14.02.2013 Views

Gruber P. Convex and Discrete Geometry

Gruber P. Convex and Discrete Geometry

Gruber P. Convex and Discrete Geometry

SHOW MORE
SHOW LESS

Create successful ePaper yourself

Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.

16 Volume of Polytopes <strong>and</strong> Hilbert’s Third Problem 285<br />

Since we have already proved the assertion (10) for m = 2, the induction assumption<br />

on m then shows that<br />

V (P) = V � (P ∩ H − ) ˙∪(P ∩ H + ) � = V (P ∩ H − ) + V (P ∩ H + )<br />

= V � P1 ˙∪··· ˙∪Pj ˙∪(Pj+1 ∩ H − ) ˙∪··· ˙∪(Pk ∩ H − ) �<br />

+ V � (Pj+1 ∩ H + ) ˙∪··· ˙∪(Pk ∩ H + ) ˙∪Pk+1 ˙∪···<br />

�<br />

˙∪Pm<br />

= V (P1) +···+V (Pj) + V (Pj+1 ∩ H − ) +···+V (Pk ∩ H − )<br />

+ V (Pj+1 ∩ H + ) +···+V (Pk ∩ H + ) + V (Pk+1) +···+V (Pm)<br />

= V (P1) +···+V (Pj) + V (Pj+1) +···+V (Pk) + V (Pk+1) +···+V (Pm).<br />

The induction on m is thus complete. Hence (10) holds for d <strong>and</strong> all m. This concludes<br />

the induction on d for (10).<br />

For the proof of the assertion (11) note first that V (P) = 0forP ∈ P with<br />

dim P ≤ d − 1. Hence V is simple. To see that it is a valuation, let P, Q ∈ P such<br />

that P ∪ Q ∈ P. The cases where at least one of P, Q, P ∩ Q is improper are easily<br />

dealt with by the simplicity of V <strong>and</strong> (10). Assume now that P, Q, P ∩ Q all are<br />

proper. Consider all hyperplanes through facets of P ∩ Q. These hyperplanes dissect<br />

P <strong>and</strong> Q into proper convex polytopes<br />

say. Then<br />

P ∩ Q, P1,...,Pm <strong>and</strong> P ∩ Q, Q1,...,Qn,<br />

P ∩ Q, P1,...,Pm, Q1,...,Qn<br />

form a dissection of P ∪ Q <strong>and</strong> it follows from (10) that<br />

V (P) + V (Q)<br />

= V (P ∩ Q) + V (P1) +···+V (Pm) + V (Q1) +···+V (Qn) + V (P ∩ Q)<br />

= V (P ∪ Q) + V (P ∩ Q).<br />

Thus V is a valuation, concluding the induction for (11).<br />

Assertion (12) easily follows from the definition of V by induction on d.<br />

If dim Z < d, the assertion (13) holds trivially. Assume now that dim Z = d. To<br />

speak more easily, call L vertical for the proof of (13). Let S be a parallel slab as in<br />

(13) <strong>and</strong> T a horizontal slab of width l. For the proof of (13) it is sufficient to show<br />

that<br />

(17) V (S ∩ Z) = V (T ∩ Z).<br />

If S <strong>and</strong> T are parallel, this is easy to see. Assume now that S <strong>and</strong> T are not parallel.<br />

By induction,<br />

(18) The contributions of the vertical facets of S ∩ Z <strong>and</strong> T ∩ Z to V (S ∩ Z) <strong>and</strong><br />

V (T ∩ Z), respectively, are the same.<br />

Next, the non-vertical facets will be considered. Let v be a unit vector parallel to<br />

L such that F, F − lv <strong>and</strong> G, G − lv are the top <strong>and</strong> bottom facets of S ∩ Z <strong>and</strong>

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!