Decomposable operators and the diagonal algebra

Written by Claude Opus 5.5 (Anthropic), September 2026. Self-checked by the writing AI. Original text: CC0 1.0.

A direct integral of Hilbert spaces carries two natural algebras of operators. A diagonal operator multiplies each fibre by a scalar that depends measurably on the point. A decomposable operator acts on each fibre by a bounded operator that depends measurably on the point. Every decomposable operator commutes with every diagonal operator. The main result of this lesson is the converse: a bounded operator that commutes with every diagonal operator is decomposable (Theorem 5.1). The standard consequences follow. The diagonal algebra and the decomposable algebra are von Neumann algebras, each is the commutant of the other, and the diagonal algebra is the centre of the decomposable one (Theorem 7.1).

In the simplest case the theorem is elementary. When the base is countable and every point has positive mass, the direct integral is an orthogonal direct sum of the fibres, and the projections onto the summands are diagonal operators. An operator that commutes with these projections maps each summand into itself, so it acts fibre by fibre (Example 9.1). In general, single points may have measure zero, and an operator on the direct integral only gives integrated information. The work is to recover operators on the fibres at almost every point from that information, making only countably many choices. These results are the starting point of reduction theory, which writes a von Neumann algebra as a direct integral over a measure space.

We assume the lesson Measurable fields of Hilbert spaces and their direct integrals and keep its notation; Section 1 recalls what we use from it. We also assume basic measure theory and Hilbert space theory. The few standard facts used without proof are listed at the end. The base is an arbitrary $\sigma$-finite measure space, and the fibres are separable.

A basic reference is [Takesaki I]; [Blackadar] gives a short survey. Direct integrals of Hilbert spaces and decomposable operators were introduced in von Neumann's reduction theory (1949); see [Blackadar, Section III.1.6].

1. Setting and notation

Throughout, $(\Gamma,\Sigma,\mu)$ is a $\sigma$-finite measure space, and measurable means $\Sigma$-measurable. We assume nothing about completeness of $\mu$, standardness of $(\Gamma,\Sigma)$ or countability properties of $\Sigma$.

Measurable fields. Over this base, $(H(\gamma)),\mathfrak M$ is a measurable field of Hilbert spaces. So $\mathfrak M$ is a vector space of sections $\gamma\mapsto\xi(\gamma)\in H(\gamma)$, called measurable sections, with three properties. The norm function $\gamma\mapsto\|\xi(\gamma)\|$ of every $\xi\in\mathfrak M$ is measurable. A section $\eta$ belongs to $\mathfrak M$ whenever $\gamma\mapsto\langle\eta(\gamma),\xi(\gamma)\rangle$ is measurable for every $\xi\in\mathfrak M$. And some sequence in $\mathfrak M$ is fundamental: at every point, its values span a dense subspace of the fibre. In particular every fibre is separable. Inner products are linear in the first variable. For sections $\xi,\eta$ we write $\langle\xi,\eta\rangle$ for the function $\gamma\mapsto\langle\xi(\gamma),\eta(\gamma)\rangle$. We use the basic closure properties of $\mathfrak M$: inner products of measurable sections are measurable functions; $f\xi\in\mathfrak M$ whenever $f$ is a measurable function and $\xi\in\mathfrak M$; pointwise limits of measurable sections are measurable; and measurable sections can be glued along a countable measurable partition of $\Gamma$.

We write $$\mathcal H=\int_\Gamma^\oplus H(\gamma)\,d\mu(\gamma)$$ for the direct integral. Let $\mathcal L^2$ be the space of measurable sections $\xi$ with $\int_\Gamma\|\xi(\gamma)\|^2\,d\mu(\gamma)<\infty$. Then $\mathcal H$ is $\mathcal L^2$ modulo equality almost everywhere, with the inner product $\int_\Gamma\langle\xi(\gamma),\eta(\gamma)\rangle\,d\mu(\gamma)$.

An orthonormal fundamental sequence. Throughout, $(e_k)_{k\geq1}$ is an orthonormal fundamental sequence, and $n(\gamma)=\dim H(\gamma)$. This means that $e_k(\gamma)\neq0$ exactly when $k\leq n(\gamma)$, and that the nonzero $e_k(\gamma)$ form an orthonormal basis of $H(\gamma)$. The dimension function $n$ is measurable. The sets $\Gamma_d=\{n=d\}$, for $d\in\{0,1,2,\ldots,\infty\}$, are the dimension strata; they form a countable measurable partition of $\Gamma$. For a finitely supported complex sequence $q=(q_1,q_2,\ldots)$ we put $v_q=\sum_lq_le_l$. This is a finite sum, so $v_q\in\mathfrak M$. Since the nonzero $e_l(\gamma)$ are orthonormal and the others vanish, $\|v_q(\gamma)\|^2=\sum_{l\leq n(\gamma)}|q_l|^2$. We write $\mathcal Q$ for the countable set of finitely supported sequences with Gaussian-rational entries.

