Characters and Character Tables

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This note introduces characters — the traces of representation matrices — which dramatically simplify representation theory. Because the trace is invariant under similarity transforms, characters capture the essential information of a representation without depending on the choice of basis. We'll build character tables, derive orthogonality relations for characters, and show how to use them to decompose any reducible representation into irreps.

1) Recap

From the previous note on representations:

Schur's Lemma: If a matrix M commutes with all matrices of a representation (M A_x = A_x M for all x), then either M = c \, \mathbb{I} (and the representation is an irrep) or the representation is reducible. Extending to two irreps: if M \, D^{(1)}(g) = D^{(2)}(g) \, M for all g, then either M = 0 (if the irreps are inequivalent) or D^{(1)} and D^{(2)} are the same irrep up to a similarity transform.

Wonderful Orthogonality Theorem (WOT): For unitary irreps,

All irreps are orthogonal! The dimensionalities \ell_j of irreps are constrained by \sum_j \ell_j^2 \leq h.

2) Character of a Representation

Irreps are defined only up to a similarity transform — any invertible matrix U gives an equally valid representation U \, D(g) \, U^{-1}. We need a quantity that is invariant under such basis changes.

2.1) The Trace Is Similarity-Invariant

The trace has a key property: \text{tr}(AB) = \text{tr}(BA). To see why:

where the last step just relabels the dummy indices. This cyclic property of the trace immediately gives us similarity invariance. If M and N are related by a similarity transform M = Q^{-1} N Q, then:

The trace is the same regardless of what basis we use!

2.2) Definition of Character

Since the trace is invariant under similarity transforms, the character does not depend on the choice of basis — it is a property of the irrep itself.

3) Character and Class

3.1) Characters Are Constant on Conjugacy Classes

Theorem: The character for each element in a conjugacy class is the same.

Proof: Recall that two elements A and B are in the same conjugacy class if A = Y^{-1} B \, Y for some group element Y. But this is exactly a similarity transform! So:

This means we only need to track characters per class, not per element — a huge simplification!

The interactive graphic below shows the irrep table for P(3). Click Take the Trace to see how the character \chi(g) = \text{tr}\, D(g) is computed from the diagonal, and how the result is constant within each conjugacy class.

3.2) The Character Table

A character table collects the characters of all irreps for each conjugacy class. Recall the irreps of P(3):

• \Gamma_1 (A_1): the trivial representation, all matrices are (1).
• \Gamma_1' (A_2): mirrors get (-1), rotations get (+1).
• \Gamma_2 (E): the 2D representation from Dresselhaus Table 1.1.

Taking the trace of each representation matrix for each class gives the character table:

You can build this table computationally in the companion notebook:

# Build the character table from irreps and conjugacy classes
char_table = rep.character_table(irreps, conj_classes)
# Rows = irreps, Columns = conjugacy classes

4) Wonderful Orthogonality Theorem for Characters

We can derive a simpler orthogonality relation that uses only characters instead of individual matrix elements.

4.1) Statement

Theorem (Dresselhaus 3.3): For irreps \Gamma_j and \Gamma_{j'},

For unitary representations (where D(R^{-1}) = D(R)^\dagger):

Since the character is constant on each class, we can also write this as a sum over classes:

where N_k = |C_k| is the number of elements in class C_k.

4.2) Proof

Start from the Wonderful Orthogonality Theorem and take diagonal elements. Setting \nu' \to \mu' and \nu \to \mu:

Now sum over \mu and \mu' to produce traces:

The \ell_j on the right cancels, giving:

For unitary representations, D(R^{-1}) = D(R)^\dagger, so \chi(R^{-1}) = \text{tr}(D(R)^\dagger) = [\chi(R)]^*. Substituting:

Finally, since the character is constant on each conjugacy class, all N_k elements in class C_k contribute the same term. Replacing the sum over group elements with a sum over classes:

4.3) Example: \Gamma_1 and \Gamma_1' of P(3)

Using the character table and summing over classes (with N_k elements per class):

As expected — different irreps give zero.

Use the interactive graphic below to explore this for all pairs of P(3) irreps:

5) What Characters Tell Us

Characters are remarkably powerful. Here are two key things they can reveal:

1. Is a representation reducible or irreducible? The orthogonality relation holds only for irreps. By comparing the characters of a representation against the character table, we can determine whether it is reducible.

2. Have we found all the irreps? Do the characters of our known irreps span the full space? (We will make this precise below.)

6) Decomposing Reducible Representations

6.1) The Decomposition Is Unique

Theorem (Dresselhaus 3.4): The reduction of a representation into irreps is unique.

