ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 23 (2023) #P2.01
https://doi.org/10.26493/1855-3974.2712.6be
(Also available at http://amc-journal.eu)
The edge-transitive polytopes that are not
vertex-transitive*
Frank Göring
Faculty of Mathematics, University of Technology, 09107 Chemnitz, Germany
Martin Winter †
Mathematics Institute, University of Warwick, Coventry CV4 7AL, United Kingdom
Received 27 October 2021, accepted 19 March 2022, published online 28 October 2022
Abstract
In 3-dimensional Euclidean space there exist two exceptional polyhedra, the rhombic
dodecahedron and the rhombic triacontahedron, the only known polytopes (besides poly-
gons) that are edge-transitive without being vertex-transitive. We show that these poly-
hedra do not have higher-dimensional analogues, that is, that in dimension d ≥ 4, edge-
transitivity of convex polytopes implies vertex-transitivity.
More generally, we give a classification of all convex polytopes which at the same time
have all edges of the same length, an edge in-sphere and a bipartite edge-graph. We show
that any such polytope in dimension d ≥ 4 is vertex-transitive.
Keywords: Convex polytopes, symmetry of polytopes, vertex-transitive, edge-transitive.
Math. Subj. Class. (2020): 52B15, 52B11
1 Introduction
A d-dimensional (convex) polytope P ⊂ Rd is the convex hull of finitely many points.
P is said to be vertex-transitive resp. edge-transitive if its (orthogonal) symmetry group
Aut(P ) ⊂ O(Rd) acts transitively on its vertices resp. edges. For a general overview over
the state of the art regarding symmetries in convex and abstract polytopes we refer to [9].
It has long been known that there are exactly nine edge-transitive polyhedra in R3 (see
e.g. [6]). These are the five Platonic solids (tetrahedron, cube, octahedron, icosahedron and
dodecahedron) together with the cuboctahedron, the icosidodecahedron, and their duals,
the rhombic dodecahedron and the rhombic triacontahedron (depicted in this order):
*This article appears as Chapter 6 in the second author’s doctoral thesis [11]. The authors thank the anonymous
referees for their careful reading and their many remarks that led to an improvement of the article in several ways.
†Corresponding author.
E-mail addresses: frank.goering@mathematik.tu-chemnitz.de (Frank Göring),
martin.h.winter@warwick.ac.uk (Martin Winter)
cb This work is licensed under https://creativecommons.org/licenses/by/4.0/
2 Ars Math. Contemp. 23 (2023) #P2.01
Little is known about analogous questions in higher dimensions. Branko Grünbaum writes
in “Convex Polytopes” [5, page 413]
No serious consideration seems to have been given to polytopes in dimension d ≥ 4
about which transitivity of the symmetry group is assumed only for faces of suitably
low dimensions, [...].
Even though families of higher-dimensional edge-transitive polytopes have been stud-
ied, to the best of our knowledge, no classification of these has been achieved so far.
Equally striking, all the known examples of such polytopes in dimension at least four are
simultaneously vertex-transitive. In dimension up to three, certain polygons (see Figure 1),
as well as the rhombic dodecahedron and rhombic triacontahedron are edge- but not vertex-
Figure 1: Some examples of edge-transitive 2n-gons with 2n ∈ {4, 6, 8} (the same works
for all n). The polygons depicted with black boundary are not vertex-transitive.
transitive. No higher dimensional example of this kind has been found. In this paper we
prove that this is not for lack of trying:
Theorem 1.1. In dimension d ≥ 4, edge-transitivity of convex polytopes implies vertex-
transitivity.
As immediate consequence, we obtain the classification of all polytopes that are edge-
but not vertex-transitive. The list is quite short:
Corollary 1.2. If P ⊂ Rd, d ≥ 2 is edge- but not vertex-transitive, then P is one of the
following:
(i) a non-regular 2k-gon (see Figure 1),
(ii) the rhombic dodecahedron, or
(iii) the rhombic triacontahedron.
Theorem 1.1 is proven by embedding the class of edge- but not vertex-transitive poly-
topes in a larger class of polytopes, defined by geometric regularities instead of symmetry.
In Theorem 2.4 we show that a polytope P ⊂ Rd which is edge- but not vertex-transitive
must have all of the following properties:
(i) all edges are of the same length,
F. Göring and M. Winter: The edge-transitive polytopes that are not vertex-transitive 3
(ii) it has a bipartite edge-graph GP = (V1 ·∪ V2, E), and
(iii) there are radii r1 ≤ r2, so that ∥v∥ = ri for all v ∈ Vi.
We compile this into a definition: a polytope that has these three properties shall be called
bipartite (cf. Definition 2.1). The edge- but not vertex-transitive polytopes then form a sub-
class of the bipartite polytopes, but the class of bipartite polytopes is much better behaved.
For example, faces of bipartite polytopes are bipartite (Proposition 2.5), something which
is not true for edge/vertex-transitive polytopes1. Our quest is then to classify all bipartite
polytopes. The surprising result: already being bipartite is very restrictive:
Theorem 1.3. If P ⊂ Rd, d ≥ 2 is bipartite, then P is one of the following:
(i) an edge-transitive 2k-gon (see Figure 1),
(ii) the rhombic dodecahedron,
(iii) the rhombic triacontahedron, or
(iv) a Γ-permutahedron for some finite reflection group Γ ⊂ O(Rd) (see Definition 2.10;
some 3-dimensional examples are shown in Figure 2).
Figure 2: From left to right: the A3-, B3 and H3-permutahedron.
The Γ-permutahedra are vertex-transitive, and all the other entries in the list are of
dimension d ≤ 3. This immediately implies Theorem 1.1.
Remarkably, despite the definition of bipartite polytope being purely geometric, all
bipartite polytopes are highly symmetric, that is, at least vertex- or facet-transitive, and
sometimes even edge-transitive.
Overview
In Section 2 we introduce the central notion of bipartite polytope and prove its most relevant
properties: that being bipartite generalizes being edge- but not vertex-transitive, and that
all faces of bipartite polytopes are again bipartite. We then investigate certain subclasses of
bipartite polytopes: bipartite polygons and inscribed bipartite polytopes. We prove that the
latter coincide with the Γ-permutahedra, a class of vertex-transitive polytopes. It therefore
remains to classify the non-inscribed cases, the so-called strictly bipartite polytopes. We
1For example, consider a vertex-transitive but not uniform antiprism. Its faces are non-regular triangles, which
are thus not vertex-transitive. Alternatively, consider the (n, n)-duoprism, n ̸= 4, that is, the cartesian product
of a regular n-gon with itself. This polytope is edge-transitive, but its facets are n-gonal prisms (the cartesian
product of a regular n-gon with an edge), which are not edge-transitive.
4 Ars Math. Contemp. 23 (2023) #P2.01
show that the classification of these reduces to the classification of bipartite polyhedra, i.e.,
the case d = 3.
From Section 3 on the investigation is focused on the class of strictly bipartite poly-
hedra. We successively determine restrictions on the structure of such, e.g. the degrees of
their vertices and the shapes of their faces. This quite elaborate process uses many clas-
sical geometric results and techniques, including spherical polyhedra, the classification of
rhombic isohedra and the realization of graphs as edge-graphs of polyhedra. As a result,
we can exclude all but two cases, namely, the rhombic dodecahedron, and the rhombic
triacontahedron. Additionally, we shall find a remarkable near-miss, that is, a polyhedron
which fails to be bipartite only by a tiny (but quantifiable) amount.
2 Bipartite polytopes
From this section on let P ⊂ Rd, d ≥ 2 denote a d-dimensional polytope of full dimension
(i.e., P is not contained in a proper affine subspace). By F(P ) we denote the face lattice
of P , and by Fδ(P ) ⊂ F(P ) the subset of δ-dimensional faces.
Definition 2.1. P is called bipartite, if
(i) all its edges are of the same length ℓ,
(ii) its edge-graph is bipartite, which we write as GP = (V1 ·∪ V2, E), and
(iii) there are radii r1 ≤ r2 so that ∥v∥ = ri for all v ∈ Vi.
If r1 < r2, then P is called strictly bipartite. A vertex v ∈ Vi is called an i-vertex. The
numbers r1, r2 and ℓ are called the parameters of a bipartite polytope.
Remark 2.2. Since P is full-dimensional by convention, Definition 2.1 only defines full-
dimensional bipartite polytopes.
To extend this notion to not necessarily full-dimensional polytopes, we shall call a
polytope bipartite even if it is just bipartite as a subset of its affine hull where we made
an appropriate choice of origin in the affine hull (note that whether a polytope is bipartite
depends on its placement relative to the origin and that there is at most one such placement
if the polytope is full-dimensional). This comes in handy when we discuss faces of bipartite
polytopes.
