2.1. Hopf cocycles

A Hopf 2-cocycle $\sigma\,:\, H\otimes H \to \Bbbk$ is a convolution-invertible linear map $x\otimes y\mapsto\sigma(x,y)$ satisfying

(2.1) \begin{align}\sigma\!\left(x_{(1)}, y_{(1)}\right) \sigma\!\left(x_{(2)} y_{(2)}, z\right) &=\sigma\!\left(y_{(1)}, z_{(1)}\right) \sigma\!\left(x, y_{(2)}z_{(2)}\right), \quad x,y,z\in H.\end{align}

It follows that $\sigma(1,1)\neq 0$ and $\sigma(x, 1) = \sigma(1, x) = \epsilon(x)\sigma(1,1)$ , $x\in H$ , so by

(2.2) \begin{align}\sigma\leftrightarrow \sigma(1,1)^{-1}\sigma\end{align}

we can always assume that $\sigma(x, 1) = \sigma(1, x) = \epsilon(x)$ . Such cocycles are called normalized. We shall denote by

\begin{equation*}Z^2(H)=\left\{\sigma\colon H\otimes H\to \Bbbk \ | \ \text{(2.1) holds}\right\}\end{equation*}

the set of normalized Hopf 2-cocycles in H.

We will use a coordinate-free version of (2.1), namely

(2.3) \begin{align}(\sigma\otimes\epsilon)\ast \sigma(m\otimes\textrm{id})&=(\epsilon\otimes\sigma)\ast \sigma(\textrm{id}\otimes\, m).\end{align}

This allows us to consider a braided version of Hopf cocycles for any bialgebra in a braided category; see [Reference Andruskiewitsch and García Iglesias1] for details.

2.1.1. Cocycle deformations

Let $\sigma\in Z^2(H)$ . Then $\cdot_{\sigma}\,:\,H\otimes H\rightarrow H$ , given by

(2.4) \begin{align}x\cdot_{\sigma}y &= \sigma\!\left(x_{(1)}, y_{(1)}\right) x_{(2)} y_{(2)}\sigma^{-1}\!\left(x_{(3)}, y_{(3)}\right), \quad x, y \in H,\end{align}

defines an associative product on the vector space H. Moreover, the collection $(H,\cdot_\sigma, \Delta)$ is a Hopf algebra with antipode $\mathcal{S}_{\sigma} = f*\mathcal{S}*f^{-1}$ , for $f =\sigma\circ(\textrm{id}\otimes\,\mathcal{S})\circ\Delta$ . This Hopf algebra is denoted $H_\sigma$ and referred to as a cocycle deformation of H.

If $\sigma\in Z^2(H)$ , then there is another deformation of the multiplication in H. Namely, $\cdot_{(\sigma)}\,:\,H\otimes H\rightarrow H$ , given by

(2.5) \begin{align}x\cdot_{(\sigma)}y &= \sigma\!\left(x_{(1)}, y_{(1)}\right) x_{(2)} y_{(2)}, \quad x, y, \in H,\end{align}

also defines an associative product on the vector space H; we denote this new algebra by $H_{(\sigma)}$ . In this case, the comultiplication $H\to H\otimes H$ defines coactions $H_{(\sigma)}\to H_{(\sigma)}\otimes H$ and $H_{(\sigma)}\to H_{\sigma}\otimes H_{(\sigma)}$ in such a way that $C=H_{(\sigma)}$ becomes a $(H_{\sigma},H)$ -bicleft object.

Moreover, any (right) cleft object C for H (and hence any cocycle deformation $H_{\sigma}$ ) arises in this way. If C is a cleft object with a normalized section $\gamma\,:\,H\to C$ , then $C\simeq H_{(\sigma)}$ ; for the cocycle $\sigma\colon H\otimes H\to \Bbbk$ determined by the formula

(2.6) \begin{align}\sigma(x,y)=\gamma\!\left(x_{(1)}\right)\!\gamma\!\left(y_{(1)}\right)\gamma^{-1}\!\left(x_{(2)}y_{(2)}\right), \quad x,y\in H.\end{align}

We stress that formula (2.6) extends to cleft objects in braided categories, see [Reference Andruskiewitsch and García Iglesias1].

Furthermore, it is possible to recover $H_{\sigma}$ completely in terms of the cleft object C, as the so-called Schauenburg left Hopf algebra L(C, H); in particular, no cocycles are explicitly involved. See [Reference Schauenburg24] for further details.

2.1.2. Cohomologous Hopf cocycles

The group $U(H^*)$ of convolution units in $H^\ast$ acts on $Z^2(H)$ via

(2.7) \begin{equation}(\alpha\rightharpoonup\sigma)(x,y)=\alpha\!\left(x_{(1)}\right)\alpha\!\left(y_{(1)}\right)\sigma\!\left(x_{(2)},y_{(2)}\right)\alpha^{-1}\!\left(x_{(3)}y_{(3)}\right).\end{equation}

We write $\sigma\sim_\alpha\sigma^{\prime}$ when $\sigma^{\prime}=\alpha\rightharpoonup\sigma$ and set

\begin{equation*}H^2(H)\,:\!=\, Z^2(H)/U(H^\ast)=\left\{[\sigma]\,:\,\sigma\in Z^2(H)\right\}.\end{equation*}

Moreover, if $\sigma^{\prime}=\alpha\rightharpoonup\sigma$ , then $H_\sigma\simeq H_{\sigma^{\prime}}$ as Hopf algebras. However, the converse is not true, see Remark 2.1 below.

The equivalence does hold for H-cleft objects, namely $(C,\gamma)\simeq (C^{\prime},\gamma^{\prime})$ if and only if $[\sigma]=[\sigma^{\prime}]$ , for $\sigma,\sigma^{\prime}$ the Hopf 2-cocycles associated to $\gamma,\gamma^{\prime}$ , respectively, as in (2.6), so that $C\simeq H_{(\sigma)}$ and $C^{\prime}\simeq H_{(\sigma^{\prime})}$ , see [Reference Doi10, Theorem 2.2]. That is, there is an equivalence

\begin{equation*}\textrm{Cleft}(H)\longleftrightarrow H^2(H).\end{equation*}

We may always assume that such $\alpha$ as in (2.7) is such that $\alpha(1)=1$ . Indeed, $c=\alpha(1)\neq 0$ as $\alpha$ is convolution invertible and

\begin{equation*}c^{-1}\alpha\rightharpoonup \sigma=c^{-1}(\alpha\rightharpoonup \sigma)\sim_{c^{-1}\epsilon} \alpha\rightharpoonup \sigma.\end{equation*}

Remark 2.1. Two cocycle deformations $H_{\sigma}$ and $H_{\sigma^{\prime}}$ may be isomorphic for two non-equivalent cocycles $\sigma,\sigma^{\prime}\in Z^2(H)$ .

We briefly develop an example. Let $H=\Bbbk\big\langle x,g\,:\,x^2=0, g^2=1, gx=-xg\big\rangle$ be the Taft algebra and let $\sigma_\lambda$ , $\lambda\in\Bbbk$ , be the cocycle determined by $\sigma_\lambda(x,x)=\lambda$ and

\begin{equation*} \sigma_\lambda\!\left(x^a g^i, x^bg^j\right)=({-}1)^{bi}\sigma_\lambda\!\left(x^a,x^b\right), \end{equation*}

for $a,b,i,j\in\{0,1\}$ . It follows that $H_{\sigma_{\lambda}}\simeq H$ for every $\lambda\in\Bbbk$ . If $\lambda\neq 0$ , then $\sigma_{\lambda}\not\sim \sigma_1$ , but $\sigma_{1}\not\sim \sigma_0=\epsilon$ . Indeed, they determine the (non-isomorphic) cleft objects $C_\lambda=\Bbbk\big\langle y,g\,:\,y^2=\lambda, g^2=1, gy=-yg\big\rangle$ , $\lambda=0,1$ .

2.2. Yetter–Drinfeld modules and Nichols algebras

We recall that the category ${}^{H}_{H}\mathcal{YD}$ of Yetter–Drinfeld modules over H is the category of all simultaneously left modules and left comodules M such that the action $\cdot$ and the coaction $\rho$ are subject to

\begin{equation*}\rho(h\cdot m)=h_{(1)}m_{({-}1)}\mathcal{S}\!\left(h_{(3)}\right)\otimes h_{(2)}\cdot m_{(0)}, \qquad m\in M, h\in H.\end{equation*}

When the antipode $\mathcal{S}$ of H is bijective this is a braided category, with braiding $c_{M,N}\colon M\otimes N\to N\otimes M$ , $m\otimes n\mapsto m_{({-}1)}\cdot n\otimes m_{(0)}$ ; $M,N\in{}^{H}_{H}\mathcal{YD}$ , $m\in M$ , $n\in N$ .

A braided vector space is a pair (V, c), where V is a vector space together with an invertible linear map $c\colon V\otimes V\to V\otimes V$ that satisfies the braid equation $(c\otimes\textrm{id})(\textrm{id}\otimes\, c)(c\otimes\textrm{id})=(\textrm{id}\otimes\, c)(c\otimes\textrm{id})(\textrm{id}\otimes \,c)$ . In particular, any $M\in{}^{H}_{H}\mathcal{YD}$ is a braided vector space with $c=c_{M,M}$ . Conversely, a realization of (V, c) over H is a structure of Yetter–Drinfeld module on V so that $c=c_{V,V}\in{}^{H}_{H}\mathcal{YD}$ .

If $B\in{}^{H}_{H}\mathcal{YD}$ is an algebra, then one may use the braiding to define an algebra structure on $B\otimes B\in{}^{H}_{H}\mathcal{YD}$ . A bialgebra, resp. Hopf algebra, in ${}^{H}_{H}\mathcal{YD}$ is defined under this convention. In this case, the bosonization $B\# H$ becomes a (Hopf) algebra. We recall that in this case the multiplication is defined via

\begin{equation*}(b\otimes h)(c\otimes k)=b \!\left(h_{(1)}\cdot c\right)\otimes h_{(2)}k, \qquad b,c\in B, \ h,k\in H.\end{equation*}

We write bh for an element $b\#h$ of $B\# H$ . Notice that this is coherent with the multiplication just defined as

\begin{equation*}B\simeq B\otimes 1_H\subset B\# H \qquad \text{ and } \qquad H\simeq 1_B\otimes H\subset B\otimes H\end{equation*}

as algebras and $(b\otimes 1)(1\otimes h)=b\otimes h$ . As well, notice that, for $h\in H$ , $b \in B$ :

(2.8) \begin{align}hb&=h_{(1)}\cdot b h_{(2)}, & bh&=h_{(2)}\!\left(\mathcal{S}^{-1}\!\left(h_{(1)}\right)\cdot b\right), & h_{(1)}b\mathcal{S}\!\left(h_{(2)}\right)&=h\cdot b.\end{align}

In turn, the comultiplication is given by the rule

(2.9) \begin{align}\Delta(x\#h)=x_{(1)}\#x_{(2)({-}1)}h_{(1)}\otimes x_{(2)(0)}\# h_{(2)}.\end{align}

See [Reference Andruskiewitsch and Schneider2] for details.

A Nichols algebra is a connected graded Hopf algebra $B=\oplus_{n\geq 0} B_n$ in ${}^{H}_{H}\mathcal{YD}$ generated by $V=B_1$ and such that $\textrm{P}(B)=V$ . We say that such a Hopf algebra is pre-Nichols when this last condition is not satisfied, so that $V\subsetneq \textrm{P}(V)$ . We denote such Nichols algebra by $\mathfrak{B}(V)$ , as it is completely determined by the braided vector space (V, c), independently of the realization in ${}^{H}_{H}\mathcal{YD}$ .

A finite-dimensional braided vector space is of diagonal type when there is a basis $\left\{x_1,\dots,x_\theta\right\}$ of V and a matrix $\mathfrak{q}=(q_{ij})_{i,j\in\mathbb{I}_\theta}$ so that the braiding $c=c^{\mathfrak{q}}$ is given by $c\!\left(x_i\otimes x_j\right)=q_{ij}x_j\otimes x_i$ , $i,j\in \mathbb{I}_\theta$ . In this setting $\mathfrak{B}(V)$ is also denoted $\mathfrak{B}_{\mathfrak{q}}$ .

A principal YD-realization of V over H is a family $(\chi_i,g_i)_{i\in \mathbb{I}_\theta}$ with $g_i\in G(H)$ and $\chi_i\colon H\to \Bbbk$ algebra maps so that

\begin{align*}\chi_j(g_i)=q_{ij} \quad \text{ and } \quad \chi_i(h)g=\chi_i\!\left(h_{(2)}\right)h_{(1)}g\mathcal{S}\!\left(h_{(3)}\right),\end{align*}

$i,j\in \mathbb{I}_\theta$ , $g\in G(H), h\in H$ . This determines a realization $V\in{}^{H}_{H}\mathcal{YD}$ via

(2.10) \begin{align}\rho(x_i)&=g_i\otimes x_i, & h\cdot x_i&=\chi_i(h)x_i, \qquad h\in H, i\in\mathbb{I}_\theta.\end{align}

For any connected $B=\bigoplus_{n\geq 0} B_n$ bialgebra and $b\in B^+\,:\!=\,\bigoplus_{n\geq 1} B_n$ , we shall consider the restricted comultiplication:

(2.11) \begin{align}\underline{\Delta}(b)\,:\!=\,\Delta(b)-b\otimes 1-1\otimes b.\end{align}

Also, we set $\underline{\Delta}(1)=1\otimes 1$ ; and write $\underline{\Delta}(b)=b_{\underline{{(1)}}}\otimes b_{\underline{{(2)}}}$ , $b\in B.$ We will use a similar notation $\underline{\rho}$ for coactions. A restricted version of coassociativity holds:

\begin{equation*} (\underline{\Delta}\otimes \textrm{id})\underline{\Delta}=(\textrm{id}\otimes \underline{\Delta})\underline{\Delta}. \end{equation*}

As usual, we write $b_{\underline{{(1)}}{\underline{{(1)}}}}\otimes b_{\underline{{(1)}}{\underline{{(2)}}}}\otimes b_{\underline{{(2)}}}=b_{\underline{{(1)}}}\otimes b_{\underline{{(2)}}{\underline{{(1)}}}}\otimes b_{\underline{{(2)}}{\underline{{(2)}}}}=b_{\underline{{(1)}}}\otimes b_{\underline{{(2)}}}\otimes b_{\underline{{(3)}}}$ .

