Group theory – fundamentals
We have a set of operators Gr that form a group. There will be ng operators, r=[1,ng],
and the operators group into classes. The number of classes, nC, also determines and is
equal to the number of irreducible representations (irreps). Our main job is to accomplish
two goals.
I.
The reduction formula. For a given basis, we wish to determine the number
of irreps that are contained in it with that symmetry group. The answer to this
questions is contained in the “reduction formula,”,
  a11  a2 2  a33  ....anC  nC .
Here Γ is the reducible representation of the basis, and Γj is a irrep subset of
that basis. The reduction formula tells us the number (aj) of states that can be
constructed from the original basis that transform according to irrep j. The
n
1 g j
*
reduction formula is a j 
  (Gr )  (Gr ).
n g r 1
II.
The character projector. An irrep Γj has a basis associated with it that can be
constructed out of the original basis. This is accomplished by the symmetryadapted coordinate filter theorem. Given any vector contained in the basis, we

can construct a symmetry-adapted coordinate q j from it by the following
ng


formula, q j    j (Gr )Gr X .
r 1
The coordinate q transforms according to the “j-th” irrep, and χj(Gr) is the
character for operator Gr on a vector (or partners of vectors) in an the j-th irrep.
III.
The row projector. The basis vectors given by II do not transform according
to a specific row. To reduce the dynamical or Hamiltonian matrices in
physical problems, we need to couple just one row at a time. To do this, we
use a row projector,
ng

j
q   [ j (Gr )] kk Gr X . The projector projects out from an arbitrary
k
r 1
vector a vector in the k’th row of the j’th irrep. To apply this projector, we
need the k,k element of the Γ matrices for each operator and each irrep. These
are not so easy to come by.
We now set up the fundamentals of the theory and define notation that will be
used in the evaluation of viral normal modes. As a running example, to make things
concrete, we use a 2 dimensional example of the group C3, which is that of an equilateral
triangle with its center at the origin, with one of its base parallel to the x axis, and one
vertex along the y axis (with x=0).
There are 6 operators, ng=6, and 3 classes, nC=3. The classes are E (identity), C3
(rotations about z), and σ (reflections about bisector of angle at vertices). There are 2 C3
operations (120o and 240o (or -120o)) and 3 σ operations. Thus the group operators are
G1=E, G2=C3, G3=C23 , G4=σ1, G5=σ2, G6=σ3.
The character table is
E
2C3
σ
A1
1
1
1
A2
1
1
-1
E
2
-1
0
There are 3 irreps (=nC), and they j=A1, A2, and E. (Don’t confuse the identity operator
G1=E with the irrep E.) The characters of these irreps are χj(Gr). For example
ΧA1(C3)=1, ΧE(C3)=-1, and ΧA2(σ)=-1.
We next define basis vectors ê1 , ê2 ,.. êk …. eˆk max where k is the dimension of
the space. The index k is k=[1,kmax] where (usually) kmax= 3natoms where natoms is the
number of atoms in the problem. This refers to the full problem, which means that natoms
can be quite large. For our 2D example of an equilateral triangle, kmax is just 2natoms,
which is 6 for the simple example of a single atom at each vertex of the triangle.
 1  
0
   
   
 0
0




 0
0


The simplest basis to choose is ê1    , ….., ê6    . We make this choice, and
 0
0




 0
0
  0  
 1  
 
 
its obvious 3D extension for the reducible representation of our natom system. Thus Γe
refers to this representation. (It is clear (or is it?) that the trace of Γe is kmax.) Not that this
basis is orthonormal, eˆi †  eˆi   i, j . Or introducing bra-ket notation, <ei|ej>=δi,j.
We next determine how an operator Gr acts on one of the basis vectors.
Let’s say we act on ê1 . We write
Gr eˆ1  eˆ1' , r .
This means that e1 is transformed to a new vector e’. For example e’ might be
1
(eˆ3  eˆ4 ) .
2
We define a reducible representation by matrices Γe(Gr); one matrix for each operator in
the group. The matrices are defined as,
Gr eˆ1 eˆ2 eˆ3 eˆ4 eˆ5 eˆ6   eˆ1 eˆ2 eˆ3 eˆ4 eˆ5 eˆ6  e (Gr ). Eq. X
But
Gr eˆ1 eˆ2 eˆ3 eˆ4 eˆ5 eˆ6    eˆ '1, r  eˆ ' 2, r  eˆ '3, r  eˆ ' 4, r  eˆ '5, r  eˆ '6, r 







 ˆ †  
 e1  


 eˆ †  
 
We apply  2   to Eq. X to obtain




 †  
 eˆ6  
  
 ˆ † 
 e1 


 eˆ † 
 2 Gr eˆ1 eˆ2 eˆ3 eˆ4 eˆ5 eˆ6  




 † 
 eˆ6 
 
 ˆ †  
 e1  


 eˆ †  
 2   eˆ1 eˆ2 ...eˆ6   e (Gr )




