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 The Euler Poincare Characteristic for Different Dimensions of Geometric Figures

After about two thousand years that mathematicians had been contemplating polyhedra (pyramids, cubes, etc.) the Swiss mathematician Leonhard Euler (1707-1783) discovered an amazing regularity among them. If the numbers of vertices and faces are added together and the number of edges subtracted the result for all polyhedra is 2 if there are no holes in the polyhedra. Later another Swiss mathematician, Simon Lhuilier (L'Huillier) (1750-1840), found that for all polyhedra having one hole the Euler computation yields the value of 0. The value for the Euler computation for a polyhedron came to be known as the Euler characteristic of the polyhedron. It was symbolized with the Greek letter Chi, Χ. Lhuilier found a formula the Euler characteristic of a polyhedron with G holes; i.e., Χ=2(1-G).

The vertices of a polyhedron are its elements of zero dimension. Edges are the polyhedron's elements of one dimension and faces are the elements of two dimensions. If Nn denotes the number of elements of n dimensions then the Euler characteristic is defined as

#### Χ = N0 - N1 + N2

With this scheme the Euler characteristic can be generalized to

#### Χm = N0 - N1 … ± Nmor, equivalently, Χm = Σi=0m(-1)iNi

This generalization is often called the Euler-Poincaré characteristic.

The Euler computation was carried out for the two dimensional surface of the polyhedra, the hull of the three dimensional figures. If Pm stands for the m dimensional analogue of a polyhedron with no holes then Euler's result was that

#### Χ2(P3) = 2

If the three dimensional cell contained within the polyhedral hull is included in the computation then

#### Χ3(P3) = 1

The analogous computations can be carried out for geometric figures of other dimensions. A square has four vertices and four edges and one square surface. Therefore Χ1(P2) = 4 - 4 = 0 and Χ2(P2) = 4 - 4 + 1 = 1.

Likewise a line segment has two vertices and one edge. (Hereafter a line segment will be referred to simply as a line.) Thus Χ0(P1) = 2 and Χ1(P1) = 2 - 1 = 1. A point is a zero dimensional figure and has just one vertex. Thus Χ0(P0) = 1.

The only reasonable definition for the hull of a point is the elimination of the point and the creation of a null figure; i.e., no vertices, no edges, etc. The characteristic for the null figure is zero.

The four dimensional analogue of a cube is called a tessaract. It has 16 vertices, 32 edges, 24 faces and 8 cells. It also has its nameless four dimensional interior which could be called a tessaract solid. Thus

#### Χ3(P4) = 16 - 32 + 24 - 8 = 0 and Χ4(P4) = 16 - 32 + 24 - 8 + 1 = 1

The above results are compiled in the following table.

 Dimensionof Space GeometricFigure Characteristicof Hull Characteristicof Solid 0 point 0 1 1 line 2 1 2 square 0 1 3 cube 2 1 4 tessaract 0 1

The pattern is clear; i.e.,

## Holes

It is very simple to see the effect of holes in a line or a square. A hole in a line creates two additional vertices and two lines where there was only one before. Let Pm(G) denote a geometric figure of m dimensions with G holes. Then

#### Χ1(P1(1)) = 4 - 2 = 2 and Χ0(P1(1)) = 4

Stated differently

#### ΔΧ1 = 1 and ΔΧ0 = 2

For a square the creation of a hole results in a figure of 8 vertices, 8 edges and 2 faces. Thus

#### Χ2(P2(1)) = 8 - 8 + 2 = 2 and Χ1(P2(1)) = 8 - 8 = 0

For the hull of a square:

For a solid square:

Thus

#### ΔΧ2 = 1 and ΔΧ1 = 0

It was found elsewhere that for a cube

#### ΔΧ3 = -1 and ΔΧ2 = -2

Again the only reasonable definition for a hole in a point is the elimination of the point and the creation of a null figure, which of course has the characteristic zero.

These results are tabulated in the following two tables

 Dimensionof Space GeometricFigure Characteristicof SolidWith No Holes Characteristicof Solid With One Hole Increment inCharacteristic 0 point 1 0 -1 1 line 1 2 +1 2 square 1 2 +1 3 cube 1 0 -1 4 tessaract 1 ? ?

 Dimensionof Space GeometricFigure Characteristicof HullWith No Holes Characteristicof Hull With One Hole Increment inCharacteristic 0 point 0 0 0 1 line 2 4 +2 2 square 0 0 0 3 cube 2 0 -2 4 tessaract 0 ? ?

Lhuilier's formula is that Χ2(P3(G)) = 2(1 - G). Its extension to the solid polyhedra is Χ3(P3(G)) = (1 - G). For lines Χ1(P1(G)) =(1 + G) and Χ0(P1(G)) = 2(1 + G). For squares the relations are the same; i.e., Χ2(P2G)) = (1 + G) and Χ2(P2(G)) = 2(1+ G). For a point the formula would be Χ-1(P0(G)) = 0. The reduction of these to one master formula awaits the analysis of the case of the tessaract.

## The Euler Poincaré Characteristic of Tessaract with a Hole

In the case of the hulls for a prism or a polygon the creation of a pit does not alter the topologically invariant characteristic. The change in the characteristic comes only when pits become holes. When two pits in a prism come together to make a hole two faces are eliminated. Since the coefficient for the number of faces in the computation of the characteristic is +1 the elimination of two faces reduces the characteristic by two; i.e., ΔΧ = -2. In the case of a polygon hull the joining of two pits to make a hole eliminates two edges. Since the coefficient of the number of edges in the computation of the characteristic is -1 the elimination of two edges increases the characteristic for the polygon hull by two; i.e., ΔΧ = +2.

Therefore the creation of pits in a tessaract hull does change the characteristic but when the pits come together to form a hole two cells are eliminated. Since the coefficient of cells in the computation of the characteristic is -1 the elimination of two cells increases the characteristic by 2; i.e., ΔΧ = +2.

Thus

 Dimensionof Space GeometricFigure Characteristicof HullWith No Holes Characteristicof Hull With One Hole Increment inCharacteristic 0 point 0 0 0 1 line 2 4 +2 2 square 0 0 0 3 cube 2 0 -2 4 tessaract 0 2 2

(To be continued.)

For more on the Euler-Poincaré Characteristic of geometric figures see Euler.