BUCKYBALLS
Ana Quintana
771202-P240
Karlstad Universitet
INDEX

ABSTRACT
3

MOLECULE’S SYMMETRY GROUPS
4

WHAT ARE FULLERENES?
 WHY DO THEY LOOK LIKE THAT?
 APPLICATIONS
 DISCOVERY OF BUCKMINSTERFULLERENE (c60)
5
6
6
7

THE PLATONIC SOLID
 TETRAHEDRON
 CUBE
 OCTAHEDRON
 DODECAHEDRON
 ICOSAHEDRON
9
9
9
10
10
11

TRUNCATED ICOSAHEDRON
 Regular icosahedron-truncated icosahedron
 Description
 PROPER SYMMETRY GROUP G
 IN SEARCH OF THE PROPER SYMMETRY GROUP
 First step: The Icosahedron
 Galois’s Letter
 Second Step: WHAT ABOUT THE
13
12
13
15
15
15
17
GRAPH
OF
ICOSAHEDRON ?
THE
TRUNCATED
18

SPECTROSCOPY
20

BIBLIOGRAPHY
24
2
ABSTRACT
Buckminsterfullerene is a natural form, or allotrope, of carbon. For many
years it was supposed that the element carbon only took two natural forms-diamond and graphite. Diamond is a solid in which each carbon atom is linked
to four neighbors, and this arrangement extends throughout the crystal to give a
rigid, hard solid. In graphite, the carbon atoms are linked to form hexagonal
rings that form flat sheets that lie on top of one another, and the result is a
slippery, soft solid. Because carbon is one of the most extensively investigated
of all elements, it came as a great surprise in 1985 when a whole family of
different allotropes was discovered. Of these, the most famous is
buckminsterfullerene, also called BuckyBall.
Buckminsterfullerene consists of 60 carbon atoms linked together to form
an almost spherical C60 molecule of joined hexagons and pentagons. The
bonds have the same arrangement as the panels on a soccer ball. The name
was given to the molecule because its structure resembles the elaborate
geometrical structures invented by the American architect Richard Buckminster
Fuller. It is now known that buckminsterfullerene is likely to be formed in sooty
flames, and there is even the possibility that it is highly abundant in the
universe, particularly near red-giant stars
The C60 “BuckyBall” molecule
3
1. MOLECULE’S SYMMETRY GROUPS
The use of group theory to determine certain properties of suitable
molecules is a well-established procedure.
For a single molecule, the group involved is the molecule’s symmetry
group which, up to conjugacy, can be considered as a subgroup of O3.
The most significant part of the symmetry group is its intersection, G, with
SO(3). In this way, G  SO(3) and will be referred to as the molecule’s proper
symmetry group.
These groups G are not very interesting from the point of view of
mathematics, with a unique exception, since they are easily constructed
solvable groups.
The unique exception is a group which is isomorphic to the alternating
group A5. As a subgroup of SO(3), it is the proper symmetry group of the
icosahedron (and the dodecahedron). It is the unique finite subgroup of SO(3)
which equals its own commutator subgroup, actually, it is the unique nonabelian simple finite subgroup of SO(3).
A group isomorphic to A5 will be referred to as an icosahedral group. Any
structure admitting such a group as a symmetry group is said to have an
icosahedral symmetry.
4
2. WHAT ARE FULLERENES?
Natural carbon can exist in several forms. The most known ones are graphite
and diamond, but there is a third type—Fullerenes, also called BuckyBalls. They
aren’t a “new form of pure solid carbon”; in fact, Fullerenes have been found to
exist in interstellar dust as well as in geological formations on Earth.
Graphite and diamonds are crystalline materials built up of extended arrays
of carbon atoms. The bonds between their atoms exhibit hexagonal and
tetrahedral structures, respectively.
Fullerenes show carbon atoms arranged in relatively small clusters of atoms.
They are large carbon-cage molecules, built up of hexagons and pentagons.
These cages are about 7—15 Å in diameter (6—10 times the diameter of a
typical atom).
Graphite nor diamonds exhibit isolated molecules of pure carbon, while in
Fullerene one finds, for the first time, a pure carbon crystalline solid with welldefined carbon molecules.
 C60 in numbers
- Molecular weight – 720
- Density - 1.72 g/cm3
- Molecular diameter - 7 Å
- The 60 carbon atoms of C60 only give rise to four absorption lines in the
infrared region, which confirms its high symmetry (IR lines at 1429, 1183, 577
and 528 cm-1). All carbon atoms are equivalent (13C NMR-signal at 143 ppm
in benzene)
5
A. WHY DO THEY LOOK LIKE THAT?
Mathematically these molecules can be described as convex polyhedrons
where the faces are either hexagons or pentagons and each vertex (carbon
atom) is the end point of the edges (carbon bonds).
Whereas the number of hexagons varies from one type of fullerene to
another, every fullerene has exactly the same number of pentagons, 12. This is
not an accident. It’s a consequence of a Theorem of Euler.

