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I am in an automorphism quest.

Let $H$ be a quasitrianglar Hopf Algebra with R-matrix $\mathcal{R} \in H \otimes H$. I know that $\mathcal{R}_{21}^{-1 }$ is a solution of the Yang Baxter equation and $$\mathcal{R}_{21}^{-1} \Delta \mathcal{R}_{21} = \Delta^{op}.$$ Let $\cdotp$ be multiplication law on $H^{\star}$ defined as : \begin{equation} \label{eq1} \alpha \cdotp \beta = \langle \alpha_{(1)} \otimes \beta_{(1)}, \mathcal{R} \rangle \alpha_{(2)} \star \beta_{(2)} \end{equation} With $\star$ the convolution product and $\Delta^{\star} (\alpha) = \alpha_{(1)} \otimes \beta_{(2)}$ the comultiplication on $H^{\star}$ dual to the multiplication on $H$. The following exchange relation is true : $$ \beta \cdotp \alpha = \langle \alpha_{(1)} \otimes \beta_{(1)} R\rangle \langle \alpha_{(3)} \otimes \beta_{(3)} \mathcal{R}_{21}^{-1} \rangle \alpha_{(2)} \star \beta_{(2)} $$

The preceding exchange relation is invariant by the substitution $\mathcal{R}_{12} \rightarrow \mathcal{R}_{21}^{-1}$. This strongly indicates that the algebras obtained by choosing $\mathcal{R}_{12}$ or $\mathcal{R}_{21}^{-1}$ in the definition of $\cdotp$ are isomorphic.

How can I find such an isomorphism ? I would like to have an explicit formula.

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  • $\begingroup$ If $\langle.,.\rangle$ stands for some dual pairing between $H$ and $H^*$ then, isn't $\langle \alpha_{(3)} \otimes \beta_{(3)} \mathcal{R}_{21}^{-1} \rangle$ an ambiguous notation ? $\endgroup$ Oct 23, 2017 at 23:42
  • $\begingroup$ Also, do you consider $H$ to be finite dimensional? Otherwise you 'd need the restricted dual $H^\circ$. $\endgroup$ Oct 23, 2017 at 23:43
  • $\begingroup$ I forgot the comas in the expression you mentioned. Assume H is finite dimensional. $\endgroup$
    – KraKeN
    Oct 24, 2017 at 8:33

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