Regular maps of high density

0.56 MB·English

by  R. H. Eggermont

REGULAR MAPS OF HIGH DENSITY
ROBH.EGGERMONTANDMAXIMHENDRIKS
5
1
0
2 Contents

n
a 1. Introduction 1
J 2. Background 2

4 3. The Regular Map Density Theorem 6
1 References 13

]
O Abstract. A regular map is a surface together with an embedded graph,
C havingpropertiessimilartothoseofthesurfaceandgraphofaplatonicsolid.
We analyze regular maps with reflection symmetry and a graph of density
.
h strictlyexceeding 1,andweconcludethatallregularmapsofthistypebelong
2
t toafamilyofmapsnaturallydefinedontheFermatcurvesxn+yn+zn=0,
a
exceptingtheonecorrespondingtothetetrahedron.
m
[
1. Introduction
1
v Objectsofhighsymmetryhavebeenofinterestforalongtime. Thousandsofyears
7
ago, the Greeks already studied the platonic solids, regular convex polyhedra with
7
congruent faces of regular polygons such that the same number of faces meet at
3
3 each vertex. The Greeks proved that there are only five of them: the tetrahedron,
0 the cube, the octahedron, the dodecahedron, and the icosahedron.
. One can consider the platonic solids to be orientable surfaces of genus 0 with an
1
0 embeddedgraphsatisfyingcertainproperties. Thenaturalquestionisthenwhether
5 this concept can be generalized. This leads to the notion of regular maps. Regular
1 maps have also been studied for many years (see for example [Sˇ06]).
:
v In contrast to the succinct and complete list of platonic solids, a complete classi-
i fication for regular maps seems far off, but at least a few families of regular maps
X
areknown. Investigationsintotheseobjectsisofgreatinterest, amongstotherrea-
r
a sons because of their intriguing connection to algebraic curves. Regular maps are
special cases of dessins d’enfants [Sch94]. As such, the combinatorial structure of
vertices,edges,andfacesgivesrisenotonlytoatopologicalrealisation,butevento
a unique algebraic curve! Explicitly computing an ideal that defines some complex
projective realisation of this algebraic curve for a given regular map is an ongoing
area of research.
Two happy examples of a whole family of regular maps for which the algebraic
curvesareknown,arenaturallydefinedontheFermatcurvesxn+yn+zn =0. For
agivenFermatcurveonehastoconsidertheactiononthiscomplexalgebraiccurve
(and hence a real surface) of its group of algebraic automorphisms, which turn out
to be realizable by linear maps in C3. A map presentation (see Section 2) of these
maps was described concisely by Coxeter and Moser (see [CM80]), although they
1
2 R.H.EGGERMONTANDM.HENDRIKS
did not hint at the link to the Fermat curves, and were perhaps not aware of it.
We give a more detailed description of the Fermat family in Section 2.
Regularmapscanbroadlybedividedintotwoclasses: chiralandreflexive. Wewill
deal only incidentally with chiral maps. In low genus, all regular maps have been
computed, with a list of reflexive maps up to genus 15 appearing in [CD01] and
with more recent lists at [Con], which runs up to genus 301 for reflexive maps at
the time of writing.
The combinatorial aspect of a regular map that is the focus of this paper is its
(graph) density, defined in Section 2. One discovers from studying the maps of low
genus,thathavingahighdensityisrelativelyrare. Infact,theonlyreflexiveregular
maps of low genus with simple graphs of density strictly exceeding 1 are members
2
of the Fermat family (every vertex in its graph being connected to precisely two
thirds of the vertices), the sole exception being the regular map Tet on the sphere,
correspondingtothetetrahedron. Thisnaturallyleadstothequestionwhetherthis
is true in higher genus as well. This paper answers this question by classifying all
reflexive regular maps with simple graphs of density strictly exceeding 1.
2
Theorem 1.1 (Regular map density theorem). Let R be a reflexive regular map
with simple graph of density strictly exceeding 1. Then either R is Tet or R is a
2
member of the Fermat family.
Before proving this theorem in Section 3, we will give the definition of a regular
map, show some properties of regular maps, and define the Fermat maps.
2. Background
Let us now give a more formal definition of regular maps.
