advanced31 min studyLesson 516 in Mathematics

Character tables and class functions

Assemble orthogonal character rows, check completeness, and distinguish arbitrary class functions from genuine characters.

On this page

What you’ll understand

  • Explain why the number of complex irreducibles equals the number of conjugacy classes.
  • Apply row and column orthogonality with the correct normalizations.
  • Expand a class function and test whether its coefficients describe a representation.

A character table arranges irreducible characters as rows and conjugacy classes as columns. Its compact appearance can hide two different geometries: rows are orthonormal with class-size weights, while the unweighted column inner products detect centralizer sizes. The table is square because irreducible characters form a basis for all complex class functions. That basis can expand arbitrary class functions, but only nonnegative integer coefficients correspond to actual representations. This distinction prevents several tempting but invalid character tests.

Why the table is square

A class function has one freely chosen complex value per conjugacy class, so the dimension of the class-function space equals the number of classes. To count irreducibles, inspect the center of the complex group algebra. A group-algebra element is central precisely when its coefficients are constant on conjugacy classes, giving a basis of class sums. On the other hand, the matrix-block decomposition gives one central scalar per irreducible block. These two computations of the center dimension prove the required equality.

Irreducible characters are already orthonormal, hence independent. Their number now equals the dimension of the class-function space, so they are a basis. A table built from some known irreducibles is complete when its independent rows exhaust this dimension, with the sum-of-squares formula supplying a useful extra check. Dimension data alone do not determine every character value. In particular, conjugate one-dimensional characters of a cyclic group have the same dimension but differ at nonidentity elements.

#G^=#{conjugacy classes},f=∑ρ∈G^⟨f,χρ⟩χρ(f a complex class function)\#\widehat G=\#\{\text{conjugacy classes}\},\qquad f=\sum_{\rho\in\widehat G}\langle f,\chi_\rho\rangle\chi_\rho\quad(f\text{ a complex class function})
This basis statement is for finite groups over the complex numbers. The coefficients of an arbitrary class function are complex numbers, not automatically multiplicities.↔ Scroll wide equations horizontally.

Column orthogonality uses centralizers

Multiply each table column by the square root of its class size divided by the group order. Row orthogonality says the resulting square matrix has orthonormal rows, so it is unitary and its columns are orthonormal too. Undoing those scale factors gives zero inner product between distinct class columns. The squared norm of a column is the group order divided by that class size, which is the size of the element's centralizer by orbit–stabilizer for conjugation.

For the symmetric group on three letters, the identity, transposition, and three-cycle columns have squared norms six, two, and three. The table rows are trivial, sign, and standard, with values respectively one everywhere, the permutation signs, and two/zero/minus one. The transposition column has entries one, minus one, and zero, producing norm two. The three-cycle column has entries one, one, and minus one, producing norm three. These checks catch errors that might escape a dimension calculation at the identity.

∑ρ∈G^χρ(g)χρ(h)‾={∣CG(g)∣,g,h conjugate,0,g,h not conjugate,∣CG(g)∣=∣G∣∣[g]∣\sum_{\rho\in\widehat G}\chi_\rho(g)\overline{\chi_\rho(h)}=\begin{cases}|C_G(g)|,&g,h\text{ conjugate},\\0,&g,h\text{ not conjugate},\end{cases}\qquad |C_G(g)|=\frac{|G|}{|[g]|}
The sum is over irreducible rows, rather than group elements. Conjugate elements give the same column; distinct conjugacy classes give orthogonal columns.↔ Scroll wide equations horizontally.

Not every row-shaped function is a character

The cyclic group of order three has three singleton classes and three one-dimensional characters. If its generator has primitive cube root value omega, the rows are the constant row, the successive powers of omega, and their complex conjugates. The finite geometric identity makes distinct rows orthogonal. Complex conjugation cannot be dropped: the sum of squares of the values of a nontrivial row is zero, whereas its squared Hermitian norm is one. A positive inner product requires the conjugate.

