Trace Hilbert spaces, commutation, comparison and expectations

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

This complete chapter consists of the trace-duality, trace-Hilbert-space, tracial-commutation, trace-comparison and expectation sections with their full proofs and examples. Results are referred to by their numbers below. The tracial commutation theorem is proved here from bounded-vector duality. General Tomita theory, coupling traces and standard-form theory are not premises. Selection and route notes: GPT-6.1 Sol (OpenAI), Ultra, October 2026; new notes CC0.

Exact prerequisites

Traces, part A, Sections 2, 3 and 7, prove the definition ideals, finite-trace projection nets, supports, sums and the semifinite part, the trace norm and the isometric dense map into the predual. The pairing $\tau(ax)=\tau(xa)$ for $x\in\mathfrak m_\tau,a\in M$ is equation (0.1) below wherever that original label is cited. The public predual chapter proves $M=(M_*)^*$, with weak-star topology equal to the ultraweak topology. Tomiyama positivity and bimodularity are Theorem 8.5 of The universal enveloping algebra, with no complete-positivity strengthening assumed. Normality is order normality, equivalent to ultraweak continuity for positive maps; the same chapter proves that normal representations have von Neumann range. The Riesz representation theorem for Hilbert spaces is in Hilbert spaces and compact operators. The finite Baire-measure representation and continuous-function density used in Theorem 3.3 are proved in Theorem 2.2, Proposition 2.3 and Proposition 3.1(4) of Haar measure on locally compact groups, with the commutative Gelfand theorem in C-star algebras. The scalar multiplication commutant is Decomposable operators and the diagonal algebra, Theorem 7.1(5). No closed unbounded-operator or Mackey-topology result is used in the selected sections.

1. Integrable elements and duality

Throughout this section $\tau$ is a faithful semifinite normal trace on $M$.

The prerequisite lesson introduced the norm $\|x\|_1=\tau(|x|)$ on $\mathfrak m_\tau$ and showed that $x\mapsto\omega_x$ is isometric with dense range in $M_*$. Let $L^1(M,\tau)$ be the completion of $(\mathfrak m_\tau,\|\cdot\|_1)$. The map $x\mapsto\omega_x$ extends to an isometric isomorphism of $L^1(M,\tau)$ onto $M_*$, and we use it to regard every element of $L^1(M,\tau)$ as a normal functional. For $y\in M$ and $x\in L^1(M,\tau)$ we write $$\tau(yx)=\tau(xy)=\omega_x(y). \tag{1.1}$$ For $x\in\mathfrak m_\tau$ this is the old meaning, by (0.1), and $|\tau(yx)|\le\|y\|\,\|x\|_1$.

Theorem 1.1 (Duality). The map $\Phi$ that sends $y\in M$ to the functional $x\mapsto\tau(yx)$ on $L^1(M,\tau)$ is an isometric linear bijection of $M$ onto the dual space $L^1(M,\tau)^*$. It carries the $\sigma$-weak topology of $M$ to the weak$^*$ topology of $L^1(M,\tau)^*$.

Proof. Let $j:L^1(M,\tau)\to M_*$, $j(x)=\omega_x$, be the isometric isomorphism above. The map $y\mapsto(\omega\mapsto\omega(y))$ is an isometric isomorphism of $M$ onto $(M_*)^*$ that carries the $\sigma$-weak topology to the weak$^*$ topology (background). Composing with the transpose of $j$, which is an isometric isomorphism of $(M_*)^*$ onto $L^1(M,\tau)^*$ and a homeomorphism for the weak$^*$ topologies, we get $\Phi(y)(x)=\omega_x(y)=\tau(yx)$. $\square$

So $M$ is the dual of $L^1(M,\tau)$, and we may think of $M$ as $L^\infty(M,\tau)$, with the operator norm in the role of the essential supremum. The elements of $\mathfrak m_\tau$ lie in both spaces. The next result identifies them inside $L^1(M,\tau)$: they are the integrable elements that are also bounded.

