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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:


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; – B. Bischof Mar 8 '12 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 '12 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 '12 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 '12 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 '12 at 18:20

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 '12 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 '12 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.

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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Let $I$ be the kernel of $\epsilon\colon A\to k$. Filter $A$ by the powers of $I$. If $A$ is a Hopf algebra, then the associated graded algebra $E^0A$ is a primitively generated Hopf algebra. If the characteristic of $k$ is zero, $E^0A$ is isomorphic to the universal enveloping algebra of its Lie algebra of primitive elements. By the Poincar\'e-Birkhof-Witt (PBW) theorem, the associated graded algebra of $E^0A$ with respect to its Lie filtration is isomorphic (when regraded by total degree) to the free commutative algebra on the primitive elements of $E^0A$, which is a polynomial algebra (assuming the $A$ you start out with is ungraded, or graded and concentrated in even degrees). Thus $A$ is two filtrations away from a polynomial algebra, which gives a significant restriction on the underlying vector space of $A$. If $k$ has positive characteristic $p$, the same argument applies with Lie algebra replaced by restricted Lie algebra, but of course the conclusion of the PBW theorem gives much less complete information, depending on the restriction ($p$th power operation).

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I'm a bit late, but here's a simple observation. Consider a topological version of the question: given a topological space $X$, how can we recognize when $X$ can be given the structure of a topological group? A simple necessary condition is that that $X$ must be homogeneous, and in particular each point should have homeomorphic neighborhoods, because $X$ must act transitively on itself.

An analogous statement about Hopf algebras is the following. Let $H$ be a commutative Hopf algebra over a field $k$. Then $G = \text{Spec } H$ is an affine group scheme over $k$; moreover, the group $G(k)$ acts on $G$, and this action is transitive on $k$-points. In particular the dimension of the Zariski tangent space at each $k$-point of $H$ must be the same. So any commutative algebra without this property can't be the underlying commutative algebra of a Hopf algebra. Examples are given by the ring of functions on any singular variety, such as the cuspidal cubic $k[x, y]/(y^2 - x^3)$ ($k$ a field of characteristic other than $2$ or $3$); in this case the dimension of the tangent space at any point is $1$ except at $(0, 0)$ where it's $2$.

I think this is essentially the point of David Speyer's answer, modulo some technicalities.

(A topological analogue of the observation about $\text{Ext}_A(k, k)$ being graded commutative is that if $X$ is a topological group then $\Omega X$ is an $E_2$-algebra, and in particular $\pi_1(X)$ is abelian by the Eckmann-Hilton argument.)

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Commutative finitely generated Hopf algebras over a field of characteristic zero are regular. See Oort.

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For the sake of those reading this question (including me!), could you specify the sense in which you mean "regular"? It's a pretty overloaded term... – MTS Apr 6 '14 at 2:15
Open-access paper link > closed-garden MR link: – darij grinberg Apr 6 '14 at 2:37

In the book Hopf algebras by Eiichi Abe (1980), there is a theorem on page 57-58 that reads that

Given a $k$-linear space $H$, suppose that there are $k$-linear maps $\mu:H\otimes H\rightarrow H$, $\eta:k\rightarrow H$, $\Delta:H\rightarrow H\otimes H$, $\varepsilon:H\rightarrow k$ such that $(H,\mu,\eta)$ is a $k$-algebra and $(H,\Delta,\varepsilon)$ is a $k$-coalgebra. Then the following conditions are equivalent. (i) $\mu,\eta$ are $k$-coalgebra morphisms. (ii) $\Delta,\varepsilon$ are $k$-algebra morphisms ...

and that

When a $k$-linear space $H$ together with $k$-linear maps $\mu,\eta,\Delta,\varepsilon$ satisfy one of the equivalent conditions of Theorem 2.1.1. (quote above) then $(H,\mu,\eta,\Delta,\varepsilon)$ or simply $H$ is called a $k$-bialgebra.

Since a Hopf algebra is a bialgebra with an antipodal map, if you have an algebra such that $\mu$ and $\eta$ are not coalgebra morphisms, or $\Delta,\varepsilon$ are not algebra morphisms then it would fail to be a bialgebra and thus a Hopf algebra.

It won't tell you for every algebra since there are bialgebras that are not Hopf algebras, but it is a start.

After looking into it further I found in the book Hopf Algebras: An Introduction by Sorin Dascalescu, et al. (2001), there is a remark on page 151 that reads

In a Hopf algebra, the antipode is unique, being the inverse of the element I in the algebra $Hom(H^c,H^a)$. The fact that $S:H\rightarrow H$ is the antipode is written as $S*I=I*S=\eta\varepsilon$, and using the sigma notation $\sum S(h_1)h_2=\sum h_1S(h_2)=\varepsilon(h)1$

[Note: $H^c$ is the bialgebra looking at the coalgebra structure on it, or $(H,\Delta,\varepsilon)$ from above. $H^a$ is the bialgebra looking at the algebra structure on it, or $(H,\mu,\eta)$ from above. Also I changed the names of the multiplication, comultiplication, unit, and counit maps to correspond with what I had above.]

So if you cannot find an antipodal map for your bialgebra, it fails to be a Hopf algebra. Again, this doesn't completely answer the question because I do not know how you would go about figuring this out, but it should help you take a step in the right direction.

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This still leaves open the question of finding a coassociative algebra map $\Delta$. – S. Carnahan Apr 6 '14 at 7:25
This answer might have been useful on MSE, but for MO it sounds a bit too patronising. People who ask questions about Hopf algebras here generally know the definitions... – Vladimir Dotsenko Apr 7 '14 at 7:39

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