Finite Drinfeld double: convention and candidate multiplication

Let $k$ be a field and let $H$ be a finite-dimensional Hopf algebra over $k$ with bijective antipode $S$. Write $H^{\ast }={\mathrm{Hom}}_k(H,k)$. Its dual coproduct is the unique map ${\mathrm{\Delta }}_{H^{\ast }}:H^{\ast }\to H^{\ast }\otimes H^{\ast }$ satisfying

${\mathrm{\Delta }}_{H^{\ast }}(f)(x\otimes y)=f(xy)$

for all $f\in H^{\ast }$ and $x,y\in H$; finite tensor duality ensures this is an algebraic tensor. The ordinary convolution product on $H^{\ast }$ is $(f\ast g)(x)=\sum f(x_1)g(x_2)$. We use the opposite algebra $H^{\ast \mathrm{o}\mathrm{p}}$, whose product is $f\odot g:=g\ast f$ and whose coproduct is the ordinary dual coproduct, not its opposite.

As a vector space set $D(H):=H^{\ast \mathrm{o}\mathrm{p}}\otimes H$ and write $fh$ for $f\otimes h$ in this normal order. Define the crossing map

$\chi (h\otimes f)=\sum f_1(S(h_1))f_3(h_3)f_2\otimes h_2\mathrm{.}$

The candidate product and unit are

$(f\otimes h)(g\otimes k)=\sum g_1(S(h_1))g_3(h_3)(g_2\ast f)\otimes h_{2}k,1_D={\epsilon }_H\otimes 1_H\mathrm{.}$

The appearance of $g_2\ast f$ is the $H^{\ast \mathrm{o}\mathrm{p}}$ product $f\odot g_2$. This item defines the bilinear candidate operation; associativity, the coalgebra structure and the antipode are supplied separately by Double crossing laws and associativity ↗ and Double coalgebra, multiplicativity, and antipode ↗.

The finite-dimensional and bijective-antipode assumptions are explicit. The HH-8 supplier Finite Hopf algebras have one-dimensional integral spaces and bijective antipode discharges the latter for every nonzero finite-dimensional Hopf algebra; [A1] verifies that its nonzero hypothesis holds here. The $H^{\ast \mathrm{o}\mathrm{p}}$ convention is chosen so that direct coaction evaluation gives the left-left Yetter–Drinfeld action. No $H^{\ast \mathrm{c}\mathrm{o}\mathrm{p}}$ formula is imported.