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Suppose finite-dimensional continuous group representations separate points of a compact Lie group. Choose an identity neighborhood $U$ containing no nontrivial subgroup, using Lie groups have no small subgroups. For each $g\notin U$, choose a representation whose matrix at $g$ is not the identity. The corresponding open nonidentity sets cover the compact complement of $U$, so finitely many representations suffice there. Their direct sum has kernel inside $U$ and therefore trivial kernel. Unitarization of a compact-group representation makes it a finite-dimensional faithful unitary representation.

For a compact Lie group $G$, the complex representation ring of a compact group $R(G)$ is the Grothendieck group of finite-dimensional continuous complex group representations, with the relations $[M\oplus N]=[M]+[N]$. Multiplication is $[M][N]=[M\otimes N]$, and the unit is the trivial one-dimensional representation. The ring of class functions iswith pointwise addition and multiplication. The character of $M$ is ${\chi }_M(g)=\mathrm{tr}{\rho }_M(g)$. The trace identities for direct sums and tensor products makea well-defined ring homomorphism $\chi :R(G)\to c\ell (G)$.

Normalize Haar measure by ${\int }_{G}dg=1$. Averaging a positive definite Hermitian inner product giveswhich is positive definite and $G$-invariant. This proves unitarization of a compact-group representation. In a unitary representation the orthogonal complement of an invariant subspace is invariant. Induction on dimension therefore gives complete reducibility of compact-group representations. Hence every element of $R(G)$ is a finite integral linear combination of irreducible classes.

We establish the needed character orthogonality for compact groups carefully. For irreducible unitary representations $V,W$, let $G$ act on ${\mathrm{Hom}}_{\mathbb{C}}(W,V)$ byIts average $P={\int }_G(g\cdot )dg$ is a projection onto ${\mathrm{Hom}}_G(W,V)$: averaging makes every image invariant, and it fixes every intertwiner. The trace of this action is ${\chi }_V(g)\overline{{\chi }_W(g)}$, since the inverse of a unitary matrix has conjugate trace. Taking the trace of the projection givesFor completeness, Schur's lemma follows here from invariance of the kernel and image of an intertwiner. A nonzero intertwiner between irreducibles is an isomorphism. An endomorphism of an irreducible complex representation has an eigenvalue $\lambda$; the noninvertible intertwiner $A-\lambda I$ must vanish. Thus the last dimension is one for isomorphic irreducibles and zero for inequivalent ones.

If a virtual character ${\sum }_V{n}_V{\chi }_V$ vanishes, integrating it against $\overline{{\chi }_W}$ gives $n_W=0$ for every irreducible $W$. The irreducible multiplicities therefore distinguish the classes, andThis also proves that two finite-dimensional representations with the same character are isomorphic.

For SU(2), the irreducibles are $V_k={\mathrm{Sym}}^k({\mathbb{C}}^2)$, $k\ge 0$, of dimension $k+1$. One can see this classification through the usual ${\mathfrak{sl}}_2$ operators $H,E,F$: a highest-weight vector $v$ has $Hv=kv$, $Ev=0$, and the relations $[H,E]=2E$, $[H,F]=-2F$, $[E,F]=H$ give, by induction,The torus weights are integers. If $F^{r}v\ne 0$ but $F^{r+1}v=0$, the identity at $j=r+1$ forces $k=r\ge 0$. The vectors $v,Fv,\dots ,F^{k}v$ span an invariant irreducible module, which in an irreducible representation is the whole space. This is the symmetric-power module. Conversely its successive monomial weight vectors are linked by $E,F$ with nonzero coefficients, proving its irreducibility. Thus the classification of finite-dimensional representations of SU2 gives every irreducible, rather than just a list of examples.

On the maximal torus $T=\{\mathrm{diag}(z,z^{-1}):|z|=1\}$, their characters areWith $x={\chi }_1=z+z^{-1}$, multiplication gives ${\chi }_{k+1}=x{\chi }_k-{\chi }_{k-1}$, starting with ${\chi }_0=1$. Hence each ${\chi }_k$ is a monic integral polynomial of degree $k$ in $x$. For example ${\chi }_2=x^2-1$ and ${\chi }_3=x^3-2x$. They form a basis over $\mathbb{Z}$, soThe character map realizes this ring as the finite integral polynomials in $z+z^{-1}$, interpreted as continuous class functions. Its image is not the whole infinite-dimensional ring of continuous class functions; injectivity is the assertion being proved.

A Non-Abelian gauge theory promotes an internal group representation symmetry to a spacetime-dependent transformation and introduces a gauge field so that differentiation transforms covariantly. The symmetry is a gauge redundancy: physically meaningful quantities must be invariant under local changes of field representative. We work in four-dimensional Minkowski spacetime with signature $(+---)$, and use skew-Hermitian matrix generators throughout.

