Let $1 \to N \to G \to H \to 1$ be a short exact sequence of topological groups. Such an exact sequence is said to be topologically split if $G$ is $N \times H$ as a topological space.

Can someone give me an example of a topologically split short exact sequence of non-discrete connected topological groups. Of course I want an example which is not a split exact sequence.

Edit: By a split exact sequence of topological groups, I mean an exact sequence of topological groups admitting a continuous section which is also a homomorphism. In other words, $G$ is a semi-direct product of $N$ by $H$.

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    $\begingroup$ What do you mean by "split exact sequence"? If it is "$G$ is the product $G\times H$", then any nontrivial semidirect product (such as Yemon Choi's answer) gives a "non-split" example. However, in this case there is always a section $H\to G$ which is a continuous homomorphism. $\endgroup$ – Laurent Moret-Bailly Apr 22 '11 at 10:13
  • $\begingroup$ Good point, Laurent - the thought crossed my mind, but in view of the author's definition of "topologically split" I assumed that "split exact" meant more than having a section in the appropriate category. $\endgroup$ – Yemon Choi Apr 22 '11 at 10:18
  • $\begingroup$ Thanks to all for your interest in the question. Yes, by split exact I meant that the exact sequence has a continuous section which is also a homomorphism. $\endgroup$ – user14595 Apr 22 '11 at 10:48
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    $\begingroup$ The answer you have accepted is "split exact" by this definition. Here's an example which isn't: clearly $0\to C_2\to C_4\to C_2\to 0$ works (with $C_n$ cyclic of order $n$) but it's discrete, so just multiply the outer two terms by the reals and the inner one by $\mathbf{R}^2$ for an example which satisfies all the conditions of the problem (as it currently stands). $\endgroup$ – Kevin Buzzard Apr 22 '11 at 11:57
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    $\begingroup$ @unknown: oh, but in that case my "answer" does not solve your problem. Having a continuous section which is also a homomorphism does not imply that $G=N\times H$ as groups, and the semi-direct product given in my answer shows you this. Could you please add an edit to your question, making it clear exactly what you mean by "split exact sequence of topological groups" $\endgroup$ – Yemon Choi Apr 22 '11 at 11:58

I think something like the real $ax+b$ group (a.k.a. the affine group of ${\mathbb R}$) ought to do the trick.

The following is not the full $ax+b$ group but may be easier to handle here. Take $$ G = \left\{\left( \matrix{ a & b \\ 0 & 1 } \right) \colon a >0, b \in {\mathbb R} \right\} $$

and take $N$ to be those matrices with $a=1$ (these correspond to translations if we view $G$ as acting by $x\mapsto ax+b$). Clearly $G/N \cong {\mathbb R}_{>0}$ as topological groups, and it is clear that $G\cong {\mathbb R}_{>0}\times{\mathbb R}$ as a topological space, but not as a topological group.

More generally, solvable Lie groups ought to give loads of examples, but I am very far from expert in such things.

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    $\begingroup$ Along these lines: en.wikipedia.org/wiki/Heisenberg_group $\endgroup$ – Mark Grant Apr 22 '11 at 10:11
  • $\begingroup$ I would appreciate very much if someone could provide some reference(s) for more examples as mentioned by Yemen. $\endgroup$ – user14595 Apr 22 '11 at 10:50
  • $\begingroup$ Please note that this answer does not actually answer what the original poster wanted to ask (it was written when the question had slightly ambiguous phrasing). It should therefore be "un-accepted" $\endgroup$ – Yemon Choi Apr 23 '11 at 18:37

Let me elaborate on the example of the 3-dimensional Heisenberg group, pointed out by Mark, which indeed provides the desired example. It is the group $$G=\{\left( \begin{array}{ccc} 1 & x & z \\ 0 & 1 & y \\ 0 & 0 & 1 \end{array} \right):x,y,z\in\mathbb{R}\}$$

Let $N$ be its centre: $$N=\{\left( \begin{array}{ccc} 1 & 0 & z \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right): z\in\mathbb{R}\}$$ so that $G/N\simeq\mathbb{R}^2$. The map $$\mathbb{R}^2\rightarrow G:(x,y)\mapsto \left(\begin{array}{ccc} 1 & x & 0 \\ 0 & 1 & y \\ 0 & 0 & 1 \end{array}\right) $$ is a continuous section, but there is no group-theoretic section (as this would make $G$ abelian!)