Proposition 1.2 (Bounded densities). For $x\in L^1(M,\tau)$ put $$N_\infty(x)=\sup\{|\tau(yx)|:\ y\in\mathfrak m_\tau,\ \|y\|_1\le1\}\in[0,\infty].$$ Then $x$ belongs to $\mathfrak m_\tau$ if and only if $N_\infty(x)<\infty$, and in that case $N_\infty(x)=\|x\|$. Equivalently: a normal functional $\omega$ on $M$ equals $\omega_x$ for some $x\in\mathfrak m_\tau$ exactly when $\omega$ is bounded on the $\|\cdot\|_1$-unit ball of $\mathfrak m_\tau$.

Proof. Let $x\in\mathfrak m_\tau$. For $y\in\mathfrak m_\tau$, (0.1) and the trace-norm inequality give $|\tau(yx)|=|\tau(xy)|\le\|x\|\,\|y\|_1$, so $N_\infty(x)\le\|x\|$. By Theorem 1.1, $\|x\|$ is the norm of $\Phi(x)$, the supremum of $|\tau(xy)|$ over the unit ball of $L^1(M,\tau)$; since $\mathfrak m_\tau$ is dense there, the supremum over its unit ball is the same. So $N_\infty(x)=\|x\|$.

Conversely, let $C=N_\infty(x)<\infty$. The functional $y\mapsto\tau(yx)$ on $\mathfrak m_\tau$ has norm at most $C$ for $\|\cdot\|_1$, so it extends to a bounded functional on $L^1(M,\tau)$, and Theorem 1.1 gives $z\in M$ with $\|z\|=C$ and $\tau(yx)=\tau(zy)$ for all $y\in\mathfrak m_\tau$. We show that $z\in\mathfrak m_\tau$. Let $z=u|z|$ be the polar decomposition, and choose projections $e_i\in F_\tau$ that increase to $1$ (semifiniteness). The element $e_iu^*$ lies in $\mathfrak m_\tau$, because $\mathfrak m_\tau$ is an ideal containing $e_i$. By (0.1), $$\tau(e_i|z|e_i)=\tau\bigl((e_iu^*)(ze_i)\bigr)=\tau\bigl(ze_i\,e_iu^*\bigr)=\tau(z\,e_iu^*)=\tau(e_iu^*\,x),$$ and the last number is at most $\|e_iu^*\|\,\|x\|_1\le\|x\|_1$ in absolute value. On the other hand $e_i|z|e_i=(|z|^{1/2}e_i)^*(|z|^{1/2}e_i)$, so the trace property gives $\tau(e_i|z|e_i)=\tau(|z|^{1/2}e_i|z|^{1/2})$, and these numbers increase to $\tau(|z|)$ by normality. Hence $\tau(|z|)\le\|x\|_1<\infty$, and $z\in\mathfrak m_\tau$. Now $\omega_z$ and $\omega_x$ are normal and agree on $\mathfrak m_\tau$, which is $\sigma$-weakly dense; so they are equal, and $x=z$ in $L^1(M,\tau)$. $\square$

Example 1.3 ($B(H)$ and the trace class). Let $M=B(H)$ and let $\tau=\operatorname{Tr}$ be the usual trace, $\operatorname{Tr}(x)=\sum_i\langle x\varepsilon_i,\varepsilon_i\rangle$ for an orthonormal basis. For a unit vector $\xi$ and any unit vector $\eta$, the rank-one operator $r=\langle\,\cdot\,,\xi\rangle\eta$ has norm $1$, and for $y\in\mathfrak m_{\operatorname{Tr}}$ the linear extension of the trace gives $\operatorname{Tr}(yr)=\langle y\eta,\xi\rangle$. The trace-norm inequality gives $|\langle y\eta,\xi\rangle|\le\|y\|_1$, so $$\|y\|\le\|y\|_1\qquad(y\in\mathfrak m_{\operatorname{Tr}}).$$ Consequently, for $x\in L^1(B(H),\operatorname{Tr})$ and $y\in\mathfrak m_{\operatorname{Tr}}$ with $\|y\|_1\le1$, $|\operatorname{Tr}(yx)|\le\|y\|\,\|x\|_1\le\|x\|_1$. So $N_\infty(x)\le\|x\|_1<\infty$ for every $x$, and Proposition 1.2 shows that $L^1(B(H),\operatorname{Tr})=\mathfrak m_{\operatorname{Tr}}$: no completion is needed. These are the trace-class operators, and Theorem 1.1 is the classical statement that $B(H)$ is the dual of the trace class. In contrast, for a diffuse algebra such as $L^\infty[0,1]$ the completion adds unbounded elements (Example 1.4).

