Gruber P. Convex and Discrete Geometry

18 Theorems of Alexandrov, Minkowski and Lindelöf 307
Proof. Since the implication (i)⇒(ii) is trivial, it is sufficient to show that
(ii)⇒(i) Let u1, −u1,...,um, −um be the exterior normal unit vectors of the
facets of P and α1,α1,...,αm,αm the corresponding areas. By (ii), u1, −u1,...,
um, −um and α1,α1,...,αm,αm are the exterior normal unit vectors and the areas
of the facets of −P. Hence the uniqueness statement in the existence and uniqueness
theorem implies that P is a translate of −P. This, in turn, shows that P is centrally
symmetric. ⊓⊔
A useful result is the following symmetry condition of Minkowski [736].
Theorem 18.3. Let P = P1 ˙∪··· ˙∪Pm, where P, P1,...,Pm ∈ Pp. If each of
the convex polytopes Pi is centrally symmetric, then P is also centrally symmetric
(Fig. 18.1).
Proof. By the above corollary, it is sufficient to show the following statement.
(4) Let F be a facet of P with exterior normal unit vector u, say, and G the
face of P with exterior normal unit vector −u. Then v(F) = v(G) and, in
particular, G is a facet of P.
Orient Ed by the vector u. Consider the facets of the polytopes Pi which are parallel
to aff F. If such a facet H is on the upper side of a polytope Pi, let its signed area
w(H) be equal to v(H), and equal to −v(H) otherwise. Then
�
w(H) = 0
H
since all Pi are centrally symmetric. Since the signed areas of the facets of the Pi,
which are strictly between the supporting hyperplanes of P parallel to aff F, cancel
out, �
w(H) = � w(H k ) + � w(Hl),
H
where the H k are the facets contained in F and the Hl the facets contained in G.
Clearly, the facets in F form a dissection of F and similarly for G. Hence,
v(F) − v(G) = � v(H k ) − � v(Hl) = � w(H) = 0,
concluding the proof of (4) and thus of the theorem. ⊓⊔
Fig. 18.1. Minkowski’s symmetry condition