ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 22 (2022) #P2.10 / 327–361
https://doi.org/10.26493/1855-3974.2433.ba6
(Also available at http://amc-journal.eu)
Characterization of a family of rotationally
symmetric spherical quadrangulations
Lowell Abrams
The George Washington University, Washington DC 20052, USA
Daniel Slilaty *
Wright State University, Dayton OH 45435, USA
Received 11 September 2020, accepted 29 August 2021, published online 27 May 2022
Abstract
A spherical quadrangulation is an embedding of a graph G in the sphere in which each
facial boundary walk has length four. Vertices that are not of degree four in G are called
curvature vertices. In this paper we classify all spherical quadrangulations with n-fold
rotational symmetry (n ≥ 3) that have minimum degree 3 and the least possible number
of curvature vertices, and describe all such spherical quadrangulations in terms of nets of
quadrilaterals. The description reveals that such rotationally symmetric quadrangulations
necessarily also have a pole-exchanging symmetry.
Keywords: Quadrangulation, spherical quadrangulation, rotational symmetry.
Math. Subj. Class. (2020): 05C10
1 Introduction
If S is a closed surface, a graph G embedded in S in which all facial boundary walks
have length four is called a quadrangulation of S. When S is the sphere, the graph G
is necessarily bipartite. Considering quadrilateral faces to be geometrically flat squares,
vertices of degree 4 extend this flatness to neighboring faces. Thus, when “most” of the
vertices of a spherical quadrangulation are of degree four, large areas will appear as a
portion of the geometrically-flat, infinite {4, 4}-planar lattice. The curvature is therefore
localized at vertices of degree other than 4.
In spherical triangulations where each face is considered to be a flat equilateral triangle,
vertices of degree 6 play a similar role in extending flatness, and curvature is thus localized
*Corresponding author.
E-mail addresses: labrams@gwu.edu (Lowell Abrams), daniel.slilaty@wright.edu (Daniel Slilaty)
cb This work is licensed under https://creativecommons.org/licenses/by/4.0/
328 Ars Math. Contemp. 22 (2022) #P2.10 / 327–361
Figure 1: A flat region with all internal vertices of degree 4.
at vertices of degree other than 6. Such spherical triangulations link to several well-studied
structures; for example, those for which all of the curvature is localized at 12 vertices of
degree 5 were popularized by the geodesic domes of Buckminster Fuller. Moreover, these
triangulations were later noted to be the topological dual graphs of what are now called
fullerene graphs. ([4] is a good starting source for fullerenes and other chemical graphs.)
As such, in a quadrangulation of S, a vertex of degree other than 4 is called a curva-
ture vertex. In this paper we investigate spherical quadrangulations with n-fold rotational
symmetry (n ≥ 3) that have minimum degree 3 and the least possible number of curva-
ture vertices, which is 8 when n = 4 and 2n + 2 otherwise. (See Proposition 2.1 and
Corollary 2.2.) The fact that curvature is localized at a relatively small number of vertices
suggests that G may have a description in terms of a geometric net of polygons. For ex-
ample, in Figure 2 we have a net of six congruent quadrilaterals which closes up to yield a
spherical quadrangulation with 3-fold rotational symmetry about poles p and q. Note that
there are 2n+ 2 = 8 curvature vertices of degree 3 each.
q
q
q
p
p
p
Figure 2: A quadrilateral net which describes a spherical quadrangulation with 3-fold rota-
tional symmetry.
Our main result (Theorem 1.1) is that every such spherical quadrangulation has a similar
L. Abrams and D. Slilaty: Characterization of a family of rotationally symmetric spherical . . . 329
description as a very simple net of 2n congruent quadrilaterals. The Three- and Four-
Parameter Constructions mentioned in Theorem 1.1 are stated in full detail in Section 3.
We also identify which of these nets produce quadrangulations that are overlay graphs of
self-dual planar embeddings.
Theorem 1.1. Let G be a spherical quadrangulation with minimum degree 3, having n-fold
rotational symmetry (n ≥ 3), and having the least possible number of curvature vertices.
If
(1) all white vertices have degree 4 and
(2) the poles of the rotational symmetry are at two black vertices,
then G can be obtained from either the Three- or Four-Parameter Construction of Section 3.
Remark 1.2. The reader may note that not all spherical quadrangulations with n-fold rota-
tional symmetry and the least possible number of curvature vertices satisfy Conditions (1)
and (2) of Theorem 1.1. Nevertheless, any such spherical quadrangulation can still be ob-
tained using Theorem 1.1 by making use of Proposition 1.3. In short, even if G does not
satisfy both (1) and (2), it is necessarily the case that G overlayed with its topological dual
graph will be a spherical quadrangulation which does satisfy both (1) and (2). This will be
described in more detail in Section 1.1 in the paragraph after Proposition 1.3.
The reader may note that the quadrangulation of Figure 2, and indeed all of the quad-
rangulations constructed in Section 3, not only possesses n-fold rotational symmetry at
poles p and q but also has a 2-fold symmetry which exchanges p and q. This is interesting
in that this 2-fold symmetry is not a priori implied by our hypotheses. Hence our spher-
ical quadrangulations possess, at the very least, an order-2n symmetry group. Any such
quadrangulations possessing additional symmetries will, of course, also be included in our
constructions.
1.1 Overlay graphs and other background on quadrangulations
An important fact about quadrangulations is that a spherical quadrangulation is always
bipartite while quadrangulations of other surfaces need not be. Given any graph H that is
cellularly embedded in a closed surface S, two bipartite quadrangulations that are naturally
associated with the embedding of H and its topological dual graph H∗ are the overlay
graph and the radial graph. For the sphere, in fact, any quadrangulation G is a radial graph
for some embedding H and its dual H∗. Applications of radial graphs, overlay graphs, and
the closely associated medial graph can be found in [1, 2, 3, 5, 7, 8, 9].
Consider a graph H cellularly embedded in a closed surface S and also consider its
topological dual graph H∗. Say that all of H (both vertices and edges) is colored “red” and
all of H∗ is colored “blue”. Embed H and H∗ simultaneously in S and at each edge/dual-
edge crossing point create a new vertex of degree four (which now has alternating red and
blue edges in rotation around it) and say that this new vertex is “white”. The graph obtained
is called the overlay graph O(H,H∗). Note that H is self dual if and only if O(H,H∗)
has a cellular automorphism which leaves white invariant and switches red and blue colors.
The overlay graph O(H,H∗) was used by Servatius and Servatius [8] to classify self-
dual embeddings in the sphere along with the pairing of their groups of color-preserving
cellular automorphisms of O(H,H∗) as index-2 subgroups of the groups of the white-
preserving cellular automorphisms. Graver and Hartung [5] do the same in the special case
330 Ars Math. Contemp. 22 (2022) #P2.10 / 327–361
of self-dual embeddings of graphs having four trivalent vertices and the remaining vertices
all of degree four. Their results, however, are much more explicit than those in [8].
The overlay graph O(H,H∗) is a bipartite quadrangulation of a closed surface S with
partite sets Red ∪ Blue and White. The radial graph of H and H∗ (denote it by R(H,H∗))
can be constructed from O(H,H∗) by placing a diagonal edge in each face of O(H,H∗)
which connects the red and blue vertices on that face and then erasing all of the edges
and white vertices of O(H,H∗). Thus R(H,H∗) is also a bipartite quadrangulation of
S. Conversely, if G is a bipartite quadrangulation of S with bipartition Red ∪ Blue, then
G = R(H,H∗) for some H and H∗ as follows. In each face of G, place a red edge
connecting the two red vertices and a blue edge connecting the two blue vertices. The
resulting red and blue graphs are H and H∗.
Unlike the radial graph, even if G is a bipartite quadrangulation of S in which all white
vertices have degree four it is not necessarily true that G is of the form O(H,H∗) for some
H . An additional condition that does ensure that G has the form O(H,H∗) is given in
Proposition 1.3. In Proposition 1.3, D(G) is the graph obtained from quadrangulation G
by placing a diagonal edge connecting the black corners of each face and then deleting the
white vertices of G. As mentioned in the previous paragraph, D(G) is the radial graph for
some K and K∗ in S when D(G) is bipartite.
Proposition 1.3. If G is a quadrangulation of closed surface S, then G = O(H,H∗) for
some H if and only if every white vertex of G has degree 4 and both G and D(G) are
bipartite. In the particular case that S is the sphere, G = O(H,H∗) for some H if and
only if every white vertex of G has degree 4.
Proof. First, let S be any closed surface. The one direction of the equivalence statement is
trivial. For the other direction, the fact that D(G) is bipartite allows us to properly 2-color
(red and blue) the vertices of D(G), which shows G is of the form O(H,H∗), as required.
The statement for the sphere follows from the first statement and the fact that any spher-
ical quadrangulation is automatically bipartite.
Now say H is a spherical quadrangulation with n-fold rotational symmetry and the
minimum number of curvature vertices, but does not satisfy the other two conditions in
Theorem 1.1. In this case O(H,H∗) inherits the n-fold rotational symmetry of H and
does satisfy the two additional conditions.
Quadrangulations have also been studied for other surfaces. Thomassen [10] and also
Márquez, de Mier, Noy, Revuelta [6] give explicit constructions for all 4-regular quadran-
gulations of the torus and Klein bottle. If G is a quadrangulation of S having no curvature
vertices, then in fact S must be the torus or Klein bottle. (See Proposition 2.1.)
2 Basic properties of spherical quadrangulations
Proposition 2.1 gives an arithmetic constraint on the quantities and degrees of curvature
vertices in quadrangulations, and Corollary 2.2 is an immediate consequence. We use χ(S)
to denote the Euler Characteristic of the surface S.
Proposition 2.1. If G is a quadrangulation of closed surface S with minimum degree 3
and vi vertices of degree i then,
v3 = 4χ(S) +
∑
i≥5
(i− 4)vi.
L. Abrams and D. Slilaty: Characterization of a family of rotationally symmetric spherical . . . 331
Furthermore, if χ(S) ̸= 0, then there are curvature vertices.
Proof. If G has f faces and e edges, then
∑
i ivi = 2e. Also, 4f = 2e and
(∑
i vi
)
− e+
f = χ(S) which when combined together yield 4(
∑
i vi) = 4χ(S) + 2e. Now subtracting
we obtain
∑
i(4− i)vi = 4χ(S) which yields our desired results.
Corollary 2.2. If G is a spherical quadrangulation with n-fold rotational symmetry
(n ≥ 3), minimum degree 3, and having the least possible number of curvature vertices,
then
• if n = 3, then G has eight vertices of degree 3 of which two are poles of the rotational
symmetry;
• if n = 4, then G has 8 vertices of degree 3 and the two poles of the rotational
symmetry are either both at vertices of degree 4 or both at the centers of faces; and
• if n > 4, then G has two vertices of degree n and 2n vertices of degree 3 where the
two vertices of degree n are the two poles of the rotational symmetry.
