ARS MATHEMATICA CONTEMPORANEA
ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 15 (2018) 487-497 https://doi.org/10.26493/1855-3974.1359.b33
(Also available at http://amc-journal.eu)
The isolated-pentagon rule and nice substructures in fullerenes*
*
Hao Li
Laboratoire de Recherche en Informatique, UMR 8623, C.N.R.S.-Université Paris-sud,
Received 22 March 2017, accepted 6 September 2017, published online 5 September 2018
After fullerenes were discovered, Kroto in 1987 proposed first the isolated-pentagon rule (IPR): the most stable fullerenes are those in which no two pentagons share an edge, that is, each pentagon is completely surrounded by hexagons. To now the structures of the synthesized and isolated (neutral) fullerenes meet this rule. The IPR can be justified from local strain in geometry and n-electronic resonance energy of fullerenes. If two pentagons abut in a fullerene, a 8-circuit along the perimeter of the pentalene (a pair of abutting pentagons) occurs. This paper confirms that such a 8-circuit is always a conjugated cycle of the fullerene in a graph-theoretical approach. Since conjugated circuits of length 8 destabilize the molecule in conjugated circuit theory, this result gives a basis for the IPR in n-electronic resonance. We also prove that each 6-circuit (hexagon) and each 10-circuit along the perimeter of a pair of abutting hexagons are conjugated. Two such types of conjugated circuit satisfy the (4n + 2)-rule, and thus stabilise the molecule.
Keywords: Fullerene, patch, stability, isolated pentagon rule, Kekule structure, conjugated cycle, cyclic edge-cut.
Math. Subj. Class.: 05C70, 05C10, 92E10
»This work was supported by NSFC (Grant Nos. 11371180, 11871256). t Corresponding author.
E-mail addresses: Hao.Li@lri.fr (Hao Li), zhanghp@lzu.edu.cn (Heping Zhang) ©® This work is licensed under https://creativecommons.org/licenses/by/3.0/
F-91405, Orsay, France
Heping Zhang t
School of Mathematics and Statistics, Lanzhou University, Lanzhou, Gansu 730000, P.R. China
Abstract
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1	Introduction
The fullerenes are closed carbon-cage molecules such that every carbon atom has bonds to three other atoms, and the length of each carbon ring is either 5 or 6. Ever since the first fullerene, Buckministerfullerene C6o, was discovered by Kroto et al. in 1985 [15], the stabilities of fullerenes have attracted many theorist's attentions. The simple Huckel molecular orbital model that predicts reliably the relative stabilities of planar aromatic hydrocarbons is not generally found to work so well for fullerenes. Kroto [14] in 1987 proposed first the isolated-pentagon rule (IPR): the most stable fullerenes are those in which no two pentagons share an edge, that is, each pentagon is completely surrounded by hexagons. Schmalz et al. [23] gave a more theoretical discussion of the rule in support of the fullerene hypothesis. Indeed the structures of the synthesized and isolated fullerenes meet this rule. The IPR can be justified from local strain and n-electronic resonance of fullerenes; for details, also see a book due to Fowler and Manolopoulos [7]. Pentagon adjacency leads to higher local curvature of the molecule surface and increases the strain energy. On the other hand, according to Huckel (4n+2)-rule, conjugated circuits of length 6,10,14,... stabilize the molecule, whereas conjugated circuits of length 4, 8,12,... destabilize the molecule. Here a conjugated circuit is a cycle of alternating single and double bonds within a Kekule structure. If two pentagons abut in a fullerene, the conjugated or resonant 8-circuit along the perimeter of the pentalene may occur, and this leads to resonance destabilization [22]. This is an interpretation of IPR in n-electronic resonation stabilization. However, a problem occurs: In a fullerene, is every 8-length circuit conjugated? To now we have not seen any definite answer to this problem in mathematics. In this article we investigate nice patches of a fullerene by applying some small cyclic edge-cuts of graphs and present a positive answer to the above problem (a patch of a fullerene is nice if its Kekule structure can be extended to a Kekule structure of the entire fullerene). As immediate consequences of our main theorems, we have that every 8-length circuit of a fullerene surrounds a pentalene (a pair of abutting pentagons) and is conjugated or alternating with respect to a Kekule structure (see Corollary 3.4). This confirms the destabilization of any pentalene as a nice substructure to the entire fullerene and thus gives a mathematical support for the IPR of fullerenes. Furthermore we also show that in a fullerene every hexagon is a conjugated 6-circuit (see Corollary 3.3) and the boundary along a naphthalene (i.e. a pair of abutting hexagons) is a conjugated 10-circuit (see Corollary 4.2). The former has already been proved (see [26]). In conjugated circuit theory [10, 19, 20], conjugated 6-circuits and 10-circuits contribute stabilizations of fullerenes and the small conjugated circuits have the greatest effects (positive and negative) on stability. For recent discussions on the IPR of fullerenes about steric strain factor and n-electronic resonance factor, see [1, 2, 8, 13, 21]. For mathematical aspects of fullerenes, see a recent survey [3].
