TITLE: Is a bialgebra with all group-like elements invertible a Hopf algebra?
QUESTION [14 upvotes]: We know that in a Hopf algebra all group-like elements are invertible. Is the converse also true? Here is the precise formulation of my question :
Let $B$ be a bialgebra and $GLE$ = { $g \in B ~|~ g \neq 0, \Delta (g) = g \otimes g$ } the set of group-like elements. We know that this set is a monoid and that if B has an antipode, namely if $B$ is a Hopf algebra, then $GLE$ is a group.
Now, suppose that in the bialgebra $B$, every group-like element is invertible, does $B$ then have an antipode? (it would be easy to define an antipode $S: B \rightarrow B$ on $GLE$ by $S(g) = g^{-1}$, but what about the other elements?)
If there is a known counterexample, then what would be the extra-condition required to assure the existence of the antipode?

REPLY [5 votes]: Sufficient conditions are known for the existence of an antipode on a bialgebra $A$ over a Commutative ring $R$ that do not assume that $A$ is either commutative or cocommutative. Assume that $A$ 
is graded, $A_i = 0$ for $i<0$, and $A_0$ is $R$-free.  Let $\epsilon\colon A\longrightarrow R$ be the counit (or augmentation).  Assume that $\epsilon^{-1}(1)$ is a group under the multiplication of  $A$.  For each grouplike element $g$, let $A_g = Rg \oplus \bar{A_g}$, where the set $\bar{A_g}$ of positive
degree elements of $A_g$ is the set of all elements $x\in A$ such that 
$$\psi(x) = x\otimes g + \sum x'\otimes x'' + g\otimes x$$
where the $x'$ and $x''$ are both of positive degree.  Assume that $A$ is the direct sum over $g\in GLE$ of the $A_g$.  Then $A$ has an antipode $\chi$.   If $A$ is commutative or cocommutative, 
then $\chi^2$ is the identity, but not in general otherwise.  The conditions are motivated by thinking 
about the homology of an $H$-space $X$ with coefficients in $R$.