Definition 2.1. A map M is an orientable surface Σ together with an embedded
connected finite graph Γ with non-empty vertex set V and non-empty edge set E
such that the complement of Γ in M is a finite disjoint union of open discs. Each
oftheseopendiscsiscalledaface ofΓ. Ifv V, e Eandf isafaceofΓ, wecall
∈ ∈
any pair of these incident if one of the pair is contained in the closure of the other.
A cellular homeomorphism of M is a homeomorphism of M that induces a graph
automorphism of Γ. An automorphism of M is an equivalence class of cellular
homeomorphisms under the equivalence relation of isotopy. We denote the set of
automorphismsofMbyAut(M),andwewriteAut+(M)forthesetoforientation-
preserving automorphisms of M. We call M chiral if Aut+(M)=Aut(M) and we
call M reflexive otherwise. Below we will work refer to reflexive maps with the
symbol R to stress that this property is assumed.
A map M is called regular if for all directed →−e,→−e(cid:48) E there is an orientation-
∈
preserving automorphism of M mapping →−e to →−e(cid:48).
Example 2.2. The platonic solids correspond to regular maps, simply by inter-
pretingthetraditionalterms‘vertices’and‘faces’accordingtoourdefinitionabove,
and defining the embedded graph by rereading the term ‘edges’. We denote the
regular map corresponding to the tetrahedron by Tet.
Notation 2.3. If M is a map, we denote by Σ(M), Γ(M), V(M) and E(M) the
corresponding surface, graph, vertex set and edge set. The set of faces of M is
denoted F(M).
REGULAR MAPS OF HIGH DENSITY 3
Notethatanycellularhomeomorphismisuniquelydetermineduptocellularisotopy
by the graph isomorphism it induces (although in general, not all graph isomor-
phisms are representable by a cellular homeomorphism).
Lemma 2.4. Let M be a map. Suppose φ Aut+(M) fixes some directed edge
∈
→−e. Then φ=Id.
Proof. Suppose e(cid:48) is an edge such that e and e(cid:48) share a common vertex. Let v be
a vertex incident to both e and e(cid:48) and consider a local picture around v. The only
way for φ to fix →−e and to preserve orientation is to act as the identity locally. In
particular,thismeansitmustfix→−e(cid:48) aswell. UsingconnectednessofΓ(M),itfollows
that φ fixes all directed edges and hence is the identity as a graph isomorphism.
This means φ=Id. (cid:3)
Asadirectcorollary,anyorientation-preservingautomorphismofamapisuniquely
determined by the image of a single directed edge. Similarly, any automorphism of
a map is uniquely determined by the combination of its orientation and the image
of a single directed edge.
Notation2.5. LetMbearegularmap. Letf beafaceofMwithcounterclockwise
orientation. Theboundaryoff isaunionofedges,andthenumberofedges(where
an edge is counted with multiplicity 2 if it borders on the same face twice) in such
a union does not depend on the face because M is regular. We always use the
letter p to denote this number. Let e ,e ,...,e be the edges on the boundary
1 2 p
of f (possibly containing doubles) in counterclockwise order. Orient these edges
in a way compatible with the orientation of f. There is a unique orientation-
preserving automorphism of M mapping →−e1 to →−e2. By necessity, it fixes f and
maps →−ei to −e−i+→1, taking indices modulo p if necessary. Effectively, it can be seen
as a rotation around f. We denote it by R . Clearly, it has order p. Observe that
f
any orientation-preserving automorphism of M fixing f is a power of R .
f
Similarly,givenavertexv ofM,thenumberofedgesincidenttov doesnotdepend
on the vertex. We always use the letter q to denote this number. Let e ,e ,...,e
1 2 q
be the edges incident to v (possibly containing doubles) in counterclockwise order.
Orienttheseedgeslocallyasarrowsdepartingfromv. Thereisauniqueorientation-
preserving automorphism of M mapping →−e1 to →−e2. It fixes v and also maps ei to
e , so it can be seen as a rotation around v. We denote it by S . Clearly, it has
i+1 v
order q. Observe that any orientation-preserving automorphism of M fixing v is a
power of S .
v
Let M be a reflexive regular map, and let e be an edge of M. We show that there
are two orientation-reversing automorphisms that fix e as a non-directed edge. We
will call these reflections.
