## Triangulating a Surface.

### November 17, 2010

In some cases, we’d like to be able to break down a surface nicely into triangles or things which look like triangles. There’s a (strong!) theorem which states that every compact surface has a finite triangulation and every surface has a (potentially infinite) triangulation. We’ll talk about how to triangulate a surface below.

(**Note: **this used to be a part of the Homology Primer, but I decided against using triangulations to talk about homology. Nonetheless, triangulation is a topic which comes up in topology so I decided to keep this post up as a reference.)

For now, all of our structures will be sitting in dimension three or less. This is not only because things can quickly get complicated when we go up a dimension but because these spaces will be more familiar to you and you will be able to “see” what you’re doing.

So, here’s what’s going to happen: given a surface, you’re going to fill it with triangles like this:

Subjected to the following rules of the game. Ready?

**(1.) Two simplices have to intersect at a simplex which is common to both simplices.**

This means, in particular, that if two triangles meet somewhere, that “somewhere” has to be an entire simplex, and that simplex has to be the entire “side” of both triangles. For example, we cannot have either of the following two figures:

since the former intersects at something which is a line segment without vertices, and the latter intersects at a little subset of a two-simplex, which is NOT itself a simplex (why?).

Similar to the latter picture, we cannot have our triangles “stabbing” each other, as in:

since their intersection is not a simplex.

Another way to state this rule is, **“Any two simplices intersect at a common face not at all.”**

**(2.) Any bounded set intersects at most finitely many simplices.**

This second rule is mainly to rule out things like this:

Where we have “infinitely many” triangles approximating things. Think about trying to approximate a circle with triangles by approximating the circle with a triangulated square, then a triangulated pentagon, then a triangulated 10-gon, then a triangulated 1000-gon, and so on — we’d soon get infinitely many triangles, but the triangulation didn’t really make things much easier to work with! This second condition is really just a formal one since in practice on these lower dimensional examples it will be hard to actually draw infinitely many lines because your hands get tired after finitely many.

And that’s it. Triangles meet each other at nice places, and there’s not a ton of them in one spot. I also want to mention that **when we triangulate a space, we can “bend” it to make the triangles fit on it nicely.** You’ll see exactly what I mean when I do the sphere and the hexagon below.

Let’s do these two examples of triangulations, and then we’ll give you some homework to do. (Note that I’m saving quotient spaces for the next time, so we’ll talk about triangulating the Klein bottle and the mobius band next time.)

## Triangulating the Sphere.

First, we take our sphere. Here it is!

Now, I’m basically going to “draw triangles on it”, but where should they go? Well, let’s just draw in some vertices.

Okay, so I’ve drawn in a number of vertices. It should be clear that connecting these will give the (triangulated) structure:

which is actually all we need (since we can “blow it up” to make the sphere). There is something to note here, though: we haven’t really triangulated the **sphere**, really, but more of something that’s really close to the sphere. It is actually enough to triangulate something homeomorphic to the surface you want to triangulate — if we blew this shape up to make the sphere, we’d get triangles, but they’d be “bent.” This is not a huge deal, ultimately, but it’s a bit harder to draw. For another triangulation of the sphere, see the previous post and look at the “orange.” Note now that there is usually more than one correct triangulation for any surface. This will turn out not to matter, but proving that it doesn’t matter is far from trivial. We will most likely not go into this as the proof is not all that illuminating.

Also note that these triangles are “filled in” — or, in other words, they’re legit 2-simplices. I haven’t colored them in to make the pictures clearer.

## Triangulating a Hexagon.

This should be a bit easier, but we’re going to do it in two different ways to show that we can be clever when we triangulate things. The first way will just be to triangulate it directly, the second way will be using a homeomorphism.

Here’s one of them. Kind of usual. Not a big deal. Note all the triangles are 2-simplices, so they’re “filled in” triangles. Here’s another triangulation:

This one is kind of neat. Note that we added an extra vertex in the center to make these triangles intersect at a simplex. **This is NOT a triangulation:**

Why not? Look at the triangles in the middle. Where do each of them intersect the top and bottom triangles?

Last, what if we wanted to be lazy? Well, we could actually just say that this is homeomorphic to a square (this should be relatively obvious) so let’s apply this homeomorphism and then triangulate the square.

Note that when we homeomorph our square back to the hexagon, the triangle here probably won’t look like a triangle on the hexagon; this will turn out not to matter so much, since most of the work we do in homotopy theory is up to homeomorphism of the space. A rule of thumb is that if a space has something that it’s homeomorphic to, then work with that space. For example, why didn’t we just homeomorph this hexagon into a triangle? We could have! Further, we could have *reduced it to a point*, which has a trivial triangulation.

## Homework.

- Triangulate the octagon in two ways: directly, by homeomorph-ing it.
- Triangulate the circle. We can’t do it directly, so what other options do we have?
- Can you find a different triangulation of the sphere?
- Count the number of vertices, edges, and faces for each of the triangulations we did of the hexagon. Consider the value where is the number of vertices, is the number of edges, and is the number of faces. What do you note?
- Count the number of vertices, edges, and faces in your triangulations of the octagon and the circle and consider . What do you note?
- Count the number of vertices, edges, and faces in both triangulations of the sphere (the one I’ve done above, and the one from problem 3) and compute .
- Take two of the following three dimensional figures and triangulate it: the tetrahedron (already triangulated!), the cube, the octahedron (already triangulated!), the dodecahedron, or the icosahedron (already triangulated!) and compute the value of . What do you note? Is that kind of weird?

## Note Concerning the Problems: The Euler Characteristic.

The thing that I’ve been mentioning in the problems (that thing) so much is called the E*uler Characteristic* and it became hugely important for the study of topology. For each of the platonic solids, you should get an Euler characteristic of 2. You should be able to prove (or at least hand-wave) and that because these have an Euler characteristic of 2, the sphere also has an Euler characteristic of 2. The plane figures (hexagon, square, etc.) that we’ve studied in this post should all have euler characteristic equal to 1.

Some thins that are neat but didn’t come up: a line has Euler Characteristic 1, and a point has Euler Characteristic 1. A torus has Euler Characteristic 0. A Klein bottle has Euler Characteristic 0, and so does a Mobius band. There are also higher dimensional versions of the Euler Characteristic.

As a final note, a basic question which motivated a lot of topology was, “when do two things have the same euler characteristic?” and it turns out that this question is not so bad to answer once we have the appropriate machinery. I won’t explicitly be talking about this so much in these posts, but it is nice to keep in the back of your mind.

just a quick correction: The torus has Euler Characteristic 0.

Whoops! You’re right; changed it above. Thank you!