Further fullerenes

## Further fullerenes

In [6] and [5] huge tables with numbers of isomers and of chiral isomers of substituted fullerenes from C20 to C70 are computed, but the corresponding cycle indices are not given. In [9] an even greater list of fullerene structures can be found. The cycle indices of the symmetry groups of these fullerenes can be computed by determining the permutation representation of the symmetry groups of these fullerenes acting on their sets of vertices, edges or faces. Here we want to present a list of the 3-dimensional cycle indices for these fullerenes.

The point group of C24 is D6d which leads to the following cycle indices.

Z3(R(C24))= (1)/(12)( 2 e66 f12 f62 v64+ 2 e312 f12 f34 v38+ e218 f12 f26 v212+ 6 e12 e217 f27 v212+ e136 f114 v124)
Z3(S(C24))= (1)/(2)Z3(R(C24)) + (1)/(24)( 4 e123 f2 f12 v122+ 2 e49 f2 f43 v46 + 6 e14 e216 f14 f25 v14 v210 ).
Since the C26 has a symmetry group of the form D3h we have
Z3(R(C26))= (1)/(6)( 3 e1 e219 f1 f27 v213+ 2 e313 f35 v12 v38 + e139 f115 v126 )
Z3(S(C26))= (1)/(2)Z3(R(C26)) + (1)/(12)( 2 e3 e66 f3 f62 v2 v32 v63 + 3 e15 e217 f15 f25 v14 v211 + e13 e218 f13 f26 v16 v210).
The fullerene C28 is of tetrahedral symmetry Td so the cycle indices can be computed as
Z3(R(C28))= (1)/(12)( 8 e314 f1 f35 v1 v39 + 3 e12 e220 f28 v214 + e142 f116 v128 )
Z3(S(C28))= (1)/(2)Z3(R(C28)) + (1)/(24)( 6 e2 e410 f44 v47 + 6 e14 e219 f14 f26 v16 v211 ).
The cycle index of R(C28) acting on its set of vertices can be found in [4] as well.

The C30 has D5h symmetry and the cycle indices for the action on the sets of edges, faces and vertices are

