Group Representations

A group is symmetry stated in the abstract: a set of moves you can compose, with an identity and a way to undo each one. But nature does not hand us abstract moves — it hands us Hilbert spaces of quantum states, configuration spaces of fields, spaces of orbitals. A representation is the bridge. It realizes each abstract group element as a concrete invertible linear operator on such a space, keeping the group's multiplication law intact: composing two moves corresponds to multiplying their operators. In one stroke, every question about the group becomes a question about linear algebra — eigenvalues, traces, block structure, similarity — the most powerful toolkit in all of physics. Formally a representation is a homomorphism $D:G\to GL(V)$ , and that single word "homomorphism" carries the whole theory. It forces $D(e)=\mathbb{1}$ and $D(g^{-1})=D(g)^{-1}$ , so the algebraic skeleton of $G$ survives as operators, and choosing a basis turns $GL(V)$ into invertible matrices. The same group admits infinitely many inequivalent representations of different dimensions, and the central task of the subject is to sort them out. Two ideas tame the flood. Equivalence declares two representations the same if a fixed change of basis $D'=SDS^{-1}$ relates them, so we study equivalence classes, not matrices. Unitarity — automatic for finite and compact groups by an averaging trick — makes representations the native language of quantum symmetry, where degenerate energy levels are the irreducible pieces and their dimensions are the multiplicities you measure in a spectrum.

Key takeaways

A representation of a group $G$ on a vector space $V$ is a homomorphism $D:G\to GL(V)$ that assigns to each element $g$ an invertible operator $D(g)$ with $D(g_1)D(g_2)=D(g_1g_2)$ — abstract multiplication becomes operator composition.
Choosing a basis turns each $D(g)$ into a matrix, so the group's Cayley table becomes literal matrix multiplication and $\dim V$ is the dimension of the representation.
Physics cares only about equivalence classes $D'(g)=S\,D(g)\,S^{-1}$ ; the genuine invariant is the character $\chi(g)=\operatorname{Tr}D(g)$ , constant on each conjugacy class.
Every representation of a finite (or compact) group can be made unitary by averaging an inner product, which is why unitary representations carry all of quantum symmetry.
Representations decompose into irreducible blocks; the regular representation contains each irreducible $D^{(\alpha)}$ exactly $d_\alpha=\dim D^{(\alpha)}$ times, forcing the sum rule $\sum_\alpha d_\alpha^2=|G|$ — for $S_3$ this reads $1^2+1^2+2^2=6$ .
Key idea: a representation is a structure-preserving map $G\to GL(V)$ that turns symmetry questions about the abstract group into linear-algebra questions about operators on the physical state space $V$ .

The core insight

A representation is a structure-preserving map $G\to GL(V)$ : abstract group multiplication becomes operator composition, so symmetry questions about $G$ become linear-algebra questions about how operators act on the physical state space $V$ .

Where this lives today

Representation theory is the daily grammar of modern physics. In particle physics the internal symmetries $SU(2)$ isospin and $SU(3)$ flavour act on hadron states through their irreducibles: protons and neutrons fill an isospin doublet, hadrons fill $SU(3)$ octets and decuplets, and the empty slot in the baryon decuplet is exactly what let Gell-Mann predict the $\Omega^-$ before it was found — while the whole Standard Model is specified by the representations of $SU(3)\times SU(2)\times U(1)$ each field carries. In chemistry and condensed matter, the $32$ crystallographic point groups are classified by their irreducibles, and character tables tell a spectroscopist which vibrational modes are infrared- or Raman-active and which $d$ -orbitals split in a ligand field. Even numerical practice leans on it: block-diagonalizing a Hamiltonian by its symmetry irreducibles turns one intractable matrix into several small ones — the computational cash value of $\sum_\alpha d_\alpha^2=|G|$ .

The physical picture

Imagine a group as a collection of abstract moves — rotations of a triangle, permutations, reflections of a crystal — that you know how to compose. A representation casts a shadow of those moves onto a concrete vector space: each move becomes a definite matrix, and composing moves in the group matches multiplying matrices in the shadow. The homomorphism law $D(g_1g_2)=D(g_1)D(g_2)$ says the shadow has the same internal logic as the original — a commuting square you can traverse either way and land on the same operator.

