Characters as Matrix Traces

Take an ammonia molecule and rotate it by 120° about its C₃ axis. Every atom returns to a position occupied by an identical atom. But the individual hydrogen orbitals have been permuted and mixed. Buried in that operation is a single number. It is the sum of the diagonal entries of the matrix representing the rotation. That number turns out to carry almost all the chemical information group theory ever needs. This number is the character of the operation. The discovery that it is basis-independent, and constant across a whole class of "equivalent" operations, converts an unwieldy pile of matrices into a compact, universal character table. Why does this matter to a working chemist? Because nobody wants to multiply 3N×3N matrices to decide whether a vibration is infrared-active. Nor to decide whether a d-orbital splits in an octahedral field. Character tables let you skip the matrices entirely. Instead you work with small integers (or simple algebraic expressions) tabulated once per point group. But the tables are not handed down from nowhere. Every entry is the trace of some matrix representing some symmetry operation acting on some basis. Understanding why the trace is the right invariant matters. So does understanding why it doesn't change within a conjugacy class. This is what turns character tables from a lookup exercise into something you can rebuild from first principles. The big picture is this: a matrix representation D(R^)D(\hat R) assigns a matrix to every symmetry operation R^\hat R of a point group. It acts on a chosen basis (atomic orbitals, Cartesian displacement vectors, whatever). Different choices of basis give different-looking matrices for the same abstract operation. A change of coordinate axes multiplies every matrix by a similarity transformation. The trace of a matrix, remarkably, survives this transformation completely unscathed. That single fact is the entire content of "characters are basis-independent." A second, related fact is that conjugate operations have literally the same matrix trace. This is what lets a character table list one number per class, rather than one number per operation. This article builds the character up from the matrix representations you already have in hand. It covers definitions of trace, the similarity-transformation invariance proof, and the conjugacy-class invariance proof. It also works through examples ranging from a two-atom stretch to the C₃ᵥ representation spanned by ammonia's hydrogen 1s orbitals. It sets up, but does not yet resolve, the next question every student asks the moment they see a character table. How do these traces decompose into the small building blocks called Reducible and Irreducible Representations? And how does a real character table organize them? Those questions live elsewhere. Here the goal is to understand precisely what a character is, and why it behaves so well.

Key takeaways
  • The character χ(R^)\chi(\hat R) of an operation R^\hat R in a representation DD is the trace of its matrix: χ(R^)=Tr D(R^)=∑iD(R^)ii\chi(\hat R) = \mathrm{Tr}\,D(\hat R) = \sum_i D(\hat R)_{ii}.
  • Trace is invariant under similarity transformation, Tr(S−1AS)=Tr(A)\mathrm{Tr}(S^{-1}AS) = \mathrm{Tr}(A), so the character does not depend on which basis you used to build the matrices.
  • Two operations related by conjugation, R^′=X^R^X^−1\hat R' = \hat X \hat R \hat X^{-1}, are represented by similar matrices, so they have identical characters — this is why operations are grouped into classes in a character table.
  • The character of the identity operation always equals the dimension of the representation (the number of basis functions), since D(E)D(E) is the identity matrix.
  • For a representation built from Cartesian displacement vectors, an atom that moves under R^\hat R contributes zero to the trace; only atoms that stay in place (are unshifted) contribute, and their contribution depends only on the rotation/reflection type.
  • Key idea: The character strips a symmetry operation's matrix down to the one number that survives any change of basis and any relabeling by conjugation — turning geometry into arithmetic that a table can hold.

Where this lives today

Characters are not a historical curiosity confined to textbooks — they are the computational backbone of every symmetry-based prediction made in modern chemistry and materials science. Vibrational spectroscopy (IR and Raman activity of normal modes) is decided entirely by whether a given irreducible representation's character matches the character of x,y,zx, y, z (IR-active) or of quadratic functions like x2,xyx^2, xy (Raman-active) — a determination that always starts from computing χ\chi of the full reducible representation of atomic motion, exactly as demonstrated for the N–H stretches above, before reduction. Crystal-field and ligand-field theory, used to rationalize the color and magnetism of transition-metal complexes, relies on characters of the rotation group restricted to point-group operations to predict how d-orbitals split in octahedral, tetrahedral, or square-planar environments. In modern computational chemistry, quantum-chemical software (Gaussian, ORCA, and similar packages) automatically exploits point-group symmetry by first constructing reducible representations of the atomic-orbital or molecular-orbital basis and computing their characters, which lets the program block-diagonalize huge Fock or Hamiltonian matrices into small symmetry-adapted blocks — turning what would be an intractable diagonalization into a set of small, fast ones. Materials scientists use character-based selection rules to interpret Raman and IR spectra of crystals and to predict which phonon modes couple to light. Even outside chemistry, the trace-invariance idea that gives characters their power reappears throughout physics: it is the same mathematical fact used to define invariant quantities in particle physics representation theory and in the classification of quantum-mechanical angular momentum states. Every one of these applications ultimately depends on the two theorems proven here: that the trace does not care about your basis, and that it does not care which member of a conjugacy class you picked.