Diagonal operators. For a bounded measurable function $f$, the diagonal operator $m_f$ is multiplication by $f$: $(m_f\xi)(\gamma)=f(\gamma)\xi(\gamma)$. It depends only on the class of $f$ in $L^\infty(\Gamma,\mu)$, and $\|m_f\|\leq\|f\|_\infty$. The map $f\mapsto m_f$ is a unital $*$-homomorphism: $m_{fg}=m_fm_g$, $m_{\bar f}=m_f^*$ and $m_1=1$.

Operator fields. A measurable field of bounded operators on $(H(\gamma))$ is a family $x=(x(\gamma))$ with $x(\gamma)\in B(H(\gamma))$ such that the section $x\xi:\gamma\mapsto x(\gamma)\xi(\gamma)$ is measurable for every $\xi\in\mathfrak M$. Fields of operators between two measurable fields are defined in the same way. The norm function $\gamma\mapsto\|x(\gamma)\|$ of a measurable field is measurable. Sums, scalar multiples, composites and adjoints of measurable fields are measurable, and so are their products with measurable functions. Measurability can be tested on one fundamental sequence: $x$ is measurable as soon as $x\xi_j\in\mathfrak M$ for every member $\xi_j$ of a fundamental sequence.

Decomposable operators. If the norm function of $x$ is essentially bounded, then $x$ acts on $\mathcal H$ by $(x\xi)(\gamma)=x(\gamma)\xi(\gamma)$. This bounded operator is written $\int^\oplus x=\int_\Gamma^\oplus x(\gamma)\,d\mu(\gamma)$, and operators of this form are called decomposable. We use four facts about them.

• (Norm formula) $\|\int^\oplus x\|=\operatorname*{ess\,sup}_\gamma\|x(\gamma)\|$.
• (Algebra) The map $x\mapsto\int^\oplus x$ is linear and multiplicative, and $(\int^\oplus x)^*=\int^\oplus x^*$.
• (Uniqueness) Two fields define the same operator if and only if they agree almost everywhere.
• (Commutation) Every decomposable operator commutes with every diagonal operator.

Commutants. For a set $\mathcal S\subseteq B(\mathcal H)$, its commutant $\mathcal S'$ is the set of bounded operators that commute with every element of $\mathcal S$. A von Neumann algebra on $\mathcal H$ is a $*$-subalgebra $M\subseteq B(\mathcal H)$ with $M''=M$.

Lemma 1.1. Let $\mathcal S\subseteq B(\mathcal H)$ be closed under adjoints. Then $\mathcal S'$ is a $*$-subalgebra of $B(\mathcal H)$ that contains $1$ and is closed in the weak operator topology. In particular, every von Neumann algebra is weakly closed.

Proof. Clearly $\mathcal S'$ is a subalgebra that contains $1$. Let $T\in\mathcal S'$ and $y\in\mathcal S$. Since $y^*\in\mathcal S$, we have $Ty^*=y^*T$, and taking adjoints gives $yT^*=T^*y$. So $T^*\in\mathcal S'$. Next, $Ty=yT$ holds exactly when $\langle T(y\xi),\eta\rangle=\langle T\xi,y^*\eta\rangle$ for all $\xi,\eta\in\mathcal H$. Both sides are weakly continuous functions of $T$, so $\mathcal S'$ is weakly closed. Finally, let $M$ be a von Neumann algebra. Applying what we proved to $\mathcal S=M$ shows that $M'$ is closed under adjoints. Applying it to $\mathcal S=M'$ shows that $M=(M')'$ is weakly closed. $\square$

We never need the bicommutant theorem.

2. The diagonal algebra and the decomposable algebra

Definition 2.1. The diagonal algebra is $\mathcal A=\{m_f:\ f\in L^\infty(\Gamma,\mu)\}$. The decomposable algebra $\mathcal D$ is the set of decomposable operators $\int^\oplus x$ on $\mathcal H$. Here $x$ runs over the measurable fields of bounded operators on $(H(\gamma))$ whose norm function $\gamma\mapsto\|x(\gamma)\|$ is essentially bounded.

Proposition 2.2.

