3.36. Groups

These are notes of Karel Výborný and Ondřej Čertík on the group theory as a result of the first VDNK (Výprava do neznámých krajů) held between October 30 and November 2, 2006 in Prague. So that the next time we look at it we should be able to quickly recover our forgotten ideas.

3.36.1. Theory

Definition of a group:

  • I1: x,y \in G \Rightarrow xy \in G

  • I2: there exist e such that ex=xe=x for each x\in G

  • I3: there exist x^{-1} such that xx^{-1}=x^{-1}x=e for each x\in G

  • I4: (xy)z=x(yz) for each x,y,z\in G

Every finite group is isomorphic to a subgroup of S_n (permutations).

Representation

Set of linear operators T(x) (for each x\in G there is one T(x))

T(x)T(y)=T(xy),\quad T(e)={\hbox{\dsrom 1}}\,.

T(x) fulfills all the group axioms I1, I2, I3, I4. The requirement T(e)=1 is non-trivial, consider for example the following 4 matrices

T(\bsigma)=\begin{pmatrix} \bsigma & 0 \\ 0 & 0 \\ \end{pmatrix}, \quad T(e)=\begin{pmatrix} \hbox{\dsrom 1} & 0 \\ 0 & 0 \\ \end{pmatrix}\,,

that fulfill T(x)T(y)=T(xy) but not T(e)={\hbox{\dsrom 1}}.

The representation T(x) is said to be faithful if there is a one-to-one relationship between T(x) and x (an isomorphism).

Equivalent representations T_1 and T_2: there exist S such that T_2=ST_1(x)S^{-1} for each x\in G.

Reducible representation T(x): there exist an equivalent representation that is diagonal:

(3.36.1.1)T'(x)=ST(x)S^{-1}=\begin{pmatrix} T_1' & 0 \\ 0 & T_2' \\ \end{pmatrix}\,, \qquad \forall x\in G\,.

We say that T' is a direct sum of T_1' and T_2': T'=T_1'\oplus T_2'.

Irreducible representation: is not reducible.

Conjugate element: x is conjugate to y (x\sim y) if there exist c such that:

x=cyc^{-1}

if x\sim y and y\sim z then x\sim z.

Conjugate class: elements which are all conjugate to each other

No element may belong to more than one class \Rightarrow every group may be broken up into separate classes.

Character \chi of the representation T(x): set of numbers \chi(x)=\Tr T(x) as the group element x runs through the group.

Equivalent representations have the same character:

\chi'(x)=\Tr T'(x)=\Tr ST(x)S^{-1}=\Tr T(x)=\chi(x)

Representations having the same character are equivalent.

Proof: Characters can be thought of as elements of a q-dimensional vector space where q is the number of conjugacy classes. Using the orthogonality relations derived above, we find that the q characters for the q inequivalent irreducible representations forms a basis set. Also, according to Maschke’s theorem, both representations can be expressed as the direct sum of irreducible representations. Since the character of the direct sum of representations is the sum of their characters, from linear algebra, we see they are equivalent.

All the elements in the same class have the same character.

Maschke’s theorem: for finite groups, every class of equivalent representations contains unitary representations. The theorem is also true for most infinite groups of interest in physics.

Let T be a reducible representation, then:

T=m_1T^{(1)} \oplus m_2T^{(2)} \oplus m_3T^{(3)}\oplus \cdots

where T^{(1)}, T^{(2)}, T^{(3)} dots are all the inequivalent irreducible representations and m_\alpha (\alpha=1,2,3,\dots) gives the number of times that the irreducible representation T^{(\alpha)} occurs in the reduction.

Similar relation holds for group characters:

\chi=m_1\chi^{(1)} + m_2\chi^{(2)} + m_3\chi^{(3)} + \cdots

and it can be shown [Elliott] (eq. 4.28, page 63):

m_\alpha={1\over g}\sum_{x\in G} \chi^{(\alpha)*}(x)\chi(x)= = {1\over g}\sum_{p} c_p\chi^{(\alpha)*}_p\chi_p

where c_p is the number of elements in the class p, g is the number of elements in G (the order of the group).

Example

Consider the S_3 permutation group. The character table is:

\begin{array}{c|ccc} S_3 & e & 3 (1 2) & 2 (1 2 3) \\ \hline A_1 & 1 & 1 & 1 \\ A_2 & 1 & -1 & 1 \\ E & 2 & 0 & -1 \\ \end{array}

From the table we read c_1 = 1, c_2 = 3, c_3 = 2 and g = c_1 + c_2 + c_3 = 1 + 3 + 2 = 6. There are 3 classes and 3 irreducible representations.

Case I

We are given a representation given by the following matrices:

e = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix},\quad\quad a = \half\begin{pmatrix} 1 & -\sqrt{3} \\ -\sqrt{3} & -1 \end{pmatrix}, \quad\quad b = \half\begin{pmatrix} 1 & \sqrt{3} \\ \sqrt{3} & -1 \end{pmatrix}, c = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix},\quad\quad d = \half\begin{pmatrix} -1 & \sqrt{3} \\ -\sqrt{3} & -1 \end{pmatrix}, \quad\quad f = \half\begin{pmatrix} -1 & -\sqrt{3} \\ \sqrt{3} & -1 \end{pmatrix}.