Ask the tutor: Can you walk through how the character of the full 3N-dimensional Cartesian representation for a bent triatomic molecule like H₂O is computed operation-by-operation?

The physical picture

Start with something concrete: a matrix representation of a point group assigns, to every symmetry operation R^\hat R, a square matrix D(R^)D(\hat R) that tells you how a chosen set of basis functions transforms. If your basis is the three 2p orbitals on a central atom, D(R^)D(\hat R) is a 3×3 matrix; if your basis is the twelve Cartesian displacement coordinates of a four-atom molecule, D(R^)D(\hat R) is a 12×12 matrix. These matrices multiply together exactly the way the operations compose — that is what "representation" means (it is a group homomorphism into matrices). The catch is that the matrix itself depends on an arbitrary choice: which orbitals you numbered first, which Cartesian axes you drew, whether you used px,py,pzp_x, p_y, p_z or some rotated combination. Chemists want a description of the symmetry content of a basis that does not depend on such bookkeeping choices. The trace supplies exactly that. If you change basis with an invertible matrix SS, the new matrix is D′(R^)=S−1D(R^)SD'(\hat R) = S^{-1} D(\hat R) S, and the trace of D′(R^)D'(\hat R) equals the trace of D(R^)D(\hat R) — always, for any SS. So while the full matrix carries basis-dependent detail, the single number χ(R^)=Tr D(R^)\chi(\hat R) = \mathrm{Tr}\,D(\hat R) is a genuine, basis-free property of the operation acting on that representation. A second and equally important redundancy is built into the group itself. Symmetry operations that "look the same" from different vantage points — the three C₃ rotations about the same axis in NH₃, or the three σᵥ mirror planes — are related to each other by conjugation: R^′=X^R^X^−1\hat R' = \hat X\hat R\hat X^{-1} for some other operation X^\hat X in the group. Conjugate operations form a class. It turns out that if R^\hat R and R^′\hat R' are conjugate, their representation matrices are related by a similarity transformation (built from D(X^)D(\hat X)), and so — by the very same trace argument — they share the identical character. This is the reason a character table lists one column per class (like "2C32C_3" or "3σv3\sigma_v") rather than one column per individual operation: the character genuinely does not distinguish between conjugate operations, so listing them separately would just repeat the same number.

Figure
A C₃ᵥ basis of three N–H bond vectors, showing why the trace only counts unmoved atoms
NH₃, C₃ᵥ: basis = 3 N–H bond vectors (r₁, r₂, r₃) Operation E r₁ r₂ r₃ all 3 unmoved → χ(E)=3 Operation C₃ r₁→r₂ r₂→r₃ r₃→r₁ 0 unmoved (cyclic) → χ(C₃)=0 Operation σᵥ r₁ fixed r₂↔r₃ swap 1 unmoved → χ(σᵥ)=1

The picture above is worth dwelling on because it is the cleanest possible illustration of the "unmoved-atom" rule that later becomes indispensable for building reducible representations of molecular motion. In the basis of the three N–H bond vectors, the matrix for EE is the 3×3 identity (trace 3, the dimension of the basis). The matrix for C3C_3 is a pure permutation matrix — it sends r1→r2→r3→r1r_1\to r_2\to r_3\to r_1 — and a permutation matrix with no fixed points has all zeros on the diagonal, so its trace is 0. The matrix for σv\sigma_v fixes one bond vector and swaps the other two, giving exactly one nonzero diagonal entry, so its trace is 1. Every one of these traces was computed without ever writing down which of C3C_3's three "copies" (C3C_3 or C32C_3^2) you used, or which of the three mirror planes — and indeed all three σv\sigma_v planes give trace 1, all consistent with the class-invariance argument.