3.1. Some remarks on $\alpha\rightharpoonup\sigma$

Let $\sigma\in Z^2_0(A)$ , $\alpha\in U(A^*)$ , so that $\sigma^{\prime}\,:\!=\, \alpha\rightharpoonup\sigma$ as in (2.7) is a Hopf cocycle cohomologous to $\sigma$ . Moreover, we assume that $\sigma^{\prime}\in Z^2_0(A)$ . Recall that we may assume $\alpha(1)=1$ .

Proof. The cocycle $\sigma_\alpha$ associated to $\gamma_\alpha$ is $\sigma^{\prime}$ , which satisfies $\sigma^{\prime}(h,k)=\epsilon(hk)$ . But

\begin{equation*}\epsilon(hk)=\sigma_\alpha(h,k)=\alpha\!\left(h_{(1)}\right)\alpha\!\left(k_{(1)}\right)\alpha^{-1}\!\left(h_{(2)}k_{(2)}\right),\end{equation*}

which shows that $\alpha(hk)=\alpha(h)\alpha(k)$ , by uniqueness of the convolution inverse.

In particular, $\alpha_{|B}\,:\,B\to \Bbbk$ is H-linear. That is,

\begin{equation*} \alpha=\alpha^{\prime}\#\varphi, \quad\text{for } \alpha^{\prime}\in\textrm{Hom}_H(B,\Bbbk) \text{ and } \varphi\in\textrm{Alg}(H,\Bbbk). \end{equation*}

Proof. Notice that

(3.4) \begin{align}\epsilon(xh)=\alpha\rightharpoonup\sigma(x,h)=\alpha\!\left(x_{(1)}\right)\alpha\!\left(h_{(1)}\right)\alpha^{-1}\!\left(x_{(2)}h_{(2)}\right).\end{align}

Thus $\psi(xh)=\alpha(x)\alpha(h)$ is a left convolution inverse to $\alpha^{-1}$ ; hence $\psi=\alpha$ . The other equality is similar, considering $\alpha\rightharpoonup\sigma(h,x)$ .

Now, $\alpha_{|B}(h\cdot x)=\alpha(h_{(1)}x \mathcal{S}(h_{2}))=\alpha\!\left(h_{(1)}\right)\alpha(x)\alpha(\mathcal{S}(h_{2}))=\epsilon(h)\alpha(x)$ .

Proof. Set $\varphi\,:\!=\,\alpha_{|H}$ ; if $\varphi\neq \epsilon$ , consider $\tau=\epsilon\#\varphi^{-1}$ and set $\tilde{\alpha}\,:\!=\,\alpha\ast\tau$ . On the one hand, $\tilde\alpha_{|H}=\epsilon$ , and $\tilde{\alpha}(xh)=\alpha(x)\epsilon(h)$ , $\tilde{\alpha}^{-1}(xh)=\alpha^{-1}(x)\epsilon(h)$ . Now, observe that

\begin{align*}\tilde\alpha\rightharpoonup\sigma(xh,yk)& =\tilde\alpha\!\left(x_{(1)}h_{(1)}\right)\tilde\alpha\!\left(y_{(1)}k_{(1)}\right)\sigma\!\left(x_{(2)}h_{(2)}, y_{(2)}k_{(2)}\right)\tilde\alpha^{-1}\!\left(x_{(3)}h_{(3)} y_{(3)}k_{(3)}\right)\\ &=\alpha\!\left(x_{(1)}\right)\alpha(y_{(1)})\sigma\!\left(x_{(2)}h_{(1)}, y_{(2)}\right)\tilde\alpha^{-1}\!\left(x_{(3)}h_{(2)}\cdot y_{(3)}\right)\epsilon(k)\\ &=\alpha\!\left(x_{(1)}\right)\alpha(y_{(1)})\sigma\!\left(x_{(2)}h_{(1)}, y_{(2)}\right)\alpha^{-1}\!\left(x_{(3)}h_{(2)}\cdot y_{(3)}\right)\epsilon(k)\\ &=\alpha\!\left(x_{(1)}\right)\alpha\!\left(h_{(1)}\cdot y_{(1)}\right)\sigma\!\left(x_{(2)}, h_{(2)}\cdot y_{(2)}\right)\alpha^{-1}\!\left(x_{(3)}h_{(3)}\cdot y_{(3)}\right)\epsilon(k)\\ &=\sigma^{\prime}(x,h\cdot y)\epsilon(k)=\sigma^{\prime}(xh,yk).\end{align*}

Hence, we can replace $\alpha$ by $\tilde{\alpha}$ .

Proof. (a) We have that $\alpha^{\prime}\rightharpoonup\sigma_B=\sigma^{\prime}_B$ , that is for $x,y\in B$ ,

\begin{equation*}\alpha^{\prime}\!\left(x_{(1)}\right)\alpha^{\prime}\!\left(x_{(3)({-}2)}x_{(2)({-}1)}\cdot y_{(1)}\right)\sigma_B\!\left(x_{(2)(0)},x_{(3)({-}1)}y_{(2)}\right)\alpha^{\prime}{}^{-1}\!\left(x_{(3)(0)}y_{(3)}\right)=\sigma^{\prime}_B(x,y).\end{equation*}

Here $x_{(1)}\otimes x_{(2)}=\Delta_B(x)$ stands for the coproduct in B. Recall that the coproduct $\Delta(xh)=x_{(1)}x_{(2)({-}1)}h_{(1)}\otimes x_{(2)(0)}h_{(2)}$ in A from (2.9) so

\begin{align*}\Delta^{(2)}(xh)&=x_{(1)}x_{(2)({-}1)}x_{(3)({-}2)}h_{(1)} \otimes x_{(2)(0)}x_{(3)({-}1)}h_{(2)}\otimes x_{(3)(0)}h_{(3)}.\end{align*}

Now, $\alpha\rightharpoonup \sigma(xh,yk)$ gives, using that that $\alpha^{\prime}$ is H-linear:

\begin{align*}&\alpha\!\left(x_{(1)} x_{(2)({-}1)}x_{(3)({-}2)} h_{(1)}\right)\alpha\!\left(y_{(1)}y_{(2)({-}1)}y_{(3)({-}2)} k_{(1)}\right)\\ & \quad \times\sigma\!\left(x_{(2)(0)}x_{(3)({-}1)}h_{(2)},y_{(2)(0)}y_{(3)({-}1)}k_{(2)}\right)\alpha^{-1}\!\left(x_{(3)(0)}h_{(3)}y_{(3)(0)}k_{(3)}\right)\\ &=\alpha^{\prime}\!\left(x_{(1)}\right)\alpha^{\prime}(y_{(1)})\sigma_B\!\left(x_{(2)},x_{(3)({-}1)}h_{(1)}\cdot y_{(2)}\right) \alpha^{\prime}{}^{-1}\!\left(x_{(3)(0)}h_{(2)}y_{(3)}\right)\epsilon(k)\\ &=\alpha^{\prime}\!\left(x_{(1)}\right)\alpha^{\prime}\!\left(x_{(3)({-}2)}x_{(2)({-}1)}h_{(1)}\cdot y_{(1)}\right)\\ & \quad \times\sigma_B\!\left(x_{(2)(0)},x_{(3)({-}1)}h_{(2)}y_{(2)}\right)\alpha^{\prime}{}^{-1}\!\left(x_{(3)(0)}h_{(3)}y_{(3)}\right)\epsilon(k)\\ &=\sigma^{\prime}_B(x,h\cdot y)\epsilon(k)=\sigma^{\prime}(xh,yk).\end{align*}

(b) Follows as (a), using that $\alpha^{\prime}$ is H-linear by Lemma 3.8 and the compatibility between $\Delta_B(x)$ and $\Delta(x\#1)$ , see (2.9).

Next lemma will also allow us to reduce computations when dealing with Hopf cocycles arising as exponentials of Hochschild cocycles.

Proof. (1) Let $x\in V$ and $b\in \mathbb{B}$ , then

\begin{align*} (\textrm{id}\otimes \Delta)\Delta(x\otimes b)=& \, x\otimes b_{(1)}\otimes 1\otimes b_{(2)}\otimes 1\otimes b_{(3)}\\&+1\otimes x_{({-}1)}\cdot b_{(1)}\otimes x_{(0)}\otimes b_{(2)}\otimes 1\otimes b_{(3)}\\&+1\otimes x_{({-}2)}\cdot b_{(1)}\otimes1\otimes x_{({-}1)}\cdot b_{(2)}\otimes x_{(0)}\otimes b_{(3)}. \end{align*}

When we act by $\alpha$ , we get

\begin{align*} \alpha\rightharpoonup\tau\!\left(x,b\right)=& \,\alpha\!\left(x\right)\alpha\!\left(b_{(1)}\right)\tau\!\left(1,b_{(2)}\right)\alpha^{-1}\!\left(b_{(3)}\right)+\alpha\!\left(x_{({-}1)}\cdot b_{(1)}\right)\tau\!\left(x_{(0)}, b_{(2)}\right)\alpha^{-1}\!\left(b_{(3)}\right)\\ &+\alpha\!\left(x_{\left({-}2\right)}\cdot b_{(1)}\right)\tau\!\left(1, x_{({-}1)}\cdot b_{(2)}\right)\alpha^{-1}\!\left(x_{(0)}b_{(3)}\right)\\ =& \,\alpha\!\left(x\right)\alpha\!\left(b_{(1)}\right)\epsilon\!\left(b_{(2)}\right)\alpha^{-1}\!\left(b_{(2)}\right)+\alpha\!\left(x_{({-}1)}\cdot b_{(1)}\right)\tau\!\left(x_{(0)}, b_{(2)}\right)\alpha^{-1}\!\left(b_{(3)}\right)\\ &+\alpha\!\left(x_{({-}1)}\cdot b_{(1)}\right)\alpha^{-1}\!\left(x_{(0)}b_{(2)}\right)\\ =&\,\alpha\!\left(x_{({-}1)}\cdot b_{(1)}\right)\tau\!\left(x_{(0)}, b_{(2)}\right)\alpha^{-1}\!\left(b_{(3)}\right)+\alpha\!\left(x_{({-}1)}\cdot b_{(1)}\right)\alpha^{-1}\!\left(x_{(0)}b_{(2)}\right). \end{align*}

(2) Use the formula in (1) to compute $\alpha\rightharpoonup \tau(x,y)=\sigma(x,y)$ , then

\begin{equation*}\alpha(x_{({-}1)}\cdot y)\alpha^{-1}\!\left(x_{(0)}\right)+\alpha^{-1}(xy)=\sigma(x,y)-\tau(x,y).\end{equation*}

Now, as

(3.6) \begin{align}\alpha^{-1}(xy)=\alpha(x)\alpha(y)-\alpha\!\left(x_{({-}1)}\cdot y\right)\alpha^{-1}\!\left(x_{(0)}\right)-\alpha(xy),\end{align}

we obtain (3.5).

4.1. Hochschild 2-cocycles with coefficients in $\Bbbk$

Let $B^+=\bigoplus_{n > 0} B_n$ . We begin with an alternative description of the set $\textrm{Z}^2(B,\Bbbk)$ .

Lemma 4.1. Let $\eta\,:\,B\otimes B\to \Bbbk$ be a linear map. The following are equivalent.

1. (a) $\eta\in \textrm{Z}^2(B,\Bbbk)$ .

2. (b) $\eta\,:\,B\otimes B\to \Bbbk$ satisfies, for $a,c\in B^+$ ,

   (4.1) \begin{equation}\eta(a,1) = \eta(1,c)=0,\end{equation}
   (4.2) \begin{equation}\eta(a,bc) = \eta(ab,c), \quad b\in B.\end{equation}

Similarly, $\eta\in \textrm{B}^2(B,\Bbbk)$ if and only if (4.1) holds and there is a linear map $f\colon B\to \Bbbk$ such that $\eta(1,1)=f(1)$ and

\begin{equation*}\eta(a,b)=-f(ab), \quad a,b\in B^+.\end{equation*}

Proof. Recall that $\eta\in\textrm{Z}^2(B,\Bbbk)$ if and only if, for $a,b,c\in B$ :

(4.3) \begin{equation}\epsilon(a)\eta(b,c)+\eta(a,bc)=\eta(ab,c)+\eta(a,b)\epsilon(c).\end{equation}

Now, if $a,c\in B^+$ , then this becomes (4.2). On the other hand, if $a=\epsilon(a)\in B_0=\Bbbk$ and $c\in B^+$ (or viceversa), then this is

\begin{align*}\epsilon(a)\eta(b,c)+\epsilon(a)\eta(1,bc)=\epsilon(a)\eta(b,c)\Rightarrow \eta(1,bc)=0 \qquad (\text{or }\eta(ab,1)=0).\end{align*}

If $a,c\in B_0=\Bbbk$ then (4.3) is tautological.

