Gruber P. Convex and Discrete Geometry

268 Convex Polytopes
consists precisely of those facets of Fi which are not visible from pi. By induction,
these facets come last in a shelling of Fi. The earlier proposition then says that these
facets come first in a different shelling of Fi. Hence F1,...,Fm, in fact, is a shelling
of P, concluding the proof of (5). Having proved (4) and (5), the induction and thus
the proof of (3) is complete. ⊓⊔
The Euler Characteristic of Polytopal Complexes
In Sect. 7.1 we have defined the Euler characteristic χ on the lattice of polytopes, i.e.
for finite unions of convex polytopes. It is the unique valuation which is 1 for convex
polytopes.
Here, the Euler characteristic χ(C) of a polytopal complex C is defined by
χ(C) = f0 − f1 +−··· ,
where fi = fi(C), i = 0, 1,...,d, is the number of convex polytopes in C of
dimension i. f0 is the number of vertices, f1 the number of edges, etc. If C and
D are complexes such that C ∪ D is also a complex, that is, P ∩ Q ∈ C ∩ D for
P ∈ C and Q ∈ D, then the following additivity property holds, as can be shown
easily,
χ(C ∪ D) + χ(C ∩ D) = χ(C) + χ(D).
In other words, χ is a valuation on the family of polytopal complexes.
It can be shown that the Euler characteristic (in the present sense) of a polytopal
complex and the Euler characteristic (in the sense of Sect. 7.1) of the underlying
polytope coincide.
The Euler polytope formula for general d
shows that the Euler characteristic of the boundary complex of a convex polytope
can be expressed in a very simple form.
Theorem 15.5. Let P ∈ Pp. Then
χ � C(bd P) � = f0 − f1 +−···+(−1) d−1 fd−1 = 1 − (−1) d ,χ � C(P) � = 1.
Proof. It is sufficient to show the following, where C(F1 ∪···∪Fi) is the polytopal
complex consisting of the facets F1,...,Fi and all their faces:
(6) Let P ∈ Pp and let F1,...,Fm, m = fd−1, be a shelling of P. Then
(i) χ � C(Fi) � = 1fori = 1,...,m.
(ii) χ � C(F1 ∪···∪ Fi) � �
1 for i = 1,...,m − 1.
=
1 − (−1) d for i = m.
(iii) χ � C(P) � = 1.
The proof of (6) is by induction. (6) is clear for d = 1. Assume now that d > 1 and
that (6) holds for d − 1. By induction, (iii) implies that