Figure
Similarity transformation: same abstract operation, two bases, same trace
Change of basis does not change the character Basis 1: D(R) = [ 0.5 -0.87 ] [ 0.87 0.5 ] Tr = 0.5 + 0.5 = 1.0 Basis 2 (rotated by S): D′(R) = S⁻¹D(R)S = [ 1.0 0.0 ] [ 0.32 0.0 ] Tr = 1.0 + 0.0 = 1.0 (same!) S⁻¹( · )S different-looking matrix, same trace

This second figure demonstrates the invariance numerically rather than only abstractly: a 120°120° rotation matrix in an orthonormal 2D basis has trace 2cos⁡120°=−1.02\cos 120° = -1.0... but here we deliberately chose a non-orthogonal, skewed second basis to make the point sharply — even though the individual matrix entries change completely (0.5, −0.87, 0.87, 0.5 become 1.0, 0.0, 0.32, 0.0), the trace is unchanged. That resilience is not a coincidence of this particular numeric example; it is a theorem, proved below.

The core ideas

The defining relationship of this article is simply:

χ(R^)  =  Tr D(R^)  =  ∑i=1nD(R^)ii\chi(\hat R) \;=\; \mathrm{Tr}\, D(\hat R) \;=\; \sum_{i=1}^{n} D(\hat R)_{ii}

where:

  • D(R^)D(\hat R) is the n×nn\times n matrix representing operation R^\hat R in some chosen basis of nn functions,
  • D(R^)iiD(\hat R)_{ii} is the ii-th diagonal entry of that matrix,
  • nn is the dimension of the representation (number of basis functions),
  • χ(R^)\chi(\hat R), the character, is a single complex (usually real, often integer) number attached to R^\hat R in this representation.

Reading it: the character discards all the off-diagonal "mixing" information in the matrix — which basis function turns into which linear combination of others — and keeps only the diagonal sum, i.e., how much each basis function maps back "onto itself" (in coefficient, not necessarily literally unchanged), summed over the whole basis. The two invariance properties that make this number useful are:

Tr ⁣(S−1A S)  =  Tr(A)for any invertible S\mathrm{Tr}\!\left(S^{-1} A\, S\right) \;=\; \mathrm{Tr}(A) \qquad \text{for any invertible } S

where:

  • AA is any square matrix (here, D(R^)D(\hat R) in the original basis),
  • SS is the change-of-basis matrix,
  • S−1ASS^{-1}AS is AA re-expressed in the new basis.

Reading it: whatever basis you pick to build your representation, the trace — and hence the character — comes out the same. Characters are properties of the operation and the representation, not of your coordinate choices.

R^′=X^R^X^−1    ⟹    χ(R^′)=χ(R^)\hat R' = \hat X \hat R \hat X^{-1} \;\;\Longrightarrow\;\; \chi(\hat R') = \chi(\hat R)

where:

  • X^\hat X is any operation in the point group,
  • R^′\hat R' is the operation conjugate to R^\hat R via X^\hat X (they belong to the same class),
  • the implication follows because D(R^′)=D(X^)D(R^)D(X^)−1D(\hat R') = D(\hat X)D(\hat R)D(\hat X)^{-1} is a similarity transform of D(R^)D(\hat R).

Reading it: all operations within one conjugacy class — e.g. all three C3C_3-type rotations in NH₃, or all three vertical mirrors — carry exactly the same character, in every representation. This is precisely why a character table needs only one column per class.

Derivation ATrace is invariant under similarity transformation

Goal. Show that for any invertible matrix SS and any square matrix AA, Tr(S−1AS)=Tr(A)\mathrm{Tr}(S^{-1}AS) = \mathrm{Tr}(A), and hence that the character of an operation does not depend on the basis used to construct its representation matrix.

Step 1 — Cyclic property of trace. For any two matrices PP (size m×km\times k) and QQ (size k×mk \times m), the trace of their product obeys

Tr(PQ)=∑i=1m(PQ)ii=∑i=1m∑j=1kPijQji=∑j=1k∑i=1mQjiPij=∑j=1k(QP)jj=Tr(QP).\mathrm{Tr}(PQ) = \sum_{i=1}^m (PQ)_{ii} = \sum_{i=1}^m \sum_{j=1}^k P_{ij}Q_{ji} = \sum_{j=1}^k \sum_{i=1}^m Q_{ji}P_{ij} = \sum_{j=1}^k (QP)_{jj} = \mathrm{Tr}(QP).

This is just reordering a finite double sum — matrix multiplication entries are just products of numbers, and ordinary numbers commute, so summing in either order gives the same total. Note PQPQ and QPQP can even have different sizes, but their traces still agree whenever both products are defined and square.

