The full dual of an infinite group algebra has no transpose coproduct

Every linear functional on the free vector space $kG$ is determined by arbitrary values on the basis $G$, so $H^{\ast }={\mathrm{Hom}}_k(H,k)$ identifies with all functions $G\to k$. The standard maps $\mathrm{\Delta }(g)=g\otimes g$, $\epsilon (g)=1$ and $S(g)=g^{-1}$ make $kG$ a Hopf algebra: the coproduct is multiplicative on basis products, coassociative and counital on each group element, and the two antipode products equal $e$. Multiplication $\mu :H\otimes H\to H$ therefore has transpose ${\mu }^{\ast }:H^{\ast }\to (H\otimes H{)}^{\ast }$, and ${\mu }^{\ast }({\delta }_e)(a,b)={\delta }_e(ab)$.

Suppose ${\mu }^{\ast }({\delta }_e)$ came from a tensor ${\sum }_{i=1}^n{F}_i\otimes G_i\in H^{\ast }\otimes H^{\ast }$. For each fixed $a\in G$, its row function of $b$ would be ${\sum }_{i=1}^n{F}_i(a)G_i(b)$, hence all row functions would lie in the finite-dimensional span of $G_1,\dots ,G_n$.

The actual row at $a$ is $b\mapsto {\delta }_e(ab)={\mathbf{1}}_{b=a^{-1}}$, the point mass at $a^{-1}$. As $a$ ranges over infinite $G$, these are infinitely many distinct point masses; any finite subfamily is linearly independent by evaluation at its respective support point. Their span is therefore infinite-dimensional, contradicting step 1.2. Thus ${\mu }^{\ast }({\delta }_e)\mathrm{\notin }{H}^{\ast }\otimes H^{\ast }$ and the asserted full-dual coproduct does not exist in the algebraic tensor category. This does not rule out the restricted dual $H^{\circ }$ of [F2].