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Let $A$ be any bialgebra (associative, unital, etc.) over a ring $k$. Then among other things it has a counit $\epsilon : A \to k$, and hence an augmentation ideal $I = \ker \epsilon$, which is a Hopf ideal. Any ideal determines a filtration $$ A \supseteq I \supseteq I^2 \supseteq \dots$$ and hence an associated graded vector space $$ \operatorname{gr} A = \frac A I \oplus \frac I {I^2} \oplus \frac {I^2} {I^3} \oplus \dots $$ Since $\epsilon: A \to A/I = k$ is a morphism of bialgebras, $I$ is a "Hopf ideal", and hence $\operatorname{gr} A$ is a bialgebra. But it is graded with zero part a Hopf algebra, hence $\operatorname{gr} A$ is Hopf.

Moreover, $\operatorname{gr} A$ is generated by $I/I^2$ as an algebra, and each element of $I/I^2$ is primitive. So in particular $\operatorname{gr} A$ is generated by its primitive part, and hence is a quotient of some universal enveloping algebra of some graded Lie algebra $\mathfrak a$; moreover, $\mathfrak a$ is generated as a Lie algebra by its degree-$1$ part, which is precisely $I/I^2$. (Generically the surjection is not an isomorphism, as $A$ might be finite-dimensional but ${\rm U}\mathfrak a$ never is.)

For now, I am interested in the following special case. The ground ring $k$ is a field of characteristic $0$. $G$ is a discrete group, and $A = k\cdot G$ is its group algebra. Then I can calculate $I / I^2$ has as its basis the non-identity elements of the abelianization $G_{\rm ab}$ of $G$. $I^2/I^3$ is spanned by pairs $(g,h) \in G\times G$, modulo $(g,1) = 0$ and $(g,hk) = (g,h) + (g,k)$, and the same relations on the other side (and maybe more relations?), and the multiplication $(I/I^2)^{\otimes 2} \to (I^2/I^3)$ is $gh = (g,h)$.

Anyway, $\operatorname{gr} (k\cdot G)$ feels a lot like some homological construction, but I don't know much homology theory. So:

Question: What's a hands-on description of $\operatorname{gr} (k\cdot G)$? How does it relate to other constructions I might have met?

Update: I definitely made an error in the above, which just means that I understand less than I thought. I'd like to explain an example, and then ask a second, more precise version of the above question.

Suppose that $G$ is a abelian. Then $k\cdot G$ is a commutative cocommutative Hopf algebra, and so is the algebra of functions on some affine algebraic group, which for want of a better name I'll call $G^\vee$ --- it's the "dual group" to $G$. The augmentation ideal then corresponds to the identity element $e\in G^\vee$, and the filtration is the filtration of the algebra of functions on $G^\vee$ in Taylor series. Then $\operatorname{gr}(k G)$ is the symmetric algebra of ${\rm T}^*_e G^\vee$. (If you complete at the augmentation ideal, you're writing down "formal power series near $e$".)

For example, when $G$ is the group with two elements and $\operatorname{char}(k) = 0$, then $I^n = I$ for $n>0$, and so $\operatorname{gr}(kG) = k$ is one-dimensional, not two-dimensional like $kG$. I was confused in my original question, because there are two kinds of filtrations on a vector space --- going up and going down --- and in one of them $\operatorname{gr}$ preserves dimensions, and in the other it may not.

So this shows that I may have been wrong when I wrote "Generically the surjection is not an isomorphism". Note that in the abelian case, $\operatorname{gr}(kG)$ is a symmetric algebra, and in particular is a universal enveloping algebra (the Hopf structure is the right one).

In the nonabelian case, I can't use quite as much geometric language, but I expect something similar should still be true:

Updated question: Is $\operatorname{gr}(kG)$ a universal enveloping algebra? If so, how is the corresponding Lie algebra related to $G$ (which remember is just a discrete group)?