Remark 2.3. An alternative definition of bipartite polytope would replace (iii) by the con-
dition that P has an edge in-sphere, that is, a sphere that touches each edge of P in a single
point (this definition was used in the abstract). The configuration depicted below (an edge
of P connecting two vertices v1 ∈ V1 and v2 ∈ V2) shows how any one of the four quanti-
ties r1, r2, ℓ and ρ (the radius of the edge in-sphere) is determined from the other three by
solving the given set of equations:
ρ2 + ℓ21 = r
2
1
ρ2 + ℓ22 = r
2
2
ℓ1 + ℓ2 = ℓ
F. Göring and M. Winter: The edge-transitive polytopes that are not vertex-transitive 5
There is a subtlety: for the edge in-sphere to actually touch the edge (rather than only
its affine hull outside of the edge) it is necessary that the perpendicular projection of the
origin onto the edge ends up inside the edge (equivalently, that the triangle conv{0, v1, v2}
is acute at v1 and v2). One might regard this as intuitively clear since we are working with
convex polytopes, but this will also follows formally as part of our proof of Proposition 3.7
(as we shall mention there in a footnote).
This alternative characterization of bipartite polytopes via edge in-spheres will become
relevant towards the end of the classification (in Section 3.9). Still, for the larger part of
our investigation, Definition 2.1(iii) is the more convenient version to work with.
2.1 General obsevations
Proposition 2.4. If P is edge- but not vertex-transitive, then P is bipartite.
This is a geometric analogue to the well known fact that every edge- but not vertex-
transitive graph is bipartite. A proof of the graph version can be found in [4]. The following
proof can be seen as a geometric analogue:
Proof of Proposition 2.4. Clearly, all edges of P are of the same length.
Fix some edge e ∈ F1(P ) with end vertices v1, v2 ∈ F0(P ). Let Vi be the orbit of vi
under Aut(P ). We prove that V1 ∪ V2 = F0(P ), V1 ∩ V2 = ∅ and that the edge graph GP
is bipartite with partition V1 ·∪ V2.
Let v ∈ F0(P ) be some vertex and ẽ ∈ F1(P ) an incident edge. By edge-transitivity,
there is a symmetry T ∈ Aut(P ) that maps ẽ onto e, and therefore maps v onto vi for
some i ∈ {1, 2}. Thus, v is in the orbit Vi. This holds for all vertices of P , and therefore
V1 ∪ V2 = F0(P ).
The orbits of v1 and v2 must either be identical or disjoint. Since V1 ∪ V2 = F0(P ),
from V1 = V2 it would follow V1 = F0(P ), stating that P has a single orbit of vertices.
But since P is not vertex-transitive, this cannot be. Thus, V1 ∩ V2 = ∅, and therefore
V1 ·∪ V2 = F0(P ).
Let ẽ ∈ F1(P ) be an edge with end vertices ṽ1 and ṽ2. By edge-transitivity, ẽ can be
mapped onto e by some symmetry T ∈ Aut(P ). Equivalently {T ṽ1, T ṽ2} = {v1, v2}.
Since v1 and v2 belong to different orbits under Aut(P ), so do ṽ1 and ṽ2. Hence ẽ has
one end vertex in V1 and one end vertex in V2. This holds for all edges, and thus, GP is
bipartite with partition V1 ·∪ V2.
It remains to determine the radii r1 ≤ r2. Set ri := ∥vi∥ (assuming w.l.o.g. that
∥v1∥ ≤ ∥v2∥). Then for every v ∈ Vi there is a symmetry T ∈ Aut(P ) ⊂ O(Rd) so that
Tvi = v, and thus
∥v∥ = ∥Tvi∥ = ∥vi∥ = ri.
Bipartite polytopes are more comfortable to work with than edge- but not vertex-
transitive polytopes because their faces are again bipartite polytopes (in the sense as ex-
plained in Remark 2.2). Later, this will enable us to reduce the problem to an investigation
in lower dimensions.
Proposition 2.5. Let σ ∈ F(P ) be a face of P . Then it holds
(i) if P is bipartite, so is σ.
6 Ars Math. Contemp. 23 (2023) #P2.01
(ii) if P is strictly bipartite, then so is σ, and v ∈ F0(σ) ⊆ F0(P ) is an i-vertex in P if
and only if it is an i-vertex in σ.
(iii) if r1 ≤ r2 are the radii of P and ρ1 ≤ ρ2 are the radii of σ, then there holds
h2 + ρ2i = r
2
i ,
where h is the height of σ, that is, the distance of aff(σ) from the origin.
Proof. Properties clearly inherited by σ are that all edges are of the same length and that
the edge graph is bipartite. It remains to show the existence of the radii ρ1 ≤ ρ2 compatible
with the bipartition of the edge-graph of σ.
Let c ∈ aff(σ) be the orthogonal projection of 0 onto aff(σ). Then ∥c∥ = h, the height
of σ as defined in (iii). For any vertex v ∈ F0(σ) which is an i-vertex in P , the triangle
∆ := conv{0, c, v} has a right angle at c. Set ρi := ∥v − c∥ and observe
ρ2i := ∥v − c∥2 = ∥v∥2 − ∥c∥2 = r2i − h2. (∗)
In particular, the value ρi does only depend on i. In other words, σ is a bipartite poly-
tope when considered as a subset of its affine hull, where the origin is chosen to be c (cf.
Remark 2.2). This proves (i), and (∗) is equivalent to the equation in (iii). From (∗) also
follows r1 < r2 ⇔ ρ1 < ρ2, which proves (ii).
The following observation will be of use later on.
Observation 2.6. Given two adjacent vertices v1, v2 ∈ F0(P ) with vi ∈ Vi, and if P has
parameters r1, r2 and ℓ, then
ℓ2 = ∥v1 − v2∥2 = ∥v1∥2 + ∥v2∥2 − 2⟨v1, v2⟩ = r21 + r22 − 2r1r2 cos∡(v1, v2),
This can be rearranged for cos∡(v1, v2). While the exact value of this expression is not
of relevance to us, this shows that this angle is determined by the parameters and does not
depend on the choice of the adjacent vertices v1 and v2.
2.2 Bipartite polygons
The easiest to describe (and to explicitly construct) are the bipartite polygons.
Foremost, the edge-graph is bipartite, and thus, a bipartite polygon must be a 2k-gon
for some k ≥ 2. One can show that the bipartite polygons are exactly the edge-transitive
2k-gons (cf. Figure 1), and that such one is strictly bipartite if and only if it is not vertex-
transitive (or equivalently, not regular). We will not make use of these symmetry properties
of bipartite polygons.
The parameters r1, r2 and ℓ uniquely determine a bipartite polygon, as can be seen by
explicit construction:
F. Göring and M. Winter: The edge-transitive polytopes that are not vertex-transitive 7
One starts with an arbitrary 1-vertex v ∈ R2 placed on the circle Sr1(0). Its neighboring
vertices are then uniquely determined as the intersections Sr2(0) ∩ Sℓ(v). The procedure
is repeated with the new vertices until the edge cycle closes (which only happens if the
parameters are chosen appropriately).
The procedure also makes clear that the interior angle αi ∈ (0, π) at an i-vertex only
depends on i, but not on the chosen vertex v ∈ Vi.
Corollary 2.7. A bipartite polygon P ⊂ R2 is a 2k-gon with alternating interior angles
α1, α2 ∈ (0, π) (αi being the interior angle at an i-vertex), and its shape is uniquely
determined by its parameters (up to congruence).
The exact values for the interior angles are not of relevance. Instead, we only need the
following properties:
Proposition 2.8. The interior angles α1, α2 ∈ (0, π) satisfy
α1 + α2 = 2α
k
reg and α2 ≤ αkreg ≤ α1, (2.1)
where αkreg := (1 − 1/k)π is the interior angle of a regular 2k-gon, and the inequalities
are satisfied with equality if and only r1 = r2.
Proof. The sum of interior angles of a 2k-gon is 2(k−1)π, and thus kα1+kα2 = 2(k−1)π,
which, after division by k, yields the first part of (2.1).
For two adjacent vertices v1, v2 ∈ F0(P ) (where vi ∈ Vi), consider the triangle ∆ :=
conv{0, v1, v2} whose edge lengths are r1, r2 and ℓ, and whose interior angles at v1 resp.
v2 are α1/2 resp. α2/2. From r1 ≤ r2 (resp. r1 < r2) and the law of sine follows α1 ≥ α2
(resp. α1 > α2). With α1 + α2 = 2αkreg this yields the second part of (2.1).
Observation 2.9. For later use (in Corollary 3.18), consider Proposition 2.8 with 2k = 4.
In this case we find,
α2 ≤
π
2
≤ α1,
that is, α1 is never acute, and α2 is never obtuse.
2.3 The case r1 = r2
We classify the inscribed bipartite polytopes, that is, those with coinciding radii r1 = r2.
This case is made especially easy by a classification result from [10]. We need the following
definition:
Definition 2.10. Let Γ ⊂ O(Rd) be a finite reflection group and v ∈ Rd a generic point
w.r.t. Γ (i.e., v is not fixed by a non-identity element of Γ). The orbit polytope
Orb(Γ, v) := conv{Tv | T ∈ Γ} ⊂ Rd
is called a Γ-permutahedron.
The relevant result then reads
Theorem 2.11 (Corollary 4.6. in [10]). If P has only centrally symmetric 2-dimensional
faces (that is, it is a zonotope), has all vertices on a common sphere and all edges of the
same length, then P is a Γ-permutahedron.
8 Ars Math. Contemp. 23 (2023) #P2.01
This provides a classification of bipartite polytopes with r1 = r2.
Theorem 2.12. If P ⊂ Rd is bipartite with r1 = r2, then it is a Γ-permutahedron.
