The Unbreakable Oneness : Euler Poincare Theorem

Take a stick (a line segment) $AB$.  Take a point $P$ on $AB$:

 An extra point $P$ turns one line segment into two line segments $AP$  and $PB$. If we take two points $P_1$ and $P_2$ on $AB$ the line  segment $AB$ turns into three line segments.

                                           
Two new points $P_1$ and $P_2$ increase the number of line segments by two. Choose eight extra points, and observe that the number of line segments go up by eight.
                                      
Two hundred new points placed anyway increase the number of segments exactly by two hundred. The new points's number is matched by the  number of extra segments. All through the process of breaking of  $AB$, we may write: 

    The number of extra points in this breaking = The number of extra line segments obtained.

Let us start afresh by taking a rectangular (or a quadrilateral) plane $ABCD$. The line segment $BD$ (or $AC$) replaces the face $ABCD$ by two faces, triangles $ABD$ and $BCD$.

                                                   
By taking pentagon, hexagon, (regular or not) one extra edge $AC$ replaces one given face by two.

   Increase in number of faces

This fact remains unaltered, even if the line segment $AC$ is replaced by another one.

Furthermore, three extra edges increase the number of faces by 3 and five extra edges increase the number of faces by 5.

 Increase in number of faces

 New situations: Observe the cases below.

---------------------------------------------------------------------------

New edges = 4,  Increase in  number of faces = 3    

New edges = 6,  Increase in  number of faces = 5   

New edges = 3,  Increase in  number of faces = 2   

---------------------------------------------------------------------------

The number of extra edges in these cases exceeds the number of extra faces by one. How this excess can be accounted for?  To explore, let us look at the following examples: 
Case: extra edges exceeding extra faces by two.

---------------------------------------------------------------------------

New edges = 7,  Increase in  number of faces = 5    

New edges = 5,  Increase in  number of faces = 3   

New edges = 5,  Increase in  number of faces = 2

--------------------------------------------------------------------------   

Case: extra edges exceeding extra faces by three.

-----------------------------------------------------------------------

New edges = 7,  Increase in  number of faces = 4    

New edges = 10,  Increase in  number of faces = 7   

-----------------------------------------------------------------------

Case : extra edges exceeding extra faces by five 

------------------------------------------------------------------- 

New edges=12 , Increase in number of faces = 8

New edges=14,  Increase in number of faces=9 

---------------------------------------------------------------------

A keen observation of the different kinds above reveals the difference in the number of new edges over the increase in number of faces is exactly compensated by the number of new vertices emerging. The different kind of breaking of a plane figure, in each of the above  example shows that number of new edges exceeds by the increased number of faces by the number of new vertices.
If $n(E)$ = number of new edges, $n(F)$ = increase in number of faces and $n(V)$ = number of new vertices emerging; then                         $ n(E) $ $-$ $n( F)$ $=$ $n(V)$  Or  $n(V)$ $-$ $n(E)$ $+$ $n(F)$ $=$ $0.$ 

An interesting game is being played. I remember how one evening my naughty daughter told me, it was more of a warning, that anyone displeasing her would be taken to task by her grandfather (i.e., my father). We are witness to our children making alliance with our parents to match against us. This biological principle of alternate generations teaming together seems to be true in Geometry of the physical world as well. The extra number of faces, in the process of division is equalled by the extra number of edges coming up; and more remarkable is the fact that, in the division process if new vertices are coming up then the number of new vertices gets aligned to the number of new faces to match equally with the number of new edges added.

Thus there remains  no change, or balancing of the new vertices and new faces together  against the new edges. It is worth noting that in all the examples discussed above the initial (undivided, single-piece) state, we have: 

For zero-dimensional objects (there is only one such) \[ \mathrm{Number\ of\ vertices} = 1.  \] For one-dimensional objects, \[ \mathrm{ Number\ of\ vertices - number\ of\ edges} = 1. \]  For two-dimensional objects,  \[ \mathrm{Number\ of\ vertices - number\ of\ edges + number\ of\ faces} = 1. \] Symbolically we may choose to write: \[V - E+ F=1.\] We shall find that for three dimensional one-piece objects, \[ \mathrm{ Number\ of\ vertices - number\ of\ edges + number\ of\ faces} - \mathrm{ number\ of\ solids} = 1. \] We can, at least, make analogous conjecture for things in higher dimension.

Foldings

Consider the following figure (made of cardboards).

 Easily verified is $V - E + F = 1$. Don't you feel tempted to turn (fold) it like in below Figure:

Folded cardboard

It  no longer lies in two dimensional Euclidean space. What about $V - E+ F$? 

Even now one! Strange, but wonderful. Further look at the cardboard in below figure.

                                  

Don't you feel tempted to turn it as in figure. 

Folded Cardboard

Yet, $V - E + F$ remains one! 

