Spectral multiplicity model for separably acting abelian von Neumann algebras

By [F1], choose the bounded self-adjoint generator $S$ with $\mathcal{A}=W^{\ast }(S)$. By [F2], $K=\sigma (S)$ is nonempty compact and contained in $\mathbb{R}$.

Regard $S$ as a bounded normal operator. Apply [F3] and enumerate the nonzero cyclic reducing summands as $H_j$, where the index set is either $\{1,\dots ,N\}$ or $\mathbb{N}$. Choose a cyclic vector $x_j\mathrm{\ne }0$ in each summand, and apply [F4] to obtain finite nonzero regular Borel measures ${\mu }_j$ on $K$ and cyclic unitaries $V_j:L^2(K,{\mu }_j)\to H_j$. Their direct sum is a unitary from ${⨁}_j{L}^2(K,{\mu }_j)$ onto $H$, and it intertwines every bounded Borel function of $S$ componentwise. AC supplies the cyclic decomposition and the cyclic vectors; no zero summand is used.

Put $a_j=\frac{2^{-j}}{1+{\mu }_j(K)},\mu ={\sum }_j{a}_j{\mu }_j\mathrm{.}$ By [F5] this is a Borel measure, and $0<\mu (K)={\sum }_j\frac{2^{-j}{\mu }_j(K)}{1+{\mu }_j(K)}\le {\sum }_j{2}^{-j}<\mathrm{\infty }\mathrm{.}$ It dominates each ${\mu }_j$ because $\mu (B)\ge a_j{\mu }_j(B)$ for every Borel $B$. It is regular by [F8]: it is finite and hence locally finite on compact $K$, and every open subset of the compact metric space $K$ is sigma-compact. For an open $O\mathrm{\ne }K$, compact sets $C_n=\{t\in K:d(t,K\setminus O)\ge 1\mathrm{/}n\}$ increase to $O$; the cases $O=\mathrm{\varnothing }$ and $O=K$ are immediate.

By [F5], take Borel Radon--Nikodym densities $h_j=d{\mu }_j\mathrm{/}d\mu$. Each $h_j$ is nonnegative almost everywhere: if $E_n=\{h_j\le -1\mathrm{/}n\}$, then $0\le {\mu }_j(E_n)={\int }_{E_n}{h}_{j}d\mu \le -\mu (E_n)\mathrm{/}n,$ so $\mu (E_n)=0$; the negative set is their countable union. Replace $h_j$ by zero there. Since ${\mu }_j(K)<\mathrm{\infty }$, the set formula also implies that $h_j$ is finite almost everywhere. Replace it by zero on any Borel null set where it is nonfinite, so each chosen $h_j$ is Borel, finite and nonnegative. Let $F_j=\{t:h_j(t)>0\},K_0={\bigcup }_j{F}_j\mathrm{.}$ The set-integral formula in [F5] gives ${\mu }_j(K\setminus F_j)=0$ and then ${\mu }_j(K\setminus K_0)=0$ for every $j$. The weighted-sum formula yields $\mu (K\setminus K_0)=0$.

Define $m_0(t)={\sum }_j{\mathbf{1}}_{F_j}(t)$ on $K_0$, and set $m(t)=m_0(t)$ there and $m(t)=1$ on $K\setminus K_0$. Countable sums of Borel indicators make $m$ Borel; it takes values in $\{1,2,\dots \}\cup \{\mathrm{\infty }\}$ and is the number of active cyclic coordinates for $\mu$-almost every $t$. For $A_r=\{t:m(t)\ge r\}$, the $r$-th active index $k_r(t)=\min \{j:t\in F_j,1+\mathrm{\#}\{i<j:t\in F_i\}=r\}$ is Borel on $A_r\cap K_0$: its level set at $j$ is $F_j\cap \{{\sum }_{i<j}{\mathbf{1}}_{F_i}=r-1\}$. The rank $r_j(t)=1+{\sum }_{i<j}{\mathbf{1}}_{F_i}(t)$ is Borel on $F_j$. These least-index definitions introduce no choice.

