Weyl systems on arbitrary locally compact abelian groups

CC0-1.0 to the extent of rights held.

Let $G$ be any locally compact Hausdorff abelian group, written additively, and $\Gamma=\widehat G$ its compact-open dual. Fix the earlier Haar convention and a Haar measure $m$ on $G$; write $L=L^2(G,m)$. Choose the earlier dual Haar measure on $\Gamma$. Inner products are linear in the first variable. Hilbert spaces, groups and multiplicities have no separability, second-countability or sigma-compactness assumption. All vector integrals use the proved finite-exponent Radon convention, with sigma-compact carriers for integrable functions.

The exact earlier inputs are L24 integration, recovery and intertwiners, L24's full group completion, H1 Fourier completion, H3 topological biduality, L25 covariant integration and vector-domain recovery, and L25 Section10's complete compact-kernel argument. The finite matrix norm/multiplicity argument is also written in L138 Section6. We use CF1,6,8 and H0's Hilbert tensor construction. The arbitrary-multiplicity extension and all vector domains needed here are proved below.

WY0. The exact assertion and signs

A Weyl system consists of strongly continuous unitary representations $U:G\to\mathcal U(H)$ and $V:\Gamma\to\mathcal U(H)$ satisfying

$U_sV_\chi=\chi(s)V_\chi U_s. \tag{WY1}$ The theorem is that there are a Hilbert space $K$ and an onto unitary $W:L\otimes K\longrightarrow H$ such that

$U_sW=W(\rho_s\otimes1_K),\qquad V_\chi W=W(M_\chi\otimes1_K), \tag{WY2}$ where

$(\rho_s\xi)(t)=\xi(t+s),\qquad (M_\chi\xi)(t)=\chi(t)\xi(t). \tag{WY3}$ Conversely these formulas give a Weyl system for every $K$. Its multiplicity is determined up to unitary isomorphism, and every bounded intertwiner is identified below. The zero system is included by $K=0$.

The sign is fixed directly: $\rho_sM_\chi\rho_s^*=\chi(s)M_\chi$. If instead the relation in (WY1) has $\overline{\chi(s)}$, replace $U_s$ by $U_{-s}$ while applying the theorem; the original $U_s$ then becomes left translation $\lambda_s\xi(t)=\xi(t-s)$. No sign is hidden in the phrase “standard representation.”

The space $L$ is nonzero: a nonzero compact continuous bump has positive finite squared integral by the earlier Haar full-support and compact-finiteness proofs. The $\rho_s$ are strongly continuous unitaries by L24's translation theorem and abelian unimodularity. The $M_\chi$ are unitary. On $\xi\in C_c(G)$,

$\|(M_\chi-M_\eta)\xi\|_2 \leq \sup_{t\in\operatorname{supp}\xi}|\chi(t)-\eta(t)|\,\|\xi\|_2, \tag{WY4}$ so compact-open convergence gives strong continuity there. Density and the uniform bound two extend it to every $L^2$ vector. Tensoring these representations with an arbitrary identity preserves unitarity and strong continuity: first test finite tensor sums, then approximate an arbitrary tensor vector. Thus the converse assertion and all representations later used on the model space are well defined.

WY1. Integrating both actions and constructing the coefficient representation

L24 supplies the nondegenerate star representations

$U(f)\xi=\int_G f(s)U_s\xi\,dm(s),\qquad V(g)\xi=\int_\Gamma g(\chi)V_\chi\xi\,d\widehat m(\chi), \tag{WY5}$ for every $f\in L^1(G)$, $g\in L^1(\Gamma)$. These are vector integrals on all of $H$, with operator bounds $\|f\|_1$ and $\|g\|_1$. A continuous vector orbit has separable image on each compact carrier piece. Approximation by strongly measurable simple functions and the integrable scalar norm bound therefore justify the integrals without assuming that $H$ is separable. Normalized shrinking compact bumps have integrated operators converging strongly to the identity, as proved in L24.

For $g\in L^1(\Gamma)$, set

$Qg(t)=\int_\Gamma g(\chi)\chi(t)\,d\widehat m(\chi). \tag{WY6}$ This is the positive inverse-sign integral on the dual group. Under the earlier negative Fourier transform and bidual map $j(t)(\chi)=\chi(t)$, it is

$Qg(t)=\widehat g(j(-t)). \tag{WY7}$ H1 applied to $\Gamma$ gives $C^*(\Gamma)=C_0(\widehat\Gamma)$, and H3 makes $t\mapsto j(-t)$ a homeomorphism onto $\widehat\Gamma$. Consequently $Q$ extends to an isometric star isomorphism $C^*(\Gamma)\to C_0(G)$; in particular its $L^1$ image is uniformly dense in $C_0(G)$. No vector-valued spectral theorem is used here.

The representation $V(g)$ extends to $C^*(\Gamma)$ by L24's actual universal norm, and hence defines a nondegenerate star representation

$\pi:C_0(G)\longrightarrow B(H),\qquad \pi(Qg)=V(g). \tag{WY8}$ Nondegeneracy follows already from the strongly convergent dual-group compact bumps in (WY5).

Let $a\in C_0(G)$, and let $\beta_s(a)(t)=a(t+s)$. For $g_s(\chi)=\chi(s)g(\chi)$, the Weyl relation and bounded maps through vector integrals give

$U_sV(g)U_s^*=V(g_s),\qquad Qg_s(t)=Qg(t+s). \tag{WY9}$ Thus $U_s\pi(a)U_s^*=\pi(\beta_s(a))$ first on the dense $Q(L^1(\Gamma))$, then on all $C_0(G)$ by the C* contractivity bound. This is precisely L25's right-translation covariant pair. L25 proves point-norm continuity of $\beta$ by compact cutoffs and finite-cover translation continuity; no regularity hypothesis on a spectral measure has been added.