Theorem (Euler) For any simple polyhedral surface (built up of
polygons) with 3 edges emanating from each vertex, we have
 (6-n) fn =12
where fn denotes the number of n-gons.
Pentagons n=5 and hexagons n=6,
(6-5) f5+ (6-6) f6=12  f5=12
Euler’s theorem implies that the number of pentagonal faces is necessarily
12, but it doesn’t give any restriction on the number of hexagonal faces.
B. APPLICATIONS
Fullerenes exhibit remarkable chemical and physical
superconductivity, ferromagnetism, tremendous stability, …
properties:
Chemically, they are quite stable; breaking the walls requires temperatures
of over 1000°C. At much lower temperatures (a few hundred °C) fullerenes will
“sublime”; which means that vapor will form directly from the solid. The balls
don’t break, they just separate from the solid intact. This property is used in
growing crystals and thin films of fullerenes.
All these characteristics, their uncommon shape and their unusual properties
for carbon materials, made scientist to speculate about amazing and
revolutionary applications.
It was speculated that BuckyBalls would make great lubricants, rolling like
little ball bearings between other molecules, or perhaps, open a new path of
investigation for that so much needed fuel for rockets. Most of these
“extraordinary” ideas have already been rejected outright. Others didn’t show
sufficient economic potential… Ultimately, we can say that no commercial
applications of round fullerene cage molecules have appeared yet.
However, this is not the end. BuckyBalls are still a new reality and scientists
have a long way to walk. Indeed, there is a very promising area of investigation
which involve modification of the C60 (or other) cage. Because of their hollow
6
shape, drugs could be trapped inside the cages, then released slowly by a
triggering mechanism that could break open the cages inside the body. Drug
delivery is still a hope, once we learn how to attach the appropriate ligands onto
the outside of the cage. BuckyBalls have also shown the ability to block the HIV
virus from attacking healthy cells under certain conditions. But, can it be
controlled?
It also turns out that vaporizing C60 leads to a smoother diamond film than
vaporizing graphite. Maybe, this will help grow tough protective coating with
better properties than what we have now.
But almost the direct outcome has been the discovery of carbon-based
nanotubes in 1997. These are structures in which crystalline arrays of carbon
atoms form tiny, hollow cylinders.
C. DISCOVERY OF BUCKMINSTERFULLERENE (c60)
In the early 1980’s, Harold W. Kroto of the University of Sussex in England
was using microwave spectroscopy to analyze the composition of carbon richstars. The analysis indicated that the atmosphere of these stars contained
cyanopolynes, which are composed of chains of alternating carbon and nitrogen
atoms. Prof. Kroto wanted to study how these chains could be formed. He
contacted Robert F. Curl and Richard E. Smalley at Rice University in Texas,
because they had been using microwave spectroscopy to analyze clusters of
metals formed in Prof. Smalley’s lab. Prof. Smalley had an apparatus that could
vaporize nearly any material into a plasma. In 1985, Kroto joined Curl and
Smalley to study the products of carbon vaporization.
They fired a high-energy laser beam at a graphite surface and used a
stream of helium gas to carry the fragments into a mass spectrometer. The
mass spectrometer revealed the masses of the fragments of graphite ejected
from the surface. These fragments varied from several atoms up to about 190
atoms. The distribution of fragments depended on the pressure of helium in the
carrier stream. As the pressure increased from several torr to about 1atm, the
distribution of fragments changed, and the fragment containing 60 carbon
atoms became by far the dominant one. Because the laser pulse and graphite
surface had not changed, they reasoned that the fragments that broke off were
not changing, but instead that the way these fragments interacted on their way
into the mass spectrometer changed. At higher helium pressures, the fragments
would be jostled together more than at lower pressures. This jostling leads to
the formation of the most stable for of small carbon atom cluster, namely C60.
7
Smalley cluster
These scientists wondered how the atoms in this cluster are arranged to
make it more stable than other clusters. They believed that its stability came
from an arrangement in which all bonding capacity of the atoms is satisfied. In a
small fragment of carbon-atom sheet ejected from a graphite surface, the atoms
around the edge of the sheet would not be fully bonded. If, however, the sheet
were to form into a ball so the edges would meet, the bonding capacity of all
atoms would be satisfied.
In thinking of how the atoms are arranged in this ball, the scientists
considered the geodesic domes designed by the architect-engineer, R.
Buckmisnter Fuller. These domes led them to suspect a structure of interlocking
hexagons and pentagons, identical to those of a soccer ball. Because this idea
was inspired by the geodesic dome, they named this C60 allotrope of carbon
buckminsterfullerene.
Building designed by Buckminster Fuller
In 1996, H. Kroto, R. Smalley, and R. Curl shared the Nobel Prize in
Chemistry for their discovery of buckminsterfullerene.
8
3. THE PLATONIC SOLIDS
The Platonic solids are made up of regular polygons. A regular polygon must
have at least three sides, so the triangle is the simplest polygon. The number of
sides can theoretically increment indefinitely.
Thus, we can see that the Platonic solids are the only five regular polyhedra:

TETRAHEDRON
It has 4 vertices, 6 edges and 4 faces, each of which is an equilateral
triangle. There are 6 planes of reflectional symmetry. Each such plane contains
one edge and bisects the opposite edge (this gives one plane for each edge,
hence 6 planes). Reflection in a plane fixes two of the vertices and exchanges
the other two, so the corresponding vertex permutation is a transposition. There
are 4 lines of 3-fold rotational symmetry, each of which passes through a vertex
and the centre of the opposite face. The corresponding vertex permutations are
3-cycles. There are also 3 lines of 2-fold rotational symmetry. Each one joins
the centers of an opposite pair of edges. The corresponding edge permutations
are transposition pairs, in other words they have the form (a b)(c d) where a, b,
c and d are all different.
If we start with a tetrahedron and find the centers of all the faces we get the
vertices of a new tetrahedron. Thus, the tetrahedron is self-dual.

CUBE
There are 8 vertices, 12 edges and 6 faces, each of which is a square.
There are rotational symmetries of order 2, order 3 and order 4. The cube is
also invariant under multiplication by -1, so
Dir(Cube) = Symm(Cube) x {1,-1}
There are 4 long diagonals which are permuted by the action of the
symmetry group, giving rise to a homomorphism
f: Symm(Cube) → S4.
The rotations of orders 2, 3 and 4 are sent to 2-cycles, 3-cycles and 4-cycles
respectively. It turns out that f is an isomorphism.
9
If we start with a cube and find the centers of all the faces we get the
vertices of an octahedron. In other words, the dual of a cube is an octahedron.

OCTAHEDRON
It has 6 vertices, 12 edges and 8 faces, each of which is an equilateral
triangle. The dual of an octahedron is a cube. As dual polyhedra have the same
symmetry groups, we have
Symm(Oct) = Symm(Cube) = S4

DODECAHEDRON
It has 20 vertices, 30 edges and 12 faces, each of which is a regular
pentagon. To construct a dodecahedron whose sides have length 1, we start
1 5
with a cube of side  
(golden number) and attach to each face a "tent"
2
with edges of length 1.
It is not obvious that the trapezia line up properly with the triangles to give
flat pentagons. We can actually inscribe 5 different cubes in a dodecahedron.
The symmetry group acts on the set of these cubes, giving a homomorphism
f: Symm (Dodec)→ S5
10
By joining each vertex to the opposite one, we obtain 10 different lines of 3fold rotational symmetry. We can twist around each of these axes by an angle
of 2p/3 or 4p/3, giving 20 different rotations of order 3, each of which gives a 3cycle. By joining the centre of each face to the centre of the opposite face, we
obtain 6 different lines of 5-fold rotational symmetry, giving 24 rotations of order
5, each of which gives a 5-cycle. By joining the centre of each edge to the
centre of the opposite edge, we obtain 15 different lines of 2-fold rotational
symmetry, each of which gives a transposition pair. It can be shown using this,
that f is actually an isomorphism
f : Symm (Dodec)→ A5
The dual of a dodecahedron is an icosahedron.

