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A group representation is a homomorphism $D:G\to \mathrm{GL}(V)$ into invertible linear maps. It is a faithful representation when it is injective, equivalently when its kernel contains only the identity. A faithful real two-dimensional representation of the cyclic group of order four isThese are rotations of the plane through $0,\pi /2,\pi ,3\pi /2$. Since $J^2=-I$ and $J^4=I$, their multiplication agrees with the group, and all four matrices are distinct.

For any invertible $S$, set $\tilde{D}(g)=SD(g)S^{-1}$. Thenso this also is a group representation. The character of a representation is ${\chi }_D(g)=\mathrm{tr}D(g)$. For the displayed quarter-turn representation, its values on $1,i,-1,-i$ are $2,0,-2,0$.

A group representation is a homomorphism $D:G\to \mathrm{GL}(V)$. It is a reducible representation if there is a nonzero proper invariant subspace of $V$ preserved by every $D(g)$, and an irreducible representation otherwise. With $S$ invertible,so changing basis gives another representation. Its character of a representation is $\chi (g)=\mathrm{tr}D(g)$, and cyclicity of the trace gives $\tilde{\chi }(g)=\mathrm{tr}(SD(g)S^{-1})=\chi (g)$.

A conjugacy class consists of $hg{h}^{-1}$ as $h$ varies. Since $(hg{h}^{-1}{)}^k=h{g}^k{h}^{-1}$, a power equals the identity exactly when $g^k$ does. Thus conjugation preserves element order, including infinite order.

For the square, let $r$ be rotation by $\pi /2$ and $s$ reflection in a horizontal symmetry axis. The dihedral group has the eight elementsThe conjugacy classes of a finite dihedral group specialize toThe two central rotations are singletons. Reflection conjugation reverses rotations, while conjugation by $r$ shifts a reflection exponent by two, giving the two reflection classes.

For finite groups over $\mathbb{C}$, character orthogonality and completeness make the irreducible characters a basis of the class functions. The latter space has one independent value per conjugacy class. Hence the square group has five inequivalent irreducible complex representations. The sum of squares of irreducible representation dimensions is eight, forcing four one-dimensional representations and one two-dimensional representation. The four scalar choices $r\mapsto \pm 1$, $s\mapsto \pm 1$ supply the former; the ordinary action on the plane supplies the latter.

A real two-dimensional group representation is a group homomorphism $D:G\to GL(2,\mathbb{R})$, so $D(AB)=D(A)D(B)$ and $D(I)=I$.

Substitute $\tilde{x}=\alpha x+\beta y$ and $\tilde{y}=\gamma x+\delta y$ into the quadratic expression. Its coefficients of $x^2$ and $y^2$ are ${\alpha }^2\hat{a}+{\gamma }^2\hat{b}$ and ${\beta }^2\hat{a}+{\delta }^2\hat{b}$. The cross term is zero for every matrix in this particular cyclic group, because each has one nonzero entry per row. ThusLet $J=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ generate the group. For $A=\pm I$, $M=I$; for $A=\pm J$, $M=S:=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. Since $S^2=I$, each $M$ is its own inverse and $D(J^k)=S^k$. The group law is respected: $S^{k+l}=S^k{S}^l$, and $S^4=I$ makes it independent of the representative modulo four. This proves that $D=M^{-1}$ is a group representation.

This quarter-turn action on diagonal quadratic forms hasso it is not faithful. Quadratic forms cannot distinguish a simultaneous sign reversal of both coordinates. The invariant coefficient directions $(1,1)$ and $(1,-1)$ correspond to the fixed form $x^2+y^2$ and the form $x^2-y^2$ whose sign changes under a quarter-turn.

The Klein four-group has multiplication tableSince it is abelian, every conjugacy class is a singleton: $\{1\},\{a\},\{b\},\{c\}$. An irreducible representation is a nonzero representation whose only invariant subspaces are zero and the whole representation space. For a finite group, the number of inequivalent complex irreducible representations equals the number of conjugacy classes, so there are four here.

