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Yes. For example, let $M$ be the figure-eight knot complement. So $M$ is a hyperbolic manifold with volume a bit more than 2. The manifold $M$ has a two-fold symmetry $\tau$ that fixes, pointwise, a …

Here is an expansion of Ian's answer. Dehn surgery on a hyperbolic knot or link generically gives another hyperbolic manifold. This follows from the Dehn surgery theorem; see Theorem 5.8.2 in chapt …

Before you try to fool SnapPea, remember that you'll almost certainly have to go above 16 (17?) crossings to do so - see https://doi.org/10.1007/BF03025227 for the tale of the tabulation of knots by H …

Here is a cool fact about $\mathrm{SL}(2, \mathbb{R}) / \mathrm{SL}(2, \mathbb{Z})$; it is homeomorphic to the complement of the trefoil knot in the three-sphere. Apparently this was first proved by …

NEW ANSWER: As there has been much confusion on this point (some of it mine...): Definition: A Riemannian 2-manifold $S$ is of hyperbolic type if the universal cover of $S$ is conformally equiva …

Consider the ideal triangle with vertices at infinity, zero, and one. Let $C$ be the semicircle perpendicular to the vertical line $[0, \infty]$ and meeting $1/2 + i/2$ (ie the midpoint of the semici …

One answer to your question comes from the paper The Weber-Seifert dodecahedral space is non-Haken by Burton, Rubinstein, and Tillmann. An earlier example is (say) the $(1, 2)$-Dehn filling of the fig …

This three-manifold can also be constructed by taking a genus two surface $S$, crossing with the interval $I$ to get $S \times I$, and attaching a pair of one-handles both of which connect $S \times \ …

$\newcommand{\HH}{\mathbb{H}}$Here is an expansion of what Anton is saying. Suppose that $M$ is a closed hyperbolic three-manifold. It follows that the universal cover of $M$ is $\HH^3$: hyperboli …

This is called the "bigon criterion". For a discussion, see Section 1.2.4 (and in particular Proposition 1.7) of the "Primer on mapping class groups" by Farb and Margalit. The Google search "bigon cr …

Here are two examples of exceptional hyperbolic fillings. In the first the core curve becomes parabolic after filling (so cannot be isotopic to a geodesic). In the second, the core curve is homotopic …

Suppose that $c = \gamma^n$ in $\pi_1(X)$. Note that, as $\pi_1(X)$ is torsion free and $c$ is assumed to be non-trivial, the element $\gamma$ generates an infinite cyclic subgroup $\langle \gamma \r …

$U$ and $L$ generate. Thus so do $AUA^{-1}$ and $ALA^{-1}$, for any element $A$ in $\mathrm{SL}(2, \mathbb{Z})$. These are the only pairs of conjugates which generate. The shortest proof I know is …

If I add the assumption that the given graph has bounded degree, and the same holds for the dual graph, then the answer to your question is no. A positive Cheeger constant implies a linear isoperimet …

See the paper "Effects of a change of pants decompositions on their Fenchel-Nielsen coordinates" by Takayuki Okai, published in Kobe J. Math. You can also find a version where some of the boundary co …