Also available at http://amc.imfm.si
ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 3 (2010) 109–120
Fullerene patches I
Jack E. Graver
Syracuse University, Department of Mathematics, Syracuse, NY 13244-1150, USA
Christina M. Graves
The University of Texas at Tyler, Department of Mathematics, Tyler, TX 75799, USA
Received 10 December 2009, accepted 17 April 2010, published online 10 May 2010
Abstract
Large carbon molecules, discovered at the end of the last century, are called fullerenes.
The most famous of these, C60 has the structure of the soccer ball: the seams represent
chemical bonds and the points where three seams come together represent the carbon
atoms. We use the term fullerene to represent all mathematically possible structures for
the chemical fullerenes: trivalent plane graphs with only hexagonal and pentagonal faces.
A simple consequence of Euler’s formula is that each fullerene has exactly 12 pentago-
nal faces; the only restriction on the number of hexagonal faces is that it not be 1. The
chemical motivation for this paper is to answer the question: When is it possible to alter
a fullerene by changing the structure inside a region of the fullerene bounded by a simple
closed circuit? Such a region or patch is said to be ambiguous if alterations may be made
to its interior without disturbing the structure of the fullerene outside of the region. In this
paper, we show that, relative to the minimum distance between pentagonal faces, there are
no small ambiguous patches.
Keywords: Fullerenes, fullerene patches, ambiguous patches, graphite patches.
Math. Subj. Class.: 05C10, 05C75, 92E10
1 Introduction
Let Γ = (V,E, F ) be a fullerene, a trivalent plane graph with only hexagonal and pentag-
onal faces. Let B = (V (B), E(B)) be an elementary circuit in Γ. The circuit B partitions
the sphere into two open hemispheres. A patch Π of Γ is constructed by deleting all ver-
tices, edges and faces interior to one of the hemispheres and replacing them by a single face
bounded by B. Our notational convention Π = (VΠ, EΠ, FΠ, B) should be interpreted as
follows: (VΠ, EΠ) is the graph remaining after the deletion; FΠ is the collection of faces
E-mail addresses: jegraver@syr.edu (Jack E. Graver), cgraves@uttyler.edu (Christina M. Graves)
Copyright c© 2010 DMFA Slovenije
110 Ars Math. Contemp. 3 (2010) 109–120
of Γ remaining after the deletion. Hence, Π̂ = (VΠ, EΠ, F̂Π), where F̂Π is the collec-
tion of faces FΠ along with the outside face, is a plane graph. The circuit B is called the
boundary of the patch. Graphite patches are defined to be plane graphs with all hexagonal
faces save one outside face and with all vertices on the boundary of the outside face having
degree 2 or 3 and with all other (internal) vertices having degree 3. While investigating
graphite patches, Guo, Hansen and Zheng [6] produced two nonisomorphic patches with
the same boundary code, that is the same sequence of vertex degrees around the boundary.
A fullerene or graphite patch that is not uniquely determined by its boundary is said to be
ambiguous.
When the boundary curve of the Guo, Hansen and Zheng ambiguous patches is traced
in the hexagonal tessellation of the plane or on a fullerene it intersects itself, that is, its
boundary is not an elementary circuit. In fact, it was shown in [5] that an ambiguous
graphite patch must triple cover some hexagon when projected onto the hexagonal tessel-
lation of the plane. Our patches have elementary circuits for boundaries when drawn on
the fullerene. However, the boundary curve of a fullerene patch with only hexagonal faces
may have self intersections when projected onto the hexagonal tessellation of the plane. In
spite of this, it seems to be the case that fullerene patches with only hexagonal faces or at
most one pentagonal face cannot be ambiguous. But this has not been proved. In this paper,
we show that ambiguous fullerene patches containing at most one pentagonal face, if they
exist, must be relatively large - roughly as large or larger than the smallest patch of the
fullerene containing two pentagonal faces. In the papers [1], by Brinkmann, Caporossi and
Hansen, [2], by Brinkmann, Friedrichs and von Nathusius, and [3], by Brinkmann, Graver
and Justus, it is shown that the number of faces in an ambiguous patch containing at most
one pentagonal face is uniquely determined.
The smallest example of an ambiguous fullerene patch is due to Endo and Kroto [4]
and is pictured in Figure 1. These two patches have the same bounding circuit and contain
two pentagons. However, the relative positions of the pentagons are different in these two
patches. In fact, most of the patches enclosing two or more pentagonal faces are ambiguous.
