Csaba Vincze (2013)

University of Debrecen

Euler's theorem can be considered as the first obstruction of building convex polyhedra. It contains a famous relationship among the numbers of vertices, edges and facets.

**Theorem 9.2.1 **
*(Euler, Leonard) The numbers v, e and f of vertices, edges and facets of a convex polyhedron are related by Euler's formula *

$v-e+f=2.$ |
(9.11) |

**Proof **The affine hull (plane) of any facet divides the space into two half-spaces such that the polyhedron is entirely contained in exactly one of them. This half-space will be called positive. After choosing a facet F consider a point c in the negative half-space with respect to F in such a way that c is contained in the positive half-spaces with respect to the facets except the
distinguished one. Use a central projection through c to present a connected finite graph in the plane of F by the projections of the vertices and edges of the polyhedron. This graph has v vertices, e edges and the projections of the facets except the distinguished one appear as convex polygons without common interior points. The number of these convex polygons is just f - 1. Consider the complement of F in
the plane instead of the missing facet. Thus we have a partition of the plane of F into f domains without common interior points. Take the next steps.** Steps of first type**. If we have a circle in the graph delete one of its edges. The output is a connected graph (the missing edge is avoidable along the complement "arc" of the circle) with the same number of vertices. The number of domains in the
partition decreases together with the number of edges. Therefore their difference and, consequently, the Euler characteristic 9.11 remains invariant. Repeat steps of first type as far as possible.** Steps of second type**. If we have no circles in the graph then there must be vertices of degree 1 (such a vertex can be found as the starting or the endpoint of the longest walk in the graph). Delete a vertex of degree 1 together with the corresponding edge. The output is a connected graph because the connectedness does not reflect to vertices with only one way to reach. The numbers of vertices andedges are reduced in this way but their difference and, consequently, the Euler characteristic 9.11 remains invariant. Repeat steps of second type as far as possible. The end of the algorithm is a graph containing only one vertex without edges and we have only one domain in the partition of the plane of F. Therefore v - e+f=1 - 0+1=2 as was to be stated.

**Remark **The higher dimensional Euler-formula is

$\sum _{k=0}^{n-1}(-1{)}^{k}{f}_{k}=1-(-1{)}^{n},$ |
(9.12) |

where n is the dimension of the space and f(k) denotes the number of k-dimensional facets (k=0, ..., n - 1).

**Excercise 9.2.2 **
*Find all convex polyhedra with f=5. *

Hint. Since the total sum of the facets is five we have only p triangular and q rectangular facets, where p+q=5. We have that

$2e=3p+4q$ |

and the possible values are p=2 and q=3 (a triangular based prism) or p=4 and q=1 (a rectangular based pyramid).

**Excercise 9.2.3 **
*Find all convex polyhedra with f=6. *

**Definition **The defect of a vertex V of a convex polyhedron is 2π minus the sum of the angles at the corners of the facets at V.

The second obstruction of building convex polyhedra is that each defect must be positive. Another result related to the defects of a convex polyhedra is Descartes's theorem. It will be applied to the characterization of regular polyhedra.

**Theorem 9.2.4 **
*(Descartes, René) The sum of the defects of vertices of a convex polyhedra is just 4π. *

**Proof **Let f(m) be the number of convex m - gons among the facets of the polyhedra. Since the sum of the
measures of the internal angles is (m - 2)π we have that the sum of the defects is

$2\pi v-\sum _{m}(m-2)\pi f\left(m\right)=2\pi (v-\sum _{m}\frac{mf\left(m\right)}{2}+\sum _{m}f(m\left)\right),$ |

where

$f=\sum _{m}f\left(m\right)$ |

and

$e=\sum _{m}\frac{mf\left(m\right)}{2}$ |

because each edge belongs to exactly two facets. Using Euler's theorem we have that the sum of the defects

$2\pi v-\sum _{m}(m-2)\pi f\left(m\right)=2\pi (v-e+f)=4\pi $ |

as was to be stated. ▮

**Excercise 9.2.5 **
*Find all convex polyhedra with regular triangles as facets. *

Hint. For the so-called Deltahedron's problem see [6] and [32] Theorem 45.6.

