Integral representations of states

Originally written by Claude Opus 5.5 (Anthropic), September 2026, with a separate historical AI spot-check of that text. GPT-6.1 Sol (OpenAI), at the Ultra setting, read and self-checked the full lesson and all its solutions, completed the operator-topology tools and supplied the exact programme prerequisites, October 2026. Public domain (CC0).

The states of a unital C$^*$-algebra $A$ form a compact convex set $\mathfrak S$, and its extreme points are the pure states. This lesson shows how to write an arbitrary state as an average of pure states, that is, as the barycentre of a probability measure on $\mathfrak S$ that lives on the pure states, as far as that makes sense. The choice of such a measure is governed by the GNS representation $\pi_\varphi$ of the state: the well-behaved representing measures, called orthogonal, correspond one-to-one to the abelian von Neumann subalgebras of the commutant $\pi_\varphi(A)'$.

Such decompositions are rarely unique. The tracial state of $M_2(\mathbb C)$ is an average of pure states in infinitely many ways, just as the centre of a square is the average of the four vertices in more than one way. A probability measure on a compact space $X$, seen as a state of $C(X)$, is an average of point masses in exactly one way, like a point of a triangle. The lesson explains the difference: a state has exactly one maximal representing measure precisely when $\pi_\varphi(A)'$ is abelian.

The first part, Sections 1–7, is Choquet's theory of compact convex sets; it uses no operator algebras. It defines barycentres and the Choquet order on measures. It shows that maximal measures live on the extreme points when the set is metrizable, and that in general they vanish on every Baire set that misses the extreme points. It ends with the Choquet–Meyer theorem: maximal representing measures are unique exactly on simplices. The second part, Sections 8–24, applies this to states. It builds the operator map of a representing measure, characterizes the orthogonal measures, matches them with the abelian subalgebras of $\pi_\varphi(A)'$, and compares them. It then decomposes every state into pure states along a maximal abelian subalgebra, and every invariant state into ergodic states. It introduces the central measure, which vanishes on every Baire set that misses the factorial states, and shows that the factorial states of a separable algebra form a Borel set. The last sections treat the face generated by a state, $G$-abelian systems and large groups of automorphisms.

Prerequisites are the full proofs of Hahn–Banach and convex separation, weak duality, Banach–Alaoglu and Krein–Milman, Hilbert-space representation, Riesz representation and elementary integration, positive functionals, states and GNS, continuous functional calculus, the spectral theorem, the double commutant theorem and Kaplansky density. The Stone–Weierstrass lesson supplies Urysohn functions and the lattice and algebra approximation lemmas. The background section below states the precise versions and their proof locations; the references identify the versions used for comparison.

The complete proofs are included in this lesson. Fremlin’s treatment of barycentres and maximal measures provides the freely readable comparison for Sections 1–7; the operator-algebra constructions in Sections 8–24 are proved here with their exact programme prerequisites.

Conventions

Convex sets. $E$ is a real locally convex Hausdorff space and $E^*$ its dual, the continuous linear functionals $E\to\mathbb R$. By the Hahn–Banach theorem $E^*$ separates the points of $E$. $K\subseteq E$ is a nonempty compact convex set. Functions are real-valued unless we say otherwise. $C(K)$ is the space of continuous functions on $K$ with the sup norm $\|\cdot\|$. We write

$\operatorname{Aff}(K)$ for the continuous affine functions on $K$;
$\begin{gathered} \operatorname{Aff}_E(K)\\ =\{p|_K+c:\ p\in E^*,\ c\in\mathbb R\}\\ \subseteq\operatorname{Aff}(K) \end{gathered}$ ;
$\mathcal P(K)$ for the continuous convex functions on $K$;
$\mathcal L(K)$ for the lower semicontinuous (lsc) convex functions $K\to(-\infty,+\infty]$.

$\partial_eK$ is the set of extreme points of $K$. A face of $K$ is a convex set $F\subseteq K$ such that $x,y\in K$, $0<t<1$ and $tx+(1-t)y\in F$ imply $x,y\in F$. A face of a face of $K$ is a face of $K$; in particular an extreme point of a face is an extreme point of $K$. If $\Phi$ is an affine map on $K$ and $y'$ is an extreme point of $\Phi(K)$, then $K\cap\Phi^{-1}(y')$ is a face of $K$.

Measures. For a compact Hausdorff space $X$, $M(X)=C(X)^*$ is the space of real Radon measures, with the dual norm. By the Riesz representation theorem each $\mu\in M(X)$ is integration against a unique regular real Borel measure, again written $\mu$, with $\|\mu\|=|\mu|(X)$. Here $|\mu|$ is the variation and $\mu=\mu^+-\mu^-$ the Jordan decomposition. For a bounded Borel function $g$ we write $\mu(g)=\int g\,d\mu$. $M^+(X)$ is the cone of positive measures and $M^+_1(X)$ the set of probability measures, a weak$^*$-compact convex set. $\delta_x$ is the point mass at $x$. A measure $\mu$ is concentrated on a Borel set $B$ if $|\mu|(X\setminus B)=0$.

Operator algebras. $A$ is a nonzero unital C$^*$-algebra over $\mathbb C$, $A_h$ its self-adjoint part, and $\mathfrak S=\mathfrak S(A)$ its state space with the weak$^*$ topology. A state $\varphi$ has its GNS representation $(\pi_\varphi,H_\varphi,\xi_\varphi)$: $\xi_\varphi$ is cyclic and $\varphi(a)=\langle\pi_\varphi(a)\xi_\varphi,\xi_\varphi\rangle$. Inner products are linear in the first variable. For a set $\mathcal S\subseteq B(H)$ closed under adjoints, $\mathcal S'$ is its commutant. It is a $*$-algebra, contains $1$, and is closed in the weak operator topology, since $Ty=yT$ says $\langle T(y\xi),\eta\rangle=\langle T\xi,y^*\eta\rangle$ for all $\xi,\eta$. A von Neumann algebra is a $*$-algebra $M\subseteq B(H)$ with $M=M''$; so every commutant of a self-adjoint set is one. If $\xi$ is cyclic for a set $\mathcal S$, it is separating for $\mathcal S'$: $x\in\mathcal S'$ and $x\xi=0$ give $x\,y\xi=y\,x\xi=0$ for $y$ in the algebra generated by $\mathcal S$, so $x=0$. A vector $\eta$ is used in the vector functional $\omega_\eta(x)=\langle x\eta,\eta\rangle$.

1. Affine functions and envelopes

We start with three lemmas. Each is proved by separating a point, or a compact convex set, from a closed convex set in $E\times\mathbb R$.

Lemma 1.1 (Affine minorants). Let $f\in\mathcal L(K)$. For every $x\in K$, $\begin{gathered} f(x)\\ =\sup\{a(x):\\ \ \\ a\in\operatorname{Aff}_E(K),\ a\\ \leq f\ \text{on }K\}. \end{gathered} \tag{1.1}$

The value $+\infty$ is allowed; Lemma 4.2 needs this case.

Proof. The right side is at most $f(x)$. Fix a real $\beta<f(x)$; we find $a\in\operatorname{Aff}_E(K)$ with $a\leq f$ and $a(x)>\beta$. First, some affine function lies below $f$: an lsc function on a compact set attains its infimum, which is $>-\infty$, so a constant works. If $f\equiv+\infty$, every constant is a minorant and we are done. Otherwise let $M=\{(y,t)\in K\times\mathbb R:\ f(y)\leq t\}\subseteq E\times\mathbb R .$ $M$ is convex because $f$ is convex. It is closed: if $(y_i,t_i)\to(y,t)$ with $(y_i,t_i)\in M$, then $y\in K$ and $f(y)\leq\liminf f(y_i)\leq t$. Also $M\neq\varnothing$ and $(x,\beta)\notin M$. The dual of $E\times\mathbb R$ consists of the maps $(y,t)\mapsto p(y)+\lambda t$ with $p\in E^*$, $\lambda\in\mathbb R$. By the separation theorem, applied to the compact set $\{(x,\beta)\}$ and the closed convex set $M$, there are such $p,\lambda$ and a real $\gamma$ with $\begin{gathered} p(x)+\lambda\beta<\gamma<p(y)+\lambda t\\ \text{for all }(y,t)\in M . \end{gathered}$ Since $(y,t)\in M$ implies $(y,t+s)\in M$ for $s\geq0$, we get $\lambda\geq0$.

If $\lambda>0$, put $a=(\gamma-p)/\lambda$ on $K$. For $y$ with $f(y)<\infty$, the point $(y,f(y))$ lies in $M$, so $f(y)>a(y)$; where $f=+\infty$ there is nothing to check. And $a(x)>\beta$.

If $\lambda=0$, then $p(x)<\gamma<p(y)$ for every $y$ with $f(y)<\infty$. Take any affine minorant $a_0\in\operatorname{Aff}_E(K)$ of $f$, and put $a_s=a_0+s(\gamma-p)$ for $s>0$. Where $f<\infty$ we have $\gamma-p<0$, so $a_s\leq a_0\leq f$; elsewhere $f=+\infty$. Since $\gamma-p(x)>0$, a large $s$ gives $a_s(x)>\beta$. $\square$

Lemma 1.2 (Density of $\operatorname{Aff}_E(K)$). $\operatorname{Aff}_E(K)$ is uniformly dense in $\operatorname{Aff}(K)$.

Proof. Let $a\in\operatorname{Aff}(K)$ and $\varepsilon>0$. The graphs $G_1=\{(y,a(y)):y\in K\}$ and $G_2=\{(y,a(y)-\varepsilon):y\in K\}$ are compact convex subsets of $E\times\mathbb R$, since $a$ is continuous and affine, and they are disjoint. Separating them gives $p\in E^*$, $\lambda\in\mathbb R$ and $\gamma$ with $p(y)+\lambda(a(y)-\varepsilon)<\gamma<p(y)+\lambda a(y)$ for all $y\in K$. Subtracting gives $\lambda\varepsilon>0$, so $\lambda>0$, and then $a(y)-\varepsilon<(\gamma-p(y))/\lambda<a(y)$. So $b=(\gamma-p)/\lambda$ on $K$ lies in $\operatorname{Aff}_E(K)$ and $\|a-b\|\leq\varepsilon$. $\square$

Lemma 1.3 (Increasing nets of affine functions). Let $f:K\to\mathbb R$ be lsc and affine. Then $I=\{g\in\operatorname{Aff}_E(K):\ g<f\ \text{on }K\}$ is directed upward, and $\sup I=f$ pointwise. So $f$ is the pointwise limit of an increasing net in $\operatorname{Aff}_E(K)$.

Proof. By Lemma 1.1, for $\varepsilon>0$ every affine minorant $a$ of $f$ gives $a-\varepsilon\in I$; so $\sup I=f$. Let $g_1,g_2\in I$. After adding a constant to $f,g_1,g_2$ we may assume that $g_1,g_2>0$. Put $\begin{gathered} M\\ =\{(y,t):\\ \ y\in K,\ f(y)\\ \leq t\},\\ M_i\\ =\{(y,t):\\ \ y\in K,\ 0\\ \leq t\\ \leq g_i(y)\}\\ (i\\ =1,2). \end{gathered}$ $M$ is closed and convex, and $M_1,M_2$ are compact and convex, so the convex hull $D$ of $M_1\cup M_2$ is compact (it is the image of $M_1\times M_2\times[0,1]$). $D$ misses $M$: a point of $D$ is $s(y_1,t_1)+(1-s)(y_2,t_2)$ with $t_i\leq g_i(y_i)<f(y_i)$, and since $f$ is affine, $st_1+(1-s)t_2<f(sy_1+(1-s)y_2)$. Separation gives $p$, $\lambda$, $\gamma$ with $p(y)+\lambda t<\gamma$ on $D$ and $p(y)+\lambda t>\gamma$ on $M$. As in the proof of Lemma 1.1, $\lambda\geq0$. If $\lambda=0$, then $p<\gamma$ on $K$ (use $(y,0)\in M_1$) and $p>\gamma$ on $K$ (use $(y,f(y))\in M$), which is absurd. So $\lambda>0$, and $g=(\gamma-p)/\lambda$ on $K$ satisfies $g_1,g_2<g<f$. $\square$

Definition 1.4 (Envelopes). Let $X\subseteq K$ be nonempty. For $g:X\to[-\infty,+\infty]$ bounded above, the upper envelope is $\begin{gathered} \overline g(x)\\ =\inf\{a(x):\\ \ a\in\operatorname{Aff}(K),\ a\\ \geq g\ \text{on }X\}\\ (x\in K), \end{gathered} \tag{1.2}$ and for $g$ bounded below the lower envelope is $\underline g(x)=\sup\{a(x):\ a\in\operatorname{Aff}(K),\ a\leq g\ \text{on }X\}$. Both are defined on all of $K$. The set $X$ is arbitrary; we mostly use $X=K$. By Lemma 1.2 the same envelopes arise if $a$ runs only through $\operatorname{Aff}_E(K)$: an $a\in\operatorname{Aff}(K)$ with $a\geq g$ is within $\varepsilon$ of some $b\in\operatorname{Aff}_E(K)$, and then $b+\varepsilon\geq g$ and $b+\varepsilon\leq a+2\varepsilon$.

Proposition 1.5 (Properties of envelopes).

$\underline g\in\mathcal L(K)$, and $-\overline g\in\mathcal L(K)$. That is, $\overline g$ is concave and upper semicontinuous (usc) with values in $[-\infty,+\infty)$.
For $g$ bounded on $X=K$: $\underline g\leq g\leq\overline g$. For $g\in C(K)$: $-\|g\|\leq\underline g\leq g\leq\overline g\leq\|g\|$.
For bounded $g,h$ on $K$: $g\leq h$ implies $\overline g\leq\overline h$; $\overline{g+h}\leq\overline g+\overline h$; $\overline{tg}=t\overline g$ for $t\geq0$; $\overline{g+b}=\overline g+b$ for $b\in\operatorname{Aff}(K)$; and $\overline{-g}=-\underline g$.
If $h\in\mathcal L(K)$ then $\underline h=h$. If $g:K\to[-\infty,\infty)$ and $-g\in\mathcal L(K)$, then $\overline g=g$. In particular $\overline f=f$ for $f\in-\mathcal P(K)$ and $\overline a=a$ for $a\in\operatorname{Aff}(K)$.

Proof. (1) A supremum of continuous affine functions is convex and lsc; the family is nonempty because $g$ is bounded below, so $\underline g>-\infty$. The statement for $\overline g$ is the same applied to $-g$, since $\overline g=-\underline{(-g)}$ directly from the definitions ($a\geq g$ on $X$ exactly when $-a\leq-g$ there). (2) The first claim is immediate from the definitions, and the constants $\pm\|g\|$ are affine. (3) If $a\geq g$ and $b'\geq h$ are affine, then $a+b'\geq g+h$; take infima. For $t>0$, $a\geq g$ exactly when $ta\geq tg$. For $t=0$: the constant $0$ is an affine majorant of the function $0$, and every affine majorant of $0$ is $\geq0$, so $\overline0=0$. The last two claims are direct. (4) The first claim is Lemma 1.1. For the second apply Lemma 1.1 to $-g$ and use (3). $\square$

Remarks 1.6.

($K$ inside $\operatorname{Aff}(K)^*$.) The map $x\mapsto\delta_x$, $\delta_x(a)=a(x)$, maps $K$ affinely and homeomorphically onto a convex set in $\operatorname{Aff}(K)^*$ that is weak$^*$-compact. It is one-to-one because $\operatorname{Aff}_E(K)$ separates points, continuous because each $a$ is continuous, and a homeomorphism because $K$ is compact. So nothing is lost if $E$ is replaced by $\operatorname{Aff}(K)^*$ with its weak$^*$ topology.
(Krein–Milman.) A nonempty compact convex set has an extreme point, and is the closed convex hull of its extreme points.

2. Measures on a compact space

The following facts about measures are used throughout. Here $X$ is a compact Hausdorff space and $\mu\in M^+(X)$.

Proposition 2.1.

(Integrals of lsc functions.) For lsc $f:X\to(-\infty,+\infty]$, $\int f\,d\mu=\sup\{\mu(g):\ g\in C(X),\ g\leq f\}$. Dually, for usc $h:X\to[-\infty,\infty)$, $\int h\,d\mu=\inf\{\mu(g):\ g\in C(X),\ g\geq h\}$.
(Monotone nets.) If $(f_i)$ is an increasing net of lsc functions $X\to(-\infty,+\infty]$ with pointwise supremum $f$, then $\int f\,d\mu=\sup_i\int f_i\,d\mu$.
(Support.) There is a largest open $\mu$-null set. Its complement $\operatorname{supp}\mu$ is closed, carries $\mu$, and every open set that meets it has positive measure. If $F$ is closed and $\mu(X\setminus F)=0$, then $\operatorname{supp}\mu\subseteq F$.
(Baire sets.) A zero set is $Z(h)=h^{-1}(0)$ with $h\in C(X)$; its complement is a cozero set. The Baire $\sigma$-algebra $\mathfrak B_0(X)$ is generated by the zero sets.
- (a) The zero sets are exactly the closed $G_\delta$ sets (equivalently, the compact $G_\delta$ sets). Finite unions and countable intersections of zero sets are zero sets. Countable unions of cozero sets are cozero sets.
- (b) If $\Phi:X\to Y$ is continuous and $Y$ is compact Hausdorff, then $\Phi^{-1}$ maps Baire sets to Baire sets.
- (c) (Regularity.) For every Baire set $B$ and $\varepsilon>0$ there are a zero set $Z\subseteq B$ and a cozero set $V\supseteq B$ with $\mu(V\setminus Z)<\varepsilon$.
- (d) If $X$ is metrizable, every closed set is a zero set, so the Baire sets are the Borel sets.
- (e) Every finite positive measure on $\mathfrak B_0(X)$ is the restriction of a unique regular Borel measure.
(Bounded Radon–Nikodym.) If $\nu\in M^+(X)$ and $\nu\leq c\mu$, then $\nu=g\mu$ for a Borel $g$ with $0\leq g\leq c$.
(Weak$^*$ density.) $C(X)$ is weak$^*$-dense in $L^\infty(X,\mu)$ (real or complex scalars).
(Image measures.) Let $\Phi:X\to Y$ be continuous, $Y$ compact Hausdorff, and $\Phi_*\mu(g)=\mu(g\circ\Phi)$. Then $\Phi_*\mu(N)=\mu(\Phi^{-1}(N))$ for every Borel set $N\subseteq Y$, so $\int g\,d\Phi_*\mu=\int g\circ\Phi\,d\mu$ for bounded Borel $g$. If $\Phi$ is one-to-one, then $G\mapsto G\circ\Phi$ is an isometric $*$-isomorphism of $L^\infty(Y,\Phi_*\mu)$ onto $L^\infty(X,\mu)$.

