Full text of DOI:10.1007/s00454-001-0008-0

Discrete Comput Geom 26:429–461 (2001)
Discrete & Computational
DOI: 10.1007/s00454-001-0008-0
Geometry
2001 Springer-Verlag New York Inc.
©
Equivelar Polyhedra with Few Vertices
B. Datta and N. Nilakantan
Department of Mathematics, Indian Institute of Science,
Bangalore 560 012, India
{dattab,nandini}@math.iisc.ernet.in
Abstract. We know that the polyhedra corresponding to the Platonic solids are equivelar.
In this article we have classified completely all the simplicial equivelar polyhedra on 11
vertices. There are exactly 27 such polyhedra. For each n
4, we have classified all the
(p, q)suchthatthereexistsanequivelarpolyhedronoftype{p, q}andofEulercharacteristic
n. We have also constructed five types of equivelar polyhedra of Euler characteristic 2m,
for each m 2.
1. Introduction
A finite collection K of cycles, edges and vertices of a complete graph is called a complex
(of dimension 2) if (i) each edge of a cycle in K is in K, (ii) each vertex of each edge
in K is in K and (iii) any two cycles have at most one common edge. The cycles,
edges and vertices in a complex are called the faces, edges and vertices in that complex,
respectively. We denote a face u1
umu1 by u1
um also.
For a complex K, the edge graph EG(K) of K is the graph whose vertices and edges
are the vertices and edges of K, respectively. EG(K), is also called the 1-skeleton of K.
The graph theoretic complement of EG(K) is called the non-edge graph of K and is
denoted by NEG(K). So, e is an edge in NEG(K) if and only if e is not an edge in K.
See [2] for the graph-theoretic terms used in this paper.
If K is a complex, then we associate another graph 3(K) with K as follows. The
vertices of 3(K) are the faces in K and for faces F1, F2 ∈ K, F1 F2 is an edge in 3(K)
whenever F1 and F2 have a common edge. For a vertex u in K let Fu be the set of faces
containing u. A complex K is called an abstract polyhedron (or simply a polyhedron) (of
dimension 2) if (iv) for each vertex v there is a face F containing v, (v) each edge is in
exactly two faces, (vi) the induced subgraph L(u) = 3(K)[Fu] is a cycle for each vertex

430
B. Datta and N. Nilakantan
u in K and (vii) the graph 3(K) is connected. Since all the polyhedra considered in this
paper are two-dimensional, we drop the qualification “two-dimensional”. Clearly, the
faces of a polyhedron determine the polyhedron. Because of this we identify a polyhedron
with the set of faces in it.
A complex may be thought of as a prescription for the construction of a topological
space by pasting together plane polygons. The topological space thus obtained from a
complex K is called the geometric carrier of K and is denoted by |K|. It is easy to see
that the geometric carrier of a polyhedron is a connected two-dimensional manifold.
Two complexes K and L are called isomorphic (denoted by K
a bijective map ϕ from the vertex-set of K to the vertex-set of L such that v1
L) if there exists
=
vk is a
face in K if and only if ϕ(v1)
ϕ(vk) is a face in L. We identify two complexes if they
are isomorphic.
If uv is an edge in a complex K, then we say u and v are adjacent in K. For a vertex
v in a complex K, the number of edges through v is called the degree of v in K. If
f0(K), f1(K) and f2(K) are the number of vertices, edges and faces, respectively, of a
polyhedron K, then the number (K) := f0(K) f1(K) + f2(K) is called the Euler
characteristic of K.
A polyhedron K is called equivelar of type {p, q} (or {p, q}-equivelar) if each face
is a p-gon (i.e., 3(K) is a p-regular graph) and the degree of each vertex is q (see [4]).
A polyhedron is called equivelar if it is equivelar of type {p, q} for some p and q.
A complex is called simplicial if each face consists of three vertices. If u is a vertex of
a simplicial complex K, then the link of u in K (denoted by LkK (u)) is the graph whose
vertices are those vertices of K which are adjacent to u and whose edges are those edges
vw in K such that uvw is a face in K. A simplicial complex with properties (iv)–(vi) is
called a combinatorial 2-manifold. Observe that in this case the link of any vertex is a
cycle. So, a connected combinatorial 2-manifold in which the degree of each vertex is the
same is a simplicial equivelar polyhedron and hence is called an equivelar combinatorial
2-manifold.
In [18]–[20] McMullen et al. considered equivelar polyhedra with geometric carriers
in R3 (and hence orientable). We consider both the orientable and non-orientable cases.
Example 1. Some equivelar polyhedra:
S42 = {abc, abd, acd, bcd},
O = {ai bj ck: 1 i, j, k 2},
C = {a1b2c1d2, a1b2d1c2, a1c2b1d2, a2b1c2d1, a2b1d2c1, a2c1b2d1},
I = {uui ui+1, ui ui+1vi+3, vi vi+1
u
i+3, vvi vi+1: 1
i+1vi,i+1ui , vi,i+1
5},
i
5},
D = {v1v2v3v4v5, vi vi+1
u
u
i+1vi+1,i+2ui+1,i+2ui,i+1
,
u12u23u34u45u51: 1
i
RP62 = {uui ui+1, ui ui+1ui+3: 1
i
5},
R = {u1,2u2,3u3,4u4,5
u
5,1, ui,i+1ui,i+1,i+3ui+2,i+3,i ui+5,i,i+2ui+5,i : 1
7+p, ui+3puj+3puk+3p: (i, j, k)
∈ {(1, 2, 5), (1, 3, 5), (1, 3, 4), (1, 8, 9), (1, 6, 8), (1, 2, 6), (2, 3, 6)},
2},
i
5},
M1 = {u1+pu4+p
u
0
p

Equivelar Polyhedra with Few Vertices
431
N1 = {uui ui+1, ui ui+1
u
i+4, ui ui+2
u
i+4, ui ui+3ui+6: 1
i
9}.
(Additions in the subscripts are modulo 5 in I, D, RP62, R and
are modulo 9 in M1, N1.)
Here S42 is equivelar of type {3, 3}, O is equivelar of type {3, 4}, C is equivelar of type
{4, 3}, I and RP62 are equivelar of type {3, 5}, D and R are equivelar of type {5, 3}, M1
is equivelar of type {3, 8} and N1 is equivelar of type {3, 9}.
The geometric carrier of each of S42, O, C, I and D is the 2-sphere and they correspond
to the Platonic solids [5], [6], [12], [8], [24], namely, tetrahedron, octahedron, cube,
icosahedron and dodecahedron, respectively. The polyhedron RP62 [1] is called the hemi-
icosahedron and the polyhedron R is called the hemi-dodecahedron. The geometric
carrier of each of RP62 and R is the real projective plane. The geometric carrier of M1
is the non-orientable surface of Euler characteristic 3. The geometric carrier of N1 is
the non-orientable surface of Euler characteristic 5.
e
Let K be a polyhedron with faces F1, . . . , Fm. Consider a complex K with vertex-set
e
{w1, . . . , wm} as: wi
wi is a face in K if and only if there exists a vertex u in K such
1
k
e
e
that Fi
Fi Fi is the cycle L(u) defined above. Then K is a polyhedron. K is called
1
k
1
e
e
the dual of K. It is easy to show that the dual of K is isomorphic to K and (K) = (K).
2
e
2
It is also not difficult to see that S42 S , C
O, I
D and RP
R. Observe that
g
e
e
=
=
=
=
6
4
e
the graph 3(K) is isomorphic to EG(K). Because of this, for a polyhedron K, 3(K) is
called the dual 1-skeleton of K.
A pattern on a connected surface M is a non-empty, connected locally finite graph 0
contained in M, such that each component of M\0 is simply connected and has compact
closure. The closure of a component of M\0 is called a face of 0. A pattern decomposes
the surface into faces. Such a decomposition is called a map (see [7], [15] and [16]).
A pattern 0 is called non-singular if each edge of 0 lies in two faces. A pattern 0 is
called equivelar of type {p, q} (or a {p, q}-pattern) if each face contains p (counted
with multiplicity) edges and each vertex has degree q.
Let 8(n) = {(p, q): there exists a {p, q}-equivelar polyhedron of Euler characteristic
n} and 6(n) denote the set of all equivelar polyhedra of Euler characteristics n. Clearly,
8(m) = ∅ for m 3. It is known (e.g., see [17]) that if (p, q) ∈ 8(n) for some n < 0,
then (p, q) ∈ 8(n) for infinitely many negative n. Here we prove:
Theorem 1. If 8(n) and 6(n) are as above, then
(i) 8(n) ∩ 8( m) = ∅ for all n 0 and m 1,
(ii) 8(2) = {(3, 3), (3, 4), (4, 3), (3, 5), (5, 3)},
(iii) 8(1) = {(3, 5), (5, 3)},
(iv) 8(0) = {(3, 6), (6, 3), (4, 4)},
(v) 8( 1) = ∅,
(vi) 8( 2) = {(3, 7), (7, 3), (4, 5), (5, 4)},
(vii) 8( 3) = {(3, 7), (7, 3), (3, 8), (8, 3), (4, 5), (5, 4), (5, 5)},
(viii) 8( 4) = {(3, 7), (7, 3), (3, 8), (8, 3), (4, 5), (5, 4), (4, 6), (6, 4), (5, 5)}
8( 2m), for all m 3,
(ix) (3, 3k 1) ∈ 8( k(3k 7)/2), (3, 3k) ∈ 8(1 k(3k 5)/2), for all k 3,

432
B. Datta and N. Nilakantan
(x) 8(n) is a finite set for each integer n and
(xi) for each n 7, there exists an n-vertex {3, 6}-equivelar polyhedron in 6(0).
Corollary 2. For each n = 0, there exist only finitely many equivelar polyhedra of
Euler characteristic n.
In [11] (also see in [7]) Edmonds et al. proved the existence and uniqueness of a
{p, q}-pattern on surfaces. Clearly, a {p, q}-equivelar polyhedron K gives a non-singular
{p, q}-pattern on |K| with the property that any two faces have at most one common
edge. So, the existence of a {p, q}-equivelar polyhedron implies the existence of a {p, q}-
pattern but not conversely. For example, by Theorem 2.4(i) of [11], a {3, 7}-pattern and
a {4, 6}-pattern exist on a non-orientable surface of Euler characteristic 1 but, from
Theorem 1(v) above, an equivelar polyhedron of Euler characteristic 1 does not exist.
Also, by the same theorem in [11], a {5, 6}-pattern exists on an orientable surface of Euler
characteristic 4 but, from Theorem 1(viii) above a {5, 6}-equivelar polyhedron of Euler
characteristic 4 does not exist. There are five choices of (p, q) for {p, q}-equivelar
polyhedra of Euler characteristic 2, where as there are infinitely many choices of (p, q)
for {p, q}-patterns (e.g., {p, 2}-patterns exist for all p 3) on the 2-sphere (see [11] and
[7]). However, in each case unique pattern exists (see Classification in [11]). Similarly
for Euler characteristic 1. From these (Classification in [11]) and Theorem 1(iii) and (iv)
above we get:
Corollary 3. If the Euler characteristic of an equivelar polyhedron is positive, then
the polyhedron is S42, C, O, I, D, RP62 or R defined in Example 1.
Corollary 2 says that 6(n) is a finite set for n = 0 and Theorem 1(xi) shows that
6(0) is an infinite set. If the Euler characteristic is 0, then (unlike when the Euler
characteristic is > 0) it is in general difficult to classify all the non-singular {p, q}-
patterns. In particular, it is very difficult to classify all the {p, q}-equivelar polyhedra of
a given non-positive Euler characteristic. Even for a negative Euler characteristic, there
can exist more than one {p, q}-equivelar polyhedra of the same Euler characteristic (e.g.,
N1, . . . , N14 in Examples 1 and 8). For simplicial polyhedra on few vertices we have:
Theorem 4. Let K be an n-vertex simplicial equivelar polyhedron of Euler character-
istic 0. If n 11, then K is isomorphic to T7, . . . , T11, A3,3, B3,3 or Q defined below.
Theorem 5. If M1 and M2 are as in Examples 1 and 7, then M1 M and any 9-vertex
=
2
neighbourly simplicial equivelar polyhedron is isomorphic to M1 or M2.
Theorem 6. Let N1, . . . , N14 be as in Examples 1 and 8. We have the following:
(i) Ni is not isomorphic to Nj for 1 i = j 14.
(ii) If M is a 10-vertex neighbourly simplicial equivelar polyhedron, then M is iso-
morphic to Ni for some i ∈ {1, . . . , 14}.