**Theorem 9.2.6 **
*(Cauchy's rigidity theorem, 1813) If there exists an edge- and facet- preserving correspondence between the vertices of two convex polyhedra such that the corresponding facets are congruent then they are congruent. *

**Proof **The rigorous proof of
the theorem involves some tools of combinatorics and (colored) graph theory [1], see also [21] and [39]. An elementary proof can be found in [32]. We summarize its basic steps. Let P(1) and P(2) be two convex polyhedra satisfying the conditions in the rigidity theorem. Cauchy's original idea is to study how the dihedral angles compare along corresponding edges. If all the dihedral angles are the same then we can build P(1)and P(2) step by step into congruent figures. Let each edge of P(1) be labelled by +, - or no mark according as its dihedral angle is less than, greater than or equal to the corresponding dihedral angle in P(2). At each vertex we intersect the polyhedra with sufficiently small spheres. This produces spherical polygons called vertex figure whose interior angles are just the dihedral angles. Therefore its vertices inherit markings+ or - from the edges. By our conditions the corresponding spherical polygons have equal sides. The basic question is that what about the interior angles. Using Steinitz's comparison lemma [32] (lemma 45.3) one can prove that either all corresponding angles are equal or, as we make a circuit of the polygon ignoring unmarked vertices, the sign must change at least four times[10]. To finish the proof we count the
total number of changes of sign in two different ways. For the sake of simplicity suppose that all edges are marked + or - and let N be the sum over all the vertices of the number of changes of sign of edges around that vertex. It follows that N is at least 4v, where v is the number of vertices. On the other hand let us count by facets. On a triangular face
two adjacent edges must have the same sign. Therefore such a face can contribute at most two changes of sign to its three vertices. Facets of n sides can contribute at most n changes of sign if n is even or n - 1 if n is odd. Especially

$4v=N\le 2{f}_{3}+\sum _{n\ge 4}n{f}_{n}$ |

Euler's theorem says that

$4v=4(e-f+2)=4\left(\frac{1}{2}\sum _{n\ge 3}n{f}_{n}-\sum _{n\ge 3}{f}_{n}+2\right)\le 2{f}_{3}+\sum _{n\ge 4}n{f}_{n},$ |

i.e.

$2\sum _{n\ge 3}(n-2){f}_{n}+8\le 2{f}_{3}+\sum _{n\ge 4}n{f}_{n}$ |

and, consequently,

$\sum _{n\ge 4}(n-4){f}_{n}+8\le 0$ |

which is a contradiction. Therefore we have no any change of sign and all corresponding dihedral angles are equal. ▮

**Remark **If there are some marked and some unmarked edges we can imitate the previous proof using only those vertices and edges that are labelled. Such a configuration is called a net. A net-face is formed by any maximal union of facets of the polyhedron which are not separated by
edges of the net. The Euler's formula for nets is

$v-e+f\ge 2$ |

and the argument of the proof still works.

**Excercise 9.2.7 **
*Find an authentic proof of the rigidity theorem for tetrahedra. *

Hint. Let

${T}_{1}:=conv\{{p}_{1},{p}_{2},{p}_{3},{p}_{4}\}\text{}and\text{}{T}_{2}=conv\{{q}_{1},{q}_{2},{q}_{3},{q}_{4}\}$ |

be two tetrahedra. By our assumption the correspondence

$f\left({p}_{i}\right)={q}_{i}(i=\mathrm{1,2},\mathrm{3,4})$ |
(9.13) |

of the vertices gives facets with congruent edges. Since 9.13 can be uniquely extended to an affine transformation

$f\left(p\right)=\phi \left(p\right)+v$ |

and the translation part is obviously an isometry we have that the linear part is also an isometry. Therefore T(2) is the image of T(1) under an isometry. In other words they are congruent.

**Excercise 9.2.8 **
*How to generalize the argumentation of the proof of Cauchy's rigidity theorem to more general polyhedra? *

Hint. Construct a class of convex polyhedra admitting a kind of inductive proof starting from tetrahedrons.