Recall that $\eta\in\textrm{B}^2(B,\Bbbk)$ if and only if there is $f\colon B\to \Bbbk$ in such a way that

\begin{equation*}\eta(a,b)=\epsilon(a)f(b)-f(ab)+f(a)\epsilon(b).\end{equation*}

This is in turn equivalent to $\eta(a,b)=-f(ab)$ , $a,b\in B^+$ .

We compute an easy example in (4.4) to fix some notation. We will apply Lemma 4.1 to write down in Section 6.2 two more non-trivial examples in (6.8) and (6.9).

Lemma 4.2. For each $i,j\in I_\theta$ , let $\xi_j^i\colon B\otimes B\to \Bbbk$ be the linear map defined by

(4.4) \begin{align}\xi_j^i(b,b^{\prime})=\delta_{b,x_j}\delta_{x_i,b^{\prime}}, \ b,b^{\prime}\in\mathbb{B}.\end{align}

Then $\xi_j^i\in \textrm{Z}^2(B,\Bbbk)$ .

4.2. Ştefan's Theorem revisited

The following result, in general terms is due to Ştefan [Reference Ştefan25], and it is of central importance in our work. It is in fact a compendium of several results in loc.cit., which we have adapted to our setting, and this allows us to find explicit formulas that are implicit in Stefan's work. It is our belief that it is in this shape that might evoke more interest in the scope of this article. In particular, we state it in terms of left actions because of this idea.

Theorem 4.4. Let H be a Hopf algebra with bijective antipode and let $B\in{}^{H}_{H}\mathcal{YD}$ . Let $A=B\# H$ and let M be an A-bimodule.

Then there is a left H-action $\rightharpoonup$ on $\textrm{H}^q(B,M)$ , $q\geq 0$ .

• For $q=0$ , this is

  (4.5) \begin{align} h\rightharpoonup m=h_{(1)}\cdot m\cdot \mathcal{S}(h_{(2)}), \end{align}
  for $h\in H$ and $m\in \textrm{H}^0(B,M)=\{m\,:\,b\cdot m=m\cdot b, \forall\,b\in B\}\subseteq M$ .

• For $q>0$ , this is

  (4.6) \begin{align} (h\rightharpoonup f)({\textbf{x}})=h_{(1)}\cdot f\!\left(\mathcal{S}(h_{(2)})\cdot {\textbf{x}}\right)\cdot \mathcal{S}\!\left(h_{(3)}\right), \end{align}
  for $h\in H$ and $f\in \textrm{Z}^q(B,M)\subset\textrm{Hom}_\Bbbk(B^{\otimes\,q}, M)$ , ${\textbf{x}}=(x_1,\cdots, x_q)\in B^{\otimes\,q}$ .

If H is semisimple, then

(4.7) \begin{equation} \textrm{H}^n(A,\Bbbk)\simeq \textrm{H}^n(B,\Bbbk)^H. \end{equation}

Proof. Following [Reference Ştefan25] we work, equivalently, with a right action. We update our equations and need to show that there is a right H-action $\leftharpoonup$ on $\textrm{H}^q(B,M)$ so that

• For $q=0$ , this is

  (4.5’) \begin{align}m\leftharpoonup h=\mathcal{S}\!\left(h_{(1)}\right)\cdot m \cdot h_{(2)}, \quad m\in \textrm{H}^0(B,M), h\in H; \end{align}

• For $q>0$ , this is

  (4.6’) \begin{align}(f\leftharpoonup h)({\textbf{x}})=\mathcal{S}\!\left(h_{(1)}\right)\cdot f(h_{(2)}\cdot {\textbf{x}})\cdot h_{(3)}, \quad f\in \textrm{Z}^q(B,M), {\textbf{x}}\in B^{\otimes\,q}. \end{align}
  Observe that the left action $h\cdot b$ can also be interpreted as $b\cdot \mathcal{S}^{-1}(h)$ .

First, recall that A is a right H-Galois object with coaction $A\to A\otimes H$ given by $b\# h\mapsto b\# h_{(1)}\otimes h_{(2)}$ , with $A^{\textrm{co} H}\simeq B$ by [Reference Schauenburg24, Theorem 2.2.7].

We let $\textrm{can}\colon A\otimes_B A\to A\otimes H$ be the canonical isomorphism; in particular,

\begin{equation*}\sum \ell_i(h)\otimes r_i(h)\,:\!=\, \textrm{can}^{-1}(1\otimes h)=\mathcal{S}\!\left(h_{(1)}\right)\otimes h_{(2)}.\end{equation*}

For $M\in A\textrm{-bimod}$ , we write $\cdot$ both for the left and right action. By [Reference Ştefan25], the H-action $\leftharpoonup$ on $H^0(B,M)=M^{B}$ is given by $m\leftharpoonup h=\sum \ell_i(h)\cdot m\cdot r_i(h)$ , which in this case translates to (4.5’):

\begin{equation*}m\leftharpoonup h=\mathcal{S}\!\left(h_{(1)}\right)\cdot m\cdot h_{(2)}.\end{equation*}

For left actions, this is (4.5).

Assume this action (4.5’) has been extended to $\textrm{H}^q(B,M)$ (here we admit $q=0$ ).

Next paragraph is almost verbatim from the proof of [Reference Ştefan25, Proposition 2.4]. We refer the reader to loc.cit. and references therein for fuller details.

If $R\colon A\textrm{-bimod}\to B\textrm{-bimod}$ is the functor given by restriction of scalars, then $\textrm{H}^{\bullet}(A,-)\circ R$ is a cohomological and coeffaceable $\delta$ -functor. Hence the morphisms $\rho^h_0\colon \textrm{H}(A,-)\circ R\to \textrm{H}(A,-)\circ R$ , $h\in H$ , given by $\rho^h_0(M)(m)=m\leftharpoonup h$ can be extended univocally to morphisms of $\delta$ -functors $\rho^h_{\bullet}\colon \textrm{H}^{\bullet}(A,-)\circ R\to \textrm{H}^{\bullet}(A,-)\circ R$ .

We will describe this action explicitly in our context. For the content of this proof, we shall write, for shortness, $H^\bullet(A,-)\circ R(M)=H^\bullet(M)$ , $M\in\mathcal{A}\textrm{-bimod}$ . We denote by $d^q_{M}\colon \textrm{Hom}_\Bbbk(B^{\otimes q},M)\to \textrm{Hom}_\Bbbk(B^{\otimes q+1},M)$ the differential.

Let us fix a short exact sequence

\begin{equation*}0\to M^{\prime}\stackrel{\iota}{\longrightarrow} M \stackrel{\pi}{\longrightarrow} M^{\prime\prime}\to 0\end{equation*}

in $A\textrm{-bimod}$ ; we write $M\ni m\twoheadrightarrow\bar{m}\in M^{\prime\prime}$ for the projection.

First, the fact that $\rho^h_{\bullet}$ is a morphism of $\delta$ -functors establishes that

\begin{equation*}\rho^h_{q+1}(\delta(\textrm{H}^q(M^{\prime\prime})))=\delta\!\left(\rho^h_{q}\!\left(\textrm{H}^q\!\left(M^{\prime\prime}\right)\right)\right).\end{equation*}

In other words this is the recipe to extend the action, as it gives

\begin{equation*}\delta([f])\leftharpoonup h=\delta([f\leftharpoonup h]), \qquad f\colon B^{\otimes q}\to M^{\prime\prime}\in\textrm{Z}^q(M^{\prime\prime}).\end{equation*}

We shall write, wlog, $\delta(f)\colon B^{\otimes q}\to M^{\prime}$ for an element in $\textrm{Z}^{q+1}(M^{\prime})$ such that $[\delta(f)]=\delta([f])$ . Now, the morphism $\delta\colon \textrm{H}^q(M^{\prime\prime})\to \textrm{H}^{q+1}(M^{\prime})$ is the usual connecting homomorphism for the long exact sequence in cohomology:

\begin{equation*}\dots \to \textrm{H}^q(M) \stackrel{\textrm{H}^q(\pi)}{\longrightarrow} \textrm{H}^q(M^{\prime\prime})\stackrel{\delta}{\longrightarrow}\textrm{H}^{q+1}(M^{\prime}) \stackrel{\textrm{H}^{q+1}(\iota)}{\longrightarrow} \textrm{H}^{q+1}(M) \to \dots\end{equation*}

Explicitly, for $f\in\textrm{Z}^q(M^{\prime\prime})=\ker d^q_{M^{\prime\prime}}$ , let $\tilde{f}\colon B^{\otimes q}\to M$ be such that $\pi(\tilde{f})=f$ , in the sense that $f\!\left(b_1,\dots,b_q\right)(\pi(m))=\tilde{f}\!\left(b_1,\dots,b_q\right)(m)$ . Now we consider $d^q_M\!\left({\tilde{f}}\right)\in \textrm{Z}^{q+1}(M^{\prime\prime})$ and see that $\textrm{H}^{q+1}(\pi)\left(\left[d^q_M\!\left({\tilde{f}}\right)\right]\right)=d^q_{M^{\prime\prime}}(\textrm{H}^{q}(\pi)(\tilde{f}))=d^q_{M^{\prime\prime}}(f)=0$ . That is $\left[d^q_M\!\left({\tilde{f}}\right)\right]\in\ker \textrm{H}^{q+1}(\pi)=\textrm{img} \textrm{H}^{q+1}(\iota)$ and we can consider $\left[d^q_M\!\left({\tilde{f}}\right)\right]$ as an element $\delta([f])\in \textrm{H}^{q+1}(M^{\prime})$ . By abuse of notation we shall write $\delta(f)=df$ .

In our setting, if ${\textbf{b}}=\left(b_1,\dots,b_{q+1}\right)$

\begin{align*}(\delta(f)\leftharpoonup h)({\textbf{b}})=d(f \leftharpoonup h)({\textbf{b}})\end{align*}

If $q=0$ , $m\in M, b\in B$ , then $dm(b)=b\cdot m-m\cdot b$ . Also, recall the commutations (2.8). The connection reads as

\begin{align*}(\delta(m)\leftharpoonup h)(b)&=d(m \leftharpoonup h)(b)=\left(d\!\left(\mathcal{S}\!\left(h_{(1)}\right)\cdot m \cdot h_{(2)}\right)(b)\right)\\ &=b\cdot \mathcal{S}\!\left(h_{(1)}\right)\cdot m\cdot h_{(2)} -\mathcal{S}\!\left(h_{(1)}\right)\cdot m \cdot h_{(2)}\cdot b\\ &=\left(b\mathcal{S}\left(h_{(1)}\right)\right)\cdot m\cdot h_{(2)} -\mathcal{S}\!\left(h_{(1)}\right)\cdot m \cdot \left(h_{(2)}b\right)\\ &=\mathcal{S}\!\left(h_{(1)}\right)\left(h_{(2)}\cdot b\right)\cdot m \cdot h_{(3)}-\mathcal{S}\!\left(h_{(1)}\right)\cdot m \cdot \left(h_{(2)}\cdot b\right)h_{(3)}\\ &=\mathcal{S}\!\left(h_{(1)}\right)\cdot \left(h_{(2)}\cdot b\cdot m - m\cdot h_{(2)}\cdot b\right)\cdot h_{(3)}\\ &=\mathcal{S}\!\left(h_{(1)}\right)\cdot dm\!\left(h_{(2)}\cdot b\right)\cdot h_{(3)},\end{align*}

so $dm\leftharpoonup h$ is as in (4.5’).

Now, we turn to $q>0$ . If ${\textbf{b}}=b_1\otimes \cdots \otimes b_{q+1}\in B^{\otimes\,q+1}$ , then recall that we write ${\textbf{b}}=b_1\otimes {\textbf{b}}^2={\textbf{b}}^q\otimes b_{q+1}$ and ${\textbf{b}}_{i\,i+1}=b_1\otimes \cdots \otimes b_ib_{i+1}\otimes \cdots\otimes b_{q+1}\in B^{\otimes\,q}$ , $i\in\mathbb{I}_q$ .