In [5], Graver and Hartung give a complete construction of spherical quadrangulations
of the form G = O(H,H∗) where H ∼= H∗ is a planar graph with four vertices of degree
3 and all other vertices of degree 4. (They do not assume any rotational symmetry). Here,
for n = 3, we assume only that G is a spherical quadrangulation with 3-fold rotational
symmetry and discover structures not found in [5].
Given a vertex v in a graph G and a subgraph H of G, the difference dG(v) − dH(v)
(that is, the degree of v in G minus the degree of v in H) is called the codegree of v with
respect to H and G. We say that v is saturated by a subgraph H when v has codegree zero
with respect to H and G.
Consider a spherical quadrangulation G with topological dual graph G∗. A collection
X of faces of G corresponds to a collection of vertices X∗ of G∗. If the induced subgraph
of G∗ on vertex set X∗ is connected, then the union of the faces in X along their incident
edges and vertices is called a face-connected subsurface. Let F be the face-connected
subsurface corresponding to X . The boundary of F , call it ∂F , is the collection of edges
(and their endpoints) incident to exactly one face in X . Topologically speaking, a face-
connected subsurface F is a sphere with holes (including the possibility of no holes) where,
of course, if there is exactly one hole, then F is topologically a disk. The boundary ∂F
is now an edge-disjoint union of cycles in G bounding the holes of F . The total length of
the union of boundary cycles is called the circumference or total circumference of F . An
interior vertex of F is a vertex not on ∂F while a boundary vertex is a vertex on ∂F .
The distance between two vertices u and v in a graph G is the length (edge length) of a
shortest uv-path in G. Denote this distance by dG(u, v). Of course, dG(u, v) is even iff u
and v are either both black or both white. Given a vertex v in a spherical quadrangulation G,
consider a face F with white vertices w1 and w2 and black vertices b1 and b2. For any vertex
v in G, evidently, |dG(v, wi)− dG(v, bj)| = 1 for each i and j ∈ {1, 2}. Additionally, the
following three possibilities may occur for F with respect to v: dG(v, w1) = dG(v, w2)
and |dG(v, b1)− dG(v, b2)| = 2; |dG(v, w1)− dG(v, w2)| = 2 and dG(v, b1) = dG(v, b2);
and dG(v, w1) = dG(v, w2) and dG(v, b1) = dG(v, b2).
Given a vertex v and a face f in a spherical quadrangulation G, let u be a vertex on f
of smallest distance from v, say distance t. The vertices in cyclic ordering around f now
332 Ars Math. Contemp. 22 (2022) #P2.10 / 327–361
have distances t, t + 1, t + d, t + 1 from v where d ∈ {0, 2}. Given integer k ≥ 0, define
Xk(v) to be the set of faces f for which t + 1 ≤ k. The face-connected subsurface of G
given by Xk(v) in Proposition 2.3 is called the k-ball centered at v and is denoted Bk(v).
Proposition 2.3. The faces in Xk(v) define a face-connected subsurface of G.
Proof. Certainly X1(v) is a face-connected subsurface. Using induction, assume that
Xk−1(v) is a face-connected subsurface and consider a face f in Xk(v) that is not in
Xk−1(v). We will complete the proof by showing that there is a facial path (that is, a path
in G∗) from f to a face in Xk−1(v). Let u be a vertex on f whose distance from v is the
smallest and let x1 and x2 be the neighbors of u on f . If we write dG(v, u) = t, then
dG(v, x1) = dG(v, x2) = t+ 1. It must be that t+ 1 = k or else f would be in Xk−1(v).
Now consider the neighbors of u in G, say x1, x2, . . . , xm. Since dG(v, u) = k − 1,
we get that dG(v, xi) ∈ {k−2, k} for each i which implies that all faces of G incident to u
are in Xk(v). Now there must be some i for which dG(v, xi) = k− 2 and so there is some
face f ′ incident to u which is in Xk−1(v). The rotation of faces in G around u contains a
path of adjacent faces from f to f ′.
The reader can easily verify that Proposition 2.4 follows directly from definitions.
Proposition 2.4. If v is a vertex in a spherical quadrangulation G and k ≥ 1, then the
following hold.
(1) If e is an edge of G whose vertices have distances t and t+1 from v with t+1 ≤ k,
then both faces of G incident to e are in Bk(v).
(2) If u is a vertex of G having distance t ≤ k from v, then u is in Bk(v).
(3) If f is a face of Bk(v) sharing an edge with ∂Bk(v), then the vertices of f have
distances k−1, k, k+1, k from v; furthermore, an edge of f on ∂Bk(v) has endpoints
with distances k and k + 1 from v.
(4) If f ′ is a face that is not in Bk(v) and which shares an edge with ∂Bk(v), then the
vertices of f ′ have distances k, k + 1, k + d, k + 1 from v where d ∈ {0, 2}.
(5) The vertices of a cycle C on ∂Bk(v) have distance from v alternating k and k + 1.
k or k + 2
k
k + 1
k + 1
f ′f
k − 1
k
Figure 3: Proposition 2.4(4).
A standard k-disk with n-fold rotational symmetry around a fixed black vertex is con-
structed as follows. Consider the standard {4, 4} planar quadrangulation and designate
one black vertex as an origin, and then label perpendicular x- and y-axes. Consider the
part of the quadrangulation in the first quadrant, with coordinate axes included, consisting
of the union of the closed faces whose interiors are either beneath or intersecting the line
L. Abrams and D. Slilaty: Characterization of a family of rotationally symmetric spherical . . . 333
x+ y = k (see Figure 4). Call this planar graph a k-wedge. Taking n copies of a k-wedge,
W1, . . . ,Wn, identify the x-axis of Wi with the y-axis of Wi+1 (subscript addition taken
modulo n) with origin vertex identified to origin vertex to obtain the standard k-disk. In
Figure 5 we show the standard 3- and 4-disks with a central vertex of degree n = 5.
y
x
Figure 4: The 11-wedge.
Figure 5: Standard 3- and 4-disks.
Consider a degree-4 vertex in G with incident edges e1, e2, e3, e4 in rotational order.
We call each of the pairs e1, e3 and e2, e4 transverse. A path P in a planar graph G is said
to be a transverse if each of its interior vertices has degree four in G with consecutive edges
on P forming a transverse pair. In a standard k-disk, there are n distinct transverse paths
of length k emanating from the origin (call it p) which we call the central rays. Note that
every vertex of distance k from p in a standard k-disk K has degree 4 in K aside from the
n endpoints of the central rays which have degree 3 in K. The vertices of distance k + 1
from p in K have degree 2 in K.
Proposition 2.5. The circumference of the standard k-disk is 2nk and the number of faces
in the standard k-disk is n
(
k+1
2
)
.
334 Ars Math. Contemp. 22 (2022) #P2.10 / 327–361
Proof. The number of faces in the first quadrant of the coordinatized {4, 4} planar quad-
rangulation that intersect the line x+ y = k is k. Therefore the number of boundary edges
of the k-wedge is 2k and the number of faces in the k-wedge is a triangular sum. The result
follows.
Proposition 2.6. If G is a spherical quadrangulation and Bk(p) is a standard k-disk in G,
then every vertex of distance k from p is on ∂Bk(p). Furthermore, any vertex of distance
k + 1 from p is either on ∂Bk(p) or is a neighbor of some vertex of distance k from p on
∂Bk(p) that is not saturated by Bk(p).
Proof. Here ∂Bk(p) is a single cycle separating the sphere into regions R1 and R2 where
without loss of generality p is in R1. Since Bk(p) is a standard k-disk, we know that every
interior vertex of Bk(p) has distance less than k from p. Given a vertex v in R2, a shortest
pv-path in G must pass through ∂Bk(p) and so v has distance strictly larger than k. This
implies our desired result.
Proposition 2.7. If D is a disk in a spherical quadrangulation G defined by a set of faces
X with |X| ≥ 2, then there are two distinct faces f1, f2 ∈ X such that each X−fi defines
a disk D′ for which the intersection of the boundary of face fi with ∂D′ is a path.
Proof. Consider a face f ∈ X whose boundary shares an edge with ∂D, call this a bound-
ary face. Using the fact that D is a disk, the reader can confirm that the following are
equivalent. (Here G∗[Y ∗] is the induced subgraph of G∗ on vertices Y ∗ where Y is a
collection of faces in G.)
• The faces X − f define a disk.
• The faces X − f define a face-connected subsurface of G, that is, G∗[X∗ − f∗] is
connected.
• The intersection of f with ∂D is a single path.
Assume by way of contradiction that G∗[X∗ − f∗] is disconnected for each boundary
face f ∈ X . By disconnectedness, the degree of f∗ in G∗[X∗] is not 1; furthermore, since
f is a boundary face, the degree of f∗ in G∗[X∗] is either 2 or 3. The number of connected
components of G∗[X∗ − f∗] is two when the degree is 2 and is two or three when the
degree is 3. We get 3 connected components precisely when f intersects ∂D in three paths
of lengths 1, 0, and 0. Each connected component of G∗[X∗ − f∗] must contain a vertex
corresponding to a boundary face of D.
Let f ∈ X be a boundary face for which the induced subgraph of G∗[X∗−f∗] contains
a connected component on vertex set C ⊆ X∗ − f∗ with |C| as small as possible. Pick
f∗0 ∈ C that is a boundary face of D. By assumption, G∗[X∗ − f∗0 ] is disconnected;
however, by planarity, one of its connected components has vertex set which is a proper
subset of C, a contradiction of minimality.
Given a face f1 ∈ X such that G∗[X∗ − f∗1 ] is connected, we will now find a face
f2 ̸= f1 such that G∗[X∗− f∗2 ] is connected. Since |X| ≥ 2, there must be boundary faces
in D other than f1. By way of contradiction, assume that for every boundary face f2 ̸= f1
we have that G∗[X∗−f∗2 ] is disconnected. Pick f2 such that G∗[X∗−f∗2 ] has a connected
component on vertices C with f∗1 /∈ C and |C| as small as possible. Following the same
argument as above, we will contradict the minimality of C.
L. Abrams and D. Slilaty: Characterization of a family of rotationally symmetric spherical . . . 335
Proposition 2.8 is surely an expected outcome; however, there are subtleties that require
verification. Proposition 2.8 provides a standard model for iteratively building up disks in
G.
Proposition 2.8. Let D be a disk in a spherical quadrangulation G such that D contains
a black vertex p having degree n ≥ 3 in G and all other vertices of D have degree 4 in G.
Then for k large enough, D is isomorphic to a face-connected subsurface of the standard
k-disk with n-fold rotational symmetry around a black central vertex p0. Furthermore, p is
identified with p0.