Example 1.4 (Measure spaces). Let $(X_k,\mu_k)_{k\in K}$ be finite measure spaces and let $(X,\mu)$ be their disjoint union: a function on $X$ is measurable when its restriction to each $X_k$ is, and $\int_Xf\,d\mu=\sum_k\int_{X_k}f\,d\mu_k$ for $f\ge0$. Put $L^p(X,\mu)=\{f:\|f\|_p<\infty\}$ modulo null functions, for $p=1,2$, and let $L^\infty(X,\mu)$ be the bounded families $(f_k)$ with $f_k\in L^\infty(X_k,\mu_k)$. Every $\sigma$-finite measure space has this form, with $K$ countable.

Let $M=L^\infty(X,\mu)$ act on $L^2(X,\mu)=\bigoplus_kL^2(X_k,\mu_k)$ by multiplication. Each summand is a maximal abelian von Neumann algebra on $L^2(X_k,\mu_k)$ (background), so $M$ is a von Neumann algebra, the direct sum of these. Put $\tau(f)=\int_Xf\,d\mu$ for $f\in M_+$. Then:

• $\tau$ is a trace, since $M$ is commutative. It is faithful. It is normal: $\tau(f)=\sum_k\langle f1_{X_k},1_{X_k}\rangle$ is a sum of vector functionals. It is semifinite: the projections $\sum_{k\in F}1_{X_k}$, $F$ finite, have finite trace and increase to $1$.
• $F_\tau$ consists of the bounded integrable functions $f\ge0$, so $\mathfrak m_\tau=L^\infty\cap L^1$, and $\|f\|_1=\int|f|\,d\mu$ is the $L^1$-norm. Likewise $\mathfrak n_\tau=L^\infty\cap L^2$ with the $L^2$-norm.
• $L^\infty\cap L^1$ is dense in $L^1(X,\mu)$: for $f\in L^1$, the functions $f\,1_{\{|f|\le n\}}\sum_{k\in F}1_{X_k}$ converge to $f$ in $L^1$ by dominated convergence, along $n\to\infty$ and finite sets $F$ increasing to $K$. As $L^1(X,\mu)$ is complete, $L^1(M,\tau)=L^1(X,\mu)$, and the pairing (1.1) is $\tau(gf)=\int_Xgf\,d\mu$.

Theorem 1.1 now says that $L^\infty(X,\mu)$ is the dual of $L^1(X,\mu)$ through $g\mapsto(f\mapsto\int gf\,d\mu)$, and Proposition 1.2 says that an integrable $f$ is essentially bounded exactly when $g\mapsto\int gf\,d\mu$ is bounded on the unit ball of $L^\infty\cap L^1$ for the $L^1$-norm, with bound $\|f\|_\infty$. By Theorem 3.3, every commutative von Neumann algebra that carries a faithful semifinite normal trace is of this form.

Remark 1.5. Faithfulness is used to make $\|\cdot\|_1$ a norm. If $\tau$ is only semifinite and normal, $\|\cdot\|_1$ vanishes exactly on $\mathfrak m_\tau(1-s(\tau))=\mathfrak m_\tau\cap M(1-s(\tau))$, and the same results hold for the algebra $Ms(\tau)$, whose predual is the set of normal functionals on $M$ that vanish on $M(1-s(\tau))$.

2. The Hilbert space of a trace

Throughout this section $\tau$ is a faithful semifinite normal trace on $M$. A reference for Sections 1–3 is [Kostecki, Section 5.2].