Give $H_t$ the span of the first $m(t)$ standard vectors in ${\mathrm{\ell }}^2$; when $m(t)=\mathrm{\infty }$ this is all of ${\mathrm{\ell }}^2$. The sections $e_r(t)={\mathbf{1}}_{A_r}(t)e_r$ form a countable fundamental family because their Gram coefficients are ${\delta }_{rs}{\mathbf{1}}_{A_r\cap A_s}$ and their span is dense in every fibre. Thus this is a measurable Hilbert field over the standard-Borel finite-measure space $(K,\mu )$. Its direct integral identifies with ${⨁}_{r\ge 1}{L}^2(A_r,\mu )$: a section has measurable coordinates $f_r$ supported on $A_r$, its squared fibre norm is ${\sum }_r\mathrm{\mid }{f}_r{\mathrm{\mid }}^2$, and monotone convergence of the finite coordinate sums gives ${\int }_K{\sum }_r\mathrm{\mid }{f}_r{\mathrm{\mid }}^{2}d\mu ={\sum }_r{\int }_{A_r}\mathrm{\mid }{f}_r{\mathrm{\mid }}^{2}d\mu \mathrm{.}$ This also proves the identification in both directions, including the countably infinite fibre.

On $K_0$, define a map from the cyclic direct sum by $f_r(t)=\sqrt{h_{k_r(t)}(t)}{g}_{k_r(t)}(t)(t\in A_r\cap K_0),$ and put $f_r=0$ on the remaining null set. The rank partitions and [F15] make every coordinate Borel. The identity [F6] gives ${\sum }_r{\int }_{A_r}\mathrm{\mid }{f}_r{\mathrm{\mid }}^{2}d\mu ={\sum }_j{\int }_{F_j}{h}_j\mathrm{\mid }{g}_j{\mathrm{\mid }}^{2}d\mu ={\sum }_j{\int }_K\mathrm{\mid }{g}_j{\mathrm{\mid }}^{2}d{\mu }_j\mathrm{.}$ For each finite partial sum these equalities follow by applying [F6] on each rank piece; monotone convergence passes to the countable sums, and the active rank enumeration is a bijection from the active $j$ coordinates to the $r=1,\dots ,m(t)$ coordinates at every $t\in K_0$. Hence the map is well-defined and isometric. It is onto: for a field $(f_r)$, set $g_j(t)=\{\begin{array}{ll}{f}_{r_j(t)}(t)\mathrm{/}\sqrt{h_j(t)},& t\in F_j,\\ 0,& t\mathrm{\notin }{F}_j\mathrm{.}\end{array}$ The Borel rank partition makes $g_j$ measurable, and the same integral identity shows that $(g_j)$ lies in the cyclic Hilbert sum and maps back to $(f_r)$ almost everywhere. Thus the constructed map $W$ is unitary and commutes with all bounded Borel scalar multipliers.

Let $\mathcal{D}=\{M_f:f\in L^{\mathrm{\infty }}(K,\mu )\}$ on this direct integral. By [F9], $\mathcal{D}$ is a WOT-closed unital $\ast$-algebra and contains $M_t$ because $K$ is compact. Therefore $W^{\ast }(M_t)\subseteq \mathcal{D}\mathrm{.}$