Proof. (1) If $g\in C(X)$ and $g\leq f$, then $\mu(g)\leq\int f\,d\mu$. For the converse, if $f\equiv+\infty$, both sides are $+\infty$ (both $0$ when $\mu=0$). Otherwise subtract the minimum of $f$ and assume $f\geq0$. Fix $N\geq1$ and $0<\delta<1$, let $m=\lfloor N/\delta\rfloor$, and let $U_k=\{f>k\delta\}$, an open set. The function $s=\delta\sum_{k=1}^m1_{U_k}$ counts the multiples $k\delta$ below $f$ up to $m\delta$, so $\min(f,N)-\delta\leq s\leq f$. Inner regularity on open sets and Urysohn's lemma give $g_k\in C(X)$ with $0\leq g_k\leq1_{U_k}$ and $\mu(g_k)\geq\mu(U_k)-\delta/m$. Then $g=\delta\sum_kg_k\leq s\leq f$ and $\begin{gathered} \mu(g)\\ \geq\int s\,d\mu-\delta^2\\ \geq\int\min(f,N)\,d\mu-\delta\mu(X)-\delta \end{gathered}$ . Let $\delta\to0$, then $N\to\infty$ (monotone convergence). The usc statement follows by applying this to $-h$.

(2) Clearly $\int f_i\leq\int f$. Let $g\in C(X)$ with $g\leq f$, and $\varepsilon>0$. The sets $\{f_i>g-\varepsilon\}$ are open, increase with $i$, and cover $X$, since $\sup_if_i(x)=f(x)>g(x)-\varepsilon$. By compactness one of them is $X$, so $\int f_i\geq\mu(g)-\varepsilon\mu(X)$. Now use (1).

(3) Let $U$ be the union of all open null sets. A compact subset of $U$ is covered by finitely many of them, so it is null, and inner regularity gives $\mu(U)=0$. The rest is immediate.

(4)(a) $Z(h)=\bigcap_n\{|h|<1/n\}$ is a closed $G_\delta$. Conversely, if $C=\bigcap_nU_n$ is closed with $U_n$ open, take Urysohn functions $h_n:X\to[0,1]$ with $h_n=0$ on $C$ and $h_n=1$ off $U_n$; then $C=Z(\sum_n2^{-n}h_n)$. Also $Z(h_1)\cup Z(h_2)=Z(h_1h_2)$ and $\bigcap_nZ(h_n)=Z(\sum_n2^{-n}\min(|h_n|,1))$; take complements for cozero sets. (b) $\Phi^{-1}(Z(h))=Z(h\circ\Phi)$, and the sets whose preimage is Baire form a $\sigma$-algebra. (c) The sets $B$ with the property form a $\sigma$-algebra containing the zero sets. A zero set $Z(h)$ is the decreasing intersection of the cozero sets $\{|h|<1/n\}$, so continuity from above works. Complements: if $Z\subseteq B\subseteq V$, then $V^c\subseteq B^c\subseteq Z^c$ with the same difference. Countable unions: choose $Z_k\subseteq B_k\subseteq V_k$ with $\mu(V_k\setminus Z_k)<\varepsilon2^{-k-1}$, put $V=\bigcup V_k$, and $Z=Z_1\cup\dots\cup Z_n$ with $n$ so large that $\mu(\bigcup_kZ_k\setminus Z)<\varepsilon/2$. (d) A closed $F$ is $Z(\operatorname{dist}(\cdot,F))$. (e) Let $\mu_0$ be such a measure. Continuous functions are Baire measurable, so $g\mapsto\int g\,d\mu_0$ lies in $M^+(X)$ and has a Radon measure $\mu$. For a zero set $Z(h)$, the functions $\max(0,1-n|h|)\in C(X)$ decrease to $1_{Z(h)}$, so $\mu(Z)=\mu_0(Z)$ by dominated convergence. Zero sets are closed under finite intersections and generate $\mathfrak B_0(X)$, so $\mu=\mu_0$ on $\mathfrak B_0(X)$ by the $\pi$–$\lambda$ theorem. Uniqueness is part of the Riesz theorem.

(5) The functional $g\mapsto\int g\,d\nu$ on $L^2(X,\mu)$ is well defined and bounded, since $\int|g|\,d\nu\leq c\int|g|\,d\mu\leq c\,\mu(X)^{1/2}\|g\|_2$. By the Riesz lemma in Hilbert space it is $g\mapsto\int g\bar u\,d\mu$ with $u\in L^2(\mu)$. With $w=\bar u$ and indicator functions, $\nu(B)=\int_Bw\,d\mu$ for every Borel set $B$. Since $0\leq\nu(B)\leq c\mu(B)$ for all $B$, we get $0\leq w\leq c$ almost everywhere.

(6) The weak$^*$ topology on $L^\infty(\mu)=L^1(\mu)^*$ is locally convex, and its continuous linear functionals are the maps $f\mapsto\int fu\,d\mu$ with $u\in L^1(\mu)$; both facts are recalled in the background section. If some $f$ were outside the weak$^*$ closure of $C(X)$, separation would give $u\in L^1(\mu)$ with $\int gu\,d\mu=0$ for all $g\in C(X)$ and $\int fu\,d\mu\neq0$. But the measure $u\mu$ has a finite Radon variation measure $|u|\mu$, by the small-set integral lemma and its regularity consequence, and it annihilates $C(X)$, so $u\mu=0$ by the uniqueness in the Riesz theorem, and $u=0$ almost everywhere. This is a contradiction.

(7) The Borel measure $\nu(N)=\mu(\Phi^{-1}(N))$ on $Y$ is regular. Given $N$ and $\varepsilon>0$, regularity of $\mu$ gives a compact $C\subseteq\Phi^{-1}(N)$ with $\mu(\Phi^{-1}(N)\setminus C)<\varepsilon$. Then $\Phi(C)\subseteq N$ is compact, and $\nu(N\setminus\Phi(C))\leq\mu(\Phi^{-1}(N)\setminus C)<\varepsilon$ because $C\subseteq\Phi^{-1}(\Phi(C))$. Outer regularity follows by passing to complements. Since $\nu$ integrates $g\in C(Y)$ as $\mu(g\circ\Phi)$, the uniqueness in the Riesz theorem gives $\nu=\Phi_*\mu$. The integral formula follows for simple functions, and then for uniform limits of them. For one-to-one $\Phi$, the map $G\mapsto G\circ\Phi$ is therefore well defined, injective and isometric on classes. It is onto: $\Phi$ is a homeomorphism of $X$ onto the compact set $\Phi(X)$, so for a bounded Borel $g$ on $X$ the function $G=g\circ\Phi^{-1}$ on $\Phi(X)$, extended by $0$, is Borel, and $G\circ\Phi=g$. $\square$

3. Barycentres

A probability measure on $K$ has a centre of mass in $K$, its barycentre. This section constructs it and collects its basic properties.

Lemma 3.1 (The barycentre exists). For every $\mu\in M_1^+(K)$ there is exactly one $y\in K$ with $p(y)=\int_Kp\,d\mu\qquad\text{for all }p\in E^*.$

Proof. Uniqueness holds because $E^*$ separates points. For a finite set $F\subseteq E^*$ let $K_F=\{y\in K:\ p(y)=\mu(p)\text{ for }p\in F\}$, a closed subset of $K$. It is nonempty. Indeed, let $T(y)=(p(y))_{p\in F}\in\mathbb R^F$; $T(K)$ is compact and convex. If $m=(\mu(p))_{p\in F}$ were not in $T(K)$, separation in $\mathbb R^F$ would give $c\in\mathbb R^F$ with $\sum_pc_pm_p>\max_{y\in K}\sum_pc_pp(y)$. But $\sum_pc_pm_p=\mu(\sum_pc_pp)$ is at most that maximum, since $\mu$ is a probability measure. Since $K_{F_1}\cap K_{F_2}=K_{F_1\cup F_2}$, the family $(K_F)$ has the finite intersection property, and compactness gives a point in all of them. $\square$

Definition 3.2 (Barycentre). This $y$ is the barycentre (or resultant) $r(\mu)$ of $\mu$, and $\mu$ represents $y$. We write $M_y(K)$ for the set of probability measures on $K$ that represent $y$.

Both sides of the next formula are continuous in $a$ for the uniform norm and agree on $\operatorname{Aff}_E(K)$ (use $\mu(1)=1$). By Lemma 1.2 they agree on $\operatorname{Aff}(K)$: $\begin{gathered} a(r(\mu))\\ =\int_Ka\,d\mu\\ (a\in\operatorname{Aff}(K)). \end{gathered} \tag{3.1}$

Proposition 3.3 (Properties of barycentres). Let $\mu\in M_1^+(K)$ and $x=r(\mu)$.

If $C\subseteq K$ is closed and convex and $\mu(K\setminus C)=0$, then $x\in C$.
(Jensen's inequality.) $g(x)\leq\int g\,d\mu$ for every $g\in\mathcal L(K)$, and $h(x)\geq\int h\,d\mu$ whenever $-h\in\mathcal L(K)$.
(Bauer's criterion.) A point $x\in K$ is extreme if and only if $M_x(K)=\{\delta_x\}$.
$r:M_1^+(K)\to K$ is affine, continuous and onto.

Proof. (1) If $x\notin C$, separation gives $p\in E^*$ with $p(x)>\sup_Cp$. But $p(x)=\int_Cp\,d\mu\leq\sup_Cp$.

(2) For each $a\in\operatorname{Aff}_E(K)$ with $a\leq g$, (3.1) gives $a(x)=\mu(a)\leq\int g\,d\mu$. Take the supremum and use (1.1). The second claim is the first for $-h$.

(3) If $x=ty+(1-t)z$ with $y\neq z$ and $0<t<1$, then $t\delta_y+(1-t)\delta_z$ is a second measure in $M_x(K)$. Conversely let $x$ be extreme, $\mu\in M_x(K)$, and suppose $\mu\neq\delta_x$. The support of $\mu$ carries $\mu$ (Proposition 2.1(3)), and $\mu$ is a probability measure other than $\delta_x$; so $\operatorname{supp}\mu$ is not contained in $\{x\}$, and there is $y\in\operatorname{supp}\mu$ with $y\neq x$. Pick a closed convex neighbourhood $W$ of $y$ in $E$ with $x\notin W$, and put $C=W\cap K$ and $t=\mu(C)>0$. If $t=1$, then $x\in C$ by (1), which is false. If $0<t<1$, the probability measures $\mu_1=t^{-1}\mu|_C$ and $\mu_2=(1-t)^{-1}\mu|_{K\setminus C}$ have barycentres $x_1\in C$ (by (1)) and $x_2\in K$, and $x=tx_1+(1-t)x_2$ by (3.1), because $E^*$ separates points. Extremality gives $x=x_1\in C$, again false.

(4) $r$ is affine by (3.1), and $r(\delta_x)=x$. For $p\in E^*$, $p\circ r(\mu)=\mu(p|_K)$ is weak$^*$-continuous in $\mu$. So $r$ is continuous into $K$ with the weak topology $\sigma(E,E^*)$. On the compact set $K$ this topology coincides with the given one, since the identity from $K$ to $(K,\sigma(E,E^*))$ is a continuous bijection from a compact space onto a Hausdorff space. $\square$

4. The Choquet order

The Choquet order compares two measures through their integrals of continuous convex functions. We first need two facts about convex functions.

Lemma 4.1 (Differences of convex functions). $\mathcal P(K)-\mathcal P(K)$ is a linear subspace of $C(K)$, dense for the sup norm, and it is closed under $\vee$ and $\wedge$.

Proof. $\mathcal P(K)$ is a convex cone that is closed under $\vee$. So $\mathcal P(K)-\mathcal P(K)$ is a linear subspace. If $f_i=g_i-h_i$ with $g_i,h_i\in\mathcal P(K)$, then $\begin{gathered} f_1\vee f_2\\ =\bigl((g_1+h_2)\vee(g_2+h_1)\bigr)\\ -(h_1+h_2)\\ \in\mathcal P(K)-\mathcal P(K), \end{gathered}$ and $f_1\wedge f_2=-((-f_1)\vee(-f_2))$. The subspace contains the constants and $\operatorname{Aff}_E(K)$, which separates the points of $K$. By Stone's lattice lemma, a linear subspace of $C(K)$ that is closed under $\vee$ and $\wedge$, contains the constants and separates points is dense. $\square$

Lemma 4.2 (Lower semicontinuous convex functions as limits). Let $f\in\mathcal L(K)$. The set $D_f$ of functions $a_1\vee\dots\vee a_n$, with $n\geq1$ and $a_i\in\operatorname{Aff}_E(K)$, $a_i\leq f$, is a subset of $\mathcal P(K)$ that is directed upward and has pointwise supremum $f$. So $f$ is the pointwise limit of an increasing net in $\mathcal P(K)$.

Proof. $D_f$ is nonempty, because some constant lies below $f$ (see the proof of Lemma 1.1), and it is directed by $\vee$. Its supremum is $f$ by (1.1). $\square$

Corollary 4.3. Let $\mu,\nu\in M^+(K)$ with $\mu(f)\leq\nu(f)$ for every $f\in\mathcal P(K)$. Then $\int g\,d\mu\leq\int g\,d\nu$ for every $g\in\mathcal L(K)$, and $\int h\,d\mu\geq\int h\,d\nu$ whenever $-h\in\mathcal L(K)$.

Proof. Let $g\in\mathcal L(K)$, and index the increasing net of Lemma 4.2 by $D_g$ itself. Its members are continuous, so the monotone-net property, Proposition 2.1(2), applies to both measures and gives $\int g\,d\mu=\sup_{d\in D_g}\mu(d)\leq\sup_{d\in D_g}\nu(d)=\int g\,d\nu$. The second claim is the first for $g=-h$. $\square$

Definition 4.4 (The Choquet order). For $\mu,\nu\in M(K)$ we write $\mu\prec\nu$ ($\nu$ majorizes $\mu$) if $\mu(f)\leq\nu(f)$ for every $f\in\mathcal P(K)$. We write $\mu\sim\nu$ if $\mu(a)=\nu(a)$ for every $a\in\operatorname{Aff}(K)$. A measure is maximal if it is maximal for $\prec$ in $M^+(K)$.

Proposition 4.5.

$\prec$ is a partial order on $M(K)$, and $\mu\prec\nu$ implies $\mu\sim\nu$.
If $\mu,\nu\in M^+(K)$ and $\mu\prec\nu$, then $\mu(1)=\nu(1)$; if $\mu$ is a probability measure, so is $\nu$, and $r(\mu)=r(\nu)$.
$\delta_{r(\mu)}\prec\mu$ for every $\mu\in M_1^+(K)$.

Proof. (1) Reflexivity and transitivity are clear. If $\mu\prec\nu\prec\mu$, the two measures agree on $\mathcal P(K)$, hence on $C(K)$, because $\mathcal P(K)-\mathcal P(K)$ is dense (Lemma 4.1). If $\mu\prec\nu$ and $a\in\operatorname{Aff}(K)$, then $a$ and $-a$ are both convex, so $\mu(a)=\nu(a)$. (2) $\pm1\in\mathcal P(K)$, so $\mu(1)=\nu(1)$; and barycentres are determined by the values on $\operatorname{Aff}(K)$, which agree by (1). (3) is Jensen's inequality, Proposition 3.3(2). $\square$

Lemma 4.6 (Maximal measures exist). Every chain in $(M^+(K),\prec)$ has an upper bound. Hence every $\nu\in M^+(K)$ is majorized by a maximal measure.

Proof. Let $\mathcal T$ be a nonempty chain, indexed by itself as a directed set. All its members have the same mass $c$, by Proposition 4.5(2). For $f\in\mathcal P(K)$, the net $(\mu(f))_{\mu\in\mathcal T}$ increases and is bounded by $c\|f\|$, so it converges. By linearity the net converges for $f\in\mathcal P(K)-\mathcal P(K)$. For $f\in C(K)$ and $\varepsilon>0$ pick such a $g$ with $\|f-g\|<\varepsilon$; then $|\mu(f)-\mu'(f)|\leq2c\varepsilon+|\mu(g)-\mu'(g)|$, so the net is Cauchy. The limit $\lambda(f)$ is a positive linear functional, so $\lambda\in M^+(K)$, and $\mu\prec\lambda$ for every $\mu\in\mathcal T$. Zorn's lemma, applied to $\{\mu:\nu\prec\mu\}$, gives a maximal element there. It is maximal in $M^+(K)$, because anything above it is above $\nu$. $\square$

Lemma 4.7. For every $\mu\in M^+(K)$ and $f\in C(K)$ there is $\nu\in M^+(K)$ with $\mu\prec\nu$ and $\nu(f)=\mu(\overline f)$.