Definition 2.1. For $x,y\in\mathfrak n_\tau$ put $\langle x,y\rangle_2=\tau(y^*x)$; this makes sense because $y^*x\in\mathfrak m_\tau$. It is a sesquilinear form, positive, and definite because $\tau$ is faithful: $\langle x,x\rangle_2=\tau(x^*x)=0$ forces $x=0$. The completion of $\mathfrak n_\tau$ is a Hilbert space $L^2(M,\tau)$, and $\mathfrak n_\tau$ is a dense subspace of it. The norm is $\|x\|_2$.

Lemma 2.2 (The two actions). Let $a\in M$ and $x\in\mathfrak n_\tau$.

1. $\|ax\|_2\le\|a\|\,\|x\|_2$, $\|xa\|_2\le\|a\|\,\|x\|_2$ and $\|x^*\|_2=\|x\|_2$.
2. There are bounded operators $\pi_\tau(a)$ and $\rho_\tau(a)$ on $L^2(M,\tau)$ with $\pi_\tau(a)x=ax$ and $\rho_\tau(a)x=xa$. The map $\pi_\tau$ is a unital $*$-homomorphism. The map $\rho_\tau$ is linear and unital, with $\rho_\tau(ab)=\rho_\tau(b)\rho_\tau(a)$ and $\rho_\tau(a^*)=\rho_\tau(a)^*$. Every $\pi_\tau(a)$ commutes with every $\rho_\tau(b)$.
3. The map $x\mapsto x^*$ extends to a conjugate-linear isometry $J$ of $L^2(M,\tau)$ with $J^2=1$, $\langle J\xi,J\eta\rangle_2=\langle\eta,\xi\rangle_2$, and $J\pi_\tau(a)J=\rho_\tau(a^*)$.
4. For $x,y\in\mathfrak n_\tau$, $\langle\pi_\tau(a)x,y\rangle_2=\tau(a\,xy^*)=\omega_{xy^*}(a)$. The maps $\pi_\tau$ and $\rho_\tau$ are injective and normal, in the sense that $a\mapsto\langle\pi_\tau(a)\xi,\eta\rangle_2$ and $a\mapsto\langle\rho_\tau(a)\xi,\eta\rangle_2$ lie in $M_*$ for all vectors $\xi,\eta$.
5. $\mathfrak m_\tau$ is dense in $L^2(M,\tau)$. If $e_i\in F_\tau$ are projections that increase to $1$, then $\pi_\tau(e_i)\to1$ and $\rho_\tau(e_i)\to1$ strongly.

We call $\pi_\tau$ the standard representation of $M$ defined by $\tau$, $\rho_\tau$ the right representation and $J$ the conjugation. When no confusion arises we write $\pi,\rho$.

Proof. (1) Since $x^*a^*ax\le\|a\|^2x^*x$, monotonicity of $\tau$ gives the first inequality. By the trace identity for $z=xa$, $\tau(a^*x^*xa)=\tau(xaa^*x^*)\le\|a\|^2\tau(xx^*)=\|a\|^2\tau(x^*x)$. Finally $\tau(xx^*)=\tau(x^*x)$.

(2) By (1), left and right multiplication by $a$ are bounded on $\mathfrak n_\tau$ for $\|\cdot\|_2$, with norm at most $\|a\|$, and extend to $L^2(M,\tau)$. The algebraic rules hold on $\mathfrak n_\tau$ and pass to the closure. For adjoints, let $x,y\in\mathfrak n_\tau$. Then $\langle ax,y\rangle_2=\tau(y^*ax)=\tau((a^*y)^*x)=\langle x,a^*y\rangle_2$. Also $\langle xa,y\rangle_2=\tau(y^*x\,a)=\tau(a\,y^*x)$ by (0.1), since $y^*x\in\mathfrak m_\tau$, and $\tau(a\,y^*x)=\tau((ya^*)^*x)=\langle x,ya^*\rangle_2$. The commutation is associativity: $(ax)b=a(xb)$.

(3) By (1), $J$ is isometric on $\mathfrak n_\tau$, so it extends; $J^2=1$ is clear. For $x,y\in\mathfrak n_\tau$, $\langle x^*,y^*\rangle_2=\tau(yx^*)=\tau(x^*y)=\langle y,x\rangle_2$ by (0.1), and this passes to limits. Finally $J\pi_\tau(a)Jx=(ax^*)^*=xa^*=\rho_\tau(a^*)x$.