2	Preliminary
To obtain the above end we now start our arguments in a graph-theoretical approach. As a molecular graph of a fullerene, a fullerene graph is a 3-connected planar cubic graph with only pentagonal and hexagonal faces. It is well known that a fullerene graph on n vertices exists for every even n > 20 except n = 22 [9]. By Euler's polyhedron formula, every fullerene graph with n vertices has exactly 12 pentagonal faces and (n/2 - 10) hexagonal faces.
Let G be a graph with vertex-set V(G) and edge-set E(G). An edge set M of a graph
H. Li and H. Zhang: The isolated-pentagon rule and nice substructures infullerenes 489
G is called a matching if no two edges in M have a common endvertex. A matching M of G is perfect if every vertex of G is incident with one edge in M. In organic molecular graphs, perfect matchings correspond to Kekule structures, playing an important role in analysis of the resonance energy and stability of polycyclic aromatic hydrocarbons.
The following classical theorem is Tutte's 1-factor theorem on the existence of perfect matching of a graph [24]. For detailed monograph on matching theory, see Lovasz and Plummer [17].
Theorem 2.1. A graph G has a perfect matching if and only if odd(G — S) < |S | for each S C V (G), where odd(G — S) denotes the number of odd components in subgraph G — S.
Subgraph G' of a graph G is called nice if G — V(G') has a perfect matching. In particular, an even cycle C of a graph G is nice if G has a perfect matching M such that C is an M-alternating cycle, i.e. the edges of C alternate in M and E(G) \ M .A nice even cycle is also called resonant or conjugated cycle (or circuit) in chemical literature. For convenience, a cycle of length k is said to be a k-cycle or k-circuit.
For nonempty subsets X, Y of V(G), let [X, Y] denote the set of edges of G that each has one end-vertex in X and the other in Y .If X = V (G) \ X = 0, then V(X) := [X, X] is called an edge-cut of G, and k-edge-cut whenever | [X, X] | = k. The edges incident with a single vertex form a trivial edge-cut. For a subgraph H of G, let H := G — V(H). We simply write V(H) for V(V(H)).
Lemma 2.2 ([25]). Every 3-edge-cut ofafullerene graph is trivial.
Lemma 2.3 ([25]). Every 4-edge-cut ofafullerene graph isolates an edge.
An edge-cut S = V(X) of G is cyclic if at least two components of G—S each contains a cycle. The minimum size of cyclic edge-cuts of G is called cyclic edge-connectivity of G, denoted by cA(G).
Theorem 2.4 ([6, 12, 18]). Let F be anyfullerene graph. Then cA(F) = 5.
From the definition with the above properties we know that each fullerene graph has the girth 5 (the minimum length of cycles), and each of its 5-cycles and 6-cycles bounds a face. A cyclic k-edge-cut of a graph isolating just a k-cycle will be called trivial.
Theorem 2.5 ([12, 16]). A fullerene graph with a non-trivial cyclic 5-edge-cut is a nan-otube with two disjoint pentacaps (see Figure 1), and each non-trivial cyclic 5-edge-cut must be an edge set between two consecutive concentric cycles of length 10.