Let σ Aut(M) Aut+(M). There is an automorphism g Aut+(M) such that
g(σ(→−e∈)) = →−e by\definition of a regular map. Moreover, t∈here is h Aut+(M)
∈
thatreverseseasadirectededge. Theautomorphismsgσ andhgσ donotpreserve
orientation and are two distinct automorphisms that fix e as a non-directed edge.
The first one fixes →−e as a directed edge. This shows existence. For uniqueness, if
σ,σ(cid:48) do not preserve orientation and if both either fix →−e or reverse it, then σ−1σ(cid:48)
is orientation-preserving and fixes →−e, and hence is Id by Lemma 2.4.
A visualization of rotations and reflections can be seen in Figure 1.
As an example, suppose v and w are adjacent vertices, and after fixing an orien-
tation, let f be the face to the left of the oriented edge −(v−,−w→). The rotation S
v
4 R.H.EGGERMONTANDM.HENDRIKS
e2
f1 f2 f2 f1
f v Sv e e
e1
ep Rf e2 eq f1 f2 f1 f2
e1 e e
Rotationaroundf Rotationaroundv Reflectionsine
Figure 1. Rotations and reflections. An indication of the action
is given by curved arrows in the left and middle figure (rotations),
and by the two-way arrows in the right figure (reflection).
around v maps −(v−,−w→) to −(−v−,S−−(−w−→)), and the rotation R around f maps the latter
v f
to −(w−−,→v). This means the automorphism R S inverts the oriented edge −(v−,−w→).
f v
◦
As a consequence, it has order 2. In particular, for any regular map M, if f is a
face adjacent to a vertex v, then (R S )2 =Id.
f v
◦
Lemma 2.6. Let M be a regular map, let v be a vertex of M, and let f be a face
of M incident to v. Then S and R generate Aut+(M).
v f
This lemma can be proved by showing that S and R can map a given directed
v f
edge to any other directed edge. A proof can be found in [Hen13, Lemma 1.1.6].
For convenience, when we have v V(M) and f a face of M incident to v, we
∈
occasionally use S and R instead of S and R . An important consequence of the
v f
lemma is that we can pin down a regular map M without explicitly talking about
its topology. We can give a presentation of Aut+(M) in the generator pair (R,S).
We call this a standard map presentation of M.
Example 2.7. We have Aut+(Tet) = 12, the cardinality of the set of directed
| |
edges of Tet. Both the number of edges incident to a vertex and the number of
edges incident to a face are 3. This means q = 3 and p = 3. Let v V(Tet)
∈
and f a face incident to v. We find the obvious relations S3 = Id and R3 = Id.
Moreover, we have the relation (SR)2 =1, from the fact that SR reverses the edge
(v,R−1(v)). Itturnsthesearetheonlyrelatorsnecessarytodefineastandardmap
presentation: Aut+(Tet)= R,S R3,S3,(RS)2 .
(cid:104) | (cid:105)
Remark 2.8. A regular map is completely determined by a standard map presen-
tation. Knowing just the isomorphism type of Aut+(M) is insufficient, however.
For example, the regular maps R3.1 and R10.9 both have Aut+(M) ∼= PSL(2,7).
Even fixing the triple (Aut+(M),p,q), and thereby also the genus, does not neces-
sarilydetermineauniqueregularmap. Thecounterexamples,tuplets,areofspecial
interest (a different story altogether). The first examples occur in genus 8 (the
twins R and R ) and genus 14 (the first Hurwitz triplet, R , R , R ).
8.1 8.2 14.1 14.2 14.3
The subscripts indicate the precise maps as listed on [Con].
A small theory of Aut+(M)-equivariant cellular morphisms between regular maps
can be developed, as described in [Hen13, Section 1.6]. The notion happily coin-
cides with taking certain quotients of Aut+(M), and a result we will use later in
Lemma 3.14 and Proposition 3.15 is the following.
REGULAR MAPS OF HIGH DENSITY 5
Proposition 2.9. Suppose M is a regular map and H a normal subgroup of
Aut(M) that is contained in Aut+(M) and does not contain an automorphism that
reverses some edge of M. Then Aut(M)/H is the automorphism group of a regular
map M satisfying Γ(M) = Γ(M)/H and F(M) = F(M)/H. There is a branched
cellular covering M M with the fiber of a cell of M a coset of H. Each cell of
→
M contains at most one ramification point, and each cell of M at most one branch
point. These numbers only depend on the dimension of the cell.