A one-dimensional representation sends the involutions $a,b$ to signs, independently. All four choices areThey are distinct irreducible representations, exhausting the four possibilities. The character table of the Klein four-group isThe specified diagonal matrices obey $D(a{)}^2=D(b{)}^2=I$, commute, and satisfy $D(a)D(b)=D(c)$, so the group relations prove that $D$ is a group representation. Its coordinate axes are invariant lines carrying respectively ${\chi }_{01},{\chi }_{10},{\chi }_{11}$. ThereforeThe trivial representation ${\chi }_{00}$ is absent. If $H$ denotes the four-row character matrix above, direct row products give $H{H}^T=4I$. This verifies character orthogonality:The character of $D$ is $(3,-1,-1,-1)$, whose inner products with the four rows are $0,1,1,1$, independently confirming the decomposition. All these irreducibles are real as well: commuting real involutions can be simultaneously diagonalized into their real sign eigenspaces.

A group representation on a vector space $W$ is a group homomorphism $\rho :G\to GL(W)$: $\rho (gh)=\rho (g)\rho (h)$ and $\rho (e)=I$. An invariant subspace $U$ obeys $\rho (g)U\subseteq U$ for every $g$. A nonzero representation is irreducible if its only invariant subspaces are $0$ and $W$. It is a faithful representation if it is injective, equivalently if its group kernel is $\{e\}$. Its character is ${\chi }_{\rho }(g)=\mathrm{tr}\rho (g)$. For inequivalent complex irreducible representations of a finite group, character orthogonality is

Use right-to-left composition of permutations and assign the matrix labels consistently byThe resulting Cayley table, with row multiplying column, isConjugation relabels the entries of a cycle, preserving its length. Thus the conjugacy classes are

Define the permutation representation by $P_{\sigma }{e}_j=e_{\sigma (j)}$, or $(P_{\sigma }{)}_{ij}={\delta }_{i,\sigma (j)}$. Applying it to a basis vector proves $P_{\sigma }{P}_{\tau }=P_{\sigma \tau }$. The six assignments above produce exactly the listed matrices. They are distinct, and $P_{\sigma }=I$ forces $\sigma (j)=j$ for every $j$, so the representation is faithful.

Every permutation fixes $(1,1,1)$, hence fixes every vector $(c,c,c)$. Therefore $V_0$ is a one-dimensional invariant subspace carrying the trivial representation. Its orthogonal complement issince the inner product with $(c,c,c)$ is $c(v_1+v_2+v_3)$. Permuting coordinates preserves their sum, so this plane is invariant too. Every real vector has the unique decompositionThus the permutation representation is the direct sum of group representations on the line and the plane.

To find the eigenvectors of $A$, note that $A(v_1,v_2,v_3)=(v_3,v_1,v_2)$. With $\zeta =e^{2\pi i/3}$, its three eigenvalues and eigenspaces areThe only real eigenvalue is one, whose eigenline is $V_0$, so $A$ has no nonzero real eigenvector in $V_0^{\perp }$. A proper nonzero real invariant subspace of that two-dimensional plane would be a line, giving such an eigenvector. Hence the plane representation is irreducible over $\mathbb{R}$.

For the later complex dimension-count theorem one should also establish irreducibility over $\mathbb{C}$; real irreducibility alone would not imply it in general. In the complexified plane, the only candidate $A$-invariant lines are its $\zeta$ and ${\zeta }^2$ eigenlines. But $C$ exchanges them:Neither is invariant under the whole group, proving complex irreducibility as well. The required decomposition is thereforewhere $W$ is the standard representation of the symmetric group for three letters. Its character is obtained by subtracting one from the number of fixed coordinates: on the three classes it is $(2,-1,0)$.

Finally, the sign representation gives a second one-dimensional irreducible, with value $+1$ on $I,A,B$ and $-1$ on the transpositions. These are exactly the two one-dimensional possibilities: if $r=A,s=C$, a one-dimensional representation sends them to commuting nonzero scalars $v,u$. The relations $r^3=s^2=e$ and $srs=r^{-1}$ imply $v^3=u^2=1$ and $v=v^{-1}$. Hence $v=1$ and $u=\pm 1$, the trivial and sign choices. Together with the two-dimensional $W$, these giveThe sum of squares of irreducible representation dimensions leaves room for no other irreducible. Thus there are exactly two one-dimensional and one two-dimensional complex irreducible representations.

A group homomorphism $D:G_1\to G_2$ satisfies $D(gh)=D(g)D(h)$ for all $g,h\in G_1$. Its group kernel is

For the matrices, put $X=R(x)$, $Y=R(y)$ and $z=e^{2\pi i/3}$. Since $z^3=1$, $z\ne 1$ and $1+z+z^2=0$, direct matrix multiplication givesThese verify the listed matrices and the defining relations, with $R(e)=I_2$. To check every product explicitly through a single calculation, write elements in the unique form $x^{ϵ}{y}^j$, with $ϵ=0,1$ and $j=0,1,2$. The relations giveThe displayed matrices obey $Y^{j}X=X{Y}^{-j}$, so their product has exactly the same form:All exponents are reduced modulo two or three respectively. Thus this verifies all 36 products, and $R$ is a two-dimensional complex group representation. It is given to be an irreducible representation.