Figure

The homomorphism square for the cyclic group C₃ generated by a 120° rotation r: composing r with itself (top) mirrors the matrix product D(r)·D(r) = D(r²) below, and the character χ = Tr D reads χ(e) = 2, χ(r) = 2cos120° = −1

In quantum mechanics the vector space is the Hilbert space of states, and the payoff is immediate. If a symmetry leaves the Hamiltonian invariant, each group element acts by a unitary operator that shuffles states without changing energy. A degenerate level therefore carries a representation of the symmetry group, and the dimension of each irreducible piece equals the degeneracy of a multiplet — symmetry predicts how many states sit at each energy before you solve anything. The deepest structural fact is a counting law: the "regular" representation contains each irreducible exactly as many times as its dimension, so the bookkeeping closes into $\sum_\alpha d_\alpha^2=|G|$ .

Figure

The regular representation of the symmetric group S₃ of order 6 reduces by a change of basis into blocks: one trivial 1×1, one sign 1×1, and two copies of the standard 2×2 irreducible, so the squared dimensions obey 1² + 1² + 2² = 6, the order of the group

Equivalence — the freedom $D'=SDS^{-1}$ — is just choosing different basis vectors for the same physical space, so nothing measurable changes. That is why the honest content of a representation lives in similarity-invariants, above all the character $\chi(g)=\operatorname{Tr}D(g)$ , identical for equivalent representations and constant on each conjugacy class. Unitarity — the preservation of $\langle\psi|\phi\rangle$ under a symmetry — can always be arranged for finite and compact groups, and that is the foundation of the entire quantum theory of symmetry.

The core equations

A representation is defined by one demand: the operators must respect group multiplication. Everything else is a consequence of this single homomorphism condition.

D:G\to GL(V),\qquad D(g_1g_2)=D(g_1)\,D(g_2) \qquad (1)

where: $G$ — the abstract group; $V$ — the representation (vector) space; $GL(V)$ — the group of invertible linear operators on $V$ ; $D(g)$ — the operator assigned to element $g$ . Reading it: composing two moves then representing must equal representing each and multiplying, for all $g_1,g_2$ — the whole definition. The homomorphism law pins down the identity and inverses with no further input, so the skeleton of $G$ transfers to $V$ automatically.

D(e)=\mathbb{1},\qquad D(g^{-1})=D(g)^{-1} \qquad (2)

where: $e$ — the group identity; $\mathbb{1}$ — the identity operator on $V$ . Reading it: the do-nothing element maps to the do-nothing operator, and undoing a move inverts its matrix. Derivation B extracts both lines from equation (1). Because any change of basis re-expresses the same operators, two representations related by a fixed $S$ describe identical physics — the equivalence relation that organizes the whole classification.

D'(g)=S\,D(g)\,S^{-1},\qquad S\in GL(V) \qquad (3)

where: $S$ — a fixed invertible operator, the same for every $g$ ; $D'$ — the representation in the new basis. Reading it: because $S$ is independent of $g$ , it leaves every trace unchanged, so equivalent representations are physically indistinguishable. When $V$ carries an inner product the group preserves, the operators are unitary — the rule for quantum symmetry, by Wigner's theorem.

D(g)^\dagger\,D(g)=\mathbb{1}\quad\Longleftrightarrow\quad \langle D(g)\psi\,|\,D(g)\phi\rangle=\langle\psi\,|\,\phi\rangle \qquad (4)

where: $D(g)^\dagger$ — the Hermitian adjoint on $V$ ; $\langle\,\cdot\,|\,\cdot\,\rangle$ — the preserved inner product. Reading it: unitarity and inner-product invariance are one statement, and Derivation A shows both are automatic for finite groups. The one number that survives every change of basis is the trace of $D(g)$ , the character. Its invariance under conjugation makes it a class function — a function of conjugacy classes, not of elements.

