\cdot(a_{4}\bullet b_{2}\bullet c)\\&\kern-3pt\overset{(\dagger)}{=}S(a_{1}\bullet b_{1})\cdot a_{2}\cdot S(a_{3})\cdot(a_{4}\bullet b_{2}\bullet c)\\&=S(a_{1}\bullet b_{1})\cdot (a_{2}\bullet b_{2}\bullet c)\\&=(a\bullet b)\rightharpoonup c, \end{align*} $$

where the equality marked with $(\dagger )$ follows from the relation $S(a_{1}\bullet b)\cdot a_{2}=S(a_{1})\cdot (a_{2}\bullet S(b))$ , whose proof is the same as in [Reference Angiono, Galindo and Vendramin5, Lemma 1.7] and simply follows from (hbc). Moreover, we have

$$\begin{align*}a\rightharpoonup(b\cdot c)=S(a_{1})\cdot(a_{2}\bullet(b\cdot c))=S(a_{1})\cdot(a_{2}\bullet b)\cdot S(a_{3})\cdot (a_{4}\bullet c)=(a_{1}\rightharpoonup b)\cdot(a_{2}\rightharpoonup c) \end{align*}$$

and

(3.16) $$ \begin{align}a\rightharpoonup1=S(a_{1})\cdot(a_{2}\bullet1)=S(a_{1})\cdot a_{2}=\epsilon(a)1,\end{align} $$

hence $(H,\cdot ,1)$ is a left $H^{\bullet }$ -module algebra. Finally, we have

$$\begin{align*}\epsilon(a\rightharpoonup b)=\epsilon(S(a_{1})\cdot(a_{2}\bullet b))\overset{(\ddagger)}{=}\epsilon(a_{1})\epsilon(a_{2})\epsilon(b)=\epsilon(a)\epsilon(b), \end{align*}$$

where $(\ddagger )$ follows from the fact that $\epsilon $ is a morphism of algebras with respect to $\cdot $ .

3.2 A general example

We now construct an easy example of a Yetter–Drinfeld brace on a Hopf algebra H. This construction depends on the assumption that the adjoint action on H is compatible with the comultiplication.

Proof Notice that $H^{\bullet }$ is automatically a right H-module coalgebra. We verify (mp.1)–(mp.4) and (⋆). Clearly $a\rightharpoonup 1=\epsilon (a)1$ holds, while (mp.2) is automatically true. Moreover, we compute

$$ \begin{align*} (a_{1}\rightharpoonup b_{1})\bullet((a_{2}\leftharpoonup b_{2})\rightharpoonup c)&=a_{1}\bullet b\bullet T(a_{2})\bullet(a_{3}\rightharpoonup c)\\ &=a_{1}\bullet b\bullet T(a_{2})\bullet a_{3}\bullet c\bullet T(a_{4})=a\rightharpoonup(b\bullet c), \end{align*} $$

and

$$\begin{align*}(a\leftharpoonup(b_{1}\rightharpoonup c_{1}))\bullet(b_{2}\leftharpoonup c_{2})=a\,\epsilon(b_{1}\rightharpoonup c_{1})\bullet b_{2}\,\epsilon(c_{2})=(a\bullet b)\leftharpoonup c. \end{align*}$$

Finally,

$$\begin{align*}(a_{1}\rightharpoonup b_{1})\bullet(a_{2}\leftharpoonup b_{2})=a_{1}\bullet b_{1}\bullet T(a_{2})\bullet a_{3}\epsilon(b_{2})=a\bullet b. \end{align*}$$

Thus, $(H,\rightharpoonup ,\leftharpoonup )$ is a matched pair of actions if and only if $H^{\bullet }$ is a left H-module coalgebra. Since $\epsilon (a\rightharpoonup b)=\epsilon (a)\epsilon (b)$ is automatically satisfied, $H^\bullet $ is a left H-module coalgebra if and only if $\Delta $ is left linear with respect to $\rightharpoonup $ ; as desired.

Remark 3.20 The linearity condition of $\Delta $ with respect to $\rightharpoonup $ , i.e., $\Delta (a\rightharpoonup b)=(a_{1}\rightharpoonup b_{1})\otimes (a_{2}\rightharpoonup b_{2})$ , explicitly reads:

$$\begin{align*}a_{1}\bullet b_{1}\bullet T(a_{4})\otimes a_{2}\bullet b_{2}\bullet T(a_{3})=a_{1}\bullet b_{1}\bullet T(a_{2})\otimes a_{3}\bullet b_{2}\bullet T(a_{4}). \end{align*}$$

The latter is equivalent to $T(a_{3})\otimes a_{1}\bullet b\bullet T(a_{2})=T(a_{1})\otimes a_{2}\bullet b\bullet T(a_{3})$ , and then to

(3.17) $$ \begin{align} a_{1}\bullet T(a_{4})\otimes a_{2}\bullet b\bullet T(a_{3})=1\otimes a_{1}\bullet b\bullet T(a_{2}), \end{align} $$

i.e., $(\mathrm {Id}\otimes \rightharpoonup )(\mathrm {Ad}_{L}\otimes \mathrm {Id})=(u_{H}\otimes \mathrm {\rightharpoonup })$ , where $\mathrm {Ad}_{L}$ denotes again the left adjoint coaction on H. From the matched pair of actions $(H,\rightharpoonup ,\leftharpoonup )$ of Lemma 3.19, we can define $\cdot $ as in (3.1), and S as in (3.2), obtaining

(3.18) $$ \begin{align} a\cdot b=a_{1}\bullet T(a_{3})\bullet b\bullet T(T(a_{2})),\qquad S(a)=a_{1}\bullet T(a_{3})\bullet T(a_{2}). \end{align} $$

Notice that (3.17) clearly holds in the commutative and cocommutative cases (where the latter was already considered, e.g., in Masuoka [Reference Masuoka32, Remark p. 302]). If $H^{\bullet }$ is commutative, then $(H,\bullet ,\bullet )$ is an example of a Yetter–Drinfeld brace. On the other hand, if $H^{\bullet }$ is cocommutative, then the previous construction reduces to the almost trivial Hopf brace $(H,\bullet ^{\mathrm {op}},\bullet )$ (the terminology is adapted from [Reference Guarnieri and Vendramin18]).

In Section 5, we shall see other explicit examples coming from coquasitriangular structures. In particular, the first example will be the Sweedler’s Hopf algebra $H_4$ . Generalizations of $H_4$ are given by the Taft algebras $H_{n^{2}}$ , which do not admit any (co)quasitriangular structure for $n>2$ (see Gelaki [Reference Gelaki17]). Matched pairs $(H_{n^{2}},H_{n^{2}})$ on $H_{n^{2}}$ are classified by Agore [Reference Agore1, Theorem 2.1], and it is not difficult to see that none of these matched pairs satisfy (⋆). This explicitly shows that matched pairs of actions $(H,\rightharpoonup ,\leftharpoonup )$ correspond to a proper subclass of the matched pairs $(H,H)$ on H.

3.3 A converse connection

In order to show that our definition of Yetter–Drinfeld braces is optimal, we now prove that the correspondence of Theorem 3.10 has a converse. Therefore, a Yetter–Drinfeld brace and a matched pair of actions are essentially the same thing. This is in fact, as we shall point out, an isomorphism of categories.

Theorem 3.21 Let $(H,\cdot , \bullet , 1, \Delta , \epsilon , S, T)$ be a Yetter–Drinfeld brace. Define

$$\begin{align*}a\leftharpoonup b:=T(a_{1}\rightharpoonup b_{1})\bullet a_{2}\bullet b_{2}. \end{align*}$$

Then, $(H,\rightharpoonup ,\leftharpoonup )$ is a matched pair of actions.

