ISEGORIA / MATH ENCYCLOPEDIA
Representation theory: groups acting as matrices
A group is an abstract list of symmetries; a representation makes each symmetry a matrix, so that composing symmetries becomes multiplying matrices. Every representation of a finite group splits into irreducible pieces, and a single row of numbers per piece, its character, decides how. The dihedral groups as rotation and permutation matrices, character tables and the inner product that decomposes anything, and the vibrations of a symmetric ring, whose frequencies and degeneracies are dictated by its symmetry.
Before you begin: Group theory and linear algebra
Predict, manipulate, then check your reasoning against the example and question. Graphs illustrate the mathematics; they do not replace a proof.
1. Symmetries as matrices
The symmetries of a regular n-gon form the dihedral group Dₙ: n rotations rᵏ and n reflections srᵏ. Each one moves the plane linearly, so it is a 2 × 2 matrix, and the matrix of a composite is the product of the matrices: that is a representation, ρ(gh) = ρ(g)ρ(h). The same group also shuffles the n vertices, which gives a second representation by n × n permutation matrices. Representations of one group can look nothing alike, but each has a character, the trace of its matrices, and the trace of a permutation matrix simply counts the vertices that stay put. Choose an element, press Play to watch it act, and drag the vector to see the matrix at work.
Worked example. In \(D_4\), the quarter turn \(r\) has matrix \(\begin{pmatrix}0&-1\\1&0\end{pmatrix}\) and trace 0, the half turn \(r^2=-I\) has trace \(-2\), and every reflection has trace 0. On the four vertices, \(r\) and \(r^2\) fix nothing, the reflections \(s\) and \(sr^2\) through opposite corners fix two vertices each, and \(sr\), \(sr^3\) through edge midpoints fix none. So the permutation character is \((4,0,0,2,0)\) on the classes \(e,\ r^2,\ \{r,r^3\},\ \{s,sr^2\},\ \{sr,sr^3\}\), and the 2 × 2 character is \((2,-2,0,0,0)\).
Watch out. The reflection animation folds the plane through a third dimension to show which way it turns over; a reflection itself is instantaneous and has no continuous path to the identity inside the plane, since its determinant is \(-1\). The labels \(r^k\) and \(sr^k\) depend on a convention for \(s\): here \(s\) is the reflection in the \(x\)-axis, which passes through vertex 1, and \(sr^k\) means "rotate by \(r^k\), then reflect".
Why must two representations with different characters be genuinely different, and not just the same one written in other coordinates?
Changing coordinates replaces every matrix \(\rho(g)\) by \(P\rho(g)P^{-1}\) for one fixed invertible \(P\), and the trace is unchanged: \(\operatorname{tr}(PAP^{-1})=\operatorname{tr}A\). So equivalent representations have equal characters. The surprise, proved by the orthogonality relations of the next experiment, is the converse for finite groups over \(\mathbb C\): equal characters force equivalent representations. The characters, one number per conjugacy class, carry everything.
2. Character tables and how to decompose
A finite group has exactly as many irreducible representations as conjugacy classes, and their characters form a square table whose rows are orthonormal under the inner product ⟨χ, ψ⟩ = (1/|G|) Σ χ(g) ψ(g)*. That orthonormality is the whole algorithm: to find how often an irreducible χᵢ occurs in any representation, take the inner product of its character with row i. No invariant subspace has to be found. Pick a group and a representation; the bars show the multiplicities, and for the regular representation, the group acting on functions on itself, every irreducible appears as often as its dimension.
