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I'm in the process of learning Lie theory.

Simply connected Lie groups correspond to finite dimensional real Lie algebras. Finite dimensional semisimple real Lie algebras correspond to Satake diagrams.

So, simply connected semisimple Lie groups correspond to Satake diagrams.

A Lie group is a geometric object and is a group. So, the most natural classification task is to classify Lie groups by geometric, topological and abstract group theoretic properties (i.e. "compact", or "simple as an abstract group").

I know, in fact, that such characterization exists: A connected Lie group is semisimple if and only if its radical (largest connected solvable normal subgroup) is trivial.

So, Satake diagrams correspond to simply connected Lie groups with trivial radical.

While this is this is the type of classification I was looking for, the property of being "simply connected with trivial radical" seems a little artbitrary.

This is my question:

What I'm trying to understand is whether the focus on semisimple Lie groups is only because "that's where we have a good understanding", or also because of some natural mathematical domains where "among Lie groups, the semisimple ones are all that matter".

Terry
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  • These are multiple questions. To the "the main question" a good example is that semisimple Lie groups are the ones we need for the classification of Irreducible symmetric spaces. – Dietrich Burde Aug 31 '16 at 09:42
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    If $G$ is compact, then its Lie algebra splits as a direct sum $\mathfrak{g}=Z_{\mathfrak{g}}\oplus [\mathfrak{g},\mathfrak{g}]$, where $[\mathfrak{g},\mathfrak{g}]$ is semisimple and $Z_{\mathfrak{g}}$ is abelian. Moreover, the connected subgroup $G_{ss}$ with Lie algebra $[\mathfrak{g},\mathfrak{g}]$ is closed and $G$ is the commuting product $G=(Z_G)0G{ss}$ where $(Z_G)_0$ is the identity component of the centre (which is also closed). Hence, every compact connected Lie group is of the form $G=(T\times H)/D$ where $T$ is a torus, $H$ is semisimple and $D$ is a finite normal subgroup. – Spenser Aug 31 '16 at 09:43
  • @DietrichBurde: Thanks for the example. In my view, I asked only one question, and I edited my question to reflect this more explicitly. – Terry Aug 31 '16 at 09:45
  • @DietrichBurde: By the Wikipedia article you linked to, it looks like a very convincing example. Of course, it would be nice to have further examples. – Terry Aug 31 '16 at 09:52
  • If I'm not mistaken, by the Levi decomposition every Lie algebra is a semidirect sum of a solvable one and a semisimple one (and every simply connected Lie group is a semidirect product of a solvable group and a semisimple group). Therefore it makes sense to divide the research to solvable groups and semisimple groups. – Cronus Sep 01 '16 at 15:56

2 Answers2

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Simple groups and simple algebras are "the ones all that matter" for an intrinsic reason. They are the atoms of this universe, so to speak, and hence are interesting. In case of Lie groups, there are furthermore reasons from geometry, why we are in particular interested in semisimple ones. However, even there, the reason is similar. We want to understand the basic "irreducible" components. One example illustrating this is the classification of Riemannian symmetric spaces. It can then be shown that any simply connected Riemannian symmetric space is a Riemannian product of irreducible ones. Therefore, it makes sense to further restrict oneself to classifying the irreducible, simply connected Riemannian symmetric spaces. We end up with simple Lie groups in Case $A$, and with compact Lie groups in case $B$. But even for compact Lie groups we are very close to semisimple Lie groups, and one could argue that also there "the ones all that matter" are semisimple ones.

Dietrich Burde
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From the standpoint of representation theory, the larger class of reductive groups are "the ones that matter" (this is an oversimplification, as finite-dimensional representations of non-reductive groups are crucial to understand, even if you only care about representations of reductive groups). But you can't understand reductive groups without understanding semisimple groups first.

hunter
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  • Thanks. Reductive are "the ones that matter" in what sense? To solve what problem(s)? – Terry Aug 31 '16 at 10:23
  • @Terry I was hoping you wouldn't ask that :-) – hunter Aug 31 '16 at 10:26
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    @Terry One big motivation is Harish-Chandra's theorem that representations of real reductive groups are induced by finite-dimensional representations from parabolics. But I am far from an expert, so you should ask someone else! – hunter Aug 31 '16 at 10:28
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    @Terry See http://mathoverflow.net/questions/223584/which-philosophy-for-reductive-groups/223895#223895 for part of the reason. – Tobias Kildetoft Aug 31 '16 at 10:29