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:

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


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?

Thank you in advance for your help!

  • 5
    $\begingroup$ You just use $S^*:H^*\to H^*$ given by $(S^*(f))(x)=f(S(x))$. $\endgroup$
    – David Hill
    Aug 16, 2019 at 17:34

1 Answer 1


For the sake of being precise, from an algebraic point of view the "dual Hopf algebra" is not constructed over the full linear dual in general, but more properly over the so called finite dual (have a look at Sweedler's Hopf algebra book).

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.

  • 1
    $\begingroup$ Thank you very much for your detailed answer! This is great. $\endgroup$
    – Lullaby
    Aug 29, 2019 at 9:53
  • $\begingroup$ $\dim(A/I)$ means vector space dimension, not Krull dimension, right? $\endgroup$ Nov 25, 2021 at 8:08
  • $\begingroup$ @CarlosEsparza Yes, it is vector space dimension. $\endgroup$ Nov 25, 2021 at 19:56

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