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For finite-dimensional unitary representations of a compact group, average the action on $\mathrm{Hom}(W,V)$ against normalized Haar measure. This is the orthogonal projection onto the intertwining operators, and its trace is the displayed integral. The Schur lemma therefore makes distinct irreducible characters orthogonal and each irreducible character of norm one.

For a strongly continuous unitary representation of a compact group, this Bochner integral exists in the Hilbert-Schmidt norm and is a positive compact operator commuting with the representation. It is nonzero if $\xi \ne 0$, because $⟨K_{\xi }\xi ,\xi ⟩={\int }_G|⟨U_g\xi ,\xi ⟩{|}^{2}dg>0$. A nonzero eigenspace is finite-dimensional and invariant. Applying this on orthogonal complements yields an orthogonal direct sum of finite-dimensional irreducible representations, with no separability assumption on the represented Hilbert space.

The Peter-Weyl theorem makes finite-dimensional unitary representations separate elements of a compact group. Intersecting their closed kernels gives the identity subgroup. If the descending chain condition holds for closed subgroups, choose finitely many representations whose kernel intersection is already trivial: otherwise repeatedly exclude a surviving nonidentity element to obtain an infinite strict chain. Their direct sum is a faithful representation into a finite-dimensional unitary group. Compactness makes its image closed, and the closed-subgroup theorem gives the original group a compatible Lie group structure.

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.

A continuous symmetric approximate identity on a compact group gives a compact self-adjoint convolution operator $T$ on $L^2(G)$ which maps into $C(G)$ and commutes with left translations. Its nonzero eigenspaces are finite-dimensional invariant spaces of continuous functions, hence consist of matrix coefficients of finite-dimensional unitary representations. If $S_N$ projects onto increasing finite sums of these eigenspaces, $S_{N}Tf\to Tf$ in $L^2$, and the bound $\Vert Th{\Vert }_{\infty }\le \Vert k{\Vert }_2\Vert h{\Vert }_2$ gives $T{S}_{N}Tf\to T^{2}f$ uniformly. Choosing the approximate identity narrow makes $T^{2}f$ uniformly close to $f$. This supplies uniform approximation without confusing an $L^2$ limit with a uniform limit.

In a unitary representation of a finite group, if a cyclic vector $v$ satisfies $\rho (g)v=av$ for some $a\ne 1$, then there are no invariant vectors. Indeed the averaging orthogonal projection $P=|G{|}^{-1}{\sum }_h\rho (h)$ satisfies $P\rho (g)=P$, so $Pv=aPv=0$. All translates of $v$ also project to zero, and they span the representation.

An equivariant Fredholm operator between unitary representations of a compact group has finite-dimensional invariant kernel and cokernel. Their formal difference is its index in the representation ring of a compact group. If domain and target decompose into irreducibles with finite multiplicities $m_{\pi },m_{\pi }^{\prime }$, Schur's lemma makes each intertwining block a matrix on the multiplicity spaces. Its rank cancels between kernel and cokernel, giving ${\mathrm{ind}}_{G}T={\sum }_{\pi }(m_{\pi }-m_{\pi }^{\prime })[\pi ]$, with finitely many nonzero terms.

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 strongly continuous unitary representation, integration against $f\in L^1(G)$ defines an operator satisfying $\Vert \pi (f)\Vert \le \Vert f{\Vert }_1$. Fubini's theorem gives $\pi (f\ast g)=\pi (f)\pi (g)$, and the group-algebra involution gives $\pi (f^{\ast })=\pi (f{)}^{\ast }$. On an Abelian group, the operator-norm closure is a commutative C-star algebra. Its algebra characters produce continuous unitary group characters, allowing the Commutative Gelfand--Naimark theorem and Riesz-Markov-Kakutani representation theorem to turn a positive vector functional into a finite spectral measure.

A maximal torus is a connected compact abelian Lie subgroup maximal among such subgroups. Restricting a finite-dimensional unitary representation to it splits into one-dimensional weight characters. For special unitary group $\mathrm{SU}(2)$ it is the diagonal circle $\mathrm{diag}(z,z^{-1})$.