Figure 1: The Endo-Kroto ambiguous patches.
Given any patch consisting of two non-adjacent pentagons joined by a simple path of
hexagons, we may construct a different patch with the same boundary. This generalization
of the Endo-Kroto construction is illustrated in Figure 2.
Let Π = (VΠ, EΠ, FΠ, B) be a patch of a fullerene Γ. If f is a face of Π, we denote the
boundary circuit of f by B(f). The vertices of the boundary, B, of Π have degree 2 or 3 in
Π and one easily sees that every patch of size greater than 1 admits at least two boundary
vertices of degree 3. The paths in the circuit B joining two consecutive vertices of degree
3 are called the segments of B. The proof of the following lemma is straight forward.
J. E. Graver and C. Graves: Fullerene Patches I 111
Π Π∗
Figure 2: Two ambiguous patches using the generalized Endo-Kroto construction.
Lemma 1.1. Let Π = (VΠ, EΠ, FΠ, B) be a patch of size at least 2. The path S =
x0, e0, x1, . . . , ek, xk+1 is a segment of B if and only if S is a connected component of
B ∩B(f) for some face f of Π.
A face f is called an i-face if B∩B(f) consists of a single segment and that segment has
length i. By the distance between two faces in a fullerene we mean the distance between
the corresponding vertices in the dual graph. By the spread of a fullerene we mean the
minimum distance between pentagonal faces taken over all pairs of pentagonal faces. By
a disk of radius r, we mean a patch consisting of all faces of distance r or less from a
central face (the center of the disk) as long as its boundary is an elementary circuit; see
Figure 3. For any patch, its diameter is the maximum of the distances (measured in the
entire fullerene) between pairs of faces in the patch. Finally, the width of a patch is the
diameter of the disk of least diameter containing it.
Figure 3: Three disks of radius 4.
The main result of this paper is:
Theorem 1.2. Ambiguous disks contain at least two pentagonal faces. In particular, all
disks with diameter less than the spread of the fullerene are unambiguous.
As a simple corollary of the main result we have:
112 Ars Math. Contemp. 3 (2010) 109–120
Corollary 1.3. In a fullerene, any patch with width less than the spread of the fullerene is
unambiguous.
2 Proof of the main result
We start this section by proving several basic lemmas. All patches that we investigate will
be considered as stand-alone patches. Hence they can be extended or altered even though
the extension or alteration cannot be carried out in the fullerene from which they came. For
the patch Π = (VΠ, EΠ, FΠ, B), p(Π) will denote the number of pentagonal faces in the
patch and si, for i = 1 to 5, will denote the number of its i-faces.
Lemma 2.1. The number of pentagonal faces in a patch Π, p(Π), is given by p(Π) =
6 + s1 − s3 − 2s4 − 3s5.
Proof. Let Π = (VΠ, EΠ, FΠ, B) be a patch in the fullerene Γ = (V,E, F ). Denote the
number of vertices on the boundary of Π that have degree j (j = 2, 3) in Π by dj . Summing
the vertex degrees, we have 3|VΠ| − d2 = 2|EΠ| or 6|VΠ| − 4|EΠ| = 2d2. Summing the
face degrees, we have 6h+5p(Π)+|B| = 2|EΠ|, where h denotes the number of hexagonal
faces of Π. Combining these two equations gives:
6|VΠ| − 6|EΠ| = 2d2 − (6h + 5p(Π) + |B|) = d2 − d3 − 6h− 5p(Π).
The number of face of the plane graph Π̂ is h + p(Π) + 1 and we then have 6|F̂Π| =
6h + 6p(Π) + 6. Adding this and the previous equation gives:
6|VΠ| − 6|EΠ|+ 6|F̂Π| = d2 − d3 + p(Π) + 6.
By Euler’s formula, the left-hand side equals 12; so, we have p(Π) = 6− d2 + d3. Finally,
one easily writes d2 and d3 in terms of s1, . . . , s5: d3 =
∑5
i=1 si while d2 =
∑5
i=1(i−1)si
giving
p(Π) = 6 +
5∑
i=1
si −
5∑
i=1
(i− 1)si = 6−
5∑
i=1
(i− 2)si
The next lemma then follows at once.
