# The Geometric Viewpoint

geometric and topological excursions by and for undergraduates

• ## Curvy Television

Television screen size became a status symbol in the American household soon after TV’s adoption
in the 1940’s, and remains so today. There are many equivalent ways to characterize screen size; today’s standard is by the diagonal length of the screen. Everyone understands that this length is related in a familiar way to a screen’s area. With today common 16:9 aspect ratio (horizontal length:vertical length), the screen area ${A(l)=\alpha l^2}$ where ${l}$ is the diagonal length and ${\alpha=337/144}$. In other words, the area varies with the square of the diagonal length.

This polynomial relationship, however, is a feature of euclidean space. TV screens placed on hyperbolic and spherical surfaces lead to surprising differences in area relative to diagonal length. Curvature will play an important role in illustrating the discrepancy between euclidean, spherical, and hyperbolic geometries. To see this, let us imagine covering our television screen with a piece of paper. Using Ilmari Karonen’s imagery from Stack Exchange we shall lay a large sheet of paper over our euclidean television, and cut it down to size.

Now imagine that the surface of our TV is lying on a sphere. We will try to lay our cut-to-size sheet over the screen, a curved surface. To anyone who has used paper-mache, it’s immediately apparent that the paper will need to wrinkle and fold up around itself to fit to the surface. There is simply too much paper and too little sphere to go around. What is happening? The intuitive answer is that there is simply not enough space on our spherical surface. This simple fact hints at a sobering idea for cartographers; there exists no flat map that is locally isometric to the Earth at any point. No matter how carefully you restrict your vision, a flat map will be always be wrong. Gauss formally proves this in his Theorema Egregium. It turns out to be a consequence of the constant positive curvature of spherical space, as opposed to zero curvature in euclidean space.

Tangent plane intersecting a surface to create a curve.

Curvature, also know as Gaussian curvature ${K}$ at a point ${\alpha}$ on a surface, is defined as the product ${\kappa_1 \cdot \kappa_2 }$, where ${\kappa_1}$ and ${\kappa_2 }$ are the principal curvatures of ${\alpha}$. Finding the principal curvatures at ${\alpha}$ can be visualized by the following method: First, find the normal vector to the surface at ${\alpha}$. Next, look at the intersection of a plane containing the normal vector with the surface. This intersection is a curve on the surface. As you rotate the plane around the normal vector, the intersection curve will change. The principal curvatures are the maximum and minimum curvature of the intersection curve at ${\alpha}$ as you rotate the plane.

On a plane, the intersection of a normal plane to a point on the surface will be a line. Since the curvature of any line is zero, ${k_1 \cdot k_2 = 0}$ at all points, resulting in constant zero curvature as we have already mentioned. Similarly, on a sphere, any plane containing a normal vector will intersect the sphere along a great circle (geodesic) of the sphere. Since any two great circles have the same (non-zero) curvature, we see that ${\kappa_1 \cdot \kappa_2 = \beta > 0}$ at all points, i.e. a sphere has constant positive Gaussian curvature.

Hyperbolic Spaces locally look like a saddle point.

Hyperbolic space challenges this definition of curvature. How can we use our method on a surface which cannot be embedded in euclidean 3-space? One way to visualize hyperbolic space is as a collection of points each of which locally looks like a saddle point. If we take a plane containing a normal vector to a saddle point, and intersect it with the saddle, we see that the curve on the saddle radically changes as we rotate the plane. The maximal and minimal curvatures of the intersection curves will be positive and negative respectively. Therefore, ${K= \kappa_1 \cdot \kappa_2 < 0}$. Since each point of hyperbolic space locally looks like an identical saddle, we see that hyperbolic space has constant negative curvature.

Largest rectangle possible in hyperbolic space

What happens when we embed our television in a hyperbolic surface? The first thing to realize is that ${2\pi}$ is the maximum area of a rectangular screen. This is similar to how a spherical rectangle is bounded by the size of the sphere. However, we do not want to restrict the size of our television. Bigger TV’s are always better! Instead, we will investigate the area of circular televisions, or hyperbolic disks, which have no upper bound. If we calculate the circumference of a hyperbolic circle in terms of its radius ${r}$, we find that ${Circ(r) = 2\pi\sinh(r)}$. Now we can simply integrate this value from ${0}$ to ${R}$ to find the area ${A(R)}$ of a hyperbolic disk, ${\int_0^R 2\pi\sinh(r) dr = 4\pi\sinh^2(\frac{1}{2}R)}$. This can be rewritten as ${A(R) = \pi(e^{R}+e^{-R})-2\pi}$, where ${R}$ is the radius of the disk.

Notice that in hyperbolic geometry, circular area grows exponentially with the radius of a disk. Our TV’s area, therefore, will grow much faster as we increase its radius than it would in euclidian space. In other words, there is more space in a hyperbolic disk than in a euclidean disk of the same radius. This is the opposite of our result from spherical geometry, and is a good example of the connection between Gaussian Curvature and the “amount of space” in a given surface. The book Crocheting Adventures with Hyperbolic Planes is a good reference for more information about curvature and the hyperbolic plane.