Proof Since H is in $\mathrm {Comon}({}^{H^\bullet }_{H^\bullet }\mathcal {YD})$ , one has that $\Delta $ and $\epsilon $ are left $H^\bullet $ -linear with respect to $\rightharpoonup $ , and hence H is a left $H^\bullet $ -module coalgebra. From the definition of $\rightharpoonup $ we immediately have $a\bullet b = a_1\cdot S(a_2)\cdot (a_3\bullet b) = a_1\cdot (a_2\rightharpoonup b)$ , whence also

$$\begin{align*}a\cdot b= a_1\cdot (a_2\rightharpoonup(T(a_3)\rightharpoonup b))=a_1\bullet (T(a_2)\rightharpoonup b). \end{align*}$$

We therefore compute

$$ \begin{align*} a\leftharpoonup(b\bullet c)&=T(a_{1}\rightharpoonup(b_{1}\bullet c_{1}))\bullet a_{2}\bullet(b_{2}\bullet c_{2})\\ \nonumber &=T(S(a_{1})\cdot(a_{2}\bullet b_{1}\bullet c_{1}))\bullet a_{3}\bullet b_{2}\bullet c_{2}\\ \nonumber &=T(S(a_{1})\cdot(a_{2}\bullet b_{1})\cdot S(a_{3}\bullet b_{2})\cdot(a_{4}\bullet b_{3}\bullet c_{1}))\bullet a_{5}\bullet b_{4}\bullet c_{2}\\ \nonumber &=T\big((a_{1}\rightharpoonup b_{1})\cdot((a_{2}\bullet b_{2})\rightharpoonup c_{1})\big)\bullet a_{3}\bullet b_{3}\bullet c_{2}\\ \nonumber &=T\Big((a_{1}\rightharpoonup b_{1})\bullet\big(T(a_{2}\rightharpoonup b_{2})\rightharpoonup((a_{3}\bullet b_{3})\rightharpoonup c_{1})\big)\Big)\bullet a_{4}\bullet b_{4}\bullet c_{2}\\ \nonumber &= T\Big(\big(T(a_{2}\rightharpoonup b_{2})\bullet a_{3}\bullet b_{3}\big)\rightharpoonup c_{1}\Big)\bullet T(a_{1}\rightharpoonup b_{1})\bullet a_{4}\bullet b_{4}\bullet c_{2}\\ \nonumber &=(T(a_{1}\rightharpoonup b_{1})\bullet a_{2}\bullet b_{2})\leftharpoonup c\\ \nonumber &=(a\leftharpoonup b)\leftharpoonup c. \end{align*} $$

Moreover, we have $a\leftharpoonup 1=T(a_{1}\rightharpoonup 1)\bullet a_{2}\bullet 1=T(\epsilon (a_{1})1)\bullet a_{2}=a$ , thus $\leftharpoonup $ is a right $H^{\bullet }$ -action. By Remark 3.16, we already know that $\Delta $ is right $H^{\bullet }$ -linear with respect to $\leftharpoonup $ , i.e., $\Delta (a\leftharpoonup b) = (a_1\leftharpoonup b_1)\otimes (a_2\leftharpoonup b_2)$ ; that H is a right $H^\bullet $ -module coalgebra, and that (⋆) is satisfied. We already know that (mp.1) is satisfied. Moreover, (mp.2) also holds:

(3.19) $$ \begin{align}1\leftharpoonup a=T(1\rightharpoonup a_{1})\bullet1\bullet a_{2}=T(a_{1})\bullet a_{2}=\epsilon(a)1. \end{align} $$

In order to show that $(H,\rightharpoonup ,\leftharpoonup )$ is a matched pair of actions on H, it remains to prove that (mp.3) and (mp.4) are satisfied. We thereby compute

$$ \begin{align*} (a_1\rightharpoonup b_1)\bullet \Big( \big( a_2\leftharpoonup b_2\big)\rightharpoonup c\Big)&= (a_1\rightharpoonup b_1)\bullet \Big( \big( T(a_2\rightharpoonup b_2)\bullet a_3\bullet b_3\big)\rightharpoonup c\Big)\\ \nonumber &= (a_1\rightharpoonup b_1)\bullet \Big(T(a_2\rightharpoonup b_2)\rightharpoonup \big(a_3\bullet b_3\rightharpoonup c\big)\Big)\\ \nonumber &= (a_1\rightharpoonup b_1)\cdot \big(a_2\bullet b_2\rightharpoonup c\big)\\ \nonumber &= (a_1\rightharpoonup b_1)\cdot (a_2\rightharpoonup(b_2\rightharpoonup c))\\ \nonumber &\kern-3pt\overset{(\dagger)}{=} a\rightharpoonup (b_1\cdot (b_2\rightharpoonup c))\\ \nonumber &= a\rightharpoonup (b\bullet c), \end{align*} $$

where $(\dagger )$ follows since $(H,\cdot ,1)$ is a left $H^{\bullet }$ -module algebra. Finally, one has

$$ \begin{align*} (a\leftharpoonup(b_1\rightharpoonup c_1))\bullet (b_2\leftharpoonup c_2)&= T(a_1\rightharpoonup (b_1\rightharpoonup c_1))\bullet a_2\bullet (b_2\rightharpoonup c_2)\bullet (b_3\leftharpoonup c_3)\\ \nonumber &\kern-3pt\overset{({\star})}{=} T(a_1\rightharpoonup (b_1\rightharpoonup c_1))\bullet a_2\bullet b_2\bullet c_2\\ \nonumber &= T(a_1\bullet b_1\rightharpoonup c_1)\bullet a_2\bullet b_2\bullet c_2\\ \nonumber &= (a\bullet b)\leftharpoonup c.\\[-35pt] \end{align*} $$

The previous construction is converse to the one of Theorem 3.10.

Proof Since the constructions of Theorems 3.10 and 3.21 are converse to each other, if we prove that these constructions are functorial then they automatically yield an isomorphism of categories.

Let be the assignment defined in Theorem 3.10. We define the assignment on morphisms as follows: if $f\colon (H,\rightharpoonup _H,\leftharpoonup _H)\to (K,\rightharpoonup _K,\leftharpoonup _K)$ is a morphism in , the corresponding morphism in is $f\colon H\to K$ read as a morphism of Hopf algebras. Since $m_{\cdot _{H}} = m_{\bullet _{H}}(\mathrm {Id}_{H}\otimes \rightharpoonup _{H})(\mathrm {Id}_{H}\otimes T_{H}\otimes \mathrm {Id}_{H})(\Delta _{H}\otimes \mathrm {Id}_{H})$ (and a similar formula holds for $m_{\cdot _{K}}$ ), one has

$$ \begin{align*} m_{\cdot_{K}}(f\otimes f) &=m_{\bullet_K}(\mathrm{Id}_{K}\otimes\rightharpoonup_{K})(\mathrm{Id}_{K}\otimes T_{K}\otimes\mathrm{Id}_{K})(\Delta_{K}\otimes\mathrm{Id}_{K})(f\otimes f)\\&=m_{\bullet_K}(\mathrm{Id}_{K}\otimes\rightharpoonup_{K})(\mathrm{Id}_{K}\otimes T_{K}\otimes\mathrm{Id}_{K})(f\otimes f\otimes f)(\Delta_{H}\otimes\mathrm{Id}_{H})\\&=m_{\bullet_K}(\mathrm{Id}_{K}\otimes\rightharpoonup_{K})(f\otimes f\otimes f)(\mathrm{Id}_{H}\otimes T_{H}\otimes\mathrm{Id}_{H})(\Delta_{H}\otimes\mathrm{Id}_{H})\\&=m_{\bullet_K}(f\otimes f)(\mathrm{Id}_{H}\otimes\rightharpoonup_{H})(\mathrm{Id}_{H}\otimes T_{H}\otimes\mathrm{Id}_{H})(\Delta_{H}\otimes\mathrm{Id}_{H})\\&=f m_{\bullet_H}(\mathrm{Id}_{H}\otimes\rightharpoonup_{H})(\mathrm{Id}_{H}\otimes T_{H}\otimes\mathrm{Id}_{H})(\Delta_{H}\otimes\mathrm{Id}_{H})\\&= f m_{\cdot_{H}} , \end{align*} $$

thus f is indeed a morphism in . Moreover, it is clear that F sends the identity morphism into the identity morphism. Finally, let $f\colon (H,\rightharpoonup _H,\leftharpoonup _H)\to (K,\rightharpoonup _K,\leftharpoonup _K)$ and $g\colon (K,\rightharpoonup _K,\leftharpoonup _K)\to (L,\rightharpoonup _L,\leftharpoonup _L) $ be morphisms in , then F sends the composition $gf$ in into the set-theoretic composition of maps $gf$ , which is the composition of morphisms in .

Analogously, one proves that the assignment defined in Theorem 3.21 is a functor.

5.1 Coquasitriangular bialgebras

The notion of quasitriangular bialgebra was introduced by Drinfeld [Reference Drinfeld11], to relax the hypothesis of cocommutativity. Here we shall operate with the dual notion, which appeared in several works in between 1990 and 1993, such as in Schauenburg [Reference Schauenburg36], Doi [Reference Doi10], Larson and Towber [Reference Larson and Towber23], and Majid [Reference Majid27].