(4) By (0.1) with the elements $y^*$ and $ax$ of $\mathfrak n_\tau$, $\tau(y^*\,ax)=\tau(ax\,y^*)$, and $xy^*\in\mathfrak m_\tau$. So $a\mapsto\langle\pi_\tau(a)x,y\rangle_2$ is $\omega_{xy^*}$, which is normal. For arbitrary vectors $\xi,\eta$, approximate them by $x,y\in\mathfrak n_\tau$; the functionals converge in norm, and $M_*$ is norm closed. For $\rho_\tau$, part (3) gives $\langle\rho_\tau(a)\xi,\eta\rangle_2=\langle J\pi_\tau(a^*)J\xi,\eta\rangle_2=\langle\pi_\tau(a)J\eta,J\xi\rangle_2$, which is normal in $a$. If $\pi_\tau(a)=0$, then $ae_i=\pi_\tau(a)e_i=0$ for projections $e_i\in F_\tau$ increasing to $1$, so $a=0$; and $\rho_\tau(a)=J\pi_\tau(a^*)J$.

(5) For $y\in\mathfrak n_\tau$, $e_iy\in\mathfrak m_\tau$, and $\|y-e_iy\|_2^2=\tau(y^*(1-e_i)y)=\tau(y^*y)-\tau(y^*e_iy)\to0$ because $y^*e_iy\uparrow y^*y$ and $\tau$ is normal. So $\mathfrak m_\tau$ is dense, and $\pi_\tau(e_i)\to1$ on the dense set $\mathfrak n_\tau$; being contractions, they converge strongly. Then $\rho_\tau(e_i)=J\pi_\tau(e_i)J\to1$ strongly. $\square$

The next two results decide when a vector of $L^2(M,\tau)$, or an element of $M$, comes from $\mathfrak n_\tau$. They are the Hilbert-space analogues of Proposition 1.2.

Proposition 2.3 (Square-integrable elements). For $x\in M$ put $$N_2(x)=\sup\{|\tau(y^*x)|:\ y\in\mathfrak m_\tau,\ \|y\|_2\le1\}\in[0,\infty].$$ Then $x\in\mathfrak n_\tau$ if and only if $N_2(x)<\infty$, and then $N_2(x)=\|x\|_2$.

Proof. The number $\tau(y^*x)$ is defined because $\mathfrak m_\tau$ is an ideal. If $x\in\mathfrak n_\tau$, then $\tau(y^*x)=\langle x,y\rangle_2$, so $N_2(x)\le\|x\|_2$ by Cauchy–Schwarz, and equality holds because $\mathfrak m_\tau$ is dense in $L^2(M,\tau)$.

Conversely let $N_2(x)<\infty$. The map $y\mapsto\overline{\tau(y^*x)}$ is linear and bounded on the dense subspace $\mathfrak m_\tau$, so the Riesz representation theorem gives $\xi\in L^2(M,\tau)$ with $\tau(y^*x)=\langle\xi,y\rangle_2$ for all $y\in\mathfrak m_\tau$. Let $e_i\in F_\tau$ be projections increasing to $1$. For $y\in\mathfrak m_\tau$, $$\langle e_ix,y\rangle_2=\tau(y^*e_ix)=\tau((e_iy)^*x)=\langle\xi,e_iy\rangle_2=\langle\pi_\tau(e_i)\xi,y\rangle_2 ,$$ where $e_ix\in\mathfrak m_\tau$. By density, $e_ix=\pi_\tau(e_i)\xi$ in $L^2(M,\tau)$. Hence $\tau(x^*e_ix)=\|e_ix\|_2^2\le\|\xi\|_2^2$. As $x^*e_ix\uparrow x^*x$, normality gives $\tau(x^*x)\le\|\xi\|_2^2<\infty$. $\square$

Proposition 2.4 (Bounded vectors). For $\xi\in L^2(M,\tau)$ the following are equivalent:

• (i) $\xi\in\mathfrak n_\tau$;
• (ii) $N_\infty(\xi)=\sup\{|\langle\xi,y\rangle_2|:\ y\in\mathfrak m_\tau,\ \|y\|_1\le1\}<\infty$;
• (iii) $N_\ell(\xi)=\sup\{\|\pi_\tau(a)\xi\|_2:\ a\in\mathfrak n_\tau,\ \|a\|_2\le1\}<\infty$;
• (iv) $N_r(\xi)=\sup\{\|\rho_\tau(a)\xi\|_2:\ a\in\mathfrak n_\tau,\ \|a\|_2\le1\}<\infty$.

