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What does the classification of Complex Semi-simple Lie algebras buy us in terms of classifying Lie groups? Certainly it classifies complex semi-simple lie groups but can we get any better? I know we can take compact real forms of the semi-simple algebras and there are several theorems about topological similarities for Lie groups with the same Lie algebra. How far can we take this? What is the biggest class of Lie groups we can rope in by this method?

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2 Answers 2

The point is, if $G$ is a real semisimple Lie group, then its Lie algebra $\mathfrak{g}$ is also semisimple and so is the complexification $\mathfrak{g} \otimes \mathbb{C}$. Or, given a complex $\mathfrak{g}$, any real form $\mathfrak{g}_\mathbb{R}$ (meaning, a pre-complexification) integrates to a real semisimple Lie group. The real forms have been classified and they are described by Satake diagrams, which are Dynkin diagrams with an extra decoration to describe the real form. So together with finding the real forms, the complex classification "buys" you everything about semi-simple real Lie groups. The great Élie Cartan not only reorganized the complex classification, he also did the real classification.

See also the Wikipedia page for real forms. As it points out, there are always two special real forms, the compact form and the completely split form, and often also others.

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Excellent. I read somewhere that this could be extended to a classification of compact, connected Lie groups. I can't find the book though. Is this true? –  Travis Oct 25 '10 at 18:12
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@Travis That's right, but not much more happens. Every compact, connected Lie group has a canonical central torus subgroup and a canonical semisimple subgroup that intersect in a finite subgroup. In other words, every example is a finite cover, or equivalently a central extension, of a torus cross a semisimple Lie group. –  Greg Kuperberg Oct 25 '10 at 18:53
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Greg's post is exactly correct. I'll add a warning, though, that whereas semisimplicity of a real Lie algebra survives complexification, simplicity does not. In particular, $\mathfrak{sl}(2,\mathbb C)$ is simple as a real Lie algebra, but its complexification $\mathfrak{sl}(2,\mathbb C) \otimes_{\mathbb R} \mathbb C = \mathfrak{sl}(2,\mathbb C) \oplus \mathfrak{sl}(2,\mathbb C)$ is not simple as a complex Lie algebra. So Cartan's classification of simple real Lie algebras does require thinking about non-simple (but semisimple) complex Lie algebras. –  Theo Johnson-Freyd Oct 25 '10 at 22:19
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@Theo But, a real Lie algebra is simple with a non-simple complexification if and only if it is secretly already complex, right? –  Greg Kuperberg Oct 25 '10 at 22:28
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@Greg: Indeed, and then the complexification is two copies, as in Theo's example. –  Allen Knutson Oct 26 '10 at 0:45

Here is an answer from a different point of view : every connected Lie group is almost a semidirect product of a semisimple Lie group (Levi factor) and a connected solvable normal Lie subgroup (its solvalble radical). See Onishchik and Vinberg's Lie Groups and Lie Algebras III for a more precise statement. In that sense, the classification of real Lie groups is reduced to that of semisimple and solvable ones. Although somewhat crude, this classification helps for example in the study of unitary representations. (See papers by M. Duflo on the orbit method.)

Another subtle issue with the classification of semisimple Lie groups is that over $\mathbb C$, the simply connected group corresponding to a semisimple Lie algebra is always linear, whereas over $\mathbb R$, this is not the case. For example, the simply connected Lie group corresponding to $\mathfrak{sl}_2(\mathbb C)$ is $\mathrm{SL}_2(\mathbb C)$, but the simply connected Lie group corresponding to $\mathfrak{sl}_2(\mathbb R)$ is an infinite-sheeted covering of $\mathrm{SL}_2(\mathbb R)$. In other words, $\pi_1(\mathrm{SL}_2(\mathbb C))=\{e\}$ whereas $\pi_1(\mathrm{SL}_2(\mathbb R))=\mathbb Z$. (Here $\pi_1$ means the first fundamental group.)

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