Every commutative $C^*$algebra is isomorphic to the set of continuous functions, that vanish at infinity, of a locally compact Hausdorff space. Every commutative finite dimensional Hopf algebra is the group algebra of some finite group. Does there exist a characterisation of the finitely generated commutative Hopf algebras that arise as group algebras?

$\begingroup$ One could say that having a basis consisting on grouplike elements is precisely what you want, but I reckon you are after a more refined characterization... $\endgroup$– javierDec 15, 2009 at 22:20

1$\begingroup$ Don't you mean "Every cocommutative finite dimensional Hopf algebra is the group algebra of some finite group."? $\endgroup$– Ben Webster ♦Dec 15, 2009 at 23:54

1$\begingroup$ @Kevin: In context, John may have meant finitedimensional Hopf algebras over $\mathbb{C}$. And I suspect that Ben was not tripped up by anything other than the abbreviated framing of the question. For one, the question's formalism is indeed dual to what I usually take to be a group algebra. $\endgroup$– Greg KuperbergDec 16, 2009 at 1:29

1$\begingroup$ Kevin You'll note, I made no claims about whether what I wrote was true. But with "commutative" in it, that claim is quite false, whereas with "cocommutative" it's morally correct, if with a caveat or two. $\endgroup$– Ben Webster ♦Dec 16, 2009 at 16:05

4$\begingroup$ I'm with Ben. "Group algebra" definitely means the algebra of finite linear combinations of group elements; functions on the group is something else. The way to measure the difference is whether the corresponding functor to Algebras is co or contravariant. $\endgroup$– Theo JohnsonFreydDec 17, 2009 at 5:53
6 Answers
Since the questioner starts off asking about $C^*$ algebras, I am going to assume that he only cares about Hopf algebras over $\mathbb{C}$. Every finitely generated, commutative $\mathbb{C}$Hopfalgebra is the polynomial functions on an algebraic group $G$. As Ben says, we can just take $G$ to be Spec of the Hopf algebra, and then the comultiplication gives a group structure on this Spec.
There are several ways that working over $\mathbb{C}$ makes things nicer than working over arbitrary fields. Other answers have pointed them out but, IMO, have done so in a way that makes things sound more confusing. Let me instead point out how nice algebraic groups over $\mathbb{C}$ are:
 You might worry that $G$ would not be reduced. This happens over fields of characteristic $p$, but not characteristic zero. See my earlier question. Also, you might worry that $G$ has singularities, but it doesn't. In short, $G$ is a complex Lie group.
 Over nonalgebraically closed fields $k$, the behavior of the $k$points of $G$ may not determine the behavior of $G$. For example, let $X = \operatorname{Spec} \mathbb{R}[x]/(x^n1)$. Then $X$ can be equipped with the structure of an algebraic group of order $n$; the coproduct is $x \mapsto x \otimes x$. Intuitively, you should think $x_1 \times x_2 \mapsto x_1 x_2$, so this group is the $n$th roots of $1$ under multiplication. By the Nullstellensatz, this can't happen over $\mathbb{C}$. The $\mathbb{C}$points will be dense (in both the ordinary and Zariski topologies) and any map will be determined by what it does to the $\mathbb{C}$points.

