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6

Examples in dimension $3$ follow from hyperbolization theorems for 3-manifolds combined with Mostow rigidity. Suppose that you have a compact 3-manifold $M$ with connected boundary $\partial M$ of genus $\ge 2$, such that $M$ is irreducible, atoroidal, and acylindrical and $\partial M$ is incompressible --- in brief this says $M$ has no "badly embedded" ...


5

Here's a specific example of such a manifold, known as the Thurston tripus or knotted Y. It is described in Example 3.3.12 of Thurston's Three-Dimensional Topology and Geometry, Volume 1. The following picture shows a hyperbolic polyhedron in $\mathbb{H}^3$ under the Poincaré ball model. It has the combinatorial structure of a truncated tetrahedron. ...


3

Your answer is almost correct. You have that straight lines are the critical points of the arc-lenght functional. These lines minimize arc-lenght if they are timelike or lightlike, but they maximize arc-lenght if spacelike. This is sort of expected, since the connection of this Lorentz plane is the same as the connection from $\Bbb R^2$ and the Christoffel ...


3

Start with the right triangle with corner angles $$ \alpha=\frac\pi7,\quad\beta=\frac\pi3,\quad\gamma=\frac\pi2 $$ It is a building block for a tiling with regular heptagons, three meeting at each vertex. The hyperbolic law of cosines gives you \begin{align*} \cosh a &= \frac{\cos\alpha + \cos\beta\cos\gamma}{\sin\beta\sin\gamma} = ...


2

I think that confusion arose because the OP of the other question used the term "hyperbolic plane" loosely, and not in a precise way. Mainly, we denote by $\Bbb L^2$ the pair $(\Bbb R^2, {\rm d}s^2)$, where $${\rm d}s^2 = -{\rm d}x^2 + {\rm d}y^2.$$ This is another way of saying that the scalar product (which is not positive definite anymore) between two ...


2

A circle orthogonal to the unit circle through the points $P,Q$ must go through $P^{-1}$ and $Q^{-1}$, too, where $P^{-1},Q^{-1}$ are the circular inverses of $P$ and $Q$ with respect to the unit circle. That simply follows from the Tangent-Secant theorem. Hence the hyperbolic line through $P=(x_1,y_1)$ and $Q=(x_2,y_2)$ is just an arc of the circle ...


2

Yes, it is. Let $\Phi$ be an isotopy. That is $\Phi:I\times M\to M$ is smooth, such that $\Phi(0,p)=p$ and $\Phi(1,p)=\phi(p)$. Let $\overline{\Phi}:I\times \overline{M}\to\overline{M}$ denote the lift of $\Phi$ equal to the identity at time $0$. By compactness, there is some $\lambda>0$ such that for every $(t,p)\in I\times M$ we have ...


2

alright, made a jpeg. First method: start with point hy. Central projection to the Beltrami-Klein model at point k. Vertical projection to the upper hemisphere at point he. Stereographic projection from the South Pole $-1$ to point p in the Poincare disk. Second method: project hy towards the South Pole directly to the Poincare Disk at point p. The ...


2

different problems may be easiest in different models. This one is best in the upper half plane. You need to know that any Mobius transformation preserves angles (including right angles) and maps any line or circle into either a line or circle. To take the unit disc to the upper half plane, use $$ f(z) = \frac{z + i}{iz+1}. $$ To take the upper half ...


2

Defining Conic Sections The first question, of course, is how to even define conic sections on the hyperbolic plane. The usual method is to use the hyperboloid model, which identifies the hyperbolic plane with the hyperboloid $$ x^2 + y^2 - z^2 = -1,\qquad z\geq 1. $$ In this model, isometries of the hyperbolic plane correspond to linear transformations of ...


2

For a curve $(x,y) = \bigl(x(t),y(t)\bigr)$ in $\mathbb{H}$, the parallel transport of a vector $\textbf{v}$ is determined by the following differential equation: $$ \frac{d\textbf{v}}{dt} \;=\; \frac{1}{y}\frac{dy}{dt}\,\textbf{v} \,-\, \frac{1}{y} \frac{dx}{dt}\,\textbf{v}^\perp $$ where $\textbf{v}^\perp$ is the vector obtained by rotating $\textbf{v}$ ...


1

I am not sure where the problem is but let me present my solution. In order to find the transport of $(0,1)$ at point $a(t)$ we need to find parallel vector field $V$ along $a$ with $V(0)=(0,1)$ and compute $V(t)$. Let: $$V(t)=V_1(t) \partial_1(a(t)) + V_2(t) \partial_2(a(t))$$ Than: $$D_tV=\sum_{k}(V_k'+\sum_{i,j}\Gamma^k_{ij}a_i'V_j)\partial_k$$ It is ...


1

Classification by points of intersection If you think about the three non-degenerate types ellipse, parabola and hyperbola, then these are classified by their points of intersection with the line at infinity. You either have two distinct points of intersection for a hyperbola, or a single point of tangency with algebraic multiplicity two for the parabola, ...


1

$\newcommand{\Reals}{\mathbf{R}}$For what it's worth, here's a plot of two such families of "equally-spaced" lines in the Poincaré model. (That is, the spacing along the "horizontal axis" of two adjacent "vertical" lines is the same for each pair, and similarly for the second family.) As expected, a line in one family fails to intersect some lines from the ...


1

Let $P$ and $Q$ be the points. Tee "line" you want is a circle $C$ containing $P$ and $Q$ and perpendicular to the unit circle, meeting it at points $A$ and $B$. (Draw that picture for yourself, please). If $P$ and $Q$ lie on a diameter of the unit circle (i.e., if $PQ$ contains the origin), then the solution is the (euclidean) line $PQ$. Else: Draw the ...


1

As user23142 mentioned there is no such model (in every model there should be unique lines between any two points on the boundary , and with a square that is not possible. Maybe an alternative: There is the Gans model ( https://en.wikipedia.org/wiki/Hyperbolic_geometry#The_Gans_Model ) which is more or less the hyperboloid model but then without the ...



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