Character Tables from Orthogonality Relations

1. The Complete Toolkit for Character Table Construction

The "row orthogonality" relation we studied previously is just one of several powerful rules that govern the structure of character tables. When combined, these rules are remarkably restrictive, often uniquely determining the entire table.

First, let's watch a video that derives the remaining essential theorems. Given your mathematical background, you should be able to follow the arguments, which are based on analyzing the group algebra and the regular representation.

Let's consolidate these insights into a definitive list of rules for constructing a character table. For a finite group $G$ with order $N_G$:

1. The number of irreducible representations (irreps) is equal to the number of conjugacy classes. This means the character table is always a square matrix.

2. The sum of the squares of the dimensions of the irreps is equal to the order of the group. Let $d_i={\chi }_i(E)$ be the dimension of the $i$-th irrep. Then:

   $$\sum _i{d}_i^2=N_G$$

   Since the dimensions $d_i$ must be positive integers, this Diophantine equation often has a unique solution, allowing us to determine the dimensions of all irreps.

3. The first row is always the trivial representation, $A_1$ or ${\Gamma }_1$, whose characters are all 1. This corresponds to a 1D representation where every group element is mapped to the number 1.

4. Row Orthogonality: The rows of the character table are orthogonal.

   $$\sum _C{N}_C{\chi }_i(C{)}^{\ast }{\chi }_j(C)=N_G{\delta }_{ij}$$

   Here, the sum is over the conjugacy classes $C$, $N_C$ is the number of elements in class $C$, and ${\chi }_i(C)$ is the character of the $i$-th irrep for that class. For $i=j$, this gives the "normalization" condition: ${\sum }_C{N}_C|{\chi }_i(C){|}^2=N_G$.

5. Column Orthogonality: The columns of the character table are also orthogonal.

   $$\sum _i{\chi }_i(C_k{)}^{\ast }{\chi }_i(C_l)=\frac{N_G}{N_{C_k}}{\delta }_{kl}$$

   Here, the sum is over the irreps $i$, and $C_k$ and $C_l$ are two different classes. This relation is especially useful for finding unknown character values down a column.

These five rules form a complete logical framework for our construction.

2. A Step-by-Step Construction: The $C_{3v}$ Group

Let's apply these rules to build the character table for the group $C_{3v}$, which is isomorphic to $S_3$. You are already familiar with the result from previous lessons; now we will derive it from scratch.

The following video provides an excellent visual walkthrough of this exact process. We will follow its steps, formalizing the logic as we go.

Step 1: Identify Classes and Group Order

The group $C_{3v}$ has 6 elements, so $N_G=6$. As we found in Module 1, these elements are partitioned into three conjugacy classes:

• $C_1=\{E\}$, so $N_{C_1}=1$.
• $C_2=\{C_3,C_3^2\}$, so $N_{C_2}=2$.
• $C_3=\{{\sigma }_v,{\sigma }_v^{\prime },{\sigma }_v^{\prime \prime }\}$, so $N_{C_3}=3$.

Step 2: Determine Number and Dimensions of Irreps

• Rule 1: There are 3 classes, so there must be 3 irreps.
• Rule 2: The dimensions $d_1,d_2,d_3$ must satisfy $d_1^2+d_2^2+d_3^2=6$.
• Rule 3: One of these irreps is the 1D trivial representation, so let $d_1=1$. This leaves $1^2+d_2^2+d_3^2=6$, or $d_2^2+d_3^2=5$. The only solution in positive integers is $d_2=1$ and $d_3=2$ (or vice-versa).

So, $C_{3v}$ has two 1-dimensional irreps ($A_1$ and $A_2$) and one 2-dimensional irrep ($E$).

Step 3: Set up the Initial Table

We can now fill in the first column with the dimensions and the first row with the characters of the trivial irrep $A_1$. Let the unknown characters be variables.

(The number of elements in each class, $N_C$, is shown in parentheses for convenience).

Step 4: Apply Orthogonality Relations

We now use the orthogonality rules to solve for $a,b,c,d$.

• Row Orthogonality for $A_1$ and $A_2$:
  ${\sum }_C{N}_C{\chi }_{A_1}(C{)}^{\ast }{\chi }_{A_2}(C)=(1)(1)(1)+(2)(1)(a)+(3)(1)(b)=0$
  This gives the equation: $1+2a+3b=0$.
  For a 1D representation, the characters are roots of unity. The orders of the elements in the classes $2C_3$ and $3{\sigma }_v$ are 3 and 2, respectively. Thus, $a^3=1$ and $b^2=1$, so $b=\pm 1$.

