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Let $G=G_1.G_2$ be a Lie Subgroup of $SO(k) \times SO(2) \subset SO(k,2)$, where $G_1=SU(k/2) \subset SO(k)$ and $G_2$ is a Lie subgroup of $SO(k) \times SO(2)$ isomorphic to $SO(2)$. Let $G_1 \cap G_2$ is discrete. We may assume that $G_1$ and $G_2$ commute in $G$. (Note that under these conditions $G$ can written as $(G_1 \times G_2)/H$ where $H$ is a discrete central subgroup of $G$).
Consider the natural action of $G$ on the anti de sitter space $H^{k+1}_1\approx SO(k,2)/SO(k,1)$ and let there exist an orbit of codimension one (so the action of $G$ on $\mathbb{R}^{k+2}_1$ has an orbit of codimension two).
Now, my question is that, can we deduce that $G_1 \cap G_2$ is trivial, i.e. $G=G_1 \times G_2$ ?

Thanks in advance.

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1 Answer 1

A. L. Onishchik has classified decompositions $G=G_1G_2$ of redcutive Lie groups (e.g., see his article "Decompositions of reductive Lie groups" in Math. USSR-Sbornik, Vol. 9 (1969), No. 4). We have $G=G_1G_2$ with proper Lie subgroups $G_1,G_2$ if and only if $G_1$ is transitive on the homogeneous space $G/G_2$, or vice versa. This is equivalent to the fact that the corresponding Lie algebras satisfy $g=g_1+g_2$, as a sum of vector spaces, not necessarily a direct sum. For example, if the groups and Lie algebras are simple, one of the examples is $A_{2n-1}=A_{2n-2}+C_n$ for all $n\ge 2$, where the sum is not direct ($A_n$ is $sl (n+1)$ etc.). This is typical. For the groups it means, that we cannot deduce that the intersection $G_1\cap G_2$ is trivial. The classification shows, I believe, that this also applies to your case.

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@Burde: Thanks for your comments. Indeed, I reduced my problem to what you see here, by the use of Onishchik's results (from his book). I guess, further conditions here (specially existence of an orbit of codimension one), forces the intersection to be trivial. –  Manuel-Velajic Apr 5 '13 at 15:41

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