# Direct integrals of von Neumann algebras *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 \(\mathcal H=\int_\Gamma^\oplus H(\gamma)\,d\mu(\gamma)\) comes with two von Neumann algebras of its own. The *diagonal algebra* \(\mathcal A\) multiplies every fibre by a scalar that varies measurably with the point. The *decomposable algebra* \(\mathcal D\) acts on every fibre by a bounded operator. Each of the two is the commutant of the other. Many algebras lie between them. Pick, at each point \(\gamma\), a von Neumann algebra \(M(\gamma)\) on the fibre \(H(\gamma)\), and keep the decomposable operators whose fibres lie in \(M(\gamma)\) at almost every point. The resulting set is written \(\int_\Gamma^\oplus M(\gamma)\,d\mu(\gamma)\). This lesson studies that construction. When the family \(\gamma\mapsto M(\gamma)\) is measurable, in the sense of Section 1, the set is a von Neumann algebra. Its commutant is obtained by integrating the commutants \(M(\gamma)'\), and its centre by integrating the centres \(M(\gamma)\cap M(\gamma)'\) (Theorems 3.2 and 4.3). So the centre equals the diagonal algebra exactly when almost all the \(M(\gamma)\) are factors. The family can be recovered from its integral, up to a null set (Theorem 4.1). Conversely, every von Neumann algebra that contains \(\mathcal A\) and is contained in \(\mathcal D\) arises in this way (Theorem 5.3). Measurable families, taken up to null sets, therefore match the von Neumann algebras between \(\mathcal A\) and \(\mathcal D\) one to one. These facts form the core of von Neumann's reduction theory (1949); see [Blackadar, Section III.1.6]. After a von Neumann algebra has been written as an integral over an abelian subalgebra of its centre, many questions about it become questions about the fibres. When the whole centre is used, the fibres are factors. The delicate point is measurability. That the fields of commutants and of centres are again measurable rests on the Borel space of von Neumann algebras introduced in [Effros 1965]. The base is an arbitrary \(\sigma\)-finite measure space throughout. We need no standard Borel structure, no completeness of the measure and no separability of \(\mathcal H\). Section 6 explains where standard bases will matter later. It also gives a base on which a harmless-looking family is not measurable, and on which the commutant formula fails for it. We assume the lessons [Measurable fields of Hilbert spaces and their direct integrals](../reader/supplements/measurable-fields-direct-integrals.html), [Decomposable operators and the diagonal algebra](../reader/supplements/decomposable-operators-diagonal-algebra.html) and [The Effros Borel structure](../reader/supplements/effros-borel-structure.html). Every fact we take from them is stated in full in the section *Background used without proof*. Example 7.2 also uses [Spatial tensor products of von Neumann algebras](../reader/supplements/spatial-tensor-products.html). Basic references are [Takesaki I, Chapter IV] and [Blackadar, Section III.1.6]. ## Conventions 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 \(\Sigma\). A *null set* is a set \(N\in\Sigma\) with \(\mu(N)=0\). A statement holds *almost everywhere*, or *for almost every* \(\gamma\), if it holds for all \(\gamma\) outside some null set. A subset \(E\subseteq\Gamma\) is *measurable up to a null set* if some \(F\in\Sigma\) has \(E\,\triangle\,F\) contained in a null set. When \(\Sigma\) is complete for \(\mu\), these are the familiar notions. Hilbert spaces are complex, inner products are linear in the first variable, and the zero space is allowed. For a Hilbert space \(K\), \(B(K)\) is the algebra of bounded operators on \(K\), and \(S'\) is the commutant of a set \(S\subseteq B(K)\). A *von Neumann algebra* on \(K\) is a \(*\)-subalgebra \(M\subseteq B(K)\) with \(M''=M\). The von Neumann algebra *generated* by a set \(S\) is \((S\cup S^*)''\). The *centre* of \(M\) is \(M\cap M'\), and \(M\) is a *factor* if its centre is \(\mathbb C1\). For von Neumann algebras \(M_1,M_2\) on \(K\) we put \(M_1\vee M_2=(M_1\cup M_2)''\). On the zero space, \(B(0)=\{0\}=\mathbb C1\) is the only von Neumann algebra, and it is a factor by this definition. We use four elementary facts about commutants. Let \(S,T\subseteq B(K)\). - **(C1)** If \(S\subseteq T\), then \(T'\subseteq S'\). Also \(S\subseteq S''\), and hence \(S'''=S'\). - **(C2)** \(S'\) is a subalgebra of \(B(K)\) containing \(1\). If \(S^*=S\), then \(S'\) is a von Neumann algebra. - **(C3)** The von Neumann algebra generated by \(S\) is the smallest von Neumann algebra containing \(S\), and its commutant is \((S\cup S^*)'\). - **(C4)** If \(M_1,M_2\) are von Neumann algebras, then \(M_1\cap M_2=(M_1'\cup M_2')'\) is a von Neumann algebra. In particular the centre of a von Neumann algebra is a von Neumann algebra. Moreover \(M_1\vee M_2\) is the von Neumann algebra generated by \(M_1\cup M_2\), and \((M_1\vee M_2)'=M_1'\cap M_2'\). **Proof.** (C1) An operator that commutes with all of \(T\) commutes with all of \(S\). Every element of \(S\) commutes with every element of \(S'\), so \(S\subseteq S''\). Applying the first statement to \(S\subseteq S''\) gives \(S'''\subseteq S'\), and applying the second to \(S'\) gives \(S'\subseteq S'''\). (C2) Sums and products of operators that commute with \(S\) commute with \(S\), and so does \(1\). Let \(S^*=S\) and \(x\in S'\). For \(s\in S\) we have \(xs^*=s^*x\), and taking adjoints gives \(sx^*=x^*s\). So \(x^*\in S'\), and \(S'\) is a \(*\)-algebra. It equals its own bicommutant by (C1). (C3) The set \(S\cup S^*\) is self-adjoint, so \((S\cup S^*)'\) is a self-adjoint set by (C2), and \((S\cup S^*)''\) is a von Neumann algebra by (C2) again. It contains \(S\) by (C1). If a von Neumann algebra \(P\) contains \(S\), it contains \(S\cup S^*\), and then \(P=P''\supseteq(S\cup S^*)''\) by (C1) used twice. The commutant is \((S\cup S^*)'''=(S\cup S^*)'\) by (C1). (C4) An operator commutes with every element of \(M_1'\cup M_2'\) exactly when it lies in \(M_1''\cap M_2''=M_1\cap M_2\). The set \(M_1'\cup M_2'\) is self-adjoint, so \(M_1\cap M_2\) is a von Neumann algebra by (C2). The set \(M_1\cup M_2\) is self-adjoint too, so it generates \((M_1\cup M_2)''=M_1\vee M_2\), whose commutant is \((M_1\cup M_2)'=M_1'\cap M_2'\) by (C3). \(\square\) The bicommutant theorem is not needed anywhere in this lesson. ## Background used without proof Throughout, \((H(\gamma))_{\gamma\in\Gamma}\), with its space \(\mathfrak M\) of measurable sections, is a fixed measurable field of Hilbert spaces over \((\Gamma,\Sigma,\mu)\), and \(\mathcal H=\int_\Gamma^\oplus H(\gamma)\,d\mu(\gamma)\) is its direct integral. The following facts are proved in the lessons named. From [Measurable fields of Hilbert spaces and their direct integrals](../reader/supplements/measurable-fields-direct-integrals.html): - **(B1) Fields.** \(\mathfrak M\) is a vector space of sections \(\xi\), with \(\xi(\gamma)\in H(\gamma)\), with three properties. The function \(\gamma\mapsto\|\xi(\gamma)\|\) is measurable for \(\xi\in\mathfrak M\). A section \(\eta\) belongs to \(\mathfrak M\) as soon as \(\gamma\mapsto\langle\eta(\gamma),\xi(\gamma)\rangle\) is measurable for every \(\xi\in\mathfrak M\). Some sequence in \(\mathfrak M\), called *fundamental*, has values that span a dense subspace of every fibre; so every fibre is separable. Inner products of measurable sections are measurable functions, and \(f\xi\in\mathfrak M\) for measurable \(f\) and \(\xi\in\mathfrak M\). Conversely, let \((\xi_j)\) be sections of a family of Hilbert spaces such that every function \(\gamma\mapsto\langle\xi_j(\gamma),\xi_i(\gamma)\rangle\) is measurable and \((\xi_j(\gamma))_j\) is total in every fibre. Then exactly one measurable field contains all the \(\xi_j\). Its measurable sections are the sections \(\eta\) for which every \(\gamma\mapsto\langle\eta(\gamma),\xi_j(\gamma)\rangle\) is measurable. (Definition 2.1, Lemma 2.2 and Theorem 3.1.) - **(B2) Dimension strata.** There are sections \(e_k\in\mathfrak M\), \(k\geq1\), such that \(e_k(\gamma)\neq0\) exactly when \(k\leq n(\gamma)=\dim H(\gamma)\), and the nonzero \(e_k(\gamma)\) form an orthonormal basis of \(H(\gamma)\). The function \(n\) is measurable, so the sets \(\Gamma_d=\{n=d\}\), \(d\in\{0,1,2,\ldots,\infty\}\), form a countable measurable partition of \(\Gamma\). Let \(\ell^2_d\) be \(\mathbb C^d\), or \(\ell^2(\mathbb N)\) if \(d=\infty\), with standard basis \((\varepsilon_k)\). For \(\gamma\in\Gamma_d\), the formula \(U(\gamma)v=\sum_{k\leq d}\langle v,e_k(\gamma)\rangle\varepsilon_k\) defines a unitary \(U(\gamma):H(\gamma)\to\ell^2_d\). (Theorems 3.1 and 5.1.) - **(B3) Operator fields.