# 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:

I2: there exist such that for each

I3: there exist such that for each

I4: for each

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

### Representation¶

Set of linear operators (for each there is one )

fulfills all the group axioms I1, I2, I3, I4. The requirement is non-trivial, consider for example the following 4 matrices

that fulfill but not .

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

Equivalent representations and : there exist such that for each .

Reducible representation : there exist an equivalent representation that is diagonal:

(3.36.1.1)¶

We say that is a direct sum of and : .

Irreducible representation: is not reducible.

Conjugate element: is conjugate to () if there exist such that:

if and then .

Conjugate class: elements which are all conjugate to each other

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

Character of the representation : set of numbers as the group element runs through the group.

Equivalent representations have the same character:

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 be a reducible representation, then:

where , , dots are all the inequivalent irreducible representations and () gives the number of times that the irreducible representation occurs in the reduction.

Similar relation holds for group characters:

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

where is the number of elements in the class , is the number of elements in (the order of the group).

#### Example¶

Consider the permutation group. The character table is:

From the table we read , , and . There are 3 classes and 3 irreducible representations.

- Case I
We are given a representation given by the following matrices:

These 6 matrices belong to the following three classes , , and the corresponding characters for each class are:

and we get:

So this representation is irreducible and it is equivalent to .

- Case II
We are given a representation given by the following matrices:

These 6 matrices belong to the following three classes , , and the corresponding characters for each class are:

and we get:

So this representation is reducible and it is equivalent to . The matrices are equivalent to:

### Other facts¶

Number of irreducible representations the number of classes.

- Regular representation of :
Take with and assign a canonical basis to the elements of . A matrix assigned to now describes the mapping , i.e. in if , then the fifth element of the first row is one and others of that row are zero in . Each IR of the reg. rep. occurs in its decomposition with the multiplicity equal to its dimension. Thus (p. 65, [Sternberg])

- Element Order
The order of an element is the least integer for which . The order can be determined from the group multiplication table for example. Theorem: 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: .

- Schur’s lemma
(a) Be an IR of . If , , then .

(b) Be , two inequivalent IRs of . Then valid implies . 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, , it lives on the direct sum of the two vector spaces of and .

Take an IR of . Then will also be a rep. of any subgroup
, 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
are simultaneously block diagonal only for , not for all . This is called *restriction* and it is denoted by .

*Induced representation*, denoted by , is an opposite of the
restriction. It works as follows: if , then
, when .

*Product of representations*, 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¶

and (see Crystallographic Point Groups) are isomorphic to (p. 35 in [Sternberg]). Written as matrices in 3D, they are 3D representations. Since has only matrices unlike , they are inequivalent.

Homeomorphism of into the Lorentz group [or into ], p. 7 [Sternberg]}. Start with the following correspondence between and :

For any matrix of take . Decode into and the relation between and defines uniquely a Lorentz transformation; thus was mapped into some Lorentz group element. If this gives a mapping from into . The mapping is because and give the same .

is not simply connected. Consider matrices , . These map into rotations by around the –axis. These matrices in form a closed loop, . If were simply connected it would be possible to contract this loop into a point while keeping and unchanged. But then the same would have to happen with the original curve of matrices while keeping and at their place. Since , this curve is not closed and such a contraction is not possible.

All IRs of 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 , which leave a monoatomic crystal lattice invariant. Those can be symmetries of an infinite crystal (e.g. is excluded since pentagons cannot cover the plane).

There are only 7 crystallographic point groups: (triclinic), (monoclinic), (orthorhombic), (rhombohedral), (tetragonal), (hexagonal) and (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]):

There are 37 subgroups together. is a subgroup of (so all 5 subgroups of are also subgroups of ). Together we get 37-5 = 32 distinct subgroups. Groups, which might at first sight appear to be missing from the list are , , , , and , but these are the same as , , , and respectively.

The following groups are isomorphic:

, ,

,

, ,

, ,

,

, ,

, , ,

The way to derive the above lists is the following.

Procedure:

Find all finite crystallographic subgroups of called

*rotation subgroups*Take each subgroup from 1) and add and close the subgroup (‘non-rot containing ’)

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

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

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

### Zoology¶

Schönflies notation: is an –fold rotation (, …) group (planar polygon), is a diedric group, i.e. plus turn-the-page two-fold rotations (e.g. , ), , and (= ) 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). Some atypical notation: , , , .

Hermann–Mauguin (HM, international) notation: 2,3,4 means , means rotation-inversion axis (rotation followed by ), is a vertical mirror plane, is a horizontal mirror plane.

Symmetry operations (in Table 3 of [Birss]): like HM, means a two-fold rotation around –axis, means some other axis in the plane than or (diagonal), is a rotation followed by . means three different two-fold axes .

### Construction and usage of the character table¶

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

The sum of the squares of the dimensions of the irreducible representations is equal to the order of the point group:

The dimension is given by the character of the identity matrix (first column) , so the sum of squares of the first column is . It is customary to put the characters of the one dimensional representation () into the first row, so the first row is filled with 1s. Also, must divide .