A fullerene patch is a 2-connected plane graph with all faces pentagonal or hexagonal except one external face, all internal vertices (not incident with the external face) of degree 3 and those incident with the external face having degree 2 or 3. The cycle bounding the external face is the boundary of the patch. We can count the pentagons of a fullerene patch as internal faces as follows.
Lemma 2.6 ([4]). For fullerene patch G, let p5 denote the number of pentagonal faces other than the external face. Then
P5 = 6 + k3 — k2 =6 + 2k3 — l,	(2.1)
where k2 and k3 denote the number of vertices of degree 2 and 3 on the boundary of G, respectively, and l is the boundary length.
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For T C V(G), the induced subgraph of G by T consists of T and all edges whose endvertices are contained in T, denoted by G[T].
In the next two sections we will investigate nice patches of fullerene graphs in cyclic 6-edge-cut and 8-edge-cut cases, respectively.
3 Cyclic 6-edge-cut
We first consider a more general case than fullerene patches.
Theorem 3.1. Let F0 be a connected induced subgraph of a fullerene graph F such that interior faces of F0 exist and each one is a pentagon or hexagon. If F has exactly six edges from F0 to the outside F0 = F — V(F0), then F0 has a perfect matching.
Proof. Let n0 and e0 denote the numbers of vertices and edges of F0 respectively. Then 3n0 = 2e0 + 6, which implies that n0 is even, i.e. F0 has an even number of vertices.
We will prove that Fo has a perfect matching by Tutte's theorem. To the contrary suppose that F0 has no perfect matchings. By Theorem 2.1, there exists a subset X0 c V(Fo) such that
odd(Fo — Xo) > |Xo|.	(3.1)
For the sake of convenience, let a := odd(F0 — X0). Since a and |X0| have the same parity, we have
a > |Xo| + 2.	(3.2)
Let Gi,..., Ga and Ga+i,..., Ga+^ denote respectively the odd components and the even components of F0 — X0, where ft denotes the number of even components of F0 — X0. For i = 1, 2,..., a + ft, let denote the number of edges of F0 which are sent to X0 from Gj, and Yi (resp. yo) the number of edges of F from Gi (resp. X0) to F0. Since V(F0) is a 6-edge-cut of F, we have
a+P
|V(F0)| = ^ Yi = 6.	(3.3)
i=0
Since F is 3-connected, for i = 1,..., a,..., a + ^ we have
|V(Gj)| = mj + Yi > 3.
(3.4)
H. Li and H. Zhang: The isolated-pentagon rule and nice substructures infullerenes 491
Figure 2: Illustration for the proof of Theorem 3.1.
By taking the number of edges of F from the components Gi to F0 and X0 into account and by using Equation (3.3) and Inequalities (3.2) and (3.4) we have
a+p
3(a + £) < ^ (mi + Yj)
i=1
a+p
<	3|Xo|- yo + ^ Yi	(3.5)
i=l
= 3|xO| +6 - 2yO
<	3a — 2yo,
which implies that ft = 0, y0 =0 and equalities always hold. Hence J2a=1 Yi = 6, and a = |X0| + 2. Further, the second equality in (3.5) implies that X0 is an independent set of F0. The first equality in (3.5) implies that mi + Yi = 3 for each 1 < i < a, that is, V(Gi) is a 3-edge-cut of F. So by Lemma 2.2 it is a trivial edge-cut and each Gi is a singleton. Let Y0 denote the set of all singletons Gi. Then F0 is a bipartite graph with partite sets X0 and Y0.
If F0 has no vertices of degree one, then F0 is 2-connected. Otherwise, F0 has a bridge, the deletion of which results in two components each containing a cycle. So the bridge together with at most three edges in V(F0) form a cyclic edge-cut, contradicting that cA(F) = 5 (Theorem 2.4). Hence F0 is a fullerene patch. Since k2 = |V(F0)| = 6, by Lemma 2.6 we have that the number p5 of pentagons contained in F0 is equal to the number k3 of vertices of degree three lying on the boundary of F0. Since F0 is bipartite, k3 = p5 = 0, which implies that F0 is just a hexagon, contradicting that a = | Y01 = |X01 +2.
