Representation theory

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].

The representation ring is the Grothendieck group of finite-dimensional complex unitary representations of a compact group, with addition induced by direct sum and multiplication by tensor product. As an abelian group it is freely generated by the irreducible unitary representations. Taking dimension gives a ring homomorphism $R(G)\to \mathbb{Z}$.

For the block-rotation maximal torus in the special orthogonal group $SO(2m+1)$, the Weyl group acts by permutations and independent inversions of $z_i$. Its invariant Laurent polynomials are the symmetric polynomials in $x_i=z_i+z_i^{-1}$, hence the polynomial ring in their elementary symmetric polynomials $e_1,\dots ,e_m$. Restriction of the group representation ring of a compact group is injective because every group element is conjugate into the torus and characters distinguish virtual representations. For the defining complexified representation $V$, the exterior-power characters satisfy ${\sum }_k{\chi }_{{\Lambda }^{k}V}{t}^k=(1+t){\prod }_i(1+x_{i}t+t^2)$. For $k\le m$, its coefficient is $e_k$ plus an integral linear combination of $e_j$ for $j<k$. Induction puts every $e_k$ in the restriction image, proving equality with the invariant ring and giving polynomial generators $[{\Lambda }^{1}V],\dots ,[{\Lambda }^{m}V]$.

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

For a subrepresentation $W\subseteq V$, the quotient vector space $V/W$ inherits the action. For a Lie algebra representation, the formula is $x(v+W)=xv+W$, well-defined because $W$ is invariant.

A group representation is a group homomorphism from a group to the general linear group of a vector space. It realizes abstract group elements as invertible linear transformations.

The defining identities are $\rho (e)=I$ and $\rho (gh)=\rho (g)\rho (h)$, so an inverse element is represented by a matrix inverse. A choice of basis gives a matrix representation, but the underlying action does not depend on that choice. The action need not distinguish every group element: the trivial representation sends all of them to $I$.

An eight-dimensional representation. For SU(3), the adjoint octet is realized by traceless matrices, and decomposing $\bar{3}\otimes 3$ separates this traceless part from the scalar singlet.

A three-dimensional representation. The isospin-one irreducible representation of SU(2) is a triplet with third components $-1,0,+1$; the fundamental SU(3) quark representation is also a triplet but belongs to a different group.

A one-dimensional group representation. For connected semisimple groups such as SU(2) or SU(3), its continuous irreducible singlet is trivial, so all Lie-algebra generators act as zero. A singlet of one subgroup need not be a singlet of a larger group.

The map $(z,U)\mapsto zU$ identifies the unitary group with $(U(1)\times SU(n))/K$, where $K=\{({\zeta }^k,{\zeta }^{-k}{I}_n):\zeta =e^{2\pi i/n}\}$. If an irreducible SU(n) tensor transformation has $b$ boxes in its Young diagram, the central scalar $\zeta I_n$ acts as ${\zeta }^b$. Therefore, for integer $q$, $(z,U)\mapsto z^{q}R(U)$ is trivial on $K$ precisely when ${\zeta }^{q-b}=1$, proving the criterion. Adding a full $n$-box column changes $b$ by $n$ and leaves the condition unchanged. In particular a lift with trivial first-factor U(1) charge descends only if its central action is trivial; this does not prevent other lifts of the same special unitary group representation from descending.

For group representations $R:G\to GL(V)$ and $S:G\to GL(W)$, their direct sum acts on the direct sum of vector spaces by $(R\oplus S)(g)(v,w)=(R(g)v,S(g)w)$. Its character and Dynkin index, when defined, are the sums of those of the two summands.

A multiplet is a collection of components transformed into one another by a specified group representation. In particle it often means the states of one irreducible representation of a symmetry group; the group and representation must be specified, since dimension alone does not identify them.