For a normalized state in a unitary representation of the ordinary four-dimensional Super-Poincaré algebra, define $M_{ab}=⟨\{Q_a,Q_b\}⟩/2$ in a real Hermitian charge basis. Then $\Vert u\cdot Q\psi {\Vert }^2=u^{T}Mu$, so its annihilator is precisely the kernel of a linear map represented by the positive matrix $M$. In a suitable real basis its eigenvalues are the two eigenvalues of $⟨P^0⟩I+⟨\mathbf{P}⟩\cdot \mathbf{\sigma }$, each repeated twice. A nonzero kernel therefore has dimension two at nonzero null momentum expectation or four at zero expectation. Complex non-Hermitian lowering operators and algebras with domain-wall extensions are not covered by this argument.

An irreducible spin representation is a nonzero Hilbert space with no proper nonzero closed invariant subspace under all three angular momentum operators. Equivalently, the associated unitary representation of SU(2) cannot be decomposed into two nonzero orthogonal invariant pieces. The irreducible possibilities are labelled by a spin $j\in \{0,\frac{1}{2},1,\frac{3}{2},\dots \}$ and have dimension $2j+1$.

Choose an orthonormal basis $|j,m_s⟩$, with $m_s=-j,-j+1,\dots ,j$. The angular momentum commutation relations imply $[{\mathbf{S}}^2,S_i]=0$, $[S_3,S_{\pm }]=\pm \hslash S_{\pm }$, $[S_{+},S_{-}]=2\hslash S_3$, and $S_{+}^{†}=S_{-}$. The complete action isIn particular, the spin raising operator annihilates $|j,j⟩$ and the spin lowering operator annihilates $|j,-j⟩$. The coefficient follows by taking the squared norm and using $S_{\mp }{S}_{\pm }={\mathbf{S}}^2-S_3^2\mp \hslash S_3$. Thus the ladder operators connect every member of the basis, while terminating at its two endpoints.

Use upper indices for the defining representation of SU(6) and lower indices for its dual representation. The Young tableaux below display tensor-slot labels; their shapes determine the Young diagrams. Full columns of height six are determinant representations and can be removed for SU(6).Here all indices run from one to six. The highest weight Dynkin labels follow from the differences of successive row lengths. In the same order the row lengths are $(1)$, $(1,1,1,1,1)$, $(2)$, $(1,1)$ and $(2,1,1,1,1)$.

The defining vector transforms as $v^a\mapsto U^a{}_b{v}^b$. For a unitary representation, the conjugate representation is equivalent to the dual representation, so $\overline{6}$ can be written as a covector with $w_a\mapsto w_b(U^{-1}{)}^b{}_a$. Its five-box column is equivalently a totally antisymmetric tensor $A^{a_1\cdots a_5}$, related by the invariant alternating tensor $w_a={ϵ}_{a{a}_1\cdots a_5}{A}^{a_1\cdots a_5}/5!$.

The symmetric tensor transforms as $S^{ab}\mapsto U^a{}_c{U}^b{}_d{S}^{cd}$, with $6(6+1)/2=21$ independent components. The antisymmetric tensor has the same transformation law but obeys $A^{ab}=-A^{ba}$, giving $6(6-1)/2=15$ independent components. For the Adjoint representation, $T\mapsto UT{U}^{-1}$; the trace constraint removes one component from the $36$ entries of a mixed tensor, leaving $35$. Its six-box Young diagram is obtained by converting the five-index column to a lower index and selecting the traceless mixed tensor.

Thus the requested dimensions, in the order above, are

First choose the same orthonormal basis for the left and right copies of the two-dimensional unitary representation $D$. Then the family representation is $F=D\oplus \epsilon$. Because the barred fermion transforms with $F(g{)}^{†}$ and the scalar field $H$ is a trivial representation, invariance of the Yukawa interaction requiresWrite $\lambda$ in a $2+1$ block form. The off-diagonal blocks would be intertwiners between inequivalent irreducible representations $D$ and $\epsilon$, so the Schur lemma makes them zero. The doublet block must be a scalar matrix, while the singlet block is arbitrary. ThereforeEquivalently, the allowed Yukawa interaction is $H[a({\bar{\psi }}_L^1{\psi }_R^1+{\bar{\psi }}_L^2{\psi }_R^2)+b{\bar{\psi }}_L^3{\psi }_R^3]$, together with its Hermitian conjugate when constructing a real Lagrangian. The tensor product rules explain the two free parameters: both $D^{\ast }\otimes D$ and ${\epsilon }^{\ast }\otimes \epsilon$ contain one trivial representation, while the mixed products contain none.