For later use, $V_\chi$ commutes with every $\pi(a)$. Indeed it commutes with $V(g)$, because $\Gamma$ is abelian and the equality passes through each vector integral; then use uniform density and contractivity. Joint strong continuity of $(s,\chi)\mapsto V_\chi U_s$ follows by expanding a difference and using the unitary bounds. All products of integrals in this proof may also be computed as iterated vector integrals with norm majorant $|f(s)g(\chi)|\|\xi\|$; L24's qualified Radon-product integration on the product of sigma-compact carriers applies. No statement about unrestricted product Borel sigma algebras is required.

WY2. The actual nondegenerate compact-operator representation

Write $\mathcal B=L^1(G,C_0(G))$, with L25's twisted convolution for $\beta$. The proved covariant integration theorem gives

$\Phi(F)\xi=\int_G\pi(F(s))U_s\xi\,dm(s),\qquad \|\Phi(F)\|\leq\|F\|_1. \tag{WY10}$ It is a nondegenerate star representation on all of $H$. In detail, if $(e_i)$ is a contractive coefficient approximate identity and $a_N$ is a compact mass-one group bump supported near zero, then

$\|\Phi(a_N e_i)\xi-\xi\| \leq\sup_{s\in N}\|U_s\xi-\xi\|+\|\pi(e_i)\xi-\xi\|\to0 \tag{WY11}$ on the product net. The coefficient convergence follows on $\pi(C_0(G))H$ from the approximate identity and then on $H$ by nondegeneracy and the contraction bound.

For $p,q\in C_c(G)$, put

$F_{p,q}(s)(t)=p(t)\overline{q(t+s)},\qquad \theta_{p,q}\xi=p\langle\xi,q\rangle. \tag{WY12}$ The support condition $t\in\operatorname{supp}p$, $t+s\in\operatorname{supp}q$ gives a joint compact support and compact support in $s$; the coefficient map is norm continuous. L25 Section10 proves the following entire algebraic and density statement:

$F_{p,q}*F_{r,z}=\langle r,q\rangle F_{p,z},\quad F_{p,q}^*=F_{q,p},\quad F_{p,q}\longleftrightarrow\theta_{p,q}. \tag{WY13}$ The correspondence is a star isomorphism between these linear spans, and the first span is $L^1(G,C_0(G))$-dense. Its proof first approximates continuous coefficient functions on one compact support, then approximates their continuous kernels on one compact rectangle by products. The substitution $y=t+s$ changes the kernel into the coefficient function with an $L^1$ error at most the uniform kernel error times the finite measure of the compact difference set. In particular this density is valid for the present arbitrary group, not just a countable exhaustion.

Here is explicitly the norm step needed for our representation. On the finite-rank span define $\kappa_0(\theta_{p,q})=\Phi(F_{p,q})$. The algebraic isomorphism makes this well defined. Any finite list of the vectors involved lies in a finite-dimensional subspace $E\subset C_c(G)$. Its full matrix algebra $\mathcal K(E)$ is contained in this span. The restriction of $\kappa_0$ is an algebraic star homomorphism of this complete matrix C* algebra, so L24's proved automatic contractivity gives

$\|\kappa_0(T)\|\leq\|T\|\quad(T\in\mathcal K(E)). \tag{WY14}$ Finite-dimensional orthonormalization uses only finite linear combinations, so the basis vectors remain in $C_c(G)$. This is the finite-corner norm mechanism of L138 Section6. Every element of the span lies in such a corner. Approximation of arbitrary Hilbert vectors by $C_c(G)$, and $\|\theta_{p,q}\|=\|p\|\|q\|$, show that the span is norm dense in $\mathcal K(L)$. The displayed rank-one norm follows directly from Cauchy–Schwarz, with equality at $q/\|q\|$ when neither vector is zero. Therefore $\kappa_0$ extends uniquely to a contractive star representation

$\kappa:\mathcal K(L)\longrightarrow B(H). \tag{WY15}$ This step needs no assumed irreducibility, simplicity of the compact algebra, or full crossed-product classification.

Let $\Phi_0$ be the actual integrated pair $(M_a,\rho_s)$ on $L$. On compact coefficients its kernel is

$K_F(t,y)=F(y-t)(t). \tag{WY16}$ Its norm is at most $\|F\|_1$, and $\Phi_0(F_{p,q})=\theta_{p,q}$. Approximate any $F\in\mathcal B$ in $L^1$ by the span in (WY13). Both integrated maps converge in operator norm, and the image of $\Phi_0$ is compact by that same approximation. Contractivity of $\kappa$ then proves

$\kappa(\Phi_0(F))=\Phi(F)\quad(F\in\mathcal B). \tag{WY17}$ Equation (WY11) implies that $\kappa$ is nondegenerate. Thus no unobserved zero representation summand can remain in the later multiplicity construction.

For $f\in L^1(G)$, $g\in L^1(\Gamma)$, the coefficient function $F(s)=f(s)Qg$ is in $\mathcal B$, and

$\Phi(F)=V(g)U(f),\qquad \Phi_0(F)=M_{Qg}\rho(f). \tag{WY18}$ Finite sums of these $F$'s are $L^1$-dense: compact continuous coefficient functions are approximated by finite scalar compact functions times fixed $C_0$ coefficients, and those coefficients by $Qg$ in uniform norm. This identifies the norm-closed integrated algebra generated linearly by these ordered products with $\overline{\kappa(\mathcal K(L))}$. WY3 will show that this image is already closed.

