The Geometry Junkyard

Euler's Formula, Proof 17: Valuations

This proof comes from Klain and Rota, who develop it as part of a more general theory of invariant measures on Euclidean spaces. Thanks also to Robin Chapman for helping explain it to me.

Let \(U^d\) be the vector space of functions from \(\mathbb{R}^d\) to \(\mathbb{R}\), and \(V^d\) be the subspace of \(U^d\) generated by characteristic functions of compact convex sets in \(\mathbb{R}^d\). We define a function \(\chi_d\) that maps \(V^d\) to real numbers, as follows:

Because \(\operatorname{D}\) and \(\sum\) are linear operators, \(\chi\) is a linear function from \(V^d\) to \(\mathbb{R}\). It can also be shown by a simple induction on dimension that, if \(f\) is the characteristic function of a compact convex set \(K\), then \(\chi(f)=1\): By induction, \(\chi(f_z)\) is the characteristic function of a closed interval, from which it is clear from the definition that \(\sum\operatorname{D}\chi(f_z)\) is \(1\). Thus also (since \(V^d\) has the characteristic functions of convex compact sets as its basis) \(\chi\) is well defined independently of the choice of coordinates for \(\mathbb{R}^d\).

If \(g\) is the characteristic function of the relative interior of a polytope \(K\), it is in \(V^d\) (by inclusion-exclusion of lower-dimensional faces) and a similar induction shows that \(\chi(g)=(-1)^m\), where \(m\) is the dimension of \(K\): Because of the independence of \(\chi\) from the coordinate system, we can assume \(m=d\). Each cross-section of \(K\) is itself a polytope, so by induction \(\chi(g_z)\) is \((-1)^{m-1}\) times the characteristic function of an open interval, and the effect of the \(\sum\) and \(\operatorname{D}\) operators is to multiply by another factor of \(-1\).

Now we let \(K\) be a convex polytope, and evaluate \(\chi\) on its characteristic function in two different ways, by partitioning \(K\) into the disjoint union of the relative interiors of its faces. By linearity of \(\chi\), the sum of its values on these relative interiors must be the same as its value on all of \(K\), that is, \[ 1 = \sum (-1)^m \] where the sum is over the set of faces of \(K\) (including \(K\) itself). Grouping the faces by their dimensions, we get a form of Euler's formula: \[ 1 = \sum_{i=0}^d (-1)^i f_i. \]

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