Definition 5.1 A coquasitriangular bialgebra, denoted by $(H,m,u,\Delta ,\epsilon , \mathcal {R})$ or simply by $(H,\mathcal {R})$ , is the datum of a bialgebra $(H,m,u,\Delta ,\epsilon )$ and of a convolution-invertible morphism , satisfying:

(coqt.1) $$ \begin{align} &\mathcal{R}(a_1\otimes b_{1})a_2 b_2 = b_1 a_1 \mathcal{R}(a_2\otimes b_{2}); \end{align} $$

(coqt.2) $$ \begin{align} &\mathcal{R}(a\otimes bc)=\mathcal{R}(a_1\otimes c)\mathcal{R}(a_2\otimes b); \end{align} $$

(coqt.3) $$ \begin{align} &\mathcal{R}(ab\otimes c)=\mathcal{R}(a\otimes c_1)\mathcal{R}(b\otimes c_2). \end{align} $$

Moreover, $(H,\mathcal {R})$ is called cotriangular if $\mathcal {R}^{-1}=\mathcal {R}^{\mathrm {op}}$ , where $\mathcal {R}^{-1}$ denotes the convolution inverse of $\mathcal {R}$ .

If $\mathcal {R}$ provides a coquasitriangular structure, then the convolution inverse $\mathcal {R}^{-1}$ satisfies:

(coqt.1′) $$\begin{align} &a_1b_1\mathcal{R}^{-1}(a_2\otimes b_2) = \mathcal{R}^{-1}(a_1\otimes b_1)b_2 a_2; \end{align} $$

(coqt.2′) $$\begin{align} &\mathcal{R}^{-1}(a\otimes bc)=\mathcal{R}^{-1}(a_{1}\otimes b)\mathcal{R}^{-1}(a_{2}\otimes c); \end{align} $$

(coqt.3′) $$\begin{align} &\mathcal{R}^{-1}(ab\otimes c)=\mathcal{R}^{-1}(b\otimes c_{1})\mathcal{R}^{-1}(a\otimes c_{2}). \end{align} $$

Let $\mathcal {R}$ be a coquasitriangular structure on a bialgebra H. Then, it is known that $\mathcal {R}$ and its convolution inverse $\mathcal {R}^{-1}$ satisfy the quantum YBE (see [Reference Majid30, Lemma 2.2.3]):

(ybe) $$ \begin{align} & \mathcal{R}^{\pm 1}(a_1\otimes b_1)\mathcal{R}^{\pm 1}(a_2\otimes c_1)\mathcal{R}^{\pm 1}(b_2\otimes c_2)\\ &\hspace{2em} = \mathcal{R}^{\pm 1}(b_1\otimes c_1)\mathcal{R}^{\pm 1}(a_1\otimes c_2)\mathcal{R}^{\pm 1}(a_2\otimes b_2).\nonumber \end{align} $$

Moreover, if H is a Hopf algebra with antipode S, then the following relations hold true:

$$\begin{align*}\mathcal{R}(S(a)\otimes b)=\mathcal{R}^{-1}(a\otimes b),\quad \mathcal{R}^{-1}(a\otimes S(b))=\mathcal{R}(a\otimes b),\end{align*}$$

$$\begin{align*}\mathcal{R}^{\pm1}(S(a)\otimes S(b))=\mathcal{R}(a\otimes b), \end{align*}$$

see [Reference Majid30, Lemma 2.2.2]. It is well known that a bialgebra H is coquasitriangular if and only if the category ${\mathfrak {M}}^{H}$ of right H-comodules is braided (see [Reference Majid30, Exercise 9.2.9] and the considerations following [Reference Majid30, Theorem 9.2.4]). Indeed, given a coquasitriangular structure $\mathcal {R}$ , the bialgebra H becomes a braided object with the braiding operator

$$\begin{align*}\sigma^{\mathcal{R}}_{H,H}(a\otimes b):= b_1\otimes a_1 \mathcal{R}(a_2\otimes b_2), \end{align*}$$

and a braiding on ${\mathfrak {M}}^{H}$ is defined for all $X,Y$ in ${\mathfrak {M}}^{H}$ by setting

(5.1) $$ \begin{align} \sigma^{\mathcal{R}}_{X,Y}:X\otimes Y\to Y\otimes X,\quad x\otimes y\mapsto y_{0}\otimes x_{0}\mathcal{R}(x_{1}\otimes y_{1}). \end{align} $$

Moreover, H is cotriangular if and only if the category ${\mathfrak {M}}^{H}$ is symmetric, i.e., $(\sigma ^{\mathcal {R}}_{X,Y})^{-1}=\sigma ^{\mathcal {R}}_{Y,X}$ for all $X,Y$ in ${\mathfrak {M}}^{H}$ .

In the rest of this section, however, we shall use another braiding operator on H. Let $(H,\mathcal {R})$ be a coquasitriangular bialgebra, then the map $\sigma _{H,H}\colon H\otimes H\to H\otimes H$ defined by

(5.2) $$ \begin{align} \sigma_{H,H}:H\otimes H\to H\otimes H,\ a\otimes b\mapsto\mathcal{R}^{-1}(a_{1}\otimes b_{1})b_{2}\otimes a_{2}\mathcal{R}(a_{3}\otimes b_{3}), \end{align} $$

is known to be a braiding operator on H, with inverse given by

$$\begin{align*}\sigma^{-1}_{H,H}:H\otimes H\to H\otimes H,\ a\otimes b\mapsto\mathcal{R}(b_{1}\otimes a_{1})b_{2}\otimes a_{2}\mathcal{R}^{-1}(b_{3}\otimes a_{3}). \end{align*}$$

Moreover, $m_{H}\sigma _{H,H}= m_{H}$ holds.

Remark 5.1 The braiding operator (5.2) extends to a braiding

$$\begin{align*}x\otimes y\mapsto \mathcal{R}^{-1}(x_{-1}\otimes y_{-1}) y_0 \otimes x_0 \mathcal{R}(x_1\otimes y_1)\end{align*}$$

on the category of bicomodules $^H{\mathfrak {M}}^H$ . However, the braidings on are not yet classified, and the problem is seemingly hard. When the category of bimodules ${}_H{\mathfrak {M}}_H$ is endowed with the tensor product $\otimes _H$ instead of , however, a classification was found by Agore, Caenepeel, and Militaru [Reference Agore, Caenepeel and Militaru2], which could be dualized for the category $^{H}{\mathfrak {M}}^{H}$ .

Remark 5.2 Let $\sigma _{H,H}$ be the braiding operator defined in (5.2). Notice that

$$\begin{align*}\sigma_{H,H}^{2}=\mathcal{R}^{-1}*(\mathcal{R}^{-1})^{\mathrm{op}}*\mathrm{Id}_{H\otimes H}*\mathcal{R}^{\mathrm{op}}*\mathcal{R}=(\mathcal{R}^{\mathrm{op}}*\mathcal{R})^{-1}*\mathrm{Id}_{H\otimes H}*(\mathcal{R}^{\mathrm{op}}*\mathcal{R}), \end{align*}$$

hence we have that $\sigma _{H,H}^{2}=\mathrm {Id}_{H\otimes H}$ if and only if $\mathrm {Id}_{H\otimes H}*(\mathcal {R}^{\mathrm {op}}*\mathcal {R})=(\mathcal {R}^{\mathrm {op}}*\mathcal {R})*\mathrm {Id}_{H\otimes H}$ . The element $\mathcal {R}^{\mathrm {op}} *\mathcal {R}$ is also known as the quantum Killing form. In particular, if $\mathcal {R}^{\mathrm {op}}=\mathcal {R}^{-1}$ , one clearly has $\sigma ^{-1}_{H,H}=\sigma _{H,H}$ .

5.2 Coquasitriangular Hopf algebras and matched pairs of actions

Suppose given a coquasitriangular Hopf algebra $(H,\bullet , 1,\Delta ,\epsilon , T,\mathcal {R})$ . We shall see that the braiding operator (5.2) yields a matched pair satisfying (⋆), and hence a Yetter–Drinfeld brace. The product and the antipode of the corresponding Hopf algebra structure in $^H_H\mathcal {YD}$ coincide with those obtained by transmutation of H, introduced by Majid [Reference Majid28]; thus providing a Yetter–Drinfeld brace-theoretic interpretation of the theory of transmutation. At the end of this section, we shall also observe that the same construction cannot be carried out with the braiding (5.1).

