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

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

+s1<br />

−sn<br />

b<br />

+s2<br />

−sk+2<br />

a<br />

+sk<br />

−sk+1<br />

Fig. 17.3. Proof of Cauchy’s rigidity theorem<br />

17 Rigidity 295<br />

vertices a, sk+1,...,sn, b <strong>and</strong> c, tk+1,...,tn, d, we obtain the contradiction that<br />

ab > cd > ab. Thus there are more than two changes of sign.<br />

Since the number of changes of sign is even, it follows that there are at least four<br />

changes of sign. ⊓⊔<br />

The next step in our proof is Cauchy’s combinatorial lemma:<br />

Lemma 17.2. Let P be a proper convex polytope in E 3 . Assume that the edges of P<br />

are marked by +, by−, or are unmarked in such a way that the following statement<br />

holds. On circling in bd P around a vertex of P which is an endpoint of a marked<br />

edge, there are at least four changes of sign of the marks encountered (omitting the<br />

unmarked edges). Then there are no marked edges.<br />

Proof. Assume that there are marked edges. Let G be a graph consisting of a maximal<br />

connected set of marked edges <strong>and</strong> their endpoints. Omitting from bd P the vertices<br />

<strong>and</strong> edges of G leaves a family of open connected sets in bd P, the countries of G.<br />

Since G is connected, the countries are simply connected. Roughly speaking, each<br />

of these countries is a union of certain facets of P. Letv,e, f be the numbers of<br />

vertices, edges <strong>and</strong> countries, respectively, determined by the connected planar graph<br />

G. Euler’s formula for graphs in E 2 , see Theorem 15.2, then shows that<br />

v − e + f = 2.<br />

For i = 2, 3,...,let fi be the number of countries with a boundary consisting of i<br />

edges, where an edge is counted twice if, on circling the boundary of the country, it<br />

appears twice. The case f2 �= 0 cannot hold. Then,<br />

f = f3 + f4 +···<br />

Since an edge of G is on the boundary of two countries or is counted twice if it is on<br />

the boundary of only one country, we see that<br />

Hence<br />

2e = 3 f3 + 4 f4 +···<br />

4v = 8 + 4e − 4 f = 8 + 2 f3 + 4 f4 + 6 f5 + 8 f6 + 10 f7 +···

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

Saved successfully!

Ooh no, something went wrong!