** Let \((K(\gamma)),\mathfrak N\) be a second measurable field over the same base. A family of bounded operators \(x(\gamma):H(\gamma)\to K(\gamma)\) is a *measurable operator field* if \(\gamma\mapsto x(\gamma)\xi(\gamma)\) lies in \(\mathfrak N\) for every \(\xi\in\mathfrak M\). No bound on \(\|x(\gamma)\|\) is assumed. It suffices that this holds for the members of one fundamental sequence of \(\mathfrak M\). The norm function \(\gamma\mapsto\|x(\gamma)\|\) is measurable. Sums, composites and adjoints of measurable operator fields are measurable, and so are their products with measurable scalar functions. (Definition 6.1 and Theorem 6.2.) - **(B4) Decomposable operators.** For a bounded measurable function \(f\), \(m_f\) is the operator of multiplication by \(f\) on \(\mathcal H\). A measurable operator field \(x\) is *essentially bounded* if its norm function is. Then \((\int^\oplus x\,\xi)(\gamma)=x(\gamma)\xi(\gamma)\) defines a bounded operator \(\int^\oplus x=\int^\oplus_\Gamma x(\gamma)\,d\mu(\gamma)\), whose norm is \(\operatorname*{ess\,sup}_\gamma\|x(\gamma)\|\). The same holds for fields between two measurable fields. The map \(x\mapsto\int^\oplus x\) is linear and multiplicative on composable fields, and \((\int^\oplus x)^*=\int^\oplus x^*\). Two fields define the same operator exactly when they agree almost everywhere. Every \(\int^\oplus x\) commutes with every \(m_f\). If \(\Sigma\) is countably generated up to null sets, for instance if it is the Borel \(\sigma\)-algebra of a standard Borel space, then \(\mathcal H\) is separable. (Theorems 9.1 and 10.1.) - **(B5) Constant fields.** Let \(K_0\) be a separable Hilbert space with orthonormal basis \((\varepsilon_k)\). By (B1), the constant sections \(\varepsilon_k\) generate a measurable field whose fibres all equal \(K_0\): the *constant field* \(K_0\). Its measurable sections are the maps \(\xi:\Gamma\to K_0\) whose coordinates \(\gamma\mapsto\langle\xi(\gamma),\varepsilon_k\rangle\) are all measurable. The map \(f\otimes v\mapsto(\gamma\mapsto f(\gamma)v)\) extends to a unitary \(W\) from \(L^2(\Gamma,\mu)\otimes K_0\) onto its direct integral. (Example 11.2.) From [Decomposable operators and the diagonal algebra](../reader/supplements/decomposable-operators-diagonal-algebra.html): - **(B6) The two algebras.** Let \(\mathcal A=\{m_f\}\) be the *diagonal algebra*, and let the *decomposable algebra* \(\mathcal D\) be the set of all \(\int^\oplus x\) with \(x\) an essentially bounded measurable operator field on \((H(\gamma))\). Both are von Neumann algebras on \(\mathcal H\), \(\mathcal A'=\mathcal D\) and \(\mathcal D'=\mathcal A\). More precisely, every \(T\in\mathcal A'\) equals \(\int^\oplus x\) for a measurable operator field \(x\) with \(\|x(\gamma)\|\leq\|T\|\) for every \(\gamma\). We have \(m_f=\int^\oplus f(\gamma)1\,d\mu(\gamma)\). If \(x\) is essentially bounded and \(x(\gamma)\) is a scalar for almost every \(\gamma\), then \(\int^\oplus x\in\mathcal A\). Finally, \(m_f=0\) exactly when \(f=0\) almost everywhere on \(\{n\geq1\}\). (Proposition 2.2, Theorems 5.1 and 7.1.) - **(B7) Matrix units.** For \(k,l\geq1\), the formula \(e_{kl}(\gamma)v=\langle v,e_l(\gamma)\rangle e_k(\gamma)\) defines a measurable operator field with \(\|e_{kl}(\gamma)\|\leq1\) and \(e_{kl}(\gamma)^*=e_{lk}(\gamma)\). (Proposition 2.2.) - **(B8) Unitary fields and matrix entries.** Let \((K(\gamma)),\mathfrak N\) be a second measurable field over the same base, with direct integral \(\mathcal K\). If \(v\) is a measurable field of unitaries \(v(\gamma):H(\gamma)\to K(\gamma)\), then \(V=\int^\oplus v\) is a unitary from \(\mathcal H\) onto \(\mathcal K\). On a constant field \(K_0\) as in (B5), a family \(y\) of bounded operators is a measurable operator field exactly when every matrix entry \(\gamma\mapsto\langle y(\gamma)\varepsilon_l,\varepsilon_k\rangle\) is measurable. (Proposition 8.1.) - **(B9) A finite equivalent measure.** There is a measurable \(w:\Gamma\to(0,1]\) with \(\int_\Gamma w\,d\mu<\infty\). The finite measure \(\lambda(E)=\int_Ew\,d\mu\) has the same null sets as \(\mu\). (Lemma 3.1.) From [The Effros Borel structure](../reader/supplements/effros-borel-structure.html): - **(B10) The Borel space of von Neumann algebras.** Let \(K\) be a separable Hilbert space, and \(B(K)_*\) the Banach space of \(\sigma\)-weakly continuous linear functionals on \(B(K)\). It is separable. The closed unit ball of \(B(K)\) is compact and metrizable in the \(\sigma\)-weak topology, and every commutant is \(\sigma\)-weakly closed. For a \(\sigma\)-weakly closed subspace \(P\subseteq B(K)\) and \(\psi\in B(K)_*\), put \(p_P(\psi)=\sup\{|\psi(x)|:x\in P,\ \|x\|\leq1\}\). Then \(p_P\) is a seminorm with \(p_P(\psi)\leq\|\psi\|\), and \(P\) is determined by the values of \(p_P\) on any dense subset of \(B(K)_*\). The set \(\mathrm{vN}(K)\) of von Neumann algebras on \(K\), with the \(\sigma\)-algebra generated by the functions \(P\mapsto p_P(\psi)\), \(\psi\in B(K)_*\), is a standard Borel space. (Conventions, Definition 3.1, Proposition 3.2, Lemma 8.1 and Theorem 9.1.) - **(B11) Measurably generated families.** A family \((L(\gamma))\) of von Neumann algebras on the fibres \(H(\gamma)\) is *measurably generated* if there are measurable operator fields \(x_j\), \(j\geq1\), such that \(L(\gamma)\) is generated by \(\{x_j(\gamma):j\geq1\}\) for *every* \(\gamma\). Let \((L(\gamma))\) and \((L_1(\gamma))\) be measurably generated. Then: - (a) the generating fields can be chosen with \(\|x_j(\gamma)\|\leq1\), and with \(\{x_j(\gamma):j\geq1\}\) \(\sigma\)-weakly dense in the unit ball of \(L(\gamma)\), for every \(\gamma\); - (b) the families \((L(\gamma)')\), \((L(\gamma)\cap L_1(\gamma))\), \((L(\gamma)\vee L_1(\gamma))\) and \((L(\gamma)\cap L(\gamma)')\) are measurably generated; - (c) the sets of those \(\gamma\) for which \(L(\gamma)\) is a factor, is abelian, or equals \(B(H(\gamma))\) belong to \(\Sigma\); - (d) a family \((L(\gamma))\) of von Neumann algebras on the fibres is measurably generated if and only if, for every \(d\), the map \(\gamma\mapsto U(\gamma)L(\gamma)U(\gamma)^*\) from \(\Gamma_d\) to \(\mathrm{vN}(\ell^2_d)\) is measurable, with \(U(\gamma)\) as in (B2). None of this uses the measure. (Definition 10.1 and Theorem 10.3.) From [Spatial tensor products of von Neumann algebras](../reader/supplements/spatial-tensor-products.html), for Example 7.2 only: - **(B12) Tensor products.** For von Neumann algebras \(M\subseteq B(H)\) and \(N\subseteq B(K)\), the spatial tensor product \(M\bar\otimes N\) is the von Neumann algebra generated by the operators \(x\otimes y\) with \(x\in M\) and \(y\in N\). If \(S\subseteq B(H)\) and \(T\subseteq B(K)\) are self-adjoint sets with \(S''=M\) and \(T''=N\), then \(M\bar\otimes N=(S\otimes1\cup1\otimes T)''\). The commutation theorem says that \((M\bar\otimes N)'=M'\bar\otimes N'\). (Definition 5.1, Theorems 5.2 and 11.4.) ## 1. Measurable fields of von Neumann algebras **Definition 1.1.** A *field of von Neumann algebras* on \((H(\gamma))\) is a family \(M=(M(\gamma))_{\gamma\in\Gamma}\) in which each \(M(\gamma)\) is a von Neumann algebra acting on \(H(\gamma)\). The field is *measurable* if there are measurable operator fields \(x_1,x_2,\ldots\) on \((H(\gamma))\) such that, for almost every \(\gamma\), the algebra \(M(\gamma)\) is generated by \(\{x_j(\gamma):j\geq1\}\). Two fields are *equivalent* if they agree almost everywhere. The generating fields need not be essentially bounded; their norms may grow without bound. Finitely many generators are allowed: repeat one of them. For a field \(M\) we write \(M'\) for the field \((M(\gamma)')\) of commutants and \(Z_M\) for the field \((M(\gamma)\cap M(\gamma)')\) of centres. For two fields \(M_1,M_2\) we write \(M_1\cap M_2\) and \(M_1\vee M_2\) for the fields \((M_1(\gamma)\cap M_2(\gamma))\) and \((M_1(\gamma)\vee M_2(\gamma))\). By (C2) and (C4), these are again fields of von Neumann algebras. A measurably generated family in the sense of (B11) is a measurable field. The only difference is the null set: Proposition 1.3(1) shows that every measurable field is equivalent to a measurably generated family. **Example 1.2.** 1. The field \(\mathbb C1=(\mathbb C1_{H(\gamma)})\) is measurable. It is generated by the identity field, since \(\{1\}''=B(H(\gamma))'=\mathbb C1\); the last equality is the computation in (2). 