Equivelar Polyhedra with Few Vertices
433
Theorem 7. There are exactly 27 (up to isomorphism) simplicial equivelar polyhedra
on 11 vertices, namely, S42, O, RP62, A3,3, B3,3, T7, . . . , T11, M1, M2, N1, . . . , N14
and Q, defined in Examples 1–8.
Remark 1. Observe that M1 and M2 are non-isomorphic but they have the same 1-
skeleton. Similarly, N1, . . . , N14 have the same 1-skeleton but they are pairwise non-
isomorphic combinatorial 2-manifolds.
Remark 2. Corollary 3 is classically known. We have added it here as an immediate
consequence of Theorem 1. Corollary 2 is also known (e.g., see [25]). We have added it
for the sake of completeness.
Remark 3. In this article we consider polyhedra from a combinatorial point of view.
For some polyhedra (e.g., L2, L3, Q, G, Dn’s, Hn’s, . . .) we have given their geometric
realizations in Section 2. A polyhedron K is also called a polyhedral 2-manifold (see [4]).
Remark 4. Property (vii), in the definition of a polyhedron, implies that the geometric
carrier of a polyhedron is connected. A complex with properties (iv)–(vi) is said to be a
weak polyhedron. Similarly, we can define {p, q}-equivelar and equivelar weak polyhe-
dra. Clearly, an equivelar weak polyhedron is the disjoint union of equivelar polyhedra
whose Euler characteristic is the sum of the Euler characteristics of the components. So,
it is sufficient to consider only equivelar polyhedra.
2. Examples
In this section we construct infinitely many equivelar polyhedra. Some of them have
already been mentioned in the previous section. We use others in the next section. Recall
that we identify a polyhedron with the set of faces in it. At the end of this section we
give the geometric realizations of some of the polyhedra.
Example 2. Some equivelar polyhedra of Euler characteristic 0:
Am,n = {ui, j ui+1, j
Bm,n = {ui, j ui+1, j
u
u
i+1, j+1, ui, j ui, j+1ui+1, j+1: 1
i+1, j+1, ui, j ui, j+1ui+1, j+1: 1
i
i
m, 1
j
1, 1
j
n}, m, n 3.
n}
m, n 3.
m
j
∪ {um, j u1,n+2 j u1,n+1 j , um, j um, j+1u1,n+1 j : 1
m, 1
n},
Cm,n = {ui, j ui+1, j ui+1, j+1ui, j+1: 1
i
j
n},
m, n 3.
(Additions in the first and second subscripts are modulo m and n,
respectively.)
Tn = {ui ui+1
u
i+3, ui ui+2ui+3: 1
i
n},
n
7.
(Additions in the subscripts are modulo n.)
Q = {012, 023, 034, 045, 056, 016, 127, 136, 138, 178, 236, 269, 279,
348, 457, 479, 489, 569, 578, 589}.

434
B. Datta and N. Nilakantan
Am,n, Bm,n, Tn and Q are equivelar of type {3, 6} and Cm,n is equivelar of type {4, 4}.
The geometric carriers of Am,n, Cm,n and Tn are the torus and the geometric carriers of
Bm,n and Q are the Klein bottle.
Example 3. Some equivelar polyhedra of type {4, 5}:
Dn = [{ai, j,kai, j+1,kai, j+1,k+1ai, j,k+1: 1 j, k 4, 1
{ai, j,kai, j+1,kai, j+1,k+1ai, j,k+1: ( j, k)
∈ {(1, 1), (1, 3), (3, 1), (3, 3)}, 1
∪ {ai,3,kai,3,k+1ai+1,2,k+1 i+1,2,k, ai,3,k+1ai,4,k+1ai+1,1,k+1ai+1,2,k+1
ai,4,k+1ai,4,kai+1,1,k i+1,1,k+1, ai,4,kai,3,kai+1,2,kai+1,1,k
k ∈ {1, 3}, 1 n}, for 2.
En = [{ai, j,kai, j+1,kai, j+1,k+1ai, j,k+1: 1 j, k 4,
{ai, j,kai, j+1,kai, j+1,k+1ai, j,k+1: ( j, k)
∈ {(1, 1), (1, 3), (3, 2), (3, 4)}, 1
∪ {ai,3,kai,3,k+1ai+1,2,k i+1,2,k+3, ai,3,k+1ai,4,k+1ai+1,1,kai+1,2,k
ai,4,k+1ai,4,kai+1,1,k+3 i+1,1,k, ai,4,kai,3,kai+1,2,k+3ai+1,1,k+3
k ∈ {2, 4}, 1 n}. for 1.
i
n}\
i
n}]
a
,
a
:
i
n
1
i
n}\
i
n}]
a
,
a
:
i
n
(Additions in the first subscripts are modulo n and in the second and third
subscripts are modulo 4.)
F1 = {ai ci di+1di , ai+1ai di bi+1, ai bi bi+1
2}.
(Additions in the subscripts are modulo 2.)
Fn = [{ai, j ai, j+1ai+1, j+1ai+1, j : 1 4, 1
{a2k 1, j a2k 1, j+1a2k, j+1a2k, j : j ∈ {1, 3}, 1
∪ {a2k 1,1a2k 1,2a2k+2,4 2k+2,3, a2k 1,2a2k,2a2k+1,4a2k+2,4
a2k,2a2k,1a2k+1,3 2k+1,4, a2k,1a2k 1,1a2k+2,3a2k+1,3
n},
Gn = [{ai, j ai, j+1ai+1, j+1ai+1, j : 1
ci+1, bi di ci+1ci , a1c1a2c2, b1d1b2d2:
1
i
j
i
2n}\
n}]
k
a
,
,
a
:
1
k
for
n
2.
j
4, 1
i
2n}\
n}]
2k+1,3, a2k 1,2a2k,2a2k+2,4a2k+1,4
2k+2,4, a2k,1a2k 1,1a2k+1,3a2k+2,3
for 2.
{a2k 1, j a2k 1, j+1a2k, j+1a2k, j : j ∈ {1, 3}, 1
∪ {a2k 1,1a2k 1,2a2k+1,4
a2k,2a2k,1a2k+2,3
n},
k
a
a
:
1
k
n
(Additions in the first and second subscripts are modulo 2n and 4,
respectively.)
G = {abcd, adef, bagh, cbij, dckl, feij, hgkl, aljg, alhf, bkeh, bkgi, ciek, cigj, dfhe,
dfjl}.

Equivelar Polyhedra with Few Vertices
435
(Dn) = 16n 40n + 20n = 4n and the geometric carrier of Dn is the orientable
surface of genus 2n + 1 for all n 2. (En) = 4n and the geometric carrier of En is
the orientable surface of genus 2n + 1 for all n 1. (Fn) = 8n 20n + 10n = 2n
and the geometric carrier of Fn is the orientable surface of genus n + 1 for all n
1.
(Gn) = 2n and the geometric carrier of Gn is a non-orientable surface for all n 2.
The geometric carrier of G is the non-orientable surface of Euler characteristic 3.
Example 4. Some equivelar polyhedra of type {3, 7}:
Hn = {ui,1vi,1ui,2, ui,2vi,1vi,2, ui,2vi,2vi,3, ui,2vi,3ui,3, ui,4vi,4ui,1, ui,1vi,4vi,1,
vi,2wi,2wi,3, vi,2wi,3vi,3, vi,3wi,3vi,4, vi,4wi,3wi,4, vi,4wi,4vi,1, vi,1wi,4wi,1,
wi,1ui,1wi,2, wi,2ui,1
ui,2, wi,2ui,2wi,3, wi,3ui,2ui,3, wi,3ui,3wi,4, wi,4ui,3ui,4,
wi,4ui,4
u
i,1, wi,4ui,1wi,1}
∪ {ui+1,3vi+1,3wi,2, ui+1,3wi,2vi,2, ui+1,3
u
i+1,4vi,2, ui+1,4vi,2vi,1,
i+1,4vi+1,4wi,1, ui+1,4vi,1wi,1, vi+1,4vi+1,3wi,1, vi+1,3wi,1wi,2:
n}, for 1.
u
1
i
n
(Additions in the subscripts are modulo n.)
H = {a1b4b1, a1b1b3, a1b3a2, a1a2c2, a1c2c1, a2b3b2, a2b2c5, a2c5c4, a2c4a3, a2a3c2,
a3c4b5, a3b5b6, a3b6a4, a3a4c3, a3c3c2, a4b6b4, a4b4b1, a4b1a5, a4a5c1, a4c1c3,
a5b1b2, a5b2c5, a5c5c6, a5c6a6, a5a6c1, a6c6c4, a6c4b5, a6b5b4, a6b4a1, a6a1c1,
b1b2b6, b1b3b6, b2b3b5, b2b5b6, b3b4b5, b3b4b6, c1c2c6, c1c3c6, c2c3c5, c2c5c6,
c3c4c5, c3c4c6}.
The geometric carrier of H is the non-orientable surface of Euler characteristic 3. The
geometric carrier of Hn is an orientable surface and (Hn) = 12n 42n + 28n = 2n
for all n 1.
Example 5. Some sequences of equivelar polyhedra of type {3, 8}:
Jn = {ui, j ui+1, j
u
i+1, j+1, ui, j ui, j+1ui+1, j+1: 1
i
2n, 1
j
3}\
{u2k 1, j u2k 1, j+1u2k, j+1: 1
k
n, j ∈ {1, 3}}
∪ {u2k 1,1u2k+1,3u2k 1,2, u2k 1,2u2k+1,3
u2k,2u2k+2,3
n},
u
2k+2,3, u2k 1,2u2k+2,3u2k,2
,
u
2k+2,1, u2k,2u2k+2,1u2k 1,1, u2k 1,1u2k+2,1u2k+1,3
for 3.
:
1
k
n
(Additions in the first and second subscripts are modulo 2n and 3,
respectively.)
Kn = [{ui, j,kui, j+1,k
ui,l,k+1ui,l+1,kui,l+1,k+1: 1
i,1,2, ui,1,3ui,2,3
u
i, j+1,k+1, ui, j,kui, j,k+1
2, 3
i,2,1, ui,3,2ui,4,1u
u
i, j+1,k+1, ui,l,kvi,l+1,kui,l,k+1
4, 1 3, 1
i,4,2, ui,3,3ui,4,3ui,3,1}]
i,1,2, ui,1,2ui,3,3 i,3,1, ui,1,2ui,3,1ui,2,2
i,1,1, ui,4,1ui+1,2,1ui+1,2,3
,
j
l
k
i
n}\
{ui,1,1ui,2,2
u
u
∪ {ui,1,1ui,4,3
u
u
i,3,3, ui,1,1ui,3,3
i,4,3, ui,2,2ui,4,3
u
u
u
,
ui,2,2ui,3,1
,

436
B. Datta and N. Nilakantan
ui,4,1ui+1,2,3
ui,4,2ui+1,1,3
u
u
i,3,2, ui,3,2ui+1,2,3
i+1,2,1, ui,4,2ui+1,2,1ui,4,1: 1
u
i+1,1,3, ui,3,2ui+1,1,3ui,4,2
n},
,
i
for
n
1.
(Additions in the first and second subscripts are modulo n and 3,
respectively.)
(Jn) = 6n 24n + 16n = 2n and the geometric carrier of Jn is the orientable
surface of genus n + 1 for all n
3. (Kn) = 12n 48n + 32n = 4n and the
geometric carrier of Kn is a non-orientable surface for all n 1.
Example 6. Some sequences of equivelar polyhedra of type {4, 6}:
L2 = {1263, 1374, 1425, 1536, 1647, 1752, 8273, 8346, 8657, 8724, 8435, 8562}.
Ln = {ai,1ai+1,1ai+1,2ai,2, ai,2ai+1,2bi+1,3bi,3, bi,3bi+1,3bi+1,4bi,4, bi,4ai+1,4ai+1,1ai,1:
1
i
2n}\{a2k 1,1a2k,1a2k,2a2k 1,2, b2k 1,3b2k,3b2k,4b2k 1,4: 1
k
n},
where
a2k,1 = b2k+1,3
for 3.
a2k
= b2k+1,4
,
a2k
= b2k+2,4
,
a2k,2 = b2k+2,3
,
1,1
1,2
,
n
(Additions in the subscripts are modulo 2n.)
Pn = {ai, j,kai, j+1,kai, j+1,k+1ai, j,k+1: 1 j, k 4, 1
{ai, j,kai, j+1,kai, j+1,k+1ai, j,k+1: ( j, k)
i
n}\
∈ {(1, 1), (1, 3), (3, 1), (3, 3)}, 1
where
ai, j,k = ai+1, j+2,k
for ( j, k) ∈ {(1, 1), (1, 2), (2, 1), (2, 2), (3, 3), (3, 4), (4, 3), (4, 4)}
and for 3.
i
n},
,
1
i
n
n
(Additions in the first subscripts are modulo n and in the second and third
subscripts are modulo 4.)
(Ln) = 4n 12n + 6n = 2n for all n
2. The geometric carrier of L2 is the
orientable surface of genus 3 and the geometric carrier of Ln is a non-orientable surface
for all n 3. (Pn) = 8n 24n + 12n = 4n and the geometric carrier of Pn is the
orientable surface of genus 2n + 1 for all n 3. (Note that Ln, for n 3, is obtained
from an 8n-vertex ({ai, j , bi,k: 1
i
2n, 1
j
2, 3
k
4}) complex (torus with
2n holes) by identifying bl,k’s with ai, j ’s. Pn is also obtained by some identifications.)
Example 7. Another 9-vertex neighbourly simplicial equivelar polyhedra:
M2 = {129, 239, . . . , 789, 189, 124, 136, 138, 147, 156, 157,
237, 245, 258, 267, 268, 346, 357, 358, 468, 478},
where the vertex set of M2 is {0, 1, . . . , 8}. Clearly, (M2) = 3.
Example 8. Thirteen more 10-vertex neighbourly simplicial equivelar polyhedra:
N2 = A ∪ {134, 136, 156, 158, 178, 179, 237, 248, 257, 259, 268,
269, 358, 359, 368, 379, 457, 467, 469, 489},

