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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. This kind of stuff can be found, But the best I believe, in Bill 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 $\Gamma \to SL(2,\mathbb{R}))$, but I do which is not know a published referencequite the same thing. For an exampleTo 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 $i : \Gamma \to \textrm{Isom}^{+}(\mathbb{H}^{2})$. It is possible to perturb $i$ to become injectiveSL(2,R)$. But the Euler number of a representation is continuous under this perturbationBy Theorem 3.19, and the representation$i$itself has Euler number zero, so the may be perturbed representation does tooto be injective. It must therefore The perturbed representation cannot be indiscretediscrete, because since (as pointed out by Misha in his comment) discrete faithful representations of are a closed subset. EDITS: Added/corrected some references. And I corrected the original version which asserted the stronger result that$\Gamma$all have Euler number i$ can be perturbed in $\pm 2$\textrm{Isom}^{+}(\mathbb{H}^{2})$. And as Kelso points out, one canNOT perturb$i$so that all nonidentity elements are hyperbolic. Goldman's thesis gives references for this result in Siegel's "Topics in Complex Function Theory II" and Jorgensen's "A note on subgroups of SL(2,C)", and also attributes it originally to Nielsen but without reference. 2 deleted 80 characters in body If$\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$. The boundary of the octagon maps to a 1-complex consisting of four based closed geodesics$\hat\alpha,\hat\beta,\hat\gamma,\hat\delta$, , all with the same base point$p \in \Sigma$but otherwise disjoint. Each of these geodesics has a$\pi/2$angle at$p$, though. If you pull them tight, you get four totally geodesic embedded circles$\alpha, \beta, \gamma, \delta$. C_a, C_b, C_c, C_d$. The circles $\alpha,\beta$ C_a,C_b$intersect each other transversely in 1 point, as do the circles$\gamma,\delta$, C_c,C_d$, and the $\alpha,\beta$ C_a,C_b$pair is disjoint from the$\gamma,\delta$C_c,C_d$ pair. The complement $S - (\alpha\cup\beta\cup\gamma\cup\delta)$ 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$, that is, any discrete faithful representation$\Gamma C_c \to \textrm{Isom}^{+}(\mathbb{H}^{2})$. All nontrivial elements are hyperbolic. When you straighten the four circles$\alpha,\beta,\gamma,\delta$in this C_a,C_b,C_c,C_d$ on the new quotient hyperbolic structuresurface, 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. This kind of stuff can be found, I believe, in Bill Goldman's thesis, but I do not know a published reference. For an example, 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 $i : \Gamma \to \textrm{Isom}^{+}(\mathbb{H}^{2})$. It is possible to perturb $i$ to become injective. But the Euler number of a representation is continuous under this perturbation, and the representation $i$ itself has Euler number zero, so the perturbed representation does too. It must therefore be indiscrete, because discrete faithful representations of $\Gamma$ all have Euler number $\pm 2$.
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 surface $\Sigma = \mathbb{H}^2 / \Gamma$. The boundary of the octagon maps to a 1-complex consisting of four based closed geodesics $\hat\alpha,\hat\beta,\hat\gamma,\hat\delta$, 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 $\alpha, \beta, \gamma, \delta$. The circles $\alpha,\beta$ intersect each other transversely in 1 point, as do the circles $\gamma,\delta$, and the $\alpha,\beta$ pair is disjoint from the $\gamma,\delta$ pair. The complement $S - (\alpha\cup\beta\cup\gamma\cup\delta)$ 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 $\Gamma \to \textrm{Isom}^{+}(\mathbb{H}^{2})$. All nontrivial elements are hyperbolic. When you straighten the four circles $\alpha,\beta,\gamma,\delta$ in this new hyperbolic structure, 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. This kind of stuff can be found, I believe, in Bill Goldman's thesis, but I do not know a published reference. For an example, 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 $i : \Gamma \to \textrm{Isom}^{+}(\mathbb{H}^{2})$. It is possible to perturb $i$ to become injective. But the Euler number of a representation is continuous under this perturbation, and the representation $i$ itself has Euler number zero, so the perturbed representation does too. It must therefore be indiscrete, because discrete faithful representations of $\Gamma$ all have Euler number $\pm 2$.