  • If $b=1$, then $1+2a+3=0⟹a=-2$. This is not a root of unity, so this is impossible for a 1D representation.
  • If $b=-1$, then $1+2a-3=0⟹a=1$. This is a valid root of unity.
    So, the second row must be $(1,1,-1)$.

• Column Orthogonality (or more Row Orthogonality):
  We can find $c$ and $d$ by making the third row ($E$) orthogonal to the first two.

  • Orthogonality with $A_1$: $(1)(1)(2)+(2)(1)(c)+(3)(1)(d)=0⟹2+2c+3d=0$.
  • Orthogonality with $A_2$: $(1)(1)(2)+(2)(1)(c)+(3)(-1)(d)=0⟹2+2c-3d=0$.
    This is a simple system of two linear equations. Adding them gives $4+4c=0⟹c=-1$. Substituting this back into the first equation gives $2-2+3d=0⟹d=0$.

Step 5: The Final Character Table

We have now determined all entries:

You can verify that this table satisfies all orthogonality relations, including those for the columns. For example, check the orthogonality of the second and third columns:
${\sum }_i{\chi }_i(C_3{)}^{\ast }{\chi }_i({\sigma }_v)=(1{)}^{\ast }(1)+(1{)}^{\ast }(-1)+(-1{)}^{\ast }(0)=1-1+0=0$
This confirms our construction is consistent.

For a more formal, text-based walkthrough of this same process, you can consult the following resource.

3. An Example with Complex Characters: The Group $C_3$

The characters for $C_{3v}$ all turned out to be real numbers. This is common, but not guaranteed. Let's consider the cyclic group $C_3=\{E,R,R^2\}$, where $R$ is a rotation by $120^{\circ }$.

• Classes: $C_3$ is an abelian group, so every element is in its own class. There are 3 classes: $\{E\},\{R\},\{R^2\}$.
• Irreps: By Rule 1, there must be 3 irreps.
• Dimensions: By Rule 2, $d_1^2+d_2^2+d_3^2=3$. The only integer solution is $d_1=d_2=d_3=1$. All irreps are one-dimensional.
• Characters as Roots of Unity: For any 1D representation $\Gamma$, the character $\chi (g)$ is simply the 1x1 matrix (a complex number) representing the group element $g$. Since $R^3=E$, it must be that $\chi (R{)}^3=\chi (E)=1$. Therefore, the character $\chi (R)$ must be a cube root of unity. The three cube roots of unity are $1,e^{2\pi i/3},e^{4\pi i/3}$. Let's call $\omega =e^{2\pi i/3}$.
• Construction:
  • The first irrep is the trivial one: ${\chi }_1(E)=1,{\chi }_1(R)=1,{\chi }_1(R^2)=1$.
  • For the second irrep, let's choose ${\chi }_2(R)=\omega$. Then ${\chi }_2(R^2)=({\chi }_2(R){)}^2={\omega }^2$.
  • For the third irrep, we must choose the remaining root of unity: ${\chi }_3(R)={\omega }^2$. This implies ${\chi }_3(R^2)=({\chi }_3(R){)}^2={\omega }^4=\omega$.

This gives us the complete character table for $C_3$:

Notice that the characters for ${\Gamma }_2$ and ${\Gamma }_3$ are complex conjugates of each other. This is a general feature: if a representation has complex characters, its complex conjugate representation will also be an irrep of the group. You can find this table and others, like for the quaternion group $Q$ and dihedral group $D_4$, in the resource LINK from the previous lesson (section 2.4.6, Table 2.2).

Conclusion

In this lesson, we have seen how a few fundamental theorems of representation theory provide a powerful and systematic framework for constructing character tables.

Key Takeaways:

• A character table is highly constrained by a set of rules: the number of irreps equals the number of classes, the sum of squared dimensions equals the group order, and the rows and columns must be orthogonal.
• The construction process is an algorithmic procedure: determine dimensions, fill in the trivial representation, and use orthogonality relations to solve for the remaining unknown characters.
• This procedure works for any finite group, revealing the complete structure of its irreducible representations—the basic building blocks of any physical theory based on that symmetry.

We now have the tools to both construct character tables and use them to decompose representations. In our next lesson, we will learn how to handle a common situation in physics: a system composed of multiple parts. We will see how to find the representation for the composite system by taking the tensor product of the individual representations and how to use characters to decompose this product into its irreducible parts. This is essential for understanding interactions, selection rules, and multi-particle quantum states.