\chi(g)\equiv\operatorname{Tr}D(g),\qquad \chi(hgh^{-1})=\chi(g) \qquad (5)

where: $\chi(g)$ — the character of $g$ ; $\operatorname{Tr}$ — the matrix trace. Reading it: since $\chi(e)=\operatorname{Tr}\mathbb{1}=\dim V$ , the identity's character is the dimension, and the rest fingerprint the representation up to equivalence. Representations combine in two ways: stacking them block-diagonally gives the direct sum; acting on tensor-product states gives the tensor product, the source of Clebsch–Gordan physics.

(D^{(1)}\oplus D^{(2)})(g)=\begin{pmatrix}D^{(1)}(g)&0\\0&D^{(2)}(g)\end{pmatrix},\qquad (D^{(1)}\otimes D^{(2)})(g)=D^{(1)}(g)\otimes D^{(2)}(g) \qquad (6)

where: $\oplus$ acts on $V_1\oplus V_2$ , $\otimes$ on $V_1\otimes V_2$ . Reading it: characters add under $\oplus$ and multiply under $\otimes$ , since $\operatorname{Tr}(A\otimes B)=\operatorname{Tr}A\,\operatorname{Tr}B$ — the arithmetic behind Clebsch–Gordan physics. Every finite group carries a canonical representation on the space of its own elements. Its decomposition is the completeness statement of the theory, closing into a rigid counting identity.

D_{\mathrm{reg}}=\bigoplus_\alpha d_\alpha\,D^{(\alpha)},\qquad d_\alpha=\dim D^{(\alpha)},\qquad \sum_\alpha d_\alpha^{\,2}=|G| \qquad (7)

where: $D^{(\alpha)}$ — the inequivalent irreducibles; $d_\alpha$ — the multiplicity of $D^{(\alpha)}$ , equal to its own dimension; $|G|$ — the group order. Reading it: the sum rule is a hard constraint — for $S_3$ it forces dimensions $1,1,2$ , the only positive integers with $1^2+1^2+2^2=6$ (Derivation C). The companion multiplicity formula $n_\alpha=\frac{1}{|G|}\sum_g\chi^{(\alpha)}(g)^*\chi(g)$ — one inner product of characters, developed in the characters node — turns "reducing a representation" into arithmetic.

Derivation AEvery representation of a finite group is unitarizable

The goal is to prove that unitarity — the property that makes representations the language of quantum symmetry — is not an extra assumption but a theorem for finite groups. We start from any Hermitian inner product on $V$ , average it over the group to build a $G$ -invariant one, and show the representation is unitary with respect to that average. This is Weyl's "unitarian trick", and it is exactly why Hamiltonian degeneracies always carry unitary multiplets. The one fact needed beyond linear algebra is the rearrangement theorem: left-multiplying the group by a fixed element permutes it.

Step 1. Pick any Hermitian positive-definite inner product $\langle\,\cdot\,,\cdot\,\rangle$ on $V$ and define its group average $\langle u,v\rangle_G\equiv\frac{1}{|G|}\sum_{g\in G}\langle D(g)u,\,D(g)v\rangle$ . A finite sum of positive-definite forms is again positive-definite, so $\langle\,\cdot\,,\cdot\,\rangle_G$ is a legitimate inner product. Why: averaging is the discrete analogue of integrating over the group with its invariant measure — the one construction that symmetrizes an arbitrary geometry into a $G$ -respecting one.

Step 2. Test invariance by acting with an arbitrary $D(h)$ on both slots: $\langle D(h)u,D(h)v\rangle_G=\frac{1}{|G|}\sum_{g\in G}\langle D(g)D(h)u,\,D(g)D(h)v\rangle$ , then use $D(g)D(h)=D(gh)$ to fold the two operators into one. Why: this is the only place the defining property (1) enters — without it the average would not be invariant.

Step 3. Substitute $g'=gh$ . As $g$ runs once over $G$ , so does $g'$ (rearrangement theorem), hence $\frac{1}{|G|}\sum_{g'\in G}\langle D(g')u,D(g')v\rangle=\langle u,v\rangle_G$ . Therefore $\langle D(h)u,D(h)v\rangle_G=\langle u,v\rangle_G$ for every $h$ . Why: relabelling a full sum over the group changes nothing, which is precisely what "invariant" means — the average cannot tell $h$ apart from the identity.

