4

Consider the flat surface $(X,\omega)$ obtained identifying pairs of parallel sides of the regular octagon of area 1 by translation. I’m trying to understand what its universal cover (with the induced metric) looks like. I’ve read that “it’s composed of an infinite number of copies of the octagon glued together with eight around each vertex to create cone points of angle $6\pi$”.

What exactly does it mean “glued together with eight around each vertex”?

Chosen a base point $x_0\in X$, the universal cover of $(X,\omega)$ should be such that the preimages of a point $x\in X$ are given by all homotopy classes of paths from $x_0$ to $x$, am I right? I'm trying to understand why these two definitions of the universal cover of $(X,\omega)$ are equivalent, but I can't see why for each homotopy class of paths between $x_0$ and $x$ one should have a point in a different copy of the octagon. Why is that?

user00169
  • 325
  • just out of curiosity i would like to ask: isn't the identified space homeomorphic to a sphere? – user300 Feb 20 '17 at 18:23
  • What exactly do you mean by "identifying the sides"? There's many ways to identify them with each other that yield different quotient spaces. – Nephry Feb 20 '17 at 18:25
  • sorry, now I corrected what I mean – user00169 Feb 20 '17 at 18:42
  • The idea is very similar to the construction of $\mathbb{R}^2$ as the universal cover for the torus. In the torus case, you start with a square with identified sides as in your case and glue together infinitely many of them side by side to get $\mathbb{R}^2$. Now ask yourself what happens to sum of the angles of each corner if you try to glue octagons "side by side". Also if you're not aware how this relates to the torus, check out this question: http://math.stackexchange.com/questions/1239104/identifying-the-two-hole-torus-with-an-octagon – Nephry Feb 20 '17 at 19:51
  • thank you, but I know the case of the torus. I wanted to know what does it mean to glue the octagons in eights around a vertex and what happens with the cone angles – user00169 Feb 20 '17 at 21:03

1 Answers1

2

To construct a model of the vertex, cut eight regular octagons from a piece of paper. Mark a vertex on each octagon, and number the octagons (mentally or physically) from $1$ to $8$. Place octagons $1$ and $2$ next to each other so that their marked vertices touch, and they meet along one side. Tape these sides together. Now repeat the process with octagon $3$ and the free edge of octagon $2$ that is incident on the marked vertex. Continue attaching octagons sequentially until all eight octagons are attached in a "chain" around the vertex. You should have a spiraling, triple-ply stack of octagons. With care, you can "unwrap" the three layers, bringing the free edge of octagon $1$ that is incident on the marked vertex next to the free edge of octagon $8$. Tape those together. You now have eight octagons surrounding a single marked vertex.

The total incident angle at this vertex is $\theta = 8 \times \frac{3}{4}\pi = 6\pi$, so the angular defect (a.k.a., the integral of the Gaussian curvature over the entire surface) is $2\pi - \theta = -4\pi$. Since the surface is orientable (edges are identified by orientation-preserving translations) and boundaryless, Gauss-Bonnet says $-4\pi = 2\pi(2 - 2g)$, so the genus is $g = 2$.

As for the universal cover: The octagon itself is a fundamental domain, and each octagon is attached to neighbors along each of its sides. To see how many octagons surround each vertex, draw a small circle around the vertex: Each time the arc hits a side (i.e., leaves the fundamental domain), continue the arc on the opposite side of the octagon. A sketch (or drawing on a physical octagon) should convince you that all eight vertices of the octagon are identified, so there are eight octagons around each vertex of the universal cover.

If you haven't experimented with cone points of incident angle greater than $2\pi$, I recommend taking two paper disks, slit radially, and joining the "top edge" of the slit in one disk to the "bottom edge" of the slit in the other. Laid flat, you have a two-ply disk with (up to) $4\pi$ worth of angle at the vertex. By "sliding the surface over itself" to increase or decrease the incident angle at the vertex, you can make a conventional cone (vertex angle smaller than $2\pi$), or a "saddle cone" with vertex angle greater than $2\pi$.

On a cone with incident angle between $0$ and $\pi$, you can draw a "monogon", a geodesic that bounds a neighborhood of the vertex. On a cone with angle less than $2\pi$, you can draw a "digon", a pair of geodesics that bound a neighborhood of the vertex. One a cone of incident angle greater than $3\pi$, no geodesic triangle encloses the vertex.

Andrew D. Hwang
  • 78,195