51

A nice summary of the status of these problems may be found here:
Otal, Jean-Pierre, William P. Thurston: ``Three-dimensional manifolds, Kleinian groups and hyperbolic geometry", Jahresber. Dtsch. Math.-Ver. 116, No. 1, 3-20 (2014). ZBL1301.00035.
I would charaterize it this way: Problems 1-18 have been completely answered, although some of the answers are ...

36

By chance I wrote, not long ago, the following applet (HTML5+JS+WebGL) that works at least on Firefox and Chrome.
https://www.math.univ-toulouse.fr/~cheritat/AppletsDivers/Escher/
This work is CC-BY-SA, including the code, but NOT the image by Escher, for which I have not asked permission: you can probably use it a few times in conferences (fair use) but ...

32

Binary Tiling
In fact, one can tile the hyperbolic plane with arbitrarily small tiles. There is a tiling of the hyperbolic plane (apparently due to Boroczky) by pentagons.
The horizontal edges are horocycles in the upper half-space model of the hyperbolic plane, and the vertical lines geodesics. The edge at the top of each tile is half the length of the ...

32

Yes, it immerses isometrically into certain solvmanifolds. Take an Anosov map of $T^2$, such as $\left[\begin{array}{cc}2 & 1 \\1 & 1\end{array}\right]$. The mapping torus admits a locally homogeneous metric modeled on the 3-dimensional unimodular solvable Lie group. The matrix has two eigenspaces with eigenvalues $\frac{3\pm\sqrt{5}}{2}$, and the ...

24

Live isometries of a hyperbolic ornament
As part of my dissertation Creating Hyperbolic Ornaments I wrote some Java software which might serve your needs. It's called morenaments conform (as of now). In particular, I can input any Euclidean ornament, hyperbolize that and then perform isometric transformations of the resulting hyperbolic ornament using ...

23

They have all been resolved in some fashion with the exception of problem 23, which asserts that the volumes of all hyperbolic 3-manifolds are not rationally related. We know basically nothing about the diophantine properties of hyperbolic volumes.

22

In genus 2, every surface has a symmetry, namely a hyperelliptic involution. In higher genus, generic surfaces will not have any symmetries. If a surface has a non-trivial symmetry group, then the quotient by the symmetry group will be an orbifold, and the moduli space of hyperbolic structures on this orbifold will have strictly smaller dimension. Hence ...

21

Edit: In a comment below, coudy asks for the following source to be included as the first in English:
Anosov, D. V. Geodesic flows on closed Riemann manifolds with negative curvature. Proceedings of the Steklov Institute of Mathematics, No. 90 (1967). Translated from the Russian by S. Feder American Mathematical Society, Providence, R.I. 1969 iv+235 pp.
...

answered Dec 26 '13 at 6:20

Benjamin Dickman

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21

Things will simplify if you just consider the circles on the Riemann sphere $S^2 = \mathbb{C}\cup\{\infty\}$, for your space is simply the space of circles on the sphere (with the lines in $\mathbb{C}$ just becoming the circles through $\infty$). Thus, your extended space of circles becomes $M^3 = \mathrm{PSL}(2,\mathbb{C})/H$ where $H\simeq\mathrm{PSL}(2,\...

answered Dec 11 '14 at 16:13

Robert Bryant

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21

About the relation to fractals: for Fuchsian groups of the first kind, the limit set has Hausdorff dimension 1, i.e., it is not fractal.
However, for all other quasifuchsian groups of the first kind, the limit set has Hausdorff dimension strictly bigger than 1, by a theorem of Rufus Bowen.

20

Shin and Strenner have shown that the conjecture is false when 3g + n > 4.
See http://arxiv.org/abs/1410.6974

19

There is no universally accepted model of random three-manifolds (or random knots/links) for that matter, however, hyperbolicity is pervasive in all known models. The most popular (but not really satisfying) model of three-manifolds is the Dunfield-Thurston model:
Finite covers of random 3-manifolds.
Nathan M. Dunfield, William P. Thurston.
Invent. Math. ...

answered Feb 8 '16 at 16:42

Igor Rivin

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19

The tilings mentioned by Ian Agol are related to an action of a Baumslag-Solitar group $\{ a,b \bigg| b^{-1}a^2b=a \}$ on the hyperbolic plane. They have arbitrarily small area, but diameter uniformly bounded away from $0$. It is possible to tesselate the hyperbolic plane with a single tile with arbitrarily small diameter, too. Let there be $n$ arcs on top ...

19

All closed hyperbolic 3-manifold groups embed into a compact Lie group.
To prove this, note first of all that given a hyperbolic 3-manifold $M$, it suffices to show that a finite-index subgroup $G\leq \pi_1(M)$ of index $m$ embeds into a compact Lie group. Then the representation $\rho: G\hookrightarrow O(n)$ will induce a represenation $Ind_G^{\pi_1(M)} \...