WY3. An onto multiplicity unitary on every Hilbert space

The following construction applies to any nondegenerate star representation $\kappa:\mathcal K(L)\to B(H)$ on arbitrary $H$, with $L\ne0$. Choose a unit vector $e\in L$, set $E=\theta_{e,e}$, and put

$K=\kappa(E)H,\qquad W_0(\xi\otimes\eta)=\kappa(\theta_{\xi,e})\eta \quad(\xi\in L,\ \eta\in K). \tag{WY19}$ The projection $\kappa(E)$ is bounded and selfadjoint, so $K$ is a closed Hilbert subspace. The formula is bilinear and hence defines a map on the algebraic tensor product. The rank-one product gives

$\begin{aligned} \langle W_0(\xi\otimes\eta),W_0(\zeta\otimes\nu)\rangle &=\langle\kappa(\theta_{e,\zeta}\theta_{\xi,e})\eta,\nu\rangle\\ &=\langle\xi,\zeta\rangle\langle\eta,\nu\rangle. \end{aligned} \tag{WY20}$ Finite sums give exactly H0's Hilbert tensor inner product. Thus $W_0$ extends to an isometry $W:L\otimes K\to H$, whose range is closed by completeness.

For every $\xi,\zeta\in L$, $h\in H$,

$\kappa(\theta_{\xi,\zeta})h =W\bigl(\xi\otimes\kappa(\theta_{e,\zeta})h\bigr), \qquad \kappa(\theta_{e,\zeta})h\in K. \tag{WY21}$ The second assertion follows by left multiplying with $\kappa(E)$. Finite ranks are norm dense in $\mathcal K(L)$, and $\kappa$ is nondegenerate, so the vectors on the left have dense span in $H$. The closed range of $W$ is therefore all of $H$. This proves surjectivity on the complete Hilbert spaces, not merely on a selected separable cyclic subspace.

For $T\in\mathcal K(L)$, the identity $T\theta_{\xi,e}=\theta_{T\xi,e}$ gives on elementary tensors, then by density,

$\kappa(T)W=W(T\otimes1_K). \tag{WY22}$ If $H\ne0$, surjectivity makes $K\ne0$. H0's tensor norm identity then shows $\|\kappa(T)\|=\|T\|$, so $\kappa$ is faithful and has closed image. When $H=0$, $K=0$ and its image is the closed zero algebra. This also completes the image assertion after (WY18).

WY4. Recovering both original Weyl actions

Apply WY3 to the representation constructed in WY2. The pair $(M_a\otimes1_K,\rho_s\otimes1_K)$ is nondegenerate and covariant on $L\otimes K$. Nondegeneracy follows from compact cutoffs on the dense compact continuous vectors of $L$, first on finite tensor sums and then everywhere. Bounded maps pass through the vector integrals on these sums and their limits, so its integrated representation is $\Phi_0(F)\otimes1_K$. Equations (WY17) and (WY22) therefore give

$\Phi(F)W=W(\Phi_0(F)\otimes1_K)\quad(F\in\mathcal B). \tag{WY23}$ L25 Section3 proves on actual finite generating vectors that a bounded intertwiner of nondegenerate integrated representations intertwines both recovered actions. Applied to (WY23), it yields

$\pi(a)W=W(M_a\otimes1_K),\qquad U_sW=W(\rho_s\otimes1_K). \tag{WY24}$ The first equality with $a=Qg$ says

$V(g)W=W(M_{Qg}\otimes1_K) =W\int_\Gamma g(\chi)(M_\chi\otimes1_K)\,d\widehat m(\chi). \tag{WY25}$ The last equality holds on every vector by the pointwise scalar integral on compact continuous vectors, its absolute integrable bound, and norm approximation; alternatively test all vector inner products and use L24's vector-integral uniqueness. Since both dual-group representations are strongly continuous and nondegenerate, L24's full integral-intertwiner theorem applied to $\Gamma$ proves $V_\chi W=W(M_\chi\otimes1_K)$. This proves (WY2) for every $s,\chi$. It does not replace a group element by an L1 Dirac mass.

WY5. Arbitrary matrix units, all intertwiners and exact multiplicity

For an arbitrary orthonormal basis $(e_i)_{i\in I}$ of $L$, choose $i_0\in I$ and take $e=e_{i_0}$ in WY3. Such a basis exists by CF1's maximal-chain/choice principle: a maximal orthonormal family has dense closed span, since the Hilbert projection theorem would otherwise produce another unit orthogonal vector. Put $E_{ij}=\theta_{e_i,e_j}$ and $P_i=\kappa(E_{ii})$. Then

$P_iP_j=0\ (i\ne j),\qquad \kappa(E_{i i_0})^*\kappa(E_{i i_0})=P_{i_0},\quad \kappa(E_{i i_0})\kappa(E_{i i_0})^*=P_i. \tag{WY26}$ Consequently $\kappa(E_{i i_0}):K\to P_iH$ is onto unitary, and the restriction of $W$ to $e_i\otimes K$ is exactly that map. Moreover

$\sum_{i\in I}P_i=1_H\quad\hbox{strongly over finite subsets},\qquad H=\bigoplus_{i\in I}P_iH. \tag{WY27}$ Indeed finite coordinate projections converge strongly to the identity on $L$: approximate a vector by its finite linear span coordinates and use contraction bounds. Their tensor amplifications converge on finite tensor sums and then on every tensor vector. Conjugate by the already onto $W$. This proves (WY27) without a countable index set or an unproved normal extension of $\kappa$.