Let $\sigma _{H,H}$ be defined as in (5.2), and define

(5.3) $$ \begin{align} &a\rightharpoonup b=(\mathrm{Id}\otimes \epsilon)\sigma_{H,H}(a\otimes b) = \mathcal{R}^{-1}(a_{1}\otimes b_{1})b_{2}\mathcal{R}(a_{2}\otimes b_{3}); \end{align} $$

(5.4) $$ \begin{align} &a\leftharpoonup b=(\epsilon\otimes \mathrm{Id})\sigma_{H,H}(a\otimes b) = \mathcal{R}^{-1}(a_{1}\otimes b_{1})a_{2}\mathcal{R}(a_{3}\otimes b_{2}). \end{align} $$

These are a left and a right action respectively, by Lemma 2.6, and H becomes a left and right module coalgebra with the respective actions. Moreover, these actions provide a matched pair $(H,H)$ , see [Reference Majid30, Example 7.2.7]. Observe that the braiding (5.2) is retrieved as

$$\begin{align*}\sigma_{H,H}(a\otimes b) = (a_1\rightharpoonup b_1)\otimes (a_2\leftharpoonup b_2).\end{align*}$$

Hence the fact that $(H,\rightharpoonup ,\leftharpoonup )$ satisfies (mp.1)–(mp.4) can also be obtained using Lemmas 2.2 and 2.3, while (⋆) is immediate.

Remark 5.3 The bialgebra $H \bowtie H$ described in Section 2, when the matched pair is derived as in (5.3) and (5.4), is denoted in Majid [Reference Majid30] by $H\bowtie _{\mathcal {R}}H$ . This has product given by

$$\begin{align*}(a\otimes h)(b\otimes g)=\mathcal{R}^{-1}(h_{1}\otimes b_{1})a\bullet b_{2}\otimes h_{2}\bullet g\mathcal{R}(h_{3}\otimes b_{3}).\end{align*}$$

Moreover, we have

$$ \begin{align*} a\leftharpoonup b=\mathcal{R}^{-1}(a_{1}\otimes b_{1})a_{2}\mathcal{R}(a_{3}\otimes b_{2})&=\mathcal{R}(T(a_{1})\otimes b_{1})a_{2}\mathcal{R}(a_{3}\otimes b_{2})\\&=a_{2}\mathcal{R}(T(a_{1})\bullet a_{3}\otimes b), \end{align*} $$

so this situation falls under the hypotheses of [Reference Majid30, Lemma 7.4.8] where, in our case, the coaction is given by $\mathrm {Ad}_{R}\colon a\mapsto a_{2}\otimes T(a_{1})\bullet a_{3}$ . Thus [Reference Majid30, Lemma 7.4.8] ensures us that H is also a right H-module with respect to $\leftharpoonup $ . Hence, using $\leftharpoonup $ and $\mathrm {Ad}_{R}$ one can construct the bialgebra (we borrow the notation from [Reference Majid30]), where $\underline {H}$ is the (right) transmutation of H [Reference Majid30, Theorem 7.4.1] (see also [Reference Majid30, Example 9.4.10]), and the symbol denotes the bialgebra with smash product algebra structure, and smash coproduct coalgebra structure. So is exactly the bosonization $H\#\underline {H}$ , built using the right action and the right coaction. By [Reference Majid30, Theorem 7.4.10], one also obtains an isomorphism . In Remark 5.6, we shall relate with a left version $\overline {H}$ of Majid’s transmutation.

Notice that

$$ \begin{align*} a\rightharpoonup b=\mathcal{R}^{-1}(a_{1}\otimes b_{1})b_{2}\mathcal{R}(a_{2}\otimes b_{3})&=\mathcal{R}^{-1}(a_{1}\otimes b_{1})b_{2}\mathcal{R}^{-1}(a_{2}\otimes T(b_{3}))\\&=b_{2}\mathcal{R}^{-1}(a\otimes b_{1}\bullet T(b_{3})). \end{align*} $$

In this case, (3.1) takes the following form:

(5.5) $$ \begin{align} a\cdot b = a_1\bullet b_2\mathcal{R}^{-1}(T(a_2)\otimes (b_1\bullet T(b_3))), \end{align} $$

and also

(5.6) $$ \begin{align}a\cdot b &= \mathcal{R}^{-1}(T(a_{3})\otimes b_{1})a_{1}\bullet b_{2}\mathcal{R}^{-1}(T(a_{2})\otimes T(b_{3}))\nonumber\\&=\mathcal{R}^{-1}(T(a_{3})\otimes b_{1})a_{1}\bullet b_{2}\mathcal{R}^{-1}(a_{2}\otimes b_{3})\nonumber\\&\kern-12pt\overset{{\rm (coqt.1')}}{=} \mathcal{R}^{-1}(T(a_3)\otimes b_1)\mathcal{R}^{-1}(a_1\otimes b_2)b_3\bullet a_2\nonumber\\ &=\mathcal{R}^{-1}((a_1\bullet T(a_3))\otimes b_1)b_2\bullet a_2, \end{align} $$

while (3.2) has the following form:

(5.7) $$ \begin{align} S(a)&=a_{1}\rightharpoonup T(a_{2})=T(a_{3})\mathcal{R}^{-1}(a_{1}\otimes T(T(a_{2})\bullet a_{4}))\nonumber\\ &=T(a_{3})\mathcal{R}(a_{1}\otimes T(a_{2})\bullet a_{4})\nonumber\\ &=T(a_{4})\mathcal{R}(a_{1}\otimes a_{5})\mathcal{R}(a_{2}\otimes T(a_{3})). \end{align} $$

Proof We already know that $(H,\rightharpoonup ,\leftharpoonup )$ satisfies (mp.1)–(mp.4) [Reference Majid30, Example 7.2.7]. Moreover, it follows from (coqt.1) that H is braided-commutative, i.e., $m_\bullet \sigma _{H,H} = m_\bullet $ , which translates into condition (⋆). Corollary 3.15 yields that $(H,\cdot ,\bullet ,1,\Delta ,\epsilon ,S,T)$ is a Yetter–Drinfeld brace.

Suppose now that $\mathcal {R}$ is cotriangular. The braided commutativity of $\cdot $ with respect to the braiding of $^{H}_{H}\mathcal {YD}$ is expressed by the condition $a\cdot b=(a_{1}\bullet T(a_{3})\rightharpoonup b)\cdot a_{2}$ , which in view of (5.6) is rewritten as

$$\begin{align*}b_3\bullet a_2 \mathcal{R}^{-1}(T(a_3)\otimes b_1)\mathcal{R}^{-1}(a_1\otimes b_2) = b_2\cdot a_3\mathcal{R}^{-1}((a_1\bullet T(a_5))\otimes b_1)\mathcal{R}((a_2\bullet T(a_4))\otimes b_3).\end{align*}$$

Suppose $\mathcal {R}^{-1} = \mathcal {R}^{\mathrm {op}}$ . Then

$$ \begin{align*} &\hspace{-2em}b_2 \cdot a_3\mathcal{R}^{-1}((a_1\bullet T(a_5))\otimes b_1)\mathcal{R}((a_2\bullet T(a_4))\otimes b_3)\\ &= b_3\bullet a_4 \mathcal{R}^{-1}(T(a_7)\otimes b_1)\mathcal{R}^{-1}(a_1\otimes b_2)\mathcal{R}^{-1}(T(b_5)\otimes a_3)\\ &\hspace{1em}\;\mathcal{R}^{-1}(b_4\otimes a_5)\mathcal{R}(a_2\otimes b_6)\mathcal{R}(T(a_6)\otimes b_7)\\&= b_3\bullet a_4 \mathcal{R}^{-1}(T(a_7)\otimes b_1) \mathcal{R}^{-1}(a_1\otimes b_2)\mathcal{R}(a_3\otimes T(b_5))\\ &\hspace{1em}\;\mathcal{R}(a_5\otimes b_4)\mathcal{R}(a_2\otimes b_6)\mathcal{R}(T(a_6)\otimes b_7)\\&= b_3\bullet a_2 \mathcal{R}^{-1}(T(a_5)\otimes b_1)\mathcal{R}^{-1}(a_1\otimes b_2) \mathcal{R}(a_3\otimes b_4)\mathcal{R}(T(a_4)\otimes b_5)\\ &= b_3\bullet a_2 \mathcal{R}^{-1}(T(a_3)\otimes b_1)\mathcal{R}^{-1}(a_1\otimes b_2), \end{align*} $$

as desired. Given a Hopf algebra $(H,\cdot , 1, \Delta , \epsilon , S)$ in a braided monoidal category $\mathcal {M}$ , it is known that, if $\cdot $ is braided-commutative in $\mathcal {M}$ , then S is involutive.

