Characters and Reducible Representations

Introduction

Writing a matrix for every operation becomes cumbersome when a molecule has many atomic orbitals or displacement coordinates. The trace of each matrix, called its character, compresses the representation while retaining the information needed to decompose it into symmetry species. Characters are especially convenient because they are unchanged by a change of basis and constant across conjugacy classes. The calculation still needs a clearly chosen basis: different function spaces produce different reducible characters for the same point group.

Core explanation

For a matrix D(R) representing operation R, χ(R) = tr D(R), the sum of its diagonal entries. The character of identity is the representation dimension n, because D(E) is the n-dimensional identity. If two operations are conjugate, their matrices are similar and have equal traces, so one character can be written for their entire class. Characters of a direct sum add: if a representation consists of blocks Γ₁ and Γ₂, then χΓ(R) = χ₁(R) + χ₂(R). This additivity makes it possible to infer which irreducible symmetry species occur in a larger basis.

For an orthonormal set of equivalent local s orbitals that an operation merely permutes, its matrix is a permutation matrix. Each orbital left on itself contributes 1 to the trace; each orbital sent to a different site contributes 0. In the three H 1s basis of pyramidal NH₃, E fixes all three and has character 3, C₃ fixes none and has character 0, and a vertical mirror fixes one while swapping two and has character 1. The resulting reducible representation has class characters (3,0,1) for E, 2C₃ and 3σv.

The fixed-function counting shortcut requires care when functions have orientation or phase. A p orbital centred on an atom may remain at that atom yet reverse sign under reflection, contributing −1 rather than +1. A vector displacement at a fixed atom contributes the trace of the three-dimensional coordinate transformation, which can be −1, 0, 1 or 3 for common operations. A moved atom's local displacement coordinates contribute zero to the diagonal trace in a simple atom-centred basis. Thus a reliable character calculation first asks what happens to each basis function, not merely which nuclei stay fixed.

A representation is reducible when one can choose a basis making every group matrix block diagonal in the same partition. The total character is then the sum of irreducible characters, with a nonnegative integer multiplicity for each species. For the NH₃ H basis, the characters match A₁ + E: one totally symmetric combination and a two-dimensional pair. At identity, dimensions add as 1 + 2 = 3, matching the three original functions. The character pattern across the other classes confirms the decomposition.

Characters do not preserve every detail of an arbitrary single matrix. Two unrelated matrices can share a trace. Their strength is group-wide: the full character set over all classes, combined with orthogonality relations, determines the irreducible content of a finite-group representation. The reduction formula later uses class sizes and irreducible character rows to find multiplicities. Before reducing, one should verify that χ(E) equals the basis dimension and that any class character is the same for all operations in that class.

Step-by-step reasoning

Choose and list the basis functions. For each operation class, transform every basis function and count its signed diagonal contribution. Sum to obtain χ for one representative, verifying class equivalence. Check χ(E) against basis size. Then compare the full class-character pattern with sums of irreducible rows, rather than assigning symmetry from one operation alone.

Visual explanation

Draw three labelled H orbitals around NH₃. Show a C₃ rotation cycling all three, so no arrow returns to its own starting circle and the trace is zero. Show a mirror fixing H₁ and exchanging H₂ with H₃, giving one diagonal contribution. Beside it, sketch a p orbital that stays on its atom but reverses its lobe phases, contributing −1.

Real-world analogy

A large team's full seating rearrangement can be described partly by how many members stay in their own seats. For plain s-orbital permutations, that count is the character. The analogy fails for oriented functions because a p orbital may remain at one site yet flip sign; a signed response, not just a fixed-seat count, is needed.

Real-world example

To build ligand group orbitals around a metal, a chemist may use one donor orbital from each equivalent ligand. Character counting quickly describes how this set transforms. Reducing the resulting representation reveals combinations that can interact with metal s, p or d orbitals of matching symmetry, saving an arbitrary trial-and-error construction.

Why?

Why is trace so useful despite discarding most matrix entries? It is invariant under a basis change and additive across blocks. Therefore the character set can be compared directly with irreducible character-table rows even when the starting local-orbital basis is not symmetry adapted. The method reveals hidden invariant subspaces without first calculating every transformed orbital explicitly.

Common misconception

Counting fixed atoms is not always the same as calculating a character. The basis might be p orbitals or displacement vectors, for which signs and coordinate rotation matter. Also, χ(E) is not necessarily 1; it equals the full dimension of the chosen representation. A three-orbital ligand basis has χ(E) = 3 even if the group contains a one-dimensional totally symmetric species.

Worked example

For three equivalent NH₃ hydrogen 1s orbitals, obtain characters χ(E)=3, χ(C₃)=0 and χ(σv)=1. The C₃v irreducible A₁ characters are (1,1,1) and E characters are (2,−1,0). Their sum is (3,0,1), exactly the reducible pattern. Thus ΓH = A₁ ⊕ E. One H combination is totally symmetric; the remaining two form an E pair.

Quick check

1. What is χ(E) for a basis of four independent ligand orbitals? Answer: Four, because the identity matrix on that basis has four diagonal ones. 2. Can a p orbital fixed at its atomic centre contribute −1 to a character? Answer: Yes. If the operation reverses its phase, its diagonal transformation coefficient is −1.

Exam focus

State the basis before reporting characters. Use class sizes when reducing later, and verify the identity character equals basis dimension. For displacement representations, use the coordinate transformation trace at each fixed atom rather than a simple fixed-atom count.

Advanced insight

The trace of a representation matrix is invariant under similarity, which is why two different orbital coordinate systems give the same character row. Full character equivalence across all classes implies equivalent decompositions into irreducible representations for a finite group, even when individual matrices in the two starting bases look very different.

Summary

Characters condense the transformation of a chosen basis into one trace per operation class. They add when representations split into blocks and are independent of basis choice. Fixed-function counting works for simple permutation bases, while oriented orbitals and displacement vectors require signed transformation coefficients. The complete character pattern reveals irreducible symmetry content.

Practice questions

1. A four-function basis has χ(E)=3 in a proposed calculation. What is wrong? Answer: The identity matrix on four independent functions has trace four. Either a basis function was omitted, the functions are not independent as claimed, or the character was miscalculated. 2. A reflection fixes one p orbital at its centre but reverses its phase. What does it contribute to the trace? Answer: It contributes −1, because character counts the diagonal transformation coefficient of the function, not only whether its atom stays at the same position. 3. Why can the NH₃ hydrogen representation be described as A₁ + E? Answer: Adding those irreducible character rows gives (3,0,1), matching the three-H reducible representation for every C₃v class, and their dimensions sum to three.