The Geometry Junkyard

Euler's Formula, Proof 6: Electrical Charge

This proof is due to Thurston. He writes:

Arrange the polyhedron in space so that no edge is horizontal – in particular, so there is exactly one uppermost vertex \(U\) and lowermost vertex \(L\).

Put a unit \(+\) charge at each vertex, a unit \(-\) charge at the center of each edge, and a unit \(+\) charge in the middle of each face. We will show that the charges all cancel except for those at \(L\) and at \(U\). To do this, we displace all the vertex and edge charges into a neighboring face, and then group together all the charges in each face. The direction of movement is determined by the rule that each charge moves horizontally, counterclockwise as viewed from above.

Polyhedron with
charges on its faces, rotated clockwise around the z-axis into a
neighboring face

In this way, each face receives the net charge from an open interval along its boundary. This open interval is decomposed into edges and vertices, which alternate. Since the first and last are edges, there is a surplus of one \(-\); therefore, the total charge in each face is zero. All that is left is \(+2\), for \(L\) and for \(U\).

Thurston goes on to generalize this idea to a proof that the Euler characteristic is an invariant of any triangulated differentiable manifold.

Proofs of Euler's Formula.
From the Geometry Junkyard, computational and recreational geometry pointers.
David Eppstein, Theory Group, ICS, UC Irvine.