Spatial tensor products: complete proof supplement through the commutation theorem

Original text by Claude Opus 5.5 (Anthropic), September 2026; public domain (CC0). Scoped selection and prerequisite bindings by GPT-6.1 Sol (OpenAI), Ultra, October 2026; original contributions CC0. The mathematical text of the selected proofs is retained.

Sections 1–11 give the complete construction of spatial tensor products and the tensor commutation theorem. Hilbert spaces and orthogonal families can have arbitrary cardinality. Countable vector series describe one normal functional and impose no separability hypothesis.

The proofs use Hilbert completion, projection and Riesz representation over the real and complex fields, bounded adjoints, continuous functional calculus and positive-functional Cauchy–Schwarz. Their full published providers are the Hilbert-space, continuous-calculus and GNS chapters of Foundations of von Neumann algebras. Section 4 supplies its own bicommutant proof.

For Theorem 10.1(3), use the published Kaplansky chapter, Theorem 7.1, and the published trace-class/predual chapter, Theorem 9.4(a). For Theorem 10.1(4), use the bounded normal-functional support proof in the universal-enveloping chapter, “State and support tools used below” and Lemma 11.1. For the comments following Corollaries 8.3–8.4, the universal-enveloping chapter, Corollary 11.4, proves normality of isomorphisms; the tensor-isomorphism proof itself explicitly tensors the given normal inverse. These four incoming statements are full proofs in the published foundation reader.

Section 11 proves tensor commutation by real orthogonal complements and cyclic corners. General Tomita–Takesaki theory, standard-form modular implementers, weights and semifiniteness are not prerequisites of this proof.

Conventions

Hilbert spaces are complex. Inner products are linear in the first variable. No separability or dimension condition is imposed anywhere, and the zero space is allowed. A sum over an arbitrary index set is the limit of the net of its finite partial sums. Every Hilbert space has an orthonormal basis.

$B(H)$ is the algebra of bounded operators on $H$ . For $\xi,\eta\in H$ put $\omega_{\xi,\eta}(x)=\langle x\xi,\eta\rangle$ and $\omega_\xi=\omega_{\xi,\xi}$ . For a set $S\subseteq B(H)$ , $S'$ is its commutant. It is a unital algebra, closed in the weak operator topology, and a $*$ -algebra when $S^*=S$ . We call a $*$ -algebra $M\subseteq B(H)$ with $M''=M$ a von Neumann algebra on $H$ . It contains $1$ and is weakly closed. On the zero space, $B(0)=\{0\}$ is a von Neumann algebra.

1. The Hilbert tensor product

Let $H\odot K$ be the algebraic tensor product. The universal property of $\odot$ , used once in each variable, gives a unique sesquilinear form on $H\odot K$ with $\langle\xi\otimes\eta,\ \xi'\otimes\eta'\rangle=\langle\xi,\xi'\rangle\langle\eta,\eta'\rangle . \tag{1.1}$

Lemma 1.1. The form (1.1) is an inner product.

Proof. Let $\zeta=\sum_{i=1}^n\xi_i\otimes\eta_i$ . Choose an orthonormal basis $u_1,\ldots,u_r$ of the span of the $\eta_i$ , and write $\eta_i=\sum_lc_{il}u_l$ . Then $\zeta=\sum_l\xi_l'\otimes u_l$ with $\xi_l'=\sum_ic_{il}\xi_i$ , and (1.1) gives $\langle\zeta,\zeta\rangle=\sum_{l,m}\langle\xi_l',\xi_m'\rangle\langle u_l,u_m\rangle=\sum_l\|\xi_l'\|^2\ge0$ . If this is $0$ , every $\xi_l'$ is $0$ . The $u_l$ are linearly independent, so $\zeta=0$ . $\square$

Definition 1.2. The Hilbert tensor product $H\otimes K$ is the completion of $H\odot K$ for the inner product (1.1).

Proposition 1.3.

(Continuity and totality.) $\|\xi\otimes\eta\|=\|\xi\|\|\eta\|$ and $\|\xi\otimes\eta-\xi'\otimes\eta'\|\le\|\xi-\xi'\|\|\eta\|+\|\xi'\|\|\eta-\eta'\|$ . If $D\subseteq H$ and $E\subseteq K$ are total, the vectors $\xi\otimes\eta$ with $\xi\in D$ , $\eta\in E$ are total in $H\otimes K$ . If $(e_i)$ and $(f_j)$ are orthonormal bases, $(e_i\otimes f_j)$ is an orthonormal basis.
(Vector maps.) For $\eta\in K$ and $\xi\in H$ , the maps $R_\eta\xi'=\xi'\otimes\eta$ and $S_\xi\eta'=\xi\otimes\eta'$ are bounded operators $R_\eta:H\to H\otimes K$ and $S_\xi:K\to H\otimes K$ , with $\|R_\eta\xi'\|=\|\xi'\|\|\eta\|$ and $\|S_\xi\eta'\|=\|\xi\|\|\eta'\|$ . Their adjoints act by $R_\eta^*(\xi'\otimes\eta')=\langle\eta',\eta\rangle\xi'$ and $S_\xi^*(\xi'\otimes\eta')=\langle\xi',\xi\rangle\eta'$ . $R_\eta$ and $S_\xi$ are linear in $\eta$ and $\xi$ , so $R_\eta^*$ and $S_\xi^*$ are conjugate-linear in them.
(Columns.) Let $(f_j)_{j\in J}$ be an orthonormal basis of $K$ and put $R_j=R_{f_j}$ . Then $R_j^*R_k=\delta_{jk}1_H$ , and for every $\zeta\in H\otimes K$ $\zeta=\sum_jR_jR_j^*\zeta,\qquad \|\zeta\|^2=\sum_j\|R_j^*\zeta\|^2 . \tag{1.2}$ Conversely, $\sum_jR_j\zeta_j$ converges whenever $\sum_j\|\zeta_j\|^2<\infty$ . Rows $S_i=S_{e_i}$ , for an orthonormal basis $(e_i)$ of $H$ , behave in the same way. If $K=0$ , then $J=\varnothing$ and $H\otimes K=0$ .
(Flip and associativity.) There are unitaries $H\otimes K\to K\otimes H$ , $\xi\otimes\eta\mapsto\eta\otimes\xi$ , and $(H\otimes K)\otimes L\to H\otimes(K\otimes L)$ , $(\xi\otimes\eta)\otimes\theta\mapsto\xi\otimes(\eta\otimes\theta)$ . Also $\xi\otimes c\mapsto c\xi$ is a unitary $H\otimes\mathbb C\to H$ .

We identify the two triple products through the associativity unitary, and write $H\otimes K\otimes L$ and $\xi\otimes\eta\otimes\theta$ . We call (1.2) the column expansion.

Proof. (1) The norm identity is (1.1). The estimate follows from $\xi\otimes\eta-\xi'\otimes\eta'=(\xi-\xi')\otimes\eta+\xi'\otimes(\eta-\eta')$ . By bilinearity, the closed span of the given vectors contains $\xi\otimes\eta$ for $\xi$ in the span of $D$ and $\eta$ in the span of $E$ . By the estimate it then contains every $\xi\otimes\eta$ , hence the dense subspace $H\odot K$ . Orthonormality of $(e_i\otimes f_j)$ follows from (1.1), and totality from what was just shown.

(2) Boundedness is the norm identity. For the adjoint, $\langle\xi'\otimes\eta',R_\eta\xi''\rangle=\langle\xi',\xi''\rangle\langle\eta',\eta\rangle=\langle\langle\eta',\eta\rangle\xi',\xi''\rangle$ . The same computation works for $S_\xi$ .

(3) $\langle R_k\xi,R_j\xi'\rangle=\langle\xi,\xi'\rangle\langle f_k,f_j\rangle$ gives $R_j^*R_k=\delta_{jk}1$ . So the ranges $R_j(H)$ are mutually orthogonal closed subspaces, and $R_jR_j^*$ is the projection onto $R_j(H)$ . Their closed span contains every $\xi\otimes f_j$ , hence all of $H\otimes K$ by (1). So $H\otimes K$ is the orthogonal sum of the $R_j(H)$ , which is (1.2). The converse is convergence of orthogonal series.

(4) The maps are defined on the algebraic tensor products by the universal property. They preserve (1.1) on elementary tensors, hence inner products of finite sums. For associativity, $(H\odot K)\odot L$ is dense in $(H\otimes K)\otimes L$ by (1), since $H\odot K$ is dense in $H\otimes K$ . The ranges are dense by (1). Isometries between dense subspaces with dense range extend to unitaries. Finally $\langle\xi\otimes c,\xi'\otimes c'\rangle=c\bar c'\langle\xi,\xi'\rangle=\langle c\xi,c'\xi'\rangle$ , and the map is onto. $\square$

2. Tensor products of operators and operator matrices

Proposition 2.1.

For $a\in B(H)$ and $b\in B(K)$ there is a unique $a\otimes b\in B(H\otimes K)$ with $(a\otimes b)(\xi\otimes\eta)=a\xi\otimes b\eta$ . Moreover $\|a\otimes b\|=\|a\|\|b\|$ , the map $(a,b)\mapsto a\otimes b$ is bilinear, $(a\otimes b)(c\otimes d)=ac\otimes bd$ , $(a\otimes b)^*=a^*\otimes b^*$ , and $1\otimes1=1$ .
(Matrices.) Fix an orthonormal basis $(f_j)_{j\in J}$ of $K$ and the columns $R_j$ of Proposition 1.3(3). For $X\in B(H\otimes K)$ put $X_{jk}=R_j^*XR_k\in B(H)$ . Then $R_j^*X\zeta=\sum_kX_{jk}R_k^*\zeta$ , a norm-convergent sum, so $X$ is determined by its matrix. Further $(a\otimes b)_{jk}=\langle bf_k,f_j\rangle a$ , $R_j^*(a\otimes1)=aR_j^*$ and $(a\otimes1)R_k=R_ka$ .
(Commutation criterion.) $X$ commutes with $a\otimes1$ if and only if every $X_{jk}$ commutes with $a$ . Hence, for every $S\subseteq B(H)$ , $(S\otimes1)'=\{X\in B(H\otimes K):\ X_{jk}\in S'\ \text{for all }j,k\},\qquad S\otimes1=\{s\otimes1:s\in S\}. \tag{2.1}$
(Commuting with $1\otimes B(K)$ .) Let $K\neq0$ . An operator $X$ commutes with every $1\otimes b$ , $b\in B(K)$ , if and only if $X=x\otimes1$ for some $x\in B(H)$ . Then $x=X_{jj}$ for every $j$ , and $\|x\|=\|X\|$ .
(Truncations.) For a finite set $F\subseteq J$ , let $p_F$ be the projection onto the span of $\{f_j:j\in F\}$ , and let $e_{jk}\in B(K)$ be the operator $\eta\mapsto\langle\eta,f_k\rangle f_j$ . Then $(1\otimes p_F)X(1\otimes p_F)=\sum_{j,k\in F}X_{jk}\otimes e_{jk}, \tag{2.2}$ its norm is at most $\|X\|$ , and it converges strongly to $X$ as $F$ increases.

Proof. (1) Two bounded operators that agree on elementary tensors are equal, because elementary tensors are total (Proposition 1.3(1)); this gives uniqueness. For existence, let $a\odot1$ be the linear map $\xi\otimes\eta\mapsto a\xi\otimes\eta$ on $H\odot K$ . On elementary tensors $R_j^*(a\xi\otimes\eta)=\langle\eta,f_j\rangle a\xi=aR_j^*(\xi\otimes\eta)$ , so $R_j^*(a\odot1)\zeta=aR_j^*\zeta$ for $\zeta\in H\odot K$ . By the column expansion (1.2), $\|(a\odot1)\zeta\|^2=\sum_j\|aR_j^*\zeta\|^2\le\|a\|^2\sum_j\|R_j^*\zeta\|^2=\|a\|^2\|\zeta\|^2 .$ So $a\odot1$ extends to an operator $a\otimes1$ with $\|a\otimes1\|\le\|a\|$ . Rows give $1\otimes b$ with $\|1\otimes b\|\le\|b\|$ . The two operators commute on elementary tensors, hence everywhere, and $a\otimes b:=(a\otimes1)(1\otimes b)$ has the required values. So $\|a\otimes b\|\le\|a\|\|b\|$ . Conversely $\|a\otimes b\|\ge\|a\xi\|\|b\eta\|$ for unit vectors $\xi,\eta$ , and the supremum of the right side is $\|a\|\|b\|$ ; if $H$ or $K$ is $0$ , both sides are $0$ . The algebraic rules hold on elementary tensors. For the adjoint, $\langle(a\otimes b)(\xi\otimes\eta),\xi'\otimes\eta'\rangle=\langle a\xi,\xi'\rangle\langle b\eta,\eta'\rangle=\langle\xi\otimes\eta,a^*\xi'\otimes b^*\eta'\rangle$ ; sesquilinearity and density finish.

(2) By (1.2) and continuity, $X\zeta=\sum_kXR_kR_k^*\zeta$ ; apply $R_j^*$ . Next, $R_j^*(a\otimes b)R_k\xi=R_j^*(a\xi\otimes bf_k)=\langle bf_k,f_j\rangle a\xi$ . The last two identities are the case $b=1$ and its adjoint.

(3) By (2), $(X(a\otimes1))_{jk}=R_j^*XR_ka=X_{jk}a$ and $((a\otimes1)X)_{jk}=aX_{jk}$ . Operators are determined by their matrices.

(4) One direction is clear. For the other, note that $1\otimes e_{jk}=R_jR_k^*$ , since both send $\xi\otimes\eta$ to $\langle\eta,f_k\rangle\,\xi\otimes f_j$ . If $X$ commutes with every $R_jR_k^*$ , then, using $R_k^*R_k=1$ , $X_{jk}=R_j^*X(R_kR_k^*)R_k=R_j^*(R_kR_k^*)XR_k=\delta_{jk}X_{kk},\qquad X_{jj}=R_j^*X(R_jR_k^*)R_k=R_j^*(R_jR_k^*)XR_k=X_{kk}.$ So $X$ has the matrix of $x\otimes1$ with $x=X_{kk}$ , and $X=x\otimes1$ by (2). By (1), $\|x\otimes1\|=\|x\|$ .