For the reverse inclusion, fix $f\in L^{\mathrm{\infty }}(K,\mu )$ and let $M=\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$. By [F12] choose a Borel representative and change it on a Borel $\mu$-null set so that $\mathrm{\mid }f(t)\mathrm{\mid }\le M$ everywhere. If $M=0$, $M_f=0$. Assume $M>0$. Choose a countable dense sequence ${\eta }_j$ in the nonzero direct-integral Hilbert space and measurable representatives; [F7] makes $t\mapsto \mathrm{\parallel }{\eta }_j(t){\mathrm{\parallel }}^2$ Borel. Define ${\nu }_j(B)={\int }_B\mathrm{\parallel }{\eta }_j(t){\mathrm{\parallel }}^{2}d\mu (t),b_j=\frac{2^{-j}}{1+\mathrm{\parallel }{\eta }_j{\mathrm{\parallel }}^2},\nu ={\sum }_j{b}_j{\nu }_j\mathrm{.}$ Each ${\nu }_j\ll \mu$, so changing $f$ on a Borel $\mu$-null set does not alter any $L^2({\nu }_j)$ class or the $L^2(\nu )$ class. By [F13] each ${\nu }_j$ and $\nu$ is a Borel measure, and $\nu (K)\le 2$ since ${\nu }_j(K)=\mathrm{\parallel }{\eta }_j{\mathrm{\parallel }}^2$ and ${\sum }_{j\in \mathbb{N}}{2}^{-j}=2$. It is nonzero because a dense sequence in a nonzero Hilbert space cannot consist entirely of zero vectors. By [F8] $\nu$ is regular, and [F16] supplies DC for [F8]'s hypotheses and the $p=2$ case of the $C_c(K)$-density theorem. Write $f=u+iv$ with real-valued $u,v$. Both belong to real $L^2(\nu )$ because $\mathrm{\mid }u\mathrm{\mid },\mathrm{\mid }v\mathrm{\mid }\le \mathrm{\mid }f\mathrm{\mid }$. Since the library's $C_c(K)$ is real-valued and $K$ is compact, $C_c(K)=C(K;\mathbb{R})$; apply the density theorem separately to choose $u_n,v_n\in C(K;\mathbb{R})$ converging to $u,v$ in real $L^2(\nu )$. Then ${\psi }_n=u_n+i{v}_n\in C(K;\mathbb{C})$ and $\mathrm{\parallel }{\psi }_n-f{\mathrm{\parallel }}_{L^2(\nu )}\to 0$. Radially clip each ${\psi }_n$ to the disc of radius $M$: since $\mathrm{\mid }f\mathrm{\mid }\le M$, the clipped functions ${\psi }_n^{(c)}$ satisfy $\mathrm{\mid }{\psi }_n^{(c)}(t)-f(t)\mathrm{\mid }\le \mathrm{\mid }{\psi }_n(t)-f(t)\mathrm{\mid }$ and $\mathrm{\parallel }{\psi }_n^{(c)}{\mathrm{\parallel }}_{\mathrm{\infty }}\le M$, and they still converge to $f$ in $L^2(\nu )$. Fix $\xi$ in the direct integral and a member ${\eta }_j$ of the dense sequence. Cauchy--Schwarz in $L^2(\mu )$ and the pointwise bound $\mathrm{\mid }⟨\xi (t),{\eta }_j(t)⟩\mathrm{\mid }\le \mathrm{\parallel }\xi (t)\mathrm{\parallel }\mathrm{\parallel }{\eta }_j(t)\mathrm{\parallel }$ give $\mathrm{\mid }⟨(M_{{\psi }_n^{(c)}}-M_f)\xi ,{\eta }_j⟩\mathrm{\mid }\le \mathrm{\parallel }\xi {\mathrm{\parallel }}_{\mathcal{H}}({\int }_K\mathrm{\mid }{\psi }_n^{(c)}-f{\mathrm{\mid }}^2\mathrm{\parallel }{\eta }_j{\mathrm{\parallel }}^{2}d\mu {)}^{1\mathrm{/}2}$, and the integral equals $\mathrm{\parallel }{\psi }_n^{(c)}-f{\mathrm{\parallel }}_{L^2({\nu }_j)}^2\le b_j^{-1}\mathrm{\parallel }{\psi }_n^{(c)}-f{\mathrm{\parallel }}_{L^2(\nu )}^2$ because $\nu ={\sum }_k{b}_k{\nu }_k$ with $b_j>0$. Hence $⟨(M_{{\psi }_n^{(c)}}-M_f)\xi ,{\eta }_j⟩\to 0$ for every $j$. The clipped approximants are uniformly bounded by $M$, so $\mathrm{\mid }⟨(M_{{\psi }_n^{(c)}}-M_f)\xi ,\eta -{\eta }_j⟩\mathrm{\mid }\le 2M\mathrm{\parallel }\xi {\mathrm{\parallel }}_{\mathcal{H}}\mathrm{\parallel }\eta -{\eta }_j{\mathrm{\parallel }}_{\mathcal{H}}$ extends this convergence from the dense family to every $\eta$; thus $M_{{\psi }_n^{(c)}}\to M_f$ in the weak operator topology. Each $M_{{\psi }_n^{(c)}}={\psi }_n^{(c)}(M_t)$ lies in $C^{\ast }(I,M_t)\subseteq W^{\ast }(M_t)$ by the continuous functional calculus, and $W^{\ast }(M_t)$ is WOT closed, so $M_f\in W^{\ast }(M_t)$.

Fix $S$ and compare any two models $(\mu ,m,U_1)$ and $(\nu ,n,U_2)$ on $K$. Choose a countable dense sequence $(z_j)$ in $H$ and define $\rho (B)={\sum }_j\frac{2^{-j}}{1+\mathrm{\parallel }{z}_j{\mathrm{\parallel }}^2}⟨E(B)z_j,z_j⟩\mathrm{.}$ By [F5, F10, F17] this is a finite nonzero Borel measure. For a Borel set $B$, $\rho (B)=0$ exactly when $E(B)z_j=0$ for every $j$, which by density and boundedness of the projection $E(B)$ is equivalent to $E(B)=0$. In either model, [F10] identifies $U_{i}E(B)U_i^{-1}$ with $M_{{\mathbf{1}}_B}$, and this operator is zero exactly when its model measure of $B$ is zero: if the measure is positive, the section ${\mathbf{1}}_B{e}_1$ is a nonzero square-integrable section because the fibre dimension is at least one almost everywhere. Therefore $\mu (B)=0⟺E(B)=0⟺\nu (B)=0,$ so the two measures have the same measure class.