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I've seen lots of notations for the group algebra ($kG$, $k(G)$, $k[G]$, $k*G$, all with their problems) This is the first time I've seen $K\cdot G$ :) Nice question, btw. –  Mariano Suárez-Alvarez Feb 14 '11 at 17:01
Do you know about Jennings's work? He gives a combinatorial description of the ranks of the graded pieces in the case that G is a p-group. I suspect this is not the case you're interested in, but it may provide some leads. –  Cam McLeman Feb 14 '11 at 17:23
@Cam: No, I don't. There seem to be lots of Jennings s --- can you give a more detailed reference? @Mariano: I have no objection to $kG$. $k\ast G$ is too melodramatic for me ($\ast$ is a fancy symbol). I don't like $k(G)$ or $k[G]$, because half the time they mean the algebra of functions on $G$, and I want the covariant one. In general, I use $\cdot$ for copower constructions, and this is $\coprod_{g\in G} k$, so it's like "$k$ times $G$". –  Theo Johnson-Freyd Feb 15 '11 at 2:37
@Theo: "The structure of the group ring of a p-group over a modular field" by S. A. Jennings (1940) is the original, but it took me a long time to get over the notation (capital $G$ is an element of a group...oy vey!). A more modern account of the same material, where the results just to by "Jennings's Theorem", is given in Dixon et al, "Analytic pro-p Groups." –  Cam McLeman Feb 16 '11 at 13:49
@Theo: in characteristic zero you can use the structure theorem proved by Sweedler in his Chapter 13 stating that a cocommutative pointed Hopf algebra over a field of characteristic zero is the smash product $U(P(H))\# kG(H)$, with $P(H)$ the lie algebra of primitive elements, and $G(H)$ the group of grouplikes. In your case with $H=\operatorname{gr}kG$, $G(H)$ is trivial. –  Mariano Suárez-Alvarez Feb 17 '11 at 4:10

2 Answers 2

up vote 5 down vote accepted

Daniel Quillen has answered this in

Quillen, Daniel G., On the associated graded ring of a group ring. J. Algebra 10 1968 411–418.

From the Mathematical Reviews by J. Knopfmacher:

"Let $KG$ denote the group algebra of a group $G$ over a field $K$ of characteristic $p$, and let $KG$ be filtered by the powers of its augmentation ideal. Then the author's main theorem states that the associated graded algebra of $KG$ is isomorphic to the universal enveloping algebra of the $p$-Lie algebra $\text{gr}^pG\otimes_ZK$, where $\text{gr}^pG$ is the graded $p$-Lie algebra defined by the $p$-lower central series of $G$. The proof of this interesting result is not obvious, and involves theorems of M. Lazard [Ann. Sci. École Norm. Sup. (3) 71 (1954), 101--190; MR0088496 (19,529b)]. As one corollary, it follows that, if $G$ and $G'$ are two finite $p$-groups whose group algebras over $Z/pZ$ are isomorphic, then $\text{gr}^pG\cong\text{gr}^pG'$."

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Awesome. For my and other readers' benefit, I'll mention that in the last 40 years, some vocabulary has changed --- Quillen writes "characteristic 1" when I would say "characteristic 0"; and the paper does cover this case. –  Theo Johnson-Freyd Feb 18 '11 at 19:38

One explicit result is the following: if $G$ is finitely generated torsion-free nilpotent, then there is an associated Lie ring $\Gamma$ (the associated graded object to $G$ and its lower central series, which is a Lie ring with commutator induced from the commutator of $G$) Then the completion of $\mathbb CG$ at the augmentation ideal is isomorphic to the completion of the enveloping algebra of $\mathbb C\otimes_\mathbb Z\Gamma$ (a complex Lie algebra, now) at its augmentation ideal.

On the other hand, in characteristic zero you can use the structure theorem proved by Sweedler in his Chapter 13 stating that a cocommutative pointed Hopf algebra $H$ over a field of characteristic zero is the smash product $\mathcal U(P(H))\\# kG(H)$, with $P(H)$ the Lie algebra of primitive elements, and $G(H)$ the group of grouplikes. In your case with $\operatorname{gr}kG$, $G(H)$ is trivial.

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What about the non nilpotent case? presumably we're getting [a quotient of, in the non-torsion-free case] the enveloping algebra of the Lie algebra of the pro-algebraic completion of your group? –  David Ben-Zvi Feb 14 '11 at 18:24

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