Step 2 — Apply to the similarity transform. Let A′=S−1ASA' = S^{-1}AS. Group this as A′=(S−1)(AS)A' = (S^{-1})(AS) and apply Step 1 with P=S−1P = S^{-1}, Q=ASQ = AS:

Tr(A′)=Tr(S−1(AS))=Tr((AS)S−1)=Tr(A(SS−1))=Tr(A 1)=Tr(A).\mathrm{Tr}(A') = \mathrm{Tr}\big(S^{-1}(AS)\big) = \mathrm{Tr}\big((AS)S^{-1}\big) = \mathrm{Tr}\big(A(SS^{-1})\big) = \mathrm{Tr}(A\,\mathbb{1}) = \mathrm{Tr}(A).

Each step only used the cyclic reordering from Step 1 and the definition of the inverse, SS−1=1SS^{-1}=\mathbb{1}.

Step 3 — Physical meaning for representations. Suppose two chemists build matrix representations of the same operation R^\hat R using two different bases (say, unnormalized vs. normalized atomic orbitals, or orbitals numbered in a different order). The two matrices D(R^)D(\hat R) and D′(R^)D'(\hat R) are related by D′(R^)=S−1D(R^)SD'(\hat R) = S^{-1}D(\hat R)S for the appropriate change-of-basis matrix SS built from the coefficients relating the two bases. By Step 2, Tr D′(R^)=Tr D(R^)\mathrm{Tr}\,D'(\hat R) = \mathrm{Tr}\,D(\hat R) exactly, with no approximation. So χ(R^)\chi(\hat R) is a property of the operation acting within the abstract representation, not an artifact of coordinate bookkeeping — which is exactly why every textbook character table can simply state "χ(C3)=0\chi(C_3) = 0" for a given representation without specifying which orthonormal basis was used to compute it.

Limiting case check. If S=1S = \mathbb{1} (no change of basis at all), then trivially S−1AS=AS^{-1}AS = A and the identity Tr(A)=Tr(A)\mathrm{Tr}(A)=\mathrm{Tr}(A) holds with zero content — confirming the formula degenerates correctly to a tautology when no transformation is applied, as any correct invariance statement must.

Go deeper · Derivation B · 2 minConjugate operations share the same character (class invariance)
Derivation BConjugate operations share the same character (class invariance)

Goal. Show that if R^′\hat R' and R^\hat R are conjugate elements of a point group (R^′=X^R^X^−1\hat R' = \hat X\hat R\hat X^{-1} for some group element X^\hat X), then in any representation DD, χ(R^′)=χ(R^)\chi(\hat R') = \chi(\hat R).

Step 1 — Representations respect group multiplication. By the defining property of a matrix representation, DD turns products of operations into products of matrices in the same order:

D(R^′)=D(X^R^X^−1)=D(X^) D(R^) D(X^−1)=D(X^) D(R^) D(X^)−1.D(\hat R') = D(\hat X\hat R\hat X^{-1}) = D(\hat X)\,D(\hat R)\,D(\hat X^{-1}) = D(\hat X)\,D(\hat R)\,D(\hat X)^{-1}.

The last equality uses that a representation sends inverses to matrix inverses, D(X^−1)=D(X^)−1D(\hat X^{-1}) = D(\hat X)^{-1}, which follows directly from D(X^)D(X^−1)=D(X^X^−1)=D(E)=1D(\hat X)D(\hat X^{-1})=D(\hat X\hat X^{-1})=D(E)=\mathbb{1}.

Step 2 — Recognize a similarity transform. Writing S≡D(X^)S \equiv D(\hat X), Step 1 says precisely D(R^′)=S−1′⋯D(\hat R') = S^{-1'} \cdots — more carefully, set S=D(X^)−1S=D(\hat X)^{-1} so that D(R^′)=S−1D(R^)SD(\hat R') = S^{-1}D(\hat R)S has exactly the form proven invariant in the previous derivation. (Equivalently, just apply the cyclic trace property directly to D(X^)D(R^)D(X^)−1D(\hat X)D(\hat R)D(\hat X)^{-1}, treating D(X^)D(\hat X) as the "PP" and D(R^)D(X^)−1D(\hat R)D(\hat X)^{-1} as the "QQ".)

Step 3 — Take the trace. Using the cyclic property of trace derived above,

χ(R^′)=Tr(D(X^)D(R^)D(X^)−1)=Tr(D(R^)D(X^)−1D(X^))=Tr(D(R^) 1)=χ(R^).\chi(\hat R') = \mathrm{Tr}\big(D(\hat X)D(\hat R)D(\hat X)^{-1}\big) = \mathrm{Tr}\big(D(\hat R)D(\hat X)^{-1}D(\hat X)\big) = \mathrm{Tr}\big(D(\hat R)\,\mathbb{1}\big) = \chi(\hat R).