Proof. If r1 = r2, then all vertices are on a common sphere (that is, P is inscribed). By
definition, all edges are of the same length. Both statements then also hold for the faces of
P , in particular, the 2-dimensional faces. An inscribed polygon with a unique edge length
is necessarily regular. With Corollary 2.7 the 2-faces are then regular 2k-gons, therefore
centrally symmetric.
Summarizing, P is inscribed, has all edges of the same length, and all 2-dimensional
faces of P are centrally symmetric. By Theorem 2.11, P is a Γ-permutahedron.
Γ-permutahedra are vertex-transitive by definition, hence do not provide examples of
edge- but not vertex-transitive polytopes.
2.4 Strictly bipartite polytopes
It remains to classify the strictly bipartite polytopes. This problem is divided into two
independent cases: dimension d = 3, and dimension d ≥ 4. The detailed study of the case
d = 3 (which turns out to be the actual hard work) is postponed until Section 3, the result
of which is the following theorem:
Theorem 2.13. If P ⊂ R3 is strictly bipartite, then P is the rhombic dodecahedron or the
rhombic triacontahedron.
Presupposing Theorem 2.13, the case d ≥ 4 is done quickly.
Theorem 2.14. There are no strictly bipartite polytopes in dimension d ≥ 4.
Proof. It suffices to show that there are no strictly bipartite polytopes in dimension d = 4,
as any higher-dimensional example has a strictly bipartite 4-face (by Proposition 2.5).
Let P ⊂ R4 be a strictly bipartite 4-polytope. Let e ∈ F1(P ) be an edge of P .
Then there are s ≥ 3 cells (aka. 3-faces) σ1, ..., σs ∈ F3(P ) incident to e, each of which
is again strictly bipartite (by Proposition 2.5). By Theorem 2.13 each σi is a rhombic
dodecahedron or rhombic triacontahedron.
The dihedral angle of the rhombic dodecahedron resp. triacontahedron is 120◦ resp.
144◦ at every edge [3]. However, the dihedral angles meeting at e must sum up to less than
2π. With the given dihedral angles this is impossible.
3 Strictly bipartite polyhedra
In this section we derive the classification of strictly bipartite polyhedra. The main goal is to
show that there are only two: the rhombic dodecahedron and the rhombic triacontahedron.
From this section on, let P ⊂ R3 denote a fixed strictly bipartite polyhedron with radii
r1 < r2 and edge length ℓ. The 2-faces of P will be shortly referred to as just faces of P .
Since they are bipartite, they are necessarily 2k-gons.
Definition 3.1. We use the following terminology:
(i) a face of P is of type 2k (or called a 2k-face) if it is a 2k-gonal polygon.
F. Göring and M. Winter: The edge-transitive polytopes that are not vertex-transitive 9
(ii) an edge of P is of type (2k1, 2k2) (or called a (2k1, 2k2)-edge) if the two incident
faces are of type 2k1 and 2k2 respectively.
(iii) a vertex of P is of type (2k1, ..., 2ks) (or called a (2k1, ..., 2ks)-vertex) if its incident
faces can be enumerated as σ1, ..., σs so that σi is a 2ki-face (note, the order of the
numbers does not matter).
We write τ(v) for the type of a vertex v ∈ F0(P ).
3.1 General observations
In a given bipartite polyhedron, the type of a vertex, edge or face already determines much
of its metric properties. We prove this for faces:
Proposition 3.2. For some face σ ∈ F2(P ), any of the following properties of σ determines
the other two:
(i) its type 2k,
(ii) its interior angles α1 > α2.
(iii) its height h (that is, the distance of aff(σ) from the origin).
Corollary 3.3. Any two faces of P of the same height, or the same type, or the same interior
angles, are congruent.
Proof of Proposition 3.2. Fix a face σ ∈ F2(P ).
Suppose that the height h of σ is known. By Proposition 2.5, a face of P of height h is
bipartite with radii ρ2i := r
2
i − h2 and edge length ℓ. By Corollary 2.7, these parameters
then uniquely determine the shape of σ, which includes its type and its interior angles. This
shows (iii) =⇒ (i), (ii).
Suppose now that we know the interior angles α1 > α2 of σ (it actually suffices to
know one of these, say α1). Fix a 1-vertex v ∈ V1 of σ and let w1, w2 ∈ V2 be its two
adjacent 2-vertices in σ. Consider the simplex S := conv{0, v, w1, w2}. The length of
each edge of S is already determined, either by the parameters alone, or by additionally
using the known interior angles via
∥w1 − w2∥2 = ∥w1 − v∥2 + ∥w2 − v∥2 − 2⟨w1 − v, w2 − v⟩
= 2ℓ2(1− cos∡(w1 − v, w2 − v)︸ ︷︷ ︸
α1
).
Thus, the shape of S is determined. In particular, this determines the height of the face
conv{v, w1, w2} ⊂ S over the vertex 0 ∈ S. Since aff{v, w1, w2} = aff(σ), this deter-
mines the height of σ in P . This proves (ii) =⇒ (iii).
Finally, suppose that the type 2k is known. We then want to show that the height h
is uniquely determined.2 For the sake of contradiction, suppose that the type 2k does not
2The reader motivated to prove this himself should know the following: it is indeed possible to write down
a polynomial in h of degree four whose coefficients involve only r1, r2, ℓ and cos(π/k), and whose zeroes
include all possible heights of any 2k-face of P . However, it turns out to be quite tricky to work out which zeroes
correspond to feasible solutions. For certain values of the coefficients there are multiple positive solutions for h,
some of which correspond to non-convex 2k-faces. There seems to be no easy way to tell them apart.
10 Ars Math. Contemp. 23 (2023) #P2.01
uniquely determine the height of the face. Then there is another 2k-face σ′ ∈ F2(P ) of
some height h′ ̸= h. W.l.o.g. assume h > h′.
Visualize both faces embedded in R2, on top of each other and centered at the origin as
shown in the figure below:
The vertices in both polygons are equally spaced by an angle of π/k (cf. Observation 2.6)
and we can therefore assume that the vertex vi of σ (resp. v′i of σ
′) is a positive multiple
of (sin(iπ/k), cos(iπ/k)) ∈ R2 for i ∈ {1, ..., 2k}. There are then factors δi ∈ R+ with
v′i = δvi.
The norms of vectors v1, v2, δ1v1 and δ2v2 are the radii of the bipartite polygons σ and
σ′. With Proposition 2.5(iii) from h > h′ follows ∥v1∥ < ∥δ1v1∥ and ∥v2∥ < ∥δ2v2∥, and
thus, (∗) δ1, δ2 > 1. W.l.o.g. assume δ1 ≤ δ2.
Since both faces have edge length ℓ, we have ∥v1 − v2∥ = ∥δ1v1 − δ2v2∥ = ℓ. Our
goal is to derive the following contradiction:
ℓ = ∥v1 − v2∥
(∗)
< δ1∥v1 − v2∥ = ∥δ1v1 − δ1v2∥
(∗∗)
< ∥δ1v1 − δ2v2∥ = ℓ,
To prove (∗∗), consider the triangle ∆ with vertices δ1v1, δ2v2 and δ1v2:
Since σ is convex, the angle α is smaller than 90◦. It follows that the interior angle of ∆ at
δ1v2 is obtuse (here we are using δ1 ≤ δ2). Hence, by the sine law, the edge of ∆ opposite
to δ1v2 is the longest, which translates to (∗∗).
As a consequence of Proposition 3.2, the interior angles of a face of P do only depend
on the type of the face (and the parameters), and so we can introduce the notion of the
interior angle αki ∈ (0, π) of a 2k-face at an i-vertex. Furthermore, set ϵk := (αk1−αk2)/2π.
By Proposition 2.8 we have ϵk > 0 and
αk1 =
(
1− 1
k
+ ϵk
)
π, αk2 =
(
1− 1
k
− ϵk
)
π.
F. Göring and M. Winter: The edge-transitive polytopes that are not vertex-transitive 11
Definition 3.4. If τ = (2k1, ..., 2ks) is the type of a vertex, then define
K(τ) :=
s∑
i=1
1
ki
, E(τ) :=
s∑
i=1
ϵki .
Both quantities are strictly positive.
Proposition 3.5. Let v ∈ F0(P ) be a vertex of type τ = (2k1, ..., 2ks).
(i) If v ∈ V1, then E(τ) < K(τ)− 1 and s = 3.
(ii) If v ∈ V2, then E(τ) > s− 2−K(τ).
Proof. Let σ1, ..., σs ∈ F2(P ) be the faces incident to v, so that σj is a 2kj-face. The
interior angle of σj at v is αkji , and the sum of these must be smaller than 2π. In formulas
2π >
s∑
j=1
αkji =
s∑
j=1
(
1− 1
kj
± ϵkj
)
π = (s−K(τ)± E(τ))π,
where ± is the plus sign for i = 1, and the minus sign for i = 2. Rearranging for E(v)
yields (∗) ∓E(τ) > s − 2 −K(τ). If i = 2, this proves (ii). If i = 1, note that from the
implication kj ≥ 2 =⇒ K(τ) ≤ s/2 follows
s
(∗)
< −E(τ) +K(τ) + 2 ≤ 0 + s
2
+ 2 =⇒ s < 4.