 †  
 eˆ6  
  
The orthogonality of the basis states on the right hand side yields
 ˆ † 
 e1 


 eˆ † 
 
 e (Gr )   2 Gr eˆ1 eˆ2 eˆ3 eˆ4 eˆ5 eˆ6 




 † 
 eˆ6 
 
Or in braket notation, the i,j’th element of the matrix is
ie, j (Gr )  ei | Gr | e j >.
Generically, the matrix Γe(Gr) is referred to as the reducible representation in the basis
of e for the operator Gr.
Symmetry-adapted basis.
We now wish to change from the êi -basis to a symmetry-adapted basis q̂i .
†
This new basis is orthonormal, qˆi  qˆ j   i, j , or <qi|qj>=δi,j.
We write the transformation as
qˆ i   U j , i eˆ j , or
j
qˆ 1 qˆ 2 qˆ 3 qˆ 4 qˆ 5 qˆ 6  eˆ1 eˆ2 eˆ3 eˆ4 eˆ5 eˆ6 U .
This can also be written as qˆi   U †i, j eˆ j , or simply q=U†e.
j
      
With this change of basis, the matrices Γ(Gr) change from Γe(Gr) in the e-rep. to Γq(Gr)
in the q-rep. The matrices transform as
U 1 e (Gr )U   q (Gr ) .
Proof.
 q (G )
 qˆi | Gr | qˆ j 
See definition from Eq. XXX above.
r 

i, j
Transformation properties.
qˆi   U †i, j eˆ j .
j
Combine these two,
 q (G )
  U ki  eˆk |Gr  U lj | eˆl  , which is
r 

i, j k
l
 e (G )
  U †   e (Gr ) U 
, we arrive at
r 

 k , l   lj
i, j k , l   
i, k
 q (Gr )  U †  e (Gr )U . QED.


Our job is to find at least one set of q’s and one transformation U. There is no
unique answer, and finding just one answer is what the symmetry-adapted coordinate
filer theorem is designed to do. Our choice of q’s will be one in which we take the
reducible representation fo an irreducible representation. In the irreducible representation,
the Γq(Gr) matrices are block diagonal, where the dimension of each block is the
dimension of the representation. In our equilateral triangle C3 example, the
representations A1, A2, and E are 1 dimensional, 1 dimensional, and 2 dimensional
respectively.
The reduction formula:
The matrices Γq(Gr) are now irreducible. We now can apply the reduction formula
to Γe,
 e  a11  a2 2  a33  ....anC  nC .
For our C3 example, the Γ matrices are composed of blocks of A1, A2, and E symmetry.
This means that the reducible representation (in the basis e) has be reduced to block
diagonal form using a symmetry-adapted q basis. Thus for C3,
 q  U †U  a1 A1  a2 A2  a3 E .
This means that the matrix representation of the Γ matrices, they can be reduced to a1
blocks of A1 symmetry (each 1×1), a2 of A2 symmetry (each 1×1 ), and a3 blocks of E
symmetry (each of 2×2).
We can easily find the coefficients aj. Since Γq and Γe are related by a similarity
transformation, their traces are the same. That is why only character tables, which only
involve traces, are so useful. We choose one operator from each class. For C3, they are E,
C3, and Sigma. Thus we arrive at three equations in 3 unknowns
 ( E )  Tr e ( E )  a1 A1 ( E )  a2  A2 ( E )  a3  E ( E )
 (C3 )  Tr e (C3 )  a1 A1 (C3 )  a2  A2 (C3 )  a3  E (C3 )
 ( )  Tr e ( )  a1 A1 ( )  a2  A2 ( )  a3  E ( ).
The character tables list χj(Gr) (e.g. χA1(E) ). The left side needs to be determined. They
are tedious to compute, but it is straightforward. The program 3C-2D.f90 computes them
for the equilateral triangle example. Here is how they are computed. We apply the
symmetry operator Gr to each basis state. If under Gr atom “i” is transformed to atom “j”,
then this atom “i” contributes nothing to the χ(Gr). (Actually we are referring to χe(Gr);
since the trace is invariant, the reference to the basis is unnecessary). Only operations in
which an atom “i” transforms to itself can there be a contribution. For example atom 1
transforms to atom 1, 2, 3, 1, 3, 2 by E, C3,C32, σ1, σ2, σ3 respectively. So only E and σ1
have atom 1 contributions to the trace. For atom 2 they are E and σ2, and for atom 3 it is
 1/ 2
 3 / 2
E and σ3. Since σ1 is 
 , we see that it is traceless. Thus