For a one-dimensional complex group representation $U$, let $u=U(x)$ and $v=U(y)$, which are nonzero complex numbers. The relations requireSince scalar multiplication commutes and $u\ne 0$, the last relation gives $v=v^2$, hence $v=1$. Then $u=\pm 1$. The choice $u=1$ is the trivial representation; the unique nontrivial choice is the sign representation:Its group kernel isA faithful representation is injective and has trivial kernel. Here the kernel has three elements, so $U$ is not faithful.

Both one-dimensional representations are irreducible representations, and they are inequivalent to each other and to the given two-dimensional one. The sum of squares of irreducible representation dimensions givesThese exhaust all complex irreducible representations. Their number equals the number of conjugacy classes, so there are three conjugacy classes. Explicitly they areThe last two sets consist of the 3-cycles and transpositions respectively.

For a group representation $D:G\to \mathrm{GL}(V)$, a faithful representation is injective, equivalently $D(g)=I$ only for $g=e$. Thus no distinct group elements are represented by the same linear map.

Two equivalent representations are related by one fixed invertible change of basis: $E(g)=S^{-1}D(g)S$ for every $g$. The same $S$ must work simultaneously for all group elements.

An irreducible representation acts on a nonzero vector space and has no invariant subspace other than $\{0\}$ and the whole space. An invariant subspace is one carried into itself by every representing matrix.

The character of a representation isThe cyclic property of the trace makes it invariant under equivalence and constant on conjugacy classes. Its value at the identity is the dimension of the representation.

A group representation assigns each $g\in G$ an invertible linear map $D(g)$ on a fixed vector space, preserving multiplication:Choosing a basis turns those maps into matrices. A faithful representation is injective, equivalently its kernel is $\{e\}$, so distinct elements have distinct matrices.

Two group representations on spaces of the same dimension are called equivalent representations if one fixed invertible change of basis conjugates every representing matrix: $E(g)=S^{-1}D(g)S$ for all $g$. Choosing a different $S$ separately for each element does not establish equivalence.

The character of a representation isThe cyclic property of the trace makes it invariant under a change of basis and constant on conjugacy classes. Thus different characters at even one group element prove inequivalence.

A vector subspace $W\subseteq {\mathbb{C}}^n$ is an invariant subspace for the group representation $D$ ifBecause $g^{-1}$ is also in the group and $D(g)$ is invertible, the inclusion is equivalently $D(g)W=W$ for every $g$.

For $n>0$, the representation is an irreducible representation when its only invariant vector subspaces areThus no nonzero proper subspace is preserved by all representing matrices.

Every $D_2(g)$ is invertible because both inversion and the conjugate transpose preserve invertibility. For $g,h\in G$, the group homomorphism property of $D_1$ and reversal of multiplication under the adjoint giveAlso $D_2(e)=I$. Hence $D_2$ is a group representation. It is the inverse-adjoint representation of $D_1$; the inverse and adjoint each reverse product order, so together they preserve it.

The symmetric group $S_n$ consists of all bijections of $\{1,\dots ,n\}$, with composition as its operation; its group order is $n!$.

An $m$-dimensional complex group representation of $G$ is a group homomorphism $\rho :G\to GL(V)$, where $V$ is an $m$-dimensional complex vector space. After a choice of basis, each $\rho (g)$ is an invertible $m\times m$ complex matrix, with $\rho (gh)=\rho (g)\rho (h)$.

An irreducible representation has $V\ne 0$ and no invariant vector subspace other than $\{0\}$ and $V$. Two group representations $\rho$ on $V$ and $\sigma$ on $W$ are isomorphic representations if there exists an invertible linear map $S:V\to W$ such that $S\rho (g)=\sigma (g)S$ for every $g$. They are inequivalent if no such map exists; in equal-dimensional matrix form this says no fixed change of basis conjugates every $\rho (g)$ to $\sigma (g)$.

Work over the complex numbers, the usual scalar field for the finite-group character theory used below. A group representation on a nonzero vector space $V$ is a homomorphism $D:G\to GL(V)$, so $D(gh)=D(g)D(h)$ and $D(e)=I$.