Equivelar Polyhedra with Few Vertices
437
N3 = A ∪ {137, 139, 146, 156, 158, 178, 237, 248, 257, 259, 268,
269, 346, 358, 359, 368, 457, 479, 489, 679},
N4 = A ∪ {136, 137, 145, 158, 168, 179, 238, 249, 256, 257, 269,
278, 347, 358, 359, 369, 467, 468, 489, 579},
N5 = A ∪ {134, 136, 157, 158, 168, 179, 237, 248, 256, 259, 268,
279, 358, 359, 369, 378, 457, 467, 469, 489},
N6 = A ∪ {134, 136, 157, 158, 168, 179, 238, 247, 256, 259, 268,
279, 357, 359, 369, 378, 458, 467, 469, 489},
N7 = A ∪ {134, 138, 156, 157, 168, 179, 236, 247, 258, 259, 269,
278, 357, 359, 367, 389, 458, 468, 469, 479},
N8 = A ∪ {134, 138, 156, 157, 168, 179, 236, 247, 257, 258, 269,
289, 358, 359, 367, 379, 459, 468, 469, 478},
N9 = A ∪ {138, 139, 146, 157, 158, 167, 236, 245, 258, 269, 278,
279, 347, 357, 359, 368, 468, 479, 489, 569},
N10 = A ∪ {136, 138, 145, 158, 167, 179, 238, 249, 256, 257, 268,
279, 347, 357, 359, 369, 468, 469, 478, 589},
N11 = A ∪ {134, 138, 156, 157, 168, 179, 237, 246, 257, 259, 268,
289, 358, 359, 367, 369, 458, 469, 478, 479},
N12 = A ∪ {134, 138, 156, 157, 168, 179, 236, 247, 258, 259, 268,
279, 357, 359, 369, 378, 458, 467, 469, 489},
N13 = A ∪ {136, 137, 145, 158, 168, 179, 238, 246, 257, 259, 267,
289, 349, 357, 358, 369, 468, 478, 479, 569},
N14 = A ∪ {134, 138, 156, 157, 168, 179, 237, 246, 257, 258, 269,
289, 358, 359, 367, 369, 459, 468, 478, 479},
where the vertex set of Ni (2
i
14) is {0, 1, . . . , 9} and A = {012, . . . , 089, 019,
14. Thus, all of them triangulate the same
124}. Clearly, (Ni ) = 5 for 1
i
non-orientable surface of Euler characteristic 5.
Ringel and Jungerman [13], [21]–[23], [14] have shown that there exist neighbourly
simplicial polyhedra on 3k and 3k + 1 vertices, for each k 3, i.e.,
Proposition 1. For k 2, if n = 3k or 3k + 1, then there exists an n-vertex equivelar
polyhedron of type {3, n 1}.
Thus, if m = k(3k 7)/2 or 1 k(3k 5)/2, for k
3, then there exists an
equivelar polyhedron of type {3, f0 1} of Euler characteristic m. In particular, there
exist neighbourly equivelar polyhedra on nine and ten vertices (M1 and N1, respectively,
in Example 1).

438
B. Datta and N. Nilakantan

Equivelar Polyhedra with Few Vertices
439

440
B. Datta and N. Nilakantan

Equivelar Polyhedra with Few Vertices
441
3. Proofs
Proof of Theorem 1. First observe that if K is an equivelar polyhedron of type {p, q},
e
then K is an equivelar polyhedron of type {q, p}. Thus, (p, q) ∈ 8(m) implies (q, p) ∈
8(m).
If K is a {p, q}-equivelar polyhedron with f0 vertices, f1 edges and f2 faces, then
q f0 = 2 f1 = pf2.
(1)
This gives
1
1
1
2
(K)
2 f1
(K)
+
=
=
.
(2)
p
q
q f0
1
2
Thus, (p, q) ∈ 8(n), for n 0, implies 1/p + 1/q
0 and (p, q) ∈ 8( m),
1
2
for m > 0, implies 1/p + 1/q
< 0. This implies (i).
1
If there esists a {p, q}-equivelar polyhedron K with (K) > 0, then 1/p + 1/q >
or (p 2)(q 2) < 4. This implies (p, q) ∈ {(3, 3), (3, 4), (4, 3), (3, 5), (5, 3)}. Thu2s
8(1), 8(2) {(3, 3), (3, 4), (4, 3), (3, 5), (5, 3)}.
(3)
From Example 1 we have (3, 3), (3, 4), (4, 3), (3, 5), (5, 3) ∈ 8(2). This proves (ii).
Let K be an equivelar polyhedron of type {p, q} and (K) = 1. If (p, q) = (3, 3),
then, from (2), we get f1 = 3, which is not possible. If (p, q) = (3, 4), then, from (2),
f0 = 3, which is again not possible. Similarly, (p, q) = (4, 3). Therefore, from (3),
(p, q) = (3, 5) or (5, 3). This and Example 1 (RP62 and R) imply (iii).
If there exists a {p, q}-equivelar polyhedron K with (K) = 0, then, by (2), 1/p +
1/q = 12 . Since p, q 3, (p, q) = (3, 6), (6, 3) or (4, 4). This together with Example
2 proves (iv).
Let K be an equivelar polyhedron of type {p, q} and (K) = 1. Since, the existence
ofanequivelarpolyhedronoftype{p, q}impliestheexistenceofanequivelarpolyhedron
of type {q, p}, therefore we may assume q p. Also p, q 3 and f0 p + 1.
If p = 3, then from (1) we get 1 = f0 f1/3 = f0 q f0/6 or (q 6) f0 = 6. This
implies f0 = 6andq = 7, acontradiction. If p = 4, then 1 = f0 f1/2 = f0 q f0/4or
(q 4) f0 = 4, which is not possible. If p 5, then f0 6 and hence f1 (6 3)/2 = 9.
1
2
1
1
5
1
18
Then, from (2), 1/q =
1/p 1/(2 f1) >
>
15 . This implies q < 5, a
contradiction to the assumption that p q. Thi2s proves (v).
e e
Observe that F1, H1, F1, H1 ∈ 6( 2) and F1 and H1 are equivelar of type {4, 5} and
{3, 7}, respectively. Therefore, 8( 2) {(5, 4), (4, 5), (7, 3), (3, 7)}.
Let K be an equivelar polyhedron of type {p, q} and (K) = 2. In this case also
we may assume that 3
p
q
f0 1.
1
2
If p
5, then f0
q + 1
6. Then, from (2), = 1/p + 1/q + 2/(q f0)
1
5
1
2
14
+
+
=
, a contradiction.
If5p = 4, then 2 = f0
f1/2 = f0 q f0/4 or (q 4) f0 = 8, which implies
5
6
30
( f0, q) = (8, 5).
If p = 3, then from (1) we get 2 = f0
f1/3 = f0 q f0/6 or (q 6) f0 = 12.
This implies ( f0, q) = (12, 7). These imply 8( 2)
{(5, 4), (4, 5), (7, 3), (3, 7)}.
This proves (vi).

442
B. Datta and N. Nilakantan
Observe that H, M1, G ∈ 6( 3) and H, M1, G are equivelar of type (3, 7), (3, 8)
and (4, 5), respectively. Thus, {(3, 7), (7, 3), (3, 8), (8, 3), (4, 5), (5, 4)} 8( 3).
Let K be an equivelar polyhedron of type {p, q} and (K) = 3. Assume p
q
f0 1.
If p = 3, then, by (1), (q 6) f0 = 18. This implies ( f0, q) = (9, 8) or (18, 7).
If p = 4, then, by (1), (q 4) f0 = 12, which implies ( f0, q) = (12, 5).
If p = q = 5, then f0 = 6. Now, S1 = {12345, 15264, 14536, 16423, 13652, 24356}
is a {5, 5}-equivelar polyhedron of Euler characteristic 3. So, (5, 5) ∈ 8( 3).
1
2
1
5
1
6
3
92
210
Finally, assume that p
contradiction. This proves (vii).
5 and q
6. Then, by (2),
+
+
=
, a
6
7
Hm (in Example 4) is an equivelar polyhedron of type {3, 7} and of Euler characteristic
2m for all m 1, therefore (3, 7) (and hence (7, 3)) ∈ 8( 2m) for all m 1.
K1 (in Example 5) is an equivelar polyhedron of type {3, 8} of Euler characteristic
4
and Jm (in Example 5) is an equivelar polyhedron of type {3, 8} and of Euler characteristic
2m for all m 3. Therefore (3, 8) (and hence (8, 3)) ∈ 8( 2m) for all m 2.
Fm (in Example 3) is an equivelar polyhedron of type {4, 5} of Euler characteristic
2m for all m 2. Therefore (4, 5) (and hence (5, 4)) ∈ 8( 2m) for all m 2.
Lm (in Example 6) is an equivelar polyhedron of type {4, 6} of Euler characteristic
2m for all m 2. Therefore (4, 6) (and hence (6, 4)) ∈ 8( 2m) for all m 2. Thus,
{(3, 7), (7, 3), (3, 8), (8, 3), (4, 5), (5, 4), (4, 6), (6, 4)}
8( 2m),
for all
m
2.
(4)
Let K be an f0-vertex equivelar polyhedron of type {p, q} and (K) = 4. Assume
q, i.e., 3 f0 1.
p
p
q
1
If p 5 and q 6, then f0 q + 1 7 and hence = 1/p + 1/q + 4/(q f0)
2
1
5
1
4
97
210
+
+
=
, a contradiction.
If6p = q = 5, then, by (2), ( f0, f1) = (8, 20) and hence f2 = 8. For each m
2,
6
7
S2m = {ai ai+1bi+m+1bi+mbi+m 1, ai ai+1bi+1ai+m+1bi : 1
i
2m, additions in the
subscripts are modulo 2m} is {5, 5}-equivelar and belongs to 6( 2m). So, (5, 5) ∈
8( 2m) for all m 2.
If p = 3, then, from (1), 4 = f0
f1/3 = f0 q f0/6 or (q 6) f0 = 24. This
implies ( f0, q) = (12, 8) or (24, 7) and hence (p, q) = (3, 8) or (3, 7).
If p = 4, then, from (1), 4 = f0 q f0/4 or (q 4) f0 = 16. This implies ( f0, q) =
(16, 5) or (8, 6) and hence (p, q) = (4, 5) or (4, 6). Thus, (p, q) ∈ 8( 4) and p
q
imply (p, q) = (3, 8), (3, 7), (4, 5), (5, 5) or (4, 6). So, 8( 4) {(3, 7), (7, 3), (3, 8),
(8, 3), (4, 5), (5, 4), (4, 6), (6, 4), (5, 5)}. This, the examples S2m’s and (4) imply (viii).
(ix) follows from the proposition stated in the previous section.
Let K be an equivelar polyhedron of type {p, q} and (K) = m, where m > 0. In
this case also, we may first assume that p q, i.e., 3
p
q
f0 1. Then, by (2),
1
1/p = 1/q + m/(q f0) 1/q + m/(q(q + 1)). This gives
2
2p(q + 1 + m)
q(q + 1)(p 2) p(q + 1)(p 2) or
(q + 1)(p 4).
(5)
(6)
2m
Clearly, if p > 4, then there are only finitely many q such that (p, q) satisfies (6). So,
there are only finitely many (p, q) ∈ 8( m) such that p, q > 4.

Equivelar Polyhedra with Few Vertices
443
If p = 3, then from (5) we get 6(q + 1 + m) 3q(q + 1) or
6m (q 6)(q + 1).
(7)
Clearly, given m > 0, only finitely many q ( 3) satisfy (7). This shows that there exists
only finitely many q such that (3, q) ∈ 8( m).
If q p = 4, then from (5) we get 4(q + 1 + m) q(q + 1) or
4m (q 4)(q + 1).
(8)
Clearly, given m > 0, only finitely many q ( 4) satisfy (8). This shows that there exists
only finitely many q
4 such that (4, q) ∈ 8( m). Therefore there are only finitely
many (p, q) ∈ 8( m) such that p or q is 4. So, 8( m) is finite for all m > 0. This
together with (ii), (iii) and (iv) imply (x).
For each n 7, Tn is an n-vertex {3, 6}-equivelar polyhedron. This proves (xi).
Proof of Corollary 2. Let (p, q) ∈ 8(n). Let K be an equivelar polyhedron of type
{p, q} and (K) = n. If n = 0, then, by (2) and (1), ( f0(K), f1(K), f2(K)) is uniquely
determined by (p, q). Since, for each f0, there exist finitely many polyhedra on f0
vertices, therefore for a given (p, q) ∈ 8(n) there are only finitely many equivelar
polyhedra of type {p, q}. The corollary now follows from Theorem 1(x).
Lemma 1. If K is a 9-vertex {3, 6}-equivelar polyhedron, then K is isomorphic to T9,
A3,3 or B3,3 defined in Example 2.
Proof. Let K bea9-vertex{3, 6}-equivelarpolyhedron. Then f2(K) = 18andNEG(K)
is a 2-regular graph and hence is either a cycle or disjoint union of cycles on nine vertices.
So, NEG(K) is isomorphic to C9, C6 t C3, C5 t C4 or 3C3 := C3 t C3 t C3. (Here
Cn denotes the cycle with n vertices. A cycle with edges v1v2, . . . , vn 1vn, vnv1 is also
denoted by Cn(v1, . . . , vn).)
If NEG(K) = C5 tC4, then there exist four vertices, a, b, c, d say, such that ac and bd
are edges in K but ab, bc, cd and da are not edges in K. Consider the following six sets of
faces, Sac = { : a, c ∈ }, Sbd = { : b, d ∈ }, Sa = { : a ∈ , c ∈ }, Sb = { : b ∈
, d ∈ }, Sc = { : c ∈ , a ∈ } and Sd = { : d ∈ , b ∈ }. Clearly, these six sets
are pairwise disjoint and #(Sac) = #(Sbd) = 2 and #(Sa) = #(Sb) = #(Sc) = #(Sd) = 4.
This implies that f2(K) 20, a contradiction. So, NEG(K) is C9, C6 t C3 or 3C3.
First consider the case when the non-edge graph consists of three 3-cycles, i.e.,
NEG(K) = C (1, 2, 3) t C (4, 5, 6) t C (7, 8, 9)
NEG(A3,3). Then, up to an iso-
=
3
3
3
morphism, LkK (1) = C6(4, 7, 5, 9, 6, 8) and hence we may assume, without loss,
LkK (4) = C6(1, 7, 2, 9, 3, 8). These imply LkK (7) = C6(1, 4, 2, 6, 3, 5), LkK (8) =
C6(1, 6, 2, 5, 3, 4). These imply LkK (2) = C6(4, 9, 5, 8, 6, 7), LkK (3) = C6(4, 8, 5, 7,
6, 9), LkK (5) = C6(1, 9, 2, 8, 3, 7), LkK (6) = C6(1, 8, 2, 7, 3, 9) and hence LkK (9) =
C6(1, 5, 2, 4, 3, 6). Clearly, these imply that K is unique, up to an isomorphism, in this
case and hence is isomorphic to A3,3
.
Now assume NEG(K) = C (1, . . . , 6) t C (7, 8, 9)
NEG(B3,3). Then, we may
assume, LkK (1) = C6(3, 8, 4, 9, 5, 7). This implies LkK (7) = C6(1, 5, 2, 4, 6, 3). Then
=
6
3
LkK (3) = C6(1, 8, 5, 9, 6, 7) and LkK (5) = C6(1, 7, 2, 8, 3, 9). These imply LkK (8) =