Lemma 2.2. If Π and Π∗ are a pair of ambiguous patches then p(Π) = p(Π∗).
By a side of a patch Π we mean a set of boundary faces between two i-faces with i ≥ 3.
Note that s3 + s4 + s5 is then the number of sides of Π. We define the length of a side
in terms of the distance between the end faces in the dual. Specifically, if all of the faces
of a side have bounding segments of length 2 and if there are ` − 1 of them, we say the
side has length ` or is an `-side. We will adopt the convention that a bounding 4-face is
a side of length 0; this will eliminate the need for special cases of the formulas that we
will develop later. (5-faces will soon be eliminated from consideration, hence we adopt
no special convention for them.) See Figure 4. A side is an h|k-side (h and k positive
integers) if when rotating in the clockwise direction its first h− 1 faces and last k− 1 faces
are 2-faces and they are separated by a single 1-face.
J. E. Graver and C. Graves: Fullerene Patches I 113
Side of length 3 1-side
0-side
4|2-side
Figure 4: Side lengths.
A disk of radius r that consists solely of hexagons will be called an H-disk. A disk
of radius r that contains exactly one pentagonal face is a P-disk. Notice that the only H-
disk of radius 0 consists of a single hexagon, and the only P -disk of radius 0 is a single
pentagon.
Proposition 2.3. An H-disk with radius r greater than or equal to 1 has six r-sides. A
P -disk with radius r greater than or equal to 1 has either
(ia) five r-sides,
(ib) four r-sides and one t-side with r < t ≤ 2r or
(ii) five r-sides and one h|k-side with h + k ≤ r.
Furthermore, the boundary faces of an H- or P -disk of radius r are precisely the faces of
distance r from the center face of the disk.
Proof. The only H- and P -disks of radius 1 are easily constructed and are pictured in
Figure 5. It is clear that these disks all satisfy the conclusions of the proposition. (The
simplest P -patch of type (ii) is pictured in Figure 7.)
Figure 5: One H-disk and two P -disks of radius one.
Let Π be an H- or P -disk of radius r. We first show that all of the boundary faces are
a distance r from the center. Suppose that f is a boundary face of Π at a distance r′ from
114 Ars Math. Contemp. 3 (2010) 109–120
the center where r′ < r. Since, f is a boundary face, there must be a face f ′ of Γ not in Π
but adjacent to f . But then the distance of f ′ from the center is less than or equal to r and,
by the definition of a disk, f ′ must be a face of the patch; contradiction.
We proceed by induction on the radius: assume that Π is an H- or P -disk of radius
r > 1 and that all H-disks and P -disks of radius less than r are accurately described by
the proposition. Delete all boundary faces of Π and any other faces that are a distance r
from the center face c to get Π′. Since the faces of Π′ are all the faces of Π at a distance of
r − 1 or less from c and the faces of Π include all faces of Γ at a distance of r or less from
c, the faces of Π′ are all faces of Γ at a distance of r − 1 or less from c. Thus Π′ is an H-
or P -disk of radius r − 1 and, by induction, it has the boundary structure described above.
Now consider a face f of Γ that is a distance r from c. It must be a face of Π and it
must share an edge with a boundary face of Π′. Hence, adding all of the faces of Γ not in
Π′ but adjacent to a boundary face of Π′, reconstructs Π. In reconstructing Π, we observe
that when adding only hexagons to the boundary of an `-side produces an ` + 1-side and
adding only hexagons to the boundary of an h|k-side produces another h|k-side.
Thus, if Π′ is an H-disk of radius r − 1, adding only hexagons creates another H-disk
Π with six r-sides. If Π′ is a P -disk, we may add only hexagons and we see immediately
that: if Π′ is of type (ia) with five r−1-sides, Π is a P -disk of type (ia) with five r-sides; if
Π′ is of type (ib) with four r− 1-sides and one t′-side (r− 1 < t′ < 2r− 2), Π is a P -disk
of type (ib) with four r-sides and one t = t′ + 1-side (r < t < 2r − 1 < 2r); finally if Π′
is of type (ii) with five r− 1-sides and one h|k-side (h + k ≤ r− 1), Π is a P -disk of type
(ii) with five r-sides and one h|k-side (h + k ≤ r − 1 ≤ r).
The only remaining case is that of adding a boundary of hexagons and exactly one
pentagon to an H-disk of radius r − 1. If the pentagon is adjacent to the central edge of a
segment of length 3, the result is a P -disk of type (ib) with four r-sides and one 2r-side.
