Gruber P. Convex and Discrete Geometry

274 Convex Polytopes
The first condition makes sure that the points (xi1,...,xid) are vertices of a convex
polytope, in the second condition it is tested whether G is the edge graph of this
polytope and the third condition makes sure that it is proper.
The theorem is now a consequence of (1) and (2). ⊓⊔
Remark. Two convex polytopes are combinatorially equivalent if there is a bijection
between the sets of vertices of these polytopes which maps (the sets of vertices of)
faces onto (the sets of vertices of) faces in both directions. A convex polytope is 2neighbourly,
if any two of its vertices are connected by an edge. A special case of
2-neighbourly polytopes are cyclic polytopes in E d where d ≥ 4. Cyclic polytopes
are the convex hulls of n ≥ d + 1 points on the moment curve
{(t, t 2 ,...,t d ) : t ∈ R} (d ≥ 3).
(Sometimes, convex polytopes are called cyclic, if they are combinatorially equivalent
to cyclic polytopes as defined here.) There are 2-neighbourly polytopes which
are not combinatorially equivalent to a cyclic polytope. See, e.g. Grünbaum [453],
p. 124. A complete graph, that is a graph in which any two vertices are connected by
an edge, thus can be realized as the edge graph of combinatorially different proper
convex polytopes. It clearly can always be realized as the edge graph of a simplex.
In spite of the above theorem, the problem to characterize, in a simple way, the
graphs which can be realized as edge graphs of convex polytopes remains open.
Edge Graphs of Convex Polytopes are d-Connected
Recall that a graph is d-connected if the deletion of any d − 1 of its vertices and
of the edges incident with these vertices leaves it connected. We show the following
theorem of Balinski [?].
Theorem 15.8. Let P be a proper convex polytope in E d . Then the edge graph of P
is d-connected.
The following proof is taken from Grünbaum [453]. For references to other proofs,
see Ziegler [1045].
Proof (by induction on d). The theorem is trivial for d = 1, 2. Assume then that
d > 2 and that it holds for 1, 2,...,d − 1.
Let v1,...,vd−1 ∈ V = V(P). We have to show that the graph arising from the
edge graph of P by deleting v1,...,vd−1 and the edges incident with these vertices,
is still connected.
Since P is the disjoint union of all relative interiors of its faces (including the
1
improper face P) , the point d−1 (v1 +···+vd−1) is in the relative interior of a face
of P, sayF. We distinguish two cases.
First, F �= P. Choose u ∈ Sd−1 and α ∈ R such that F ={x ∈ P : u · x = α}
and P ⊆{x : u·x ≤ α}. Since 1
d−1 (v1+···+···+vd−1) ∈ F and v1,...,vd−1 ∈ P,
we see that v1,...,vd−1 ∈ F. Choose β