The Hopf algebra SL _q (2) with $q=-1$ will be an example of a coquasitriangular Hopf algebra with involutive antipode S, which is not cotriangular.

Remark 5.6 Let us show how the Hopf algebra $(H,\cdot ,1,\Delta ,\epsilon ,S)$ is related to the transmutation of the Hopf algebra $(H,\bullet ,1,\Delta ,\epsilon ,T)$ . For more details about transmutation theory we redirect the reader to Majid [Reference Majid28, Reference Majid30]. Let $(H,\bullet ,1,\Delta ,\epsilon ,T,\mathcal {R})$ be a coquasitriangular Hopf algebra, so that $\mathcal {R}^{-1}$ induces a braiding on the category $^{H}\mathfrak {M}$ . It is possible to define a second product operation $\overline {\bullet }$ on H, and a second antipode $\overline {T}$ , such that they provide, on H, a Hopf algebra structure $\overline {H}$ in $^{H}\mathfrak {M}$ (with same comultiplication $\Delta $ and counit $\epsilon $ ), where the comodule structure is given by $\mathrm {Ad}_L$ . This is called the transmutation of H (see dual of [Reference Majid30, Example 9.4.10]). The transmuted product and the transmuted antipode are, respectively,

$$ \begin{align*}&a\,\overline{\bullet}\, b:= \mathcal{R}^{-1}(T(a_2)\otimes(b_1\bullet T(b_3))) a_1\bullet b_2,\\ &\overline{T}(a):= \mathcal{R}^{-1}(a_1\otimes T(T(a_2)\bullet a_4)) T(a_3).\end{align*} $$

By (5.5) we have $a\,\overline {\bullet }\, b=a\cdot b$ and by (5.7) we have $\overline {T}(a)=S(a)$ , so $(H,\cdot ,1,\Delta ,\epsilon ,S)$ coincides with the transmutation of $(H,\bullet ,1,\Delta ,\epsilon ,T)$ . Hence the transmutation of coquasitriangular Hopf algebras gives us examples of Yetter–Drinfeld braces. Theorem 3.10 implies, in particular, that $\overline {H}$ lies in $\mathrm {Hopf}({}^{H^{\bullet }}_{H^{\bullet }}\mathcal {YD})$ , with action given by $\rightharpoonup $ . Observe from duals of [Reference Zhu41, Theorem 4.8 and Corollary 4.9] (with $f=\mathrm {Id}_{H^{\bullet }}$ ) that there is a braided monoidal equivalence between and ${}^{H^{\bullet }}_{H^{\bullet }}\mathcal {YD}$ , hence one can also recover an object in .

Remark 5.7 We know from Counterexample 2.7 that the following two actions do not form a matched pair:

$$\begin{align*}a\rightharpoonup b=\mathcal{R}(a\otimes b_{2})b_{1},\qquad a\leftharpoonup b=a_{1}\mathcal{R}(a_{2}\otimes b). \end{align*}$$

We may define $\cdot $ as in (3.1):

$$\begin{align*}a\cdot b :=a_{1}\bullet(T(a_{2})\rightharpoonup b)=a_{1}\bullet \mathcal{R}(T(a_{2})\otimes b_{2})b_{1}=a_{1}\bullet b_{1}\mathcal{R}^{-1}(a_{2}\otimes b_{2}). \end{align*}$$

With this product, $(H,\cdot , 1)$ is an algebra. But now observe that, since $\epsilon (a\cdot b)=\mathcal {R}^{-1}(a\otimes b)$ , $\epsilon $ is not a morphism of algebras with respect to $\cdot $ , unless $\mathcal {R}^{-1}=\epsilon \otimes \epsilon $ and hence $\mathcal {R}=\epsilon \otimes \epsilon $ : in this case, by (coqt.1), H is commutative.

In the case of cocommutative Hopf braces, it is known (see [Reference Angiono, Galindo and Vendramin5, Corollary 2.5]) that the additional operation (3.1) is commutative if and only if the braiding operator ${\sigma (a\otimes b):= (a_1\rightharpoonup b_1)\otimes (a_2\leftharpoonup b_2)}$ is involutive. An analogous result for Yetter–Drinfed braces may possibly hold, substituting commutativity with braided commutativity in $^H_H\mathcal {YD}$ . However, we attempted mimicking the proof of [Reference Angiono, Galindo and Vendramin5, Theorem 2.3] in several straightforward ways, and all of them failed; thus leaving the following problem open:.

6.1 The Sweedler’s Hopf algebra

Let us suppose . The Sweedler’s Hopf algebra $H_4 = E(1)$ is the free -algebra generated by two elements g and x modulo the relations

$$\begin{align*}g^{2}=1,\ x^{2}=0,\ \text{and}\ xg=-gx, \end{align*}$$

where $1$ denotes the unit of $H_4$ . It becomes a Hopf algebra with comultiplication, counit and antipode given by $\Delta (g)=g\otimes g$ , $\Delta (x)=x\otimes 1+g\otimes x$ , $\epsilon (g)=1$ , ${\epsilon (x)=0}$ , ${S(g)=g}$ and $S(x)=xg$ , respectively. In particular, $H_4$ is neither commutative nor cocommutative. We recall that $H_4$ has a one-parameter family of cotriangular structures given by

$$ \begin{align*} \mathcal{R}\begin{pmatrix} 1\otimes1 & 1\otimes g &1\otimes x & 1\otimes xg\\ g\otimes1 & g\otimes g & g\otimes x & g\otimes xg\\ x\otimes1 & x\otimes g & x\otimes x & x\otimes xg\\ xg\otimes1 & xg\otimes g & xg\otimes x & xg\otimes xg \end{pmatrix}=\begin{pmatrix} 1 & 1 & 0 & 0\\ 1 & -1 & 0 & 0\\ 0 & 0 & k & -k\\ 0 & 0 & k & k \end{pmatrix}, \end{align*} $$

where k is a parameter in (see [Reference Majid30, Example 2.2.6]). The convolution inverse of $\mathcal {R}$ is $\mathcal {R}^{-1}=\mathcal {R}^{\mathrm {op}}$ given by the transpose matrix

$$ \begin{align*} \mathcal{R}^{-1}\begin{pmatrix} 1\otimes1 & 1\otimes g &1\otimes x & 1\otimes xg\\ g\otimes1 & g\otimes g & g\otimes x & g\otimes xg\\ x\otimes1 & x\otimes g & x\otimes x & x\otimes xg\\ xg\otimes1 & xg\otimes g & xg\otimes x & xg\otimes xg \end{pmatrix}=\begin{pmatrix} 1 & 1 & 0 & 0\\ 1 & -1 & 0 & 0\\ 0 & 0 & k & k\\ 0 & 0 & -k & k \end{pmatrix}. \end{align*} $$

Given $\alpha :=a_{1}1+a_{2}g+a_{3}x+a_{4}xg$ in $H_4$ , one has

$$ \begin{align*} (\Delta\otimes\mathrm{Id})\Delta(\alpha)&=a_{1}1\otimes1\otimes1+a_{2}g\otimes g\otimes g+a_{3}x\otimes1\otimes1+a_{3}g\otimes x\otimes1\\&\hspace{1em}\;+a_{3}g\otimes g\otimes x+a_{4}xg\otimes g\otimes g+a_{4}1\otimes xg\otimes g+a_{4}1\otimes1\otimes xg. \end{align*} $$

Hence, given $\beta :=b_{1}1+b_{2}g+b_{3}x+b_{4}xg$ in $H_4$ , one has

$$ \begin{align*} \alpha\rightharpoonup\beta&=(a_{1}b_{1}+a_{2}b_{1}+k(a_{3}b_{3}-a_{3}b_{4}-a_{4}b_{3}+a_{4}b_{4}))1\\&\hspace{1em}\;+(a_{1}b_{2}+a_{2}b_{2}+k(a_{3}b_{4}-a_{3}b_{3}-a_{4}b_{4}+a_{4}b_{3}))g\\&\hspace{1em}\;+(a_{1}b_{3}-a_{2}b_{3})x+(a_{1}b_{4}-a_{2}b_{4})xg. \end{align*} $$

Analogously, for $\leftharpoonup $ one has:

$$ \begin{align*}\alpha\leftharpoonup \beta &= (a_1b_1+a_1b_2+a_3b_3 k+a_3b_4 k+a_4 b_3 k+a_4b_4 k)1 \\ &\hspace{1em}\;+(a_2b_1+a_2b_2-a_3b_3 k-a_3b_4 k-a_4b_3k -a_4b_4 k)g\\&\hspace{1em}\; +(a_3b_1-a_3b_2)x+(a_4b_1-a_4b_2)xg.\end{align*} $$

Table 1:

Left and right actions for $H_4$ .

