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 The Euler Characteristic in Algebraic Topology

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 millennia apparently no one 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; i.e., Χ. Let Χ(P) denote the value of (V−E+F) for a polyhedron P. 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 characteristic 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 analysis and proof of the Euler and Lhuilier formulas. 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.

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

Χ(prism) = V - E + F = 2N - 3N + (N+2) = 2

Thus the Euler formula is verified for prisms. The reason for choosing to work with prisms is that it is simple to deal with holes for them. Consider first what happens if a prismatic pit is created down from the top but does not go through 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 one was before. These along with N rectangular sides of the pit and the polygon at the bottom of the pit means that the number of faces is increased by (N + N +1 -1)=2N. The number of edges is increased by N around the top of the pit plus N from each vertex to the corresponding vertex around the top edge of the prism. There are N edges down the sides of the pit and N edges at the bottom of the pit. Thus the increase in the number of edges created for the prism by the pit is (2N + N + N)=4N. The number of vertices is increased by N around the top of the pit and N around the bottom of the pit for a total increase in the number of vertices of 2N.

The change in the Euler characteristic for a prism with a pit is therefore

ΔΧ = 2N - 4N + 2N = 0

The same construction could be carried out from the bottom face and the same results in terms of faces, edges and vertices would prevail.

Now consider what happens if the pits from the top and bottom are extended until they meet and create a hole in the prism. The only thing that changes is that the two polygons at the ends of the pits disappear and thus the number of faces is reduced by 2. Thus

ΔΧ = −2

For G holes then clearly

Thus

Χ(prism with G holes) = Χ(prism) − 2G = 2 − 2G = 2(1 − G)

This same result would occur if a smaller prism were extruded from the top or bottom face of the prism. If such extrusions were extended from the top and bottom and bent to meet the effect on the Euler characteristic would be the same as in the creation of a hole in the prism; i.e., the elimination of the faces at the ends of the extrusions. This process creates a handle and hence

Χ(prism with a handle) = Χ(prism) − 2 = 2 − 2 = 0 and thus Χ(prism with G handles) = 2 − 2G = 2(1 − G)

The creation of a handle simultaneously creates a topological hole, so naturally the effects are the same.

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