ICOSAHEDRON.
It has 12 vertices, 30 edges and 20 faces, each of which is an equilateral
triangle. The dual is a dodecahedron, so we have
Symm (Icos) = Symm (Dodec) = A5
11
Briefly,
Platonic Solids
Tetrahedron
Octahedron and
cube
Icosahedron
and
dodecahedron
Proper symmetry
Mckay
group
A4
correspondence
group E6
S4
group E7
A5
group E8
Simple
Lie
Groups
A4: alternating group consisting of all even permutations of 4 things (rotating
the tetrahedron we can achieve any even permutation of its 4-vertices).
S4: group of all permutations of 4 things--we can draw 4 line segments
connecting the opposite vertices of the cube.
A5: alternating group of all even permutations of 5 things.
12
4. TRUNCATED ICOSAHEDRON

Regular icosahedron-truncated icosahedron
A truncated icosahedron is a regular icosahedron in which every vertex has
been, in a symmetric fashion, chopped off. This procedure has the effect of
replacing each vertex by a hexagonal face. In this way, we have added one
face, deleted one vertex, and added five new edges and five new vertices, for
every time we do it.
Regularicosahedron- Truncated icosahedron

Description
Among the many fullerene molecules, the most prominent and studied is
C60. Its corresponding polyhedron is the truncated icosahedron, which is seen
on the surface of a soccer ball.
The truncated icosahedron has 32 faces. That is, besides the 12 pentagons,
there are 20 hexagons. It’s of significance that the pentagons are isolated from
one another; in fact, this could be a requirement for the stability of a fullerene
molecule. C60 is the smallest fullerene molecule in which this isolation occurs.
There are 90 edges in the truncated icosahedron, 60 of which bound the 12
pentagons and separate hexagons from pentagons. These edges are called
PENTAGONAL EDGES. The remaining 30 edges, HEXAGONAL EDGES,
separate hexagons from hexagons. The pentagonal edges are single carbon
bonds and the hexagonal edges are double carbon bonds. In this way, each
hexagon has alternating single and double bonds. These hexagonal faces
remind a benzene ring, although, unlike benzene, these single and double
bonds remain fixed in the case of C60.
The truncated icosahedron has 60 vertices. Because of the isolation of the
pentagonal faces, each vertex lies on a unique pentagon. Therefore, these
polygons define a natural equivalence relation on the set of vertices, partitioning
the set of vertices into 12 equivalence classes where each class is the 5element set of vertices of one of the pentagons.
Each vertex gathers 3 edges as well, two pentagonal and one hexagonal,
implying the idea of a deep relationship between them.
13
In terms of its symmetry, a truncated icosahedron has 3 defining
dimensions: the two dihedral angles, 138 11’ (hexagon-hexagon) and 142 37’
(hexagon-pentagon), and the length of an edge. All other dimensions can be
derived from these three using trigonometry.
The sixty-member carbon “bucky ball” molecule
Fullerene molecules: C60 (soccer ball), C70 (rugby ball) and C80
Another remarkable characteristic of this polyhedron is its high symmetry. It
has 2-, 3-, and 5-fold symmetry.
Truncated Icosahedral Symmetry
14
A. PROPER SYMMETRY GROUP G
The structure of the truncated icosahedron is completely determined by the
graph  of its vertices and edges.
The proper symmetry group G of the buckyball is a 60-element icosahedral
group. This group G operates in a simple transitive way on the set V of vertices.
In this manner, given a pair of ordered vertices there is a unique proper
symmetry which carries the first to the second. The action of G on V is
equivalent to the action of G on itself by left translations. In particular, the action
of G on V doesn’t even perceive the edge structure in .
This behavior of G doesn’t satisfy the most important requirement for a
symmetry group, the capability of being able to describe faithfully the nature of
the molecule. Hence, we would like to find a group that can explain the full
structure of .
The edge structure in  determines a 6060 adjacency matrix H and,
consequently H would be expressed group theoretically. The eigenvalues of this
matrix, via a “Hunckel approximation”, enter into the determination of the
molecular energy levels of C60.
B. IN SEARCH OF THE PROPER SYMMETRY GROUP
The search of an appropriate symmetry group is not an easy task,
considering the difficulty in describing faithfully the structure of the truncated
icosahedron, where we must be really careful with the tricky description of its
edges.
If p is a prime number, let Fp denote the finite field of p elements.
The group SL (2, p) is the group of all 22 matrices with entries in Fp having
determinant equal to 1.
PSL (2, p) is SL (2, p) modulo its (2-element if p is odd) center, and it’s
named “Projective Special Linear group over F ”
The group PSL (2, p) is simple if p  5 and PSL (2,5) is an ICOSAHEDRAL
GROUP.

