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Let $J$ be the plane rotation matrix through $\pi /2$, generating a cyclic group of order four. A quadratic form $a{x}^2+b{y}^2$ has its two coefficients exchanged by a quarter-turn and unchanged by a half-turn. Thus the pullback action on this two-dimensional space gives the displayed group representation: $D(J^k)=D(J{)}^k$, with $D(J{)}^2=I$. It is not a faithful representation, because its kernel of a group homomorphism is $\{I,J^2\}$. Its two invariant coefficient lines correspond to $x^2+y^2$ and $x^2-y^2$, on which a quarter-turn acts by $+1$ and $-1$, respectively. Restricting to this diagonal space is valid for these signed coordinate permutations; a general linear transformation can introduce an $xy$ term.

A complex group representation $\rho :G\to \mathrm{GL}(n,\mathbb{C})$ defines a real representation on ${\mathbb{R}}^{2n}$ by viewing each complex coordinate as its real and imaginary parts. In grouped coordinates its matrices are the realification of a complex matrix shown in the title. Direct block multiplication gives $\mathcal{R}(AB)=\mathcal{R}(A)\mathcal{R}(B)$ and $\mathcal{R}(I_n)=I_{2n}$, proving the representation property and preservation of invertibility.

Taking the trace gives its character ${\chi }_{\mathbb{R}}(g)=2\mathrm{Re}{\chi }_{\mathbb{C}}(g)$. After complexification of a real vector space, this realification splits as the original representation and its conjugate. Realification need not preserve irreducibility: the answer depends on whether the irreducible complex representation has real, complex or quaternionic representation type. The standard two-dimensional complex representation of the quaternion group gives a four-dimensional real example with character four at the identity, minus four at its central negative identity, and zero at the other six elements.

A representation over the rational numbers is a group representation on a finite-dimensional rational vector space. Choosing a basis gives rational matrices. It differs from a rational representation of an algebraic group, where rational refers to regular matrix-coefficient functions rather than the ground field. Scalar extension yields representations over every characteristic-zero field.

Representation theory describes abstract symmetries by linear maps on vector spaces. A group representation assigns an invertible linear map to each group element, respecting multiplication. The payoff is that a symmetry problem becomes a problem about matrices, invariant subspaces, and eigenvalues. Decomposing the action can reveal combinations of variables that a symmetric system cannot mix.

Take the cyclic group $C_3=\{e,g,g^2\}$, with $g^3=e$. On three real coordinates, let $g$ send $(a,b,c)$ to $(c,a,b)$:This permutation representation does more than relabel a list: it acts linearly on every possible assignment of values to the three positions. The vector $(1,1,1)$ stays fixed. The vector subspace $W=\{(a,b,c):a+b+c=0\}$ is also preserved, since a permutation does not change the sum.

Every vector separates uniquely into these two types:For example, $(4,1,1)=2(1,1,1)+(2,-1,-1)$. The first term belongs to a trivial representation; the second is a contrast with zero coordinate sum. This is a direct sum decomposition ${\mathbb{R}}^3=\mathbb{R}(1,1,1)\oplus W$. The symmetry preserves the mean and rotates the contrasts.

The rotation is literal after choosing the orthonormal basis $u=(2,-1,-1)/\sqrt{6}$, $v=(0,1,-1)/\sqrt{2}$ of $W$. In this basis,It is a rotation through $120$ degrees, and three applications return to the starting point. There is no nonzero invariant real line, because this rotation has no real eigenvector. Thus the two-dimensional action is an irreducible representation over the real numbers. Over the complex numbers it instead splits into two eigenlines with eigenvalues $e^{2\pi i/3}$ and $e^{-2\pi i/3}$. The coefficient field is part of an irreducibility claim.

Why would this decomposition matter for a physical calculation? Consider three equal masses with scalar displacements $x_i$ and the symmetric quadratic coupling energyThe equations of motion are $m\ddot{x}=-kLx$, whereis a graph Laplacian and $J$ is the all-ones matrix. The constant vector has eigenvalue zero: translating all coordinates equally costs no coupling energy. On $W$, $J$ vanishes and $L=3I$, so both independent contrast directions have angular frequency $\sqrt{3k/m}$. A three-coordinate coupled problem reduces to a free mean and two identical harmonic oscillators.

For initial displacement $x(0)=(4,1,1)$ and initial velocity zero, the exact solution is $x(t)=2(1,1,1)+\cos (\sqrt{3k/m}t)(2,-1,-1)$. The fixed mean and oscillating contrast can be read directly from the decomposition. More generally, a linear operator commuting with the group action carries its fixed-vector subspace into itself; with an invariant inner product, it preserves the corresponding orthogonal complement too.

For a finite group, averaging over the action constructs an invariant inner product over the real numbers or complex numbers. The resulting invariant complements explain Maschke's theorem: a finite-dimensional group representation decomposes into irreducible representations. Characters compress the action into traces and help identify those pieces. The assumptions matter; in modular representation theory the coefficient field can have characteristic dividing the group order, and such decompositions can fail. Chapter 4 of Pavel Etingof and coauthors' notes develops these mechanisms. ^[ref].

For a group representation of $G$ and a subgroup $H$, restriction keeps the vector space and lets only the elements of $H$ act. It is denoted ${\mathrm{Res}}_H^{G}V$.

