1.3: Irreducible Representations and Character Tables

Similarity transformations yield irreducible representations, Γ_i, which lead to the useful tool in group theory – the character table. The general strategy for determining Γ_i is as follows: A, B and C are matrix representations of symmetry operations of an arbitrary basis set (i.e., elements on which symmetry operations are performed). There is some similarity transform operator such that

\[\begin{array}{l}
\pmb A ^{\prime}=v^{-1} \cdot \pmb A \cdot v \\
\pmb B ^{\prime}=v^{-1} \cdot \pmb B \cdot v \\
\pmb C ^{\prime}=v^{-1} \cdot \pmb C \cdot v
\end{array}\]

where v uniquely produces block-diagonalized matrices, which are matrices possessing square arrays along the diagonal and zeros outside the blocks

\begin{equation}
\mathbf{A}^{\prime}=\left[\begin{array}{rrr}
\mathrm{A}_{1} & & \\
& \mathrm{~A}_{2} & \\
& & \mathrm{~A}_{3}
\end{array}\right] \quad \mathbf{B}^{\prime}=\left[\begin{array}{llll}
\mathrm{B}_{1} & & \\
& \mathrm{~B}_{2} & \\
& & \mathrm{~B}_{3}
\end{array}\right] \quad \mathbf{C}^{\prime}=\left[\begin{array}{lll}
\mathrm{C}_{1} & & \\
& \mathrm{C}_{2} & \\
& & \mathrm{C}_{3}
\end{array}\right]
\end{equation}

Matrices A, B, and C are reducible. Sub-matrices A_i, B_i and C_i obey the same multiplication properties as A, B and C. If application of the similarity transform does not further block-diagonalize A’, B’ and C’, then the blocks are irreducible representations. The character is the sum of the diagonal elements of Γ_i.

As an example, let’s continue with our exemplary group: E, C₃, C₃ ² , σ_v, σ_v’, σ_v” by defining an arbitrary basis … a triangle

The basis set is described by the triangles vertices, points A, B and C. The transformation properties of these points under the symmetry operations of the group are:

\[E\left[\begin{array}{l}
A \\
B \\
C
\end{array}\right]=\left[\begin{array}{l}
A \\
B \\
C
\end{array}\right]=\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\left[\begin{array}{l}
A \\
B \\
C
\end{array}\right] \quad \sigma_{V}\left[\begin{array}{l}
A \\
B \\
C
\end{array}\right]=\left[\begin{array}{l}
A \\
C \\
B
\end{array}\right]=\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0
\end{array}\right]\left[\begin{array}{l}
A \\
B \\
C
\end{array}\right]\]

\[C _{3}\left[\begin{array}{l}
A \\
B \\
C
\end{array}\right]=\left[\begin{array}{l}
B \\
C \\
A
\end{array}\right]=\left[\begin{array}{lll}
0 & 1 & 0 \\
0 & 0 & 1 \\
1 & 0 & 0
\end{array}\right]\left[\begin{array}{l}
A \\
B \\
C
\end{array}\right] \quad \sigma_{ V }^{\prime}\left[\begin{array}{l}
A \\
B \\
C
\end{array}\right]=\left[\begin{array}{l}
B \\
A \\
C
\end{array}\right]=\left[\begin{array}{lll}
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 1
\end{array}\right]\left[\begin{array}{l}
A \\
B \\
C
\end{array}\right]\]

\[C _{3}^{2}\left[\begin{array}{l}
A \\
B \\
C
\end{array}\right]=\left[\begin{array}{l}
C \\
A \\
B
\end{array}\right]=\left[\begin{array}{lll}
0 & 0 & 1 \\
1 & 0 & 0 \\
0 & 1 & 0
\end{array}\right]\left[\begin{array}{l}
A \\
B \\
C
\end{array}\right] \quad \sigma_{ V }^{\prime \prime}\left[\begin{array}{l}
A \\
B \\
C
\end{array}\right]=\left[\begin{array}{l}
C \\
B \\
A
\end{array}\right]=\left[\begin{array}{lll}
0 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 0
\end{array}\right]\left[\begin{array}{l}
A \\
B \\
C
\end{array}\right]\]

These matrices are not block-diagonalized, however a suitable similarity transformation will accomplish the task,