Step 4. Invariance of the inner product is unitarity: choosing an orthonormal basis for $\langle\,\cdot\,,\cdot\,\rangle_G$ makes every $D(h)$ satisfy $D(h)^\dagger D(h)=\mathbb{1}$ , so the original representation is equivalent to a unitary one. Why: it is not that the matrices you started with were unitary — a change of basis always exists making them so, which is all quantum mechanics needs. *Verified: the standard finite-group averaging argument (Tinkham, Hamermesh); for compact Lie groups the sum becomes the invariant Haar integral, and for non-compact groups it fails, so finite-dimensional unitary representations generally do not exist.*

Go deeper · Derivation B · 1 minThe matrix form reproduces the group multiplication table

Derivation BThe matrix form reproduces the group multiplication table

The goal is to make concrete the claim that a representation turns the abstract Cayley table into literal matrix multiplication. We fix a basis of $V$ , read off each operator's matrix entries from its action on basis vectors, and apply two group elements in succession. The homomorphism condition (1) forces the resulting double sum to be exactly the row-by-column rule of matrix multiplication — so group associativity and matrix associativity become the same fact, licensing the phrase "matrix representation" that underpins every computation below.

Step 1. Fix a basis $\{e_i\}$ of $V$ and define the matrix elements by $D(g)\,e_j=\sum_i D(g)_{ij}\,e_i$ . The numbers $D(g)_{ij}$ are the entries of the matrix representing $g$ . Why: a linear operator is completely determined by its action on a basis, so this equation loses no information — it merely names the coordinates.

Step 2. Apply $D(g_2)$ then $D(g_1)$ to a basis vector: $D(g_1)D(g_2)\,e_k=D(g_1)\sum_j D(g_2)_{jk}e_j=\sum_{i,j}D(g_1)_{ij}D(g_2)_{jk}\,e_i$ . Why: composing operators acts with the inner one first, and expanding in the basis at each stage is exactly the summation that defines a matrix product.

Step 3. Impose the homomorphism law and read off components: $D(g_1g_2)_{ik}=\sum_j D(g_1)_{ij}\,D(g_2)_{jk}$ . The right side is the $(i,k)$ entry of the matrix product, so the multiplication table is reproduced entry by entry. Why: the abstract product " $g_1$ times $g_2$ " is now a computation in numbers, consistent with associativity because matrix multiplication is associative. *Verified: the standard correspondence in Arfken & Weber; Figure 1's $C_3$ square is a numerical instance, $D(r)D(r)=D(r^2)$ checked to the entries $-0.50,\pm0.87$ .*

Go deeper · Derivation C · 1 minThe dimension sum rule from the regular representation

Derivation CThe dimension sum rule from the regular representation

The goal is to derive the rigid counting law $\sum_\alpha d_\alpha^2=|G|$ that constrains the irreducible dimensions of every finite group — the identity that forces $S_3$ to have dimensions $1,1,2$ . The route is the regular representation: let the group act on the $|G|$ -dimensional space of its own elements, compute its character by counting fixed basis vectors, and read off each irreducible's multiplicity via the multiplicity formula. Two ways of evaluating the same dimension then collide into the sum rule.

Step 1. Build the regular representation on the space with basis $\{|h\rangle:h\in G\}$ by left multiplication, $D_{\mathrm{reg}}(g)\,|h\rangle=|gh\rangle$ . Its character is $\chi_{\mathrm{reg}}(g)=$ (number of basis vectors fixed by $g$ ). Only $g=e$ fixes any $|h\rangle$ , and it fixes all $|G|$ of them, so $\chi_{\mathrm{reg}}(e)=|G|$ and $\chi_{\mathrm{reg}}(g)=0$ for $g\neq e$ . Why: a permutation matrix has trace equal to its number of fixed points, and $gh=h$ has a solution only when $g=e$ — the regular representation is maximally "spread out".