1. $\mathcal A$ is a commutative $*$-subalgebra of $B(\mathcal H)$, and $\mathcal D$ is a $*$-subalgebra. Both contain $1$.
2. $m_f=\int^\oplus f(\gamma)1_{H(\gamma)}\,d\mu(\gamma)$. Hence $\mathcal A\subseteq\mathcal D$ and $\mathcal A\subseteq\mathcal D'$.
3. (Scalar fields) Let $x$ be a measurable field with essentially bounded norm, and suppose that $x(\gamma)$ is a multiple of $1_{H(\gamma)}$ for almost every $\gamma$. Then $\int^\oplus x=m_c$ with $c=\langle xe_1,e_1\rangle$. So the decomposable operators with scalar fibres are exactly the diagonal operators $m_f$.
4. $\|m_f\|=\operatorname*{ess\,sup}_{\{n\geq1\}}|f|$. In particular $m_f=0$ if and only if $f=0$ almost everywhere on $\{n\geq1\}$. Extend each $g\in L^\infty(\{n\geq1\},\mu)$ by $0$ to a function $\tilde g$ on $\Gamma$. Then $g\mapsto m_{\tilde g}$ is an isometric $*$-isomorphism of $L^\infty(\{n\geq1\},\mu)$ onto $\mathcal A$.
5. (Matrix units) For $k,l\geq1$, put $e_{kl}(\gamma)v=\langle v,e_l(\gamma)\rangle e_k(\gamma)$. This is a measurable field with $\|e_{kl}(\gamma)\|\leq1$ and $e_{kl}(\gamma)^*=e_{lk}(\gamma)$. We write $E_{kl}=\int^\oplus e_{kl}\in\mathcal D$.

Proof. (1) The map $f\mapsto m_f$ is a unital $*$-homomorphism defined on the commutative algebra $L^\infty(\Gamma,\mu)$. So its image $\mathcal A$ is a commutative $*$-subalgebra of $B(\mathcal H)$ that contains $m_1=1$. Sums, scalar multiples, composites and adjoints of measurable fields are measurable, and their norm functions stay essentially bounded. Since $x\mapsto\int^\oplus x$ is linear and multiplicative and $(\int^\oplus x)^*=\int^\oplus x^*$, the set $\mathcal D$ is a $*$-subalgebra. It contains $1$, the operator of the identity field.

(2) The field $f1=(f(\gamma)1_{H(\gamma)})$ is measurable: for $\xi\in\mathfrak M$, the section $(f1)\xi=f\xi$ is measurable, because measurable sections are closed under multiplication by measurable functions (measurable fields). Its norm function is $|f|\,1_{\{n\geq1\}}$, which is essentially bounded. The two operators $m_f$ and $\int^\oplus f1$ agree on every vector by definition. So $\mathcal A\subseteq\mathcal D$. Since decomposable operators commute with diagonal operators, every $m_f$ commutes with $\mathcal D$, that is, $\mathcal A\subseteq\mathcal D'$.

(3) Since $xe_1\in\mathfrak M$, the function $c$ is measurable, because inner products of measurable sections are measurable. As $\|e_1(\gamma)\|\leq1$, we have $|c(\gamma)|\leq\|x(\gamma)\|$, so $c$ is essentially bounded. If $x(\gamma)=a1$ and $n(\gamma)\geq1$, then $c(\gamma)=a\langle e_1(\gamma),e_1(\gamma)\rangle=a$. If $n(\gamma)=0$, then $x(\gamma)$ and $c(\gamma)1$ are both the zero operator on the zero space. So $x(\gamma)=c(\gamma)1_{H(\gamma)}$ for almost every $\gamma$. Fields that agree almost everywhere define the same operator, so (2) gives $\int^\oplus x=\int^\oplus c1=m_c$. Conversely, every $m_f$ is decomposable with scalar fibres, by (2).

(4) Apply the norm formula to the field $f1$ of (2). Its norm function is $|f|\,1_{\{n\geq1\}}$, so $\|m_f\|=\operatorname*{ess\,sup}_{\{n\geq1\}}|f|$, and $m_f=0$ exactly when $f=0$ almost everywhere on $\{n\geq1\}$. Every section vanishes on $\{n=0\}$, where the fibres are zero, so $m_f=m_{f1_{\{n\geq1\}}}$. Moreover $f1_{\{n\geq1\}}=\tilde g$ for the restriction $g$ of $f$ to $\{n\geq1\}$. So the map $g\mapsto m_{\tilde g}$ is onto $\mathcal A$. It is a $*$-homomorphism, because $f\mapsto m_f$ is one and extension by $0$ respects products and complex conjugation. It is unital, because $m_{1_{\{n\geq1\}}}=m_1=1$. It is isometric by the norm formula, since $\|m_{\tilde g}\|=\operatorname*{ess\,sup}_{\{n\geq1\}}|g|=\|g\|_\infty$.

