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Let $G$ be a simple Lie group, let $T^a$ be Hermitian matrices for a finite-dimensional unitary representation $R$, and normalizeA matter field $\psi$ transforms locally as ${\psi }^{\prime }(x)=U(x)\psi (x)$, where $U(x)=e^{ig{ϵ}^a(x)T^a}$. An ordinary derivative of $\psi$ does not transform covariantly because it differentiates $U$. Introduce a gauge field $A_{\mu }=A_{\mu }^a{T}^a$ and the gauge covariant derivativeDemanding $D_{\mu }^{\prime }{\psi }^{\prime }=U{D}_{\mu }\psi$ determines the Yang-Mills gauge transformationTo first order in $ϵ$,

The gauge field strength is defined by $[D_{\mu },D_{\nu }]=-ig{F}_{\mu \nu }$:or, in components,Covariance of the commutator givesThe commutator term distinguishes Yang-Mills theory from an Abelian gauge theory and produces cubic and quartic gauge-boson interactions.

For a Dirac field of mass $m$ in $R$, the Lagrangian isEquivalently, the gauge term is proportional to $-\mathrm{tr}(F_{\mu \nu }{F}^{\mu \nu })$. The cyclic property of the trace and $F_{\mu \nu }^{\prime }=U{F}_{\mu \nu }{U}^{-1}$ make it invariant. Unitarity gives ${\bar{\psi }}^{\prime }=\bar{\psi }{U}^{-1}$, while $D_{\mu }^{\prime }{\psi }^{\prime }=U{D}_{\mu }\psi$, so both the matter kinetic term and mass term are invariant. A complex scalar $\varphi$ in a unitary representation may instead be coupled throughprovided the scalar potential $V$ is $G$-invariant.

The simplicity assumption means that the Lie algebra $\mathfrak{g}$ is nonabelian and has no proper nonzero ideal of a Lie algebra. Its Adjoint representation is therefore irreducible, and every invariant symmetric bilinear form is proportional to the Killing form. Consequently the pure gauge kinetic term has one overall gauge coupling for a simple factor. The theory has no independent Abelian gauge direction; if the gauge algebra were a direct sum of simple and Abelian ideals, each factor could instead carry its own coupling. A simple group may still have a discrete center, but this does not add a gauge boson because gauge bosons are indexed by the Lie algebra.

Schur lemma says that an intertwining linear map between irreducible finite-dimensional complex representations is either zero or an isomorphism. In particular, every endomorphism of an irreducible complex representation is a scalar multiple of the identity.

A continuous representation of a topological group is a continuous homomorphismfor a finite-dimensional complex vector space $V$. It is a unitary representation if $V$ has a positive-definite Hermitian inner product for whichfor every $g\in G$ and $v,w\in V$.

For $G=S^1$, start with any positive-definite Hermitian form and average it using normalized Haar measure:Translation invariance makes this form $S^1$-invariant, and positivity is preserved, proving unitarity. Since $S^1$ is abelian, the operators $\rho (z)$ commute; since they are unitary, they are normal. Simultaneous diagonalization therefore decomposes $V$ into common one-dimensional eigenspaces. Thus every representation of the circle group is a direct sum of one-dimensional representations.

WriteThe group law and inverse areA calculation givesEvery element of the centre $Z$ occurs by taking, for example, $x^{\prime }=y=0$ and $y^{\prime }=1$. Hence the commutator subgroup is exactly $Z$. The image of a one-dimensional representation is abelian, so the one-dimensional representation kills the commutator subgroup and its kernel contains $Z$.

Now let $(\rho ,V)$ be a complex representation of $G/Z_0$. Its restriction to the central subgroupis a representation of the circle group. Decompose it into its distinct weight spaces:Because $Z/Z_0$ is central, every $\rho (g)$ commutes with its action and preserves every common eigenspace. Thus the $V_i$ are $G/Z_0$-subrepresentations, as asserted by the central circle weight-space decomposition.

Let $n_i=\dim V_i$. The mapis a one-dimensional representation of $G/Z_0$; after composition with $G\to G/Z_0$, its kernel contains $Z$. On $z\in Z/Z_0$ its value isso ${\theta }_i^{n_i}=1$. A continuous character of $S^1$ has the form $z\mapsto z^m$; the displayed identity forces $m=0$. Therefore every ${\theta }_i$ is trivial. Since the ${\theta }_i$ were distinct, $d=1$ and ${\theta }_1=1$.

It follows that every finite-dimensional complex representation of $G/Z_0$ kills the entire nontrivial central circle $Z/Z_0$. Its kernel is therefore nontrivial, so no such representation is faithful. This is the real Heisenberg quotient has no faithful finite-dimensional representation.