For a complex group representation with matrices $R(g)$, its conjugate representation has matrices $\overline{R(g)}$ on the conjugate vector space. In a unitary representation it is equivalent to the dual representation, since $(R(g{)}^{-1}{)}^T=\overline{R(g)}$. On a diagonal maximal torus, its weights are the negatives of the original weights.

For a finite group $G$ represented unitarily by $U(g)$, the central projector onto the isotypic subspace of an irreducible representation $\lambda$ is $P_{\lambda }=(d_{\lambda }/|G|){\sum }_g{\chi }_{\lambda }(g{)}^{\ast }U(g)$, where $d_{\lambda }$ is its dimension. Character orthogonality implies that the operator is identity on copies of $\lambda$ and zero on other irreducibles. For a one-dimensional irreducible, applying it to an atomic orbital gives either zero or a symmetry-adapted linear combination of that symmetry, to be normalized afterward. A multidimensional central projector selects an entire isotypic component rather than one uniquely chosen component function.

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.

Two finite-dimensional group representations $D,E$ over the same field are equivalent if there is one invertible linear map $S$ intertwining all their group actions, equivalently $E(g)=S^{-1}D(g)S$ for every $g$. The same change of basis must work simultaneously for all group elements. The cyclic property of the trace implies that equivalent representations have identical characters: $\mathrm{tr}(S^{-1}D(g)S)=\mathrm{tr}D(g)$. Consequently a difference at a single group element proves inequivalence. This implication does not need the group to be finite, and the reverse implication is not asserted for arbitrary groups or fields.

For a complex finite-dimensional group representation, the displayed matrices give another group representation: taking inverses and taking the conjugate transpose each reverse multiplication order, so their composition preserves it. This operation is involutive. With the standard Hermitian inner product, an invariant vector subspace $W$ for $D^{♯}$ has orthogonal complement invariant for $D$, because $D(g{)}^{†}W=D^{♯}(g^{-1})W\subseteq W$. Thus $⟨w,D(g)v⟩=⟨D(g{)}^{†}w,v⟩=0$ for $v\in W^{\perp },w\in W$. It follows that $D^{♯}$ is an irreducible representation exactly when $D$ is. No finiteness assumption on the group or unitarity assumption on the group representation is needed.

A one-dimensional complex group representation acts on a one-dimensional complex vector space, and is equivalently a group homomorphism into the multiplicative group ${\mathbb{C}}^{\times }$. It is automatically an irreducible representation. Its character of a representation is its scalar value, a one-dimensional character. A change of basis cannot change those scalar values, so two such group representations are equivalent exactly when their homomorphisms agree. For a symmetric group $S_n$ with $n\ge 2$, all transpositions are conjugate, square to the identity and generate the group. Their common scalar image is therefore $1$ or $-1$, giving exactly the trivial representation and the sign representation.

The top exterior power of an $n$-dimensional group representation is one-dimensional, and $g$ acts on it by $\det R(g)$. It is trivial for the defining representation of $SU(n)$. This is why a full $n$-box column in an $SU(n)$ Young diagram can be removed without changing the representation.

A projective representation is a group homomorphism into the projective linear group of a vector space. Choosing invertible linear representatives makes multiplication hold up to nonzero scalar factors. A projective unitary representation chooses unitary operators as representatives, whose scalar factors have modulus one.

Let $A=\mathrm{diag}(1,-1)$ and $B=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. They generate the dihedral group of a square, realized by $\{\pm I,\pm A,\pm B,\pm AB\}$. Its image in the projective linear group has four elements, since opposite matrices become equal. The projective classes of $A,B$ commute because $AB=-BA$, and each squares to the identity. Sending $[A^a{B}^b]$ to $(a,b)$ gives the Klein four-group. This is an example in which a nontrivial scalar commutator disappears after passing to a projective representation.

An orthogonal representation is a real group representation preserving a positive-definite inner product, equivalently a group homomorphism into an orthogonal group. Every finite-dimensional continuous real representation of a compact group has this form after averaging an inner product over Haar measure, by orthogonalization of a compact-group representation.

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.