These 6 matrices belong to the following three classes \{e\}, \{a, b, c\}, \{d, f\} and the corresponding characters for each class are:

\chi_1 &= 2 \\ \chi_2 &= 0 \\ \chi_3 &= -1

and we get:

m_1 &= {1\over 6} (1\cdot 1\cdot 2 + 3\cdot1\cdot 0 + 2\cdot1\cdot (-1)) = 0 \\ m_2 &= {1\over 6} (1\cdot 1\cdot 2 + 3\cdot(-1)\cdot 0 + 2\cdot1\cdot (-1)) = 0\\ m_3 &= {1\over 6} (1\cdot 2\cdot 2 + 3\cdot 0\cdot 0 + 2\cdot(-1)\cdot (-1)) = 1\\

So this representation is irreducible and it is equivalent to m_1 A_1 \oplus m_2 A_2 \oplus m_3 E = E.

Case II

We are given a representation given by the following matrices:

e = d = f = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix},\quad\quad a = b = c = \half\begin{pmatrix} -1 & -\sqrt{3} \\ -\sqrt{3} & 1 \end{pmatrix}.

These 6 matrices belong to the following three classes \{e\}, \{a, b, c\}, \{d, f\} and the corresponding characters for each class are:

\chi_1 &= 2 \\ \chi_2 &= 0 \\ \chi_3 &= 2

and we get:

m_1 &= {1\over 6} (1\cdot 1\cdot 2 + 3\cdot1\cdot 0 + 2\cdot1\cdot 2) = 1 \\ m_2 &= {1\over 6} (1\cdot 1\cdot 2 + 3\cdot(-1)\cdot 0 + 2\cdot1\cdot 2) = 1\\ m_3 &= {1\over 6} (1\cdot 2\cdot 2 + 3\cdot 0\cdot 0 + 2\cdot(-1)\cdot 2) = 0\\

So this representation is reducible and it is equivalent to m_1 A_1 \oplus m_2 A_2 \oplus m_3 E = A_1 \oplus A_2. The matrices are equivalent to:

e = d = f = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix},\quad\quad a = b = c = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.

Other facts

Number of irreducible representations = the number of classes.

Regular representation of G:

Take R^n with n=\# G and assign a canonical basis to the elements g_i of G. A matrix A_a assigned to a\in G now describes the mapping (g_1,g_2,\ldots)\mapsto (ag_1,ag_2,\ldots), i.e. in if ag_1=g_5, then the fifth element of the first row is one and others of that row are zero in A_a. Each IR of the reg. rep. occurs in its decomposition with the multiplicity equal to its dimension. Thus (p. 65, [Sternberg])

\# G = \sum p_i^2\,.

Element Order

The order n of an element g is the least integer for which g^n=e. The order n can be determined from the group multiplication table for example. Theorem: n must divide the size (order) of the group (for finite groups). Example: For a group of six elements, the only possible orders are 1, 2, 3 and 6. Note: the element order is the same for the whole conjugacy class because: x^n = \left(c y c^{-1}\right)^n = c y^n c^{-1} = c e c^{-1} = e.

Schur’s lemma

(a) Be r an IR of G. If [r(a),T]=0, \forall a\in G, then T=cI.

(b) Be r_1, r_2 two inequivalent IRs of G. Then r_1(a)T=Tr_2(a) valid \forall a\in G implies T=0. See p. 55 in [Sternberg]. This can be used to derive the orthogonality relations for characters.

Complete reducibility

Every rep can be decomposed into IRs: true for finite (p. 52) and compact (p. 179 in [Sternberg]) groups. Counterexample for larger groups, p. 53.

Sum of reps.

Opposite process to reduction, \rho\oplus\sigma, it lives on the direct sum of the two vector spaces of \rho and \sigma.

Take an IR \rho of G. Then \rho will also be a rep. of any subgroup H\subset G, but it need not be an IR, because the condition for reducibility, Eq. (3.36.1.1), is less strict: it suffices if the matrices T(g) are simultaneously block diagonal only for g\in H, not for all g\in G. This is called restriction and it is denoted by \downarrow.

Induced representation, denoted by \uparrow, is an opposite of the restriction. It works as follows: if F=G\otimes H, then \rho(f)=\rho(g), when f=g\otimes h.

Product of representations, \rho\otimes\sigma lives on the direct product of the two vector spaces. Product of IRs need not be an IR. Most prominent example: adding of angular momenta.

Interesting examples

O and T_d (see Crystallographic Point Groups) are isomorphic to S_4 (p. 35 in [Sternberg]). Written as matrices in 3D, they are 3D representations. Since O has only \det A=1 matrices unlike T_d, they are inequivalent.

