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 The Euler Poincare Characteristic  for Polyhedral Solids

## Euler's Formula for Polyhedra

The regular polyhedra were known at least since the time of the ancient Greeks. The names of the more complex ones are purely Greek. But despite their being known for close to two millenia no one apparently noticed the fact that the sum of the number of faces F and the number of vertices V less the number of edges E is equal to two for all of them; i.e.,

#### V - E + F = 2

The value of (V-E+F) is usually denoted by the Greek letter Chi (Χ). Thus Χ(cube)=2.

It was the Swiss mathematician Leonhard Euler who recognized and published this fact. The value of two is said to be the Euler characteristic of each of the polyhedra. This value is not changed by stretching or shrinking any side or face or even shrinking a side or face to zero. This means that the Euler characteric is a topological invariant because it is not altered by any continuous mapping.

Another French-Swiss mathematician, Simon Lhuilier (1750-1840), found a slight generalization of Euler's formula to take into account polyhedra having holes. Lhuilier's formula is

#### V - E + F = 2 − 2G = 2(1− G)

where G is the number of holes in the polyhedron. Thus the Euler characteristic is 2 for a regular polyhedron but 0 for a torus-like polyhedron.

This is elegantly simple result. The following material is an extension of the Euler and Lhuilier formulas to polyhedral solids. The language is ambiguous in this matter. The term cube in ordinary discourse refers to a cubic solid whereas for the Euler formula it refers to the surface hull of the cubic solid. For the case of sphere there is an alternate term, ball to use for the solid leaving sphere to apply for the surface. For the other figures there are no convenient alternate terms and therefore terms like cubical solid and cubical surface must be used in stead.

The Euler Poincare characteristic for a polyhedral solid will be denoted as Χ3 and defined as

#### Χ3 = V - E + F - C

where C is the number of solid cell components in the figure.

It is convenient to work with prisms; i.e., polyhedra which have top and bottom faces which are the same polygon. Let N be the order of the polygon. This means that the number of edges and the number of vertices of the top and bottom faces are both N. Each edge of the polygon corresponds to a rectangular face on the side of the prism. Thus the number of faces on the sides of the prism is N. These together with the top and bottom polygons means that the number of faces is N+2. There is a side edge for each vertex of the top polygon and 2N edges for the top and bottom polygon. Therefore the number of edges is 3N. The vertices of the prism are the same as the vertices of the top and bottom polygon. Therefore the number of vertices of the prism is 2N. There is only one cell for the prism.

When these values are substituted into the formula Χ3 the result is

## A Single Hole

The reason for choosing to work with prisms is that it is simple to deal with holes for them. Consider what happens if a prismatic hole is created down from the top to the bottom. The subdivision of the top of a square prism is shown below.

For a polygon with N sides the top of the subdivision creates N faces where only one was before. Each one of these subdivisions corresponds to a cell component below it. Thus there are N cells where there was only one before. This means the net increase in the number of cells is (N-1). The N subdivision faces along with N rectangular sides of the hole and the N subdivision faces at the bottom of the prism plus the N faces separating the N subdivision solids means that the number of faces is increased by ((N-1) + N + N + (N-1))=4N-2. The number of edges is increased by N around the top of the hole plus N from each vertex of the hole to the corresponging vertex around the top face of the prism. There are N edges down the sides of the hole and N edges of the hole at the bottom face plus the N edges from the vertices of the hole to the vertices of the bottom face. Thus the increase in the number of edges created for the prism by the hole is (2N + N + 2N)=5N. The number of vertices is increased by N around the top of the hole and N around the bottom of the hole for a total increase in the number of vertices of 2N.

The change in the Euler Poincare characteristic for a prismatic solid with a hole is therefore

This means that

#### Χ3(prismatic solid with one hole) = 1 - 1 = 0

There is a simpler construction that works just as well but which might not be as convincing as the above one. First the concept of simply connected must be defined. A figure is simply connected if any loop created within the figure can be shrunk to a point without any part of it passing out of the figure. All that has to be done is create an edge from one vertex of the top and bottom poygons of the hole to the corresponding vertex of the top and bottom faces of the prism. This makes the top and bottom faces of the prism simply connected. There is one face created in the interior of the prism and this face makes the interior cell simply connected. Thus the increases created by the prismatic hole are then: 1. +2 for the hole itself less the two top and bottom faces of the hole for a net change of 0, 2. the two additional edges on the top and bottom faces 3. the additional face in the interior of the prism. Thus ΔΧ3 = (2-2) - 2 + 1 = -1.

## Multiple Holes

If a prismatic hole is created in one of the subdivision polygons of a prismatic solid with one hole the same accounting applies and the Euler Poincare characteristic would again be reduced by 1.

Thus for G holes

Thus

#### Χ3(prism with G holes) = Χ3(prism) - G = 1 - G

Since the Euler Poincare characteristic is a topological invariant the general formula is

#### Χ3(polyhedral solid with G holes) = Χ3(polyhedral solid) - G = 1 - G

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