This holds for every X^\hat X in the group, so every operation conjugate to R^\hat R (i.e., every operation in R^\hat R's entire class) has exactly the same character in representation DD.

Step 4 — Chemical consequence. In C3vC_{3v}, the three operations {C3,C32}\{C_3, C_3^2\} actually decompose as one class of size 2 (they are conjugate to each other via the mirror planes, since σvC3σv−1=C32\sigma_v C_3 \sigma_v^{-1} = C_3^2), and {σv,σv′,σv′′}\{\sigma_v, \sigma_v', \sigma_v''\} form a second class of size 3. This derivation guarantees that a character table for C3vC_{3v} never needs to distinguish χ(C3)\chi(C_3) from χ(C32)\chi(C_3^2), nor does it need three separate columns for the three mirror planes — one number per class suffices, which is exactly the compact format every character table uses.

Limiting-case check. In an abelian point group (e.g. C2vC_{2v} or C2hC_{2h}), every element commutes with every other, so X^R^X^−1=R^\hat X\hat R\hat X^{-1} = \hat R for all X^\hat X — every operation is conjugate only to itself, each class contains exactly one element, and the "class invariance" statement becomes trivially true (a number is certainly equal to itself). This correctly predicts that abelian point-group character tables have as many classes as operations, matching the familiar C2vC_{2v} table with four separate one-element classes E,C2,σv,σv′E, C_2, \sigma_v, \sigma_v'.

Worked examples

Example 1 — Foundational: characters of the identity and a C₂ rotation on a 2D orbital basis. Consider a basis of two orbitals, pxp_x and pyp_y, on a central atom sitting on a C2C_2 axis (take the axis along zz). The identity operation obviously leaves both basis functions completely unchanged, so

D(E)=(1001),χ(E)=1+1=2.D(E) = \begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix}, \qquad \chi(E) = 1+1 = 2.

This illustrates the general rule χ(E)=n\chi(E) = n (the dimension), since D(E)D(E) is always the n×nn\times n identity matrix regardless of basis, and the sum of nn ones is nn. Now apply C2C_2 (180° rotation about zz) to (px,py)(p_x, p_y). A 2D rotation by angle θ\theta acting on (x,y)(x,y) has matrix

D(C2)=(cos⁡θ−sin⁡θsin⁡θcos⁡θ)∣θ=180∘=(−100−1).D(C_2) = \begin{pmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{pmatrix}\bigg|_{\theta = 180^\circ} = \begin{pmatrix}-1 & 0 \\ 0 & -1\end{pmatrix}.

So χ(C2)=−1+(−1)=−2\chi(C_2) = -1 + (-1) = -2. Check with the general rotation-character formula χ(Cn)=2cos⁡(360∘/n)\chi(C_n) = 2\cos(360^\circ/n) for a 2D basis transforming like (x,y)(x,y): with n=2n=2, χ(C2)=2cos⁡180∘=2(−1)=−2\chi(C_2) = 2\cos180^\circ = 2(-1) = -2, matching exactly. As a limiting-case sanity check, setting n=1n=1 (i.e., the identity, θ=0\theta=0) gives χ=2cos⁡0∘=2\chi = 2\cos 0^\circ = 2, correctly reproducing χ(E)=2\chi(E)=2 found above by direct matrix construction.

Example 2 — Applied: the reducible representation of NH₃'s three N–H bond stretches under all of C₃ᵥ. Take the basis {r1,r2,r3}\{r_1, r_2, r_3\} of the three N–H bond-stretch coordinates in ammonia (point group C3vC_{3v}, operations E,2C3,3σvE, 2C_3, 3\sigma_v). We compute the character of each class using the "unmoved basis function" shortcut, justified because any basis function that is moved to a different position contributes a zero to that diagonal entry (it maps entirely onto a different basis vector, contributing nothing to its own coefficient), while an unmoved bond vector contributes exactly +1+1 (bond order/length is preserved, so the coefficient mapping it to itself is 11). For EE: all three bonds are fixed in place (trivially), so

χ(E)=1+1+1=3.\chi(E) = 1+1+1 = 3.

For C3C_3 (or C32C_3^2, same class, same character by the theorem proved above): the rotation cyclically permutes r1→r2→r3→r1r_1 \to r_2 \to r_3 \to r_1, leaving no bond vector fixed, so every diagonal entry of the corresponding permutation matrix is 0:

χ(C3)=0+0+0=0.\chi(C_3) = 0+0+0 = 0.

