I've believed for a long time that every Lie group can be decomposed as the semidirect product of a connected Lie group and a discrete Lie group. However, in this Math Overflow thread, it is mentioned that this is not true, and that there exist obstructions to this decomposition. Unfortunately, the level of complexity in the answers goes a bit over my head.

This type of decomposition is also addressed in this thread.

With this loss of faith in my intuition, I have three questions:

  • Is it true that Lie groups do not necessarily decompose into the semidirect product of a connected group and a discrete group?
  • If this decomposition doesn't always work, then is there a particularly nice argument as to why it doesn't?
  • If this decomposition doesn't always work, then is there a nice example I can use to convince myself that my intuition is wrong?

Edit: As much as I love Qiaochu Yuan's answer, which is thorough and (as usual) a role model among Stack Exchange answers, this is not what I'm looking for. I'm aware of the general extension problem, and that not every short exact sequence splits, but my intuition has led me to think that each Lie group can be decomposed this way. I will certainly play around with his outline for a construction, but I would much prefer a more concrete answer. Once again, thank you!

Edit: As of yet, playing with the construction suggested by Qiaochu hasn't yielded anything interesting. This question is still open to answers!

  • $\begingroup$ "discrete Lie group" is redundant; every (say, countable) discrete group is canonically a ($0$-dimensional) Lie group. $\endgroup$ Sep 22, 2014 at 5:47
  • $\begingroup$ I was attempting to make the distinction between a countable discrete group and, say, $\mathbb{R}$ with the discrete topology. But yes, it is redundant if you aren't talking to a pedant. $\endgroup$ Sep 22, 2014 at 10:03
  • $\begingroup$ @Qiaochu -- What's redundant about it? A "discrete group" need not be a Lie groaup, because it might not be countable. And a "Lie group" need not have the discrete topology. $\endgroup$
    – Jack Lee
    Sep 22, 2014 at 14:30
  • $\begingroup$ @JackLee: Some people will define smooth manifolds without requiring second-countability. In Manfredo do Carmo's Riemannian Geometry, they aren't even required to be Hausdorff. Since I was first guided into differential geometry by Spivak's books, which use metric spaces, and your Introduction to Smooth Manifolds, which is awesome, I tend to reject these definitions. I believe Qiaochu subscribes to the other doctrine. $\endgroup$ Sep 22, 2014 at 15:11

3 Answers 3


This is an extended comment Qiaochu answer; so extended indeed that it is larger than its progenitor. (Warning: I largely made it up on the go, so there may be mistakes.)

First, the concrete. Consider the "mixed Heisenberg group": namely, the set of "matrices"

$\begin{pmatrix} 1 & a & c\\ 0 & 1 & b\\ 0 & 0 & 1\\ \end{pmatrix}$ with $a, b \in \mathbb{Z}$ and $c \in \mathbb{R}$. Since $\mathbb{Z} \subset \mathbb{R}$ we can consider this a subset of the real Heisenberg group. The formulas

$\begin{pmatrix} 1 & a & c\\ 0 & 1 & b\\ 0 & 0 & 1\\ \end{pmatrix} \begin{pmatrix} 1 & a' & c'\\ 0 & 1 & b'\\ 0 & 0 & 1\\ \end{pmatrix}= \begin{pmatrix} 1 & a+a' & c+c'+ab'\\ 0 & 1 & b+b'\\ 0 & 0 & 1\\ \end{pmatrix}$ and $\begin{pmatrix} 1 & a & c\\ 0 & 1 & b\\ 0 & 0 & 1\\ \end{pmatrix}^{-1}= \begin{pmatrix} 1 & -a & ab-c\\ 0 & 1 & -b\\ 0 & 0 & 1\\ \end{pmatrix}$ convince us that this is a subgroup.

Now this is pretty clearly a Lie subgroup (it's a closed submanifold; you can check things explicitly or apply Cartan's theorem (an overkill in this case)).

It is a central extension of $\mathbb{Z}^2$ by $\mathbb{R}$ and is not a direct product, as in Quiaochu's answer.

Let's spell out the check: the element of $H^2(\pi_0(G), G_0)$ Qiaochu referred to has a representative constructed by choosing a set-theoretic section $s$ (in this case there is an obvious one, sending $(a, b)$ to $\begin{pmatrix} 1 & a & 0\\ 0 & 1 & b\\ 0 & 0 & 1\\ \end{pmatrix}$), and measuring its failure to be a group homomorphism $\phi((a,b), (a', b'))=s((a,b)) s((a',b'))s((a+a', b+b'))^{-1} \in G_0=\mathbb{R}$

In our case this is equal to $ab'$. One can fairly easily check that this $\phi$ is not $dh((a,b), (a', b'))=h((a,b))+h((a',b')-h((a+a',b+b'))$ for any $h:\mathbb{Z}^2 \to \mathbb{R}$ (plug in $a=b'=0$ and then $a'=b=0$ and compare the results), hence the class is non-trivial, so the extension is not split (the map $\phi$ changes when we change a section $s$ to $t$ by exactly $d(st^{-1})$ where $(st^{-1})$ is viewed as map $\mathbb{Z}^2 \to \mathbb{R}$; thus the class is well defined - this is what one checks to see that $H^2(Q, N)$ classifies extensions).