Homeomorphism of SL(2,C) into the Lorentz group [or SU(2) into SO(3)], p. 7 [Sternberg]}. Start with the following 1-1 correspondence between \vec{x} and x:

\vec{x}=(x_0,x_1,x_2,x_3)^T\,,\qquad x=\begin{pmatrix} x_0+x_3 & x_1-ix_2 \\ x_1+ix_2 & x_0-x_3 \end{pmatrix}\,.

For any matrix of A\in SL(2,C) take AxA^*=x'. Decode x' into \vec{x}' and the relation between \vec{x} and \vec{x}' defines uniquely a Lorentz transformation; thus A was mapped into some Lorentz group element. If x_0=0 this gives a mapping from SU(2) into SO(3). The mapping is 2-1 because A and -A give the same x'.

SO(3) is not simply connected. Consider matrices U_{\theta}=diag(e^{-i\theta}, e^{i\theta})\in SU(2), \theta\in[0,\pi]. These map into SO(3) rotations by 2\theta around the z–axis. These matrices A_\theta=R_{z,2\theta} in SO(3) form a closed loop, R_{z,0}=R_{z,2\pi}. If SO(3) were simply connected it would be possible to contract this loop into a point while keeping A_0 and A_\pi unchanged. But then the same would have to happen with the original curve of matrices U_\theta while keeping U_0 and U_\pi at their place. Since U_{\pi}=-I\not= U_0=I, this curve is not closed and such a contraction is not possible.

All IRs of S_3 are in [Sternberg], p. 57.

3.36.2. Crystallographic Point Groups

Point group is a subgroup of O(3).

Crystallographic point groups are all subgroups of O(3), which leave a monoatomic crystal lattice invariant. Those can be symmetries of an infinite crystal (e.g. C_5 is excluded since pentagons cannot cover the plane).

There are only 7 crystallographic point groups: S_2 (triclinic), C_{2h} (monoclinic), D_{2h} (orthorhombic), D_{3d} (rhombohedral), D_{4h} (tetragonal), D_{6h} (hexagonal) and O_{h} (cubic).

For simple monoatomic crystals with one atom per unit cell these seven are the only possible crystallographic point groups. For more complicated crystals with a molecule or an arrangement of atoms in the unit cell, the symmetry will be reduced to the subgroup which leaves not only the lattice but also the unit cell invariant.

The complete list of all possible crystallographic point groups will therefore be given by the above seven together with all their subgroups (Tab. 3 in [Birss] or Tab. 4 in [Sternberg]):

\begin{array}{cc} S_2 & C_{1h}, S_2 \\ C_{2h} & C_2, C_{1h}, C_{2h} \\ D_{2h} & D_2, C_{2v}, D_{2h} \\ D_{3d} & C_3, S_6, D_3, C_{3v}, D_{3d} \\ D_{4h} & C_4, S_4, C_{4h}, D_4, C_{4v}, D_{2d}, D_{4h} \\ D_{6h} & C_3, S_6, D_3, C_{3v}, D_{3d}, C_6, C_{3h}, C_{6h}, D_6, C_{6v}, D_{3h}, D_{6h} \\ O_h & T, T_h, O, T_d, O_h \\ \end{array}

There are 37 subgroups together. D_{3d} is a subgroup of D_{6h} (so all 5 subgroups of D_{3d} are also subgroups of D_{6h}). Together we get 37-5 = 32 distinct subgroups. Groups, which might at first sight appear to be missing from the list are C_{1v}, D_1, D_{1h}, S_1, and S_3, but these are the same as C_{1h}, C_2, C_{2v}, C_{1h} and C_{3h} respectively.

The following groups are isomorphic:

C_{1h}, S_2, C_2

S_4, C_4

S_6, C_{3h}, C_6

C_{2h}, C_{2v}, D_2

C_{3v}, D_3

D_{2d}, C_{4v}, D_4

D_{3d}, D_{3h}, C_{6v}, D_6

T_d, O

The way to derive the above lists is the following.

Procedure:

  1. Find all finite crystallographic subgroups of SO(3) called rotation subgroups

  2. Take each subgroup from 1) and add -I and close the subgroup (‘non-rot containing -I’)

  3. for each subgroup G^\wedge in 1), find whether it has some normal subgroups G^+ of index 2 (half a size of G^\wedge) and construct G^+\cup (-I)aG^+, where a\notin G^+ and a\in G^\wedge; this will be a ‘non-rot not containing -I’ (for each G^\wedge there can be zero, one or more such G^+).

The sum of 1.,2.,3. are all finite crystallographic groups of O(3). The procedure is described in [Sternberg], p. 28-40.

An example: O (all rot. symm. of a cube, i.e. no mirroring) is 1), O^h (all symm. of a cube) is 2) made of O and T_d (all symm. of a tetrahedron) is 3) made of 1).

Zoology

Schönflies notation: C_n is an n–fold rotation (2_z, 3_z …) group (planar polygon), D_n is a diedric group, i.e. C_n plus turn-the-page two-fold rotations (e.g. 2_x, 2_\perp), T, O and I (= Y) are the rotational symmetries of a tetrahedron, octahedron (identical to those of a cube) and icosahedron (identical to those of a dodecahedron), respectively. Additional indexes mean reflection planes, horizontal, vertical, diagonal (h,v,d) or -I (i). Some atypical notation: S_2=C_i, S_6=C_{3i}, S_4=C_{2i}, C_s=C_{1h}.

