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Topic: Rigidity of Convex Polytopes
Replies: 6   Last Post: Jun 20, 1994 9:29 AM

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 John Sullivan Posts: 14 Registered: 12/6/04
Re: Rigidity of Convex Polytopes
Posted: Jun 20, 1994 9:20 AM

I haven't had a chance this morning to look up the reference
someone suggested yesterday, but over the weekend I was thinking
about the problem of rigidity for convex polytopes.
It seems to me that although rigidity in higher dimensions is not a corollary
of Cauchy's theorem about polyhedra, the two lemmas used for that argument
can also be used for the higher dimensional case.

Cauchy's theorem is a consequence of two lemmas. The first is geometric
and characterizes the kind of nonrigidity found at a single vertex.
If there are \$k\$ triangles around a vertex, then of course there are
\$k-3\$ degrees of freedom. But the lemma says that if you compare
two (convex) configurations of the vertex, and mark the edges
incident to the vertex '+', '-', or '0' depending on the sign change
(from one configuration to the other) of the dihedral angle,
then there are at least four changes of sign in these marks
unless all marks are '0'. Thus for \$k=4\$ we must have '+-+-' (or '0000'),
for \$k=5\$, '+-+-?' (or '00000'), etc.

The topological lemma then says that if we have any triangulation
of the sphere with edges marked '+', '-', '0', such that the
number of sign changes around any vertex is 0 or >=4, then in
fact all marks are 0.

In fact, for both lemmas, I prefer to think of the dual picture,
obtained by applying the Gauss map to your polyhedron.
So the first talks about a single spherical \$k\$-gon with
fixed internal angles; it has \$k-3\$ degrees of freedom in edge
lengths, but if we compare two configurations we find the
'+...-...+...-...' pattern. The second lemma deals with a spherical
net with marked edges such that the count of sign changes in the
ring of edges around any face is 0 or >=4.

Now let's see how these lemmas, for instance, imply rigidity for
the tetraplex, a 4d polytope made of 600 regular tetrahedra.
Again look at the Gauss map, or spherical dual. Each tetrahedron
corresponds to a point in \$S^3\$, its normal vector, each triangle to
a great circle arc, each edge to a pentagonal region on some great \$S^2\$.
These pentagons have equal angles arccos(-1/3) (the exterior dihedral
angle of the tetrahedron), but of course have two degrees of freedom
for edge lengths (the dihedral angles between the five corresponding
tetrahedra joined at an edge). However, the geometric lemma applies
to show how these edge length can vary.

Now the dual of the tetraplex divides the three-sphere into 120
regions, combinatorially each a dodecahedron. Even though the
12 pentagons bounding each dodecahedron are not geometrically
in one S^2, they still form topologically a spherical net,
so we can apply the topological lemma to show that each dodecahedron
is rigid. That means all edge lengths are equal, and the tetraplex
only exists in its regular configuration.

I learned about the structure of Cauchy's argument from an
article by JJStoker in Proc NAS 55(1966) p1398.
I assume that this application to higher dimensional cases was
first done by Alexandrov, or maybe by Servatius et al, but
as I said, I haven't yet tracked down any reference mentioning
higher dimensional rigidity. I'll try to do that to make sure
my argument is the right one :-)

-John Sullivan

Date Subject Author
6/16/94 John Sullivan
6/16/94 Joseph O'Rourke
6/17/94 John Conway
6/19/94 joe malkevitch
6/19/94 John Conway
6/20/94 John Sullivan
6/20/94 Walter Whiteley