(5) For $\xi\in\mathfrak M$, the section $e_{kl}\xi=\langle\xi,e_l\rangle e_k$ is measurable: the function $\langle\xi,e_l\rangle$ is measurable, and multiplying the measurable section $e_k$ by a measurable function gives a measurable section (measurable fields). A rank-one operator $v\mapsto\langle v,b\rangle a$ has norm $\|a\|\,\|b\|$, so $\|e_{kl}(\gamma)\|=\|e_k(\gamma)\|\,\|e_l(\gamma)\|\leq1$. Finally $\langle e_{kl}v,v'\rangle=\langle v,e_l\rangle\langle e_k,v'\rangle=\langle v,e_{lk}v'\rangle$, so $e_{kl}(\gamma)^*=e_{lk}(\gamma)$. $\square$

3. Test vectors and coefficient densities

This section and the next prepare the proof of Theorem 5.1. Let $T$ commute with the diagonal algebra. Its matrix coefficients $\langle Tm_f\,\cdot,\cdot\rangle$, as functions of $f$, are integrals against $f$. Their densities will be the matrix entries of the fibre operators.

Lemma 3.1. Let $T\in\mathcal A'$.

1. (Weight) There is a measurable $w:\Gamma\to(0,1]$ with $\int_\Gamma w\,d\mu<\infty$. Fix one, and put $\lambda(B)=\int_Bw\,d\mu$ for $B\in\Sigma$. Then $\lambda$ is a finite measure with the same null sets as $\mu$.
2. (Test vectors) The sections $\tilde e_k=w^{1/2}e_k$ lie in $\mathcal L^2$. The vectors $m_f\tilde e_k$, for bounded measurable $f$ and $k\geq1$, span a dense subspace of $\mathcal H$.
3. (Densities) For all $k,l\geq1$ there is a measurable, $\lambda$-integrable function $t_{kl}$ that vanishes at every point of $\{n<\max(k,l)\}$ and satisfies $$\langle Tm_f\tilde e_l,\tilde e_k\rangle=\int_\Gamma f\,t_{kl}\,d\lambda \qquad\text{for every bounded measurable }f. \tag{3.1}$$

Proof. (1) Since $\mu$ is $\sigma$-finite, $\Gamma$ is a disjoint union of measurable sets $F_1,F_2,\ldots$ of finite measure. Put $w=\sum_j2^{-j}(1+\mu(F_j))^{-1}1_{F_j}$. Each point lies in exactly one $F_j$, so $0<w\leq\tfrac12$. Also $\int w\,d\mu=\sum_j2^{-j}\mu(F_j)(1+\mu(F_j))^{-1}\leq\sum_j2^{-j}=1$. Since $w>0$ everywhere, $\lambda(B)=0$ exactly when $\mu(B)=0$.

(2) The section $\tilde e_k$ is measurable, since $w^{1/2}$ is a measurable function. Moreover $\|\tilde e_k(\gamma)\|^2\leq w(\gamma)$, which is integrable, so $\tilde e_k\in\mathcal L^2$. Now let $\eta\in\mathcal H$ be orthogonal to every $m_f\tilde e_k$. Put $g_k=w^{1/2}\langle e_k,\eta\rangle$. It is measurable, and $|g_k|\leq w^{1/2}\|\eta(\cdot)\|$. The right side is integrable by the Cauchy–Schwarz inequality in $L^2(\Gamma,\mu)$, since $w^{1/2}$ and $\|\eta(\cdot)\|$ are square integrable. The hypothesis says that $\int fg_k\,d\mu=\langle m_f\tilde e_k,\eta\rangle=0$ for every bounded measurable $f$. Take $f=\overline{g_k}/|g_k|$ on $\{g_k\neq0\}$ and $f=0$ elsewhere. Then $\int|g_k|\,d\mu=0$, so $g_k=0$ almost everywhere. As $w>0$, $\langle e_k(\gamma),\eta(\gamma)\rangle=0$ for almost every $\gamma$. Discarding one null set for each $k$, we find that at almost every $\gamma$ the vector $\eta(\gamma)$ is orthogonal to all $e_k(\gamma)$. These vectors contain an orthonormal basis of $H(\gamma)$, so $\eta(\gamma)=0$ almost everywhere, and $\eta=0$. Hence the vectors $m_f\tilde e_k$ span a dense subspace.

(3) Fix $k,l$. For $B\in\Sigma$ put $\nu_{kl}(B)=\langle Tm_{1_B}\tilde e_l,\tilde e_k\rangle$.