Here, by Γ(M)/H we mean the graph obtained by identifying vertices respectively
edges of Γ(M) if they lie in the same H-orbit, and by F(M)/H we mean the set
obtained by identifying faces of F(M) if they lie in the same H-orbit.
Example 2.10 (Fermat maps). For n Z , let
>0
∈
G = R,S R3,S2n,(RS)2,[R,S]3 .
n
(cid:104) | (cid:105)
For each n, the group G is the group of orientation-preserving automorphisms of
n
a regular map that we call the Fermat map Fer(n), obtained by considering the
solutions of xn +yn +zn = 0, acted upon by its algebraic automorphism group.
We omit the proof for this claim, but we do note that Fer(n) is a reflexive regular
map, andthatthisisaremarkableproperty. Thegroupstructurecanbedescribed
as Gn ∼=Z2n(cid:111)Sym3. The graph Γ(Fer(n)) is a simple graph, has 3n vertices and is
of genus (cid:0)n−1(cid:1). Each of the faces of Fer(n) is a triangle, since R has order 3.
2
Figure 2. The first Fermat maps (n = 1,2,3,4). The visualisa-
tion of Fer(4) was constructed in 1987 by Ulrich Brehm [Bre87].
Definition 2.11. LetMbearegularmapandletv,v(cid:48) V(M). Thenthedistance
∈
betweenv andv(cid:48) istheminimallengthofapathinΓ(M)fromv tov(cid:48);itisdenoted
by d(v,v(cid:48)). The set of vertices at distance at most i from v is denoted D(v,i). The
set of vertices at distance precisely i from v is denoted by ∂D(v,i).
The density δ(M) of M, which is the central notion for the rest of this paper, is
defined as
∂D(v,1)
δ(M):= | |.
V(M)
| |
Note that δ(M) does not depend on the choice of v by the regularity of the map.
Example 2.12. The Fermat maps all have density 2. The tetrahedron Tet has
3
density 3.
4
Our main objective from here on will be to show that the Fermat maps and the
tetrahedronaretheonlyreflexiveregularmapswithsimplegraphofdensitystrictly
exceeding 1, as announced in our regular map density theorem 1.1.
2
6 R.H.EGGERMONTANDM.HENDRIKS
3. The Regular Map Density Theorem
LetuswriteV=V(R)forconvenience. Westartwitharathertechnicalbutcrucial
lemma:
Lemma 3.1. Let v,v(cid:48) V. Suppose we have j Z such that Sj fixes v(cid:48). Then the
∈ ∈ v
following claims hold.
1: There is k Z such that Sj = Skj. Moreover, k is well-defined and
invertible mo∈dulo q . v v(cid:48)
gcd(j,q)
2: Let g Aut(R). Suppose Sj = Skj. Then Sj = Skj . Moreover, Sj
∈ v v(cid:48) g(v) g(v(cid:48)) v
fixes g(v) if and only if it fixes g(v(cid:48)) and in this case, if Sj = Sl , then
v g(v)
Sj =Skl .
v g(v(cid:48))
3: Let k Z such that Sj = Skj. For all i Z, we have Sj = Skj and
∈ v v(cid:48) ∈ v Svi(v(cid:48))
Sj =Sj .
v Si (v)
v(cid:48)
4: Let g ,...,g Aut(R) and suppose there is v V such that Sj fixes both
1 n ∈ i ∈ v
v and g (v ) for all i 1,2,...,n . Then Sj fixes g g ...g (v(cid:48)).
i i i ∈{ } v n n−1 1
Proof. For Claim 1, note that any orientation-preserving element that fixes v(cid:48) is
a rotation around v(cid:48), say Sj = Si . The order of these rotations must be equal,
v v(cid:48)
andinthiscaseisequaltothesmallestpositiveintegerxsuchthatxj 0 mod q,
which is q , using the fact that (Sj)x = 1 precisely if xj is a m≡ultiple of q.
gcd(j,q) v
Analogously, we observe that the order of these rotations must be equal to q ,
gcd(i,q)
meaning gcd(i,q) = gcd(j,q). In particular, both i and j are integer multiples of
gcd(j,q), and moreover, these multiples must be invertible modulo q (if not,
gcd(j,q)
the order of i or j modulo q would be strictly smaller than q ). Clearly, there
gcd(j,q)
exists k, well-defined and invertible modulo q , such that jk i mod q. This
gcd(j,q) ≡
shows Claim 1.
