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I have a 3-dimensional real Lie algebra given by the relations: \begin{equation} [e_1,e_2] = e_2-2e_3\end{equation} \begin{equation} [e_1,e_3] = 2e_2+e_3\end{equation} \begin{equation} [e_2,e_3] = 0\end{equation} I need to describe some Lie group with such Lie algebra as a subgroup in $GL(3,\mathbb{R})$ In the first point of the task, it was necessary to find all the ideals and I did it, but I don't know how this can help in solving the second point.

AndrewGap
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    It may be easiest to solve your problem using the Bianchi classification in dimension $3$. It looks like your case is Type VII. – kabenyuk Oct 24 '23 at 11:03
  • I think I can't use this classification and I have to find another solution. – AndrewGap Oct 24 '23 at 12:39
  • You can have a look at some posts here, e.g. How to obtain a Lie group from a Lie algebra. You Lie algebra is a matrix Lie algebra, and you can easily find matrix Lie groups with this Lie algebra. – Dietrich Burde Oct 24 '23 at 13:46
  • So, I was able to find the group I needed. It can be written as a semidirect product of $\mathbb{R}^{2}$ and $G_1$, where $G_1$ is generated by the matrix $\begin{pmatrix} e^t & 0 & 0 \ 0 & cos(2t) & sin(2t) \ 0 & -sin(2t) & cos(2t) \end{pmatrix}$. Now I want to describe all connected Lie groups with such a Lie algebra. To do this, I use the third Lie theorem, that is, I need to find a universal cover of my semidirect product and factorize over all discrete subgroups of the center. I will assume that this cover will be isomorphic $\mathbb{R}^{3}$, but I'm not sure. Am I right? – AndrewGap Oct 25 '23 at 12:13
  • No, that's not correct. The simply connected covering of a non-Abelian group is non-Abelian. – kabenyuk Oct 25 '23 at 15:12

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