• $\nu_{kl}$ is a complex measure. Let $B$ be the disjoint union of $B_1,B_2,\ldots$ in $\Sigma$. Then $\|1_B\tilde e_l-\sum_{i\leq m}1_{B_i}\tilde e_l\|^2=\int_{\bigcup_{i>m}B_i}\|\tilde e_l\|^2\,d\mu$, which tends to $0$ by dominated convergence. Since $T$ and the inner product are continuous, $\nu_{kl}(B)=\sum_i\nu_{kl}(B_i)$.
• A bound. $T$ commutes with $m_{1_B}$, and $m_{1_B}$ is a self-adjoint idempotent. Hence $$\nu_{kl}(B)=\langle Tm_{1_B}m_{1_B}\tilde e_l,\tilde e_k\rangle=\langle Tm_{1_B}\tilde e_l,\,m_{1_B}\tilde e_k\rangle , \tag{3.2}$$ so $|\nu_{kl}(B)|\leq\|T\|\,\|1_B\tilde e_l\|\,\|1_B\tilde e_k\|$. Here $\|1_B\tilde e_k\|^2=\int_Bw\,\|e_k\|^2\,d\mu\leq\lambda(B)$, and likewise for $l$, so $|\nu_{kl}(B)|\leq\|T\|\lambda(B)$. In particular $\nu_{kl}(B)=0$ whenever $\lambda(B)=0$. By the Radon–Nikodym theorem there is a measurable $\lambda$-integrable $t_{kl}$ with $\nu_{kl}(B)=\int_Bt_{kl}\,d\lambda$ for all $B\in\Sigma$.
• Vanishing off the strata where both vectors live. If $B\subseteq\{n<l\}$, then $1_B\tilde e_l=0$, because $e_l(\gamma)=0$ when $l>n(\gamma)$. If $B\subseteq\{n<k\}$, then $1_B\tilde e_k=0$. In both cases (3.2) gives $\nu_{kl}(B)=0$. So $\int_Bt_{kl}\,d\lambda=0$ for every measurable $B\subseteq\{n<\max(k,l)\}$. Taking for $B$ the parts of that set where the real or imaginary part of $t_{kl}$ is positive or negative shows $t_{kl}=0$ almost everywhere on it. Replacing $t_{kl}$ by $1_{\{n\geq\max(k,l)\}}t_{kl}$ changes no integral, and we make this choice.
• All bounded $f$. (3.1) holds for $f=1_B$ by construction, hence for simple $f$ by linearity. A bounded measurable $f$ is a uniform limit of simple functions $f_j$. Then $|\langle Tm_{f-f_j}\tilde e_l,\tilde e_k\rangle|\leq\|T\|\sup|f-f_j|\,\|\tilde e_l\|\,\|\tilde e_k\|\to0$, and $|\int(f-f_j)t_{kl}\,d\lambda|\leq\sup|f-f_j|\int|t_{kl}|\,d\lambda\to0$. So (3.1) holds for $f$. $\square$

The vanishing clause is where the dimension strata enter. The coefficient $t_{kl}$ lives only where both $e_k$ and $e_l$ are nonzero.

4. A pointwise bound

Lemma 4.1. Let $T\in\mathcal A'$, with $t_{kl}$ as in Lemma 3.1(3). For finitely supported complex sequences $p,q$, put $$\tau_{q,p}=\sum_{k,l}\overline{p_k}\,q_l\,t_{kl},$$ a finite sum of measurable functions. Then $$|\tau_{q,p}(\gamma)|\leq\|T\|\,\|v_q(\gamma)\|\,\|v_p(\gamma)\|\qquad\text{for almost every }\gamma . \tag{4.1}$$

In the proof of Theorem 5.1, $\tau_{q,p}(\gamma)$ becomes $\langle x(\gamma)v_q(\gamma),v_p(\gamma)\rangle$ for the field $x$ that represents $T$. So (4.1) is the bound $\|x(\gamma)\|\leq\|T\|$, tested on one pair of vectors. The difficulty is that $T$ only gives integrated information. The proof localizes to the set where (4.1) fails, with weights chosen so that the two sides of the Cauchy–Schwarz bound can be compared pointwise.

Proof. Put $r=\|v_q(\cdot)\|$ and $s=\|v_p(\cdot)\|$. These are bounded measurable functions. Note that $w^{1/2}v_q=\sum_lq_l\tilde e_l$, and similarly for $p$.