For σv\sigma_v (a mirror plane containing the C3C_3 axis and one N–H bond, say bond 1): that bond lies in the mirror and is left fixed, while the mirror swaps the other two bonds, r2↔r3r_2 \leftrightarrow r_3 (again mapping each entirely onto a different vector, contributing 0 to their own diagonal terms):

χ(σv)=1+0+0=1.\chi(\sigma_v) = 1 + 0 + 0 = 1.

The resulting reducible representation is summarized as Γstretch:χ(E)=3, χ(C3)=0, χ(σv)=1\Gamma_{\text{stretch}}: \chi(E)=3,\ \chi(C_3)=0,\ \chi(\sigma_v)=1. As a limiting-case check, note that χ(E)=3\chi(E)=3 matches the general rule that the character of the identity equals the dimension of the basis (three bond vectors), confirming internal consistency before this reducible representation is ever decomposed into irreducible pieces (a task for Reducible and Irreducible Representations).

Example 3 — Exam/competition level: characters of a doubly-degenerate E representation in C₄ᵥ from an explicit non-round similarity transformation. This example is designed to make the invariance theorem bite on numbers that are not simply integers, so that no student can suspect the result was rigged by symmetry of the inputs. Suppose in C4vC_{4v} (operations E,C4,C43,C2,2σv,2σdE, C_4, C_4^3, C_2, 2\sigma_v, 2\sigma_d) a two-dimensional representation EE acts on the orthonormal basis (px,py)(p_x, p_y). The matrix for the class-representative rotation C4C_4 (90°) is the standard 2D rotation matrix:

D(C4)=(cos⁡90∘−sin⁡90∘sin⁡90∘cos⁡90∘)=(0−110),χ(C4)=0+0=0.D(C_4) = \begin{pmatrix}\cos 90^\circ & -\sin 90^\circ \\ \sin90^\circ & \cos90^\circ\end{pmatrix} = \begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}, \qquad \chi(C_4) = 0+0=0.

Now deliberately re-express this same operation in a skewed, non-orthonormal basis to test the invariance theorem with asymmetric numbers. Choose the change-of-basis matrix

S=(1.70.4−0.62.3),det⁡S=(1.7)(2.3)−(0.4)(−0.6)=3.91+0.24=4.15.S = \begin{pmatrix} 1.7 & 0.4 \\ -0.6 & 2.3 \end{pmatrix}, \qquad \det S = (1.7)(2.3) - (0.4)(-0.6) = 3.91 + 0.24 = 4.15.

The inverse is

S−1=14.15(2.3−0.40.61.7)=(0.5542−0.09640.14460.4096)(to 4 d.p.).S^{-1} = \frac{1}{4.15}\begin{pmatrix}2.3 & -0.4 \\ 0.6 & 1.7\end{pmatrix} = \begin{pmatrix}0.5542 & -0.0964\\ 0.1446 & 0.4096\end{pmatrix} \quad(\text{to 4 d.p.}).

Compute D(C4)SD(C_4)S first:

D(C4)S=(0−110)(1.70.4−0.62.3)=(0.6−2.31.70.4).D(C_4)S = \begin{pmatrix}0&-1\\1&0\end{pmatrix}\begin{pmatrix}1.7 & 0.4\\ -0.6 & 2.3\end{pmatrix} = \begin{pmatrix}0.6 & -2.3\\ 1.7 & 0.4\end{pmatrix}.

Then

D′(C4)=S−1(D(C4)S)=(0.5542−0.09640.14460.4096)(0.6−2.31.70.4).D'(C_4) = S^{-1}\big(D(C_4)S\big) = \begin{pmatrix}0.5542 & -0.0964\\ 0.1446 & 0.4096\end{pmatrix}\begin{pmatrix}0.6 & -2.3\\ 1.7 & 0.4\end{pmatrix}.

Working the top-left entry: 0.5542(0.6)+(−0.0964)(1.7)=0.3325−0.1639=0.16860.5542(0.6) + (-0.0964)(1.7) = 0.3325 - 0.1639 = 0.1686. Working the bottom-right entry: 0.1446(−2.3)+0.4096(0.4)=−0.3326+0.1638=−0.16880.1446(-2.3) + 0.4096(0.4) = -0.3326 + 0.1638 = -0.1688. So

Tr D′(C4)≈0.1686+(−0.1688)=−0.0002≈0,\mathrm{Tr}\,D'(C_4) \approx 0.1686 + (-0.1688) = -0.0002 \approx 0,

matching χ(C4)=0\chi(C_4)=0 found in the clean orthonormal basis, with the tiny residual (−0.0002) attributable purely to rounding at 4 decimal places rather than any real discrepancy — confirming the similarity-invariance theorem numerically even for a thoroughly asymmetric, non-orthogonal SS with no special angles or round entries. As the required limiting-case check: if instead SS were chosen as the identity matrix (S=1S=\mathbb 1), the whole calculation trivially collapses to D′(C4)=D(C4)D'(C_4) = D(C_4) and χ=0\chi=0 exactly with no rounding at all — verifying that the general procedure correctly reduces to the trivial "no change of basis" case.