Hence, we obtain, with commutations (2.8) as above:

\begin{align*}d(f \leftharpoonup h)({\textbf{b}})& = b_1\cdot (f \leftharpoonup h)\left({\textbf{b}}^2\right)-(f \leftharpoonup h)({\textbf{b}}_{12})+(f \leftharpoonup h)({\textbf{b}}_{23})\\ &\quad +\cdots +({-}1)^{n-1}(f \leftharpoonup h)({\textbf{b}}_{q-1q})+({-}1)^n(f \leftharpoonup h)({\textbf{b}}^q)\cdot b_{q+1}\\ &=b_1\cdot \mathcal{S}\!\left(h_{(1)}\right)\cdot f\!\left(h_{(2)}\cdot {\textbf{b}}^2\right) \cdot h_{(3)}-\mathcal{S}\!\left(h_{(1)}\right)\cdot f\!\left(h_{(2)}\cdot {\textbf{b}}_{12}\right)h_{(3)}\\ &\quad +\cdots +({-}1)^n\mathcal{S}\!\left(h_{(1)}\right)\cdot f \!\left(h_{(2)}\cdot {\textbf{b}}^q\right)\cdot h_{(3)}\cdot b_{q+1}\\ &=\mathcal{S}(h_{(1)} ) \cdot h_{(2)}\cdot b_1 \cdot f\!\left(h_{(3)}\cdot {\textbf{b}}^2\right) \cdot h_{(3)}-\mathcal{S}\!\left(h_{(1)}\right)\cdot f\!\left(h_{(2)}\cdot {\textbf{b}}_{12}\right)h_{(3)}\\ &\quad +\cdots +({-}1)^n\mathcal{S}\!\left(h_{(1)}\right)\cdot f \!\left(h_{(2)}\cdot {\textbf{b}}^q\right)\cdot h_{(3)}\cdot b_{q+1}\cdot h_{(4)}\\ &=\mathcal{S}(h_{(1)} ) \cdot \left(h_{(2)}\cdot {\textbf{b}}\right)_1 \cdot f\!\left(\left(h_{(2)}\cdot {\textbf{b}}\right)^2\right) \cdot h_{(3)}-\mathcal{S}\!\left(h_{(1)}\right)\cdot f\!\left(\left(h_{(2)}\cdot {\textbf{b}}\right)_{12}\right)h_{(3)}\\ &\quad +\cdots +({-}1)^n\mathcal{S}\!\left(h_{(1)}\right)\cdot f \!\left(\left(h_{(2)}\cdot {\textbf{b}}\right)^q\right)\cdot \left(h_{(2)}\cdot {\textbf{b}}\right)_{q+1}\cdot h_{(3)}\\ &=\mathcal{S}\!\left(h_{(1)}\right)df\!\left(h_{(2)}\cdot {\textbf{b}}\right)h_{(3)},\end{align*}

which shows (4.6’).

By uniqueness of the extension; this shows that (4.6') is indeed the action announced on [Reference Ştefan25, Proposition 2.4]; which implies (4.6) for the case of left actions.

The identity (4.7) is [Reference Ştefan25, (3.6.1) in p.229].

4.3. Invariant Hochschild 2-cocycles with coefficients in $\Bbbk$

In this setting, the action $\rightharpoonup$ in (4.6) from Theorem 4.4 becomes

\begin{equation*}h\rightharpoonup \eta(x,y)= \eta\!\left(\mathcal{S}(h_{(2)})\cdot x,\mathcal{S}\!\left(h_{(1)}\right)\cdot y\right).\end{equation*}

Since the antipode $\mathcal{S}$ is bijective, a cocycle $\eta\in \textrm{Z}^2(B)$ is H-invariant when

(4.8) \begin{align}\eta(h_{(1)}\cdot a,h_{(2)}\cdot b)=\epsilon(h)\eta(a,b), \ a,b\in B.\end{align}

Similarly, if $\eta\in Z^2(B)^H$ , then $\eta$ extends to $\eta\in \textrm{Z}^2(B\# H)$ via the formula

(4.9) \begin{align}\eta (a\#h,y\#k)=\eta(a,h\cdot b)\epsilon(k), \ a,b\in B, h,k\in H.\end{align}

Conversely, if $\eta\in Z^2(B\# H)$ then the restriction

\begin{equation*}\eta_{|B}\,:\!=\, \eta_{|B\otimes B}\colon B\otimes B\to \Bbbk\end{equation*}

defines a cocycle $\eta_{|B}\in Z^2(B)$ ; this is not necessarily H-invariant. If $\int\in H$ is an integral with $\epsilon({\int})=1$ , then we can project $\eta\in Z^2(B)$ onto $\eta^{\int}\in Z^2(B)^H$ by

\begin{equation*}\eta^{\int}(a,b)=\eta\!\left(\!\smallint{}_{(1)}\cdot a,\smallint{}_{(2)}\cdot b\right), \qquad a,b\in B.\end{equation*}

5.1. Exponential and invariance

Let $H,B\in{}^{H}_{H}\mathcal{YD}$ , $A=B\# H$ be as in Section 1.1.

We start with some simple though meaningful observations in Lemma 5.3. Let $\eta\in \textrm{Z}^2(B,\Bbbk)^H$ be an H-invariant Hochschild 2-cocycle; and set $\eta\in \textrm{Z}^2(A,\Bbbk)$ the extension as in (4.9). Notice that

\begin{equation*}\eta(x,h)=\eta(x,1)\epsilon(h)=0, \qquad \eta(h,x)=\eta(1,h\cdot x)=0,\end{equation*}

so $\eta_{|A^+\otimes A_0+A_0\otimes A^+}=0$ . Hence if $\eta_{|A_0\otimes A_0}=0$ , then the exponential $\sigma\,:\!=\, e^\eta$ is well-defined. Moreover, as in our case $\eta(h,k)=\epsilon(hk)\eta(1,1)$ , $h,k\in H=A_0$ , it is enough to require

(5.4) \begin{align}\eta(1,1)=0.\end{align}

Next lemma shows that we can always assume (5.4), without loss of generality.

Proof. Let $\theta=\eta(1,1)\epsilon$ , then $\theta\in \textrm{B}^2(B,\Bbbk)$ and (a) holds. Also, $e^\theta=e^{\eta(1,1)}\epsilon$ . (b) is clear and it implies that $e^{\eta^{\prime}}\colon B\otimes B\to \Bbbk$ is well-defined normalized convolution invertible linear map. As $\eta^{\prime}\ast\theta=\theta\ast\eta^{\prime}$ , then $e^{\theta}\ast e^{\eta^{\prime}}=e^{\eta}$ , which proves (c). (d) is straightforward, cf. (2.2).

Remark 5.4. If $\eta(1,1)=0$ , then

\begin{equation*}e^{\eta}(x,y)=\sum\limits_{k=0}^{m_{x,y}}\dfrac{1}{k!}\eta^{\ast k}(x,y), \qquad m_{x,y}=\min\{\deg(x),\deg(y)\}, \ x,y\in B.\end{equation*}

In particular,

(5.5) \begin{align}e^\eta(x,x_i)=\eta(x,x_i), \qquad x\in B, \ i\in\mathbb{I}_\theta.\end{align}

In our setting, we require that $\sigma=e^\eta$ satisfies (1.2). Next proposition shows that this is to require that $\eta\in \textrm{Z}^2(B,\Bbbk)^H$ .

Proof. We show that $e^\eta\in Z^2(B)^H\Rightarrow\eta\in \textrm{Z}^2(B,\Bbbk)^H$ , the other implication is clear. We need to show that $\eta\!\left(h_{(1)}\cdot x,h_{(2)}\cdot y\right)=\epsilon(h)\eta(x,y)$ . If $\deg y=0$ , then this is automatic. If $\deg y=1$ , say $y=x_i$ , then by (5.5) we have that

\begin{align*}\eta\!\left(h_{(1)}\cdot x,h_{(2)}\cdot x_i\right)=e^{\eta}\!\left(h_{(1)}\cdot x,h_{(2)}\cdot x_i\right)=\epsilon(h)e^\eta(x,x_i)=\epsilon(h)\eta(x,x_i).\end{align*}

Now, for the general case, assume $\deg y=n>1$ and let us write $y=y^{\prime}x_i$ , $\deg y^{\prime}=n-1$ . Then

\begin{align*}\eta\!\left(h_{(1)}\cdot x, h_{(2)}\cdot y\right)&=\eta\!\left(h_{(1)}\cdot x,h_{(2)}\cdot y^{\prime} h_{(3)}\cdot x_i\right)=\eta\!\left(h_{(1)}\cdot x h_{(2)}\cdot y^{\prime}, h_{(3)}\cdot x_i\right)\\ &=\eta\!\left(h_{(1)}\cdot (x y^{\prime}), h_{(2)}\cdot x_i\right)=\epsilon(h)\eta\!\left(x y^{\prime}, x_i\right)=\epsilon(h)\eta(x, y).\end{align*}

Hence the result follows.

Remark 5.6. Set $\tau=e^{\eta}$ in Lemma 3.11 so $\tau(x,y)=e^{\eta}(x,y)=\eta(x,y)$ . In particular, (3.5) becomes

(5.6) \begin{align}\alpha(x)\alpha(y)-\alpha(xy)=\sigma(x,y)-\eta(x,y).\end{align}

If $x=y$ , then this is

\begin{equation*}\alpha(x)^2-\alpha\!\left(x^2\right)=\sigma(x,x)-\eta(x,x).\end{equation*}

Assume that $c(x\otimes x)=-x\otimes x$ , thus $x^2$ is primitive and $\alpha^{-1}\!\left(x^2\right)=-\alpha\!\left(x^2\right)$ .

If, moreover, $\alpha$ is H-linear, then

\begin{align*} -\alpha\!\left(x^2\right)&=\alpha^{-1}\!\left(x^2\right)\stackrel{(3.6)}{=}\alpha(x)^2-\alpha\!\left(x_{({-}1)}\cdot x\right)\alpha^{-1}\!\left(x_{(0)}\right)-\alpha\!\left(x^2\right)\\ &=\alpha(x)^2-\epsilon\!\left(x_{({-}1)}\right)\alpha( x)\alpha^{-1}\!\left(x_{(0)}\right)-\alpha\!\left(x^2\right)\\ &=\alpha(x)^2-\alpha( x)\alpha^{-1}(x)-\alpha\!\left(x^2\right)=2\alpha(x)^2-\alpha\!\left(x^2\right). \end{align*}

That is, $\alpha(x)=0$ and (5.6) becomes

(5.7) \begin{align} \alpha(xy)=\eta(x,y)-\sigma(x,y),\qquad x,y\in V. \end{align}

In particular, if $x^2=0$ (v.g. when B is a Nichols algebra) we obtain

(5.8) \begin{align} \sigma(x,x)=\eta(x,x). \end{align}

In other words, $\eta(x,x)=\sigma(x,x)$ is a necessary condition to $e^{\eta}\sim\sigma$ .

5.2. Equivalence of the commutation conditions

In [Reference García and Mastnak13], for a Hopf algebra $A=\mathfrak{B}(V)\# H$ , the authors work with cocycles $\eta$ concentrated in degree (1,1), that is $\eta(A^n,A^m)=0$ if $(n,m)\neq (1,1)$ and show in [Reference García and Mastnak13, Lemma 2.3] that (5.1) and (5.2) are equivalent to each other and in turn to two other identities in $V^{\otimes 4}$ :

(5.9) \begin{align}\begin{split}({\eta}\otimes {\eta})(\textrm{id}\otimes c\otimes \textrm{id})&= ({\eta}\otimes {\eta})(\textrm{id}\otimes c\otimes \textrm{id})(c\otimes c),\\({\eta}\otimes {\eta})(\textrm{id}\otimes c\otimes \textrm{id})(\textrm{id}\otimes \textrm{id}\otimes c)&= ({\eta}\otimes {\eta})(\textrm{id}\otimes c\otimes \textrm{id})(c\otimes \textrm{id}\otimes \textrm{id}).\end{split}\end{align}

We are interested in finding more general contexts where this equivalence holds, to reduce the number of identities to check. We obtain some answers in Lemmas 5.7 and 5.8 below.

Proof. Interchange a and c in one of the lines from (5.3) to obtain the other.

The following lemma is rather technical, see (6.17) for a concrete and detailed application.

Lemma 5.8. Let A be a Hopf algebra and fix a linear basis $\mathbb{B}$ of A. Let $\eta\in\textrm{Z}^2(A,\Bbbk)$ be a Hochschild 2-cocycle. Fix $T=\Bbbk[\tau_x\,:\,x\in\mathbb{B}]$ the polynomial algebra on $\dim A$ variables given by the symbols $\tau_x\,:\!=\, \eta({-},x)\colon A\to \Bbbk$ .

In particular, for every $y,z\in A$ , there are families of scalars $\left(u_x=u_x(y,z)\right)_{x\in\mathbb{B}}$ , $\left(d_x=d_x(y,z)\right)_{x\in\mathbb{B}}\in\Bbbk$ such that the following identities hold in T:

\begin{align*} \eta\!\left({-},y_{(1)}z_{(1)}\right)\eta\!\left(y_{(2)},z_{(2)}\right)&=\sum_{x\in\mathbb{B}} u_x(y,z)\tau_x, \\ \eta\!\left(y_{(1)},z_{(1)}\right)\eta\!\left({-},y_{(2)}z_{(2)}\right)&=\sum_{x\in\mathbb{B}} d_x(y,z)\tau_x. \end{align*}

Assume that

(5.10) \begin{align} u_x(y,z)=d_x(y,z), \qquad \forall x,y,z\in \mathbb{B}. \end{align}

Then (5.1) implies (5.2).

The converse is also true, mutatis mutandis.

Proof. Assume that (5.1) and (5.10) hold. Let us set $\omega_z\,:\!=\, \eta({-},z)\colon A\to \Bbbk$ , for each $z\in\mathbb{B}$ . Now, if $x,y\in A$ , then

\begin{align*}\eta\!\left(x_{(1)}y_{(1)}, - \right)\eta\!\left(x_{(2)},y_{(2)}\right)&=\sum_{z\in\mathbb{B}}u_z(x,y)\omega_z\stackrel{(5.10)}{=}\sum_{z\in\mathbb{B}}d_z(x,y)\omega_z\\&=\eta\!\left(x_{(1)},y_{(1)}\right)\eta(x_{(2)}y_{(2)},-).\end{align*}

Which shows that (5.2) also holds.

