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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?

edited Jan 20 2012 at 16:16

asked Jan 20 2012 at 10:53

SB
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We really need a "Counterexamples in Hopf algebras" book... – darij grinberg Jan 20 2012 at 13:36

Ralph · Accepted Answer · 2012-01-20 16:50:18Z UTC

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

$\quad\quad$On Bialgebras which are simple Hopf Modules, Amer. Math. Soc. 80(1980),563-568

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)

Commutative bialgebras: That's a result of Takeuchi (can be found in this paper), also reproved by Radford in the cited paper.
Cocummutative pointed bialgebras (see Sweedler's book "Hopf Algebras", Prop. 9.2.5).

edited Jan 20 2012 at 16:50

answered Jan 20 2012 at 16:43

Ralph
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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 2012 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 2012 at 17:50

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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 2012 at 17:55

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... 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 2012 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 2012 at 18:23

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Peter May · Answer 2 · 2012-01-21 04:27:57Z UTC

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 $R$.

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

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