# Euler's Formula, Proof 18: Hyperplane Arrangements

This proof comes from a 1997 paper by Jim Lawrence. It applies to convex polytopes in Rd, and resembles in some ways the valuation proof.

If A is a finite set of hyperplanes in Rd, it partitions Rd into faces, sets of points that are all contained within the same set of hyperplanes, and that are on the same side of all hyperplanes that do not contain them. Lawrence defines an A-polyhedron to be any union of faces, and he defines a function χ from A-polyhedra to integers, where if f is a face then χ(f)=(-1)dim f, and if P is an A-polyhedron composed of faces fi then χ(P)=sum χ(fi).

It is not difficult to see that χ has the same value for P in an arrangement A for which P is an A-polytope; for, adding any superfluous hyperplane to the arrangement merely splits some faces into two same-dimension faces and one face of one lesser dimension, which does not change the total value of χ for those faces, and removing any hyerplane merely reverses this process. Consequentially, if P is any nonempty intersection of finitely many open halfspaces then &chi(P)=(-1)dim P, for A can be assumed to be the arrangement of boundary hyperplanes of the halfspaces defining P, for which P is a face. In particular, from the intersection of zero halfspaces we obtain χ(Rd)=(-1)d. Also, it is obvious from the construction that the value of χ on a disjoint union of polyhedra is simply the sums of its values on the individual polyhedra making up the union.

We can now calculate the Euler characteristic of a closed d-dimensional convex polyhedron P, by embedding P on a hyperplane missing the origin in Rd+1, forming the infinite cone C of positive scalar multiples of points in P, and computing X(P)=χ(C). This valuation (when viewed in terms of the arrangement of hyperplanes through the facets of this cone) includes a term of the form (-1)k for every k-dimensional face of P, including 1 for the empty face (corresponding to the apex of the cone), -1 for the vertices (rays forming edges of the cone, which are 1-dimensional faces of the arrangement), etc.

X(P) = χ(Rd+1) - χ(Rd+1\C) = (-1)d+1 - χ(union Hi),

where Hi is the open halfspace bounded by the i'th hyperplane on the side of that hyperplane away from C. We already know the value of the first of these two terms, and the second can be calculated by inclusion-exclusion as

-sumS subset [1,n] (-1)|S|χ(intersectioni in S H_i)

summed over all nonempty subsets of the halfspaces Hi. But all such subsets have nonempty intersection (they contain the cone complementary to C) and are open convex polyhedra, so we can simplify the sum above to

-sumi=1n (-1)i (n choose i) (-1)d+1 = (-1)d+1,

which exactly cancels the χ(Rd+1) term leaving X(C)=0.

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

Semi-automatically filtered from a common source file. Last update: .