Double coalgebra, multiplicativity, and antipode

The dual coproduct is coassociative because for all $x,y,z\in H$, both iterated coproducts of $f$ evaluate to $f((xy)z)=f(x(yz))$, and its counit is $f\mapsto f(1_H)$; these are also the dual coalgebra maps in [F3]. The tensor coalgebra formula for ${\mathrm{\Delta }}_D$ is therefore coassociative with counit ${\epsilon }_D$.

By [F3], the dual Hopf algebra has antipode $S^{\ast }(f)=f\circ S$. Since $S$ is bijective, $S^{\ast }$ is bijective with inverse $\bar{f}=f\circ S^{-1}$. The same coproduct makes the opposite product a bialgebra: $\mathrm{\Delta }(f\odot g)=\mathrm{\Delta }(g\ast f)=\mathrm{\Delta }(g)\mathrm{\Delta }(f)=\sum (g_1\ast f_1)\otimes (g_2\ast f_2)=\sum (f_1\odot g_1)\otimes (f_2\odot g_2)$. The opposite unit is still ${\epsilon }_H$, and $\mathrm{\Delta }({\epsilon }_H)={\epsilon }_H\otimes {\epsilon }_H$ because ${\epsilon }_H(xy)={\epsilon }_H(x){\epsilon }_H(y)$; its counit $f\mapsto f(1_H)$ is multiplicative for $\odot$ since $(f\odot g)(1_H)=(g\ast f)(1_H)=g(1_H)f(1_H)=f(1_H)g(1_H)$ using ${\mathrm{\Delta }}_H(1_H)=1_H\otimes 1_H$. Pairing the two proposed antipode products with $x\in H$ gives $\sum ({\bar{f}}_1\odot f_2)(x)=f(S^{-1}(x_2)x_1)=\epsilon (x)f(1_H)$ and $\sum (f_1\odot {\bar{f}}_2)(x)=f(x_2{S}^{-1}(x_1))=\epsilon (x)f(1_H)$, using the inverse antipode identities obtained by applying $S^{-1}$ to the equations in [F6] and reversing products by [F4]. Thus $\bar{f}$ is the antipode of $H^{\ast \mathrm{o}\mathrm{p}}$. The tensor coalgebra restricts to these Hopf structures on the two embedded factors.

Define $\psi :H^{\ast }\otimes H\to H\otimes H^{\ast }$ by $\psi (f\otimes h)=\sum f_1(h_1)f_3(S(h_3))h_2\otimes f_2$. Pairing the dual factor of $\chi \psi (f\otimes h)$ with $x$ gives $\sum f(h_{1}S(h_2)x{h}_{4}S(h_5))h_3=f(x)h$, by the two contractions $h_{1}S(h_2)=\epsilon (h)1$ and $h_{4}S(h_5)=\epsilon (h)1$. Pairing the dual factor of $\psi \chi (h\otimes f)$ with $x$ gives $\sum f(S(h_1)h_{2}xS(h_4)h_5)h_3=f(x)h$ by the other two antipode contractions. Finite dual separation [F2] proves $\psi ={\chi }^{-1}$; equivalently, the inverse crossing relation is $fh=\sum f_1(h_1)f_3(S(h_3))h_2{f}_2$.

The coproduct is multiplicative on each embedded factor by step 1.2 and the bialgebra equations [F1]. For the mixed crossing, applying $⟨-,x⟩\otimes {\mathrm{id}}_H\otimes ⟨-,y⟩\otimes {\mathrm{id}}_H$ to ${\mathrm{\Delta }}_D(hf)$ and expanding $hf$ gives $\sum f(S(h_1)xy{h}_4)h_2\otimes h_3$. Applying the same evaluation to ${\mathrm{\Delta }}_D(h){\mathrm{\Delta }}_D(f)=\sum (h_1{f}_1)\otimes (h_2{f}_2)$ and expanding both crossings gives $\sum f(S(h_1)x{h}_{3}S(h_4)y{h}_6)h_2\otimes h_5$. The antipode identity [F6] contracts the adjacent factors $h_{3}S(h_4)$, leaving $\sum f(S(h_1)xy{h}_4)h_2\otimes h_3$, the same tensor. By [F2], finite dual evaluations separate tensors, so ${\mathrm{\Delta }}_D(hf)={\mathrm{\Delta }}_D(h){\mathrm{\Delta }}_D(f)$.

