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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.

Normalized Haar measure makes this averaging a norm-one projection onto continuous class functions. Averaging a trace coefficient replaces its coefficient endomorphism by $\int \theta (h{)}^{-1}\alpha \theta (h)dh$. On an irreducible summand this average is $(\mathrm{tr}\alpha /\dim V)I$ by Schur lemma. It follows that averaging finite-dimensional matrix coefficients yields finite linear combinations of irreducible characters. Together with the Peter-Weyl theorem, this proves uniform density of those characters among continuous class functions.

On $S^2=SU(2)/U(1)$ a degree-$q$ equivariant line bundle corresponds to isotropy character $\mathrm{diag}(e^{it},e^{-it})\mapsto e^{-iqt}$. The Peter-Weyl theorem computes its section multiplicities by the occurrence of weight $-q$ in the SU(2) representation ${\mathrm{Sym}}^n{\mathbb{C}}^2$. Its weights $n,n-2,\dots ,-n$ give the displayed multiplicity-one decomposition. The spinor lines have degrees $-1,+1$ and thus contain all odd $n$, once each.

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.

Interpret $\theta$ as a continuous finite-dimensional complex group representation, so its values lie in ${\mathrm{GL}}_{\mathbb{C}}(V)$, and interpret approximation in the uniform norm. Assume the usual Hausdorff convention for the compact topological group. The following argument proves the required form of the Peter-Weyl theorem without first assuming that $G$ is a Lie group.

Start with normalized Haar measure. It is both left- and right-invariant on a compact group. Average any Hermitian inner product to make a finite-dimensional representation unitary. The left regular action $L_{a}h(x)=h(a^{-1}x)$ is already unitary on $L^2(G)$.

Choose a continuous nonnegative function $k$ supported in a small symmetric identity neighbourhood, with $k(z^{-1})=k(z)$ and $\int k=1$. Such functions are obtained from a nonzero continuous bump near the identity, symmetrizing and normalizing. DefineThe convolution operator $T_k$ commutes with left translations. Its continuous kernel is Hermitian, hence it is self-adjoint. It is compact on $L^2(G)$: approximate the continuous kernel uniformly by finite sums of separated functions of $x$ and $y$, using the Stone-Weierstrass theorem on the compact product. The corresponding operators have finite rank and converge in operator norm. It also maps into $C(G)$ and satisfiesby the Cauchy-Schwarz inequality; continuity follows from the uniform continuity of the translated kernel.

The spectral theorem for compact self-adjoint operators decomposes the orthogonal complement of its kernel into finite-dimensional eigenspaces with nonzero eigenvalues. Every such eigenspace is left-translation invariant, and all its elements are continuous, since $h=T_{k}h/\lambda$ there. If $W\subset C(G)$ is one of these invariant spaces, write $\rho (a)=L_a{|}_W$. Evaluation at the identity givesThus $h$ is a matrix coefficient of the dual representation. This identifies each finite spectral sum with finite-dimensional representation data.

To get uniform rather than just $L^2$ approximation, let $S_N$ project onto increasing finite sums of the nonzero eigenspaces. For $f\in C(G)$, $T_{k}f$ is orthogonal to the kernel, so $S_N{T}_{k}f\to T_{k}f$ in $L^2$. Applying the displayed smoothing bound givesMeanwhile, choosing the support of $k$ sufficiently small makes $T_{k}f$ uniformly close to $f$, by uniform continuity under small translations. Since $\Vert T_k{\Vert }_{C\to C}\le 1$, it also makes $T_k^{2}f$ uniformly close to $f$. Every $T_k{S}_N{T}_{k}f$ belongs to a finite sum of invariant eigenspaces, hence is a finite sum of matrix coefficients. This completes the convolution proof of uniform Peter-Weyl approximation.

