# monoid bialgebra is a Hopf algebra if and only if monoid is a group

Assume that $H$ is a Hopf algebra with comultiplication $\Delta$, counit $\varepsilon$ and antipode $S$. It is well known, that if $c\in H$ and $\Delta(c)=\sum\limits_{i=1}^{n}a_{i}\otimes b_{i}$, then $\sum\limits_{i=1}^{n}S(a_{i})b_{i}=\varepsilon(c)1=\sum\limits_{i=1}^{n}a_{i}S% (b_{i})$ (actualy, this condition defines the antipode), where on the left and right side we have multiplication in $H$.

Now let $G$ be a monoid and $k$ a field. It is well known that $kG$ is a bialgebra (please, see parent object for details), but one may ask, when $kG$ is a Hopf algebra? We will try to answer this question.

A monoid bialgebra $kG$ is a Hopf algebra if and only if $G$ is a group.

Proof. ,,$\Leftarrow$” If $G$ is a group, then define $S:kG\to kG$ by $S(g)=g^{-1}$. It is easy to check, that $S$ is the antipode, thus $kG$ is a Hopf algebra.

,,$\Rightarrow$” Assume that $kG$ is a Hopf algebra, i.e. we have the antipode $S:kG\to kG$. Then, for any $g\in G$ we have $S(g)g=gS(g)=1$ (because $\Delta(g)=g\otimes g$ and $\varepsilon(g)=1$). Here $1$ is the identity in both $G$ and $kG$. Of course $S(g)\in kG$, so

 $S(g)=\sum_{h\in G}\lambda_{h}h.$

Thus we have

 $1=\big{(}\sum_{h\in G}\lambda_{h}h\big{)}g=\sum_{h\in G}\lambda_{h}hg.$

Of course $G$ is a basis, so this decomposition is unique. Therefore, there exists $g^{\prime}\in G$ such that $\lambda_{g^{\prime}}=1$ and $\lambda_{h^{\prime}}=0$ for $h^{\prime}\neq g^{\prime}$. We obtain, that $1=g^{\prime}g$, thus $g$ is left invertible. Since $g$ was arbitrary it implies that $g$ is invertible. Thus, we’ve shown that $G$ is a group. $\square$

Title monoid bialgebra is a Hopf algebra if and only if monoid is a group MonoidBialgebraIsAHopfAlgebraIfAndOnlyIfMonoidIsAGroup 2013-03-22 18:58:51 2013-03-22 18:58:51 joking (16130) joking (16130) 4 joking (16130) Theorem msc 16W30