If the pentagon is not adjacent to the central edge of a length 3 segment, then it will have
just one edge on the boundary of the enlarged disk. Number the faces on the side of the
enlarged disk that includes the pentagon f1, . . . , fr+1, where f1 and fr+1 have bounding
segments of length 3. If the pentagon is face fh+1, the P -disk resulting from adding the
bounding faces will be of type (ii) with five r-sides and one h|k-side where k + h = r.
Corollary 2.4. H-disks are unambiguous.
Proof. When an H-disk is projected onto the hexagonal tessellation of the plane its bound-
ary curve is an elementary circuit. Hence it is unambiguous.
To prove that P -disks are unambiguous, we are going to need to know just which
boundary curves are possible for certain classes of hexagonal patches - patches with only
hexagonal faces. By Lemma 2.1, when a hexagonal patch has s1 = s4 = s5 = 0 it has
exactly six sides. If the lengths of these sides are a, b, c, d, e, f in cyclic order (counter-
clockwise), we say the patch has side parameters, or simply parameters, (a, b, c, d, e, f).
If a hexagonal patch has s1 = s5 = 0, s4 ≥ 1, then s3 = 6 − 2s4 and it will have 6 − s4
sides of positive length. By our convention, each 4-face is a side of length 0. Hence such
a patch can also be described by six side parameters (a, b, c, d, e, f) where non-adjacent
variables could take on the value 0; for example, the patch with parameters (a, 0, c, d, e, f)
where a, c, d, e, f > 0 is a 5-sided hexagonal patch with the a and c sides separated by a
4-face.
J. E. Graver and C. Graves: Fullerene Patches I 115
Lemma 2.5. Let Π be a hexagonal patch with s1 = s5 = 0 and side parameters (a, b, c, d,
e, f). Then there are no consecutive sides of length 0 and the following three equalities
must hold: a + b = d + e, b + c = e + f and c + d = f + a.
Proof. If we project Π onto the hexagonal tessellation, then the boundary of the patch
can be represented in the dual (triangular) tessellation by the circuit joining consecutive
bounding faces. In this representation, a side of length x is represented as straight lines of
x dual edges; these straight lines make 120◦ angles at 3-faces and 60◦ angles at 4-faces.
Since s1 = s5 = 0 and s3 + 2s4 = 6 (by Lemma 2.1), we see that Π has at most three
4-faces. If Π has three 4-faces, then Π is a “triangular” patch. It is clear that this triangle
must be an equilateral triangle and that all sides of Π are the same length. Thus, the side
parameters of Π are (a, 0, a, 0, a, 0) which do satisfy the above equalities. (See Figure 6.)
d
d
b=0
b+c
a+b
f
d
c
b
bb
a
e+f
d+e
b+c
a+b
f
ee e d
c
b
bb
a
e=0
e=0
a
c
d=a
f=ca=c+d
e=c
c
d d
c
f=0 0
b=0b=0
d=0f=0
c=a
e=a
a
Figure 6: Hexagonal Patch Boundaries.
If Π has exactly two 4-faces, called f1 and f2, then it must have two 3-faces, g1 and
g2. If f1 and f2 are separated by only 2-faces, then projecting the patch onto the hexagonal
tessellation has as its diagram a trapezoid with sides of length (a, 0, c, d, e, 0) where a is
the side between f1 and f2. It is apparent that sides a and d are parallel and that b + c =
c = e = e+f . We can enlarge the patch by extending sides c and e to sides of length c+d.
This new patch is an equilateral triangle, and thus a+b = a = e+d and a+f = a = c+d.
On the other hand, if f1 and f2 are separated by 2-faces and one 3-face, then projecting Π
onto the hexagonal tessellation yields a parallelogram with side parameters (a, 0, c, d, 0, f)
which must satisfy a = d and c = f giving a + f = d + c, a + b = a = d = e + d and
b + c = c = f = e + f .
If Π has exactly one 4-face, then projecting Π onto the hexagonal tessellation yields a
patch with side parameters (a, b, c, d, 0, f) where sides f and c are parallel and sides a and
d are parallel. By extending sides a and c by b hexagons, we get a parallelogram which
must satisfy a+b = d and b+c = f . It follows that a+b = d = e+d and b+c = f = e+f .