2. The field \((B(H(\gamma)))\) is measurable: it is generated by the matrix-unit fields \(e_{kl}\) of (B7). Indeed, fix \(\gamma\) with \(n(\gamma)\geq1\), and let \(s\in B(H(\gamma))\) commute with every \(e_{kl}(\gamma)\). For \(k\leq n(\gamma)\) we have \(e_{k1}(\gamma)e_1(\gamma)=e_k(\gamma)\), so \[ se_k(\gamma)=se_{k1}(\gamma)e_1(\gamma)=e_{k1}(\gamma)se_1(\gamma)=\langle se_1(\gamma),e_1(\gamma)\rangle\,e_k(\gamma). \] So \(s\) agrees with the scalar \(c=\langle se_1(\gamma),e_1(\gamma)\rangle\) on an orthonormal basis, and \(s=c1\). The set \(\{e_{kl}(\gamma)\}\) is self-adjoint and its commutant is \(\mathbb C1\), so by (C3) the von Neumann algebra it generates is \((\mathbb C1)'=B(H(\gamma))\). If \(n(\gamma)=0\), then \(B(H(\gamma))=\mathbb C1\), and there is nothing to prove. **Proposition 1.3.** Let \(M\), \(M_1\) and \(M_2\) be measurable fields of von Neumann algebras. 1. (*Repair on a null set*) There are a null set \(N\) and a measurably generated family \(\tilde M\) with \(\tilde M(\gamma)=M(\gamma)\) for all \(\gamma\notin N\). 2. (*Generators in the unit ball*) There are measurable operator fields \(y_k\), \(k\geq1\), with \(\|y_k(\gamma)\|\leq1\) for every \(\gamma\), such that for almost every \(\gamma\) the operators \(y_k(\gamma)\) lie in \(M(\gamma)\), are \(\sigma\)-weakly dense in its unit ball, and generate it. 3. (*Operations*) The fields \(M'\), \(Z_M\), \(M_1\cap M_2\) and \(M_1\vee M_2\) are measurable. 4. (*Fibrewise properties*) The set of \(\gamma\) for which \(M(\gamma)\) is a factor is measurable up to a null set. So are the set where \(M(\gamma)\) is abelian and the set where \(M(\gamma)=B(H(\gamma))\). 5. (*Effros form*) A field \(L\) of von Neumann algebras on \((H(\gamma))\) is measurable exactly when, for every \(d\), some measurable map \(\Phi_d:\Gamma_d\to\mathrm{vN}(\ell^2_d)\) satisfies \(\Phi_d(\gamma)=U(\gamma)L(\gamma)U(\gamma)^*\) for almost every \(\gamma\in\Gamma_d\). **Proof.** (1) Choose generating fields \(x_j\) and a null set \(N\) such that \(M(\gamma)\) is generated by \(\{x_j(\gamma)\}\) for \(\gamma\notin N\). For every \(\gamma\), let \(\tilde M(\gamma)\) be the von Neumann algebra generated by \(\{x_j(\gamma)\}\). Then \(\tilde M\) is measurably generated, and it agrees with \(M\) off \(N\). (2) Apply (B11)(a) to \(\tilde M\). It gives fields that, for every \(\gamma\), lie in \(\tilde M(\gamma)\), are dense in its unit ball and generate it. Off \(N\), \(\tilde M(\gamma)=M(\gamma)\). (3) Take repairs \(\tilde M,\tilde M_1,\tilde M_2\) as in (1), with null sets \(N,N_1,N_2\). By (B11)(b), the families \(\tilde M'\), \(Z_{\tilde M}\), \(\tilde M_1\cap\tilde M_2\) and \(\tilde M_1\vee\tilde M_2\) are measurably generated. Off the null set \(N\cup N_1\cup N_2\) they agree with \(M'\), \(Z_M\), \(M_1\cap M_2\) and \(M_1\vee M_2\). The generating fields of a family that agrees with a given field off a null set generate that field almost everywhere. So the four fields are measurable. (4) By (B11)(c), the set \(F\) of \(\gamma\) for which \(\tilde M(\gamma)\) is a factor lies in \(\Sigma\). The set of \(\gamma\) for which \(M(\gamma)\) is a factor differs from \(F\) only inside \(N\). The same argument applies to the other two properties. (5) Let \(L\) be measurable, with a repair \(\tilde L\) and a null set \(N\) as in (1). By (B11)(d), the map \(\Phi_d(\gamma)=U(\gamma)\tilde L(\gamma)U(\gamma)^*\) is measurable on \(\Gamma_d\), and it equals \(U(\gamma)L(\gamma)U(\gamma)^*\) for \(\gamma\in\Gamma_d\setminus N\). Conversely, let maps \(\Phi_d\) be given, and let \(N_d\subseteq\Gamma_d\) be a null set off which \(\Phi_d(\gamma)=U(\gamma)L(\gamma)U(\gamma)^*\). For \(\gamma\in\Gamma_d\) put \(\tilde L(\gamma)=U(\gamma)^*\Phi_d(\gamma)U(\gamma)\). The map \(a\mapsto U(\gamma)^*aU(\gamma)\) is a \(*\)-isomorphism of \(B(\ell^2_d)\) onto \(B(H(\gamma))\) that carries commutants to commutants, so \(\tilde L(\gamma)\) is a von Neumann algebra acting on \(H(\gamma)\). Since \(U(\gamma)\tilde L(\gamma)U(\gamma)^*=\Phi_d(\gamma)\), the family \(\tilde L\) is measurably generated by (B11)(d). It agrees with \(L\) off the null set \(\bigcup_dN_d\), so \(L\) is measurable. \(\square\) **Remark 1.4.** 1. Every von Neumann algebra \(P\) acting on a separable space \(K\) is generated by a countable set. Indeed, by (B10) the unit ball of \(P\) lies in a compact metrizable space, so it has a countable \(\sigma\)-weakly dense subset \(S\). The von Neumann algebra generated by \(S\) is a commutant, hence \(\sigma\)-weakly closed (B10). It contains \(S\), hence the unit ball of \(P\), hence \(P\); and it lies in \(P\) by (C3). So measurability is not about the number of generators at one point. It asks that the generators be chosen measurably in \(\gamma\). 2. A field that is equivalent to a measurable field is measurable, since the same generating fields serve outside a null set. 3. By (B10), the maps \(\Phi_d\) of Proposition 1.3(5) take values in standard Borel spaces. So a measurable field is, up to a null set, a measurable map from each dimension stratum into a standard Borel space of von Neumann algebras. ## 2. The direct integral of a field of von Neumann algebras **Definition 2.1.** Let \(M\) be a field of von Neumann algebras on \((H(\gamma))\), measurable or not. Put \[ \int_\Gamma^\oplus M(\gamma)\,d\mu(\gamma)=\Big\{\int^\oplus x:\ x\ \text{an essentially bounded measurable operator field with}\ x(\gamma)\in M(\gamma)\ \text{for almost every}\ \gamma\Big\}, \tag{2.1} \] and abbreviate it to \(\int^\oplus M\). When \(M\) is measurable, we call \(\int^\oplus M\) the *direct integral* of the field \(M\). By definition it is a subset of \(\mathcal D\). **Proposition 2.2.** Let \(M\), \(M_1\) and \(M_2\) be fields of von Neumann algebras on \((H(\gamma))\), measurable or not. 1. (*Representing fields*) Let \(T\in\int^\oplus M\). Every essentially bounded measurable operator field \(x\) with \(T=\int^\oplus x\) satisfies \(x(\gamma)\in M(\gamma)\) for almost every \(\gamma\). Moreover, \(T=\int^\oplus x\) for a measurable operator field \(x\) with \(x(\gamma)\in M(\gamma)\) and \(\|x(\gamma)\|\leq\|T\|\) for every \(\gamma\). 2. (*Algebra*) \(\int^\oplus M\) is a \(*\)-subalgebra of \(\mathcal D\) that contains \(\mathcal A\), and \(\mathcal A\subseteq(\int^\oplus M)'\). 3. (*Monotonicity*) If \(M_1(\gamma)\subseteq M_2(\gamma)\) for almost every \(\gamma\), then \(\int^\oplus M_1\subseteq\int^\oplus M_2\). In particular, equivalent fields have the same direct integral. 4. (*Intersections*) \(\int^\oplus(M_1\cap M_2)=\int^\oplus M_1\cap\int^\oplus M_2\). 5. (*The extreme fields*) \(\int^\oplus B(H(\gamma))\,d\mu(\gamma)=\mathcal D\) and \(\int^\oplus\mathbb C1\,d\mu(\gamma)=\mathcal A\). **Proof.** (1) Choose an essentially bounded \(x_0\) with \(T=\int^\oplus x_0\) and \(x_0(\gamma)\in M(\gamma)\) off a null set \(N_0\). If \(T=\int^\oplus x\), then \(x=x_0\) off a null set \(N_1\) by (B4), so \(x(\gamma)\in M(\gamma)\) off \(N_0\cup N_1\). For the second claim, \(T\) lies in \(\mathcal D=\mathcal A'\), so (B6) gives a measurable operator field \(y\) with \(T=\int^\oplus y\) and \(\|y(\gamma)\|\leq\|T\|\) for every \(\gamma\). By the first claim, \(y(\gamma)\in M(\gamma)\) off a null set \(N_2\). The field \(x=1_{\Gamma\setminus N_2}\,y\) is measurable by (B3), and it agrees with \(y\) almost everywhere, so \(\int^\oplus x=T\). At each \(\gamma\), \(x(\gamma)\) is either \(y(\gamma)\in M(\gamma)\) or \(0\in M(\gamma)\), and \(\|x(\gamma)\|\leq\|T\|\). (2) Let \(S=\int^\oplus x\) and \(T=\int^\oplus y\) lie in \(\int^\oplus M\), and let \(c\in\mathbb C\). By (B4), \(S+cT=\int^\oplus(x+cy)\), \(ST=\int^\oplus xy\) and \(S^*=\int^\oplus x^*\). The fields \(x+cy\), \(xy\) and \(x^*\) are measurable and essentially bounded by (B3). Their values lie in \(M(\gamma)\) for almost every \(\gamma\), because \(M(\gamma)\) is a \(*\)-algebra. So \(\int^\oplus M\) is a \(*\)-subalgebra of \(\mathcal D\). For bounded measurable \(f\), \(m_f=\int^\oplus f(\gamma)1\,d\mu(\gamma)\) by (B6), and \(f(\gamma)1\in M(\gamma)\); so \(\mathcal A\subseteq\int^\oplus M\). Every element of \(\int^\oplus M\) is decomposable, so it commutes with \(\mathcal A\) by (B4). (3) An essentially bounded field whose values lie in \(M_1(\gamma)\) almost everywhere has its values in \(M_2(\gamma)\) almost everywhere, because a union of two null sets is null. (4) The inclusion \(\subseteq\) follows from (3). Let \(T\in\int^\oplus M_1\cap\int^\oplus M_2\), and write \(T=\int^\oplus x\). By (1), applied to \(M_1\) and to \(M_2\), the operator \(x(\gamma)\) lies in \(M_1(\gamma)\) and in \(M_2(\gamma)\) for almost every \(\gamma\). So \(T\in\int^\oplus(M_1\cap M_2)\). (5) The first equality is the definition of \(\mathcal D\). For the second, every \(m_f=\int^\oplus f(\gamma)1\,d\mu(\gamma)\) lies in \(\int^\oplus\mathbb C1\). Conversely, a decomposable operator whose field is scalar almost everywhere lies in \(\mathcal A\) by (B6). \(\square\) **Proposition 2.3 (Unitary transfer).