§3. Gluing of surfaces. A problem of topological classification.

L et ₁ and ₂ are two surfaces with boundary and suppose, that each of the surfaces has an edge homeomorphic to a circle. We join (glue, paste) the edges and we get a new surface. One can say that a hole in the first surface is pasted by the second one or vise versa. If both surfaces ₁ and ₂ are closed, the new surface is closed as well.

E xample 4. If we paste the sphere and the torus we get the surface ‘a sphere with a hand’, which is homeomorphic to the torus (pic.32). The sphere with two hands is homeomorphic to the surface, which can be obtained by pasting of two toras.

Exercise 31. Consider the following evolvent (pic.34). What surface we get, if we past the sides indicated by the same letter taking into account its direction?

A problem of topological classification of surfaces means the following. We should submit a list of closed surfaces, which are not homeomorphic to each other, and any closed surface is homeomorhic to one of the surfaces from the list.

Denote P_k the sphere with k hands. For example, P_o is the sphere and P₁ is the torus.

Theorem. P_o, P₁, P₂, …, P_k , … is a full topological classification of the closed orientable surfaces.

We will submit the proof of the theorem later.

§4. Eulerian characteristics of a surface.

Let  be a surface with boundary or without boundary, having 1 or 2 sides. Suppose, that  admits decomposition on polygonal domains, it means that we can draw a graph G on the surface, which divides the surface on a finite number of domains each of them is homeomorphic to the circle. Denote V – the number of vertexes of the graph, E – the number of edges and F – the number of polygonal domains (faces). The number

()=VE+F (5)

is called Eulerian characteristics of the surface . Precisely speaking, this number is determined not by the surface itself, but by the graph G. We proved for the sphere that this number doesn’t depend on the graph and it is equal to 2. We will show now that it is true for any surface.

Let two graphs G₁ and G₂ be drawn on the surface . Each of them determines a decomposition of the surface on polygonal domains. Let V_i be the number of vertexes of the graph G_i, E_i – the number of edges and F_i – the number of faces. The graphs G₁ and G₂ could have infinite number of intersection points. But we can move the graph G₁ a bit so that two graphs would have finite number of intersection points. If the graph G₁G₂ is not connected, we can move G₁ and G₂ so, that they would have common points.

We add all the common points to the set of the vertexes of the graph G₁G₂. Denote V – the number of vertexes of the graph G₁G₂, E – the number of edges and F – the number of faces. We are going to prove two equalities

(6)

and they imply V₁E₁+F₁= V₂E₂+F₂. Both equalities (6) have the same proof. So we will prove only the first one.

L et M be an arbitrary face determined by the graph G₁ (pic.35). Denote V, E the number of vertexes and the number of edges of the graph G₁G₂ located inside M (not on the border) and let this graph decomposes M on F faces.

Denote q the number of vertexes and the number of edges at the same time located on the border of M. Let M_o be the same polygon, but without any decomposition. Let us cut the polygon M out of the surface  and paste it by the border with M_o. We get the surface homeomorphic to the sphere (pic. 36). Each of the polygons M and M_o is homeomorphic to the hemisphere. We have a connected graph drawn on the sphere, which has V+q vertexes, E+q edges and F+1 faces, because F faces of M are added by one more face M_o. According to the Euler theorem we have

( V+q)(E+q)+(F+1)=2.

Therefore

VE+F=1. (7)

Let us return to the surface , where the graph G₁G₂ is drawn. We delete from the graph its part located inside M. Instead of V vertexes, E edges and F faces we get 0 vertexes, 0 edges and 1 face (M itself). I.e. the number VE+F will be replaced by the number 00+1=1. According to (7) nothing will be changed!

We delete now all the parts of the graph G₁G₂ located inside all the faces of the graph G₁. We get a new graph G* and the number V*E*+F* remains the same, i.e. V*E*+F*=VE+F.

What is the difference between G₁ and G*? G* has some additional vertexes on the edges of G₁. If one adds a vertex on the edge of a graph, he adds 1 more edge at the same time. So this action does not change the number VE+F. Therefore V*E*+F*=V₁E₁+F₁ and the first of the equalities is true.

We proved, that Eulerian characteristics does not depend on a decomposition of the surface on polygonal domains and it is determined by the surface itself. Moreover, it is topological invariant.

Really, let a finite graph G be drawn on the closed surface ₁ and let f:₁₂ be a homeomorphism. Then G=f(G) is the graph on the surface ₂ which has the same number of edges and the same number of vertexes as G. And G decomposes ₂ on the same number of domains as G decomposes ₁. Therefore (₁)=(₂).

N ow it is easy to prove, that the sphere is not homeomorphic to the torus. Consider the following graph on the torus. It has 1 vertex, 1 edge and there is only one face. Therefore (T²)=12+1=0. And we know, that (S²)=2.

Exercises 32. Prove, that the sphere with q holes has the Eulerian characteristics 2q.

33. Let ₁ and ₂ be two surfaces, that have boundary homeomorphic to the circle. Prove, that if we paste them by there’s boundaries, we will get the surface with the Eulerian characteristics (₁)+(₁).

34. Find the Eulerian characteristics of circle, of a hand and of the Mőbius band.

35. Prove, that the Eulerian characteristics of the surface P_k is equal to 2k.

34. A graph is drawn on a surface and it has V vertexes, E edges. Suppose, that the graph decomposes the surface on F domains, but some of the domains are not homeomorphic to a circle. Prove, that VE+F().