Adding a + b = d and f = b + c gives a + b + f = b + c + d and canceling the b gives
c + d = f + a.
Finally, if Π has no 4-faces, then Π is a patch with side parameters (a, b, c, d, e, f)
116 Ars Math. Contemp. 3 (2010) 109–120
where sides a and d are parallel, sides b and e are parallel, and sides f and c are parallel.
By extending sides a and c by d hexagons, and extending sides f and d by e hexagons, we
create a parallelogram (see Figure 6). This parallelogram gives a+b = d+e, b+c = e+f
and a + b + e + f = b + c + d + e directly and a + f = c + d by canceling the b and e in
the last equation.
If a P -disk Π were ambiguous, there exists a patch Π∗ with the same boundary. So
we introduce a class of patches that will include all of these possible alternatives. By an
r-patch, we mean a patch of one of the following two types. An r-patch of the first type
is a patch with s1 = s4 = s5 = 0, s3 = 5 and with 4 sides of length r and one side
of length t (t ≥ r). That is, an r-patch of the first type has side parameters (r, r, r, r, t),
where t ≥ r. An r-patch of the second type is a patch with s4 = s5 = 0, s1 = 1 and
s3 = 6, with five r-sides and one h|k-side. That is, an r-patch of the second type has
side parameters (r, r, r, r, r, h|k), where h and k are positive integers. It is an immediate
corollary of Lemma 2.1 that an r-patch contains exactly one pentagon. One easily checks
that there are only two 1-patches: the two P -disks of radius 1 with parameters (1, 1, 1, 1, 1)
and (1, 1, 1, 1, 2). These are both r-patches of the first type and are pictured in Figure 5.
The simplest r-patch of the second type has parameters (2, 2, 2, 2, 2, 1|1) and is pictured in
Figure 7; note that it is a P -disk of radius 2.
Figure 7: The r-patch of the second type with parameters (2, 2, 2, 2, 2, 1|1).
The following lemma gives another property of r-patches — namely that we can delete
all bounding faces of an r-patch (r > 1) and still get a legitimate patch. It will always be
possible to remove all of the faces on a side unless one of those faces actually meets the
boundary in two distinct segments — its removal would then leave a “disconnected” patch.
It will always be possible to remove all of the faces on the boundary unless two faces on
different sides share an edge. See Figure 8.
Lemma 2.6. Let Π = (VΠ, EΠ, FΠ, B) be an r-patch r > 1.
i. Let f be a face of Π. Then, B(f) ∩B consists of at most one path.
ii. Let f and g be faces of Π that are not consecutive as one cyclically traverses the
boundary. Then, B(f) ∩B(g) = ∅.
J. E. Graver and C. Graves: Fullerene Patches I 117
Figure 8: Impossible r-patch boundaries.
Proof. (i) Assume there exists a face f in Π consisting of 2 disjoint segments, a cut-face.
Clearly both segments of f have length 2 or less. If both segments have length 2, then both
faces adjacent to f are also cut-faces. Extending to the left, the string of cut-faces must
terminate. Terminating with a single face is impossible since this is an r-patch and such a
face would be a 4-face or 5-face. Hence the leftmost cut-face must meet two faces on its
left and must have at least one bounding segment of length 1. If f meets two faces on its
right, then it would have two bounding segments of length 1. If f meets only one face on
its right then the right-most cut face in the string of cut-faces to the right must also have at
least one bounding segment of length 1. Hence this patch must have at least two bounding
segments of length 1 contradicting the assumption that it is an r-patch.
(ii) Consider any r-patch Π and suppose that it has a pentagonal 3-face f on its bound-
ary. If the patch is of type (r, r, r, r, t), replacing f by a hexagon will result in a hexagonal
patch with boundary lengths (r, 0, r, r, r, t), (r, r, 0, r, r, t) or (r, r, r, r, 0, t); all of which
violate the conditions of Lemma 2.5. If the patch is of type (r, r, r, r, r, h|k), replacing f
by a hexagon will result in a hexagonal patch with boundary lengths (r, 0, r, r, r, r, h|k),
(r, r, 0, r, r, r, h|k) or (r, r, r, r, r, 0, h|k). Now we can eliminate the segment of length 1
in the h|k-side by attaching an h× k parallelogram of hexagons (see the left-hand diagram
in Figure 9, also see Figure 10). The resulting six-sided hexagonal patches will then have
boundary lengths (h + r, 0, r, r, r, r + k), (h + r, r, r, 0, r, r + k) or (h + r, r, r, r, r, k); all
of which violate the conditions of Lemma 2.5. Hence Π has no pentagonal 3-face f on its
boundary.