Proof. For $g\in C(K)$, $\overline g$ is a bounded usc function (Proposition 1.5), so $\Psi(g)=\int\overline g\,d\mu$ is defined. By Proposition 1.5(3), $\Psi$ is sublinear: $\Psi(g+h)\leq\Psi(g)+\Psi(h)$ and $\Psi(tg)=t\Psi(g)$ for $t\geq0$. On the line $\mathbb Rf$ put $\nu_0(tf)=t\mu(\overline f)$. Then $\nu_0\leq\Psi$ there. For $t\geq0$ this is equality. For $t<0$, $0=\overline{tf+|t|f}\leq\overline{tf}+|t|\overline f$ gives $t\overline f\leq\overline{tf}$, and we integrate. By the Hahn–Banach extension theorem, $\nu_0$ extends to a linear $\nu$ on $C(K)$ with $\nu\leq\Psi$. If $g\leq0$, then $\overline g\leq0$, so $\nu(g)\leq0$: $\nu$ is positive. If $g\in\mathcal P(K)$, then $\overline{-g}=-g$ (Proposition 1.5(4)), so $-\nu(g)=\nu(-g)\leq\mu(\overline{-g})=-\mu(g)$, that is, $\mu(g)\leq\nu(g)$. $\square$

Corollary 4.8 (The upper envelope as a maximum). For $x\in K$ and $f\in C(K)$, $\overline f(x)=\max\{\nu(f):\ \nu\in M_x(K)\}.$ If $f\in\mathcal P(K)$, the maximum is attained at a maximal measure.

Proof. For $\nu\in M_x(K)$, Jensen's inequality for the usc concave $\overline f$ gives $\nu(f)\leq\nu(\overline f)\leq\overline f(x)$. Lemma 4.7 with $\mu=\delta_x$ gives $\nu\succ\delta_x$ with $\nu(f)=\overline f(x)$, and $\nu\in M_x(K)$ by Proposition 4.5(2). If $f$ is convex, a maximal $\nu'\succ\nu$ (Lemma 4.6) lies in $M_x(K)$ and has $\nu'(f)\geq\nu(f)$. $\square$

Remark 4.9. For nonconvex $f$ the maximum need not be attained at a maximal measure. Let $K$ be a square, $x$ its centre, and $f\in C(K)$ a bump with $f(x)=1$ that vanishes near the four vertices. Then $\overline f(x)\geq f(x)=1$, but every maximal measure in $M_x(K)$ lives on the vertices (Example 7.5) and gives $f$ the value $0$.

The next lemma describes the Choquet order through decompositions of measures. It is used in Sections 14, 15 and 17.

Lemma 4.10 (The order through decompositions). For $\mu,\nu\in M^+(K)$ the following are equivalent.

$\mu\prec\nu$.
Whenever $\mu=\sum_{i=1}^n\mu_i$ with $\mu_i\in M^+(K)$, there are $\nu_i\in M^+(K)$ with $\nu=\sum_i\nu_i$ and $\nu_i\sim\mu_i$ for each $i$.

If (1) holds, the $\nu_i$ in (2) can be chosen with $\mu_i\prec\nu_i$.

Proof. (1)$\Rightarrow$(2): On $C(K)^n$ with the norm $\|\vec f\|=\max_i\|f_i\|$ put $\Psi(\vec f)=\sum_i\mu_i(\overline{f_i})$. It is sublinear by Proposition 1.5(3), as in the proof of Lemma 4.7. On the diagonal $\Delta=\{(f,\dots,f)\}$ put $\Psi_0(f,\dots,f)=\nu(f)$. Then $\begin{gathered} \Psi_0(f,\dots,f)\\ =\nu(f)\\ \leq\nu(\overline f)\\ \leq\mu(\overline f)\\ =\Psi(f,\dots,f), \end{gathered}$ where the middle step is Corollary 4.3 applied to $-\overline f\in\mathcal L(K)$, and the last step uses $\sum_i\mu_i=\mu$. The Hahn–Banach theorem gives a linear extension $\Lambda$ of $\Psi_0$ to $C(K)^n$ with $\Lambda\leq\Psi$. Since $\|\overline{f_i}\|_\infty\leq\|f_i\|$, we get $\Lambda(\vec f)\leq\sum_i\|\mu_i\|\|f_i\|\leq\|\mu\|\|\vec f\|$; with $-\vec f$, $|\Lambda(\vec f)|\leq\|\mu\|\|\vec f\|$. So $\Lambda$ is bounded, each map $f\mapsto\Lambda(0,\dots,f,\dots,0)$ is a measure $\nu_i\in M(K)$, and $\Lambda(\vec f)=\sum_i\nu_i(f_i)$. Moreover $\sum_i\|\nu_i\|=\|\Lambda\|\leq\|\mu\|$ (choose $f_i$ of norm $\leq1$ with $\nu_i(f_i)$ close to $\|\nu_i\|$). With $\vec1=(1,\dots,1)$, and $\nu(1)=\mu(1)$ by Proposition 4.5(2): $\begin{gathered} \sum_i\nu_i(1)\\ =\Lambda(\vec1)\\ =\nu(1)\\ =\mu(1)\\ =\|\mu\|\\ \geq\sum_i\|\nu_i\| \end{gathered}$ . Since $\nu_i(1)\leq\|\nu_i\|$ for each $i$, all these are equalities, and a real measure with $\nu_i(1)=\|\nu_i\|$ is positive. For $f\in C(K)$, $\nu_i(f)=\Lambda(0,\dots,f,\dots,0)\leq\mu_i(\overline f)$ (note $\overline0=0$). For $f\in\mathcal P(K)$, applying this to $-f$ and using $\overline{-f}=-f$ gives $\mu_i(f)\leq\nu_i(f)$. So $\mu_i\prec\nu_i$, hence $\mu_i\sim\nu_i$. Finally $\sum_i\nu_i=\nu$ because $\Lambda$ extends $\Psi_0$.

(2)$\Rightarrow$(1): Let $f\in\mathcal P(K)$ and $\varepsilon>0$. Each point of $K$ has a closed convex neighbourhood in $E$ on which $f$ varies by less than $\varepsilon$ (continuity and local convexity). By compactness there are closed convex $G_1,\dots,G_n\subseteq K$ that cover $K$ with $|f(y)-f(z)|<\varepsilon$ for $y,z\in G_i$. Let $A_i=G_i\setminus\bigcup_{j<i}G_j$ and $\mu_i=\mu|_{A_i}$. By (2), $\nu=\sum_i\nu_i$ with $\nu_i\sim\mu_i$; so $\nu_i(K)=\mu_i(K)$. For $\mu_i\neq0$, the normalized $\mu_i$ and $\nu_i$ have the same barycentre $x_i$, and $x_i\in G_i$ by Proposition 3.3(1). Then $\mu_i(f)\leq\mu_i(K)(f(x_i)+\varepsilon)=\nu_i(K)(f(x_i)+\varepsilon)$, and Jensen's inequality gives $\nu_i(K)f(x_i)\leq\nu_i(f)$. Summing, $\mu(f)\leq\nu(f)+\varepsilon\nu(K)$. $\square$

5. Boundary measures

Maximal measures are the measures that have been "pushed out" as far as the Choquet order allows. This section identifies them as the measures carried by the sets on which the upper envelopes of continuous functions agree with the functions.

Definition 5.1 (Boundary sets). For $f\in C(K)$ the boundary set is $B_f=\{x\in K:\ \overline f(x)=f(x)\}$, with the upper envelope $\overline f$ of (1.2).

Since $\overline f-f\geq0$ is usc, $B_f=\bigcap_n\{\overline f-f<1/n\}$ is a $G_\delta$ set.

Lemma 5.2. $\partial_eK=\bigcap_{f\in C(K)}B_f$.

Proof. Let $x\in\partial_eK$ and $f\in C(K)$. Corollary 4.8 gives $\nu\in M_x(K)$ with $\nu(f)=\overline f(x)$. By Bauer's criterion, Proposition 3.3(3), $\nu=\delta_x$; so $\overline f(x)=f(x)$.

Conversely, let $x$ lie in every $B_f$, and let $\nu\in M_x(K)$. For $f\in C(K)$ we have $x\in B_f\cap B_{-f}$, and $\overline{-f}=-\underline f$, so $\underline f(x)=f(x)=\overline f(x)$. Jensen's inequality for the lsc convex $\underline f$ and the usc concave $\overline f$ gives $\begin{gathered} f(x)\\ =\underline f(x)\\ \leq\nu(\underline f)\\ \leq\nu(f)\\ \leq\nu(\overline f)\\ \leq\overline f(x)\\ =f(x). \end{gathered}$ So $\nu(f)=f(x)$ for all $f$, that is, $\nu=\delta_x$. By Bauer's criterion, $x$ is extreme. $\square$

Definition 5.3 (Boundary measures). $\mu\in M(K)$ is a boundary measure if $|\mu|(K\setminus B_f)=0$ for every $f\in C(K)$.

Theorem 5.4.

A measure $\mu\in M^+(K)$ is maximal if and only if it is a boundary measure.
Every point of $K$ is the barycentre of a boundary probability measure.

Proof. (1) Let $\mu$ be maximal and $f\in C(K)$. Lemma 4.7 gives $\nu\succ\mu$ with $\nu(f)=\mu(\overline f)$; maximality gives $\nu=\mu$. So $\int(\overline f-f)\,d\mu=0$ with $\overline f-f\geq0$, and $\mu(K\setminus B_f)=0$.

Conversely let $\mu$ be a boundary measure and $\nu\in M^+(K)$ with $\mu\prec\nu$. For $f\in\mathcal P(K)$: $\nu(f)\leq\nu(\overline f)\leq\mu(\overline f)=\mu(f)\leq\nu(f).$ The first step is $f\leq\overline f$. The second is Corollary 4.3, applied to $-\overline f\in\mathcal L(K)$. The third holds because $\overline f=f$ $\mu$-almost everywhere, and the fourth is $\mu\prec\nu$. So $\mu$ and $\nu$ agree on $\mathcal P(K)$, and $\mu=\nu$ because $\mathcal P(K)-\mathcal P(K)$ is dense (Lemma 4.1).

(2) By Lemma 4.6, $\delta_x$ is majorized by a maximal $\mu$. By (1) it is a boundary measure, and by Proposition 4.5(2) it represents $x$. $\square$

Proposition 5.5 (The boundary measures).

The positive boundary measures form a convex cone. If $0\leq\nu\leq\mu$ and $\mu$ is a boundary measure, so is $\nu$. A measure $\mu$ is a boundary measure exactly when $\mu^+$ and $\mu^-$ are, and the boundary measures form a linear subspace of $M(K)$ that is closed under $\vee$ and $\wedge$.
If $|\mu|$ is concentrated on a Borel set $B\subseteq\partial_eK$, then $\mu$ is a boundary measure.

Proof. (1) $|\mu+\nu|\leq|\mu|+|\nu|$, $|t\mu|=|t||\mu|$, and $\mu^\pm\leq|\mu|=\mu^++\mu^-$. The lattice operations of $M(K)$ satisfy $\mu\vee\nu=\tfrac12(\mu+\nu+|\mu-\nu|)$. (2) $B\subseteq\partial_eK\subseteq B_f$ by Lemma 5.2. $\square$

6. The metrizable case and Baire sets

A boundary measure is carried by every set $B_f$. If one of these sets equals $\partial_eK$, boundary measures are carried by the extreme points; this happens when $K$ is metrizable. For general $K$ they still vanish on every Baire set that misses $\partial_eK$.

Definition 6.1 (Strictly convex functions). $f:K\to\mathbb R$ is strictly convex if $f(tx+(1-t)y)<tf(x)+(1-t)f(y)$ whenever $x\neq y$ and $0<t<1$.

Lemma 6.2. If $f\in\mathcal P(K)$ is strictly convex, then $\partial_eK=B_f$. Hence if a strictly convex continuous function exists, $\partial_eK$ is a $G_\delta$ set.

Proof. $\partial_eK\subseteq B_f$ by Lemma 5.2. If $x=ty+(1-t)z$ with $y\neq z$ and $0<t<1$, then, since $\overline f$ is concave and $\overline f\geq f$, $\begin{gathered} \overline f(x)\\ \geq t\overline f(y)+(1-t)\overline f(z)\\ \geq tf(y)+(1-t)f(z)>f(x), \end{gathered}$ so $x\notin B_f$. $\square$

Lemma 6.3. If $K$ is metrizable, there is a strictly convex $f\in\mathcal P(K)$.

Proof. A compact metric $K$ has a dense sequence: choose finite $1/n$-nets and take their countable union, as in Lemma 5.0 of the Hilbert-space lesson. First, $C(K)$ is separable. Let $d$ be a metric and $(x_n)$ a dense sequence. The functions $d(\cdot,x_n)$ separate points, so the algebra they generate together with $1$ is dense, by the real Stone–Weierstrass theorem; its combinations with rational coefficients form a countable dense set. A separable metric space has a countable base of balls about a dense sequence with positive rational radii. Intersect these balls with a subset and choose a point in every nonempty intersection; the chosen countable set is dense in that subset. Thus $\mathcal P(K)\setminus\{0\}$ contains a sequence $(f_n)$ that is dense in $\mathcal P(K)$ (the constant $1$ shows that $\mathcal P(K)\neq\{0\}$). Put $f=\sum_n2^{-n}\|f_n\|^{-1}f_n$, a uniformly convergent series of convex functions, so $f\in\mathcal P(K)$.

Suppose $f(x)=tf(y)+(1-t)f(z)$ with $x=ty+(1-t)z$, $y\neq z$, $0<t<1$. Each $f_n$ satisfies "$\leq$", and the weighted sum is an equality, so $f_n(x)=tf_n(y)+(1-t)f_n(z)$ for every $n$. By density this holds for every $g\in\mathcal P(K)$. Take $a\in\operatorname{Aff}_E(K)$ with $a(y)\neq a(z)$. Then $a$ and $a^2$ are in $\mathcal P(K)$, and $\begin{gathered} a(x)^2\\ =ta(y)^2+(1-t)a(z)^2,\\ a(x)\\ =ta(y)+(1-t)a(z), \end{gathered}$ which contradicts the strict convexity of $s\mapsto s^2$. $\square$

Theorem 6.4 (Choquet's theorem). Let $K$ be metrizable. Then $\partial_eK$ is a $G_\delta$ set, and $|\mu|(K\setminus\partial_eK)=0$ for each boundary measure $\mu$. Hence each point of $K$ has a representing probability measure that is carried by $\partial_eK$.

Proof. Take $f$ as in Lemma 6.3. By Lemma 6.2, $B_f=\partial_eK$, and a boundary measure $\mu$ has $|\mu|(K\setminus B_f)=0$. The second claim follows from Theorem 5.4(2). $\square$

Lemma 6.5. Let $K$ be any compact convex set, and $(f_n)$ a sequence in $\mathcal L(K)$ with $\sup_n\sup_Kf_n<\infty$. Put $f=\limsup_nf_n$. Then $\sup_{\partial_eK}f=\sup_Kf$.

Here $K$ need not be metrizable. Metrizability enters the proof only through an auxiliary set $K'$.

Proof. Let $\alpha=\sup_{\partial_eK}f$ and $x\in K$. The $f_n$ are real-valued, since they are bounded above. By Lemma 1.1 there are $a_n\in\operatorname{Aff}_E(K)$ with $a_n\leq f_n$ and $a_n(x)>f_n(x)-1/n$. Each $a_n$ is the restriction of a continuous affine function on $E$, again written $a_n$. Let $\Phi(y)=(a_n(y))_{n\geq1}\in\mathbb R^{\mathbb N}$. With the product topology, $\mathbb R^{\mathbb N}$ is locally convex and has a compatible metric. The map $\Phi$ is affine and continuous, so $K'=\Phi(K)$ is compact, convex and metrizable. Let $p_n$ be the $n$-th coordinate, so $a_n=p_n\circ\Phi$.

Let $y'\in\partial_eK'$. The set $K\cap\Phi^{-1}(y')$ is a nonempty closed face of $K$ (see the Conventions), so by the Krein–Milman theorem it has an extreme point $y$, and $y$ is extreme in $K$, because an extreme point of a face is an extreme point of $K$. Hence $\begin{gathered} \limsup_np_n(y')\\ =\limsup_na_n(y)\\ \leq\limsup_nf_n(y)\\ =f(y)\\ \leq\alpha . \end{gathered}$ By Theorem 6.4, applied to the metrizable set $K'$, there is $\mu\in M_1^+(K')$ with $r(\mu)=\Phi(x)$ and $\mu(K'\setminus\partial_eK')=0$. The $p_n$ are bounded above on $K'$, uniformly in $n$, because the $a_n$ are bounded above by the $f_n$. By Fatou's lemma applied to $C-p_n\geq0$, $\begin{gathered} f(x)\\ =\limsup_na_n(x)\\ =\limsup_n\int_{K'}p_n\,d\mu\\ \leq\int_{\partial_eK'}\limsup_np_n\,d\mu\\ \leq\alpha . \end{gathered}$ The first equality uses $f_n(x)-1/n<a_n(x)\leq f_n(x)$, and the second uses $p_n(\Phi(x))=\int p_n\,d\mu$, since $p_n$ is linear. $\square$

Theorem 6.6. If $\mu$ is a boundary measure on $K$, then $|\mu|(B)=0$ for every Baire set $B\subseteq K$ with $B\cap\partial_eK=\varnothing$.