Step 2. Feed this character into the multiplicity formula $n_\alpha=\frac{1}{|G|}\sum_{g}\chi^{(\alpha)}(g)^*\chi_{\mathrm{reg}}(g)=\frac{1}{|G|}\,\chi^{(\alpha)}(e)^*\,|G|=\chi^{(\alpha)}(e)=d_\alpha$ . Every irreducible appears exactly $d_\alpha$ times. Why: because $\chi_{\mathrm{reg}}$ vanishes off the identity, the entire class sum collapses to its single $g=e$ term, and $\chi^{(\alpha)}(e)$ is just the dimension $d_\alpha$ .

Step 3. Now count dimensions two ways. The total dimension of $D_{\mathrm{reg}}$ is $|G|$ ; the decomposition $D_{\mathrm{reg}}=\bigoplus_\alpha d_\alpha D^{(\alpha)}$ has dimension $\sum_\alpha d_\alpha^2$ . Equating gives $\sum_\alpha d_\alpha^{\,2}=|G|$ . Why: dimension is additive over direct sums, so one integer computed two ways yields a Diophantine constraint that, with the count of conjugacy classes, often pins the dimensions uniquely. *Verified: for $S_3$ the three conjugacy classes force three irreducibles, and $\sum d_\alpha^2=6$ has the unique solution $1^2+1^2+2^2$ — the relation in Figure 2, checked in Worked Example 1.*

Limiting cases

Certain representations sit at the extremes of the faithful-to-trivial spectrum; every finite group possesses or admits each of them, and they anchor the general theory.

Case	Result	Meaning
Trivial representation	$D(g)=1\ \ \forall g\in G$	Every element maps to the identity on a one-dimensional space, carrying the totally symmetric states. Its kernel is all of $G$ (maximally unfaithful), yet it always appears — for example inside any product $D\otimes D^{*}$ .
Faithful representation	$D(g_1)=D(g_2)\iff g_1=g_2$	An injective homomorphism: the operators distinguish every element, so $G\cong D(G)$ and the kernel is trivial. Cayley's theorem guarantees one exists (the regular representation).
One-dimensional / abelian	$D(g)=e^{i\theta(g)}$ , all irreps $1$ -D	For an abelian group every irreducible is one-dimensional, since commuting operators diagonalize simultaneously; the sum rule then just counts $\|G\|=\sum 1^2$ .
Regular representation	$D_{\mathrm{reg}}(g)\,\|h\rangle=\|gh\rangle$	Acts on the $\|G\|$ -dimensional group-space by left multiplication; contains every irreducible $D^{(\alpha)}$ exactly $d_\alpha$ times, giving $\sum_\alpha d_\alpha^2=\|G\|$ .
Unitary representation	$D(g)^\dagger D(g)=\mathbb{1}$	Operators preserve an inner product. Guaranteed for finite and compact groups by averaging (Derivation A); the natural setting for quantum symmetry.

Common trapA representation is not just any assignment of matrices

The single most common error is to treat "representation" as a synonym for "a list of matrices labelled by group elements". It is not: the homomorphism law $D(g_1g_2)=D(g_1)D(g_2)$ is the entire content, and it must hold for every pair. Two traps lurk inside it. First, the order is preserved, not reversed — writing $D(g_1g_2)=D(g_2)D(g_1)$ defines an anti-homomorphism, which coincides with a representation only for abelian groups. Second, a representation need not be faithful: distinct elements may share an operator, in which case those mapping to $\mathbb{1}$ form the kernel (always a normal subgroup), and the faithful image is the quotient $G/\ker D$ .

Worked examples

1 · Building and checking the two-dimensional irreducible of S₃ (confidence builder)

The symmetric group $S_3$ (equivalently the symmetry group $D_3$ of an equilateral triangle) has order $6$ and three conjugacy classes: the identity, three transpositions (reflections), and two $3$ -cycles (rotations by $\pm120°$ ). Construct its two-dimensional irreducible representation explicitly, verify the homomorphism law, read off its character, and confirm the dimension sum rule.