Let f and g be adjacent, non-consecutative boundary faces - a cut-pair. Since they are
non-consecutive, the face labeled y in the right-hand diagram of Figure 9 must belong to
the patch. If neither of the faces labeled x and z belong to the patch, then y would be a
hexagonal 4-face, a pentagonal 3-face or a cut face. None of these is possible in an r-patch.
Hence we may assume that z belongs to the patch. We may also assume without loss of
generality that f and g are the leftmost cut-pair and so x is also in the patch. By a similar
argument v and one of u and w on the right must belong to the patch - say w. Hence g
is a 1-face. If u is a face of the patch, then f is also a 1-face. If u is not a face of the
patch, then v and w form a cut pair and the right-most cut pair in this chain of cut-pairs
must also include a 1-face. Hence this patch must have at least two 1-faces contradicting
the assumption that it is an r-patch.
The core of a patch Π is the plane subgraph obtained by removing all the boundary
faces of Π. By the construction of P -disks in Lemma 2.3, it is clear that a P -disk of radius
r has either an H-disk or a P -disk of radius r− 1 as its core. By Lemma 2.6, all r-patches
have a legitimate patch as their core. We use this fact in carrying out the next step to
118 Ars Math. Contemp. 3 (2010) 109–120
r
r
r
r
r
h
h
k
k
f
g
x
y
z
u
v
w
Figure 9: Diagrams for the proof of Lemma 5
eventually show that the only r-patches are the P -disks. The next step is to show that the
parameters of any r-patches actually match the parameters of some P -disk.
Lemma 2.7. Let Π be an r-patch of the first type with parameters (r, r, r, r, t) Then t ≤ 2r
and
i. if t = 2r, Π has a pentagon as the central face of the t-side;
ii. if t < 2r, Π has only hexagons on the boundary.
Proof. Let Π have parameters (r, r, r, r, t). In the proof of the previous lemma, we showed
that a pentagon in the boundary of an r-patch cannot be a 3-face. Suppose that the ith face
(0 < i < r) in the first r-side is a pentagon. Replace this face by a hexagon. This new
hexagonal patch has parameters (i, r− i, r, r, r, t); which is impossible since i + (r− i) 6=
r + r. We get the same kind of contradiction if we assume that any face on any of the
r-sides are pentagons. Now assume that the ith face (0 < i < t) in the t-side is a pentagon.
Again, we replace the pentagon by a hexagon to get a hexagonal patch with parameters
(r, r, r, r, i, t− i). We must have r +r = r + i and r +r = i+(t− i). This is possible only
if i = r and t = 2r. Thus, the only way a pentagon can be on the boundary is if t = 2r and
the central face on the t-side is a pentagon.
To prove that t < 2r when Π has only hexagonal boundary faces, we’ll go by induction
on r. This inequality holds for r = 1 since the only 1-patch with only hexagonal boundary
faces has parameters (1, 1, 1, 1, 1). Let Π be an r-patch with parameters (r, r, r, r, t) where
r > 1 and with only hexagons on its boundary. We can then remove the boundary and
consider its core. If a `-side has only hexagons, removing them produces an (` − 1)-side.
Thus, the core must be an (r−1)-patch with side parameters (r−1, r−1, r−1, r−1, t−1).
By induction, we have that (t− 1) ≤ 2(r − 1) or t < 2r.
Lemma 2.8. Let Π be an r-patch of the second type with parameters (r, r, r, r, r, h|k).
Then h + k ≤ r and
i. if h + k = r, then Π has a pentagon as the face with bounding segment of length 1;
ii. if h + k < r, then Π has only hexagons on the boundary.
Proof. Let Π have parameters (r, r, r, r, r, h|k). Suppose that the ith face in the h|k-side,
i < h, is a pentagon. We replace the pentagon by a hexagon to get a hexagonal patch with
J. E. Graver and C. Graves: Fullerene Patches I 119
parameters (r, r, r, r, r, i, (h−i)|k). If we attach the hexagons of a (h−i)×k parallelogram
(see Figure 10), we get a 6-sided hexagonal patch with parameters (r, r, r, r, i+k, r+h−i).