The minimum degree of a vertex in a polyhedron is at least three, hence s = 3, and (∗)
becomes (i).
This allows us to exclude all but a manageable list of types for 1-vertices. Note that a
vertex v ∈ V1 has a type of some form (2k1, 2k2, 2k3).
Corollary 3.6. For a 1-vertex v ∈ V1 of type τ holds K(τ) > 1 + E(τ) > 1. One checks
that this leaves exactly the options in Table 1.
τ K(τ) Γ
(4, 4, 4) 3/2 I1 ⊕ I1 ⊕ I1
(4, 4, 6) 4/3 I1 ⊕ I2(3)
(4, 4, 8) 5/4 I1 ⊕ I2(4)
...
...
...
(4, 4, 2k) 1 + 1/k I1 ⊕ I2(k)
(4, 6, 6) 7/6 A3 = D3
(4, 6, 8) 13/12 B3
(4, 6, 10) 31/30 H3
Table 1: Possible types of 1-vertices, their K-values and the Γ of the Γ-permutahedron in
which all vertices have this type.
12 Ars Math. Contemp. 23 (2023) #P2.01
The types in Table 1 are called the possible types of 1-vertices. Each of the possi-
ble types is realizable in the sense that there exists a bipartite polyhedron in which all
1-vertices have this type. Examples are provided by the Γ-permutahedra (the Γ of that
Γ-permutahedron is listed in the right column of Table 1). These are not strictly bipartite
though.
The convenient thing about Γ-permutahedra is that all their vertices are of the same
type. We cannot assume this for general strictly bipartite polyhedra, not even for all 1-
vertices.
3.2 Spherical polyhedra
The purpose of this section is to define a second notion of interior angle for each face. These
angles can be defined in several equivalent ways, one of which is via spherical polyhedra.
A spherical polyhedron is an embedding of a planar graph into the unit sphere, so that
all edges are embedded as great circle arcs, and all regions are convex3. If 0 ∈ int(P ), we
can associate to P a spherical polyhedron PS by applying central projection
R3 \ {0} → S1(0), x 7→
x
∥x∥
to all its vertices and edges (this process is visualized below).
The vertices, edges and faces of P have spherical counterparts in PS obtained as pro-
jections onto the unit sphere. Those will be denoted with a superscript “S ”. For example,
if e ∈ F1(P ) is an edge of P , then eS denotes the corresponding “spherical edge”, which
is a great circle arc obtained as the projection of e onto the sphere.
We still need to justify that the spherical polyhedron of P is well-defined, by proving
that P contains the origin:
Proposition 3.7. 0 ∈ int(P ).
Proof. The proof proceeds in several steps.
Step 1: Fix a 1-vertex v ∈ V1 with neighbors w1, w2, w3 ∈ V2, and let ui := wi − v
be the direction of the edge conv{v, wi} emanating from v. Let σij ∈ F2(P ) denote the
2k-face containing v, wi and wj . The interior angle of σij at v is then ∡(ui, uj), which by
Proposition 2.8 and k ≥ 2 satisfies
∡(ui, uj) >
(
1− 1
k
)
π ≥ π
2
=⇒ ⟨ui, uj⟩ < 0.
Step 2: Besides v, the polyhedron P contains another 1-vertex v′ ∈ V1. It then
holds v′ ∈ v + cone{u1, u2, u3}, which means that there are non-negative coefficients
3Convexity on the sphere means that the shortest great circle arc connecting any two points in the region is
also contained in the region.
F. Göring and M. Winter: The edge-transitive polytopes that are not vertex-transitive 13
a1, a2, a3 ≥ 0, at least one positive, so that v + a1u1 + a2u2 + a3u3 = v′. Rearranging
and applying ⟨v, ·⟩ yields
a1⟨v, u1⟩+ a2⟨v, u2⟩+ a3⟨v, u3⟩ = ⟨v, v′⟩ − ⟨v, v⟩ (∗)
= r21 cos∡(v, v
′)− r21 < 0.
The value ⟨v, ui⟩ is independent of i (see Observation 2.6). Since there is at least one
positive coefficient ai, from (∗) follows ⟨v, ui⟩ < 0.4
Step 3: By the previous steps, {v, u1, u2, u3} is a set of four vectors with pair-wise neg-
ative inner product. The convex hull of such an arrangement in 3-dimensional Euclidean
space does necessarily contain the origin in its interior, or equivalently, there are positive
coefficients a0, ..., a3 > 0 with a0v + a1u1 + a2u2 + a3u3 = 0 (for a proof, see Proposi-
tion A.1). In other words: 0 ∈ v + int(cone{u1, u2, u3}).
Step 4: If H(σ) denotes the half-space associated with the face σ ∈ F2(P ), then
0 ∈ v + int(cone{u1, u2, u3}) =
⋂
σ∼v
int(H(σ)).
Thus, 0 ∈ int(H(σ)) for all faces σ incident to v. But since every face is incident to a
1-vertex, we obtain 0 ∈ int(H(σ)) for all σ ∈ F2(P ), and thus 0 ∈ int(P ) as well.
The main reason for introducing spherical polyhedra is that we can talk about the spher-
ical interior angles of their faces.
Let σ ∈ F2(P ) be a face, and v ∈ F0(σ) one of its vertices. Let α(σ, v) denote the
interior angle of σ at v, and β(σ, v) the spherical interior angle of σS at vS . It only needs
a straight-forward computation (involving some spherical geometry) to establish a direct
relation between these angles: e.g. if v is a 1-vertex, then
sin2(ℓS) · (1− cosβ(σ, v)) =
( ℓ
r2
)2
· (1− cosα(σ, v)),
where ℓS denotes the arc-length of an edge of PS (indeed, all edges are of the same length).
An equivalent formula exists for 2-vertices. The details of the computation are not of
relevance, but can be found in Appendix A.2.
The core message is that the value of α(σ, v) uniquely determines the value of β(σ, v)
and vice versa. In particular, since the value of α(σ, v) = αki does only depend on the
type of the face and the partition class of the vertex, so does β(σ, v), and it makes sense to
introduce the notion βki for the spherical interior angle of a 2k-gonal spherical face of P
S
at (the projection of) an i-vertex. Thus, we have
βk1i = β
k2
i ⇐⇒ α
k1
i = α
k2
i
3.2⇐⇒ k1 = k2, (3.1)
where we use Proposition 3.2 for the last equivalence.
Observation 3.8. The spherical interior angles βki have the following properties:
(i) The spherical interior angles surrounding a vertex add up to exactly 2π. That is, for
an i-vertex v ∈ F0(P ) of type (2k1, ..., 2ks) holds
βk1i + · · ·+ β
ks
i = 2π.
4Note that this provides the formal proof mentioned in Remark 2.3, namely, that the triangle conv{0, v1, v2}
is acute at v1 and v2.
14 Ars Math. Contemp. 23 (2023) #P2.01
(ii) The sum of interior angles of a spherical polygon always exceed the interior angle
sum of a respective flat polygon. That is, it holds
kβk1 + kβ
k
2 > 2(k − 1)π =⇒ βk1 + βk2 > 2
(
1− 1
k
)
π.
This has some consequences for the strictly bipartite polyhedron P :
Corollary 3.9. P contains at most two different types of 1-vertices, and if there are two,
then one is of the form (4, 4, 2k), and the other one is of the form (4, 6, 2k′) for distinct
k ̸= k′ and 2k′ ∈ {6, 8, 10}.
Proof. Each possible type listed in Table 1 is either of the form (4, 4, 2k) or of the form
(4, 6, 2k′) for some 2k ≥ 4 or 2k′ ∈ {6, 8, 10}.
If P contains simultaneously 1-vertices of type (4, 4, 2k1) and (4, 4, 2k2), apply Ob-
servation 3.8(i) to see
β21 + β
2
1 + β
k1
1
(i)
= β21 + β
2
1 + β
k2
1 =⇒ β
k1
1 = β
k2
1
(3.1)
=⇒ k1 = k2.
If P contains simultaneously 1-vertices of type (4, 6, 2k′1) and (4, 6, 2k
′
2), then
β21 + β
3
1 + β
k′1
1
(i)
= β21 + β
3
1 + β
k′2
1 =⇒ β
k′1
1 = β
k′2
1
(3.1)
=⇒ k′1 = k′2.
Finally, if P contains simultaneously 1-vertices of type (4, 4, 2k) and (4, 6, 2k′), then
β21 + β
2
1 + β
k
1
(i)
= β21 + β
3
1 + β
k′
1 =⇒ βk1 − βk
′
1 = β
3
1 − β21︸ ︷︷ ︸
̸= 0 by (3.1)
(3.1)
=⇒ k ̸= k′.
Since each edge of P is incident to a 1-vertex, we obtain
Observation 3.10. If P has only 1-vertices of types (4, 4, 2k) and (4, 6, 2k′), then each
edge of P is of one of the types
(4, 4), (4, 2k)︸ ︷︷ ︸
from a (4, 4, 2k)-vertex
, (4, 6), (4, 2k′) or (6, 2k′)︸ ︷︷ ︸
from a (4, 6, 2k′)-vertex
.
Corollary 3.11. The dihedral angle of an edge e ∈ F1(P ) of P only depends on its type.