Worked example. For \(S_3\) permuting three points, \(\chi=(3,1,0)\) on the classes \(e\), transpositions (3 of them), 3-cycles (2). Then \(m_{\rm triv}=(3+3\cdot1+0)/6=1\), \(m_{\rm sign}=(3-3+0)/6=0\), \(m_{\rm std}=(3\cdot2+0+0)/6=1\): the permutation representation is the trivial line (the all-ones vector) plus the two-dimensional standard representation (vectors summing to zero). As a check, \(\langle\chi,\chi\rangle=(9+3+0)/6=2=1^2+1^2\). For \(A_4\), whose characters involve \(\omega=e^{2\pi i/3}\), the 3-dimensional irreducible tensored with itself gives \((9,1,0,0)\), which is the trivial representation plus \(\chi_\omega\) plus \(\chi_{\omega^2}\) plus two copies of the three-dimensional one.
Watch out. Everything here is over the complex numbers. Over the reals, an irreducible can fail to split further even though its complexification does: the quaternion group acting on \(\mathbb H=\mathbb R^4\) is irreducible over \(\mathbb R\), yet its character \((4,-4,0,0,0)\) is twice the two-dimensional complex irreducible. For infinite groups the sum becomes an integral over the group, and that works only for compact groups; the tables shown are complete for these four small groups.
Why does every irreducible representation appear in the regular representation exactly as many times as its dimension?
The regular representation permutes the basis vectors \(e_h\) by \(g:e_h\mapsto e_{gh}\), and a nontrivial \(g\) moves every one of them, so its character is \(|G|\) at the identity and \(0\) everywhere else. The inner product with \(\chi_i\) then keeps only the identity term: \(m_i=\frac{1}{|G|}\,|G|\,\overline{\chi_i(e)}=d_i\). Counting dimensions on both sides gives \(|G|=\sum_i m_id_i=\sum_i d_i^2\), the identity in the readout.
3. Symmetry sets the frequencies
Put n equal beads on a closed string and let each move along its spoke. The equations of motion commute with every symmetry of the ring, so by Schur’s lemma the normal modes can be chosen inside the irreducible representations of Dₙ. Those are Fourier waves around the ring, and a wave and its mirror image, q and n − q, belong to one two-dimensional irreducible, so they must share a frequency: the degeneracy is forced, not accidental. Make one bead heavier and only the mirror through it survives. Each pair then splits into a mode that is even under that mirror and one that is odd; the odd one has a node at the heavy bead, so it never feels the extra mass.
Worked example. For \(n=6\) the frequencies are \(2\omega_0|\sin(\pi q/6)|\): \(0\) for \(q=0\) (all beads shifted together, the trivial irreducible \(A_1\)), \(\omega_0\) twice (\(q=1,5\), the irreducible \(E_1\)), \(\sqrt3\,\omega_0\) twice (\(q=2,4\), \(E_2\)), and \(2\omega_0\) once (\(q=3\), neighbours in antiphase, a one-dimensional \(B\)). That is \(1+2+2+1=6\) modes. Make bead 1 two and a half times heavier: the sum of the squared frequencies drops from \(12\omega_0^2\) to \((10+0.8)\,\omega_0^2\), as the trace of \(M^{-1}K\) requires, while the two odd modes stay at exactly \(\omega_0\) and \(\sqrt3\,\omega_0\).
Watch out. The model is the simplest one with the right symmetry: small displacements, a string of tension \(F\) and spacing \(a\), each bead coupled only to its two neighbours. The \(q=0\) mode has zero frequency because nothing resists a uniform shift; in a real molecule that role is played by a rigid translation or rotation. In molecules the same reasoning, with the point group in place of \(D_n\), predicts how many vibrational frequencies appear and which are degenerate, before any force constant is known.
Why does the odd mode keep its frequency exactly, not just approximately, when bead 1 is made heavier?
Odd under the mirror through bead 1 means \(u_{-j}=-u_j\), and in particular \(u_0=-u_0=0\): bead 1 does not move. The equation of motion of bead 1 then reads \(m_0\cdot0=(F/a)(u_1+u_{-1})=0\), which holds whatever \(m_0\) is, and every other bead obeys the same equation as before. So the old odd eigenvector, with its old frequency, is still an exact solution. In the language of representations, the perturbation commutes with the mirror, so it cannot mix the odd subspace with the even one, and on the odd subspace it is zero.