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5

There are at least four relevant articles of Beltrami. In the first one (Ann. Mat. Pura App.I (1865), no. 7, 185-204) he computes all two dimensional Riemannian metrics for which a coordinate system exists where all geodesics are represented as straight lines: he finds out that only surfaces of constant curvature have this property. I have little doubts that ...


4

If a Mobius transformation $$\phi: z \mapsto \frac{a z + b}{c z + d}$$ preserves the upper half-plane, then by continuity (as a self-map of the Riemann sphere) it must map $\mathbb{R} \cup \{\infty\}$ to itself, and some algebra forces that (possibly after canceling a common nonreal factor) $a, b, c, d$ are all real. Any such map maps the u.h.p. either to ...


2

In the first place, $\cos(z) = \dfrac{e^{iz}+e^{-iz}}{2} = \cosh(iz)$ and $\sin(z) = \dfrac{e^{iz}-e^{-iz}}{2i}=-i\sinh(iz)$. So why don't people stick to the exponential function and forget about all trigonometric or hyperbolic functions? In some cases it is indeed better to go straight to the complex exponential function, but in other cases one might ...


2

These functions, all tightly related to each other, nonetheless have different mathematical applications. In particular, the hyperbolic trigonometric functions have applications to hyperbolic geometry. For instance, in the hyperbolic plane the circumference of a circle of radius $r$ equals $2\sinh(r)$.


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First let me correct a mistake in your question: is it WRONG to think of this as an astroid. It is just 4 arcs of circles, NOT an astroid, though it looks somewhat similar. Second. The conformal map is certainly NOT fractional-linear (which you call Mobius). This conformal map is written explicitly in the paper I already referred to on MO: arXiv:1110.2696. ...


1

Actually, one of the very few formulas in Riemann's Habilitation provides a metric $\sqrt{\sum dx^2}/(1+\alpha/4\sum x^2)$ which, for various values of $\alpha$, gives hyperbolic, spherical, and Euclidean metric.


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There is a section “The Geometry of Commutators” on pg 184 in Alan Beardon’s book “The Geometry of Discrete Groups” that investigates just this kind of thing. This kind of reasoning can be found in that section: Suppose both $f,g$ are hyperbolic and the axis of $f$ and $g(\mbox{axis of f})$ do not meet. To classify $[f,g]$ as elliptic, parabolic or ...


1

A homothetic transformation of the Euclidean plane preserves the absolute values of angles but scales areas by $\lambda^2$. In hyperbolic geometry, the area of a polygon is equal to the its angle deficit. So as long as its angles don't change, its area can't change either. For this reason, the identity transformation with $\lambda=1$ is the only homothetic ...


1

I wasn't sure whether you saw this in chat so I'm posting as an answer here: Even under the stronger hypothesis that $M$ be simply connected of nonpositive curvature and a visibility manifold, there can exist pairs of bi-asymptotic geodesics. To construct a counterexample, let $N$ be a negatively curved surface with a single cusp. Chop off the cusp at ...


1

I don't think it has a special name. the line is not so special. (every point has one, two of these lines from different points but the same axis can intersect, while if they are perpendiculatr to the same line they don't) Maybe best is to describe it as: "the line trough $P$ that is perpendicular to the perpendicular from $P$ to $l$" or "the line ...


1

Avoid squaring whenever possible as it immediately introduces extraneous root(s) We have $\displaystyle\frac{e^{iz}+e^{-iz}}2=\sqrt2$ $$\iff(e^{iz})^2-2\sqrt2(e^{iz})+1=0$$ $$\implies e^{iz}=\dfrac{2\sqrt2\pm\sqrt{8-4}}2=\sqrt2\pm1$$ $$\iff iz=\log(\sqrt2\pm1)$$ $$\iff z=-i\log(\sqrt2\pm1)$$


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First separate to cases $y=0$ and $y \neq 0$. You will find that the first has no solution. The latter implies $x=k\pi$ that implies $y = (-1)^k \textrm{arcosh}(\sqrt{2}) = (-1)^k \ln(\sqrt{2} + 1)$.


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Question 1: The constant-curvature metric on $B$ is unique. To see this, note that equipping $B$ with any metric of constant curvature $-1$ gives us a biholomorphim (aka a conformal isomorphism) between the universal covering surface of $B$ and the unit disk in the complex plane. Combining two such metrics would give us a biholomorphism from the disk to ...


1

Here are some hints about how to do this constructively. Suppose $p,q$ are points in the hyperbolic plane such that the length of the arc of the horocycle between $p,q$ is equal to $\ell$. Then we may assume that, in the upper half plane, $p=(0,1)$ and $q=(\ell,1)$. To get a formula for the length of the geodesic segment between $p,q$, construct a ...



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