There is also an exact operator form of uniqueness. If two Weyl systems have the above models $L\otimes K$ and $L\otimes K'$, every bounded simultaneous intertwiner $B:L\otimes K\to L\otimes K'$ has the form

$B=1_L\otimes b\quad\hbox{for a unique }b\in B(K,K'). \tag{WY28}$ To prove it, integration, uniform density of $Qg$, and the compact-kernel construction show that $B$ intertwines every $T\otimes1$, $T\in\mathcal K(L)$. The rank-one projection $E$ forces $B(e\otimes\eta)=e\otimes b\eta$ for a bounded linear $b$, with $\|b\|\leq\|B\|$. Commuting the rank-one map $\theta_{\xi,e}$ across $B$ gives (WY28) on elementary tensors and then everywhere. Conversely (WY28) intertwines the model actions by their formulas; its boundedness follows from the tensor norm estimate. Thus a simultaneous unitary intertwiner exists exactly when $K,K'$ are unitarily isomorphic. This includes zero multiplicities.

In particular the joint commutant is $1_L\otimes B(K)$. A nonzero Weyl system is irreducible exactly when $\dim K=1$. If $K$ has dimension greater than one, the projection onto any proper nonzero closed subspace of $K$ gives a nontrivial reducing projection. If $\dim K=1$, the commutant consists of scalars, so any joint invariant closed subspace, which reduces both unitary groups because their inverses are present, has scalar projection and is trivial. This proves the irreducible uniqueness statement as a consequence of the complete multiplicity construction.

WY6. Exact finite reconstruction of two interlaced copies

Let $G=\mathbb Z/5\mathbb Z$ with counting Haar, $\omega=e^{2\pi i/5}$, $L=\mathbb C^5$ and $K=\mathbb C^2$. On $L$, set $S_s\delta_j=\delta_{j-s}$ and $D_n\delta_j=\omega^{nj}\delta_j$. Then $S_sD_n=\omega^{ns}D_nS_s$, agreeing with (WY1). To display a non-diagonal presentation of the multiplicity, index a physical orthonormal basis $(h_k)_{k=0}^9$ and let

$R(\delta_i\otimes f_a)=h_{\tau(2i+a)},\qquad \tau(k)=3k+1\pmod {10},\quad a\in\{0,1\}. \tag{WY29}$ The map $\tau$ is a permutation since 3 is invertible modulo10. Thus $R$ is unitary. Put $U_s=R(S_s\otimes1)R^*$, $V_n=R(D_n\otimes1)R^*$. For $i\in\mathbb Z/5\mathbb Z$, exact geometric sums give

$P_i=\frac15\sum_{n=0}^4\omega^{-ni}V_n =R(E_{ii}\otimes1)R^*,\qquad \kappa(E_{ij})=P_iU_{j-i}. \tag{WY30}$ The sum is one at label $i$ and zero at every other label because $\sum_{n=0}^4z^n=0$ for a nontrivial fifth root and equals5 for $z=1$. For the second identity, $S_{j-i}\delta_k=\delta_{k-j+i}$, and projection onto $\delta_i$ keeps exactly $k=j$.

The five support fibres in physical coordinates are respectively

$\{1,4\},\quad\{7,0\},\quad\{3,6\},\quad\{9,2\},\quad\{5,8\}. \tag{WY31}$ Take the multiplicity space $P_0H=\operatorname{span}\{h_1,h_4\}$, with the displayed ordered basis. Then the onto map of (WY19) is

$W(\delta_i\otimes h_{\tau(a)}) =\kappa(E_{i0})h_{\tau(a)} =U_{-i}h_{\tau(a)}=h_{\tau(2i+a)}. \tag{WY32}$ This explicitly recovers two copies from interlaced coordinate fibres. The reconstruction and character phases are exact; the finite figure is not a reduction of the arbitrary-Hilbert theorem.

The complete primary source used for comparison is Marc A. Rieffel, Induced representations of C*-algebras (1974), author-hosted edition, printed179–180 and231–232 for the rank-one/module mechanism and239–240 for its broader equivalence context. Here the compact representation, the onto tensor map and all intertwiner assertions have been proved locally. No imprimitivity, disintegration or Stone–von Neumann theorem is imported; normal crossed-product commutants, general duality and weights remain separate obligations.

Recovering multiplicity from one projection

Exact Weyl fibres and the onto tensor construction

The first two panels are an exact finite example, not a claim that the group has ten elements. The group is $G=\mathbb Z/5\mathbb Z$, its Haar measure is counting measure, its dual characters are $\chi_n(s)=\omega^{ns}$, $\omega=e^{2\pi i/5}$, and each dual point has the corresponding Haar mass $1/5$. The Hilbert space has dimension ten because the multiplicity space has dimension two. All indices $i,j,n,s$ for the group are modulo five; only the physical scrambling index is modulo ten.

Let $S_s\delta_j=\delta_{j-s}$, $D_n\delta_j=\omega^{nj}\delta_j$, and define the unitary permutation $R(\delta_i\otimes f_a)=h_{3(2i+a)+1\bmod10}$, $a\in\{0,1\}$. Then $U_s=R(S_s\otimes1)R^*,\qquad V_n=R(D_n\otimes1)R^*,\qquad U_sV_n=\omega^{ns}V_nU_s.$ For the Weyl sign, apply the two sides to $\delta_j$: the exponents are $nj$ and $ns+n(j-s)$, equal modulo five. This is WY0's positive-character convention; using the conjugate scalar instead would require the stated left-translation model.