Hermann–Mauguin (HM, international) notation: 2,3,4 means C_n, \bar{4} means rotation-inversion axis (rotation followed by -I), m is a vertical mirror plane, /m is a horizontal mirror plane.

Symmetry operations (in Table 3 of [Birss]): like HM, 2_x means a two-fold rotation around x–axis, 2_\perp means some other axis in the xy plane than x,y or xy (diagonal), \bar{3}_z is a rotation followed by -I. 3(2_\perp) means three different two-fold axes 2_\perp.

Construction and usage of the character table

For simpler groups the character tables can be fully constructed by the following rules:

  1. The sum of the squares of the dimensions n_i of the irreducible representations is equal to the order g of the point group:

    \sum_{\mu=1}^k n_\mu^2 = g

    The dimension n_\mu is given by the character of the identity matrix (first column) n_\mu = \chi^\mu(E), so the sum of squares of the first column is g. It is customary to put the characters of the one dimensional representation (\chi^1(C_i)=1) into the first row, so the first row is filled with 1s. Also, n_i must divide g.

  2. The number of irreducible representations r (rows) is equal to the number of classes k (columns)

  3. The rows must satisfy

    \sum_{i=1}^k g_i \chi^\mu(C_i) \chi^\nu(C_i)^* = g \delta_{\mu\nu}

  4. The columns must satisfy

    \sum_{\nu=1}^k \chi^\nu(C_i) \chi^\nu(C_j)^* = {g\over g_i} \delta_{ij}

  5. Characters of all one-dimensional representation must be roots of unity, equal to \chi = e^{2\pi i k\over n}, where n is the element order, which must divide the group size g (and it is the same for the whole conjugacy class). In general, k is any integer (for faithful representation it would be k=1). This follows from the fact, that the character is also the one-dimensional representation matrix and they all commute, thus the group is Abelian. Also, the characters (=representation matrices) must respect the group operations, so for example if for two group elements g_1^2=g_2, then their characters must also obey \chi_1^2 = \chi_2.

  6. Character of an element is the complex conjugate of its inverse. If they both belong to the same conjugacy class, the character must be real. If the character is complex, it means that its inverse is not from the same conjugacy class and then there must be a complex conjugate for another conjugacy class from the same irreducible representation.

  7. Characters come in complex conjugate pairs, since complex conjugate of a representation is also a representation. If there is only one representation of the dimension d, then it must be real (since it is its own conjugate). If there are two representations of dimension d and one is complex, then the other one must be its complex conjugate. Another way to look at this is that if we conjugate each entry of the character table, then we must get the same character table (up to a possible reordering of rows within the same dimension).

  8. If there is one dimensional representation A_1 (with characters \chi_1) and any other representation T of dimension d (with characters \chi), then there must be a representation of dimension d with characters \chi_1\chi (corresponding to the tensor product A_1 \otimes T).

There exists a systematic approach that works for any group, but it is complicated (see for example [Dixon67], [Blokker72], [Cannon69] and [Chillag86]).

The notation for irreducible representation: One-dimensional irreducible representations are labeled either A or B according to whether the character of a 2\pi\over n (proper or improper) rotation about the symmetry axis of highest order n is +1 or -1. If there is no symmetry axis, all one-dimensional representations are labeled A.

For general information, see [Elliott] (sec. 4.15, page 67) and [Bishop], page 128.

Example I

Let’s take the group C_{3v}, which has three classes E (1 element), C_3 (2 elements) and \sigma_v (3 elements).

So g_1=1, g_2=2 and g_3=3 and the order is g=g_1+g_2+g_3=6. Therefore it has three irreducible representations, whose dimensions must satisfy:

n_1^2 + n_2^2 + n_3^2 = 6

The only integer solution (up to a permutation) is n_1=n_2=1 and n_3=2. So we immediately have:

\begin{array}{c|ccc} C_{3v} & E & 2 C_3 & 3 \sigma_v \\ \hline & 1 & 1 & 1 \\ & 1 & a & b \\ & 2 & c & d \\ \end{array}

The rule 3. generates the following equations for all \mu and \nu:

\begin{array}{cc|c} \mu & \nu & \sum_{i=1}^k g_i \chi^\mu(C_i) \chi^\nu(C_i)^* = g \delta_{\mu\nu} \\ \hline 1 & 1 & 6 = 6 \\ 1 & 2 & 1 + 2a + 3b = 0 \\ 1 & 3 & 2 + 2c + 3d = 0 \\ 2 & 2 & 1 + 2a^2 + 3b^2 = 6 \\ 2 & 3 & 2 + 2ac + 3bd = 0 \\ 3 & 3 & 4 + 2c^2 + 3d^2 = 6 \end{array}

