This small note is an introduction to Topology that follows the introductory arguments of (Armstrong 2013).

Euler’s Theorem

We will start our journey in topology following a classical example in the history of Mathematics the relation:

$$ v - e + f = 2 $$

Valid for classical Polyhedrons.

Basic definitions

Polyhedron

It’s a collection of plane polygons (see Programmazione lineare#Poliedro) such that:

1. Every polygon shares each of its edges with exactly another polygon
2. We have vertexes that can be shared by many polygons. Informally we have a piece of surface with a vertex.

Theorem statement

If we have a Polygon $P$ such that

1. Any two vertices of $P$ are connected through a path in the edges. (connected principle)
2. Any loop on $P$ made of its vertices made of straight line segments (we don’t need to pass through the edges) separates $P$ in two pieces. (informally if we cut there then the surface does not separate completely). Then we have $v - e + f = 2$, where $v$ are the number of vertices, $e$ the number of edges and $f$ the number of faces.

The two conditions above are justified by the counter example observations: . Which clearly do not follow the theorem

Simple proof

Here, I will just sketch the general proof (there will be some interesting observation, but that is left for later). We know that vertices + edges are just a graph, and for every graph there is a easy way to find a tree $T$ with all of the graph nodes. By a property of the trees, we know that every tree has $v - e = 1$, easily provable with induction on the number of nodes. The we construct some sort of a dual tree which we call $\Gamma$ built in this way.

1. Vertices are random points on every face of $P$
2. Two vertices are connected if their corresponding faces share an edge, and this edge does not belong to $T$. It can be proven that $\Gamma$ is indeed a tree, thanks to the 2 initial hypothesis, and then we know that $$ v - e + f = v(T) - \left[ e(T) + e(\Gamma)\right] - v(\Gamma) = \left[ v(T) - e(T) \right] + \left[ v(\Gamma) - e(\Gamma) \right] = 1 + 1 = 2 $$

The beautiful observation about this proof, is that every polygon that satisfies those hypothesis can be divided in two parts, which imply it is topologically equivalent to split a sphere (this is why it’s 2!) Which means that it is just a deformed sphere from the faces point of view.

Legendre’s Proof

Before we can understand this proof, we need to understand how to calculate the area of a spherical polygon which is just a polygon mapped to the surface of a unit sphere.

Spherical Polygon Area

Lemma: given angles $\alpha_{1}, \dots \alpha_{k}$ and $n$ edges then the area of the spherical polygon is

$$ \sum_{i = 1}^{k} \alpha_{i} + 2\pi - n\pi $$

This value is known as the excess in spherical geometry, and one can prove that the area is equal to $A = E\cdot R^{2}$ where $R$ is the radius of the sphere. This is also known as Girard’s theorem.

Proof: If you have Girard’s theorem (advised to check the link for a visualization), useful for simple triangles, then it’s easy to recompose everything and have that theorem, in this case we will just prove Girard’s theorem:

Consider a Spherical Lune, it’s easy to see that it’s area, given an angle $\alpha$, is given by this proportion

$$ 2\pi : \alpha = 4\pi : \text{ Area} \implies \text{Area} = 2\alpha $$

If we sum all the lunes of a triangle, we have the area of the entire sphere and three times our triangle. Let’s all $A$ the area of the triangle, we have (because we count the antipodal triangle, and use lunes to cover the entire sphere):

$$ 2(2\alpha + 2\beta + 2\gamma) = 4\pi + 2A + 2A \implies A = \alpha + \beta + \gamma - \pi = E $$

The proof

So now we know that a specific spherical polygon’s area is $\sum_{i = 1}^{k} \alpha_{i} - (2 - n) \pi$ We now sum over all the polygons and we obtain $2\pi v - 2n\pi e + 2\pi f = 4\pi$ where $v$ is the number of vertices, $e$ of edges and $f$ faces, that is the number of polygons. This is true because every vertex at the end is counted as $2\pi$ in radians, and every edge is counted twice, and we are summing $f$ polygons so we have the last term. Simplifying the $2\pi$ we obtain $v - e + f = 2$ which is the Euler’s theorem. This is a nice proof, and quite easy after you know the Spherical Polygon area lemma.

References

[1] Armstrong “Basic Topology” Springer Science & Business Media 2013