18

I think Scott's argument is that the lengths of $6g-6$ curves can't form coordinates for Teichmuller space. If one has $6g-6$ geodesics which parameterize, then they must be filling (they meet every simple closed curve). But the length of a filling (immersed) curve is proper in Teichmuller space, and hence the infimum of the length is achieved (in fact, at a ...

18

Check out Indra's Pearls. (Mumford, Series, Wright).

answered Jan 21 '17 at 22:11

Igor Rivin

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17

Dehn gave at least three solutions of the conjugacy problem for surface
groups, which can be found in my translation in the book Papers on Group
Theory and Topology (Springer 1986), Papers 2, 4, and 5.
The first is based on an idea of PoincarÃ©: lifting a curve to the universal
cover, which is the disk model of the hyperbolic plane, replacing it by the
...

answered Jul 17 '13 at 6:09

John Stillwell

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17

The Nash-Kuiper embedding theorem applies here, as the obvious smooth topological embedding of the hyperbolic plane as the unit disk in 3-dimensional Euclidean space decreases distances. Therefore there is a $C^1$ isometric embedding. I don't know of a stronger result.

dg.differential-geometry ap.analysis-of-pdes riemannian-geometry hyperbolic-geometry isometric-immersion

17

I'm assuming that your intention is that negatively curved means having negative sectional curvature.
Your question, with regard to uniqueness up to homeomorphism, is a special case of the Borel Conjecture (topological uniqueness of aspherical manifolds) which is still unsolved despite much progress. In the special case of negative curvature, there is a ...

17

I don't understand your reference to the model (since the geometry of the hyperbolic plane does not depend on any model), but, in fact, the Beltrami-Klein model demonstrates that any qualitative statement about convex sets in the Euclidean plane holds in the Hyperbolic plane and vice versa, since the model maps convex sets to convex sets.
EDIT This has (...

answered Feb 2 '18 at 20:01

Igor Rivin

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16

I would suggest John Barrett's essay on The Hyperbolic Theory of Special Relativity as a comprehensive answer.
The principle of relativity corresponds to the hypothesis that the kinematic space is a space of constant negative curvature. The value of the radius of curvature is the speed of light.
The relativistic law of combination of velocities can be ...

gt.geometric-topology dg.differential-geometry mp.mathematical-physics hyperbolic-geometry special-relativity

answered Mar 15 '13 at 12:41

Carlo Beenakker

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15

Sullivan proved that every closed hyperbolic manifold has a stably parallelizable finite cover. This is not true for say complex hyperbolic manifolds (of real dimension $>2$). See Farrell's "Lectures on Surgical Methods in Rigidity".
Real Pontryagin classes of complete hyperbolic (or more generally conformally flat) manifolds vanish.
Orientable closed ...

answered Jun 8 '13 at 14:16

Igor Belegradek

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15

To answer Joseph's questions:
First, it's not impossible to integrate the geodesic flow of the hyperbolic plane in these coordinates, but the formulae I got aren't very nice, so I'm not going to type them in unless I can find a better way to express them. It's probably easier than I got on a first pass through, but I don't have time to work on simplifying ...

answered Nov 25 '13 at 14:09

Robert Bryant

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15

The simplest answer is that the universal cover together with the deck group action contain a lot of information about the manifold, and the representations of the group provide one way to extract it.
Representation of the group correspond to either locally constant sheaves or flat connections.
The acyclic representations, i.e., those for which the ...

rt.representation-theory gt.geometric-topology hyperbolic-geometry quantum-topology character-varieties

answered Mar 27 '16 at 17:51

Liviu Nicolaescu

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14

For general $p$, the only known method is to construct a Dirichlet fundamental domain and read off the group presentation from it. The procedure for computation of a fundamental domain is called "Jorgenesen's algorithm: List elements of $\Gamma$ (using the embedding to $GL(2(n+1),Z)$ via restriction of scalars). For each $\gamma\in \Gamma$ construct the ...

14

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 Sierpinski carpet. (This appears in the Kapovich and Kleiner paper mentioned below, but must be standard---the point is the topological fact that any planar ...

14

The free discrete subgroups consist of the closure of the Riley slice of Schottky space (notice that one may assume $Re(\alpha)\geq 0, Im(\alpha)\geq 0$, since $\alpha \mapsto -\alpha, \overline{\alpha}$ preserves discreteness).
Here's a picture of the first quadrant of the Riley slice:
The exterior of the black fractal represents free discrete groups that ...

14

I do not know of any specific reference, but I will try to explain the situation a bit. For most of the stuff, one does not actually need references besides the paper of Borel-Serre defining the Borel-Serre compactification. Maybe the book of Elstrodt, Grunewald, Mennicke "Groups acting on hyperbolic space" contains some useful information, I currently do ...

14

Yes. the fundamental theorem is that the constructible angles in the non-Euclidean plane are exactly the constructible angle in the Euclidean plane. Lengths come from the two laws of cosines and the law of sines. In particular, length 1 is not constructible. As i recall, positive length $x$ is constructible if and only if $\sinh x$ is constructible in the ...

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