Proof. Suppose that e is a (2k1, 2k2)-edge. Then e is incident to a 1-vertex v ∈ V1 of
type (2k1, 2k2, 2k3). By Observation 3.8(i) holds βk31 = 2π − βk11 − βk21 , which further
determines k3. By Proposition 3.2 we have uniquely determined interior angles αk11 , α
k2
1
and αk31 .
It is known that for a simple vertex (that is, a vertex of degree three) the interior angles
of the incident faces already determine the dihedral angles at the incident edges (for a
proof, see the Appendix, Proposition A.2). Consequently, the dihedral angle at e is already
determined.
The next result shows that Γ-permutahedra are the only bipartite polytopes in which a
1-vertex and a 2-vertex can have the same type.
F. Göring and M. Winter: The edge-transitive polytopes that are not vertex-transitive 15
Corollary 3.12. P cannot contain a 1-vertex and a 2-vertex of the same type.
Proof. Let v ∈ F0(P ) be a vertex of type (2k1, 2k2, 2k3). The incident edges are of type
(2k1, 2k2), (2k2, 2k3) and (2k3, 2k1) respectively. By Corollary 3.11 the dihedral angles
of these edges are uniquely determined, and since v is simple (that is, has degree three), the
interior angles of the incident faces are also uniquely determined (cf. Appendix, Proposition
A.2). In particular, we obtain the same angles independent of whether v is a 1-vertex or a
2-vertex.
A 1-vertex is always simple, and thus, a 1-vertex and a 2-vertex of the same type would
have the same interior angles at all incident faces, that is, αk1 = α
k
2 for each incident
2k-face. But this is not possible if P is strictly bipartite (by Proposition 2.5(ii) and Propo-
sition 2.8).
3.3 Adjacent pairs
Given a 1-vertex v ∈ V1 of type τ1 = (2k1, 2k2, 2k3), for any two distinct i, j ∈ {1, 2, 3},
v has a neighbor w ∈ V2 of type τ2 = (2ki, 2kj , ∗, ..., ∗), where ∗ are placeholders for
unknown entries. The pair of types
(τ1, τ2) = ((2k1, 2k2, 2k3), (2ki, 2kj , ∗, ..., ∗))
is called an adjacent pair of P . It is the purpose of this section to show that certain adjacent
pairs cannot occur in P . Excluding enough adjacent pairs for fixed τ1 then proves that the
type τ1 cannot occur as the type of a 1-vertex.
Our main tools for achieving this will be the inequalities established in Proposition 3.5 (i)
and (ii), that is
E(τ1)
(i)
< K(τ1)− 1 and E(τ2)
(ii)
> s− 2−K(τ2),
where s is the number of elements in τ2. For a warmup, and as a template for further
calculations, we prove that the adjacent pair (τ1, τ2) = ((4, 6, 8), (6, 8, 8)) will not occur
in P .
Example 3.13. By Proposition 3.5(i) we have
(∗) ϵ2 + ϵ3 + ϵ4 = E(τ1)
(i)
< K(τ1)− 1 =
1
2
+
1
3
+
1
4
− 1 = 1
12
.
On the other hand, by Proposition 3.5(ii) we have
(∗∗) 2
12
= 3− 2−
(1
3
+
1
4
+
1
4
)
= s− 2−K(τ2)
(ii)
< E(τ2) = ϵ3 + ϵ4︸ ︷︷ ︸
<1/12
+ ϵ4︸︷︷︸
<1/12
<
2
12
,
which is a contradiction. Hence this adjacent pair cannot occur. Note that we used (∗) to
upperbound certain sums of ϵi in (∗∗).
An adjacent pair excluded by using the inequalities from Proposition 3.5(i) and (ii) as
demonstrated in Example 3.13 will be called infeasible.
16 Ars Math. Contemp. 23 (2023) #P2.01
The argument applied in Example 3.13 will be repeated many times for many different
adjacent pairs in the upcoming Sections 3.5, 3.4, 3.6, 3.8, and we shall therefore use a
tabular form to abbreviate it. After fixing, τ1 = (4, 6, 8), the argument to refute the adjacent
pair (τ1, τ2) = ((4, 6, 8), (6, 8, 8)) is abbreviated in the first row of the following table:
τ2 s− 2−K(τ2)
?
< E(τ2)
(6, 8, 8) 2/12 ̸< (ϵ3 + ϵ4) + ϵ4 < 2/12
(6, 8, 6, 6) 9/12 ̸< (ϵ3 + ϵ4) + ϵ3 + ϵ3 < 3/12
The second row displays the analogue argument for another example, namely, the pair
((4, 6, 8), (6, 8, 6, 6)), showing that it is infeasible as well. Both rows will reappear in the
table of Section 3.5 where we exclude (4, 6, 8) as a type for 1-vertices entirely. Note that
the terms in the column below E(τ2) are grouped by parenthesis to indicate which subsums
are upper bounded via Proposition 3.5(i). In this example, if there are n groups, then the
sum is upper bounded by n/12.
The placeholders in an adjacent pair ((2k1, 2k2, 2k3), (2ki, 2kj , ∗, ..., ∗)) can, in theory,
be replaced by an arbitrary sequence of even numbers, and each such pair has to be refuted
separately. The following fact will make this task tractable: write τ ⊂ τ ′ if τ is a subtype
of τ ′, that is, a vertex type that can be obtained from τ ′ by removing some of its entries.
We then can prove
Proposition 3.14. If (τ1, τ2) is an infeasible adjacent pair, then the pair (τ1, τ ′2) is infea-
sible as well, for every τ ′2 ⊃ τ2.
Proof. Suppose τ2 = (2k1, ..., 2ks), τ ′2 = (2k1, ..., 2ks, 2ks+1, ..., 2ks′) ⊃ τ2, and that the
pair (τ1, τ ′2) is not infeasible. Then τ
′
2 satisfies Proposition 3.5(ii)
E(τ ′2) > s
′ − 2−K(τ ′2)
=⇒ E(τ2) > s− 2−K(τ2) +
s′∑
i=s+1
α
ki
2 /π>0︷ ︸︸ ︷(
1− 1
ki
− ϵki
)
> s− 2−K(τ2).
But this is exactly the statement that τ2 satisfies Proposition 3.5(ii) as well, i.e., that the pair
(τ1, τ2) is also not infeasible.
By Proposition 3.14 it is sufficient to exclude so-called minimal infeasible adjacent
pairs, that is, infeasible adjacent pairs (τ1, τ2) for which (τ1, τ ′2) is not infeasible for any
τ ′2 ⊂ τ2.
A second potential problem is, that we know little about the values that might replace
the placeholders in τ2 = (2ki, 2kj , ∗, ..., ∗). For our immediate goal, dealing with the
following special case is sufficient:
Proposition 3.15. The placeholders in an adjacent pair ((4, 6, 2k′), (6, 2k′, ∗, ..., ∗)) can
only contain 4, 6 and 2k′.
Proof. Suppose that P contains an adjacent pair
(τ1, τ2) = ((4, 6, 2k
′), (6, 2k′, 2k, ∗, ..., ∗))
induced by a 1-vertex v ∈ V1 of type τ1 with neighbor w ∈ V2 of type τ2. Suppose further
that 2k ̸∈ {4, 6, 2k′}. The vertex w is then incident to a 2k-face, and therefore also to a
1-vertex u ∈ V1 of type (4, 4, 2k) (u cannot be of type (4, 6, 2k) because of k ̸= k′ and
Corollary 3.9). This configuration is depicted below:
F. Göring and M. Winter: The edge-transitive polytopes that are not vertex-transitive 17
Note that w is also incident to a 4-face, and thus (6, 2k′, 2k, 4) ⊆ τ2.
By Proposition 3.5(i) the existence of 1-vertices of type (4, 4, 2k) and (4, 6, 2k′) yields
inequalities
ϵ2 + ϵ2 + ϵk <
1
k
and ϵ2 + ϵ3 + ϵk′ <
1
k′
− 1
6
. (3.2)
Since τ2 has τ := (6, 2k′, 2k, 4) as a subtype, by Proposition 3.14 it suffices to show that
the pair ((4, 6, 2k′), (6, 2k′, 2k, 4)) is infeasible. This follows via Proposition 3.5(ii):
7
6
− 1
k
− 1
k′
= 4− 2−K(τ)
(ii)
< E(τ) = ϵ2 + ϵ3 + ϵk′︸ ︷︷ ︸
<1/k′−1/6
+ ϵk︸︷︷︸
<1/k
(3.2)
<
1
k
+
1
k′
− 1
6
,
which rearranges to 1/k + 1/k′ > 2/3. Recalling 2k′ ∈ {6, 8, 10} =⇒ k′ ≥ 3 (from
Corollary 3.9) and 2k ̸∈ {4, 6, 2k′} =⇒ k ≥ 4 shows that this is not possible.
3.4 The case τ1 = (4, 6, 10)
If P contains a 1-vertex of type (4, 6, 10), then it contains an adjacent pair of the form
(τ1, τ2) = ((4, 6, 10), (6, 10, ∗, ..., ∗)).
We proceed as demonstrated in Example 3.13. Proposition 3.5(i) yields ϵ2+ϵ3+ϵ5 < 1/30.