If F0 has a vertex x of degree one, let xy be the edge of F0, and xy1 and xy2 be the other two edges in F incident with x. Then V(F0 — x) = (V(F0) \ {xy1, xy2}) U {xy} forms a cyclic 5-edge-cut of F since F0 — x contains all cycles of F0 and F0 — x can be obtained from F — F0 by adding a 2-length path y1xy2 and contains at least seven pentagons. Since F0 — x is bipartite, cyclic 5-edge-cut V(F0 — x) is not trivial, and F0 — x
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is always 2-connected from Theorem 2.5. By Lemma 2.6 we have p5 = k3 + 1 for the fullerene patch F0 — x, which implies that F0 has at least one pentagon, contradicting that F0 is bipartite.	□
Corollary 3.2. For each cyclic 6-edge cut E0 of a fullerene graph F, both components of F — E0 have a perfect matching.
Proof. It follows that F — E0 has exactly two components from Lemma 2.2 and 3-edge-connectedness of F. Such two components fulfil the conditions of Theorem 3.1 and thus each has a perfect matching.	□
0 <£>
Figure 3: Some nice substructures of fullerene graphs.
Figure 4: Some nice patches of fullerene graphs with six 2-degree vertices.
From Corollary 3.2 we can find many nice substructures of fullerene graphs, examples of which are shown in Figures 3 and 4. It should be mentioned that the third nice substructure fulvene in Figure 3 has been discovered by Doslic applying 2-extendability of fullerenes [5, 27], and the first one has been proved in investigating k-resonance [26, 11]; see the following.
Corollary 3.3 ([26]). Each hexagon of a fullerene graph is resonant.
Corollary 3.4. Each 8-length cycle (if exists) of a fullerene graph bounds a pentalene (a pair of abutting pentagons) and is thus resonant.
Proof. Let C be a 8-length cycle of a fullerene graph F .If F has an edge e whose endver-tices both lie in C but e ^ E(C), then e is called a chord of C. If C has no chords, then the
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eight edges issuing from C can be classified into two edge-cuts of size from 3 to 5, which lie in the interior and the exterior of C respectively. If one is a 3-edge-cut, then Lemma 2.2 implies that it is trivial, and thus a triangle or quadrilateral appear, a contradiction. If both are 4-edge-cuts, then Lemma 2.3 implies that F has only 12 vertices, also a contradiction. So C must have a chord. Further, this chord and C form a pair of 5-length cycles sharing this chord, which must bound pentagonal faces of F by Theorem 2.4. That is, C bounds a pentalene and is resonant from Corollary 3.2.	□
4 Cyclic 8-edge-cut
Theorem 4.1. If E0 is a cyclic 8-edge-cut of a fullerene graph F and E0 is a matching, then F — E0 has a perfect matching.
Proof. There exists a nonempty and proper subset X of vertex set V(F) such that E0 = V(X) = [X,X]. Let Fo := F[X] and Fo := F[X]. We claim that both Fo and Fo are connected and E0 is a minimal edge-cut. If not, then one of F0 and F0, say F0, is disconnected. Then F0 has exactly two components since F is 3-connected. Since E0 is a matching, F0 and each component of F0 have the minimum degree 2 and contain a cycle. So a cyclic edge-cut of at most four edges occurs in F, a contradiction. So the claim is verified. Hence each of F0 and F0 has exactly one face of size more than six, which has exactly 8 two-degree vertices on its boundary.
We only show that F0 has a perfect matching (the same for F0). If F0 has a bridge, then it follows that F0 can be obtained from two pentagons by adding one edge between them by Theorems 2.4 and 2.5. In this case F0 has a perfect matching. So in the following we always suppose that F0 is a patch of F. We adopt similar arguments and notations as in the proof of Theorem 3.1 (see Figure 2). It is known that F0 has an even number of vertices. Suppose to the contrary that F0 has no perfect matchings. By Tutte's theorem we can choose a minimal subset X0 c V(F0) satisfying a := odd(F0 — X0) > |X0| + 2.
