TITLE: How to recognize a Hopf algebra?
QUESTION [30 upvotes]: 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?

REPLY [7 votes]: 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.)