6.1. Hopf cocycles

The (braided) cleft objects for the Nichols algebra $\mathfrak{B}_{\mathfrak{q}}$ are given by the family of algebras $\mathcal{E}(\boldsymbol\lambda)$ , $\boldsymbol\lambda=(\lambda_1,\lambda_2,\lambda_{12})\in\Bbbk^3$ , generated by $y_1,y_2$ and relations

(6.4) \begin{align}y_1^2&=\lambda_1, & y_2^2&=\lambda_2, & y_{12}^2&=\lambda_{12};\end{align}

again, $y_{12}\,:\!=\, y_1y_2 - q_{12}y_2x_1$ . The scalars $\lambda_1,\lambda_2,\lambda_{12}\in\Bbbk$ are subject to:

(6.5) \begin{align}\lambda_i&\neq 0 \text{ only if } \chi_i^2=\epsilon, i\in\mathbb{I}_2; & \lambda_{12}&\neq 0 \text{ only if } \left(\chi_{1}\chi_{2}\right)^2=\epsilon.\end{align}

This conditions can also be written as

\begin{align*}\lambda_jq_{ij}^2&=\lambda_j, \, i,j\in\mathbb{I}_2; & \lambda_{12}q_{ij}^2&=\lambda_{12}, \ i\neq j\in\mathbb{I}_2.\end{align*}

This algebra inherits a basis as in (6.2).

For each triple $\boldsymbol\lambda=(\lambda_1,\lambda_2,\lambda_{12})$ , the corresponding deformation of $A=\mathfrak{B}_{\mathfrak{q}}\# H$ is the quotient $\mathfrak{u}_{\mathfrak{q}}(\boldsymbol\lambda)$ of $T(V)\# H$ modulo the relations

(6.6) \begin{align}a_1^2=\lambda_1\!\left(1-g_1^2\right), a_2^2=\lambda_2\!\left(1-g_2^2\right), a_{12}^2=\lambda_{12}\!\left(1-g_1^2g_2^2\right)+4q_{12}\lambda_1\lambda_2g_2^2\!\left(1-g_1^2\right).\end{align}

Here $V=\Bbbk\{a_1,a_2\}$ and $a_{12}\,:\!=\, a_1a_2 - q_{12}a_2a_1$ .

Henceforward, we will assume that

(6.7) \begin{equation}q_{12}=\pm 1,\end{equation}

as we shall investigate Hopf 2-cocycles $\sigma_{\boldsymbol\lambda}$ associated to non-trivial deformations $\mathcal{E}(\boldsymbol\lambda)$ and $\mathfrak{u}_{\mathfrak{q}}(\boldsymbol\lambda)$ .

Lemma 6.3. The section $\gamma\colon\mathfrak{B}_{\mathfrak{q}}\to\mathcal{E}(\boldsymbol\lambda)$ is given by

\begin{align*}\gamma(x_i)&=y_i,\ i\in\mathbb{I}_2, & \gamma(x_{12})&=y_{12}, & \gamma(x_{2}x_1)&=y_{2}y_1,\\\gamma(x_{12}x_1)&=y_{12}y_{1} - 2q_{21}\lambda_1 y_2, & \gamma(x_2x_{12})&=y_{2}y_{12},& \gamma(x_2x_{12}x_1)&=y_2y_{12}y_1.\end{align*}

The convolution inverse for $\gamma$ is determined by

\begin{align*}\gamma^{-1}(x_i)&=-y_i,\ i\in\mathbb{I}_2, & \gamma^{-1}(x_{12})&=y_{12}+2q_{12}y_2y_1, \\\gamma^{-1}(x_{2}x_1)&=q_{21}y_{12}-y_2y_1, &\gamma^{-1}(x_{12}x_1)&=-y_{12}y_1+2q_{21}\lambda_1y_2, \\ \gamma^{-1}(x_2x_{12})&=-y_2y_{12},& \gamma^{-1}(x_2x_{12}x_1)&=y_2y_{12}y_1-\lambda_{12}.\end{align*}

Proof. The values for $\gamma(x_i), \gamma(x_{12}), \gamma(x_{2}x_1)$ are straightforward. We see that

\begin{align*}\underline{\Delta}(x_{12}x_1)&=-q_{21}x_1\otimes x_{12}+x_{12}\otimes x_1+2x_1\otimes x_2x_1,\\\underline{\rho}(y_{12}y_1)&=2q_{21}\lambda_1\,1\otimes x_2 -q_{21}y_1\otimes x_{12}+x_{12}\otimes x_1+2y_1\otimes x_2x_1\end{align*}

and thus setting $\gamma(x_{12}x_1)=(y_{12}y_{1} - 2q_{21}\lambda_1 y_2 )$ gives $(\gamma\otimes \textrm{id})\Delta(x_{12}x_1)=\rho\gamma(x_{12}x_1)$ as expected. On the other hand, it is easy to see $\gamma(x_2x_{12})=y_{2}y_{12}$ . Finally,

\begin{align*}\underline{\Delta}(x_2x_{12}x_1)&=2x_2x_1\otimes x_2x_1 - q_{21}^{2}x_1\otimes x_2x_{12} - q_{21}x_{12}\otimes x_2x_1 \\ &\quad - q_{21}^{2}x_{12}x_1\otimes x_2+ x_2\otimes x_{12}x_1 - q_{21}x_2x_1\otimes x_{12} + x_2x_{12}\otimes x_1,\\ \underline{\rho}(y_2y_{12}y_1)&=2q_{21}\lambda_1 y_2\otimes x_2+2y_2y_1\otimes x_2x_1 - q_{21}^{2}y_1\otimes x_2x_{12} - q_{21}y_{12}\otimes x_2x_1 \\ &\quad - q_{21}^{2}y_{12}y_1\otimes x_2+ y_2\otimes x_{12}x_1 - q_{21}y_2y_1\otimes x_{12} + y_2y_{12}\otimes x_1.\end{align*}

So $(\gamma\otimes\textrm{id})\Delta(x_2x_{12}x_1)=\rho(y_2y_{12}y_1)$ as (recall $\lambda_1q_{21}^3=\lambda_1 q_{21}$ ):

\begin{align*}(\gamma\otimes\textrm{id})\underline{\Delta}(x_2x_{12}x_1) & = 2y_2y_1\otimes x_2x_1 - q_{21}^{2}y_1\otimes x_2x_{12} - q_{21}y_{12}\otimes x_2x_1 \\ &\quad - q_{21}^{2}(y_{12}y_{1} - 2q_{21}\lambda_1 y_2 )\otimes x_2+ y_2\otimes x_{12}x_1 - q_{21}y_2y_1\otimes x_{12} + y_2y_{12}\otimes x_1.\end{align*}

Now, to see that $\gamma^{-1}$ is the convolution inverse of $\gamma$ , we only need to verify that $\gamma\ast\gamma^{-1}(b)=0=\gamma^{-1}\ast\gamma(b)$ , for each $b\in\mathbb{B}_{\mathfrak{q}}$ . To illustrate this, we expose some calculations below.

\begin{align*}\gamma\ast\gamma^{-1}(x_{12}x_1)&=\gamma(x_{12}x_1)-q_{21}y_1\!\left(y_{12}+2q_{12}y_2y_1\right)-y_{12}y_1+2y_1(q_{21}y_{12}-y_2y_1)+\gamma^{-1}(x_{12}x_1)\\ &=-y_{12}y_1+2y_1y_2y_1-y_{12}y_1+2y_{12}y_1-2y_1y_2y_1=0.\\ \gamma\ast\gamma^{-1}(x_{2}x_{12}x_1) & = y_2y_{12}y_1+2y_2y_1(q_{21}y_{12}-y_2y_1)+q_{21}^2y_1y_2y_{12}-y_2y_{12}y_1\\ &\quad -q_{21}y_{12}(q_{21}y_{12}-y_2y_1)+q_{21}^2(y_{12}y_1-2q_{21}\lambda_1y_2)y_2+y_2y_{12}y_1\\ &\quad -q_{21}^2\lambda_{12}+y_2({-}y_{12}y_1+2q_{21}\lambda_1y_2)-q_{21}y_2y_1\!\left(y_{12}+2q_{12}y_2y_1\right)\\ & =2q_{21}y_2y_1y_{12}+q_{21}^2(y_{12}+q_{12}y_2y_1)y_{12}-q_{21}^2\lambda_{12}+q_{21}y_{12}y_2y_1\\ & \quad +q_{21}^2y_{12}y_1y_2-2q_{21}\lambda_1\lambda_2-q_{21}^2\lambda_{12}+2q_{21}\lambda_1\lambda_2-q_{21}y_2y_1y_{12}\\ & =2y_2y_{12}y_1-q_{21}y_2y_1y_{12}+q_{21}y_{12}y_2y_1+q_{21}^2y_{12}(y_{12}+q_{12}y_2y_1)\\ &\quad -q_{21}^2\lambda_{12}-y_2y_{12}y_1\\ & =y_2y_{12}y_1-y_2y_{12}y_1=0.\end{align*}

The rest of the computations, which are simpler, are left to the reader.

Remark 6.4. Notice that $\gamma\colon\mathfrak{B}_{\mathfrak{q}}\to\mathcal{E}(\boldsymbol\lambda)$ as in Lemma 6.3 is H-linear. Indeed, it suffices to check this on $\gamma(x_{12}x_1)$ and this follows since

\begin{equation*}\lambda_1 h\cdot y_2=\lambda_1\chi_2(h)y_2= \lambda_1\chi_1^2(h)\chi_2(h)y_2, \quad h\in H.\end{equation*}

Theorem 6.5. Let $\boldsymbol\lambda=(\lambda_1,\lambda_2,\lambda_{12})$ . The Hopf cocycle $\sigma_{\boldsymbol\lambda}\,:\!=\,\sigma$ associated to the cleft object $\mathcal{E}(\boldsymbol\lambda)$ is given by

Proof. In first place, we use Lemma 3.1 to calculate the elements of $S_\sigma$ as in (3.1). Indeed, let $i,j\in\mathbb{I}_2$ , then

\begin{align*} \sigma\!\left(x_i,x_j\right)=-q_{ij}y_jy_i+\gamma^{-1}\!\left(x_ix_j\right)=\delta_{ij}\lambda_i,\end{align*}

here we use that $\gamma^{-1}\!\left(x_ix_j\right)=q_{ij}y_jy_i$ . It is easy to see that

\begin{equation*}\sigma(x_i,x_{12})=\sigma(x_i,x_2x_1)=0.\end{equation*}

For the other hand, we show $\sigma(x_1,x_2x_{12})$ and $\sigma(x_2,x_{12}x_1)$ .

\begin{align*}\sigma(x_1, x_2x_{12}) & =y_2y_{12}y_2-2q_{12}y_2y_1\gamma^{-1}(x_1x_2)-y_2\gamma^{-1}(x_{12}x_1)-y_{12}\gamma^{-1}(x_1x_2) +\gamma^{-1}(x_1x_2x_{12})\\ &=y_2y_{12}y_2-2y_2y_1y_2y_1+y_2(y_{12}-2q_{21}\lambda_1y_2)-q_{12}y_{12}y_2y_1-y_1y_{12}y_2+\lambda_{12}\\ &=y_2y_{12}y_1-2y_2y_{12}y_1-2q_{12}\lambda_1\lambda_2+y_2y_{12}y_1+2q_{12}\lambda_1\lambda_2+\lambda_{12}=\lambda_{12}.\\ \sigma(x_2, x_{12}x_1) & =q_{21}^2(y_{12}y_1-2q_{21}\lambda_1y_2)y_2+y_1y_2y_{12} -q_{21}y_{12}\gamma^{-1}(x_2x_1)+\gamma^{-1}(x_2x_{12}x_1)\\ & =y_{12}y_1y_2-2q_{21}\lambda_1\lambda_2+\lambda_{12}-y_2y_{12}y_1-y_{12}y_1y_2+y_2y_{12}y_1+\lambda_{12}=2q_{12}\lambda_1\lambda_2.\end{align*}

The reader will have no difficulty to see that

\begin{equation*}\sigma(x_1,x_{12}x_1)=\sigma(x_2,x_2x_{12})=\sigma(x_i,x_1x_{12}x_2)=0.\end{equation*}

Thus, $S_\sigma=\{0,\lambda_1,\lambda_2,\lambda_{12},2q_{12}\lambda_1\lambda_2\}$ . From now on, we use the formulas of decomposition in Example 3.4 to get the others values of $\sigma$ , following Lemma 3.2. In this case, the matrix coefficients of the braiding $\left(c_{i,j}^{p,q}\right)_{ \begin{smallmatrix} i,j,\\ p,q \end{smallmatrix} \in\mathbb{I}}$ are given by $c_{i,j}^{p,q}=\delta_{i,q}\delta_{j,p}q_{ij}$ so $\mathbb{Z}\!\left[c_{i,j}^{p,q}\right]_{\begin{smallmatrix} i,j,\\ p,q \end{smallmatrix} \in\mathbb{I}}=\mathbb{Z}\!\left[q_{12}^{\pm 1}\right]\simeq \mathbb{Z}$ , as $q_{11}=q_{22}=-1$ and $q_{21}=$ $-q_{12}^{-1}=\mp1$ .

