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For Question 2: In addition to Kellerhals' lower bounds on $\epsilon(n)$, there is an absolute constant $C>0$ such that $$ \epsilon(n)\le \frac{C}{\sqrt{n}}, $$ see Proposition 5.2 in Belolipetsky, Mi …

There is such a proof given by Hjelmslev; it is based on a clever application of central symmetries (point-reflections). You can find it on pages 102-104 of Ф. Бахман, Построение геометрии на основе …

The thing is that every compact Riemannian surface admits a $C^\infty$ isometric embedding in ${\mathbb E}^5$, see Michael Albanese's answer here; the result is a version of the Nash isometric embeddi …

Let $M$ be a compact hyperbolic 3-manifold and let $C(M)$ be the moduli space of flat conformal structures on $M$. The hyperbolic metric on $M$ defines a distinguished point $h\in C(M)$. One says tha …

First of all, I would use a different name for what you call geodesics on $X$, let's call them regular geodesics. The reason is that geodesics in Riemannian geometry are curves satisfying the equation …

I am sure this was known earlier (check Ivanov's book, most likely, it is there), but you can refer to the following theorem of Lindenstraus and Mirzakhani (see their paper Ergodic theory of the space …

This story is somewhat reminiscent of a (probably apocryphal) story of somebody writing a telegram to a mathematical research institute (Steklov Institute in Moscow, if I remember it correctly): "Sol …

To make things interesting, I will consider two classes of manifolds (one strictly larger than the other) without compactness assumption: connected manifolds (dimension is finite but not fixed) admi …

You definition is closely related to the notion of the "cone type" introduced by Jim Cannon. As Yves noted, $U(x,y)$ is compact. It is also connected. Compactness part is immediate from the Arzela- …

Let us say that an abstract group $\Gamma$ is self-contained (I just made up this terminology) if $\Gamma$ is isomorphic to its proper subgroup of finite index. Next, each discrete group of isometri …

B.Bowditch, Geometrical finiteness for hyperbolic groups. J. Funct. Anal. 113 (1993), no. 2, 245–317. The statement that you need is a corollary of his Proposition 5.6 in conjunction with finiteness …

Since the fiber is contractible, there is a global section. The details are in the papers Earle, Clifford J.; Eells, James, A fibre bundle description of Teichmüller theory. J. Differential Geometr …

These are difficult questions and very little in general is known about this. Mostly, what's known is ad hoc results for specific classes of manifolds (say, take some surgeries on 2-bridge knots...). …

I assume that $\varphi$ has finite order; a similar construction works in the infinite order case, just it is a bit harder. Consider $D$, the complement to the fixed point of $\varphi$. Then $S'=D/<\v …

For the second question the answer is obviously negative since distinct Euclidean lines cannot converge to each other (in the sense that the distance function goes to zero). For the first question, I …