The Drinfeld double of a finite group algebra

The group-like maps are those of [F1]; multiplicativity, coassociativity and both counit equations hold on each group basis vector, and $S(g)g=gS(g)=e$. The dual coproduct is $\mathrm{\Delta }({\delta }_x)={\sum }_{ab=x}{\delta }_a\otimes {\delta }_b$, since both sides evaluate on $(a,b)$ as ${\mathbf{1}}_{ab=x}$.

For $h=g\in G$ and dual factors ${\delta }_y$ and ${\delta }_x$, the iterated dual coproduct is ${\mathrm{\Delta }}^{(2)}({\delta }_y)={\sum }_{abc=y}{\delta }_a\otimes {\delta }_b\otimes {\delta }_c$. In [L1], the scalar factors require $a=g^{-1}$ and $c=g$, so $b=gy{g}^{-1}$; the remaining convolution ${\delta }_b\ast {\delta }_x$ is pointwise multiplication and is zero unless $b=x$. This gives the stated double product.

The dual bases of $kG$ are $(g{)}_{g\in G}$ and $({\delta }_g{)}_{g\in G}$. Substitution in [L3] gives the displayed $R$ and inverse, since $S^{-1}(g)=g^{-1}$.

Substituting $\mathrm{\Delta }({\delta }_x)$ and $\mathrm{\Delta }(g)$ into [L2] gives the stated coproduct and counit. The antipode formula becomes $S_D({\delta }_{x}g)=g^{-1}{\delta }_{x^{-1}}$; reordering by the product of step 1.2 gives ${\delta }_{g^{-1}{x}^{-1}g}{g}^{-1}$.

For any left $kG$-coaction write $\delta (v)={\sum }_{g}g\otimes v_g$, a finite sum. Coassociativity and comparison of coefficients in the group basis imply $\delta (v_g)=g\otimes v_g$, while counitality gives $v={\sum }_g{v}_g$; uniqueness of these coefficients makes the sum of the subspaces $V_g$ direct. Conversely such a grading defines a coaction. For homogeneous $v_x\in V_x$, the Yetter–Drinfeld equation at $g\in G$ is exactly $\delta (g\cdot v_x)=gx{g}^{-1}\otimes g\cdot v_x$, so compatibility is equivalent to conjugation compatibility. The braiding [L4] is therefore $c(v_x\otimes w)=x\cdot w\otimes v_x$.