Suppose you're given some closed surface, like the one on the right. How can you tell which of the surfaces in the classification it's topologically equivalent to? One way is to try deforming it until you can make it look like an n-holed torus or a connected sum of projective planes. While in this case doing so isn't terribly difficult, in general it can be quite a puzzle. Yet there's another, much simpler method, relying on a single number. That number is the Euler characteristic of a surface.
Calculating the Euler characteristic of a surface is traditionally done using a triangulation of that surface. A triangulation is just what it sounds like: the division of a surface into triangles in a "nice" way. Nice here means that the intersection of any two triangles is either a single point which is a vertex of each triangle or a single edge which is a side of each of the triangles. Here's an illustration.
The Euler characteristic is equal to the number of vertices minus the number of edges plus the number of triangles in a triangulation. Normally it's denoted by the Greek letter χ, chi (pronounced kai); algebraically, χ=v-e+f, where f stands for number of faces, in our case, triangles.
From the above activity you should notice a certain pattern: the Euler characteristic of two triangulations of the same surface is the same! This is a theorem, part of which we'll prove.
Suppose we have a sphere of radius 1. Recall that the area of a sphere is equal to 4πr2, so that our unit radius sphere has an area of 4π.
The following steps, coupled with the illustration below, will guide you in finding the area of a triangle on a sphere in terms of its angles.
If everything went according to plan, you found that in fact 4π=4x+4y+4z-4A, where A is the area of the triangle, so that A=x+y+z-π. This confirms that, as we've seen in the last section, the angles of a triangle on a sphere add up to at least pi, since having negative area is impossible. How does this relate to the Euler characteristic? Let's find out.
So you got 4π=2πv-πf. What now? Well, we would like to
get the Euler characteristic χ=v-e+f in here, so why not add and
subtract 3πf on the right, getting
by part 1 in the activity above. Canceling 2π on both sides, we get
Congratulations! We've just shown that no matter what triangulation you take, the Euler characteristic of a sphere is 2. Note that knowing that in a plane the sum of a triangle's angles is exactly π, we can apply the same analysis to our flat surfaces, the Torus and Klein bottle.
From the above you should see that the Euler characteristic of both the Klein bottle and the torus is 0.
The projective plane, like the sphere, has spherical geometry. Modify our analysis of the sphere to find the Euler characteristic of the projective plane. (Hint: almost everything stays the same, except the projective plane has half the area of a sphere.)
Now we have all the tools needed to calculate the Euler characteristic of any surface!
We've done it! Now, simply knowing a surface's Euler characteristic, and whether it's oriented or not, we can tell what it is. Here's a pictorial summary of what we've discovered over the last five sections.