How to think of the group ring as a Hopf algebra?

Question

Given a finite group $G$ and a field $K$, one can form the group ring $K[G]$ as the free vector space on $G$ with the obvious multiplication. This is very useful when studying the representation theory of $G$ over $K$, as for instance if $K=\mathbb{C}$, by Maschke's Theorem and Wedderburn's Theorem we can write $\mathbb{C}[G] = \bigoplus_i \mathrm{M}_{n_i}(\mathbb{C})$, and each factor corresponds to an $n_i$-dimensional irreducible representation of $G$.

However, this decomposition of the group ring doesn't remember as much as one would initially hope, for instance one has that $\mathbb{C}[D_4] \cong \mathbb{C}[Q_8]$, where $D_4$ is the dihedral group of order $8$ and $Q_8$ is the quaternion group. So one can't recover the group from the group ring in general.

One way to remedy this is by imposing more structure on the group ring $K[G]$. For instance, it is a cocommutative Hopf algebra, and one can recover the group as the set of group-like elements in $K[G]$.

Given that we have more information here to keep track of, I'm not sure what the Hopf algebra "looks like". Is there some structure theorem that tells us what the group ring looks like as a Hopf algebra, especially in terms of the representation theory of $G$?

(Any answers providing general intuition about how to think of Hopf algebras in general are more than welcome.)

I don't understand what "looks like" means here. If I remember correctly, Hopf algebraists are often happy to say that something "looks like" a group Hopf algebra, so group Hopf algebras are considered as something rather basic already. — darij grinberg, Oct 20 '11 at 21:34
What I can say is that *as a coalgebra*, a group Hopf algebra looks extremely simply: It is a direct sum of $1$-dimensional coalgebras, each of which is just given by $\Delta g=g\otimes g$ for some generator $g$ (this is, of course, the only possible form of a $1$-dimensional coalgebra). This direct sum decomposition is unique (up to the order of the addends) and gives you your group $G$ back, at least as a set (you then get the group operation from the algebra structure). — darij grinberg, Oct 20 '11 at 21:39
As Darij says, Hopf theorists consider group algebras as *building blocks*, so it is not quite clear what you want here. — Mariano Suárez-Álvarez, Oct 20 '11 at 22:45
Yes, well that's fair enough. It's just that, as someone who might be scared of general group rings might be reassured that its structure as an algebra can be easily understood with Wedderburn's Theorem, someone who's a bit scared of Hopf algebras could be reassured by some similar simple classification result. But it's become rather clear that the information will flow the other way: one can't get a good classification as finite groups are hard to classify, so in the end you instead just get to understand these particularly simple Hopf algebras in terms of the group ring, and not conversely. — Will, Oct 21 '11 at 00:38
Finite dimensional Hopf algebras are (much!) harder to classify than finite groups — Mariano Suárez-Álvarez, Oct 21 '11 at 14:03
In response to your last paragraph, I learned that one way to think about Hopf algebras in general is to think of them as those algebras whose representations behave like the representations of a finite group in certain ways. Namely, having a coproduct means that you can define the tensor product of representations and having an antipode means that you can define the dual of a representation. So Hopf algebras can be thought of loosely as "those algebras whose categories of representations have tensor products and duals". — Flounderer, Oct 25 '11 at 22:35
One way to be able to think of any finite dimensional cocommutative Hopf algebra as something that "looks like" a group algebra is by taking the dual instead. This is then a commutative Hopf algebra, which means that it is the algebra of regular functions of an affine group scheme. — Tobias Kildetoft, Sep 26 '12 at 15:53

score 11 · Answer 1 · answered Oct 28 '13 at 23:30

So first of all the group algebra $K[G]$ has, in addition to its algebra structure, a coproduct,

$$ \Delta: k[G] \rightarrow k[G] \otimes k[G] \\ g \mapsto g \otimes g $$ an antipode, $$ S: k[G] \rightarrow k[G]\\ g \mapsto g^{-1} $$ and a counit $\epsilon:k[G] \rightarrow k$ sending $g \mapsto 1$ for all $g \in G$. It is these maps that give $k[G]$ the structure of a Hopf algebgra.

As far as structure theorems go, there are several. These are particularly striking in the case of the cohomology of a H-space $X$ over a field $k$ of characteristic 0, when Hopf proved that the Hopf algebra $H^\bullet(X; k)$ is:

An exterior algebra generated by homogeneous elements of odd degree if $H^\bullet(X;k)$ is finite dimensional
A free graded-commutative algebra if each $H^n(x;k)$ is finite dimensional.

