Lie (and topological) group extensions of $\mathbb{R}^2$ by $\mathbb{R}$

What are all the non-split Lie (and topological) group extensions $0 \to \mathbb{R} \to G \to \mathbb{R}^2 \to 0$? Here, $\mathbb{R}$ and $\mathbb{R}^2$ are regarded as Lie (and topological) groups with respect to the usual addition. One example of a non-split extension is the Heisenberg group $H_3(\mathbb{R})$ (Please see a post by Alain Valette at https://mathoverflow.net/questions/63630). Since, every abelian topological extension of $\mathbb{R}^n$ by a locally compact abelian group is trivial, we have that every abelian topological extension of $\mathbb{R}^2$ by $\mathbb{R}$ is trivial. Hence, we need to see only non-abelian extensions.

• Well, this means that the Lie group is of dimension 3, right? I believe there is a classification of three-dimensional Lie algebras (by Bianchi, no?), which should then give a classification of 3-dimensional (connected, simply connected) Lie groups. Then you can check the above case-by-case. – Mark Jul 4 '11 at 12:02
• Thanks Mark. Could you (or someone) please explain this classification in brief. It might be of interest to general public. – jap Jul 4 '11 at 12:54

Central extensions $$0 \to \mathbb{R} \to G \to \mathbb{R}^2 \to 0$$ in which $G$ is a principal $\mathbb{R}$-bundle over $\mathbb{R}^2$ (I suppose you mean that by "topological") are classified by continuous maps $$f: \mathbb{R}^2 \times \mathbb{R}^2 \to \mathbb{R}$$ satisfying $$f(x,y)f(y,z) = f(x,z).$$ The abelian ones are those corresponding to maps with $f(x,y) = f(y,x)$.

This follows from a general theory for topological central extensions described in J.-L. Brylinksi's "Differentiable cohomology of gauge groups" (for the smooth case, but that is not relevant) combined with the fact that every principal bundle over $\mathbb{R}^2$ is trivializable.

EDIT: From a map $f$, you get the extension $G$ explicitly as the topological space $G = \mathbb{R} \times \mathbb{R}^2$ with the multiplication given by $$(a_1,x_1)(a_2,x_2) = (a_1 + a_2 + f(x_1,x_1^{-1}),x_1 + x_2).$$

• Thanks Konrad. I guess this is the condition of the 2-cocycle considered by Brylinksi. I am looking for the extensions explicitly as topological groups (Lie groups). – jap Jul 5 '11 at 5:09
• @jap: well, see my edit! – Konrad Waldorf Jul 5 '11 at 13:40

This is just an answer to the request for the Bianchi classification, not to the original question. I'm putting it as an answer because it's too long for a comment.

A 3-dimensional Lie algebra $L$ is either semi-simple, in which case it is isomorphic to either ${\frak{so}}(3)$ or ${\frak{sl}}(2,\mathbb{R})$, or else it has a basis $x_1,x_2,x_3$ such that $$[x_1,x_2]=0\qquad [x_2,x_3] = b_{11} x_1 + b_{12}x_2\qquad [x_3,x_1] = b_{21} x_1 + b_{22}x_2$$ where the $2$-by-$2$ matrix $B = (b_{ij})$ is equal to one of the following $$\begin{pmatrix}0&0\cr 0&0\end{pmatrix},\ \begin{pmatrix}1&0\cr 0&0\end{pmatrix},\ \begin{pmatrix}1&0\cr 0&1\end{pmatrix},\ \begin{pmatrix}1&0\cr 0&-1\end{pmatrix}$$ or $$\begin{pmatrix}0&1\cr -1&0\end{pmatrix},\ \begin{pmatrix}1&1\cr -1&0\end{pmatrix},\ \begin{pmatrix}\sigma&1\cr -1&\sigma\end{pmatrix},\ \begin{pmatrix}\sigma&1\cr-1&-\sigma\end{pmatrix}$$ where $\sigma>0$ is a real number. These are all pairwise non-isomorphic.

The proof is fairly straightforward and can be found in many places.