Proof. $|\mu|$ is a positive boundary measure, since Definition 5.3 depends only on $|\mu|$. First let $C$ be a zero set disjoint from $\partial_eK$, say $C=Z(h)$ with $h\geq0$. The functions $f_n=\max(0,1-nh)\in C(K)$ satisfy $0\leq f_n\leq1$, $f_n=1$ on $C$, and $f_n\to0$ pointwise off $C$. The lower envelopes $\underline{f_n}$ lie in $\mathcal L(K)$ and are bounded by $1$. Since $\overline{-f_n}=-\underline{f_n}$, the boundary set $B_{-f_n}$ is $\{\underline{f_n}=f_n\}$; it contains $\partial_eK$ (Lemma 5.2) and carries $|\mu|$. So on $\partial_eK$ we have $\limsup_n\underline{f_n}=\limsup_nf_n=0$, and Lemma 6.5 gives $\limsup_n\underline{f_n}\leq0$ on $K$. By Fatou's lemma, $\begin{gathered} |\mu|(C)\\ \leq\limsup_n|\mu|(f_n)\\ =\limsup_n|\mu|(\underline{f_n})\\ \leq|\mu|\bigl(\limsup_n\underline{f_n}\bigr)\\ \leq0 . \end{gathered}$ For a general Baire set $B$ disjoint from $\partial_eK$, every zero set $Z\subseteq B$ is null, and $|\mu|(B)=0$ by the regularity in Proposition 2.1(4)(c). $\square$

Remark 6.7 (The Bishop–de Leeuw theorem). By Theorems 5.4 and 6.6, every maximal probability measure vanishes on every Baire set that misses $\partial_eK$; one says that it is pseudoconcentrated on $\partial_eK$. So every point of $K$ is the barycentre of a probability measure that is pseudoconcentrated on $\partial_eK$. For metrizable $K$ this is the same as being carried by $\partial_eK$ (Theorem 6.4), because then $\partial_eK$ is a $G_\delta$ set and every Borel set is a Baire set. For nonmetrizable $K$ the set $\partial_eK$ need not be a Borel set, even when $K$ is a simplex, so the Baire form is the natural statement.

For a non-Borel extreme boundary, the Baire pseudo-support statement above is the precise conclusion.

7. Simplices

In a triangle every point is a unique convex combination of the vertices; in a square it is not. Simplices are the compact convex sets that behave like the triangle, and the Choquet–Meyer theorem characterizes them by the uniqueness of maximal measures.

Definition 7.1 (Simplex). $K$ is a simplex if $\operatorname{Aff}(K)^*$ is a vector lattice for the dual order: $\psi\geq0$ when $\psi(a)\geq0$ for every $a\geq0$ in $\operatorname{Aff}(K)$.

Lemma 7.2 (Positive functionals on $\operatorname{Aff}(K)$).

$\psi\in\operatorname{Aff}(K)^*$ is positive if and only if $\psi=t\delta_x$ with $t\geq0$ and $x\in K$.
The restriction $R(\mu)=\mu|_{\operatorname{Aff}(K)}$ is a positive linear map of $M(K)$ onto $\operatorname{Aff}(K)^*$, and $R(\mu)=\mu(1)\delta_{r(\mu)}$ for $0\neq\mu\in M^+(K)$.

Proof. (1) Let $\psi\geq0$. From $-\|a\|\leq a\leq\|a\|$, $|\psi(a)|\leq\psi(1)\|a\|$. If $\psi(1)=0$, then $\psi=0$. Otherwise extend $\psi$ by the Hahn–Banach theorem to $\tilde\psi\in C(K)^*$ with $\|\tilde\psi\|=\psi(1)=\tilde\psi(1)$. Such a functional is positive: for $0\leq g\leq1$, $\|1-2g\|\leq1$ gives $\tilde\psi(1)-2\tilde\psi(g)\leq\tilde\psi(1)$. So $\tilde\psi/\psi(1)$ is a probability measure with some barycentre $x$, and $\psi(a)=\psi(1)a(x)$ by (3.1). (2) $R$ is onto because every functional on $\operatorname{Aff}(K)$ extends to $C(K)$ (Hahn–Banach); the formula is (3.1). $\square$

Lemma 7.3 (Riesz decomposition). In a vector lattice, let $p_1,\dots,p_m,q_1,q_2\geq0$ with $\sum_jp_j=q_1+q_2$. Then there are $p_{jk}\geq0$ with $p_j=p_{j1}+p_{j2}$ and $\sum_jp_{jk}=q_k$.

Proof. Induction on $m$; $m=1$ is trivial. Put $P'=\sum_{j\geq2}p_j$, $p_{11}=p_1\wedge q_1$, $p_{12}=p_1-p_{11}$, $r_1=q_1-p_{11}$ and $r_2=q_2-p_{12}$. Then $p_{11},p_{12},r_1\geq0$ and $r_1+r_2=P'$. Also $p_{12}=(p_1-q_1)\vee0$, and $p_1-q_1=q_2-P'\leq q_2$, so $p_{12}\leq q_2\vee0=q_2$ and $r_2\geq0$. Apply the induction hypothesis to $P'=r_1+r_2$. $\square$

Theorem 7.4 (Choquet–Meyer). The following are equivalent.

(a) $K$ is a simplex.
(b) Every $x\in K$ is the barycentre of exactly one boundary probability measure (equivalently, $M_x(K)$ has exactly one maximal element).
(c) $\overline f$ is affine for every $f\in\mathcal P(K)$.

Proof. (a)$\Rightarrow$(c). Let $f\in\mathcal P(K)$ and $x=ty+(1-t)z$ with $0<t<1$. Since $\overline f$ is concave, we must show $\overline f(x)\leq t\overline f(y)+(1-t)\overline f(z)$. By Corollary 4.8, $\overline f(x)=\sup\{\nu(f):\ \nu\in M_x(K)\}$. Discrete measures suffice: given $\nu\in M_x(K)$ and $\varepsilon>0$, cover $K$ by closed convex $G_1,\dots,G_n$ on each of which $f$ varies by less than $\varepsilon$, put $A_i=G_i\setminus\bigcup_{j<i}G_j$, and let $x_i\in G_i$ be the barycentre of $\nu|_{A_i}/\nu(A_i)$ (when $\nu(A_i)>0$). Then $\nu'=\sum_i\nu(A_i)\delta_{x_i}\in M_x(K)$ and $|\nu(f)-\nu'(f)|\leq\varepsilon$. So let $\nu'=\sum_jc_j\delta_{x_j}\in M_x(K)$. In $\operatorname{Aff}(K)^*$ we have $\sum_jc_j\delta_{x_j}=\delta_x=t\delta_y+(1-t)\delta_z$. By the Riesz decomposition (Lemma 7.3) and Lemma 7.2(1), $c_j\delta_{x_j}=c_{j1}\delta_{x_{j1}}+c_{j2}\delta_{x_{j2}}$ with $\sum_jc_{j1}\delta_{x_{j1}}=t\delta_y$ and $\sum_jc_{j2}\delta_{x_{j2}}=(1-t)\delta_z$. Evaluating at $1$ and at affine functions gives $c_j=c_{j1}+c_{j2}$ and $c_jx_j=c_{j1}x_{j1}+c_{j2}x_{j2}$, so convexity of $f$ gives $c_jf(x_j)\leq c_{j1}f(x_{j1})+c_{j2}f(x_{j2})$. The measure $t^{-1}\sum_jc_{j1}\delta_{x_{j1}}$ lies in $M_y(K)$, so $\sum_jc_{j1}f(x_{j1})\leq t\overline f(y)$, and likewise for $z$. Hence $\nu'(f)\leq t\overline f(y)+(1-t)\overline f(z)$.

(c)$\Rightarrow$(b). Existence is Theorem 5.4(2). Let $\mu,\nu$ be boundary probability measures with barycentre $x$, and $f\in\mathcal P(K)$. Then $\mu(f)=\mu(\overline f)$, since $\mu$ is carried by $B_f$, where $\overline f=f$. The function $-\overline f$ is real-valued, lsc and affine, so by Lemma 1.3 it is the limit of an increasing net $(a_i)$ in $\operatorname{Aff}_E(K)$. By the monotone-net property, Proposition 2.1(2), and by (3.1), $\mu(-\overline f)=\sup_i\mu(a_i)=\sup_ia_i(x)=-\overline f(x)$. So $\mu(f)=\overline f(x)=\nu(f)$, and $\mu=\nu$ by Lemma 4.1.

(b)$\Rightarrow$(a). Let $\mathcal B$ be the set of boundary measures, a linear subspace of $M(K)$ closed under $\vee$ and $\wedge$ (Proposition 5.5(1)). So $\mathcal B$ is a vector lattice, and it suffices to show that $R|_{\mathcal B}$ is an order isomorphism onto $\operatorname{Aff}(K)^*$. Onto: given $\psi=R(\mu)$, majorize $\mu^\pm$ by maximal measures $m^\pm$ (Lemma 4.6), which are boundary measures by Theorem 5.4; then $R(m^\pm)=R(\mu^\pm)$ by Proposition 4.5(1), and $\psi=R(m^+-m^-)$. One-to-one: if $\mu\in\mathcal B$ and $R(\mu)=0$, then $\mu^+$ and $\mu^-$ are positive boundary measures with the same mass $c$ and, if $c>0$, the same barycentre; by (b), $\mu^+=\mu^-$, so $\mu=0$. Order: $R$ is positive. If $\mu\in\mathcal B$ and $R(\mu)\geq0$, then $R(\mu)=c\delta_x$ (Lemma 7.2(1)), and $c\,m_x$, with $m_x$ the boundary probability measure at $x$, satisfies $R(cm_x)=R(\mu)$; so $\mu=cm_x\geq0$. $\square$

Example 7.5 (The square and the triangle). Let $K$ be the square with vertices $v_1=(0,0)$, $v_2=(1,0)$, $v_3=(1,1)$, $v_4=(0,1)$, and $c$ its centre. $K$ is metrizable, so maximal measures live on the four vertices (Theorem 6.4), and every measure on the vertices is a boundary measure (Proposition 5.5(2)). A measure $\sum_ip_i\delta_{v_i}$ represents $c$ exactly when $p_1=p_3$, $p_2=p_4$ and $p_1+p_2=\frac12$. So the maximal measures in $M_c(K)$ form the segment $t\cdot\frac12(\delta_{v_1}+\delta_{v_3})+(1-t)\cdot\frac12(\delta_{v_2}+\delta_{v_4})$, $t\in[0,1]$, and the square is not a simplex. For $f(x,y)=(x-y)^2$, which is convex, $\overline f(v_1)=\overline f(v_3)=0$, while $\overline f(c)=\max_t(1-t)=1$ by Corollary 4.8; so $\overline f$ is not affine on the diagonal, as condition (c) of Theorem 7.4 predicts. In a triangle, barycentric coordinates are unique, every point has one maximal measure, and the triangle is a simplex.

Exercise 7.6 (medium; The Choquet order on an interval). Let $K=[0,1]$. Show that for $\mu,\nu\in M_1^+(K)$, $\mu\prec\nu$ if and only if $r(\mu)=r(\nu)$ and $\int(t-s)^+d\mu(t)\leq\int(t-s)^+d\nu(t)$ for every $s\in[0,1]$. Deduce that $(1-x)\delta_0+x\delta_1$ is the only maximal measure in $M_x(K)$.

Solution. Necessity: $t\mapsto(t-s)^+$ is convex, and $\mu\prec\nu$ forces equal barycentres (Proposition 4.5(2)). Sufficiency: a continuous convex $f$ on $[0,1]$ is the uniform limit of its piecewise linear interpolants at the points $k/N$, which are convex and have the form $\alpha+\beta t+\sum_kc_k(t-s_k)^+$ with $c_k\geq0$ (the jumps of the slope). Integrate: the affine part has the same integral against $\mu$ and $\nu$, and each $c_k(t-s_k)^+$ has the larger integral against $\nu$. For the last claim, $\partial_eK=\{0,1\}$, and the only probability measure on $\{0,1\}$ with barycentre $x$ is $(1-x)\delta_0+x\delta_1$; every maximal measure lives on $\{0,1\}$ (Theorem 6.4). So $[0,1]$ is a simplex, by Theorem 7.4.

8. The state space

From now on $A$ is a nonzero unital C$^*$-algebra; for a nonunital algebra a nondegenerate representation extends uniquely to a unital representation of its unitization by $\widetilde\pi(a+\lambda1)=\pi(a)+\lambda I$. Expanding the product and adjoint identities proves that this is a $*$-homomorphism; it is bounded by Theorem 4.2 of the continuous-functional-calculus lesson. A unital extension is unique because the adjoined unit must act as $I$. We work with the nonzero unital algebra throughout. We apply Sections 1–7 with $E=A^*_h$, the real space of hermitian functionals, $\psi(a^*)=\overline{\psi(a)}$, with the weak topology $\sigma(A^*_h,A_h)$. The continuous linear functionals for this topology are the maps $\psi\mapsto\psi(h)$ with $h\in A_h$, by the duality theorem for weak topologies. States are hermitian, because positive functionals are. For $a\in A$ we write $\hat a(\omega)=\omega(a)$, a continuous complex function on $\mathfrak S$, and $\mathcal A_{\mathbb C}=\{\hat a:\ a\in A\}$.

Proposition 8.1.

$\mathfrak S$ is compact and convex in $E$, and $\operatorname{Aff}_E(\mathfrak S)=\{\hat h:\ h\in A_h\}$. A measure $\mu\in M_1^+(\mathfrak S)$ has barycentre $\varphi$ exactly when $\varphi(a)=\int\omega(a)\,d\mu(\omega)$ for all $a\in A$.
(Kadison's representation.) For $h\in A_h$, $\|\hat h\|=\|h\|$, and $h\geq0$ if and only if $\hat h\geq0$. The map $h\mapsto\hat h$ is an isometric order isomorphism of $A_h$ onto $\operatorname{Aff}(\mathfrak S)$.
(Radon–Nikodym map.) Let $\varphi\in\mathfrak S$, and write $\pi,H,\xi$ for $\pi_\varphi,H_\varphi,\xi_\varphi$. Put $\begin{gathered} \Theta_\varphi(x)(a)\\ =\langle\pi(a)x\xi,\xi\rangle\\ (x\in\pi(A)',\ a\in A). \end{gathered}$ Then $\Theta_\varphi:\pi(A)'\to A^*$ is linear and one-to-one, $\Theta_\varphi(x^*)=\Theta_\varphi(x)^*$, and a self-adjoint $x$ is positive if and only if $\Theta_\varphi(x)$ is. It maps $\{x\in\pi(A)':\ 0\leq x\leq1\}$ onto $\{\psi\in A^*:\ 0\leq\psi\leq\varphi\}$.
(Pure states.) For $\varphi\in\mathfrak S$ the following are equivalent: (a) $\varphi\in\partial_e\mathfrak S$; (b) every positive functional $\psi\leq\varphi$ is a multiple of $\varphi$; (c) $\pi_\varphi(A)'=\mathbb C1$. We call such states pure and write $P(A)=\partial_e\mathfrak S$.
Every state $\varphi$ is the barycentre of a maximal probability measure $\mu$ on $\mathfrak S$: $\begin{gathered} \varphi(a)\\ =\int_{\mathfrak S}\omega(a)\,d\mu(\omega)\\ (a\in A), \end{gathered} \tag{8.1}$ and $\mu(B)=0$ for every Baire set $B\subseteq\mathfrak S$ that misses $P(A)$. If $A$ is separable, $\mathfrak S$ is metrizable, $P(A)$ is a $G_\delta$ set, and $\mu$ is concentrated on it: $\begin{gathered} \varphi(a)\\ =\int_{P(A)}\omega(a)\,d\mu(\omega)\\ (a\in A). \end{gathered} \tag{8.2}$

Proof. (1) $\mathfrak S$ is a weak$^*$-closed subset of the unit ball of $A^*$, so it is weak$^*$-compact by the Banach–Alaoglu theorem. On hermitian functionals the topologies $\sigma(A^*,A)$ and $\sigma(A^*_h,A_h)$ agree, since $\psi(h+ik)=\psi(h)+i\psi(k)$. The constant $c$ is $\widehat{c1}$. The last claim follows by complex linearity from the case $a\in A_h$.

(2) Let $\pi$ be a faithful representation of $A$ (the Gelfand–Naimark theorem); it is isometric, being an injective $*$-homomorphism. The norm of the self-adjoint operator $\pi(h)$ equals $\sup_{\|\eta\|=1}|\langle\pi(h)\eta,\eta\rangle|$, and each $a\mapsto\langle\pi(a)\eta,\eta\rangle$ is a state. So $\|\hat h\|\geq\|h\|$; the reverse holds because states have norm one. If $\hat h\geq0$, then $0\leq\omega(h)\leq\|h\|$ for every state, so $\|\,\|h\|1-h\,\|=\sup_\omega|\omega(\|h\|1-h)|\leq\|h\|$. The spectrum of $h$ therefore lies in $[0,2\|h\|]$, and $h\geq0$. The image $\{\hat h\}$ is complete, hence closed, and it is dense by Lemma 1.2. So it is all of $\operatorname{Aff}(\mathfrak S)$.

(3) Linearity is clear. If $\Theta_\varphi(x)=0$, then $\langle x\pi(a)\xi,\pi(b)\xi\rangle=\Theta_\varphi(x)(b^*a)=0$ for all $a,b$, so $x=0$. The adjoint formula is a direct computation. For $x\geq0$, $\Theta_\varphi(x)(a^*a)=\langle x\pi(a)\xi,\pi(a)\xi\rangle\geq0$, and $\Theta_\varphi(x)(a^*a)\leq\|x\|\varphi(a^*a)$. Conversely, if $x=x^*$ and $\Theta_\varphi(x)\geq0$, then $\langle x\eta,\eta\rangle\geq0$ on the dense set $\pi(A)\xi$, so $x\geq0$. Now let $0\leq\psi\leq\varphi$. By the Cauchy–Schwarz inequality $|\psi(b^*a)|^2\leq\psi(a^*a)\psi(b^*b)\leq\|\pi(a)\xi\|^2\|\pi(b)\xi\|^2$. So $(\pi(a)\xi,\pi(b)\xi)\mapsto\psi(b^*a)$ is a well-defined bounded sesquilinear form on $\pi(A)\xi$, and there is an operator $x$ with $0\leq x\leq1$ and $\langle x\pi(a)\xi,\pi(b)\xi\rangle=\psi(b^*a)$. For $c\in A$, $\begin{gathered} \langle x\pi(c)\pi(a)\xi,\pi(b)\xi\rangle\\ =\psi((c^*b)^*a)\\ =\langle\pi(c)x\pi(a)\xi,\pi(b)\xi\rangle \end{gathered}$ , so $x\in\pi(A)'$. With $b=1$ we get $\Theta_\varphi(x)=\psi$.