I include these results as they provide both a motivation for what a structure theorem for Hopf algebras looks like, and some historical context, as these objects were the motivation for the definition of a Hopf algebra. As far as more general structure theorems for Hopf algebras go there are some nice results of Cartier, Gabriel and Milnor-Moore. Here are two theorems:

For the first theorem note that $U(\mathfrak{g})$ denotes the universal enveloping algebra. I remark here that $k[G]$ is an example of a cocommutative Hopf algebra.

Theorem (Cartier-Gabriel) Assume that $k$ is algebraically closed, that $A$ is a cocommutative Hopf algebra. Let $\mathfrak{g}$ be the space of primitive elements, and $\Gamma$ the group of group like elements in $A$. Then there is an isomorphism of $\Gamma \ltimes U(\mathfrak{g})$ onto $A$ as Hopf algebras, inducing the identity on $\Gamma$ and on $\mathfrak{g}$.

Theorem (Milnor-Moore) Let $A = \bigoplus_{n \geq0}A_n$ be a graded Hopf algebra over $k$. Assume

$A_0=k$ (we say $A$ is connected in this case [HINT: think about motivating example above!])
The product in $A$ is commutative

Then $A$ is a free commutative algebra (a polynomial algebra) generated by homogeneous elements.

Reference: Pierre Cartier has a survey paper called "A primer of Hopf algebras" which has all of this stuff and more. There are further structure theorems for Hopf algebras but they are too hard to state without the addition of more definitions (e.g. conilpotent), which would just take too long here.

This answer is great. But the reader should be careful to understand what $g \mapsto g \otimes g$ means.. It means that $\sum_{i=1}^n a_i g_i \mapsto \sum_{i=1}^n a_i (g_i \otimes g_i)$. It does NOT mean that $\sum_{i=1}^n a_i g_i \mapsto (\sum_{i=1}^n a_i g_i) \otimes (\sum_{i=1}^n a_i g_i)$. — Maithreya Sitaraman, Sep 14 '18 at 04:44

score 1 · Answer 2 · answered May 21 '19 at 14:02

If you have a Hopf algebra $H$, then the set $G(H):=\{0\neq g\in H : \Delta(g)=g\otimes g\}$ is always a group (the operation is the product in $H$, the inverse $g^{-1}$ is $S(g)$) (*). So, if you have a Hopf algebra admiting a basis with elements $x_i$ such that $\Delta(x_i)=x_i\otimes x_i$ (that is, it "looks like a group algebra"), then $H=k[G(H)]$ IS a group algebra. The elements in $G(H)$ are called the "group-like" elements.

If in $H$ there is an element $x$ such that $\Delta x=x\otimes 1+1\otimes x$, then $x$ is called PRIMITIVE. The set of primitive elements, denoted $P(H)$ is always a Lie subalgebra of $H$ (where the bracket is the commutator, $[x,y]=xy-yx$). Thus, you have a Hopf algebra map $U(P(H))\to H$ (where $U(P(H)$ = the universal envelopping algebra of the Lie algebra $P(H)$).

An "intuitive" way of looking at Cartier-Gabriel theorem is that cocommutativity of $H$ implies that $H$ is generated by group-likes and primitives.

A generalization of primitive elements is the following: if $g_1$ and $g_2$ are group-likes, a SKEW $g_1$-$g_2$-primitive is an element $x$ such that $\Delta x=g_1\otimes x+x\otimes g_2$.

A very interesting question is when a Hopf algebra is generated by group-likes and skew primitives. This can be though as a generalization of Cartier-Gabriel theorem for the non cocomutative case. In the finite dimensional case, it is an open conjecture that all Hopf algebras are of this form (say $k=\overline k$).

(*) if $H=O(G)$, the reglar functions on an affine group $G$, then the group-like elements in $H^*$ are the algebra maps $H\to k$. (In general $H^*$ is not a Hopf algebra -only when $H$ is finite dimensional- but the notion of group-likes in $H^*$ makes sense anyway, that is, elements in $H^*$ satisfying $m^*(g)=g\otimes g\in H^*\otimes H^*\subset (H\otimes H)^*$.

score 0 · Answer 3 · answered Aug 23 '13 at 12:21

If you want to have a nice intuition about Hopf algebras, I suggest to look at the spectrum of an affine algebraic group. Every affine variety $V$ corresponds to an algebra $A$. A Hopf algebra structure on $A$ equips the variety $V$ then with a multiplication, identity and inversion.

Sideremark: I see another question implicitly here. You can recover a compact group from its representation theory. This is called Tannaka Krein duality.

How to think of the group ring as a Hopf algebra?

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