Fullerenes

From CanisiusmathWiki

C60 or the Bucky Ball

Table of contents

1 Abstract

2 Some General Background Information

3 Infinite Fullerenes

4 Covers of Fullerenes

5 Toridal Fullerenes

6 Observations about Eigenvalues

7 Ihara Zeta Function For Finite Fullerenes

8 Ihara Zeta Function For Infinite Fullerenes

9 Allowing Heptagons

[edit]

Abstract

Fullerenes determine graphs where the primitive loops are pentagons or hexagons. We consider spherical (classical), toroidal and kleinian fullerenes, and we study their spectral properties. As an application, we write infinite fullerenes as limits of finite toroidal fullerenes, and we use this information to compute the Ihara zeta function.

[edit]

Some General Background Information

Fullerenes are geometric structures that are composed of, in the classical sense, 12 pentagons and n = 0, 2, 3, 4, ... hexagons which preserve 3-regularity across the entire structure. They are of particular interest to chemists since they model the bonding of carbon molecules in spherical and tubular form. The Bucky Ball or $C 60$ is the first fullerene that has isolated pentagons, a necessary condition for the construction of spherical carbon molecules. We hope to investigate the mathematical properties of fullerenes, infinite fullerenes, fullerenes on other topological structures, and the Ihara Zeta function.

[edit]

Infinite Fullerenes

Infinite fullerenes come from tiling the infinite plane with hexagons and less than or equal to 6 pentagons. When there is zero or one pentagon, there is a unique solution to the tiling but when there are 2 to 6 pentagons there will be infinite solutions. For an infinite tilling, choose a vertex and count all vertices that are a walk n away; let $\varphi_n$ denote the number of vertices that can be reached with a walk of length n. With zero pentagons, a graphite sheet is produced, and using the method of walks produces the equation $\varphi_n = 1 + \sum_{j=0}^n 3j$ . For the other tilings an equation is not possible to be derived due to the lack of symmetry; the method however allows for infinite fullerenes to be compared to finite ones.

If we let the number of vertices, n, approach infinity, then we can use tilings of the plane as an isomorphism of finite fullerenes. For construction of infinite fullerenes assume that IPR standards are necessary; pentagons spaced out symmetrically along the fullerene are non-adjacent. Using this assumption and the fact that since the pentagonal faces are an infinite distance apart, then each region can be made isomorphic to the plane tiled with hexagons and 1 pentagon. These fullerenes that are produced can be labeled as $C_n\$ as the other IPR fullerenes are labeled; but they are not constructible in the lab due to their instability.

If we instead force the pentagons to lie in pairs where members of the pairs are a finite distance apart then this region will be isomorphic to the tiled plane with 2 pentagons that are a finite distance apart.

[edit]

Covers of Fullerenes

Statement: Fullerenes can be embedded in the sphere, $S^2\$ , the torus, $T^2\$ , the klein bottle, $K^2\$ , and the projective plane $P^2\$ ; the number of pentagons on each surface is 12, 0, 6, and 0, respectively.

Proof: Using the euler characteristic equation, $\chi = v\ - e\ + f\$ , where $v\$ is the number of vertices, $e\$ is the number of edges, and $f\$ is the number of faces, the number of pentagons for each surface can be determined. First, note that, $3v\ =2e\ =5f_5\ +6f_6\$ due to fullerenes being 3 regular.

$\chi = v\ -\ e\ + f\ \Rightarrow 6\chi = 6v\ -\ 6e\ +\ 6f\ = 4e\ -\ 6e\ +\ 6f\ = -2e\ +\ 6f\ = -\left(5f_5\ +\ 6f_6\right) +\ \left(6f_5\ +\ 6f_6\right) \Rightarrow 6\chi = f_5 \Rightarrow \chi = \frac{f_5}{6}$

Now let $\chi = 2 - 2g\$ for orientable surface and $\chi = 2 - g\$ for non-orientable surfaces.
For $S^2 , \chi = 2 - 2(0) = 2 \Rightarrow 2 = \frac{f_5}{6} \Rightarrow f_5 = 12$ .
For $T^2 , \chi = 2 - 2(1) = 0 \Rightarrow 0 = \frac{f_5}{6} \Rightarrow f_5 = 0$ .
For $P^2 , \chi = 2 - 1 = 1 \Rightarrow 1 = \frac{f_5}{6} \Rightarrow f_5 = 6$ .
For $K^2 , \chi = 2 - 2 = 0 \Rightarrow 0 = \frac{f_5}{6} \Rightarrow f_5 = 0$ . $\Box$

Since it is at least very difficult for a fullerene to cover another fullerene in $S^2\$ or $P^2\$ , due to the presence of 12 and 6 pentagons in all $S^2\$ and $P^2\$ respectively and the restriction of three valency, we will focus on fullerenes in $T^2\$ and $K^2\$ .