Let Gi,..., Ga and Ga+i,..., Ga+p denote respectively the odd components and the even components of F0 — X0. For i = 1, 2,..., a + ft, let denote the number of edges of F0 which are sent to X0 from Gj, and 7i (resp. 70) the number of edges of F from Gi (resp. X0) to the patch F0. By |V(F0) | = Y,0=cf Yi =8 and Inequality (3.4), we have
a+p
3(a + ft) < Y (mi + Yi) i=i
a+p
<	3|X0|— 70 + Y Yi	(4.1)
i=i
= 3|X0| + 8 — 270
<	3a + 2 — 2y0,
which implies that ft = 0, 0 < 70 < 1, and |X0| + 2 = a. So the forth equality in Inequality (4.1) holds.
If 70 = 1, then |[X — X0,X]| = ^jf Yi = 7 and all equalities in Inequality (4.1) hold. Like the proof of Theorem 3.1 we have that X0 is an independent set, mi + Yi = 3 for each 1 < i < a and each Gi is a singleton. Hence F0 is a bipartite graph. By Lemma 2.6 we have that F0 has two three-degree vertices on the boundary of F0. That implies that
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F0 is just the graph obtained by gluing two hexagons along an edge. So F0 has the same cardinalities of two partite sets, which contradicts that |X0| + 2 = a.
From now on we suppose that y0 = 0. That is, each vertex of X0 has degree 3 in F0. We claim that second equality in Inequality (4.1) must hold. Otherwise, F0 [X0] has exactly one edge, say uv, and the first equality holds, so each Gi is a singleton. Without loss of generality, suppose that y1 and y2 are two neighbors of u other than v, and V(Gi) = jyi} and V(G2) = {y2}. Let X0 := X0 \ {u}, and Xi := {u,yi,ys}. Then Gi := F0[Xi] is a 3-vertex path obtained by combining Gi and G2 with vertex u. Hence F0 - X0 has the odd components Gi, G3,..., Ga, and odd(F0 - X0) = a - 1 = |X01 + 2, contradicting the minimality of X0.
Hence X0 is an independent set of F0, and the first inequality is strict. Since for each 1 < i < a, mi + Yj is always odd, there exists an i0 such that mio + Yio =5 and mi + Yi = 3 for all i = i0. For convenience, we may suppose that i0 = 1. So Gi is an odd component with at least three vertices and G2,..., Ga are all singletons. Let Y0 denote the set of all singletons Gi (2 < i < a). Then H := (X0, Y0) is a bipartite graph as the induced subgraph of fullerene graph F.
If Gi is a tree, then it is a 2-length path, say xyz, since V(Gi) has exactly five edges. For F0, by Lemma 2.6 we have p5 = k3 - 2. Since E0 is a matching, x and z both have neighbors in X0, so y1 < 3. The latter implies J2a=2 Yi > 5. That is, the boundary of F0 contains at least 5 two-degree vertices belonging to Y0.
We assert that p5 < 2. Since H is bipartite, any pentagon P of F0 must intersect G1. If P only intersects a vertex of G1, say z, then P - z is a path of length 3 in H which connects two vertices of X0, contradicting that any path between two vertices in the same partite set of a bipartite graph has an even length. Similarly we have that P cannot contain both edges of G1. If F0 has two distinct pentagons sharing the same edge of G1, then one pentagon must have two edges G1, a contradiction. So the assertion holds.
By the assertion and p5 = k3 - 2 we have k3 < 4. This implies that the boundary of F0 has at most 4 vertices in X0. Let C be the boundary of F0. Then C - V(C) n X0 has at most |V(C) n X0| components. On the other hand, C - V(C) n X0 has all singletons in V(C) n Y0 as components. But |V(C) n Y0| > 5, contradicting |V(C) n X0| < 4.
