Gruber P. Convex and Discrete Geometry

14 Preliminaries and the Face Lattice 257
the contrary, there is a hyperplane HFG which contains all vertices of F and G and
thus F and G. LetHG be a support hyperplane of P with G = HG ∩ P. Clearly
HG �= HFG. Rotate HG keeping the (d − 2)-dimensional plane HG ∩ HFG fixed to
the first position, say H, where it contains a further vertex v of P, not in HFG. Then
H is a support hyperplane of P and H ∩ P is a face E of P with E � G and such
that E ∩ HFG = G. Hence E ∩ F =∅. This contradicts the maximality of G. ⊓⊔
14.4 Convex Polytopes and Simplicial Complexes
A (finite) simplicial complex C in E d is a family of finitely many simplices in E d such
that for any S ∈ C each face of S is also in C and for any S, T ∈ C, the intersection
S ∩ T is a face of both S and T . In topology a (convex or non-convex) polytope is
defined to be the union of all simplices of a simplicial complex in E d , see Alexandroff
and Hopf [9] or Maunder [698]. The problem arises, whether a convex polytope P in
the sense of convex geometry can be obtained in this way. In some cases even more is
demanded, the vertices of the simplices all should be vertices of P. If the latter holds,
we speak of a simplification of P. That simplifications always exist seems to be well
known. Anyhow, this result is used by several authors, including Macdonald [675],
Ehrhart [292, 293] and Betke and Kneser [108] without further comment. Thus it is
important in our context, see Sects. 19.1, 19.2 and 19.4. For convex polytopes a proof
was communicated to the author by Peter Mani-Levitska [685] and only then we
found the proof of Edmonds [286] in the literature. Edmonds’s idea is also described
by Lee [635, 636]. Since these proofs are essentially different, and each contains an
interesting idea, both are presented.
Convex Polytopes have Simplifications
Our aim is to prove the following result.
Theorem 14.9. Let P ∈ P. Then P has a simplification.
The k-skeleton k-skel P of a convex polytope P in E d is the union of its faces of
dimension ≤ k. Byasimplification of k-skel P we mean simplifications of the faces
of P of dimension ≤ k which fit together at common sub-faces. Let f0(P) denote
the number of vertices of P.
Proof (by Edmonds). We may assume that P is proper. By induction, the following
will be shown.
(6) Let k ∈{0, 1,...,d}. Then each face of P of dimension ≤ k has a simplification.
These simplifications together form a simplification of k-skel P.
Consider a linear ordering of the set of vertices of P. Clearly, (6) holds for k = 0.
Assume now, that k > 0 and that (6) holds for k − 1. We construct a simplification
of k-skel P as follows. Let F be a face of P with dim F = k. By the induction
assumption, all proper faces of F have a simplification. Let p be the first vertex of