(4) (a)$\Rightarrow$(c): Let $x\in\pi(A)'$ with $0\leq x\leq1$, and $t=\langle x\xi,\xi\rangle$. If $0<t<1$, then $\varphi=t\,\Theta_\varphi(x)/t+(1-t)\,\Theta_\varphi(1-x)/(1-t)$ is a convex combination of states, so $\Theta_\varphi(x)=t\varphi=\Theta_\varphi(t1)$ and $x=t1$. If $t=0$, then $x^{1/2}\xi=0$; as $\xi$ is separating for $\pi(A)'$, $x=0$. If $t=1$, apply this to $1-x$. So every positive contraction in $\pi(A)'$ is a scalar, and $\pi(A)'=\mathbb C1$. (c)$\Rightarrow$(b) follows from (3). (b)$\Rightarrow$(a): if $\varphi=t\psi_1+(1-t)\psi_2$ with states $\psi_i$ and $0<t<1$, then $t\psi_1\leq\varphi$, so $t\psi_1=\lambda\varphi$, and $\lambda=t$ by evaluating at $1$.

(5) Apply Theorem 5.4(2) and Theorem 6.6 to $K=\mathfrak S$, and use (1) to write the barycentre condition as (8.1). If $(a_n)$ is dense in the unit ball of $A$, then $d(\omega,\omega')=\sum_n2^{-n}|\omega(a_n)-\omega'(a_n)|$ is a metric on $\mathfrak S$ whose topology is weaker than the weak$^*$ topology. A weaker Hausdorff topology on a compact space coincides with it. Now use Theorem 6.4. $\square$

Condition (c) of part (4) says that $\pi_\varphi$ is irreducible. Indeed, a closed subspace is invariant under $\pi_\varphi(A)$ exactly when its projection lies in $\pi_\varphi(A)'$, and a von Neumann algebra whose only projections are $0$ and $1$ is $\mathbb C1$, by the spectral theorem. So a state is pure exactly when its GNS representation is irreducible.

9. The operator map of a representing measure

A representing measure of a state $\varphi$ acts on the GNS space of $\varphi$ through a positive map into the commutant $\pi_\varphi(A)'$. We first give a test for the continuity of such maps.

Operator topology tools

The weak operator topology tests $T\mapsto\langle T\eta,\zeta\rangle$, and the strong operator topology tests $T\mapsto\|T\eta\|$. We will use the following facts for arbitrary Hilbert spaces.

Lemma 9.0.

The closed unit ball of $B(H)$ is compact in the weak operator topology. Consequently every bounded weakly closed set of operators is weakly compact.
The weakly continuous and strongly continuous linear functionals on $B(H)$ are exactly the finite sums of vector functionals. Every convex set has the same weak and strong closures.
On a norm-bounded set, the weak and $\sigma$-weak topologies agree. Multiplication by a fixed operator on either side is continuous for the $\sigma$-weak topology. Multiplication by a fixed $g\in L^\infty(\mu)$ is continuous for $\sigma(L^\infty,L^1)$.

Proof. (1) Map an operator of norm at most one to all its coefficients $\langle T\eta,\zeta\rangle$, in the product of the closed discs of radii $\|\eta\|\|\zeta\|$, indexed by $(\eta,\zeta)\in H^2$. The product is compact by Tychonoff's theorem. Its points belonging to the image are exactly the coefficient arrays that are linear in $\eta$ and conjugate-linear in $\zeta$. These identities are closed conditions. Every such array is a bounded sesquilinear form, with bound one supplied by the discs, and is therefore the coefficient array of an operator of norm at most one by Hilbert representation for bounded forms, Theorem 3.1. The coefficient map is injective, and the weak operator topology is precisely the topology it induces from the product. Its closed image is compact. A bounded weakly closed set lies in a scalar multiple of this compact ball and is closed there.

(2) By the duality theorem for weak topologies, Theorem 1.2, the weakly continuous linear functionals are the finite sums of vector functionals. They are strongly continuous by Cauchy–Schwarz. Conversely, strong continuity of a linear functional $F$ gives finitely many vectors $\eta_1,\dots,\eta_n$ and $C>0$ with $|F(T)|\leq C\max_i\|T\eta_i\|$: use a basic neighbourhood on which $|F|<1$ and scale $T$; when all $T\eta_i=0$, scaling forces $F(T)=0$. Thus $F$ factors as a bounded linear functional on the subspace of $H^n$ consisting of $R(T)=(T\eta_i)_i$. Its bound for the Hilbert product norm follows from $\max_i\|T\eta_i\|\leq\|R(T)\|$. Extend it to $H^n$ by Hahn–Banach, and apply the Riesz–Fréchet theorem, Theorem 2.3, to obtain vectors $\zeta_i$ with $F(T)=\sum_{i=1}^n\langle T\eta_i,\zeta_i\rangle .$ The real continuous linear functionals are the real parts of these: for a real-linear $G$, use the complex-linear functional $F(T)=G(T)-iG(iT)$. Hence the two locally convex topologies also have the same real continuous functionals. Theorem 4.2 on closures of convex sets, whose proof separates a point from a closed convex set, gives the closure equality.

(3) A $\sigma$-weak test is $T\mapsto\sum_n\langle T\eta_n,\zeta_n\rangle$, with the two vector sequences square summable. On a ball of radius $M$, its tail is bounded uniformly by $M\sum_{n>N}\|\eta_n\|\|\zeta_n\|$, which tends to zero by Cauchy–Schwarz. It is a uniform limit there of weakly continuous finite sums, proving the topology equality. For fixed $a,b$, the test after $T\mapsto aTb$ has vectors $b\eta_n,a^*\zeta_n$, again square summable; this proves continuity. Finally, $\int (gf)u\,d\mu=\int f(gu)\,d\mu$, with $gu\in L^1$, proves the last assertion. These are also the bounded-net and multiplication arguments of Lemma 1.2 of the double-commutant lesson. $\square$

Lemma 9.1 (Normality test). Let $\mu\in M^+(X)$ and let $\kappa:L^\infty(X,\mu)\to B(H)$ be a bounded linear map. Suppose that for $\eta,\zeta$ in a dense subspace $D\subseteq H$ the functional $f\mapsto\langle\kappa(f)\eta,\zeta\rangle$ has the form $f\mapsto\int fu\,d\mu$ with $u\in L^1(\mu)$. Then the same holds for all $\eta,\zeta\in H$, and for all sums $f\mapsto\sum_n\langle\kappa(f)\eta_n,\zeta_n\rangle$ with $\sum_n\|\eta_n\|^2<\infty$ and $\sum_n\|\zeta_n\|^2<\infty$. So $\kappa$ is continuous from the weak$^*$ topology of $L^\infty(\mu)$ to the $\sigma$-weak topology; we call it normal.

Proof. The map sending $u\in L^1(\mu)$ to the functional $f\mapsto\int fu\,d\mu$ is isometric (test with $f=\bar u/|u|$ where $u\neq0$), so its range $N$ is norm-closed in $L^\infty(\mu)^*$. If $\eta_k\to\eta$ and $\zeta_k\to\zeta$ with $\eta_k,\zeta_k\in D$, then $\begin{gathered} |\langle\kappa(f)\eta,\zeta\rangle\\ -\langle\kappa(f)\eta_k,\zeta_k\rangle|\\ \leq\|\kappa\|\|f\|(\|\eta-\eta_k\|\|\zeta\|\\ +\|\eta_k\|\|\zeta-\zeta_k\|) \end{gathered}$ , a norm convergence of functionals. So the limit lies in $N$. The series converges in norm because $\sum_n\|\eta_n\|\|\zeta_n\|<\infty$. $\square$

Proposition 9.2. Let $\mu\in M^+_1(\mathfrak S)$ have barycentre $\varphi$, and write $\pi,H,\xi$ for its GNS triple. There is exactly one linear map $\kappa_\mu:L^\infty(\mathfrak S,\mu)\to\pi(A)'$ with $\begin{gathered} \langle\kappa_\mu(f)\pi(a)\xi,\xi\rangle\\ =\int_{\mathfrak S}f(\omega)\,\omega(a)\,d\mu(\omega)\\ (f\in L^\infty(\mathfrak S,\mu),\ a\in A). \end{gathered} \tag{9.1}$ It is positive, $\kappa_\mu(1)=1$, $\|\kappa_\mu(f)\|\leq\|f\|_\infty$ for real $f$, and $\kappa_\mu$ is normal.

Proof. An operator $x\in\pi(A)'$ is determined by the functional $\Theta_\varphi(x)(a)=\langle\pi(a)x\xi,\xi\rangle=\langle x\pi(a)\xi,\xi\rangle$ (Proposition 8.1(3)). This gives uniqueness. For existence let $\kappa'(f)(a)=\int f\hat a\,d\mu$. If $0\leq f\leq c$, then $\kappa'(f)$ is a positive functional and $\kappa'(f)\leq c\varphi$, because $\omega(a^*a)\geq0$. By Proposition 8.1(3) there is a unique $x_f\in\pi(A)'$ with $\Theta_\varphi(x_f)=\kappa'(f)$, and $0\leq x_f\leq c$. Every $f\in L^\infty$ is a combination $\sum_{k=0}^3i^kf_k$ with $f_k\geq0$, so $\kappa'(f)$ lies in the range of $\Theta_\varphi$, and $\kappa_\mu=\Theta_\varphi^{-1}\circ\kappa'$ is linear, positive and satisfies (9.1). $\kappa_\mu(1)=1$ because $\kappa'(1)=\varphi=\Theta_\varphi(1)$. For real $f$, $-\|f\|\leq f\leq\|f\|$ gives $-\|f\|\leq\kappa_\mu(f)\leq\|f\|$. Finally, for $\eta=\pi(a)\xi$ and $\zeta=\pi(b)\xi$, $\langle\kappa_\mu(f)\eta,\zeta\rangle=\langle\kappa_\mu(f)\pi(b^*a)\xi,\xi\rangle=\int f\,\widehat{b^*a}\,d\mu ,$ and $\widehat{b^*a}$ is bounded and continuous. The normality test, Lemma 9.1, applies with $D=\pi(A)\xi$. $\square$

10. Orthogonal measures

A representing measure is orthogonal when its operator map is multiplicative. This section gives equivalent conditions and shows that the image of the operator map is then an abelian von Neumann algebra.

Lemma 10.1 (Positive contractions and projections). Let $p,q\in B(H)$.

If $p$ is a projection and $0\leq x\leq p$, $0\leq x\leq1-p$, then $x=0$.
If $p,q$ are projections and $p+q$ is a projection, then $pq=0$.
If $p,q\geq0$ and $p+q=1$, then $p$ and $q$ commute and $0\leq pq\leq p$, $0\leq pq\leq q$.

Proof. (1) $(1-p)x(1-p)\leq(1-p)p(1-p)=0$, so $x^{1/2}(1-p)=0$ and $x=xp=px$. Likewise $pxp\leq p(1-p)p=0$ gives $xp=0$. So $x=0$. (2) $(p+q)^2=p+q$ gives $pq+qp=0$. Multiply by $p$ on the left, and separately on the right: $pq+pqp=0=pqp+qp$. So $pq=qp$, and then $2pq=0$. (3) $q=1-p$ commutes with $p$, and $pq=p^{1/2}qp^{1/2}\geq0$. Also $p-pq=p(1-q)=p^2\geq0$ and $q-pq=q^2\geq0$. $\square$

Theorem 10.2. Let $\mu\in M_1^+(\mathfrak S)$ with barycentre $\varphi$. The following are equivalent.

$\kappa_\mu$ is multiplicative.
$\kappa_\mu(\chi_E)\kappa_\mu(1-\chi_E)=0$ for every Borel set $E\subseteq\mathfrak S$.
For every Borel set $E$, the functionals $\varphi_E(a)=\int_E\omega(a)\,d\mu(\omega)$ and $\varphi_{E^c}=\varphi-\varphi_E$ are orthogonal: a positive $\psi$ with $\psi\leq\varphi_E$ and $\psi\leq\varphi_{E^c}$ is $0$.

In this case $\kappa_\mu$ is a $*$-isomorphism of $L^\infty(\mathfrak S,\mu)$ onto its range, $\|\kappa_\mu(f)\|=\|f\|_\infty$, and $\|\kappa_\mu(f)\xi_\varphi\|^2=\int|f|^2\,d\mu$.

Proof. (1)$\Rightarrow$(2): $\chi_E(1-\chi_E)=0$.

(2)$\Rightarrow$(3): Put $p=\kappa_\mu(\chi_E)$; then $\kappa_\mu(1-\chi_E)=1-p$ and $p(1-p)=0$, so $p$ is a projection. We have $\varphi_E=\Theta_\varphi(p)$ and $\varphi_{E^c}=\Theta_\varphi(1-p)$. If $0\leq\psi\leq\varphi_E$ and $\psi\leq\varphi_{E^c}$, then $\psi\leq\varphi$, so $\psi=\Theta_\varphi(x)$ with $x\geq0$ (Proposition 8.1(3)). Since $\Theta_\varphi$ reflects order, $x\leq p$ and $x\leq1-p$. By Lemma 10.1(1), $x=0$.

(3)$\Rightarrow$(1): Fix $E$, and let $p=\kappa_\mu(\chi_E)$, $q=\kappa_\mu(1-\chi_E)$. If $x\in\pi(A)'$ and $0\leq x\leq p$, $0\leq x\leq q$, then $\Theta_\varphi(x)$ lies below $\varphi_E$ and $\varphi_{E^c}$, so $x=0$. By Lemma 10.1(3), $x=pq$ is such an element, so $pq=0$, and $p=p(p+q)=p^2$. Thus $\kappa_\mu(\chi_E)$ is a projection for every Borel $E$. If $G\cap H=\varnothing$, then $\kappa_\mu(\chi_G)+\kappa_\mu(\chi_H)=\kappa_\mu(\chi_{G\cup H})$ is a projection, so the two are orthogonal (Lemma 10.1(2)). For Borel $E,F$, write $\chi_E=\chi_{E\cap F}+\chi_{E\setminus F}$ and $\chi_F=\chi_{E\cap F}+\chi_{F\setminus E}$, and expand: $\kappa_\mu(\chi_E)\kappa_\mu(\chi_F)=\kappa_\mu(\chi_{E\cap F})=\kappa_\mu(\chi_E\chi_F).$ So $\kappa_\mu$ is multiplicative on simple functions, which are norm-dense in $L^\infty$; and $\kappa_\mu$ is bounded.

Now let $\kappa_\mu$ be multiplicative. Positivity gives $\kappa_\mu(\bar f)=\kappa_\mu(f)^*$, so $\kappa_\mu$ is a $*$-homomorphism, and $\|\kappa_\mu(f)\xi\|^2=\langle\kappa_\mu(|f|^2)\xi,\xi\rangle=\int|f|^2d\mu$ by (9.1) with $a=1$. Hence $\kappa_\mu$ is one-to-one. For the norm: $\|\kappa_\mu(f)\|^2=\|\kappa_\mu(|f|^2)\|\leq\|f\|_\infty^2$. Conversely, for $\varepsilon>0$ the set $E=\{|f|\geq\|f\|_\infty-\varepsilon\}$ has positive measure, $p=\kappa_\mu(\chi_E)$ is a nonzero projection, and $p\kappa_\mu(|f|^2)p=\kappa_\mu(\chi_E|f|^2)\geq(\|f\|_\infty-\varepsilon)^2p$. So $\|\kappa_\mu(f)p\|\geq\|f\|_\infty-\varepsilon$. $\square$

Definition 10.3. $\mu\in M_1^+(\mathfrak S)$ is orthogonal if the conditions of Theorem 10.2 hold. Its associated abelian algebra is $\mathcal C_\mu=\kappa_\mu(L^\infty(\mathfrak S,\mu))\subseteq\pi_\varphi(A)'$.

The algebra $\mathcal C_\mu$ is commutative, because $\kappa_\mu$ is a $*$-homomorphism on the commutative algebra $L^\infty$. The next proposition shows that it is weakly closed, and it collects tools used later.

Proposition 10.4 (Structure of an orthogonal measure). Let $\mu$ be orthogonal with barycentre $\varphi$, and write $\pi,H,\xi$ for its GNS triple.

The map $V_\mu:f\mapsto\kappa_\mu(f)\xi$, $f\in L^\infty(\mu)$, extends to an isometry $V_\mu:L^2(\mathfrak S,\mu)\to H$, and $\kappa_\mu(g)V_\mu=V_\mu M_g$ for $g\in L^\infty(\mu)$, where $M_g$ is multiplication by $g$.
Let $e_\mu$ be the projection onto $V_\mu(L^2)$. Then $V_\mu(L^2)=[\mathcal C_\mu\xi]$, the closed span of $\mathcal C_\mu\xi$, and $e_\mu\in\mathcal C_\mu'$.
(The $L^2$ criterion.) $\mathcal C_\mu=\{y\in\pi(A)':\ y\xi\in e_\mu H\}$.
$\mathcal C_\mu$ is a von Neumann algebra (commutative), and $\xi$ is separating for it.
For every $a\in A$, $\begin{gathered} \kappa_\mu(\hat a)\xi\\ =e_\mu\pi(a)\xi\\ \text{and}\\ e_\mu\pi(a)e_\mu\\ =\kappa_\mu(\hat a)e_\mu . \end{gathered} \tag{10.1}$

Proof. (1) The isometry is the last claim of Theorem 10.2. The intertwining holds on $L^\infty$ by multiplicativity, and both sides are continuous on $L^2$. (2) $\kappa_\mu(L^\infty)\xi$ is dense in the range of $V_\mu$, which is closed. The range is invariant under the self-adjoint set $\mathcal C_\mu$, so $e_\mu$ commutes with it.

