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Sam Nead's answer does it, but perhaps I can offer a slightly different perspective on Question 1. No complicated hyperbolic gluing results are needed. I assume we are satisfied with the characterisat …

I think your questions are answered by this paper of Leininger, McReynolds, Neumann and Reid. In their terminology, your first question is asking about manifolds with equal length sets, and your seco …

There are plenty of other possibilities. Here are a few examples: The boundary of the fundamental group of an acylindrical hyperbolic 3-manifold with totally geodesic boundary is homeomorphic to a S …

Your group is the fundamental group of a 2-dimensional orbifold with underlying surface the 2-sphere and 3 cone points of order p. It follows that it acts properly discontinuously and cocompactly by …

[17/4/17: edited to correct proof.] This is true. First, we need a lemma which builds a related cover. Throughout, $\alpha$ is a simple closed geodesic of length $\ell$, and $\beta_1,\ldots,\beta_n$ …

This question is certainly open in general. I don't know if anyone has formally expressed an 'expectation' in print, but you might be interested in the following pieces of positive evidence. I believ …

Proposition 9.6 of M. Kapovich's 'Homological dimension and critical exponent of Kleinian groups' (it's Proposition 9.5 of the arXiv version) asserts that the cohomological dimension of the pair $(G,\ …

The answer to your Question 1 is certainly 'yes', for the reason you already explained: $\mathbb{H}^2/K_4$ is a Riemann surface, and its quotient by the finite group $B(2,4)$ is compact. I don't know …

It does hold - Wise proved that finite-volume non-compact hyperbolic 3-manifolds are virtually special, hence virtually RFRS and so virtually fibred by one of Agol's results. Details and references a …

This sort of thing was worked out by Coornaert, who constructed Patterson--Sullivan measures on the boundaries of word-hyperbolic groups: Michel Coornaert, Mesures de Patterson-Sullivan sur le bord d …

I think Lee's and Steve's comments pretty much answer this question. Let me try to summarize, and clear up a couple of misconceptions that seem to be lurking. For convenience, I'll denote your group …

As this is too long for a comment, I'll post it as an answer. As you seem to be interested in CAT(0) cube complexes, I want to point out that there is, in fact, a ninth type of quasiconvexity, which …

Agol's original paper on RFRS gives a nice short proof that the fundamental group of any manifold which also happens to be a finite-index subgroup of your favourite right-angled reflection group is RF …

One can also see it using the theory of ends. If $\pi_1M$ were freely decomposable, then it would follow from the easy direction of Stallings' Ends Theorem that $\pi_1M$ had two or infinitely many en …

Theorem(Long and Niblo): If $M$ is a 3-manifold and $S$ is an incompressible component of $\partial M$ then $\pi_1 S$ is separable in $\pi_1 M$ (pick a base point in $S$ to make sense of this). This …