Solving all these equations simultaneously, we get two independent solutions. One is:

a &= 1 \\ b &= -1 \\ c &= -1 \\ d &= 0

and the other is:

a &= -{7\over 5} \\ b &= {3\over 5} \\ c &= {1\over 5} \\ d &= -{4\over 5}

The rule 4. generates the following equations for all i and j:

\begin{array}{cc|c} i & j & \sum_{\nu=1}^k \chi^\nu(C_i) \chi^\nu(C_j)^* = {g\over g_i} \delta_{ij} \\ \hline 1 & 1 & 6 = 6 \\ 1 & 2 & 1 + a + 2c = 0 \\ 1 & 3 & 1 + b + 2d = 0 \\ 2 & 2 & 1 + a^2 + c^2 = 3 \\ 2 & 3 & 1 + ab + cd = 0 \\ 3 & 3 & 1 + b^2 + d^2 = 2 \end{array}

Both of the above solutions for (a, b, c, d) satisfy all of these equations, so the column equations are redundant.

Now we use the rule 5. and see that the second solution is not a root of unity, so we discard it. The final character table is:

\begin{array}{c|ccc} C_{3v} & E & 2 C_3 & 3 \sigma_v \\ \hline A_1 & 1 & 1 & 1 \\ A_2 & 1 & 1 & -1 \\ E & 2 & -1 & 0 \\ \end{array}

Code:

from sympy import var, solve, Matrix
var("a, b, c, d")
g = [1, 2, 3]
M = Matrix([
    [1, 1, 1],
    [1, a, b],
    [2, c, d]])

def rows(mu, nu, M, g):
    eq = 0
    for i in range(len(g)):
        eq += g[i] * M[mu, i] * M[nu, i]
    if mu == nu:
        eq -= sum(g)
    return eq

def cols(i, j, M, g):
    eq = 0
    for nu in range(len(g)):
        eq += M[nu, i] * M[nu, j]
    if i == j:
        eq -= sum(g) / g[i]
    return eq

print "Character table:"
print M
print "Rows conditions:"
eqs = []
for mu in range(3):
    for nu in range(mu, 3):
        eq = rows(mu, nu, M, g)
        eqs.append(eq)
        print mu+1, nu+1, ":    ", eq
print "-"*40
print "Columns conditions:"
eqs2 = []
for i in range(3):
    for j in range(i, 3):
        eq = cols(i, j, M, g)
        eqs2.append(eq)
        print i+1, j+1, ":    ", eq
print "-"*40
print "Solving the 1, 2, 4, 5 equations out of 0..5 from the rows conditions"
s = solve(eqs[1:3]+eqs[4:], [a, b, c, d])
print s
print "Test that all the solutions satisfy the rest of the equations:"
for a, b, c, d in s:
    print
    print "Solution:", a, b, c, d
    r = eqs[3].subs({
        "a": a,
        "b": b,
        "c": c,
        "d": d,
        })
    print "Equation 3 from rows conditions, result: ", r
    assert r == 0
    print "Columns conditions:"
    for i, eq in enumerate(eqs2):
        r = eq.subs({
                "a": a,
                "b": b,
                "c": c,
                "d": d,
                })
        print "Equation %i from columns conditions, result: %r" % (i, r)
        assert r == 0

This prints:

Character table:
[1, 1, 1]
[1, a, b]
[2, c, d]
Rows conditions:
1 1 :     0
1 2 :     2*a + 3*b + 1
1 3 :     2*c + 3*d + 2
2 2 :     2*a**2 + 3*b**2 - 5
2 3 :     2*a*c + 3*b*d + 2
3 3 :     2*c**2 + 3*d**2 - 2
----------------------------------------
Columns conditions:
1 1 :     0
1 2 :     a + 2*c + 1
1 3 :     b + 2*d + 1
2 2 :     a**2 + c**2 - 2
2 3 :     a*b + c*d + 1
3 3 :     b**2 + d**2 - 1
----------------------------------------
Solving the 1, 2, 4, 5 equations out of 0..5 from the rows conditions
[(-7/5, 3/5, 1/5, -4/5), (1, -1, -1, 0)]
Test that all the solutions satisfy the rest of the equations:

Solution: -7/5 3/5 1/5 -4/5
Equation 3 from rows conditions, result:  0
Columns conditions:
Equation 0 from columns conditions, result: 0
Equation 1 from columns conditions, result: 0
Equation 2 from columns conditions, result: 0
Equation 3 from columns conditions, result: 0
Equation 4 from columns conditions, result: 0
Equation 5 from columns conditions, result: 0

Solution: 1 -1 -1 0
Equation 3 from rows conditions, result:  0
Columns conditions:
Equation 0 from columns conditions, result: 0
Equation 1 from columns conditions, result: 0
Equation 2 from columns conditions, result: 0
Equation 3 from columns conditions, result: 0
Equation 4 from columns conditions, result: 0
Equation 5 from columns conditions, result: 0

Example II

We derive the character table for C_{3v} again, using another approach. First we determine the element orders, that must divide the size of the group (possible values are 1, 2, 3, 6). Element order of the class E is 1, because E^2=1. The element order of C_3 is 3, because C_3^3 = 1. Finally, the element order of \sigma_v is 2, because \sigma_v^2=1.

\begin{array}{c|ccc} C_{3v} & E & 2 C_3 & 3 \sigma_v \\ \mbox{class sizes} & 1 & 2 & 3 \\ \mbox{element orders} & 1 & 3 & 2 \\ \hline A_1 & 1 & 1 & 1 \\ A_2 & 1 & a & b \\ E & 2 & c & d \\ \end{array}

Rule 7: The characters of the representation A_2 must be real, because otherwise A_1 would have to be a complex conjugate. E is the only representation of dimension 2, so it must be real as well.

