research30 min studyLesson 91 in Mathematics

Representation theory and symmetry

Turn abstract composition into linear maps, then distinguish invariant pieces, faithful actions, and irreducibility.

On this page

What you’ll understand

  • Verify the homomorphism law for a concrete group representation.
  • Find invariant subspaces and explain why faithful does not mean irreducible.
  • Use a character without confusing it with a complete invariant in every setting.

A symmetry can be described without choosing coordinates: compose two rotations, reverse a permutation, or apply an identity operation. Representation theory asks how such an abstract rule can act linearly on a vector space. This translation makes eigenvectors, traces, and decomposition available, but the translation has obligations. Arbitrary matrices indexed by group elements are not automatically a representation. We will inspect a two-element symmetry completely before explaining which conclusions depend on the field, dimension, and group being studied.

Composition must survive the translation

A group G consists of elements with an associative composition, an identity, and inverses. A representation over a field sends each element to an invertible linear map on a vector space V, preserving multiplication and identity. If applying h and then g is written gh, the corresponding matrix product must use the same order. Coordinates only describe the linear maps; changing a basis conjugates every matrix by the same invertible matrix. Two such descriptions are equivalent representations, even if individual entries look different.

The kernel records group elements acting as the identity on every vector. A faithful representation has trivial kernel, so it distinguishes group elements. That property says nothing about whether the vector space splits into simpler invariant pieces. For the group with two elements, a reflection matrix acting on the plane is faithful because the nonidentity element changes some vectors. Nevertheless its fixed axis and reversed axis are separate one-dimensional invariant subspaces. Recovering the symmetry and decomposing its action are distinct questions with distinct tests.

ρ:G⟶GL(V),ρ(gh)=ρ(g)ρ(h),ker⁡ρ={g:ρ(g)=I}\rho:G\longrightarrow\mathrm{GL}(V),\qquad \rho(gh)=\rho(g)\rho(h),\qquad \ker\rho=\{g:\rho(g)=I\}
A group representation is a homomorphism into invertible linear maps. Faithfulness means that the displayed kernel contains only the group identity.↔ Scroll wide equations horizontally.

Invariant subspaces reveal simpler actions

A subspace W is invariant when every represented group element sends W into itself. The zero subspace and the whole space always qualify. A nonzero representation is irreducible when there are no other invariant subspaces. This is stronger than asking whether one chosen matrix has distinct eigenvalues, because the same subspace must work for the entire group. For commuting complex matrices, common invariant directions often help; for noncommuting actions, an eigenline of one element may be moved away by another element.

A decomposition into invariant summands lets each operation be written in blocks. For finite groups over the complex numbers, averaging an inner product over all group elements makes it invariant. The orthogonal complement of an invariant subspace is then invariant too, yielding complete reducibility in finite dimension. This argument divides by the group order and relies on a positive definite complex inner product. In a field whose characteristic divides that order, analogous splitting can fail. Likewise representations of infinite groups do not inherit this finite averaging proof merely because their matrices are small.

⟨v,w⟩G=1∣G∣∑g∈G⟨ρ(g)v,ρ(g)w⟩,V=W⊕W⊥G\langle v,w\rangle_G=\frac1{|G|}\sum_{g\in G}\langle\rho(g)v,\rho(g)w\rangle,\qquad V=W\oplus W^{\perp_G}
For a finite group acting on a finite-dimensional complex space, the averaged inner product is positive definite and invariant. An invariant W then has an invariant orthogonal complement.↔ Scroll wide equations horizontally.

Traces summarize, but assumptions govern what they determine

The character assigns to a group element the trace of its action. Trace is unchanged by simultaneous basis conjugation and is constant on conjugacy classes. It adds under direct sums and multiplies under tensor products. For finite-dimensional complex representations of finite groups, character theory determines multiplicities of irreducible summands. This powerful classification statement is not a universal assertion about every field, every infinite group, or arbitrary nonsemisimple algebra representations. Always keep the setting attached to the theorem rather than treating a useful numerical summary as complete by definition.

To study a new symmetry action, first write the group relations and check them on the proposed matrices. Then locate invariant subspaces and test whether a direct complement exists under the available hypotheses. Finally use traces to compare decompositions without performing a fresh basis change for every element. The vector lab below supplies only one rotating vector and its projections; it helps distinguish a vector from its coordinate description. It does not enumerate a group, identify irreducibles, or certify that a whole family of matrices respects group multiplication.

χV⊕W(g)=χV(g)+χW(g),χV⊗W(g)=χV(g)χW(g)\chi_{V\oplus W}(g)=\chi_V(g)+\chi_W(g),\qquad \chi_{V\otimes W}(g)=\chi_V(g)\chi_W(g)
These finite-dimensional trace identities hold for direct-sum and tensor-product representations. Classification by characters needs additional representation-theoretic hypotheses.↔ Scroll wide equations horizontally.

A faithful reflection that is reducible

Let G={e,s} with s2=es^2=e. On R2\mathbb R^2, set ρ(s)=(0110)\rho(s)=\begin{pmatrix}0&1\\1&0\end{pmatrix}. Verify and decompose the action.

  1. Multiplication gives ρ(s)2=I\rho(s)^2=I, so the defining relation holds; set ρ(e)=I. The two matrices differ, making the representation faithful.

  2. For u=(1,1), swapping coordinates gives u. For v=(1,−1), swapping coordinates gives −v. These nonzero independent vectors span invariant lines.

  3. In the basis (u,v), the matrix is diag(1,−1). The original two-dimensional action is the direct sum of the trivial and sign representations, so it is reducible.

  4. The character is χ(e)=2 and χ(s)=0, equal to the sum of the characters (1,1) and (1,−1). Trace zero does not mean the represented map is zero.

The result

The action faithfully distinguishes both group elements while splitting into two irreducible one-dimensional real actions.

Explore the model

Interactive lab

Interactive lab · vectors

A vector and its coordinate projections

Predict which component vanishes at 90°, then change angle and magnitude and observe the vector through its horizontal and vertical components. This supports the distinction between an action on a space and its coordinates. The lab contains one vector, not a group representation or an irreducible-decomposition algorithm.

A vector and its coordinate projectionsPlayback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Current readings: x component: 1.393; y component: 0.975; Magnitude: 1.7. All parameters are adjustable with the labeled controls below.−7.46−3.7303.737.46−3.5−1.7501.753.5xyvvₓ = 1.39
Vectorx / y projectionsTrace marker
x component
1.393
y component
0.975
Magnitude
1.7
t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forCoordinates describe vectors; a representation must additionally preserve every composition relation.

Check your understanding

Which statement about a faithful representation is justified?

Try it yourself

For the cyclic group generated by r with r4=er^4=e, consider the real plane rotation R=(0−110)R=\begin{pmatrix}0&-1\\1&0\end{pmatrix}. Is this real representation irreducible? What changes over C\mathbb C?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

A proper nonzero real subspace of the plane is a line, and an invariant line would supply a real eigenvector.

Compare with the solution

R has characteristic polynomial λ²+1, with no real eigenvalue, so it has no invariant real line and the two-dimensional real representation is irreducible. Over the complex numbers its eigenvalues i and −i give invariant complex eigenlines, so its complexification splits. Irreducibility depends on the scalar field.

Take it with you

  • A representation preserves composition; a list of matrices needs those relations checked.
  • Faithfulness concerns the kernel, whereas irreducibility concerns invariant subspaces.
  • Decomposition and character theorems must retain their field and group hypotheses.

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