Table 1: Long description

The table is split into two primary sections: Left action and Right action. Each section has a header row and four data rows. The columns for both sections are labeled with the elements 1, g, x, and x g.

Left Action section:

* Row 1 (element 1): The results are 1, g, x, x g.

* Row 2 (element g): The results are 1, g, negative x, minus x g.

* Row 3 (element x): The results are 0, 0, k 1 minus k g, k g minus k 1.

* Row 4 (element x g): The results are 0, 0, k g minus k 1, k 1 minus k g.

Right Action section:

* Row 1 (element 1): The results are 1, 1, 0, 0.

* Row 2 (element g): The results are g, g, 0, 0.

* Row 3 (element x): The results are x, negative x, k 1 minus k g, k 1 minus k g.

* Row 4 (element x g): The results are x g, minus x g, k 1 minus k g, k 1 minus k g.

Table 2:

Multiplication table for $\,\overline {\cdot }\,$ in $H_4$ .

Table 2: Long description

The table defines the operation dot with an overline.

Header Row and Column: The elements are 1, g, x, and x g.

Row 1 (Element 1):

- 1 times 1 is 1.

- 1 times g is g.

- 1 times x is x.

- 1 times x g is x g.

Row 2 (Element g):

- g times 1 is g.

- g times g is 1.

- g times x is x g.

- g times x g is x.

Row 3 (Element x):

- x times 1 is x.

- x times g is x g.

- x times x is k 1 minus k g.

- x times x g is k g minus k 1.

Row 4 (Element x g):

- x g times 1 is x g.

- x g times g is x.

- x g times x is k g minus k 1.

- x g times x g is k 1 minus k g.

The two actions are summarized in Table 1. We denote by $\,\overline {\cdot }\,$ the transmuted multiplicative structure on $H_4$ , defined according to (5.6). This is explicitly given by

$$ \begin{align*} \alpha\,\overline{\cdot}\, \beta&= \alpha_1 (S(\alpha_2)\rightharpoonup\beta)\\ &= a_1(S(1)\rightharpoonup \beta)+a_2g(S(g)\rightharpoonup \beta)+a_3 x(S(1)\rightharpoonup \beta)\\&\hspace{1em}\;+a_3 g(S(x)\rightharpoonup \beta)+a_4 xg(S(g)\rightharpoonup \beta)+a_4(S(xg)\rightharpoonup \beta)\\ &= a_1\beta +a_2 g (b_11+b_2 g -b_3 x-b_4 xg)+a_3 x\beta +a_3 g((kb_4-kb_3)1\\&\hspace{1em}\;+(kb_3-kb_4)g)+a_4 xg(b_11+b_2 g -b_3 x-b_4 xg)\\&\hspace{1em}\;+a_4((kb_4-kb_3)1+(kb_3-kb_4)g)\\ &=(a_1b_1+a_2b_2+ka_3 b_3-ka_3b_4+ka_4b_4-ka_4 b_3)1\\&\hspace{1em}\; +(a_1b_2+a_2b_1+ka_3b_4-ka_3 b_3+ka_4b_3-ka_4b_4)g\\&\hspace{1em}\; +(a_1b_3+a_2b_4+a_3b_1+a_4b_2)x\\&\hspace{1em}\;+(a_1b_4+a_2b_3+a_3b_2+a_4b_1)xg. \end{align*} $$

The multiplication $\,\overline {\cdot }\,$ is described in Table 2. The algebra $(H_4, \,\overline {\cdot }\, ,1)$ is generated by $g,x$ modulo the relations $g \,\overline {\cdot }\, g =1$ , $x\,\overline {\cdot }\, x = k1-kg$ , $x\,\overline {\cdot }\, g= g\,\overline {\cdot }\, x$ . The new antipode $\overline {S}$ is defined by $\overline {S}(\alpha ):=\alpha _{1}\rightharpoonup S(\alpha _{2})$ as in (5.7), whence $\overline {S}(1)=1$ , $\overline {S}(g)=g$ , $\overline {S}(x)= -xg=-x\overline {\cdot }g$ , and $\overline {S}(xg) = \overline {S}(x\,\overline {\cdot }\, g)=\overline {S}(g)\,\overline {\cdot }\, \overline {S}(x) = -x$ . Notice that $\,\overline {\cdot }\,$ is braided-commutative and $\overline {S}$ is involutive, as we know from Theorem 5.5. Remark 5.2 implies that the braiding $\sigma _{H_4,H_4}\colon \alpha \otimes \beta \mapsto (\alpha _1\rightharpoonup \beta _1)\otimes (\alpha _2\leftharpoonup \beta _2)$ is involutive.

Observe that $\,\overline {\cdot }\,$ is also commutative, even if $\sigma ^{\mathcal {YD}}_{H_4,H_4}$ does not coincide with the flip map $\tau $ (for instance, it is easy to see that $\sigma ^{\mathcal {YD}}_{H_4,H_4}$ and $\tau $ differ on $x\otimes x$ ). Notice moreover that $\,\overline {\cdot }\,$ is compatible with $\Delta $ in the braided sense, but not in the classical sense: indeed $(x\,\overline {\cdot }\, x)_1\otimes (x\,\overline {\cdot }\, x)_2 = (k1-kg)\otimes (k1-kg) \neq (k1-kg)\otimes 1+2xg\otimes g +1\otimes (k1-kg) = (x_1\,\overline {\cdot }\, x_1)\otimes (x_2\,\overline {\cdot }\, x_2)$ . In particular, this Yetter–Drinfeld brace is not a Hopf brace.

We have, as a byproduct of Remark 6.1, the following result.

6.2 The Hopf algebras E(n)

Let us suppose . A generalization of the Sweedler’s Hopf algebra is provided by the Hopf algebras $E(n)$ generated as algebras by $g, x_i$ for $i = 1,\ldots , n$ , with relations

$$\begin{align*}g^2 = 1,\;\; x_i^2 = 0,\;\; x_ig = -g x_i,\;\; x_ix_j = -x_j x_i,\;\;\text{ for all }i=1,\ldots, n,\end{align*}$$

endowed with the comultiplication $\Delta (1) = 1\otimes 1$ , $\Delta (g) = g\otimes g$ , $\Delta (x_i) = x_i\otimes 1+ g \otimes x_i$ , the counit $\epsilon (g) = 1$ , $\epsilon (x_i) =0$ , and the antipode $S(g)=g$ , $S(x_i) = x_i g$ . The quasitriangular structures on $E(n)$ are classified by Panaite and Van Oystaeyen [Reference Panaite and Van Oystaeyen34], and the coquasitriangular structures are obtained by duality (see e.g., Carnovale and Cuadra [Reference Carnovale and Cuadra9, p. 256]):

$$ \begin{align*} &\mathcal{R}_A\begin{pmatrix} 1\otimes 1& 1\otimes g & 1\otimes x_j & 1\otimes x_j g \\ g\otimes 1& g\otimes g & g\otimes x_j & g\otimes x_j g \\ x_i\otimes 1& x_i\otimes g & x_i\otimes x_j & x_i\otimes x_j g \\ x_ig\otimes 1& x_ig\otimes g & x_ig\otimes x_j & x_ig\otimes x_j g \end{pmatrix}= \begin{pmatrix}1 & 1 & 0 & 0 \\ 1 & -1 & 0 & 0 \\ 0 & 0 & A_{ij}& -A_{ij}\\ 0 & 0 & A_{ij}& A_{ij} \end{pmatrix}, \end{align*} $$

where $A=(A_{ij})_{ij}$ is an $n\times n$ matrix with entries in . It is easy to see that $\mathcal {R}_A$ is cotriangular if and only if A is symmetric. Panaite and Van Oystaeyen [Reference Panaite and Van Oystaeyen34] conjectured that $R_A$ is triangular if and only if A is symmetric; which was later proven by Carnovale and Cuadra [Reference Carnovale and Cuadra9, Proposition 2.1].

