MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is there analog of Lie algebra for the case of topological groups which are not necessarily differentiable manifolds, and in particular for finite groups? here by "analog" i mean that it should have similar kind of relations to given group as of a finite dimensional Lie algebra with its corresponding Lie group. for example there should be some analog of "Exponential map"; Every linear representation of group should also be a linear representation of its algebra, and this should be in some sense compatible with the exponential map between the two; etc


share|cite|improve this question
Something like a local group? – Qiaochu Yuan Apr 22 '12 at 17:46
There are finite Lie algebras associated with finite $p$-groups (used, for example, in the solution of restricted Burnside problem,…), and finite Lie algebras associated with finite simple Lie groups, . – Mark Sapir Apr 22 '12 at 18:17
Perhaps you want to look – Mark Sapir Apr 22 '12 at 18:22
There is the notion of a Lie algebra for an arbitrary group scheme (this includes linear algebraic groups over arbitrary fields): see – Kevin Ventullo Apr 22 '12 at 18:23
@ Qiaochu Yuan. Anything like a Lie group :-) – dushya Apr 22 '12 at 20:32

For finite $p$-groups one has the Lazard correspondance. This gives an equivalence of categories between $p$-groups of class less than $p$ ("Lazard groups") and Lie rings of class less than $p$ whose additive groups are $p$-groups. Given such a $p$-group $G$, its Lazard corespondent is a Lie ring defined on the same underlying set as $G$, with addition and bracket product defined in terms of the group operations as follows:

$$ x +_L y = x y [x,y]^{-1/2} [y,[x,y]]^{7/12}[x,[y,x]]^{5/12} [y,[y,[y,x]]]^{5/8} [y,[x,[y,x]]]^{1/2}[x,[x,[y,x]]] ^{3/8}\ldots $$

$$[x,y]_L = [x,y][y,[x,y]]^{-1/2}[x,[y,x]]^{1/2}[y,[y,[y,x]]]^{-1/3}[y,[x,[y,x]]]^{-1/4}[x,[x,[y,x]]]^{-1/3}\ldots $$

(these are related to the Baker-Campbell-Hausdorff formula). Example: for $G$ the extra-special $p$-group of order $p^3$ and exponent $p>2$, the corresponding Lie ring is the 3-dimensional Heisenberg Lie algebra over $\mathbb{F}_p$. See for example Khukhro's book on $p$-automorphisms of $p$-groups or Lazard's Sur les groups nilpotents et les anneaux de Lie.

There is a correspondence of representations here, but only in small dimension. Any characteristic $p$ rep can be thought of as a homomorphism into the group of upper unitriangular $n\times n$ matrices. For $n\leq p$ this has class $\leq p-1$, and its associated Lie ring is isomorphic, via the log map, to the Lie algebra of strictly upper triangular matrices. So functoriality gives a correspondence of representations of dimension $\leq p$.

It is worth pointing out that in characteristic p, it is not usually possible to find a Lie algebra with the same representation theory as a given group. This is because all Lie algebras have their cohomology finitely generated over the subring generated by degree two elements,while few groups have this property. However there are miraculous special cases such as the dihedral group of order eight whose modular group algebra is actually isomorphic to a universal enveloping algebra.

share|cite|improve this answer
$-1/2, 7/12,5/12,5/8,1/2,3/8.\cdots$ How does the sequence continue ? That's not obvious at all. – Todd Leason Apr 23 '12 at 8:41
I don't know if there's a known closed formula. Even for the simpler BCH formulas at it is not so easy to give one. – M T Apr 23 '12 at 8:54

This is studied by Helge Glockner in this nice paper (which has good references, too).

share|cite|improve this answer
Thanks for your answer, – dushya Apr 22 '12 at 20:39

Every locally compact group $G$ contains an open, closed, almost connected subgroup. This is a variant of van Dantzig's theorem. Pull back an open, compact subgroup in the totally disconnected group $G/G_0$ along the surjection $G \twoheadrightarrow G/G_0$, where $G_0$ is the connected component of the identity.

An almost connected, locally compact group $G'$ admits a net of normal compact subgroups $\cap N = \{ 1 \}$ and such that $G'/N$ is isomorphic to a Lie group, i.e. $G'$ is a projective limit of Lie groups:

$$ G' = \lim\limits_{\leftarrow} G'/N.$$

This is the solution to Hilbert's 5th problem.

Now, one is able to define everything via inductive and projective limits.

Edit: As far as I understand, these are the systems of proejctive Lie groups also considered in the refernce of Igor Rivin. Terrence Tao has written something about this on his blog: One personal note: You do not necessarily need to understand the proofs, to apply the theorems.

share|cite|improve this answer
In fact, this is where the terminology of smooth functions on totally disconnected groups or smooth representation of these originates from. – Marc Palm Apr 22 '12 at 19:32
Can you please suggest some reference(s) where this projective limit relation between almost connected, locally compact groups and Lie groups has been explained/studied/used. – dushya Apr 22 '12 at 20:23
… There is a chapter appeoximation through Lie groups or the like. I think they use invariant subgroup instead of normal subgroup. Hope that helps, Best Marc – Marc Palm Apr 22 '12 at 20:32
Thank you Marc, – dushya Apr 22 '12 at 20:38

If $G$ is a finitely generated group which is torsion free nilpotent of class $n$, then $G$ is the Lie group of some $\mathbb{Z}$-Lie algebra $\mathfrak g$ which is also nilpotent of class $n$. Hence you can define an algebraic group $G(k)$ for any field $k$ by taking the exponential of $\mathfrak g\otimes_{\mathbb{Z}} k$.

