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Euler’s Polyhedron Theorem

The next result was at least in an implicit form known to R.Descartes in 1620:
Euler’s theorem . For a convex polyhedron we have: (number of vertices) - (number of edges) + (number of faces) = 2.

For instance, for a tetrahedron we have .

Euler’s theorem extends to a sphere with an arbitrary number of handles:

Theorem. If , , and are the number of vertices, edges, and faces of any trinagulation of a sphere with handles, then .

Sketch of the proof: A sphere with handles can be represented as series of cubes  each with a smaller cube punctured from it, and glued along the non-punctured face.

Now convert the polygonal division on this chain on glued together cubes into a triangular one by drawing diagonals in the existing faces. Every new diagonal increases the number of edges by one, and also the number of faces and so their difference remains unchanged. A direct count gives


,
and the result follows.

Note that the last result is a special case of a basic result in topology, where the value of the expression depends on the topological character of the surface. A standard way to prove this is to show that where the integer is called th Betti number of the surface, and it is a topological invariant of the surface. The sum is also called Euler characteristic of the surface.

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