444
B. Datta and N. Nilakantan
C6(1, 3, 5, 2, 6, 4), LkK (9) = C6(1, 4, 2, 6, 3, 5), LkK (2) = C6(4, 7, 5, 8, 6, 9), LkK (6)
= C6(2, 9, 3, 7, 4, 8). So, LkK (4) = C6(1, 9, 2, 7, 6, 8). These determine the polyhe-
dron uniquely and therefore K is isomorphic to B3,3
.
Finally assume NEG(K) = C9(1, . . . , 9). Then LkK (1) = C6(4, 6, 8, 3, 5, 7), C6(4,
6, 8, 5, 3, 7), C6(4, 6, 3, 8, 5, 7), C6(4, 7, 5, 3, 6, 8) or C6(4, 6, 3, 7, 5, 8).
If LkK (1) = C6(4, 6, 8, 3, 5, 7), then LkK (3) = C6(1, 5, 7, 9, 6, 8). These imply
LkK (5) = C3(1, 3, 7), a contradiction. If LkK (1) = C6(4, 6, 3, 8, 5, 7), then LkK (3) =
C6(1, 6, 9, 7, 5, 8). These give LkK (5) = C4(1, 7, 3, 8), a contradiction. If LkK (1) =
C6(4, 7, 5, 3, 6, 8), then 15, 16 are edges and 67, 78, 89 are non-edges in LkK (3). It is
then not possible to construct LkK (3), which is a 6-cycle with vertex-set {1, 5, 6, 7, 8, 9}.
Similarly, it is not possible to construct LkK (8) when LkK (1) = C6(4, 6, 8, 5, 3, 7).
Thus, LkK (1) = C6(4, 6, 3, 7, 5, 8).
Similarly, replacing 1 by i, LkK (i) = C6(i + 3, i + 5, i + 2, i + 6, i + 4, i +
7), for 1
i
9 (additions in the subscripts are modulo 9). This shows that the
simplicial polyhedron is unique, up to an isomorphism, with the non-edge graph a 9-
cycle. Therefore, K is isomorphic to T9. This completes the proof.
Lemma 2. If K is a 10-vertex {3, 6}-equivelar polyhedron, then K is isomorphic to
T10 or Q defined in Example 2.
Proof. Let K be a 10-vertex {3, 6}-equivelar polyhedron. Choose a vertex, say u0. Let
the link of u0 be C6(u1, . . . , u6). Since the link of each vertex is a 6-cycle, ui ui+1ui+2
is not a face in K for i = 1, . . . , 6 (additions in the subscripts are modulo 6). If either
ui ui+1ui+3 or ui ui+1ui+4 is a face for each i, then we get 12 faces and hence the number
of faces through the remaining three vertices is 8, a contradiction. So, assume u1u2u7
is a face, where u7 is one of the remaining three vertices. Then u1u6u7 and u2u3u7 cannot
be faces.
Case 1: u1u3u6 or u2u3u6 is a face. Assume, without loss, that u1u3u6 is a face.
If u5 is in the link of u1, then u1u3u5, u1u5u7 are faces and hence V (Lk(u3)) =
{u0, u1, u2, u4, u5, u6}. This gives 12 faces through u0, u1 and u3. Hence, the number
of faces through the remaining two vertices is 8, a contradiction. If u4 is in the link
of u1, then u1u3u4 and u1u4u7 are faces. Here, five vertices of each of Lk(u3) and
Lk(u4) are known. This implies V (Lk(u8)) = {u2, u3, u5, u6, u7, u9} and V (Lk(u9)) =
{u2, u4, u5, u6, u7, u8}, where u8 and u9 are the remaining two vertices. Therefore,
Lk(u4) = C6(u7, u1, u3, u0, u5, u9) and Lk(u3) = C6(u2, u0, u4, u1, u6, u8). Then
Lk(u2) = C6(u7, u1, u0, u3, u8, u9). However, now Lk(u7) contains C4(u1, u4, u9, u2),
which is impossible. Therefore, the sixth vertex in the link of u1 is one of the remaining
two vertices, say u8.
Clearly, u1u3u8 and u1u7u8 are faces. Thus, V (Lk(u3)) = {u0, u1, u2, u4, u6, u8}.
Let u9 be the remaining vertex. Then V (Lk(u9)) = {u2, u4, u5, u6, u7, u8}. The link
of u2 shows that u2u7u9 is a face. Now, from the link of u3, either u2u3u8 and u3u4u6
are faces or u2u3u6 and u3u4u8 are faces. In either case, u2u5 is not an edge and hence
V (Lk(u5)) = {u0, u4, u6, u7, u8, u9}.
Subcase 1.1: u2u3u6 and u3u4u8 are faces. Then, by considering Lk(u2), Lk(u6) and
Lk(u5) successively, u2u6u9, u5u6u9 are faces and u4u5u9 is not a face. Finally, by

Equivelar Polyhedra with Few Vertices
445
completing Lk(u9) and Lk(u5) successively, the other faces are u4u7u9, u4u8u9, u5u8u9,
u4u5u7 and u5u7u8. In this case K is isomorphic, via the map f : ui → i, to Q.
Subcase 1.2: u2u3u8 and u3u4u6 are faces. Then, by considering Lk(u2), Lk(u8), Lk(u5)
and Lk(u4) successively, the other faces are u2u8u9, u5u7u8, u5u8u9, u4u5u7, u5u6u9,
u4u6u9, u4u7u9. So, K isisomorphic, viathecompositionof f and(1, 2, 3)(4, 8)(5, 7)(6,
9), to T10.
Case 2: u1u4u6 or u2u3u5 is a face. Assume, without loss, that u1u4u6 is a face. Let the
remaining two vertices be u8 and u9.
If either u1u8 or u1u9 is an edge, say u1u8 is an edge, then, by considering Lk(u1) and
Lk(u4), successively, u1u4u8, u1u7u8, u3u4u6 andu4u5u8 arefaces. Clearly, V (Lk(u9)) =
{u2, u3, u5, u6, u7, u8}. By considering Lk(u6) and Lk(u5), u3u6u9 and u5u6u9 are faces
and u2u3u5 is not a face. Then, from Lk(u3), u2u3u8 and u3u8u9 are faces. This gives
deg(u8) > 6, a contradiction. So, the sixth vertex in the link of u1 is either u3 or u5.
If neither u8 nor u9 is in the link of u4, then from the links of u0, u1 and u4 we get
12 faces. Therefore, the number of faces containing u8 or u9 is 8, a contradiction. So,
Lk(u4) contains one of u8 or u9, say u8. Then V (Lk(u9)) = {u2, u3, u5, u6, u7, u8}.
Subcase 2.1: u3 is in the link of u1. The links of u1 and u4 show that u1u3u4, u1u3u7,
u4u5u8 and u4u6u8 are faces. It is easy to see, by considering the link of u3, that u2u3u9
and u3u7u9 are faces. However, this implies V (Lk(u8)) = {u2, u4, u5, u6, u7, u9}. Then,
by completing Lk(u2), Lk(u6) and Lk(u5) successively, the other faces are u2u7u8,
u2u8u9, u5u6u9, u6u8u9, u5u7u8 and u5u7u9. In this case K is isomorphic, via the
composition of f and (1, 9, 5, 3, 8, 4)(2, 7, 6), to T10.
Subcase 2.2: u5 is in the link of u1. Clearly, u1u4u5 and u1u5u7 are faces. In this case,
by considering Lk(u4), Lk(u5), Lk(u6), Lk(u7) and Lk(u2), u3u4u8, u4u6u8, u5u6u9,
u5u7u9, u6u8u9, u2u7u8, u3u7u8, u3u7u9, u2u3u9 and u2u8u9 are faces. Then K is
isomorphic, via the map (u1, u6)(u2, u5)(u3, u4)(u7, u9), to Q.
Case 3: None of u1u3u6, u1u4u6, u2u3u5 or u2u3u6 is a face. In this case we can assume
that u1u6u8 is a face, where u8 is one of the remaining two vertices. Let u9 be the
remaining vertex.
If u3 ∈ Lk(u1), then, by considering the links of u1 and u3, u1u3u7, u1u3u8, u2u3u8
and u3u4u7 are faces in K. Clearly, V (Lk(u9)) = {u2, u4, u5, u6, u7, u8}. The links of
u2 and u8 show that deg(u7) > 6, a contradiction. Similarly, we get a contradiction if
u5 ∈ Lk(u1).
If u4 ∈ Lk(u1), then u1u4u7 and u1u4u8 are faces in K. Here, V (Lk(u9)) =
{u2, u3, u5, u6, u7, u8}. To complete Lk(u4), either u3u4u7 or u3u4u8 has to be a face.
In both cases we see that u6u8u9 ∈ K. In the first case, the links of u4 and u5 show that
u4u5u8 and u5u8u9 are faces, which imply that deg(u8) < 6, a contradiction. In the sec-
ond case, Lk(u4), Lk(u5) and Lk(u7) show that u4u5u7, u5u7u9, u2u7u8 and u7u8u9 are
faces. This implies deg(u8) > 6, a contradiction. Hence, the sixth vertex in Lk(u1) is u9.
Clearly, the edge u2u3 belongs to either u2u3u8 or u2u3u9.
If u2u3u8 is a face, the sixth vertex, y, in Lk(u2), is one of u4, u5 or u6. If y = u4
or u5, the links of u2 and y show that deg(u8) > 6, a contradiction. If y = u6, by
considering the links of u2 and u6, we see that u2u6u7, u2u6u8 and u5u6u7 are faces.

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B. Datta and N. Nilakantan
Clearly, u3u4u9 ∈ K. The sixth vertex in Lk(u7) has to be u3. However, this implies that
deg(u3) > 6, a contradiction. Hence, u2u3u9 ∈ K.
The sixth vertex, say x, in Lk(u2), is one of u4, u5 or u8.
If x = u4 or u5, by considering the links of x and u9, we observe that deg(u9) > 6, a
contradiction. Hence, x = u8.
The links of u2, u8, u4, u7 and u3 show that u2u7u8, u2u8u9, u4u6u8, u4u7u8, u3u4u6,
u4u5u7, u5u7u9, u3u5u6 and u3u5u9 are faces. Here, K is isomorphic, via the map
(1, 4)(2, 5) (3, 6), to Q. This completes the proof.
Lemma 3. If K is an 11-vertex {3, 6}-equivelar polyhedron, then K is isomorphic to
T11 defined in Example 2.
Proof. Let K be an 11-vertex {3, 6}-equivelar polyhedron. Choose a vertex, say u0,
and let the link of u0 be C6(u1, . . . , u6). By an argument similar to that in the previous
lemma, we assume that u1u2u7 is a face.
Claim. One of u1u3u6, u1u4u6, u2u3u5 or u2u3u6 has to be a face.
If not assume u1u6u8 is a face, where u8 is one of the remaining three vertices. Let
u9 and u10 be the remaining two vertices.
If u3 ∈ Lk(u1), then the links of u1 and u3 show that u1u3u7, u1u3u8, u2u3u8 and
u3u4u7 are faces in K. It is clear that Lk(u2) contains either u9 or u10, say u9. Then
V (Lk(u10)) = {u4, u5, u6, u7, u8, u9}. Thelinksofu8 andu7 nowshowthatdeg(u9) < 6,
a contradiction.
If u4 ∈ Lk(u1), then it is clear that the edge u5u6 belongs to u5u6u7 or u5u6u9 (or
u5u6u10). If u5u6u7 ∈ K, then deg(u9), deg(u10) < 6. If u5u6u9 (or u5u6u10) is a face,
then, by considering Lk(u4), either u3u4u7 or u3u4u8 ∈ K. In both cases we see that the
link of u5 has only five vertices, a contradiction.
Finally, assume (without loss) that u9 ∈ Lk(u1). The edge u2u3 can belong to one of
u2u3u8, u2u3u9 or u2u3u10.
If u2u3u8 ∈ K, then the sixth vertex, say y, in Lk(u2) is u4, u5, u6 or u10. If y = u4 or
u5, then V (Lk(u10)) = {u3, u4, u5, u6, u7, u9}\{y} and hence deg(u10) < 6. If y = u6,
we can easily see that V (Lk(u10)) = {u3, u4, u5, u7, u8, u9}. Then, by completing the
links of u6, u7 and u8, we see that deg(u9) = 4, a contradiction. If y = u10, then we see
that either u3u6u8 or u6u8u10 has to be a face to complete Lk(u8). In both the cases we
see that there exists no vertex x (= u0) such that u3u4x is a face in K, a contradiction.
If u2u3u9 is a face in K, then, by an argument similar to the one above, we see that
u4, u5 ∈/ Lk(u2). Hence either u8 or u10 ∈ Lk(u2). If u8 ∈ Lk(u2), it is easy to see that
u10 ∈ Lk(u8). The last vertex in Lk(u9) has to be u5. Completing the link of u7, we
observe that deg(u5) > 6, a contradiction. If u10 ∈ Lk(u2), then u2u7u10 and u2u9u10
are simplices. To complete Lk(u9), u3u7u9 and u8u9u10 have to be faces. It is clear that
u4u6u8 ∈ K. The last vertex in Lk(u7) is one of u4, u5 or u6, all of which are impossible.
If u2u3u10 ∈ K, then the sixth vertex, z, in Lk(u2) is u4, u5 or u8. If z = u4, the links
of u2 and u4 show that u2u4u7, u2u4u10, u4u5u10 and u3u4u7 have to be faces. It is easy
to see that Lk(u3) has only five vertices, a contradiction. If z = u8, considering the links
of u2, u8, u9, u7, u3 and u6, we see that u4u6u7 and u4u5u6 are faces, a contradiction. If