Strip

Now suppose we begin with a strip as shown in above figure and turn it by merging $AB$ to $A'B'$, resulting in figure:

New formation

The merger of $AB$ into $A' B'$ is a formidably new thing. It changes the structure, of the original piece. It has given birth to a hollow see-through, a tunnel; something unconceivable of the original strip. And $V - E + F$ is no more one. Mergers can create many new situations one such is the double tunnel as seen in figure:

Double tunnel

Polygons to Containers

 To escape from the myriad of situations, that the merger can create, let me take refuge in the school days play of hollow paper balls roughly spherical. Let us try  such mergers leading to some sort of cyclic shapes only. 

As a preparation towards cyclic shapes, let us first make lidless containers. Consider the first shape in Figure \ref{27-28}, easily available by breaking/cutting cardboards; not forgetting to verify $V - E + F$ is one:

Cardboard and Cuboidal Container

Let us first turn along $DC$ and $DA$. What change it brings to $V - E+ F$ of the figure? Two points $E$ and $E'$ have merged into one reducing the value of $V$ by $1$; while the two edges $DE$ and $DE'$ have merged into one reducing the value of $E$ by $1$; and thus $V - E+ F$ remains unaltered.  Similarly, turning along $AB$ and $BC$, we observe that $AF$ merges to $AF'$, $BG$ merges to $BG'$ and $CH$ merges to $CH'$, keeping $ V - E+ F$ unaltered at one.

Lidless cuboid as in above figure has $ V - E+ F = 1$, like the plane figure it is based upon.

Next, consider the following: turning about $AB$ and $AC$, so that $AD$ merges with $AE$. As a result, we obtain a joker's cap, hollow at triangle $BCD$ results.

Joker's cap

Yet  $V - E + F$ remains one.

Experiment with more and more polygons, divided into polygons; folding them up into single-opening lidless containers we can ensure the oneness of $V - E + F$. 

It would be equally interesting to flatten many ready made lidless cardboard containers you come across, and ensure the oneness of their $V - E + F$ at each stage of the process.

We may recall how attempts to fracture a line segment failed to alter $V - E$; and then attempt to fracture a single plane polygon failed to alter $V - E + F$ for the polygon. You would do well in going ahead by taking solid polyhedra (convex) verifying oneness of $V - E + F - S$, and failing to alter it by making fractures that is plane cuts, denoting the number of solids by $S$.  Take care not to cut it in two.
                                                      

Fractured Bread (cut but not separated)

Observe:
 New vertices emerging = 6. New edges emerging = 7.
 New faces emerging =2.  Increase in the number of solids = 1.
 Change in $V - E + F - S = 6 - 7 +2 - 1 = 0$. 

This oneness of the objects can be changed, as observed earlier, by making surgeries (cutting or joining). This oneness, twoness, ... is a specialty, a peculiarity of the objects and is called its  Euler's Characteristic.

Placing the Lid on
We don't have choice about the shape of the lid. It has to be of the  same size and shape as that of the hole. So the number and size of the  vertices and edges of the lid has to just fit upon the hole.  

Placing the lid on an open cuboid

Placing the lid on an open container

Placing the lid on, on the lidless polyhedron does not leave any  freedom to add or to subtract the number of vertices or edges. Placing the lid amounts to only adding an extra face to the lidless solid. So obviously the value of number of faces for the closed polygon is one  greater than for the lidless polyhedron; the number of vertices and edges remaining the same.  Thus for a closed polyhedron $V - E+ F = 2$; a great historical result first observed by Euler, the Great. This  is probably the first result of a Geometry, which did not depend on the measures of lengths or angles of objects, suggesting the birth of a new Geometry.

An astonishing discovery this is that no factory can manufacture a regular solid with six (or more) sided polygonal base; and just one with pentagonal base.  This is easily derivable from Euler's result that there are only five regular  olyhedrons:  Tetrahedron, Cube,   Octahedron, Icosahedron and Dodecahedron.
 

Five regular polyhedrons

The above five figures can be obtained by folding up the following plane cut-outs:

Tetrahedron cutout 

Cube cutout

Isocahedron cutout

Octahedron cutout

Dodecahedron cutout

Have you noticed  the shapes of the patches a football is made of? Not all are the same. It is interesting to learn that there lies a lot of Mathematics (Chemistry and Physics too) behind the shape of the football. Take an Icosahedron, mark its twelve vertices. Chop off all the vertices, trisecting each set of the five edges culminating at the vertices. And you get the football.  It is good fun kicking a football. Hold its shape reverentially. It is greater fun discovering its symmetries. The symmetries have three types of axis of rotations: axes joining  midpoints of opposite pairs of edges of hexagons, axes joining the  centres of the opposite pairs of hexagons and axes joining the centres of opposite pairs of pentagons; giving birth to Icosahedral group.