A faithful representation is injective: distinct group elements have distinct representing maps. Equivalently, its kernel is just $\{e\}$.

An irreducible representation has no nonzero proper invariant subspace: if $W\subseteq V$ and $D(g)W\subseteq W$ for every $g\in G$, then $W=\{0\}$ or $W=V$. Irreducibility concerns invariant subspaces, whereas faithfulness concerns whether any group element is represented by the identity.

The homomorphism law and $g{g}^{-1}=g^{-1}g=e$ giveThus $D(g^{-1})$ is both a left and right inverse of $D(g)$, provingIf only multiplicativity is initially stated, $D(e{)}^2=D(e)$ together with invertibility of representation matrices first implies $D(e)=I$. This explains why a group representation takes values in the general linear group rather than arbitrary square matrices.

For $G_1=⟨X⟩$, define the one-dimensional complex group representationExplicitly $W\mapsto 1$, $X\mapsto i$, $\hat{W}\mapsto -1$ and $\hat{X}\mapsto -i$. Since $i^4=1$, the map is well-defined on exponents modulo four and preserves products. Its four images are distinct, so it is also a faithful representation. The same construction applies to either of the other cyclic subgroups.

Here $a^2=-I$, $a^3=-a$ and $a^4=I$, so $K$ is a cyclic group of order four. Its multiplication table isWrite the specified image of $a$ as $A$. Direct matrix multiplication givesA group representation is a group homomorphism to a general linear group, so the images of all powers of $a$ are forced:All four images are invertible. Defining $\Phi (a^j)=A^j$ is well-defined modulo four because $A^4=I_2$, and $A^{j+\ell }=A^j{A}^{\ell }$ verifies every product in the table. These observations prove both existence and uniqueness of this representation with the prescribed image.

Distinguish the quaternion group generators $\mathbf{i},\mathbf{j},\mathbf{k}$ from the complex scalar $i$. Denote their proposed image matrices byDirect multiplication givesFor example $AB=\mathrm{diag}(i,-i)=C$ and $ABC=C^2=-I_2$. Sending $I$ to $I_2$ and each negative quaternion element to the negative of its positive image now reproduces every quaternion product. The matrices all have determinant one and are invertible, so this is a group representation in $GL(2,\mathbb{C})$.

The eight matrices $\pm I_2,\pm A,\pm B,\pm C$ are distinct. Therefore the group homomorphism is injective, and the representation is faithful; equivalently its kernel is $\{I\}$.

The character of a representation is the trace of its representing matrix. ConsequentlyThese are constant on the five conjugacy classes $\{I\}$, $\{-I\}$, $\{\pm \mathbf{i}\}$, $\{\pm \mathbf{j}\}$, $\{\pm \mathbf{k}\}$.

The complete list of subgroups isTo see completeness, Lagrange's theorem allows only orders $1,2,4,8$. The only element of order two is $-I$, which fixes the order-two subgroup. Every noncentral element has order four and generates the corresponding listed cyclic subgroup. Any order-four subgroup contains such an element, because there are only two central elements; it must equal the cyclic subgroup generated by that element. This accounts for all possibilities.

Definition and equations. A normal mode is a motion in which every component of a linear system has the same time dependence with fixed amplitude ratios and phases. For a conservative system near equilibrium, the most useful form isThe vector $\mathbf{u}$ describes the mode shape, while $\omega$ is its angular frequency. Linear combinations of modes describe arbitrary sufficiently small motions when the modes form a complete basis.

Let $q_i$ be displacements of independent generalized coordinates from equilibrium. Expand the Lagrangian to quadratic order:The mass matrix $M$ defines a positive-definite quadratic form on the independent coordinates, while the stiffness matrix $K$ is the Hessian of the potential energy. Substituting a mode gives the generalized eigenvalue problem for small oscillationsA practical calculation identifies the equilibrium, chooses independent coordinates, computes the quadratic kinetic and potential energies, solves for the eigenvalues and then obtains their eigenvectors. One can equivalently diagonalize the real symmetric matrix $M^{-1/2}K{M}^{-1/2}$. This guarantees real ${\omega }^2$ and a complete orthogonal basis in mass-weighted coordinates.