First step: The Icosahedron
Let V be the 12-element set of vertices of an icosahedron P where {c,d}  V
define an edge of P if the scalar product <cd>= (the golden number). Let A 
SO(3) be the group of all rotations which stabilize the group V. Then A is an
icosahedral group. Let Aj  A, for any integer j>1, denote the set of all elements
in A of order j. The Aj  0 only if j=2, 3 or 5, then
15
|A(2)|=15
|A(3)|=20
|A(5)|=24
For any prime p, the group PSL(2,p) naturally operates (transitively) on p+1
points. More concretely, it operates on the projective line Fp  {} over Fp as
the group of fractional transformations
a  b

c  d
Since A  PSL (2,5), it means that A naturally operates on six points. But, as
we mentioned before AA5, the symmetry group of the icosahedron, so A
operates on a set F of just five objects. This is not very easy to see if we just
look at the icosahedron, but algebraically it is quite clear. Define a relation on
the 15-element set A2, where if ,   A2 then    if  commutes with . The
relation between these two elements of A2 is an equivalence relation and there
are 5 equivalence classes each of which has 3 elements. Algebraically, we can
take F= {the set of maximal commuting subsets of A2}.
However, we still have a question, where in A can we find the graph of
vertices and edges of an icosahedron? In order to answer this question we
need to use the concept of conjugacy class. The sets A 2 and A3 are conjugacy
classes in A. On the other hand, A5 decomposes into a union of 2 conjugacy
classes, C and C’, each with 12 elements. Both of them are A-sets with respect
to the action of conjugation and closed under inversion. The map CC’, 2
is an A-bijection
Consider one of this two conjugacy classes of elements of order 5 in A, C 
A5. We will define a graph  whose set of vertices is C and, for u,vC, {u,v} is
an edge of  if uvC. This is well defined as result of the definition of u and v as
conjugate elements. It is also remarkable the fact of that any graph we have
defined is necessary invariant under the icosahedral group.
The next theorem, which can be proved, gives the solution to the previous
question,
Theorem 1. The graph  is isomorphic to the graph of vertices
and edges of an icosahedron. With respect to such an isomorphism
for u C the vertex corresponding to u-1 is the antipode of the
vertex corresponding to u.
Therefore, we can conclude saying that it‘s possible to find in A the graph of
the vertices and edges of an icosahedron. Starting with the edge defined by u
and v, and choosing the right orientation of the icosahedron, the five neighbors
of the vertex u can be expressed in terms of u and v, exhibiting the five faces of
the icosahedron which have u as a vertex.
16
So, what we are supposed to do now is replace the icosahedron by the
truncated icosahedron and, using the same method, find its graph. But this is
not so easy, and we should make use of a few extra tools.