Step 1: an integrated bound. Let $g_1,g_2$ be bounded measurable functions. Since $m_{g_2}^*=m_{\overline{g_2}}$ commutes with $T$, we have $\langle Tm_{g_1}a,m_{g_2}b\rangle=\langle Tm_{g_1\overline{g_2}}a,b\rangle$ for all vectors $a,b$. Expanding $w^{1/2}v_q$ and $w^{1/2}v_p$ and applying (3.1) with $f=g_1\overline{g_2}$ gives $$\langle Tm_{g_1}(w^{1/2}v_q),\,m_{g_2}(w^{1/2}v_p)\rangle =\sum_{k,l}\overline{p_k}\,q_l\,\langle Tm_{g_1\overline{g_2}}\tilde e_l,\tilde e_k\rangle =\int_\Gamma g_1\overline{g_2}\,\tau_{q,p}\,d\lambda .$$ Also $\|m_{g_1}(w^{1/2}v_q)\|^2=\int|g_1|^2r^2\,d\lambda$ and $\|m_{g_2}(w^{1/2}v_p)\|^2=\int|g_2|^2s^2\,d\lambda$. Since $|\langle T a,b\rangle|\leq\|T\|\,\|a\|\,\|b\|$, we get $$\Big|\int_\Gamma g_1\overline{g_2}\,\tau_{q,p}\,d\lambda\Big| \leq\|T\|\Big(\int_\Gamma|g_1|^2r^2\,d\lambda\Big)^{1/2}\Big(\int_\Gamma|g_2|^2s^2\,d\lambda\Big)^{1/2}. \tag{4.2}$$

Step 2: weights supported where (4.1) fails. Let $B=\{|\tau_{q,p}|>\|T\|rs\}$. Let $\varphi$ be the measurable function with $|\varphi|=1$ and $\varphi\,\tau_{q,p}=|\tau_{q,p}|$; take $\varphi=1$ where $\tau_{q,p}=0$. Choose $g_1=1_Bs\varphi$ and $g_2=1_Br$. Then $g_1\overline{g_2}\,\tau_{q,p}=1_B\,rs\,|\tau_{q,p}|$, so the left side of (4.2) is $\int_Brs\,|\tau_{q,p}|\,d\lambda$. Both integrals on the right of (4.2) equal $\int_Br^2s^2\,d\lambda$, so the right side is $\|T\|\int_Br^2s^2\,d\lambda$. Hence $$\int_B rs\,\big(|\tau_{q,p}|-\|T\|rs\big)\,d\lambda\leq0 .$$ On $B$ the integrand is $\geq0$, and its second factor is $>0$. So the integrand vanishes almost everywhere on $B$, and therefore $rs=0$ almost everywhere on $B$.

Step 3: $\tau_{q,p}$ vanishes wherever $rs=0$. Suppose $r(\gamma)=0$. Then $q_l=0$ for every $l\leq n(\gamma)$, and for $l>n(\gamma)$ we have $t_{kl}(\gamma)=0$ by Lemma 3.1(3). So every term of $\tau_{q,p}(\gamma)$ is zero. The case $s(\gamma)=0$ is the same, with $k$ in place of $l$. On $B$ we have $\tau_{q,p}\neq0$, so $rs>0$ at every point of $B$. With Step 2, $\lambda(B)=0$, so $\mu(B)=0$. $\square$

5. Decomposable means commuting with the diagonal algebra

Theorem 5.1 (Commutant of the diagonal algebra). A bounded operator $T$ on $\mathcal H$ commutes with every diagonal operator if and only if $T$ is decomposable. In that case $T=\int^\oplus x$ for a measurable field $x$ with $\|x(\gamma)\|\leq\|T\|$ for every $\gamma$. Then $\|T\|=\operatorname*{ess\,sup}_\gamma\|x(\gamma)\|$, and $x$ is unique up to equality almost everywhere.

Reference: [Takesaki I, Corollary IV.8.16]. For a constant fibre, [Takesaki I, Theorem IV.7.10] allows the fibre to be nonseparable but needs a locally compact base with a Radon measure; here the fibres are separable and the base is any $\sigma$-finite measure space.

Proof. If $T$ is decomposable, it commutes with every $m_f$, since decomposable operators commute with diagonal operators. Conversely, let $T\in\mathcal A'$. Take $w$, $\tilde e_k$ and $t_{kl}$ from Lemma 3.1 and $\tau_{q,p}$ from Lemma 4.1.

Step 1: one null set. For $p,q\in\mathcal Q$, the set $\{|\tau_{q,p}|>\|T\|\,\|v_q\|\,\|v_p\|\}$ is measurable, and it is null by Lemma 4.1. Let $N$ be the union of these countably many sets. It is a measurable null set.