Panel A displays the physical coordinate order $h_0,\ldots,h_9$. The fibre label $i$ under each coordinate is its $V_n$ exponent: $V_n$ acts there as $\omega^{ni}$. Blue and orange distinguish the multiplicity coordinates $a=0,1$, not two different group characters. The exact selectors are $P_i=\frac15\sum_{n=0}^4\omega^{-ni}V_n, \qquad \kappa(E_{ij})=P_iU_{j-i}.$ The geometric-sum proof is given completely in WY6. It yields the five ordered fibre pairs $(1,4),(7,0),(3,6),(9,2),(5,8)$. Thus the extracted space is $K=P_0H=\operatorname{span}\{h_1,h_4\}$. The projections $P_i$ have rank two; $E_{ii}$ is rank one on $\mathbb C^5$. These two ranks refer to different spaces.

Panel B sorts the physical coordinates by the onto map $W(\delta_i\otimes h_{3a+1})=\kappa(E_{i0})h_{3a+1} =U_{-i}h_{3a+1}=h_{3(2i+a)+1\bmod10}.$ Here $a=0,1$, so the starting coordinates are exactly $h_1,h_4$. Each solid arrow shows $U_1$ sending fibre $i$ to fibre $i-1$; each dashed arrow is the same shift from fibre zero to fibre four. The two rows carry identical $U$ and $V$ actions. This is an exact coordinate reconstruction, not a numerical eigenspace fit.

Panel C states the arbitrary-space mechanism proved in WY2–WY5. For a nonzero Hilbert space $L=L^2(G)$, choose a unit vector $e$ and the rank-one projection $E=\theta_{e,e}$ on $L$. The nondegenerate representation $\kappa$ of $\mathcal K(L)$ is constructed from the complete integrated kernel algebra. Set $K=\kappa(E)H$. The rule $W(\xi\otimes\eta)=\kappa(\theta_{\xi,e})\eta, \qquad \eta\in K,$ has the exact tensor inner product, because $\theta_{e,\zeta}\theta_{\xi,e}=\langle\xi,\zeta\rangle E.$ It extends by completeness to an isometry on $L\otimes K$. Every rank-one image satisfies $\kappa(\theta_{\xi,\zeta})h =W\bigl(\xi\otimes\kappa(\theta_{e,\zeta})h\bigr), \qquad \kappa(\theta_{e,\zeta})h\in K.$ Nondegeneracy and finite-rank density make the range dense, while isometry makes it closed; hence $W$ is onto. Subsequent vector-domain recovery intertwines both original unitary actions. For an arbitrary orthonormal basis the resulting coordinate projections sum strongly over finite subsets, not through an assumed countable sequence. The zero system is represented by $K=0$.

The original renderer and exact checker verifies all 625 finite matrix-unit products, all 25 adjoints, all 125 modular phase identities and all ten basis images. The stored floating-point selector error is a cross-check of the exact geometric-sum proof. The PNG, SVG and integer data are reproducible. The diagram and caption are original CC0 expression. The human-source comparison is Rieffel's complete author-hosted 1974 paper, printed231–232 and239–240, for the rank-one/module mechanism; every assertion used in the diagram is proved in the linked local body.

WY7. Left translations, conjugate characters and multiplier recovery

There is a useful second realization of the same positive-sign relation. On $L=L^2(G)$, put $(L_s\xi)(r)=\xi(r-s),\qquad (Q_\chi\xi)(r)=\overline{\chi(r)}\xi(r),\qquad (J\xi)(r)=\xi(-r). \tag{WM1}$ Here $J$ is complex linear. Inversion preserves Haar measure on an abelian group, by L24, so $J$ is a selfadjoint unitary. Direct substitution gives $J\rho_sJ=L_s,\qquad JM_\chi J=Q_\chi,\qquad L_sQ_\chi L_s^*=\chi(s)Q_\chi . \tag{WM2}$ Indeed the last multiplier is $\overline{\chi(r-s)}=\chi(s)\overline{\chi(r)}$. Strong continuity follows from WY0 by conjugation; it also follows directly from L24's translation estimate and the compact-support character estimate (WY4). No inversion of a group parameter is implicit in (WM2).

For a Weyl pair $(U,V)$, let $\pi_+$ be the representation (WY8), and define $\pi_-(a)=\pi_+(a(-\,\cdot)),\qquad q_-g(r)=\int_{\widehat G}g(\chi)\overline{\chi(r)}\,d\chi . \tag{WM3}$ Reflection is an isometric star automorphism of $C_0(G)$. Thus $\pi_-$ is nondegenerate, $\pi_-(q_-g)=V(g)$, and the $q_-g$'s are uniformly dense in $C_0(G)$, by WY1. Its exact covariance is $U_s\pi_-(a)U_s^*=\pi_-(\beta_s a),\qquad (\beta_s a)(r)=a(r-s). \tag{WM4}$ To check the sign, the right side transported through reflection is $a(-r-s)$, which is the right-translation covariance of $\pi_+$ applied to $a(-r)$.

The character functions need not vanish at infinity. We therefore construct their multiplier action, rather than treat them as elements of $C_0(G)$. Let $0\leq e_i\leq1$ be compact cutoffs forming a $C_0(G)$ approximate identity. For $a\in C_b(G)$, the operators $\pi_-(ae_i)$ have norm at most $\|a\|_\infty$. On each generating vector, $\pi_-(ae_i)\pi_-(b)h =\pi_-(ae_i b)h\longrightarrow\pi_-(ab)h \qquad(b\in C_0(G)). \tag{WM5}$ They converge on the dense span of these vectors and hence strongly on all of $H$: approximate an arbitrary vector by one in that span and use the common norm bound. Write the limit as $\widetilde\pi_-(a)$. Formula (WM5) proves independence of the cutoffs and uniqueness among bounded operators with this action on the generating vectors. It also proves multiplication and linearity. Applying the same construction to $\overline a$, and passing to inner products in $\pi_-(ae_i)^*=\pi_-(\overline a e_i)$, proves the adjoint law. Nondegeneracy gives $\widetilde\pi_-(1)=1_H$. This is the unique unital star extension to $C_b(G)$, since any extension must satisfy (WM5).

