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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.)

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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
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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
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As Darij says, Hopf theorists consider group algebras as building blocks, so it is not quite clear what you want here. –  Mariano Suárez-Alvarez Oct 20 '11 at 22:45
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Finite dimensional Hopf algebras are (much!) harder to classify than finite groups –  Mariano Suárez-Alvarez Oct 21 '11 at 14:03
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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

3 Answers 3

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:

  1. An exterior algebra generated by homogeneous elements of odd degree if $H^\bullet(X;k)$ is finite dimensional
  2. 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

  1. $A_0=k$ (we say $A$ is connected in this case [HINT: think about motivating example above!])
  2. 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.

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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.

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Honestly I think that Wolfram/MathWorld has an excellent and comprehensive record on this. I checked a couple of times to be sure that it covers your question clearly. Indeed (1) to (6) is great intro to set the scene for Hopf Algebra then directly go to (9) to (11) which in confirmation to darij's comment is perfectly simple presented. Not sure what more one can say to this.

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