Proof. Let X be the collection of faces defining D. If |X| = 1, then the result is clearly
true. If |X| ≥ 2, then by Proposition 2.7, there is an ordering f1, . . . , fm of the elements of
X , such that p is on f1 and Xi = {f1, . . . , fi} defines a disk Di such that fi+1 intersects
∂Di in a path. Inductively Di is a face-connected subsurface of a standard k-disk, Sk, for
some large enough value of k. If ∂Di intersects ∂Sk, then increase k by 2 so that ∂Di no
longer intersects the boundary of the k-disk.
Let Pi be the path of intersection of fi+1 with ∂Di (the length of Pi being 1, 2, or 3).
Every internal vertex of Pi must be saturated by Di as a subgraph of G and each endpoint
of Pi is not saturated. Additionally, every vertex of D other than p has degree 4 in G, so
the codegree of any vertex of Di is the same with respect to being a subgraph of G or Sk.
We now have that every internal vertex of Pi is saturated by Di as a subgraph of Sk and
each endpoint of Pi is not saturated. Thus Pi is incident to a unique face f ′ of Sk that is
not in Di. The face f ′ may now be identified with fi+1 and we have Di+1 as a subgraph
of the standard k-disk Sk.
3 The two constructions
We will define two families of spherical quadrangulations, one defined with three indepen-
dent parameters and the other with four. Each spherical quadrangulation is described as a
net of 2n congruent convex quadrilaterals with vertices on the 2-dimensional integer lattice
in which two sides of the quadrilateral are perpendicular and of the same length. For lack
of a more specific term, we will call such a quadrilateral a special integer quadrilateral.
3.1 Three Parameters
Choose positive even integer a, non-negative integer s, and l ∈ {0, . . . , a − 1}. Consider
the special integer quadrilateral of Figure 6. By reflecting along the line y = x we may
assume that l ∈ {0, . . . , a2}
We assemble a net of 2n such quadrilaterals as indicated in Figure 7 to obtain a spherical
quadrangulation with n-fold rotational symmetry with poles at black vertices p and q and
with all white vertices of degree 4. Note that the arrangement of the quadrilaterals in
this construction will always yield a quadrangulation with an order-2 rotational symmetry
which exchanges p and q.
In Proposition 3.2 we characterize when the Three-Parameter Construction yields a
spherical grid which is the overlay graph of a self-dual embedding. Proposition 3.1 gives
us a necessary and sufficient condition for making this characterization.
336 Ars Math. Contemp. 22 (2022) #P2.10 / 327–361
(0, 0)
(0, a)
(a, 0)
(s+ l, a+ s− l)
Figure 6: The Three-Parameter Construction with (a, s, l) = (8, 3, 1).
p
p
p
q
q
q
Figure 7: A net constructed from 6 copies of the quadrilateral in Figure 6.
Proposition 3.1. If O(H,H∗) has n-fold rotational symmetry with poles at black vertices,
the minimum possible number of curvature vertices, and an additional symmetry that ex-
changes the poles, then H ∼= H∗ if and only if one pole is in H and the other in H∗.
Proof. Let p and q be the poles of the n-fold rotational symmetry of O(H,H∗). If one of
p and q is in H and the other in H∗, then because R(H,H∗) is bipartite, the symmetry
which exchanges poles must exchange all of H and H∗ and so is an isomorphism between
H and H∗.
Conversely, assume that H ∼= H∗ and say that H is red and H∗ is blue. As such, the
rotational symmetry, which fixes p and q, preserves the red and blue colors whereas the
symmetry which exchanges p and q exchanges red and blue colors.
If n = 3, then four of the degree-3 vertices are red and four are blue and p and q
both have degree 3. Furthermore, the six degree-3 vertices other than p and q are therefore
divided into two orbits under the rotational symmetry of three vertices each, one being red
L. Abrams and D. Slilaty: Characterization of a family of rotationally symmetric spherical . . . 337
and the other blue. Therefore the fourth red degree-3 vertex is one of p and q and the fourth
blue degree-3 vertex is the other.
If n ≥ 4, then the rotational symmetry divides the 2n degree-3 vertices into two orbits
of n vertices each; one must be red and the other blue. Call these Or and Ob. Without loss
of generality, say that p is red. Thus the distance in R(H,H∗) from p to the vertices in Or
is even while the distance to the vertices in Ob is odd. Now the symmetry which exchanges
p and q must therefore exchange Or and Ob, more generally, exchange red and blue colors.
Thus p ∈ H and q ∈ H∗.
Proposition 3.2. A spherical quadrangulation constructed from the Three-Parameter Con-
struction is the overlay graph of a self-dual graph if and only if s and l have different
parities.
Proof. Let G be a quadrangulation constructed using the Three-Parameter Construction
and let p and q be the poles of the rotational symmetry. By Proposition 1.3, G ∼= O(H,H∗)
for some H . By Proposition 3.1, a necessary and sufficient condition for H ∼= H∗ would
without loss of generality be that p ∈ H and q ∈ H∗. This is true if and only if the distance
from p to q in D(G) is odd.
Consider one quadrilateral of the construction. In this quadrilateral, there is a path in
the graph D(G), from (0, 0) to (0, a) of length a. From (0, a) to (s + l, a + s − l), there
is a path in D(G) of length (s + l) + (s − l) = 2s when s ≡ l mod 2 and of length
(s + l − 1) + (s − l − 1) + 1 = 2s − 1 when s ̸≡ l mod 2. Thus there is a path from p
to q in D(G) of length 2a+ 2s when s ≡ l mod 2 and of length 2a+ 2s− 1 when s ̸≡ l
mod 2, as required.
3.2 Four parameters
Choose positive integers a and b of the same parity. Assume that a ≥ b. Choose non-
negative integers h and w of the same parity, not both zero, and which satisfy
− h
w
≤ b− a
b+ a
and − a
b
(a− w) ≤ b+ h.
(In the case that w = 0 say that − hw = −∞.) Consider the special integer quadrilateral of
Figure 8.
(0, 0)
(a, b)
(−b, a)
(a− w, b+ h)
Figure 8: A special integer quadrilateral with (a, b, h, w) = (7, 3, 2, 2).
Given a choice of a and b, the constraints placed on non-negative integers h and w guarantee
that the quadrilateral defined will indeed be convex. The first inequality guarantees that the
338 Ars Math. Contemp. 22 (2022) #P2.10 / 327–361
point (a−w, b+h) lies above the line containing (−b, a) and (a, b). The second inequality
guarantees that at x = a−w, the y-coordinate b+ h is greater than the y-coordinate of the
line of slope − ba . (See Figure 9.)
(0, 0)
(a, b)
(−b, a)
(a− w, b+ h)
Figure 9: Constraints on (a, b, h, w) guarantee convexity.
We assemble a net of 2n such quadrilaterals as indicated in Figure 10 to obtain a spher-
ical quadrangulation with n-fold rotational symmetry with poles at black vertices p and q
and with every white vertex of degree 4. As with the three-parameter construction, the four-
parameter construction always yields a quadrangulation with an order-2 rotational symme-
try which exchanges p and q.
q
q
q
p
p
p
Figure 10: A net constructed from 6 copies of the quadrilateral in Figure 8.
Proposition 3.3. If G is constructed from the four-parameter construction with parameters
(a, b, h, w), then G is the overlay graph of a self-dual graph if and only if h and w are both
odd.
Proof. Say G is constructed using the Four-Parameter Construction and let p and q be the
poles of the rotational symmetry. As in the proof of Proposition 3.2, G ∼= O(H,H∗) for
some H (say H is red and H∗ is blue) and a necessary and sufficient condition for H ∼= H∗
would be that the distance from p to q in D(G) is odd.
L. Abrams and D. Slilaty: Characterization of a family of rotationally symmetric spherical . . . 339
Consider one quadrilateral of the construction with p at (0, 0). There is a path in D(G)
from p to the vertex at (a, b) of length a+b when a and b are both even and a+b−1 when a
and b are both odd. Call this length x. There is a path in D(G) from (a, b) to (a−w, b+h)
of length w + h when w and h are both even and of length w + h − 1 when h and w are
both odd. Call this length y. Thus there is a pq-path in D(G) of length 2x + y which is
even when h and w are both even and is odd when h and w are both odd, as required.
4 The two constructions are sufficient
In this section we prove Theorem 1.1. So, throughout this section, let n ≥ 3 be a fixed in-
teger and G a spherical quadrangulation satisfying the hypothesis of Theorem 1.1. Recall
that making use of Proposition 1.3 allows this theorem to cover all spherical quadrangu-
lations with minimum degree three and the minimum number of curvature vertices. Our
first step is to prove Proposition 4.2, which separates the remainder of this proof into two
distinct cases. In Section 4.3 we find that all graphs in the first case are given by the Three-
Parameter Construction. In Section 4.4 we find that all graphs in the second case are given
by the Four-Parameter Construction.
4.1 Initial k-balls are standard disks
Proposition 4.1. If every vertex of Bk(p) aside from p has degree 4 in G, then Bk+1(p) is
a standard (k + 1)-disk.
Proof of Proposition 4.1. First we prove that B1(p) is a standard 1-disk. The n white
neighbors of p are all distinct because G is simple. Now the only way in which B1(p)
is not a disk would be if the black vertices along the boundary walk are not all distinct. If,
by way of contradiction, we assume that these black vertices are not all distinct, then rota-
tional symmetry implies that there is only one black vertex on ∂B1(p). This black vertex
must therefore have degree 2n, a contradiction.
Inductively assume that the result holds up to some k − 1 ≥ 0 in G. Assume, by way
of contradiction, that every vertex v ̸= p of Bk(p) has degree 4 in G and yet Bk+1(p) is not
a standard (k + 1)-disk. Since every vertex v ̸= p of Bk(p) has degree 4 in G, we get that
every vertex v ̸= p of Bk−1(p) has degree 4 in G and so inductively Bk(p) is a standard
k-disk.
By Proposition 2.4, every face f that is not in Bk(p) but shares an edge e with ∂Bk(p)
is in Bk+1(p). Each vertex of distance k from p is in Bk(p) (again by Proposition 2.4)
and so has degree 4 in G. Therefore each edge e on ∂Bk(p) satisfies the following: if e is
incident to a central ray then the two edges of f incident to e are not on ∂Bk(p) and if e
is not incident to a central ray, then there are two consecutive edges of f on ∂Bk(p) (see
Figure 11).
The former type of face we will call a “radial” face and the latter a “notch” face. For a
notch face there cannot be 3 edges of f that are on ∂Bk(v) because this would either force
a vertex on ∂Bk(p) to have degree less that 3 in G, a contradiction, or force two vertices
on ∂Bk(p) to be identified, a contradiction because Bk(p) is a standard k-disk. In Case 1
assume that there is a radial face f in Bk+1(p) having opposing edges that are both on
∂Bk(p). In Case 2, every radial face has exactly one of its edges on ∂Bk(p).