The map $S_D(fh)=S(h)\bar{f}$ reverses products on each embedded factor. Applying the inverse crossing of step 1.3 to $\bar{f}\otimes S(h)$, and using the anti-multiplicative and anti-comultiplicative identities [F4] together with $\bar{f}=f\circ S^{-1}$, gives $\bar{f}S(h)=\sum f_1(S(h_1))f_3(h_3)S(h_2){\bar{f}}_2$. This is exactly the reversal of the defining crossing $hf=\sum f_1(S(h_1))f_3(h_3)f_2{h}_2$. Hence $S_D$ reverses the mixed relation and is an anti-algebra map on the crossed product.

For simple normal tensors $fh$ and $gk$, the product is $f(hg)k$ by [L2]. The defining tensor coalgebra formula gives ${\mathrm{\Delta }}_D(f(hg)k)={\mathrm{\Delta }}_D(f){\mathrm{\Delta }}_D(hg){\mathrm{\Delta }}_D(k)$, because each displayed term is in normal order. The mixed calculation of step 2.1 gives ${\mathrm{\Delta }}_D(hg)={\mathrm{\Delta }}_D(h){\mathrm{\Delta }}_D(g)$, and the factor restrictions of step 1.2 give the remaining products. Hence ${\mathrm{\Delta }}_D((fh)(gk))={\mathrm{\Delta }}_D(fh){\mathrm{\Delta }}_D(gk)$; linearity proves multiplicativity on all of $D(H)$.

Applying ${\epsilon }_D$ to the crossing gives $\sum f_1(S(h_1))f_2(1_H)f_3(h_3){\epsilon }_H(h_2)=\sum f(S(h_1)h_3){\epsilon }_H(h_2)=f(1_H){\epsilon }_H(h)$: the dual coproduct combines the first three evaluations, and applying $\mathrm{id}\otimes \epsilon \otimes \mathrm{id}$ to ${\mathrm{\Delta }}^{(2)}(h)$ gives $\mathrm{\Delta }(h)$ before the antipode contraction [F6]. Thus the counit is multiplicative on the crossing and on both factors; the normal-product calculation of step 3.1 makes it multiplicative on all products. Also ${\mathrm{\Delta }}_D(1_D)=1_D\otimes 1_D$, since ${\mathrm{\Delta }}_{H^{\ast }}({\epsilon }_H)={\epsilon }_H\otimes {\epsilon }_H$ and ${\mathrm{\Delta }}_H(1_H)=1_H\otimes 1_H$, and ${\epsilon }_D(1_D)=1$. Therefore $D(H)$ is a bialgebra.

For ${\mathrm{\Delta }}_D(fh)=\sum f_1{h}_1\otimes f_2{h}_2$, the left antipode contraction is $\sum S(h_1)({\bar{f}}_1\odot f_2)h_2$. By step 1.2 the dual factor contracts to $f(1_H){\epsilon }_H$, and the remaining $S(h_1)h_2$ contracts to ${\epsilon }_H(h)1_D$ by [F6]. The right contraction is $\sum f_1{h}_{1}S(h_2){\bar{f}}_2$; first contract $h_{1}S(h_2)$ by [F6] and then use the other dual antipode identity of step 1.2. Both sides equal $f(1_H){\epsilon }_H(h)1_D={\epsilon }_D(fh)1_D$. Thus $S_D$ is an antipode.