## How to recognize a Hopf algebra?

Suppose we are handed an algebra $A$ over a field $k$. What should we look at if we want to determine whether $A$ can or cannot be equipped with structure maps to make it a Hopf algebra?

I guess in order to narrow it down a bit, I'll phrase it like this: what are some necessary conditions on an algebra for it to be a Hopf algebra?

Thoughts so far:

The first obvious condition is that $A$ must be augmented, i.e. there must be a nontrivial character $\varepsilon : A \to k$. Since this is generally not that hard to determine if we are given the algebra in some fairly concrete way, let's suppose that $A$ is given to us with an augmentation map.

If $A$ is finite-dimensional, then $A$ must be a Frobenius algebra. But not every finite-dimensional Frobenius algebra is a Hopf algebra, e.g. $\Lambda^\bullet(k^2)$ is not a Hopf algebra if the characteristic of $k$ is not 2. And generally I am more interested in the infinite-dimensional case.

All I can come up with is this: the category of finite-dimensional $A$-modules must be a (left) rigid monoidal category. But I don't know if that is a helpful observation: given a category with a forgetful functor to finite-dimensional vector spaces over some field, how can one prove that it can't be given the structure of a braided rigid monoidal category?

And perhaps there are some homological invariants that one can look at?

To sum up, the question is:

### Question

Given a $k$-algebra $A$ and a nonzero character $\varepsilon : A \to k$, are there invariants we can look at in order to show that $A$ cannot be given the structure of a Hopf algebra?

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Actually, I learned on MO a while back that EVERY algebra has a hopf structure! Here is the paper by Agore; arxiv.org/abs/0905.2613 – B. Bischof Mar 8 2012 at 18:18
@Bischof, that 's not quite true. You cannot turn $k[x]/(x^2)$ into a Hopf algebra if the characteristic of $k$ is not two. – Mariano Suárez-Alvarez Mar 8 2012 at 18:52
@Bischof, although that paper is quite interesting, I think you have misinterpreted the results. What it says is that there is a right adjoint to the forgetful functor from Hopf algebras to algebras, i.e. there is a cofree Hopf algebra on any algebra. If $A$ is the algebra and $H(A)$ denotes the cofree Hopf algebra on $A$, then there is a natural algebra map from $H(A)$ to $A$ (corresponding through the adjunction to the identity map of $H(A)$), but this doesn't mean that $A$ itself can be given a Hopf structure. – MTS Mar 8 2012 at 21:35
You are both right, I was being fast and loose in my comment. Sorry if I mislead you. – B. Bischof Mar 9 2012 at 15:58
No worries, it inspired me to look more closely at the paper. Do you happen to know of any place where examples of these cofree Hopf algebras are described? – MTS Mar 9 2012 at 18:20
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A trivial consequence of what Vladimir says is that if $A$ is a Hopf algebra and $k$ is the trivial module (via an augmentation map $\epsilon$, then $\operatorname{Ext}_A(k,k)$ is graded commutative. It's possible to give necessary conditions for this, for example the degree one elements are graded commutative iff you can find a map $f: I^2/I^3 \to S^2(I/I^2)$ (the symmetric square) such that $fm = p$ where $I$ is the augmentation ideal, $m: (I/I^2)^{\otimes 2} \to I^2/I^3$ is multiplication and $p: (I/I^2)^{\otimes 2} \to S^2(I/I^2)$ is the natural quotient.

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Nice observation! This works for the exterior algebra: since $\Lambda(V)$ is Koszul, the Yoneda algebra is the Koszul dual, which is of course the symmetric algebra $S(V^*)$. – MTS Mar 8 2012 at 17:54
Commutativity of the Yoneda algebra cannot tell a non-Hopf bialgebra (or even a quasi-Hopf algebra) from a Hopf algebra, though. – Mariano Suárez-Alvarez Mar 8 2012 at 19:01

A homological condition that might be useful: in the Hopf case, the Yoneda algebra $Ext_A^\bullet(k,k)$ embeds into the Hochschild cohomology $HH^\bullet(A,A)$, moreover, there is a Gerstenhaber algebra structure on the Yoneda algebra, and this embedding is an embedding of Gerstenhaber algebras.