By Proposition 3.15 the placeholders can only take on values 4, 6 or 10. The following table
lists the minimally infeasible adjacent pairs and proves their infeasibility.
τ2 s− 2−K(τ2)
?
< E(τ2)
(6, 10, 6) 4/30 ̸< (ϵ3 + ϵ5) + ϵ3 < 2/30
(6, 10, 10) 8/30 ̸< (ϵ3 + ϵ5) + ϵ5 < 2/30
(6, 10, 4, 4) 14/30 ̸< (ϵ2 + ϵ3 + ϵ5) + ϵ2 < 2/30
By Proposition 3.14 we conclude: the placeholder in τ2 = (6, 10, ∗, ..., ∗) can contain
no 6 or 10, and at most one 4. This leaves us with the option τ2 = (4, 6, 10), which is the
same as τ1 and therefore not possible by Corollary 3.12. Therefore, P cannot contain a
1-vertex of type (4, 6, 10).
3.5 The case τ1 = (4, 6, 8)
If P contains a 1-vertex of type (4, 6, 8), then it also contains an adjacent pair of the form
(τ1, τ2) = ((4, 6, 8), (6, 8, ∗, ..., ∗)).
We proceed as demonstrated in Example 3.13. Proposition 3.5(i) yields ϵ2+ϵ3+ϵ4 < 1/12.
By Proposition 3.15 the placeholders can only take on values 4, 6 or 8. The following table
lists the minimally infeasible adjacent pairs and proves their infeasibility.
18 Ars Math. Contemp. 23 (2023) #P2.01
τ2 s− 2−K(τ2)
?
< E(τ2)
(6, 8, 8) 2/12 ̸< (ϵ3 + ϵ4) + ϵ3 < 2/12
(6, 8, 4, 4) 5/12 ̸< (ϵ2 + ϵ3 + ϵ4) + ϵ2 < 2/12
(6, 8, 4, 6) 7/12 ̸< (ϵ2 + ϵ3 + ϵ4) + ϵ3 < 2/12
(6, 8, 6, 6) 9/12 ̸< (ϵ2 + ϵ3 + ϵ4) + ϵ3 + ϵ3 < 3/12
By Proposition 3.14 we conclude: the placeholder in τ2 = (6, 8, ∗, ..., ∗) can contain no 8,
and at most one 4 or 6, but not both at the same time.
This leaves us with the options τ2 = (4, 6, 8) and τ2 = (6, 6, 8). In the first case,
τ1 = τ2 which not possible by Corollary 3.12. In the second case, there would be two
adjacent 6-faces, but P does not contain (6, 6)-edges by Observation 3.10 with 2k′ = 8.
Therefore, P cannot contain a 1-vertex of type (4, 6, 8).
3.6 The case τ1 = (4, 6, 6)
If P contains a 1-vertex of type (4, 6, 6), then it also contains an adjacent pair of the form
(τ1, τ2) = ((4, 6, 6), (6, 6, ∗, ..., ∗)).
We proceed as demonstrated in Example 3.13. Proposition 3.5(i) yields ϵ2+ϵ3+ϵ3 < 1/6.
By Proposition 3.15 the placeholders can only take on values 4 or 6. The following table
lists the minimally infeasible adjacent pairs and proves their infeasibility.
τ2 s− 2−K(τ2)
?
< E(τ2)
(6, 6, 4, 4) 2/6 ̸< (ϵ2 + ϵ3 + ϵ3) + ϵ2 < 2/6
(6, 6, 6, 4) 3/6 ̸< (ϵ2 + ϵ3 + ϵ3) + ϵ3 < 2/6
(6, 6, 6, 6) 4/6 ̸< (ϵ3 + ϵ3) + (ϵ3 + ϵ3) < 2/6
By Proposition 3.14 we conclude: the placeholder in τ2 = (6, 6, ∗, ..., ∗) can contain at
most one 4 or 6, but not both at the same time.
This leaves us with the options τ2 = (4, 6, 6) and τ2 = (6, 6, 6). In the first case we
have τ1 = τ2, which is not possible by Corollary 3.12. Excluding (6, 6, 6) needs more
work: fix a 6-gon σ ∈ F2(P ). Each edge of σ is either of type (4, 6) or of type (6, 6) (by
Observation 3.10). Each 1-vertex of σ (which must be of type (4, 6, 6)) is then incident to
exactly one of these (6, 6)-edges of σ. Thus, there are exactly three (6, 6)-edges incident
to σ (see Figure 3). On the other hand, each 2-vertex of σ is incident to an even number
of (6, 6)-edges of σ (since if a 2-vertex is incident to at least one (6, 6)-edge, then we
have previously shown that its type must be (6, 6, 6), implying another incident (6, 6)-
edge). Therefore the number of (6, 6)-edges incident to σ must be even (see Figure 3), in
contradiction to the previously obtained number three of such edges.
Consequently, P cannot contain a 1-vertex of type (4, 6, 6).
Observation 3.16. It is a consequence of Sections 3.6, 3.5, 3.4 that P cannot have a 1-
vertex of a type (4, 6, 2k′) for a 2k′ ∈ {6, 8, 10}. By Corollary 3.9 this means that all
1-vertices of P are of the same type τ1 = (4, 4, 2k) for some fixed 2k ≥ 4.
It is worth to distinguish the case (4, 4, 4) from the cases (4, 4, 2k) with 2k ≥ 6.
F. Göring and M. Winter: The edge-transitive polytopes that are not vertex-transitive 19
Figure 3: Possible distributions of (4, 6)-edges (gray) and (6, 6)-edges (thick) around a
6-gon as discussed in Section 3.6. The top row shows configurations compatible with the
conditions set by 1-vertices (black), and the bottom row shows the configurations compat-
ible with the conditions set by the 2-vertices (white).
3.7 The case τ1 = (4, 4, 4)
In this case, all 2-faces are 4-gons, and all 4-gons are congruent by Proposition 3.2. A
4-gon with all edges of the same length is known as a rhombus, and the polyhedra with
congruent rhombic faces are known as rhombic isohedra (from german Rhombenisoeder).
These have a known classification:
Theorem 3.17 (S. Bilinksi, 1960 [2]). If P is a polyhedron with congruent rhombic faces,
then P is one of the following:
(i) a member of the infinite family of rhombic hexahedra, i.e., P can be obtained from a
cube by stretching or squeezing it along a long diagonal,
(ii) the rhombic dodecahedron,
(iii) the Bilinski dodecahedron,
(iv) the rhombic icosahedron, or
(v) the rhombic triacontahedron.
The figure below depicts these polyhedra in the given order (from left to right; including
only one instance from the family (i)):
The rhombic dodecahedron and triacontahedron are known edge- but not vertex-transitive
polytopes. We show that the others are not even strictly bipartite.
Corollary 3.18. If P is strictly bipartite with all 1-vertices of type (4, 4, 4), then P is one
of the following:
(i) the rhombic dodecahedron,
20 Ars Math. Contemp. 23 (2023) #P2.01
(ii) the rhombic triacontahedron.
Proof. The listed ones are edge-transitive but not vertex-transitive. Also they are not in-
scribed. By Proposition 2.4 they are therefore strictly bipartite.
We then have to exclude the other polyhedra listed in Theorem 3.17. The rhombic
hexahedra include the cube, which is inscribed, hence not strictly bipartite. In all the other
cases, there exist vertices where acute and obtuse angles meet (see the figure). So this vertex
cannot be assigned to either V1 or V2 (cf. Observation 2.9), and the polyhedron cannot be
bipartite.
These are the only strictly bipartite polyhedra we will find, and both are edge-transitive
without being vertex-transitive.
3.8 The case τ1 = (4, 4, 2k), 2k ≥ 6
If P contains a 1-vertex of type (4, 4, 2k) with 2k ≥ 6, then it also has an adjacent pair of
the form
(τ1, τ2) = ((4, 4, 2k), (4, 2k, ∗, ..., ∗)).
We proceed as demonstrated in Example 3.13. Proposition 3.5(i) yields ϵ2+ϵ2+ϵk < 1/k.
Since (4, 4, 2k) is the only type of 1-vertex of P , there are only 4-faces and 2k-faces and the
placeholders can only take on the values 4 and 2k (note that we do not use Proposition 3.15
for this). The following table lists some inequalities derived for infeasible pairs:
τ2 s− 2−K(τ2)
?
< E(τ2)
(4, 2k, 4, 4, 4) 1− 1/k < (ϵ2 + ϵ2 + ϵk) + (ϵ2 + ϵ2) < 2/k
(4, 2k, 4, 4, 2k) 3/2− 2/k < (ϵ2 + ϵ2 + ϵk) + (ϵ2 + ϵk) < 2/k
One checks that these inequalities are not satisfied for 2k ≥ 6. Proposition 3.14 then states
that the placeholders can contain at most two 4-s, and if exactly two, then nothing else.
Moreover, τ2 must contain at least as many 4-s as it contains 2k-s, as otherwise we would
find two adjacent 2k-faces while P cannot contain a (2k, 2k)-edge by Observation 3.10.
We are therefore left with the following options for τ2:
(4, 4, 2k), (4, 4, 4, 2k) and (4, 2k, 4, 2k).