Equivelar Polyhedra with Few Vertices
447
z = u5, to complete Lk(u5), we see that either u4u5u7 or u4u5u10 is a face. In the first
case the links of u7 and u4 show that deg(u10) > 6. In the second case we see that there
exists no vertex x (= u1) such that u7u9x is a face. This proves the claim.
By the claim we can assume without loss that u1u3u6 or u1u4u6 is a face.
Case 1: u1u3u6 is a face in K. Let u be the sixth vertex in Lk(u1).
If u = u5, we see from the links of u1 and u3 that deg(u5) > 6, a contradiction.
If u = u4, the links of u1 and u3 show that u1u3u4, u1u4u7, u2u3u8 and u3u6u8 are
faces, where u8 is one of the remaining three vertices. Let u9 and u10 be the other two
vertices. Now, one of u9 or u10, say u9, has to be in the link of u2. Hence, V (Lk(u10)) =
{u4, u5, u6, u7, u8, u9}. Then it is easy to see that V (Lk(u9)) = {u2, u5, u7, u8, u10}, a
contradiction. So, u is one of the remaining three vertices, say u8.
Clearly, u1u3u8 and u1u7u8 are faces. Let u9 and u10 be the other two vertices. To
complete Lk(u3) either u2u3u6 or u2u3u8 ∈ K. If u2u3u6 ∈ K, we see that the sixth
vertex in both Lk(u2) and Lk(u6) is the same, say u9. Then deg(u10) < 6, a contradiction.
So, u2u3u8 and u3u4u6 ∈ K.
Clearly, V (Lk(u9)) = {u2, u4, u5, u7, u8, u10} and V (Lk(u10)) = {u4, u5, u6, u7, u8,
u9}. Then, by considering the links of u2, u6, u4, u8 and u5, u2u7u9, u2u8u9, u4u6u10,
u5u6u10, u4u5u9, u4u9u10, u7u8u10, u8u9u10, u5u7u9 and u5u7u10 are faces in K. Here,
K is isomorphic, via the map (1, 2, 3)(4, 9, 6, 10, 7, 5, 8), to T11.
Case 2: u1u4u6 is a face in K. Let u be the sixth vertex in Lk(u1).
If u = u5, then V (Lk(u4)) = {u0, u1, u3, u5, u6, u8} and V (Lk(u6)) = {u0, u1, u4,
u5, u8, u9} (otherwise f2 < 22), where u8 and u9 are two of the remaining three vertices.
Then the vertex set of the link of the remaining vertex is a subset of {u2, u3, u7, u8, u9},
a contradiction. If u is one of the remaining three vertices, say u8, then the links of u1
and u4 show that u1u4u8, u1u7u8, u3u4u6 and u4u5u8 are faces in K. The face other
that u0u5u6 having u5u6 as an edge has to be u5u6u9 (in all other cases f2 < 22),
where u9 is one of the remaining two vertices. If u10 is the remaining vertex, then
V (Lk(u10)) = {u2, u3, u5, u7, u8, u9}. The links of u3, u5 and u10 imply that deg(u2) =
5, a contradiction. So, u = u3.
Clearly, u1u3u4 and u1u3u7 are faces. The sixth vertex in Lk(u4) has to be one of the
three remaining vertices, say u8. Then u4u5u8 and u4u6u8 ∈ K. If u9 and u10 are the
remaining two vertices, then V (Lk(u9)) = {u2, u3, u5, u7, u8, u10} and V (Lk(u10)) =
{u2, u5, u6, u7, u8, u9}. The links of u3, u2, u6, u5 and u7 show that u2u3u9, u3u7u9,
u2u7u10, u2u9u10, u5u6u10, u6u8u10, u5u8u9, u5u9u10, u7u8u9 and u7u8u10 are faces in
K. In this case, K is isomorphic, via the map (0, 1)(2, 9, 7, 8, 5, 4)(3, 10, 6), to T11. This
completes the proof of the lemma.
Proof of Theorem 4. Let K be a {3, q}-equivelar simplicial polyhedron on n ( 11)
vertices.
Since (K) = 0 we have q = 6 and hence n 7.
It is not difficult to show (also see in [9]) that T7 is the only (up to isomorphism)
7-vertex neighbourly combinatorial 2-manifold. Hence, if n = 7, then K is isomorphic
to T7.
The case n = 8 follows from the classification of combinatorial 2-manifolds on eight
vertices in [10]. (It is also not difficult, by a similar argument as in the proof of Lemma 1,

448
B. Datta and N. Nilakantan
to show that there exists a unique combinatorial 2-manifold with non-edge graph 4K2
(the disjoint union of four edges). This implies that K is isomorphic to T8.)
If n = 9, then, by Lemma 1, K is isomorphic to A3,3, B3,3 or T9.
If n = 10, then, by Lemma 2, K is isomorphic to or T10 or Q.
If n = 11, then, by Lemma 3, K is isomorphic to T11. This completes the proof.
Lemma 4. If M1 and M2 are as in Examples 1 and 7, then M1 M .
=
2
Proof. For 1
i
2, let A(3(Mi )) denote the adjacency matrix of the graph 3(Mi ).
Let Pi (x) denote the characteristic polynomial of A(3(Mi )). Then
P1(x) = (x 3)(x 2)6(x 1)3x4(x + 1)3(x + 2)6(x + 3),
P2(x) = (x 3)(x 2)4(x 1)x4(x + 1)2(x + 2)2(x2 3)(x2 + 2x 1)
(x3 + 2x2 4x 6)2.
If M1 and M2 are isomorphic, then 3(M1) and 3(M2) are isomorphic as graphs and
hence P1(x) = P2(x). Clearly, P1(x) = P2(x). Hence, M1 M .
=
2
Lemma 5. If N1, . . . , N14 are as in Examples 1 and 8, then Ni
14.
N for 1 i = j
j
=
Proof. For 1
i
14, let A(3(Ni )) denote the adjacency matrix of the graph 3(Ni ).
Let Pi (x) denote the characteristic polynomial of A(3(Ni )). Then
P1(x) = (x 3)(x 2)(x 1)(x + 1)2(x + 2)(x3 + x2 5x 3)2
(x9 12x7 + 2x6 + 45x5 12x4 52x3 + 9x2 + 15x + 1)2,
P2(x) = (x 3)(x 1)9(x + 2)4(x2 + 3x + 1)3(x2
x
3)5,
P3(x) = (x 3)(x 1)3(x + 2)(x2
x
3)2(x2 + x 1)3
(x5 + x4 8x3 5x2 + 13x + 6)3,
P4(x) = (x 3)(x 1)2x2(x13 20x11 + 154x9 + 8x8 576x7 82x6 + 1073x5
+ 272x4 893x3 316x2 + 235x + 96)
(x12 + 5x11 7x10 67x9 25x8 + 315x7 + 315x6 599x5 850x4
+ 358x3 + 781x2 + 57x 144),
P5(x) = (x 3)(x 1)x3(x12 20x10 + 154x8 + 4x7 570x6 38x5 + 1015x4
+ 110x3 723x2 90 x + 93)
(x13 + 4x12 12x11 60x10 + 42x9 + 340x8
+ 2x7 910x6 269x5 + 1170x4 + 453x3 650x2 207x + 112),
P6(x) = (x 3)(x 1)x(x2 5)(x2 + 3x + 1)(x2
x
1)
(x11 + 2x10 14x9 25x8 + 72x7 + 111x6 162x5 210x4 + 145x3
+ 156x2 28x 24)
(x10 14x8 x7 + 68x6 + 11x5 132x4 30x3 + 81x2 + 12x 8),

Equivelar Polyhedra with Few Vertices
449
P7(x) = (x 3)(x 1)(x + 1)
(x27 + 3x26 35x25 105x24 + 541x23 + 1621x22 4851x21
14513x20 + 27857x19 + 83335x18 106646x17 320442x16
+ 274411x15 + 836409x14 466133x13 1470839x12 + 497081x11
+ 1695261x10 297277x9 1217051x8 + 70531x7 + 501773x6
+ 9836x5 105408x4 6268x3 + 9416x2 + 568x 228),
P8(x) = (x 3)x2(x + 1)5(x2 5)3(x2
x
1)2
(x6 10x4 + 5x3 + 25x2 25x + 5)2,
P9(x) = (x 3)x2(x3 + x2 5x 1)(x3 + x2 5x 4)2
(x9 12x7 + x6 + 48x5 6x4 74x3 + 12x2 + 36x 10)2,
P10(x) = (x 3)x2(x27 + 3x26 36x25 108x24 + 576x23 + 1728x22 5384x21
16184x20 + 32482x19 + 98344x18 131768x17 406048x16
+ 362570x15 + 1159262x14 664274x13 2283392x12
+ 766277x11 + 3040125x10 481011x9 2623483x8
+ 78215x7 + 1363257x6 + 75039x5 373481x4
37464x3 + 40010x2 + 4277x 493),
P11(x) = (x 3)x(x + 1)(x27 + 2x26 38x25 70x24 + 646x23 + 1080x22
6472x21 9660x20 + 42370x19 + 55418x18
189975x17 213030x16 + 594709x15 + 556564x14
1301093x13 981238x12 +1959990x11 +1134608x10
1967022x9 812454x8 + 1239001x7 + 322970x6
439776x5 56142x4 + 71163x3 + 1616x2
3360x + 160),
P12(x) = (x 3)x(x2 5)(x2
x
1)(x4 + 2x3 4x2 5x + 2)
(x10 14x8 x7 + 68x6 + 11x5 132x4 30x3 + 81x2 + 12x 8)
(x10 + 2x9 12x8 21x7 + 50x6 + 71x5 86x4 84x3 + 57x2
+ 22x 8),
P13(x) = (x 3)(x + 1)(x2 3)(x3 x2 3x + 1)(x3 + 3x2
x
5)
(x10 14x8 + 2x7 + 65x6 14x5 114x4 + 25x3 + 66x2 12x 9)2,
P14(x) = (x 3)(x29 + 3x28 36x27 108x26 + 576x25 + 1726x24 5388x23
16128x22 + 32588x21 + 97682x20 132955x19 401795x18
+ 369861x17 + 1143465x16 690891x15 2250991x14
+ 824419x13 + 3013939x12 552352x11 2648342x10
+ 117160x9 + 1435760x8 + 77028x7 433570x6 45664x5
+ 61804x4 + 5564x3 3748x2 119x + 63).
Now, Ni
N implies 3(N ) and 3(N ) are isomorphic as graphs and hence P (x) =
=
j
i
j
i

450
B. Datta and N. Nilakantan
Pj (x). Since Pi (x) and Pj (x) have different prime factorizations Pi (x) = Pj (x), for
i = j 14. Hence, for 1 i = j 14, Ni N .
1
=
j
Proof of Theorem 5. Let K be a 9-vertex neighbourly simplicial equivelar polyhedron.
Choose a vertex, say u9, and let the link of u9 be C8(u1, . . . , u8).
Claim. There exists a face of the form ui ui+1ui+3 or ui ui+1ui+6 for some i ∈ {1,
. . . , 8} (additions in the subscripts are modulo 8).
If possible, assume that either ui ui+1ui+4 or ui ui+1ui+5 is a face for each i. Then
we can assume without loss that u1u2u5 is a face in K. Then, by repeated use of the
assumption, it is clear that u1u4u8, u4u5u8, u3u7u8 and u3u4u7 are faces in K. To
complete Lk(u4) and Lk(u7), u2u4u6, u2u4u7, u1u4u6 and u1u6u7 have to be faces, a
contradiction. This proves the claim.
By the claim, we can assume without loss that u1u2u4 ∈ K. We observe that the edge
u1u8 belongs to one of u1u3u8, u1u5u8 or u1u6u8.
Case 1: u1u3u8 is a face. By considering Lk(u1), one of u1u3u5, u1u3u6 or u1u3u7 is a
face.
If u1u3u7 ∈ K, then, by considering the links of u1 and u5, u1u5u6, u1u4u6 and
u1u5u7 are faces and hence, by considering Lk(u7), u3u5u7 ∈/ K. Then, by considering
the links of u5, u8 and u6, u2u3u5, u3u5u8 ∈ K, u5u7u8 ∈ K and hence u2u5u7, u4u5u8,
u2u7u8, u2u6u8, u4u6u8, u2u3u6 and u3u6u7 are faces. Then u2u3u5, u2u3u6 and u2u3u9
are the faces through u2u3, a contradiction.
If u1u3u5 ∈ K, then, by considering Lk(u1) and Lk(u6), u1u6u7, u1u4u6, u1u5u7
are faces. The edge u4u6, belongs to either u3u4u6 or u4u6u8. In the first case, Lk(u3)
and Lk(u6) show that u2u3u6 ∈/ K and u2u5u6, u2u6u8, u3u6u8, u2u3u7, u3u5u7 ∈ K.
However, this implies that Lk(u5) contains C3(u3, u1, u7), a contradiction. In the sec-
ond case u2u3 ∈ Lk(u6). Considering Lk(u7) and Lk(u5), u2u4u7, u3u4u7, u2u5u7 and
u3u7u8 ∈ K. Then Lk(u8) contains C4(u7, u9, u1, u3), a contradiction. So, u1u3u6 ∈ K.
From Lk(u1), u1u5u7 is a face and either u1u4u5 or u1u4u7 is a face.
If u1u4u5 ∈ K, then so is u1u6u7. To complete Lk(u7), either u2u4u7 and u3u4u7 or
u2u4u7 and u2u3u7 have to be faces. This shows that deg(u4) < 8 in the first case and
deg(u2) < 8 in the second case, a contradiction. Thus, u1u4u7 ∈ K.
Clearly, u1u5u6 is a face. By considering the links of u3 and u6, u3u6u8 ∈/ K and
u4u6u8, u2u6u8 ∈ K. Then, by considering the links of u6, u4 and u8, u2u4u6, u2u4u7
and u2u4u8 are not faces and hence u2u4u5 is a face. Now, by considering the links of
u2, u5 and u4, u2u5u7 is not a face and u2u5u8, u2u6u7, u2u3u7, u3u5u7, u3u5u8 u3u4u6,
u4u7u8 are faces. Here, K is M2.
Case 2: u1u5u8 is a face. Clearly, one of u1u3u5, u1u5u6 or u1u5u7 is a face.
Subcase 2.1: u1u3u5 is a face. If u1u3u7 is a face, then the link of u1 shows that u1u4u6
and u1u6u7 are faces. By considering the links of u3, u8, u7, u4 and u2, u2u5u7, u4u5u7,
u2u3u7, u4u7u8, u3u4u6, u2u4u8 and u2u5u6 ∈ K. Then Lk(u5) is C5(u7, u4, u9, u6, u2),
a contradiction. So, by considering Lk(u1), u1u3u6, u1u6u7 and u1u4u7 ∈ K.
Clearly, either u3u4 or u4u5 ∈ Lk(u7). If u4u5u7 ∈ K, then u2u3u7, u2u7u8 and
u3u5u7 are faces (if u3u7u8 ∈ K, then, to complete Lk(u3), u3u5u8 ∈ K, which implies