Rule 5: A_2 is Abelian, with element orders 1, 3 and 2. As such, we must have:

a = e^{2\pi i k\over 3} \\ b = e^{2\pi i l\over 2} \\

Where k and l are unknown integers. However, since both a and b is real, the only solution is k=0 (corresponding to a=1) and l=0, 1 (corresponding to b=\pm 1).

Rule 3 gives:

1 + 2a + 3b = 0

And plugging in a=1 this implies b=-1, consistent with the previous paragraph.

Rule 8: multiplying A_2 by E must give characters of dimension 2, which is E, so we get:

+1 \cdot c &= c \\ -1 \cdot d &= d \\

From which d=0. Rule 3 gives:

2 + 2c + 3d = 0

Where we use d=0 and we get c=-1. The final character table is:

\begin{array}{c|ccc} C_{3v} & E & 2 C_3 & 3 \sigma_v \\ \hline A_1 & 1 & 1 & 1 \\ A_2 & 1 & 1 & -1 \\ E & 2 & -1 & 0 \\ \end{array}

Example III

We derive the character table for C_2.

\begin{array}{c|ccc} C_2 & E & C_2 \\ \mbox{class sizes} & 1 & 1 \\ \mbox{element orders} & 1 & 2 \\ \hline A_1 & 1 & 1 \\ A_2 & 1 & a \\ \end{array}

We have two classes, group order is 2, so we must have two representations of dimension 1. Using the rule 3. we get:

1 + a = 0

so a=-1 and the final character table is:

\begin{array}{c|ccc} C_2 & E & C_2 \\ \hline A_1 & 1 & 1 \\ A_2 & 1 & -1 \\ \end{array}

Example IV

We derive the character table for C_3.

\begin{array}{c|ccc} C_3 & E & C_3 & C_3^2 \\ \mbox{class sizes} & 1 & 1 & 1 \\ \mbox{element orders} & 1 & 3 & 3 \\ \hline \end{array}

We have 3 classes and representations, group order is 3, so they must be one dimensional:

\begin{array}{c|ccc} C_3 & E & C_3 & C_3^2 \\ \mbox{class sizes} & 1 & 1 & 1 \\ \mbox{element orders} & 1 & 3 & 3 \\ \hline A_1 & 1 & 1 & 1 \\ A_2 & 1 & a & b \\ A_3 & 1 & c & d \\ \end{array}

Rule 3 says:

1 + a + b = 0

Rule 5 says:

a &= \omega^k \\ b &= \omega^l \\

where \omega = e^{2\pi i \over 3}, so:

1 + \omega^k + \omega^l = 0

Which has only two solutions: k=1, l=2 and k=2, l=1. If we choose the first solution, we get a=\omega and b=\omega^2=\bar\omega. Using the rule 7. it follows that c=\bar a = \bar\omega=\omega^2 and d=\bar b = \omega. If we choose the second solution, we get the pairs a, b and c, d interchanged, however, we can reorder the rows, so these two options are equivalent. The final character table is:

\begin{array}{c|ccc} C_3 & E & C_3 & C_3^2 \\ \hline A_1 & 1 & 1 & 1 \\ A_2 & 1 & \omega & \omega^2 \\ A_3 & 1 & \omega^2 & \omega \\ \end{array} \omega = e^{2\pi i \over 3} = {-1+i\sqrt 3 \over 2}

Example V

Group C_4:

\begin{array}{c|cccc} C_4 & E & C_4 & C_4^2 & C_4^3 \\ \mbox{class sizes} & 1 & 1 & 1 & 1\\ \mbox{element orders} & 1 & 4 & 2 & 4 \\ \hline A_1 & 1 & 1 & 1 & 1 \\ A_2 & 1 & a & b & c \\ A_3 & 1 & & & \\ A_4 & 1 & & & \\ \end{array}

Rule 5 gives:

a &= i^k \\ b &= (-1)^l \\ c &= i^m \\

Rule 3 gives:

(3.36.2.1)1 + a + b + c = 0

Using the rule 7. we know that at least one of A_2, A_3 and A_4 must be real, so let it be A_2. The only real solutions of the equation (3.36.2.1) are a=1, b=-1, c=-1 and permutations. The representation however must be isomorphic to the C_4 group, so in particular a^2 = b, from which b=1 and then a=-1 and c=-1.