Galois’s Letter
Galois’ result, reported in his famous letter to Chevalier, written the night
before his life-ending duel, is that if p>11 then PSL(2,p) can not operate, nontrivially, on a set with fewer than p+1 elements. In particular, it can not operate
non-trivially on a set with p elements when p>11.
Implicit in his statement is certainly the knowledge that his affirmation is not
true for the 3 cases of a simple PSL (2, p), where p11- i.e. p=5, 7, 11.
The situations p = 5, 7,11, where this happens satisfies that PSL (2, p) is
simple, as well. In each case, it is very amusing to look at how PSL (2, p) acts
non-trivially on a set S with p elements and consider the subgroup that doesn’t
move a particular element of S. It’s explained as follows,
The fact that PSL(2,5) operates on 5 points, as we discussed before,
establishes the isomorphism of PSL(2,5) with A5.
The group PSL(2, 7) is isomorphic to PSL(3,2). This group operates on a
projective plane, which is indicated with the 3, and the plane is over the field of
two elements, pointed out with the 2. This plane has 7 lines and 7 points,
showing seven things on which PSL(2,7) operates.
PSL (2, 11) is the most interesting case. It operates on eleven points and
the amazing of its symmetry is due to its unique and notable geometry. The
eleven points will be the field F11 itself-the integers from 0 to 11. Calculating the
elements belonging to the group, we obtain eleven 5-elements sets. Analyzing
these sets-lines, we observe the presence of repeated points, exactly 2, in
every pair of chosen lines. This means that the intersection of every two of the
lines establishes a bijection between the 55-element set of all pairs of distinct
points and the 55-element set of all pairs of distinct lines. This is named biplane
geometry.
Let’s identify the symmetric group S11 with the permutation group of the set
F11, the subgroup of S11 which stabilizes the set of these 11 lines is isomorphic
to PSL(2,11). Therefore, this relation develops an embedding of PSL(2,11) into
S11 or an action of PSL(2,11) on eleven objects. The isotropy subgroup of
PSL(2,11) at a point in F11 is isomorphic to PSL(2,5). The second embedding of
PSL(2,5) in PSL(2,11) is obtained by taking the isotropy subgroup of a line
instead of a point.
After all the explanation about the way these simple groups operate, we can
make some conclusions. For example, when p = 5, we have PSL (2, 5) = A5,
and if we look at the subgroup of even permutations of 5 things that leaves a
17
particular thing alone, we get A4. If we play this game with PSL (2, 7) we get S 4,
and if we do the same with PSL (2,11), we get A5.
The three groups obtained, A4, S4 and A5 are, up to conjugacy, the only finite
subgroups of SO(3) which operate irreducibly on 3. They are the proper
symmetry groups of the five Platonic Solids!
A5 is both PSL (2, 5) and the subgroup of PSL (2, 11) that fixes a point of S.
This gives a lot of relationships between A5, PSL (2, 5) and PSL (2, 11). This is
what we were looking for. One can start thinking of A5 as the vertices of the
buckyball, and some of them are connected by an edge using the embedding of
A5 in PSL(2,11).
 Second Step: WHAT ABOUT THE GRAPH OF THE TRUNCATED
ICOSAHEDRON ?
Using an inference from Galois’ report, we can consider a more appropriate
group for our purpose. Let’s think about a 60-element PSL(2,11) set-PSL(2,11)/H, where the subgroup H is chosen to be Z11, a cyclic group of order
11. Is possible that it has a natural structure of the graph of the truncated
icosahedron? A quick answer to this question would be, yes, of course. As a
consequence of the way it has been defined, it’s easy to see a canonical
PSL(2,11) invariant decomposition of this group, X, into a union of 12
“pentagons”. Making use of the known Borel groups, B, one can reach the
conclusion that B/Z11  Z5.
The orbits of B/Z11 are the 12 pentagons. Also, from Galois’ statement, you
can infer that X is the principal homogeneous space for A 5 which is exactly how
A5 operates on the vertices of a TI.
But, what happens with the most intricate part of the graph, the hexagonal
edges? The answer yields on a kind of picture, Cayley graph which is defined
on a group by choosing generators. In the case we are considering now, the
most convenient is make use of the different choices for an icosahedral group.
Hence, non-trivial elements ,  in an icosahedral group A’ define a Cayley
graph on A’. The following theorem establishes that this Cayley graph is the
graph of the TI.
Theorem 2. Let M be any 60-element set and let S60 be the full
group of permutations of M. Assume that ,S60 satisfy the
relations 5=1, 2=1, ()3=1, and that ,  and  have no fixed
points in M. Then the subgroup A’S60 generated by  and  is an
icosahedral group and M is a principal homogeneous space for A’ .
Let A be the centralizer of A’ in S60 so that A is also an icosahedral
group (A and A’ are each other’s centralizer) and M is also a
principal homogeneous space for A.
Let  be the graph on M so that for any x M the edges (three of
them) containing x are {x, x}, {x, -1x} and {x, x}. Then  is
18
isomorphic to the graph of a truncated icosahedron where the two
pentagonal edges containing x are {x,  x} and {x, -1x} and the
unique hexagonal edge containing x is {x, x}.
Finally, A is the proper symmetry group of .
We are almost ready to give an answer to the question in trouble, but we still
have to fix some details. A conjugacy class M of elements of order 11 in