Step 2: the fibre operators. Fix $\gamma\notin N$. Let $D(\gamma)$ be the linear span of the $e_k(\gamma)$ with $k\leq n(\gamma)$; it is dense in $H(\gamma)$. Every $v\in D(\gamma)$ is the finite sum $\sum_{l\leq n(\gamma)}\langle v,e_l(\gamma)\rangle e_l(\gamma)$. On $D(\gamma)$ define $$\beta_\gamma(v,v')=\sum_{k,l\leq n(\gamma)}\overline{\langle v',e_k(\gamma)\rangle}\,\langle v,e_l(\gamma)\rangle\,t_{kl}(\gamma),$$ a finite sum. It is linear in $v$ and conjugate-linear in $v'$. Suppose all coordinates of $v$ and $v'$ are Gaussian rationals. Then $v=v_q(\gamma)$ and $v'=v_p(\gamma)$ for some $p,q\in\mathcal Q$ supported in $\{k\leq n(\gamma)\}$, and $\beta_\gamma(v,v')=\tau_{q,p}(\gamma)$. As $\gamma\notin N$, this gives $|\beta_\gamma(v,v')|\leq\|T\|\,\|v\|\,\|v'\|$. Now fix a finite $m\leq n(\gamma)$. On the span of $e_1(\gamma),\ldots,e_m(\gamma)$, both sides of the inequality are continuous functions of the $2m$ coordinates. Gaussian-rational coordinates are dense, so the inequality holds on the whole span. Every pair of vectors of $D(\gamma)$ lies in such a span. Hence $\beta_\gamma$ is bounded by $\|T\|$ on $D(\gamma)$. A bounded sesquilinear form on a dense subspace extends uniquely to the whole space with the same bound, and a bounded sesquilinear form on a Hilbert space is given by a unique bounded operator (see Background used without proof). So there is a unique $x(\gamma)\in B(H(\gamma))$ with $\langle x(\gamma)v,v'\rangle=\beta_\gamma(v,v')$ for $v,v'\in D(\gamma)$, and $\|x(\gamma)\|\leq\|T\|$. For $\gamma\in N$, put $x(\gamma)=0$.

Step 3: measurability. For all $k,l$ and every $\gamma$, $$\langle x(\gamma)e_l(\gamma),e_k(\gamma)\rangle=1_{\Gamma\setminus N}(\gamma)\,t_{kl}(\gamma). \tag{5.1}$$ Indeed, for $\gamma\notin N$ and $k,l\leq n(\gamma)$ this is the definition of $\beta_\gamma$. If $k>n(\gamma)$ or $l>n(\gamma)$, the left side is $0$ because $e_k(\gamma)=0$ or $e_l(\gamma)=0$, and $t_{kl}(\gamma)=0$ by Lemma 3.1(3). For $\gamma\in N$ both sides are $0$. The right side of (5.1) is measurable, because $N$ is measurable. A section is measurable as soon as its inner products with the members of one fundamental sequence are measurable (the testing criterion). Applied to the section $xe_l$ and the fundamental sequence $(e_k)$, this gives $xe_l\in\mathfrak M$ for every $l$. An operator field that maps every member of a fundamental sequence to a measurable section is measurable (testing operator fields). So $x$ is a measurable field.

Step 4: $T=\int^\oplus x$. Put $X=\int^\oplus x$. Since $\|x(\gamma)\|\leq\|T\|$ for every $\gamma$, the norm formula shows that $X$ is bounded with $\|X\|\leq\|T\|$. For bounded measurable $f$, (5.1), the fact that $N$ is null, and (3.1) give $$\langle Xm_f\tilde e_l,\tilde e_k\rangle=\int_\Gamma f\,w\,\langle xe_l,e_k\rangle\,d\mu=\int_\Gamma f\,t_{kl}\,d\lambda=\langle Tm_f\tilde e_l,\tilde e_k\rangle .$$ Both $X$ and $T$ commute with every $m_g$: $X$ because it is decomposable, $T$ by hypothesis. So for bounded measurable $f,g$, $$\langle Xm_f\tilde e_l,m_g\tilde e_k\rangle=\langle Xm_{f\bar g}\tilde e_l,\tilde e_k\rangle=\langle Tm_{f\bar g}\tilde e_l,\tilde e_k\rangle=\langle Tm_f\tilde e_l,m_g\tilde e_k\rangle .$$ By Lemma 3.1(2), the vectors $m_f\tilde e_l$ span a dense subspace. Sesquilinearity and continuity give $\langle X\xi,\eta\rangle=\langle T\xi,\eta\rangle$ for all $\xi,\eta$, so $T=X$. The identity $\|T\|=\operatorname*{ess\,sup}_\gamma\|x(\gamma)\|$ is the norm formula. The field $x$ is unique up to equality almost everywhere, because two fields that define the same operator agree almost everywhere. $\square$