Equivelar Polyhedra with Few Vertices
451
deg(u5) < 8). Lk(u5) shows that u2u5u6 and u2u5u8 are faces. The link of u8 now
contains C5(u2, u7, u9, u1, u5), a contradiction. So, u3u4 ∈ Lk(u7).
From the links of u7, u3, u8 and u2, u3u5u7, u2u5u7, u2u7u8, u2u3u8, u3u6u8,
u4u5u8, u4u6u8, u2u4u6 and u2u5u6 are faces. Here, K is isomorphic, via the map
(1, 3, 5, 7)(2, 4, 6, 8), to M2.
Subcase 2.2: u1u5u6 is a face. To complete Lk(u1), u1u3u6, u1u3u7 and u1u4u7 have to
be faces. Lk(u5) shows that the edge u4u5 belongs to either u2u4u5 or u4u5u7.
If u4u5u7 ∈ K, then, by considering Lk(u5) and Lk(u7), u2u3u5 ∈ K, u3u5u7 ∈/ K
and hence u2u5u7 and u3u5u8 ∈ K. To complete Lk(u3), u3u7u8 and u3u4u6 have to be
faces (since Lk(u4) shows that u3u4u7 ∈/ K). This implies that the link of u8 contains
C5(u3, u7, u9, u1, u5), a contradiction. So, u2u4u5 ∈ K.
The links of u5, u2, u8 and u4 show that u3u5u7, u2u5u7, u3u5u8, u2u6u8 and u4u6u8
are faces. Then, from Lk(u2), either u2u3u6 ∈ K or u2u3u8 ∈ K.
In the first case the links of u2, u6 and u8 show that u2u7u8, u4u6u7 and u3u4u8 are
faces. Here, K is isomorphic, via the map (1, 7, 2, 5, 8, 4, 3, 6, 9), to M2.
In the second case the links of u2, u6 and u8 show that u2u6u7, u3u4u6 and u4u7u8
are faces. Here, K is isomorphic, via the map (1, 2, 7, 5, 3)(4, 8), to M1.
Subcase 2.3: u1u5u7 is a face. From Lk(u1), u1u3u6 and one of u1u3u4 or u1u4u6 are
faces.
If u1u3u4 ∈ K, then so is u1u6u7. The links of u3, u6 and u4 show that u3u5u6,
u2u3u6 ∈/ K, u2u4u6, u4u6u8, u3u6u8, u4u5u7 and u4u7u8 are faces. This shows that the
link of u7 contains C6(u4, u5, u1, u6, u9, u8), a contradiction. So, u1u4u6 is a face.
Clearly, u1u3u7 ∈ K. We now observe that either u3u4u7 or u3u4u8 is a face (since,
Lk(u6) shows that u3u4u6 ∈/ K).
In the first case the links of u3, u4, u5 and u6 show that u3u5u8 ∈ K, u4u5u7 ∈/ K and
hence u2u5u7, u2u4u5, u3u5u6, u2u3u8, u4u6u8, u4u7u8, u2u6u8 and u2u6u7 are faces.
Here, K is isomorphic, via the map (1, 4, 6, 3, 9, 7, 5, 2, 8), to M2.
In the second case u4u5u7 has to be a face (if u2u4u5 ∈ K, then, from Lk(u4)
and Lk(u7), deg(u7) < 8). Considering the links of u5, u3, u7 and u6, u2u3u5 ∈ K,
u3u5u8 ∈/ K and u2u5u8, u3u5u6, u3u7u8, u2u4u7, u2u6u7, u2u6u8 and u4u6u8 are faces.
Here, K is isomorphic, via the map (1, 3, 6, 4, 9, 7)(5, 8), to M2.
Case 3: u1u6u8 is a face. Clearly, u3u6, u5u6 or u6u7 ∈ Lk(u1).
If u1u6u7 ∈ K, then u1u3u5 ∈ K. The links of u6, u2 and u1 show that u2u4u6,
u3u4u6, u1u3u7 and u1u4u5 are faces. Here, Lk(u4) contains C6(u6, u3, u9, u5, u1, u2),
a contradiction.
If u1u5u6 ∈ K, then u1u3u7 ∈ K. It is clear from Lk(u6) that u2u4u6 and u3u4u6
are faces. Considering the links of u1, u4, u8 and u5, u1u4u7, u1u3u5, u4u5u8, u4u7u8,
u3u5u7 ∈ K, a contradiction. So, u3u6 ∈ Lk(u1).
Clearly, u5u7 ∈ Lk(u1). Now, either u1u4u5 or u1u4u7 ∈ K.
If u1u4u5 ∈ K, then u1u3u7 ∈ K. The links of u4, u8, u6 and u3 show that u4u6u8,
u4u6u7, u2u5u6, u2u3u6 and u3u5u7 which imply deg(u7) < 8. So, u1u4u7 is a face.
The link of u1 shows that u1u3u5 is a face. Now, u4u5u8 (= u4u5u9) is the face having
u4u5 as an edge. (If u2u4u5 ∈ K, considering the links of u4, u7 and u2, we observe that
u3u5 ∈ Lk(u7), a contradiction. Lk(u7) shows that u4u5u7 ∈/ K.) The links of u3 and u5

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show that u2u5u6 is a face. To complete Lk(u5), u2u5u7 and u3u5u8 have to be faces (if
u2u3u5 ∈ K, then u2u3 ∈ Lk(u8), a contradiction). Now, the second face through u6u8
is u2u6u8 or u4u6u8.
In the first case the links of u6, u4 and u7 show that u3u4u6, u4u6u7, u2u4u8, u2u3u7
and u3u7u8 are faces. Here, K is isomorphic, via the map (2, 5)(3, 9, 6, 8)(4, 7), to M2.
In the second case the links of u8, u7 and u6 show that u2u3u8, u2u7u8, u3u4u7, u3u6u7
and u2u4u6 are faces. Here, K is isomorphic, via the map (2, 7, 3), to M1.
The theorem now follows from Lemma 4.
Proof of Theorem 6. Let K be a 10-vertex neighbourly simplicial equivelar polyhedron.
Choose a vertex, say u, and let the link of u be C9(u1, . . . , u9).
Case 1: There exists no face of the form ui ui+1ui+3 or ui ui+1ui+7. If ui ui+1ui+5 is a face
for each i ∈ {1, . . . , 9} (additions in the subscript are modulo 9), then u1u2u6, u1u5u6
and u1u5u9 ∈ K. This implies that C5(u2, u6, u5, u9, u) is in Lk(u1), a contradiction.
So assume, without loss, that u1u2u5 ∈ K.
Claim 1. u1u4u9 is a face.
Since u1u8u9 is not a face, the second face through u1u9 is u1u4u9, u1u6u9 or u1u7u9.
However, by the assumption, u1u7u9 is not a face.
If u1u6u9 ∈ K, then u1u5 is in one of u1u3u5, u1u4u5, u1u5u7 or u1u5u8.
If u1u3u5 ∈ K, then u1u4u7 and u1u4u8 ∈ K (since u3u4 and u7u8 ∈/ Lk(u1)). The
edge u5u6 belongs to either u2u5u6 or u5u6u9. If u2u5u6 ∈ K, the links of u5, u1, u3
and u6 show that u4u5u8, u5u8u9, u3u5u7, u1u3u8, u1u6u7, u3u4u9, u4u6u9, u2u4u6 and
u2u4u7 ∈ K. The link of u6 contains C7(u2, u5, u, u7, u1, u9, u4), a contradiction. If
u5u6u9 ∈ K, then the links of u6, u1 and u8 imply u1u7 ∈/ Lk(u6), u1u3u7, u1u6u8 ∈ K,
u3u4u6, u4u6u8 ∈/ K and hence u4u6u7 ∈ K, a contradiction to the assumption.
If u1u4u5 ∈ K, then, by considering the links of u5, u1, u6, u4 and u3, u1u3u7, u1u3u8,
u5u6u9, u1u4u7, u1u6u8, u4u6u8, u2u4u6, u2u3u6, u3u6u7 (since u3u4, u3u8 ∈/ Lk(u6)),
u3u4u9, u4u7u9, u4u7u8, u3u5u8 and u5u7u8 are faces. This gives a contradiction to the
assumption.
If u1u5u7 ∈ K, then u1u3u8 and u1u4u8 ∈ K. If u2u5u6 ∈ K, we see that u8u9 ∈
Lk(u5) (if not, Lk(u5) shows that u3u5u8 and u4u5u8 ∈ K which implies that C4(u3, u1,
u4, u5) is in Lk(u8)). Hence, the links of u5, u1, u6, u7 and u9 show that u3u5u7, u1u4u7,
u1u3u6, u2u6u8, u4u6u8, u4u6u9, u4u7u9, u2u7u9, u2u7u8, u2u3u9, u3u5u9 and u4u5u8
are faces. Here, u4u8 is an edge in u1u4u8, u4u5u8 and u4u6u8, a contradiction. If
u5u6u9 ∈ K, then u4u8 ∈ Lk(u5), u3u5 ∈/ Lk(u8). The links of u5, u1 and u9 show
that u3u5u7, u3u5u9, u1u4u7, u1u3u6, u3u4u9 and u4u6u9 ∈ K. Here u6u9 is an edge in
u1u6u9, u5u6u9 and u4u6u9, a contradiction.
Finally, let u1u5u8 ∈ K. Then u1u3u7, u1u4u7 ∈ K. If u2u5u6 ∈ K, then the links
of u5, u1, u6, u9 and u2 show that u4u5u9, u5u7u9, u3u5u7, u3u5u8, u1u4u8, u1u3u6,
u3u6u7, u4u6u9, u4u6u8, u2u6u8, u3u8u9, u2u3u9, u2u7u9 and u2u4u8 ∈ K. Hence
Lk(u8) contains C3(u2, u6, u4), a contradiction. If u5u6u9 ∈ K, then Lk(u5) shows that
u2u4u5, u3u5u7, u3u5u8 and u5u8u9 ∈ K. Hence Lk(u9) contains C5(u5, u8, u, u1, u6),
a contradiction. This proves Claim 1.