Strictly, the normalized functions form the unit sphere rather than a vector space; let $V=L^2(\mathbb{R})$ and restrict to normalized states when interpreting wavefunctions. The group law givesTranslation preserves Lebesgue measure and both exponential factors have unit modulus, so $D(g)$ preserves the inner product and is a unitary representation. This is the Schrödinger representation of the Heisenberg group: $q$ translates position, $r$ translates momentum, and $s$ contributes the physically irrelevant overall phase.

For a compact group, the matrix coefficients of its finite-dimensional irreducible unitary representations span a dense subspace of $C(G)$ and form a complete orthogonal decomposition of $L^2(G)$ with Haar measure. If $K$ is a closed subgroup and $W$ is a finite-dimensional unitary $K$-representation, the space of L2 sections of the associated homogeneous bundle has multiplicity $\dim {\mathrm{Hom}}_K(E{|}_K,W)$ for each irreducible $G$-representation $E$. This is the compact-group form of Frobenius reciprocity and determines homogeneous-bundle multiplicities from isotropy weights.

A strongly continuous unitary representation of the circle has integer spectral weights: $H$ is the orthogonal sum of spaces on which $U_z=z^m$. Positive energy means that these weights are bounded below, equivalently its self-adjoint generator has spectrum bounded below. Multiplication by an integer character shifts the lower bound to zero. A unitary operator with a nonzero rotation weight cannot exist in its operator algebra: repeated application of it or its inverse would move a nonzero vector to arbitrarily low energy.

Writing complex coordinates as $z=x+iy$ converts the unitary representation $U=A+iB$ into the indicated real matrix. It preserves the Euclidean metric because $U^{†}U=I$, and its real determinant is $|{\det }_{\mathbb{C}}U{|}^2=1$. Thus even $U(n)$ embeds in $SO(2n)$. The complexified defining real representation restricts as $\mathbf{n}\oplus \bar{\mathbf{n}}$, whereas the underlying real space is the realification of a complex representation of dimension $n$.

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 $R(G)\to \mathbb{Z}$.

A spin rotation acts on a quantum state through the unitary representation generated by its angular momentum operators. With $\hslash =1$, rotation by angle $\theta$ about axis $\hat{\mathbf{n}}$ is $e^{-i\theta \hat{\mathbf{n}}\cdot \mathbf{S}}$. Simultaneously rotating every site is a global symmetry of an isotropic Heisenberg model.

A strongly continuous unitary representation of the circle group splits as an orthogonal sum of character subspaces. The projections are $P_n={\int }_{\mathbb{T}}{z}^{-n}{U}_{z}dz$. Character orthogonality makes them mutually orthogonal, and Fejér sum shows their ranges span the represented Hilbert space. The self-adjoint generator has integer eigenvalues, with domain given by ${\sum }_n{n}^2\Vert P_n\xi {\Vert }^2<\infty$.

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.

A unitary representation of the fundamental group of a connected complex manifold gives a bundle $(\tilde{X}\times V)/{\pi }_{1}X$ with locally constant unitary transition matrices. Its induced flat unitary connection has $(0,1)$ part equal to its Dolbeault operator, so it is the Chern connection for the induced Hermitian metric. On a compact Riemann surface, zero curvature gives degree zero; the subbundle curvature inequality makes the bundle semistable.

For Hermitian $J_3$ and adjoint angular momentum ladder operators $J_{+}^{†}=J_{-}$ on a positive-definite inner-product space, a highest-weight vector of weight $j$ obeys $J_{+}|j,j⟩=0$. The commutation relations give $J_{+}(J_{-}{)}^{\ell }|j,j⟩=\ell (2j-\ell +1)(J_{-}{)}^{\ell -1}|j,j⟩$. Taking inner products forces $2j$ to be a nonnegative integer: otherwise the first negative factor would give a negative squared norm. The norm first vanishes at $\ell =2j+1$, so the generated unitary representation has dimension $2j+1$. Without positivity an abstract highest-weight module can remain infinite-dimensional even at an integral highest weight.

A unitary irreducible representation is both unitary and irreducible. Such representations are the building blocks in the harmonic analysis of groups and in quantum symmetry classifications.

Normalizing the tableau lines in a compatible real phase convention gives the orthonormal basis formAdmissible swaps have $|d|\ge 2$. Nonadmissible swaps act by $+1$ within a row and $-1$ within a column. The symmetric matrix is an orthogonal involution and is useful for making the unitary representation and its phases explicit.