By Lemma 2.5, we have (i + k) + (r + h− i) = 2r; from which we conclude that we must
have h + k = r. Considering another adjacent pair, we have (r + h − i) + r = 2r; from
which we conclude that i = h. Thus the only case in which one of the faces on the h|k-side
could be a pentagon is the case where h + k = r and the pentagonal face is the face with
boundary segment of length 1.
r i
h-i k r
r i
h-i
k
r
r i+k
r+h-i
Figure 10: Diagrams for the proof of Lemma 2.8
Next suppose that the ith face in first r-side is a pentagon. Replace it in the patch by a
hexagon. This results in a hexagonal patch with parameters (i, r − i, r, r, r, r, h|k). Filling
in to eliminate the length 1 segment then yields a 6-sided hexagonal patch with parameters
(i + h, r − i, r, r, r, r + k); which is impossible since (r − i) + r = r + (r + k) > 2r.
We get the same kind of contradiction if we assume that any face on any of the r-sides are
pentagons. Thus, if Π has a pentagon on the boundary, h + k = r and the pentagon is the
face with bounding segment of length 1.
To prove that h + k < r when Π has only hexagonal boundary faces, we proceed by
induction on r. One easily checks that there are no 1-patches of this second type and the
only 2-patch of the second type is pictured in Figure 7 and has parameters (2, 2, 2, 2, 2, 1|1).
Hence, the lemma holds for all r-patches with r ≤ 2. Let Π be an r-patch with parameters
(r, r, r, r, r, h|k) for r > 2 and with only hexagons on its boundary. We can remove the
boundary and consider its core. If an r-side has only hexagons, removing them produces an
(r − 1)-side; if an h|k-side has only hexagons, removing them produces another h|k-side.
Thus, the core must be an (r − 1)-patch with sides (r − 1, r − 1, r − 1, r − 1, r − 1, h|k).
By induction, we see that h + k ≤ r − 1 or h + k < r.
This final proposition concludes our proof of the main result.
Proposition 2.9. P -disks are unambiguous.
Proof. This proof will proceed by induction on r. It is easy to check that the only two
P -disks of radius 1 are unambiguous. Hence we assume that Π is a P -disk of radius r > 1
and that P -disks of radius r − 1 are unambiguous. Let Π∗ be another patch with the same
boundary as Π. By Proposition 2.3, Π∗ must be an r-patch with the same parameters as Π.
We will show that Π∗ is in fact the same patch as Π.
For the first case, we assume that Π has parameters (r, r, r, r, t) with t < 2r. By
Lemma 2.7, we see that Π and Π∗ have only hexagons on their boundaries. Stripping off
120 Ars Math. Contemp. 3 (2010) 109–120
the hexagonal boundary faces of both Π and Π∗ yields in each case a core that is an (r−1)-
patch with parameters (r−1, r−1, r−1, r−1, t−1). By Proposition 2.3, there is a P -disk
with these parameters and by the induction hypothesis it must be the only (r − 1)-patch
with these parameters. Hence, Π and Π∗ have the same P -disk as core. There is only one
way to add hexagons around the boundary of a P -disk, so Π∗ must be the same patch as Π.
Now, assume that Π has parameters (r, r, r, r, 2r). By Lemma 2.7, both Π and Π∗ have
a pentagon as the central face of the 2r-side. Thus the core of both Π and Π∗ is the H-disk
of radius r − 1. Replacing the boundary faces, we see that Π∗ = Π.
If Π has parameters (r, r, r, r, r, h|k) with h+k < r, both Π and Π∗ have only hexagons
on the boundary. Thus, the core of Π is the P -disk with radius r − 1 and parameters
(r − 1, r − 1, r − 1, r − 1, r − 1, h|k). Since the core of Π∗ is an r-patch with the same
parameters, it is by induction the same P -disk as the core of Π. Again, replacing the
boundary faces, we see that Π∗ = Π.
Finally, assume Π has parameters (r, r, r, r, r, h|k) with h + k = r. By Lemma 2.8,
Π and Π∗ must have a pentagon on the boundary and that pentagon is the boundary face
with bounding segment of length 1. Thus, the core of both Π and Π∗ is the H-disk of
radius r − 1. As above Π and Π∗ have the same core and the same boundary faces and are
therefore equal.
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