# How to obtain a Lie group from a Lie algebra

Please pardon me, if the question is too simple.

How can we obtain a Lie group form a given Lie algebra (say in 3 dimension) in practise? Can someone illustrate it in 3 dimension?

Is there a complete listing (up to isomorphism ) of all 3 dimensional Lie groups?

-
Pete's answer here may be of interest to you. – t.b. Jul 31 '11 at 12:18
I'd recommend that you work through Lecture 10 of Fulton-Harris, Representation theory (see also lectures 7 and 8). Also, there's little hope of listing them all (without assuming connectedness) as $\Gamma \times G$ is a Lie group for every discrete group $\Gamma$. – t.b. Jul 31 '11 at 14:08
Thanks Theo. Alright, is there any source where I can find a complete listing of connected 3-dimensional Lie groups. – ronny Jul 31 '11 at 15:00

In a comment, you ask for a list of three dimensional connected Lie groups. It can be obtained in two steps: first, classify all three dimensional Lie algebras; from each of them, construct the corresponding connected, simply connected Lie group and look for discrete central subgroups, and list all the corresponding quotients.

The first step is easy to do.

You can find the classification of complex three dimensional Lie algeras in many places. I remember the list to be at least in [Nuss, Philippe. L'homologie cyclique des algèbres enveloppantes des algèbres de Lie de dimension trois. J. Pure Appl. Algebra 73 (1991), no. 1, 39--71. MR1121630 (92i:19004)], and there is a much more technological approach in [Otto, Carolyn; Penkava, Michael. The moduli space of three-dimensional Lie algebras. Algebra and its applications, 255--268, Contemp. Math., 419, Amer. Math. Soc., Providence, RI, 2006. MR2279122 (2008d:17007)] Obtaining the list is a good exercise, though.

I assume, though, that you want real three-dimensional groups. That takes a little more work. You can either proceed directly, much as in the complex situation (there is just aa couple of places where the non-algebraic-closedness of $\mathbb R$ is felt) or reuse the complex classification by complexifying your algebra, using the classification of complex algebras, and then looking for possible real forms.

Next: for most of the Lie algebras it is very easy to find the simply connected Lie group. Only $\mathfrak{sl}_2$ is complicated (it is not a linear group). The others are nilpotent or solvable.

I have never tried listing discrete central subgroups in those groups. The centers are one-dimensional in most cases, so it should not be too hard

-
Many Thanks Mariano. I think, I can proceed following your answer. Thanks to all. – ronny Aug 1 '11 at 5:20