There is a useful basis qualification. The basis used in solution is not orthonormal: its Gram matrix is $K=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$, and $D(g{)}^{†}KD(g)=K$. If the fields are expressed in exactly those coordinates, with the barred components defined by ordinary conjugate transpose, the invariant doublet coefficient is $aK$, rather than $a{I}_2$. Canonically normalizing the kinetic term returns the boxed form. Independently chosen left and right orthonormal bases replace $a{I}_2$ by $aS$, where $S$ is their fixed unitary intertwiner; the physical degeneracy is unchanged.

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.

We prove uniform approximation, not just approximation in $L^2$. The analytic facts used are: a continuous square-integrable kernel gives a compact integral operator on $L^2$; the spectral theorem for compact self-adjoint operators decomposes the closure of its range into its finite-dimensional nonzero eigenspaces; and continuous functions are dense in $L^2$ for normalized Haar measure on a compact Lie group. Normalized Haar measure on a compact group is both left- and right-invariant and is preserved by inversion.

Choose a continuous nonnegative function $k$ of integral one, supported in a sufficiently small symmetric neighborhood of the identity, with $k(g^{-1})=k(g)$. Such functions are obtained from a local bump and its inverted bump followed by normalization. Define the convolution operatorIts kernel is continuous on $G\times G$, so $T$ is compact. The symmetry of $k$ makes it self-adjoint. It commutes with the left translations $(L_{g}h)(x)=h(g^{-1}x)$ by a change of variable $y=gz$.

By Cauchy-Schwarz inequality and invariance of Haar measure,The same integral formula and continuity of the kernel show $Th$ is continuous for every $h\in L^2(G)$. Moreover, rewriting $y=x{z}^{-1}$ gives $(Tf)(x)=\int f(x{z}^{-1})k(z)dz$. Uniform continuity on the compact group therefore makes $\Vert Tf-f{\Vert }_{\infty }$ arbitrarily small when the support of $k$ is small. Also $\Vert T{\Vert }_{\infty \to \infty }\le 1$ because $k\ge 0$ and has integral one.

Fix $\epsilon >0$ and take $k$ so $\Vert Tf-f{\Vert }_{\infty }<\epsilon /3$. ThenLet $P_N$ be the orthogonal projection onto increasing finite sums of the nonzero eigenspaces of $T$. Since $Tf$ lies in the closure of the range, the spectral theorem gives $P_{N}Tf\to Tf$ in $L^2$. Applying the smoothing estimate givesThus some $h=T{P}_{N}Tf$ satisfies $\Vert h-f{\Vert }_{\infty }<\epsilon$.

Each nonzero eigenspace is finite dimensional, invariant under $L_g$, and consists of continuous functions: if $Tv=\lambda v$ with $\lambda \ne 0$, then $v=Tv/\lambda$. The finite sum $M$ containing $h$ therefore carries the finite-dimensional unitary representation $\rho (g)=L_g{|}_M$. It is continuous, since translating each of its finitely many continuous basis functions depends continuously on $g$ in the uniform and hence the $L^2$ norm.

Let $\ell$ be evaluation at the identity on $M$, and let the dual representation on $M^{\ast }$ be $\theta (g)\ell =\ell \circ \rho (g^{-1})$. For the linear functional $L_h$ on $M^{\ast }$ defined by $L_h(u)=u(h)$,The rank-one endomorphism $\alpha =\ell \otimes L_h$ of $M^{\ast }$ satisfies $\mathrm{tr}(\alpha \theta (g))=L_h(\theta (g)\ell )$. ThereforeThis is the requested Peter-Weyl theorem, proved via the convolution proof of uniform Peter-Weyl approximation. The extra convolution after the spectral truncation is what upgrades the $L^2$ estimate to a uniform one.