Compose $W$ with the inverse of the direct sum of the cyclic unitaries $V_j$. This gives a unitary $U:H⟶{\int }_K^{\oplus }{H}_{t}d\mu (t)$ that intertwines every bounded Borel function of $S$ with the same scalar multiplier; in particular $US{U}^{-1}=M_t$.

Let $g=d\nu \mathrm{/}d\mu$ and $q=d\mu \mathrm{/}d\nu$. Since both measures are finite, their Radon--Nikodym set formulas make these densities finite almost everywhere. As in Step 1.4, the threshold-set argument gives nonnegative Borel versions; replace them by zero on the Borel null sets where they are negative or nonfinite. If $g=0$ on a set of positive $\mu$-measure, the set formula gives zero $\nu$-measure there, contradicting $\mu \ll \nu$; similarly $q>0$ $\nu$-almost everywhere. Applying [F6] to the set formula for $g$ and the nonnegative function $q$ gives ${\int }_{B}qgd\mu ={\int }_{B}qd\nu =\mu (B)(B\subseteq K\text{ Borel})\mathrm{.}$ Uniqueness in [F5] therefore gives $qg=1$ almost everywhere. The map $J:L^2(K,\nu ;{\mathbb{C}}^{n(t)})⟶L^2(K,\mu ;{\mathbb{C}}^{n(t)}),J\xi =\sqrt{g}\xi ,$ is an isometry by [F6]; multiplication by $\sqrt{q}$ is its inverse, so it is unitary. It commutes with every bounded Borel scalar multiplier. Hence $R=J{U}_2{U}_1^{-1}$ is a unitary from the first model to the second model written over $\mu$, and it intertwines every scalar multiplier.

Since $\mathcal{A}=W^{\ast }(S)$, unitary conjugation and steps 1.8--1.9 yield $U\mathcal{A}{U}^{-1}=W^{\ast }(US{U}^{-1})=W^{\ast }(M_t)=\mathcal{D}\mathrm{.}$ These are both inclusions: WOT closedness gives $W^{\ast }(M_t)\subseteq \mathcal{D}$, and the bounded-continuous approximation gives the reverse inclusion.

Over $(K,\mu )$ form the measurable direct-sum field $Z_t={\mathbb{C}}^{m(t)}\oplus {\mathbb{C}}^{n(t)}$ using the interleaved fundamental families: their Gram coefficients are the two Borel Gram matrices in diagonal blocks, so the field is measurable and has dense fundamental span. Its direct integral identifies with the Hilbert sum of the two model spaces by pairing section coordinates and adding their squared norms. In that sum, define the off-diagonal operator $\widehat{R}(x,y)=(0,Rx)\mathrm{.}$ It commutes with every scalar multiplier. The commutant theorem [F9] makes it decomposable, say $\widehat{R}={\int }^{\oplus }{T}_{t}d\mu (t)$ on $Z_t$. Let $P_F,P_G$ be the global projections onto the first and second summands, induced by the pointwise projections $Q_F(t),Q_G(t)$. Globally, $\widehat{R}=P_G\widehat{R}{P}_F,{\widehat{R}}^{\ast }\widehat{R}=P_F,\widehat{R}{\widehat{R}}^{\ast }=P_G\mathrm{.}$ The adjoint/product and exact-norm clauses in [F9] imply, outside a common Borel null set, $T_t=Q_G(t)T_t{Q}_F(t),T_t^{\ast }{T}_t=Q_F(t),T_t{T}_t^{\ast }=Q_G(t)\mathrm{.}$ Indeed each difference field induces the zero operator, so its essential supremum norm is zero; the finitely many exceptional null sets can be united. On this common conull set, the off-diagonal block of $T_t$ is a unitary map ${\mathbb{C}}^{m(t)}\to {\mathbb{C}}^{n(t)}$. The dimensions of unitarily isomorphic finite or countably infinite Hilbert spaces agree, so $m(t)=n(t)$ almost everywhere. This argument includes both different finite ranks and the finite/infinite and infinite/infinite cases; it uses no published intertwiner-dimension lemma.

Steps 1.1--1.9, 2.1, and 3.1 prove existence of the generator, finite regular measure, measurable multiplicity field, unitary spectral model and algebra identity. Steps 1.10, 2.2, and 3.2 prove fixed-generator measure-class and almost-everywhere multiplicity uniqueness. No uniqueness is asserted across different generators. [step 1.1, step 1.2, step 1.7, step 2.1, step 3.1, step 1.10, step 2.2, step 3.2] $\mathrm{\square }$