# Is a bialgebra with all group-like elements invertible a Hopf algebra?

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?

-
We really need a "Counterexamples in Hopf algebras" book... –  darij grinberg Jan 20 '12 at 13:36

The answer is "no". A counterexample is given by Radford in Example 2 (p. 567) in the paper

He takes the coalgera $C = \mathbb{C}^\ast = Hom_{\mathbb{R}}(\mathbb{C},\mathbb{R})$ over $\mathbb{R}$. Then the tensor algebra $T(C) = \mathbb{R} \oplus C \oplus (C \otimes C) \oplus ...$ is a cocommutative bialgebra that is no Hopf algebra and has $1$ as the only group-like element (thus forming a group).

Additional conditions that imply the existence of an antipode are (over fields)

1. Commutative bialgebras: That's a result of Takeuchi (can be found in this paper), also reproved by Radford in the cited paper.

2. Cocummutative pointed bialgebras (see Sweedler's book "Hopf Algebras", Prop. 9.2.5).

-
I don't understand what is the bialgebra structure on $T(C)$. How is the identification $C = \mathbb C^*$ used? Currently, it seems that $C$ is nothing more than a 2-dimensional vector space, in which case the standard cocommutative bialgebra structure on $T(C)$ makes it into the universal enveloping of the free Lie algebra on two generators. This is certainly Hopf. –  Theo Johnson-Freyd Jan 20 '12 at 17:45
I went and looked at the paper, and so can answer my question. The algebra structure on $T(C)$ is the usual associative one. The coalgebra structure, though, is not graded. Rather, one gives $C^{\otimes n}$ the tensor coalgebra structure, and $T(C)$ is a direct sum of coalgebras. –  Theo Johnson-Freyd Jan 20 '12 at 17:50
Incidentally, there is a cute geometric interpretation of this construction. Let $X$ be a space. Then there is a monoid of "words in $X$", given by $T(X) = \bigcup_n X^{\times n}$, with "concatenation" as the monoid operation. This is clearly the construction here, because $\oplus$ of coalgebras is the categorical coproduct, and $\otimes$ is the categorical product. And because the category of cocommutative coalgebras is very much like a category of "spaces". The coalgebra $C$ corresponds to what the algebraic geometers call a "$\operatorname{Spec}(\mathbb C)$". The grouplikes ... –  Theo Johnson-Freyd Jan 20 '12 at 17:55
... in a coalgebra are precisely the "$\mathbb R$-points" of the corresponding space. So here the idea is that if $X$ has no $\mathbb R$-points, then $T(X)$ has only one $\mathbb R$-point (the identity). On the other hand, if you test it against $X$-points, then you can tell it is not a group object. Just as cocommutative coalgebras are a type of "space", cocommutative Hopf algebras are precisely the "group objects" in the category of cocommutative coalgebras. –  Theo Johnson-Freyd Jan 20 '12 at 17:57
Theo, the tensorproduct of coalgebras has a natural coalgebra structure. By iteration $C \otimes \cdots \otimes C$ is a coalgebra. Then extend linearly to the direct sum. This construction is sometimes called "the free bialgebra on C". –  Ralph Jan 20 '12 at 18:23

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 $R$.

-
Isn't the freeness assumption for $A_0$ already a consequence of $A=\oplus_g A_g$ and the invertibility of the group-likes, i.e. $A_0=R[G]$ where $G$ is the group of group-like elements ? –  Ralph Jan 21 '12 at 10:10
So this generalizes the well-known fact that a connected $\mathbb{N}$-graded bialgebra is a Hopf algebra. Nice. –  Ralph Jan 21 '12 at 10:33
Sure. I was specializing a more general result. Let $C$ be a coalgebra, $A$ an algebra. $Hom_R(C,A)$ is an $R$-algebra. Let $G(C,A)$ be the submonoid of those $f$ such that $f\eta = \eta$ and $\epsilon f = \epsilon$. One looks for sufficient conditions on $C$ and $A$ for when G(C,A)$is a group. Then one must assume$C_0$is$R$-free, since the group hypothesis refers to$A\$. –  Peter May Jan 21 '12 at 21:14