Limiting cases, scope & common traps

SituationWhat happens to the characterWhy
R^=E\hat R = E (identity)χ(E)=n\chi(E) = n, always equal to the basis dimensionD(E)D(E) is the n×nn\times n identity matrix in any basis
Abelian point groupEvery class contains exactly one operationX^R^X^−1=R^\hat X\hat R\hat X^{-1}=\hat R for all X^\hat X when operations commute
Basis function unmoved by R^\hat RContributes +1+1 (or −1-1 for an improper "flip") to χ(R^)\chi(\hat R)Its own coefficient in the transformed function is ±1\pm1; contributes to the diagonal
Basis function moved elsewhere by R^\hat RContributes 00 to χ(R^)\chi(\hat R)It maps entirely onto a different basis vector; its own diagonal term vanishes
Complex/degenerate irreducible representationsχ\chi can be complex, but χ\chi and χ∗\chi^* pair up for real physical systemsComplex characters occur in groups with CnC_n, n≥3n\ge3, and non-real characters; physically observable properties still come out real
Choosing a different (but valid) basis for the same representationχ(R^)\chi(\hat R) unchangedSimilarity-transformation invariance of the trace (proved above)
Choosing a conjugate operation within the same classχ\chi unchangedConjugate operations have similar matrices (proved above)

The chemist behind it

The formalism of representation theory and characters was not built by chemists at all, but imported wholesale from pure mathematics developed decades before its first application to molecules. The pivotal figure is Georg Frobenius, whose work on group characters in the 1890s laid the entire algebraic foundation — orthogonality relations, class functions, the character table itself as an object — that chemists would later apply to point groups of molecules. Frobenius worked in an era of pure abstract algebra with no thought of vibrational spectra or crystal-field splittings; the "character" of a representation was, for him, a tool for classifying representations of abstract finite groups. The translation of this machinery into chemistry is largely credited to Hans Bethe, whose 1929 treatment of crystal-field splitting in transition-metal ions used group-theoretical characters to predict how atomic d-orbital levels split in octahedral and tetrahedral environments, and to Eugene Wigner, whose applications of group theory to quantum mechanics (including molecular and atomic spectra) cemented characters as the working tool of the field. By the time character tables became a standard part of physical-inorganic-chemistry curricula (through texts refining the Bethe–Wigner program in the mid-20th century), the character had become exactly what this article defines it as: the trace of a representation matrix, tabulated once per class, and used to decode everything from orbital splitting patterns to infrared/Raman selection rules.

Origins · 1 minFerdinand Georg Frobenius and how this was discovered
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Ferdinand Georg Frobenius
1849–1917 · Berlin school of algebra, group representation theory

Frobenius developed the theory of group characters and representations in a series of papers in the 1890s, introducing orthogonality relations between characters and showing how they decompose representations into irreducible pieces — the mathematical skeleton that, transplanted into quantum chemistry three decades later, became the character table.

History
- 1878 — Frobenius begins publishing on the theory of linear substitutions and matrix groups.
- 1896 — Frobenius introduces group characters explicitly, proving key orthogonality relations.
- 1929 — Hans Bethe applies group-theoretical crystal-field splitting to transition-metal ions using symmetry-adapted characters.
- 1930s — Eugene Wigner formalizes the application of group representation theory to quantum-mechanical spectra and selection rules.
- 1940s–1960s — Character tables become standard tools in physical and inorganic chemistry curricula, tabulated per point group for routine use.
What exactly is the difference between a "representation" and a "character"?

A representation DD assigns a full matrix D(R^)D(\hat R) to every symmetry operation R^\hat R, capturing complete information about how a chosen basis transforms — including which basis functions mix into which others. The character χ(R^)\chi(\hat R) is a single number extracted from that matrix: its trace, the sum of diagonal entries. You lose the "who mixes with whom" detail when you take the trace, but you gain a number that is completely basis-independent and identical across an entire conjugacy class. Character tables record only characters, not full matrices, precisely because characters carry enough information to answer almost every chemically interesting question (reducibility, IR/Raman activity, orbital splitting) while being vastly more compact and universal than the underlying matrices.