340 Ars Math. Contemp. 22 (2022) #P2.10 / 327–361
k
k
k
k
f
k f
e′
Figure 11: A radial face and a notch face.
Case 1: Let e and e′ be opposing edges of radial face f that are both on ∂Bk(p). Since e
and e′ are nonconsecutive on f , they must also be non-consecutive along ∂Bk(p) (see the
left in Figure 11). Since e and e′ are the only edges of f on ∂Bk(p), it must also be the
case that e′ is incident to a central ray of Bk(p) as well (which cannot be the same ray).
Let O be the orbit of e under the n-fold rotational symmetry. (Note that |O| = n.) The
black and white coloring of the vertices, along with orientability, forces edges e and e′ to
both point in the same direction along the disk from their central rays, which implies that
e′ ∈ O. Therefore the orbit of f under the n-fold rotational symmetry does not have order
n and so f contains a pseudofixed point in its interior, a contradiction of the fact that p and
q are the only pseudofixed points of the rotational symmetry of the sphere.
Case 2: Let B denote the face-connected subsurface consisting of Bk(p) along with the
faces not in Bk(p) that share an edge with ∂Bk(p). Note that B ⊆ Bk+1(p). Take two
faces f and f ′ of B that are not in Bk(p) which are consecutive with respect to their edges
on ∂Bk(p). Because no vertex v ̸= p of B ⊆ Bk+1(p) has degree other than 4, if one of
f and f ′ is a notch face then f and f ′ share no edges in common, and if f and f ′ are both
radial faces then they must share one edge in common. Thus B is obtained from a standard
(k + 1)-disk, call it K, after perhaps making identifications along ∂K. (See Figure 12.)
Bk(p) B Bk+1(p)
K
Figure 12: The relationships between B, K, Bk(p) and Bk+1(p).
Note that there can be no facial identifications in obtaining B from K because that forces
the identified face to have two opposite edges on ∂Bk(p) which would put us back into
Case 1. In Case 2.1 there are no identifications, in Case 2.2 there is an edge identification,
and in Case 2.3 there are no edge identifications but there are vertex identifications.
Case 2.1: Here we have that B = K is a standard (k + 1)-disk. In this case, the vertices
on the cycle ∂B in G must have distances from p in G alternating between k+1 and k+2.
If Bk+1(p) = B, then we are done. If not, then there is a face f in Bk+1(p) that is not in
B and shares an edge with ∂B. Because f is in Bk+1(p), the distances of the vertices on f
from p must be k+1, k+2, k+1, k; however, ∂B separates p from f in the sphere which
means that no vertex on f can have distance less than k + 1 from p.
L. Abrams and D. Slilaty: Characterization of a family of rotationally symmetric spherical . . . 341
Case 2.2: Let e and e′ be two edges on ∂K that are identified in obtaining B from K. If e
and e′ are consecutive along ∂K with e having endpoints u and v and e′ having endpoints
v and w, then the distances in K from p to u, v, and w are either k + 1, k + 2, k + 1 or
k+2, k+1, and k+2. In the first case, the identification of e and e′ would create a vertex
in G of degree 1, a contradiction. In the second case, the identification of e and e′ would
create either a vertex of degree 2 in G (again a contradiction) or a vertex of degree 3 in G
that is in Bk(p), a contradiction of our inductive hypothesis.
If e and e′ are in the same orbit under the rotational symmetry of K, then the interior of
e would contain a pseudofixed point of the rotational symmetry; however, p and q are the
only pseudofixed points in G of the rotational symmetry, a contradiction.
Now, given that e and e′ are not consecutive along ∂K and not in the same orbit,
rotational symmetry yields an orbit of n distinct edges e1, . . . , en identified to an orbit of
n distinct edges e′1, . . . , e
′
n. Planarity of G forces the cyclic ordering of these edges along
∂K to be e1, e′1, . . . , en, e
′
n.
If either e1 or e′1 (say e1) is not incident to a central ray of K, then e has an endpoint v
of degree 4 in K. Thus v has distance k+1 from p in K and say without loss of generality
that v is black. The identification of e1 and e′1 would force v to have degree at least five in
B because the black endpoint of e′1 has degree at least 3 in K. This is impossible unless
the vertex resulting from the identification of e1 and e′1 is the other pole q; however, this
implies that the corresponding endpoints of e2, . . . , en and e′2, . . . , e
′
n are also the pole q.
This is impossible because q has degree n in G and such identifications would force q to
have degree at least 2n, which contradicts the fact that G has maximum degree n.
If both e1 and e′1 are incident to central rays, then these two orbits of edges account for
all of the 2n edges on ∂K that are incident to the central rays. In Figure 13 the edges with
the same numbers are identified and therefore, by the rotational symmetry, the endpoints
of the central rays must be identified to the other pole of the rotational symmetry, call it q.
Making these edge identifications results in a surface K ′ that is topologically a sphere
with n holes; that is, ∂K ′ consists of vertex-disjoint cycles C1, . . . , Cn. As discussed
above, there can be no further edge identifications in going from K ′ to B. If there are
vertex identifications in going from K ′ to B, then each identification is between two white
degree-2 vertices on ∂K ′. These white vertices must be on the same cycle Ci. The reason
for this is as follows. If vertex x on Ci is identified to vertex y on Cj , then let Q be a simple
xy-path in K ′ whose interior avoids ∂K ′ (not necessarily a path in the graph). Now any
cycle on K ′ which avoids its boundary (again, not necessarily a cycle in the graph itself)
and separates Ci from Cj must transversely intersect Q an odd number of times. Thus the
spherical embedding G would have two cycles drawn on it which intersect transversely an
odd number of times, a contradiction.
Now say that two white vertices on Ci (call them x and y) are identified in going from
K ′ to B. Note that the black vertices on Ci all have degree 4 in K ′ and the white vertices
on Ci all have degree 2 in K ′ save for one which has degree 3. Let Pi be the xy-path on
Ci which avoids the degree-3 white vertex. After identifying x and y, facial boundaries
of length four and saturated black vertices forces the identification of the adjacent pair of
white vertices on Pi, and so on. These identifications will eventually result either in a white
vertex being forced to have degree 2 in G (a contradiction) or a face in G being forced to
have length 2 (again a contradiction).
Lastly, assuming there are no further vertex or edge identifications, we have K ′ = B.
Let Di be the disk in G bounded by Ci whose faces are not in B. The black vertices on
342 Ars Math. Contemp. 22 (2022) #P2.10 / 327–361
p
q q
q
q
q
1 1
2
2
3
34
4
5
5
Figure 13: Edges with the same numbers are identified which then implies that the vertex
q is the other pole.
Ci have degree 2 in Di and the white vertices on Ci have degree 4 in Di save for one
white vertex which has degree 3 in Di. By rotational symmetry, there must be two black
degree-3 vertices in the interior of Di, with the remaining interior vertices having degree
4. This forces Di to have exactly three vertices of odd degree, a contradiction.
Case 2.3: If two identified vertices on ∂K are in the same orbit under the rotational sym-
metry, then the resulting vertex will be pseudofixed and so the identified vertex is the pole
q. However, q will now be forced to have degree at least 2n, a contradiction. So now
take an orbit of n distinct vertices v1, . . . , vn on ∂K that are pairwise identified to the n
distinct vertices v′1, . . . , v
′
n in going from K to B. There can be no additional identifica-
tions among these vertices. Planarity now forces these 2n vertices to have cyclic ordering
v1, v
′
1, . . . , vn, v
′
n along ∂K. Thus the vertex v1 = v
′
1 in G has degree 4 in G. So now
if K ′ is obtained from K by making these n identifications only, then K ′ is obtained as
shown in Figure 14 (for n = 5).
v1 v′1
v2
v′2
v3
v′3v4
v′4
v5
v′5
Figure 14: The surface K ′ from Case 2.3.
L. Abrams and D. Slilaty: Characterization of a family of rotationally symmetric spherical . . . 343
Rechoose the identified vertex pairs v1, v′1, . . . , vn, v
′
n so that vi and v
′
i are as close together
along ∂K as possible. Thus the orbit of n holes in K ′ have boundaries that are cycles in
G, say C1, . . . , Cn. Now either the endpoint of a central ray of K ′ is on C1 and is adjacent
to v1 = v′1 or not. Let these be Cases 2.3.1 and 2.3.2.
Case 2.3.1: Consider the disk H in G with ∂H = C1 whose faces are not in K. By
rotational symmetry either 0, 1, or 2 of the degree-3 vertices of G appear in H and the pole
q does not appear in H . Assume for the moment that 0 of the degree-3 vertices of G appear
in H . Say that C1 has length 2m. The degrees in H of the vertices on C1 are therefore
2, 3, 4, 2, 4, 2, . . . , 4, 2 in which the first degree-2 vertex is v1 = v′1 and the degree-3 vertex
is the endpoint of the central ray of K ′. The remaining vertices of H all have degree 4.
Thus the sum of the degrees of the vertices in H is
4i+ 3 + 4(m− 1) + 2m = 4i+ 6m− 1
where i is the number of interior vertices. So now if ϵ is the number of degree-3 vertices
of G appearing in H , then the sum of the degrees of the vertices in H is 4i+ 6m− 1− ϵ.
Now if e is the number of edges in H , we obtain
4i+ 6m− 1− ϵ = 2e.
If f is the number of quadrilateral faces in H , then
4f + 2m = 2e.
Now Euler’s Formula implies that
1 = i+ 2m− e+ f = 1
4
(2e+ ϵ+ 1− 6m) + 2m− e+ 1
4
(2e− 2m) = 1
4
(1 + ϵ) ≤ 3
4
,
which is a contradiction.
Case 2.3.2: Let u1 and w1 be the neighbors of v1 = v′1 on C1. Note that these three vertices
all have degree 4 in K ′ and so have no edges extending into the interior of H . This forces
these three vertices to be on the same quadrilateral face of G and this face is in H . Because
v1 and v′1 are chosen to be as close together as possible along ∂K, it must be that v1 and v
′
1
are at distance 4 apart along ∂K. Hence C1 has length four and H has only one face. Let
x be the fourth vertex of C1. Since u1 and w1 both have degree 4 in K ′ and K, it must be
that x has degree 2 in K ′ which implies that x also has degree 2 in G, a contradiction.
Proposition 4.2. If every vertex v ̸= p of Bk−1(p) has degree 4 in G but there are curvature
vertices of G in Bk(p)− p, then Bk(p) is a standard k-disk and the following hold.
(1) If k is even, then the n endpoints of the central rays of Bk(p) have degree 3 in G and
all other vertices of Bk(p)− p have degree 4 in G.
(2) If k is odd, then there are either n or 2n degree-3 vertices of G which have distance
k + 1 from p on ∂Bk(p) and all other vertices of Bk(p) − p have degree 4 in G,
including the endpoints of the central rays.