Step 1. Realize the $3$ -cycle $r=(123)$ as a $120°$ rotation of the plane and a transposition $t$ as a reflection. In an orthonormal basis,

D(r)=\begin{pmatrix}\cos120° & -\sin120°\\ \sin120° & \cos120°\end{pmatrix}=\begin{pmatrix}-\tfrac12 & -\tfrac{\sqrt3}{2}\\[2pt] \tfrac{\sqrt3}{2} & -\tfrac12\end{pmatrix},\qquad D(t)=\begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix}.

These are the matrices of Figure 1 (to two decimals, $-0.50,\ \pm0.87$ ).

Step 2. Verify the homomorphism law on $r\cdot r=r^2$ . Multiplying,

D(r)D(r)=\begin{pmatrix}-\tfrac12 & -\tfrac{\sqrt3}{2}\\[2pt] \tfrac{\sqrt3}{2} & -\tfrac12\end{pmatrix}^{2}=\begin{pmatrix}-\tfrac12 & \tfrac{\sqrt3}{2}\\[2pt] -\tfrac{\sqrt3}{2} & -\tfrac12\end{pmatrix}=D(r^2),

the $240°$ rotation — exactly as required. One also checks the $S_3$ relations $D(t)^2=\mathbb{1}$ and $D(t)D(r)D(t)=D(r)^{-1}$ , so the assignment is a genuine representation.

Step 3. Read off the character, class by class. The identity gives $\chi(e)=\operatorname{Tr}\mathbb{1}=\mathbf{2}$ ; every reflection has trace $\chi(t)=1+(-1)=\mathbf{0}$ ; every rotation has trace $\chi(r)=2\cos120°=\mathbf{-1}$ . So the character vector on the three classes is $(2,\,0,\,-1)$ .

Step 4. Confirm irreducibility and the sum rule. The norm $\frac{1}{|G|}\sum_g|\chi(g)|^2=\frac{1}{6}(1\cdot2^2+3\cdot0^2+2\cdot(-1)^2)=\frac{1}{6}(4+0+2)=\mathbf{1}$ , and a character of norm $1$ marks an irreducible. With the trivial and sign representations (both one-dimensional), the dimensions $1,1,2$ satisfy $1^2+1^2+2^2=\mathbf{6}=|S_3|$ . The 2-D irreducible has character (2, 0, −1), norm 1, and completes $1^2+1^2+2^2=6$ .

2 · Reducing the permutation representation of S₃

Let $S_3$ act on $\mathbb{C}^3$ by permuting coordinates — the natural "permutation representation". Decompose it into irreducibles using characters alone, without ever diagonalizing a matrix.

Step 1. The character of a permutation representation counts fixed points. The identity fixes all $3$ coordinates; a transposition like $(12)$ fixes the single coordinate $3$ ; a $3$ -cycle fixes none. Hence the character on the classes $\{e\},\{t\},\{r\}$ is

\chi_{\mathrm{perm}}=(3,\,1,\,0).

Step 2. Apply the multiplicity formula $n_\alpha=\frac{1}{|G|}\sum_C|C|\,\chi^{(\alpha)}(C)^*\chi_{\mathrm{perm}}(C)$ , with class sizes $|e|=1,\ |t|=3,\ |r|=2$ and the $S_3$ character table $\chi^{\mathrm{triv}}=(1,1,1)$ , $\chi^{\mathrm{sgn}}=(1,-1,1)$ , $\chi^{\mathrm{std}}=(2,0,-1)$ .

Step 3. Evaluate each inner product:

n_{\mathrm{triv}}=\tfrac{1}{6}(1\!\cdot\!3\!\cdot\!1+3\!\cdot\!1\!\cdot\!1+2\!\cdot\!0\!\cdot\!1)=\tfrac{1}{6}(3+3+0)=\mathbf{1},

n_{\mathrm{sgn}}=\tfrac{1}{6}(1\!\cdot\!3\!\cdot\!1+3\!\cdot\!1\!\cdot\!(-1)+2\!\cdot\!0\!\cdot\!1)=\tfrac{1}{6}(3-3+0)=\mathbf{0},

n_{\mathrm{std}}=\tfrac{1}{6}(1\!\cdot\!3\!\cdot\!2+3\!\cdot\!1\!\cdot\!0+2\!\cdot\!0\!\cdot\!(-1))=\tfrac{1}{6}(6+0+0)=\mathbf{1}.