For a fixed character $\chi$, set $g_\chi(\eta)=g(\chi^{-1}\eta)$ on the dual group. Haar substitution and the representation law give $q_-g_\chi=\overline\chi\,q_-g,\qquad V(g_\chi)=V_\chi V(g). \tag{WM6}$ First use $b=q_-g$ in (WM5), then its uniform density and nondegeneracy. It follows that $\widetilde\pi_-(\overline\chi)=V_\chi. \tag{WM7}$ This also verifies the multiplier convention without a spectral-measure uniqueness theorem.

In the canonical model, integrating the $Q_\chi$'s gives $M_{q_-g}$. Any operator commuting with all $Q_\chi$ commutes with these vector integrals, and their uniform closure contains every $M_a$, $a\in C_0(G)$. Conversely $M_{\overline\chi e_i}\to Q_\chi$ strongly, since compact cutoffs converge strongly on $L^2(G)$, first on $C_c(G)$ and then by density. Consequently $\{Q_\chi:\chi\in\widehat G\}'' =\{M_a:a\in C_0(G)\}'' . \tag{WM8}$ These arguments use only finite-exponent Haar integration and bounded cutoff nets.

Finally, if $W_+$ is the onto unitary of WY3–WY4, put $W_-=W_+(J\otimes1_K).$ Then $U_sW_-=W_-(L_s\otimes1_K),\quad V_\chi W_-=W_-(Q_\chi\otimes1_K),\quad \pi_-(a)W_-=W_-(M_a\otimes1_K). \tag{WM9}$ All three equalities follow by substituting (WM2)–(WM3) into (WY24)–(WY25). The map is onto, and $K$ is unchanged. Thus the left-translation and right-translation realizations have exactly the same multiplicity, including $K=0$.

WY8. Compact kernels, both operator orders and the generated algebra

Continue with (WM3)–(WM4). For $F\in C_c(G\times G)$, define $a_s^F(r)=F(r,r-s),\qquad b_s^F(q)=F(q+s,q). \tag{WM10}$ If the two coordinate projections of $\operatorname{supp}F$ are contained in compact sets $A,B$, then $a_s^F=0$ for $s\notin A-B$, and every slice is supported in $A$. The map $s\mapsto a_s^F$ is continuous in the $C_0$ norm. To see this at $s_0$, restrict $s$ to a compact neighbourhood of $s_0$. The variables $(r,r-s)$, with $r\in A$, then range over a compact set. Continuity at each point and a finite subcover give uniform convergence in $r$ as $s\to s_0$; outside $A$ every slice is zero. This is the compact-parameter argument of L25 Section10.

The integrand $\pi_-(a_s^F)U_s h$ is norm continuous for each $h$, has compact support in $s$, and has norm at most $\|a_s^F\|_\infty\|h\|$. L24's vector integral therefore defines a bounded operator $S_Fh=\int_G\pi_-(a_s^F)U_s h\,ds,\qquad \|S_F\|\leq\int_G\|a_s^F\|_\infty\,ds \leq m(A-B)\|F\|_\infty . \tag{WM11}$ No operator-norm continuity of $U_s$ is required. Since $\beta_s(b_s^F)=a_s^F$, covariance gives the second, exactly equal order $S_Fh=\int_G U_s\pi_-(b_s^F)h\,ds. \tag{WM12}$

Define kernel composition and involution by $(F\star H)(r,q)=\int_G F(r,z)H(z,q)\,dz,\qquad F^\sharp(r,q)=\overline{F(q,r)}. \tag{WM13}$ The product has support in the product of the first projection for $F$ and the second projection for $H$. Its integration variable lies in the compact intersection of their other two projections. The same finite-cover continuity estimate, followed by multiplication by the finite measure of that compact intersection, proves continuity of $F\star H$. Hence $F\star H\in C_c(G\times G)$. The involution plainly preserves compact support and continuity. In a triple product, both intermediate variables lie in fixed compact sets; the product of the three sup norms is an integrable majorant there. L24's qualified Radon-product Fubini theorem gives associativity. Conjugation of (WM13) gives $(F\star H)^\sharp=H^\sharp\star F^\sharp$ and $(F^\sharp)^\sharp=F$.

Here is the full represented product calculation. Write the second translation variable as $t$, and then set $v=s+t$. Covariance makes the coefficient of $U_v$ $\begin{aligned} \int_G a_s^F(r)(\beta_s a_{v-s}^H)(r)\,ds &=\int_G F(r,r-s)H(r-s,r-v)\,ds\\ &=\int_G F(r,z)H(z,r-v)\,dz\\ &=(F\star H)(r,r-v). \end{aligned} \tag{WM14}$ The substitution $z=r-s$ preserves Haar measure because $G$ is abelian. On each vector, the original double integral has a norm majorant supported on a product of compact difference sets; therefore vector Fubini and the substitution are legitimate. For adjoints, first test an inner product against two vectors, use the scalar integral, and set $v=-s$. The resulting coefficient is $(\beta_v\overline{a_{-v}^F})(r) =\overline{F(r-v,r)} =a_v^{F^\sharp}(r).$ Thus the actual bounded operators satisfy $S_FS_H=S_{F\star H},\qquad S_F^*=S_{F^\sharp}. \tag{WM15}$ All product integrals here are Radon integrals on compact, hence sigma-finite, carriers; no equality of uncompleted product Borel sigma algebras is assumed.