(3) The inclusion "$\subseteq$" is clear. Let $y\in\pi(A)'$ with $y\xi\in e_\mu H$; so $y\xi=V_\mu f$ with $f\in L^2(\mu)$. Let $E_n=\{|f|\leq n\}$. By (1), $\kappa_\mu(\chi_{E_n})y\xi=V_\mu(\chi_{E_n}f)=\kappa_\mu(\chi_{E_n}f)\xi$. Both $\kappa_\mu(\chi_{E_n})y$ and $\kappa_\mu(\chi_{E_n}f)$ lie in $\pi(A)'$, for which $\xi$ is separating, so they are equal. Hence $\|\chi_{E_n}f\|_\infty=\|\kappa_\mu(\chi_{E_n}f)\|\leq\|y\|$ for every $n$, so $|f|\leq\|y\|$ almost everywhere. Then $f\in L^\infty$, $\kappa_\mu(f)\xi=V_\mu f=y\xi$, and $y=\kappa_\mu(f)$.

(4) $\mathcal C_\mu$ is a commutative $*$-algebra. Let $y$ be in its weak closure. Then $y\in\pi(A)'$, and $y\xi$ lies in the weak closure of $\mathcal C_\mu\xi$, hence in $e_\mu H$. By (3), $y\in\mathcal C_\mu$. So $\mathcal C_\mu$ is weakly closed and contains $1$; by von Neumann's bicommutant theorem it is a von Neumann algebra. $\xi$ is separating because $\mathcal C_\mu\subseteq\pi(A)'$.

(5) For $f\in L^\infty$, $\begin{gathered} \langle\kappa_\mu(\hat a)\xi,\kappa_\mu(f)\xi\rangle\\ =\int\bar f\hat a\,d\mu\\ =\langle\kappa_\mu(\bar f)\pi(a)\xi,\xi\rangle\\ =\langle\pi(a)\xi,\kappa_\mu(f)\xi\rangle . \end{gathered}$ So $\kappa_\mu(\hat a)\xi-\pi(a)\xi$ is orthogonal to $e_\mu H$, while $\kappa_\mu(\hat a)\xi\in e_\mu H$. This is the first formula. For the second, let $x\in\mathcal C_\mu$. Then $\begin{gathered} e_\mu\pi(a)e_\mu x\xi\\ =e_\mu x\pi(a)\xi\\ =x\,e_\mu\pi(a)\xi\\ =x\kappa_\mu(\hat a)\xi\\ =\kappa_\mu(\hat a)e_\mu x\xi \end{gathered}$ . The vectors $x\xi$ are dense in $e_\mu H$, and both sides vanish on $(1-e_\mu)H$. $\square$

11. Simplicial measures

Definition 11.1. A measure $\mu\in M_x(K)$ is simplicial if it is an extreme point of the convex set $M_x(K)$.

Proposition 11.2 (Any compact convex set). For $\mu\in M_x(K)$, $\mu$ is simplicial if and only if $\operatorname{Aff}(K)$ is dense in the real space $L^1(K,\mu)$.

Proof. Suppose $\operatorname{Aff}(K)$ is dense, and $\mu=\frac12(\mu_1+\mu_2)$ with $\mu_i\in M_x(K)$. Then $\mu_i\leq2\mu$, so $\mu_i=h_i\mu$ with $0\leq h_i\leq2$ (Proposition 2.1(5)). For $a\in\operatorname{Aff}(K)$, (3.1) gives $\int a(h_i-1)\,d\mu=a(x)-a(x)=0$. The functional $g\mapsto\int g(h_i-1)\,d\mu$ is continuous on $L^1$ and vanishes on a dense set, so $h_i=1$ and $\mu_i=\mu$.

Suppose $\mu$ is simplicial and $\operatorname{Aff}(K)$ is not dense. By the Hahn–Banach theorem and the duality $L^1(\mu)^*=L^\infty(\mu)$ there is a real $h\in L^\infty(\mu)$, $h\neq0$, with $\int ah\,d\mu=0$ for all $a\in\operatorname{Aff}(K)$; scale it so that $\|h\|_\infty\leq1$. Then $\mu\pm h\mu$ are positive, and they lie in $M_x(K)$ because $\int a(1\pm h)\,d\mu=a(x)$. Since $\mu=\frac12((\mu+h\mu)+(\mu-h\mu))$, extremality gives $h\mu=0$, which is false. $\square$

Proposition 11.3. Let $\mu\in M_\varphi(\mathfrak S)$, and consider: (i) $\mu$ is orthogonal; (ii) $\mathcal A_{\mathbb C}=\{\hat a:\ a\in A\}$ is dense in $L^1(\mathfrak S,\mu)$ (complex scalars); (iii) $\mu$ is simplicial. Then (i)$\Rightarrow$(ii)$\Leftrightarrow$(iii). Under (i), $\mathcal A_{\mathbb C}$ is even dense in $L^2(\mathfrak S,\mu)$.

Proof. (i)$\Rightarrow$(ii): By (10.1), $V_\mu(\hat a)=\kappa_\mu(\hat a)\xi=e_\mu\pi(a)\xi$. These vectors are dense in $e_\mu H=V_\mu(L^2)$, because $\pi(A)\xi$ is dense in $H$. $V_\mu$ is an isometry onto $e_\mu H$, so $\mathcal A_{\mathbb C}$ is dense in $L^2(\mu)$, and hence in $L^1(\mu)$, since $\|g\|_1\leq\|g\|_2$ for a probability measure.

(ii)$\Leftrightarrow$(iii): By Proposition 8.1(2), $\operatorname{Aff}(\mathfrak S)=\{\hat h:\ h\in A_h\}$, and $\mathcal A_{\mathbb C}=\operatorname{Aff}(\mathfrak S)+i\operatorname{Aff}(\mathfrak S)$. The complex span of a set of real functions is dense in complex $L^1$ exactly when the set is dense in real $L^1$: approximate real and imaginary parts separately, and conversely note $\|g-\operatorname{Re}u\|_1\leq\|g-u\|_1$ for real $g$. Now apply Proposition 11.2. $\square$

Remark 11.4. The equivalence (ii)$\Leftrightarrow$(iii) uses nothing about C$^*$-algebras: it is Proposition 11.2, which holds for every compact convex set. The implication (i)$\Rightarrow$(iii) cannot be reversed. Here is an explicit simplicial measure that is not orthogonal.

Example 11.5 (Simplicial but not orthogonal). Let $A=M_2(\mathbb C)$. For $v\in\mathbb R^3$ with $|v|\leq1$ let $\rho_v=\frac12(1+v_1\sigma_1+v_2\sigma_2+v_3\sigma_3)$, with the Pauli matrices $\sigma_i$, and $\omega_v(a)=\operatorname{tr}(\rho_va)$. Every state is some $\omega_v$, and $\omega_v$ is pure exactly when $|v|=1$. For density matrices, $\omega_\rho\leq\omega_{\rho'}$ if and only if $\rho\leq\rho'$. Let $v^{(1)},v^{(2)},v^{(3)}$ be unit vectors in the $v_1v_3$-plane at mutual angles of $120^\circ$, so $\sum_kv^{(k)}=0$, and let $\mu=\frac13\sum_k\delta_{\omega_{v^{(k)}}}$. It represents the tracial state $\tau=\omega_0$.

$\mu$ is simplicial. $L^1(\mu)$ is $\mathbb C^3$. The functions $\hat1,\hat\sigma_1,\hat\sigma_3$ give the vectors $(1,1,1)$, $(v^{(k)}_1)_k$, $(v^{(k)}_3)_k$, which are independent because the three points are not collinear. So (ii) holds.
$\mu$ is not orthogonal. Take $E=\{\omega_{v^{(1)}}\}$. The density of $3\varphi_{E^c}=\omega_{v^{(2)}}+\omega_{v^{(3)}}$ is $1-\frac12v^{(1)}\cdot\sigma$, with eigenvalues $\frac12$ and $\frac32$. Let $\psi=\frac16\omega_{v^{(1)}}$, with density $\frac1{12}(1+v^{(1)}\cdot\sigma)$. In the eigenbasis of $v^{(1)}\cdot\sigma$ the densities of $\psi$, $\varphi_E$ and $\varphi_{E^c}$ are $\operatorname{diag}(\frac16,0)$, $\operatorname{diag}(\frac13,0)$ and $\operatorname{diag}(\frac16,\frac12)$. So $0\neq\psi\leq\varphi_E$ and $\psi\leq\varphi_{E^c}$.

By contrast, $\frac12(\delta_{\omega_v}+\delta_{\omega_{-v}})$ with $|v|=1$ is orthogonal: $\rho_v$ and $\rho_{-v}$ are orthogonal rank-one projections, and a positive matrix below multiples of both is $0$. The normalized surface measure on the sphere $\{\omega_v:|v|=1\}$ also represents $\tau$, by symmetry. It is maximal, since it lives on $\partial_e\mathfrak S$ and is therefore a boundary measure (Proposition 5.5(2) and Theorem 5.4). But it is not orthogonal: $\kappa$ would embed the infinite-dimensional $L^\infty$ of the sphere into the four-dimensional $\pi_\tau(A)'$.

12. Abelian algebras with a cyclic vector

This section proves three facts about commutants that the rest of the lesson uses: an abelian algebra with a cyclic vector has an abelian commutant, compressions of von Neumann algebras by projections behave well, and abelian subalgebras of a commutant match certain projections.

Lemma 12.1. Let $\mathfrak A\subseteq B(H)$ be a commutative $*$-algebra (not necessarily closed) with a cyclic vector $\xi$. Then $\mathfrak A'$ is commutative and $\mathfrak A'=\mathfrak A''$. If $\mathfrak A$ is a von Neumann algebra, then $\mathfrak A'=\mathfrak A$: it is maximal abelian.

Proof. For $a\in\mathfrak A$, $\|a^*\xi\|^2=\langle aa^*\xi,\xi\rangle=\langle a^*a\xi,\xi\rangle=\|a\xi\|^2$. So $J(a\xi)=a^*\xi$ is a well-defined conjugate-linear isometry on $\mathfrak A\xi$. It extends to a conjugate-linear isometry $J$ of $H$ onto $H$ with $J^2=1$, and $\langle J\eta,J\zeta\rangle=\langle\zeta,\eta\rangle$. For $c,b\in\mathfrak A$, $Jc^*(b\xi)=b^*c\xi=cb^*\xi=cJ(b\xi)$, so $Jc^*=cJ$. Let $T\in\mathfrak A'$. Then $JTJ(b\xi)=J(b^*T\xi)=bJT\xi$ and $T^*(b\xi)=bT^*\xi$. Moreover $JT\xi=T^*\xi$, since for $c\in\mathfrak A$ $\begin{gathered} \langle JT\xi,c\xi\rangle\\ =\langle JT\xi,J(c^*\xi)\rangle\\ =\langle c^*\xi,T\xi\rangle\\ =\langle c^*T^*\xi,\xi\rangle\\ =\langle T^*\xi,c\xi\rangle . \end{gathered}$ So $JTJ=T^*$ on a dense set, hence everywhere. For $S,T\in\mathfrak A'$: $\begin{gathered} T^*S^*\\ =(ST)^*\\ =J(ST)J\\ =(JSJ)(JTJ)\\ =S^*T^* \end{gathered}$ , and $\mathfrak A'$ is commutative. Then $\mathfrak A'\subseteq\mathfrak A''$; and $\mathfrak A\subseteq\mathfrak A'$ gives $\mathfrak A''\subseteq\mathfrak A'$. $\square$

Lemma 12.2 (Reduction). Let $\mathcal N$ be a von Neumann algebra on $H$, $e$ a projection, and $\mathcal N_e=\{exe|_{eH}:\ x\in\mathcal N\}\subseteq B(eH)$.

If $e\in\mathcal N'$, then $(\mathcal N_e)'=(\mathcal N')_e$, and $x\mapsto x|_{eH}$ is a $*$-homomorphism of $\mathcal N$ onto $\mathcal N_e$.
If $e\in\mathcal N$, then $(\mathcal N_e)'=(\mathcal N')_e$.
In both cases $\mathcal N_e$ is a von Neumann algebra on $eH$.

Proof. (1) Here $exe=xe$, so $x\mapsto x|_{eH}$ is a $*$-homomorphism. If $y\in\mathcal N'$, then $eye|_{eH}$ commutes with every $xe|_{eH}$. Conversely, let $T\in B(eH)$ commute with $\mathcal N_e$, and let $\tilde T=Te$ (zero on $(1-e)H$). Each $x\in\mathcal N$ leaves $eH$ and $(1-e)H$ invariant, so $\tilde T\in\mathcal N'$ and $T=e\tilde Te|_{eH}$.

(2) The inclusion "$\supseteq$" is direct. Let $T\in B(eH)$ commute with every $exe|_{eH}$. For $x_1,\dots,x_n\in\mathcal N$ and $\eta_1,\dots,\eta_n\in eH$, the matrix $X=[ex_j^*x_ie]_{j,i}$ acts on $(eH)^n$ and is positive, since $\langle X\vec\eta,\vec\eta\rangle=\|\sum_ix_i\eta_i\|^2$. The operator $T^{(n)}=T\oplus\dots\oplus T$ commutes with $X$, hence with $X^{1/2}$. So $\begin{gathered} \Bigl\|\sum_ix_iT\eta_i\Bigr\|^2\\ =\langle T^{(n)*}T^{(n)}X\vec\eta,\vec\eta\rangle\\ =\|T^{(n)}X^{1/2}\vec\eta\|^2\\ \leq\|T\|^2\Bigl\|\sum_ix_i\eta_i\Bigr\|^2 . \end{gathered}$ So $y(\sum_ix_i\eta_i)=\sum_ix_iT\eta_i$ defines a bounded operator on the closed span $[\mathcal NeH]$; extend it by $0$ on the orthogonal complement. For $z\in\mathcal N$ the formula gives $yz=zy$ on $[\mathcal NeH]$; the complement is $\mathcal N$-invariant and $y$ vanishes there. So $y$ commutes with $\mathcal N$. For $\eta\in eH$, $y\eta=y(e\eta)=eT\eta=T\eta$. So $T=eye|_{eH}$ with $y\in\mathcal N'$.

(3) If $e\in\mathcal N'$, apply (2) to the von Neumann algebra $\mathcal N'$, which contains $e$: $((\mathcal N')_e)'=(\mathcal N'')_e=\mathcal N_e$. If $e\in\mathcal N$, apply (1) to $\mathcal N'$ in the same way. So $\mathcal N_e$ is a commutant. $\square$

Proposition 12.3. Let $\mathcal M$ be a unital $*$-algebra of operators on $H$ with a cyclic vector $\xi_0$.

(a) If $\mathcal A\subseteq\mathcal M'$ is a commutative $*$-algebra and $e$ is the projection onto $[\mathcal A\xi_0]$, then $e\mathcal Me$ is commutative, that is, $e\mathcal Me\subseteq(e\mathcal Me)'$.
(b) If $e$ is a projection with $e\xi_0=\xi_0$ and $e\mathcal Me$ is commutative, then $\mathcal A=\mathcal M'\cap\{e\}'$ is a commutative von Neumann algebra, $e$ is the projection onto $[\mathcal A\xi_0]$, and $x\mapsto x|_{eH}$ is a $*$-isomorphism of $\mathcal A$ onto the maximal abelian algebra $(e\mathcal Me|_{eH})'$ of $eH$.
(c) The maps $\mathcal A\mapsto[\mathcal A\xi_0]$ and $e\mapsto\mathcal M'\cap\{e\}'$ are inverse bijections. One side is the set of commutative von Neumann algebras contained in $\mathcal M'$; the other is the set of projections $e$ with $e\xi_0=\xi_0$ and $e\mathcal Me$ commutative.

Proof. (a) $e\in\mathcal A'$, because $[\mathcal A\xi_0]$ is invariant under the self-adjoint set $\mathcal A$. The restrictions $\mathcal A_e=\{a|_{eH}\}$ form a commutative $*$-algebra on $eH$ with cyclic vector $e\xi_0$: since $a(1-e)\xi_0=(1-e)a\xi_0=0$, we have $a\xi_0=ae\xi_0$, so $\mathcal A_e(e\xi_0)=\mathcal A\xi_0$ is dense in $eH$. (If $1\in\mathcal A$, then $e\xi_0=\xi_0$.) By Lemma 12.1, $(\mathcal A_e)'$ is commutative. For $x\in\mathcal M$ and $a\in\mathcal A$, $(exe)(ae)=exae=eaxe=(ae)(exe)$. So $e\mathcal Me|_{eH}\subseteq(\mathcal A_e)'$, which is commutative; and $e\mathcal Me$ vanishes on $(1-e)H$.

(b) Let $S=\{exe|_{eH}:\ x\in\mathcal M\}$, a commuting self-adjoint set that contains $1_{eH}$. The vector $\xi_0$ is cyclic for $S$ in $eH$, because $e\mathcal M\xi_0$ is dense in $eH$ and $exe\xi_0=ex\xi_0$. By Lemma 12.1, applied to the algebra generated by $S$, $S'$ is commutative and $S'=S''$; so $S'$ is maximal abelian on $eH$. The set $\mathcal A$ is the commutant of $\mathcal M\cup\{e\}$, a von Neumann algebra. Restriction $r(x)=x|_{eH}$ is a $*$-homomorphism of $\mathcal A$ into $S'$. It is one-to-one: $r(x)=0$ gives $x\xi_0=0$, and $\xi_0$ is separating for $\mathcal M'$. It is onto $S'$: let $y\in S'$. For $m\in\mathcal M$, the positive operator $s=em^*me|_{eH}$ lies in $S$, so $y^*y$ commutes with $s$ and with $s^{1/2}$; also $y\xi_0\in eH$. Hence $\begin{gathered} \|my\xi_0\|^2\\ =\langle em^*me\,y\xi_0,y\xi_0\rangle\\ =\langle y^*y\,(em^*me)\xi_0,\xi_0\rangle\\ \leq\|y\|^2\langle em^*me\xi_0,\xi_0\rangle\\ =\|y\|^2\|m\xi_0\|^2 . \end{gathered}$ So $x_y(m\xi_0)=my\xi_0$ defines a bounded operator on $H$, and $x_y\in\mathcal M'$. Also $x_{y^*}=x_y^*$: both sides of $\langle x_ym\xi_0,m'\xi_0\rangle=\langle m\xi_0,x_{y^*}m'\xi_0\rangle$ equal $\langle y\,(em'^*me)\xi_0,\xi_0\rangle$. For $m_1,m_2\in\mathcal M$, $\begin{gathered} \langle x_ym_1\xi_0,em_2\xi_0\rangle\\ =\langle(em_2^*e)(em_1e)\,y\xi_0,\xi_0\rangle\\ =\langle y\,em_1\xi_0,em_2\xi_0\rangle , \end{gathered}$ because $y$ commutes with the elements $em_ie|_{eH}$ of $S$. So $ex_y=ye$ as operators on $H$. Applying this to $y^*$ and taking adjoints gives $x_ye=ey$, where $ey$ means $y$ on $eH$ and $0$ on $(1-e)H$. So $x_y$ commutes with $e$, lies in $\mathcal A$, and $r(x_y)=y$. Finally $[\mathcal A\xi_0]=[S'\xi_0]\supseteq[S\xi_0]=eH$, and $\mathcal A\xi_0=\mathcal Ae\xi_0\subseteq eH$.