(5) $1\otimes p_F=\sum_{j\in F}R_jR_j^*$ , and $R_jX_{jk}R_k^*=X_{jk}\otimes e_{jk}$ , since both send $\xi\otimes\eta$ to $\langle\eta,f_k\rangle X_{jk}\xi\otimes f_j$ . This gives (2.2). Put $Q_F=1\otimes p_F$ . Then $Q_F\to1$ strongly by (1.2), and $Q_FXQ_F\zeta-X\zeta=Q_FX(Q_F\zeta-\zeta)+(Q_F-1)X\zeta\to0$ . $\square$

3. Normal functionals and the amplification

The ultraweak topology. Let $B(H)_*$ be the set of functionals $\rho(x)=\sum_n\langle x\xi_n,\eta_n\rangle,\qquad \sum_n\|\xi_n\|^2<\infty,\quad \sum_n\|\eta_n\|^2<\infty , \tag{3.1}$ with countable or finite index sets. A square-summable family of any size has only countably many nonzero members, so larger index sets give nothing new. The series converges absolutely, and $|\rho(x)|\leq\|x\|\,(\sum\|\xi_n\|^2)^{1/2}(\sum\|\eta_n\|^2)^{1/2}$ . $B(H)_*$ is a linear space: merge two families to add, and scale one sequence to multiply by a scalar. The ultraweak topology, also called the $\sigma$ -weak topology, is the weakest topology in which every $\rho\in B(H)_*$ is continuous. It is finer than the weak operator topology, so every weakly closed set, in particular every commutant, is ultraweakly closed.

A linear map $\Phi$ from a subspace of $B(H)$ to $B(L)$ is normal if it is ultraweakly continuous, that is, if $\rho\circ\Phi$ is ultraweakly continuous for every $\rho\in B(L)_*$ . For a subspace $Y\subseteq B(H)$ , $Y_*$ denotes its normal functionals. Compositions of normal maps are normal. For fixed $A\in B(L,L')$ and $C\in B(L',L)$ , the map $X\mapsto AXC$ from $B(L)$ to $B(L')$ is normal, since $\sum_n\langle AXC\zeta_n,\zeta_n'\rangle=\sum_n\langle X(C\zeta_n),A^*\zeta_n'\rangle$ . For positive maps, normality is often expressed instead through suprema of bounded increasing nets; part (3) of the next proposition gives this order form for the amplification.

Proposition 3.1.

(Normal functionals.) Let $Y\subseteq B(H)$ be any linear subspace. A linear functional on $Y$ is ultraweakly continuous if and only if it is the restriction of some $\rho\in B(H)_*$ . In particular, every normal functional on $Y$ is bounded.
(Normality of the amplification.) The map $a\mapsto a\otimes1_K$ , $B(H)\to B(H\otimes K)$ , is a unital $*$ -homomorphism. It is isometric if $K\ne0$ and zero if $K=0$ . It is normal: for $\rho=\sum_n\omega_{\zeta_n,\zeta_n'}\in B(H\otimes K)_*$ , $\rho(a\otimes1)=\sum_{n}\sum_j\langle aR_j^*\zeta_n,R_j^*\zeta_n'\rangle,\qquad \sum_{n,j}\|R_j^*\zeta_n\|^2=\sum_n\|\zeta_n\|^2 , \tag{3.2}$ so $\rho(\,\cdot\otimes1)\in B(H)_*$ . It is strongly continuous on bounded sets. The same holds for $b\mapsto1_H\otimes b$ , and for the restriction of either map to any subspace, such as a von Neumann algebra.
(Order form.) Let $(a_\alpha)$ be a bounded increasing net of self-adjoint operators on $H$ . It converges strongly to its least upper bound $a$ among self-adjoint operators. The net $(a_\alpha\otimes1)$ is increasing, converges strongly to $a\otimes1$ , and $a\otimes1$ is its least upper bound. If all $a_\alpha$ lie in a von Neumann algebra $M$ , then $a\in M$ , and $x\mapsto x\otimes1$ carries the supremum in $M$ to the supremum in every von Neumann algebra that contains $M\otimes1$ .
(Normal functionals as vector functionals.) Let $M\subseteq B(H)$ be a $*$ -subalgebra with $1\in M$ , let $\varphi\in M_*$ , and let $R$ be an infinite-dimensional Hilbert space. There are $\zeta,\zeta'\in H\otimes R$ with $\varphi(x)=\langle(x\otimes1)\zeta,\zeta'\rangle$ for $x\in M$ . If $\varphi$ is positive, one can take $\zeta'=\zeta$ . Then $\varphi=\sum_n\omega_{\xi_n}|_M$ with $\sum_n\|\xi_n\|^2=\varphi(1)$ .

Reference: Part (4) extends [Takesaki I, Lemma IV.5.4], which assumes $R$ separable and gives two vectors. The proof of [Takesaki I, Theorem IV.5.5] needs a single vector at that point; the positive case of (4) supplies it.

Proof. (1) Restrictions are continuous by definition. Conversely, let $\varphi$ be continuous at $0$ . There are $\rho_1,\ldots,\rho_m\in B(H)_*$ and $\delta>0$ with $|\varphi(y)|<1$ whenever $y\in Y$ and $\max_i|\rho_i(y)|<\delta$ . If $\rho_i(y)=0$ for all $i$ , the same holds for $ty$ , $t>0$ ; so $|\varphi(y)|<1/t$ for every $t$ , and $\varphi(y)=0$ . Hence $\varphi$ vanishes on the kernel of $y\mapsto(\rho_1(y),\ldots,\rho_m(y))\in\mathbb C^m$ . So $\varphi$ factors through the image of this map. Extending the factor linearly to $\mathbb C^m$ gives constants $c_i$ with $\varphi=\sum_ic_i\rho_i$ on $Y$ , and $\sum_ic_i\rho_i\in B(H)_*$ . Boundedness follows from the estimate after (3.1).

(2) The algebraic properties and the norm are Proposition 2.1(1). By the column expansion (1.2) and polarization, $\langle(a\otimes1)\zeta,\zeta'\rangle=\sum_j\langle R_j^*(a\otimes1)\zeta,R_j^*\zeta'\rangle=\sum_j\langle aR_j^*\zeta,R_j^*\zeta'\rangle$ , using $R_j^*(a\otimes1)=aR_j^*$ from Proposition 2.1(2). This is (3.2); the double family $(R_j^*\zeta_n)$ is square-summable, and so is $(R_j^*\zeta_n')$ . Now let $a_\alpha\to a$ strongly with $\|a_\alpha\|\le C$ . Then $\|((a_\alpha-a)\otimes1)\zeta\|^2=\sum_j\|(a_\alpha-a)R_j^*\zeta\|^2$ . Each term tends to $0$ and is at most $4C^2\|R_j^*\zeta\|^2$ . Given $\varepsilon>0$ , a finite set of indices carries all of $\sum_j\|R_j^*\zeta\|^2$ except $\varepsilon$ , so the sum tends to $0$ . Rows handle $1\otimes b$ . Restrictions of continuous maps are continuous.

(3) For $\xi\in H$ , $\langle a_\alpha\xi,\xi\rangle$ is increasing and bounded, so it converges. By polarization $\langle a_\alpha\xi,\eta\rangle$ converges for all $\xi,\eta$ . The limit is a bounded sesquilinear form, so it equals $\langle a\xi,\eta\rangle$ for a self-adjoint $a$ . Clearly $a\ge a_\alpha$ for every $\alpha$ . A self-adjoint upper bound $b$ satisfies $\langle b\xi,\xi\rangle\ge\lim_\alpha\langle a_\alpha\xi,\xi\rangle=\langle a\xi,\xi\rangle$ , so $a$ is the least upper bound. For $T=a-a_\alpha\ge0$ , the Cauchy–Schwarz inequality for the positive form $\langle T\cdot,\cdot\rangle$ gives $\|T\xi\|^4=\langle T\xi,T\xi\rangle^2\le\langle T\xi,\xi\rangle\langle T^2\xi,T\xi\rangle\le\langle T\xi,\xi\rangle\,\|T\|\,\|T\xi\|^2$ . So $\|T\xi\|^2\le\|T\|\langle T\xi,\xi\rangle\to0$ , and $a_\alpha\to a$ strongly. For $c\ge0$ , $\langle(c\otimes1)\zeta,\zeta\rangle=\sum_j\langle cR_j^*\zeta,R_j^*\zeta\rangle\ge0$ ; hence $(a_\alpha\otimes1)$ is increasing. It is bounded and converges strongly to $a\otimes1$ by (2). By the first part its least upper bound is its strong limit $a\otimes1$ . A von Neumann algebra is strongly closed, so $a\in M$ , and $a$ is then also the supremum in $M$ . If $P\supseteq M\otimes1$ is a von Neumann algebra, $a\otimes1\in P$ is the least upper bound in all of $B(H\otimes K)$ , hence in $P$ .

(4) By (1), $\varphi=\sum_n\omega_{\xi_n,\eta_n}|_M$ with square-summable families indexed by a set $\mathbb N_0\subseteq\mathbb N$ . Choose an orthonormal family $(g_n)_{n\in\mathbb N_0}$ in $R$ , and put $\zeta=\sum_n\xi_n\otimes g_n$ and $\zeta'=\sum_n\eta_n\otimes g_n$ . By the column expansion (1.2), applied with a basis of $R$ that contains the $g_n$ , $\langle(x\otimes1)\zeta,\zeta'\rangle=\sum_n\langle x\xi_n,\eta_n\rangle=\varphi(x)$ .

Now let $\varphi$ be positive, and write $\pi(x)=x\otimes1_R$ . A positive functional on a unital $*$ -algebra is hermitian: a self-adjoint $x$ equals $y^*y-z^*z$ with $y=(1+x)/2$ and $z=(1-x)/2$ , so $\varphi(x)$ is real. Hence, for $x=x^*$ , with $\theta=\zeta+\zeta'$ and $\theta'=\zeta-\zeta'$ , $4\varphi(x)=2\langle\pi(x)\zeta,\zeta'\rangle+2\langle\pi(x)\zeta',\zeta\rangle=\langle\pi(x)\theta,\theta\rangle-\langle\pi(x)\theta',\theta'\rangle .$ So $4\varphi(y^*y)\le\|\pi(y)\theta\|^2$ for $y\in M$ . Let $L_0$ be the closure of $\pi(M)\theta$ . The Cauchy–Schwarz inequality for $\varphi$ gives $|\varphi(y^*x)|^2\le\varphi(x^*x)\varphi(y^*y)\le\tfrac1{16}\|\pi(x)\theta\|^2\|\pi(y)\theta\|^2$ . So $B(\pi(x)\theta,\pi(y)\theta):=\varphi(y^*x)$ is a well-defined positive sesquilinear form on $\pi(M)\theta$ , bounded by $\tfrac14$ . It extends to $L_0$ , where it is given by a positive operator $T_0\in B(L_0)$ . For $a\in M$ we have $B(\pi(a)u,v)=B(u,\pi(a^*)v)$ on $\pi(M)\theta$ , since both sides are $\varphi(y^*ax)$ . So $T_0$ commutes with the restrictions of $\pi(M)$ to the invariant subspace $L_0$ . The projection $P_0$ onto $L_0$ commutes with $\pi(M)$ , because $L_0$ is invariant under the self-adjoint set $\pi(M)$ . Hence $T=T_0P_0$ is positive and commutes with $\pi(M)$ , and so does $T^{1/2}$ , a norm limit of polynomials in $T$ . With $\xi''=T^{1/2}\theta$ , $\varphi(x)=B(\pi(x)\theta,\theta)=\langle T\pi(x)\theta,\theta\rangle=\langle\pi(x)\xi'',\xi''\rangle .$ Expanding $\xi''=\sum_k\xi_k\otimes g_k$ along an orthonormal basis $(g_k)$ of $R$ gives $\varphi=\sum_k\omega_{\xi_k}|_M$ and $\varphi(1)=\|\xi''\|^2=\sum_k\|\xi_k\|^2$ . $\square$

Remarks. Part (1) is finite-dimensional linear algebra, and the boundedness of a normal functional is a consequence, not a hypothesis. In part (4) any infinite-dimensional $R$ works, while a finite-dimensional one does not in general (Exercise 3.4). The one-vector form for positive $\varphi$ is used in the proof of Theorem 8.2. Part (2) is not strong continuity on all of $B(H)$ : Example 3.2 shows that it fails for unbounded nets when $H$ and $K$ are both infinite-dimensional.

Example 3.2 (the amplification is not strongly continuous). Let $H=K=\ell^2(\mathbb N)$ and $\zeta_0=\sum_kk^{-1}\delta_k\otimes\delta_k\in H\otimes K$ . For each finite set $F\subseteq H$ , choose a unit vector $u_F$ orthogonal to $F$ , put $c_F=(\sum_kk^{-2}|\langle\delta_k,u_F\rangle|^2)^{-1/2}$ , and let $x_F\xi=c_F\langle\xi,u_F\rangle\delta_1$ . Then $x_F\xi=0$ for $\xi\in F$ , so the net $(x_F)$ , directed by inclusion, tends to $0$ strongly. But $(x_F\otimes1)\zeta_0=c_F\sum_kk^{-1}\langle\delta_k,u_F\rangle\,\delta_1\otimes\delta_k$ has norm $1$ . So $x\mapsto x\otimes1$ is strongly continuous only on bounded sets, as in Proposition 3.1(2). Both dimensions matter. If $\dim K=d<\infty$ , it is strongly continuous everywhere, since $\|(x\otimes1)\zeta\|^2=\sum_{j\le d}\|xR_j^*\zeta\|^2$ is a finite sum. If $\dim H<\infty$ , the strong topology on $B(H)$ is the norm topology, and continuity is automatic.

