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Archive for **February 27th, 2022**

### Field of Topology in Math? Geodesic, Great Circles, Polyhedron? And what the Mathematical Giant Euler has to do with them?

Posted February 27, 2022

on:**Note**: I will skip** Legendre Ingenious Proof **of Euler’s Polyhedron Formula and refer you to the link. Or maybe Not.

https://atonu-roy-chowdhury.medium.com/legendres-ingenious-proof-of-euler-s-polyhedron-formula-68bccd2a006

Atonu Roy Chowdhury

Euler’s** polyhedron formula**

V-E+F=2

This formula will be explained later and is often referred as *The Second Most Beautiful Math Equation*, second to none other than another identity** (e^{iπ}+1=0)** by **The German Mathematical Giant** Euler who demonstrated by observation that planets circle the Sun in **elliptical fashion**.

(Like why Not simplify this identity as **e^{iπ}**= -1? I had no idea that complex numbers i were invented during Euler period. And how can **e^{iπ}** be computed manually? Try it on you calculator. I can conjecture that Euler reached this formula through a deductive process?)

Euler’s Polyhedron Formula basically gives us a fundamental and elegant result about Polyhedrons (Greek suffix *poly *means many, and *hedra* means face). **Polyhedron is a three-dimensional shape that consists of multiple flat polygonal faces. **

A Football (or a Soccer Ball if you prefer that name), is a polyhedron that contains 12 pentagonal face and 20 hexagonal face.

By definition, a polyhedron has some polygonal **Faces**. Here the yellow triangles and red pentagons are the faces.

A face is polygon so by definitely it has some *gon-*s or sides. You can see here that sometimes two faces share some common side (that’s why it looks like a closed box, otherwise just a random collection of polygons wouldn’t be considered as polygons).

These common sides are called **Edges**. In this example, the edges are marked grey. Also, three or more faces has some common endpoints.

These common endpoints are called **Vertices** (plural of **Vertex**).

The formal definition of **convex polyhedron **deals with convex hull, so I’m not gonna explain that now.

If you take any two points on or inside the polyhedron, the line segment through that two points will be strictly on or inside the polyhedron. The examples we’ve seen till now, every single one of them was convex polyhedrons.

Below are some examples of **non-convex Polyhedrons**.

Euler’s formula establishes a relation between the number of Vertices, number of Edges, number of Faces in a convex Polyhedron.

Let V, E, F respectively denotes the number of Vertices, Edges, Faces in a convex polyhedron. Then Euler’s formula states that

How simple a formula, yet so elegant and beautiful! But to me, Legendre’s proof is somewhat more beautiful than the formula itself.

Before diving into the proof, we need some background knowledge. First we need to know about ** Great Circles **(Great Circles are often called

*) and*

**Geodesic***Geodesic Polygons*. These are related to spherical geometry.

A *Great Circle* is any circle on the sphere that** has the same radius as the sphere.** For example, if we assume the Earth to be a perfect sphere, then the equator line is a great circle.

Also, the **lines of longitude** are also great circles. Oftentimes, you may find the definition that great circle is the circle that cuts the sphere into two congruent hemispheres. That definition is equivalent to the definition I stated here.

Now, about *Geodesic Polygons*, we shall define * Geodesic Triangles* first. Then we will expand the definition to higher polygons.

A *Geodesic Triangle *is a region on the surface of the sphere, which is bounded by three geodesics (or great circles). Some books state it as **Spherical Triangle **too.

Now, Euclidean Geometry insists that the sum of the angles in a Triangle must be 180 degrees, or π radian (throughout this blog, I shall prefer radians more than degrees; unless specified, the unit of angle is radian). That’s because of Eulid 5th Postulate. But this **postulate doesn’t hold in spherical geometry**. A Geodesic triangle can have angle sum more than 2π.

How do we find the angle between two geodesics?

We shall extend our notion of angle between two sides to angles between two intersecting circles.

Suppose two great circles intersect at some point B. Then we** draw the tangents at point B** to both of the circles. We know how to calculate angle between two lines, so we find out the angle between these two tangent lines. This angle is defined as the angle between those two circles.

A natural question arises, two circles can intersect each other at two points. Is the angle equal in both points? The answer is yes. The proof is left as an exercise for the reader.

A Geodesic triangle can have **angle sum more than π**. For example, the larger triangle in the image below is a geodesic triangle with 3 right angles

**The angle sum of geodesic triangle is closely related to the area**.

Seems like, the larger the angle sum, the larger the area. This, in fact, is true.

There is a theorem called **Harriot-Girard Theorem **that gives us a precise relation between sum of the angles and the area.

The theorem states that: if a geodesic triangle on a unit sphere (sphere with radius 1) has angles a, b, c; then the area of that triangle will be **a+b+c–π**.