From now on suppose that G1 contains a cycle. Then V(G1) is a cyclic 5-edge-cut of F. By Theorem 2.5 V(G1) is a matching and G1 is also a patch (precisely, G1 is a pentagon or contains a pentacap according as the cyclic 5-edge-cut V(G1) is trivial or not), so each vertex of H has degree at least two, and each component of H contains a cycle. If H is disconnected, then H has exactly two components H1 and H2 since F0 is 2-edge-connected and V(G1) has exactly five edges. Further, between G1 and each Hi has at least two edges. So V(G1) has two consecutive edges along the boundary of G1 separately from G1 to H1 and H2. These two edges must be contained in a cycle of length at least 8 bounding a face of F, a contradiction. Hence H is connected.
Since G1 and F0 are two connected subgraphs of F with exactly one face of size more than six, there are two possible cases to be considered.
Case 1. G1 and F0 lie in different faces of H. Suppose that G1 lies in a bounded face f of H and F0 does in the exterior face of H. Then the boundary df of f is a 10-length cycle since 5 neighbors of G1 in H belong to X0 and are separated by 5 vertices in Y0. Hence F is a nanotube with two pentacaps and F0 has exactly 6 pentagons. By Lemma 2.6 the boundary of F0 has exactly 8 vertices of degree 3 in F0. Hence the boundary of F0 is an alternating cycle of three-degree and two-degree vertices. But in this nanotube there is only
H. Li and H. Zhang: The isolated-pentagon rule and nice substructures infullerenes 495 10-length cycle as such boundary of a patch, a contradiction.
Figure 5: Illustration for Case 2 in the proof of Theorem 4.1 (the vertices in X0 are colored white and other vertices black).
Case 2. G1 and F0 lie in the exterior face of H. Then the boundary of F0 is formed by a path P of H and a path P1 of Gi and two edges between them. So 0 < y1 < 3, and there are 8 - y1 two-degree vertices lying on P, which belong to Y0 and are thus non-adjacent mutually. So there are at least 7 - y1 three-degree vertices in X0 on P that can separate them. Since the four end-vertices of P and P1 are all of degree three in F0, there are at least 11 - y1 vertices of degree three of F0 on the boundary. That is, for F0, k3 > 11 - y1. On the other hand, if G1 is a pentagon, then F0 has at most 5 - y1 pentagons, so k3 < 7 - y1 by Lemma 2.6, a contradiction. Otherwise, V(G1) is a non-trivial cyclic 5-edge-cut and F0 has exactly 6 pentagons. Hence, by Lemma 2.6 we have that for F0, k3 = 8. So y1 = 3. Take two consecutive edges e and f of V(G1) along the boundary of G1 separately from G1 to F0 and H. Since V(G1) is a non-trivial cyclic 5-edge-cut, by Theorem 2.5 we have that e and f have non-adjacent end-vertices in G1. So these two edges belong to a cycle of length at least 7 bounding a face of F (see Figure 5). But this is impossible.	□
From Theorem 4.1 we further find many nice substructures of fullerene graphs, which are listed in Figure 6. In particular, the first one is the naphthalene (a pair of abutting hexagons), whose boundary is a resonant cycle of length 10.
Corollary 4.2. Any adjacent hexagons of a fullerene graph form a nice substructure, and the boundary (10-length cycle) is thus resonant.
However, not all 10-length cycles of fullerene graphs are resonant. For example, see Figure 1. The following corollary gives a criterion for a 10-length cycle of a fullerene graph to be resonant.
Corollary 4.3. A 10-length cycle C of a fullerene graph F is resonant if and only if it bounds either the naphthalene or the second patch in Figure 4.
Proof. The sufficiency is immediate from Corollaries 3.2 and 4.2. So we only consider the necessity. Suppose that 10-length cycle C of a fullerene graph F is resonant. Let F0 be
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Figure 6: Some nice patches of fullerene graphs with eight 2-degree vertices.
the patch of F bounded by 10-length cycle C with p5 < 6. So F0 has an even number of vertices, and we can have that k3 and k2 both are even. By Lemma 2.6 we have p5 = 2k3 -4 and 2 < k3 < 5. The possible values of k3 are 2 and 4. If k3 = 2, then C bounds a pair of adjacent hexagons. If k3 = 4, then F0 has exactly two vertices in the interior of C which are adjacent by Lemma 2.3. In fact, F0 is the second patch in Figure 4.	□
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