# Decomposing maximal compact subgroups of SO(n,1)

Let $G=SO(n,1)$ and let $G=KAN$ be an Iwasawa decomposition of $G$. Let $M$ be the centralizer of $A$ in $K$. In this case, we have $K≃SO(n)$, $A≃\Bbb R$(this is the maximal diagonalizable subgroup), $N≃\Bbb{R}^{n−1}$ and $M≃SO(n−1)$. Let $\mathfrak k$ be the Lie algebra of $K$ and $\mathfrak m\subseteq\mathfrak k$ be the lie algebra of $M$. Let $\mathfrak h$ be the orthocomplement to $\mathfrak m$ in $\mathfrak k$. That is $\mathfrak k= \mathfrak m\oplus \mathfrak h$. Let $H$ be the Lie hgroup associated to generate by $exp(\mathfrak h)$. What can we say about $HM$? When is it true that $HM=K$?

I believe $K/HM$ is discrete and finite. In this case there exists $e=\omega_1,\omega_2,\dots,\omega_l$ s.t. $K=HM\sqcup\omega_2 HM\sqcup\dots\sqcup\omega_l HM$. Can I explicitly calculate these $\omega_i$?

• ${\mathfrak h}$ is not a Lie algebra, what do you mean by $H$? – Misha Feb 22 '13 at 4:18
• the issue is one of connected components; so one must be careful about the groups $K$ and $M$. your $K$ is not $SO(n)$ but $O(n)$. Your $M$ is not $SO(n-1)$ but $O(n-1)$. In this case $K/HM$ is connected. – Venkataramana Feb 22 '13 at 4:27
• this is independent of what $\mathfrak h$ may be. As Misha says, $\mathfrak h$ is not a Lie algebra. – Venkataramana Feb 22 '13 at 4:51
• @Misha. Thank you. Indeed, $\mathfrak h$ is not a Lie Algebra. For my purpose it is enough to consider the group generated by $exp(\mathfrak h)$. In this case, what can I say about the cosets of $K/HM$? – Davis Feb 22 '13 at 5:35

I believe that your set $\exp(\mathfrak{h})$ will always act transitively on $S^n$ and will contain a stabilizer of at least one point of $S^n$, which should imply that $\exp(\mathfrak{h})$ alone generates $K$. Or am I missing something?