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Assignmnet 12

Problems

Getting Used to Hyperbolic Space

Exercise 1 (Hyperbolic Circle Area) In this exercise you will go through the transition-style arguments we use to turn formulas on the sphere into thier analogs in hyperbolic geometry, much like we did in class.

The starting point is the formula for the area of a circle of radius \(r\) on the sphere of radius \(R\):

\[A(r)=\int_0^r C(r)dr=\int_0^r 2\pi\sin\left(\frac{r}{R}\right)dr=2\pi R^2\left(1-\cos\frac{r}{R}\right)\]

  • Re-express this in terms of curvature
  • Convert to the Taylor series
  • Plug in \(\kappa=-1\)
  • Convert back to hyperbolic trigonometric functions

What is the correct hyperbolic formula? (We wrote it down without doing the full derivation in class, so you can confirm)

Exercise 2 (Hyperbolic Pizza) One way to try and develop intuition for the strange behavior of circles is to think about the type of circles we see in daily life: pizzas! One major factor determining how good a pizza is is its crust percentage which we will define as \[\mathrm{CrustPercent}=\frac{\area(\mathrm{Crust})}{\area(\mathrm{Pizza})}\]

In this probelm we will consider pizzas which have 1 inch crusts: meaning a 10 inch (radius) pizza has a 9inch radius center of toppings, surrounded by a 1 inch thick circle of crust.

  • Show the CrustPercent for Euclidean pizza is \[\frac{2}{r}-\frac{1}{r^2}.\] From this we see that as \(r\to \infty\) the crust percent drops to zero: this makes sense, if you imagine an extremely large pizza with only a 1inch thick crust, it’s totally reasonable that most of the pizza is not crust!
  • What is the CrustPercent for a hyperbolic pizza of radius \(r\)? Show that when \(r\) is large, this limits to the constant \[\mathrm{CrustPercent}\to 1-\frac{1}{e}\approx 63\%\] Thus crust is an inevitable part of life in hyperbolic space: no matter what size pizza you make it will always be well over half crust!

Exercise 3 (Hyperbolic Pizza II) In this problem, we will imagine our unit to be inches (so, the radius appearing in formulas for space of curvature \(-1\) is measured in inches).

You are at a pizzaria and are trying to decide if the 5 inch radius pizza they sell is large enough for you and your friends. They also sell a six inch (radius) pizza, but it costs twice as much. At first, you think this sounds crazy! But is this actually a good deal, or not?

Your friend is feeling very hungry, and jokingly asks the pizzamaker how large of a pizza he would need to order so that its areas is the same as an american football field (\(100\times 50\) yards). The pizzamaker says “I think I have room for that in my oven, coming right up!” How big of a pizza is he going to make?

Hint: invert the formula for area in terms of radius, to get radius in terms of area, then plug into a calculator!

Working with the Models

Exercise 4 (Hyperbolic space is homogeneous) We proved in class that the horizontal translations \(T(x,y)=(x+t,y)\) are isometries of the Half Plane model, and we also proved that the similarities \(S(x,y)=(sx,sy)\) are also isometries.

Combining these two, prove that \(\HH^2\) is homogeneous: that is, that for any two points \(p,q\in\HH^2\), there exists an isometry that takes \(p\) to \(q\).

Hint: can you first show that you can build an isometry that takes \((0,1)\) to any point in the plane? Then combine two of these to get what you want?

Exercise 5 (The Circumference of Circles) In the Disk Model, if \(a<1\) the Euclidean circle \(x^2+y^2=a^2\) centered at \(O\) is also a hyperbolic circle. In the text, we compute that its hyperbolic circumference is

\[C = \frac{4\pi a}{1-a^2}\] and that its hyperbolic radius is

\[r=2\mathrm{arctanh}(a)\]

Using these two, prove that \(C(r)=2\pi\sinh(r)\). Hint: solve for \(a\) in terms of \(r\) and plug into circumference. Then use hyperbolic trigonometric identites to simplify!

Together these two arguments prove that the geometry modeled by the Half Plane and the Disk has the circumference function \(C(r)=2\pi\sinh(r)\) for circles based at any point (the second problem establishes this for circles at a special point, and hte first problem establishes that space is homogeneous, so its the same at all points). Thus, this space truly is hyperbolic geometry, and has curvature \(-1\) (we proved in class, any space with this circumference function has constant curvature \(-1\)).