# Example of a bialgebra which is Frobenius but not Hopf

Let $$\Bbbk$$ be a field. A well-known result (see Larson, Sweedler, An Associative Orthogonal Bilinear Form for Hopf Algebras and Pareigis, When Hopf algebras are Frobenius algebras) states that a bialgebra $$B$$ over $$\Bbbk$$ is a finite-dimensional Hopf algebra if and only if it is Frobenius as an algebra and the Frobenius morphism $$\psi:B\to\Bbbk$$ is a (left) integral on $$B$$ (i.e. $$\sum b_1\psi(b_2)=\psi(b)1_B$$ for all $$b\in B$$).

For didactical reasons, I would like to have an example of a (necessarily finite-dimensional) bialgebra $$B$$ which is Frobenius as an algebra but that is not a Hopf algebra. However, I didn't manage to construct one. Does anybody know one? Even a non-elementary one would be fine.

Alternatively, if somebody knows a proof of the fact that a Frobenius bialgebra is automatically a Hopf algebra, without further assumptions on the Frobenius homomorphism, then I would be very glad to see it.

Here is an example. Take k a field and S the multiplicative semigroup {0,1}. For notational convenience denote it S={1,x}. Consider B=k[S] the semigroup algebra.

B is bialgebra and not Hopf because S is a semigroup and nota a group.

$$B\cong k[x]/(x^2-x)\cong k\times k$$ is semisimple. In particular is a Frobenius algebra.

• Thanks, that's a nice example. For personal future reference, an explicit Frobenius coordinate system is given by $e= 1\otimes 1 - 1\otimes x - x\otimes 1$ and $\psi:B\to\Bbbk: 1\mapsto 0, x\mapsto -1$. – Ender Wiggins May 26 '19 at 16:55

Ussually finite dimension is part of the definition of a Frobenius algebra. So, what definition of "Frobenius" algebra do you have in mind?

In wikipedia (https://en.wikipedia.org/wiki/Frobenius_algebra#Category-theoretical_definition) a categorical definition is given. If you take a "Frobenius object" in the category of vector spaces, then Frobenius impies finite dimensionality.

• This is exactly the reason why I stressed finite-dimensional: I was trying to avoid un-helpful (non-)answers because I would like to see a genuine, ring-theoretic example of a bialgebra over a field that (as an algebra) is Frobenius (Casimir+Frobenius morphism or non-degenerate bilinear associative form, choose the description you prefer) – Ender Wiggins May 17 '19 at 7:26
• Just to be sure, when I read your post I thought of an infinite dimensional example, but now I read carefully and I see that what you wrote is "necessarily finite dimensional". But then, what's wrong with Larson-Sweedler-Pareigis result? You are lookin for finite or infinite dimensional examples? – Marco Farinati May 17 '19 at 11:59
• Since the additional condition on the Frobenius morphism is required to have a Hopf algebra structure, I would like to see an example of a bialgebra which is Frobenius but doesn't satisfy that condition, so that it is not a Hopf algebra. In particular, being Frobenius, it has to be finite-dimensional. – Ender Wiggins May 18 '19 at 12:08
• I think you misunderstood the statement. Larson-Sweedler proves that a finitely generated bialgebra over a p.i. ring, (e.g. a finite dimensional bialgebra over a field) a Frobenius structure provides an atipode (and viceversa). So, if a finite dimensional bialgebra $B$ is Frobenius as algebra, then it is necesarily Hopf. – Marco Farinati May 21 '19 at 13:28
• I am missing the converse: why a bialgebra which is Frobenius as an algebra should admit an antipode? Larson-Sweedler proved only one implication, from Hopf to Frobenius. Pareigis proved the converse, but under the additional assumption that the Frobenius morphism is an integral. I would like to see a proof of $B$ Frobenius $\Rightarrow$ $B$ fd Hopf or a counterexample of a Frobenius bialgebra which is not Hopf – Ender Wiggins May 21 '19 at 17:36