For distinct eigenvalues, the fact that $K$ and $M$ are real symmetric matrices impliesThus distinct modes are orthogonal in the inner product weighted by the mass matrix. Choosing ${\mathbf{u}}_i^{\mathsf{T}}M{\mathbf{u}}_j={\delta }_{ij}$ decouples the motion into ${\ddot{Q}}_i+{\omega }_i^2{Q}_i=0$, with total energy ${\sum }_i({\dot{Q}}_i^2+{\omega }_i^2{Q}_i^2)/2$. A positive-definite matrix $K$ gives positive eigenvalues and oscillatory stability. Zero modes can describe rigid translations or rotations without restoring force; a negative eigenvalue gives exponential instability rather than an oscillatory mode.

Symmetry. A symmetry preserving both quadratic energies maps modes into modes of the same frequency. Coordinates adapted to symmetry divide the eigenvalue problem into smaller independent blocks. For two identical subsystems, exchanging them gives symmetric and antisymmetric modes. The degeneracy of normal modes means that independent modes share one frequency. Higher-dimensional group representations of a symmetry can enforce such degeneracies; symmetry does not require every pair of modes to be degenerate, and accidental degeneracies can also occur. Within a degenerate eigenspace, orthogonal linear combinations are equally valid mode shapes.

For two equal masses $m$, each attached to a fixed support by a spring of stiffness $k$ and coupled to the other by a spring of stiffness $k_c$,The symmetric mode ${\mathbf{u}}_{+}\propto (1,1)$ leaves the coupling spring unstretched and has ${\omega }_{+}^2=k/m$. The antisymmetric mode ${\mathbf{u}}_{-}\propto (1,-1)$ stretches it and has ${\omega }_{-}^2=(k+2k_c)/m$. Superposing the two gives beats when their frequencies are close.

Continuous systems. The same idea replaces matrices by differential operators and mode vectors by functions satisfying boundary conditions. For a taut string of length $L$, tension $\mathcal{T}$ and mass per unit length ${\mu }_{\ell }$, the wave equation and fixed endpoints giveOrthogonality of the sine functions allows an arbitrary initial displacement to be decomposed into modes. Changing the endpoints changes the spectrum: boundary conditions are part of the physical mode problem.

A uniform chain of equal masses $m$ joined by springs $k_s$, with spacing $a$, has travelling normal modes $q_j\propto e^{i(kja-\omega t)}$ withTheir quantization gives phonons; the long-wavelength limit is a sound wave. Rectangular electromagnetic cavity modes satisfy Maxwell's equations and conductor boundary conditions, with $\omega =c|\mathbf{k}|$ in vacuum. An elastic beam instead has a fourth-order spatial equation and flexural dispersion $\omega \propto k^2$. These examples show how restoring forces and geometry determine the frequency spectrum.

Real systems. Small-amplitude motion is often approximately linear even when the full equations are nonlinear. Large amplitudes produce frequency shifts, mode coupling and transfer of energy between modes. Friction broadens resonances and makes free oscillations decay. A damping matrix proportional to $M$ or $K$ can retain the undamped mode shapes; general damping need not, and the appropriate eigenfrequencies and modes can be complex. External periodic forcing predominantly excites modes near resonance, with amplitude set by coupling to the forcing and dissipation. Imperfections split ideal degeneracies, while radiation or coupling to a surrounding fluid can remove energy from the chosen subsystem. Normal modes remain useful when linearization, the boundary conditions and the treatment of damping accurately describe the motion being measured.

For a group representation on $V$, a polynomial invariant is a polynomial $p$ satisfying $p(gv)=p(v)$ for every $g\in G$. Homogeneous degree-$d$ invariants form $({\mathrm{Sym}}^d{V}^{\ast }{)}^G$. Using the symmetric power matters: an alternating invariant tensor contracted with repeated copies of the same commuting vector can vanish.

A nonabelian simple group has no nonidentity conjugacy class of prime-power size. Coprime-degree irreducible characters vanish on such an element: the conjugacy-class sum makes the character-to-degree ratio an algebraic integer, and the Kronecker theorem on algebraic integers in the unit disk makes a nonzero ratio a root of unity, forcing a scalar in a faithful group representation. Column character orthogonality then makes $1/p$ an algebraic integer, impossible. This is the character-theoretic ingredient in Burnside's theorem.

A pseudoreal complex representation is isomorphic to its conjugate but has an invariant antilinear map squaring to minus the identity. Such group representations admit antisymmetric invariant bilinear forms and can support half-hypermultiplets in four-dimensional extended supersymmetry.