Let $\mathcal{H}\_2 = \{(x,t) \in \mathbf{R}^2: t > 0\}$ be the upper half-plane, and let $\mathcal{H}\_3$ be the hyperbolic 3-space $\{(x,t) \in \mathbf{C} \times \mathbf{R}: t > 0\}$. Clearly $\mathcal{H}\_2$ embeds in $\mathcal{H}\_3$. There is an action of $PSL(2, \mathbf{C})$ on $\mathcal{H}\_3$, extending the familiar action of $PSL(2, \mathbf{R})$ on $\mathcal{H}\_2$.



> 
> If $\gamma \in PSL(2, \mathcal{O}\_K)$, where $K$ is an imaginary quadratic field which is *not* $\mathbf{Q}(\zeta\_3)$, is it the case that we have must have either $\gamma \cdot \mathcal{H}\_2 = \mathcal{H}\_2$ or $\mathcal{H}\_2 \cap \gamma \cdot\mathcal{H}\_2 = \varnothing$? 
> 
> 
> 


For a general $PSL(2, \mathbf{C})$, or for non-integral $\gamma \in PGL(2, K)$, the intersection $\mathcal{H}\_2 \cap \gamma \cdot\mathcal{H}\_2$ can certainly be a nonempty proper subspace of $\mathcal{H}\_2$ (in which case it's a geodesic arc). This case also seems to occur for some $\gamma \in PSL(2, \mathbf{Z}[\zeta\_3])$, but I have found no other examples for any other quadratic fields. Is there a reason for this, or have I just not looked hard enough?



Hyperbolic planes inside hyperbolic 3-space quotients
Let $\Gamma = PSL\_2(\mathcal{O}\_K)$. Note that complex conjugation acts as an orientation-reversing isometry on $\mathcal{H}\_3$, the action being compatible with that on $\Gamma$.


Let $\gamma\in \Gamma$ be such that $a = \mathcal{H\_2}\cap \gamma^{-1}\mathcal{H}\_2$ is a nontrivial geodesic.
For all $x\in a$ we have $\bar{x}=x$ and $\gamma x = \overline{\gamma x} = \bar{\gamma}\bar{x} = \bar{\gamma}x$, so $\sigma = \gamma^{-1}\bar{\gamma}\in \Gamma$ fixes $a$ pointwise, so $\sigma$ is elliptic. Elementary considerations on the entries of the matrix $\sigma$, using


* $\sigma \in \Gamma$
* $\bar{\sigma} = \sigma^{-1}$
* $tr(\sigma)\in\mathcal{O}\_K\cap (-2,2) = \{-1,0,1\}$


quickly rule out all cases where $K \neq \mathbb{Q}(\zeta\_3)$.