In the canonical left model, the substitution $q=r-s$ gives $(S_F\xi)(r)=\int_G F(r,q)\xi(q)\,dq . \tag{WM16}$ This identity is first checked on compact continuous vectors, with the preceding Fubini bounds, and then extended to every $L^2$ vector by (WM11). In particular, for $p,q\in C_c(G)$, $F_{p,q}(r,t)=p(r)\overline{q(t)},\qquad S_{F_{p,q}}=\theta_{p,q},\qquad \theta_{p,q}\xi=p\langle\xi,q\rangle . \tag{WM17}$ In the arbitrary Weyl system, conjugating (WM11) by the already constructed $W_-$ gives instead $S_F=W_-(T_F\otimes1_K)W_-^*,\qquad T_{p,q}:=S_{F_{p,q}} =W_-(\theta_{p,q}\otimes1_K)W_-^* , \tag{WM18}$ where $T_F$ is the canonical kernel operator (WM16). The finite-rank norm argument and extension in WY2–WY3 therefore apply with exactly this kernel convention. Approximation in $L^2$ extends $T_{p,q}$ to arbitrary $p,q\in L^2(G)$. In particular $\begin{split} T_{p,q}T_{u,v}&=\langle u,q\rangle T_{p,v},\qquad T_{p,q}^*=T_{q,p},\\ U_sT_{p,q}&=T_{L_sp,q},\qquad V_\chi T_{p,q}=T_{Q_\chi p,q}. \end{split} \tag{WM19}$ These are identities on all of $H$, not just on an algebraic tensor domain.

For completeness, the compact kernels generate precisely $N=\{U_s,V_\chi\}''$. First $S_F\in N$: an operator in $N'$ commutes with $\pi_-(a)$, by the integrated formula and uniform density in WY7, and then commutes with (WM11). Conversely put $\mathcal N=\{S_F:F\in C_c(G\times G)\}''$. For $a,h\in C_c(G)$, the function $F(r,q)=a(r)h(r-q)$ is compactly supported and gives $S_F=\pi_-(a)U(h)$. The bounds on integrated representations and density extend membership in $\mathcal N$ to $a\in C_0(G)$, $h\in L^1(G)$. Shrinking Haar bumps make $U(h_i)\to1$ strongly, so $\pi_-(a)\in\mathcal N$. Coefficient cutoffs then give $U(h)\in\mathcal N$. Translated Haar bumps give $U_s\in\mathcal N$ for every $s$: their integrals are $U_sU(h_i)$ and converge strongly to $U_s$. Finally $\pi_-(\overline\chi e_i)\to V_\chi$ strongly by (WM5)–(WM7). Thus $N=\mathcal N$.

This argument does not infer generation merely from the algebraic rank-one relations. It uses the actual nondegenerate integrated operators and their strong recovery limits. Uniform approximation of compact kernels by finite product kernels on one compact rectangle, proved in L25 Section10, also gives $\|S_F-S_{F_n}\| \leq m(A-B)\|F-F_n\|_\infty\longrightarrow0 \tag{WM20}$ whenever their common rectangle is $A\times B$. Hence the represented finite-rank algebra has the same von Neumann closure.

WY9. Generated-algebra corners and the multiplicity projection

Assume $H\ne0$, so that $K\ne0$. Work through $W_-$ on $L\otimes K$. WY5 gives the joint commutant $1_L\otimes B(K)$. Its commutant is exactly $B(L)\otimes1_K$, with no countable-basis hypothesis. Here is a direct proof. Fix a unit vector $\eta\in K$. An operator $T$ commuting with $1\otimes B(K)$ preserves $L\otimes\mathbb C\eta$, because it commutes with the projection onto that space. Therefore $T(\xi\otimes\eta)=A\xi\otimes\eta$ for one $A\in B(L)$, with $\|A\|\leq\|T\|$. Commutation with $1\otimes\theta_{\zeta,\eta}$ gives $T(\xi\otimes\zeta)=A\xi\otimes\zeta$ for every $\zeta\in K$. Finite tensor sums are dense, so $T=A\otimes1$. The reverse inclusion follows by multiplication of elementary tensors. Consequently the full generated algebra is $W_-^*NW_-=B(L)\otimes1_K,\qquad W_-^*N'W_-=1_L\otimes B(K). \tag{WM21}$ In the canonical system $K=\mathbb C$, this proves that the Weyl pair generates all of $B(L)$.

For a unit vector $e\in L$, the projection $P_e=\theta_{e,e}\otimes1_K \tag{WM22}$ belongs to the generated algebra and satisfies $P_e(B(L)\otimes1)P_e=\mathbb CP_e$. Indeed $\theta_{e,e}A\theta_{e,e}=\langle Ae,e\rangle\theta_{e,e}$, and every scalar occurs by choosing $A$ scalar. Thus $P_e$ is a nonzero minimal projection of the generated algebra, even when its range has large Hilbert dimension. Its range $e\otimes K$ is unitarily $K$. By contrast, for a unit vector $\eta\in K$, $Q_\eta=1_L\otimes\theta_{\eta,\eta} \tag{WM23}$ is minimal in the commutant, and its range $L\otimes\eta$ is unitarily $L$. These two projection choices recover the two different tensor factors.