** Let \((K(\gamma)),\mathfrak N\) be a second measurable field over \((\Gamma,\Sigma,\mu)\), with direct integral \(\mathcal K\). Let \(v\) be a measurable field of unitaries \(v(\gamma):H(\gamma)\to K(\gamma)\), and put \(V=\int^\oplus v\). For a field \(M\) on \((H(\gamma))\), let \(vMv^*\) be the field \((v(\gamma)M(\gamma)v(\gamma)^*)\) on \((K(\gamma))\). Then \[ V\Big(\int^\oplus M\Big)V^*=\int^\oplus vMv^* . \] If \(M\) is measurable, so is \(vMv^*\). **Proof.** The map \(a\mapsto v(\gamma)av(\gamma)^*\) is a \(*\)-isomorphism of \(B(H(\gamma))\) onto \(B(K(\gamma))\) that carries commutants to commutants. So \(vMv^*\) is a field of von Neumann algebras, and the map carries the von Neumann algebra generated by a set \(S\) to the one generated by \(v(\gamma)Sv(\gamma)^*\). The operator \(V\) is unitary by (B8). Let \(T\in\int^\oplus M\), and write \(T=\int^\oplus x\) with \(x(\gamma)\in M(\gamma)\) for every \(\gamma\) (Proposition 2.2(1)). By (B4), \(VTV^*=\int^\oplus vxv^*\). The field \(vxv^*\) is measurable by (B3). It has the same norm function as \(x\), and its values lie in \(v(\gamma)M(\gamma)v(\gamma)^*\). So \(V(\int^\oplus M)V^*\subseteq\int^\oplus vMv^*\). The adjoint field \(v^*\) is a measurable field of unitaries from \((K(\gamma))\) to \((H(\gamma))\), and \(\int^\oplus v^*=V^*\). So the same argument gives \(V^*(\int^\oplus vMv^*)V\subseteq\int^\oplus v^*vMv^*v=\int^\oplus M\). Conjugating by \(V\) gives the reverse inclusion. Finally, if measurable fields \(x_j\) generate \(M(\gamma)\) off a null set, then the measurable fields \(vx_jv^*\) generate \(v(\gamma)M(\gamma)v(\gamma)^*\) off the same null set. \(\square\) ## 3. The commutant theorem **Lemma 3.1.** 1. Let \(x\) be an essentially bounded measurable operator field, and let \(T=\int^\oplus t\in\mathcal D\). Then \(T\) commutes with \(\int^\oplus x\) if and only if \(t(\gamma)x(\gamma)=x(\gamma)t(\gamma)\) for almost every \(\gamma\). 2. (*Truncation*) Let \(x\) be a measurable operator field and \(m\geq1\). Then \(x^{[m]}=1_{\{\|x\|\leq m\}}\,x\) is a measurable operator field with \(\|x^{[m]}(\gamma)\|\leq m\) for every \(\gamma\), and \(x^{[m]}(\gamma)=x(\gamma)\) whenever \(\|x(\gamma)\|\leq m\). If \(M\) is a field of von Neumann algebras with \(x(\gamma)\in M(\gamma)\) for almost every \(\gamma\), then \(\int^\oplus x^{[m]}\) and its adjoint lie in \(\int^\oplus M\). **Proof.** (1) By (B4), \(T\int^\oplus x-(\int^\oplus x)T=\int^\oplus(tx-xt)\). The field \(tx-xt\) is measurable and essentially bounded by (B3). By (B4), it defines the zero operator exactly when it vanishes almost everywhere. (2) The norm function of \(x\) is measurable (B3), so \(1_{\{\|x\|\leq m\}}\) is a measurable function and \(x^{[m]}\) is a measurable operator field (B3). The bound and the equality are clear. Where \(x(\gamma)\in M(\gamma)\), the operator \(x^{[m]}(\gamma)\), which is \(x(\gamma)\) or \(0\), lies in \(M(\gamma)\). So \(\int^\oplus x^{[m]}\in\int^\oplus M\), and its adjoint lies there too by Proposition 2.2(2). \(\square\) **Theorem 3.2 (Commutant theorem).** Let \(M\) be a measurable field on \((H(\gamma))\), as in Definition 1.1. 1. \(\big(\int^\oplus M\big)'=\int^\oplus M'\). 2. \(\int^\oplus M\) is a von Neumann algebra acting on \(\mathcal H\), and its centre contains the diagonal algebra \(\mathcal A\). 3. \(\int^\oplus M'\) is also a von Neumann algebra, with commutant \(\big(\int^\oplus M'\big)'=\int^\oplus M\). **Proof.** (1) *The inclusion \(\supseteq\).* Let \(S\in\int^\oplus M'\) and \(T\in\int^\oplus M\). Write \(S=\int^\oplus s\) and \(T=\int^\oplus t\), with \(s(\gamma)\in M(\gamma)'\) and \(t(\gamma)\in M(\gamma)\) for almost every \(\gamma\). Then \(s(\gamma)t(\gamma)=t(\gamma)s(\gamma)\) for almost every \(\gamma\), and Lemma 3.1(1) gives \(ST=TS\). This part holds for every field \(M\). *The inclusion \(\subseteq\).* Let \(T\in(\int^\oplus M)'\). Since \(\mathcal A\subseteq\int^\oplus M\) (Proposition 2.2(2)), \(T\) commutes with \(\mathcal A\), so \(T\in\mathcal D\) by (B6); write \(T=\int^\oplus t\). Choose measurable operator fields \(x_j\) and a null set \(N_0\) such that \(M(\gamma)\) is generated by \(\{x_j(\gamma)\}\) for \(\gamma\notin N_0\). For all \(j,m\geq1\), the operators \(\int^\oplus x_j^{[m]}\) and \((\int^\oplus x_j^{[m]})^*=\int^\oplus(x_j^{[m]})^*\) lie in \(\int^\oplus M\) (Lemma 3.1(2)), so \(T\) commutes with them. By Lemma 3.1(1), there is a null set \(N_{j,m}\) outside which \(t(\gamma)\) commutes with \(x_j^{[m]}(\gamma)\) and with \(x_j^{[m]}(\gamma)^*\). Let \(N\) be the union of \(N_0\) and all the \(N_{j,m}\); it is a null set. Fix \(\gamma\notin N\) and \(j\), and take \(m\geq\|x_j(\gamma)\|\). Then \(x_j^{[m]}(\gamma)=x_j(\gamma)\), so \(t(\gamma)\) commutes with \(x_j(\gamma)\) and \(x_j(\gamma)^*\). Hence \(t(\gamma)\) lies in \(\{x_j(\gamma),x_j(\gamma)^*:j\geq1\}'\), which is \(M(\gamma)'\) by (C3). Therefore \(T\in\int^\oplus M'\). (2) The field \(M'\) is measurable by Proposition 1.3(3). Applying (1) to it gives \((\int^\oplus M')'=\int^\oplus M''=\int^\oplus M\). Together with (1) for \(M\) itself, \[ \Big(\int^\oplus M\Big)''=\Big(\int^\oplus M'\Big)'=\int^\oplus M . \] Since \(\int^\oplus M\) is a \(*\)-algebra (Proposition 2.2(2)), it is a von Neumann algebra. By Proposition 2.2(2), \(\mathcal A\) lies both in \(\int^\oplus M\) and in its commutant, hence in its centre. (3) By (1), \(\int^\oplus M'\) is the commutant of the self-adjoint set \(\int^\oplus M\), so it is a von Neumann algebra by (C2). The identity \((\int^\oplus M')'=\int^\oplus M\) was shown in (2). \(\square\) **Remark 3.3 (What the proof uses).** 1. The inclusion \(\int^\oplus M'\subseteq(\int^\oplus M)'\) holds for every field. The reverse inclusion uses only the generating fields of \(M\), cut down by truncation. The Effros Borel structure enters once, through Proposition 1.3(3): it tells us that \(M'\) is measurable, so that part (1) can be applied to \(M'\) in the proof of part (2). 2. The base enters only through (B4), which gives the norm formula and the uniqueness of representing fields, and through the identity \(\mathcal A'=\mathcal D\) of (B6). Both use the \(\sigma\)-finiteness of \(\mu\) and the separability of the fibres. No completeness of \(\mu\), no standard Borel structure, no countable generation of \(\Sigma\) and no separability of \(\mathcal H\) is used. For instance, take the product of uncountably many copies of \(\{0,1\}\), with the product of the fair-coin measures. There the direct integral of the constant field \(\mathbb C\) is an \(L^2\)-space that is not separable ([Decomposable operators and the diagonal algebra](../reader/supplements/decomposable-operators-diagonal-algebra.html), Remark 5.2), and the theorem applies. 3. For \(M(\gamma)=B(H(\gamma))\), part (1) says \(\mathcal D'=\int^\oplus\mathbb C1=\mathcal A\), which is part of (B6). 4. The measurability of \(M\) cannot be dropped from parts (1) and (3): see Example 6.3. ## 4. Uniqueness of the field, centres and factors **Theorem 4.1 (Uniqueness of the field).** Let \(M\) be a measurable field and \(L\) any field of von Neumann algebras on \((H(\gamma))\). Then \(\int^\oplus M\subseteq\int^\oplus L\) if and only if \(M(\gamma)\subseteq L(\gamma)\) for almost every \(\gamma\). In particular, two measurable fields have the same direct integral if and only if they are equivalent. **Proof.** If \(M(\gamma)\subseteq L(\gamma)\) almost everywhere, the inclusion follows from Proposition 2.2(3). Conversely, assume \(\int^\oplus M\subseteq\int^\oplus L\). Choose generating fields \(x_j\) of \(M\) and a null set \(N_0\) as in Definition 1.1. For all \(j,m\), the operator \(\int^\oplus x_j^{[m]}\) lies in \(\int^\oplus M\) (Lemma 3.1(2)), hence in \(\int^\oplus L\). By Proposition 2.2(1), applied to \(L\) and to the representing field \(x_j^{[m]}\), there is a null set \(N_{j,m}\) outside which \(x_j^{[m]}(\gamma)\in L(\gamma)\). Outside the null set \(N=N_0\cup\bigcup_{j,m}N_{j,m}\), every \(x_j(\gamma)\) lies in \(L(\gamma)\): take \(m\geq\|x_j(\gamma)\|\). Then \(L(\gamma)\), a von Neumann algebra that contains all the \(x_j(\gamma)\), contains the algebra \(M(\gamma)\) that they generate, by (C3). The last sentence follows by using the first one in both directions. \(\square\) Only \(M\) has to be measurable in the first statement. Example 6.3 shows that the uniqueness fails for a field that is not measurable. **Corollary 4.2.** Let \(M\) be a measurable field. 1. \(\int^\oplus M=\mathcal D\) if and only if \(M(\gamma)=B(H(\gamma))\) for almost every \(\gamma\). 2. \(\int^\oplus M=\mathcal A\) if and only if \(M(\gamma)=\mathbb C1\) for almost every \(\gamma\). **Proof.