## Girard’s Theorem

### Consider the white triangle \(\sf T \) on the sphere shown above. Girard’s Theorem gives a formula for the area of…

I’m not gonna show the rigorous proof of this theorem here. A visual demonstration is given in the Wolfram link above. Also a rigorous proof is given in the Princeton blog.

Now that we know what is Geodesic Triangle, can we extend the notion of triangle to any polygon? The answer is: yes, we can.

In an analogous manner, we define a * Geodesic Polygon *as the area bounded by three or more geodesics. Needless to say, the sides of a Geodesic Polygon are parts of some great circles.

In this photo, ABCDE is a geodesic pentagon. The sides AB, BC, CD, DE, EA are actually parts of some geodesics.

In the same way as before, we can now generalize the **Harriot-Girard Theorem** for polygons. The generalized Harriot-Girard Theorem is: if an n-sided geodesic polygon has angles a₁, a₂, a₃, … , aₙ ; then the area of this polygon is: **a₁ + a₂ + a₃ + … + aₙ –( n–2 )π = sum of angles–nπ+2π.**

The proof to the generalization is just using the previous result. For any n-sided geodesic polygons, we can divide it into n-2 geodesic triangles. Just choose any vertex, then connect the other vertices to this chosen vertex using some great circles.

Thus we obtain n-2 geodesic triangles. Notice that, the sum of the areas of these triangles is precisely the area of the polygon. Furthermore, the sum of all the angles of the triangles is equal to the sum of the angles of the polygon. Now, if we apply Harriot-Girard Theorem on all the n-2 triangles, we obtain the result.

Alright, enough about spherical geometry. How are these even related to our original problem, Euler’s Polyhedron Formula?

What Legendre does next is, he connects all these seemingly unrelated ideas from spherical geometry to solve a combinatorial problem.

Let’s consider a convex polyhedron with V vertices, E edges and F faces. Take any point X inside the polyhedron. Then construct a hollow sphere centered at X that surrounds the polyhedron completely. The units are not really important here. So we can assume without loss of generality that the sphere is a unit sphere.

We project the polyhedron onto the sphere through the point X. We can think of this projection as light and shadow. Assume that there is a lightbulb at the point X. Then we shall mark the shadows of the edges onto the sphere. The shadows will make some geodesic polygons.

The next trick Legendre pulls out of his sleeve is ** Counting in two ways.** This means, if we calculate the same quantity in two different ways, the values we find using the two ways must be the same. Because we are essentially calculating the same thing. And how a thing can have multiple different values?

What Legendre calculates here is the surface area of the sphere. One possible way to calculate surface area is: we know the formula surface area =4*πr*². Here the radius is 1, so the surface area is **4 π**

*.*

We can calculate the same thing by adding the areas of the geodesic polygons we got after projecting. By the generalized Harriot-Girard Theorem, **area=sum of angles–nπ+2π**, where n is the number of edges of that geodesic polygon. Notice that, each face of the polyhedron corresponds to exactly one geodesic polygon. That means, there are total F polygons. If we take the sum of areas over all polygons, we get

If we choose a vertex of the polyhedron, the projected vertex makes some angles across some geodesic polygons. The sum of these angles is 2π (again, the proof is left as an exercise for the reader)**. (**Σsum of angles) is basically sum of angles across all the vertices. So this quantity is 2πV.

Now, what was n? n was the number of edges in the polygon. From the definition of edge in a polyhedron, each edge is shared by exactly two faces. Hence, when we calculate Σn, we count each edge twice. So Σnπ = πΣn=π(2E)=2πE.

We showed earlier that there are total F polygons. And we are summing over these F polygons. So Σ2π=2πF. Σarea is the surface area of the sphere, which is 4π. Plugging in values, we get

Hence our desired formula is proved.

Now what’s so great about this proof?

This proof connects seemingly unrelated concepts — spherical geometry, angles, areas— to prove a **combinatorial theorem**.

The ideas are really very clever and beautiful. this proof suggests that Euler’s Polyhedron Formula is more than just a fundamental formula about polyhedrons, and there must be a **close relation between this formula and geometry.**

And that indication eventually lead to the discovery of a **new mathem**atical branch, Topology.

I started this with a quote, so it’s ideal to end with another quote. If you have any suggestions about new articles/improvements on this article/any questions, don’t hesitate to ask. You can comment here (I’m not really sure how commenting on Medium works), or email me, or send me a message on instagram, or send me a message on twitter, or send me a message on discord (atonu_#1514).

**Note:**

**Leonhard Euler** lived 76 years, during this period he wrote more than **850 books in Mathematics,**

If we collect his books, we will make 75 volumes with 600 pages for each volume.

On top of all this, he has more than **4000 mathematical correspondences with other scholars.**

He wrote almost in all branches of mathematics and he created two new branches, **Calculus of variations** and** Functions of complex numbers**.

“A not good math paper has zero ideas, it’s just pushing through things that everybody already knows but nobody bothered to do. Then there are good math papers that have one new idea which is really shocking.”

–Alex Kontorovich