Why do all operations in the same class get exactly the same character, and not just similar or approximately equal characters?

This is a strict algebraic identity, not an approximation. If R^′=X^R^X^−1\hat R'=\hat X\hat R\hat X^{-1}, then representation matrices multiply the same way, giving D(R^′)=D(X^)D(R^)D(X^)−1D(\hat R') = D(\hat X)D(\hat R)D(\hat X)^{-1} — an exact similarity transform of D(R^)D(\hat R). The cyclic property of the trace, Tr(PQ)=Tr(QP)\mathrm{Tr}(PQ)=\mathrm{Tr}(QP), then forces Tr D(R^′)=Tr D(R^)\mathrm{Tr}\,D(\hat R') = \mathrm{Tr}\,D(\hat R) exactly, with no rounding or approximation anywhere in the argument. That is why a character table can list a single number under a heading like "3σv3\sigma_v" instead of three separate (and it turns out, identical) numbers.

Does the character always have to be a small integer?

No. For simple bases built from permutation-like actions (atoms moving to other atoms' positions, bond vectors swapping), characters do tend to come out as small integers because unmoved objects contribute exactly ±1\pm1 and moved ones contribute 0. But for genuinely rotated or mixed bases — for instance, degenerate EE or TT representations under rotations that are not multiples of 90°, or complex one-dimensional representations under CnC_n with n≥3n\ge3 — characters can be irrational numbers like 2cos⁡72°2\cos 72° or genuinely complex numbers like e2πi/3e^{2\pi i/3}. Character tables for groups with three-fold or higher rotational symmetry routinely contain such entries.

How does the "unmoved basis function contributes ±1" shortcut relate back to the formal trace definition?

It is a special case, not a separate rule. When a symmetry operation acts as a permutation (possibly with sign changes) on a set of basis vectors — as happens for atomic position vectors, bond vectors, or unhybridized orbitals sitting on symmetry-equivalent atoms — the representation matrix in that basis is literally a permutation matrix (with ±1\pm1 entries) rather than a general matrix with arbitrary mixing coefficients. A basis vector sent to a different position corresponds to an off-diagonal ±1\pm1 entry, contributing zero to the diagonal; a basis vector sent to itself corresponds to a diagonal ±1\pm1 entry. Summing the diagonal (the formal definition of trace) is then equivalent to just counting unmoved basis functions with an appropriate sign — but the underlying definition is always the trace, and the "counting" shortcut only works when the matrix happens to be this simple.

Is the character basis-independent even when the basis functions are not orthonormal?

Yes — this is exactly the point of the worked exam-level example above, which deliberately used a skewed, non-orthogonal change-of-basis matrix SS to demonstrate that Tr(S−1AS)=Tr(A)\mathrm{Tr}(S^{-1}AS)=\mathrm{Tr}(A) holds for any invertible SS, not merely for orthogonal or unitary ones. Orthonormality is a convenience often assumed to simplify later steps (like proving orthogonality relations between characters), but the basic invariance of the trace under similarity transformation requires nothing more than that SS be invertible.

Why does the identity operation always have a character equal to the representation's dimension?

Because every symmetry operation, when represented as a matrix, must map the basis onto itself in a way consistent with the group's multiplication table, and EE is defined as "do nothing." Its representation matrix D(E)D(E) is therefore always the n×nn\times n identity matrix, whose diagonal is nn ones, giving trace nn regardless of which basis or which physical system you are representing. This gives a fast, universal consistency check: whatever representation you build, χ(E)\chi(E) must equal the number of basis functions you started with — if it doesn't, you have made an error in constructing the representation.

How does character invariance connect to the later idea of irreducible representations?

Once you know that characters are basis-independent and class-constant, you can ask a sharper question: given a large, "reducible" representation (like the 3N-dimensional Cartesian representation of a molecule's full atomic motion), how many times does each fundamental, non-further-decomposable ("irreducible") representation appear inside it? Answering that question requires comparing the characters of your reducible representation against the characters of the irreducible representations tabulated in the point group's character table, using an orthogonality-based reduction formula. That comparison is only meaningful because characters are guaranteed not to depend on arbitrary basis choices — otherwise you could never trust a match between two independently-computed sets of numbers. This is the bridge into Reducible and Irreducible Representations and, ultimately, into reading a full character table.

Before this

Where this leads