# Similarity and a Duality for Fullerenes

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## Abstract

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## 1. Introduction

**Figure 2.**Fullerene on an icosahedron. Top right: signature graph of fullerene. Bottom left: one possible shortest spanning tree. Right: flat map of fullerene.

## 2. Similarity

**Lemma 1.**Let O be the center of a hexagon in Λ. Dilating by an integer factor from O sends centers of faces onto centers of faces.

**Proof.**In our coordinate system for Λ, the centers of faces all have integer coordinates. Consider the center of a face given by coordinates $(a,b)$. Let Λ be fixed. Dilating the plane about O by $m\in \mathbb{Z}$ will send $(a,b)$ to $(ma,mb)$, which is the center of a new face. ☐

**Lemma 2.**Let O be the center of a hexagon in Λ. Dilating by an integer factor from O sends segments to segments.

**Proof.**Let Λ be fixed and $(a,b)$ be the coordinates of the segment that connects faces f and g. By Lemma 1, we know that the dilation sends the centers of f and g to the centers of new faces, which we denote by $mf$ and $mg$. As such, the dilated segment connects the centers of $mf$ and $mg$ and has coordinates $(ma,mb)$. ☐

**Lemma 3.**Let O be the center of a hexagon in Λ. Dilating adjacent segments by an integer factor from O yields adjacent segments with the same angle measure.

**Proof.**Dilations preserve angle measures. ☐

**Theorem 4.**Let $\mathcal{S}$ be the signature of a reduced fullerene Γ. If all Coxeter coordinates in $\mathcal{S}$ are multiplied by $m\in \mathbb{Z}$, then the resulting structure is the signature of a new fullerene.

**Proof.**Consider a spanning tree of Γ with edge and angle labels. Scaling the spanning tree by multiplying all edge coordinates by m yields a new labeled tree. We take a flat map of the original spanning tree, and using the three lemmas, we see that the image of the boundary of the flat map under dilation by m corresponds to the dilated spanning tree. This shows that we do indeed have the flat map of a new fullerene.

**Case 1: $\theta =1$**

**Figure 4.**Calculating the length of the third side for adjacent segments with an enclosed angle of one.

**Case 2: $\theta =2$**

**Case 3: $\theta =3$**

**Case 4: $\theta =4$**

## 3. Leapfrog Fullerenes

**Lemma 5.**Let σ be a segment in Λ, and let ${\Lambda}^{\prime}$ the leapfrog tessellation. The Coxeter coordinates of σ in ${\Lambda}^{\prime}$ are calculated from the Coxeter coordinates of σ in Λ as described in the following table:

## 4. Duality

**Theorem 6.**Dual similarity classes cannot both be of the leapfrog type.

**Proof.**Let Γ be a reduced leapfrog fullerene, and let ${\Gamma}^{\prime}$ be its leapfrog dual. Assume by way of contradiction that both fullerenes are in leapfrog-type classes. Since the Coxeter coordinates of the segments in the signature of Γ must be congruent modulo three, it follows that the Coxeter coordinates are all of the form $\left(p+r,p\right)$, $\left(p,p+r\right)$, $\left(p,p\right)$ or $\left(r\right)$, where $r=3s$ for some $s\in \mathbb{N}$, but s and p are not a multiple of three. By the previous lemma, we get the following table.

## 5. Chains and the Clar and Fries Numbers

**Lemma 7.**Let K be a Kekulé structure for fullerene Γ, and let ${\varphi}_{\Gamma}\left(K\right)$ denote the number of benzene rings in K, then:

**Proof.**$3{\varphi}_{\Gamma}\left(K\right)$ counts the Kekulé edges that lie on a benzene ring: it counts those that lie on two benzene rings twice, those that lie on one benzene ring just once and those that lie on no benzene rings not at all. Hence, $3{\varphi}_{\Gamma}\left(K\right)+2\left|{A}_{0}\left(K\right)\right|+\left|{A}_{1}\left(K\right)\right|$ double counts the edges in K. We get:

**Lemma 8.**Let K be a Kekulé structure for Γ, and let ${\gamma}_{\Gamma}\left(K\right)$ denote the size of the largest Clar set C in K, then:

**Proof.**The faces in C and the edges in A cover every vertex exactly once. Hence, $6\left|C\right|+2\left|A\right|=\left|V\right|$. Solving for $\left|C\right|$ yields the desired result. ☐

- for all i, ${e}_{i}$ has exactly one endpoint in ${f}_{i-1}$, while the other endpoint is in ${f}_{i}$;
- if ${f}_{i}$ is a hexagon, the endpoints of ${e}_{i}$ and ${e}_{i+1}$ are either opposite or adjacent around the boundary of ${f}_{i}$;
- either ${f}_{n}={f}_{0}$ and all faces are distinct hexagons or ${f}_{0}$ and ${f}_{n}$ are distinct pentagons, and all other faces on the chain are distinct hexagons.

- the chains in one equivalence class have length $max\{p,q\}$ and a color of incompatibility different from the color of the faces of the chain;
- the chains in the other equivalence class have length $p+q$ and the third color as the color of incompatibility.

**Figure 13.**The best chain decomposition for the reduce icosahedral fullerene with coordinates $(2,1)$.

**Figure 14.**The scaled up chain decomposition for the similar icosahedral fullerene with coordinates $(4,2)$.

## Author Contributions

## Conflicts of Interest

## References

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**MDPI and ACS Style**

Edmond, J.J.; Graver, J.E.
Similarity and a Duality for Fullerenes. *Symmetry* **2015**, *7*, 2047-2061.
https://doi.org/10.3390/sym7042047

**AMA Style**

Edmond JJ, Graver JE.
Similarity and a Duality for Fullerenes. *Symmetry*. 2015; 7(4):2047-2061.
https://doi.org/10.3390/sym7042047

**Chicago/Turabian Style**

Edmond, Jennifer J., and Jack E. Graver.
2015. "Similarity and a Duality for Fullerenes" *Symmetry* 7, no. 4: 2047-2061.
https://doi.org/10.3390/sym7042047