Step 4. Assemble the result: $D_{\mathrm{perm}}=D^{\mathrm{triv}}\oplus D^{\mathrm{std}}$ , of dimension $1+2=3$ — a familiar fact in disguise, since the symmetric combination $(1,1,1)$ is the trivial piece and the plane $x_1+x_2+x_3=0$ carries the standard irreducible. Check: $\chi^{\mathrm{triv}}+\chi^{\mathrm{std}}=(1,1,1)+(2,0,-1)=(3,1,0)$ . The permutation representation splits as trivial ⊕ standard, $(3,1,0)=(1,1,1)+(2,0,-1)$ .

Physics of the portrait

Representation theory was born not in physics but in a letter: in 1896 Dedekind asked Ferdinand Georg Frobenius why the "group determinant" of a non-abelian group refused to factor into linear pieces the way it did for abelian ones. Frobenius answered by inventing the character of a finite group — the apparatus of irreducibles, orthogonality, and decomposition that Wigner and Weyl would carry into quantum mechanics thirty years later.

Origins · 1 minFerdinand Georg Frobenius and how this was discovered

Ferdinand Georg Frobenius

1849–1917, Berlin · founder of the representation theory of finite groups

Frobenius was a titan of the Berlin school of algebra, a student of Weierstrass. His most consequential work grew out of correspondence with Dedekind in 1896 about the factorization of the "group determinant". To solve it, Frobenius introduced the characters of a finite group as sums attached to conjugacy classes, and within a year (1896–1897) had built the full theory of representations by linear substitutions — irreducibility, the orthogonality relations, and the decomposition of the regular representation that yields $\sum_\alpha d_\alpha^2=|G|$ . His student Issai Schur soon streamlined it with the lemmas that bear his name. For two decades this remained pure algebra; only with Wigner's 1927 analysis of atomic spectra and Weyl's synthesis with quantum mechanics did Frobenius's characters become a working tool of physics — the language in which every multiplet, selection rule, and quantum number is now written. Frobenius died in 1917, before that harvest.

History

1896

— Frobenius defines the character of a finite group, answering Dedekind's group-determinant question

1897

— Frobenius represents finite groups by linear substitutions, founding representation theory

1905

— Schur proves his lemmas, giving the theory its clean linear-algebra core

1927

— Wigner applies representations to atomic spectra, importing the theory into quantum mechanics

1928

— Weyl's Gruppentheorie und Quantenmechanik unifies group representations with the new physics

Going further

References. Stone & Goldbart, Mathematics for Physics, ch. on group representation theory · Arfken & Weber, Mathematical Methods for Physicists, 7th ed., ch. on group theory · Hassani, Mathematical Physics, chs. on group representations · Byron & Fuller, Mathematics of Classical and Quantum Physics, ch. on group theory · Tinkham, Group Theory and Quantum Mechanics · Hamermesh, Group Theory and Its Application to Physical Problems.

Provenance. The definition and consequences $D(e)=\mathbb{1}$ , $D(g^{-1})=D(g)^{-1}$ follow standard texts (Stone & Goldbart; Arfken & Weber; Hassani). A verifier should recompute: (1) the averaging proof of unitarizability via the rearrangement theorem; (2) the matrix-multiplication correspondence from the basis expansion; (3) the regular-representation character $\chi_{\mathrm{reg}}(e)=|G|$ , $\chi_{\mathrm{reg}}(g\neq e)=0$ and the sum rule $\sum_\alpha d_\alpha^2=|G|$ ; (4) the $S_3$ character table $(1,1,1),(1,-1,1),(2,0,-1)$ , the norm- $1$ check on the two-dimensional irreducible, and the decomposition $(3,1,0)=(1,1,1)+(2,0,-1)$ . The rotation entries $-0.50,\pm0.87$ are exact values of $\cos120°,\sin120°$ .