Proof. By Proposition 4.1, Bk(v) is a standard k-disk. By Proposition 2.4, the vertices of
Bk(p) that are not in Bk−1(p) come in two types: those on ∂Bk(p) having distance k + 1
344 Ars Math. Contemp. 22 (2022) #P2.10 / 327–361
from p and the endpoints of the central rays of Bk(p), which have distance k from p. If k is
odd, then because the curvature vertices of G are black, Bk(p) is of Type (2). If k is even,
then because the curvature vertices of G are black, Bk(p) is of Type (1) and there are either
n or 2n such curvature vertices.
Proposition 4.2 gives us two cases for G. In Section 4.3 we will discuss the case for G
in Part (1) of Proposition 4.2 and in Section 4.4 the case for Part (2) of Proposition 4.2.
4.2 Necklaces
Take quadrilaterals q1, . . . , qn whose vertices are properly colored black and white. Label
the black vertices of qi with bi,1 and bi,2. A diamond necklace of length n with a black
diagonal is the graph obtained from q1, . . . , qn by identifying bi,2 with bi+1,1 for each
i ∈ {1, . . . , n} where addition in subscripts is taken modulo n so as to obtain a cyclic
arrangement of these quadrilaterals. The top of Figure 15 shows a diamond necklace with a
black diagonal. A diamond necklace of length n with a white diagonal is defined similarly.
When t diamond necklaces of the same length with diagonals of alternating colors are
stacked together as on the bottom of Figure 15, we obtain a straight thorax of thickness t.
xx
x
y
z
x
y
z
Figure 15: A diamond necklace with a black diagonal and a straight thorax of thickness 5.
Consider a diamond necklace of length n(k + 1) for some k ≥ 1 with black-diagonal
vertices b0, . . . , bn(k+1)−1 and choose a positive integer 1 ≤ l ≤ k. If we embed our
necklace in the plane with b0, . . . , bn(k+1)−1 oriented in the clockwise direction, then there
are well-defined inner and outer boundary cycles of length 2n(k + 1) each. For each
vertex i(k + 1) ∈ {0, k + 1, . . . , (n − 1)(k + 1)} identify the two inner-boundary edges
incident to bi(k+1) and identify the two outer-boundary edges incident to bi(k+1)+l. The
resulting graph with n-fold rotational symmetry is called a (k, l)-zig-zag necklace with a
black diagonal. Note that the lengths of each of the two boundary cycles of the (k, l)-zig-
zag necklace is 2nk. A (k, l)-zig-zag necklace with a white diagonal is defined similarly.
The graph in Figure 16 shows a portion of a (k, 6)-zig-zag necklace.
Any number of (k, l)-zig-zag necklaces with diagonals of alternating colors may be
stacked as shown in Figure 17.
L. Abrams and D. Slilaty: Characterization of a family of rotationally symmetric spherical . . . 345
b0
b1
b2
b3
b4
b5
b6
Figure 16: A portion of a (k, 6)-zig-zag necklace.
Figure 17: Stacking zig-zag necklaces.
4.3 Curvature vertices on the ends of central rays
Proposition 4.3. Let Bk(p) be a standard k-disk in G as described in Proposition 4.2(1). If
the only curvature vertices of Bk+t(p) are p, v1, . . . , vn, then Bk+t+1(p) is a disk obtained
from the standard disk Bk(p) by adding a straight thorax with thickness t+1. Furthermore,
the circumferences of Bk(p) and Bk+t(p) are the same.
Proof. First we observe that the statement about the circumference of Bk+t(p) is evident
by the structure of necklaces. We now proceed with the rest of the proof. Certainly for
t = 0, Bk+t(p) is a disk obtained from the standard disk Bk(p) by adding a straight thorax
with thickness t = 0. So now assume that this same statement holds for some t ≥ 0
and the only curvature vertices of Bk+t(p) are p, v1, . . . , vn. Also as part of the induction
hypothesis we include that ∂Bk+t(p) has vertices in one color class (those of distance k+ t
from p) saturated by Bk+t(p) and vertices in the other color class (those of distance k+t+1
from p) having degree 2 in Bk+t(p).
Now consider Bk+t+1(p). Every face f of Bk+t+1(p) that is not in Bk+t(p) yet shares
an edge with ∂Bk+t(p) must share two consecutive edges with ∂Bk+t(p) because the ver-
tices in one color class of ∂Bk+t(p) are saturated by Bk+t(p). Furthermore, since Bk+t(p)
is a disk, f cannot share three edges with ∂Bk+t(p) because if it did, then f would have
two vertices that are saturated by Bk+t(p) which would force f to share all four of its edges
with ∂Bk+t(p). Since ∂Bk+t(p) is a cycle, this would imply that the length of ∂Bk+t(p) is
four; however, it must have length at least 2n ≥ 6.
346 Ars Math. Contemp. 22 (2022) #P2.10 / 327–361
Let B be the face-connected subsurface obtained from Bk+t(p) by adding to it the faces
of Bk+t+1(p) that share an edge with ∂Bk+t(p). We claim that B is a disk. If B is not a
disk, then B is obtained as follows. Let B′ be the disk obtained from Bk+t(p) by adding to
it a diamond necklace of faces. Now B is obtained from B′ by identifying edges or vertices
on ∂B′. Note that it is not possible to identify faces because ∂Bk+t(p) is a cycle.
It is not possible to identify two non-consecutive edges of ∂B′ because any two such
edges on ∂B′ contain endpoints that are on ∂Bk+t(p), which is a disk, and therefore are
distinct. Now suppose that e1 and e2 are consecutive edges on ∂B′ whose common end-
point is v. The degree of v in B′ is either 2 or 4. It is not possible that v has degree 2 in B′
because identifying e1 and e2 would then yield a vertex of degree 1 in G. It is not possible
that v has degree 4 in B′ because identifying e1 and e2 would create a vertex of degree 3
in Bk+t(p) that is not among p, v1, . . . , vn.
So it must be that B is obtained from B′ by identifications of vertices along ∂B′. Note
that the degree in B′ of the vertices of ∂B′ alternate between 2 and 4 where the degree-2
vertices have distance k+t+2 from p and the degree-4 vertices have distance k+t+1 from
p. Thus identification of any two degree-2 vertices on ∂B′ will then force the identification
of another pair of degree-2 vertices on ∂B′. These identifications will continue until we
force G to contain either a facial cycle of length 2 (a contradiction) or a vertex of degree 2
(again a contradiction).
Thus B′ = B is a disk and the vertices of ∂B alternate with distances k + t + 1 and
k+ t+2 from p. Thus B = Bk+t+1(p) because if there is a face f in Bk+t+1(p) but not in
B, its closest vertex to p has distance at least k+ t+1 and so this face is not in Bk+t+1(p),
a contradiction.
Proposition 4.4. If Bk+t(p) is a disk in a spherical quadrangulation G as given in Proposi-
tion 4.3, Bk+t(p) contains no curvature vertices of G other than p, v1, . . . , vn, but
Bk+t+1(p) contains an additional curvature vertex of G, then Bk+t+1(p) is also a disk
as given in Proposition 4.3, contains curvature vertices u1, . . . , un on its boundary cycle,
and each ui has distance k + t+ 2 from p.
Proof. This follows from Proposition 4.3 and the fact that the only vertices in Bk+t+1(p)
that are not in Bk+t(p) are the outer vertices of the new diamond-necklace layer of the
thorax.
Proposition 4.5. If Bk+t(p) is a disk as given in Proposition 4.4 which contains curvature
vertices p, v1, . . . , vn in its interior and curvature vertices u1, . . . un on its boundary, then
G is obtained from Bk+t(p) by identifying ∂Bk+t(p) with the boundary of a standard k-disk
D such that u1, . . . , un are identified with the endpoints of the central rays of D.
Proof. By Proposition 4.4, each ui has distance k+t+1 from p. This implies that k+t+1
is even, and since k is even, we must have that t is odd.
Say that l is the smallest distance in G from the pole q to any vertex on ∂Bk+t(p) and
let u be such a vertex. It must be that d(u, p) = k + t + 1 rather than k + t, because the
vertices of ∂Bk+t(p) of distance k + t from p are saturated by Bk+t(p) and so any path
from q to one of these vertices of distance k + t from p must go through the vertices of
distance k + t + 1 from p. Therefore u is black and has degree 2 in Bk+t(p). Also, since
u is black, l must be even.
First suppose that u can be chosen to be in Bl−1(q). The intersection of Bl−1(q) and
Bk+t(p) may only consist of a collection of black vertices because the white vertices of
L. Abrams and D. Slilaty: Characterization of a family of rotationally symmetric spherical . . . 347
Bl−1(q) have distance at most l − 1 from q. Therefore Bl−2(q) contains no curvature
vertices aside from q and Proposition 4.1 implies that Bl−1(q) is a standard (l − 1)-disk.
Thus u has degree 2 in Bl−1(q) and must have degree 4 in G. The neighbor wp of u on
∂Bk+t(p) in the direction of rotation is saturated by Bk+t(p) and the neighbor wq of u
on ∂Bl−1(q) in the rotational direction is either saturated or, alternatively, is the endpoint
of a central ray and, because it is white, has codegree 1 with respect to Bl−1(q). Let
up be the next vertex in rotational order along ∂Bk+t(p) and let uq be the next vertex in
rotational order along ∂Bl−1(q). If wq is saturated by Bl−1(q), then because every face of
G has length four we must have that up = uq and this vertex has degree 4 in G. (See the
left configuration in Figure 18). If wq is the end of a central ray of Bl−1(q) (which has
codegree 1), then again, the fact all faces have length four implies that up is adjacent to wq
and so up is a curvature vertex of G and uq = u′p where u
′
p is the next black vertex in the
rotational direction on ∂Bk+t(p). (See the right configuration in Figure 18).
u
wp
wq
up
uq
Face of
length 4 u
wp
wq
up Face of
length 4
uq
Central
Ray
w′p
u′p
Figure 18: Forced vertex identifications on the boundaries of Bk+1(p) and a standard k-
disk.
This process of identifying vertices and adjacencies continues all the way around
∂Bk+t(p) and ∂Bl−1(q) so that the black vertices on ∂Bk+t(p) correspond to the black
vertices and endpoints of the central rays of Bl−1(q). Therefore l = k and G is obtained as
stated in the proposition.
Next suppose that u cannot be chosen to be in Bl−1(q). Proposition 2.6 implies that
u is the endpoint of a central ray of Bl(q). Let w be this endpoint of the central ray of
Bl−1(q) that is adjacent to u. Thus u has codegree 1 or 2 with respect to Bk+t(p). In either
case there is a face f incident to the uw-edge that contains an edge wb1 of ∂Bl−1(q) and
an edge uw1 of ∂Bk+t(p). So now a fourth edge for f would be w1b1; however, w1 is
saturated by Bk+t(p) and b1 /∈ Bk+t(p), a contradiction (see Figure 19).
