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Gruber P. Convex and Discrete Geometry

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286 <strong>Convex</strong> Polytopes<br />

T ∩ Z, respectively. Let ±u be the exterior normal unit vectors of S ∩ Z at the facets<br />

F, F − lv. Clearly, ±v are the exterior normal unit vectors of T ∩ Z at the facets<br />

G, G − lv. Choose p ∈ F <strong>and</strong> q ∈ G arbitrarily. Then<br />

(19) The contributions of the facets F, F − lv of S ∩ Z to V (S ∩ Z) <strong>and</strong> of the<br />

facets G, G − lv of T ∩ Z to V (T ∩ Z) are<br />

1 � � l<br />

p · u v(F) + (p − lv) · (−u) v(F−lv) = u · vv(F),<br />

d<br />

d<br />

1 � � l<br />

q · vv(G) + (q − lv) · (−v) v(G − lv) =<br />

d<br />

d v(G).<br />

Here, the translation invariance of v was used. Clearly, G = F ′ , where “ ′ ”is<br />

the orthogonal projection of the hyperplane HF = aff F onto the hyperplane HG =<br />

aff G. Since S, T are not parallel, HF, HG are also not parallel. Let w : P(HF) → R<br />

be defined by:<br />

w(Q) = v(Q ′ ) for Q ∈ P(HF).<br />

Since, by induction, v is a simple translation invariant, monotone valuation on<br />

P(HG), it is easy to see, that this also holds for w on P(HF). Thus, there are two<br />

simple, translation invariant valuations v <strong>and</strong> w on P(HF). The uniqueness theorem<br />

16.1 then yields<br />

w(Q) = αv(Q) for Q ∈ P(HF),<br />

where α ≥ 0 is a suitable constant. To determine its value, choose a (d − 1)dimensional<br />

cube K in HF of edge-length 1 such that a (d − 2)-dimensional facet of<br />

K is contained in HF ∩ HG. Then K ′ is a box in HG of edge-lengths 1,...,1, u · v.<br />

Thus<br />

w(K ) = v(K ′ ) = u · v = αv(K ) = α,<br />

i.e. α = u · v. Here we have used property (12) for v. Hence<br />

By (19), this shows that<br />

v(G) = v(F ′ ) = αv(F) = u · vv(F).<br />

(20) The contributions of the facets F, F − lv of S ∩ Z to V (S ∩ Z) <strong>and</strong> of the<br />

facets G, G − lv of T ∩ Z to V (T ∩ Z) are the same.<br />

Having proved (18) <strong>and</strong> (20), the proof of (17) is complete, concluding the induction<br />

for (13).<br />

For the proof of (14), let P ∈ P <strong>and</strong> t ∈ E d .IfP is improper, V (P) = 0 =<br />

V (P + t) since V is simple by (11), <strong>and</strong> we are done. Assume now that P is proper.<br />

Let L be a line through o with direction t <strong>and</strong> call it vertical. Let H be the (d − 1)dimensional<br />

subspace of E d orthogonal to L. LetF1,...,Fm be the facets of P on<br />

the upper side <strong>and</strong> G1,...,Gn the facets on the lower side of P. The translate P + t<br />

of P may be obtained as follows. Dissect each facet Fi into pieces Fi1,...,Fin <strong>and</strong><br />

each facet G j into pieces G1 j,...,Gmj such that

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