The group operations give:

a^2 &= b \\ a b &= c \\ a c &= 1 \\

which gives:

2k &= l \\ k +l &= m \\ k + m &= 0, 4, 8, 12, ... \\

The possible solutions are:

\begin{array}{ccc} k & l & m \\ \hline 2 & 0 & 2 \\ 1 & 2 & 3 \\ 3 & 2 & 1 \\ \end{array}

The first solution is real and it is equal to A_2. The other two solutions are complex conjugate and they must be solutions of A_3 and A_4, because A_3 and A_4 cannot be real (otherwise they would have to be equal to A_2 and the orthogonality relation for columns would not hold). The final character table is:

\begin{array}{c|cccc} C_4 & E & C_4 & C_4^2 & C_4^3 \\ \hline A_1 & 1 & 1 & 1 & 1 \\ A_2 & 1 & -1 & 1 & -1 \\ A_3 & 1 & i & -1 & -i \\ A_4 & 1 & -i & -1 & i \\ \end{array}

Example VI

Group T:

\begin{array}{c|cccc} T & E & 4 C_3 & 4 C_3^2 & 3 C_2 \\ \mbox{class sizes} & 1 & 4 & 4 & 3\\ \mbox{element orders} & 1 & 3 & 3 & 2 \\ \hline A_1 & 1 & 1 & 1 & 1 \\ A_2 & 1 & a & b & c \\ A_3 & 1 & & & \\ T & 3 & d & e & f \\ \end{array}

The group size is 1 + 4 + 4 + 3 = 12, so the only possible option for dimensions of the 4 representations is 1, 1, 1 and 3.

Rule 5 gives:

a &= \omega^k \\ b &= \omega^l \\ c &= (-1)^m \\

where \omega = e^{2\pi i \over 3}. Rule 3 gives:

1 + 4 \omega^k + 4 \omega^l + 3 (-1)^m = 0

The only solution is m=0, k=1 and l=2 (and k with l interchanged). This fully determines A_2 and A_3. The last row is determined from column orthogonality conditions (we compare the given column with the first column):

1 + \omega + \omega^2 + 3 d & = 0 \\ 1 + \omega^2 + \omega + 3 e & = 0 \\ 1 + 1 + 1 + 3 f & = 0 \\

Using the relation 1 + \omega + \omega^2=0 we get d = 0, e=0 and f=-1.

The final character table is:

\begin{array}{c|cccc} T & E & 4 C_3 & 4 C_3^2 & 3 C_2 \\ \hline A_1 & 1 & 1 & 1 & 1 \\ A_2 & 1 & \omega & \omega^2 & 1 \\ A_3 & 1 & \omega^2 & \omega & 1 \\ T & 3 & 0 & 0 & -1 \\ \end{array}

3.36.3. Applications of finite groups

Distinct energy levels (‘vibrations’)

Assume that we know number of atoms in a molecule and measure the number of its distinct vibrational modes (frequencies) in a multiplet. We want to know its symmetry.

We go through the list of all possible (point) symmetries S of such a molecule and look at what all reps. S has. If an n–tuplet was observed among the vibrational modes and there is no n-dimensional IR of S, then can be excluded.

This procedure assumes that (a) the original symmetry S is slightly disturbed because of something and (b) two multiplets (m and n dimensional) do not accidentally happen to have the same frequencies (‘accidental degeneracy’).

Selection rules (‘transitions’)

According to [Pilar], p. 572.

Probability of an optical transition is proportional to

(3.36.3.1)\braket{i| H_1 |f}\,,

where \ket{i}, \ket{f} are the initial and final states and H_1 is the operator of the interaction causing the transition. This is the Fermi golden rule (first order time dep. perturbation theory).

The integral ((3.36.3.1)) may vanish because of the symmetry. A simple 1D example is that \ket{f} is an even function f(x), \ket{i} is an odd function i(x) and H_1 is an even function h_1(x). Then i^*(x)h_1(x)f(x) is odd and thus the integral over (-\infty,\infty) vanishes. The group theory only generalizes this observation.

The procedure is: find the IRs \rho_i, \rho_f to which \ket{i}, \ket{f} belong and also \rho, the regular rep of H_1 in order to catch all IRs of H_1 (is this procedure correct?). Then construct \rho_i\otimes\rho\otimes\rho_f, decompose it into IRs and see if the trivial rep is present. If not, the integral ((3.36.3.1)) vanishes. This procedure is claimed to be equivalent to checking whether \rho_i\otimes \rho_f and \rho contain at least one common IR.

The infrared absorption (IRa) is described by H_1\propto x (or y, z, depending probably on the polarization of light), the Raman scattering has H_1\propto x^2 (it comes from the second order perturbation theory?).

Zoology

Todo:

  • Describe the representations A_1, A_2, B_1, E etc.

  • Reps are specified by the generating functions f(x,y,z) and the symmetry operations T acting on these functions f(x,y,z)\mapsto f(x',y',z') then transform the arguments, (x,y,z)\mapsto (x',y',z')=T(x,y,z). Explain what functions are commonly used (x, R_x,\ldots) and give maybe some examples.

  • Further reading: Davydov, p. 318, 195. Joe Penrose: Symmetry in Science.