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453
Clearly, the second face through u1u4 is u1u4u6, u1u4u7 or u1u4u8.
Claim 2. u1u4u7 is a face.
If u1u4u6 ∈ K, then u1u3u7, u1u3u8 are faces. We see that u4u5 is an edge in either
u4u5u8 or u4u5u9. If u4u5u9 ∈ K, then the links of u5, u1 and u4 imply that u2u5u6,
u3u5u7, u3u5u8, u1u5u8, u5u7u9, u1u6u7, u3u4u7 and u2u4u7 ∈ K. This implies u3u7 is
in u1u3u7, u3u5u7 and u3u4u7, a contradiction. Hence, u4u5u8 ∈ K and either u1u5u8
or u1u5u7 ∈ K. In the former case the links of u1, u4 and u6 show that u1u6u8 ∈ K,
u4u6u8 and u4u6u9 ∈/ K and hence u2u4u6 is a face. It is clear that u1u6u8 is the only
face having u6u8 as an edge, a contradiction. In the latter case the links of u1, u5, u7 and
u4 show that u1u6u7, u2u5u7, u3u5u7, u3u5u9, u2u7u9, u4u7u9, u4u8u9 and u3u4u5 ∈ K,
a contradiction.
If u1u4u8 ∈ K, either u4u5u9 or u4u5u8 ∈ K. In either case Lk(u4) implies that
u3u4u7, u2u4u7 and u2u4u6 are faces. Hence u1u6u7 ∈ K. To complete Lk(u1), either
u1u3u8 or u1u6u8 ∈ K. If u1u3u8 ∈ K, then to complete Lk(u6), either u6u8u9 or
u5u6u8 ∈ K, a contradiction. Hence u1u6u8 and therefore u1u3u5 and u1u3u7 ∈ K.
Clearly, u2u7u8 ∈ K. This implies deg(u7) < 9, a contradiction. This proves Claim 2.
Again, either u4u5u9 or u4u5u8 ∈ K. If u4u5u9 ∈ K, then it is clear from Lk(u4) that
u3u4u8, u4u6u8, u2u4u6 and u2u4u7 ∈ K. It is easy to see that u3u6u7, u2u5u6, u2u7u8,
u2u3u9 and u5u8u9 ∈ K. Hence deg(u9) < 9, a contradiction. Thus, u4u5u8 ∈ K.
Clearly, u3u4u7 or u3u4u9 is a face through u3u4. If u3u4u9 ∈ K, the links of u4, u1
and u6 show that u2u4u7, u4u6u8, u1u3u7, u1u6u8 and u3u6u7 ∈ K. To complete Lk(u7),
either u5u7u8 or u7u8u9 ∈ K, a contradiction. So, u3u4u7 ∈ K.
The links of u4, u1 and u7 show that u1u6u7, u1u3u8, u2u4u6, u2u7u8 and u5u7u9 are
faces.
Now, either u2u5u8 or u5u8u9 ∈ K. If u5u8u9 ∈ K, the links of u5, u1, u4, u6 show
that C3(u3, u1, u6) is in Lk(u8), a contradiction. So, u2u5u8 ∈ K.
Now, the links of u1, u5, u6, u8 and u2 show that u1u3u5, u1u6u8, u3u5u7, u5u6u9,
u4u6u8, u2u3u6, u3u6u9, u3u8u9, u2u7u9 and u2u4u9 are faces. Here, K is N1.
Case 2: There exists a face of the form ui ui+1ui+3 or ui ui+1ui+7. We can assume without
loss that u1u2u4 ∈ K. From Lk(u1) we see that u1u9 is an edge in one of u1u3u9, u1u5u9,
u1u6u9 or u1u7u9.
Subcase 2.1: u1u7u9 ∈ K. The edge u1u4 belongs to one of u1u3u4, u1u4u5, u1u4u6 or
u1u4u8.
2.1.1: u1u3u4 ∈ K. From Lk(u1) we see that one of u1u5u7, u1u6u7 or u1u7u8 ∈ K.
If u1u6u7 ∈ K, we see from Lk(u1) and Lk(u6) that u1u6u8, u1u5u8 and u1u3u5 ∈ K.
The link of u4 shows that one of u4u5u7, u4u5u8 or u4u5u9 ∈ K.
If u4u5u7 is a face, the links of u4 and u7 show that either u4u7u8 or u4u7u9 ∈ K.
In both cases, after completing Lk(u4), we see from Lk(u6) and Lk(u7) that u2u3u6 and
u2u3u7 are faces, a contradiction (since uu2u3 is already a face).
If u4u5u8 is a face, then it is clear that u4u6 ∈/ Lk(u8) and u4u8u9 ∈/ K (since, if
u4u8u9 ∈ K, then Lk(u4) implies that u4u6u7 ∈ K, a contradiction). The links of u4
and u7 show that u4u7u8 and u4u7u9 ∈ K, which imply that C6(u9, u1, u6, u, u8, u4) is
in Lk(u7). So, u4u5u9 ∈ K.

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Since u6u7 ∈/ Lk(u4), we see on completing Lk(u4) that u4u6u8 and u4u7u8 ∈ K.
The links of u7, u4, u8 and u6 show that u4u6u9, u2u4u7, u2u3u8 and u2u3u6 ∈ K, a
contradiction. Therefore, either u1u5u7 or u1u7u8 ∈ K.
2.1.1.1: u1u5u7 ∈ K. The link of u1 shows that either u1u5u6 or u1u5u8 is a face.
2.1.1.1.1: u1u5u6 isaface. From Lk(u1) we see that u1u6u8, u1u3u8 ∈ K. Considering
the links of u4 and u5, either u4u5u8 or u4u5u9 ∈ K.
Subcase A: u4u5u8 is a face. The links of u5 and u7 show that either u2u5u7 or
u3u5u7 ∈ K.
A.1: u2u5u7 is a face. If u2u3u5 ∈ K, Lk(u5), Lk(u3) and Lk(u8) show that u3u5u9,
u3u6u7 and u3u6u8 are faces. This implies that C3(u3, u1, u6) is in Lk(u8).
To complete Lk(u5), u2u5u9, u3u5u8 and u3u5u9 ∈ K. The links of u3 and u6 show
that u3u6u7, u4u6u9 and u2u4u6 ∈ K (since, u4u6 ∈/ L(u8) and Lk(u3)). The links of
u4, u8, u7 and u6 show that u4u7u9, u4u7u8, u2u6u8, u2u8u9, u2u3u7 and u3u6u9 are
faces. In this case K is N11 (more precisely, K is isomorphic to N11 by the map ϕ, where
ϕ(u) = 0 and ϕ(ui ) = i, for 1
i
9).
A.2: u3u5u7 ∈ K. If u2u3u5 ∈ K, the links of u5 and u3 show that u5u8u9 and
u3u8u9 ∈ K (since u3u6 ∈/ L(u8)), a contradiction. Hence, u2u5u9, u2u5u8 and u3u5u9
arethefacesrequiredtocompleteLk(u5). Ifu2u3u7 ∈ K, Lk(u3)showsthatu3u5u6 ∈ K,
a contradiction.
We now observe from the links of u3, u5 and u9 that the edge u2u3 belongs to u2u3u6.
The edge u3u6 is in either u3u6u7 or u3u6u9.
In the first case the links of u3, u7, u4 and u2 show that u3u8u9, u2u4u7, u4u7u9,
u4u6u9, u4u6u8, u2u7u8 and u2u6u9 are faces. Here, K is N7.
In the second case the links of u3, u7, u2 and u4 show that u3u7u8, u2u4u7, u4u6u7,
u2u7u9, u2u6u8, u4u6u9 and u4u8u9 are faces. Now, K is N12.
Subcase B: u4u5u9 ∈ K. If u2u4u8 ∈ K, the links of u4 and u7 show that u4u6u7,
u4u6u8 and u4u7u9 ∈ K. Here C6(u8, u4, u7, u, u5, u1) is in Lk(u6). We now observe
from Lk(u4) that either u2u4u6 or u2u4u7 ∈ K.
In the first case the links of u4, u8, u7, u6 and u9 show that u4u6u8, u4u7u8, u4u7u9,
u2u5u8, u2u3u7, u3u5u8 u2u8u9, u3u6u7, u3u6u9, u2u6u9, u2u5u7 and u3u5u9 are faces.
Now, K is N14.
In the second case the links of u4, u6, u7, u2 and u3 show that u4u6u8, u4u7u8, u4u6u9,
u2u3u6, u3u6u7, u2u6u9, u2u5u7, u3u7u9, u2u5u8, u2u8u9, u3u5u9 and u3u5u8 are faces.
Here, K is N8.
2.1.1.1.2: u1u5u8 ∈ K. The links of u1 and u4 show that u1u6u8, u1u3u6 ∈ K and
one of u4u5u7, u4u5u8 or u4u5u9 ∈ K.
Subcase A: u4u5u7 ∈ K. If u2u4u6 ∈ K, then Lk(u4) and Lk(u8) show that u4u7u9
and u4u8u9 ∈ K, thereby showing that deg(u9) < 9. In the case when u2u4u9 ∈ K,
Lk(u4) and Lk(u7) show that u4u6u8 and u2u3u7 are faces (since, from Lk(u4), either
u4u7u8 or u4u6u7 ∈ K). Now, the links of u5 and u2 show that u2u5u9, u3u5u9 and
u2u6u8 are faces. This implies that u6u8 is an edge in u1u6u8, u2u6u8 and u4u6u8, a
contradiction. Thus, from Lk(u4), u2u4u8 ∈ K.
The links of u4, u7, u8, u2 and u9 show that u4u6u9, u4u6u7, u4u8u9, u2u3u7,
u2u7u9, u3u7u8, u2u5u6, u3u5u9, u2u5u9, u3u6u9, u2u6u8 and u3u5u8 are faces. Here,
K is N5.

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Subcase B: u4u5u8 ∈ K. It is clear that C4(u4, u5, u1, u6) is in Lk(u8) if u4u6u8 ∈ K.
If u4u7u8 ∈ K, considering the links of u4, u7 and u8 we see that u2u3u7 and u2u3u8 are
faces, a contradiction. Hence, from Lk(u4), u4u8u9 ∈ K. Now, on considering Lk(u9)
also, u4u6u9, u4u6u7 and u2u4u7 ∈ K. The links of u8, u2 and u9 show that u2u3u8,
u3u7u8, u2u6u8, u2u5u9 and u3u5u9 are faces. To complete Lk(u5), either u2u5u6 or
u3u5u6 is a face.
In the first case the links of u2, u5 and u3 show that u2u7u9, u3u5u7 and u3u6u9 are
faces. Clearly, K is N6.
In the second case the links of u3, u9 and u5 show that u3u7u9, u2u6u9 and u2u5u7
are faces and therefore K is isomorphic, via the map (0, 4, 2, 3, 1)(5, 7, 6, 9), to N7.
Subcase C: u4u5u9 ∈ K. Considering the link of u4, we see that one of u4u6u9,
u4u7u9 or u4u8u9 is a face.
C.1: u4u6u9 ∈ K. The links of u4 and u8 show that u4u7u8, u2u4u7, u4u6u8 and
u2u3u8 ∈ K. If u6u7u9 ∈ K, Lk(u7) shows that u2u3u7 ∈ K, a contradiction. Hence,
we can conclude from Lk(u7) that u3u6u7 ∈ K. From Lk(u6) and Lk(u3), we see that
u2u6u9, u2u5u6 and u3u5u9 ∈ K. To complete Lk(u8), either u2u8u9 or u3u8u9 is a face.
In the first case the links of u8, u2 and u7 show that u3u5u8, u2u5u7 and u3u7u9 are
faces. Here, K is isomorphic, via the map (0, 1, 3, 2, 4)(5, 9, 8, 6, 7), to N14.
In the second case the links of u8, u3 and u2 show that u2u5u8, u3u5u7 and u2u7u9
are faces. Here, K is isomorphic, via the map, (0, 1, 3, 2, 4)(5, 9, 8, 6, 7), to N11.
C.2: u4u7u9 ∈ K. The link of u4 shows that either u4u6u7 or u4u7u8 ∈ K. In both
cases u2u3 ∈ Lk(u7). If u3u5u6 ∈ K, the link of u3 and u8 show that u3u7u9 and
u3u8u9 ∈ K, a contradiction. Hence, u2u5u6 ∈ K (the links of u8 and u5 show that
u5u6u8 and u5u6u9 ∈/ K). Considering the links of u5 and u9, we see that u3u5u9,
u2u6u9, u2u8u9 and u3u6u9 ∈ K. The links of u6, u4, u2 and u8 show that u4u6u8,
u4u6u7, u2u4u8, u2u5u7, u3u7u8 and u3u5u8 are faces. In this case K is isomorphic, via
the map (0, 2, 1, 4, 3)(5, 7, 9, 6), to N11.
C.3: u4u8u9 ∈ K. The links of u4, u6, u3, u8, u2 and u5 imply that u4u6u8, u4u6u7,
u2u4u7, u2u5u6, u2u6u9, u3u6u9, u2u3u8, u2u5u8, u2u7u9, u3u5u7, u3u7u8 and u3u5u9
are faces. Here, K is isomorphic, via the map (0, 2, 1, 4, 3)(5, 7, 8, 9, 6), to N14.
2.1.1.2: u1u7u8 ∈ K. The link of u1 shows that either u1u5u8 or u1u6u8 is a face.
Claim 3. u1u5u8 is a face.
If u1u6u8 ∈ K, the links of u1 and u4 show that u1u5u6, u1u3u5 and one of u4u5u7,
u4u5u8, u4u5u9 ∈ K. If u4u5u7 ∈ K, from Lk(u4), either u4u6u7 or u4u6u9 ∈ K
(since u4u7u8 ∈/ K). In both the cases the links of u2, u4, u6 and u7 show that uu2u3,
u2u3u6, u2u3u7 ∈ K, a contradiction. If u4u5u8 ∈ K, Lk(u4) and Lk(u8) show that
either u4u6u8 or u4u8u9 ∈ K. Again, in both cases, considering Lk(u4), Lk(u9) and
Lk(u8) we see that u2u3u6, u2u3u8 ∈ K. So, u4u5u9 ∈ K. Since u7u8 already belongs
to two faces, u4u7u8 ∈/ K. Hence, u4u6u7 and u4u6u8 ∈ K, thereby showing that
C6(u8, u4, u7, u, u5, u1) is in Lk(u6). This proves Claim 3.
The links of u1, u4 and u5 show that u1u3u6, u1u5u6 and either u4u5u7 or u4u5u9 ∈ K.
Subcase A: u4u5u7 ∈ K. Considering Lk(u4), either u4u6u7 or u4u7u9 ∈ K. In both
cases u2u3 ∈ Lk(u7) and therefore, from Lk(u5), u2u5u9 and u3u5u9 ∈ K.