If a representation has one-dimensional endomorphism algebra, it cannot be a direct sum of two nonzero invariant subspaces: the two coordinate projections would be independent endomorphisms. This obstruction does not require either summand to be irreducible, and does not assert that scalar endomorphisms alone imply irreducibility.

A two-dimensional representation of an abelian group fitting into an extension of two distinct linear characters splits. Choose an element on which the linear characters differ. Its matrix has two distinct eigenvalues, and all commuting group operators preserve its two eigenlines. Those eigenlines supply the two invariant linear character summands. The argument does not apply when the two linear characters coincide.

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.

A smooth vector in any abstract group representation of a locally profinite group has an open stabilizer, equivalently is fixed by some compact-open subgroup. Smooth vectors form an invariant linear subspace: use intersections for addition and conjugation for translation. A smooth representation is one in which every vector has this property.

A smooth linear character is a one-dimensional smooth representation, equivalently a homomorphism whose kernel is open. On a local multiplicative group, each smooth linear character is trivial on some sufficiently deep principal-unit subgroup. Its inverse has the same kernel.

The unnormalized Jacquet module is the space of coinvariants of a unipotent subgroup $U$. If a subgroup $T$ normalizes $U$, it acts on this quotient. Every linear functional transforming by a linear character trivial on $U$ factors through it. Normalized Jacquet functors include an additional modulus twist; that convention must be specified.

The right action on compactly supported locally constant functions is a smooth representation. For a compact-open subgroup $K$, its fixed-vector space has basis the characteristic functions of cosets in $G/K$, since compact support on that discrete quotient is finite support. Consequently this representation is admissible exactly when $G$ is compact.

An admissible representation is a smooth representation for which every fixed-vector space $V^K$ for a compact-open subgroup has finite dimension. Admissibility can hold for an infinite-dimensional representation; it bounds each local fixed level, not their union.

For the upper-triangular subgroup of ${\mathrm{GL}}_n(F)$, conjugation by $\mathrm{diag}({\varpi }^{-(n-1)},\dots ,1)$ expands the upper-unipotent coordinates. For a decomposed compact-open subgroup $K=T_0{U}_0$, its conjugates eventually contain $K$, while their fixed-vector spaces are isomorphic. Finite dimension therefore forces those spaces to be equal. Inverse conjugation contracts any fixed unipotent element into $U_0$, proving that the unipotent radical acts trivially in every admissible representation of this triangular group.

The fixed-vector space is $V^K=\{v:\pi (k)v=v\text{ for all }k\in K\}$. Conjugation gives an isomorphism $V^K\to V^{gK{g}^{-1}}$, $v\mapsto \pi (g)v$. If $K\subseteq L$, then $V^L\subseteq V^K$.

A Galois representation is a continuous group representation of an absolute Galois group, usually on a vector space or lattice over a local field. Examples include the cyclotomic character and the representations attached to modular forms. The coefficient topology is part of the definition.

For an abelian variety $A$ and a prime number $\ell$ different from the characteristic, the inverse limit of its geometric ${\ell }^n$-torsion, with transition multiplication by $\ell$, is its Tate module. For an elliptic curve, it is free of rank two over the p-adic integers and satisfies $T_{\ell }(E)/{\ell }^n{T}_{\ell }(E)\simeq E[{\ell }^n]$. The absolute Galois group acts continuously and commutes with all base-defined endomorphisms.

Let $R=K\cap {\mathrm{End}}_k(E)$. If $\alpha \in R$ maps the Tate module into ${\ell }^n{T}_{\ell }(E)$, it kills $E[{\ell }^n]$ and factors through the quotient isogeny $[{\ell }^n]$. The resulting endomorphism is $\alpha /{\ell }^n$, belongs to $K$ and is defined over $k$, hence belongs to $R$. Approximation in $R\otimes {\mathbb{Z}}_{\ell }$ now identifies the displayed stabilizer. This does not assume that the integral Tate module is free over a possibly nonmaximal CM order.