Remark 5.2 (What the proof uses). In the arguments of this lesson, $\sigma$-finiteness enters once, through the weight $w$ of Lemma 3.1(1). It makes $\lambda$ finite and the Radon–Nikodym step available. The norm formula and the uniqueness of fields, which we take from the lesson on direct integrals, also use it. Separability of the fibres enters through the countable sequence $(e_k)$ and the countable set $\mathcal Q$. Nothing else is used: no completeness of $\mu$, no standard Borel structure, no countably generated $\sigma$-algebra, and no separability of $\mathcal H$. For example, let $\Gamma=\{0,1\}^I$ for an uncountable set $I$, with the product $\sigma$-algebra and the product of the fair-coin measures, and let $H(\gamma)=\mathbb C$. The functions $\gamma\mapsto2\gamma_i-1$, $i\in I$, are orthonormal in $L^2(\Gamma,\mu)$, so $\mathcal H=L^2(\Gamma,\mu)$ is not separable. The theorem still applies. Every choice in the proof is countable: the fibre operators are defined outside one null set $N$, a countable union of null sets. In particular the proof needs no lifting of $L^\infty(\Gamma,\mu)$, that is, no linear and multiplicative choice of one representative in every class.

6. Intertwiners between two direct integrals

Proposition 6.1. Let $(H(\gamma)),\mathfrak M$ and $(K(\gamma)),\mathfrak N$ be measurable fields over the same base, with direct integrals $\mathcal H$ and $\mathcal K$. Write $m^H_f$ and $m^K_f$ for their diagonal operators. A bounded operator $T:\mathcal H\to\mathcal K$ satisfies $Tm^H_f=m^K_fT$ for every bounded measurable $f$ if and only if $T=\int^\oplus x$ for a measurable field $x$ of bounded operators from $H(\gamma)$ to $K(\gamma)$ with essentially bounded norm. The field can be chosen with $\|x(\gamma)\|\leq\|T\|$ for every $\gamma$.

Proof. If $T=\int^\oplus x$, then $x(\gamma)(f(\gamma)v)=f(\gamma)x(\gamma)v$ pointwise, so $T$ intertwines. For the converse, let $L(\gamma)=H(\gamma)\oplus K(\gamma)$ be the direct-sum field, whose measurable sections are the pairs $(\xi,\eta)$ with $\xi\in\mathfrak M$ and $\eta\in\mathfrak N$. Let $\mathcal L$ be its direct integral. A measurable pair $(\xi,\eta)$ is square integrable exactly when both components are, since $\|(\xi,\eta)\|^2=\|\xi\|^2+\|\eta\|^2$. It vanishes almost everywhere exactly when both components do. So $W(\xi,\eta)=(\xi,\eta)$ is a unitary from $\mathcal L$ onto $\mathcal H\oplus\mathcal K$, and $Wm^L_fW^*=m^H_f\oplus m^K_f$. Put $$\tilde T=W^*\begin{pmatrix}0&0\\ T&0\end{pmatrix}W .$$ Both $\tilde Tm^L_f$ and $m^L_f\tilde T$ send $W^*(\xi,\eta)$ to $W^*(0,Tm^H_f\xi)=W^*(0,m^K_fT\xi)$. So $\tilde T$ commutes with the diagonal algebra of $\mathcal L$. By Theorem 5.1, $\tilde T=\int^\oplus z$ with $\|z(\gamma)\|\leq\|\tilde T\|=\|T\|$ for every $\gamma$. Let $x(\gamma)$ be the lower-left block of $z(\gamma)$: $x(\gamma)v$ is the second component of $z(\gamma)(v,0)$. It is a measurable field, because a field of operators on a direct-sum field is measurable exactly when its four matrix blocks are. Moreover $\|x(\gamma)\|\leq\|z(\gamma)\|\leq\|T\|$. For $\xi\in\mathcal H$, $T\xi$ is the second component of $W\tilde TW^*(\xi,0)$, which is $x\xi$. So $T=\int^\oplus x$. $\square$

7. Commutants, von Neumann algebras and the centre

Theorem 7.1.

1. $\mathcal A'=\mathcal D$ and $\mathcal D'=\mathcal A$.
2. $\mathcal A$ and $\mathcal D$ are von Neumann algebras on $\mathcal H$: $\mathcal A''=\mathcal A$ and $\mathcal D''=\mathcal D$. In particular, both are closed in the weak operator topology.
3. The centre $\mathcal D\cap\mathcal D'$ of $\mathcal D$ is $\mathcal A$.
4. The following are equivalent: (a) $\mathcal A$ is maximal abelian, that is, $\mathcal A'=\mathcal A$; (b) $\mathcal D$ is abelian; (c) $\dim H(\gamma)\leq1$ for almost every $\gamma$.
5. For every $\sigma$-finite measure space $(\Gamma,\Sigma,\mu)$, the multiplication operators by functions in $L^\infty(\Gamma,\mu)$ form a von Neumann algebra on $L^2(\Gamma,\mu)$ that is maximal abelian.