This question is motivated by a previous one: Conditions for a smooth manifold to admit the structure of a Lie group, and wants to be a sort of "converse".

Here I am taking an abstract group $G$ and looking for necessary conditions for it to admit the structure of a Lie group.

Of course I am thinking of structures with underlying manifold not $0$-dimensional (since every group can be given the discrete topology this wouldn't be interesting).

In the mentioned question they take a manifold, put a group structure on it and look at the constraints this imposes on its "geometric structure" (homotopy, co-homology, bundles...); so here I want to take a group, put a manifold structure over it and find constraints on its "algebraic structure" (subgroups, quotients, representation, whatever...).

In this question: Given a group $ G $, how many topological/Lie group structures does $ G $ have? there is a wonderful answer which addresses the problem of existence and uniqueness of Lie structure on a topological group.

But this is not what I'm looking for, because there it is assumed a priori that $G$ admits the structure of a topological group; then this structure is "smoothened" to obtain a Lie structure. What I want to do is to start with a $G$ without any topological structure. So my question could also be rephrased as:

Which algebraic properties distinguish Lie groups from abstract groups?

To be explicit, the no-small-subgroups property quoted in the second question is the kind of things I'm looking for.

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  • I may have an explanation for this. Now that it's been 4 years, have you come up with an explanation yourself? I'm surprised that there haven't been any answers to this. – Chickenmancer May 06 '17 at 16:16
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    @Chickenmancer Please write an answer if you can add to this question. – JoeyBF Jun 15 '17 at 23:55

1 Answers1


The question is actually quite hard. Below is a partial answer.

First of all, I am using the "textbook" definition of a manifold (including Hausdorff and 2-nd countable). Thus, for instance, every countable group has a Lie group structure (when equipped with the discrete topology). In view of this, it makes sense to restrict to connected Lie groups: Once you gave an algebraic description of connected Lie groups, a general group $G$ is isomorphic to a Lie group if and only if $G$ contains a normal subgroup $H< G$ of (at most) countable index such that $H$ is isomorphic to a connected Lie groups.

I will also work with real Lie groups.

With this in mind, I will start with few examples.

Lemma 1. ${\mathbb Q}$ (the additive group of rational numbers) can be abstractly defined as a nontrivial torsion-free divisible rank 1 abelian group.

(Recall that an abelian group $A$ is said to be divisible if for every natural number $n$ and every $a\in A$, there exists $b\in A$ such that $nb=a$ (where $nb$ means $b+b+...+b$, sum is taken $n$ times). An abelian group $A$ is said to have rank one if for any two non-neutral elements $a, b\in A$ there exist natural numbers $m, n$ such that $ma=nb$ or $ma=-nb$.)

Lemma 2. ${\mathbb R}$ (the additive group of real numbers) can be abstractly defined as the direct sum of $c$ copies of ${\mathbb Q}$ where $c$ has cardinality of continuum.

Thus, we have a definition of the additive group ${\mathbb R}^n$ as the $n$-fold direct sum of ${\mathbb R}$, but as an abstract group, it is isomorphic to ${\mathbb R}$.

Corollary. 1. The group ${\mathbb S}^1$ (the multiplicative group of unit complex numbers) can be abstractly defined as the as the quotient of ${\mathbb R}$ by an infinite cyclic subgroup. Equivalently, it is isomorphic to ${\mathbb R}\times ({\mathbb Q}/{\mathbb Z})$.

  1. A (nontrivial) group $A$ is isomorphic to a connected abelian Lie group if and only if $A$ is isomorphic to the direct product of finitely many copies of ${\mathbb S}^1$ and ${\mathbb R}$.

For instance, ${\mathbb C}^\times$, the multiplicative group of complex numbers, can be abstractly defined as the direct product ${\mathbb S}^1\times {\mathbb R}$.

I will also need a description of ${\mathbb R}$ as a field. Let $F$ be a field. Define the relation $\le$ on $F$ by: $x\le y \iff \exists z\in F, (y-x)=z^2$, i.e. $y-x$ is the square of an element of $F$. Then $F$ is isomorphic to ${\mathbb R}$ if and only if the following hold:

a. $\le$ is a total order on $F$.

b. $(F, \le)$ is an ordered field.

c. $(F, \le)$ satisfies the completeness axiom: For every nonempty subset $E\subset F$ which is bounded above, there exists the least upper bound.

Now, the natural thing to do would be to proceed to nilpotent and then solvable Lie groups since these can be described as iterated central extensions and semidirect products (starting with abelian Lie groups). However, I do not know how to do this and I suspect that this is already quite hard. The nilpotent case hinges on the following:

Let $A$ be an abelian topological group and $G$ be a topological group. Central extensions (in the category of topological groups) of the form $$ 1\to A\to \tilde{G}\to G\to 1 $$ (where $A$ embeds in $G$ as a central subgroup) are classified by continuous cohomology classes, elements of $H_{c}^2(G; A)$. Abstract central extensions are described as elements of $H^2(G; A)$. There is a natural homomorphism $$ H_{c}^2(G; A)\to H^2(G; A) $$ and the question is to understand its image (as topology of $G$ and $A$ varies!) in the case when $G$ is a nilpotent Lie group and $A$ is an abelian Lie group. It is conceivable that (as long as we do not insist on having connected Lie groups) that every such extension is realizable.

