Tracial GNS representations and finite von Neumann algebras

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

Definition 11.1 and Proposition 11.2(1)–(2) develop tracial GNS representations, with complete proofs. The auxiliary conjugate-linear isometry is used only to establish separation. The equality $JMJ=M'$ and vector-state/spatial-automorphism theorems are separate assertions. Selection and route notes: GPT-6.1 Sol (OpenAI), Ultra, October 2026; new notes CC0.

Prerequisites and notation

Inner products are linear in the first variable. Normal vector functionals, bounded separate continuity of multiplication, and ultraweak density in the bicommutant are proved in The double commutant theorem. The full GNS construction, including a nonunital algebra and positive contractive approximate units, is in Representations and positive functionals. The equivalence of order normality and ultraweak continuity is in The universal enveloping algebra. The references to Fact 2.4(4)-(5) below mean these precise density and GNS inputs. Fact 2.5(1) means the finite-trace extension and trace identity proved in Proposition 2.2 of Traces, part A.

Definition 11.1. A positive linear functional $\varphi$ on a $C^*$-algebra $A$ is tracial, or central, if $\varphi(x^*x)=\varphi(xx^*)$ for all $x\in A$.

Then $\varphi(xy)=\varphi(yx)$ for all $x,y\in A$. Indeed, in any $*$-algebra $4y^*x=\sum_{k=0}^3i^k(x+i^ky)^*(x+i^ky)$ and $4xy^*=\sum_{k=0}^3i^k(x+i^ky)(x+i^ky)^*$; applying $\varphi$ gives $\varphi(y^*x)=\varphi(xy^*)$, and we replace $y$ by $y^*$.

Proposition 11.2. Let $A$ be a $C^*$-algebra, with or without a unit, let $\varphi$ be a positive linear functional on $A$ that is tracial, and let $(\pi,K,\xi)$ be a cyclic representation with $\varphi(a)=\langle\pi(a)\xi,\xi\rangle$ (Fact 2.4(5)). Let $M=\pi(A)''$.

1. $\tilde\varphi(x)=\langle x\xi,\xi\rangle$ is a faithful normal finite trace on $M$.
2. $\xi$ is cyclic and separating for $M$.

Proof. The trace property. For $a,b\in A$, $\tilde\varphi(\pi(a)\pi(b))=\varphi(ab)=\varphi(ba)=\tilde\varphi(\pi(b)\pi(a))$. Fix $y\in\pi(A)$. The functional $x\mapsto\tilde\varphi(xy)-\tilde\varphi(yx)$ is $\sigma$-weakly continuous and vanishes on $\pi(A)$, which is $\sigma$-weakly dense in $M$ because $\pi$ is nondegenerate (Fact 2.4(4)); so it vanishes on $M$. Now fix $x\in M$: the functional $y\mapsto\tilde\varphi(xy)-\tilde\varphi(yx)$ vanishes on $\pi(A)$, hence on $M$. So $\tilde\varphi(xy)=\tilde\varphi(yx)$ on $M$, and $\tilde\varphi$, a positive normal functional, is a finite normal trace (Fact 2.5(1)).

The involution. For $a\in A$, $\|\pi(a^*)\xi\|^2=\varphi(aa^*)=\varphi(a^*a)=\|\pi(a)\xi\|^2$. So $J_0(\pi(a)\xi)=\pi(a^*)\xi$ is a well-defined conjugate-linear isometry of the dense subspace $\pi(A)\xi$ onto itself, with $J_0^2=1$. It extends to a conjugate-linear isometry $J$ of $K$ with $J^2=1$; so $J$ is onto. Polarization gives $\langle J\eta,J\zeta\rangle=\langle\zeta,\eta\rangle$: the real parts agree because $\|J(\eta+\zeta)\|=\|\eta+\zeta\|$, and replacing $\zeta$ by $i\zeta$ handles the imaginary parts. If $A$ has a unit, $J\xi=J\pi(1)\xi=\xi$. In general, let $(u_i)$ be an approximate unit of positive contractions (Fact 2.4(5)). Then $\pi(u_i)\pi(a)\xi=\pi(u_ia)\xi\to\pi(a)\xi$, so the bounded net $\pi(u_i)$ tends to $1$ strongly on the dense subspace $\pi(A)\xi$, hence on $K$, and $\xi=\lim_i\pi(u_i)\xi$. As $J\pi(u_i)\xi=\pi(u_i)\xi$, $J\xi=\xi$.

For $a,b,x\in A$, $J\pi(a)J\pi(x)\xi=J\pi(a)\pi(x^*)\xi=J\pi(ax^*)\xi=\pi(xa^*)\xi$. Hence $$J\pi(a)J\,\pi(b)\pi(x)\xi=\pi(bxa^*)\xi=\pi(b)\,J\pi(a)J\,\pi(x)\xi .$$ So $J\pi(a)J$ commutes with $\pi(b)$ on a dense subspace, hence everywhere, and $J\pi(A)J\subseteq\pi(A)'=M'$. The map $x\mapsto JxJ$ is continuous for the weak operator topology, since $\langle JxJ\eta,\zeta\rangle=\langle J\zeta,xJ\eta\rangle$, and $M$ is the weak closure of $\pi(A)$; so $JMJ\subseteq M'$.

For $x\in M$ and $a\in A$, using $\langle J\alpha,\beta\rangle=\langle J\beta,\alpha\rangle$ and the trace property, $$\langle J(x\xi),\pi(a)\xi\rangle=\langle\pi(a^*)\xi,x\xi\rangle=\tilde\varphi\bigl(x^*\pi(a^*)\bigr)=\tilde\varphi\bigl(\pi(a^*)x^*\bigr)=\langle x^*\xi,\pi(a)\xi\rangle .$$ So $J(x\xi)=x^*\xi$.

Cyclic and separating. $\xi$ is cyclic for $M\supseteq\pi(A)$. And $M'\xi\supseteq JMJ\xi=JM\xi=M^*\xi=M\xi$, which is dense; so $\xi$ is cyclic for $M'$, that is, separating for $M$. Hence $\tilde\varphi(x^*x)=\|x\xi\|^2=0$ forces $x=0$: $\tilde\varphi$ is faithful. $\square$

Corollary. The algebra $M$ of Proposition 11.2 is finite.

Proof. If $v^*v=1$, traciality gives $\tilde\varphi(1-vv^*)=0$. Faithfulness gives $vv^*=1$. Thus the identity is a finite projection. $\square$

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