For the compact Simple Lie group $G$, its real Lie algebra has a negative-definite Killing form. Hence $B=-\kappa$ is a positive invariant bilinear form on a Lie algebra. Choose a $B$-orthonormal basis $T_a$, $a=1,\dots ,\dim G$, with $[T_a,T_b]=f_{ab}{}^c{T}_c$. Invariance makes $f_{abc}=B([T_a,T_b],T_c)$ fully antisymmetric. This construction uses the Adjoint representation and works independently of any chosen scalar representation.

Let ${\mathcal{A}}_{\mu }={\mathcal{A}}_{\mu }^a{T}_a$ be a Lie-algebra-valued gauge potential, with the coupling absorbed into this connection. If $R$ is the given finite-dimensional irreducible representation of $G$, its Lie algebra representation $\rho =dR$ definesFor a finite local transformation $U(x)\in G$, require ${\Phi }^{\prime }=R(U)\Phi$ and $D_{\mu }^{\prime }{\Phi }^{\prime }=R(U)D_{\mu }\Phi$. Differentiating the first equation forces the gauge-field transformation lawThus the gauge potential transforms inhomogeneously; an ordinary derivative alone would leave an uncancelled derivative of $U$.

The commutator of gauge covariant derivatives determines the gauge field strength:The Lie algebra representation preserves the bracket, so this formula holds for any $R$, including a trivial scalar representation. The gauge field strength itself is defined in the gauge algebra. The transformation of the connection impliesIt is therefore covariant, rather than invariant as a Lie-algebra-valued quantity.

For infinitesimal $U=1+ϵ+O({ϵ}^2)$, the same equations give the infinitesimal gauge transformation with an anti-Hermitian connection:Substitution into $\delta (D_{\mu }\Phi )$ cancels the terms containing ${\partial }_{\mu }ϵ$ and gives $\delta (D_{\mu }\Phi )=\rho (ϵ)D_{\mu }\Phi$, the gauge covariance of a scalar covariant derivative.

The Yang-Mills theory has Lagrangian densityThis is the Killing-form Yang-Mills Lagrangian. Its sign uses a positive internal form $B$ and the stated spacetime signature. In a $B$-orthonormal basis it is $-{\mathcal{F}}_{\mu \nu }^a{\mathcal{F}}^{a\mu \nu }/(4g_{\mathrm{YM}}^2)$. Gauge invariance follows because $B$ is invariant under the adjoint group action. Infinitesimally, the variation is proportional toAfter rescaling ${\mathcal{A}}_{\mu }=g_{\mathrm{YM}}{A}_{\mu }$, the gauge coupling appears in $D_{\mu }={\partial }_{\mu }+g_{\mathrm{YM}}\rho (A_{\mu })$ and in the non-Abelian part of the canonically normalized field strength. Squaring that field strength gives cubic and quartic gauge-field self-interactions. An ordinary local quadratic mass term in the gauge potential is not invariant because its transformation contains ${\partial }_{\mu }ϵ$.

For the scalar field kinetic term, compactness supplies a positive invariant Hermitian form $h$ on its representation space. One can obtain it by averaging any positive form with normalized Haar measure, the unitarization of a compact-group representation. Then $R(U)$ is unitary for $h$, and $\rho (ϵ)$ is anti-Hermitian. The Yang-Mills theory coupled to an arbitrary scalar representation hasHere $V$ is any real gauge-invariant scalar potential, so $V(R(U)\Phi )=V(\Phi )$. For example,is available for every unitary representation, with $\lambda \ge 0$ giving a bounded quartic contribution. Other invariant interactions may exist for particular representations; the potential is not required to depend only on $h(\Phi ,\Phi )$.

The scalar kinetic term is gauge-invariant because both covariant derivatives transform by the same unitary $R(U)$. Infinitesimally its variation vanishes byThe potential is invariant by its defining condition. For a real irreducible representation, use the averaged positive symmetric form instead and write the kinetic term as $\frac{1}{2}h(D_{\mu }\Phi ,D^{\mu }\Phi )$; all transformation and invariance arguments remain valid. Expansion of the covariant kinetic term produces both the linear gauge-scalar interaction and the quadratic term needed by local gauge invariance.

The invariant inner products, gauge covariant derivatives and gauge field strength together construct the full theory for every compact Simple Lie group and every supplied finite-dimensional irreducible representation of the scalar fields. Both finite and infinitesimal transformation laws have been included, and every sign follows the same anti-Hermitian convention.

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.