A coefficient $\ell (\theta (g)v)$ equals $\mathrm{tr}(\alpha \theta (g))$ for the rank-one endomorphism $\alpha =v\otimes \ell$. Finite sums of coefficients can be combined using a direct-sum representation and a block-diagonal $\alpha$. Therefore the conclusion has exactly the requested form:

For a class function $f$, apply conjugation averaging on a compact group,It fixes $f$ and is a contraction in the uniform norm, so averaging an approximation preserves its error bound. A trace coefficient averages to $\mathrm{tr}(\overline{\alpha }\theta (g))$, whereThis endomorphism commutes with the representation. A finite-dimensional unitary representation splits into irreducibles by repeatedly taking invariant orthogonal complements. On each irreducible diagonal block, Schur lemma makes its average a scalar multiple of the identity, namely its trace divided by the block dimension. Off-diagonal blocks do not contribute to the total trace. Thus the averaged coefficient is a finite linear combination of irreducible characters. Consequently irreducible complex characters span a uniformly dense subspace of the continuous class functions.

Use the round unit sphere with its complex orientation and unique spin structure. Its chiral spinor bundles are $S^{+}=\mathcal{O}(-1)$ and $S^{-}={\Lambda }^{0,1}\otimes \mathcal{O}(-1)$, the latter a smooth line bundle of degree $+1$. The Dirac operator isHere Clifford multiplication has $c(v{)}^2=-|v{|}^2$, and $D$ acts self-adjointly on the L2 spinors with first-order Sobolev domain. The geometric spin connection is invariant under the lifted $SU(2)$ action, hence $D$ is equivariant.

We justify the spectral properties needed for its phase. The Peter-Weyl theorem and Frobenius reciprocity for compact groups give the decomposition of sections of an isotropy-weight-$q$ line bundle: $E_n$ occurs once exactly when $n\ge |q|$ and $n\equiv q(\mathrm{mod}2)$, because its torus weights are $n,n-2,\dots ,-n$. The spinor isotropy weights are $+1,-1$, so each chirality contains $E_1,E_3,E_5,\dots$, each once. The horizontal connection Laplacian is the total Casimir minus the vertical generator square, ${\nabla }^{\ast }\nabla =C-1/4$ on these weight-one bundles. Expanding the Dirac square in a normal orthonormal frame gives the diagonal connection-Laplacian terms and the off-diagonal curvature contraction. The Clifford relations contract spin curvature to $\mathrm{Scal}/4$, the Lichnerowicz formula for a twisted Dirac operator with no twisting bundle. Since the scalar curvature here is two,Thus there is no kernel, the eigenvalues of $D$ are $\pm 1,\pm 2,\dots$ with finite multiplicities, and the resolvent is compact by spectral truncation. These round-sphere spectral facts agree with the explicit calculation in Abrikosov's paper.

Define the chiral phase of the round-sphere Dirac operatorIt is unitary because both chiral kernels vanish; equivariance follows from that of $D$. The full phase is $F=D|D{|}^{-1}=\left(\begin{array}{cc}0& V^{\ast }\\ V& 0\end{array}\right)$.

For a smooth scalar $f$, the Leibniz rule gives $[D,M_f]=c(df)$, a bounded bundle endomorphism. The spectral integral for the sign operator gives, in the strong topology,Write $R_{\pm }=(D\mp it{)}^{-1}$. Since $2D(D^2+t^2{)}^{-1}=R_{+}+R_{-}$ and $[R_{\pm },M_f]=-R_{\pm }[D,M_f]R_{\pm }$, the commutator integral is norm convergent: its integrands are compact and bounded by $\Vert [D,M_f]\Vert /(1+t^2)$. A norm limit of their integrals is compact. Thus $[F,M_f]$ is compact, and its off-diagonal blocks say $V{M}_f^{+}-M_f^{-}V$ is compact. Uniform approximation of continuous $f$ by smooth functions extends this to all $f\in C(S^2)$, using the bound $2\Vert f-g{\Vert }_{\infty }$ for the change of commutator.

The chiral phase is an equivariant unitary which intertwines the two multiplication representations modulo compacts. This specifies the grading implicit in the subsequent index question; treating the full self-adjoint phase as an endomorphism would yield a different, zero, compressed index.