The following is a generic example. Every topological space is also a topological group. Let $0\to Y\to Z\to X\to 0$ be an exact sequence of Banach spaces. If you prefer take a Banach space $Z$, then a closed subspace $Y$ and set $X=Z/Y$. By the so-called Michael selection theorem the quotient map $\pi: Z\to X$ has a continuous section $\sigma:X\to Z$, so certainly these sequences split ``topologically'' in the sense you used. However the splitting of the sequence through a continuous homomorphism (=bounded linear operator if the underlying field is that of real numbers) depends on the concrete examples one want to use, see, for instance, the lovely survey

Plichko, Anatolij M., and Yost, David. "Complemented and uncomplemented subspaces of Banach spaces." Extracta Mathematicae 15.2 (2000): 335--371. https://www.eweb.unex.es/eweb/extracta/ http://eudml.org/doc/38635.


Here's another example that I think should work, but in the p-adic world.

Let $F$ be a p-adic field, and consider the group $G = SL(2,F)$. Inside $G$, we various elliptic tori as follows : Let $E$ be a quadratic extension of $F$. Then $T := E^1$, the set of norm $1$ elements in $E$, embeds in $SL(2,F)$ as $a + b \delta$ maps to the $2 \times 2$ matrix $(a,b, b \Delta, a)$ where $E = F(\sqrt{\Delta})$, $\delta = \sqrt{\Delta}$.

i.e. a in upper left, b in upper right, $b \Delta$ in lower left, $a$ in lower right (sorry, I don't seem to be using the array command correctly here).

Now, $G$ has a canonical $2$-fold cover $\widetilde{G} = \widetilde{SL(2,F)}$, the metaplectic cover. It sits in an exact sequence $$1 \rightarrow \mathbb{Z} / 2 \mathbb{Z} \rightarrow \widetilde{G} \rightarrow G \rightarrow 1$$ This is a topological central extension. Moreover, the 2-cocycle of this extension can be written down explicitly, but we won't need the full cocycle. For more information, see page 7 of


This extension restricts to an extension $$1 \rightarrow \mathbb{Z} / 2 \mathbb{Z} \rightarrow \widetilde{T} \rightarrow T \rightarrow 1$$ where $\widetilde{T} := \pi^{-1}(T)$, where $\pi$ is the projection $\pi : \widetilde{G} \rightarrow G$. It's a fact that the 2-cocycle of this extension is given by $c(x,y) = (x,y)_F$, where $( \cdot, \cdot)_F$ is the Hilbert symbol of $F$ (see loc. cit.)

In order to determine whether this extension splits as abstract groups, we just need to show whether any primage of $-1 \in T$ has order $2$ (for the proof of this, see page 8 loc. cit.). The definition of multiplication in an extension says that $(\pm 1, -1)^2 = (1, (-1,-1)_F)$. So at least if $F$ has residual characteristic $\neq 2$, we have that $(-1,-1)_F = 1$. Therefore, the above sequence splits as abstract groups.

However, by Remark 4.1 of loc. cit., this sequence also splits topologically.

After reading Kevin's comment, here is a response that hopefully will answer the question.

A general class of examples that answer the question can be gleamed from Moore's paper "Group extensions of p-adic and adelic linear groups". In this paper, Moore defined cohomology groups that take into account topology. That is, (I quote from the second paragraph of his paper) "If $G$ and $A$ are locally compact separable topological groups, and if $G$ acts on $A$ as a topological transformation group of automorphisms, one may modify the definitions and arrive at cohomology groups $H^n(G,A)$ which take into account the topology".

Let $G$ be a locally compact separable group, and $A$ a locally compact separable topological $G$-module. As in Moore's notation, we let $G^a$ and $A^a$ denote the underlying abstract groups of $G$ and $A$ respectively, considered without their topologies. Denote $H^n(G^a,A)$ the ordinary group cohomology (no topologies considered. Again this is Moore's notation). Then we have the natural homomorphism $$ H^n(G,A) \rightarrow H^n(G^a,A)$$

$\mathbf{Theorem \ 2.3}$ (of Moore's paper) : If $G$ is perfect, then the natural map $H^2(G,A) \rightarrow H^2(G^a,A)$ is injective.

Thus, with the given assumptions on $G$ and $A$, take any topological extension $$1 \rightarrow A \rightarrow E \rightarrow G \rightarrow 1$$ that splits algebraically. Then it splits topologically.

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    $\begingroup$ I interpreted "non-discrete" as meaning "none of the groups are allowed to be discrete". $\endgroup$ – Kevin Buzzard Apr 22 '11 at 20:26
  • $\begingroup$ Thanks Kevin, I edited my response and hopefully this works. $\endgroup$ – Moshe Adrian Apr 22 '11 at 22:29
  • $\begingroup$ Yes Kevin. By non-discrete, I mean that none of the groups are allowed to be discrete. I have modified my original question by adding connectedness also. I apologize I should have written it while posting the question. $\endgroup$ – user14595 Apr 23 '11 at 4:45

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