One example of application is shown below.

\begin{align*} \sigma(x_{12}x_1,x_{12}x_1)&=\sigma(x_1,x_2x_1x_{12}x_1)+\sigma(x_2,g_1\cdot (x_{12}x_1))\sigma(x_1,x_1)\\ &\quad+\sigma(g_2\cdot x_1,g_2\cdot (x_{12}x_1))\sigma(x_1,x_2)-\sigma(x_1,x_2)\sigma(x_1,x_{12}x_1)\\ &\quad-\sigma(x_1,g_2\cdot x_1)\sigma(x_2,x_{12}x_1)+\sigma(x_2x_1,2x_1)\sigma(x_1,x_2x_1)\\ &\quad-\sigma(x_2x_1,q_{21}x_1)\sigma(x_1,x_{12})+\sigma(x_2x_1,x_{12})\sigma(x_1,x_1)\\ &\quad+\sigma(x_2,g_1\cdot(2x_1))\sigma(x_1,x_1x_2x_1)-\sigma(x_2,g_1\cdot(q_{21}x_1))\sigma(x_1,x_1x_{12}x_1)\\ &\quad+\sigma(x_2,g_1\cdot x_{12})\sigma(x_1,x_1^2)+\sigma(g_2\cdot x_1,g_2\cdot (2x_1))\sigma\!\left(x_1,x_2^2x_1\right)\\ &\quad-\sigma(g_2\cdot x_1,g_2\cdot (q_{21}x_1))\sigma(x_1,x_2x_{12})+\sigma(g_2\cdot x_1,g_2\cdot x_{12})\sigma(x_1,x_2x_1)\\ &=q_{12}\sigma(x_2, x_{12}x_1)\sigma(x_1,x_1)-q_{21}\sigma(x_1, x_1)\sigma(x_2,x_{12}x_1) -q_{21}\sigma( x_1,x_1)\sigma(x_1,x_2x_{12})\\ &=2q_{12}\sigma(x_2, x_{12}x_1)\sigma(x_1,x_1)+q_{12}\sigma( x_1,x_1)\sigma(x_1,x_2x_{12})\\ &=4\lambda_1^2\lambda_2+q_{12}\lambda_{12}\lambda_1.\end{align*}

The other values are computed analogously.

6.2. Hochschild cocycles

Recall the cocycles $\xi_j^i\in \textrm{Z}^2\!\left(\mathfrak{B}_{\mathfrak{q}},\Bbbk\right)$ , $i,j\in\mathbb{I}_2$ , from Example 4.2. Additionally, we denote $\xi_0=\epsilon\otimes\epsilon\in \textrm{Z}^2\!\left(\mathfrak{B}_{\mathfrak{q}},\Bbbk\right)$ , the cocycle such that $\xi_0(1,1)=1$ , and zero elsewhere.

We shall consider two more Hochschild 2-cocycles $\xi_{121}^2,\xi_{212}^1$ in $\mathfrak{B}_{\mathfrak{q}}$ . Namely, let $\xi_{121}^2\colon \mathfrak{B}_{\mathfrak{q}}\otimes \mathfrak{B}_{\mathfrak{q}}\to \Bbbk$ be the linear map given by

(6.8) \begin{align}\begin{split}\xi_{121}^2(x_2,x_{12}x_1)&=\xi_{121}^2(x_2x_{12},x_1)=\xi_{121}^2(x_2x_1,x_2x_1)=\xi_{121}^2(x_{12},x_{12})=1,\\\xi_{121}^2(x_{12},x_2x_1)&=\xi_{121}^2(x_2x_1,x_{12})=-q_{12}.\end{split}\end{align}

and zero elsewhere in the basis $\mathbb{B}\otimes \mathbb{B}$ . Following Lemma 4.1, it is an easy computation to check that $\xi_{121}^2\in \textrm{Z}^2\!\left(\mathfrak{B}_{\mathfrak{q}},\Bbbk\right)$ . As well, we define $\xi_{212}^1\in \textrm{Z}^2\!\left(\mathfrak{B}_{\mathfrak{q}},\Bbbk\right)$ via

(6.9) \begin{align}\xi_{212}^1(x_1,x_2x_{12})=\xi_{212}^1(x_{12},x_{12})=\xi_{212}^1(x_{12}x_{1},x_2)=1\end{align}

and zero elsewhere in $\mathbb{B}\otimes \mathbb{B}$ .

We further note that $\xi_{121}^2+\xi_{212}^1\in\textrm{B}^2(B,\Bbbk)$ . Indeed, let $f\colon B\to \Bbbk$ be the linear map defined on the basis $\mathbb{B}$ by $f(x_2x_{12}x_1)=1$ and zero elsewhere. Then

(6.10) \begin{align} \left(\xi_{121}^2-\xi_{212}^1\right)(b,b^{\prime})=f(bb^{\prime}), \qquad b,b^{\prime}\in\mathbb{B}. \end{align}

Recall the group $\Gamma\leq G(H)$ from (6.3).

Proof. By Lemma 4.1, see Remark 4.3, $\eta\in \textrm{Z}^2\!\left(\mathfrak{B}_{\mathfrak{q}},\Bbbk\right)$ is determined by the values

\begin{align*}\eta&(x_1,x_1), & \eta&(x_1,x_2), & \eta&(x_{1},x_{12}), & \eta&(x_1,x_2x_{12}), \\\eta&(x_1,x_2x_1), & \eta&(x_2,x_2), & \eta&(x_{2},x_1), & \eta&(x_2,x_{12}), \\\eta&(x_2,x_{12}x_1) & \eta&(1,1).\end{align*}

Furthermore, as $x_1x_2x_1=x_{12}x_1$ and $x_2x_1x_2=x_2x_{12}$ , we get

\begin{align*}\eta(x_1,x_{12}x_1)=\eta(x_2,x_2x_{12})=0.\end{align*}

Also, it is easy to see that

\begin{align*}\eta(x_1,x_2x_{12}x_1)&=\eta(x_2,x_2x_{12}x_1)=\eta(x_2,x_2x_{12})=\eta(x_2,x_2x_1)=0,\\\eta(x_1,x_2x_1)&=q_{21}\eta(x_1,x_{12}).\end{align*}

Now we let $\Gamma$ act and obtain

\begin{equation*}\eta(x_1,x_2)=\eta(g_1\cdot x_1,g_1\cdot x_2)=-q_{12}\eta(x_1,x_2)\end{equation*}

which forces $q_{12}=-1$ but also

\begin{equation*}\eta(x_1,x_2)=\eta(g_2\cdot x_1,g_2\cdot x_2)=-q_{21}\eta(x_1,x_2),\end{equation*}

which also gives $q_{21}=-1$ , a contradiction to $q_{12}q_{21}=-1$ . Thus $\eta(x_1,x_2)=0$ , similarly $\eta(x_2,x_1)=0$ . Similarly, we need to have

\begin{equation*}-q_{21}^2\eta(x_{1},x_{12})=\eta(x_{1},x_{12}),\end{equation*}

which gives $\eta(x_{1},x_{12})=0$ ; and similarly

\begin{equation*}-q_{12}^2\eta(x_{2},x_{12})=\eta(x_{2},x_{12}), \ -q_{12}^2\eta(x_2x_1,x_2)=\eta(x_2x_1,x_2)\end{equation*}

so $\eta(x_{12},x_2)=0$ . That is, we are left with the values

\begin{align*}\eta&(1,1), &\eta&(x_1,x_1), & \eta&(x_2,x_2), & \eta&(x_2,x_{12}x_1), & \eta&(x_1,x_2x_{12})\end{align*}

and therefore $\eta\in\Bbbk\left\{\xi_0, \xi_1^1,\xi_2^2,\xi_{121}^2, \xi_{212}^1\right\}$ .

The second statement is straightforward; recall $h\cdot x_i=\chi_i(h)x_i$ , $h\in H, i\in\mathbb{I}_\theta$ .

6.3. Exponentials

Let $\eta\in \textrm{Z}^2\!\left(\mathfrak{B}_{\mathfrak{q}},\Bbbk\right)^H$ , we are interested in finding the conditions for $\eta$ such that $e^\eta$ is a Hopf cocycle; set

(6.11) \begin{align}\eta=\eta_1\xi_1^1+ \eta_2\xi^2_2+\eta_{121}\xi_{121}^2+\eta_{212}\xi_{212}^1.\end{align}

In particular, $\eta(x_i,x_i)=\eta_i$ , $i\in\mathbb{I}_2$ , $\eta(x_{2},x_{12}x_1)=\eta_{121}$ and $\eta(x_{1},x_2x_{12})=\eta_{212}$ .

Next proposition characterizes the cocycles $\eta$ that satisfy (5.1) and (5.2); so in particular $e^{\eta}$ is a Hopf cocycle. It is enough to assume that $\eta\in \textrm{Z}^2\!\left(\mathfrak{B}_{\mathfrak{q}},\Bbbk\right)^\Gamma$ .

Proposition 6.7. A cocycle $\eta\in\textrm{Z}^2\!\left(\mathfrak{B}_{\mathfrak{q}},\Bbbk\right)^\Gamma$ satisfies (5.1) and (5.2) if and only if

\begin{equation*}\eta \in \textrm{C}\,:\!=\,\left\{\eta_1\xi_1^1,\eta_2\xi_2^2, \eta_{121}\xi_{121}^2+\eta_{212}\xi_{212}^1\,:\,\eta_1,\eta_2,\eta_{121},\eta_{212}\in\Bbbk\right\}.\end{equation*}

In particular if $\eta\in \textrm{C}$ , then $e^\eta$ is a Hopf 2-cocycle.

We will show in Lemma 6.11 that conditions (5.1) and (5.2) are not necessary for $e^\eta$ to be a Hopf 2-cocycle.

Remark 6.8. Let $\eta=\eta_1\xi_1^1+ \eta_2\xi^2_2+\eta_{121}\xi_{121}^2+\eta_{212}\xi_{212}^1\in \textrm{Z}^2\!\left(\mathfrak{B}_{\mathfrak{q}},\Bbbk\right)^\Gamma$ . We can rephrase Proposition 6.7 as follows: $\eta\in \textrm{C}$ if and only if

(6.12) \begin{align} \eta_i\eta_j=\eta_i\eta_{iji}=\eta_i\eta_{jij}=0,\qquad i\neq j\in \mathbb{I}_2. \end{align}

Proof. Suppose that $\eta$ verifies (5.1) and (5.2). That is for any $x,y,z\in \mathfrak{B}_{\mathfrak{q}}$ ,

(6.13) \begin{equation}\eta\!\left(x,y_{(1)}z_{(1)}\right)\eta\!\left(y_{(2)},z_{(2)}\right) =\eta\!\left(x,y_{(2)}z_{(2)}\right)\eta\!\left(y_{(1)},z_{(1)}\right),\end{equation}

(6.14) \begin{equation}\eta\!\left(x_{(1)}y_{(1)},z\right)\eta\!\left(x_{(2)},y_{(2)}\right) = \eta\!\left(x_{(2)}y_{(2)},z\right)\eta\!\left(x_{(1)},y_{(1)}\right).\end{equation}

Now, we see that if $y=x_2x_1$ and $z=x_2$ , then (6.13) gives

(6.15) \begin{align}q_{21}\eta(x,x_1)\eta(x_2,x_2)=q_{12}\eta(x,x_1)\eta(x_2,x_2), \quad\text{for any $x\in\mathfrak{B}_{\mathfrak{q}}$.}\end{align}

Similarly, if $y=x_1x_2$ and $z=x_1$ , then (6.13) gives

(6.16) \begin{align}q_{21}\eta(x,x_2)\eta(x_1,x_1)=q_{12}\eta(x,x_2)\eta(x_1,x_1),\quad\text{for any $x\in \mathfrak{B}_{\mathfrak{q}}$.}\end{align}

Thus, from (6.15) and (6.16), we obtain

\begin{align*} \eta(x_1,x_1)\eta(x_2,x_2)=\eta(x_2,x_{12}x_1)\eta(x_2,x_2)=\eta(x_1,x_2x_{12})\eta(x_1,x_1)=0.\end{align*}

If we suppose that $\eta(x_1,x_1)\neq 0$ , then $\eta(x_2,x_2)=\eta(x_1,x_2x_{12})=0$ . Furthermore, if $y=x_{12}x_1$ and $z=x_1$ , then we obtain from (6.13) that

\begin{align*}\eta(x,x_{12})\eta(x_1,x_1)=-2q_{21}\eta(x_1,x_1)\eta(x,x_2x_1)+\eta(x_1,x_1)\eta(x,x_{12}),\end{align*}

for any $x\in\mathfrak{B}_{\mathfrak{q}}$ . Hence, we get that $\eta(x_2,x_{12}x_1)=0$ . Through the same calculations for $\eta(x_2,x_2)\neq 0$ , we get that $\eta(x_1,x_1)=\eta(x_2,x_{12}x_1)=\eta(x_1,x_2x_{12})=0$ . Thus, $\eta\in \textrm{C}$ .

Conversely, suppose that $\eta\in \textrm{C}$ . On one hand, if $\eta=\eta_1\xi_i^i$ , $i\in\mathbb{I}_2$ , then by [Reference García and Mastnak13, Lemma 2.3.], we have the conclusion as $\eta$ clearly satisfies (5.9).

