# Rethinking Pythagoras - 4a. Stretching the Truth (Poincaré Disc Model)

Author(s):
Daniel J. Heath (Pacific Lutheran University)

We define distances and circles in hyperbolic space analogously to how we did on the sphere, and assume that the reader is familiar with the Poincaré Disc Model of hyperbolic geometry, in particular, its (distance-inducing) differential $$ds=2(dx^2+dy^2)^{1/2}/(1-x^2-y^2)$$. For those who need a refresher, Wikipedia has a reference page for the Poincaré Disc Model , Wolfram Math World  has another, and the Geometry Center  has an archive containing descriptions, pictures, equations, and an applet concerning it.

### 4a.1 Using Differential Calculus

Let $$A$$ and $$C$$ represent the area and circumference, respectively, of a circle in the hyperbolic plane. Repeating an idea from Subsections 2.1 and 3.1, we note that $$dA/dr=C$$. It is harder to conjecture a formula for circumference in hyperbolic geometry. Using Geometry Playground, we note that for small $$r$$, $$A\approx \pi r^2$$ and $$C\approx \pi r^2$$, but for large $$r$$, both $$A$$ and $$C$$ seem to approach $$\pi e^r$$. Those familiar with hyperbolic trigonometric functions are probably already beginning to see the "parallel," and are conjecturing that $$C=2\pi\sinh r$$, where $$\sinh r=(e^r-e^{-r})/2$$ (and similarly $$\cosh r=(e^r+e^{-r})/2$$). If this is true, then a quick repeat of the steps of Subsection 3.1 shows that $$A=2\pi\cosh r-2\pi$$, a remarkable resemblance to the spherical case.

It is elementary to verify that $$\frac{d}{dx}\sinh x=\cosh x$$ and $$\frac{d}{dx}\cosh x=\sinh x$$, so that calculating MacLaurin series for each is a straightforward matter. Then we can see that

$A=2\pi\cosh r-2\pi=2\pi\left(1+\frac1{2!}r^2+\frac1{4!}r^4+\dots\right)-2\pi\approx\pi r^2$

$C=2\pi\sinh r=2\pi\left(\frac1{1!}r^1+\frac1{3!}r^3+\frac1{5!}r^5+\dots\right)\approx 2\pi r$

the latter approximations true for small $$r$$ . Again it is clear that both expressions are asymptotic to their Euclidean counterparts when distances are small. In addition, they imply hyperbolic trigonometric analogs for the familiar limit formulas from calculus:

$\lim_{r\to 0}\left(\frac{\cosh r-1}{r^2}\right)=\frac12$

implied by the former, and

$\lim_{r\to 0}\left(\frac{\sinh r}{r}\right)=1$

implied by the latter. It would be difficult to determine whether your universe were Euclidean, spherical, or hyperbolic if you were trapped in only one small corner of it, since in any of the three geometries, measurements on a small scale are either Euclidean or almost Euclidean. Indeed, physicists have not yet agreed on the overall shape of our universe , , .

### 4a.2 Using Integral Calculus

We take a circle in the Poincaré Disc model, and for convenience center it at the origin. In the model, we assume that it passes through the point $$(x_0,0)$$. We can find its area by integrating the square of the line element, $$ds^2=4(dx^2+dy^2)/(1-x^2-y^2)^2$$ over the region inside the circle. This is easier in polar coordinates, where

$A=\int_0^{2\pi}\int_0^{x_0} \frac4{(1-r^2)^2} rdrd\theta$

This is relatively simple as double integrals go. After substitution $$u=1-r^2$$, $$du=-2rdr$$, we can move quickly to the solution

(3)$A=(4\pi x_0^2)/(1-x_0^2)$

However, we want to know the area in terms of the radius. To find the radius, we need to calculate the distance between the origin and the point $$(x_0,0)$$. This requires a hyperbolic arclength integral. Thus:

$r=\int_0^{x_0} ds$

$=\int_0^{x_0} \frac{2(dx^2+dy^2)}{(1-x^2-y^2)}$

$=\int_0^{x_0}\frac{2dx}{1-x^2}$

the latter because $$y=0$$  on the geodesic joining the two points. This integral can be evaluated using the method of partial fractions, but we prefer to use the hyperbolic trigonometric substitution $$x=\tanh\theta$$ , $$dx=\mathrm{sech}^2\theta d\theta$$, giving us, after simplification:

$r=\int_0^{\tanh^{-1}(x_0)}\frac{2\mathrm{sech}^2\theta d\theta}{\mathrm{sech}^2\theta}$

$=2\tanh^{-1}(x_0)$

Then we have $$x_0=\tanh(r/2)$$ . Substitution into Equation (3) yields:

$A=4\pi\tanh^2(r/2)/(1-\tanh^2(r/2))$

$=4\pi\tanh^2(r/2)/(\mathrm{sech}^2(r/2))$

$=4\pi\sinh^2(r/2)$

(4)$A=2\pi\cosh r -2\pi$

the latter equality via hyperbolic trigonometric identity $$\sinh(r/2)=\frac12(\cosh r -1)$$. This is precisely what we conjectured in Subsection 4a.1. From here we can differentiate to verify our circumference conjecture of the same subsection.

### 4a.3 The Pythagorean Theorem

After seeing the similarities between spherical and hyperbolic area, we begin with a headstart; our initial conjecture at a Hyperbolic Pythagorean Theorem is $$\cosh a\cdot\cosh b=\cosh c$$. In Geometry Playground , we choose hyperbolic geometry and construct a right triangle and its edge-circles as in Figure 1b, or simply open Geometry Playground with a pre-constructed hyperbolic right triangle in the Poincaré Disc Model, and then construct the edge-circles on that right triangle as in Figure 1b.

Measuring the area $$A$$ of each circle, and calculating $$\cosh r=(A+2\pi)/(2\pi)$$ from Equation (4), it shouldn't take long for the relationship $$\cosh a\cdot\cosh b=\cosh c$$ to stand out, as expected. In fact, though we leave it without proof, this is the "Hyperbolic Pythagorean Theorem."

We again note that as $$a$$, $$b$$, and $$c$$ decrease, the Hyperbolic Pythagorean Theorem approaches the Euclidean one. Substituting MacLauren series $$\cosh r=1+r^2/2!+r^4/4!+\dots$$ for $$\cosh a$$, $$\cosh b$$, and $$\cosh c$$, we obtain

$\left(1+\frac1{2!}a^2+\frac1{4!}a^4+\dots\right)\left(1+\frac1{2!}b^2+\frac1{4!}b^4+\dots\right)=1+\frac1{2!}c^2+\frac1{4!}c^4+\dots$

Distributing the left, we arrive at

$1+\frac1{2!}a^2+\frac1{2!}b^2+\dots=1+\frac1{2!}c^2+\dots$

with all remaining terms of degree 4 or higher. Since we are assuming $$a$$, $$b$$, and $$c$$ are all small, we ignore the higher order terms. Canceling the constant terms and multiplying by 2, we obtain the Euclidean Pythagorean Theorem.