A class function is a genuine complex representation character exactly when its expansion in irreducible rows has nonnegative integer coefficients. The necessity follows from decomposition, and the sufficiency follows by taking the indicated direct sum. A norm-one class function alone need not be genuine: the negative of an irreducible character also has norm one. Similarly, a class indicator is a useful function but usually has fractional or negative expansion coefficients. The irreducibility norm criterion assumes at the outset that the function is already an actual character.

Expand a cyclic class function with complex rows

On C3=(e,r,r2)C_3=(e,r,r^2) let ω=e2πi/3\omega=e^{2\pi i/3}, with rows χ0=(1,1,1)\chi_0=(1,1,1), χ1=(1,ω,ω2)\chi_1=(1,\omega,\omega^2), and χ2=(1,ω2,ω)\chi_2=(1,\omega^2,\omega). Expand f=(2,−1,−1)f=(2,-1,-1).

  1. The constant-row coefficient is (2−1−1)/3=0(2-1-1)/3=0. For χ1\chi_1, conjugation gives coefficient (2−ω2−ω)/3=1(2-\omega^2-\omega)/3=1, since ω+ω2=−1\omega+\omega^2=-1.

  2. The coefficient for χ2\chi_2 is likewise one. Adding the nontrivial rows gives (2,ω+ω2,ω2+ω)=(2,−1,−1)(2,\omega+\omega^2,\omega^2+\omega)=(2,-1,-1).

  3. Both coefficients are nonnegative integers, so this is a two-dimensional genuine character. Its norm is (4+1+1)/3=2(4+1+1)/3=2, agreeing with two inequivalent summands. In contrast, −χ1-\chi_1 has norm one but identity value minus one and is not a representation character.

The result

The class function is the sum of the two conjugate nontrivial cyclic characters. Positivity and integrality of all expansion coefficients establish genuineness.

Explore the model

Interactive lab

Interactive lab · waves

Complex phase changes the sum

Pause and Reset, then set both phase difference 120120 degrees and amplitude ratio 11. Predict resultant amplitude 11, because ∣1+e2πi/3∣=1|1+e^{2\pi i/3}|=1. Reset and set both phase 00 degrees and ratio 11, predicting amplitude 22. The instrument adds two equal-frequency waves and exposes their phase cancellation. It does not generate a character table or add all three cube roots; compute 1+ω+ω2=01+\omega+\omega^2=0 separately. Control changes pause and retain time.

Complex phase changes the sumTwo waves share the same frequency and wave number. Playback advances t in sin(2x − t) + a sin(2x − t + phase). Current readings: Sum at x = 0: 0.866; Resultant amplitude: 1.732; Phase difference: 60°. All parameters are adjustable with the labeled controls below.−3.14−1.5701.573.14−3.4−1.701.73.4x · radiansAmplitude
SumFirst componentSecond component
Sum at x = 0
0.866
Resultant amplitude
1.732
Phase difference
60°
t = 0 s

Two waves share the same frequency and wave number. Playback advances t in sin(2x − t) + a sin(2x − t + phase). Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forCharacter values can be complex phases, and their inner products need complex conjugation.

Check your understanding

What additional condition makes a class function a genuine complex representation character?

Try it yourself

On the three classes of S3S_3, expand the indicator f=(0,1,0)f=(0,1,0) of the transpositions in the trivial, sign, and standard character basis. Explain why it is not an actual character.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Only the transposition term survives each inner product, and that class has three elements.

Compare with the solution

The coefficients are 3/6=1/23/6=1/2 for trivial, −3/6=−1/2-3/6=-1/2 for sign, and zero for standard. Hence f=(1−ε)/2f=(\mathbf1-\varepsilon)/2. The coefficients are neither nonnegative nor integral. Also its identity value is zero, whereas the only zero-dimensional representation has the zero character on every class. Being an ordinary nonnegative indicator function does not make it a representation character.

Take it with you

  • The center counts both conjugacy classes and irreducible types.
  • Column norms measure centralizer sizes.
  • Genuine characters have nonnegative integer irreducible coefficients.

Read further

Original sources and deeper explanations to follow at your own pace.

Optional progress, saved in this browser when available. Revisit any lesson at any time.

Lesson not marked complete.