Now, we show that the hypothesis of Lemma 5.8 hold when

(6.17) \begin{align} \eta=\eta_{121}\xi_{121}^2+\eta_{212}\xi_{212}^1. \end{align}

Let $x,y,z\in\mathbb{B}_\mathfrak{q}$ , such that $\deg(y), \deg(z)$ are positive. In first place, note that if $\deg(y)+\deg(z)<4$ , then we obtain zero in both sides of (5.1) and thus $u_x(y,z)=d_x(y,z)=0$ . Now, if $\deg(y)+\deg(z)=4$ , then (5.1) becomes $\eta(x,1)\eta(y,z)$ and so, $u_x(y,z)=0=d_x(y,z)$ when $x\neq 1$ and $u_1(y,z)=\eta(y,z)=d_1(y,z)$ . On the other hand, we explicitly show the most representative calculations below.

deg(y) + deg(z) = 5:

1. (i) $y=x_{12}$ and $z=x_{12}x_1$ :

   \begin{align*}\eta({-},y_1z_1)\eta(y_2,z_2)=&2\eta({-},x_1)\eta(x_2,x_{12}x_1)-2q_{21}\eta({-},x_1)\eta(x_{12},x_2x_1) +\eta({-},x_1)\eta(x_{12},x_{12}),\\ \eta(y_1,z_1)\eta({-},y_2z_2)=&\eta(x_{12},x_{12})\eta({-},x_1).\end{align*}
   Here

• $u_{x_1}\!\left(x_{12},x_{12}x_1\right)=\eta(x_{12},x_{12})=d_{x_1}\!\left(x_{12},x_{12}x_1\right)$ .

• $u_x\!\left(x_{12},x_{12}x_1\right)=d_x\!\left(x_{12},x_{12}x_1\right)=0$ for $x\neq x_1$ .

1. (ii) $y=x_{12}$ and $z=x_2x_{12}$ :

   \begin{align*}\eta({-},y_1z_1)\eta(y_2,z_2)=&\eta({-},x_2)\eta(x_1,x_2x_{12})-q_{12}\eta({-},x_2)\eta(x_2x_1,x_{12}),\\\eta(y_1,z_1)\eta({-},y_2z_2)=&-q_{21}\eta(x_2x_1,x_{12})\eta({-},x_2)+\eta(x_1,x_2x_{12})\eta({-},x_2)\\&+2\eta(x_2x_1,x_2x_1)\eta({-},x_2).\end{align*}
   Then

• $u_{x_2}(x_{12},x_2x_{12})=\eta(x_{12},x_{12})=d_{x_2}(x_{12},x_2x_{12})$ .

• $u_x(x_{12},x_2x_{12})=d_x(x_{12},x_2x_{12})=0$ for $x\neq x_2$ .

1. (iii) $y=x_2x_1$ and $z=x_{12}x_{1}$ :

   \begin{align*}\eta({-},y_1z_1)\eta(y_2,z_2)=&q_{21}\eta({-},x_1)\eta(x_2,x_{12}x_1)-2q_{21}\eta({-},x_1)\eta(x_2x_1,x_2x_1)\\&+\eta({-},x_1)\eta(x_2x_1,x_{12}),\\\eta(y_1,z_1)\eta({-},y_2z_2)=&\eta(x_2x_1,x_{12})\eta({-},x_1)+q_{12}\eta(x_2,x_{12}x_1)\eta({-},x_1).\end{align*}
   In this case $u_{x}(x_{12},x_2x_{12})=0=d_{x}(x_{12},x_2x_{12})$ for any $x\in \mathbb{B}_\mathfrak{q}$ .

2. (iv) $y=x_{2}x_1$ and $z=x_{2}x_{12}$ :

   \begin{align*}\eta({-},y_1z_1)\eta(y_2,z_2)=&\eta({-},x_2)\eta(x_1,x_2x_{12})-q_{12}\eta({-},x_2)\eta(x_2x_1,x_{12}),\\\eta(y_1,z_1)\eta({-},y_2z_2)=&-q_{21}\eta(x_2x_1,x_{12})\eta({-},x_2)+\eta(x_1,x_2x_{12})\eta({-},x_2)\\&+2\eta(x_2x_1,x_2x_1)\eta({-},x_2).\end{align*}
   Then,

• $u_{x_2}(x_{2}x_1,x_2x_{12})=\eta(x_{12},x_{12})=d_{x_2}(x_{2}x_1,x_2x_{12})$ .

• $u_x(x_{2}x_1,x_2x_{12})=d_x(x_{2}x_1,x_2x_{12})=0$ for $x\neq x_2$ .

1. (v) $y=x_{2}$ and $z=x_{2}x_{12}x_1$ :

   \begin{align*}\eta(x,y_1z_1)\eta(y_2,z_2)=&-\eta(x,x_2)\eta(x_2,x_{12}x_1),\\\eta(y_1,z_1)\eta(x,y_2z_2)=&-\eta(x_2,x_{12}x_1)\eta(x,x_2).\end{align*}
   Then,

• $u_{x_2}(x_{2},x_2x_{12}x_1)=-\eta(x_{2},x_{12}x_1)=d_{x_2}(x_{2},x_2x_{12}x_1)$ .

• $u_x(x_{2},x_2x_{12}x_1)=d_x(x_{2},x_2x_{12}x_1)=0$ for $x\neq x_2$ .

deg(y) + deg(z) = 6:

1. (i) $y=x_{12}$ and $z=x_2x_{12}x_1$ :

   \begin{align*}\eta({-},y_1z_1)\eta(y_2,z_2)=&q_{21}\eta({-},x_2x_{1})\eta(x_{12},x_{12}),\\\eta(y_1,z_1)\eta({-},y_2z_2)=&\eta(x_{12},x_2x_1)\eta({-},x_2x_1)+q_{21}\eta(x_1,x_2x_{12})\eta({-},x_2x_1).\end{align*}
   Then,

• $u_{x_2x_1}(x_{12},x_2x_{12}x_1)=q_{21}\eta(x_{12},x_{12})=d_{x_2x_1}(x_{12},x_2x_{12}x_1)$ .

• $u_x(x_{12},x_2x_{12}x_1)=d_x(x_{12},x_2x_{12}x_1)=0$ for $x\neq x_2x_1$ .

1. (ii) $y=x_{2}x_1$ and $z=x_2x_{12}x_1$ :

   \begin{align*}\eta({-},y_1z_1)\eta(y_2,z_2)=&\eta({-},x_2x_1)\eta(x_1,x_2x_{12})-q_{21}\eta({-},x_{12})\eta(x_2,x_{12}x_1)\\&+q_{21}\eta({-},x_{12})\eta(x_2x_1,x_2x_1),\\\eta(y_1,z_1)\eta({-},y_2z_2)=&\eta(x_1,x_2x_{12})\eta({-},x_2x_1).\end{align*}
   Thus,

• $u_{x_2x_1}(x_{2}x_1,x_2x_{12}x_1)=\eta(x_{1},x_2x_{12})=d_{x_2x_1}(x_{2}x_1,x_2x_{12}x_1)$ .

• $u_x(x_{2}x_1,x_2x_{12}x_1)=d_x(x_{2}x_1,x_2x_{12}x_1)=0$ for $x\neq x_2x_1$ .

1. (iii) $y=x_{12}x_1$ and $z=x_2x_{12}$ :

   \begin{align*}\eta({-},y_1z_1)\eta(y_2,z_2)=&-2q_{12}\eta({-},x_1x_2)\eta(x_2x_1,x_{12})-\eta({-},x_1x_2)\eta(x_{12},x_{12})\\&+\eta({-},x_{12})\eta(x_1,x_2x_{12}),\\\eta(y_1,z_1)\eta({-},y_2z_2)=&-2q_{12}\eta(x_{12},x_2x_1)\eta({-},x_1x_2)-\eta(x_{12},x_{12})\eta({-},x_{1}x_2)\\&+\eta(x_1,x_2x_{12})\eta({-},x_{12}).\end{align*}
   Then,

• $u_{x_{12}}(x_{12}x_1,x_2x_{12})=-q_{12}\eta(x_2x_1,x_{12})=d_{x_{12}}(x_{12}x_1,x_2x_{12})$ .

• $u_{x_{2}x_1}(x_{12}x_1,x_2x_{12})=-\eta(x_2x_1,x_{12})=d_{x_{2}x_1}(x_{12}x_1,x_2x_{12})$ .

• $u_x(x_{12}x_1,x_2x_{12})=d_x(x_{12}x_1,x_2x_{12})=0$ for $x\neq x_{12},x_2x_1$ .

1. (iv) $y=x_{12}x_1=z$ or $y=x_2x_{12}=z$ . In these cases, we obtain zero in both sides of (5.1) and so $u_x(y,z)=0=d_x(y,z)$ .

deg(y) + deg(z) = 7:

1. (i) $y=x_{12}x_1$ and $z=x_2x_{12}x_1$ :

   \begin{align*}\eta({-},y_1z_1)\eta(y_2,z_2) & = -4\eta({-},x_1x_2x_1)\eta(x_2x_1,x_2x_1)+2q_{21}\eta({-},x_1x_{12})\eta(x_2x_1,x_2x_1)\\ & \quad +2q_{21}\eta({-},x_1x_2x_1)\eta(x_2x_1,x_{12})-\eta({-},x_1x_{12})\eta(x_{12},x_2x_1)\\ &\quad +2q_{21}\eta({-},x_1x_2x_1)\eta(x_{12},x_2x_1)-\eta({-},x_1x_2x_1)\eta(x_{12},x_{12})\\ &\quad +\eta({-},x_{12}x_1)\eta(x_1,x_2x_{12})+\eta({-},x_{12}x_1)\eta(x_{12}x_1,x_2),\\ \eta(y_1,z_1)\eta({-},y_2z_2) & =\eta(x_{12}x_1,x_2)\eta({-},x_{12}x_1)+\eta(x_1,x_2x_{12})\eta({-},x_{12}x_1)\\ &\quad -2q_{12}\eta(x_{12},x_2x_1)\eta({-},x_1x_2x_1)-\eta(x_{12},x_{12})\eta({-},x_1x_2x_1)\\ &\quad -\eta(x_{12},x_2x_1)\eta({-},x_1x_{12}).\end{align*}
   Here we have

• $u_{x_{12}x_1}(x_{12}x_1,x_2x_{12}x_1)=2\eta(x_1,x_2x_{12})=d_{x_{12}x_1}(x_{12}x_1,x_2x_{12}x_1)$ .

• $u_x(x_{12}x_1,x_2x_{12}x_1)=d_x(x_{12}x_1,x_2x_{12}x_1)=0$ for $x\neq x_{12}x_1$ .

1. (ii) $y=x_2x_{12}$ and $z=x_2x_{12}x_1$ :

   \begin{align*}\eta({-},y_1z_1)\eta(y_2,z_2) & =q_{21}\eta({-},x_{12}x_2)\eta(x_2,x_{12}x_1)-2\eta({-},x_2x_1x_2)\eta(x_2,x_{12}x_1)\\ &\quad +q_{21}\eta({-},x_2x_{12})\eta(x_{12},x_2x_1)-\eta({-},x_2x_{12})\eta(x_2x_{12},x_1),\\ \eta(y_1,z_1)\eta({-},y_2z_2) & =-\eta(x_2x_{12},x_1)\eta({-},x_2x_{12})-q_{21}\eta(x_{12},x_2x_1)\eta({-},x_2x_{12})\\ &\quad +2\eta(x_2x_1,x_2x_1)\eta({-},x_2x_{12})-q_{21}\eta(x_{12},x_{2}x_1)\eta({-},x_2x_{12}).\end{align*}
   Then,

• $u_{x_2x_{12}}(x_{2}x_{12},x_2x_{12}x_1)=-\eta(x_2,x_{12}x_1)=d_{x_2x_{12}}(x_2x_{12},x_2x_{12}x_1)$ .

• $u_x(x_{2}x_{12},x_2x_{12}x_1)=d_x(x_{2}x_{12},x_2x_{12}x_1)=0$ for $x\neq x_2x_{12}$ .

$gr(y)+gr(z)=8$ :

(i) We obtain zero in both sides of (5.1), hence $u_x(y,z)=d_x(y,z)=0$ .

The other cases are very similar. Now, we have that $\eta$ verifies (5.1) and the hypothesis of the Lemma 5.8; hence $\eta$ verifies (5.2) and $e^{\eta}$ is multiplicative.

Lemma 6.9. Let $\sigma=\sigma_{\boldsymbol\lambda}$ be as in Theorem 6.5 and $\eta\in \textrm{Z}^2\!\left(\mathfrak{B}_{\mathfrak{q}},\Bbbk\right)^H$ as in (6.11).

Assume that $e^{\eta}$ is a Hopf cocycle, cohomologous to $\sigma$ . Then

(6.18) \begin{align} \lambda_1=\eta_1,\quad\lambda_2=\eta_2,\quad\eta_{121}+\eta_{212}=\lambda_{12}+2q_{12}\lambda_1\lambda_2.\end{align}

Moreover, if $\alpha\rightharpoonup\sigma=e^{\eta}$ , $\alpha\in U\!\left(\mathfrak{B}_{\mathfrak{q}}^{\ast}\right)$ , then $\alpha(1)=1$ and

(6.19) \begin{align}\begin{split}\alpha(x_1)&=\alpha(x_2)=\alpha(x_{12})=\alpha(x_2x_1)=\alpha(x_{12}x_1)=\alpha(x_2x_{12})=0,\\\alpha(x_2x_{12}x_1)&=\eta_{212}-\lambda_{12}.\end{split}\end{align}

Proof. Suppose that there exist $\alpha\in U(\mathfrak{B}_{\mathfrak{q}}^\ast)$ such that $\alpha\rightharpoonup \sigma=e^\eta$ . In first place, we may define $\alpha(1)=1$ and then Remark 5.6 gives us $\lambda_1=\eta_1$ and $\lambda_2=\eta_2$ . In particular, it follows that

• $\alpha(x_1)=\alpha(x_2)=0$ .