Statement: In $T^2\$ there will always be a fullerene that covers another fullerene.
Proof: For $T^2\$ we proved above that there were no pentagonal faces, so a fullerene, $F\in T^2\$ is composed entirely of hexagons. Since $F\$ is made entirely of hexagons it can just be considered as a tiling. Now allow $F\$ to have an arbitrary number of hexagons. One can now cut $F\$ once horizontally and once vertically such that no edges are broken. This produces a tilling of the rectangle $\Phi\$ which recovers $F\$ when the sides are identified. Now make a copy of $\Phi\$ and label it ${\Phi}'\$ . Glue the right side of $\Phi\$ to the left side of ${\Phi}'\$ . Because of the properties of $\Phi\$ and ${\Phi}'\$ , we obtain a tilling ${\Phi}_2\$ of the parallelogram with hexagons, except at the edges. Again by the properties of $\Phi\$ and ${\Phi}'\$ , we obtain a tilling of the torus when we identify the sides appropriately. The map

${\pi} = \mathrm{id}{\cup}\mathrm{id}: {\Phi}_2 = {\Phi}{\cup}{\Phi}' \to {\Phi}$

induces a map $p\$ between a fullerene $F_2\$ , which is produced by ${\Phi}_2\$ , and $F\$ . We will show that $p\$ is a graph covering map. It is obvious that $p\$ is a graph map. We need to check the neighborhood condition: For each vertex $v\$ in $F_2\ \mbox{, } p\$ , induces a bijection between $N(v)\$ and $N(p(v))\$ . This is obvious for the vertices on the interior of $\Phi\$ and ${\Phi}'\$ . Also it is obvious for vertices on the `horizontal' sides of the squares. So we need to check the condition for $v\$ on the vertical sides of $\Phi\$ and ${\Phi}'\$ . Now let $a\$ be a glued vertex on the right edge of $\Phi\$ and the left edge of $\Phi^'\$ . Because of the way that ${\Phi}_2\$ is formed, the neighborhood of $a\$ in $F_2\$ is isomorphic to the neighborhood of $a\$ in $F\$ . Thus $N(a) \approx N(p(a))$ . Now let $b\$ be another glued vertex that lies to the right of $\Phi^'\$ and the left of $\Phi\$ . The same argument works in this case because of the way that $F_2\$ is formed. Thus $p\$ is a double covering map. $\Box$

Statement: In $K^2\$ there will always be a fullerene that covers another fullerene.
Proof: The proof for this is similar to the one above except for the fact that the copy of $\Phi\$ , ${\Phi}'\$ , will be mirrored left to right along the axis of similar orientation and then glued. $\Box$

[edit]

Toridal Fullerenes

Fullerenes on the torus are analogous to tiling a plane with all hexagons; to facilitate our exploration of this subject the following code for Mathematica (which can be extended to other languages with little work) was produced which gives the adjacency matrix for toroidal fullerenes.

In[1]: 
(*user input, number of hexagons vertical and horizontal*)
vertical = ???; 
horizontal = ???;
(*end of user input*)

modding = vertical*4; 
tempTable := Flatten[ 
             {(*1*)    Mod[i, modding] == 1 && (i == j - 1 || i == j - (modding/2 - 1)),
             (*mid+1*) Mod[i, modding] == ((modding/2) + 1) && (i == j - 1 ||i == j - ((modding/2) - 1) || i == j - (modding/2)),
             (*mid*)   Mod[i, modding] == modding/2 && i == j - modding/2,
             (*endc*)  Table[i == z && j == horizontal*vertical*2 - (modding/2 - z), {z, 1, modding/2, 2}],
             (*col1*)  Table[Mod[i, modding] == z && i == j - 1, {z, 3, modding/2, 2}],
             (*col2*)  Table[Mod[i, modding] == z && (i == j - modding/2 || i == j - 1), {z, 2, (modding/2) - 2,2}],
             (*col3*)  Table[Mod[i, modding] == z && i == j - 1, {z, modding/2 + 2, modding - 1, 2}],
             (*col4*)  Table[Mod[i, modding] == z && (i == j - 1 || i == j - modding/2), {z, modding/2 + 1, modding - 1, 2}]}, 3];

J[i_, j_] := If[Apply[Or, tempTable], True, False];

A = Table[If[J[i, j], 1, 0], {i, 1, horizontal*vertical*2}, {j, 1, horizontal*vertical*2}]; A = A + Transpose[A];