3
/
2

1
/
2


 ( E )  Tr e ( E )  6 ,  (C3 )  Tr e (C3 )  0 , thus  ( )  Tr e ( )  0 .
The three equations become
6  a1  a 2  2a3
0  a1  a 2  a3
0  a1  a 2  0a3
Which has solution (a1, a2, a3)=(1, 1, 2).
The reduction formula performs the same calculation in one shot. The reduction
formula is,
n
1 g j
*
aj 
  (Gr )  (Gr ).
n g r 1
The quantities χj(Gr) is the character of the operator Gr (or any operator in its class) in the
symmetry-adapted basis of symmetry j (e.g. j=A1, A2, or E in C3) The character table
gives the characters χj(Gr) of the various irreps. For example in C3, χA1(E)=1, χE(σ)=0 etc.
To use the reduction formula, we need to know the trace of the various operators,
χ(Gr). The trace is independent of representation, so we can use either the e- or qrepresentations. Since we do not yet know the q-rep, we use the e-rep.
An example is offered by C3. The program C3-2D.f90 does this example and obtains
(a1, a2, a3)=(1, 1, 2).
The character projector.
The character projector is used to find the Nj (=aj·nj) basis vectors that span the
space of representation “j” which occurs aj times anNd has nj rows. (For example, Hg has
5 rows and occurs 8 times for C60, so NHg is 40.) It must be noted that the basis found is
complete and spans the space, but the vectors do not transform as a specific row or the
irrep. As such, this is only an intermediate step.
ng
j
The character projector, Pjχ, is P    j (Gr )Gr . We apply Pχ to a sequence
r 1



of vectors X 1 , X 2 , … X n . The vectors may be (1000..), (0100…) etc., or they may be
a sequence of vectors generated by a random number generator. We apply Pχ to these


vectors on at a time to generate the sequence v m  P X m . At each step m, an overalp
matrix is computed with that we have so far in our sequence, Si, j (m)  vi | v j  . If the

overlap maxtrix has a zero eigenvalue, that last vector v m is skipped over. A zero

eigenvalue indicates that v m is not linearly independent of precious vectors. This is
unlikely to happen with random vectors, but happens often when we use a simple
sequence of vectors. The process is repeated until m= aj·nj linearly independent vectors
are created. This is the dimension of the space Nj. The final Nj×Nj overlap matrix has all

 
positive non-zero eigenvalues. The basis vectors v1 , v2 , …, v N span the space but arre
j
 
not orthonormal. A Gram-Schmidt process is used to find an orthonormal basis q1 , q2 ,


…, q N from the vi basis.
j
The row projector.
A key quantity we need is a basis for a each row in an irrep. Suppose irrep “j” has
nj rows. For example the Hg irrep of Ih has 5 rows.
We can apply the row projector onto a random vector to generate a basis for the
rows of representation “j.” This requires the Γ matrices. Is there a more automatic
method? Eric has derived the “Great Overlap Theorem” which is a method to numerically
generate a complete basis.

Suppose we are given a vector q j from the character projector starting from a

random vector X . It is a vector in the space irrep “j” but it does not transform as any
specific row. But from this vector we can generate an ng×ng matrix which we call the

“Great Overlap” matrix S. We operate on the vector q (for simplicity, we drop the j



superscript, q j ) by all r operators of the group, qr  Gr q , or | qr | Gr q  . The
overlap matrix is, S r, r '  qr | qr '  Gr q | Gr 'q  . The matrix is Hermitian, and we
solve for the eigenvalue/vector spectrum. We find that the eigenvalue spectrum has nj×nj
non-zero eigenvalues, and the remainder (ng-nj2) are all zero. The zero eigenvalues is to
be interpreted to mean that the in the space of vectors |qr>, there are only nj2 linearly
independent vectors and the remaining ng-nj2 are linearly dependent on the others and
hence give zero eigenvalue.
The important behavior of the Great Overlap matrix is the existence of nj×nj nonzero eigenvalues. Each eigenvalue is repeated nj times. Thus for a apace of dimension 4,
the eigenvalue spectrum of S is (λ1, λ1, λ1, λ1), (λ2, λ2, λ2, λ2), (λ3, λ3, λ3, λ3), and (λ4, λ4,
λ4, λ4). The values of the eigenvalues depend on the vector |q> used, however the sum of
the eigenvalues is ng, the size of the group (e.g. λ1+ λ2+ λ3+ λ4 = ng/nj=30 for Hg).
References
Group Theory for Atoms, Molecules and Solids, Brian S. Wherrett
(Prentice Hall International, Englewood Cliffs, 1986) Chapter 5, pp 47-63.
Chemical Applications of Group Theory, F. Albert Cotton, (John Wiley and Sons, NY,
1990, 3rd edition).
Rotation-vibration spectra of icosahedral molecules. II. Icoshedral symmetry, vibrational
eigenfrequencies, and normal modes of Buckminsterfullerene, David E. Weeks and
William G. Harter, J. Chem. Phys. 90, 4744-4771 (1989). This paper lists the icosahedral
j
generator irrep matrices, mn (Gr ) . See Fig. 21.
Point group theory tables, Simon L. Altmann and Peter Herzig (Oxford Science
publications, Clarendon Press, Oxford, 1994). Tables list the basis of the irrep in
spherical harmonics. From these it is possible to generate the Γ matrices, from which the
row projector can be built. (See for example, p652, Table 74.6a for group I.)
Additional thoughts
Phi_ij=gammadagger(G)ii’ phii’j’ gamma(G)j’j
Vvec=sum_i v^e_i e_i sum_i v^q_i q_i.
v^q_i = sum_j Udag_i,j v^e_j
phi^q = Udag phi^e U = a1*phi^1+a2*phi^2+…
phi^a1=a1Xa2 matrix.