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Gruber P. Convex and Discrete Geometry

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296 <strong>Convex</strong> Polytopes<br />

Call the index of a vertex of G the number of changes of sign of the marks encountered<br />

on circling the vertex once. Let I be the sum of all indices. By the assumption<br />

of the lemma,<br />

4v ≤ I.<br />

We now count the indices in a different way. If the boundary of a country determined<br />

by G consists of i edges, circling it there are at most i changes of sign if i is even <strong>and</strong><br />

at most i − 1ifi is odd. Thus<br />

<strong>and</strong> we obtain the contradiction that<br />

I ≤ 2 f3 + 4 f4 + 4 f5 + 6 f6 + 6 f7 +···<br />

8 + 2 f5 + 2 f6 +···≤0. ⊓⊔<br />

Proof of Cauchy’s rigidity theorem. The given homeomorphism maps the facets,<br />

edges <strong>and</strong> vertices of P onto the facets, edges <strong>and</strong> vertices of Q, respectively, where<br />

corresponding facets are congruent <strong>and</strong> corresponding edges have equal length. For<br />

the proof of the congruence of P <strong>and</strong> Q it is sufficient to show that the dihedral<br />

angles of P <strong>and</strong> Q at corresponding edges are equal.<br />

Mark an edge of P by +, −, or leave it unmarked, if the dihedral angle of P<br />

at this edge is greater than, less than, or equal to the dihedral angle of Q at the<br />

corresponding edge. We have to show that no edge of P is marked. To see the latter,<br />

we first show the following:<br />

(4) Let p be a vertex of P which is incident with at least one marked edge.<br />

Then, circling around p, we encounter at least four changes of sign.<br />

Consider spheres with centres at p <strong>and</strong> q, where q is the vertex of Q corresponding<br />

to p, <strong>and</strong> radius ϱ>0. Choose ϱ so small that the sphere with centre p meets only<br />

those facets <strong>and</strong> edges of P which contain p <strong>and</strong> similarly for q. Intersecting these<br />

spheres with P <strong>and</strong> Q, gives convex spherical polygons S <strong>and</strong> T , respectively. Call<br />

vertices of S <strong>and</strong> T corresponding if they are determined by corresponding edges of<br />

P <strong>and</strong> Q. The interior angles of S <strong>and</strong> T at their vertices are simply the dihedral<br />

angles of P <strong>and</strong> Q at their edges with endpoints p <strong>and</strong> q. Mark a vertex of S by<br />

+, −, or leave it unmarked if this holds for the edge of P which determines this<br />

vertex. If a vertex of S is marked by +, −, or is left unmarked, then the interior<br />

angle of S at this vertex is greater than, less than, or equal to the interior angle of T<br />

at the corresponding vertex. An application of Corollary 17.1 then shows that circling<br />

around S we have at least four changes of sign. Translating this back to P, we thus<br />

see that circling around the vertex p, there are at least four changes of sign of the<br />

marks at the edges of P with endpoint p. The proof of (4) is complete.<br />

Cauchy’s combinatorial lemma <strong>and</strong> Proposition (4) finally show that there are no<br />

marked edges at all. ⊓⊔<br />

Higher Dimensions<br />

Assume that one can prove Cauchy’s theorem in E k . Then it holds also in S k by<br />

the same arguments as for E k . By intersecting a closed convex polyhedral surface

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