(c) By (a) and (b) both maps land in the right sets, and $e\mapsto\mathcal M'\cap\{e\}'\mapsto e$ is the identity by (b). Let $\mathcal A\subseteq\mathcal M'$ be an abelian von Neumann algebra and $e=[\mathcal A\xi_0]$. Clearly $\mathcal A\subseteq\mathcal B=\mathcal M'\cap\{e\}'$, and $\mathcal B$ is abelian by (b). By Lemma 12.2(1) and (3), $\mathcal A_e$ is a von Neumann algebra on $eH$, and it has the cyclic vector $\xi_0$, so it is maximal abelian (Lemma 12.1). For $x\in\mathcal B$, $x|_{eH}$ commutes with $\mathcal A_e$, so $x|_{eH}=a|_{eH}$ for some $a\in\mathcal A$, and $x=a$ because restriction is one-to-one on $\mathcal B$. $\square$

13. Orthogonal measures and abelian subalgebras

The main structural fact about orthogonal measures: they are in one-to-one correspondence with the abelian von Neumann subalgebras of the commutant.

Theorem 13.1. Let $\varphi\in\mathfrak S$, and let $\mathcal C\subseteq\pi_\varphi(A)'$ be a commutative von Neumann algebra. Then exactly one orthogonal $\mu\in M_\varphi(\mathfrak S)$ has $\mathcal C_\mu=\mathcal C$. So $\mu\mapsto\mathcal C_\mu$ maps the orthogonal measures in $M_\varphi(\mathfrak S)$ bijectively onto the commutative von Neumann algebras contained in $\pi_\varphi(A)'$.

Proof. Write $\pi,H,\xi$, and let $e$ be the projection onto $[\mathcal C\xi]$; $e\in\mathcal C'$, and $\xi$ is separating for $\mathcal C\subseteq\pi(A)'$.

Step 1: the compression. By Lemma 12.2, $\mathcal C_e$ is a von Neumann algebra on $eH$, and $x\mapsto x_e=x|_{eH}$ is a $*$-homomorphism of $\mathcal C$ onto it; it is one-to-one because $\xi$ is separating. $\mathcal C_e$ is abelian with cyclic vector $\xi$, hence maximal abelian by Lemma 12.1: $(\mathcal C_e)'=\mathcal C_e$. Also $(\mathcal C_e)'=(\mathcal C')_e$ by Lemma 12.2(1). Since $\pi(A)\subseteq\mathcal C'$, each $e\pi(a)e|_{eH}$ lies in $\mathcal C_e$. So there is a unique $\theta(a)\in\mathcal C$ with $\theta(a)e=e\pi(a)e$; in particular $\theta(a)\xi=e\pi(a)\xi$. The map $\theta:A\to\mathcal C$ is linear, unital and positive: $x\mapsto x_e$ is an isometric $*$-isomorphism of $\mathcal C$ onto $\mathcal C_e$, so it reflects positivity.

Step 2: the measure. By the Gelfand representation (the commutative Gelfand–Naimark theorem), $\mathcal C\cong C(\Omega)$ for its spectrum $\Omega$; write $\Gamma$ for this isomorphism. For a character $\chi\in\Omega$, $\chi\circ\theta$ is a state, and $t:\chi\mapsto\chi\circ\theta$ is continuous from $\Omega$ to $\mathfrak S$. So $\bar\theta(g)=\Gamma^{-1}(g\circ t)$ is a unital $*$-homomorphism $C(\mathfrak S)\to\mathcal C$, and $\bar\theta(\hat a)=\theta(a)$. Let $\mu\in M_1^+(\mathfrak S)$ be the measure $g\mapsto\langle\bar\theta(g)\xi,\xi\rangle$. Then $\int\hat a\,d\mu=\langle\theta(a)\xi,\xi\rangle=\langle e\pi(a)\xi,\xi\rangle=\varphi(a)$, so $\mu\in M_\varphi(\mathfrak S)$.

Step 3: $\kappa_\mu=\bar\theta$ on $C(\mathfrak S)$. For $g\in C(\mathfrak S)$ and $a\in A$, using $\bar\theta(\bar g)\xi\in eH$, $\begin{gathered} \langle\bar\theta(g)\pi(a)\xi,\xi\rangle\\ =\langle\pi(a)\xi,e\bar\theta(\bar g)\xi\rangle\\ =\langle\theta(a)\xi,\bar\theta(\bar g)\xi\rangle\\ =\langle\bar\theta(g\hat a)\xi,\xi\rangle\\ =\int g\hat a\,d\mu . \end{gathered}$ By the uniqueness in Proposition 9.2, $\kappa_\mu(g)=\bar\theta(g)$. So $\kappa_\mu$ is multiplicative on $C(\mathfrak S)$. It is normal, $C(\mathfrak S)$ is weak$^*$-dense in $L^\infty(\mu)$ (Proposition 2.1(6)), and multiplication is separately continuous in both topologies; so $\kappa_\mu$ is multiplicative on $L^\infty$. Thus $\mu$ is orthogonal, and $\mathcal C_\mu\subseteq\mathcal C$, since $\mathcal C$ is weakly closed.

Step 4: $\mathcal C_\mu=\mathcal C$. By (10.1), $[\mathcal C_\mu\xi]\supseteq[\kappa_\mu(\mathcal A_{\mathbb C})\xi]=[e\pi(A)\xi]=eH$, so $e_\mu=e$. For $y\in\mathcal C$, $y\xi\in eH=e_\mu H$, and the $L^2$ criterion, Proposition 10.4(3), gives $y\in\mathcal C_\mu$.

Uniqueness. Let $\mu,\nu$ be orthogonal with $\mathcal C_\mu=\mathcal C_\nu=\mathcal C$. Then $e_\mu=e_\nu=e$, and (10.1) gives $\kappa_\mu(\hat a)\xi=e\pi(a)\xi=\kappa_\nu(\hat a)\xi$, so $\kappa_\mu(\hat a)=\kappa_\nu(\hat a)$ (separating vector). The restrictions of $\kappa_\mu$ and $\kappa_\nu$ to $C(\mathfrak S)$ are $*$-homomorphisms that agree on $\mathcal A_{\mathbb C}$. This set contains $1$, is closed under conjugation ($\overline{\hat a}=\widehat{a^*}$), and separates the points of $\mathfrak S$, so it generates a dense subalgebra, by the complex Stone–Weierstrass theorem. So $\kappa_\mu=\kappa_\nu$ on $C(\mathfrak S)$, and $\mu(g)=\langle\kappa_\mu(g)\xi,\xi\rangle=\nu(g)$. $\square$

14. Comparing orthogonal measures

Under the correspondence of Theorem 13.1, the Choquet order between orthogonal measures is inclusion between their abelian algebras.

Theorem 14.1. Let $\varphi\in\mathfrak S$ and let $\mu,\nu\in M_\varphi(\mathfrak S)$ be orthogonal. The following are equivalent.

$\mu\prec\nu$.
$\int\omega(h)^2\,d\mu(\omega)\leq\int\omega(h)^2\,d\nu(\omega)$ for every $h\in A_h$.
$\mathcal C_\mu\subseteq\mathcal C_\nu$.

Proof. (1)$\Rightarrow$(2): $\hat h$ is real and affine, so $\hat h^2$ is convex and continuous.

(2)$\Rightarrow$(3): For $a=h+ik$ with $h,k\in A_h$, $|\hat a|^2=\hat h^2+\hat k^2$. By (10.1) and the isometry $V_\mu$, $\begin{gathered} \|e_\mu\pi(a)\xi\|^2\\ =\|\kappa_\mu(\hat a)\xi\|^2\\ =\int|\hat a|^2\,d\mu\\ \leq\int|\hat a|^2\,d\nu\\ =\|e_\nu\pi(a)\xi\|^2 . \end{gathered}$ So $\langle e_\mu\eta,\eta\rangle\leq\langle e_\nu\eta,\eta\rangle$ on the dense set $\pi(A)\xi$, hence $e_\mu\leq e_\nu$. For $x\in\mathcal C_\mu\subseteq\pi(A)'$, $x\xi\in e_\mu H\subseteq e_\nu H$, and the $L^2$ criterion, Proposition 10.4(3), gives $x\in\mathcal C_\nu$.

(3)$\Rightarrow$(1): We check condition (2) of Lemma 4.10. Let $\mu=\sum_i\mu_i$ with $\mu_i\geq0$. By Proposition 2.1(5), $\mu_i=f_i\mu$ with $0\leq f_i\leq1$ and $\sum_if_i=1$. Since $\kappa_\mu(f_i)\in\mathcal C_\nu$ and $\kappa_\nu$ is an isomorphism onto $\mathcal C_\nu$, there are $g_i\in L^\infty(\nu)$ with $\kappa_\nu(g_i)=\kappa_\mu(f_i)$; then $0\leq g_i\leq1$ and $\sum_ig_i=1$. Put $\nu_i=g_i\nu$. Then $\sum_i\nu_i=\nu$ and $\begin{gathered} \nu_i(\hat a)\\ =\langle\kappa_\nu(g_i)\pi(a)\xi,\xi\rangle\\ =\langle\kappa_\mu(f_i)\pi(a)\xi,\xi\rangle\\ =\mu_i(\hat a) \end{gathered}$ . Since $\operatorname{Aff}(\mathfrak S)=\{\hat h\}$ (Proposition 8.1(2)), $\nu_i\sim\mu_i$. $\square$

15. Multiplicity-free states

Lemma 15.1 (Extreme positive contractions). In a C$^*$-algebra $\mathcal B$ of operators, the extreme points of $\{x\in\mathcal B:\ 0\leq x\leq1\}$ are exactly the projections in $\mathcal B$.

Proof. Let $p$ be a projection and $p=\frac12(x+y)$ with $0\leq x,y\leq1$. Then $(1-p)x(1-p)\leq2(1-p)p(1-p)=0$, so $x=pxp\leq p$, and likewise $y\leq p$. From $x+y=2p$ we get $x=p$. If $x$ is not a projection, then $x-x^2\neq0$, and $x=\frac12\bigl((2x-x^2)+x^2\bigr)$, where $0\leq x^2\leq1$ and $0\leq2x-x^2=1-(1-x)^2\leq1$. $\square$

Theorem 15.2. For $\varphi\in\mathfrak S$ the following are equivalent.

$\pi_\varphi(A)'$ is abelian.
$M_\varphi(\mathfrak S)$ has exactly one maximal measure.

In that case the maximal measure is the orthogonal measure $\mu$ with $\mathcal C_\mu=\pi_\varphi(A)'$, and $\nu\prec\mu$ for every $\nu\in M_\varphi(\mathfrak S)$.

Proof. (1)$\Rightarrow$(2): By Theorem 13.1 there is an orthogonal $\mu$ with $\mathcal C_\mu=\pi(A)'$. Let $\nu\in M_\varphi(\mathfrak S)$; we check condition (2) of Lemma 4.10 for the pair $\nu,\mu$. If $\nu=\sum_i\nu_i$ with $\nu_i\geq0$, then $\nu_i=g_i\nu$ with $0\leq g_i\leq1$ and $\sum_ig_i=1$. The operators $\kappa_\nu(g_i)$ are positive, lie in $\pi(A)'=\mathcal C_\mu$, and add up to $1$. Since $\kappa_\mu$ is a $*$-isomorphism onto $\mathcal C_\mu$, $\kappa_\nu(g_i)=\kappa_\mu(f_i)$ with $f_i\geq0$ and $\sum_if_i=1$. With $\mu_i=f_i\mu$, $\mu_i(\hat a)=\langle\kappa_\mu(f_i)\pi(a)\xi,\xi\rangle=\nu_i(\hat a)$, so $\mu_i\sim\nu_i$. Lemma 4.10 gives $\nu\prec\mu$. So $\mu$ majorizes all of $M_\varphi(\mathfrak S)$. It is maximal: if $\mu\prec\lambda$, then $\lambda\in M_\varphi(\mathfrak S)$ (Proposition 4.5(2)), so $\lambda\prec\mu$ and $\lambda=\mu$. Any maximal $\nu\in M_\varphi(\mathfrak S)$ satisfies $\nu\prec\mu$, hence $\nu=\mu$.

(2)$\Rightarrow$(1): Let $\mu$ be the unique maximal measure in $M_\varphi(\mathfrak S)$. Every $\lambda\in M_\varphi(\mathfrak S)$ is majorized by a maximal measure (Lemma 4.6), which is $\mu$.

Step 1: $\kappa_\mu$ is one-to-one. If $\mu=\frac12(\mu_1+\mu_2)$ in $M_\varphi(\mathfrak S)$, then $\mu_1,\mu_2\prec\mu$, so $\mu(f)\geq\mu_i(f)$ for $f\in\mathcal P(\mathfrak S)$, while $\mu(f)$ is their average. So $\mu_1=\mu_2=\mu$ on $\mathcal P(\mathfrak S)$, and hence everywhere (Lemma 4.1). Thus $\mu$ is simplicial, and $\mathcal A_{\mathbb C}$ is dense in $L^1(\mu)$ (Proposition 11.3). If $\kappa_\mu(f)=0$, then $\int f\hat a\,d\mu=0$ for all $a$, so $f=0$.

Step 2: $\kappa_\mu$ maps $\{0\leq f\leq1\}$ onto $\{h\in\pi(A)':\ 0\leq h\leq1\}$. Positivity gives "into". Let $0\leq h\leq1$ in $\pi(A)'$. The bicommutant $\mathcal C=\{h\}''$ is abelian and lies in $\pi(A)'$. Let $\nu$ be the orthogonal measure with $\mathcal C_\nu=\mathcal C$ (Theorem 13.1). For $g\in C(\mathfrak S)$ put $\nu_1(g)=\langle\kappa_\nu(g)h\xi,\xi\rangle$ and $\nu_2(g)=\langle\kappa_\nu(g)(1-h)\xi,\xi\rangle$. These are positive measures, because $\kappa_\nu(g)\geq0$ commutes with $h$ and $1-h$ when $g\geq0$; and $\nu=\nu_1+\nu_2$. Since $\nu\prec\mu$, Lemma 4.10 gives $\mu=\mu_1+\mu_2$ with $\mu_i\sim\nu_i$, and $\mu_1=g\mu$ with $0\leq g\leq1$. For $a\in A$, using $h\xi\in[\mathcal C\xi]=e_\nu H$ and (10.1), $\begin{gathered} \langle\kappa_\mu(g)\pi(a)\xi,\xi\rangle\\ =\mu_1(\hat a)\\ =\nu_1(\hat a)\\ =\langle\kappa_\nu(\hat a)e_\nu h\xi,\xi\rangle\\ =\langle e_\nu\pi(a)e_\nu h\xi,\xi\rangle\\ =\langle h\pi(a)\xi,\xi\rangle . \end{gathered}$ So $\kappa_\mu(g)=h$.

Step 3. By Steps 1–2, $\kappa_\mu$ is an affine bijection of $\{0\leq f\leq1\}$ onto $\{0\leq h\leq1\}\subseteq\pi(A)'$. The indicator functions $\chi_E$ are extreme in the first set (if $\chi_E=\frac12(f_1+f_2)$ with $0\leq f_i\leq1$, then $f_i=\chi_E$ almost everywhere). So each $\kappa_\mu(\chi_E)$ is extreme in the second set, hence a projection by Lemma 15.1. The proof of (3)$\Rightarrow$(1) in Theorem 10.2 used only this, so $\kappa_\mu$ is multiplicative. Its range contains every positive contraction of $\pi(A)'$, and these span $\pi(A)'$; so $\pi(A)'=\kappa_\mu(L^\infty)$ is abelian. $\square$

Definition 15.3. A representation $\pi$ of $A$ is multiplicity-free if $\pi(A)'$ is abelian.

So a state is multiplicity-free (its GNS representation is) exactly when its maximal representing measure is unique. That measure is orthogonal and vanishes on Baire sets that miss $P(A)$ (Proposition 8.1(5)).

If every state is multiplicity-free, the Choquet–Meyer theorem says that $\mathfrak S$ is a simplex. This happens only for abelian algebras.

Theorem 15.4. The state space $\mathfrak S(A)$ is a simplex exactly when $A$ is abelian.

Proof. If $A$ is abelian, $A\cong C(X)$ by the commutative Gelfand–Naimark theorem, and by Proposition 8.1(2) $\operatorname{Aff}(\mathfrak S)\cong A_h=C_{\mathbb R}(X)$ as ordered spaces. So $\operatorname{Aff}(\mathfrak S)^*\cong C_{\mathbb R}(X)^*=M(X)$ with its usual order, which is a vector lattice.