Example 3.3 (the norm need not be attained by a normal functional). For $x$ in a von Neumann algebra, $\|x\|=\sup\{|\omega(x)|:\omega\ \text{normal},\ \|\omega\|\le1\}$ , because the vector functionals $\omega_{\xi,\eta}$ with unit vectors $\xi,\eta$ already give this supremum. The supremum need not be attained. Let $x$ be the diagonal operator with entries $1-1/n$ , $n\ge1$ , on $\ell^2(\mathbb N)$ , so $\|x\|=1$ , and let $\omega\in B(\ell^2)_*$ with $\|\omega\|\le1$ . For the projection $q_N$ onto the span of the first $N$ basis vectors, let $a_N$ and $b_N$ be the norms of $y\mapsto\omega(q_Nyq_N)$ and $y\mapsto\omega((1-q_N)y(1-q_N))$ . Block-diagonal test operators show that $a_N+b_N\le1$ . Indeed, let $y_1,y_2$ have norm at most one, multiplied by scalars of modulus one so that $\omega(q_Ny_1q_N)\ge0$ and $\omega((1-q_N)y_2(1-q_N))\ge0$ . The operator $q_Ny_1q_N+(1-q_N)y_2(1-q_N)$ has norm at most one, so the sum of these two numbers is at most $\|\omega\|\le1$ ; taking suprema over $y_1$ and $y_2$ gives the claim. Since $x=q_Nxq_N+(1-q_N)x(1-q_N)$ and $\|q_Nxq_N\|=1-1/N$ , we get $|\omega(x)|\le(1-1/N)a_N+b_N\le1-a_N/N$ . So $|\omega(x)|=1$ would force $a_N=0$ and $b_N=1$ for all $N$ . But $b_N\to0$ : writing $\omega=\sum_k\omega_{\xi_k,\eta_k}$ , $b_N\le(\sum_k\|(1-q_N)\xi_k\|^2)^{1/2}(\sum_k\|(1-q_N)\eta_k\|^2)^{1/2}$ , which tends to $0$ by dominated convergence. Hence no normal functional of norm at most one attains the norm of $x$ .

Exercise 3.4 (the space $R$ in Proposition 3.1(4) must be infinite-dimensional). Let $H$ be infinite-dimensional with an orthonormal sequence $(\delta_n)$ , let $\varphi=\sum_n2^{-n}\omega_{\delta_n}$ on $B(H)$ , and let $R=\mathbb C^d$ with $d<\infty$ . Show that no $\zeta,\zeta'\in H\otimes R$ satisfy $\varphi(x)=\langle(x\otimes1)\zeta,\zeta'\rangle$ for all $x\in B(H)$ .

Solution. Suppose they do. Expanding along a basis of $R$ gives $\varphi(x)=\sum_{k\le d}\langle x\xi_k,\eta_k\rangle$ for some vectors $\xi_k,\eta_k$ . The linear map $v\mapsto(\langle v,\xi_k\rangle)_{k\le d}$ on the span of $\delta_1,\ldots,\delta_{d+1}$ has a nonzero kernel, so there is a unit vector $u$ in that span with $u\perp\xi_k$ for all $k$ . Let $p_u$ be the projection onto $\mathbb Cu$ . Then $p_u\xi_k=0$ , so $\sum_k\langle p_u\xi_k,\eta_k\rangle=0$ . But $\varphi(p_u)=\sum_n2^{-n}|\langle\delta_n,u\rangle|^2>0$ , since $u\ne0$ lies in the span of the $\delta_n$ . This is a contradiction.

4. The bicommutant theorem and the normal extension principle

Lemma 4.1 (amplified density). Let $A\subseteq B(H)$ be a $*$ -subalgebra with $1\in A$ , and let $x\in A''$ . For every sequence $(\xi_n)$ in $H$ with $\sum_n\|\xi_n\|^2<\infty$ , finite families included, and every $\varepsilon>0$ , there is $a\in A$ with $\sum_n\|(x-a)\xi_n\|^2<\varepsilon^2 . \tag{4.1}$

Proof. Let $R=\ell^2(\mathbb N)$ with standard basis $(g_n)$ , and $\zeta=\sum_n\xi_n\otimes g_n\in H\otimes R$ . Let $E$ be the closure of $\{(a\otimes1)\zeta:a\in A\}$ and $P$ the projection onto $E$ . $E$ is invariant under the self-adjoint set $A\otimes1$ : for $a\in A$ we have $(a\otimes1)P=P(a\otimes1)P$ , and replacing $a$ by $a^*$ and taking adjoints gives $P(a\otimes1)=P(a\otimes1)P$ . So $P\in(A\otimes1)'$ . By the commutation criterion (2.1), every matrix entry $P_{jk}$ lies in $A'$ . As $x\in A''$ , $x$ commutes with every $P_{jk}$ , and Proposition 2.1(3) shows that $x\otimes1$ commutes with $P$ . Since $1\in A$ , $\zeta\in E$ , so $(x\otimes1)\zeta=P(x\otimes1)\zeta\in E$ . Choose $a\in A$ with $\|((x-a)\otimes1)\zeta\|<\varepsilon$ . By the column expansion (1.2), the square of the left side is $\sum_n\|(x-a)\xi_n\|^2$ . $\square$

Theorem 4.2 (bicommutant theorem). For a $*$ -subalgebra $A\subseteq B(H)$ with $1\in A$ , $A''$ is the closure of $A$ in the strong, the weak and the ultraweak topology. A $*$ -subalgebra that contains $1$ and is closed in one of these topologies is a von Neumann algebra.

Proof. Strong density is Lemma 4.1 for finite families. For ultraweak density, let $x\in A''$ and $\rho_1,\ldots,\rho_m\in B(H)_*$ with $\rho_i=\sum_n\omega_{\xi^i_n,\eta^i_n}$ . Apply Lemma 4.1 to the merged family $(\xi^i_n)_{i,n}$ . Then $|\rho_i(x-a)|\le(\sum_n\|(x-a)\xi^i_n\|^2)^{1/2}(\sum_n\|\eta^i_n\|^2)^{1/2}$ is small for all $i$ at once. Conversely, $A''$ is weakly closed and the weak topology is coarser than the other two, so each closure of $A$ lies in $A''$ . The last sentence follows. $\square$

The general form of the theorem, for $*$ -subalgebras that need not contain $1$ , is proved in the lesson The double commutation theorem. The unital case is all we need.

Proposition 4.3 (normal extension principle). Let $M$ be a von Neumann algebra on $H$ , $A\subseteq M$ a $*$ -subalgebra with $1\in A$ and $A''=M$ , and $\Phi,\Psi:M\to B(L)$ normal linear maps.

(i) If $\Phi=\Psi$ on $A$ , then $\Phi=\Psi$ .
(ii) If $P\subseteq B(L)$ is a von Neumann algebra and $\Phi(A)\subseteq P$ , then $\Phi(M)\subseteq P$ .
(iii) Let $S\subseteq M$ be a self-adjoint set with $S''=M$ , and let $\Phi,\Psi$ be $*$ -homomorphisms. If $\Phi(S)\subseteq P$ and $\Phi(1)\in P$ , then $\Phi(M)\subseteq P$ . If $\Phi=\Psi$ on $S\cup\{1\}$ , then $\Phi=\Psi$ .

Proof. (i) Let $\rho\in B(L)_*$ . The functional $\rho\circ(\Phi-\Psi)$ is normal on $M$ . By Proposition 3.1(1) it is the restriction of some $\sigma=\sum_n\omega_{\xi_n,\eta_n}\in B(H)_*$ , and $\sigma$ vanishes on $A$ . For $x\in M$ and $a\in A$ , $|\sigma(x)|=|\sigma(x-a)|\le(\sum_n\|(x-a)\xi_n\|^2)^{1/2}(\sum_n\|\eta_n\|^2)^{1/2}$ , which Lemma 4.1 makes arbitrarily small. So $\rho(\Phi(x))=\rho(\Psi(x))$ for every $\rho$ , in particular for every vector functional, and $\Phi(x)=\Psi(x)$ .

(ii) For $p'\in P'$ , the map $x\mapsto\Phi(x)p'-p'\Phi(x)$ is normal and vanishes on $A$ . By (i) it vanishes on $M$ . So $\Phi(M)\subseteq P''=P$ .

(iii) Let $A$ be the span of $1$ and all finite products of elements of $S$ . It is a $*$ -subalgebra containing $1$ and $S$ , and $A'=S'$ , so $A''=S''=M$ . Since $\Phi$ is a $*$ -homomorphism and $P$ is a unital $*$ -algebra, $\Phi(A)\subseteq P$ ; in the second case $\Phi=\Psi$ on $A$ . Apply (ii) and (i). $\square$

For example, let $G$ be a locally compact group with left regular representation $\lambda$ . The set $S=\lambda(G)$ is self-adjoint, because $\lambda(g)^*=\lambda(g^{-1})$ , and it contains $1=\lambda(e)$ . So by (iii), a normal $*$ -homomorphism on the group von Neumann algebra $\lambda(G)''$ is determined by its values on the $\lambda(g)$ , and it maps $\lambda(G)''$ into every von Neumann algebra that contains the images of the $\lambda(g)$ .

5. The spatial tensor product

Definition 5.1. Let $M\subseteq B(H)$ and $N\subseteq B(K)$ be von Neumann algebras. $M\odot N$ is the linear span of the operators $x\otimes y$ with $x\in M$ , $y\in N$ . By Proposition 2.1(1) it is a $*$ -subalgebra of $B(H\otimes K)$ containing $1$ . The spatial tensor product is $M\bar\otimes N=(M\odot N)'' , \tag{5.1}$ the smallest von Neumann algebra that contains every $x\otimes y$ . For sets $S\subseteq B(H)$ and $T\subseteq B(K)$ we write $S\otimes1=\{s\otimes1:s\in S\}$ and $1\otimes T=\{1\otimes t:t\in T\}$ .

Theorem 5.2.

(Closure.) $M\bar\otimes N$ is a von Neumann algebra on $H\otimes K$ . It is the closure of $M\odot N$ in the strong, the weak and the ultraweak topology. In particular it is strongly closed.
(Generators.) If $S\subseteq B(H)$ and $T\subseteq B(K)$ are self-adjoint sets with $S''=M$ and $T''=N$ , then $M\bar\otimes N=(S\otimes1\cup1\otimes T)''$ . In particular $M\bar\otimes N=(M\otimes1\cup1\otimes N)''$ .
(Bounded strong limits.) Let $(X_\alpha)$ be a bounded net in $B(H\otimes K)$ , let $X\in B(H\otimes K)$ , and let $D\subseteq H\otimes K$ be a total set with $X_\alpha\zeta\to X\zeta$ for every $\zeta\in D$ . Then $X_\alpha\to X$ strongly. If every $X_\alpha$ lies in $M\bar\otimes N$ , so does $X$ . One may take $D=\{\xi\otimes\eta:\xi\in D_1,\ \eta\in D_2\}$ for total sets $D_1\subseteq H$ , $D_2\subseteq K$ .
(Amplification.) For every set $S\subseteq B(H)$ , $(S\otimes1)''=S''\otimes1$ . Hence $M\otimes1$ is a von Neumann algebra, $M\bar\otimes\mathbb C1_K=M\otimes1$ and $\mathbb C1_H\bar\otimes N=1\otimes N$ . The map $x\mapsto x\otimes1$ is a normal unital $*$ -homomorphism of $M$ onto $M\otimes1\subseteq M\bar\otimes N$ . It is injective and isometric when $K\ne0$ .
(Matrices over $M$ .) With matrices as in Proposition 2.1(2), $M\bar\otimes B(K)=\{X\in B(H\otimes K):X_{jk}\in M\ \text{for all }j,k\}=(M'\otimes1)', \tag{5.2}$ and $B(H)\bar\otimes B(K)=B(H\otimes K)$ .

Proof. (1) By (5.1), $M\bar\otimes N$ is the double commutant of a self-adjoint set, hence a von Neumann algebra. The rest is the bicommutant theorem (Theorem 4.2) with $A=M\odot N$ .

(2) By the commutation criterion (2.1), $(S\otimes1)'$ consists of the operators whose entries lie in $S'=S'''=M'$ . So $(S\otimes1)'=(M\otimes1)'$ . Rows give $(1\otimes T)'=(1\otimes N)'$ . Hence $(S\otimes1\cup1\otimes T)'=(M\otimes1\cup1\otimes N)'$ . The last set equals $(M\odot N)'$ , because $M\otimes1$ and $1\otimes N$ lie in $M\odot N$ and $x\otimes y=(x\otimes1)(1\otimes y)$ . Take commutants.

(4) If $K=0$ , all sets involved are $\{0\}$ . Let $K\ne0$ . By (2.1), $(S\otimes1)'$ contains $1\otimes B(K)$ , whose matrix entries are scalars. So $(S\otimes1)''\subseteq(1\otimes B(K))'=B(H)\otimes1$ by Proposition 2.1(4). Now let $x\in B(H)$ . If $x\in S''$ , then $x$ commutes with the entries of every $X\in(S\otimes1)'$ , since they lie in $S'$ ; by Proposition 2.1(3), $x\otimes1$ commutes with $X$ . If conversely $x\otimes1\in(S\otimes1)''$ , then it commutes with $c\otimes1$ for every $c\in S'$ , so $xc=cx$ and $x\in S''$ . Thus $(S\otimes1)''=S''\otimes1$ . For $S=M$ this says $(M\otimes1)''=M\otimes1$ . Applying Proposition 2.1(4) with $H=\mathbb C$ gives $\{1_K\}''=B(K)'=\mathbb C1_K$ ; so (2) with $T=\{1\}$ gives $M\bar\otimes\mathbb C1=(M\otimes1)''=M\otimes1$ , and symmetrically for $N$ . The remaining claims are the normality of the amplification (Proposition 3.1(2)) and the norm identity of Proposition 2.1(1).

(5) The entries of $x\otimes y$ are multiples of $x$ , so $M\odot B(K)\subseteq\mathcal C:=\{X:X_{jk}\in M\}$ . By (2.1), $\mathcal C=(M'\otimes1)'$ , the commutant of a self-adjoint set; so $\mathcal C''=\mathcal C$ and $M\bar\otimes B(K)\subseteq\mathcal C$ . Conversely, for $X\in\mathcal C$ the truncations (2.2) lie in $M\odot B(K)$ , are bounded by $\|X\|$ and converge strongly to $X$ . By (3), $X\in M\bar\otimes B(K)$ . For $M=B(H)$ we have $M'=\mathbb C1$ and $(\mathbb C1\otimes1)'=B(H\otimes K)$ . $\square$

Part (3) is how membership in a tensor product is usually proved in practice: one exhibits a norm-bounded net in $M\odot N$ that converges on a total set of test vectors. Example 12.2 does this for a discrete group, and the end of Section 12 describes the same argument for a locally compact group. Without the norm bound, convergence on a total set does not give strong convergence (Example 5.5). Part (2) allows generating sets that are not von Neumann algebras themselves. For instance, let $G$ be a locally compact group, $m$ the multiplication representation of $L^\infty(G)$ on $L^2(G)$ , and $\lambda$ the left regular representation. Both $m(L^\infty(G))$ and $\lambda(G)$ are self-adjoint sets, so $m(L^\infty(G))''\bar\otimes\lambda(G)''=(m(L^\infty(G))\otimes1\cup1\otimes\lambda(G))''$ , whether or not $m(L^\infty(G))$ is already a von Neumann algebra.

