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I am trying to decide whether $spin(3)$(i.e. the unit quaternions) can be realized as a homogeneous space of $SL(3, \mathbb{R})$ but I reach nowhere. Does someone know a reference/answer for this?

Thanks!

MEEL
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1 Answers1

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Sorry if answering one owns questions is not very well seen but I want feedback on it and I think I got it.

Let $G = SL(3, \mathbb{R})$ and suppose $spin(3) = \mathbb{S}^3$ is a homogeneous space of it. Then, since $M$ is simply connected, there exists a compact subgroup $C$ of $G$ that also acts transitively on $\mathbb{S}^3$. If X is the isotropy subgroup of some point in the sphere then $C/X = \mathbb{S}^3$. Moreover, since the sphere is simply connected $X$ is connected.

On the other hand, the maximal compact subgroups of $G$ are isomorhpic to $SO(3)$ and $C$ is contained in one such group, say $K$. Then $\mbox{dim}\; C\le \mbox{dim}\; K = 3 = \dim \mathbb{S}^3$. Since the sphere is a quotient of $C$, we conclude that $\mbox{dim}\; C = 3$ and so it is open in $K$ and being also open, $C = K$ and since $X$ is connected and of dimension $0$ it is a point (it must be the identity). We deduce $SO(3) = \mathbb{S}^3$ which is false!

Hence, $spin(3)$ is not an homogeneous space of $SL(3, \mathbb{R})$.

MEEL
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    The fact that $ G $ acting transitively on a simply connected space implies that its maximal compact subgroup acts transitively is due to Montgomery 1950 see https://mathoverflow.net/questions/408900/does-the-maximal-compact-subgroup-always-act-transitively-on-a-compact-homogeneo?rq=1 – Ian Gershon Teixeira Feb 14 '22 at 17:36