Theorem 4.6. The graph described in Proposition 4.5 is given by the Three-Parameter
Construction.
Proof. Consider the part of Bk(p) between two consecutive central rays, call it Wk. Let
o1 and o2 be the curvature vertices on the central rays of Wk which have distance k from
p. Consider the black diagonal line D in Wk from o1 to o2. Now let W be the portion
of Bk+t(p) consisting of Wk along with the faces between the black diagonals emanating
from o1 and o2 which are perpendicular to D. Let o be the curvature vertex in W which
348 Ars Math. Contemp. 22 (2022) #P2.10 / 327–361
u
w b1 /∈ Bk+t(p)
w1
f
Figure 19: Final contradiction in the proof of Proposition 4.5.
has distance k + t from p. As shown in Figure 20, there is a special integer quadrilateral
for the Three-Parameter Construction contained within W .
p
o1
o
o2
Figure 20: A special integer quadrilateral within a single wedge.
By inspection, the spherical quadrangulation constructed by the special integer quadrilat-
eral from Figure 20 contains Bk+t(p) with curvature vertices positioned as shown. By
Proposition 4.5, there is only one spherical quadrangulation which contains Bk+t(p) with
curvature vertices in a given position. Thus G is given by the Three-Parameter Construc-
tion.
4.4 Curvature vertices off the central rays
Let Bk(p) be a standard k-disk in G as described in Proposition 4.2(2). The disk Bk(p)
contains at least one orbit of n degree-3 curvature vertices. Let t ≥ 0 be the smallest integer
for which Bk+t(p) contains both orbits of n degree-3 curvature vertices. In Proposition 4.8
we describe three possible structures for Bk+t(p). Finally, we show that each structure is
given by the Four-Parameter Construction.
L. Abrams and D. Slilaty: Characterization of a family of rotationally symmetric spherical . . . 349
4.4.1 Two new types of disk
Note that the length of ∂Bk(p) is 2nk. The black vertices along ∂Bk(p) have degree 2 in
Bk(p) and the white vertices along ∂Bk(p) have degree 4 in Bk(p) save for the endpoints
of the central rays which have degree 3 in Bk(p). Consider vertices w1, v1, . . . , wn, vn in
clockwise order around ∂Bk(p) in which w1, . . . , wn are the ends of the central rays and
v1, . . . , vn is one orbit of black vertices on ∂Bk(p). Say that the distance from wi to vi
along ∂Bk(p) is 2l − 1. Let T be a t-layered stack of (k, l)-zig-zag necklaces. Note that
∂T consists of two cycles. Let ∂inT be the inner cycle of ∂T and say that the necklace
along ∂inT has a black diagonal and label these black vertices as b0, . . . , bn(k+1)−1. Note
that bj for j not divisible by k + 1 appears on ∂inT and has degree four in T except when
j = i(k + 1) + l, in which case bj has degree 3 in T . Also, the white vertices on ∂inT all
have degree 2 in T save for the white vertices on ∂inT adjacent to bi(k+1)’s, which have
degree 3 in T . Thus we can identify ∂Bk(p) with ∂inT so that vi is identified with bi(k+1)+l
and wi is adjacent to bi(k+1). We call the resulting disk Zk,l,t(p). Our discussion assumes
that t ≥ 1, but as a convention we can define Zk,l,0(p) to be the standard k-disk with l
defined by either one of the two orbits of degree-3 curvature vertices on ∂Bk(p). Note that
Zk,l,t(p) is a (k + t)-ball centered at p and every vertex v ̸= p in the interior of Zk,l,t(p)
has degree 4 in Zk,l,t(p) save for v1, . . . , vn which all have degree 3. Figure 21 depicts
Z5,2,3(p) for n = 5 (ignore the shading in the outer faces for the moment).
Figure 21: The disk Z5,2,3(p). If the shaded faces are removed, then the remaining faces
define Z5,2,2(p).
Now for t ≥ 2 that is even we define a disk Ẑk,l,t(p) from Zk,l,t(p). Say that v′i is the
black vertex on ∂Zk,l,t(p) that is on the transverse path from vi emanating outwards from
Bk(p). (Call this transverse path from vi a curvature ray.) Also, say that w′i is the endpoint
of the central ray of Zk,l,t(p) that contains wi. Now since t is even, the black vertices on
∂Zk,l,t(p) all have degree 2 except for v′1, . . . , v
′
n, which have degree 3 in Zk,l,t(p). Let
l′ = min{l − 1, k − l}. Label the l′ black vertices on ∂Zk,l,t(p) in the clockwise direction
from v′i with 1, 2, . . . , l
′ and do the same for the l′ black vertices on ∂Zk,l,t(p)
350 Ars Math. Contemp. 22 (2022) #P2.10 / 327–361
in the counter-clockwise direction from v′i (see the left-hand configuration in Figure 22 in
which k = 7, l = 3, t = 2, and l′ = l − 1 = 2).
1 12 2
vi
v′i
p
vi
p
v′i
Figure 22: Constructing Ẑk,l,t(p) with k = 7, l = 3, t = 2, and l′ = l − 1 = 2.
Now identify the black vertices having the same labels as shown on the right in Fig-
ure 22. Repeat these identifications for each i. The resulting disk is Ẑk,l,t(p). Note that
Ẑk,l,t(p) is a (k+t)-ball centered at p and that every vertex v ̸= p in the interior of Ẑk,l,t(p)
has degree 4 in Ẑk,l,t(p) save for v1, . . . , vn, v′1, . . . , v
′
n which all have degree 3.
4.4.2 The three structures
Proposition 4.7. Let Bk(p) be a standard k-disk in G with k odd and with exactly n degree-
3 vertices v1, . . . , vn of G appearing on ∂Bk(p). For each t ≥ 0, if the only curvature
vertices in Bk+t−1(p) are among p, v1, . . . , vn, then Bk+t(p) is either Zk,l,t(p) or Ẑk,l,t(p)
where l is specified by the position of v1, . . . , vn on ∂Bk(p).
Proof. The proof will be by induction on t where the case for t = 0 is given by Propo-
sition 4.1. Assuming for some t ≥ 1 that the only curvature vertices in Bk+t−1(p)
are among p, v1, . . . , vn we now consider Bk+t(p). For t = 1, we already know that
Bk+t−1(p) = Bk(p) is a standard k-disk which is also Zk,l,0(p). For t ≥ 2, the induction
hypothesis assumes that Bk+t−1(p) is either Zk,l,t−1(p) or Ẑk,l,t−1(p). However, while the
only curvature vertices of Bk+t−1(p) are among p, v1, . . . , vn, in fact, Ẑk,l,t−1(p) contains
more curvature vertices than this. Hence Bk+t−1(p) = Zk,l,t−1(p).
By Proposition 2.4 every face of G not in Bk+t−1(p) but sharing an edge with
∂Bk+t−1(p) is in Bk+t(p). Consider the face-connected subsurface B ⊆ Bk+t(p) con-
sisting of Bk+t−1(p) along with the faces not in Bk+t−1(p) but sharing an edge with
∂Bk+t−1(p). We will show that B = Zk,l,t(p) or B = Ẑk,l,t(p) and that B = Bk+t(p).
Given an edge e of ∂Bk+t−1(p) = ∂Zk,l,t−1(p), let fe be the face of B that is not in
Bk+t−1(p) and is incident to e. For comparison as a standard model, consider the disk
Zk,l,t(p) (separate from G) whose subdisk Zk,l,t−1(p) is identified with Bk+t−1(p) in G.
Let f ′e be the face of Zk,l,t(p) that is not in Bk+t−1(p) = Zk,l,t−1(p) and is incident to e.
If fe (or f ′e) is incident to a central ray, then call fe (or f
′
e) a radial face; otherwise, call fe
(or f ′e) a notch face. Note that f
′
e1 = f
′
e2 if and only if f
′
e1 is a notch face with e1 and e2
both incident to a common vertex that is saturated with respect to Bk+t−1(p) and G (see
L. Abrams and D. Slilaty: Characterization of a family of rotationally symmetric spherical . . . 351
Figure 21). We must show that the corresponding necessary and sufficient condition holds
for fe1 = fe2 .
If fe is a notch face, then fe shares two consecutive edges (say e1 and e2) with
∂Bk+t−1(p) where the common endpoint of e1 and e2, call it v12, is a vertex saturated
by Bk+t−1(p) in G. It cannot be that a third edge of fe is on ∂Bk+t−1(p) because such
an edge would have to be consecutive with e1 or e2 on the cycle ∂Bk+t−1(p), whereas the
two endpoints of e1 and e2 other than v12 have codegree 1 or 2 with respect to Bk+t−1(p).
If fe is a radial face of B, then denote the edges of fe by e1, e2, e3, e4 in rotational
order along fe. Assuming that e1 is on ∂Bk+t−1(p) and is incident to a central ray of
Bk+t−1(p) (call it r) we get that each endpoint of e1 has positive codegree with respect to
Bk+t−1(p). Thus e2 and e4 are not on the cycle ∂Bk+t−1(p). Without loss of generality
assume that edge e2 is the transverse continuation of r. Assume by way of contradiction
that e3 is on ∂Bk+t−1(p). Since e2 and e4 are not on ∂Bk+t−1(p), we again must have that
e3 is also incident to a central ray of Bk+t−1(p), call it r′. Note that r ̸= r′ because r = r′
would imply that G is not simple, a contradiction. Now either e1 and e3 are in the same
orbit under the rotational symmetry or not. If so, then the orbit of fe under the rotational
symmetry consists of n/2 faces and so there is a pseudofixed point in the interior of fe, a
contradiction. If not, then when adding fe to Bk+t−1(p), the black and white bipartition
forces there to be a half twist which creates a Möbius band in G, a contradiction.
The previous two paragraphs show that fe 7→ f ′e is a one to one correspondence be-
tween the faces of B that are not in Bk+t−1(p) and the faces of Zk,l,t(p) that are not in
Bk+t−1(p). Furthermore, fe 7→ f ′e takes notch faces to notch faces and radial faces to
radial faces; also, if f1 and f2 are two consecutive faces of B, then their common vertex
along ∂Bk+t−1(p), call it v, has codegree one or two and this determines whether or not
f1 and f2 share an edge incident to v. Therefore B is obtained from Zk,l,t(p) by making
zero or more identifications along ∂Zk,l,t(p). If there are no identifications, then we have
that B = Zk,l,t(p). We also get that B = Bk+t(p) because no face outside of B can have
a vertex of distance k + t− 1 from p and so we are done. So now, in Case 1 say that there
are edges on ∂Zk,l,t(p) that are identified and in Case 2 say that no edges along ∂Zk,l,t(p)
are identified but that there are vertices that are identified.