• $\alpha\!\left(x_ix_j\right)=0$ , for all $i,j\in \mathbb{I}_2$ with $i\neq j$ .

On the other hand, from the relations

\begin{align*}\alpha\rightharpoonup \sigma(x_i,x_jx_i)&=e^{\eta}(x_i,x_jx_i)=0, & \alpha\rightharpoonup \sigma(x_i,x_{12})&=e^{\eta}(x_i,x_{12})=0,\end{align*}

we obtain that $\alpha^{-1}(x_ix_jx_i)=0$ , so $\alpha(x_ix_jx_i)=0$ . Now we compute $\alpha(x_2x_{12}x_1)$ , expanding $\alpha\rightharpoonup \sigma(x_1,x_2x_{12})=e^\eta(x_1,x_2x_{12})$ and $\alpha\rightharpoonup \sigma(x_2,x_{12}x_1)=e^\eta(x_2,x_{12}x_1)$ . For the first,

\begin{align*}\alpha\rightharpoonup \sigma(x_2,x_{12}x_1)&=\eta_{121}\\\sigma(x_2,x_{12}x_1)+\alpha^{-1}(x_2x_{12}x_1)&=\eta_{121}\\\alpha^{-1}(x_2x_{12}x_1)&=\eta_{121}-2q_{12}\lambda_1\lambda_2.\end{align*}

For the second, we obtain

\begin{align*}\alpha\rightharpoonup \sigma(x_1,x_{2}x_{12})&=\eta_{212}\\\sigma(x_1,x_{2}x_{12})-\alpha^{-1}(x_2x_{12}x_1)&=\eta_{212}\\\alpha^{-1}(x_2x_{12}x_1)&=\lambda_{12}-\eta_{212}.\end{align*}

Hence, as $\alpha^{-1}(x_2x_{12}x_1)$ exists, then $\eta_{121}+\eta_{212}=\lambda_{12}+2q_{12}\lambda_1\lambda_2$ .

Let $\overline{\textrm{C}}\subset\textrm{Z}^2\!\left(\mathfrak{B}_{\mathfrak{q}},\Bbbk\right)^\Gamma$ be the set defined by

\begin{align*}\overline{\textrm{C}}\,:\!=\,\{\eta_i\xi_i^i+\eta_{121}\xi_{121}^2-\eta_{121}\xi_{212}^1,\eta_{121}\xi_{121}^2+\eta_{212}\xi_{212}^1 \,:\,\eta_i,\eta_{121},\eta_{212}\in\Bbbk,i\in\mathbb{I}_2\}.\end{align*}

Notice that $\eta\in \overline{\textrm{C}}$ if and only if

(6.20) \begin{align}\eta_1\eta_2=\eta_1(\eta_{121}+\eta_{212})=\eta_2(\eta_{121}+\eta_{212})=0.\end{align}

Remark 6.10. In particular, $\textrm{C}\subset \overline{\textrm{C}}$ and this inclusion is strict. However, they coincide up to cobordisms, see (6.10):

\begin{equation*}\overline{\textrm{C}}=\textrm{C}+\Bbbk\left\{\xi_{121}^2-\xi_{212}^1\right\}.\end{equation*}

We have the following characterization of Hopf cocycles that are an exponential map of a Hochschild cocycle.

Proof. Let $\eta\in \textrm{Z}^2\!\left(\mathfrak{B}_{\mathfrak{q}},\Bbbk\right)^\Gamma$ be a Hochschild cocycle as in (6.11). Assume $e^\eta$ is a Hopf 2-cocycle. We compute the values of $e^\eta$ in two ways. Namely, via the decomposition formulas in Example 3.4 and via the definition of an exponential map. We obtain a list of equations that imply that $\eta \in \overline{\textrm{C}}$ . We show below some of these calculations.

$e^{\eta}(x_2x_1,x_{12})$ : Using the decomposition formula, we obtain

\begin{equation*}e^{\eta}(x_2x_1,x_{12})=2\eta_1\eta_2-q_{12}\eta_{121},\end{equation*}

but the definition of exponential map gives

\begin{equation*}e^{\eta}(x_2x_1,x_{12})=\eta_1\eta_2-q_{12}\eta_{121}.\end{equation*}

Thus, $\eta_1\eta_2=0$ . From now on, we suppose that this relations hold.

$e^{\eta}(x_{12}x_1,x_{12}x_1)$ : We obtain

\begin{equation*}e^{\eta}(x_{12}x_1,x_{12}x_1)=q_{12}\eta_1\eta_{121}+q_{12}\eta_1\eta_{212}, \qquad e^{\eta}(x_{12}x_1,x_{12}x_1)=0.\end{equation*}

So, we get the relation $\eta_1(\eta_{121}+\eta_{212})=0$ .

$e^{\eta}(x_2x_{12},x_2x_{12})$ : Here we get

\begin{equation*}\sigma(x_2x_{12},x_2x_{12})=-q_{12}\eta_2\eta_{121}-q_{12}\eta_2\eta_{212},\qquad e^{\eta}(x_2x_{12},x_2x_{12})=0.\end{equation*}

Hence $\eta_2(\eta_{121}+\eta_{212})=0$ .

Therefore, we see that $\eta\in\overline{\textrm{C}}$ .

Conversely, suppose that $\eta\in\overline{\textrm{C}}$ . If $\eta=\eta_{121}\xi_{121}^2+\eta_{212}\xi_{212}^1$ , then $e^{\eta}$ is a Hopf cocycle by Proposition 6.7. Now, suppose that $\eta=\eta_1\xi_1^1+\eta_{121}\xi_{121}^2-\eta_{121}\xi_{212}^1$ . The case $\eta=\eta_2\xi_2^2+\eta_{121}\xi_{121}^2-\eta_{121}\xi_{212}^1$ is analogous.

We claim that, by definition, $e^{\eta}$ is given by the table

Indeed, by Remark 5.4 and a degree argument, plus considering that $\eta_1\eta_2=0$ , we see that $e^{\eta}(x,y)=\eta(x,y)$ for $\deg x+\deg y\leq4$ .

On the other hand, if $\deg x+\deg y=5$ , then

\begin{align*} e^\eta(x,y)=\eta(x,y)+\eta\!\left(x_{\underline{{(1)}}},y_{\underline{{(1)}}}\right)\eta\!\left(x_{\underline{{(2)}}},y_{\underline{{(2)}}}\right),\end{align*}

so $e^\eta(x,y)=0$ in this case. Now, when $\deg x+\deg y=6$ and $\deg x\neq\deg y$ , it follows that for each of the four possible cases, there are scalars $(a_{ij})_{i,j}$ , depending on x, y, such that the following relation holds:

\begin{align*} e^\eta(x,y)=\eta(x,y)+\sum_{i,j\in\mathbb{I}_2,i\leq j}a_{ij}\eta(x_i,x_i)\eta\!\left(x_j,x_{ij}x_j\right).\end{align*}

Thus $e^\eta(x,y)=0$ as $\eta(x,y)=0$ and $\eta\!\left(x_j,x_{ij}x_j\right)=q_{ij}\eta\!\left(x_j,x_jx_{ij}\right)=0$ . If $\deg x=\deg y=3$ , then it follows that $e^\eta(x,y)=0$ .

If $\deg x+\deg y=7$ , then we see that $e^\eta(x,y)=0$ . Finally, if $x=y=x_2x_{12}x_1$ , then it is easy to see that $\eta^{\ast k}(x,y)=0$ for $k\neq 2$ and

\begin{align*}\eta^{\ast 2}(x,y)=&\, -4\eta(x_2x_1,x_2x_1)^2 +4q_{21}\eta(x_2x_1,x_{12})\eta(x_2x_1,x_2x_1)\\ &+\eta(x_1,x_2x_{12})\eta(x_2x_{12},x_1)+2q_{21}\eta(x_{12},x_2x_1)\eta(x_2x_1,x_2x_1)\\ &-\eta(x_{12},x_{12})\eta(x_2x_1,x_2x_1)-\eta(x_{12},x_2x_1)\eta(x_2x_1,x_{12}) \end{align*}

\begin{align*} &+2\eta(x_{12}x_1,x_2)\eta(x_2,x_{12}x_1)+2q_{21}\eta(x_2x_1,x_2x_1)\eta(x_{12},x_2x_1)\\ &-\eta(x_2x_1,x_{12})\eta(x_{12},x_2x_1)-\eta(x_2x_1,x_2x_1)\eta(x_{12},x_{12})\\ &+\eta(x_2x_{12},x_1)\eta(x_1,x_2x_{12})\\ =&\, -4\eta_{121}^2+4\eta_{121}^2+\eta_{212}\eta_{121}+2\eta_{121}^2-\eta_{121}^2-\eta_{212}\eta_{121}-\eta_{121}^2\\ &+2\eta_{212}\eta_{121}+2\eta_{121}^2-\eta_{121}^2-\eta_{121}^2-\eta_{212}\eta_{121}+\eta_{121}\eta_{212}\\ =&\, 2\eta_{212}\eta_{121}.\end{align*}

Hence, recalling that $\eta_{212}=-\eta_{121}$ , we obtain that $e^\eta(x,y)=\frac{1}{2}\eta^{\ast 2}(x,y)=-\eta_{121}^2$ .

Now, let $\sigma=\sigma_{\eta_1,0,0}$ as in Theorem 6.5 and $\alpha\in U\!\left(\mathfrak{B}_{\mathfrak{q}}^{\ast}\right)$ as in (6.19). Then $\alpha\rightharpoonup\sigma_{\eta_1,0,0}(x_i,b)=e^\eta(x_i,b)$ , for any $i\in\mathbb{I}_2$ , $b\in \mathbb{B}_\mathfrak{q}$ , see Lemma 6.9. Next, we apply the decomposition formulas to compute all values of the Hopf cocycle $\alpha\rightharpoonup\sigma_{\eta_1,0,0}$ and we obtain that $\alpha\rightharpoonup\sigma=e^{\eta}$ . Therefore, $e^{\eta}$ is a Hopf cocycle.

Let $\textrm{Exp}(\textrm{C}), \textrm{Exp}(\overline{\textrm{C}})\subset Z^2(\mathfrak{B}_{\mathfrak{q}})$ be the subsets of Hopf cocycles that arise as exponential maps of elements of C and $\overline{\textrm{C}}$ , respectively. Next, we show that it is enough to consider exponential cocycles in $\textrm{Exp}(\textrm{C})$ .

Proof. One implication is trivial. Assume $\sigma\sim_\alpha e^\eta$ , $\eta\in \overline{\textrm{C}}$ . That is, $\alpha\rightharpoonup\sigma=e^\eta$ .

If $\eta=\eta_{121}\xi_{121}^2+\eta_{212}\xi_{212}^1$ , then $\eta\in \textrm{C}$ and we can take $\alpha=\epsilon$ .

Fix now $\eta=\eta_1\xi_1^1+\eta_{121}\xi_{121}^2-\eta_{121}\xi_{212}^1$ . By Lemma 6.9, (notice that $\eta_{212}=-\eta_{121}$ ) if we define $\alpha^{\prime}$ by

\begin{align*}\alpha^{\prime}(1)=1,\quad \alpha^{\prime}(x)=0, \text{ for $0<\deg x\leq 3$},\quad\alpha^{\prime}(x_2x_{12}x_1)=\eta_{121},\end{align*}

then $\alpha^{\prime}\rightharpoonup e^\eta=\sigma_{\eta_1,0,0}$ . But $\sigma_{\eta_1,0,0}=e^{\eta^{\prime}}$ and $\eta^{\prime}=\eta_1\xi_1^1\in \textrm{C}$ . Hence $\sigma\sim_\beta e^{\eta^{\prime}}$ , for $\beta=\alpha^{\prime}\ast\alpha$ .

The next theorem characterizes the cocycles that are co-homologous to an exponential map of a Hochschild cocycle.

Proof. First, notice that we can assume $\eta\in \textrm{C}$ , by Lemma 6.12.

$(a)\Rightarrow (b)$ . Combining Lemma 6.9 with equations (6.12), we obtain

\begin{equation*}\lambda_1\lambda_2=0,\qquad \lambda_1\lambda_{12}=\lambda_2\lambda_{12}=0.\end{equation*}

$(b)\Rightarrow (c)$ . By looking at the tables in Theorem 6.5 and in the proof of Lemma 6.11, we see that this implication is clear as $\sigma_{(\lambda,0,0)}=e^{\lambda\xi_1^1}$ . Similarly, $\sigma_{(0,\lambda,0)}=e^{\lambda\xi_2^2}$ . Finally, if $\eta=\lambda\xi_{212}^1$ , it follows from Lemma 6.9 that $\sigma_{(0,0,\lambda)}=e^\eta$ , as $\alpha=\epsilon$ therein.

$(c)\Rightarrow (a)$ . Clear.

We remark that $e^\beta\neq \epsilon$ , as shown in the table in the proof of Lemma 6.11.