Now if $G$ is a discrete group, define the rational serie as $D_i(G)=${$x \in G, x^r \in \Gamma_i(G)$ for some $r>0$ } where $\Gamma_i(G)$ is the $i$th term of the lower central serie. Therefore, by construction $G/D_i(G)$ is torsion free nilpotent of class $i$, hence you can associate to it a Lie algebra $\mathfrak g_i(k)$. Now define $\mathfrak g(k)$ as the inverse limit of the $\mathfrak g_i(k)$. it is called the Malcev Lie algebra of $G$. It is a complete, separated pro-nilpotent Lie algebra. Set $G(k)=\exp(\mathfrak g(k))$, it is a pro-unipotent group coming with a morphism $G \rightarrow G(k)$ which is universal for this property. Note that if $\bigcap_{i\geq 0} D_i(G)$ = {1} (such a group is called residually torsion free nilpotent) this morphism is injective, so it may happen that $G(k)$ capture a lot of things about $G$.

Indeed every representation of $\mathfrak g(k)$ extends to a representation of $G(k)$ just by taking the exponential, and therefore to a representation of $G$. Conversly, every $k$-(pro-)unipotent representation of $G$ induces a representation of $\mathfrak g(k)$.

Note that $\mathfrak g(k)$ is a rather complicated object, so it doesn't seems to help a lot. But in many interesting case, $\mathfrak g(k)$ is isomorphic as a filtered Lie algebra to an "easy to handle" graded Lie algebra (namely to the associated graded of $G$, see Ralph's answer). In that case you really get something like the relation between a Lie group and its easier to handle Lie algebra.

share|cite|improve this answer

If $G$ is a group and $G = \gamma_{0}^p(G) \supseteq ... \supseteq \gamma_{k}^p(G) \supseteq \gamma_{k+1}^p(G)\supseteq ...$ its p-lower central series then we have the associated Lie algebra over $k := \mathbb{F}_p$:

$$\operatorname{gr}^p(G) = \bigoplus_{k\ge 0}\gamma_{k}^p(G) / \gamma_{k+1}^p(G)$$ those Lie bracket is given by the commutator map: $[\bar{x},\bar{y}] := \overline{[x,y]}$.

We can associate $G$ with another Lie algebra: Let $I$ be the augmentation ideal of $kG$. We have the graded ring $\operatorname{gr}(kG):= \bigoplus_{k\ge 0} I^k/I^{k+1}$. Then the map $$\operatorname{gr}^p(G) \to \operatorname{gr}(kG),\; \bar{x} \mapsto \overline{x-1}$$ is a homomorphism of Lie $k$-algebras.

Now by a celebrated theorem of Quillen the induced map $$U(\operatorname{gr}^p(G) \otimes_{\mathbb{Z}} k) \to \operatorname{gr}(kG)$$ is an isomorphism of $k$-algebras $(U$ denotes the universal envelope$)$.

As an application, one obtains $H^\ast(G;k) \cong H^\ast(\operatorname{gr}^p(G);k)$ if $G$ is a $p$-group. Hence one can compute group cohomology by Lie algebra cohomology in this case.

share|cite|improve this answer
There is a spectral sequence computing $H^*(G,k)$ from $H^*(\operatorname{gr}^p(G),k)$, but they won't be isomorphic as rings in general: the latter is finitely generated over its subring generated by degree two elements (e.g. Friedlander and Parshall "Geometry of p-unipotent Lie algebras" J.Alg. 109) but $H^*(G,k)$ may not have this property. If they did all periodic $kG$-modules would have period at most two, but examples of larger period are known. – M T Apr 25 '12 at 7:56
The isomorphism above is the cohomological analogue of Theorem 4.3 ii) in Grünenfelder: On the homology of filtered and graded rings, J. Pure and Appl. Algebra, 14(1979), 21-37. As homology and cohomology are dual to each other (over a field) the isomorphism is true at least additively. But I guess it is also an isomorphism of rings. I'll have a look later and will also look into the Friedlander-Parshall paper. – Ralph Apr 25 '12 at 8:33
I checked it. It's an isomorphism of $k$-vector spaces. More specifically, there is a filtration on the group cohomology such that $H^i(\operatorname{gr}^p(G);k)\cong \operatorname{gr}H^i(G;k)$, but I didn't check if the latter preserves products. – Ralph Apr 26 '12 at 17:53
In your answer above, I believe Quillen's isomorphism involves the restricted enveloping algebra of the (restricted) Lie algebra $\text{gr}^p(G)$. This is a finite-dimensional algebra, whereas the usual universal enveloping algebra is infinite-dimensional. The statement about cohomology then refers to the cohomology of $\text{gr}^p(G)$ as a restricted Lie algebra. – Christopher Drupieski Jun 22 '12 at 1:31
@Ralph, would you please give some reference to these constructions? – Zuriel Oct 31 '14 at 16:00

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.