Case 1: Assume that e1 and e2 are on ∂Zk,l,t(p) and are identified in going from Zk,l,t(p)
to B. If ei is not incident to a central ray of Zk,l,t(p), then ei has one endpoint that is on
Bk+t−1(p) = Zk,l,t−1(p), is not a curvature vertex, and has degree 4 in Zk,l,t(p). Any
identification with another vertex of the same color would yield a vertex of degree more
than four, a contradiction. Thus e1 and e2 are both incident to central rays of Zk,l,t(p).
Because the rotational symmetry has only two fixed points (i.e., the poles), the n endpoints
of the central rays of Zk,l,t(p) must either correspond to n distinct vertices in B or one
vertex in B that is fixed under the n-fold rotational symmetry (that is, the other pole of
the rotational symmetry, call it q). We assume the latter is true as this is the only way in
which edges of ∂Zk,l,t(p) may be identified. Now the 2n edges of ∂Zk,l,t(p) incident to
the central rays are identified as per the numbering in Figure 23 to obtain Z ′.
There are two subcases to consider here: in Case 1.1 say that 2 ≤ l ≤ k − 1 and in
Case 1.2 say that l ∈ {1, k}.
Case 1.1: Now ∂Z ′ consists of n vertex-disjoint cycles. Since q is black, the black vertices
on ∂Z ′ all have degree 4 in Z ′ and the white vertices on ∂Z ′ all have degree 2 in Z ′ except
352 Ars Math. Contemp. 22 (2022) #P2.10 / 327–361
q q
q
q
q
p
1
2
3
2
3
4
45
1
5
q5
q
q
q
q
4
4
3
3
2
2
1
1
5
Figure 23: The disk Z ′ for l = 2 and l = 1, respectively.
for exactly 2n white vertices on ∂Z ′ having degree 3 in Z ′: the n endpoints of the curvature
rays along with the n neighbors of q. As shown at the beginning of Case 1, no two edges
of ∂Z ′ may be identified and so B is obtained from Z ′ by the identification of zero or more
pairs of white vertices having degree 2 in Z ′ to obtain white vertices of degree 4 in G.
We cannot, of course, identify white vertices from two distinct cycles of ∂Z ′ because this
would create a non-separating cycle in the embedding of G in the sphere, a contradiction.
Consider a cycle C in ∂Z ′ with vertices w1, b1, w2, b2, . . . , wm, bm and say by way of
contradiction that wi and wj are identified in going from Z ′ to B. Call the resulting face-
connected subsurface after this identification Z ′′. Now wi, bi, bj−1 (and also wi, bi−1, bj)
all have degree four in Z ′′. Because 2 ≤ l ≤ k − 1, we now get that these three vertices
are on a common face of G and so wi+1 and wj−1 (and also wi−1 and wj+1) must be
identified in going from Z ′′ to B. This identification process will continue and eventually
yield a contradiction by either: trying to identify a white vertex of degree 3 with a white
L. Abrams and D. Slilaty: Characterization of a family of rotationally symmetric spherical . . . 353
vertex of degree 2 or 3, by creating a face of length four having a white vertex of positive
codegree on its boundary, or by creating a cycle of length 2.
Thus we must have that Z ′ = B; however, this is not possible for the following reason.
Consider a cycle C in ∂B and let P be the disk of G with ∂P = C. The black vertices
on C have degree 2 in P , the white vertices on C have degree 4 in P save for two of them
which have degree 3 in P . By the rotational symmetry, the interior vertices of P include
exactly one black degree-3 vertex and all other interior vertices have degree 4 in P . Thus
P has an odd number of vertices of odd degree, a contradiction.
Case 1.2: As in Case 1.1, ∂Z ′ consists of n vertex-disjoint cycles and the black vertices
on ∂Z ′ all have degree 4 in Z ′. However, in this case, the white vertices on ∂Z ′ all have
degree 2 in Z ′ except for the n white vertices adjacent to q which have degree 4 in Z ′. Let
C be one cycle in ∂Z ′. As in Case 1.1, no two edge of C may be identified in going from
Z ′ to B. Label the vertices in rotational order along C with w1, b1, . . . , wm, bm where w1
is the white vertex of C having degree 4 in Z ′. Since bm, w1, b1 all have degree 4, they
must be on the same face of G and so we must have that wm = w2. This in turn implies
that wm−1 = w3, etc. These identifications are not possible, however, because m = 2k−2
and k is even and thus these identifications would create a face of length 2, a contradiction.
Case 2: In this case, the only possible identifications along ∂Zk,l,t(p) in going from
Zk,l,t(p) to B are pairs of vertices which have degree 2 in Zk,l,t(p). Say that the ver-
tices on ∂Zk,l,t(p) of distance k + t + 1 from p have color κ and the vertices of distance
k + t have color λ. Hence {κ, λ} = {black,white} (e.g., in Figure 21 κ = white). The
vertices of color κ on ∂Zk,l,t(p) have degree 2 in Zk,l,t(p) save for the n endpoints of the
curvature rays (which have degree 3 in Zk,l,t(p)) and the vertices of color λ on ∂Zk,l,t(p)
have degree 4 in Zk,l,t(p) save for the n endpoints of the central rays (which have degree 3
in Zk,l,t(p)).
Label the vertices along ∂Zk,l,t(p) in rotational order with λ1, κ1, . . . , λm, κm in which
λ1 is the endpoint of a central ray. Say that κi and κj are identified in B, and say that
v1, . . . , vn is the orbit of κi under the rotational symmetry and u1, . . . , un the orbit of κj .
Then |{u1, . . . , un, v1, . . . , vn}| = 2n in Zk,l,t(p) and |{u1, . . . , un, v1, . . . , vn}| = n or 1
in G. It cannot be that |{u1, . . . , un, v1, . . . , vn}| = 1 in G because then these 2n degree-2
vertices in Zk,l,t(p) would then identify to one vertex of degree 4n in G, a contradiction.
Thus |{u1, . . . , un, v1, . . . , vn}| = n in G. Because of the rotational symmetry these two
orbits of vertices must alternate along the cycle ∂Zk,l,t(p) and because G is spherical,
identified pairs of vertices (e.g., κi and κj) must appear consecutively along ∂Zk,l,t(p).
Let γij be the κiκj-path along ∂Zk,l,t(p) which contains no other vertices from
v1, . . . , vn, u1, . . . , un. Again, because of the rotational symmetry, at most one endpoint
of a curvature ray and at most one endpoint of a central ray occurs on γij . Suppose that
λi+1 and λj are the neighbors of κi = κj on γij . If λi+1 and λj both have degree 4 in
Zk,l,t(p), then κi, λi+1, and λj must all be on the same face of G and so we must have
that κi+1 = κj−1 in B. Similarly if λi and λj+1 both have degree 4 in Zk,l,t(p), then
κi−1 = κj+1 in B. These identifications of degree-2, κ-colored vertices must continue
in each direction along ∂Zk,l,t(p) until either we reach the endpoint of a curvature ray or
central ray. Thus γij contains either the endpoint of a curvature ray or the endpoint of a
central ray, but not both.
We claim that γij contains the endpoint of a curvature ray and not the endpoint of a
central ray. This is because if the latter were true, then, because the endpoint of a central
354 Ars Math. Contemp. 22 (2022) #P2.10 / 327–361
ray is of color λ, this identification of κ-colored vertices along γij would end with either a
face of length two (a contradiction) or a face of length four containing a κ-colored vertex
of degree 2 (again a contradiction).
Now since the endpoint of a curvature ray is contained in γij , the identification of κ-
colored vertices along γij ends with the identification of some κa−1 and κa+1 where κa
is the endpoint of the curvature ray and so has degree 3 in G. Also note that this implies
that κa is the midpoint of γij . Identifications of κ-colored vertices from κi = κj that
are off of γij must then stop at the endpoint of the central rays. These identifications of
vertices in Zk,l,t(p) result in the disk Ẑk,l,t(p). These are the only vertex identifications
that can happen in going from Zk,l,t(p) to B because we started with an arbitrary vertex
identification. Thus B = Ẑk,l,t(p) and we must also have that B = Ẑk,l,t(p) = Bk+t(p)
because any face of G outside of Ẑk,l,t(p) cannot contain a vertex of distance k + t − 1
from p.
Proposition 4.8 gives us the three possible structures for Bk+t(p). The proof of Propo-
sition 4.8 is similar to the proof of Proposition 4.2 using Proposition 4.7 in the place of
Proposition 4.1.
Proposition 4.8. Let k ≥ 1 be odd and let O1 be one orbit of n degree-3 curvature vertices
contained in ∂Bk(p). Let O2 be the second orbit of n degree-3 curvature vertices of G. Let
t ≥ 0 be such that Bk+t−1(p) contains no curvature vertices aside from O1 ∪ {p} and
Bk+t(p) contains O2. One of the following holds.
(1) If t is odd (that is, k+ t is even), then Bk+t(p) = Zk,l,t(p) and the vertices of O2 are
the endpoints of the central rays of Bk+t.
(2) If t is even (that is, k + t is odd), then either
(a) Bk+t(p) = Ẑk,l,t(p) and the vertices of O2 are the endpoints of the curvature
rays and appear in the interior of Bk+t(p), or
(b) Bk+t(p) = Zk,l,t(p) and the vertices of O2 are on ∂Bk+t(p) but not the end-
points of the central rays or curvature rays.
Proposition 4.9. If Bk+t(p) = Zk,l,t(p) is as given in Proposition 4.8(2)(b), then G is
given by the Four-Parameter Construction with uniquely determined parameters.
Proof. Consider two consecutive central rays of Zk,l,t(p) along with the vertices, edges and
faces of Zk,l,t(p) between these central rays, that is, the closure of a fundamental region of
the rotational symmetry, call it F . A rendering of F is shown on the left in Figure 24 in
which the dashed lines have length t (with t = 0 a possibility) and are identified.
Let o1 ∈ O1 and o2 ∈ O2 be the vertices of O1 ∪O2 in F . Since o2 is not on the endpoint
of the central ray or curvature ray, o2 appears on ∂Zk,l,t(p) in one of the two circled areas
shown on the left of the figure; take the right of Figure 24 as an illustration. We may assume
without loss of generality that o2 is in the upper region because if o2 is in the lower circled
region on the left of Figure 24, then we may reflect Zk,l,t(p) to get Zk,k−l+1,t(p) and then
have o2 in the upper region.
Now take the fundamental region F ′ adjacent to and in the counterclockwise direction
from F . The rendering of F ∪ F ′ shown in Figure 25 is geometrically flat and so we coor-
dinatize in the obvious way with p at (0, 0). As a result, the grey lines shown in Figure 25
L. Abrams and D. Slilaty: Characterization of a family of rotationally symmetric spherical . . . 355
p
o1
p
o1
o2
Figure 24: The fundamental region F .
form a quadrilateral with a right angle at the origin. Evidently the quadrilateral is also con-
vex because each of the interior angles is less than 180◦. Hence the grey quadrilateral is
of the type used in the Four-Parameter Construction; furthermore, the parameters defining
this quadrilateral are uniquely determined by the positions of o1 and o2 within F . Hence
Zk,l,t(p) contains n of these special quadrilaterals at p.
o′1
o′2
p
o1
o2
F ′
F
Figure 25: Two fundamental regions F and F ′ along with a four-parameter special integer
quadrilateral.