PSL(2,11) is required. The next theorem sets up all the necessary details.
Theorem 3. Let A be any icosahedral subgroup of PSL(2,11). Let
A(2) be the set of all elements of order 2 in A. Then for ant x 
PSL(2,11) or order 11 there exists a unique element x A(2) such
that the commutator x= x-1xxx is again in A(2). Moreover, there
exists a unique choice CA of a (12- element) conjugacy class of
elements of order 5 in A and a unique choice of a (60-element)
conjugacy class, M, of elements of order 11 en PSL(2,11) such that
if x M there exists u CA which normalizes the cyclic group
19
5. SPECTROSCOPY
Let us describe now, according to the results obtained, the infra-red and
Raman spectra of a buckyball. The space of vibrational states has dimension
174 = 180 – 6. We would like to prove the following:
1) there are at most 46 distinct vibrational modes, that is, the space of
classical vibrational states can be decomposed into 46 irreducible
representations,
2) of these representations, at most four are visible in the infra-red (this
four just are equivalent to the representation of the group I h on the
ordinary 3-dimensional space ℝ3),
3) ten lines are visible in the Raman spectrum or, more precisely, the
complexification of the space S2 (ℝ3), which is 6-dimensional,
decomposes into a direct sum of the trivial representation and a 5dimesional representation, which corresponds to the traceless tensors
(this one is irreducible and occurs eight times in the space of
vibrational states, whereas the trivial one occurs twice).
The above facts concerning the spectrum of the buckyball have all been
verified experimentally. The two different kinds of Raman spectra,
corresponding to the trivial and to the 5-dimensional representation, can be
distinguished through the use of polarized light.
To begin, we must list all the irreducibles of Ih. Since Ih is the direct product
of I with ℤ2, it is enough to find the irreducibles of I and then label each with a +
or – sign according to the trivial or sign representation of the ℤ2 component.
Now I~A5, which is a subgroup of S5. The elements of A5 consist of the
identity, 20 three-cycles, 15 elements of the form (ab)(cd), and 24 five-cycles.
The 20 three-cycles form a single conjugacy class in A5, since any two threecycles are conjugate by an even permutation—the odd permutation, (de),
commutes with (abc). In the same way, the 15 elements of the form (ab)(cd)
form a single conjugacy class, as the odd permutation (ab) commutes with
(ab)(cd). On the other hand, a five-cycle, =(12345), does not commute with
any odd element, since it cannot carry the elements of any two-cycle or any
four-cycle into itself. But 2 =(13524) is conjugate to  in S5 by the odd element
(2354) and hence by no even element. So, the 24 five-cycles split up in A5 into
two conjugacy classes, those conjugate to  and those conjugate to 2. Thus
there are five conjugacy classes in all, and so five inequivalent irreducible
representations.
Let’s restrict the irreducible representations of S5 to A5. Some computations
show that the first three lines restrict to irreducibles of A 5. The fourth line, the
six-dimensional representations of S5, does not remain irreducible; the
character satisfies  ,   A5  2 . So it splits into two irreducibles of dimensions,
say, d1 and d2. Since
20
1  16  25  d12  d 22  60,
we must have
d1  d2  3.
One of these three-dimensional representations we already know—the
complexification of the representation identifying A5 with I acting as rotational
symmetries of the icosahedron. Since all elements are rotations, having trace
1+2 cos, we see that, for this representation,
 abc  0
and
 abcd   1.
If  corresponds to rotation trough angle 2/5, then
2
4
    1  2 cos ,
  2   1  2 cos
5
5
This is an irreducible character, and interchanging the roles of  and 2 in
this last equation gives the character of the other three-dimensional
representation.
We may label the irreducibles of I as 1, 4, 5, 3 and 3’. Therefore the
regular representation of I decomposes into
1 + 4 + 5 + 3 + 3 = 16
irreducible representations.
Let E B denote the vector bundle describing the displacements of the
carbon atoms in the buckyball from their equilibrium positions. It is a
homogeneous vector bundle with respect to the groups I and I h. The group I has
a trivial isotropy group at any point of B. Hence the space (E) transforms as
the direct sum of three copies of the regular representation of I. It decomposes
into 48 irreducibles. We remove two copies of 3 when we subtract off overall
translations and overall rotations. Therefore there are, at most, 46 distinct
vibrational modes.
Let X be a vertex of the buckyball and let H be the isotropy group of X in
I. So H consists of two elements, the identity and the reflection, r x, in the plane
passing through X and bisecting the buckyball. The representation of I h given by
its action on three-dimensional space is labeled as 3-, since the inversion
operator ( the generator of the ℤ2 component) acts as –Id. The vector bundle E
is induced from the restriction of this representation to H. The group H has only
two irreducible representations, both one-dimensional: the trivial representation,
1+, and the “sign” representation, 1-, which assigns the value –1 to rx. Thus the
three-dimensional representation 3-  of H must decompose into a sum of these
one-dimensional representations. Let’s choose coordinates so that r x is
reflection in the y, z plane in order to see what this decomposition is. The matrix
representing rx is then,
  1 0 0