The case τ2 = (4, 4, 2k) is impossible by Corollary 3.12. We show that τ2 = (4, 4, 4, 2k)
is also not possible: consider the local neighborhood of a (4, 4, 4, 2k)-vertex (the high-
lighted vertex) in the following figure:
F. Göring and M. Winter: The edge-transitive polytopes that are not vertex-transitive 21
Since the 1-vertices (black dots) are of type (4, 4, 6), this configuration forces on us the ex-
istence of the two gray 6-faces. These two faces intersect in a 2-vertex, which is then
incident to two 2k-faces and must be of type (4, 2k, 4, 2k). But we can show that the types
(4, 4, 4, 2k) and (4, 2k, 4, 2k) are incompatible by Observation 3.8(i):
β22 + β
2
2 + β
2
2 + β
k
2
(i)
= β22 + β
k
2 + β
2
2 + β
k
2 =⇒ β22 = βk2
(3.1)
=⇒ 4 = 2k ≥ 6.
Thus, (4, 4, 4, 2k) cannot occur.
We conclude that every 2-vertex incident to a 2k-face must be of type (4, 2k, 4, 2k).
Consider then the following table:
τ2 s− 2−K(τ2)
?
< E(τ2)
(4, 2k, 4, 2k) 1− 2/k < (ϵ2 + ϵ2 + ϵk) + ϵ2 < 2/k
The established inequality yields 2k ≤ 6, and hence 2k = 6. We found that then all 1-
vertices must be of type (4, 4, 6), and all 2-vertices incident to a 6-face must be of type
(4, 6, 4, 6).
3.9 The case τ1 = (4, 4, 6)
At this point we can now assume that all 1-vertices of P are of type (4, 4, 6) and that each
2-vertex of P that is incident to a 6-face is of type (4, 6, 4, 6). In particular, P contains a
2-vertex w ∈ V2 of this type. Since there is no (6, 6)-edge in P , the two 6-faces incident to
w cannot be adjacent. In other words, the faces around w must occur alternatingly of type
4 and type 6, which is the reason for writing the type (4, 6, 4, 6) with alternating entries.
On the other hand, P contains (4, 4)-edges, and none of these is incident to a (4, 6, 4, 6)-
vertex surrounded by alternating faces. Thus, there must be further 2-vertices of a type
other than (4, 6, 4, 6), necessarily not incident to any 6-face. These must then be of type
(4r) := (4, ..., 4︸ ︷︷ ︸
r
), for some r ≥ 3.
Proposition 3.19. r = 5.
Proof. If there is a (4r)-vertex, Observation 3.8(i) yields β22 = 2π/r. Analogously, from the
existence of a (4, 6, 4, 6)-vertex follows
2β22 + 2β
3
2
(i)
= 2π =⇒ β32 =
2π − 2β22
2
=
(
1− 2
r
)
π.
Recall kβk1 + kβ
k
2 > 2π(k − 1) from Observation 3.8(ii). Together with the previously
established values for β22 and β
3
2 , this yields
β21 >
2π(2− 1)− 2β22
2
=
(
1− 2
r
)
π, and
β31 >
2π(3− 1)− 3β32
3
=
(1
3
+
2
r
)
π.
(3.3)
Since the 1-vertices are of type (4, 4, 6), Observation 3.8(i) yields
2π
(i)
= 2β21 + β
3
1
(3.3)
> 2
(
1− 2
r
)
π +
(1
3
+
2
r
)
π =
(7
3
− 2
r
)
π.
22 Ars Math. Contemp. 23 (2023) #P2.01
Figure 4: The edge-graph of the final candidate polyhedron.
And one checks that this rearranges to r < 6.
This leaves us with the options r ∈ {3, 4, 5}. If r = 4, then β32 = π/2 = β22 ,
which is impossible by Equation (3.1). And if r = 3, then (3.3) yields β31 > π, which is
also impossible for a convex face of a spherical polyhedron. We are left with r = 5.
To summarize: P is a strictly bipartite polyhedron in which all 1-vertices are of type
(4, 4, 6), and all 2-vertices are of types (4, 6, 4, 6) or (45), and both types actually occur in
P . This information turns out to be sufficient to uniquely determine the edge-graph of P ,
which is shown in Figure 4.
This graph can be constructed by starting with a hexagon in the center with vertices
of alternating colors (indicating the partition classes). One then successively adds further
faces (according to the structural properties determined above), layer by layer. This process
involves no choice and thus the result is unique.
As mentioned in Remark 2.3, a bipartite polyhedron has an edge in-sphere. Thus, P is
a polyhedral realization of the graph in Figure 4 with an edge in-sphere. It is known that
any two such realizations are related by a projective transformation [8]. One representative
Q ⊂ R3 from this class (which we do not yet claim to coincide with P ) can be constructed
by applying the following operation ⋆ to each vertex of the regular icosahedron:
The operation is performed in such a way, so that
• the five new “outer” vertices of the new 4-gons are positioned in the centers of edges
of the icosahedron.
• the edges of each new 4-gon are tangent to a common sphere centered at the center
of the icosahedron
F. Göring and M. Winter: The edge-transitive polytopes that are not vertex-transitive 23
The resulting polyhedron Q looks as follows:
One can verify that Q has indeed the desired edge-graph.
It is clear from the construction that Q has an edge in-sphere, and any two of its 4-
gonal or 6-gonal faces are congruent (as we would expect from a bipartite polyhedron).
Like-wise, P has an edge in-sphere and the same edge-graph. Hence, P must be a pro-
jective transformation of Q. However, any projective transformation that is not just a re-
orientation or a uniform rescaling will inevitably destroy the property of congruent faces.
In conclusion, we can assume that P is identical to Q (up to scale and orientation).
It remains to check whether Q is indeed a bipartite polyhedron. For this, recall that any
two of the following properties imply the third (cf. Remark 2.3):
(i) Q has an edge in-sphere.
(ii) Q has all edges of the same length.
(iii) for each vertex v ∈ F0(Q), the distance ∥v∥ only depends on the partition class of
the vertex.
Now, Q satisfies (i) by construction, and it would need to satisfy both (ii) and (iii) in order
to be bipartite. The figure certainly suggests that all edges of Q are of the same length.
However, as we shall show now, Q cannot satisfy both (ii) and (iii) at the same time, and
thus, can satisfy neither. In particular, the edges must have a tiny difference in length that
cannot be spotted visually, making Q into a remarkable near-miss (we will quantify this
below).
For what follows, let us assume that (ii) holds, that is, that all edges of Q are of the same
length, in particular, that all 4-gons are rhombuses. Our goal is to show that ∥v∥ depends
on the type of the vertex v ∈ V2 (not only its partition class), establishing that (iii) does not
hold.
For this, start from the following well-known construction of the regular icosahedron
from the cube of edge-length 2 centered at the origin.
24 Ars Math. Contemp. 23 (2023) #P2.01
The construction is as follows: insert a line segment in the center of each face of the cube
as shown in the left image. Each line segment is of length 2φ, where φ ≈ 0.61803 is the
positive solution of φ2 = 1−φ (one of the numbers commonly knows as the golden ratio).
The convex hull of these line segments gives the icosahedron with edge length 2φ.
It is now sufficient to consider a single vertex of the icosahedron together with its inci-
dent faces. The image below shows this vertex after we applied ⋆.
The image on the right is the orthogonal projection of the configuration on the left onto
the yz-plane. This projection makes it especially easy to give 2D-coordinates for several
important points:
F. Göring and M. Winter: The edge-transitive polytopes that are not vertex-transitive 25
The points A and C are 2-vertices of Q of type (45) and (4, 6, 4, 6) respectively. Both
points and the origin O are contained in the yz-plane onto which we projected. Conse-
quently, distances between these points are preserved during the projection, and assuming
that Q is bipartite, we would expect to find |OA| = |OC| = r2. We shall see that this
is not the case, by explicitly computing the coordinates of A and C in the new coordinate
system (y, z).
By construction, C = (0, 1) and |OC| = 1. Other points with easily determined
coordinates are P, Q, R, S, T (the midpoint of R and S) and U (the midpoint of Q and
S).
By construction, the point B lies on the line segment QT. The parallel projection
of a rhombus is a (potentially degenerated) parallelogram, and thus, opposite edges in the
projection are still parallel. Hence, the gray edges in the figure are parallel. For that reason,
the segment UB is parallel to PQ. This information suffices to determine the coordinates
of B, which is now the intersection of QT with the parallel of PQ through U. The
coordinates are given in the figure.
The rhombus containing the vertices A, B and C degenerated to a line. Its fourth
vertex is also located at B. Therefore, the segments CB and BA are translates of each
other. Since the point B and the segment CB are known, this allows the computation of
the coordinates of A as given in the figure.
We can finally compute |OA|. For this, recall (∗)φ2n = F2n−2 − φF2n−1, where Fn
denotes the n-th Fibonacci number with initial conditions F0 = F1 = 1. Then
|OA|2 = (4φ− 3)2 + (3φ− 1)2
= 25φ2 − 30φ+ 10
(∗)
= 25(1− φ)− 30φ+ 10
= 35− 55φ
= 1 + (34− 55φ) (∗)= 1 + φ10 > 1,
and thus, Q cannot be bipartite. Remarkably, we find that
|OA| =
√
1 + φ10 ≈ 1.00405707
is only about 0.4% larger than |OC| = 1, and so while Q is not bipartite, it is a remarkable
near-miss.