When they hold, and $\xi=x\in\mathfrak n_\tau$, all three suprema equal the operator norm $\|x\|$.

Proof. (i)$\Rightarrow$(iii): for $a\in\mathfrak n_\tau$, $\pi_\tau(a)x=ax=\rho_\tau(x)a$, so $\|ax\|_2\le\|x\|\,\|a\|_2$ by Lemma 2.2(1). Thus $N_\ell(x)\le\|x\|$.

(iii)$\Rightarrow$(ii): let $y\in\mathfrak m_\tau$ with polar decomposition $y=u|y|$. Then $k=|y^*|=u|y|u^*$ satisfies $ku=u|y|=y$, and $k\in F_\tau$, so $k^{1/2}\in\mathfrak n_\tau$ with $\|k^{1/2}\|_2^2=\tau(k)=\tau(|y|)=\|y\|_1$. Write $y=k^{1/2}(k^{1/2}u)$. As $\pi_\tau(k^{1/2})$ is self-adjoint, $$|\langle\xi,y\rangle_2|=|\langle\pi_\tau(k^{1/2})\xi,\,k^{1/2}u\rangle_2|\le\|\pi_\tau(k^{1/2})\xi\|_2\,\|k^{1/2}u\|_2\le N_\ell(\xi)\,\|k^{1/2}\|_2\cdot\|k^{1/2}\|_2=N_\ell(\xi)\,\|y\|_1 ,$$ using Lemma 2.2(1) for $\|k^{1/2}u\|_2\le\|k^{1/2}\|_2$. So $N_\infty(\xi)\le N_\ell(\xi)$.

(ii)$\Rightarrow$(i): the map $w\mapsto\langle\xi,w^*\rangle_2$ is linear on $\mathfrak m_\tau$ and bounded for $\|\cdot\|_1$, since $\|w^*\|_1=\|w\|_1$. By Theorem 1.1 there is $x'\in M$ with $\|x'\|=N_\infty(\xi)$ and $\langle\xi,w^*\rangle_2=\tau(wx')$ for $w\in\mathfrak m_\tau$; with $w=y^*$ this reads $\langle\xi,y\rangle_2=\tau(y^*x')$. Let $e_i\in F_\tau$ be projections increasing to $1$. For $y\in\mathfrak m_\tau$, $$\langle e_ix',y\rangle_2=\tau(y^*e_ix')=\tau((e_iy)^*x')=\langle\xi,e_iy\rangle_2=\langle\pi_\tau(e_i)\xi,y\rangle_2 ,$$ so $e_ix'=\pi_\tau(e_i)\xi$. Then $\tau(x'^*e_ix')=\|\pi_\tau(e_i)\xi\|_2^2\le\|\xi\|_2^2$, and normality gives $\tau(x'^*x')\le\|\xi\|_2^2$: so $x'\in\mathfrak n_\tau$. Finally $\pi_\tau(e_i)\xi\to\xi$ and $e_ix'\to x'$ in $L^2(M,\tau)$ (Lemma 2.2(5)), so $\xi=x'$.

Norms: for $x\in\mathfrak n_\tau$, $N_\infty(x)=\sup\{|\tau(wx)|:w\in\mathfrak m_\tau,\|w\|_1\le1\}$, which is $\|x\|$ by Theorem 1.1 and density. With the inequalities above, $\|x\|=N_\infty(x)\le N_\ell(x)\le\|x\|$.