Z3(R(C30))= (1)/(10)( 4 e59 f12 f53 v56 + 5 e1 e222 f1 f28 v215 + e145 f117 v130)
Z3(S(C30))= (1)/(2)Z3(R(C30)) + (1)/(20)( 4 e5 e104 f2 f5 f10 v103 + 5 e15 e220 f15 f26 v16 v212 + e15 e220 f15 f26 v215 ).
The symmetry group of C32 is D3, so it consists only of rotations and we have
Z3(R(C32))= (1)/(6)( 2 e316 f36 v12 v310 + 3 e12 e223 f29 v216 + e148 f118 v132 ).
Since C34 has C3v symmetry the cycle indices are
Z3(R(C34))= (1)/(3)( 2 e317 f1 f36 v1 v311 + e151 f119 v134)
Z3(S(C34))= (1)/(2)(Z3(R(C34)) + e15 e223 f15 f27 v14 v215 ).
The point group of C36 is D6h, from which the following cycle indices can be computed.
Z3(R(C36))= (1)/(12)( 2 e69 f2 f63 v66 + 2 e318 f12 f36 v312 + 4 e227 f12 f29 v218 + 3 e12 e226 f210 v218 + e154 f120 v136 )
Z3(S(C36))= (1)/(2)Z3(R(C36)) + (1)/(24)( 2 e32 e68 f2 f32 f62 v66 + e227 f210 v218 + 2 e69 f12 f63 v66 +
3 e14 e225 f14 f28 v18 v214 + 3 e16 e224 f16 f27 v14 v216 + e16 e224 f16 f27 v218 ).
C38 is of C3v symmetry so we have
Z3(R(C38))= (1)/(3)( 2 e319 f37 v12 v312 + e157 f121 v138)
Z3(S(C38))= (1)/(6)( 2 e319 f37 v12 v312 + 3 e15 e226 f15 f28 v16 v216 + e157 f121 v138).
For the C40 two possible symmetry groups are given, namely Td and D5d. In the first case the cycle indices are
Z3(R(C40))= (1)/(12)( 3 e230 f12 f210 v220 + 8 e320 f1 f37 v1 v313 + e160 f122 v140 )
Z3(S(C40))= (1)/(2)Z3(R(C40)) + (1)/(24)( 6 e415 f2 f45 v410 + 6 e16 e227 f16 f28 v14 v218 ).
For the symmetry group D5d we compute
Z3(R(C40))= (1)/(10)( 4 e512 f12 f54 v58 + 5 e12 e229 f211 v220 + e160 f122 v140 )
Z3(S(C40))= (1)/(2)Z3(R(C40)) + (1)/(20)( 4 e106 f2 f102 v104 + e230 f211 v220 + 5 e16 e227 f16 f28 v14 v218 ).
Since the point group of C42 is D3 we have
Z3(R(C42))= (1)/(6)( 2 e321 f12 f37 v314 + 3 e1 e231 f1 f211 v221 + e163 f123 v142 ).
For the D3h symmetry group of C44 we can compute
Z3(R(C44))= (1)/(6)( 2 e322 f38 v12 v314 + 3 e12 e232 f212 v222 + e166 f124 v144)
Z3(S(C44))= (1)/(2)Z3(R(C44)) + (1)/(12)( 2 e32 e610 f32 f63 v2 v32 v66 + 4 e16 e230 f16 f29 v16 v219 ).
A second form of C44 is chiral and has T as its symmetry group which gives the following cycle index:
Z3(R(C44))= (1)/(12)( 3 e12e232 f212 v222 + 8 e322 f38 v12v314 + e166 f124 v144 ).
The symmetry group of C46 is C3 so we compute
Z3(R(C46))= (1)/(3)( 2 e323 f1 f38 v1 v315 + e169 f125 v146 ).
The C48 has D3 as its symmetry group, so we have
Z3(R(C48))= (1)/(6)( 2 e324 f12f38 v316 + 3 e12 e235 f213 v224 + e172 f126 v148 ).
For the D5h symmetry of C50 we derive
Z3(R(C50))= (1)/(10)( 4 e515 f12 f55 v510 + 5 e1 e237 f1 f213 v225 + e175 f127 v150 )
Z3(S(C50))= (1)/(2)Z3(R(C50)) + (1)/(20)( 4 e5 e107 f2 f5 f102 v52 v104 + 5 e17 e234 f17 f210 v14 v223 +
e15 e235 f15 f211 v110 v220 ) .
The symmetry group of C52 is T, which consists only of rotational symmetries.
Z3(R(C52))= (1)/(12)( 3 e12e238 f214 v226 + 8 e326 f1 f39 v1 v317 + e178 f128 v152 ).
C54 has point group D3, so we compute
Z3(R(C54))= (1)/(6)( 3 e1 e240 f1 f214 v227 + 2 e327 f12 f39 v318 + e181 f129 v154 ).
For the D3d symmetry group of C56 the cycle indices are
Z3(R(C56))= (1)/(6)( 2 e328 f310 v12 v318 + 3 e12 e241 f215 v228 + e184 f130 v156 )
Z3(S(C56))= (1)/(2)Z3(R(C56)) + (1)/(12)( 2 e614 f65 v2 v69 + 3 e16 e239 f16 f212 v18 v224 +
e242 f215 v228 ) .
The symmetry group of C58 is C3v so we have
Z3(R(C58))= (1)/(3)( 2 e329 f1 f310 v1 v319 + e187 f131 v158)
Z3(S(C58))= (1)/(2)(Z3(R(C58)) + e17 e240 f17 f212 v16 v226 ).
C80 is the next fullerene with Ih symmetry. Its cycle index is
Z3(R(C80))=(1)/(60) ( 24 e524f12f58v516+ 20 e340f314v12 v326+ 15 e260f12 f220v240+ e1120f142v180)
Z3(S(C80))=(1)/(2) Z3(R(C80)) + (1)/(120) ( e260f221v240 + 20e620f67v2v613 + e18e256f18f21715v18v236 + 24e1012f2f104v108).
The first fullerene with symmetry group I is the C140. Its cycle index is
Z3(R(C140))=(1)/(60) ( 24 e542f12f514v528+ 20 e370f324v12v346+ 15 e12e2104f236v270+ e1210f172v1140)

We could go on listing the cycle indices of many more fullerenes. In many cases it is possible to arrange the v vertices of Cv in several different ways leading to different symmetry groups and to different cycle indices. For instance for the fullerene C78 there are 4 possible isomers given in [9]. So from chemical properties we first have to determine the actual shape of the molecule. In [11] it is shown that C76 is of D2 symmetry and not of Td symmetry which would be possible as well.

harald.fripertinger@kfunigraz.ac.at,
last changed: January 23, 2001

 Further fullerenes