** The fields \((B(H(\gamma)))\) and \(\mathbb C1\) are measurable (Example 1.2), and their direct integrals are \(\mathcal D\) and \(\mathcal A\) (Proposition 2.2(5)). Apply Theorem 4.1. \(\square\) **Theorem 4.3 (Centre).** Let \(M\) be a measurable field. The field \(Z_M\) of centres is measurable, and \[ \Big(\int^\oplus M\Big)\cap\Big(\int^\oplus M\Big)'=\int^\oplus Z_M . \tag{4.1} \] In particular, the centre of \(\int^\oplus M\) equals \(\mathcal A\) exactly when almost every fibre \(M(\gamma)\) is a factor. **Proof.** The field \(Z_M\) is measurable by Proposition 1.3(3). By Theorem 3.2(1) and Proposition 2.2(4), \[ \Big(\int^\oplus M\Big)\cap\Big(\int^\oplus M\Big)'=\int^\oplus M\cap\int^\oplus M'=\int^\oplus(M\cap M')=\int^\oplus Z_M . \] If almost every \(M(\gamma)\) is a factor, then \(Z_M\) is equivalent to \(\mathbb C1\), and \(\int^\oplus Z_M=\int^\oplus\mathbb C1=\mathcal A\) by Proposition 2.2(3) and (5). Conversely, if the centre is \(\mathcal A\), then \(\int^\oplus Z_M=\mathcal A\), and Corollary 4.2(2), applied to the measurable field \(Z_M\), gives \(Z_M(\gamma)=\mathbb C1\) for almost every \(\gamma\). \(\square\) **Remark 4.4.** 1. By Proposition 1.3(4), the set of \(\gamma\) where \(M(\gamma)\) is a factor is measurable up to a null set. So the criterion of Theorem 4.3 is a condition on one set that is measurable up to a null set: its complement must be contained in a null set. 2. Over \(\Gamma_0\) every fibre is the zero space, and \(\{0\}\) counts as a factor. The diagonal algebra does not see \(\Gamma_0\) at all, since \(m_f=0\) whenever \(f\) vanishes on \(\{n\geq1\}\) (B6). If one prefers not to call the zero algebra a factor, Theorem 4.3 holds with "almost every \(\gamma\) with \(n(\gamma)\geq1\)". 3. For \(M(\gamma)=B(H(\gamma))\), Theorem 4.3 says that \(\mathcal A\) is the centre of \(\mathcal D\). ## 5. Von Neumann algebras between the diagonal and the decomposable algebra **Lemma 5.1 (Countably many generators).** Let \(S\subseteq\mathcal D\) be a countable set. For each \(s\in S\) choose an essentially bounded measurable operator field \(\beta_s\) with \(s=\int^\oplus\beta_s\), and let \(M_S(\gamma)\) be the von Neumann algebra generated by \(\{\beta_s(\gamma):s\in S\}\). Then \(M_S\) is a measurable field, and the von Neumann algebra generated by \(\mathcal A\cup S\) equals \(\int^\oplus M_S\). Another choice of the fields \(\beta_s\) changes \(M_S\) only on a null set. **Proof.** The field \(M_S\) is generated at every point by the countably many measurable fields \(\beta_s\), so it is measurable. If \(S\) is empty, then \(M_S=\mathbb C1\), which the identity field generates (Example 1.2(1)). Let \(\mathcal B_S\) be the von Neumann algebra generated by \(\mathcal A\cup S\). *The inclusion \(\mathcal B_S\subseteq\int^\oplus M_S\).* By Theorem 3.2(2), \(\int^\oplus M_S\) is a von Neumann algebra. It contains \(\mathcal A\) (Proposition 2.2(2)), and it contains every \(s\in S\), because \(\beta_s(\gamma)\in M_S(\gamma)\) for every \(\gamma\). By (C3) it contains \(\mathcal B_S\). *The inclusion \(\int^\oplus M_S\subseteq\mathcal B_S\).* By (C3), \(\mathcal B_S'=(\mathcal A\cup S\cup S^*)'\). Let \(T\in\mathcal B_S'\). Then \(T\) commutes with \(\mathcal A\), so \(T=\int^\oplus t\) by (B6). For \(s\in S\), \(T\) commutes with \(s=\int^\oplus\beta_s\) and with \(s^*=\int^\oplus\beta_s^*\). By Lemma 3.1(1) there is a null set \(N_s\) outside which \(t(\gamma)\) commutes with \(\beta_s(\gamma)\) and \(\beta_s(\gamma)^*\). Outside the null set \(\bigcup_{s\in S}N_s\), the operator \(t(\gamma)\) lies in \(\{\beta_s(\gamma),\beta_s(\gamma)^*:s\in S\}'\), which is \(M_S(\gamma)'\) by (C3). So \(T\in\int^\oplus M_S'=(\int^\oplus M_S)'\), by Theorem 3.2(1). This shows \(\mathcal B_S'\subseteq(\int^\oplus M_S)'\). Taking commutants and using (C1) and Theorem 3.2(2), \(\int^\oplus M_S=(\int^\oplus M_S)''\subseteq\mathcal B_S''=\mathcal B_S\). Finally, two representing fields of the same operator agree almost everywhere (B4). As \(S\) is countable, a new choice of all the \(\beta_s\) changes the generators only on one null set. \(\square\) **Lemma 5.2 (Countable generation over the diagonal algebra).** Let \(\mathcal B\) be a von Neumann algebra on \(\mathcal H\) with \(\mathcal A\subseteq\mathcal B\subseteq\mathcal D\). Then there is a countable set \(S_0\subseteq\mathcal B\) such that \(\mathcal B\) is generated by \(\mathcal A\cup S_0\). **Proof.** For each \(b\in\mathcal B\), fix an essentially bounded measurable operator field \(\beta_b\) with \(b=\int^\oplus\beta_b\). For every countable \(S\subseteq\mathcal B\), let \(M_S\) be the field of Lemma 5.1 built with these \(\beta_b\), and let \(\mathcal B_S\) be the von Neumann algebra generated by \(\mathcal A\cup S\). Then \(\mathcal B_S=\int^\oplus M_S\), and \(\mathcal B_S\subseteq\mathcal B\) by (C3). If \(S\subseteq T\), then \(M_S(\gamma)\subseteq M_T(\gamma)\) for every \(\gamma\), because the generating set grows. *A size function.* For each \(d\geq1\), choose a sequence \((\psi^d_i)_{i\geq1}\) that is dense in the unit ball of the separable space \(B(\ell^2_d)_*\) (B10). For a countable \(S\subseteq\mathcal B\) and \(\gamma\in\Gamma_d\) with \(d\geq1\), put \[ F_S(\gamma)=\sum_{i\geq1}2^{-i}\,p_{P_S(\gamma)}(\psi^d_i),\qquad\text{where}\quad P_S(\gamma)=U(\gamma)M_S(\gamma)U(\gamma)^*\in\mathrm{vN}(\ell^2_d), \tag{5.1} \] and put \(F_S=0\) on \(\Gamma_0\). Three properties hold. - *\(F_S\) is measurable, with values in \([0,1]\).* The field \(M_S\) is measurably generated, so \(\gamma\mapsto P_S(\gamma)\) is measurable on \(\Gamma_d\) by (B11)(d). Each function \(P\mapsto p_P(\psi)\) is measurable on \(\mathrm{vN}(\ell^2_d)\), by the definition of its Borel structure (B10). So every term of (5.1) is measurable on \(\Gamma_d\). Hence \(F_S\) is measurable on each set of the countable measurable partition \((\Gamma_d)\), and so on \(\Gamma\). The bound holds because \(0\leq p_P(\psi)\leq\|\psi\|\leq1\). - *If \(S\subseteq T\), then \(F_S\leq F_T\).* Indeed \(P_S(\gamma)\subseteq P_T(\gamma)\), and \(p_P(\psi)\) is a supremum over the unit ball of \(P\), which grows with \(P\). - *If \(S\subseteq T\) and \(F_S(\gamma)=F_T(\gamma)\), then \(M_S(\gamma)=M_T(\gamma)\).* For \(\gamma\in\Gamma_0\), both algebras are \(\{0\}\). Let \(\gamma\in\Gamma_d\) with \(d\geq1\). Each term of \(F_S(\gamma)\) is at most the corresponding term of \(F_T(\gamma)\), and the sums agree, so \(p_{P_S(\gamma)}(\psi^d_i)=p_{P_T(\gamma)}(\psi^d_i)\) for all \(i\). Both functions are seminorms bounded by the norm, so they are \(1\)-Lipschitz and positively homogeneous. Hence they agree on the set \(\{t\psi^d_i:t\geq0,\ i\geq1\}\), which is dense in \(B(\ell^2_d)_*\). By (B10), \(P_S(\gamma)=P_T(\gamma)\), and conjugating by \(U(\gamma)\) gives \(M_S(\gamma)=M_T(\gamma)\). *An essential supremum.* Let \(\lambda\) be the finite measure of (B9), and let \(c\) be the supremum of \(\int F_S\,d\lambda\) over all countable \(S\subseteq\mathcal B\). Then \(c\leq\lambda(\Gamma)<\infty\). Choose countable sets \(S_1,S_2,\ldots\) with \(\int F_{S_k}\,d\lambda\to c\), and let \(S_0=\bigcup_kS_k\), again countable. Since \(F_{S_0}\geq F_{S_k}\) for every \(k\), we get \(\int F_{S_0}\,d\lambda=c\). *Conclusion.* Let \(b\in\mathcal B\), and put \(T=S_0\cup\{b\}\). Then \(F_T\geq F_{S_0}\), while \(\int F_T\,d\lambda\leq c=\int F_{S_0}\,d\lambda\). So the nonnegative measurable function \(F_T-F_{S_0}\) has integral \(0\). It vanishes \(\lambda\)-almost everywhere, hence outside a \(\mu\)-null set (B9). By the third property, \(M_T(\gamma)=M_{S_0}(\gamma)\) outside that null set. Proposition 2.2(3) and Lemma 5.1 give \(\mathcal B_T=\int^\oplus M_T=\int^\oplus M_{S_0}=\mathcal B_{S_0}\). Hence \(b\in\mathcal B_T=\mathcal B_{S_0}\). As \(b\) was arbitrary, \(\mathcal B\subseteq\mathcal B_{S_0}\subseteq\mathcal B\). \(\square\) **Theorem 5.3 (Von Neumann algebras between \(\mathcal A\) and \(\mathcal D\)).** Let \(\mathcal B\) be a von Neumann algebra on \(\mathcal H\). Then \(\mathcal B=\int^\oplus M\) for some measurable field \(M\) if and only if \(\mathcal A\subseteq\mathcal B\subseteq\mathcal D\). In that case: 1. the field \(M\) is unique up to equivalence; 2. \(\mathcal B'=\int^\oplus M'\), and the centre of \(\mathcal B\) is \(\int^\oplus Z_M\); 3. \(M\) can be chosen so that each \(M(\gamma)\) is generated by the fibres \(\beta_s(\gamma)\) of countably many elements \(s\) of \(\mathcal B\). **Proof.** If \(\mathcal B=\int^\oplus M\) with \(M\) measurable, then \(\mathcal A\subseteq\mathcal B\subseteq\mathcal D\) by Proposition 2.2(2). Conversely, let \(\mathcal A\subseteq\mathcal B\subseteq\mathcal D\). Lemma 5.2 gives a countable \(S_0\subseteq\mathcal B\) such that \(\mathcal B\) is generated by \(\mathcal A\cup S_0\), and Lemma 5.1 gives \(\mathcal B=\int^\oplus M_{S_0}\). This proves the equivalence and (3). Part (1) is Theorem 4.1, and part (2) is Theorems 3.2(1) and 4.3. \(\square\) **Remark 5.4 (The separable case).