The number of irreducible representations (rows) is equal to the number of classes (columns)

The rows must satisfy

The columns must satisfy

Characters of all one-dimensional representation must be roots of unity, equal to , where is the element order, which must divide the group size (and it is the same for the whole conjugacy class). In general, is any integer (for faithful representation it would be ). 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 , then their characters must also obey .

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.

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 , then it must be real (since it is its own conjugate). If there are two representations of dimension 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).

If there is one dimensional representation (with characters ) and any other representation of dimension (with characters ), then there must be a representation of dimension with characters (corresponding to the tensor product ).

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 or according to whether the character of a (proper or improper) rotation about the symmetry axis of highest order is or . If there is no symmetry axis, all one-dimensional representations are labeled .

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

#### Example I¶

Let’s take the group , which has three classes (1 element), (2 elements) and (3 elements).

So , and and the order is . Therefore it has three irreducible representations, whose dimensions must satisfy:

The only integer solution (up to a permutation) is and . So we immediately have:

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

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

and the other is:

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

Both of the above solutions for 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:

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 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 is 1, because . The element order of is 3, because . Finally, the element order of is 2, because .

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

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

Where and are unknown integers. However, since both and is real, the only solution is (corresponding to ) and (corresponding to ).

Rule 3 gives:

And plugging in this implies , consistent with the previous paragraph.

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

From which . Rule 3 gives:

Where we use and we get . The final character table is:

#### Example III¶

We derive the character table for .

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

so and the final character table is:

#### Example IV¶

We derive the character table for .

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

Rule 3 says:

Rule 5 says:

where , so:

Which has only two solutions: , and , . If we choose the first solution, we get and . Using the rule 7. it follows that and . If we choose the second solution, we get the pairs and interchanged, however, we can reorder the rows, so these two options are equivalent. The final character table is:

#### Example V¶

Group :

Rule 5 gives:

Rule 3 gives:

(3.36.2.1)¶

Using the rule 7. we know that at least one of , and must be real, so let it be . The only real solutions of the equation (3.36.2.1) are , , and permutations. The representation however must be isomorphic to the group, so in particular , from which and then and .

The group operations give:

which gives:

The possible solutions are:

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

#### Example VI¶

Group :

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:

where . Rule 3 gives:

The only solution is , and (and with interchanged). This fully determines and . The last row is determined from column orthogonality conditions (we compare the given column with the first column):

Using the relation we get , and .

The final character table is:

## 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 of such a molecule and look at what all reps. has. If an –tuplet was observed among the vibrational modes and there is no -dimensional IR of , then can be excluded.

This procedure assumes that (a) the original symmetry is slightly disturbed because of something and (b) two multiplets ( and 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)¶

where , are the initial and final states and 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 is an even function , is an odd function and is an even function . Then is odd and thus the integral over vanishes. The group theory only generalizes this observation.

The procedure is: find the IRs , to which , belong and also , the regular rep of in order to catch all IRs of (is this procedure correct?). Then construct , 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 and contain at least one common IR.

The infrared absorption (IRa) is described by (or , , depending probably on the polarization of light), the Raman scattering has (it comes from the second order perturbation theory?).

### Zoology¶

Todo:

Describe the representations , , , etc.

Reps are specified by the generating functions and the symmetry operations acting on these functions then transform the arguments, . Explain what functions are commonly used (, ) 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 and are differentiable , p. 172 in [Sternberg]) usually a subgroup of 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 , p. 234 in [Sternberg]. If then where . At least in and probably in any , any element of can be written as where is a rotation around some axis. These ’s are the generators of .

Typical example: for there are three generators, , , , where is the rotation by around –axis in . The generators form a vector space (here the linear span of , , ) with an additional operation of commutation. This structure is closed and it is called the Lie algebra of the group . The commutation relations between the generators fully specify the Lie algebra. E.g. 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 . From [Georgi].

#### IRs of SU(2)¶

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 . The conjugacy classes of are exemplified by matrices and their possible characters can only be

with integer.

All the corresponding reps exist, they are defined on the space by .

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

IRs of are just those of but must be an integer.

### Young diagrams¶

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

find all conjugacy classes of

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. ) is a unique mark of each conjugation class. The Young diagram is written by rows, each row has empty boxes and . Each conjugacy class has one YD. An YD of has boxes.

A Young tabloid (YTd) is obtained by filling an YD with numbers 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 . Take an YD . On the space of all corresponding YTd’s () a rep. of the is created. It is decomposed into IRs and shown to have some ‘new’ IR compared to .

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

Comments from p. 82 of [Sternberg]: Basis of is defined (; means probably a function on which is zero for all unless ). The action of on this basis functions is described.

## 3.36.5. Literature¶

Books:

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

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

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

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

Articles:

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

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

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

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