Assume $A_{ij} = A_{ji}$ , thus $\mathcal {R}^{-1} = \mathcal {R}^{\mathrm {op}}$ . We define a second multiplicative structure $\,\overline {\cdot }\,$ as in (5.6). This is described in Table 3 (in particular, it is commutative). In terms of generators and relation, $(E(n), \,\overline {\cdot }\,,1)$ is the algebra generated by $g, x_i$ with relations $g \,\overline {\cdot }\, g = 1$ , $x_i\,\overline {\cdot }\, x_j + x_j\,\overline {\cdot }\, x_i = 2A_{ij}(1-g)$ , $x_i\,\overline {\cdot }\, g = g\,\overline {\cdot }\, x_i$ . The action $\rightharpoonup $ is described in Table 4. The antipode $\overline {S}$ , defined as in (5.7), acts as follows: $\overline {S}(1) = 1$ , $\overline {S}(g)=g$ , $\overline {S}(x_i) = -x_i g = -x_i\,\overline {\cdot }\, g$ .

Table 3:

Multiplication table for $\,\overline {\cdot }\,$ in $E(n)$ .

Table 3: Long description

The table consists of a header row and a header column defining the operands, with four rows and four columns of data.

Header Row (Top): 1, g, x sub j, x sub j g.

Header Column (Left): 1, g, x sub i, x sub i g.

Row 1 (Multiplication by 1):

- 1 times 1 yields 1

- 1 times g yields g

- 1 times x sub j yields x sub j

- 1 times x sub j g yields x sub j g

Row 2 (Multiplication by g):

- g times 1 yields g

- g times g yields 1

- g times x sub j yields x sub j g

- g times x sub j g yields x sub j

Row 3 (Multiplication by x sub i):

- x sub i times 1 yields x sub i

- x sub i times g yields x sub i g

- x sub i times x sub j yields A sub i j 1 minus A sub i j g plus x sub i x sub j

- x sub i times x sub j g yields A sub i j g minus A sub i j 1 plus x sub i x sub j g

Row 4 (Multiplication by x sub i g):

- x sub i g times 1 yields x sub i g

- x sub i g times g yields x sub i

- x sub i g times x sub j yields A sub i j g minus A sub i j 1 plus x sub i x sub j g

- x sub i g times x sub j g yields A sub i j 1 minus A sub i j g plus x sub i x sub j

Table 4:

Table for the action $\rightharpoonup $ in $E(n)$ .

Table 4: Long description

The table consists of five columns and five rows including the header.

Header Row:

- Column 1: right harpoon with barb up operator.

- Column 2: 1.

- Column 3: g.

- Column 4: x sub j.

- Column 5: x sub j g.

Row 2:

- Column 1: 1.

- Column 2: 1.

- Column 3: g.

- Column 4: x sub j.

- Column 5: x sub j g.

Row 3:

- Column 1: g.

- Column 2: 1.

- Column 3: g.

- Column 4: minus x sub j.

- Column 5: minus x sub j g.

Row 4:

- Column 1: x sub i.

- Column 2: 0.

- Column 3: 0.

- Column 4: A sub i j 1 minus A sub i j g.

- Column 5: A sub i j g minus A sub i j 1.

Row 5:

- Column 1: x sub i g.

- Column 2: 0.

- Column 3: 0.

- Column 4: A sub i j g minus A sub i j 1.

- Column 5: A sub i j 1 minus A sub i j g.

6.3 The Hopf algebra SL _q (2)

Let , and consider, following Kassel [Reference Kassel21, Section IV.6], the -algebra $\mathrm {SL}_q(2)$ generated by $a,b,c,d$ modulo the following relations:

$$ \begin{align*} &ba = qa b, && ca = qac, && db = qbd,\\ & dc = qcd,&& bc = cb,&& ad-da = (q^{-1}-q)bc,\\ & && da- qbc = 1. && \end{align*} $$

This is known to be a Hopf algebra with

$$\begin{align*}\Delta\begin{pmatrix} a &b\\ c & d \end{pmatrix} = \begin{pmatrix} a &b\\ c & d \end{pmatrix}\otimes \begin{pmatrix} a &b\\ c & d \end{pmatrix},\quad \epsilon \begin{pmatrix} a &b\\ c & d \end{pmatrix} = \begin{pmatrix} 1&0\\ 0&1 \end{pmatrix}, \quad S\begin{pmatrix} a &b\\ c & d \end{pmatrix} = \begin{pmatrix} d & -qb\\ -q^{-1}c & a \end{pmatrix}, \end{align*}$$

and, when $q\neq 1$ , it is neither commutative nor cocommutative. The Hopf algebra $\mathrm {SL}_q(2)$ is coquasitriangular [Reference Kassel21, Corollary VIII.7.2], with

$$ \begin{align*} \mathcal{R}&\begin{pmatrix} a\otimes a & b \otimes b &a \otimes b& b \otimes a\\ c\otimes c & d\otimes d& c\otimes d & d\otimes c\\ a\otimes c & b\otimes d& a\otimes d& b\otimes c\\ c\otimes a & d\otimes b & c\otimes b & d\otimes a \end{pmatrix}=q^{-1/2}\begin{pmatrix} q & 0 & 0 & 0\\ 0 & q & 0 & 0\\ 0 & 0 & 1 & q-q^{-1}\\ 0 & 0 & 0 & 1 \end{pmatrix},\\ \mathcal{R}^{-1}&\begin{pmatrix} a\otimes a & b \otimes b &a \otimes b& b \otimes a\\ c\otimes c & d\otimes d& c\otimes d & d\otimes c\\ a\otimes c & b\otimes d& a\otimes d& b\otimes c\\ c\otimes a & d\otimes b & c\otimes b & d\otimes a \end{pmatrix}=q^{1/2}\begin{pmatrix} q^{-1} & 0 & 0 & 0\\ 0 & q^{-1} & 0 & 0\\ 0 & 0 & 1 & q^{-1}-q\\ 0 & 0 & 0 & 1 \end{pmatrix}. \end{align*} $$

Notice that $\mathcal {R}^{-1}=\mathcal {R}^{\mathrm {op}}$ if and only if $q^{1/2}=q^{-1/2}$ , if and only if $q=1$ . In this case $\mathcal {R}=\mathcal {R}^{-1}=\mathcal {R}^{\mathrm {op}}$ , and the previous matrices are the identities. The explicit form of the iterated $\Delta $ is here reported:

$$ \begin{align*} (\Delta\otimes \mathrm{Id})\Delta \, (a) &= a\otimes a \otimes a +a \otimes b\otimes c + b\otimes c\otimes a+ b\otimes d\otimes c;\\ (\Delta\otimes \mathrm{Id})\Delta \, ( b) &= a\otimes a\otimes b + a\otimes b\otimes d + b\otimes c\otimes b + b\otimes d\otimes d;\\ (\Delta\otimes \mathrm{Id})\Delta \, ( c) &= c\otimes a\otimes a + c\otimes b\otimes c+ d\otimes c\otimes a+ d\otimes d\otimes c;\\ (\Delta\otimes \mathrm{Id})\Delta \, ( d) &= c\otimes a\otimes b + c\otimes b\otimes d + d\otimes c\otimes b + d\otimes d\otimes d. \end{align*} $$

We define an additional operation $\,\overline {\cdot }\,$ on $\mathrm {SL}_q(2)$ as in (5.6). Using the iterated $\Delta $ , we compute Table 5 for the multiplication $\,\overline {\cdot }\,$ , and Table 6 for the left action $\rightharpoonup $ . Hence we have that $\,\overline {\cdot }\,$ is commutative if and only if $q=q^{-1}$ , i.e., if and only if $q=\pm 1$ . In particular, if $q=-1$ , $\,\overline {\cdot }\,$ is commutative but $\mathcal {R}$ is not cotriangular.

Table 5:

Multiplication table for $\,\overline {\cdot }\,$ in $\mathrm {SL}_q(2)$ .

Table 5: Long description

The table consists of 5 columns and 5 rows. The top-left corner cell contains the operator dot-bar. The header row (top) and header column (left) both list the elements a, b, c, and d in that order.

Row 1 (element a):

* a times a yields a-squared.

* a times b yields q a b.

* a times c yields q super negative 1 a c.

* a times d yields a d.

Row 2 (element b):

* b times a yields q super negative 1 a b.

* b times b yields q super negative 1 b-squared.

* b times c yields (q-squared minus 1) a-squared plus (1 minus q-squared) a d plus q b c.

* b times d yields (q-cubed minus q) a b plus b d.

Row 3 (element c):

* c times a yields q a c.

* c times b yields q b c.

* c times c yields q super negative 1 c-squared.