** If \(\mathcal H\) is separable, Lemma 5.2 is immediate. By (B10), the unit ball of \(B(\mathcal H)\) is compact and metrizable in the \(\sigma\)-weak topology. So the unit ball of \(\mathcal B\) has a countable \(\sigma\)-weakly dense subset \(S_0\), and \(S_0\) generates \(\mathcal B\) by the argument of Remark 1.4(1). The essential supremum in the proof of Lemma 5.2 replaces the separability of \(\mathcal H\) by that of the fibres. This matters when \(\mathcal H\) is not separable, which can happen when \(\Sigma\) is not countably generated up to null sets (Remark 3.3(2)). **Corollary 5.5 (Dictionary).** The map \(M\mapsto\int^\oplus M\) induces a bijection from the equivalence classes of measurable fields of von Neumann algebras on \((H(\gamma))\) onto the von Neumann algebras \(\mathcal B\) with \(\mathcal A\subseteq\mathcal B\subseteq\mathcal D\). It preserves and reflects inclusion. It carries - the field \(M'\) to the commutant \((\int^\oplus M)'\); - the field \(M_1\cap M_2\) to \(\int^\oplus M_1\cap\int^\oplus M_2\); - the field \(Z_M\) to the centre of \(\int^\oplus M\); - the fields \((B(H(\gamma)))\) and \(\mathbb C1\) to \(\mathcal D\) and \(\mathcal A\). **Proof.** The map is well defined and preserves inclusion by Proposition 2.2(3); it is injective and reflects inclusion by Theorem 4.1; and it is onto by Theorem 5.3. The four items are Theorem 3.2(1), Proposition 2.2(4), Theorem 4.3 and Proposition 2.2(5). \(\square\) Joins are treated in Exercise 8.3. ## 6. Standard and non-standard bases **Remark 6.1.** Sections 1 to 5 hold over every \(\sigma\)-finite measure space. In particular they hold when \(\Sigma\) is the Borel \(\sigma\)-algebra of a standard Borel space, and also when \(\Sigma\) is its completion for \(\mu\), which is the setting of \(\mu\)-measurable fields. Standard bases matter for two later questions. How can a given abelian von Neumann algebra be realized as a diagonal algebra? And is the base of such a realization determined by the algebra? Corollary 6.2 collects what the standard case adds, and Example 6.3 shows what can go wrong without it. **Corollary 6.2 (Standard bases).** Let \(\Gamma\) be a standard Borel space, \(\Sigma\) its Borel \(\sigma\)-algebra, and \(\mu\) a \(\sigma\)-finite measure on \(\Sigma\). 1. \(\mathcal H\) is separable. 2. A field \(M\) of von Neumann algebras is measurable exactly when, for each \(d\), the map \(\gamma\mapsto U(\gamma)M(\gamma)U(\gamma)^*\) agrees almost everywhere on \(\Gamma_d\) with a Borel map into the standard Borel space \(\mathrm{vN}(\ell^2_d)\). 3. Let \(\mathcal B\) be a von Neumann algebra with \(\mathcal A\subseteq\mathcal B\subseteq\mathcal D\). There is a sequence \((b_k)\) that is \(\sigma\)-weakly dense in the unit ball of \(\mathcal B\). For each \(k\) choose a representing field \(\beta_k\) of \(b_k\), and let \(M(\gamma)\) be the von Neumann algebra generated by \(\{\beta_k(\gamma):k\geq1\}\). Then \(\mathcal B=\int^\oplus M\). 4. For a measurable field \(M\), the set of \(\gamma\) where \(M(\gamma)\) is a factor is a Borel set up to a null set, and the centre of \(\int^\oplus M\) is \(\mathcal A\) exactly when its complement is contained in a null set. **Proof.** (1) is (B4). (2) is Proposition 1.3(5) together with (B10). (3) By (1) and Remark 5.4, such a sequence exists and generates \(\mathcal B\). Since \(\mathcal A\subseteq\mathcal B\), the set \(\mathcal A\cup\{b_k\}\) generates \(\mathcal B\) as well, and Lemma 5.1 applies. (4) is Proposition 1.3(4) with Theorem 4.3. \(\square\) **Example 6.3 (A separating \(\sigma\)-algebra that is not countably generated).** Let \(\Gamma=[0,1]\). For \(\gamma\in\Gamma\), let \(p(\gamma)\) be the orthogonal projection of \(\mathbb C^2\) onto the line spanned by \((\cos\gamma,\sin\gamma)\), and let \[ \mathcal E(\gamma)=\mathbb Cp(\gamma)+\mathbb C\big(1-p(\gamma)\big)\subseteq M_2(\mathbb C). \] We compare the same family \(\mathcal E\) over two \(\sigma\)-algebras on \([0,1]\), in both cases on the constant field \(\mathbb C^2\). *Two facts about \(2\times2\) matrices.* First, \(\mathcal E(\gamma)'=\mathcal E(\gamma)\). An operator that commutes with \(p(\gamma)\) maps the range and the kernel of \(p(\gamma)\) into themselves. Both are lines, so the operator is a combination of \(p(\gamma)\) and \(1-p(\gamma)\). Conversely \(\mathcal E(\gamma)\) is abelian. So \(\mathcal E(\gamma)\) is a von Neumann algebra, generated by \(p(\gamma)\), and it is not a factor, since it is abelian of dimension \(2\). Second, \(\mathcal E(\gamma)\cap\mathcal E(\gamma')=\mathbb C1\) for \(\gamma\neq\gamma'\). Let \(a=\alpha p(\gamma)+\beta(1-p(\gamma))\) lie in \(\mathcal E(\gamma')\). If \(\alpha\neq\beta\), then \(p(\gamma)=(a-\beta)/(\alpha-\beta)\) commutes with \(p(\gamma')\). Then \(p(\gamma')\) maps the range of \(p(\gamma)\), a line, into itself, so that line lies in the range or in the kernel of \(p(\gamma')\). But the two lines are neither equal nor orthogonal, because \(0<|\gamma-\gamma'|\leq1<\pi/2\). So \(\alpha=\beta\), and \(a\) is a scalar. *The Borel sets.* Give \([0,1]\) its Borel \(\sigma\)-algebra and Lebesgue measure. The entries of \(p(\gamma)\) are continuous in \(\gamma\), so \(p\) is a measurable operator field on the constant field \(\mathbb C^2\) (B8). It generates \(\mathcal E(\gamma)\) at every point, so \(\mathcal E\) is measurable. Since \(\mathcal E'=\mathcal E\), Theorem 3.2 gives \((\int^\oplus\mathcal E)'=\int^\oplus\mathcal E\). So the direct integral is a maximal abelian von Neumann algebra and its own centre. No fibre is a factor. *The countable–co-countable sets.* Now let \(\Sigma_c\) consist of the countable subsets of \([0,1]\) and their complements. Let \(\mu_c(E)\) be \(0\) for countable \(E\) and \(1\) otherwise. This is a probability measure. The \(\sigma\)-algebra \(\Sigma_c\) contains every singleton, so it separates points, and its null sets are the countable sets. A function \(f:[0,1]\to\mathbb C\) is \(\Sigma_c\)-measurable exactly when it is constant outside a countable set. Such a function is clearly measurable. Conversely, let \(f\) be measurable and \(k\geq1\). Cover \(\mathbb C\) by countably many disjoint squares of side \(2^{-k}\). Their preimages are disjoint sets in \(\Sigma_c\) that cover \([0,1]\). They cannot all be countable, and two of them cannot both be co-countable, so exactly one of them, for a square \(Q_k\), is co-countable. Outside one countable set, \(f\) takes its values in every \(Q_k\) at once. Two values in \(Q_k\) differ by at most \(2^{-k}\sqrt2\), so \(f\) is constant outside that countable set. Consequently, over \(\Sigma_c\) the measurable sections of the constant field \(\mathbb C^2\) are the maps that are constant outside a countable set (B5). The direct integral is \(\mathbb C^2\) itself: every class has a constant representative \(v\), and its norm is \(\|v\|\). By (B8), a family of operators is a measurable operator field exactly when it is constant outside a countable set. So \(\mathcal D=M_2(\mathbb C)\), acting on \(\mathbb C^2\), and \(\mathcal A=\mathbb C1\). The family \(\mathcal E\) is not measurable for \(\Sigma_c\). Suppose that measurable fields \(x_j\) generated \(\mathcal E(\gamma)\) for all \(\gamma\) outside a countable set. Each \(x_j\) equals a constant \(a_j\) outside a countable set. So outside one countable set, \(\mathcal E(\gamma)\) would be the algebra generated by the \(a_j\), the same for all such \(\gamma\). This contradicts \(\mathcal E(\gamma)\neq\mathcal E(\gamma')\) for \(\gamma\neq\gamma'\). Now compute. Let \(T=\int^\oplus x\) with \(x(\gamma)\in\mathcal E(\gamma)\) outside a countable set. Then \(x\) equals a constant \(a\) outside a countable set, so \(a\in\mathcal E(\gamma)\) for all \(\gamma\) outside a countable set, and in particular for two different \(\gamma\). So \(a\) is a scalar, and \(\int^\oplus\mathcal E=\mathcal A=\mathbb C1\). Since \(\mathcal E'=\mathcal E\), also \(\int^\oplus\mathcal E'=\mathbb C1\). But \[ \Big(\int^\oplus\mathcal E\Big)'=(\mathbb C1)'=M_2(\mathbb C)\neq\mathbb C1=\int^\oplus\mathcal E' . \] So Theorem 3.2(1) fails for this field. Theorem 4.1 fails as well: \(\int^\oplus\mathcal E=\int^\oplus\mathbb C1\), although \(\mathcal E(\gamma)\neq\mathbb C1\) for every \(\gamma\). *What the base does not see.