The center in (WM21) consists of scalars. If $A\otimes1$ is central, $A$ commutes with every rank-one operator on $L$. Commutation with $\theta_{e,e}$ gives $Ae=ce$; commutation with $\theta_{\xi,e}$ then gives $A\xi=c\xi$ for every $\xi$. This proves factoriality, and (WM22) proves that it is type I. The algebra and the multiplicity are both identified; factoriality by itself would not identify $K$.

The orientation of a rank-one corner is also explicit. Let $f=\theta_{u,v}$, $g=\theta_{p,q}$, with all four vectors nonzero. For every $A\in B(L)$, $\begin{split} gAf^*&=\langle Av,q\rangle\,\theta_{p,u},\\ g^*Af&=\langle Au,p\rangle\,\theta_{q,v}. \end{split} \tag{WM24}$ These follow by applying each side to a vector and using the linear-first inner product. In the first line, take $A=\theta_{q,v}$ to obtain a nonzero coefficient; in the second take $A=\theta_{p,u}$. Thus each full corner space is exactly one dimensional. Its normalized generator is respectively $\theta_{p/\|p\|,u/\|u\|}$ or $\theta_{q/\|q\|,v/\|v\|}$. Each is a partial isometry from the second displayed unit-vector line onto the first. Tensoring with $1_K$ gives the corresponding corner of the full von Neumann algebra and an onto partial isometry between its image ranges.

For example on $\mathbb C^2$, with $E_{ij}=\theta_{e_i,e_j}$, $f=g=E_{01},\quad A=E_{00} \quad\Longrightarrow\quad g^*Af=E_{11}. \tag{WM25}$ The output line is $\mathbb Ce_1$. Every index and adjoint in this calculation determines a different initial or final line.

If one rank-one input is zero, or $H=0$, the corresponding corner is the zero space, of dimension zero. The nonzero qualifications in the one-dimensional and minimal-projection assertions are therefore necessary. The multiplicity construction itself retains the zero system as $K=0$.

WY10. Two inequivalent multiplicities on the same infinite Hilbert space

Let $G=\mathbb Z$, with counting measure, and identify its dual with the circle $\mathbb T$, with probability Haar measure. Write $\chi_z(n)=z^n$. On the single Hilbert space $\mathcal H=\ell^2(\mathbb Z)$, the first Weyl pair is $U_s^{(1)}\delta_k=\delta_{k+s},\qquad V_z^{(1)}\delta_k=z^{-k}\delta_k. \tag{WM26}$ Define the second pair by the unique decomposition $k=2n+a$, $a\in\{0,1\}$: $U_s^{(2)}\delta_{2n+a}=\delta_{2(n+s)+a},\qquad V_z^{(2)}\delta_{2n+a}=z^{-n}\delta_{2n+a}. \tag{WM27}$ The unitary $R:\ell^2(\mathbb Z)\otimes\mathbb C^2\longrightarrow\mathcal H, \qquad R(\delta_n\otimes e_a)=\delta_{2n+a} \tag{WM28}$ is onto because this indexing is a bijection of orthonormal bases. It identifies (WM27) with two canonical copies. For both pairs, evaluation on each basis vector gives $U_s^{(j)}V_z^{(j)}=z^sV_z^{(j)}U_s^{(j)}$. Continuity in the discrete variable is automatic. Continuity in $z$ holds on finite coordinate sums, and the unitary bounds extend it to every $\ell^2$ vector. Thus these are continuous Weyl systems, not just algebraic pairs.

Their ambient Hilbert spaces are identical, but their multiplicities can be recovered without cancelling infinite dimensions. Form the actual strong vector averages $P^{(j)}=\int_{\mathbb T}V_z^{(j)}\,dz . \tag{WM29}$ The scalar Haar integral of $z^{-n}$ is $1$ for $n=0$ and $0$ otherwise. For $n\ne0$, choose $w\in\mathbb T$ with $w^{-n}\ne1$; invariance makes the integral equal to $w^{-n}$ times itself, so it vanishes. Applying this to the basis vectors in (WM26)–(WM27), and using the bounded extension of the integrals, gives $\operatorname{ran}P^{(1)}=\mathbb C\delta_0,\qquad \operatorname{ran}P^{(2)} =\operatorname{span}\{\delta_0,\delta_1\}. \tag{WM30}$ A simultaneous unitary intertwiner of the two pairs would intertwine (WM29), by bounded maps through vector integration, and hence give a unitary between these ranges. Their dimensions are one and two, a contradiction. The multiplicities are exactly one and two although both underlying Hilbert spaces have the same countably infinite dimension.

In the second model, $P^{(2)}=R(\theta_{\delta_0,\delta_0}\otimes1)R^*$ is minimal in the generated algebra and has rank two on $\mathcal H$. The commutant projection $Q=R(1\otimes\theta_{e_0,e_0})R^* \tag{WM31}$ is the projection onto the even coordinates. It is minimal in the commutant but has infinite-dimensional range. This is a concrete instance of the two different factor-recovery statements (WM22)–(WM23).

One and two Weyl copies on the same integer coordinate Hilbert space, with the generated-algebra and commutant projections distinguished

The diagram displays finite windows of infinite bases. In the lower panel, the integer $n$ is the canonical group-coordinate label, while the physical coordinate is $2n+a$. Translation by one moves both rows one column to the right, or two physical coordinates. The blue column is $\operatorname{ran}P^{(2)}$, of dimension two. The orange row is a finite window of $\operatorname{ran}Q$, whose basis continues in both directions and whose dimension is infinite. No truncation of either representation is used in the proof.