For Claim 2, we have the equalities Sj = gS±jg−1 = gS±kjg−1 = Skj , where
g(v) v v(cid:48) g(v(cid:48))
the sign in the exponent corresponds to the orientation of g being positive or neg-
ative. If Sj fixes g(v) respectively g(v(cid:48)), it is a power of Sj respectively Sj
v g(v) g(v(cid:48))
(which is itself a power of Sj ). In either case, Sj fixes both g(v) and g(v(cid:48)).
g(v) v
Moreover, if Sj =Sl , we have Sj =Skl . This completes the proof of Claim 2.
v g(v) v g(v(cid:48))
The equality Sj = Skj is easily seen because of Claim 2, using g = Si. The
v Si(v(cid:48)) v
v
equality Sj =Sj follows by symmetry. This shows Claim 3.
v Si (v)
v(cid:48)
To prove Claim 4, note that by Claim 2, we have Sj fixes g (v(cid:48)) if and only if it
v 1
fixes g (v). Exchanging the roles of v(cid:48) and v , we have Sj fixes g (v) if and only if
1 1 v 1
it fixes g (v ). The latter is true by assumption, so Sj fixes g (v(cid:48)). Claim 4 now
1 1 v 1
follows by induction. (cid:3)
While we formulated the previous lemma rather technically, the statements should
be seen in a more geometrical and intuitive way. Any rotation Sj that fixes a
v0
vertex v must be a rotation around v of the same order (Claim 1). Moreover,
1 1
such a situation translates well under conjugation (Claims 2 and 3). Finally, the
set of fixed points of an orientation-preserving automorphism Sj is closed under
v
the set of automorphisms g(cid:48) that map at least one fixed point of g to another fixed
point of g (Claim 4). We will use this lemma often.
REGULAR MAPS OF HIGH DENSITY 7
From here on, we will work under the following assumption.
Assumption 3.2. R is a reflexive regular map with a simple graph.
An easy graph lemma gives us a point of departure to say something about regular
maps with high density.
Lemma 3.3. If Γ(R) has density δ(R) 1, then Γ(R) has diameter at most 2.
≥ 2
Proof. Letv ,v V. Supposethatd(v ,v ) 2. Then∂D(v ,1)and∂D(v ,1)are
0 1 1 0 0 1
∈ ≥
containedinV v ,v ,whichhascardinality V 2. Sincewehave ∂D(v ,1) +
0 1 0
−{ } | |− | |
∂D(v ,1) V > V 2, we find that v and v share at least two common
1 0 1
|neighbors.| I≥n p|ar|ticu|lar|,−this implies d(v ,v ) 2. (cid:3)
0 1
≤
In the following proposition, we will show that the faces of reflexive regular maps
of high density are triangles (p=3).
Proposition 3.4. Suppose δ(R)> 1. Then p=3.
2
Proof. Let v V and let f be a face incident to v. Consider v = R2(v) and
∈ 1 f
supposev ∂D(v,1). ThenthesameholdsforSi(v )foranyi 0,1,...,q 1 ,
1 (cid:54)∈ v 1 ∈{ − }
because it preserves distances. In particular, the size of the orbit under S of
v
v is at most V ∂D(v,1). We have V < 2q because δ(R) > 1. The(cid:104)ref(cid:105)ore,
1 | \ | | | 2
V ∂D(v,1) = V q < 2q q = q. So the size of the orbit of v under S is
1 v
| \ | | |− − (cid:104) (cid:105)
strictly smaller than q, and hence there is j 1,...,q 1 satisfying Sj(v )=v .
Note that on one hand, Sj does not fix any∈ne{ighbor o−f v,}and since δ(Rv)>1 1, w1e
v 2
find Sj has at most V q < q fixed points. On the other hand, observe that for
v(cid:48) = Rv (v), we have|v|=−S−1(v). By Claim 4 of Lemma 3.1, using g = S−1, we
f 1 v(cid:48) i v(cid:48)
find Sj fixes Si (v) for any i, so Sj fixes all neighbors of v(cid:48), meaning it has at least
v v(cid:48) v
q fixed points, a contradiction. We conclude v ∂D(v,1).
