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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
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@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
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@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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2 Answers

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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
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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
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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.

Reference: this article of Marco Farinati and Andrea Solotar.

I have a feeling that it would give some information already for exterior algebras, though I don't have time to check it carefully now. Of course, to use this observation for exterior algebras, the graded commutative product from the Gerstenhaber structure (highlighted by mt in his answer) is enough. But I think that there are cases where the Lie bracket will help to settle the answer.

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