26 Ars Math. Contemp. 23 (2023) #P2.01
Since P was assumed to be bipartite, but was also shown to be identical to Q, we
reached a contradiction, which finally proves Theorem 2.13, and the goal of the paper is
achieved.
4 Conclusions and open questions
In this paper we have shown that any edge-transitive (convex) polytope in four or more
dimensions is necessarily vertex-transitive. We have done this by classifying all polytopes
which simultaneously have all edges of the same length, an edge in-sphere and a bipartite
edge graph (which we named bipartite polytopes).
The obstructions we derived for being edge-transitive without being vertex-transitive
have been primarily geometric and less a matter of symmetry (a detailed investigation of
the Euclidean symmetry groups was not necessary, but it might be interesting to view the
problem from this perspective). We suspect that dropping convexity or considering com-
binatorial symmetries instead of geometric ones will quickly lead to examples of “just
edge-transitive structures”. For example, it is easy to find embeddings of graphs into Rd
with these properties.
Slightly stronger than being simultaneously vertex- and edge-transitive is being tran-
sitive on arcs, that is, on incident vertex-edge pairs. This additional degree of symmetry
allows an edge to be not only mapped onto any other edge, but also onto itself with reversed
orientation. While there are graphs that are vertex- and edge-transitive without being arc-
transitive (the so-called half-transitive graphs, see [7]), we believe it is unlikely that this
distinction is necessary for convex polytopes.
Question 4.1. Is there a polytope P ⊂ Rd that is edge-transitive and vertex-transitive,
but not arc-transitive?
In a different direction, the questions of this paper naturally generalize to faces of higher
dimensions. In general, the interactions between transitivities of faces of different dimen-
sions have been little investigated. For example, already the following question seems to
be open:
Question 4.2. For fixed k ∈ {2, ..., d − 3}, are there convex d-polytopes for arbitrarily
large d ∈ N that are transitive on k-dimensional faces without being transitive on either
vertices or facets?
Of course, any such question could be attacked by attempting to classify the k-face-
transitive (convex) polytopes for some k ∈ {1, ..., d− 2}. It seems to be unclear for which
k this problem is tractable (for comparison, k = 0 is intractable, see [1]), and it appears
that there are no techniques applicable to all (or many) k at the same time.
ORCID iDs
Frank Göring https://orcid.org/0000-0001-8331-2138
Martin Winter https://orcid.org/0000-0002-3817-9494
References
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doi:10.1007/bf02429904, https://doi.org/10.1007/bf02429904.
F. Göring and M. Winter: The edge-transitive polytopes that are not vertex-transitive 27
[2] S. Bilinski, Über die Rhombenisoeder, Period. Math.-Phys. Astron., II. Ser. 15 (1960), 251–
263.
[3] H. S. M. Coxeter, Regular Polytopes, Dover Publications, 1973.
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157–164, 1978, doi:10.1007/bfb0062528, https://doi.org/10.1007/bfb0062528.
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3190110204.
[7] D. F. Holt, A graph which is edge-transitive but not arc-transitive, J. Graph Theory
5 (1981), 201–204, doi:10.1002/jgt.3190050210, https://doi.org/10.1002/jgt.
3190050210.
[8] H. Sachs, Coin graphs, polyhedra, and conformal mapping, Discrete Math. 134
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tional Geometry Third Edition, Chapman and Hall/CRC, pp. 477–503, 2017.
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(2021), 1446–1462, doi:10.1007/s00454-021-00303-6, https://doi.org/10.1007/
s00454-021-00303-6.
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Transitivity, Ph.D. thesis, Technische Universität Chemnitz, 2021, https://www-user.
tu-chemnitz.de/~wimart/phd/thesis.pdf.
28 Ars Math. Contemp. 23 (2023) #P2.01
A
A.1 Geometry
Proposition A.1. Given a set x0, ..., xd ∈ Rd\{0} of d+1 vectors with pair-wise negative
inner product, then there are positive coefficients α0, ..., αd > 0 with
α0x0 + · · ·+ αdxd = 0.
Proof. We proceed by induction. The induction base d = 1 which is trivially true.
Now suppose d ≥ 2, and, W.l.o.g. assume ∥x0∥ = 1. Let π0 be the orthogonal projec-
tion onto x⊥0 , that is, π0(u) := u− x0⟨x0, u⟩. In particular, for i ̸= j and i, j > 0
⟨π0(xi), π0(xj)⟩ = ⟨xi, xj⟩︸ ︷︷ ︸
<0
−⟨x0, xi⟩︸ ︷︷ ︸
<0
⟨x0, xj⟩︸ ︷︷ ︸
<0
< 0.
Then {π(x1), ..., π0(xd)} is a set of d vectors in x⊥0 ∼= Rd−1 with pair-wise negative inner
product. By induction assumption there are positive coefficients α1, ..., αd > 0 so that
α1π0(x1) + · · ·+ αdπ0(xd) = 0.
Set α0 := −⟨x0, α1x1+ · · ·+αdxd⟩ > 0. We claim that x := x0α0+ · · ·+αdxd = 0.
Since Rd = span{x0} ⊕ x⊥0 , it suffices to check that ⟨x0, x⟩ = 0 as well as π0(x) = 0.
This follows:
⟨x0, x⟩ = α0 ⟨x0, x0⟩︸ ︷︷ ︸
=1
+ ⟨x0, α1x1 + · · ·+ αdxd⟩︸ ︷︷ ︸
=−α0
= 0,
π0(x) = α0 π0(x0)︸ ︷︷ ︸
=0
+α1π0(x1) + · · ·+ αdπ0(xd)︸ ︷︷ ︸
=0
= 0.
Proposition A.2. Let P ⊂ R3 be a polyhedron with v ∈ F0(P ) a vertex of degree three.
The interior angles of the faces incident to v determine the dihedral angles at the edges
incident to v and vice versa.
Proof. For w1, w2, w3 ∈ F0(P ) the neighbors of v, let ui := wi − v denote the direction
of the edge ei from v to wi. Let σij be the face that contains v, wi and wj . Then ∡(ui, uj)
is the interior angle of σij at v.
The set {u1, u2, u3} is uniquely determined (up to some orthogonal transformation) by
the angles ∡(ui, uj). Furthermore, since P is convex, {u1, u2, u3} forms a basis of R3, and
this uniquely determines the dual basis {n12, n23, n31} for which ⟨nij , ui⟩ = ⟨nij , uj⟩ =
0. In other words, nij is a normal vector to σij . The dihedral angle at the edge ej is
then π − ∡(nij , njk), hence uniquely determined. The other direction is analogous, via
constructing {u1, u2, u3} as the dual basis to the set of normal vectors.
A.2 Computations
The edge lengths in a spherical polyhedron are measured as angles between its end ver-
tices. Consider adjacent vertices vS1 , v
S
2 ∈ F0(PS), then the incident edge has (arc-)length
ℓS := ∡(vS1 , v
S
2 ) = ∡(v1, v2).
It follows from Observation 2.6 that these angles are completely determined by the
parameters, hence the same for all edges of PS .
F. Göring and M. Winter: The edge-transitive polytopes that are not vertex-transitive 29
Proposition A.3. For a face σ ∈ F2(P ) and a vertex v ∈ F0(σ), there is a direct relation-
ship between the value of α(σ, v) and the value of β(σ, v).
Proof. Let w1, w2 ∈ V2 be the neighbors of v in the 2k-face σ, and set ui := wi − v. Then
∡(u1, u2) = α(σ, v). W.l.o.g. assume that v is a 1-vertex (the argument is equivalent for a
2-vertex).
For convenience, we introduce the notation χ(θ) := 1− cos(θ). We find that
(∗) 2ℓ2 · χ(α(σ, v)) = ℓ2 + ℓ2 − 2ℓ2 cos(∡(u1, u2))
= ∥u1∥2 + ∥u2∥2 − 2⟨u1, u2⟩
= ∥u1 − u2∥2 = ∥w1 − w2∥2
= ∥w1∥2 + ∥w2∥2 − 2⟨w1, w2⟩
= r22 + r
2
2 − r22 cos∡(w1, w2) = 2r22 · χ(∡(w1, w2)).
The side lengths of the spherical triangle wS1 v
SwS2 are ∡(w1, w2), ℓ
S and ℓS . By the
spherical law of cosine5 we obtain
cos∡(w1, w2) = cos(ℓ
S) cos(ℓS) + sin(ℓS) sin(ℓS) cos(β(σ, v))
= cos2(ℓS) + sin2(ℓS)(cos(β(σ, v))− 1 + 1)
= [cos2(ℓS) + sin2(ℓS)] + sin2(ℓS)(cos(β(σ, v))− 1)
= 1− sin2(ℓs) · χ(β(σ, v))
=⇒ sin2(ℓS) · χ(β(σ, v)) = χ(∡(w1, w2))
(∗)
=
( ℓ
r2
)2
· χ(α(σ, v)).
5cos(c) = cos(a) cos(b) + sin(a) sin(b) cos(γ), where a, b and c are the side lengths (arc-lengths) of a
spherical triangle, and γ is the interior angle opposite to the side of length c.