8
$\begingroup$

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?

Improve this question
$\endgroup$
4
  • $\begingroup$ I don't know the answer, but the following might be useful. Assume $\gamma$ is hyperbolic. If the intersection is a geodesic arc, then $\gamma$ must fix that arc, and thus that arc is the axis for $\gamma$. Conversely, if $\gamma$ has an axis in $\mathcal{H}^2$, then $\mathcal{H}^2 \cap \gamma \cdot \mathcal{H}^2$ will be that axis. So ignoring the parabolic and elliptic cases, what you're asking is whether or not that group contains a hyperbolic element whose axis lies in $\mathcal{H}^2$, or equivalently whose two fixed points on the boundary lie in the boundary of $\mathcal{H}^2$. $\endgroup$
    – Andy Putman
    Commented May 8, 2017 at 21:48
  • 1
    $\begingroup$ @AndyPutman Why does $\gamma$ have to fix the intersection? $\endgroup$
    – David Cohen
    Commented May 8, 2017 at 21:53
  • $\begingroup$ So I thought about this a bit and I didn't come up with an answer, but I would like to note that the following obvious strategy for producing a counterexample will never work. Given $\gamma\in PSL(2,\mathcal{O}_K)$, we may view gamma as a Moebius transformation of $\mathbb{C}\cup\infty$ in the usual way (i.e., $\frac{az+b}{cz+d}$). If $\gamma$ took the points $0,1,\infty$ to some points $p,q,r$ where $Im(p)$ and $Im(q)$ are positive but $Im(r)<0$, then $\mathcal{H}_2\cap\gamma\cdot\mathcal{H}_2$ would be two points. Unfortunately, this can never happen (!) for the fields $K$ in question. $\endgroup$
    – David Cohen
    Commented May 8, 2017 at 22:46
  • $\begingroup$ @DavidCohen: You're right, that's what I get for trying to answer MO questions while massively jet lagged... $\endgroup$
    – Andy Putman
    Commented May 9, 2017 at 6:34

1 Answer 1

Reset to default
10
$\begingroup$

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 $K \neq \mathbb{Q}(\zeta_3)$.

Improve this answer
$\endgroup$
8
  • 1
    $\begingroup$ Why are the endpoints of the axis of an element of $\Gamma$ in $\mathbb{P}^1(K)$? $\endgroup$
    – David Cohen
    Commented May 8, 2017 at 23:41
  • $\begingroup$ Of course! $\gamma^{-1} \bar\gamma$ is elliptic. I can't believe I didn't think of that. Very nice :-) $\endgroup$
    – David Loeffler
    Commented May 9, 2017 at 6:43
  • $\begingroup$ @DavidCohen Ah yes, they could be in another imaginary quadratic field. But the rest of the arguments still works. I will edit the answer accordingly. $\endgroup$
    – Aurel
    Commented May 9, 2017 at 7:19
  • $\begingroup$ Hmm actually the end points could be in a real quadratic field. I will have to see if we can fix the argument. $\endgroup$
    – Aurel
    Commented May 9, 2017 at 9:39
  • 2
    $\begingroup$ This is a great solution. $\endgroup$
    – David Cohen
    Commented May 9, 2017 at 16:02

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .