I just wonder are there any Lie subgroups of $\textrm{Gl}(n;\mathbb R)$ besides $\textrm{Sl}(n;\mathbb R)$, $\textrm{O}(n;\mathbb R)$ and $\textrm{SO}(n;\mathbb R)$, and is there any classification of them?
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There are many, many, many others. There are even Lie subgroups inside the three Lie subgroups which you listed. This is a fine question, of course, but maybe take a look at the Wikipedia page on Lie groups: https://en.wikipedia.org/wiki/Lie_group – Jesse Madnick Jul 10 '17 at 07:28
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There is a subgroup of all diagonal matrices with nonnegative entries, which is isomorphic to $\Bbb R^n$ with addition. – edm Jul 10 '17 at 07:31
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Sure. Let uss see:
- the diagonal matrices with non-zero determinant;
- the upper (or lower) trianguar matrices with non-zero determinant;
- the Heisenberg group;
- $\mathrm{SU}(n,\mathbb{C})$ can be seen as a subgroup of $\mathrm{GL}(2n,\mathbb{R})$;
- the symplectic group can be seen as a subgroup of $\mathrm{GL}(4n,\mathbb{R})$
And so on... Actually, it is hard (although not impossible) to find a Lie group wich is not a subgroup of some $\mathrm{GL}(n,\mathbb{R})$. And, no, there is no classification.
José Carlos Santos
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Very true, but there is a classification of maximal connected Lie subgroups (Dynkin). – Moishe Kohan Jul 10 '17 at 08:28
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1Dynkin (1950s) classified maximal Lie subalgebras of classical simple Lie algebras $g$ (while $gl$ is not simple, $sl$ is). The classification of maximals is roughly the following. Either the subalgebra is maximal parabolic, and these are classified by looking at the nodes of Dynkin diagram, or is semisimple and has equal rank with $g$, and is obtained by removing one node from the extended Dynkin diagram of $g$. If you are interested, I can dig out the more exact description. I found this result very useful (but not well known). Later, Karpelevich extended this to real Lie algebras. – Moishe Kohan Jul 10 '17 at 08:53
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