* Over \(\Sigma_c\), the diagonal algebra is \(\mathbb C1\), and the direct integral of the constant field is \(\mathbb C^2\), exactly as over a single point of mass one. But no bijection exists between a co-countable subset of \([0,1]\) and a point. So without standardness, the base of a direct integral cannot be recovered from its diagonal algebra, even up to null sets. For standard \(\sigma\)-finite bases, an isomorphism between the \(L^\infty\)-algebras does come from a Borel isomorphism of the bases, defined outside null sets [Takesaki I, Lemma IV.8.22]. This is why the uniqueness of disintegrations is stated for standard bases. ## 7. Examples **Example 7.1 (A countable base).** Let \(\Gamma\) be countable, \(\Sigma\) the set of all subsets, and \(0<\mu(\{\gamma\})<\infty\) for every \(\gamma\). Every section is measurable. The direct integral \(\mathcal H\) is the Hilbert sum of the fibres, with inner products weighted by \(\mu(\{\gamma\})\), and \(m_{1_{\{\gamma\}}}\) is the projection onto the \(\gamma\)-th summand ([Measurable fields of Hilbert spaces and their direct integrals](../reader/supplements/measurable-fields-direct-integrals.html), Example 11.1). Every family of bounded operators is a measurable operator field, and the only null set is empty. So every field \(M\) of von Neumann algebras is measurable. By Remark 1.4(1), each \(M(\gamma)\) is generated by a sequence \((a_{\gamma,j})_j\), and the families \(x_j(\gamma)=a_{\gamma,j}\) are measurable operator fields. The direct integral \(\int^\oplus M\) is the set of uniformly bounded families \((x(\gamma))\) with \(x(\gamma)\in M(\gamma)\) for every \(\gamma\). Theorem 3.2(1) can be checked by hand here. An operator \(T\) that commutes with \(\int^\oplus M\) commutes with the projections \(m_{1_{\{\gamma\}}}\in\mathcal A\). So it maps each summand into itself and is a bounded family \((t(\gamma))\). For \(\gamma_0\in\Gamma\) and \(a\in M(\gamma_0)\), the family equal to \(a\) at \(\gamma_0\) and to \(0\) elsewhere lies in \(\int^\oplus M\). Commuting with it forces \(t(\gamma_0)a=at(\gamma_0)\). So \(t(\gamma_0)\in M(\gamma_0)'\) for every \(\gamma_0\), and \(T\in\int^\oplus M'\). The centre of \(\int^\oplus M\) consists of the bounded families of central elements. It equals \(\mathcal A\), the bounded families of scalars, exactly when every \(M(\gamma)\) is a factor. **Example 7.2 (Constant fields and tensor products).** Let \(K_0\) be a separable Hilbert space. Let \(H(\gamma)=K_0\) be the constant field of (B5), with its unitary \(W:L^2(\Gamma,\mu)\otimes K_0\to\mathcal H\), and let \(M_0\) be a von Neumann algebra on \(K_0\). Consider the constant field \(M(\gamma)=M_0\). (a) *The field is measurable.* By Remark 1.4(1), \(M_0\) is generated by a sequence \((a_j)\). A constant family \(a\) of operators has constant matrix entries, so it is a measurable operator field (B8). Hence the constant fields \(a_j\) generate \(M(\gamma)\) at every point. (b) *The operators involved.* Write \(m\) for the representation of \(L^\infty(\Gamma,\mu)\) by multiplication operators on \(L^2(\Gamma,\mu)\). For \(g\in L^2(\Gamma,\mu)\), \(v\in K_0\), a bounded measurable \(f\) and \(a\in B(K_0)\), \[ W\big((m_f\otimes1)(g\otimes v)\big)=\big(\gamma\mapsto f(\gamma)g(\gamma)v\big)=m_fW(g\otimes v),\qquad W\big((1\otimes a)(g\otimes v)\big)=\big(\gamma\mapsto g(\gamma)av\big)=\Big(\int^\oplus a\Big)W(g\otimes v). \] The elementary tensors are total and all these operators are bounded. So \(W(m_f\otimes1)W^*=m_f\), and \(W(1\otimes a)W^*=\int^\oplus a\) is the decomposable operator with constant fibre \(a\). (c) *The direct integral is a tensor product.* By Lemma 5.1 with \(S=\{\int^\oplus a_j:j\geq1\}\), the algebra \(\int^\oplus M\) is generated by \(\mathcal A\cup S\). By (b), \(W^*(\int^\oplus M)W\) is generated by \(m(L^\infty)\otimes1\) and the operators \(1\otimes a_j\). Here \(m(L^\infty)\) is a von Neumann algebra acting on \(L^2(\Gamma,\mu)\), because it is the diagonal algebra of the constant field \(\mathbb C\) (B6). The self-adjoint set \(\{a_j,a_j^*\}\) generates \(M_0\). So (B12) gives \[ W^*\Big(\int^\oplus M_0\,d\mu\Big)W=m\big(L^\infty(\Gamma,\mu)\big)\bar\otimes M_0 . \tag{7.1} \] (d) *Consequences.* Apply (7.1) also to \(M_0'\) and to the centre \(Z_0=M_0\cap M_0'\). Then Theorems 3.2(1) and 4.3 give \[ \big(m(L^\infty)\bar\otimes M_0\big)'=m(L^\infty)\bar\otimes M_0',\qquad \big(m(L^\infty)\bar\otimes M_0\big)\cap\big(m(L^\infty)\bar\otimes M_0\big)'=m(L^\infty)\bar\otimes Z_0 . \] The first identity is the commutation theorem of (B12) for this pair, obtained here without it, for a \(\sigma\)-finite measure and a separable \(K_0\). If \(M_0\) is a factor, then \(Z_0=\mathbb C1\), and by (7.1) for \(\mathbb C1\) the centre is \(W^*\mathcal AW=m(L^\infty)\otimes1\), in line with Theorem 4.3. **Example 7.3 (Varying dimension and zero fibres).** Let \(\Gamma=(0,2]\) with Lebesgue measure on its Borel sets. Put \(d(\gamma)=k\) for \(\gamma\in(\tfrac1{k+1},\tfrac1k]\), \(k\geq1\), and \(d(\gamma)=0\) for \(\gamma\in(1,2]\). Let \(H(\gamma)=\mathbb C^{d(\gamma)}\), where \(\mathbb C^0=0\). Let \(\xi_k(\gamma)\) be the \(k\)-th standard basis vector of \(\mathbb C^{d(\gamma)}\) if \(k\leq d(\gamma)\), and \(0\) otherwise. The functions \(\langle\xi_k,\xi_l\rangle=\delta_{kl}1_{\{d\geq k\}}\) are measurable, and the \(\xi_k(\gamma)\) span every fibre. So by (B1) exactly one measurable field contains them. All fibres over \((1,2]\) are zero. For \(j\geq1\), let \(q_j(\gamma)\) be the \(j\)-th diagonal matrix unit of \(M_{d(\gamma)}(\mathbb C)\) if \(j\leq d(\gamma)\), and \(0\) otherwise. Since \(q_j\xi_k=\delta_{jk}\xi_k\), each \(q_j\) is a measurable operator field (B3). Let \(M(\gamma)=\Delta_{d(\gamma)}\), the algebra of diagonal matrices in \(M_{d(\gamma)}(\mathbb C)\), with \(\Delta_0=\{0\}\). - *\(M\) is measurable.* An operator on \(\mathbb C^k\) commutes with all diagonal matrix units exactly when it maps each coordinate line into itself, that is, when it is diagonal. So \(\{q_j(\gamma)\}'=\Delta_{d(\gamma)}\). Since \(\Delta_{d(\gamma)}\) is spanned by the \(q_j(\gamma)\), it has the same commutant, so \(\Delta_{d(\gamma)}'=\Delta_{d(\gamma)}\). The \(q_j(\gamma)\) are self-adjoint, so the algebra they generate is \(\{q_j(\gamma)\}''=\Delta_{d(\gamma)}'=\Delta_{d(\gamma)}\) (C3). - *The direct integral is maximal abelian.* Since \(M'=M\), Theorem 3.2 gives \((\int^\oplus M)'=\int^\oplus M\). So \(\int^\oplus M\) is a maximal abelian von Neumann algebra on \(\mathcal H\), and it is its own centre. - *Factor fibres.* \(M(\gamma)\) is a factor exactly when \(d(\gamma)\leq1\), that is, on \((\tfrac12,2]\): on \((\tfrac12,1]\) the fibre algebra is \(\mathbb C=B(\mathbb C)\), and on \((1,2]\) it is the zero algebra. The complement \((0,\tfrac12]\) has positive measure, so the centre is strictly larger than \(\mathcal A\) (Theorem 4.3). For instance, \(\int^\oplus q_1\) is central, but it is not diagonal, because \(q_1(\gamma)\) is not a scalar on \((0,\tfrac12]\). - *The full field.* For the field \((B(H(\gamma)))\), every fibre is a factor, so the centre of \(\mathcal D\) is \(\mathcal A\). By (B6), \(m_f=0\) exactly when \(f\) vanishes almost everywhere on \((0,1]\): the diagonal algebra does not see the zero fibres over \((1,2]\). **Example 7.4 (Null sets are invisible).** Let \(\Gamma=[0,1)\) with Lebesgue measure on its Borel sets, and let \(H(\gamma)=\ell^2(\mathbb Z)\) be the constant field, with standard basis \((\varepsilon_m)_{m\in\mathbb Z}\). Let \(u_\gamma\) be the unitary with \(u_\gamma\varepsilon_m=e^{2\pi i\gamma m}\varepsilon_m\). Its matrix entries are continuous in \(\gamma\), so \(u\) is a measurable operator field (B8). Let \(M(\gamma)\) be the von Neumann algebra generated by \(u_\gamma\); the field \(M\) is measurable. Let \(\Delta\subseteq B(\ell^2(\mathbb Z))\) be the algebra of diagonal operators. An operator \(x\) commutes with \(u_\gamma\) exactly when \(\langle x\varepsilon_m,\varepsilon_{m'}\rangle\big(e^{2\pi i\gamma m}-e^{2\pi i\gamma m'}\big)=0\) for all \(m,m'\). If \(\gamma\) is irrational, the numbers \(e^{2\pi i\gamma m}\) are distinct, so \(x\) is diagonal: \(\{u_\gamma\}'=\Delta\). An operator that commutes with a unitary also commutes with its inverse, which is its adjoint, so \(\{u_\gamma,u_\gamma^*\}'=\Delta\). The same computation with the projections onto the lines \(\mathbb C\varepsilon_m\) gives \(\Delta'=\Delta\). Hence \(M(\gamma)=\Delta'=\Delta\) for irrational \(\gamma\). At rational points the fibre is different: for example \(M(0)=\mathbb C1\), and at \(\gamma=p/q\) in lowest terms with \(q\geq2\) the fibre is a \(q\)-dimensional abelian algebra ([The Effros Borel structure](../reader/supplements/effros-borel-structure.html), Example 9.4). The rational points form a null set. So \(M\) is equivalent to the constant field \(\Delta\), and \(\int^\oplus M=\int^\oplus\Delta\) by Proposition 2.2(3). Since \(\Delta'=\Delta\), Theorem 3.2(1) gives \((\int^\oplus M)'=\int^\oplus M\): the direct integral is maximal abelian and its own centre. The set of \(\gamma\) where \(M(\gamma)\) is a factor is \(\{0\}\). It is null, and the centre is not \(\mathcal A\), in line with Theorem 4.3. By Example 7.2, \(W^*(\int^\oplus\Delta)W=m(L^\infty[0,1))\bar\otimes\Delta\). ## 8. Exercises **Exercise 8.1 (Abelian and maximal abelian direct integrals).