Example 5.3 (diagonal algebras of any size). Let $I$ and $J$ be arbitrary sets, possibly uncountable, and let $\ell^\infty(I)$ act on $\ell^2(I)$ by multiplication. It is its own commutant, by the argument given below for $I\times J$ , so it is a von Neumann algebra. Counting measure on an uncountable set is not $\sigma$ -finite, so the $\sigma$ -finite result in the lesson on decomposable operators and the diagonal algebra does not apply. By Proposition 1.3(1), $\delta_i\otimes\delta_j\mapsto\delta_{(i,j)}$ is a unitary $\ell^2(I)\otimes\ell^2(J)\to\ell^2(I\times J)$ , and under it $\ell^\infty(I)\bar\otimes\ell^\infty(J)=\ell^\infty(I\times J). \tag{5.3}$ Indeed, $m_a\otimes m_b$ becomes multiplication by $(i,j)\mapsto a(i)b(j)$ . The algebra $\ell^\infty(I\times J)$ is its own commutant: an operator commuting with every rank-one projection $m_{1_{\{(i,j)\}}}$ is diagonal, and a bounded diagonal operator is a multiplication. So $\ell^\infty(I\times J)$ is a von Neumann algebra that contains every $m_a\otimes m_b$ , and $\ell^\infty(I)\bar\otimes\ell^\infty(J)\subseteq\ell^\infty(I\times J)$ . Conversely, for $f\in\ell^\infty(I\times J)$ and finite $F\subseteq I\times J$ , the operator $m_{f1_F}=\sum_{(i,j)\in F}f(i,j)\,m_{1_{\{i\}}}\otimes m_{1_{\{j\}}}$ lies in $\ell^\infty(I)\odot\ell^\infty(J)$ , has norm at most $\|f\|_\infty$ , and agrees with $m_f$ on $\delta_{(i,j)}$ once $(i,j)\in F$ . By Theorem 5.2(3), $m_f\in\ell^\infty(I)\bar\otimes\ell^\infty(J)$ .

Example 5.4 (zero spaces). If $K=0$ , then $H\otimes K=0$ and $M\bar\otimes N=\{0\}=B(0)$ . The amplification $x\mapsto x\otimes1_K$ is then the zero map, which is not injective when $H\ne0$ . This is the only way injectivity fails in Theorem 5.2(4). For instance, for a locally compact group $G$ , the amplification $x\mapsto x\otimes1$ with $K=L^2(G)$ is injective, because $L^2(G)\ne0$ : Haar measure charges nonempty open sets.

Example 5.5 (the norm bound in Theorem 5.2(3) is needed). On $\ell^2(\mathbb N)$ with basis $(\delta_k)_{k\ge1}$ , let $T_n\xi=n\langle\xi,\delta_n\rangle\delta_1$ . For each $k$ , $T_n\delta_k=0$ once $n>k$ , so $T_n\to0$ on the total set $\{\delta_k\}$ . But for $\zeta=\sum_kk^{-1}\delta_k$ we have $T_n\zeta=\delta_1$ for every $n$ . So $T_n$ does not tend to $0$ strongly, and $\|T_n\|=n$ is unbounded. Taking $K=\mathbb C$ , this is an example in $B(H)\bar\otimes B(\mathbb C)=B(H\otimes\mathbb C)$ .

6. Reduced and induced algebras

For a set $S\subseteq B(H)$ and a projection $e$ , write $S_e=\{ese|_{eH}:s\in S\}\subseteq B(eH)$ . For a von Neumann algebra $M$ the subscript is used in two cases. If $e\in M$ , $M_e$ is the reduced algebra: $eMe$ acting on $eH$ . If $e'\in M'$ , then $xe'=e'xe'$ , so $M_{e'}=\{xe'|_{e'H}:x\in M\}$ ; this is the induced algebra, and $x\mapsto x_{e'}=xe'|_{e'H}$ is the induction. Thus $(M')_e$ is induced when $e\in M$ , and $(M')_{e'}$ is reduced when $e'\in M'$ .

Proposition 6.1. Let $M$ be a von Neumann algebra on $H$ .

For a projection $e\in M$ , $M_e$ is a von Neumann algebra on $eH$ , and $(M_e)'=(M')_e$ .
For a projection $e'\in M'$ , $M_{e'}$ is a von Neumann algebra on $e'H$ with commutant $(M')_{e'}=\{e'x'e'|_{e'H}:x'\in M'\}$ . The induction is a normal unital $*$ -homomorphism of $M$ onto $M_{e'}$ . Let $c$ be the projection onto the closed span $[M'e'H]$ . Then $c\in M$ , and the induction is injective if and only if $c=1$ .
(Reductions of tensor products.) Let $N$ be a von Neumann algebra on $K$ , and $e\in M$ , $f\in N$ projections. Identify $(e\otimes f)(H\otimes K)$ with $eH\otimes fK$ , the closure of $eH\odot fK$ . Then $(M\bar\otimes N)_{e\otimes f}=M_e\bar\otimes N_f,\qquad (M'\bar\otimes N')_{e\otimes f}=(M')_e\bar\otimes(N')_f , \tag{6.1}$ the first a reduced algebra, the second an induced one. Applied to $M'$ and $N'$ , this gives the same identities for projections $e'\in M'$ , $f'\in N'$ , with reduction and induction exchanged: $(M'\bar\otimes N')_{e'\otimes f'}=(M')_{e'}\bar\otimes(N')_{f'}$ and $(M\bar\otimes N)_{e'\otimes f'}=M_{e'}\bar\otimes N_{f'}$ .

Proof. (1) We first identify the commutant of $M_e$ , and then deduce that $M_e$ is a von Neumann algebra.

$(M')_e\subseteq(M_e)'$ : each $x'\in M'$ commutes with $e$ and with every $exe$ , so $x'|_{eH}$ maps $eH$ into itself and commutes with $M_e$ .

$(M_e)'\subseteq(M')_e$ : let $T\in B(eH)$ commute with $M_e$ ; then $T^*$ does too, since $M_e$ is self-adjoint. For $a_1,\ldots,a_n\in M$ and $\xi_1,\ldots,\xi_n\in eH$ , let $\mathbf A$ be the operator matrix $[ea_k^*a_ie]_{k,i}$ acting on $eH\otimes\mathbb C^n$ , and $\vec\xi=(\xi_i)$ . Then $\langle\mathbf A\vec\xi,\vec\xi\rangle=\|\sum_ia_i\xi_i\|^2$ , so $\mathbf A\ge0$ . The entries of $\mathbf A$ commute with $T$ and $T^*$ , so by the commutation criterion (Proposition 2.1(3)) $\mathbf A$ commutes with $C=T^*T\otimes1$ , and hence with $\mathbf A^{1/2}$ . Therefore $\Big\|\sum_ia_iT\xi_i\Big\|^2=\langle C\mathbf A\vec\xi,\vec\xi\rangle=\langle C\mathbf A^{1/2}\vec\xi,\mathbf A^{1/2}\vec\xi\rangle\le\|T\|^2\Big\|\sum_ia_i\xi_i\Big\|^2 .$ (The first equality expands both sides, using $T\xi_i\in eH$ and $T(ea_k^*a_ie)=(ea_k^*a_ie)T$ .) So $\sum a_i\xi_i\mapsto\sum a_iT\xi_i$ is well defined and bounded on the span of $MeH$ . Extend it by continuity to $[MeH]$ and by $0$ on $[MeH]^\perp$ ; call the result $x'$ . Both subspaces are invariant under the self-adjoint set $M$ , and $x'b=bx'$ holds on each of them for $b\in M$ . So $x'\in M'$ . Taking $n=1$ and $a_1=1$ gives $x'|_{eH}=T$ .

$M_e$ is a von Neumann algebra: let $S\in B(eH)$ commute with $(M')_e$ , and let $\tilde S$ be the operator $\xi\mapsto S(e\xi)$ on $H$ . For $x'\in M'$ , $\tilde Sx'\xi=S(x'e\xi)=x'S(e\xi)=x'\tilde S\xi$ , using $ex'=x'e$ . So $\tilde S\in M''=M$ , and $S=e\tilde Se|_{eH}\in M_e$ . Hence $(M_e)''=((M')_e)'\subseteq M_e$ .

(2) Apply (1) to the von Neumann algebra $M'$ and $e'\in M'$ . The reduced algebra $(M')_{e'}$ is a von Neumann algebra with commutant $(M'')_{e'}=M_{e'}$ . So $M_{e'}=((M')_{e'})'$ is a von Neumann algebra, with commutant $(M')_{e'}$ . The induction is multiplicative because $e'$ commutes with $M$ , and unital. It is normal because $\sum_n\langle xe'\zeta_n,\zeta_n'\rangle$ , for $\zeta_n,\zeta_n'\in e'H$ , is an element of $B(H)_*$ evaluated at $x$ . The subspace $[M'e'H]$ is invariant under the self-adjoint set $M'$ , so $c\in M''=M$ . If $xe'=0$ with $x\in M$ , then $xa'e'=a'xe'=0$ for $a'\in M'$ , so $xc=0$ ; hence $c=1$ forces $x=0$ . If $c\ne1$ , then $1-c$ is a nonzero element of $M$ with $(1-c)e'=0$ , because $e'H\subseteq[M'e'H]$ .

(3) The range of $e\otimes f$ is the closed span of the vectors $e\xi\otimes f\eta$ , a copy of $eH\otimes fK$ . On it, $(e\otimes f)(x\otimes y)(e\otimes f)$ acts as $(exe|_{eH})\otimes(fyf|_{fK})$ . The compression $\Phi(X)=(e\otimes f)X(e\otimes f)|_{eH\otimes fK}$ is normal and maps $M\odot N$ onto $M_e\odot N_f$ . By the normal extension principle (Proposition 4.3(ii)), $\Phi(M\bar\otimes N)\subseteq M_e\bar\otimes N_f$ . Conversely, $e\otimes f\in M\bar\otimes N$ , so $\Phi(M\bar\otimes N)=(M\bar\otimes N)_{e\otimes f}$ is a von Neumann algebra by (1), and it contains $M_e\odot N_f$ ; hence it contains $M_e\bar\otimes N_f$ . For the second identity, $e\otimes f$ commutes with $M'\odot N'$ , hence lies in $(M'\bar\otimes N')'$ . The induction by $e\otimes f$ is a normal $*$ -homomorphism on $M'\bar\otimes N'$ that maps $x'\otimes y'$ to $x'_e\otimes y'_f$ . Its range is a von Neumann algebra by (2), and the same two inclusions follow. $\square$

7. Tensor products with $B(K)$ : commutants and matrix units

Proposition 7.1. For a von Neumann algebra $M\subseteq B(H)$ and any Hilbert space $K$ , $(M\otimes1)'=M'\bar\otimes B(K),\qquad (M\bar\otimes B(K))'=M'\otimes1 . \tag{7.1}$

Proof. By the commutation criterion (2.1), $(M\otimes1)'=\{X:X_{jk}\in M'\}$ . This is $M'\bar\otimes B(K)$ by (5.2) applied to $M'$ . Also by (5.2), $M\bar\otimes B(K)=(M'\otimes1)'$ . So its commutant is $(M'\otimes1)''$ , which is $M'''\otimes1=M'\otimes1$ by Theorem 5.2(4) with $S=M'$ . $\square$

These are the cases $N=\mathbb C1$ and $N=B(K)$ of the commutation theorem (Theorem 11.4). They need only the matrix calculus.

Definition 7.3. A matrix unit in a von Neumann algebra $M$ is a family $(w_{ij})_{i,j\in I}$ in $M$ with $w_{ij}^*=w_{ji}$ , $w_{ij}w_{kl}=\delta_{jk}w_{il}$ and $\sum_iw_{ii}=1$ , the sum taken strongly.

The $w_{ii}$ are mutually orthogonal projections, so the sum makes sense.

Proposition 7.4. Let $(w_{ij})$ be a matrix unit in $M\subseteq B(H)$ with $I\ne\varnothing$ . Fix $i_0\in I$ and put $e=w_{i_0i_0}$ . Let $(\varepsilon_i)$ be the standard basis of $\ell^2(I)$ . The formula $U(\xi\otimes\varepsilon_i)=w_{ii_0}\xi$ , for $\xi\in eH$ , defines a unitary $U:eH\otimes\ell^2(I)\to H$ , and $U^*MU=M_e\bar\otimes B(\ell^2(I)). \tag{7.2}$

Proof. Put $w_i=w_{ii_0}$ . Then $w_i^*w_j=w_{i_0i}w_{ji_0}=\delta_{ij}e$ and $w_iw_i^*=w_{ii}$ . So $w_i$ is a partial isometry with initial projection $e$ and final projection $w_{ii}$ , and $\langle w_i\xi,w_j\xi'\rangle=\delta_{ij}\langle\xi,\xi'\rangle$ for $\xi,\xi'\in eH$ . Hence $U$ maps each column $eH\otimes\varepsilon_i$ isometrically onto $w_{ii}H$ , and these ranges are mutually orthogonal with $\sum_iw_{ii}=1$ . With the column expansion (1.2), $U$ is unitary. Let $R_i$ be the columns of $eH\otimes\ell^2(I)$ , so that $UR_i=w_i|_{eH}$ . For $x\in M$ , the entries of $U^*xU$ are $R_j^*U^*xUR_k=(w_j^*xw_k)|_{eH}=(ew_{i_0j}xw_{ki_0}e)|_{eH}\in M_e$ . Conversely, let $X$ have entries $X_{jk}=(ey_{jk}e)|_{eH}$ with $y_{jk}\in M$ . For finite $F\subseteq I$ , $U\big(\sum_{j,k\in F}R_jX_{jk}R_k^*\big)U^*=\sum_{j,k\in F}w_{ji_0}ey_{jk}ew_{i_0k}\in M$ . These operators are bounded by $\|X\|$ and converge strongly to $UXU^*$ , by the truncation statement, Proposition 2.1(5). $M$ is strongly closed, so $UXU^*\in M$ . Hence $U^*MU=\{X:X_{jk}\in M_e\}$ , which is $M_e\bar\otimes B(\ell^2(I))$ by (5.2); $M_e$ is a von Neumann algebra by Proposition 6.1(1). $\square$

8. Maps between tensor products and normal homomorphisms

Let $M,N,P$ be von Neumann algebras on $H,K,L$ . For $X\in B(H\otimes K)$ and $Y\in B(K\otimes L)$ put $X_{12}=X\otimes1_L$ and $Y_{23}=1_H\otimes Y$ on $H\otimes K\otimes L$ . For $Z\in B(H\otimes L)$ let $F:H\otimes L\otimes K\to H\otimes K\otimes L$ be the unitary $\xi\otimes\theta\otimes\eta\mapsto\xi\otimes\eta\otimes\theta$ (Proposition 1.3(4)), and put $Z_{13}=F(Z\otimes1_K)F^*$ . Then $(x\otimes z)_{13}=x\otimes1\otimes z$ .