(*code for graphics modified from wolfram example*)

h[x_, y_] := Polygon[Table[{Cos[2 Pi k/6] + x, Sin[2 Pi k/6] + y}, {k, 6}]]; 
graph = Graphics[{EdgeForm[Opacity[.7]], LightBlue, Table[h[3 i + 3 ((-1)^j + 1)/4, Sqrt[3]/2 j], {i, vertical}, {j, horizontal}]}]; 

Print[graph]; 
Print["Characteristic Polynomial"]; Print[CharacteristicPolynomial[A, x]]; 
Print["Determinant"]; Print[Det[A]]; 
Print["Eigenvalues"]; Print[Eigenvalues[A]]; 
Print["Adjacency Matrix"]; Print[MatrixForm[A]];

The preceding code will produce adjacency matrices, and from that any number of graph theory calculations, for hexagonal lattices. The user must replace ??? with integers. The following is a sample output for vertical = 3 and horizontal = 2 and its cover vertical = 3 and horizontal = 4.

[edit]

Observations about Eigenvalues

Note: These are merely conjectures observed from a limited number of examples

Let $F_n \in T^2$ be a fullerene described by the hexagonal tiling $m \times2^n$ , where $m,n \in \mathbb{Z}, m \ge 2$ . This produces a covering sequence, $\big\{ F_n \big\}$ , where $F n + 1$ is simply the double cover of $F n$ . Then,

There will always be $\lambda = \pm 3$ . This is trivial due to the regularity of fullerenes.

If $m | 2$ , then $\lambda=\pm1$ with order $2 n + 1$ .

The $λ$ 's for $F_n \subset F_{n+1}$ . This result is also trivial due to $F n | (F n + 1$ .

$n=2, \lambda=\pm\sqrt{5}.$

$n=3, \lambda =\pm \sqrt{5 \pm 2 \sqrt{2}}.$

$n=4, \lambda = \pm \sqrt{5 \pm 2 \sqrt{2 \pm \sqrt{2}}}.$

$n=5, \lambda = \pm \sqrt{5 \pm 2 \sqrt{2 \pm \sqrt{2 \pm \sqrt{2}}}}, etc.$

If you let all operands be either plus or minus, this sequence approaches $\pm 3$ .

Statement: $\sqrt{5 + 2 \sqrt{2 + \sqrt{2 + \sqrt{2 + . . .}}}} = 3$
Proof: First let $S = \left \{ \sqrt{2} , \sqrt{2 + \sqrt{2}} , \sqrt{2 + \sqrt{2 + \sqrt {2}}} , . . . \right \}$ . S by construction is a monotone increasing function with $\forall s \in S < 2$ . To prove this let $s \in S = \sqrt{2 + \sqrt{2 + \sqrt {2 + . . . \sqrt{2}}}} < \sqrt{2 + \sqrt{2 + \sqrt {2 + . . . \left( \sqrt{2} \right) ^2 }}} = \sqrt{2 + \sqrt{2 + \sqrt {2 + . . . 2}}} = 2$ . Since $S$ is bounded and monotone increasing a limit, $L$ exists. Let $L^2 = \left( \sqrt{2 + \sqrt{2 + \sqrt {2 + . . . }}} \right )^2 = 2 + \sqrt{2 + \sqrt{2 + \sqrt {2 + . . . }}} \Rightarrow 2 + L = L^2 \Rightarrow L = 2$ . Now $\sqrt{5 + 2 \sqrt{2 + \sqrt{2 + \sqrt{2 + . . .}}}} = \sqrt{5 + 2 \left( 2 \right)} = 3$ . $\Box$

As result of the previous finding we can say that the spectral gap goes to zero.

After the first several covers in this sequence all the unique eigenvalues seem to appear, and after that only a similar pattern to the above holds.

All eigenvalues come in plus/minus pairs. This indicates that they are centered at 0 which means that the graph is bipartite. This holds since for all $F \in T^2$ there are no odd cycles.

[edit]

Ihara Zeta Function For Finite Fullerenes

First the Ihara Zeta function, an analogue for the Riemann Zeta Function, is defined as

$\zeta_X(t) = \prod_{\left[C\right]} \left(1-t^{\left|C\right|}\right)^{-1} = e \left( {\sum_{r=1}^\infty \frac{c_r t^r}{r}} \right)$

where $\left[C\right]$ is the equivalence classes of closed, prime paths and $c_r\$ is the number of closed loops of length r that are oriented (oriented meaning that loops do not overlap and the starting point of loops do not matter). Now assume that the graph will be k-regular. By Bass' Theorem, the Ihara Zeta function can expressed as