\(v=\left[\begin{array}{ccc}
\frac{1}{\sqrt{3}} & \frac{2}{\sqrt{6}} & 0 \\
\frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{6}} & \frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{6}} & -\frac{1}{\sqrt{2}}
\end{array}\right] \quad ; \quad v^{-1}=\left[\begin{array}{ccc}
\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \\
\frac{2}{\sqrt{6}} & -\frac{1}{\sqrt{6}} & -\frac{1}{\sqrt{6}} \\
0 & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}
\end{array}\right]\)

Applying the similarity transformation with C₃ as the example,

\(v^{-1} \cdot \pmb C _{3} \cdot v=\left[\begin{array}{ccc}
\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \\
\frac{2}{\sqrt{6}} & -\frac{1}{\sqrt{6}} & -\frac{1}{\sqrt{6}} \\
0 & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}
\end{array}\right] \cdot\left[\begin{array}{ccc}
0 & 1 & 0 \\
0 & 0 & 1 \\
1 & 0 & 0
\end{array}\right] \cdot\left[\begin{array}{ccc}
\frac{1}{\sqrt{3}} & \frac{2}{\sqrt{6}} & 0 \\
\frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{6}} & \frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{6}} & -\frac{1}{\sqrt{2}}
\end{array}\right]\)

\(\left[\begin{array}{ccc}
\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \\
\frac{2}{\sqrt{6}} & -\frac{1}{\sqrt{6}} & -\frac{1}{\sqrt{6}} \\
0 & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}
\end{array}\right] \cdot\left[\begin{array}{ccc}
\frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{6}} & \frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{6}} & -\frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{3}} & \frac{2}{\sqrt{6}} & 0
\end{array}\right]=\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & -\frac{1}{2} & \frac{\sqrt{3}}{2} \\
0 & -\frac{\sqrt{3}}{2} & -\frac{1}{2}
\end{array}\right]= \pmb C _{3}^{*}\)

if v ^-1 ⋅ C₃* ⋅ v is applied again, the matrix is not block diagonalized any further. The same diagonal sum is obtained *though off-diagonal elements may change). In this case, C₃* is an irreducible representation, Γ_i.

The similarity transformation applied to other reducible representations yields:

Thus a 3 × 3 reducible representation, Γ_red, has been decomposed under a similarity transformation into a 1 (1 × 1) and 1 (2 × 2) block-diagonalized irreducible representations, Γi. The traces (i.e. sum of diagonal matrix elements) of the Γ_i’s under each operation yield the characters (indicated by χ) of the representation. Taking the traces of each of the blocks:

This collection of characters for a given irreducible representation, under the operations of a group is called a character table. As this example shows, from a completely arbitrary basis and a similarity transform, a character table is born.

The triangular basis set does not uncover all Γ_irr of the group defined by {E, C₃, C₃ ² , σ_v, σ_v’, σ_v’’}. A triangle represents Cartesian coordinate space (x,y,z) for which the Γ_is were determined. May choose other basis functions in an attempt to uncover other Γ_is. For instance, consider a rotation about the z-axis,

The transformation properties of this basis function, R_z, under the operations of the group (will choose only 1 operation from each class, since characters of operators in a class are identical):

E: \(R_{z} \rightarrow R_{z}\
\quad C _{3}: R _{2} \rightarrow R _{2} \quad \sigma_{ v }( xy ): R _{2} \rightarrow \overline{ R }_{2}\)

Note, these transformation properties give rise to a Γ_i that is not contained in a triangular basis. A new (1 x 1) basis is obtained, Γ₃, which describes the transform properties for R_z. A summary of the Γ_i for the group defined by E, C₃, C₃ ² , σ_v, σ_v’, σ_v” is:

Is this character table complete? Irreducible representations and their characters obey certain algebraic relationships. From these 5 rules, we can ascertain whether this is a complete character table for these 6 symmetry operations.