The connection between curvature and space has been used by physicists to investigate our universe. A major question in the field of astronomy is whether or not our universe, on a large scale, has positive, zero, or negative curvature. One method to test this theory is to calculate the density of galaxies at different distances from our own galaxy. A constant density would be reminiscent of a flat (zero-curvature) universe. However, if we observed a growing density of galaxies as we looked further away, this would hint at negatively curved space, because we are observing “more space” with our instruments, and hence we would see more galaxies. Similarly, a shrinking density of galaxies would suggest a positively curved universe. Using this method, physicists have been able to rule out the possibility of a large curvature in our universe, however the difficulty of accurately measuring densities has restricted the ability to make a stronger claim.

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• ## Pythagoras in the hyperbolic plane

The Pythagorean theorem is surely the most famous of all mathematical theorems. The simplicity of its statement (that a right triangle with sides of length $a$ and $b$ and hypotenuse of length $c$ satisfies $a^2 + b^2 = c^2$) and the multiplicity of beautiful proofs (like the one shown at right from Byrne’s edition of Euclid’s Elements) contribute to its memorability.

Byrne’s version of Euclid’s proof of the Pythagorean theorem.

The fame of Pythagoras’ theorem in Euclidean geometry makes it natural to ask if it holds in other geometries. For instance, is the Pythagorean theorem true in hyperbolic geometry, that radical challenger to Euclid’s throne?

The naive answer is “No, the Pythagorean theorem does not hold in hyperbolic geometry”, as it is logically equivalent to Euclid’s 5th postulate (which is the defining difference between Euclidean and hyperbolic geometry). However, there is a theorem in hyperbolic geometry which is analogous to Pythagoras theorem:

The hyperbolic Pythagorean theorem
If a right triangle in the hyperbolic plane has sides of length $a$ and $b$ and a hypotenuse of length $c$, then $\cosh(a)\cosh(b) = \cosh c$.

Right triangle in the hyperbolic plane

As in the figure at left, the edges of the triangle are hyperbolic geodesics (we’ll review what those are below), with the sides of length $a$ and $b$ adjacent to the right angle and the hypotenuse (of length $c$) is the edge across from the right angle. The function $\cosh x = \frac{e^x + e^{-x}}{2}$ is hyperbolic cosine.

### The hyperbolic plane

The hyperbolic plane was discovered by Bolyai and Lobachevsky when they investigated the effect of replacing Euclid’s parallel postulate with an alternative. They operated purely deductively: they had no graphical representation of hyperbolic geometry to work with. Later mathematicians, such as Klein and Poincaré, discovered ways of representing hyperbolic geometry inside of Euclidean geometry by giving new meanings to terms such as “line”. Using calculus, we can give a succinct description (called the Poincaré disc model) of the hyperbolic plane as follows.

The entirety of hyperbolic geometry will take place inside the open unit disc (the blue disc at left) in the plane $\mathbb R^2$. The unit circle (the boundary of the disc) is not part of the world in which we do hyperbolic geometry. We refer to it as the infinity circle. A path $\gamma(t) = (x(t), y(t))$ for $t_0 \leq t \leq t_1$ in the disc has length defined by an integral similar to the integral defining path length in euclidean geometry. The length of $\gamma$ in euclidean geometry is given by $\int_{t_0}^{t_1} ||\gamma'(t)||\thinspace dt$, where $||v||$ denotes the magnitude of the vector $v$. The length of $\gamma$ in hyperbolic geometry on the other hand is given by the integral $\int_{t_0}^{t_1} \frac{||\gamma'(t)||}{1 - ||\gamma(t)||^2} \thinspace dt$. A path between points $A$ and $B$ is a geodesic if it has length no greater than the length of any other path between $A$ and $B$. Geodesics in hyperbolic geometry are the analogue of straight lines in euclidean geometry. If there were light in the hyperbolic plane, it would travel along geodesics.

A triangle is formed by three geodesics intersecting pairwise.

In the Poincaré disc model of hyperbolic geometry it turns out that the geodesics are segments of diameters of the disc and portions of circles in $\mathbb R^2$ which intersect the infinity circle at right angles. If three geodesics intersect in three points, not all lying on the same geodesic, then the three geodesics define a triangle. The image on the right shows our triangle arising from three geodesics. It may seem as though the triangle we’ve drawn is somewhat special in that two of the sides lie on diameters of the disc. However using hyperbolic isometries (the analogue of euclidean translations, rotations, and reflections) we may move (without changing lengths or angles) any hyperbolic triangle so that two of its sides lie on diameters, as we have indicated.

### Proving the hyperbolic Pythagorean theorem

Here is a sketch of the proof of the hyperbolic Pythagorean theorem. It is an abbreviated version of the proof given by Martin Greenberg in his excellent text Euclidean and non-Euclidean Geometries.