$\begingroup$ Yes, my question was about $\mathbb{C}algebras. But it's probably a little late to change it now. $\endgroup$ Dec 17, 2009 at 15:34
Every finitely generated commutative Hopf algebra is the functions on a group scheme of finite type. Just take Spec: the coproduct and antipode give you the group operations.
As Ben says, a commutative Hopf algebra $R$ is, by definition, the coordinate ring of an affine group scheme. There are also group schemes that are not affine, such as abelian varieties, so these are excluded from the question. Since the question is about algebras, let's say that the group scheme $G$ is defined over a field $k$. Then $G$ is morally, but not actually, the same as its group of $k$rational points $G(k)$. Unless $G$ is a finite group with $R = k[G(k)]^*$, the dual of the usual group algebra, there are three possible differences between $R$ and a (dual) group algebra.
First, the group elements of $G(k)$ are ideals of $R$ with residue field $k$, and their group law is given by the coproduct on $R$. If $p$ is such an ideal, viewed as a point on $G$, then in general the function that is 1 on $p$ and $0$ on the rest of $G$ is not regular; it is not an element of $R$. This is one way to tell that $G$ must be 0dimensional in order for $R$ to be a literal group algebra.
Second, if $k$ is not algebraically closed, there may be other closed points in $G$ whose residue field is a field extension of $k$. If the field extension is separable, then the group law on these points is multivalued. For instance if $G = \text{GL}(n,\mathbb{R})$, then it has complex points, which correspond to complex conjugate pairs of complex matrices. The way to multiply two of these points is to multiply conjugates in all possible ways. (This is an example of making a tensor product $E \otimes_k F$ of two fields over a field, an operation that also came up in another MO question.)
Third, in characteristic $p$, $G$ may not be reduced. The simplest example is actually finitedimensional: Take the universal enveloping algebra $U(L)$ of an abelian Lie algebra, which is to say a polynomial algebra, and divide by the ideal of $p$th powers of elements of $L$ to obtain a finitedimensional Hopf algebra $u(L)$ which is a local ring. (This is similar and related to a common construction with quantum group Hopf algebras.) However, Cartier and Oort showed that algebraic group schemes in characteristic zero are reduced. You can always reduce $G$ as a scheme, but, as in the example $u(L)$, you may be throwing away everything interesting.

$\begingroup$ I am sure your usage of "complex points" is correct in the algebraicgeometry sense, but it is not the sense in which the affinealgebraicgroups community (or at least my part of it!) usually understands it. Namely, I would say that a complex point of $G = \operatorname{GL}(n)_{/\mathbb R}$ is an element of $G(\mathbb C) = \operatorname{GL}(n, \mathbb C)$, i.e., a single complex matrix, not a conjugate pair. Just a note for anyone reading the affinealgebraicgroups literature. $\endgroup$– LSpiceMay 4 at 20:40
Nobody has mentioned the important examples of graded Hopf algebras that occur in algebraic topology. For example, the cohomology of a Lie group with complex coefficients is a commutative Hopf algebra over C that is not a group algebra, or even a finite group scheme in the usual sense. Of course, it is graded, so perhaps is not relevant to the original question, but these are so important it seems worthwhile to mention them.

1$\begingroup$ You could also call them "Hopf superalgebras", and there is the interesting phenomenon that a Hopf superalgebra is not strictly a Hopf algebra. The bialgebra axiom changes in the category of supervector spaces = $\mathbb{Z}/2$graded vector spaces. Cohomology rings are also supercommutative. Anyway, supercommutative Hopf superalgebras are a lot like nonreduced group schemes. (You probably know all this, Mark, but just to relate it to the other answers.) $\endgroup$ Dec 16, 2009 at 15:29

$\begingroup$ They're the functions on a super group scheme. $\endgroup$– Ben Webster ♦Dec 16, 2009 at 16:02
Maybe this can be of use: James S. Milne, Algebraic Groups, Lie Groups, and their Arithmetic Subgroups. Chapter I of this text considers affine algebraic groups; these are in onetoone contravariant correspondence with finitely generated commutative Hopf algebras. Hopf algebras that happen to be group algebras correspond to diagonalizable algebraic groups; these are considered in I.12. Unfortunately, I don't see any results on how to recognize a diagonalizable algebraic group, but maybe this viewpoint can help you in googling.
I am going to interpret the (last sentence of the) question as asking what group algebras are finitely generated and commutative.
So, let $G$ be a group and $k$ a field such that the group algebra $kG$ is finitely generated and commutative. Then of course $G$ is commutative. Moreover, $kG$ is of finite GelfandKirillov dimension and noetherian, so theorem 5.1.5 in [Jespers, Eric; Okniński, Jan. Noetherian semigroup algebras. Algebras and Applications, 7. Springer, Dordrecht, 2007. x+361 pp.] implies that $G$ is finitely generated. Since we know all finitely generated abelian groups, and their group algebras are plainly finitely generated commutative Hopf algebras, this provides the characterization we wanted.