3.36.4. Continuous Groups

Lie groups+algebras

A continuous group with metrics is a Lie group (more exactly a differentiable manifold and a\mapsto ag and a\mapsto a^{-1} are differentiable \forall g, p. 172 in [Sternberg]) usually a subgroup of GL(n) is meant, a linear Lie group (i.e. matrices). Peter–Weyl theorem (p. 179 in [Sternberg]) looks like that compact Lie groups are practically as nice as finite groups.

Consider G=O(n), p. 234 in [Sternberg]. If A\in G then \exp(-tA)\subset G where t\in R. At least in O(3) and probably in any O(n), any element of G can be written as \exp(-tA) where A is a \pi/2 rotation around some axis. These A’s are the generators of G.

Typical example: for G=SO(3) there are three generators, iA_x, iA_y, iA_z, where A_x is the rotation by \pi/2 around x–axis in R^3. The generators form a vector space (here the linear span of iA_x, iA_y, iA_z) with an additional operation of commutation. This structure is closed and it is called the Lie algebra of the group G. The commutation relations between the generators fully specify the Lie algebra. E.g. [iA_x,iA_y]=iA_z and the two other ones.

This is a great simplification because a continuous (infinite) group was thus mapped on a vector space, the algebra, where it suffices to look at the basis elements, the generator. The net effect is that we have to watch only three objects instead of infinitely many in the example above.

Todo: weights, roots and Dynkin diagrams. Octets and decuplets. Classification of IRs of SU(n). From [Georgi].

IRs of SU(2)

  1. 181 in [Sternberg]; alternative somewhere in [Georgi].

The Peter-Weyl theorem concerns also the orthogonality of characters and that in turn strongly restricts any possible characters of SU(2). The conjugacy classes of SU(2) are exemplified by matrices U_\theta=\diag(e^{i\theta},e^{-i\theta}) and their possible characters can only be

\chi(\theta)=\sum_{k=-s}^{s} \exp(-i2k\theta)

with 2s integer.

All the corresponding reps exist, they are defined on the space z_1^{2s},z_1^{2s-1}z_2,\ldots, z_2^{2s} by U_{-\theta}z_1^{2s-k}z_2^k\mapsto [\exp i(2s-2k)\theta]z_1^{2s-k}z_2^k.

For an IR of SU(2) the complex conjugate is just the original. For other SU(n) it is not necessarily the case, p. 182 in [Sternberg].

IRs of SO(3) are just those of SU(2) but s must be an integer.

Young diagrams

YD is a systematic method to find all IRs of any symmetric group S_n (permutations of an n-element set). The idea:

  • find all conjugacy classes of S_n

  • assign an IR to each of them

Char’n of the conjugacy classes: each permutation can be decomposed into cycles. This cycle structure (i.e. how many cycles of length 1, how many of length 2, etc. =[\nu_1,\nu_2\,\ldots,\nu_n]) is a unique mark of each conjugation class. The Young diagram is written by rows, each row has \lambda_i empty boxes and \lambda_i-\lambda_{i+i}=\nu_i\ge 0. Each conjugacy class has one YD. An YD of S_n has n boxes.

A Young tabloid (YTd) is obtained by filling an YD with numbers 1,\ldots,n where ordering in each row does not matter. A Young tableau is an YTd where all orderings (thus also in rows) matter.

The IRs of S_n. Take an YD \lambda. On the space of all corresponding YTd’s (M_\lambda) a rep. of the S_n is created. It is decomposed into IRs and shown to have some ‘new’ IR compared to \mu>\lambda.

Details are explained in [Sternberg], p. 76 or in the lecture notes of J. Niederle.

Comments from p. 82 of [Sternberg]: Basis of M_\lambda is defined (e_t; \delta_{\{t\}} means probably a function on M_\lambda which is zero for all \{y\} unless \{y\}=\{t\}). The action of a\in S_n on this basis functions is described.

3.36.5. Literature

Books:

[Birss] (1,2)

Birss, R.R. (1961). Symmetry and Magnetism. Volume III. North-Holland Publishing Company.

[Bishop]

Bishop D.M. (1993). Group theory and chemistry. Courier Dover Publications.

[Elliott] (1,2)

Elliott, J.P. and Dawber, P.G. (1979). Symmetry in Physics. The Macmillan Press Ltd.

[Georgi] (1,2)

Georgi, H. (1990). Lie Algebras in Particle Physics. Addison-Wesley.

[Pilar]

Pilar, F.L. (1990). Elementary Quantum Chemistry, McGraw–Hill.

[Sternberg] (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15)

Sternberg, S. (1994). Group theory and physics. Cambridge University Press

Articles:

[Blokker72]

Blokker, E. International journal of quantum chemistry VI, 925, (1972)

[Cannon69]

Cannon, J.J. Communications of the association for computing machinery, 12, 3 (1969)

[Chillag86]

Chillag, D. (1986). Character Values of Finite Groups as Eigenvalues Of Nonnegative Integer Matrices. Proceedings of the American Mathematical Society, 97(3), 565-567.

[Dixon67]

Dixon, J.D. Numerische Mathematik 10, 446 (1967)