To deduce a faithful finite-dimensional representation, first note that these representations separate points. For $g\ne e$, choose a continuous function taking different values at $g$ and $e$, and approximate it closely enough by a trace coefficient to retain that difference. The representation appearing in that trace must satisfy $\theta (g)\ne I$.

There is an identity neighborhood $U$ containing no nontrivial subgroup. Here is the Lie groups have no small subgroups argument: choose an injective exponential chart on a Lie-algebra ball of radius $r$, and take $U=\exp (B_{\delta })$ with $0<\delta <r/2$. If a subgroup contained a nonidentity $\exp X$ in $U$, choose the first integer $m$ with $m\Vert X\Vert \ge \delta$. Then $\delta \le m\Vert X\Vert <2\delta <r$, so its power $\exp (mX)$ lies outside $U$, a contradiction. For a zero-dimensional group, take $U=\{e\}$.

For every $g\notin U$, choose a representation nontrivial at $g$. The sets on which these representations are not the identity are open and cover the compact set $G\setminus U$. A finite subcover gives finitely many representations, whose direct sum has kernel contained in $U$. A subgroup contained in $U$ is trivial, so this direct sum is faithful. Averaging its Hermitian form makes it unitary. Thus the finite faithful representation from point-separating representations yieldsUsing the no-small-subgroups neighborhood is essential; $G\setminus \{e\}$ alone need not be compact, so a finite-subcover argument on that punctured set would not be valid.

Let $\rho :{\pi }_1(X)\to U(V)$ be the unitary representation. The associated unitary flat holomorphic vector bundle iswhere deck transformations act on the first factor and $\rho$ acts on the second. Local lifts to the universal cover give holomorphic orthonormal frames with locally constant unitary transition matrices. The induced Hermitian metric is therefore well defined, and the induced flat unitary connection is its Chern connection.

Let $s$ be a global holomorphic section. In one of these parallel orthonormal frames write $s=(f_1,\dots ,f_r)$, where every $f_j$ is holomorphic. Its squared norm $u=|s{|}^2={\sum }_j|f_j{|}^2$ is a globally defined smooth function because the transition matrices are unitary. In a local complex coordinate,Choose any smooth conformal metric on the curve, locally $\kappa (z)|dz{|}^2$. Its Laplacian satisfiesOn the compact curve without boundary, Stokes theorem gives ${\int }_X\Delta ud\mathrm{vol}=0$. A continuous nonnegative function with zero integral vanishes everywhere, so each $f_j^{\prime }=0$. Thus the section is locally constant in every parallel frame and is globally parallel. This proves that holomorphic sections of a unitary flat bundle are parallel.

A parallel section is determined by its value $v$ at a chosen base point. Following it around a loop returns the value $\rho (\gamma )v$, so a global parallel section exists exactly when $\rho (\gamma )v=v$ for every $\gamma \in {\pi }_1(X)$. Conversely, such a fixed vector gives the constant equivariant section on $\tilde{X}$, which descends holomorphically. Evaluation and this construction are inverse linear maps, giving the canonical identificationUnitarity is what makes the local squared norms glue and the integrated nonnegative identity available; no assumption that the universal cover is compact is made.

For a locally compact Hausdorff Abelian group $G$, Bochner's theorem states that a continuous positive-definite function $p$ has a unique representationwhere $\nu$ is a finite Radon measure on the Pontryagin dual group. Conversely every such measure gives a continuous positive-definite function. Here positive definiteness means ${\sum }_{i,j}{c}_i\overline{c_j}p(x_i{x}_j^{-1})\ge 0$ for all finite choices. The Abelian and continuity hypotheses are part of the theorem; the representation is not claimed for arbitrary non-Abelian groups or discontinuous functions.

