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I have a question about hyperbolic 2-torus, from now on donoted by $\Sigma_{2}$

Actually I've tried to prove that for a group $\Gamma \subset \textrm{Isom}^{+}(\mathbb{H}^{2})$ represented by $\Gamma = \left< a,b,c,d~~|~~[a,b][c,d]=1 \right>$, $a$,$b$,$c$, and $d$ are hyperbolic element in $\textrm{Isom}^{+}(\mathbb{H}^{2})$(which fix 2 points only on $\partial \mathbb{H}^{2}$.) if and only if $\mathbb{H}^{2}/\Gamma =\Sigma_{2}$. "If"-part is easy so I mainly consider "only if"-part.

It can be verified that one could make $\Sigma_{2}$ by pairing sides of a hyperbolic regular octagon which has angle $\frac{\pi}{4}$. Let the pairing isometries be $\alpha$, $\beta$, $\gamma$ and $\delta$$\in \textrm{Isom}^{+}(\mathbb{H}^{2})$. Then they generate a holonomy group, or deck transformation group of $\mathbb{H}^{2}$ with the relation $[\alpha,\beta][\gamma,\delta]=1$, where $[A,B]=ABA^{-1}B^{-1}$. Note that $\alpha$, $\beta$, $\gamma$ and $\delta$ should be hyperbolic elements.

Now the question is following :

For $\Gamma$, is there ALWAYS an 'corresponding' octagon(not necessarily regular) which is a fundamental domain for $\Sigma_{2}$?? I mean the side-pairing of the octagon induces $\Sigma_{2}$ and the paring maps are exactly $a$,$b$,$c$, and $d$.

Or, is it possible that the octagon is weird(?), for example, some of its edges intersect each other not at a vertex but at an middle of an edge? (in this case, the boundary of the octagon has self intersection.)

(I might have heard that sometimes the 'weird case' occurs but I'm not sure.)

Thanks in advance.

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I had a negative comment earlier, but I changed my mind. There is a slightly subtle point here regarding discreteness of $\Gamma$... – Lee Mosher Jun 20 '12 at 13:43
@Kenso: In order for the "easy" part of "if and only if" to be true you have to assume that the group $\Gamma$ is torsion-free (otherwise there are easy counter-examples). Are you making a similar assumption for the "hard" direction? You also have to be more precise with your definition of a "fundamental polygon". From your question, it appears that you do not use the standard definition. For instance, do you require "polygon" to be embedded (or only immersed)? Do you want the angle sum of each vertex cycle to be $2\pi$? – Misha Jun 20 '12 at 17:24
@Misha: If I'm not mistaken, I was confusing you because I did not mention that my $\Sigma_{2}$ is smooth. For easy part, I think $\Gamma$ does not have any elliptic elements(candidates of finite order) since it is the deck transformation group for $\Sigma$. For the "fundamental domain polygon", I meant that polygon whose edges are all geodesics in $\mathbb{H}^{2}$, which makes $a$~$d$ be the side-pairings to get $\Sigma_{2}$. Since $\Sigma_{2}$ should be smooth, the angle sum should be $2\pi$. – KENSO Jun 20 '12 at 18:14
To sum, my question is whether $\Sigma_{2}$ can be obtained from 'a geodesic octagon' in $\mathbb{H}^{2}$ by USING the generators $a$~$d$ of $\Gamma$ to side-pair, if $\Sigma_{2}=\mathbb{H}^{2}/ \Gamma$. – KENSO Jun 20 '12 at 18:42
Assuming that $\Gamma$ is torsion-free, the answer to your last question is yes and Lee sketched a proof in his answer. (The same result holds for quotients of any genus.) I would have given a different argument though. Note, however, that in order to get an octagonal fundamental domain bounded by geodesic segments you may have to replace your original generating set with a different one (the replacement is done via an automorphism of $\Gamma$). – Misha Jun 21 '12 at 4:55
up vote 2 down vote accepted