 0 1 0
 0 0 1


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which shows that we have the next block decomposition
3-  = 1-  1+  1+.
(1)
Frobenius reciprocity says that 3- occurs once induced from 1- and twice
in the representation induced from 1+. From the decomposition above we see
that dim Hom(3-, (E))=5. Subtracting one copy of 3- corresponding to overall
translations shows that there are exactly four vibrational lines in the infra-red.
This proves statement 2) above.
Using Frobenius reciprocity again to compute the multiplicities of all the
irreducibles in the induced representations, we see that the reflection rx is the
product of the inversion –Id with a rotation x through 180 lying in I:
rx = - x.
Let K={e, x} be the corresponding two element subgroup of I, and 1K its
one-dimensional representations. Then for any + representation of I h we have
dim HomH (1, k+ ) = dim HomK (1K, k+ ),
whereas for the – representations the signs are reversed:
dim HomH (1, k- ) = dim HomK (1K∓, k+ ).
So, for example,
dim HomIh (5+, 1+ ) = dim HomH (1+, 5+ )= dim HomH (1K+, k+ ) =
1
2
(5  1)  3.
Proceeding in this way we compute the decompositions
1+ = 1+  2 x 3-  3+  2 x 3’-  3’+  2 x 4+  2 x 4-  3 x 5+  2 x 5and
1- = 1-  2 x 3+  3-  2 x 3’+  3’-  2 x 4+  2 x 4-  3 x 5-  2 x 5+.
(2)
(3)
We can use the expressions above to determine the number of lines in
the Raman spectrum. In the Raman experiment the operator, A, transforms like
an element of S2(ℝ3)+. Using the characters of I, we obtain easily,
3  3 = 3  1  5.
The first summand on the left is just the space of anti-symmetric tensors.
So, we see that the complexification of S2(ℝ3) transforms, under Ih, like 1+  5+.
The 5+ corresponds to traceless tensors and the 1+ corresponds to multiples of
the identity. The reason for the + is that the parity operator, P, has no effect on
tensors of even degree.
Let us examine these two components separately. Notice that 1+ does
not occur at all on the right-hand side of (3), and occurs once on the right-hand
side of (2). Hence we conclude from (1) and Frobenius reciprocity that
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Dim HomIh (1+, Vib) = 2.
In other words, there should be two Raman lines corresponding to the 1+
representation of Ih. The 5+ occurs twice on the right-hand side of (3), and three
times on the right-hand of (2). Hence we conclude from (1) and Frobenius
reciprocity that
Dim HomIh (5+, Vib) = 2+ 3+ 3 = 8.
So there should be eight lines corresponding to 5+. All ten lines have
been observed. In fact that 1+ lines can be distinguished, experimentally, from
the 5+ lines through the use of polarized light.
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6. BIBLIOGRAPHY
a) This Week’s Finds in Mathematical Physics (Week 79)
John Baez
April 1, 1996
b) The Graph of the Truncated Icosahedron and the Last Letter of Galois
B. Konstant
Notices of the AMS, Sep. 1995
c) Representation theory : a first course
William Fulton, Joe Harris
New York ; Berlin : Springer-Vlg, 1991
d) Affine Lie algebras and quantum groups : an introduction, with
applications in conformal field theory
Jürgen Fuchs
Cambridge : Cambridge Univ. Press, 1992
e) Symmetries, lie algebras and representations : a graduate course for
physicists
Jürgen Fuchs, Christoph Schweigert
Cambridge : Cambridge University Press, 1997
f) The Royal Swedish Academy of Sciences
“The 1996 Nobel Prize in Chemistry”
http://www.nobel.se/chemistry/laureates/1996/press.html
g) Scientific American
A few articles
http://www.sciam.com
Fullerene Chemistry (Jos Lurie)
http://www.twr.ac.za/jos7.htm
h) Fullerenes (Kim Allen)
http://www.mindspring.com/~kimal/Fuller/index.html
Truncated Icosahedral Symmetry (Dan Busby)
http://sam.phys.1su.edu/tiga/symmetry.html
i) The Platonic Solids (Tom Gettys)
http://sam.phys.1su.edu/tiga/symmetry.html
j) Buckyballs (Professor Bassam Z. Shakhashir)
http://scifun.chem.wisc.edu/chemweek/buckball/buckball.html
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