Proposition 8.1.

(Associativity.) Under the identification of Proposition 1.3(4), $(M\bar\otimes N)\bar\otimes P=M\bar\otimes(N\bar\otimes P)=(M\otimes1\otimes1\ \cup\ 1\otimes N\otimes1\ \cup\ 1\otimes1\otimes P)'' , \tag{8.1}$ the von Neumann algebra generated by the $x\otimes y\otimes z$ . We write $M\bar\otimes N\bar\otimes P$ .
(Flip.) For the flip unitary $\Sigma:H\otimes K\to K\otimes H$ , $\Sigma(x\otimes y)\Sigma^*=y\otimes x$ and $\Sigma(M\bar\otimes N)\Sigma^*=N\bar\otimes M$ .
(Legs.) The maps $X\mapsto X_{12}$ , $Y\mapsto Y_{23}$ and $Z\mapsto Z_{13}$ are normal unital $*$ -homomorphisms, injective when the added space is nonzero. They map $M\bar\otimes N$ , $N\bar\otimes P$ and $M\bar\otimes P$ into $M\bar\otimes N\bar\otimes P$ .
(Spatial isomorphisms.) For unitaries $U_1:H\to H_1$ and $U_2:K\to K_1$ , $(U_1\otimes U_2)(M\bar\otimes N)(U_1\otimes U_2)^*=(U_1MU_1^*)\bar\otimes(U_2NU_2^*)$ .
(Tensoring an implemented map.) Let $R$ be a Hilbert space, $V:H_0\to H\otimes R$ an isometry whose range projection $e'=VV^*$ commutes with $M\otimes1_R$ , and $\pi(x)=V^*(x\otimes1_R)V$ for $x\in M$ . Then $\pi:M\to B(H_0)$ is a normal unital $*$ -homomorphism. For every von Neumann algebra $P$ on $L$ , the map $(\pi\otimes\iota)(X)=(V\otimes1_L)^*\,X_{13}\,(V\otimes1_L)\qquad(X\in M\bar\otimes P), \tag{8.2}$ with $X_{13}$ acting on $H\otimes R\otimes L$ , is a normal unital $*$ -homomorphism with $(\pi\otimes\iota)(x\otimes z)=\pi(x)\otimes z$ , and it is the only normal map on $M\bar\otimes P$ with these values. Likewise $(\iota\otimes\pi)(X)=(1_L\otimes V)^*(X\otimes1_R)(1_L\otimes V)$ on $P\bar\otimes M$ satisfies $(\iota\otimes\pi)(z\otimes x)=z\otimes\pi(x)$ . If $\pi(M)\subseteq Q$ for a von Neumann algebra $Q$ on $H_0$ , then $(\pi\otimes\iota)(M\bar\otimes P)\subseteq Q\bar\otimes P$ and $(\iota\otimes\pi)(P\bar\otimes M)\subseteq P\bar\otimes Q$ .

Proof. (1) The set $S=M\otimes1\cup1\otimes N$ is self-adjoint with $S''=M\bar\otimes N$ by Theorem 5.2(2). Applying Theorem 5.2(2) to the pair $M\bar\otimes N$ , $P$ with the generating sets $S$ and $P$ gives $(M\bar\otimes N)\bar\otimes P=(S\otimes1\cup1\otimes P)''$ , which is the right side of (8.1). The same argument with $M$ and $T=N\otimes1\cup1\otimes P$ treats $M\bar\otimes(N\bar\otimes P)$ . Products of the three kinds of generators give the $x\otimes y\otimes z$ .

(2) Conjugation by a unitary is a $*$ -isomorphism that carries commutants to commutants, and it maps $M\odot N$ onto $N\odot M$ .

(3) Each map is an amplification (Proposition 3.1(2)) followed by conjugation by a unitary, so it is a normal unital $*$ -homomorphism. It is injective when the added space is nonzero, by the norm identity of Proposition 2.1(1). The images of elementary tensors, such as $(x\otimes z)_{13}=x\otimes1\otimes z$ , lie in $M\bar\otimes N\bar\otimes P$ ; the normal extension principle (Proposition 4.3(ii)) extends this to the whole algebras.

(4) Conjugation by $U_1\otimes U_2$ maps $x\otimes y$ to $U_1xU_1^*\otimes U_2yU_2^*$ and commutants to commutants.

(5) $\pi$ is linear, preserves adjoints and is normal. As $e'$ commutes with $y\otimes1$ and $V^*V=1$ , $\pi(x)\pi(y)=V^*(x\otimes1)e'(y\otimes1)V=V^*(xy\otimes1)e'V=\pi(xy)$ , and $\pi(1)=1$ . For (8.2), $X\mapsto X_{13}$ is a normal $*$ -homomorphism by (3). For $X=x\otimes z$ , $X_{13}=(x\otimes1_R)\otimes z$ commutes with $e'\otimes1_L$ . By Proposition 4.3(ii), applied with the von Neumann algebra $\{e'\otimes1\}'$ , $X_{13}$ commutes with $e'\otimes1$ for every $X\in M\bar\otimes P$ . The computation for $\pi$ then repeats with $V\otimes1$ in place of $V$ , so $\pi\otimes\iota$ is multiplicative; it is normal as a composite of normal maps. On elementary tensors, $(V\otimes1)^*((x\otimes1_R)\otimes z)(V\otimes1)=V^*(x\otimes1_R)V\otimes z$ . Uniqueness is Proposition 4.3(i) with $A=M\odot P$ . The map $\iota\otimes\pi$ is treated in the same way, without a flip. The range statements follow from Proposition 4.3(ii), since $\pi(x)\otimes z\in Q\odot P$ . $\square$

By uniqueness, $\pi\otimes\iota$ on $M\bar\otimes P$ depends only on $\pi$ , not on the choice of $R$ and $V$ . The next theorem shows that every normal unital $*$ -homomorphism has the form in (5).

Theorem 8.2 (normal homomorphisms). Let $M\subseteq B(H)$ be a von Neumann algebra and $\pi:M\to B(L)$ a normal unital $*$ -homomorphism. There are a Hilbert space $R$ and an isometry $V:L\to H\otimes R$ such that $e'=VV^*$ lies in $(M\otimes1_R)'=M'\bar\otimes B(R)$ and $\pi(x)=V^*(x\otimes1_R)V\qquad(x\in M). \tag{8.3}$ So $\pi$ is the amplification $x\mapsto x\otimes1_R$ , followed by the induction by $e'$ , followed by the unitary $V^*:e'(H\otimes R)\to L$ . Consequently:

$\pi(M)$ is a von Neumann algebra on $L$ ;
if $L\ne0$ , $\pi$ is injective if and only if the projection onto $[(M\otimes1_R)'e'(H\otimes R)]$ is $1$ .

Reference: Compare [Takesaki I, Theorem IV.5.5].

Proof. If $L=0$ , take $R=0$ . Otherwise, by Zorn's lemma, choose nonzero vectors $\lambda_i\in L$ , $i\in I$ , whose cyclic subspaces $L_i=[\pi(M)\lambda_i]$ are pairwise orthogonal, with the index set $I$ maximal for this property. Each $L_i$ is invariant under the self-adjoint set $\pi(M)$ . If $\lambda$ is orthogonal to every $L_i$ , then so is $[\pi(M)\lambda]$ , since $\langle\pi(x)\lambda,\pi(y)\lambda_i\rangle=\langle\lambda,\pi(x^*y)\lambda_i\rangle=0$ ; maximality forces $\lambda=0$ . So $L=\bigoplus_iL_i$ . Let $R=\ell^2(I\times\mathbb N)$ , the orthogonal sum of the subspaces $R_i=\ell^2(\{i\}\times\mathbb N)$ . The functional $\psi_i(x)=\langle\pi(x)\lambda_i,\lambda_i\rangle$ is positive and normal on $M$ . By the positive case of Proposition 3.1(4), applied with the infinite-dimensional space $R_i$ , there is $\zeta_i\in H\otimes R_i\subseteq H\otimes R$ with $\psi_i(x)=\langle(x\otimes1)\zeta_i,\zeta_i\rangle$ . Then $\|(x\otimes1)\zeta_i\|^2=\psi_i(x^*x)=\|\pi(x)\lambda_i\|^2$ , so $(x\otimes1)\zeta_i\mapsto\pi(x)\lambda_i$ extends to a unitary $U_i$ of $[(M\otimes1)\zeta_i]$ onto $L_i$ . The spaces $[(M\otimes1)\zeta_i]\subseteq H\otimes R_i$ are mutually orthogonal. Let $e'$ be the projection onto their sum; it commutes with $M\otimes1$ because the sum is invariant under this self-adjoint set. Let $V:L\to H\otimes R$ be the isometry equal to $U_i^{-1}$ on each $L_i$ ; its range projection is $e'$ . For $x,y\in M$ , $V\pi(x)\pi(y)\lambda_i=V\pi(xy)\lambda_i=(xy\otimes1)\zeta_i=(x\otimes1)V\pi(y)\lambda_i$ . So $V\pi(x)=(x\otimes1)V$ on each $L_i$ , hence on $L$ , and $\pi(x)=V^*(x\otimes1)V$ . The identity $(M\otimes1_R)'=M'\bar\otimes B(R)$ is (7.1).

(1) By Theorem 5.2(4), $M\otimes1_R$ is a von Neumann algebra. By Proposition 6.1(2), so is its induced algebra by $e'$ . And $\pi(M)$ is the image of that induced algebra under the unitary $V^*$ .

(2) If $L\ne0$ , then $R\ne0$ and the amplification is injective. $V^*$ is unitary on $e'(H\otimes R)$ . Apply Proposition 6.1(2) to $M\otimes1_R$ . $\square$

Corollary 8.3 (normal isomorphisms). Let $\pi:M_1\to M_2$ be a normal $*$ -isomorphism of von Neumann algebras $M_k\subseteq B(H_k)$ , $H_1\ne0$ . With $R$ , $V$ , $e'$ as in Theorem 8.2, let $M=M_1\otimes1_R$ on $H_1\otimes R$ , $e_2'=e'$ , and $e_1'=1\otimes q$ for a rank-one projection $q$ on $R$ . Then $e_1',e_2'\in M'$ , the projections onto $[M'e_k'(H_1\otimes R)]$ are both $1$ , $M_1$ is spatially isomorphic to the induced algebra $M_{e_1'}$ , $M_2$ to $M_{e_2'}$ , and $\pi$ corresponds to $(x\otimes1)_{e_1'}\mapsto(x\otimes1)_{e_2'}$ .

The hypothesis that $\pi$ is normal is automatic, because every $*$ -isomorphism between von Neumann algebras is normal (fact (d) of the background section).

Proof. By (7.1), $M'=M_1'\bar\otimes B(R)$ . It contains $1\otimes B(R)$ , so $[M'(1\otimes q)(H_1\otimes R)]\supseteq H_1\otimes R$ . For $e_2'$ , use part (2) of Theorem 8.2; $H_2\ne0$ because $M_2\cong M_1\ne0$ . For a unit vector $g\in qR$ , the unitary $\xi\mapsto\xi\otimes g$ of $H_1$ onto $H_1\otimes qR$ carries $x$ to $(x\otimes1)_{e_1'}$ , and $V^*$ carries $(x\otimes1)_{e_2'}$ to $\pi(x)$ . $\square$

Corollary 8.4 (tensor products of normal homomorphisms). Let $\pi_k:M_k\to N_k$ , $k=1,2$ , be normal unital $*$ -homomorphisms between von Neumann algebras. There is a unique normal unital $*$ -homomorphism $\pi_1\otimes\pi_2:M_1\bar\otimes M_2\to N_1\bar\otimes N_2$ with $(\pi_1\otimes\pi_2)(x\otimes y)=\pi_1(x)\otimes\pi_2(y)$ . If $\pi_1$ and $\pi_2$ are $*$ -isomorphisms onto $N_1$ and $N_2$ with normal inverses, then $\pi_1\otimes\pi_2$ is a $*$ -isomorphism onto $N_1\bar\otimes N_2$ .

Proof. By (8.3), $\pi_1$ has the form of Proposition 8.1(5), and $\pi_1(M_1)\subseteq N_1$ . So, by the range statement there, $\pi_1\otimes\iota:M_1\bar\otimes M_2\to N_1\bar\otimes M_2$ exists, and likewise $\iota\otimes\pi_2:N_1\bar\otimes M_2\to N_1\bar\otimes N_2$ . Put $\pi_1\otimes\pi_2=(\iota\otimes\pi_2)\circ(\pi_1\otimes\iota)$ . Uniqueness is Proposition 4.3(i). In the isomorphism case, $(\pi_1^{-1}\otimes\pi_2^{-1})\circ(\pi_1\otimes\pi_2)$ is normal and is the identity on $M_1\odot M_2$ , hence on $M_1\bar\otimes M_2$ ; likewise in the other order. $\square$

By fact (d) of the background section, the inverses in the last sentence of Corollary 8.4 are automatically normal. The lesson does not use this fact.

9. Slice maps

A slice map integrates out one leg of a tensor product against a functional on that leg. Let $M\subseteq B(H)$ and $N\subseteq B(K)$ be von Neumann algebras, and let $R_\eta$ , $S_\xi$ be the vector maps of Proposition 1.3(2).

Lemma 9.1. For $X\in M\bar\otimes N$ and all vectors, $S_{\xi'}^*XS_\xi\in N$ and $R_{\eta'}^*XR_\eta\in M$ .

