Interior angles of a polygon in the hyperbolic plane

Question

This is fairly simple question about how to make sense of "angle" in the hyperbolic plane.

The hyperbolic plane can be tesselated with pentagons, four to each vertex. In this tesselation, each pentagon has interior angles of exactly 90 degrees.

Suppose you start at one of the vertices of one of these pentagons. Then you can walk around the pentagon, to get back where you started. To do so, you first walk along an edge (which is a hyperbolic geodesic), you then turn 90 degrees to the right, walk along another edge, turn 90 degrees to the right again, and repeat five times.

After doing this, you have gotten back to where you started, and at the same orientation. However, you have rotated 90 degrees five times! So you have rotated by a total of 450 degrees and preserved orientation.

How does this work, exactly? Surely we would still say that 360 degrees is a full rotation in the hyperbolic plane.

Answer 1

Certainly, if you stand in one spot and spin around, a "full rotation" of $360^\circ$ restores your orientation; this is true. However, once you move from that spot, things get interesting.

Here's perhaps a simpler thought experiment:

Start at the North Pole of the Earth. Travel directly south to the Equator. Take a left, and travel a quarter of the way around the Equator. Take another left, and travel directly north to the North Pole. One more left leaves you facing your original path. You have returned you to your original location and orientation after having made three right-angle turns. Yet, $270^\circ\neq 360^\circ$. Hmm ...

The reason this travel experience doesn't match what happens when you walk around on the plane is because ... well ... a sphere isn't a plane. A plane is flat, but a sphere is not. Not-unreasonably, the "curvature" of the very surface upon which you travel affects how much turning you need to do to restore your orientation.

Like the sphere, the hyperbolic plane is curved. A discourse on curvature is beyond the scope of this answer. (See, for instance, Wikipedia's "Gaussian curvature" entry.) However, I'll note that we assign the sphere positive curvature, and we assign the hyperbolic plane negative curvature. (Indeed, the hyperbolic plane is often called a pseudosphere.) In the very loosest possible manner of speaking, this is "why" our travel path turns less than $360^\circ$ on a sphere, and more than $360^\circ$ on the hyperbolic plane, while a travel path on a flat ("zero-curvature") plane turns exactly $360^\circ$.

Answer 2

The sum of the interior angles in the hyperbolic plane, as well as on the sphere, measure the polygon-area, when the normal angles of the polygon are subtracted.

For a sphere, one can have the faces of a dodecahedron, which gives a corner-angle of 120 degrees. In total, the pentagon on the euclidean plane has only 108 degrees, so the total angle is in excess by 60 degrees. The sphere has a total excess of 720 degrees, so there are 12 such pentagons.

In the hyperbolic case, the angle is in defect of the euclidean case, and the area is proportional to the defect. In the case above, the pentagon has rightangles, or 90, five make 450 degrees. But the pentagon in Euclidean space is 540 degrees, so it is in defect by 90 degrees.

You can make a right decagon from this, by including chords across the 10 pentagons that touch a central pentagon. This then includes into the decagon, five parts of three-sides of a pentagon, and five parts of two sides of a pentagon, and the core, all together six times the area.

A decagon in euclidean geometry has a corner-angle of 144 degrees, the total angles is 1440 degrees. The right decagon has a total angle of 900 degrees (10 right angles), so the defect is 540 degrees, or six times that of the right pentagon we made it from.