1
Label the neighbors of v by the elements of Z/q∈Z, counterclockwise and let f be
i
the face on the left of −(v−,→i) (consistent with the orientation).
Suppose S−1(v)=i. Since R is reflexive and the reflection that fixes the oriented
0
edge −(−v−,→0) maps i to i, we find S (v) = i. Observe that this means that
0
− −
S (v) = SiS S−i(v) = 0. We now see two faces on the left of −(0−,→i), being f and
i v 0 v 0
f . We conclude f = f . Since v occurs only once on each face (using the
i−1 0 i−1
assumption that Γ(R) is simple), we conclude i=1 and hence p=3, as was to be
shown. (cid:3)
0
f =f
i 1 0
fi 1 1 i − 1 i
i − i
f0 Svi f0
v 0 v 0
−(−−i−,→0)mapsto−(0−,→i)
i − i
− 1 i − 1 i
− − − −
Figure 3. Proof of Proposition 3.4
8 R.H.EGGERMONTANDM.HENDRIKS
Definition 3.5. Assume we have a regular map with p = 3, and let v V. A
∈
diagonal neighbor of v is an element of V of the form S2(v) with w a neighbor of
w
v. Let v(cid:48) V as well. We call v and v(cid:48) diagonally aligned if there is a sequence
∈
(v ,v ,...,v ) of elements of V satisfying v =v, v =v(cid:48) and for all i 1,...,n
0 1 n 0 n
∈{ }
the vertex v is a diagonal neighbor of v .
i i−1
If v V is part of a triangle vwu, then wu is part of precisely one other triangle,
∈
say wuv(cid:48). In this case, v(cid:48) is a diagonal neighbor of v. All diagonal neighbors of v
are of this form.
v1 v2
v
vn
v
0
Figure 4. Diagonal neighbors and diagonal alignment
The observation that v(cid:48) is a diagonal neighbor of v if and only if v is a diagonal
neighbor of v(cid:48) shows that being diagonally aligned is an equivalence relation. We
writev v(cid:48) ifvandv(cid:48) arediagonallyaligned. Itiseasytoseethatbeingdiagonally
∼
aligned is preserved by graph isomorphisms.
Forv V,wewriteV = v(cid:48) V:v(cid:48) v . Itisaneasyexercisetoseethatifvwu
v
∈ { ∈ ∼ }
isafaceofΓ(V),thenanyelementofVisdiagonallyalignedtoatleastoneofv,wor
u (and in fact, either V , V and V are pairwise disjoint or V =V =V =V).
v w u v w u
Definition 3.6. Let R be a regular map. We define J to be the minimal element
of 1,2,...,q suchthatSJ fixesallelementsinV . WecallJ theprimitive period
{ } v v
of R.
Note that J exists because Sq = Id. Moreover, J is independent of choice of v by
v
Claim 2 of Lemma 3.1 (using the fact that v v(cid:48) if and only if g(v) g(v(cid:48)) for all
∼ ∼
v,v(cid:48) V and all g Aut(R)), so we are justified in not using a subscript. Thirdly,
∈ ∈
J divides q, since Sq =Id fixes V pointwise.
v v
Assumption 3.7. Fromthispointon,weaddtoourpreviousassumptions(Risa
reflexive regular map with simple graph) that R satisfies δ(R)> 1. In particular,
2
we will have p=3 and q 2.
≥
Let v V. The following lemma shows that we can classify V as the set of fixed
v
∈
points of SJ.
v
Lemma 3.8. Let v V and let v(cid:48) be a diagonal neighbor of v. Then the following
∈
claims hold.
1: The element J is the minimal element in 1,...,q such that SJ fixes v(cid:48).
{ } v
REGULAR MAPS OF HIGH DENSITY 9
2: We have J =q if and only if v(cid:48) ∂D(v,1).
∈
3: We have J =q if and only if V =V.
v
4: If w V is fixed by SJ, then w V .
5: If J <∈q, then V ∂Dv(v,1)= ∈, q vis even, and δ(R) 2.
v∩ ∅ ≤ 3
6: If J =q, then q is odd.