(i)$\Leftrightarrow$(iv): $J$ is isometric, maps $\mathfrak n_\tau$ onto itself, and $\|\rho_\tau(a)\xi\|_2=\|J\rho_\tau(a)\xi\|_2=\|\pi_\tau(a^*)J\xi\|_2$ by Lemma 2.2(3). As $a\mapsto a^*$ preserves $\mathfrak n_\tau$ and $\|\cdot\|_2$, $N_r(\xi)=N_\ell(J\xi)$, and $\|x^*\|=\|x\|$. $\square$

Example 2.5. (a) $B(H)$. By Example 1.3, $\|y\|\le\|y\|_1$ for $y\in\mathfrak m_{\operatorname{Tr}}$. For $x\in\mathfrak n_{\operatorname{Tr}}$ this gives $\|x\|^2=\|x^*x\|\le\|x^*x\|_1=\|x\|_2^2$. Let $\xi\in L^2(B(H),\operatorname{Tr})$ be the limit of $x_n\in\mathfrak n_{\operatorname{Tr}}$. For $a\in\mathfrak n_{\operatorname{Tr}}$, $\|\pi(a)\xi\|_2=\lim_n\|ax_n\|_2\le\|a\|\,\|\xi\|_2\le\|a\|_2\|\xi\|_2$. So $N_\ell(\xi)\le\|\xi\|_2$, and $\xi\in\mathfrak n_{\operatorname{Tr}}$ by Proposition 2.4. Every vector is bounded: $L^2(B(H),\operatorname{Tr})$ is the space of Hilbert–Schmidt operators, with $\|x\|\le\|x\|_2$.

(b) $L^\infty[0,1]$. With Lebesgue measure, $L^2(M,\tau)=L^2[0,1]$ by Example 1.4, and for $f\in L^2$, $N_\ell(f)=\sup\{\|gf\|_2:\ g\in L^\infty,\ \|g\|_2\le1\}$. For $f(t)=t^{-1/4}$ and $g=\delta^{-1/2}1_{[0,\delta]}$, $\|gf\|_2^2=\delta^{-1}\int_0^\delta t^{-1/2}\,dt=2\delta^{-1/2}\to\infty$. So $f$ is not a bounded vector, although it lies in $L^2$. Here the bounded vectors are exactly the essentially bounded functions, and Proposition 2.4 recovers $\|f\|_\infty=\sup\{\|gf\|_2:\|g\|_2\le1\}$.

3. The commutation theorem

Theorem 3.1 (The commutation theorem). Let $\tau$ be a faithful semifinite normal trace on $M$. Then $\pi_\tau(M)$ and $\rho_\tau(M)$ are von Neumann algebras on $L^2(M,\tau)$, $$\pi_\tau(M)'=\rho_\tau(M),\qquad \rho_\tau(M)'=\pi_\tau(M),\qquad J\pi_\tau(M)J=\rho_\tau(M). \tag{3.1}$$ $\pi_\tau$ is a normal $*$-isomorphism of $M$ onto $\pi_\tau(M)$, and $\rho_\tau$ is a normal anti-isomorphism of $M$ onto $\rho_\tau(M)$. The center of $\pi_\tau(M)$ is $\pi_\tau(Z)=\rho_\tau(Z)$.

Proof. By Lemma 2.2, $\pi_\tau$ is a faithful normal representation, so its image is a von Neumann algebra (background), and $\pi_\tau$ is an isomorphism onto it. Since $J$ is a conjugate-linear isometry with $J^2=1$, the map $X\mapsto JXJ$ preserves products, adjoints and strong limits. By Lemma 2.2(3), $\rho_\tau(M)=J\pi_\tau(M)J$; so $\rho_\tau(M)$ is strongly closed, contains $1$ and is closed under adjoints: a von Neumann algebra.