$\zeta_X(t) = \left(1-t^2\right)^{-\frac{1}{2}\left(k-2\right)\left|X\right|}\mbox{det}\left(I-tkM+\left(k-1\right)It^2\right)^{-1}$

where $M\$ is the Markov operator on $x\$ and $k\$ is the degree. Because fullerenes are 3-regular, letting k = 3 yields

$\zeta_X(t) = \left(1-t^2\right)^{-\frac{1}{2}\left|X\right|}\mbox{det}\left(I-3tM+2It^2\right)^{-1}$

The Ihara zeta function can also be expresed as a product of eigenvalues:

$\zeta_X(t) = \left(1-t^2\right)^{\left|V\right|-\left|E\right|}\prod_{i=1} \left(1+2t^2-\lambda_it\right)^{-1}$

Using Mathematica, we are able to quickly find the Ihara zeta function for a number of toroidal fullerenes. Below is the Ihara zeta function for a "2,2" fullerene:

$\zeta_X(z)^{-1} = \left(1-z^2\right)^4\left(1+4z^2-2z^4-60z^6-183z^8-240z^{10}-32z^{12}+256z^{14}+256z^{16}\right)$

[edit]

Ihara Zeta Function For Infinite Fullerenes

We can extend the Ihara Zeta Function for infinite graphs and look for the limit.

Let $F_n \in T^2$ be a fullerene where $n \in \mathbb{N}$ such that $\left \{ F_n \right \}$ is a finite graph covering sequence (i.e. $F n + 1$ is a graph cover for $F n$ ). Now let $\left | F_1 \right | = \alpha$ where $α$ is the number of vertices; thus $| F n | = 2 n - 1 α$ .

We are now exploring the equation: $\zeta_X\left(t\right) = \lim_{n \rightarrow \infty} \prod_{i=1} {\left ( 1 + 2t^2 - \lambda_i \left ( F_n \right ) t \right )}^{- \frac{1}{2^{n-1} \alpha}}$

To look at the limit of the function we explored graph covering families in the form of $m \times 2^n$ , vertical by horizontal hexagons on the sliced open fullerene, where $m \geq 2$ and $n \in \mathbb{N}$ . Below we graphed some of the examples, examples include n = 1,2,3,4,5 in respect from bottom to top.

[edit]

Allowing Heptagons

Statement: The embedding of pentagons, hexagons, and heptagons in the sphere, $S^2\$ , the torus, $T^2\$ , the klein bottle, $K^2\$ , and the projective plane $P^2\$ will result in $f_5\ = f_7\ + 12$ , $f_5\ = f_7\$ , $f_5\ = f_7\ + 6$ , and $f_5\ = f_7\$ respectively.

Proof: Using the euler characteristic equation, $\chi = v\ -\ e\ +\ f\$ , where $v\$ is the number of vertices, $e\$ is the number of edges, and $f\$ is the number of faces, the number of pentagons and heptagons for each surface can be determined. First, note that $3v\ = 2e\ = 5f_5\ +\ 6f_6\ +\ 7f_7\$ due to fullerenes being 3 regular.

$\chi = v\ -\ e\ +\ f\ \Rightarrow 6\chi = 6v\ -\ 6e\ +\ 6f\ = 4e\ -\ 6e\ +\ 6f\ = -2e\ +\ 6f\ = -\left(5f_5\ +\ 6f_6\ +\ 7f_7\right)\ +\ \left(6f_5\ +\ 6f_6\ +\ 6f_7\right)$ $\Rightarrow 6\chi = f_5\ -\ f_7\ \Rightarrow \chi = \frac{f_5 - f_7}{6}$

Now let $\chi = 2 - 2g\$ for orientable surface and $\chi = 2 - g\$ for non-orientable surfaces.
For $S^2 , \chi = 2 - 2(0) = 2 \Rightarrow 2 = \frac{f_5 - f_7}{6} \Rightarrow f_5 - f_7 = 12 \Rightarrow f_5 = f_7 + 12$ .
For $T^2 , \chi = 2 - 2(1) = 0 \Rightarrow 0 = \frac{f_5 - f_7}{6} \Rightarrow f_5 - f_7 = 0 \Rightarrow f_5 = f_7$ .
For $P^2 , \chi = 2 - 1 = 1 \Rightarrow 1 = \frac{f_5 - f_7}{6} \Rightarrow f_5 - f_7= 6 \Rightarrow f_5 = f_7 + 6$ .
For $K^2 , \chi = 2 - 2 = 0 \Rightarrow 0 = \frac{f_5 - f_7}{6} \Rightarrow f_5 - f_7 = 0 \Rightarrow f_5 = f_7$ . $\Box$