Let $\mathbb H^3$ be hyperbolic $3$-space and let $\Gamma \subset PSL(2,\mathbb R) \subset PSL(2,\mathbb C) \cong Isom ^+(\mathbb H^3)$ be the (Fuchsian) fundamental group of a Riemann Surface $\Sigma$. Then $\Gamma$ also acts freely and properly discontinuously on $\mathbb H^3$, so we may look at the quotient manifold $\mathbb H^3 \backslash \Gamma$. Something that seems to be common knowledge among experts on hyperbolic geometry, but something that I cannot quite wrap my head around, is the following assertion:

There exists an isometric embedding $i: \Sigma \to \mathbb H^3 \backslash \Gamma$.

Let $p: \mathbb H^3 \to \mathbb H^3 \backslash \Gamma$ be the corresponing covering map. Assuming we could indeed find an isometrically embedded copy of $\Sigma$, then each component of $p^{-1}(\Sigma)$ would be an isometrically embedded copy of $\mathbb H^2$ in $\mathbb H^3$ (a hyperbolic plane), invariant under the action of $\Gamma$. However, it is not at all clear to me how such a copy would look like. Using the Poincaré half-plane model of $\mathbb H^3$ (so that topologically, we can identify $\mathbb H^3$ with $\mathbb C \times \mathbb R^+$), hyperbolic planes in $\mathbb H^3$ correspond (precisely?) to planes/ hemispheres perpendicular to the $x_1x_2$-plane. Moreover, according to what I've read in some sources, in this model, the action of $PSL(2,\mathbb R)$ is generated by the following isometries, where $\lambda$ ranges over $\mathbb R$: \begin{align} (z,x_3) \mapsto (\lambda z, |\lambda|x_3): \lambda \neq 0 \\ \\ (z,x_3) \mapsto (z + \lambda, x_3) \\ \\ (z,x_3) \mapsto (\frac{-\overline{z}}{|z|^2 + x_3^2},\frac{x_3}{|z|^2 + x_3^2}) \\ \end{align}

However, I fail to see how any subgroup, generated by some collection of compositions of the above elements (let alone a fuchsian subgroup), would leave any of the aforementioned hyperplanes invariant. Any help is highly appreciated

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    $\begingroup$ Hint: $\Gamma$ preserves unique hyperbolic plane in $H^3$, namely, the one whose boundary is the limit circle of $\Gamma$. $\endgroup$ – Moishe Kohan Dec 1 '16 at 5:29
  • $\begingroup$ But in order to understand the limit circle, one must somehow first understand the action of $\Gamma$ on $\mathbb H^3$, right ? $\endgroup$ – H1ghfiv3 Dec 1 '16 at 10:14

More a (very) long comment than an answer

Also I am not knowledgable on Fuchsian Group and $\mathbb H^3$ ( hyperbolic $3$-space) but I know a bit about hyperbolic $2$-space and extrapolate a bit from that. so maybe some of my comments are just plain wrong.

But still some , I hope helpful comments: (maybe I am a bit pedantic at some points but that is also to help grow my own understanding)

Be careful with Hyperbolic planes (planes in hyperbolic geometry using the Poincare half plane model) and Hyperplanes (surfaces of points equidistant to an hyperbolic plane) they are not the same, I think you everywhere mean Hyperbolic planes.

You make some mistakes with your indices I guess $ z = x_1 + x_2 i$ and $x_3 > 0 $

Hyperbolic planes in $\mathbb H^3$ correspond to planes/ hemispheres perpendicular to the boundary of the model (the euclidean $x_3 = 0 $ -plane.)

Also I think $\lambda$ ranges over $\mathbb C$ (not just $\mathbb R$ ) otherwise the 3 isometries do not cover all isometries , (for example if $\lambda$ ranges over $\mathbb R$ no isometry can be built that maps the $ z + \overline{z} = 0 $ plane to the $ z - \overline{z} = 0 $ plane)

Then there are some hyperbolic planes that are invariant.

in 1) $$(z,x_3) \mapsto (\lambda z, |\lambda|x_3): \lambda \neq 0 $$

Leaves the plane unit hemisphere invariant if $ |\lambda | =1 $

(the unit hemisphere is $z\overline{z} +x_3^2 = 1$ )

in 3) $$(z,x_3) \mapsto (\frac{-\overline{z}}{|z|^2 + x_3^2},\frac{x_3}{|z|^2 + x_3^2}) $$

Leaves the unit hemisphere, and the planes $ z + \overline{z} = 0 $ and$ z - \overline{z} = 0 $ invariant. (and maybe more important moves every point inside the unit hemisphere to outside the hemisphere and every point outside the hemisphere to inside the hemisphere)

But the real question here is why is that important? Is it not more that some combinations of isometries are able to create invariant hyperplanes not that one of the 3 (primal) isometries can do that.

Please check all I wrote, I am even less knowledgable on Fuchsian Group than you but I hope this hyperbolic geometry view helps

  • $\begingroup$ I used x_0,x_1,x_2 as notation for the coordinates. I've changed it up now to match your far more conventional notation. Sorry for the confusion. $\endgroup$ – H1ghfiv3 Dec 1 '16 at 9:58
  • $\begingroup$ Okay, I think I found some helpful information: Any element $g \in PSL(2,\mathbb C)$ leaving some hyperbolic plane $P \subset \mathbb H^3$ invariant (so $P$ is some plane/hemisphere perpendicular to $x_3 = 0$) is known as an hyperbolic element. Looking at the extended action on $\hat {\mathbb C} \subset \partial \mathbb H^3$, $g$ preserves $P$ if and only if it both fixes precisely two points $p,q$ on $\partial P \subset \partial \mathbb H^3$ and satisfies $tr(g)^2 \in \mathbb R^+ \setminus [0,4]$. The limit set $\Lambda(g)$ (set of accumulation points) of $g$ is just $\{p,q\}$.... $\endgroup$ – H1ghfiv3 Dec 1 '16 at 12:35
  • $\begingroup$ Consequently, if $\Gamma$ is the fundamental group of a surface, then $\Gamma$ is in particular a torsion-free fuchsian group of first type. This means that, if $\Gamma$ preserves some plane $P \subset \mathbb H^3$, then all elements $g \in \Gamma$ are hyperbolic in the above sense and satisfy $\Lambda(\Gamma) := \bigcup_{g \in \Gamma} \Lambda(g) = \partial P$ (the limit circle Moishe was alluding to). Thus, the only thing left to check is that the action of $\Gamma$ on $P$ is exactly what we would expect, so that $P \backslash \Gamma$ is indeed homemorphic to $\Sigma$. $\endgroup$ – H1ghfiv3 Dec 1 '16 at 12:41
  • $\begingroup$ You rate me much to high , I am not even sure if my notation is any more conventional than yours , I just wanted to make it consistent and easy to understand (to me), seeing the edit you made: In the original post just once the mistake of using $x_3$ instead of $x_2$, I should have noticed that earlier and stuck to $z$ and $x_2$ ). Also I don't understand more than 80 % of what you wrote now (I only understand a bit of hyperbolic geometry) but I hope my post helped you and maybe someday i will understand your comments :) But can you help me: does $\lambda$ range over C or R ? $\endgroup$ – Willemien Dec 1 '16 at 23:03
  • $\begingroup$ Sorry for the late response, had a busy weekend. $\lambda$ may be any complex number. I will gladly try to make the stuff I wrote above clearer and less obscure, but maybe this is better suited for a private conversation ? Also, this short overview really helped me out with understanding how isometries act on $\mathbb H^3$. $\endgroup$ – H1ghfiv3 Dec 5 '16 at 13:36

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