Reducing a Representation

Introduction

Once a basis has been converted into a list of characters, the next question is which irreducible symmetry species it contains. The character reduction formula answers this systematically. It works because irreducible character rows are orthogonal when summed over the complete group with class multiplicities. A short C₂v calculation using water's two H 1s orbitals makes the formula concrete and provides several checks against sign or column-order errors.

Core explanation

Let Γ be a reducible representation with character χΓ(R), and let i label an irreducible representation with character χᵢ(R). Its multiplicity is aᵢ = (1/h) Σclasses N R χΓ(R) χᵢ(R) . Here h is group order, N R is the number of operations in the class, and means complex conjugation if characters are complex. Most elementary molecular point-group tables have real characters, so the conjugation makes no visible difference. The result must be a nonnegative integer if the basis and characters are valid.

For C₂v, choose z as the C₂ axis and operation order E, C₂(z), σ(xz), σ(yz). Put bent water in the yz plane with H positions (0,+a,b) and (0,−a,b). The basis consists of their two 1s orbitals, h₁ and h₂. E fixes both, giving character 2. C₂ exchanges them, giving 0. σ(xz) also exchanges them, giving 0. σ(yz), the molecular plane, fixes both, giving 2. Thus ΓH has characters (2,0,0,2). All C₂v classes contain one operation, so each N R is 1 and h = 4.

The C₂v rows in this convention are A₁: (1,1,1,1), A₂: (1,1,−1,−1), B₁: (1,−1,1,−1) and B₂: (1,−1,−1,1). Multiplying ΓH termwise by A₁ and summing gives (2+0+0+2)/4 = 1. Multiplying by A₂ gives (2+0+0−2)/4 = 0. B₁ likewise gives zero, while B₂ gives (2+0+0+2)/4 = 1. Therefore ΓH = A₁ ⊕ B₂. The physical functions can be chosen as h₁+h₂ of A₁ symmetry and h₁−h₂ of B₂ symmetry, after normalising and accounting for overlap if needed.

Two checks are essential. At identity, a₁d₁ + a₂d₂ + … must equal the original basis dimension. Here 1 + 1 = 2. Reconstructing the entire character row by adding A₁ and B₂ gives (2,0,0,2), exactly ΓH. A noninteger coefficient or failed reconstruction usually signals an incorrect operation order, missing class multiplicity, wrong fixed-function count or a basis that is not closed under the group.

For a larger non-abelian group, class sizes cannot be omitted. In C₃v, the classes E, 2C₃ and 3σv have sizes 1, 2 and 3, so the formula includes corresponding weights. Treating one representative per class as if each counted once would be wrong. Character-table headings already supply the weights, which is why their leading numbers matter.

Reduction reveals symmetry content but not automatically explicit orbital coefficients. A projection operator or a physically guided sum-and-difference construction is needed to build the actual SALCs. It also does not assign an energy order: A₁ and B₂ tell us how functions transform, while orbital energies require Hamiltonian matrix elements and chemical context.

Step-by-step reasoning

Specify the geometry, basis and operation-column order. Compute χΓ by transforming every basis function, then copy group order and class sizes. For each irrep row, multiply characters class by class, include multiplicity and divide by h. Check that all coefficients are nonnegative integers, dimensions sum correctly and reconstructed characters equal the original row.

Visual explanation

Draw the two H orbitals of water and arrows for each C₂v operation: two operations exchange them and two leave both fixed. Beside the drawing, place the character row (2,0,0,2) above the four irreducible rows. Highlight the termwise multiplication with A₁ and B₂, which each yield one copy.

Real-world analogy

An unknown mixture can be resolved into known components by comparing its response under several independent tests. Each irrep row is a distinct response pattern; orthogonality lets the reduction formula isolate how much of each pattern is present. The analogy does not replace the arithmetic: every operation must be weighted according to its class size.

Real-world example

In a molecular-orbital treatment of water, the two H 1s functions are reorganised into A₁ and B₂ combinations. The A₁ combination can match O 2s and 2p z, while B₂ can match O 2p y in the stated axes. The result tells an orbital-diagram builder which mixing blocks are possible before evaluating numerical energies or overlaps.

Why?

Why do the coefficients come out as integers? A reducible representation is a direct sum containing whole copies of irreducible representation spaces; it cannot contain 0.4 of an invariant orbital pair. Orthogonality of character rows extracts the copy count exactly. A fractional result is evidence of a setup or arithmetic error rather than an exotic partial symmetry species.

Common misconception

Do not multiply only the representative characters while forgetting that a class may contain several operations. Also, a coefficient of zero means a given species is absent from the chosen basis, not absent from the whole molecule. The water H 1s basis lacks A₂, but other atomic orbitals or vibrational motions can still transform as A₂.

Worked example

Using ΓH = (2,0,0,2) and B₂ = (1,−1,−1,1), calculate a(B₂) = [2(1)+0(−1)+0(−1)+2(1)]/4 = 1. For B₁ = (1,−1,1,−1), a(B₁) = [2+0+0−2]/4 = 0. Together with a(A₁)=1 and a(A₂)=0, the result is ΓH = A₁ + B₂. Re-adding those rows reproduces all four original characters.

Quick check

1. What should a valid reduction coefficient be? Answer: A nonnegative integer counting copies of the irrep in the chosen representation. 2. Why is the class-size factor needed in a non-abelian group? Answer: A character-table column can stand for several operations, and the group sum must count each operation.

Exam focus

Write the reduction formula and define h, N R and both character terms. Show the operation order explicitly. Finish with dimension and reconstruction checks; they catch most errors more reliably than a single coefficient calculation.

Advanced insight

Character reduction is an inner-product calculation in the space of class functions. Irreducible characters form an orthonormal set under the weighted group inner product, so the multiplicity is the projection of the reducible character onto one irreducible row. This viewpoint also explains why merely matching the identity character cannot uniquely determine a decomposition.

Summary

The reduction formula decomposes a basis's character pattern into irreducible species using group order and class sizes. For water's two H 1s orbitals, Γ = (2,0,0,2) reduces to A₁ + B₂ in the stated C₂v axes. Nonnegative integer coefficients, dimension matching and complete character reconstruction verify the result.

Practice questions

1. In the water calculation, why is χ(C₂) zero rather than −2? Answer: C₂ exchanges the two distinct H 1s basis functions, so its two-by-two permutation matrix has zeros on the diagonal. Neither individual 1s function simply changes sign on itself. 2. A proposed decomposition has dimensions adding to three for a four-function basis. Can it be correct? Answer: No. The identity character of the basis is four, so the sum of irrep dimensions weighted by multiplicity must also be four. 3. How would you test whether Γ = A₁ + B₂ beyond checking identity dimension? Answer: Add their characters in every operation column and compare with the original Γ row. Matching all four characters is a stronger complete reconstruction check.