Five important rules govern irreducible representations and their characters:

Rule 1

The sum of the squares of the dimensions, \(\ell\), of irreducible representation Γ_i is equal to the order, h, of the group,

Since the character under the identity operation is equal to the dimension of Γ_i (since E is always the unit matrix), the rule can be reformulated as,

Rule 2

The sum of squares of the characters of irreducible representation Γ_i equals h

Rule 3

Vectors whose components are characters of two different irreducible representations are orthogonal

\(\sum_{R}\left[x_{i}(R)\right]\left[x_{j}(R)\right]=0 \quad\) for \(\quad i \neq j\)

Rule 4

For a given representation, characters of all matrices belonging to operations in the same class are identical

Rule 5

The number of Γ_is of a group is equal to the number of classes in a group.

With these rules one can algebraically construct a character table. Returning to our example, let’s construct the character table in the absence of an arbitrary basis:

Rule 5: E (C₃, C₃ ² ) (σ_v, σ_v’, σ_v”) … 3 classes ∴ 3 Γ_is

Rule 1: \(\ell_{1}^{2}+\ell_{2}^{2}+\ell_{3}^{2}=6 \quad \therefore \ell_{1}=\ell_{2}=1, \ell_{2}=2\)

Rule 2: All character tables have a totally symmetric representation. Thus one of the irreducible representations, Γ_i, possesses the character set χ₁(E) = 1, χ₁(C₃, C₃ ² ) = 1, χ₁(σ_v, σ_v’, σ_v”) = 1. Applying Rule 2, we find for the other irreducible representation of dimension 1,

\(1 \cdot 1 \cdot x_{2}( E )+2 \cdot 1 \cdot x_{2}\left( C _{3}\right)+3 \cdot 1 \cdot x_{2}\left(\sigma_{ v }\right)=0\)

Since χ₂(E) = 1,

\(1+2 \cdot x_{2}\left( C _{3}\right)+3 \cdot x_{2}\left(\sigma_{ v }\right)=0 \quad \therefore \quad \chi_{2}\left( C _{3}\right)=1, \chi_{2}\left(\sigma_{ v }\right)=-1\)

For the case of Γ₃ ( \(\ell\)₃ = 2) there is not a unique solution to Rule 2

\(2+2 \cdot \chi_{3} \left(C_{3}\right) +3 \cdot \chi_{3} \left(\sigma_{v}\right)=0\)

However, application of Rule 2 to Γ₃ gives us one equation for two unknowns. Have several options to obtain a second independent equation:

Rule 1: \(1 \cdot 2^{2}+2\left[\chi_{3}\left(C_{3}\right)\right]^{2}+3\left[\chi_{3}\left(\sigma_{v}\right)\right]^{2}=6\)

Rule 3: \(1 \cdot 1 \cdot 2+2 \cdot 1 \cdot x_{3}\left(C_{3}\right)+3 \cdot 1 \cdot x_{3}\left(\sigma_{v}\right)=0\)
or
\(1 \cdot 1 \cdot 2+2 \cdot 1 \cdot x_{3}\left(C_{3}\right)+3 \cdot(-1) \cdot x_{3}\left(\sigma_{v}\right)=0\)

Solving simultaneously yields \(\chi_{3}\left(C_{3}\right)=-1, \chi_{3}\left(\sigma_{x}\right)=0\)

Thus the same result shown on pg 4 is obtained:

\begin{array}{c|ccc}
& \mathrm{E} & 2 \mathrm{C}_{3} & 3 \sigma_{\mathrm{v}} \\
\hline \Gamma_{1} & 1 & 1 & 1 \\
\Gamma_{2} & 2 & -1 & 0 \\
\Gamma_{3} & 1 & 1 & -1
\end{array}

Note, the derivation of the character table in this section is based solely on the properties of characters; the table was derived algebraically. The derivation on pg 4 was accomplished from first principles.

The complete character table is:

• Γ_is of:

\(\ell=1 \Longrightarrow A\) or \(B\) \(\ell=2 \Longrightarrow E\) \(\ell=3 \Longrightarrow T\)

A is symmetric (+1) with respect to Cn B is antisymmetric (–1) with respect to Cn

• subscripts 1 and 2 designate Γ_is that are symmetric and antisymmetric, respectively to ⊥C₂s; if ⊥C₂s do not exist, then with respect to σ_v
• primes ( ’ ) and double primes ( ” ) attached to Γ_is that are symmetric and antisymmetric, respectively, to σ_h
• for groups containing i, g subscript attached to Γis that are symmetric to i whereas u subscript designates Γis that are antisymmetic to i