First prove existence without presupposing the representation. The elementary consequences of positive definiteness give $p(x^{-1})=\overline{p(x)}$, $p(e)\ge 0$ and $|p(x)|\le p(e)$. If $p(e)=0$, then $p=0$ and take $\nu =0$. Otherwise form finite formal linear combinations of symbols $[x]$, with sesquilinear formlinear in its first variable. Positive definiteness makes this form positive semidefinite. The Cauchy-Schwarz inequality for such a form shows that null vectors are orthogonal to everything, so quotienting by them gives an inner product space. Complete it to a Hilbert space $H$. Since $G$ is Abelian, translation $U_a[x]=[ax]$ preserves the form and extends to a unitary operator. Its inverse is $U_{a^{-1}}$, so $(U_a)$ is a unitary representation. With $v=[e]$,The translated copies of $v$ span a dense subspace. MoreoverFinite sums and density, together with $\Vert U_a\Vert =1$, give strong continuity on all of $H$. This constructs the cyclic unitary representation of a positive-definite function directly.

For $f\in L^1(G)$ define the integrated unitary representation $\pi (f)=\int f(x)U_{x}dx$, with the integral applied to each vector. Its norm is at most $\Vert f{\Vert }_1$. Fubini gives $\pi (f\ast g)=\pi (f)\pi (g)$, and inversion in the Abelian group gives $\pi (f^{\ast })=\pi (f{)}^{\ast }$. Thusis a commutative C-star algebra. It is nondegenerate: choose nonnegative compactly supported functions $u_V$ of Haar integral one with supports shrinking to the identity element. Strong continuity impliesfor every $w\in H$. These form an approximate identity in the representation. Neighbourhoods may form a net, since no metrizability is being assumed here.

The Commutative Gelfand--Naimark theorem identifies $\mathcal{A}$ with $C_0(X)$, where $X$ is its character space. The positive vector functional $\omega (B)=⟨Bv,v⟩$ becomes a bounded positive functional on $C_0(X)$, so the Riesz-Markov-Kakutani representation theorem supplies a finite positive Radon measure $\rho$ withThese general representation theorems are applied to the explicitly constructed algebra; they are not an assumption of Bochner's conclusion.

For each $\theta \in X$, the functional $f\mapsto \theta (\pi (f))$ is a nonzero algebra character of $L^1(G)$, since $\pi (L^1(G))$ is dense in $\mathcal{A}$. By Question 3 it has the form $\int f(x){\chi }_{\theta }(x)dx$. The resulting map $\iota :X\to \hat{G}$ is continuous: pointwise convergence of $\theta$ on $\mathcal{A}$ gives convergence on all $\pi (f)$, and the topological character identification in Question 3 gives compact-open convergence of ${\chi }_{\theta }$. Push $\rho$ forward under $\iota$ to obtain $\nu$. This finite measure is Radon: compact subsets approximating the measure of any inverse-image Borel set have compact images, giving inner regularity; finiteness and complements then give outer regularity.

To recover $p(x)$, note that $U_x\pi (u_V)=\pi (T_x{u}_V)\in \mathcal{A}$. HenceThe left side tends to $p(x)$. On the right, $|{\Phi }_{\chi }(u_V)|\le 1$ and ${\Phi }_{\chi }(u_V)\to 1$ uniformly on compact subsets of $\hat{G}$. For completeness, the evaluation $(\chi ,x)\mapsto \chi (x)$ is jointly continuous: near $x_0$ restrict $x$ to a compact neighbourhood, control the character uniformly there, and use continuity of the fixed character. A finite cover of a compact set of characters then makes those characters equicontinuous at $e$, proving the stated uniform convergence. Continuity of $\iota$ gives uniform convergence on compact subsets of $X$, and the finite Radon measure $\rho$ makes the remaining tails arbitrarily small. Thus passage to the limit is valid even for a net, andTaking $x=e$ also gives the required total mass $p(e)$.

Conversely, a finite positive Radon measure givesIts transform is continuous. Indeed, choose a compact set of characters with arbitrarily small measure outside it; joint continuity of evaluation gives uniform continuity in $x$ near the specified point on that compact set, while the outside contribution is bounded by twice its measure. This proves continuity without using an unjustified dominated-convergence assertion for arbitrary nets.