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If u4u6u7 ∈ K, the links of u4, u6, u8, u3 and u5 show that u4u8u9, u2u6u8, u3u6u8,
u2u6u9, u2u4u8, u3u5u8, u4u6u9, u3u7u9 and u2u5u7 are faces and, thus, K is N2.
If u4u7u9 ∈ K, the links of u4, u9, u6, u7, u5 and u8 show that u2u4u8, u4u6u8, u4u6u9,
u3u5u7, u2u6u7, u2u5u8, u3u6u8, u3u8u9 and u2u6u9 are faces. Here, K is isomorphic,
via the map (0, 4, 2, 3, 1)(5, 8, 7, 6, 9), to N5.
Subcase B: u4u5u9 ∈ K. The links of u4, u7, u6, u8, u5 and u9 show that u4u6u8,
u4u6u7, u2u4u7, u4u8u9, u3u5u7, u3u6u9, u2u6u9, u2u6u8, u2u3u8, u3u5u8, u2u5u7,
u2u5u9 andu3u7u9 ∈ K. Inthiscase K isisomorphic, viathemap(0, 4, 6, 8, 1)(2, 5, 9, 3,
7), to N5.
2.1.2: u1u4u5 ∈ K.
Claim 4. For a vertex x = u, u2, u3 and u4 cannot occur together in any order in the
link of any of the vertices.
Let x be any vertex in K. If u2u3x and u3u4x are faces, then C4(u2, x, u4, u) is in
Lk(u3). If u2u4x and u3u4x are faces, thenC6(u3, u, u5, u1, u2, x) isin Lk(u4) andfinally
if u2u3x and u2u4x are faces, then C5(u4, u1, u, u3, x) is in Lk(u2). Hence, Claim 4 is
proved.
The links of u1 and u5 show that u1u3u5 or u1u5u8 ∈ K.
Claim 5. u1u5u8 is a face.
If u1u3u5 ∈ K, then Lk(u1) and Lk(u5) show that u1u6u8 and one of u2u5u6, u5u6u8
or u5u6u9 ∈ K.
If u2u5u6 ∈ K, Lk(u5) shows that u5u7u8 and u5u7u9 ∈ K (if u5u8u9 and u5u7u9 are
faces, then deg(u9) < 9). Hence u7u8 ∈/ Lk(u1) and therefore u1u6u7 and u1u3u8 ∈ K.
This shows that C6(u1, u9, u5, u8, u, u6) is in Lk(u7).
If u5u6u8 ∈ K, considering the links of u6 and u1, we see that u1u3u6 and u1u7u8 are
faces, which imply that u2, u3 and u4 are together in Lk(u8), which is impossible from
Claim 4. Thus, u5u6u9 ∈ K.
It is clear that u1u7u8 and u1u3u6 are faces (since u1u6u7 ∈/ K, from Lk(u6) and
Claim 4). Since u7u8 ∈/ Lk(u5), u2u5u7 and u2u5u8 are faces. To complete Lk(u5),
either u5u7u9 or u5u8u9 ∈ K, both of which are impossible by looking at Lk(u7),
Lk(u8) and Claim 4. This proves Claim 5.
From Claim 5 and Lk(u1), u1u5u8 and therefore u1u3u6 ∈ K.
The link of u5 shows that u5u6 is an edge in one of u2u5u6, u3u5u6 or u5u6u9.
If u3u5u6 ∈ K, considering the links of u6 and u1, u1u6u8 and u1u3u7 are faces. The
links of u5, u7 and Claim 4 show that u2u3u5 ∈/ K. Hence, Lk(u5) and Lk(u3) show
that u3u5u9 and u2u5u7 ∈ K. The links of u8, u5 and Claim 4 show that u5u8u9 ∈/ K,
u5u7u9 and u2u5u8 ∈ K. Since u6u9 ∈/ Lk(u3), it is clear from Lk(u9) and Claim 4 that
u2u6u9 and u4u6u9 are faces. Since either u2u3u9 or u3u4u9 is a face, Lk(u3) implies
that u3u7u8 ∈ K. The links of u7, u6 and u4 show that u2u4u7, u4u6u7, u2u6u8, u4u8u9
and u3u4u8 are faces, thereby showing that C5(u3, u4, u9, u, u7) is in Lk(u8). Therefore,
either u2u5u6 or u5u6u9 is a face.
Subcase A: u2u5u6 ∈ K. The links of u5 and u7 and Claim 4 show that u2u3u5 ∈/ K.

Equivelar Polyhedra with Few Vertices
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If u2u5u9 ∈ K, so is u3u5u7. If u5u7u8 ∈ K, Lk(u7), Lk(u8) and Claim 4 show that
neither u1u6u7 nor u1u6u8 is a face, a contradiction. Hence, from Lk(u5), u3u5u8 and
u5u7u9 ∈ K. Considering the links of u1 and u7 we see that u1u6u7 and u1u3u8 are faces.
This is seen to be impossible by applying Claim 4 to Lk(u7). Hence, Lk(u5) shows that
u2u5u7 ∈ K.
We now observe, from Lk(u5), that either u3u5u7 or u5u7u9 is a face.
In the first case the links of u5, u8 and u1 show that u5u8u9, u3u5u9, u1u3u8 and
u1u6u7 are faces (if u1u6u8 ∈ K, the remaining three vertices in Lk(u8) are u2, u3 and
u4). Since u6u8 ∈/ Lk(u3), Lk(u8) shows that u2u6u8 and u4u6u8 are faces. The links of
u6, u9, u7 and u8 show that u3u6u9, u4u6u9, u2u4u9, u2u7u9, u3u4u7, u4u7u8 and u2u3u8
are faces and hence K is N10.
In the second case the links of u5, u8 and u1 show that u3u5u8, u3u5u9, u1u6u8 and
u1u3u7 are faces. It is easy to see from Lk(u3), Lk(u6) and Lk(u9) that u4u6u7 and u3u6u9
are faces (if either u2u3u9 or u3u4u9 is a face, then from Claim 4 and on completing
Lk(u9), we get deg(u6) < 9). The links of u9, u4, u6, u7 and u8 show that u2u4u9,
u2u6u9, u4u8u9, u4u6u8, u3u4u7, u2u7u8, and u2u3u8 are faces. Here, K is N4.
Subcase B: u5u6u9 be a face. The link of u9 and Claim 4 show that u5u7u9 ∈/ K and
hence Lk(u5) shows that either u2u5u9 or u3u5u9 is a face.
B.1: u2u5u9 ∈ K. Here, u3u7 ∈ Lk(u5) and the links of u5, u7, u8 and Claim 4 show
that u5u7u8 ∈/ K. Now, Lk(u5), Lk(u8) and Lk(u1) show that u3u5u8, u2u5u7, u1u6u8
and u1u3u7 ∈ K. Now, the links of u7 and u6 show that the edge u7u9 belongs to u4u7u9
and therefore u2u6u7 and u4u7u8 are faces (if u4u6u7 ∈ K, then considering Lk(u4),
Lk(u9) and Lk(u6), u4u6u8 and u2u6u9 are faces which show that C3(u2, u5, u6) is in
Lk(u9)). To complete Lk(u6), either u2u4u6 or u4u6u9 ∈ K.
In the first case the links of u2, u3 and u8 show that u2u3u8, u2u8u9, u3u6u9, u3u4u9
and u4u6u8 are faces. In this case, K is N13.
In the second case the links of u9, u8 and u6 show that u3u8u9, u2u3u9, u2u4u8, u2u6u8
and u3u4u6 ∈ K. Here, K is isomorphic, via the map (0, 1, 5, 7)(2, 8)(3, 6, 9, 4), to N13.
B.2: u3u5u9 ∈ K. It is clear that if u2u3u5 and u5u7u8 ∈ K, then the links of u2, u7,
u1 and u3 show that C7(u3, u6, u1, u5, u7, u, u9) is in Lk(u8). Therefore, from Lk(u5),
u2u5u8, u2u5u7 and u3u5u7 ∈ K. Now, if u3u7u8 ∈ K, the links of u7, u1, u4 and u8
show that C5(u4, u8, u, u1, u7) is in Lk(u9). Hence, u4u7u8 is a face.
In case u4u7u9 ∈ K, the links of u7, u1, u6 and u2 show that u1u3u7, u2u6u7,
u1u6u8 ∈ K and u2u6u8, u2u3u6 ∈/ K (if u2u3u6 ∈ K, to complete Lk(u6), u4u6u8 and
u4u6u9 ∈ K which implies that C4(u6, u9, u7, u8) is in Lk(u4)). Since, u3u6u9 ∈/ K
(from Lk(u9)), to complete Lk(u6), u2u4u6 and u3u4u6 ∈ K, a contradiction. Further,
u2u4u7 ∈/ K (if u2u4u7 ∈ K, completing Lk(u7), we get u2u4u9 ∈ K, a contradiction).
Hence, from Lk(u7), u3u4u7 ∈ K.
The links of u7, u1, u4, u6 and u9 show that u1u6u7, u2u7u9, u1u3u8, u4u6u8, u4u6u9,
u2u4u9, u2u3u6, u2u6u8 and u3u8u9 are faces. In this case, K is isomorphic, via the map
(2, 9)(3, 8)(4, 7)(5, 6), to N10.
2.1.3: u1u4u6 ∈ K. Using the same method as the one above, we find that K is isomorphic
to one of N2, N4, N5, N7, . . . , N12.
2.1.4: u1u4u8 ∈ K. In this case we find that K is isomorphic to one of N4, . . . , N7,
N9, . . . , N14.

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Subcase 2.2: u1u3u9 ∈ K. The link of u1 shows that one of u1u4u6, u1u4u5, u1u4u7 or
u1u4u8 ∈ K.
2.2.1: u1u4u6 ∈ K. We observe that either u7u8 ∈ Lk(u1) or u7u8 ∈/ Lk(u1)
2.2.1.1: u7u8 ∈ Lk(u1). The link of u1 shows that either u1u5u7 or u1u5u8 ∈ K.
2.2.1.1.1: u1u5u7 ∈ K. In this case, K is isomorphic to one of N1, N7, N11 or N14.
2.2.1.1.2: u1u5u8 ∈ K. The links of u1 and u7 show that u1u5u6 and u1u3u7 ∈ K.
Considering Lk(u6) and Lk(u7), we see that either u2u6u7 or u6u7u9 ∈ K.
Claim 6. u6u7u9 is a face.
If u2u6u7 ∈ K, then Lk(u7) and Lk(u3) shows that either u3u4u7 or u3u5u7 ∈ K.
If u3u4u7 ∈ K, the links of u7, u4, u5 and u3 show that u2u5u7, u5u7u9, u4u7u9,
u2u4u5, u4u6u8, u4u8u9, u3u5u8, u3u5u9, u2u3u6 and u3u6u8 ∈ K. Here, deg(u6) < 9.
Hence, u3u5u7 ∈ K.
To complete Lk(u7), u2u4u7, u4u7u9 and u5u7u9 have to be faces (if u4u5u7 ∈ K,
Lk(u5) implies that either u3u5u9 or u5u8u9 is a face, which is seen to be impossible
from the links of u3 and u8). This shows that the face (= uu4u5) having u4u5 as an edge
is u2u4u5, a contradiction. This proves Claim 6.
From Lk(u7) and Claim 6, we see that one of u2u3u7, u3u4u7 or u3u5u7 is a face.
If u3u4u7 is a face, the links of u7, u4, u8 and u5 show that u2u4u7, u2u5u7, u5u7u9,
u4u5u9 and u2u3u5 are faces which implies that deg(u2) < 9.
If u2u3u7 is a face, the links of u7, u2, u5, u3 and u6 show that u2u5u7, u4u5u7, u4u7u9,
u2u5u9, u3u5u9, u3u5u8, u3u4u6, u3u6u8, u2u6u8, u2u6u9, u2u4u8 and u4u8u9 are faces.
Here, K is N3.
In the last case the link of u7 shows that u2u4u7, u2u5u7 and u4u7u9 ∈ K (if u4u5u7 ∈
K, to complete Lk(u5), either u3u5u9 or u5u8u9 ∈ K, a contradiction). It is clear that
u4u5u9 ∈ K, from Lk(u4) and Lk(u5). The links of u4, u6, u2 and u3 show that u3u4u8,
u4u6u8, u2u3u6, u2u6u8, u3u6u9, u2u8u9, u2u5u9 and u3u5u8 are faces. Here, K is
isomorphic, via the map (0, 9, 7)(1, 2, 5, 4, 6, 8), to N5.
2.2.1.2: u7u8 ∈/ Lk(u1). Hence, u1u5u7 and u1u5u8 are faces. To complete Lk(u1),
either u1u6u7 and u1u3u8 or u1u6u8 and u1u3u7 are faces
2.2.1.2.1: u1u6u7 and u1u3u8 are faces. The edge u7u8 belongs to one of u2u7u8,
u3u7u8 or u4u7u8.
Subcase A: u2u7u8 ∈ K. If u2u3u7 ∈ K, the links of u7 and u3 show that u3u7u9,
u4u7u9, u4u5u7 and u3u5u8 are faces. This implies that C3(u5, u1, u3) is in Lk(u8).
Hence, either u2u4u7 or u2u7u9 ∈ K.
A.1: u2u7u9 ∈ K. To complete Lk(u7), u3u5u7, u3u4u7 and u4u7u9 have to be faces.
Considering Lk(u4), we see that either u4u6u9 or u4u8u9 ∈ K (since, u2u4 ∈/ Lk(u9)).
In the first case the links of u4, u8, u9 and u6 show that u2u4u8, u4u5u8, u3u6u8,
u6u8u9, u2u5u9, u3u5u9, u2u3u6 and u2u5u6 are faces and, hence, K is isomorphic, via
the map (0, 3, 1, 2, 8, 4, 6, 5, 9), to N10.
In the second case the links of u4, u6, u8, u2 and u9 show that u4u6u8, u2u4u5, u2u5u8,
u3u6u8, u2u3u6, u2u6u9, u5u6u9 and u3u5u9 are faces. Here, K is N9.
A.2: u2u4u7 ∈ K. The link of (u7) shows that u3u7u9 ∈ K. Considering the links
of u8 and u9, u2u8u9 and u2u3u9 ∈/ K and hence the link of u2 shows that u2u5u9 and
u2u6u9 are faces. To complete Lk(u7), either u3u4u7 or u4u7u9 ∈ K.