Proof. $1\otimes N'$ commutes with $M\odot N$ , so $X$ commutes with $1\otimes y'$ for $y'\in N'$ . Also $(1\otimes y')S_\xi=S_\xi y'$ and $S_{\xi'}^*(1\otimes y')=y'S_{\xi'}^*$ . Hence $S_{\xi'}^*XS_\xi y'=S_{\xi'}^*X(1\otimes y')S_\xi=y'S_{\xi'}^*XS_\xi$ , and $S_{\xi'}^*XS_\xi\in N''=N$ . The other case uses $M'\otimes1$ in the same way. $\square$

Theorem 9.2 (slice maps). Let $X\in M\bar\otimes N$ .

For every bounded linear functional $\omega$ on $N$ , normal or not, there is a unique operator $(\iota\otimes\omega)(X)\in B(H)$ with $\langle(\iota\otimes\omega)(X)\xi,\xi'\rangle=\omega(S_{\xi'}^*XS_\xi)\qquad(\xi,\xi'\in H). \tag{9.1}$ It lies in $M$ , $\|(\iota\otimes\omega)(X)\|\le\|\omega\|\|X\|$ , the map $X\mapsto(\iota\otimes\omega)(X)$ is linear, and $(\iota\otimes\omega)(x\otimes y)=\omega(y)x$ . If $H\ne0$ , the slice map $\iota\otimes\omega:M\bar\otimes N\to M$ has norm $\|\omega\|$ .
(Module rules, adjoints, positivity.) For $a,b\in M$ and $c,d\in N$ , $(\iota\otimes\omega)((a\otimes1)X(b\otimes1))=a\,(\iota\otimes\omega)(X)\,b$ and $(\iota\otimes\omega)((1\otimes c)X(1\otimes d))=(\iota\otimes\omega_{c,d}^{\flat})(X)$ , where $\omega_{c,d}^{\flat}(y)=\omega(cyd)$ . Also $(\iota\otimes\omega)(X^*)=(\iota\otimes\omega^\natural)(X)^*$ , where $\omega^\natural(y)=\overline{\omega(y^*)}$ . If $\omega$ is positive, so is $\iota\otimes\omega$ .
(Normality.) If $\omega$ is normal, so is $\iota\otimes\omega$ . More precisely, if $\omega=\sum_n\omega_{\eta_n,\eta_n'}|_N$ as in Proposition 3.1(1), then $(\iota\otimes\omega)(X)=\sum_nR_{\eta_n'}^*XR_{\eta_n}, \tag{9.2}$ a norm-convergent series whose right side is a normal map on all of $B(H\otimes K)$ . In particular $(\iota\otimes\omega_{\eta,\eta'})(X)=R_{\eta'}^*XR_\eta$ . Conversely, if $H\ne0$ and $\iota\otimes\omega$ is normal, then $\omega$ is normal.
(The other leg.) For a bounded functional $\varphi$ on $M$ , $\langle(\varphi\otimes\iota)(X)\eta,\eta'\rangle=\varphi(R_{\eta'}^*XR_\eta)$ defines $(\varphi\otimes\iota)(X)\in N$ , with the mirror images of (1)–(3). In particular $(\omega_{\xi,\xi'}\otimes\iota)(X)=S_{\xi'}^*XS_\xi$ , and $(\varphi\otimes\iota)((1\otimes c)X(1\otimes d))=c\,(\varphi\otimes\iota)(X)\,d$ for $c,d\in N$ .
(Fubini identity.) For $\varphi\in M_*$ and $\omega\in N_*$ , $\varphi\big((\iota\otimes\omega)(X)\big)=\omega\big((\varphi\otimes\iota)(X)\big). \tag{9.3}$ If $\varphi=\sum_m\omega_{\xi_m,\xi_m'}|_M$ and $\omega=\sum_n\omega_{\eta_n,\eta_n'}|_N$ , both sides equal the absolutely convergent double series $\sum_{m,n}\langle X(\xi_m\otimes\eta_n),\xi_m'\otimes\eta_n'\rangle$ .

Proof. (1) The right side of (9.1) is defined by Lemma 9.1. It is linear in $\xi$ and conjugate-linear in $\xi'$ , because $S_\xi$ is linear in $\xi$ and $S_{\xi'}^*$ is conjugate-linear in $\xi'$ . It is bounded by $\|\omega\|\|X\|\|\xi\|\|\xi'\|$ . A bounded sesquilinear form is given by a unique operator, of norm at most the bound. For $a'\in M'$ , $S_{a'\xi}=(a'\otimes1)S_\xi$ and $S_{\xi'}^*(a'\otimes1)=S_{a'^*\xi'}^*$ , and $X$ commutes with $a'\otimes1$ . So $\langle(\iota\otimes\omega)(X)a'\xi,\xi'\rangle=\omega(S_{a'^*\xi'}^*XS_\xi)=\langle(\iota\otimes\omega)(X)\xi,a'^*\xi'\rangle=\langle a'(\iota\otimes\omega)(X)\xi,\xi'\rangle$ , and $(\iota\otimes\omega)(X)\in M''=M$ . Since $S_{\xi'}^*(x\otimes y)S_\xi=\langle x\xi,\xi'\rangle y$ , we get $(\iota\otimes\omega)(x\otimes y)=\omega(y)x$ . If $H\ne0$ , then $\|(\iota\otimes\omega)(1\otimes y)\|=|\omega(y)|$ and $\|1\otimes y\|=\|y\|$ , which gives the norm.

(2) The identities $S_{\xi'}^*(a\otimes1)X(b\otimes1)S_\xi=S_{a^*\xi'}^*XS_{b\xi}$ and $S_{\xi'}^*(1\otimes c)X(1\otimes d)S_\xi=c\,S_{\xi'}^*XS_\xi\,d$ give the two module rules. For adjoints, $\omega(S_{\xi'}^*X^*S_\xi)=\omega\big((S_\xi^*XS_{\xi'})^*\big)=\overline{\omega^\natural(S_\xi^*XS_{\xi'})}=\overline{\langle(\iota\otimes\omega^\natural)(X)\xi',\xi\rangle}=\langle(\iota\otimes\omega^\natural)(X)^*\xi,\xi'\rangle$ . If $\omega\ge0$ and $X\ge0$ , then $\langle(\iota\otimes\omega)(X)\xi,\xi\rangle=\omega(S_\xi^*XS_\xi)\ge0$ .

(3) $\langle R_{\eta_n'}^*XR_{\eta_n}\xi,\xi'\rangle=\langle X(\xi\otimes\eta_n),\xi'\otimes\eta_n'\rangle=\langle S_{\xi'}^*XS_\xi\eta_n,\eta_n'\rangle$ . Summing over $n$ gives $\omega(S_{\xi'}^*XS_\xi)$ , because $S_{\xi'}^*XS_\xi\in N$ . The series converges in norm, since $\|R_{\eta_n'}^*XR_{\eta_n}\|\le\|X\|\|\eta_n\|\|\eta_n'\|$ . For $\rho=\sum_k\omega_{\zeta_k,\zeta_k'}\in B(H)_*$ , $\rho\big(\sum_nR_{\eta_n'}^*XR_{\eta_n}\big)=\sum_{k,n}\langle X(\zeta_k\otimes\eta_n),\zeta_k'\otimes\eta_n'\rangle$ , and $\sum_{k,n}\|\zeta_k\otimes\eta_n\|^2=\sum_k\|\zeta_k\|^2\sum_n\|\eta_n\|^2<\infty$ , likewise for the primed vectors. So the map is normal. Conversely, if $\iota\otimes\omega$ is normal, so is $y\mapsto(\iota\otimes\omega)(1\otimes y)=\omega(y)1$ , because the amplification $y\mapsto1\otimes y$ is normal (Proposition 3.1(2)); evaluating at a unit vector shows that $\omega$ is normal.

(4) Exchange the legs, or conjugate by the flip (Proposition 8.1(2)).

(5) By (9.1) and (9.2), $\varphi((\iota\otimes\omega)(X))=\sum_m\omega(S_{\xi_m'}^*XS_{\xi_m})=\sum_m\sum_n\langle X(\xi_m\otimes\eta_n),\xi_m'\otimes\eta_n'\rangle$ . The terms are bounded by $\|X\|\|\xi_m\|\|\xi_m'\|\|\eta_n\|\|\eta_n'\|$ , which is summable over $(m,n)$ by Cauchy–Schwarz. So the order of summation does not matter, and the same computation for the right side of (9.3) gives the same double series. $\square$

Exercise 9.3 (slices by a functional that is not normal). Let $N=\ell^\infty(\mathbb N)$ on $\ell^2(\mathbb N)$ , and let $\omega$ be a state of $N$ that vanishes on every finitely supported sequence, for instance the limit along a free ultrafilter. Let $M\subseteq B(H)$ with $H\ne0$ . Show that $\iota\otimes\omega:M\bar\otimes N\to M$ is positive, unital and of norm one, but not normal.

Solution. By Theorem 9.2(1) and (2), $\iota\otimes\omega$ is positive, has norm $\|\omega\|=1$ , and $(\iota\otimes\omega)(1)=\omega(1)1=1$ . The projections $p_n=1_{\{1,\ldots,n\}}$ increase strongly to $1$ , $\omega(p_n)=0$ and $\omega(1)=1$ . A normal functional is continuous for the strong topology on bounded sets: write it as in (3.1) and use dominated convergence, as in the proof of Proposition 3.1(2). So $\omega$ is not normal. By the converse in Theorem 9.2(3), $\iota\otimes\omega$ is not normal.

10. Product functionals and the predual

The support of a positive normal functional $\psi\ne0$ on a von Neumann algebra $Q$ is the unique projection $e\in Q$ such that $\psi(x)=\psi(exe)$ for all $x\in Q$ and $\psi$ is faithful on $eQe$ (fact (c) of the background section); put $s(0)=0$ . Applied to $x(1-e)$ and $(1-e)x$ , the first condition gives $\psi(x)=\psi(xe)=\psi(ex)$ . A positive normal $\psi$ is faithful exactly when $s(\psi)=1$ .

Theorem 10.1. Let $M\subseteq B(H)$ and $N\subseteq B(K)$ be von Neumann algebras, $\varphi\in M_*$ and $\omega\in N_*$ .

(Product functionals.) The functional $\varphi\otimes\omega:=\varphi\circ(\iota\otimes\omega)=\omega\circ(\varphi\otimes\iota) \tag{10.1}$ is normal on $M\bar\otimes N$ , and it is the only normal functional with $(\varphi\otimes\omega)(x\otimes y)=\varphi(x)\omega(y)$ . Its norm is $\|\varphi\|\|\omega\|$ . When $\varphi$ and $\omega$ are positive, $\varphi\otimes\omega$ is positive; when both are states, it is a state. For vector series $\varphi=\sum_m\omega_{\xi_m,\xi_m'}|_M$ and $\omega=\sum_n\omega_{\eta_n,\eta_n'}|_N$ , $\varphi\otimes\omega=\sum_{m,n}\omega_{\xi_m\otimes\eta_n,\ \xi_m'\otimes\eta_n'}|_{M\bar\otimes N}$ .
(Density.) The span of the functionals $\omega_{\xi,\xi'}\otimes\omega_{\eta,\eta'}=\omega_{\xi\otimes\eta,\xi'\otimes\eta'}|_{M\bar\otimes N}$ is norm-dense in $(M\bar\otimes N)_*$ . So is the span of all $\varphi\otimes\omega$ .
(Pairing with the algebraic tensor product.) For $\theta\in(M\bar\otimes N)_*$ , $\|\theta\|=\sup\{|\theta(X)|:X\in M\odot N,\ \|X\|\le1\}$ . So restriction to $M\odot N$ embeds $(M\bar\otimes N)_*$ isometrically in the dual of $M\odot N$ with the operator norm, and by (2) the image of $M_*\odot N_*$ is dense in the image. Since $(M\bar\otimes N)_*$ is norm-closed in the dual of $M\bar\otimes N$ (fact (b) of the background section), the image is the closure of $M_*\odot N_*$ . This part uses Kaplansky's density theorem (fact (a)).
(Faithfulness and supports.) If $\varphi$ and $\omega$ are faithful and positive, $\varphi\otimes\omega$ is faithful. For positive $\varphi,\omega$ , $s(\varphi\otimes\omega)=s(\varphi)\otimes s(\omega)$ . Hence, if $H\ne0\ne K$ , $\varphi\otimes\omega$ is faithful exactly when $\varphi$ and $\omega$ are.

Proof. (1) Both composites are normal by Theorem 9.2(3) and (4), and they are equal by the Fubini identity (9.3). On $x\otimes y$ the value is $\varphi(\omega(y)x)=\varphi(x)\omega(y)$ . Uniqueness is Proposition 4.3(i) with $A=M\odot N$ . By Theorem 9.2(1), $|\varphi((\iota\otimes\omega)(X))|\le\|\varphi\|\|\omega\|\|X\|$ . Conversely $|(\varphi\otimes\omega)(x\otimes y)|=|\varphi(x)||\omega(y)|$ and $\|x\otimes y\|=\|x\|\|y\|$ ; suprema over the unit balls give $\|\varphi\otimes\omega\|\ge\|\varphi\|\|\omega\|$ . Positivity follows from Theorem 9.2(2), and $(\varphi\otimes\omega)(1)=\varphi(1)\omega(1)$ . The vector formula is Theorem 9.2(5).

(2) Let $\theta\in(M\bar\otimes N)_*$ . By Proposition 3.1(1), $\theta$ is the restriction of some $\rho=\sum_k\omega_{\zeta_k,\zeta_k'}$ with $\zeta_k,\zeta_k'\in H\otimes K$ . The tail $\sum_{k>n}\omega_{\zeta_k,\zeta_k'}$ has norm at most $(\sum_{k>n}\|\zeta_k\|^2)^{1/2}(\sum_{k>n}\|\zeta_k'\|^2)^{1/2}$ , which tends to $0$ . For $k\le n$ , approximate $\zeta_k$ and $\zeta_k'$ by vectors $\alpha,\alpha'\in H\odot K$ , using $\|\omega_{\zeta,\zeta'}-\omega_{\alpha,\alpha'}\|\le\|\zeta-\alpha\|\|\zeta'\|+\|\alpha\|\|\zeta'-\alpha'\|$ . By sesquilinearity, $\omega_{\alpha,\alpha'}$ is a finite sum of functionals $\omega_{\xi\otimes\eta,\xi'\otimes\eta'}$ , and each of these equals $\omega_{\xi,\xi'}\otimes\omega_{\eta,\eta'}$ on $M\bar\otimes N$ by (1).