The problem becomes even harder for solvable group, so I will consider the complementary case case of semisimple Lie groups: Every simply connected Lie group is isomorphic to a semidirect product of a solvable Lie group and a semisimple Lie group.

Up to taking finite central extensions, every connected semisimple Lie group is a finite direct product of connected simple Lie groups. The latter are defined by the property that their Lie algebras are simple. This is "almost the same" as the simplicity condition as abstract groups (the difference comes again from central extensions by finitely generated abelian groups: mostly finite or infinite cyclic). Let me ignore the distinction in order to simplify the matter.

Simple Lie groups are completely classified (see e.g. this wikipedia article) into several infinite families and finitely many exceptional groups. An easy way out is to say that a simple Lie group is one from a list. This is, of course, completely unsatisfactory, so I will give an answer (of sorts) below.

First of all, every simple Lie group has an important numerical invariant, called rank (see for instance this discussion). For instance, $SL_n({\mathbb R})$ has rank $n-1$.

The theory then splits in three very different cases:

  1. $rank(G)=0$, i.e. $G$ is compact.

  2. $rank(G)=1$; for instance, $O(n,1)$ and $U(n,1)$.

  3. Higher rank: $rank(G)\ge 2$.

I have to say that I do not have anything interesting to say about compact groups, so below I will discuss an algebraic description of groups of higher rank. This feels like the "generic case" and it is hard enough.

The key thing to know is that each semisimple Lie group $G$ has a BN-pair structure:

This structure is given by a pair of subgroups $B< G, N<G$ and their intersection $H=B\cap N$. This data should satisfy the 5 axioms listed in the linked Wikipedia page. These axioms involve no topology and are purely group-theoretic.

Given this, one defines the Weyl group $W=N/H$ and its distinguished generating set $S$ ("simple generators"). The cardinality $n$ of $S$ is going to be the rank of $G$.

Many groups which are far from Lie groups will have such a BN-pair structure, we need extra axioms:

Axiom 6. The group $W$ is a finite Coxeter group; in particular, the cardinality $n$ of $S$ is finite.

(This axiom rules out Kac-Moody groups among other things.)

The Coxeter group $W$ has Coxeter-Dynkin diagram $D(W,S)$, which is a decorated graph encoding the relators of this finite group; the vertices of this graph are the simple generators $s\in S$. Let $n_1,...,n_k$ denote numbers of vertices of these connected components; thus, $n=n_1+...+n_k$. This number is the rank of $G$.

The case of simple Lie groups corresponds to $k=1$.

Theorem. (J.Tits) Suppose that each number $n_i$ is $\ge 3$. Then the group $G$ satisfying Axioms 1-6 is of "algebraic origin". Namely, there exists a division ring $F$ and a semisimple algebraic group ${\mathbb G}$ such that $G$ is isomorphic to the group of $F$-points of ${\mathbb G}$, ${\mathbb G}(F)$. (I am lying here a little bit but this little lie will be irrelevant.)

For instance, maybe ${\mathbb G}= SL_{n+1}$, then ${\mathbb G}(F)= SL_{n+1}(F)$: The group of matrices with entries in the field $F$ and unit determinant. As an example, one can take $F$ equal to the division ring of quaternions ${\mathbb H}$.

With a bit more work (one needs an extra "Moufang axiom" which I am not going to define here) this theorem also works for groups of rank 2. However, it utterly fails for groups of rank 1.

Furthermore, one can work out the ring structure of $F$ from the group-theoretic structure of $G$ (I am not going to do this either). The field requirement for $F$ amounts to solvability assumption on $B$. Now, we can use the characterization of ${\mathbb R}$ given earlier. Lastly, the groups ${\mathbb G}({\mathbb R})$ are exactly the connected semisimple real Lie groups (as mentioned above, up to taking some finite index subgroups and central extensions with finitely generated abelian kernels).

Like it or not, this is a characterization that you asked for, except for the fact that it leaves out the case of groups of rank $\le 1$ and restricts to semisimple Lie groups.

See this mathoverflow discussion for more references.

Update: As it turns out, the extension problem is not as bad as I thought. In Theorem 3 in

David Wigner, Cohomology of Topological Groups, Transactions of the American Mathematical Society, Vol. 178 (1973), pp. 83-93

it is proven that the natural map $H^*_c(G,A)\to H^*(G,A)$ is an isomorphism, where $G$ is a connected Lie group, $A$ is a simply connected abelian Lie group and the action of $G$ on $A$ is continuous. Hence, every extension of $G$ by $A$ (as an abstract group) is equivalent to an extension as a Lie group. Thus, indeed, the overall problem of characterizing Lie groups as abstract groups reduces to:

(a) The characterization problem for semisimple Lie groups which is mostly solved by the answer given above.

(b) Characterizing which abstract homomorphisms $\phi: G\to Aut(A)$ from connected Lie groups to abstract automorphism groups of abelian Lie groups (for simplicity, ${\mathbb R}^n$) are equivalent to continuous representations to the group of continuous automorphisms, i.e. continuous representatioins $$ G\to GL(n, {\mathbb R}) $$ This problem might be hard but at least it reduces to a problem of the finite-dimensional representation theory.

darij grinberg
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Moishe Kohan
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