Gruber P. Convex and Discrete Geometry

264 Convex Polytopes
The Converse of Euler’s Polytope Formula for d = 3 by Steinitz
Steinitz [962, 965] proved the following converse of Euler’s formula, perhaps conjectured
already by Euler. Together with Euler’s polytope formula, it yields a characterization
of the f -vectors of convex polytopes in E 3 .
Theorem 15.3. Let v,e, f be positive integers such that
v − e + f = 2, v≤ 2 f − 4, f ≤ 2v − 4.
Then there is a proper convex polytope in E 3 with v vertices, e edges and f facets.
Proof. Clearly, we have the following proposition.
(4) Let P0 be a proper convex polytope in E 3 with f -vector (v0, e0, f0).
Assume that P has at least one vertex incident with three edges and at least
one triangular facet. By cutting off this vertex, resp. by pasting a suitable
triangular pyramid to P at this facet we obtain proper polytopes with f -
vectors (v0 + 2, e0 + 3, f0 + 1), and (v0 + 1, e0 + 3, f0 + 2), respectively,
satisfying the same assumption.
A consequence of this proposition is the following.
(5) Let P0 be as in (4) with f -vector (v0, e0, f0). By repeating the cutting
process i times and the pasting process j times, we obtain a proper convex
polytope with f -vector (v0 + 2i + j, e0 + 3i + 3 j, f0 + i + 2 j).
To prove the theorem note that, by assumption,
2v − f − 4, 2 f − v − 4
are non-negative integers. Their difference is divisible by 3. Thus they give the same
remainder on division by 3, say r ∈{0, 1, 2}. Hence there are non-negative integers
i, j, such that
2v − f − 4 = 3i + r, 2 f − v − 4 = 3 j + r
and thus
(6) v = (4 + r) + 2i + j, f = (4 + r) + i + 2 j.
Take for P0 a pyramid with basis a convex (3 + r)-gon. The f -vector of P0 is the
vector
(4 + r, 6 + 2r, 4 + r).
Applying (5) to this P0 with the present values of i and j, we obtain a convex polytope
P with v vertices and f facets, see (6). By Euler’s polytope formula, P has
e = v + f − 2 edges. ⊓⊔