(3) The inequality $\ge$ is clear. $M\bar\otimes N$ is the weak closure of $M\odot N$ (Theorem 5.2(1)). By Kaplansky's density theorem, every $X\in M\bar\otimes N$ with $\|X\|\le1$ is the strong limit of a net $X_\alpha\in M\odot N$ with $\|X_\alpha\|\le1$ . With $\theta$ written as in (2), $|\theta(X_\alpha-X)|\le(\sum_k\|(X_\alpha-X)\zeta_k\|^2)^{1/2}(\sum_k\|\zeta_k'\|^2)^{1/2}$ . Each term tends to $0$ and is at most $4\|\zeta_k\|^2$ , so the sum tends to $0$ , as in the proof of Proposition 3.1(2). Hence $|\theta(X)|$ is at most the supremum over the unit ball of $M\odot N$ . The remaining sentences follow from this, from (2), and from the closedness of the predual.

(4) Faithfulness. Let $X\in M\bar\otimes N$ , $X\ge0$ , with $(\varphi\otimes\omega)(X)=0$ . Then $(\varphi\otimes\iota)(X)\ge0$ and $\omega((\varphi\otimes\iota)(X))=0$ ; as $\omega$ is faithful, $(\varphi\otimes\iota)(X)=0$ . For $\eta\in K$ , the Fubini identity (9.3) with the normal functional $\omega_\eta$ gives $\varphi((\iota\otimes\omega_\eta)(X))=\langle(\varphi\otimes\iota)(X)\eta,\eta\rangle=0$ , and $(\iota\otimes\omega_\eta)(X)\ge0$ ; so $(\iota\otimes\omega_\eta)(X)=0$ . Hence $\langle X(\xi\otimes\eta),\xi\otimes\eta\rangle=\langle(\iota\otimes\omega_\eta)(X)\xi,\xi\rangle=0$ , that is $X^{1/2}(\xi\otimes\eta)=0$ , for all $\xi,\eta$ . Since product vectors are total (Proposition 1.3(1)), $X=0$ .

Supports. Let $e=s(\varphi)$ , $f=s(\omega)$ and $p=e\otimes f\in M\bar\otimes N$ . If $\varphi=0$ or $\omega=0$ , both sides are $0$ . Otherwise:

$X\mapsto(\varphi\otimes\omega)(pXp)$ and $\varphi\otimes\omega$ are normal and agree on each $x\otimes y$ , since $\varphi(exe)\omega(fyf)=\varphi(x)\omega(y)$ . By Proposition 4.3(i) they agree on $M\bar\otimes N$ .
$\varphi\otimes\omega$ is faithful on $p(M\bar\otimes N)p$ . Let $Y\ge0$ there with $(\varphi\otimes\omega)(Y)=0$ . Since $Y=(1\otimes f)Y(1\otimes f)$ , Theorem 9.2(4) gives $(\varphi\otimes\iota)(Y)=f(\varphi\otimes\iota)(Y)f$ . This is a positive element of $fNf$ on which $\omega$ vanishes, so it is $0$ . For $\eta\in K$ , $(\iota\otimes\omega_\eta)(Y)=e(\iota\otimes\omega_\eta)(Y)e\ge0$ by Theorem 9.2(2), and $\varphi$ vanishes on it by (9.3); so it is $0$ . Then $Y=0$ as in the first paragraph.
$p\ne0$ and $\varphi\otimes\omega\ne0$ .

By the uniqueness of the support, $s(\varphi\otimes\omega)=p$ . Finally, let $H\ne0\ne K$ . If $e\ne1$ , pick $\xi\ne0$ with $e\xi=0$ and any $\eta\ne0$ ; then $(e\otimes f)(\xi\otimes\eta)=0\ne\xi\otimes\eta$ . So $e\otimes f=1$ forces $e=1$ , and likewise $f=1$ . $\square$

The proof of (4) does not use the commutation theorem of Section 11.

Example 10.2 (matrices: partial traces and product traces). Let $H=\mathbb C^m$ , $K=\mathbb C^n$ , and let $\operatorname{Tr}_n$ be the trace on $B(\mathbb C^n)$ ; it is normal, with norm $n$ . For $X\in B(\mathbb C^m\otimes\mathbb C^n)$ , (9.2) with $\operatorname{Tr}_n=\sum_j\omega_{f_j,f_j}$ gives the partial trace $(\iota\otimes\operatorname{Tr}_n)(X)=\sum_jX_{jj}$ . Its norm as a map is $n=\|\operatorname{Tr}_n\|$ , attained at $X=1$ , as Theorem 9.2(1) predicts. The product functional $\operatorname{Tr}_m\otimes\operatorname{Tr}_n$ is $\operatorname{Tr}_{mn}$ , since both are normal and agree on elementary tensors (uniqueness in Theorem 10.1(1)); its norm is $mn=\|\operatorname{Tr}_m\|\|\operatorname{Tr}_n\|$ .

11. The commutation theorem and its consequences

In this section $\langle\xi,\eta\rangle_{\mathbb R}=\operatorname{Re}\langle\xi,\eta\rangle$ . It is a real inner product with the same norm, and it makes $H$ a real Hilbert space. For a real subspace $X\subseteq H$ , $X^\perp$ is its real-orthogonal complement; $X^{\perp\perp}$ is the closure $\overline X$ , and $(iX)^\perp=iX^\perp$ . If $X\perp Y$ and $X+Y$ is dense, then $X^\perp=\overline Y$ : indeed $\overline X+\overline Y$ is closed, as an orthogonal sum of closed subspaces, and dense, so it is $H$ . For a set $S$ of operators, $S_h$ is its self-adjoint part.

Lemma 11.1. If $a,b\in B(H)$ are commuting self-adjoint operators and $\xi\in H$ , then $a\xi\perp ib\xi$ in the real sense. In particular $M_h\xi\perp iM'_h\xi$ for a von Neumann algebra $M$ .

Proof. $\langle a\xi,ib\xi\rangle=-i\langle ba\xi,\xi\rangle$ , and $ba=ab$ is self-adjoint, so $\langle ba\xi,\xi\rangle$ is real. $\square$

Lemma 11.2. Let $M$ be a von Neumann algebra on $H$ with a cyclic vector $\xi_0$ , that is, $[M\xi_0]=H$ . Let $A\subseteq M$ and $B\subseteq M'$ be $*$ -subalgebras.

(a) If $A_h\xi_0+iB_h\xi_0$ is dense in $H$ , then $A''=M$ and $B''=M'$ .
(b) (Rieffel–van Daele.) $M_h\xi_0+iM'_h\xi_0$ is dense in $H$ , and $(M_h\xi_0)^\perp$ is the closure of $iM'_h\xi_0$ .
(c) If $1\in A$ , $1\in B$ , $A''=M$ and $B''=M'$ , then $A_h\xi_0+iB_h\xi_0$ is dense in $H$ .

Proof. (a) Put $X=A_h\xi_0$ and $Y=iB_h\xi_0$ . By Lemma 11.1, $X\perp Y$ , so $X^\perp=\overline Y$ and $Y^\perp=\overline X$ .

Step 1: $A\xi_0$ is dense. By Lemma 11.1, $M_h\xi_0\perp Y$ , so $M_h\xi_0\subseteq\overline X$ . Hence $M\xi_0=M_h\xi_0+iM_h\xi_0$ lies in the closure of $A\xi_0$ .

Step 2. Let $b\in(A')_h$ . By Lemma 11.1, $ib\xi_0\in X^\perp=\overline Y$ , so there are $b_n\in B_h$ with $b_n\xi_0\to b\xi_0$ . For $c\in B'$ and $x,y\in A$ , $\langle cbx\xi_0,y\xi_0\rangle=\lim_n\langle cxb_n\xi_0,y\xi_0\rangle=\lim_n\langle b_ncx\xi_0,y\xi_0\rangle=\lim_n\langle cx\xi_0,yb_n\xi_0\rangle=\langle cx\xi_0,by\xi_0\rangle=\langle bcx\xi_0,y\xi_0\rangle .$ Here $bx=xb$ and $by=yb$ because $b\in A'$ ; $b_n\in M'$ commutes with $x,y\in M$ ; $c\in B'$ commutes with $b_n\in B$ ; and $b_n$ , $b$ are self-adjoint. By Step 1, $cb=bc$ . So $(A')_h\subseteq B''$ , and $A'\subseteq B''$ because $A'$ is spanned by its self-adjoint part.

Step 3. From $A\subseteq M$ and $B\subseteq M'$ we get $M'\subseteq A'$ and $B''\subseteq M'$ . So $A'\subseteq B''\subseteq M'\subseteq A'$ : all are equal, $A''=M''=M$ , and $B''=M'$ .

(b) Let $\eta_0$ be real-orthogonal to $M_h\xi_0+iM'_h\xi_0$ . On $H\otimes\mathbb C^2$ with basis $\varepsilon_1,\varepsilon_2$ , let $\pi(x)=x\otimes1$ . By (7.1) and (5.2), $\pi(M)'=M'\bar\otimes B(\mathbb C^2)$ consists of the $2\times2$ matrices with entries in $M'$ . Let $\zeta_0=\xi_0\otimes\varepsilon_1+\eta_0\otimes\varepsilon_2$ , and let $P\in\pi(M)'$ be the projection onto $[\pi(M)\zeta_0]$ , with entries $p,r,r^*,q\in M'$ ; then $0\le p\le1$ . The first component of $P\zeta_0=\zeta_0$ is $p\xi_0+r\eta_0=\xi_0 . \tag{$\alpha$}$ For $a\in M_h$ , $\langle a\xi_0,\eta_0\rangle$ is purely imaginary, so $\langle a\xi_0,\eta_0\rangle=-\overline{\langle a\xi_0,\eta_0\rangle}=-\langle a\eta_0,\xi_0\rangle$ . Both sides are complex-linear in $a$ , so this holds for all $a\in M$ . It says that $\eta_0\otimes\varepsilon_1+\xi_0\otimes\varepsilon_2$ is orthogonal to every $\pi(a)\zeta_0$ . So $P$ annihilates it, and the first component gives $p\eta_0+r\xi_0=0 . \tag{$\beta$}$ For $b\in M'_h$ , $\operatorname{Re}\langle ib\xi_0,\eta_0\rangle=0$ , so $\langle b\xi_0,\eta_0\rangle$ is real and equals $\langle b\eta_0,\xi_0\rangle$ . By linearity, $\langle b\xi_0,\eta_0\rangle=\langle b\eta_0,\xi_0\rangle\qquad(b\in M'). \tag{$\gamma$}$ By ( $\beta$ ), ( $\gamma$ ) for $b=r$ , and ( $\alpha$ ), $0\le\langle p\eta_0,\eta_0\rangle=-\langle r\xi_0,\eta_0\rangle=-\langle r\eta_0,\xi_0\rangle=-\langle(1-p)\xi_0,\xi_0\rangle\le0 .$ So $p^{1/2}\eta_0=0$ and $(1-p)^{1/2}\xi_0=0$ , hence $p\eta_0=0$ and $(1-p)\xi_0=0$ . A cyclic vector for $M$ is separating for $M'$ : if $b\in M'$ and $b\xi_0=0$ , then $bM\xi_0=Mb\xi_0=0$ , so $b=0$ . So $p=1$ , and $\eta_0=p\eta_0=0$ . This proves the density. The statement about $(M_h\xi_0)^\perp$ follows from Lemma 11.1 and the remark at the start of this section.

(c) By the bicommutant theorem (Theorem 4.2), $A$ is weakly dense in $M$ . If $a_\alpha\in A$ tend weakly to $x\in M_h$ , then $(a_\alpha+a_\alpha^*)/2\in A_h$ tend weakly to $x$ , and $(a_\alpha+a_\alpha^*)\xi_0/2$ tends weakly to $x\xi_0$ in $H$ . A closed real subspace $Z$ of $H$ is weakly closed: if $v\notin Z$ and $u=v-P_Zv$ , with $P_Z$ the real orthogonal projection, then $\langle v,u\rangle_{\mathbb R}=\|u\|^2>0$ while $\langle z,u\rangle_{\mathbb R}=0$ on $Z$ . Hence $M_h\xi_0$ lies in the closure of $A_h\xi_0$ . Likewise $M'_h\xi_0$ lies in the closure of $B_h\xi_0$ . Apply (b). $\square$

Remark. The hypothesis of (a) forces $A$ and $B$ to be nondegenerate. Step 1 gives $[A\xi_0]=H$ . Step 2 with $b=1$ gives $\xi_0\in\overline{B_h\xi_0}\subseteq[BH]$ . The projection onto $[BH]$ lies in $M'$ , because $[BH]$ is invariant under $M$ , and it fixes $\xi_0$ , which is separating for $M'$ ; so it is $1$ . Some condition of this kind is needed in (c): on $H=\mathbb C$ with $M=M'=\mathbb C$ , $A=\{0\}$ and $B=M'$ satisfy $A''=M$ and $B''=M'$ , but $A_h\xi_0+iB_h\xi_0=i\mathbb R\xi_0$ is not dense.

Lemma 11.3. Let $X\subseteq H$ and $Y\subseteq K$ be real subspaces such that $X+iX$ is dense in $H$ and $Y+iY$ is dense in $K$ . Let $X\odot Y$ be the real span of the $\xi\otimes\eta$ with $\xi\in X$ , $\eta\in Y$ . Then $X\odot Y+i(X^\perp\odot Y^\perp)$ is dense in $H\otimes K$ .