Conversely, by Proposition 4.7, the entire Four-Parameter Construction using this
quadrilateral must contain Zk,l,t(p) because the positions of the curvature vertices in the
Four-Parameter Construction come from the corners of the quadrilaterals. So one possi-
bility for G is given by the Four-parameter construction. Now, we will show that there is
only one spherical quadrangulation which contains Zk,l,t(p) and has O2 in this position on
∂Zk,l,t(p), which will complete our proof.
Consider the vertices on the boundary ∂Zk,l,t(p). The black vertices on this boundary
356 Ars Math. Contemp. 22 (2022) #P2.10 / 327–361
have degree 2 in Zk,l,t(p) save for the n endpoints of the curvature rays which have degree
3. The white vertices on this boundary have degree 4 in Zk,l,t(p) save for the n endpoints of
the central rays which have degree 3. Let w1, b1, c1, . . . , wn, bn, cn be vertices in clockwise
rotational order on ∂Zk,l,t(p) in which the wi’s are the endpoints of the central rays, bi’s
the vertices of O2, and ci’s the endpoints of the curvature rays.
Now let D be the disk of G constructed from the faces of G not in Zk,l,t(p). Thus D
contains the pole q ̸= p and ∂D = ∂Zk,l,t(p); furthermore, by rotational symmetry the
pole q is in the interior of D. Now,
• the wi’s, bi’s, and ci’s all have degree 3 in D,
• the remaining black vertices on ∂D have degree 4 in D, and
• the remaining white vertices on ∂D have degree 2 in D.
By Proposition 2.8, we may consider D as a subgraph of a standard r-disk Sr with a
black central vertex, call it q0, for some large enough value of r. Of course the embedding
of D must have q corresponding to q0 and, since q is in the interior of D, the central rays
of D must lie on the central rays of Sr. Thus the embedding of D in Sr is unique up to
dihedral symmetry. For the uniqueness of D as a completion of G, we need to show that
there is no other disk D′ in Sr having n-fold rotational symmetry around q with a bijection
between the vertices of ∂D′ and ∂D which respects degrees and cyclic ordering.
Consider the black vertices of Sr and connect pairs of black vertices on the same face
with an edge (say it is also black). This black graph is a quadrangulation with every internal
vertex of degree 4 aside from q which has degree n. Call any transverse path in the black
graph a black diagonal path of Sr or D. Call the n black diagonal paths of Sr or D that
originate from q the diagonal rays of Sr or D.
Now consider the boundary faces of D and the black diagonal edges in each. These
black edges form a cycle, call it C, in the black-diagonal graph and C is contained entirely
inside the disk D. Note that the cycle C consists of black diagonal paths whose endpoints
are the ci’s, bi’s, and w′i’s where w
′
i is the black neighbor of wi that is not in Zk,l,t(p).
Traversing C in Sr with q to our right, the ci’s and bi’s represent a right turn rather than a
transverse path and the w′i’s represent a left turn. We will now show that C (and hence the
boundary faces of D) is uniquely determined by the positions of ci’s, bi’s, and w′i’s on ∂D.
This will imply the uniqueness of D.
Now let V be the region of Sr between and including two consecutive diagonal rays,
call them Y1 and Y2 in the clockwise direction. The intersection of C with Y1 has one or
more connected components, each of which is either an isolated vertex or a path of positive
length. If there is no path of positive length, then let y1 be some vertex of C on Y1. If
there is a path of positive length in the intersection, then let y1 be the last vertex of some
intersection path when traversing C in the clockwise direction. Let y2 be the corresponding
vertex on Y2 under the rotational symmetry in the clockwise direction, and let P be the
y1y2-path in C in the clockwise direction. Consider the square Q in V given by the black
diagonals of Sr shown in Figure 26.
In the clockwise traversal of C, P contains two right turns and one left turn and at the rest
of the vertices of P , a transverse crossing. The sequence of turns is either left-right-right,
right-left-right, or right-right-left; however, if necessary we can reflect R around the axis
Y1 and reverse the traversal of C so that the first turn is right. In Figure 26, V is rendered
as part of the standard 4 × 4 grid in the xy-plane between the perpendicular lines y = x
L. Abrams and D. Slilaty: Characterization of a family of rotationally symmetric spherical . . . 357
Y2Y1
y2
y1
q
q0
Figure 26: The square defined by Y1, y1, Y2, and y2 in V .
and y = −x. In the traversal of P , two right turns and a left turn yield a net change of 90◦
degrees in the clockwise direction. Because of the way in which y1 is chosen, the first edge
of P is in the direction of the arrow shown in Figure 26. Now the path P is in V and is
completely determined by the placement of the two right turns in the 3-turn sequence. In
Figure 27 we have three examples of P . Since these turns are determined by the placements
of the ci’s, bi’s, and wi’s on ∂D, there is only one possibility for P and so for C and hence
for D.
Proposition 4.10. If Bk+t(p) = Ẑk,l,t(p) is as given in Proposition 4.8(2)(a), then G is
given by the Four-Parameter Construction with uniquely determined parameters.
Proof. As in the proof of Proposition 4.9, consider two adjacent fundamental regions F
and F ′ of Ẑk,l,t(p) between three consecutive central rays. These may be rendered in a
geometrically flat fashion as in Figure 28 with identically labeled vertices being identified
in G and appropriate identifications of dashed edges. Note that the path of dashed edges
has positive length. Since the rendering is flat we have a special integer quadrilateral as
used in the Four-Parameter Construction with parameters uniquely determined by k, l, and
t as shown in the figure. Therefore the Four-Parameter Construction yields one possibility
for G. In order to show that this is the only possibility for G, we will show that there is
only one possibility for the disk in G around q sharing its boundary with Ẑk,l,t(p).
Consider the dashed edge shown in Figure 29 along with its orbit of n edges under the
rotational symmetry. Let Z̃ be the disk around p consisting of Ẑk,l,t(p) along with the n
faces bounded by these n edges and Ẑk,l,t(p).
Let D be the disk defined by the faces of G not contained in Z̃. Note that D contains q
in its interior and ∂D = ∂Z̃. All of the white vertices of ∂D = ∂Z̃ have degree 4 in Z̃ and
degree 2 in D. Among the black vertices of ∂D = ∂Z̃, n have degree 3 in both Z̃ and D
and the rest have degree 2 in Z̃ and degree 4 in D.
Say that l is the smallest distance in G from the pole q to any vertex on ∂D = ∂Z̃. Let
u be one such vertex on the common boundary. It must be that d(u, p) = k + t+ 1 rather
358 Ars Math. Contemp. 22 (2022) #P2.10 / 327–361
q
q0
q
q0
q
q0
Figure 27: Three examples of the path P . In the third example, B intersects Y1 in a path.
than k + t. This is because the vertices of ∂Bk+t(p) of distance k + t from p are all white
and are saturated by Z̃ and so any path from q to one of these vertices of distance k + t
from p must go through the vertices of distance k+ t+ 1 from p. Therefore u is black and
has degree 2 or 3 in Z̃.
Since u is black, l must be even. First suppose that u can be chosen to be in Bl−1(q)
(which by Proposition 4.1 is a standard (l−1)-disk); that is, u is a vertex of degree 2 on the
boundary of Bl−1(q). Since the white vertices of ∂Bl−1(q) have distance l − 1 from q and
l is the smallest distance of a vertex from q to Z̃, these white vertices on ∂Bl−1(q) are not
in Z̃. So any black vertex on ∂Bl−1(q) which is identified to a black vertex on ∂Z̃ forces
another identification of two black boundary vertices. Eventually these identifications will
run to the degree-3 vertices of Z̃ on ∂Z̃. But this forces these black vertices to have degree
5 in G, a contradiction.
L. Abrams and D. Slilaty: Characterization of a family of rotationally symmetric spherical . . . 359
o2
o′2
o′2
p
o1
o′1
b(1,1)
o′1
b(1,1)
b(1,2)
b(1,2)
b(2,1)
b(2,1)
b(2,2)
b(2,2)
Figure 28: A flat rendering of two fundamental regions for Ẑk,l,t(p).
vi
p
v′i
Figure 29: There are n edges in G with endpoints in Ẑk,l,t(p). One such edge is shown as
a dashed curve.
Thus u is a vertex of degree 3 on the boundary of Bl(q); that is, u is the endpoint of a
central ray of Bl(q) which by Proposition 4.1 is a standard l-disk. Degree considerations
now force u and the vertices in its orbit to be identified with the degree-3 vertices of Z̃ on
∂Z̃. From here ∂Bl(q) must then be identified with ∂Z̃.
Proposition 4.11. If Bk+t(p) = Zk,l,t(p) is as given in Proposition 4.8(1), then G is given
by the Four-Parameter Construction with uniquely determined parameters.
Proof. Again, as in the proof of Proposition 4.9, consider two adjacent fundamental regions
F and F ′ of Zk,l,t(p) between three consecutive central rays rendered in a geometrically
flat fashion as in Figure 30. Again we have a special integer quadrilateral contained in
F ∪ F ′ as shown in the figure with uniquely determined parameters. Therefore the Four-
Parameter Construction yields one possibility for G, and we will now show that there is
only one possibility for the disk in G around q sharing its boundary with Zk,l,t(p).
Let m be the shortest distance from q to a vertex u on ∂Zk,l,t(p). Since all of the black
vertices on ∂Zk,l,t(p) are saturated by Zk,l,t(p), it must be that u is white and hence m is
odd. By Proposition 4.1 and the definition of m, Bm(q) is a standard m-disk. If u is not
the endpoint of a central ray of Bm(q), then u is saturated by Bm(q). Since u has degree
360 Ars Math. Contemp. 22 (2022) #P2.10 / 327–361
o2
p
o1
o′1
o′2 o
′
n
Figure 30: A flat rendering of two fundamental regions for Zk,l,t(p).
4 in G, it must be that u has degree 2 in ∂Zk,l,t(p) and that the boundary edges incident
to u in Bm(q) are identified to the boundary edges incident to u in ∂Zk,l,t(p). The only
boundary edges of both disks that are left are those incident to the central rays of Bm(q)
and the curvature rays of ∂Zk,l,t(p). Degree considerations and the fact that all faces must
have length 4 now force all edges of ∂Bm(q) to be identified to all edges of ∂Zk,l,t(p).
ORCID iDs
Lowell Abrams https://orcid.org/0000-0002-8174-5957
Daniel Slilaty https://orcid.org/0000-0002-7918-3641
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