** Let \(M\) be a measurable field. Show that \(\int^\oplus M\) is abelian if and only if \(M(\gamma)\) is abelian for almost every \(\gamma\). Show also that \((\int^\oplus M)'=\int^\oplus M\) if and only if \(M(\gamma)'=M(\gamma)\) for almost every \(\gamma\). *Solution.* A von Neumann algebra \(P\) is abelian exactly when \(P\subseteq P'\). By Theorem 3.2(1), \(\int^\oplus M\) is abelian if and only if \(\int^\oplus M\subseteq\int^\oplus M'\). By Theorem 4.1, applied to the measurable field \(M\) and to \(L=M'\), this holds if and only if \(M(\gamma)\subseteq M(\gamma)'\) for almost every \(\gamma\), that is, if and only if \(M(\gamma)\) is abelian for almost every \(\gamma\). For the second claim, \((\int^\oplus M)'=\int^\oplus M'\) by Theorem 3.2(1), and the fields \(M\) and \(M'\) are both measurable (Proposition 1.3(3)). By Theorem 4.1, \(\int^\oplus M'=\int^\oplus M\) if and only if \(M(\gamma)'=M(\gamma)\) for almost every \(\gamma\). Examples 7.3 and 7.4, and Example 6.3 over the Borel sets, are instances of the second claim. **Exercise 8.2 (When is a direct integral a factor?).** Let \(M\) be a measurable field, assume \(\mathcal H\neq0\), and put \(\Gamma_+=\{n\geq1\}\). Show that \(\int^\oplus M\) is a factor if and only if - (i) every \(E\in\Sigma\) with \(E\subseteq\Gamma_+\) satisfies \(\mu(E)=0\) or \(\mu(\Gamma_+\setminus E)=0\), and - (ii) almost every \(M(\gamma)\) is a factor. Check that (i) holds for the countable–co-countable base of Example 6.3. *Solution.* Suppose \(\int^\oplus M\) is a factor. Its centre is \(\mathbb C1\) and contains \(\mathcal A\) (Theorem 3.2(2)), so \(\mathcal A=\mathbb C1\), and the centre equals \(\mathcal A\). Theorem 4.3 gives (ii). For (i), let \(E\subseteq\Gamma_+\) be measurable. Then \(m_{1_E}\) is a projection in \(\mathbb C1\), so it is \(0\) or \(1\); these differ because \(\mathcal H\neq0\). If \(m_{1_E}=0\), then \(1_E=0\) almost everywhere on \(\Gamma_+\) by (B6), so \(\mu(E)=0\). If \(m_{1_E}=1=m_1\), then \(1_E-1=0\) almost everywhere on \(\Gamma_+\), so \(\mu(\Gamma_+\setminus E)=0\). Conversely, assume (i) and (ii). By (ii) and Theorem 4.3, the centre of \(\int^\oplus M\) is \(\mathcal A\), so it suffices to show \(\mathcal A=\mathbb C1\). Note that \(\mu(\Gamma_+)>0\), since otherwise every vector of \(\mathcal H\) would vanish almost everywhere. Let \(f\) be a bounded measurable real function, and let \(R\) be the set of rationals \(t\) with \(\mu(\Gamma_+\cap\{f\leq t\})=0\). If \(t\in R\) and \(st\) almost everywhere on \(\Gamma_+\). For rational \(t>c\), we have \(t\notin R\), so \(\Gamma_+\cap\{f\leq t\}\) is not null; by (i), its complement \(\Gamma_+\cap\{f>t\}\) in \(\Gamma_+\) is null, and \(f\leq t\) almost everywhere on \(\Gamma_+\). Using countably many rationals on each side, \(f=c\) almost everywhere on \(\Gamma_+\), and so \(m_f=m_c=c1\) by (B6). For complex \(f\), apply this to the real and imaginary parts. So \(\mathcal A=\mathbb C1\), and \(\int^\oplus M\) is a factor. In Example 6.3, every set in \(\Sigma_c\) is countable or has a countable complement, so it is null or has a null complement. Hence (i) holds for every field over that base. For instance, let \(M_0\) be a factor on a separable \(K_0\neq0\). The constant field \(M_0\) is measurable (Example 7.2(a)), so its direct integral over \(\Sigma_c\) is a factor. In fact, as in Example 6.3, every measurable section of the constant field \(K_0\) is constant outside a countable set, and the direct integral is \(M_0\) itself, acting on \(K_0\). **Exercise 8.3 (Joins).** Let \(M_1\) and \(M_2\) be measurable fields. Show that \(\int^\oplus(M_1\vee M_2)=\int^\oplus M_1\vee\int^\oplus M_2\), where the right side is the von Neumann algebra generated by the two direct integrals. *Solution.* The field \(M_1\vee M_2\) is measurable (Proposition 1.3(3)), so its direct integral is a von Neumann algebra (Theorem 3.2(2)); so are \(\int^\oplus M_1\) and \(\int^\oplus M_2\). By (C4), Theorem 3.2(1) and Proposition 2.2(4), \[ \Big(\int^\oplus M_1\vee\int^\oplus M_2\Big)'=\Big(\int^\oplus M_1\Big)'\cap\Big(\int^\oplus M_2\Big)'=\int^\oplus M_1'\cap\int^\oplus M_2'=\int^\oplus(M_1'\cap M_2'). \] By (C4) at each point, \(M_1(\gamma)'\cap M_2(\gamma)'=(M_1(\gamma)\vee M_2(\gamma))'\). So the last term is \(\int^\oplus(M_1\vee M_2)'\), which equals \((\int^\oplus(M_1\vee M_2))'\) by Theorem 3.2(1). Two von Neumann algebras with the same commutant are equal, since each is the commutant of its commutant. **Exercise 8.4 (Central projections).** Let \(M\) be a measurable field. Show that a projection \(P\) on \(\mathcal H\) lies in the centre of \(\int^\oplus M\) if and only if \(P=\int^\oplus p\) for a measurable operator field \(p\) such that \(p(\gamma)\) is a projection in the centre of \(M(\gamma)\) for every \(\gamma\). Deduce that if almost every \(M(\gamma)\) is a factor, the central projections of \(\int^\oplus M\) are exactly the operators \(m_{1_E}\), \(E\in\Sigma\). *Solution.* If \(p\) is such a field, then \(\|p(\gamma)\|\leq1\), and \(P=\int^\oplus p\) lies in \(\int^\oplus Z_M\), which is the centre (Theorem 4.3). By (B4), \(P^*=\int^\oplus p^*=P\) and \(P^2=\int^\oplus p^2=P\), so \(P\) is a projection. Conversely, let \(P\) be a central projection. By Theorem 4.3 and Proposition 2.2(1), \(P=\int^\oplus x\) with \(x(\gamma)\) in the centre of \(M(\gamma)\) for every \(\gamma\). From \(P=P^*P\) and (B4), \(x(\gamma)=x(\gamma)^*x(\gamma)\) outside a null set \(N\). Put \(p=1_{\Gamma\setminus N}\,x\). Then \(p(\gamma)=p(\gamma)^*p(\gamma)\) for every \(\gamma\). Taking adjoints gives \(p(\gamma)^*=p(\gamma)\), and then \(p(\gamma)^2=p(\gamma)\). So each \(p(\gamma)\) is a projection in the centre of \(M(\gamma)\), and \(\int^\oplus p=P\). Now let \(M(\gamma)\) be a factor outside a null set \(N'\). The function \(c=\langle pe_1,e_1\rangle\) is measurable (B1); let \(E=\{c=1\}\). For \(\gamma\notin N'\) with \(n(\gamma)\geq1\), \(p(\gamma)\) is a projection in \(\mathbb C1\), so it is \(0\) or \(1\), and it equals \(c(\gamma)1=1_E(\gamma)1\). For \(n(\gamma)=0\), both sides are the zero operator. So \(p(\gamma)=1_E(\gamma)1\) almost everywhere, and \(P=m_{1_E}\) by (B4) and (B6). Conversely, every \(m_{1_E}\) is a projection in \(\mathcal A\), which lies in the centre. ## Where this leads - *Disintegration.* Let \(\mathcal B\) be a von Neumann algebra acting on a separable space, and let \(\mathcal C\) be an abelian von Neumann algebra contained in its centre. Suppose \(\mathcal C\) has been realized as the diagonal algebra of a direct integral. Then \(\mathcal C\subseteq\mathcal B\subseteq\mathcal C'=\mathcal D\), so Theorem 5.3 writes \(\mathcal B\) as the direct integral of a measurable field. If \(\mathcal C\) is the whole centre, almost every fibre is a factor, by Theorem 4.3. How to realize \(\mathcal C\) as a diagonal algebra over a standard base, and in what sense such a realization is unique, is the subject of a later lesson on disintegration; for a summary, see [Blackadar, III.1.6.3–III.1.6.4]. - *Types.* Once an algebra is written as an integral over its centre, the type decomposition of [Projections and types of von Neumann algebras](../../foundations-of-von-neumann-algebras/projections-and-types-of-von-neumann-algebras.html) can be read fibre by fibre [Blackadar, III.1.6.4]. - *States.* For a state of a C\*-algebra, the counterpart of the decomposition over the centre is the central measure of [Integral representations of states](../../foundations-of-von-neumann-algebras/integral-representations-of-states.html), whose abelian algebra is the centre of the represented algebra. ## References - [Effros 1965] E. G. Effros, The Borel space of von Neumann algebras on a separable Hilbert space, *Pacific Journal of Mathematics* 15 (1965), 1153–1164. https://doi.org/10.2140/pjm.1965.15.1153 - [Blackadar] B. Blackadar, *Operator Algebras: Theory of C\*-Algebras and von Neumann Algebras*, [revised author edition, 8 February 2017](https://bruceblackadar.com/Mathematics/Cycr.pdf). - [Takesaki I] M. Takesaki, *Theory of Operator Algebras I*, Springer, New York, 1979; reprinted as Encyclopaedia of Mathematical Sciences 124, Springer, Berlin, 2002.