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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 http://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.

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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
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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

2 Answers 2

up vote 2 down vote accepted

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 $$ (a_1,x_1)(a_2,x_2) = (a_1 + a_2 + f(x_1,x_1^{-1}),x_1 + x_2). $$

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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
    
Many thanks Konrad. –  jap Jul 5 '11 at 15:26

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.

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