* c times d yields c d.

Row 4 (element d):

* d times a yields a d.

* d times b yields b d.

* d times c yields (q-cubed minus q) a c plus c d.

* d times d yields d-squared.

Table 6:

Table for the action $\rightharpoonup $ in $\mathrm {SL}_q(2)$ .

Table 6: Long description

The table consists of a header row and four data rows.

Header Row: The top-left cell contains a right harpoon with barb up symbol. The following columns are headed by the variables a, b, c, and d.

Row 1: Starts with variable a. The results of the action are: a, q super negative 1 b, q c, and d.

Row 2: Starts with variable b. The results of the action are: (1 minus q super negative 2) b, 0, (q minus q super negative 1) d plus (q super negative 1 minus q) a, and (1 minus q squared) b.

Row 3: Starts with variable c. All four result cells in this row contain the value 0.

Row 4: Starts with variable d. The results of the action are: a, q b, q super negative 1 c, and d.

The algebra $(H,\,\overline {\cdot }\,,1)$ is generated by $a,b,c,d$ modulo the relations

$$ \begin{align*}&c\,\overline{\cdot}\, a = q^2 a\,\overline{\cdot}\, c, && b\,\overline{\cdot}\, d - d\,\overline{\cdot}\, b=(q^2-1)a\,\overline{\cdot}\, b, \\ & a\,\overline{\cdot}\, b = q^2 b\,\overline{\cdot}\, a,&& b\,\overline{\cdot}\, c - c\,\overline{\cdot}\, b = (q^2-1)a \,\overline{\cdot}\, a - (q^2-1)a\,\overline{\cdot}\, d, \\ & a\,\overline{\cdot}\, d = d\,\overline{\cdot}\, a,&& d\,\overline{\cdot}\, c - c\,\overline{\cdot}\, d = (q^2-1) c\,\overline{\cdot}\, a, &\end{align*} $$

$$\begin{align*}a\,\bar{\cdot}\, d - q^{-2} c\,\bar{\cdot} \,b= 1. \end{align*}$$

We now define the transmuted antipode $\overline {S}(x):=x_{1}\rightharpoonup S(x_{2})$ , as in (5.7). One has

$$ \begin{align*} &\overline{S}( a ) = q^{-2} d + (1-q^{-2}) a , & \overline{S}( b ) = -q^{-2} b ,\\ & \overline{S}( c )= -q^{-2} c , & \overline{S}( d ) = a. \end{align*} $$

Notice that $\overline {S}$ is involutive if and only if $q = \pm 1$ . Hence, if $q=-1$ , $\overline {S}$ is involutive but $\mathcal {R}$ is not cotriangular.

6.4 An example in the class of Suzuki Hopf algebras

Let be an algebraically closed field of characteristic , as in Suzuki [Reference Suzuki38]. The Suzuki Hopf algebra $A^{\nu ,\lambda }_{1,2}$ , with parameters , is generated by $a,b,c,d$ subject to the relations

$$ \begin{align*}& a^{2}=d^{2},\; b^{2}=c^{2}, \; a^{2}+\nu b^{2}=1,\; cb=\lambda bc,\; ad=da,\\ & ab=ba=ac=ca=bd=db=cd=dc=0. \end{align*} $$

Let

$$\begin{align*}\mathbf{t}= \begin{pmatrix}a& b\\ c& d\end{pmatrix}, \end{align*}$$

with $\mathbf {t}_{ij}$ denoting the entry on the ith row and jth column. The comultiplication, counit and antipode are respectively given by:

$$\begin{align*}\Delta \mathbf{t} = \mathbf{t}\otimes \mathbf{t},\;\; \epsilon (\mathbf{t}_{ij}) = \delta_{ij},\;\; S (\mathbf{t}_{ij}) = (\mathbf{t}_{ji})^{3}.\end{align*}$$

The map

$$\begin{align*}\mathcal{R}_{\alpha, \beta}\begin{pmatrix} a\otimes a & a\otimes b &a\otimes c& a\otimes d\\ b\otimes a&b\otimes b&b\otimes c &b\otimes d\\ c\otimes a&c\otimes b&c\otimes c&c\otimes d\\ d\otimes a& d\otimes b&d\otimes c&d\otimes d \end{pmatrix} = \begin{pmatrix} 0&0&0&0\\ 0&\alpha &\beta&0\\ 0&\beta&\alpha&0\\ 0&0&0&0 \end{pmatrix}\end{align*}$$

defines a coquasitriangular structure whose convolution inverse is $\mathcal {R}^{-1}_{(\alpha ,\beta )} = \mathcal {R}_{(\alpha ^{-1},\beta ^{-1})}$ (see [Reference Suzuki38, proof of Proposition 2.1]), which is cotriangular if $(\alpha ,\beta ) = (-1,-1)$ or $(\alpha ,\beta )=(1,1)$ , and $(\nu ,\lambda ) = (1,1)$ (see [Reference Suzuki38, Proposition 3.10]).

We define an algebra operation $\,\overline {\cdot }\,$ as in (5.6), whose multiplication table is reported in Table 7. In particular, $\,\overline {\cdot }\,$ is commutative. The algebra $(H,\,\overline {\cdot }\,,1)$ is generated by $a,b,c,d$ modulo the relations

$$ \begin{align*} &a\,\overline{\cdot}\, a = d \,\overline{\cdot}\, d, \quad c \,\overline{\cdot}\, c = \alpha\beta^{-1}\lambda b \,\overline{\cdot}\, b,\quad c\,\overline{\cdot}\, b = b\,\overline{\cdot}\, c,\quad a\,\overline{\cdot}\, d = d\,\overline{\cdot}\, a,\\ &a\,\overline{\cdot}\, b=b\,\overline{\cdot}\, a=a\,\overline{\cdot}\, c=c\,\overline{\cdot}\, a=b\,\overline{\cdot}\, d=d\,\overline{\cdot}\, b=c\,\overline{\cdot}\, d=d\,\overline{\cdot}\, c=0.\end{align*} $$

The action $\rightharpoonup $ is reported in Table 8. The corresponding antipode $\overline {S}$ , defined as in (5.7), is identified by $\overline {S} (a) = \alpha ^2\beta ^2 d^3$ , $\overline {S}(b) = \alpha ^3\beta b^3$ , $\overline {S}(c) = \alpha ^3\beta c^3$ , $\overline {S}(d) = \alpha ^2\beta ^2 a^3$ .

Table 7:

Multiplication table for $\,\overline {\cdot }\,$ in $A_{1,2}^{\nu ,\lambda }$ .

Table 7: Long description

The table defines the multiplication of elements a, b, c, and d.

Header Row: The columns are labeled a, b, c, and d.

Row 1 (a):

- a times a is alpha super negative 2 beta super negative 2 a d.

- a times b is 0.

- a times c is 0.

- a times d is alpha super negative 2 beta super negative 2 a-squared.

Row 2 (b):

- b times a is 0.

- b times b is alpha super negative 2 beta super negative 2 b c.

- b times c is alpha super negative 3 beta super negative 1 b-squared.

- b times d is 0.

Row 3 (c):

- c times a is 0.

- c times b is alpha super negative 3 beta super negative 1 b-squared.

- c times c is alpha super negative 1 beta super negative 3 lambda b c.

- c times d is 0.

Row 4 (d):

- d times a is alpha super negative 2 beta super negative 2 a-squared.

- d times b is 0.

- d times c is 0.

- d times d is alpha super negative 2 beta super negative 2 a d.

Table 8:

Table for the action $\rightharpoonup $ in $A^{\nu ,\lambda }_{1,2}$ .

Table 8: Long description

The table consists of five columns and five rows.

Header Row: The first cell contains a right harpoon with barb up symbol. The following headers are a, b, c, and d.

Row 1: The row header is a. The corresponding values are d, alpha super negative 1 beta c, alpha beta super negative 1 b, and a.

Row 2: The row header is b. All corresponding values in this row are 0.

Row 3: The row header is c. All corresponding values in this row are 0.

Row 4: The row header is d. The corresponding values are d, alpha beta super negative 1 c, alpha super negative 1 beta b, and a.

We know by Theorem 5.5 that the additional operation (3.1) is braided commutative if $\mathcal {R}$ is cotriangular. In all the previous examples, when $\mathcal {R}$ is cotriangular (3.1) is also commutative. We could neither prove nor disprove that this is true in general, and thus the following problem is open.

Notice that the converse is false (see the example $SL_q(2)$ with $q = -1$ ).