A connected semisimple Lie subgroup of $SO(n)$ is closed in $SO(n)$ (Kobayashi and Nomizu, 1963, p. 279). Can we extend this result to all semisimple groups, put differently, is any connected semisimple Lie subgroup of a semisimple group closed in this semisimple group?

I thought that maybe I could use arguments like "any connected semisimple group is faithfully represented as subgroup of $SO(n)$" which would immediately prove the extension to all semisimple group. A push in the right direction or reference would be enough.


Yes, it is true. It is a consequence of the facts below.

  1. If $h$ is a Lie subalgebra of $gl(n,K)$ ($K = \mathbb{R}$ or $\mathbb{C}$) then $[h,h]$ is the subalgebra of an algebraic subgroup of $GL(n,K)$ hence corresponds to a (connected) closed subgroup ( an old result of Chevalley and Tuan.

  2. If $G$ has a morphism $\phi$ to a $GL(n,K)(=G_1) $ with an injective tangent map $d\phi$ and $d\phi(h)$ corresponds to a closed subgroup $H_1$ in $G_1$ then $h$ corresponds to a closed subgroup $\phi^{-1}(H_1)$.

From 1., every semisimple Lie subalgebra of a $GL(n,K)$ corresponds to an algebraic subgroup hence a closed subgroup.

For 2., map consider $\phi \colon G \to GL(g)$, $g \mapsto Ad(g)$, with differential $d\phi=ad$ injective, since $g$ has zero center.

  • $\begingroup$ Thanks! Is there also a way in applying 2. without using 1. and instead using the result from my question? $\endgroup$ – user1043065 May 24 '15 at 13:42
  • $\begingroup$ No worries... Well, you cannot map injectively lie algebras like $sl(m, \mathbb{R})$ to $su(n)$. In fact, if a semisimple Lie algebra maps into $su(n)$ then any connected Lie group having it as an algebra is compact. ( that's a theorem of Weyl on compact semisimple Lie algebras). $\endgroup$ – Orest Bucicovschi May 24 '15 at 22:44
  • $\begingroup$ So I can use my own result if the Lie algebra of the group (containing the subgroup) is itself compact (has negative definitie Killing form), right? $\endgroup$ – user1043065 May 25 '15 at 7:08
  • $\begingroup$ @user1043065:For a real Lie algebra, negative definite Killing form( equivalent to semisimple and compact) already has the group (connected piece at least) compact. The compact Lie groups are subgroups of some $SO(n)$. Now you can apply your method. I think it works even it the group you start with has a compact Lie algebra,(compact so reductive but not necessarily semisimple) since the group will cover a compact group. Now apply your method to the compact group and also use 1. Hence: your method works if the Lie algebra of the group is compact $\endgroup$ – Orest Bucicovschi May 25 '15 at 8:25

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