By Lemma 2.2(2), $\rho_\tau(M)\subseteq\pi_\tau(M)'$. For the converse, let $b\in\pi_\tau(M)'$ and $x\in\mathfrak n_\tau$. Consider the vector $\xi=bx$. For $a\in\mathfrak n_\tau$, $$\|\pi_\tau(a)\xi\|_2=\|b\,\pi_\tau(a)x\|_2=\|b(ax)\|_2\le\|b\|\,\|ax\|_2\le\|b\|\,\|x\|\,\|a\|_2 .$$ So $N_\ell(\xi)\le\|b\|\|x\|$, and Proposition 2.4 gives an element $w\in\mathfrak n_\tau$ with $bx=w$. For $a\in\mathfrak n_\tau$, $$\rho_\tau(w)a=aw=\pi_\tau(a)bx=b\,\pi_\tau(a)x=b(ax)=b\,\rho_\tau(x)a .$$ Both $\rho_\tau(w)$ and $b\rho_\tau(x)$ are bounded and agree on the dense set $\mathfrak n_\tau$, so $b\rho_\tau(x)=\rho_\tau(w)\in\rho_\tau(M)$. Now let $e_i\in F_\tau$ be projections increasing to $1$; they lie in $\mathfrak n_\tau$. By Lemma 2.2(5), $b\rho_\tau(e_i)\to b$ strongly, and $\rho_\tau(M)$ is strongly closed; so $b\in\rho_\tau(M)$. This proves $\pi_\tau(M)'=\rho_\tau(M)$, and then $\rho_\tau(M)'=\pi_\tau(M)''=\pi_\tau(M)$ by the bicommutant theorem.

$\rho_\tau$ is injective and normal by Lemma 2.2(4), and it reverses products. For $a\in Z$ and $x\in\mathfrak n_\tau$, $ax=xa$, so $\pi_\tau(a)=\rho_\tau(a)$. The center of $\pi_\tau(M)$ is $\pi_\tau(M)\cap\pi_\tau(M)'=\pi_\tau(M)\cap\rho_\tau(M)$, and $\pi_\tau$ maps $Z$ onto the center of $\pi_\tau(M)$ because it is an isomorphism. $\square$

When $\tau(1)<\infty$, the vector $1\in\mathfrak n_\tau$ is cyclic for $\pi_\tau(M)$ and for $\rho_\tau(M)$, since $\pi_\tau(x)1=x=\rho_\tau(x)1$; by (3.1) it is then also separating for both.

Example 3.2 (Matrices). Let $M=M_n(\mathbb C)$ and $\tau=\operatorname{Tr}$. Then $L^2(M,\tau)$ is $M_n(\mathbb C)$ with $\langle x,y\rangle_2=\operatorname{Tr}(y^*x)$, $\pi$ and $\rho$ are left and right multiplication, and $J$ is the adjoint. Theorem 3.1 is here elementary: if a linear map $T$ of $M_n(\mathbb C)$ commutes with every left multiplication, then $T(x)=T(x\cdot1)=xT(1)$, so $T=\rho(T(1))$.

Theorem 3.3 (Commutative algebras). Let $A$ be a commutative von Neumann algebra carrying a faithful semifinite normal trace $\tau$.

1. $\pi_\tau(A)$ is maximal abelian on $L^2(A,\tau)$: $\pi_\tau(A)'=\pi_\tau(A)$.
2. There are compact Hausdorff spaces $\Omega_k$ ($k\in K$) with finite positive Baire measures $\mu_k$, and a $*$-isomorphism $\Phi$ of $A$ onto the algebra $L^\infty(X,\mu)$ of the disjoint union $(X,\mu)$ of the $(\Omega_k,\mu_k)$, as in Example 1.4, such that $\tau(a)=\int_X\Phi(a)\,d\mu$ for $a\in A_+$.

Consequently, by Example 1.4, $\mathfrak m_\tau$, $\mathfrak n_\tau$, $L^1(A,\tau)$ and $L^2(A,\tau)$ correspond to $L^\infty\cap L^1$, $L^\infty\cap L^2$, $L^1(X,\mu)$ and $L^2(X,\mu)$. The space $X$ is locally compact, each $\Omega_k$ being open and compact in it, and $\mu$ is finite on compact sets.

Proof. (1) For $a\in A$ and $x\in\mathfrak n_\tau$, $ax=xa$, so $\pi_\tau(a)=\rho_\tau(a)$. By Theorem 3.1, $\pi_\tau(A)'=\rho_\tau(A)=\pi_\tau(A)$.