If $\Gamma$ is a discrete subgroup then the answer to your question is yes. Starting with a hyperbolic structure from a regular octagon with $\pi/4$ angles as you describe, consider the map from the octagon to the quotient hyperbolic surface $\Sigma = \mathbb{H}^2 / \Gamma$. The boundary of the octagon maps to a 1-complex consisting of four based closed geodesics, all with the same base point $p \in \Sigma$ but otherwise disjoint. Each of these geodesics has a $\pi/2$ angle at $p$, though. If you pull them tight, you get four totally geodesic embedded circles $C_a, C_b, C_c, C_d$. The circles $C_a,C_b$ intersect each other transversely in 1 point, as do the circles $C_c,C_d$, and the $C_a,C_b$ pair is disjoint from the $C_c,C_d$ pair. The complement $S - (C_a \cup C_b \cup C_c \cup C_d)$ is an annulus with ``scalloped'' boundary, each boundary component a concatenation of 4 geodesic arcs; the surface $S$ can be reconstructed by gluing these arcs in pairs in the appropriate fashion. Now pick any hyperbolic structure on $\Sigma$, that is, any discrete faithful representation $C_c \to \textrm{Isom}^{+}(\mathbb{H}^{2})$. All nontrivial elements are hyperbolic. When you straighten the four circles $C_a,C_b,C_c,C_d$ on the new quotient hyperbolic surface, you get the same intersection pattern and the same qualitative description of the complement, however the "shape" of that annulus has changed, i.e. the length of the core curve, the lengths of and angles between the scalloped sides, and possible "twisting" around the core curve. But you can still "pull in" the boundary of this annulus to give the octagon structure that you ask for. To do this, pick a corner on each boundary circle, drag that corner to the core circle of the annulus, and then drag the two corners along the core circle until they touch; when you do this dragging maneuver, no new identifications of the annulus boundary will be introduced until the two points touch and the annulus becomes an octagon with side pairings as you ask for (maybe you'll have to drag around and around the core curve in order to get the exact side pairings, rather than getting them conjugated by a power of the core curve due to twisting around the core of the annulus).

On the other hand, I believe there do exist injective homomorphisms $\Gamma \to \textrm{Isom}^{+}(\mathbb{H}^{2})$ with nondiscrete image so that $a,b,c,d$ are all hyperbolic, in which case what you ask for is hopeless. But the best I can do just by quoting known results is to use Theorem 3.19 of Goldman's thesis which says that injective homomorphisms are dense in the space of representations $\Gamma \to SL(2,\mathbb{R}))$, which is not quite the same thing. To apply this, pick one hyperbolic element of $\textrm{Isom}^{+}(\mathbb{H}^{2})$ and map all of $a,b,c,d$ to that element, which induces a non-injective homomorphism $i : \Gamma \to SL(2,R)$. By Theorem 3.19, the representation $i$ may be perturbed to be injective. The perturbed representation cannot be discrete, since (as pointed out by Misha in his comment) discrete faithful representations are a closed subset.

EDITS: Added/corrected some references. And I corrected the original version which asserted the stronger result that $i$ can be perturbed in $\textrm{Isom}^{+}(\mathbb{H}^{2})$.

And as Kelso points out, one canNOT perturb $i$ so that all nonidentity elements are hyperbolic. Goldman's thesis gives references for this result in Siegel's "Topics in Complex Function Theory II" and Jorgensen's "A note on subgroups of SL(2,C)", and also attributes it originally to Nielsen but without reference.

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@Lee: You do not need Goldman for this, it follows from the fact that the space of discrete and faithful representations is a component of the character variety $X(\pi_1(\Sigma), PSL(2, {\mathbb R})$: Limit of discrete and faithful representations is again discrete and faithful (Chuckrow-Jorgensen-et al). – Misha Jun 20 '12 at 15:32
@Lee: There are also discrete but non-faithful examples, obtained by mapping $\pi_1(\Sigma_2)$ to the fundamental group of the orbifold which is torus with two cone points of order $2$. By carefully choosing the homomorphism you can assume that generators have hyperbolic images. – Misha Jun 20 '12 at 15:36
@Kelso: You are right that one cannot find an indiscrete but faithful representation in which every nontrivial element to be hyperbolic, but it is possible to arrange (at least in SL(2,R)) for $a,b,c,d$ to be hyperbolic. – Lee Mosher Jun 20 '12 at 17:20
@Lee and @Kenso: Suppose that $\Sigma$ is a compact surface of genus $g\ge 2$ and $\rho: \pi=\pi_1(\Sigma)\to PSL(2, {\mathbb C})$ is a nonelementary representation (no discreteness assumptions here). Then there is a choice of "standard" generators $a_i, b_i, i=1,...,g$ (satisfying the product-of-commutators relation) for $\pi$, so that $\rho(a_i), \rho(b_i), i=1,...,g$, are all hyperbolic/loxodromic with pairwise disjoint fixed point sets. This follows from my paper with Gallo and Marden. – Misha Jun 20 '12 at 17:44
@Kenso: A group of isometries of hyperbolic space $H^n$ is called elementary if it either fixes a point in $H^n$ or a point on the boundary sphere of $H^n$ or preserves a geodesic in $H^n$. A representation to $Isom(H^n)$ whose image is not an elementary group is called nonelementary. – Misha Jun 21 '12 at 4:46

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