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A (non-compact) symmetric space is of the form $G/K$, where $G$ is a (non-compact) semisimple Lie group defined over $\mathbb{R}$, and $K$ is a maximal compact subgroup of $G$.

Let $M = G/K$ be a rank-one symmetric space of the noncompact type. There are only three families of rank $1$ symmetric spaces:

  1. hyperbolic $n$-space, corresponding to the Lie group $\operatorname{SO}(n,1)$.

  2. complex hyperbolic $n$-space, corresponding to the Lie group $\operatorname{SU}(n,1)$.

  3. quaternionic hyperbolic $n$-space, corresponding to the Lie group $\operatorname{Sp}(n,1)$.

There is also one exceptional example:

  1. the Cayley upper half plane, corresponding to the Lie group $F_4^{-20}$.

Now, let $G = NAK$ be the Iwasawa decomposition of $G$. Does anyone know what is the Iwasawa decomposition ($N={?}$ and $K={?}$) of the four cases 1), 2), 3) and 4)?

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    $\begingroup$ $K$ is the maximal compact subgroup, which is $\mathrm{S}(\mathrm{O}(n)\times\mathrm{O}(1)$ in the first case, $\mathrm{S}(\mathrm{U}(n)\times\mathrm{U}(1)$ in the second case, and probably similar in the third case (but I haven't doubled checked). $\endgroup$
    – YCor
    Commented Jun 3 at 14:12
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    $\begingroup$ $N$ can be identified to the sphere at infinity minus one point, and identifies to $\mathbf{R}^{n-1}$ in the first case, the $(2n-1)$-dimensional Heisenberg group in the second case,a quaternionic Heisenberg group in the 3rd case (dimension $4n-1$, 1st layer of dimension $4n-4$ and 2nd layer of dimension $3$); in the 4th case it's a $(8+7)$-dimensional 2-step-nilpotent group. In all cases $A$ is 1-dimensional. $\endgroup$
    – YCor
    Commented Jun 3 at 14:12

1 Answer 1

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For simplicity, I will work only with connected Lie groups (and, accordingly, identity components of isometry groups):

  1. Real hyperbolic space $M=\mathbb H^n$: $G=SO^+(n,1)$: $K=SO(n)$, $N\cong {\mathbb R}^{n-1}$.

  2. Complex hyperbolic space $M=\mathbb C H^n$: $G=PU(n,1)$: $K=U(n)$, $N$ is the $2n-1$-dimensional Heisenberg group, the central extension of ${\mathbb R}^{2n-2}$ by $\mathbb R$ where the 2-cocycle of the extension is given by the symplectic form on ${\mathbb R}^{2n-2}$.

  3. Quaternionic hyperbolic space $M=\mathbb H H^n$: $G=PSp(n,1)$: $K=Sp(n)$. The group $N$ is some 2-step nilpotent group, not sure if it has a name, its center is 3-dimensional, the quotient by the center is $\mathbb R^{4n-4}$.

  4. Octonionic hyperbolic plane $\mathbb O H^2$: $K=Spin(9)$. The group $N$ is again 2-step nilpotent.

You can find a detailed discussion in chapter 19 of

Mostow, G. D., Strong rigidity of locally symmetric spaces, Annals of Mathematics Studies. No. 78. Princeton, N. J.: Princeton University Press and University of Tokyo Press. V, 195 p. $ 7.00 (1973). ZBL0265.53039.

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    $\begingroup$ Why isn't it $K = \operatorname S(\operatorname O(n) \times \operatorname O(1))$ for (1) and $K = \operatorname S(\operatorname U(n) \times \operatorname U(1))$ for (2), as @YCor says? $\endgroup$
    – LSpice
    Commented Jun 3 at 17:26
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    $\begingroup$ Yes, I think @LSpice's description is more "natural", even if there is some collapsing... Similarly, for $Sp^*(n,1)$, whose maximal compact is $Sp^*(n)\times Sp^*(1)$... $\endgroup$
    – paul garrett
    Commented Jun 3 at 17:37
  • $\begingroup$ @LSpice: Thank you, I was ignoring connected components etc. $\endgroup$
    – Moishe Kohan
    Commented Jun 3 at 18:23

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