Proof. Let $\zeta$ be real-orthogonal to $X\odot Y$ and to $i(X^\perp\odot Y^\perp)$ . For fixed $\xi$ , $\eta\mapsto\langle\zeta,\xi\otimes\eta\rangle_{\mathbb R}$ is a bounded real-linear functional on $K$ . So there is a unique $t\xi\in K$ with $\langle t\xi,\eta\rangle_{\mathbb R}=\langle\zeta,\xi\otimes\eta\rangle_{\mathbb R}$ for all $\eta$ . The map $t:H\to K$ is real-linear and bounded. It is conjugate-linear: $\langle t(i\xi),\eta\rangle_{\mathbb R}=\langle\zeta,\xi\otimes i\eta\rangle_{\mathbb R}=\langle t\xi,i\eta\rangle_{\mathbb R}=\langle-it\xi,\eta\rangle_{\mathbb R}$ . Let $t'$ be its real adjoint, $\langle t'\eta,\xi\rangle_{\mathbb R}=\langle\eta,t\xi\rangle_{\mathbb R}$ ; it is conjugate-linear too. So $T=t't$ is complex-linear, and $\langle T\xi,\xi\rangle=\langle T\xi,\xi\rangle_{\mathbb R}+i\langle T\xi,i\xi\rangle_{\mathbb R}=\|t\xi\|^2+i\langle t\xi,-it\xi\rangle_{\mathbb R}=\|t\xi\|^2$ . So $T\ge0$ . The two orthogonality hypotheses say:

$\langle t\xi,\eta\rangle_{\mathbb R}=0$ for $\xi\in X$ , $\eta\in Y$ ; so $t'Y\subseteq X^\perp$ , and $t'\overline Y\subseteq X^\perp$ ;
$\langle t\xi,i\eta\rangle_{\mathbb R}=\langle\zeta,i(\xi\otimes\eta)\rangle_{\mathbb R}=0$ for $\xi\in X^\perp$ , $\eta\in Y^\perp$ ; so $tX^\perp\subseteq(iY^\perp)^\perp=i\overline Y$ .

Hence $TX^\perp\subseteq t'(i\overline Y)=-it'(\overline Y)\subseteq iX^\perp$ , and $T^2X^\perp\subseteq T(iX^\perp)=iT(X^\perp)\subseteq X^\perp$ . $T$ is a norm limit of real polynomials in $T^2$ (approximate $\sqrt s$ uniformly on $[0,\|T\|^2]$ ), and $X^\perp$ is closed, so $TX^\perp\subseteq X^\perp$ . Thus $TX^\perp\subseteq X^\perp\cap iX^\perp=\{0\}$ : a vector in both is real-orthogonal to $X$ and to $iX$ , hence to the dense set $X+iX$ . So $\|t\xi\|^2=\langle T\xi,\xi\rangle=0$ for $\xi\in X^\perp$ . Then $\langle t'\eta,\xi\rangle_{\mathbb R}=\langle\eta,t\xi\rangle_{\mathbb R}=0$ for $\eta\in K$ and $\xi\in X^\perp$ , so $t'K\subseteq X^{\perp\perp}=\overline X$ . With the first bullet, $t'Y\subseteq\overline X\cap X^\perp=\{0\}$ . Hence $\langle t\xi,\eta\rangle_{\mathbb R}=\langle\xi,t'\eta\rangle_{\mathbb R}=0$ for all $\xi\in H$ and $\eta\in Y$ , that is $tH\subseteq Y^\perp$ . As $t(i\xi)=-it\xi$ and $iH=H$ , also $tH\subseteq iY^\perp$ . By density of $Y+iY$ , $Y^\perp\cap iY^\perp=\{0\}$ , so $t=0$ . Thus $\operatorname{Re}\langle\zeta,\xi\otimes\eta\rangle=0$ for all $\xi,\eta$ ; replacing $\xi$ by $i\xi$ gives $\langle\zeta,\xi\otimes\eta\rangle=0$ , and $\zeta=0$ because product vectors are total (Proposition 1.3(1)). $\square$

Theorem 11.4 (commutation theorem). For von Neumann algebras $M\subseteq B(H)$ and $N\subseteq B(K)$ , $(M\bar\otimes N)'=M'\bar\otimes N' . \tag{11.1}$

Reference: Compare [Takesaki I, Lemmas IV.5.7–IV.5.8 and Theorem IV.5.9].

Proof. Step 0. $M'\odot N'$ commutes with $M\odot N$ . So $M'\bar\otimes N'\subseteq(M\bar\otimes N)'$ and $M\bar\otimes N\subseteq(M'\bar\otimes N')'$ .

Step 1: cyclic vectors. Let $\xi_0$ be cyclic for $M$ and $\eta_0$ cyclic for $N$ . Put $X=M_h\xi_0$ and $Y=N_h\eta_0$ ; then $X+iX=M\xi_0$ and $Y+iY=N\eta_0$ are dense. By Lemma 11.2(b), $X^\perp$ and $Y^\perp$ are the closures of $iM'_h\xi_0$ and $iN'_h\eta_0$ . By Lemma 11.3 and the continuity of $(\xi,\eta)\mapsto\xi\otimes\eta$ (Proposition 1.3(1)), the real span of $M_h\xi_0\odot N_h\eta_0$ and $i\,(iM'_h\xi_0\odot iN'_h\eta_0)=i\,(M'_h\xi_0\odot N'_h\eta_0)$ is dense in $H\otimes K$ . With $\zeta_0=\xi_0\otimes\eta_0$ , the first set lies in $(M\bar\otimes N)_h\zeta_0$ and the second in $(M'\bar\otimes N')_h\zeta_0$ , since $x\otimes y$ is self-adjoint for self-adjoint $x,y$ . So $(M\bar\otimes N)_h\zeta_0+i(M'\bar\otimes N')_h\zeta_0$ is dense. The vector $\zeta_0$ is cyclic for $M\bar\otimes N$ : $(M\odot N)\zeta_0$ contains $M\xi_0\odot N\eta_0$ , which is dense by Proposition 1.3(1). Lemma 11.2(a), for the von Neumann algebra $M\bar\otimes N$ with $A=M\bar\otimes N$ and $B=M'\bar\otimes N'$ , gives $M'\bar\otimes N'=(M'\bar\otimes N')''=(M\bar\otimes N)'$ .

Step 2: the general case. Let $Y'\in(M\bar\otimes N)'$ and $X'\in(M'\bar\otimes N')'$ . It suffices to show $X'Y'=Y'X'$ : then $(M\bar\otimes N)'\subseteq(M'\bar\otimes N')''=M'\bar\otimes N'$ , and Step 0 gives equality. It suffices in turn to show $\langle X'Y'\zeta,\zeta\rangle=\langle Y'X'\zeta,\zeta\rangle$ for $\zeta=\xi\otimes\eta$ . Indeed, if $\langle T(\xi\otimes\eta),\xi\otimes\eta\rangle=0$ for all $\xi,\eta$ , polarization in $\xi$ gives $R_\eta^*TR_\eta=0$ , so $\langle S_{\xi'}^*TS_\xi\eta,\eta\rangle=0$ for all $\xi,\xi',\eta$ ; polarization in $\eta$ gives $S_{\xi'}^*TS_\xi=0$ , and $T=0$ by Proposition 1.3(1).

Fix $\xi,\eta$ . Let $e'\in M'$ and $f'\in N'$ be the projections onto $[M\xi]$ and $[N\eta]$ , and $g'=e'\otimes f'\in M'\bar\otimes N'$ . Then $g'\zeta=\zeta$ . The induced algebras $M_{e'}$ and $N_{f'}$ have the cyclic vectors $\xi$ and $\eta$ , and by Proposition 6.1(2) their commutants are $(M')_{e'}$ and $(N')_{f'}$ . Step 1 gives $(M_{e'}\bar\otimes N_{f'})'=(M')_{e'}\bar\otimes(N')_{f'}$ on $g'(H\otimes K)=e'H\otimes f'K$ .

$g'\in(M\bar\otimes N)'$ . By Proposition 6.1(1), the reduced algebra of $(M\bar\otimes N)'$ by $g'$ is the commutant of the induced algebra $(M\bar\otimes N)_{g'}$ , which is $M_{e'}\bar\otimes N_{f'}$ by (6.1). So $B:=g'Y'g'|_{g'(H\otimes K)}\in(M_{e'}\bar\otimes N_{f'})'$ .
$g'$ lies in $M'\bar\otimes N'$ , the commutant of the von Neumann algebra $(M'\bar\otimes N')'$ . By Proposition 6.1(2), the induced algebra of $(M'\bar\otimes N')'$ by $g'$ is the commutant of the reduced algebra $(M'\bar\otimes N')_{g'}=(M')_{e'}\bar\otimes(N')_{f'}$ (6.1). By Step 1 that commutant is $M_{e'}\bar\otimes N_{f'}$ . So $A:=X'g'|_{g'(H\otimes K)}\in M_{e'}\bar\otimes N_{f'}$ .

Hence $AB=BA$ . Since $X'$ commutes with $g'$ , $g'X'Y'g'=(g'X'g')(g'Y'g')$ and $g'Y'X'g'=(g'Y'g')(g'X'g')$ . As $\zeta=g'\zeta$ , $\langle X'Y'\zeta,\zeta\rangle=\langle AB\zeta,\zeta\rangle=\langle BA\zeta,\zeta\rangle=\langle Y'X'\zeta,\zeta\rangle$ . $\square$

Corollary 11.5. Let $M_k,N_k$ be von Neumann algebras on $H_k$ ( $k=1,2$ ), and let $\vee$ denote the von Neumann algebra generated.

(Joins.) $(M_1\bar\otimes M_2)\vee(N_1\bar\otimes N_2)=(M_1\vee N_1)\bar\otimes(M_2\vee N_2)$ . This part does not use the commutation theorem.
(Intersections.) $(M_1\bar\otimes M_2)\cap(N_1\bar\otimes N_2)=(M_1\cap N_1)\bar\otimes(M_2\cap N_2)$ .
(Centres.) The centre of $M_1\bar\otimes M_2$ is $Z(M_1)\bar\otimes Z(M_2)$ .
(Slice criterion.) $X\in B(H_1\otimes H_2)$ lies in $M_1\bar\otimes M_2$ if and only if $R_{\eta'}^*XR_\eta\in M_1$ and $S_{\xi'}^*XS_\xi\in M_2$ for all vectors; equivalently, iff $(\iota\otimes\omega)(X)\in M_1$ and $(\omega'\otimes\iota)(X)\in M_2$ for all normal functionals $\omega$ on $B(H_2)$ and $\omega'$ on $B(H_1)$ . When $M_2=B(H_2)$ , the first family of conditions suffices, and the commutation theorem is not needed.

Proof. (1) By Theorem 5.2(2), both sides are generated by $M_1\otimes1$ , $1\otimes M_2$ , $N_1\otimes1$ and $1\otimes N_2$ ; on the right use the self-adjoint sets $S=M_1\cup N_1$ and $T=M_2\cup N_2$ , with $S''=M_1\vee N_1$ and $T''=M_2\vee N_2$ .

(2) Apply (1) to the commutants and take commutants, using (11.1) and $(P\vee Q)'=P'\cap Q'$ : $((M_1'\bar\otimes M_2')\vee(N_1'\bar\otimes N_2'))'=(M_1\bar\otimes M_2)\cap(N_1\bar\otimes N_2)$ and $((M_1'\vee N_1')\bar\otimes(M_2'\vee N_2'))'=(M_1\cap N_1)\bar\otimes(M_2\cap N_2)$ .

(3) Take $N_k=M_k'$ in (2), and use (11.1) for $(M_1\bar\otimes M_2)'$ .

(4) Necessity is Lemma 9.1, together with Theorem 9.2(3) for the slices of $B(H_1)\bar\otimes B(H_2)=B(H_1\otimes H_2)$ . Conversely, the matrix entries of $X$ along any basis of $H_2$ lie in $M_1$ , so $X$ commutes with $M_1'\otimes1$ by the commutation criterion (Proposition 2.1(3)). Rows give $X\in(1\otimes M_2')'$ . So $X\in(M_1'\otimes1\cup1\otimes M_2')'=(M_1'\bar\otimes M_2')'$ , which is $M_1\bar\otimes M_2$ by (11.1). For $M_2=B(H_2)$ , use (5.2) instead. $\square$

Part (4) is the "Fubini" description of the tensor product. It is the usual route to statements like $W\in L^\infty(G)\bar\otimes L(G)$ for a multiplicative unitary $W$ . The bounded strong limits of Theorem 5.2(3) give another route, which needs neither the commutation theorem nor any countability; Example 12.2 follows it.

Exercise 11.6 (factors without the commutation theorem). Let $M_1\subseteq B(H_1)$ and $M_2\subseteq B(H_2)$ be factors, with $H_1,H_2\ne0$ . Show, without the commutation theorem, that $M_1\bar\otimes M_2$ is a factor.

Solution. This is the argument of B. Blackadar, Operator Algebras, III.1.5.10. Let $Z$ be the centre of $M_1\bar\otimes M_2$ . Then $Z$ commutes with $M_1\otimes1$ , which lies in the algebra, and with $M_1'\otimes1$ , which lies in its commutant. So the von Neumann algebra $Z'$ contains $((M_1\cup M_1')\otimes1)''=(M_1\cup M_1')''\otimes1$ , by Theorem 5.2(4). As $M_1$ is a factor, $(M_1\cup M_1')'=M_1'\cap M_1=\mathbb C1$ , so $(M_1\cup M_1')''=B(H_1)$ . Likewise $Z'\supseteq1\otimes B(H_2)$ . By Theorem 5.2(2) and (5), $Z'$ contains $B(H_1)\bar\otimes B(H_2)=B(H_1\otimes H_2)$ . Hence $Z\subseteq B(H_1\otimes H_2)'=\mathbb C1$ .

Exercise 11.7 (maximal abelian subalgebras). Let $A_k\subseteq M_k$ be von Neumann subalgebras with $A_k'\cap M_k=A_k$ ( $k=1,2$ ). Show that $(A_1\bar\otimes A_2)'\cap(M_1\bar\otimes M_2)=A_1\bar\otimes A_2$ , and that $A_1\bar\otimes A_2$ is abelian.

Solution. Each $A_k$ is abelian, since $A_k=A_k'\cap M_k\subseteq A_k'$ . So $A_1\odot A_2$ is commutative, and for a commutative set $S$ , $S\subseteq S'$ gives $S''\subseteq S'=(S'')'$ ; so $A_1\bar\otimes A_2$ is abelian. By (11.1), $(A_1\bar\otimes A_2)'=A_1'\bar\otimes A_2'$ . By Corollary 11.5(2), $(A_1'\bar\otimes A_2')\cap(M_1\bar\otimes M_2)=(A_1'\cap M_1)\bar\otimes(A_2'\cap M_2)=A_1\bar\otimes A_2$ .

Editable source · Sources and component terms