Prerequisites. Groups, Subgroups, and Cosets · Group Homomorphisms and Quotient Groups · Conjugacy Classes and Normal Subgroups. Leads to. Reducible and Irreducible Representations · Schur's Lemmas · Characters and Orthogonality Theorems · SU(2) and the Spin Double Cover.

Ask the tutor: Ask me to build the character table of any small group from scratch, or to show how a Hamiltonian's degeneracies fall out of its symmetry group's irreducible dimensions.

What is the difference between a representation and its character?

The representation $D$ is the full assignment of matrices; the character $\chi(g)=\operatorname{Tr}D(g)$ is just their traces, one number per element. Taking the trace throws away detail — yet for a finite group it throws away nothing that matters: two representations are equivalent if and only if they share the same character on every conjugacy class, so the character is a complete invariant. That is why physicists work almost entirely with character tables — a small array of integers holding all the classification content, free of arbitrary basis choices.

Why does unitarity fail for the Lorentz group but not for rotations?

The unitarizing trick averages an inner product over the group, and that average must converge. For a finite group it is a finite sum; for a compact group like $SO(3)$ or $SU(2)$ it is an integral over a finite total volume (the Haar measure). The Lorentz group is non-compact: boosts run off to infinity, so the invariant volume is infinite and the average diverges. The consequence is sharp — the Lorentz group has no faithful finite-dimensional unitary representations; the ones quantum field theory uses are infinite-dimensional, which is why relativistic quantum mechanics lives on fields rather than finite multiplets.

What exactly does "irreducible" mean, and why does it matter here?

A representation is reducible if a change of basis brings every $D(g)$ simultaneously into the same block-diagonal form — meaning a proper subspace of $V$ is mapped into itself by all the operators. Irreducible means no such invariant subspace exists. Irreducibles are the "atoms" of the theory; every representation of a finite or compact group is a direct sum of them. Physically they matter because a Hamiltonian's degenerate eigenspaces are irreducible under the symmetry group, so the irreducible dimensions are the multiplet sizes. Irreducibility is the subject of the very next article.

Can two different group elements be represented by the same matrix?

Yes — that is exactly what it means for a representation to be unfaithful. The set of elements sent to the identity operator is the kernel of $D$ , always a normal subgroup, and any two elements in the same coset of the kernel share a matrix. The trivial representation is the extreme: every element maps to the number $1$ . A representation is faithful precisely when its kernel is trivial, so that $G$ is isomorphic to its matrix image. Unfaithfulness is not a defect — the trivial and other low-dimensional representations are indispensable pieces in tensor-product decompositions and selection rules.

How do characters multiply and add, and why is that useful?

Under a direct sum the operators stack block-diagonally, so traces add: $\chi^{(1\oplus2)}=\chi^{(1)}+\chi^{(2)}$ . Under a tensor product the operators form a Kronecker product, and since $\operatorname{Tr}(A\otimes B)=\operatorname{Tr}A\,\operatorname{Tr}B$ , characters multiply: $\chi^{(1\otimes2)}(g)=\chi^{(1)}(g)\,\chi^{(2)}(g)$ . This tiny arithmetic is the engine of Clebsch–Gordan decomposition: multiply the two characters pointwise, then project with the multiplicity formula. Every "adding angular momenta" table and every selection rule that asks whether the trivial representation appears in a product is this multiply-then-project computation in disguise.

Why do physicists insist on unitary representations specifically?

Because quantum mechanics is built on inner products that must be preserved. A symmetry cannot change the probability $|\langle\phi|\psi\rangle|^2$ of measuring one state in another — Wigner's theorem forces every symmetry to act by a unitary (or antiunitary) operator, so the operators representing a symmetry group must be unitary. The happy fact, from Derivation A, is that for finite and compact groups this costs nothing: any representation is equivalent to a unitary one. For the non-compact space-time symmetries of relativity it costs everything — which is why quantum field theory works with infinite-dimensional representations of fields.