The triangle ABC

Let $\triangle ABC$ be a right triangle in the hyperbolic plane with $C$ the right angle. Without loss of generality, we may assume that the vertex $A$ is the origin and that two of the edges, one of which is the hypotenuse, are portions of diameters, as in our picture. Let $d(AB)$ be the hyperbolic distance from point $A$ to point $B$ and $\overline{AB}$ the euclidean distance.

We have already defined the hyperbolic cosine function $\cosh$. The hyperbolic sine function is defined similarly: $\sinh(x) = (e^x - e^{-x})/2$ and the hyperbolic tangent function is simply $\text{tanh }(x) = \sinh(x)/\cosh(x)$.

Using our path length formula, it is straightforward to verify that $\text{tanh}(d(AB)) = \frac{2\overline{AB}}{1+\overline{AB}^2}$ and $\text{tanh}(d(AC)) = \frac{2\overline{AC}}{1+\overline{AC}^2}$.

We begin by showing:

Lemma: $\sin A = \frac{\sinh a}{\sinh c}$ and $\cos A = \frac{\text{tanh } a}{\text{tanh } c}$

Once we have those equations, the hyperbolic Pythagorean theorem can be derived from the equality $\sin^2 A + \cos^2 A = 1$ by applying identities for $\sinh$ and $\cosh$ analogous to the identities involving $\sin$ and $\cos$. We leave it as a pleasant challenge to the reader to work out those details.

Extend the geodesics into the plane outside the disc

We now set about proving the Lemma, by considering the image in $\mathbb R^2$ at right. Extend the geodesics making up the sides of $\triangle ABC$ into $\mathbb R^2$. This means that the geodesic $a$ is now part of a circle $H$ centered at a point $O$.

The circle $H$ intersects the infinity circle in two points. Join those two points by a line (drawn in red in the figure) and let $P$ and $Q$ be the points where that line intersects the extensions of the other two edges of $\triangle ABC$, as in the picture. (The points $P$ and $Q$ have a special relationship to the points $B$ and $Q$: they are the images of $P$ and $Q$ under the conversion from the Poincaré disc model of the hyperbolic plane, to the Klein model of the hyperbolic plane. But that is a story for a different day.)

The key to the whole business is to apply the definition of cosine to the euclidean triangle $\triangle PAQ$. Doing so, we obtain: $\cos A = \overline{AQ}/\overline{AP}$. Converting to hyperbolic distances, we arrive at

(*)     $\cos A = \text{tanh }b/\text{tanh }c$.

Let $R$ be the point, other than $B$, where the line $AB$ intersects the circle $H$. The point $R$ is the result of applying the inversion $(x,y) \to \frac{1}{x^2 + y^2}(x,y)$ to the point $B$. This implies that $\overline{AR} = 1/\overline{AB}$. Hence,

(**)      $\overline{BR} = \overline{AR} - \overline{AB} = 2/\sinh d(AB)$.

Recalling that $c = d(AB)$, we have $\overline{BR} = 2/c$. Letting $S$ be the other intersection point between the line $AC$ and the circle $H$, we also have $\overline{CS} = 2/\sinh b$.

The angle BOT is equal to the angle at B of triangle ABC.

Finally, as in the figure at left, let $T$ be the orthogonal projection of $O'$, the center of $H$ onto the line $AB$. Some rather easy arithmetic, using the fact that the angles of a euclidean triangle sum to $\pi$, shows that the angle $BOT$ is equal to the angle $B$. Combining this fact with (*) and (**), we conclude that $\sin B = \sinh b/ \sinh c$. Interchanging the roles of $A$ and $B$ in the preceding argument, concludes the proof of the lemma.

### Final Thoughts

The euclidean and hyperbolic planes are certainly the most important of the two-dimensional geometries. The third most important geometry is spherical geometry. There is also a version of the Pythagorean theorem for triangles on the sphere. Thurston, in his famous book Three-dimensional Geometry and Topology, sketches a strategy for giving a combined proof of the law of cosines in the hyperbolic plane and in the sphere. The corresponding Pythagorean theorems follow from that. For the hyperbolic law of cosines, Thurston uses the hyperboloid model of the hyperbolic plane, which gives a unification of the Poincaré disc model and the Klein model alluded to earlier.

Finally, many of the beautiful proofs of the Pythagorean theorem make use of dissections of a square and the fact that $a^2$ is the area of a square with sides of length $a$. Is there a dissection proof of the hyperbolic Pythagorean theorem?

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• ## Introducing: The Geometric Viewpoint

This is a blog of essays, aimed at undergraduate mathematics majors, on topological and geometric topics. The essays will require different levels of mathematical background and will appear on an ad hoc basis. If you would like to submit an essay for consideration, please email it in PDF and LaTex to Scott Taylor. Essays from Colby students are especially welcome.

The Kinoshita graph is a marvelous encapsulation of the relationship between geometry and topology in low dimensions. Perhaps it will be the subject of an essay on this blog?

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