Finite double and its canonical quasitriangular structure

By [F1], ${\sum }_i{f}_i\otimes h_i$ represents ${\mathrm{id}}_H$, so it is independent of the chosen dual bases. Applying the embeddings $H^{\ast }\to D(H)$ and $H\to D(H)$ in the two tensor factors preserves this independence and gives the displayed $R$.

Applying $S^{-1}$ to the two antipode equations, and reversing products by the anti-multiplicativity result [F5], gives $\sum x_2{S}^{-1}(x_1)=\epsilon (x)1_H=\sum S^{-1}(x_2)x_1$. Multiplying $R$ by the proposed inverse, the first double factor uses $f_i\odot f_j=f_j\ast f_i$ and the second uses $h_i{S}^{-1}(h_j)$. Pairing the first dual factor with $x\in H$ gives ${\sum }_{i,j}{f}_j(x_1)f_i(x_2)h_i{S}^{-1}(h_j)=\sum x_2{S}^{-1}(x_1)=\epsilon (x)1_H$. In the reverse product the pairing is ${\sum }_{i,j}{f}_j(x_1)f_i(x_2)S^{-1}(h_i)h_j=\sum S^{-1}(x_2)x_1=\epsilon (x)1_H$. Finite dual separation [F1] makes both products $1_D\otimes 1_D$.

On any two left $D(H)$-modules, $(\tau R)(v\otimes w)={\sum }_i{h}_i\cdot w\otimes (f_i\otimes 1_H)\cdot v={\sum }_i{f}_i(v_{-1})h_i\cdot w\otimes v_0=\sum v_{-1}\cdot w\otimes v_0=c(v\otimes w)$ by the dual-basis identity ${\sum }_i{f}_i(x)h_i=x$. Thus $\tau R$ is the natural invertible module braiding of [L3].

Apply $D(H)$-linearity of this braiding to the regular left modules $V=W=D(H)$, whose tensor actions are supplied by [F3]. Since $c=\tau \circ L_R$, pulling the target action on $W\otimes V$ back along the flip gives the operator identity $L_R{L}_{{\mathrm{\Delta }}_D(d)}=L_{{\mathrm{\Delta }}_D^{\mathrm{o}\mathrm{p}}(d)}{L}_R$ on $D(H)\otimes D(H)$. Evaluating at $1_D\otimes 1_D$ yields $R{\mathrm{\Delta }}_D(d)={\mathrm{\Delta }}_D^{\mathrm{o}\mathrm{p}}(d)R$.

Apply the hexagon $c_{V,W\otimes U}=({\mathrm{id}}_W\otimes c_{V,U})(c_{V,W}\otimes {\mathrm{id}}_U)$ from [L3] to three regular modules with their tensor actions [F3]. Substituting $c=\tau R$ and tracking the initial order $V,W,U$ makes its two crossing operators $R_{12}$ followed by $R_{13}$; hence their combined operator is $R_{13}{R}_{12}$. The left side is the block flip applied to $(\mathrm{id}\otimes {\mathrm{\Delta }}_D)R$, with the fixed reassociation, so [F2] gives ${\alpha }_{D,D,D}^{-1}(\mathrm{id}\otimes {\mathrm{\Delta }}_D)R=R_{13}{R}_{12}$. The other hexagon $c_{V\otimes W,U}=(c_{V,U}\otimes {\mathrm{id}}_W)({\mathrm{id}}_V\otimes c_{W,U})$ first crosses positions $2,3$ and then $1,3$, producing $R_{13}{R}_{23}$ and thus $({\mathrm{\Delta }}_D\otimes \mathrm{id})R=R_{13}{R}_{23}$. Evaluating both regular-module operator identities at $1_D\otimes 1_D\otimes 1_D$ proves equality in $D(H{)}^{\otimes 3}$. These are the stated leg conventions, so all three quasitriangular identities hold.