Proof. Claim 1 can be shown by induction. Suppose there is J(cid:48) such that SJ(cid:48) fixes
v
v(cid:48). Then by Claim 3 of Lemma 3.1, it fixes any diagonal neighbor of v (since all
of these are of the form Si(v(cid:48))), and analogously it fixes any diagonal neighbor of
v
any diagonal neighbor of v, etcetera. Hence SJ(cid:48) fixes V , and hence J divides J(cid:48).
v v
Clearly, SJ fixes v(cid:48), so this shows Claim 1.
v
ForClaim2,notethatS inducesbijectionsonthesets v ,∂D(v,1)and∂D(v,2).
v
{ }
Note that ∂D(v,1) has cardinality q and both v and ∂D(v,2) have cardinality
less than q since δ(R) > 1. In particular, if v{(cid:48) } ∂D(v,1), its orbit under the
2 (cid:54)∈
powers of S has cardinality strictly less than q, and hence J < q. Conversely, if
v
v(cid:48) ∂D(v,1), then it is fixed by SJ, and therefore J =q.
∈ v
ForClaim3,notethatifV =V,thenitcontainsaneighborofv,andhenceJ =q.
v
Conversely, if J =q, then v(cid:48) is a neighbor of v by the second part. Since V is the
v
equivalence class of v under , any rotation around v fixes it as a set. Hence all
neighbors of v are elements o∼f V . But then V q > 1 V, and hence V = V
v | v| ≥ 2| | v
(using the fact that either V = 1 V or V =V).
| v| 3| | v
For Claim 4, observe that if SJ fixes w, then it also fixes V . If V = V , then
v w w (cid:54) v
V contains a neighbor of v, and hence J =q. This gives a contradiction, since by
w
Claim 3, V =V and hence w V .
v v
∈
For Claim 5, suppose J < q. If V ∂D(v,1) = , then SJ fixes a neighbor of v,
v ∩ (cid:54) ∅ v
which gives a contradiction. So V ∂D(v,1)= .
v
Let w be a neighbor of v and cons∩ider the set ∅S2i(v) : i 0,1,..., q−1 of
{ w ∈ { (cid:98) 2 (cid:99)}}
cardinality 1+ q−1 . All of these elements lie in V . If q is odd, then this set
(cid:98) 2 (cid:99)} v
contains Sq−1(v), which is a neighbor of v, a contradiction. Hence q is even, and
w
this set has cardinality q. In particular, we find V ∂D(v,1) q, and hence
2 | \ | ≥ 2
δ(R) q = 2.
≤ q+q2 3
For Claim 6, suppose J = q and q is even. Then v(cid:48) ∂D(v,1). Let w ∂D(v,1)
∈ ∈
such that v(cid:48) = S2(v), and note that there is a directed edge −(−w−,−S−−(−v→)). Con-
w w
sider the (unique) reflection that maps this edge to −(−S−−(v−)−,−w→). Since q is even,
w
this reflection does not fix any neighbor of v. On the other hand, it fixes v(cid:48), a
contradiction. (cid:3)
We can now show that the only reflexive regular map of density greater than 1
2
with q odd is Tet. After showing this, we can focus on the case where q is even.
Proposition 3.9. Suppose δ(R)> 1 and q is odd. Then R is Tet, and δ(R)= 3.
2 4
Proof. Letv V. ByLemma3.8,theprimitiveperiodJ ofRisequaltoq. Number
∈
theneighborsofvby0,1,...,q 1clockwiseandletv(cid:48) =S2(v),adiagonalneighbor
− 0
of v. It is a neighbor of v by the second part of Lemma 3.8.
Now, consider the reflection that maps that maps −(−0,−q−−−→1) to −(−q−−−1−,→0). It fixes
− −
v and v(cid:48). Moreover, it maps any neighbor i of v to i 1 mod q. We conclude
v(cid:48) = q−1. − −
2
10 R.H.EGGERMONTANDM.HENDRIKS
q+1
Observe that we have a face 0,q−1,q−21. Applying the rotation Sv2 yields the
triangle q+1,q−1,0. Becauserotationspreserveorientation,weconclude q+1 q 1
2 2 2 ≡ −
mod q. Since q > 2, we conclude q+1 = q 1 and hence q = 3. We now easily
deduce R=Tet. 2 − (cid:3)
Assumption 3.10. From here on, we’ll assume R is a reflexive regular map with
simple graph satisfying δ(R) > 1 that is not equal to Tet. In particular, we will
2
have p = 3, q is even, and δ(R) 2. Moreover, we have J < q, and for all v, no
≤ 3
neighbor of v is diagonally aligned to v.
