Topologically split extensions of topological groups

Question

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 $H$.

Answer 1

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.

Answer 2

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!)

Answer 3

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.

Answer 4

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

http://citeseerx.ist.psu.edu/viewdoc/download;jsessionid=7B8F219E349E46B736984D32600EB0CB?doi=10.1.1.78.7539&rep=rep1&type=pdf

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.