In the compact picture of a principal-series group representation of the Lorentz spinor double cover, functions on SU(2) have a fixed right-torus character $e^{-in\vartheta }$. Expanding into spin matrix coefficients leaves one right weight in each allowed spin, giving the displayed multiplicity-one rotation decomposition. The boost action couples these spaces. The central element acts by $(-1{)}^n$; even $n$ yields integer rotation spins and a group representation of the connected Lorentz group.

For chosen inequivalent unitary irreducible representations $\rho ,\sigma$ of a finite group and a $d_{\rho }\times d_{\sigma }$ matrix $M$, the average ${\mathbb{E}}_x\rho (x)M\sigma (x{)}^{\ast }$ is zero when $\rho \ne \sigma$. For $\rho =\sigma$, it is $(\mathrm{tr}M/d_{\rho })I$. This follows from the Schur lemma, since the average intertwines the two group representations. Equivalent representations in different bases require the corresponding intertwiner; the identity-matrix formula assumes literally the same representative.

An unbroken subgroup $H$ must fix the vacuum expectation value $v$ of a scalar field in a group representation $V$. Thus a nonzero such $v$ exists exactly when $V^H\ne \{0\}$. For a compact group, complete reducibility of compact-group representations identifies this fixed vector subspace with the sum of the trivial representations in the restriction of a representation $V{|}_H$. Every $v\in V^H$ has $H\subseteq \mathrm{Stab}(v)$; equality requires a separate stabilizer calculation. The criterion neither determines the full unbroken group nor guarantees that a given scalar potential has its minimum at that direction.

A complex group representation $\pi :G\to \mathrm{GL}(V)$ is smooth when every vector is fixed by some compact-open subgroup. Equivalently each vector's stabilizer is open. Smoothness is local constancy of orbit maps with the vector space regarded discretely, not ordinary continuity into a finite-dimensional complex topology.

For the B2 root system with long simple root first, the four-dimensional spinor representation has highest weight $(0,1)$. Its symmetric square has highest weight $(0,2)$ and dimension ten, so it is the Adjoint representation. The exterior square has the four nonzero defining weights and two zero weights; it splits into the five-dimensional vector representation of highest weight $(1,0)$ and a trivial representation. Thus the tensor square decomposes as displayed. The invariant alternating bilinear form reflects the pseudoreal representation structure. The spinor representation is a group representation of the Spin group $\mathrm{Spin}(5)$, while its tensor square descends to $SO(5)$.

A subrepresentation is an invariant subspace of a group representation or Lie algebra representation, with the restricted action.

The regular group representation of a finite group over the complex numbers contains each irreducible representation with multiplicity equal to its dimension. Consequently the sum of the squared dimensions is the group order. Once a list of distinct irreducibles exhausts this sum, it is complete.

A supersymmetric vector multiplet contains a gauge boson and its gaugino partners, with additional scalar fields for extended supersymmetry. Its gauge group representation is the adjoint group representation.

The symmetric cube of a group representation $V$ is the symmetric part of $V^{\otimes 3}$. Over the complex numbers, its character isThe projection onto symmetric tensors averages the six permutations of the three factors. The identity permutation contributes $\chi (g{)}^3$, the three transpositions each contribute $\chi (g)\chi (g^2)$, and the two three-cycles each contribute $\chi (g^3)$. Cubic polynomial invariants are fixed vectors in ${\mathrm{Sym}}^3(V^{\ast })$, whose dimension follows by averaging this character for a finite group.

For the defining real representation of $SO(d)$, the symmetric square splits as a scalar trace plus symmetric trace-free tensors: ${\mathrm{Sym}}^2({\mathbb{R}}^d)=\mathbb{R}\oplus {\mathrm{Sym}}_0^2({\mathbb{R}}^d)$. The trace projection is $S\mapsto S-(\mathrm{tr}S)I/d$, so the second summand has dimension $d(d+1)/2-1$.

For $d\ge 3$ the trace-free summand is an irreducible real group representation. To see this, diagonalize a nonzero symmetric trace-free matrix in a nonzero invariant subspace. Two diagonal entries differ. Infinitesimal rotation in their plane yields a nonzero symmetric off-diagonal matrix; rotations carry it to every coordinate pair, and a rotation through $\pi /4$ produces a difference of two diagonal entries. These off-diagonal matrices and diagonal differences span all symmetric trace-free matrices, so the invariant subspace is the whole summand. For $d=24$, its dimension is 299, the third-level nontrivial multiplet of the lowest levels of a fully transverse ND bosonic string.

The diagonal element $\mathrm{diag}(m,m^{-1})$ conjugates $U(t)$ to $U(m^{2}t)=U(t{)}^{m^2}$. In a finite-dimensional unitary representation, conjugacy therefore makes its eigenvalue multiset invariant under every square power. Square-power rigidity of a finite nonzero spectrum makes the image of every upper unipotent the identity. Its kernel is a normal subgroup; elementary unipotent generators of SL2R then force the entire group representation to be trivial. This algebraic argument does not require continuity.

Using $I_3=\mathrm{diag}(1/2,-1/2,0)$ and flavour $Y=\mathrm{diag}(1/3,1/3,-2/3)$, the three coordinate vectors of the SU(3) group defining group representation have the displayed weights of a representation. The complex-conjugate group representation has their negatives. These coordinates use flavor hypercharge; another Cartan normalization rescales the vertical axis.

With gauge covariant derivative $D=d+A$, a Wilson loop is the matrix trace in a specified group representation of parallel transport around a closed curve. The holonomy of a connection changes by conjugation under gauge equivalence of principal connections, so its trace is gauge-invariant. It detects global connection data even when the local gauge curvature is zero.