# Dual of a Hopf algebra

Given is a Hopf algebra $$(H,m,\eta, \Delta, \epsilon, S)$$. We know that there is a dual notion of it, called the dual Hopf algebra on $$H^{*}$$ as a vector space. It has the natural structure of a Hopf algebra.

We know that the finite-dimensional algebra $$(H,m, \eta)$$ has the structure of a coalgebra, given by the maps:

$$m^{*}: H^{*} \rightarrow (H \otimes H)^{*}\cong H^{*} \otimes H^{*}, m^{*}(f)(a \otimes b)=f(ab),$$

$$u^{*}: H^{*} \rightarrow \mathbb{K}$$, $$u^{*}(f)=f(1_{H})$$,

for any $$f \in H^{*}$$ and $$a,b \in H$$.

On the other side the coalgebra $$(H, \Delta, \epsilon)$$ has the structure of an algebra, this is true even if $$H$$ is of infinite dimension. Its structural maps should be the following:

$$\Delta^{*}: H^{*} \otimes H^{*} \rightarrow H^{*}$$, $$\Delta^{*}(f \otimes g)(\Delta(h))=f(h_{(1)})g(h_{(2)}),$$

$$\epsilon^{*}=\epsilon(h)$$,

for any $$f,g \in H^{*}$$ and $$h \in H$$.

My question is how to define the dual notion of the antipode? How would the precise assignment look like?

• You just use $S^*:H^*\to H^*$ given by $(S^*(f))(x)=f(S(x))$. Aug 16 '19 at 17:34

The idea is the following: if $$(A,m,u)$$ is an algebra, then you can consider the biggest subspace $$A^\circ$$ of $$A^*$$ on which $$m^*$$ induces a comultiplication (i.e., such that $$m^*(A^\circ)\subseteq A^*\otimes A^*$$). Concretely, it turns out that $$A^\circ=\left\{f\in A^*\mid \ker(f)\supseteq I\text{ for an ideal }I \text{ such that }\mathrm{dim}(A/I)<\infty\right\}$$ i.e. the subspace of all those linear functionals that vanish over a finite-codimensional ideal. More or less by definition, $$A^\circ$$ becomes a coalgebra, called the finite dual coalgebra of $$A$$. Explicitly, for every $$f\in A^\circ$$ $$\Delta_\circ(f)=\sum f_1\otimes f_2 \quad \text{ iff } \quad \sum f_1(a)f_2(b) = f(ab) \quad \forall a,b\in A; \\ \varepsilon_\circ(f) = f(1_A).$$ The interesting fact is that if $$(B,m,u,\Delta,\varepsilon)$$ is a bialgebra, then the convolution product $$(f*g)(a)=\sum f(a_1)g(a_2) \quad \forall\,f,g\in B^*, a\in B$$ on $$B^*$$ restricts to a multiplication $$m_\circ:B^\circ\otimes B^\circ\to B^\circ$$ and the unit $$\varepsilon$$ of $$B^*$$ actually belongs to $$B^\circ$$, so that $$(B^\circ,m_\circ,\varepsilon,\Delta_\circ,\varepsilon_\circ)$$ becomes a bialgebra as well.
If, in addition, $$B$$ admits an antipode $$S$$, then $$S^*:B^*\to B^*$$ (co)restricts to a map $$S_\circ: B^\circ \to B^\circ$$ which is still an antipode for $$B^\circ$$.
Since in the finite-dimensional case, $$B^\circ = B^*$$, you recover the content of your post plus David's comment about the aspect of the antipode.
• $\dim(A/I)$ means vector space dimension, not Krull dimension, right? Nov 25 at 8:08