Let $G$ be a group and let $k$ be a field (characteristic 0 if you want). Let $L$ be the graded Lie ring associated to the lower central series of $G$, that is, $L$, as a graded abelian group is $\oplus_{i \geq 1} G_{i}/G_{i+1}$ where $G_1 = G$, $G_i = (G_{i-1}, G)$, and the Lie bracket is induced by the commutator $(a,b)$ on $G$. Tensor $L$ with $_{\mathbb Z}k$ to get a Lie algebra $\hat L$ over $k$.

Is there any relationship between the cohomology algebra of $G$, $H^{\*}(G, k)$, and the cohomology of the (graded) Lie algebra $\hat L$, $H^{*}(\hat L, k)$?

I realize I am not being precise about what category these cohomologies are being computed in, as in graded versus ungraded. I am hoping someone can tell me there is a relationship or that there is no connection. Perhaps there are conditions on $G$ that ensure these cohomology algebras are isomorphic? References?

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    Are you assuming that $G$ is a nilpotent group? For a semisimple Lie group $G,$ the Lie algebra $L$ will be 0, but the group cohomology of $G$ is isomorphic to the Lie algebra cohomology of the Lie algebra $g$ of $G.$ If, on the other hand, $G$ is a nilpotent Lie group then from the standard spectral sequence, its group cohomology is isomorphic to the cohomology of $g,$ but extra work is required to compare it with the cohomology of $L.$ – Victor Protsak Dec 4 '10 at 2:51
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    I think you should look up "Malcev completion". Roughly, a (discrete) nilpotent group can be "tensored with the rationals"; "rational" nilpotent groups G correspond precisely to nilpotent Lie algebras g over Q; and for general G by considering the limit of nilpotent quotients you get a limit of nilpotent Lie algebras that is sort of remembering as much about this discrete group as a Lie algebra can. So your graded Lie algebra can be improved to a filtered one. Sorry to be so vague. – Tom Goodwillie Dec 4 '10 at 4:30
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    The way you get the Lie alg from the group G is: Make the group algebra over Q; make it a Hopf algebra with the coalgebra structure given by $\Delta (g)=g\otimes g$; the primitive elements ($x$ such that $\Delta(x)=x\otimes 1 + 1\otimes x$) form a Lie algebra. – Tom Goodwillie Dec 4 '10 at 4:40
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    Interesting question! I cannot even start thinking because I don't see any functor from $G$-modules to $\tilde{L}$-modules. Is there a functor? – Bugs Bunny Dec 4 '10 at 12:42
up vote 17 down vote accepted

The continuous cohomology of a group $\Gamma$ is the direct limit $$H^*_{\text{cts}}(\Gamma;\mathbb Q)=\lim_{\longrightarrow}\ H^*(\Gamma/K;\mathbb Q)$$ of the cohomology of all its finitely generated nilpotent quotients $\Gamma/K$. The basic properties of continuous cohomology are established in Hain, "Algebraic cycles and extensions of variations of mixed Hodge structure", 175–221 in Complex geometry and Lie theory, Proc. Sympos. Pure Math 53.

There is an obvious comparison map $H^*_{\text{cts}}(\Gamma;\mathbb Q)\to H^*(\Gamma;\mathbb Q)$, which is always an isomorphism on $H^0$ and $H^1$, and is always injective on $H^2$. A finitely generated group $\Gamma$ is called pseudo-nilpotent if this map is an isomorphism in every degree.

Nomizu's theorem implies that for finitely generated groups, $H^*_{\text{cts}}(\Gamma;\mathbb Q)$ coincides with the continuous cohomology $H^*_{\text{cts}}(\mathfrak{g};\mathbb Q)$ of the Malcev Lie algebra $\mathfrak{g}$ of $\Gamma$. The Malcev Lie algebra is a certain pronilpotent $\mathbb Q$–Lie algebra associated to $\Gamma$ mentioned above by Tom Goodwillie. It has the property that its associated graded $\text{gr}(\mathfrak{g})$ is isomorphic to your $\hat{L}=\text{gr}(\Gamma)$, the associated graded of the group $\Gamma$. Moreover, in many (possibly all?) cases there is an isomorphism $H^*_{\text{cts}}(\mathfrak{g};\mathbb Q)\approx H^*(\hat L;\mathbb Q)$.

Some examples of pseudo-nilpotent groups—that is, groups for which the cohomology of the group and its associated Lie algebra coincide—are free groups, fundamental groups of Riemann surfaces, and pure braid groups. (This property is closely related to the property of a space $X$ being a rational $K(\pi,1)$, in the sense that the localization-at-0/rationalization of $X$ is aspherical.) Definitely not all groups have this property, however. For example, the reason that the pure braid group is pseudo-nilpotent is that it is the fundamental group of the complement of a particularly nice hyperplane arrangement. But without some condition on the arrangement, Falk showed that there are aspherical-hyperplane-complement-groups that are not pseudo-nilpotent.