Since any representation can be block-diagonalized into irreps, its character is a linear combination of irrep characters:

where a_i is the number of times irrep \Gamma_i appears in the decomposition.

6.2) The Decomposition Formula

To find the coefficients a_j, consider:

This is effectively a projection of \chi onto the irrep \Gamma_j. Substituting \chi(C_k) = \sum_{\Gamma_i} a_i \, \chi^{(\Gamma_i)}(C_k):

6.3) Example: 3D Permutation Representation of P(3)

Consider the 3D permutation matrices for P(3), where each group element permutes the coordinates (x_1, x_2, x_3). Example matrices for each class:

Taking traces gives the character for this representation. Comparing against the irrep characters:

Now apply the decomposition formula (h = 6), where each term is {\color{#b12a90}N_k} \left[{\color{...}\chi^{(\Gamma_j)}(C_k)}\right]^* {\color{#333}\chi^{\text{3D perm}}(C_k)}:

Therefore:

The 3D permutation representation decomposes into the trivial representation plus the 2D irrep, with dimensions 1 + 2 = 3, as expected.

Verify this computationally in the companion notebook:

# Check: the 3D permutation rep is reducible
perm_rep = groups.permutation_matrices(3)
rep.is_an_irrep(table, perm_rep)  # → False

# Decompose into irreps
sub_irreps = rep.decompose_rep_into_irreps(perm_rep)
# → [Γ_1 (dim 1), Γ_2 (dim 2)]

7) The Number of Irreducible Representations

Theorem (Dresselhaus 3.5): The number of irreps equals the number of conjugacy classes.

7.1) Proof

Rewrite the WOT for characters with normalized vectors. Define:

Then the orthogonality theorem becomes:

Each v^{(\Gamma_j)}_k gives the k-th component of a K-dimensional vector (where K = number of classes).

• There can be at most K such orthonormal vectors, since the (K+1)-th would be linearly dependent on the others.
• There must be at least K such vectors: if there were fewer, the irreps would not span the K-dimensional space, and we would not be able to express every reducible representation in terms of irreps.

Therefore, the number of irreps = K, the number of conjugacy classes.

Example: P(3) has 3 conjugacy classes (\{E\}, \{A, B, C\}, \{D, F\}) \Rightarrow 3 irreps (\Gamma_1, \Gamma_1', \Gamma_2).

8) Second Orthogonality Relation for Characters

Theorem (Dresselhaus 3.6): Not only are the rows of a character table orthogonal — the columns are too!

The first orthogonality relation (for rows) tells us that different irreps are orthogonal. This second relation tells us that different conjugacy classes are orthogonal. Together, they confirm that the character table is a (rescaled) unitary matrix — it has exactly as many rows as columns, i.e. the number of irreps equals the number of classes.

We won't prove this in class — see Dresselhaus Ch. 3.6 for the proof.

9) The Regular Representation Revisited

We introduced the regular representation previously. Now we can use character theory to prove a key property.

9.1) Character of the Regular Representation

Recall that D^{\text{reg}}(X) is an h \times h matrix with a 1 wherever element X appears in the (rearranged) multiplication table. By construction:

• D^{\text{reg}}(E) = \mathbb{I}_h, so \chi^{\text{reg}}(E) = h.
• For g \neq E: element g never appears on the diagonal of the multiplication table (by the rearrangement theorem), so \chi^{\text{reg}}(g) = 0.

9.2) Each Irrep Appears \ell_i Times

Theorem: The regular representation contains each irrep \Gamma_i a number of times equal to its dimensionality \ell_i.

Proof: Since D^{\text{reg}} is reducible, we can write \chi^{\text{reg}}(C_k) = \sum_{\Gamma_i} a_i \, \chi^{(\Gamma_i)}(C_k). By the decomposition formula:

But \chi^{\text{reg}}(C_k) = 0 for all classes except the identity class, and \chi^{\text{reg}}(E) = h. Since N_E = 1:

Therefore there are \ell_i copies of \Gamma_i in \Gamma^{\text{reg}}.

9.3) The Dimension Formula: \sum_j \ell_j^2 = h

Theorem: The order of a group h and the dimensionalities \ell_j of its irreps \Gamma_j are related by \sum_j \ell_j^2 = h.

Proof: The regular representation has dimension h (by construction). But each irrep \Gamma_j is contained \ell_j times, and each copy contributes \ell_j dimensions:

Additional resources: Character Tables for Point Groups used in Chemistry — an encyclopedic collection of character tables for point groups, by Gernot Katzer.

Related exercises: Representations Exercises