Finally suppose two finite Radon measures have the same transform. Multiply by $f\in L^1(G)$ and integrate over $G$. Fubini shows that the measures agree on every function $\chi \mapsto {\Phi }_{\chi }(f)$. Question 3 showed that these functions are uniformly dense in $C_0(\hat{G})$. Boundedness of both measure functionals extends equality to all of $C_0(\hat{G})$; uniqueness in the Riesz representation theorem gives equality of the measures. This proves existence, uniqueness and the converse, completing Bochner's theorem. In particular $p(e)=1$ corresponds exactly to a probability measure on the dual group.

Here a supersymmetry-preserving combination means a real combination of Hermitian Majorana spinor supercharges, in a positive-norm unitary representation of the ordinary Super-Poincaré algebra. This reality qualification is essential. Let $|\psi ⟩$ be normalized and in the charge domains, and define the real symmetric matrixFor real $u$, the Hermitian operator $q_u=u^a{Q}_a$ satisfiesTherefore $M$ is a positive semidefinite matrix, andThe reverse implication uses positivity: diagonalize $M$, so a zero quadratic form has zero component in every positive-eigenvalue direction. This converts the problem into nullity of real N=1 supercharges, not just a condition on the determinant of an operator-valued expression.

In the equivalent Weyl spinor form, $\{q_{\alpha },q_{\beta }^{†}\}=2{\sigma }_{\alpha \dot{\beta }}^m{P}_m$. After a fixed invertible change of real charge basis, $M$ is the realification of the Hermitian two-by-two matrixup to the harmless spatial-sign convention. A complex Hermitian matrix has the same eigenvalues in its realification, each repeated twice. The Pauli matrix multiplication law gives $(\mathbf{p}\cdot \mathbf{\sigma }{)}^2=|\mathbf{p}{|}^{2}I$, so these eigenvalues are $E+|\mathbf{p}|$ and $E-|\mathbf{p}|$, each twice. Positivity forces $E\ge |\mathbf{p}|$. If one nonzero real $u$ annihilates the state, $M$ has a kernel: either $E=|\mathbf{p}|>0$, with exactly two zero eigenvalues, or $E=|\mathbf{p}|=0$, with four. ThusFor a momentum eigenstate these are the null-momentum and zero-momentum cases; the expectation-matrix proof also covers states not assumed to have definite momentum. There are no massive states preserving a real charge in this unextended algebra.

If “linear combination” instead allows arbitrary complex coefficients, the unrestricted claim is false. In a massive fermionic Fock space with two lowering operators $a_1,a_2$ and a spin-half Clifford vacuum spectator, takeThen $a_1\psi =0$. The three vectors $a_2\psi$, $a_2^{†}\psi$ and $a_1^{†}\psi$ are linearly independent, by their occupation numbers and spectator spin labels. The complex annihilator is therefore only one-dimensional. Such a lowering operator is not a Hermitian generator of preserved real supersymmetry. The boxed conclusion applies to the physical real-charge interpretation; it also assumes no domain-wall charge in N=1 supersymmetry.

By part (i), finite-dimensional continuous unitary representations separate points of the compact group. More explicitly, if every representation sent some $g\ne e$ to the identity, all their matrix coefficients would take the same values at $g$ and $e$. Uniform density would force every continuous scalar function to do so, contradicting separation of points on a compact Hausdorff space. HenceEach representation kernel is a closed normal subgroup. We claim that the intersection can be reduced to finitely many kernels. Begin with $K_0=G$. If $K_j\ne \{e\}$, choose $g_j\in K_j\setminus \{e\}$ and a representation ${\theta }_{j+1}$ with ${\theta }_{j+1}(g_j)\ne I$. ThenAn indefinite continuation would violate the descending chain condition for closed subgroups. Thus after finitely many steps, $K_N=\{e\}$.

The direct sum $\rho ={\theta }_1\oplus \cdots \oplus {\theta }_N$ is consequently a finite-dimensional faithful representation, with continuous injective homomorphismCompactness of $G$ makes this a homeomorphism onto a compact, hence closed, image. The closed-subgroup theorem makes that image an embedded Lie subgroup of the finite-dimensional unitary group. Transporting its smooth structure to $G$ provesThis is the mechanism behind the assertion that compact groups with a descending chain condition are Lie groups: the chain condition turns an arbitrarily large separating family into one finite-dimensional faithful representation.