We can also take a "mixed Heisenberg group mod n" for an odd prime, meaning that $a,b \in \mathbb{Z}/n\mathbb{Z}$ and $c \in \mathbb{R}/n\mathbb{Z}$. The above construction still works, but everything is now compact/finite.

Now, the general. I claim that in general the construction described by Qiaochu naturally yields a Lie group - because the extension is central. Let $p: G\to \pi_0(G)$ be the projection map. Topologically, define a subset $U$ of $G$ contained in $G_b=p^{-1}(b)$ to be open if for some $y \in G_b$ the set $y^{-1}U$ is open. [Notes: 1) $x y^{-1}= y^{-1}(xy^{-1})y=y^{-1}x$, so it does not matter which side we multiply on 2) if $z \in G_b$ then $z^{-1}U= (z^{-1}y)y^{-1}U$ is open if and only if $y^{-1}U$ is open, so it does not matter which $y \in G_b$ we use.] Now declare any subset of $G$ open if its intersection with each $G_b$ is open. Around any $g\in G$ one gets a chart by composing left (or right, these are the same) multiplication by $g^{-1}$ with a chart $C: U_e \to \mathbb{R}^n$ in $G_0$ around identity. Now we need to check that inverses and multiplication are smooth; however, because $G_0$ is in the center, the maps on subsets in $\mathbb{R}^n$ induced via the above charts are equal to the maps for corresponding maps in $G_0$ (for example, using $inv$ to denote map inverse and ${-1}$ to denote group inverse, we have $C( g[(g C^{inv}(r))^{-1}])=C((C^{inv}(r))^{-1})$; etc.) So $G$ is indeed a Lie group.

Now if the extension is not central presumably bad things may happen if action of $Q$ on $N$ is not via Lie automorphisms (this seems fairly "easy" to imagine, just take any Hamel basis of $\mathbb{R}$ over $\mathbb{Q}$ and define additive map $f:\mathbb{R}\mapsto\mathbb{R}$ by permuting the basis vectors by arbitrary non-trivial permutation, thus $f \in Aut(\mathbb{R})$; now define semidirect product of $\mathbb{R}$ and $\mathbb{Z}$ using this (you can even take a permutation of order 2 and use $\mathbb{Z}/2\mathbb{Z}$ instead of $\mathbb{Z}$...). However, if the action is by $Aut(N)$ then the semidirect product is also a Lie group. It seems that for non-split extensions with such an action one should also get a Lie group structure on $G$, but I have not checked the details.


The natural discrete group one wants to use is the group $\pi_0(G)$ of connected components, and the natural connected group one wants to use is the connected component $G_0$ of the identity. These groups organize into a short exact sequence

$$1 \to G_0 \to G \to \pi_0(G) \to 1$$

and this short exact sequence refines to a semidirect product decomposition iff it admits a splitting; that is, iff there exists a group homomorphism $\pi_0(G) \to G$ which is a section of the natural map $G \to \pi_0(G)$.

In general there's no reason for such a thing to exist; that is, you should expect generically that most short exact sequences aren't split in this sense. (The difficulty is in the requirement that the map $\pi_0(G) \to G$ is required to be a group homomorphism; because $\pi_0(G)$ is discrete, it is of course straightforward to find a set-theoretic splitting.) The simplest example of a non-split short exact sequence of ordinary groups is

$$1 \to \mathbb{Z}_2 \to \mathbb{Z}_4 \to \mathbb{Z}_2 \to 1.$$

Here the problem is that the nontrivial element of $\mathbb{Z}_2$ has exactly two lifts to $\mathbb{Z}_4$, both of which have order $4$, so neither of these lifts can be extended to a group homomorphism.

I don't know a "natural" example of a Lie group with this property (e.g. I think all of the classical Lie groups are semidirect products), but here's a strategy for constructing one. First, construct a non-split short exact sequence

$$1 \to G_0 \to G \to \pi_0(G) \to 1$$

where $G_0$ is some random connected Lie group and $\pi_0(G)$ is some random discrete group, but where both are regarded with the discrete topology. You can find such things by constructing nontrivial central extensions; $G_0$ will be an abelian Lie group like $S^1$ and it will be central in $G$ (which you haven't constructed yet). Central extensions of this form are classified by second group cohomology $H^2(\pi_0(G), G_0)$, and it's not hard to find finite groups $\pi_0(G)$ with interesting nontrivial second group cohomology. This is particularly true if you take $G_0 = \mathbb{C}^{\times}$, where the corresponding cohomology group is well-studied as the Schur multiplier because it is related to projective representations; in particular there are many tables of Schur multipliers.

This constructs $G$ as an abstract group; then (this is the part where I'm least confident I can work out the details) use a set-theoretic splitting of the short exact sequence to equip $G$ with a compatible smooth structure.


Let $G$ be a locally compact abelian group such that $G_{0}$, the identity component, is open in $G$. Then $G\cong G_{0} \bigoplus G/G_{0}$. This shows that $G$ is a direct sum of a connected group and a discrete group.

  • $\begingroup$ This is prop. 23.28 in Stroppel, Locally Compact Groups. $\endgroup$
    – Watson
    Oct 18, 2018 at 9:57

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