Suppose vwu is a face of Γ(R). Then S maps V to V and vice versa. In
v w u
particular, S2 is a bijection of V and V (and of V of course). This motivates us
v w u v
to give the following definition:
Definition 3.11. Let j :=lcm(J,2). We call j the even period of R.
Lemma 3.12. For any v,w V, we have [Sj,Sj]=Id and [Sj,S4]=Id.
∈ v w v w
Proof. If v w, we have Sj is a rotation around w, and hence it commutes with
∼ v
S , which implies both of the relations.
w
Suppose v w. Note that Sj fixes V pointwise and hence for any v(cid:48) V , we
(cid:54)∼ v v ∈ v
have [Sj,Sj](v(cid:48)) = SjSjS−jS−j(v(cid:48)) = SjS−j(v(cid:48)) = v(cid:48), using the fact that Sj
v w v w v w w w w
fixes V as a set and Sj fixes V pointwise. Likewise, [Sj,Sj] fixes V pointwise.
v v v v w w
Thereisv(cid:48) V suchthatv(cid:48) isaneighborofw. Now[Sj,Sj]fixestheedge−(−v−(cid:48),−w→)
∈ v v w
and hence it is Id.
Also, we have [Sj,S4] = SjS−j . Let v(cid:48)(cid:48) = S2(v(cid:48)). We see that both v(cid:48) and
v w v S4(v) w
w
S4(v(cid:48)) are diagonal neighbors of v(cid:48)(cid:48). There is k such that Sj =Skj by Claim 1 of
w v(cid:48) v(cid:48)(cid:48)
Lemma 3.1. Note that k only depends on the fact that v(cid:48)(cid:48) is a diagonal neighbor of
v(cid:48). ThismeansthatSj =Skj aswell,sincev(cid:48)(cid:48) isadiagonalneighborofS4(v(cid:48)).
S4(v(cid:48)) v(cid:48)(cid:48) w
w
WeconcludeSj =Sj andhence[Sj ,S4]=Id. Thisimplies[Sj,S4]=Id. (cid:3)
S4(v(cid:48)) v(cid:48) v(cid:48) w v w
w
Let v V and let v(cid:48) be a diagonal neighbor of v. We have Sj = Skj for some k,
unique∈ly defined modulo q. By Claim 2 of Lemma 3.1, we findvk is ivn(cid:48)dependent of
j
choice of v and v(cid:48) (as long as they are diagonal neighbors). In particular, we find
Sj =Skj andhencek2 1 mod q. Wewillshowsomethingstronger, namelythe
v(cid:48) v ≡ j
following.
Lemma 3.13. Let v V and let v(cid:48) be a diagonal neighbor of v. Then Sj =Sj .
∈ v v(cid:48)
Moreover, for any face vuw of Γ(R), we have SjSjSj =Id.
v u w
Proof. Let w be a neighbor of both v and v(cid:48) such that S2(v) = v(cid:48). Label the
w
neighbors of w counterclockwise with the elements of Z/qZ with v =v.
0
We have Sj =Sj for any m Z using the fact that Sj commutes with S4 by
vi vi+4m ∈ vi w
the previous lemma.
Note that V ∂D(w,1)= v :m Z and V ∂D(w,1)= v :m Z .
v∩ { 2m ∈ } v1 ∩ { 2m+1 ∈ }
We claim that Sj fixes V ∂D(w,1) as a set. Note that it fixes V as a set since
v v1∩ v1
j is even.
Suppose Sj (v ) is not a neighbor of w for some i. Then also Sj (v ) is not a
v0 i v4m i+4m
neighbor of w (because Sj (v ) is simply S4m(Sj (v )), and rotations around
v4m i+4m w v0 i
w preserve distance to w). Since Sj = Sj, we find that Sj (v ) is not a
v4m v v0 i+4m