I recommend reading Hain's excellent article for more information; unfortunately I could never find it online, though you can read snippets on Google Books. In the interest of full disclosure: much of this answer was taken from my paper "Representation theory and homological stability", with Benson Farb (arXiv:1008.1368, pages 61-62).

Yes, there is a connection. The cohomology of the Lie algebra is connected to the cohomology of the group via a spectral sequence.

I'm going to assume $k$ is a field of characteristic $p \geq 0$. Then it is a result of Lazard (Sur les groupes nilpotents et les anneaux de Lie, Ann. Sci. Ecole Norm. Sup. (3), 1954, 71, 101-190) that your Lie algebra L is a $p$-restricted Lie algebra over the field $\mathbb{F}_p$, where $\mathbb{F}_p=\mathbb{Z}$ if $p=0$. If $p=0$, then $L$ is a Lie ring over the integers.

Now let $I$ be the augmentation ideal of the group ring $kG$. We can filter the group ring by the powers of $I$, and get the associated graded ring $\text{gr } kG = \bigoplus_{n=0}^\infty I^n/I^{n+1}$. The associated graded ring inherits from $kG$ the structure of a Hopf algebra. It is a result of Quillen (On the associated graded ring of a group ring, J. Algebra, 1968, 10, 411-418) that $\text{gr } kG$ is isomorphic as a Hopf algebra to $u(L \otimes_{\mathbb{Z}} k)$, the $p$-restricted enveloping algebra of $L \otimes_{\mathbb{Z}} k$. (If $p=0$, then it is just the usual universal enveloping algebra, I think.)

Now, there is a spectral sequence connecting the cohomology of the associated graded ring $\text{gr } kG$ to that of $kG$: $E_1^{i,j} = H^{i+j}(\text{gr }kG,k)_{(i)} \Rightarrow H^{i+j}(kG,k)$. For the construction of this spectral sequence, you can see Section 3 of the paper Complexity for modules over finite Chevalley groups and classical Lie algebras by Lin and Nakano (Invent. Math., 1999, 138 (1), 85-101). That paper also contains some applications in the special case when $G$ is a finite group of Lie type of a certain kind, or is the $p$-Sylow subgroup of such.

Addendum: This last bit is something of an attempt to address Bugs Bunny's comment. Given a $kG$-module $M$, we can form the associated graded module $\text{gr }M = \bigoplus_{n=0}^\infty (I^n.M)/(I^{n+1}.M)$. Then $\text{gr }M$ is a graded $\text{gr }kG$-module, so by restriction a module for $L \otimes_{\mathbb{Z}} k$. Then you get a spectral sequence looking like $E_1^{i,j} = H^{i+j}(\text{gr }kG,\text{gr }M)_{(i)} \Rightarrow H^{i+j}(kG,M)$.

  • Thanks very much everyone for the responses. The references look interesting. Best, Pete – Pete Goetz Dec 6 '10 at 0:11
  • Concerning the old paper by Lazard, it's freely available at www.numdam.org (a quick search for "Lazard" returns this and various other papers by him). Minor quibble: the ring of integers isn't a "field" when $p=0$. – Jim Humphreys Jan 4 '11 at 17:58
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    A correction to this answer: actually the Lie algebra in the original question is NOT what shows up here when p > 0. Rather, one must replace L (the associated graded of the lower central series) by the associated graded of the p-lower central series, which satisfies the additional axiom $x\in G_i\implies x^p\in G_{p\cdot i}$. To see that this is necessary for Quillen's theorem, it suffices to consider $\Gamma=\mathbb{Z}/p^2\mathbb{Z}$ (or to look at his paper). – Tom Church Feb 5 '17 at 17:47

The following two papers seem to go, independently, more or less in line with Tom Goodwillie's comment above:

I just happened to stumble upon this old question. The checked answer refers to a 1999 paper for a spectral sequence that was studied in a 1994 paper that addresses the question at hand:

Annetta Bajer. The May spectral sequence for a finite $p$-group stops. J. Algebra 167 (1994), no. 2, 448–459.

Here is the Math Review of that paper: ``Let $k$ be a field of characteristic $p>0$, let $G$ be a finite $p$-group, and let $gr kG$ be the graded $k$-algebra associated to the filtration of $kG$ by the powers of its augmentation ideal. The May spectral sequence of the title is a spectral sequence whose $E_2$-page is $Ext_{gr kG}(k,k)$ which converges to a filtration of $H^{*}(G,k)=Ext_{kG}(k,k)$. The author shows that this spectral sequence has only finitely many nonzero differentials, and deduces a similar result for a related spectral sequence converging to a filtration of $H^*(G,M)$ for $M$ a finitely generated $kG$-module.''

The published reference for the MSS in general goes back to 1966: The cohomology of restricted Lie algebras and of Hopf algebras. J. Algebra (1966), 123--146. ([3] on my web page).

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