Conversely let $\mathfrak S$ be a simplex. By the Choquet–Meyer theorem (Theorem 7.4), every state has exactly one maximal representing measure, so $\pi_\varphi(A)'$ is abelian for every state $\varphi$ (Theorem 15.2). Suppose $A$ is not abelian. Then some pure state $\omega$ has $\dim H_\omega\geq2$: otherwise every pure state is multiplicative, so every pure state kills every commutator $[a,b]$; by Krein–Milman every state does, and then $[a,b]=0$, because the states separate the points of $A$ (Proposition 8.1(2)). Let $\pi=\pi_\omega$ on $H$, irreducible, and let $\eta_1,\eta_2\in H$ be orthonormal. The state $\varphi(a)=\frac12(\langle\pi(a)\eta_1,\eta_1\rangle+\langle\pi(a)\eta_2,\eta_2\rangle)$ is given by $\pi\oplus\pi$ and $\zeta=(\eta_1,\eta_2)/\sqrt2$. The commutant of $\pi\oplus\pi$ consists of the $2\times2$ matrices with entries in $\pi(A)'=\mathbb C1$, that is, $M_2(\mathbb C)\otimes1$. The projection $p\otimes1$ onto $[(\pi\oplus\pi)(A)\zeta]$ lies there and fixes $\zeta$, so $p_{11}\eta_1+p_{12}\eta_2=\eta_1$ and $p_{21}\eta_1+p_{22}\eta_2=\eta_2$, which forces $p=1$. So $\zeta$ is cyclic, $\pi_\varphi\cong\pi\oplus\pi$, and $\pi_\varphi(A)'\cong M_2(\mathbb C)$ is not abelian. This is a contradiction. $\square$

16. The face generated by a state

The smallest face of $\mathfrak S$ that contains a state $\varphi$ is described by the commutant $\pi_\varphi(A)'$, through the Radon–Nikodym map.

Proposition 16.1. Let $\varphi\in\mathfrak S$ with GNS triple $\pi,H,\xi$, and let $F_\varphi$ be the smallest face of $\mathfrak S$ that contains $\varphi$.

$\begin{gathered} F_\varphi\\ =\{\psi\in\mathfrak S:\\ \ t\psi\\ \leq\varphi\text{ for some }t>0\}\\ =\{\Theta_\varphi(x):\\ \ x\in\pi(A)'_+,\ \\ \langle x\xi,\xi\rangle\\ =1\} \end{gathered}$ . If $\psi\in\mathfrak S$ has the form $\Theta_\varphi(x)$ for some $x\in\pi(A)'$, then automatically $x\geq0$.
$F_\varphi$ is closed in $\mathfrak S$ $\iff$ $\pi(A)'$ is finite-dimensional $\iff$ $F_\varphi$ is finite-dimensional.
The cone $\mathbb R_+F_\varphi$ is a lattice in its own order if and only if $\pi(A)'$ is abelian.
If the closure $\overline{F_\varphi}$ is a simplex, then $\pi(A)'$ is abelian.
There are $A$ and $\varphi$ with $\pi_\varphi(A)'$ abelian and $\overline{F_\varphi}$ not a simplex.

Parts (3)–(5) distinguish the lattice property of the generated face from the simplex property of its closure. The proof includes an explicit counterexample to the converse of (4).

Proof. (1) Call the first set $F$. It is convex and contains $\varphi$. It is a face: if $\psi=s\psi_1+(1-s)\psi_2\in F$ with $t\psi\leq\varphi$, then $ts\psi_1\leq\varphi$. It lies in every face $F'\ni\varphi$: if $t\psi\leq\varphi$ with $0<t<1$, then $\varphi=t\psi+(1-t)\psi'$ with the state $\psi'=(\varphi-t\psi)/(1-t)$, so $\psi\in F'$; and $t=1$ forces $\psi=\varphi$. The second description follows from Proposition 8.1(3): $t\psi\leq\varphi$ means $t\psi=\Theta_\varphi(y)$ with $0\leq y\leq1$. If $\Theta_\varphi(x)=\psi$ is a state, then $\Theta_\varphi(x^*)=\psi^*=\psi$, so $x=x^*$, and $x\geq0$ because $\Theta_\varphi$ reflects order.

(2) The linear span of $F_\varphi$ is $\Theta_\varphi(\pi(A)')$: every $x$ is a combination of positive elements, and a positive $x\neq0$ has $\langle x\xi,\xi\rangle>0$ (separating vector). As $\Theta_\varphi$ is one-to-one, $\dim F_\varphi<\infty$ exactly when $\dim\pi(A)'<\infty$. If $\dim\pi(A)'<\infty$, the set $\{x\geq0:\ \langle x\xi,\xi\rangle=1\}$ is compact: $x\mapsto\langle x\xi,\xi\rangle$ is positive on the compact set $\{x\geq0,\ \|x\|=1\}$, hence at least some $c>0$ there, so the set is bounded. Its continuous image $F_\varphi$ is compact, hence closed. If $\pi(A)'$ is infinite-dimensional, it has an infinite sequence of nonzero orthogonal projections $p_n$ (Lemma 16.2 below). Put $c_n=\langle p_n\xi,\xi\rangle>0$; $c_n\to0$ because $\sum c_n\leq1$. Pass to a subsequence with $c_{n_k}\leq4^{-k}$, and put $\psi=\sum_k2^{-k}\Theta_\varphi(p_{n_k})/c_{n_k}$, a norm-convergent series of states. $\psi$ is a norm limit of normalized partial sums, which lie in $F_\varphi$; so $\psi\in\overline{F_\varphi}$. If $\psi$ were in $F_\varphi$, then $\psi=\Theta_\varphi(x)$ with $x$ bounded, and $\psi\geq2^{-k}c_{n_k}^{-1}\Theta_\varphi(p_{n_k})$ would give $x\geq2^kp_{n_k}$, so $\|x\|\geq2^k$ for all $k$. So $F_\varphi$ is not closed.

(3) The real span of $F_\varphi$ is $\Theta_\varphi(\pi(A)'_h)$, and $\Theta_\varphi$ maps $\pi(A)'_+$ onto $\mathbb R_+F_\varphi$; so it is an order isomorphism. If $\pi(A)'$ is abelian, $\pi(A)'_h\cong C_{\mathbb R}(\Omega)$ is a lattice. Conversely, suppose $\mathcal N=\pi(A)'$ is not abelian. A von Neumann algebra is generated by its projections (spectral theorem), so some projection $p\in\mathcal N$ is not central, and then $p\mathcal N(1-p)\neq0$. Take $x\in p\mathcal N(1-p)$ with $0<\|x\|\leq1$. Then $x^2=0$, $(x+x^*)^2=xx^*+x^*x$, so $\|x+x^*\|=\|x\|\leq1$, and $u=\frac12(1+x+x^*)$ satisfies $0\leq u\leq1=p+(1-p)$. If $\mathcal N_h$ were a lattice, the Riesz decomposition (Lemma 7.3) would give $u=u_1+u_2$ with $0\leq u_1\leq p$ and $0\leq u_2\leq1-p$. Then $u_1=pu_1p$ and $u_2=(1-p)u_2(1-p)$ (as in the proof of Lemma 10.1(1)), so $pu(1-p)=0$. But $pu(1-p)=\frac12x\neq0$.

(4) Suppose $\pi(A)'$ is not abelian. By Theorem 15.2, $M_\varphi(\mathfrak S)$ does not have exactly one maximal measure; it has at least one (Lemma 4.6), so it has two, $\mu_1\neq\mu_2$. Every $\mu\in M_\varphi(\mathfrak S)$ lives on $L=\overline{F_\varphi}$: for $\omega\in\operatorname{supp}\mu$ and a closed convex neighbourhood $W$ of $\omega$, $C=W\cap\mathfrak S$ has $\mu(C)>0$, and the barycentre $\psi_C$ of $\mu|_C/\mu(C)$ satisfies $\mu(C)\psi_C\leq\varphi$, so $\psi_C\in F_\varphi$, and $\psi_C\in C$ by Proposition 3.3(1). As $W$ shrinks, $\psi_C\to\omega$. Both $\mu_i$ are maximal also among measures on $L$: if $\mu_i\prec\lambda$ on $L$, then $\mu_i(f)\leq\lambda(f)$ for every $f\in\mathcal P(\mathfrak S)$ (restrict $f$ to $L$), so $\lambda=\mu_i$. By the Choquet–Meyer theorem, $L$ is not a simplex.

(5) Counterexample. Let $A=C([0,1],M_2(\mathbb C))$, and let $v_0=e_1$, $v_1=(e_1+e_2)/\sqrt2$, $v_2=e_2$, $v_3=(e_1-e_2)/\sqrt2$, unit vectors whose Bloch vectors are the vertices of a square. Let $J_k=[\frac12-2^{-k},\frac12-2^{-k-1})$ for $k\geq1$; these intervals cover $[0,\frac12)$. Put $\xi(x)=v_{k\bmod4}$ for $x\in J_k$, $\xi(x)=v_0$ for $x\geq\frac12$, and $\varphi(a)=\int_0^1\langle a(x)\xi(x),\xi(x)\rangle\,dx .$ GNS. On $\mathcal H=L^2([0,1];\mathbb C^2)$ let $\pi(a)$ be multiplication by $a(\cdot)$; then $\varphi(a)=\langle\pi(a)\xi,\xi\rangle$. The vector $\xi$ is cyclic: if $\eta\perp\pi(f\otimes E_{ij})\xi$ for all $f\in C[0,1]$, then $\xi_j\overline{\eta_i}=0$ almost everywhere for all $i,j$, and since $\xi(x)\neq0$, $\eta=0$. Commutant. An operator $T\in\pi(A)'$ commutes with $1\otimes E_{ij}$, so $T=t\otimes1$ with $t\in B(L^2[0,1])$. It commutes with $\pi(f\otimes1)=M_f\otimes1$ for $f\in C[0,1]$; by Proposition 2.1(6) and the normality of $g\mapsto M_g$ it commutes with every $M_g$, $g\in L^\infty$, so $t=M_g$ for some $g\in L^\infty$, because the multiplication operators by $L^\infty$ form a maximal abelian algebra on $L^2[0,1]$. So $\pi_\varphi(A)'=\{M_g\otimes1\}$ is abelian. The face. By (1), $F_\varphi$ consists of the states $a\mapsto\int g(x)\langle a(x)\xi(x),\xi(x)\rangle dx$ with $g\geq0$ bounded and $\int g=1$. Taking $g=\chi_{J_k}/|J_k|$ with $k\equiv j\pmod4$ and letting $k\to\infty$ gives, in the closure, the four states $\rho_j(a)=\langle a(\frac12)v_j,v_j\rangle$. They are pure (vector states of the irreducible representation $a\mapsto a(\frac12)$), hence extreme in $\overline{F_\varphi}$. Since $|v_0\rangle\langle v_0|+|v_2\rangle\langle v_2|=1=|v_1\rangle\langle v_1|+|v_3\rangle\langle v_3|$, the point $\rho=\frac12(\rho_0+\rho_2)=\frac12(\rho_1+\rho_3)$ of $\overline{F_\varphi}$ is the barycentre of two different measures concentrated on extreme points. Both are boundary measures (Proposition 5.5(2)), hence maximal (Theorem 5.4). By the Choquet–Meyer theorem, $\overline{F_\varphi}$ is not a simplex. $\square$

Lemma 16.2. An infinite-dimensional von Neumann algebra $\mathcal N$ contains an infinite sequence of nonzero, pairwise orthogonal projections.

Proof. Suppose not. Then every nonzero projection dominates a minimal one, since a strictly decreasing sequence $p>p_1>p_2>\dots$ would give the orthogonal nonzero differences $p_k-p_{k+1}$. Let $\mathcal A$ be a maximal abelian $*$-subalgebra; it equals $\mathcal A'\cap\mathcal N$, a von Neumann algebra. A maximal orthogonal family of minimal projections of $\mathcal A$ is finite, say $e_1,\dots,e_n$, and $\sum e_i=1$. Each $e_i\mathcal A=\mathbb Ce_i$, by the spectral theorem. A self-adjoint $y\in e_i\mathcal Ne_i$ commutes with every $e_j$, hence with $\mathcal A$, so $y\in\mathcal A\cap e_i\mathcal Ne_i=\mathbb Ce_i$. So $e_i\mathcal Ne_i=\mathbb Ce_i$. For $v,w\in e_i\mathcal Ne_j$, $w^*v\in e_j\mathcal Ne_j=\mathbb Ce_j$ and $vv^*\in\mathbb Ce_i$; if $w\neq0$, scale it so that $ww^*=e_i$, and then $v=ww^*v\in\mathbb Cw$. So $\dim e_i\mathcal Ne_j\leq1$, and $\dim\mathcal N\leq n^2$. $\square$

17. Invariant states

Let $G$ be a group, with no topology, and $s\mapsto\alpha_s$ a homomorphism of $G$ into the automorphism group of $A$. The invariant states are $\mathfrak S^G=\{\omega\in\mathfrak S:\ \omega\circ\alpha_s=\omega\text{ for all }s\}$, and the ergodic states are the extreme points $\partial_e\mathfrak S^G$. The results of this section are used for the ergodic decomposition in Section 19, and in Sections 23 and 24.

Proposition 17.1. Let $\varphi\in\mathfrak S^G$, with GNS triple $\pi,H,\xi$.

$\mathfrak S^G$ is compact and convex.
There is exactly one unitary representation $U=U_\varphi$ of $G$ on $H$ with $U_s\pi(a)\xi=\pi(\alpha_s(a))\xi$. It satisfies $U_s\xi=\xi$ and $U_s\pi(a)U_s^*=\pi(\alpha_s(a))$.
Let $\mathfrak M_\varphi=\pi(A)'\cap U(G)'$. Then $\Theta_\varphi$ maps $\{x\in\mathfrak M_\varphi:\ 0\leq x\leq1\}$ onto the set of invariant functionals $\psi$ with $0\leq\psi\leq\varphi$.
$\varphi\in\partial_e\mathfrak S^G$ if and only if $\mathfrak M_\varphi=\mathbb C1$.
If $\mu\in M_\varphi(\mathfrak S)$ and $\kappa_\mu(L^\infty(\mu))\subseteq U(G)'$, then $\operatorname{supp}\mu\subseteq\mathfrak S^G$, so $\mu(\mathfrak S^G)=1$.
Suppose $\mathfrak M_\varphi$ is abelian, and let $\mu$ be the orthogonal measure with $\mathcal C_\mu=\mathfrak M_\varphi$ (Theorem 13.1). Then $\mu$ is concentrated on $K=\mathfrak S^G$, and, as a measure on $K$, it majorizes every probability measure on $K$ with barycentre $\varphi$. So it is the unique maximal measure on $K$ that represents $\varphi$, and $\mu(B)=0$ for every Baire subset $B$ of the compact space $K$ that misses $\partial_eK$.

Proof. (1) $\mathfrak S^G$ is the intersection of the closed convex sets $\{\omega:\ \omega(\alpha_s(a))=\omega(a)\}$.

(2) $\|\pi(\alpha_s(a))\xi\|^2=\varphi(\alpha_s(a^*a))=\|\pi(a)\xi\|^2$, so $U_s$ is a well-defined isometry on $\pi(A)\xi$ with dense range; it extends to a unitary, and $U_sU_t=U_{st}$. Then $\begin{gathered} U_s\pi(a)U_s^*\pi(b)\xi\\ =U_s\pi(a\,\alpha_{s^{-1}}(b))\xi\\ =\pi(\alpha_s(a))\pi(b)\xi \end{gathered}$ , and $U_s\xi=\pi(\alpha_s(1))\xi=\xi$.

(3) For $x\in\pi(A)'$, $U_s^*xU_s\in\pi(A)'$, and since $U_s\xi=\xi$, $\begin{gathered} \Theta_\varphi(x)(\alpha_s(a))\\ =\langle U_s\pi(a)U_s^*x\xi,\xi\rangle\\ =\langle\pi(a)\,U_s^*xU_s\,\xi,\xi\rangle\\ =\Theta_\varphi(U_s^*xU_s)(a). \end{gathered}$ So $\Theta_\varphi(x)$ is invariant exactly when $U_s^*xU_s=x$ for all $s$, because $\Theta_\varphi$ is one-to-one. Combine with Proposition 8.1(3).

(4) The proof of Proposition 8.1(4), (a)$\Leftrightarrow$(c), works word for word with $\mathfrak S^G$ in place of $\mathfrak S$ and $\mathfrak M_\varphi$ in place of $\pi(A)'$, using (3).

(5) For $f\in L^\infty(\mu)$, $s\in G$ and $a\in A$, since $\kappa_\mu(f)$ commutes with $U_s$ and $U_s^*\xi=\xi$, $\begin{gathered} \int f\,\widehat{\alpha_s(a)}\,d\mu\\ =\langle\kappa_\mu(f)U_s\pi(a)U_s^*\xi,\xi\rangle\\ =\langle U_s\kappa_\mu(f)\pi(a)\xi,\xi\rangle\\ =\int f\hat a\,d\mu . \end{gathered}$ So $\widehat{\alpha_s(a)}=\hat a$ almost everywhere. The closed set $\{\omega:\ \omega(\alpha_s(a))=\omega(a)\}$ therefore has a null complement and contains $\operatorname{supp}\mu$ (Proposition 2.1(3)). Intersect over $s$ and $a$.

(6) By (5), $\mu$ is concentrated on the closed convex set $K$, and its restriction is a probability measure on $K$ with barycentre $\varphi$. Let $\lambda\in M_1^+(K)$ have barycentre $\varphi$, and $\lambda=\sum_i\lambda_i$ with $\lambda_i\geq0$. Then $\lambda_i=g_i\lambda$ with $0\leq g_i\leq1$ and $\sum g_i=1$. The functional $\psi_i(a)=\lambda_i(\hat a)$ is positive, satisfies $\psi_i\leq\varphi$, and is invariant, because every point of $K$ is. By (3), $\psi_i=\Theta_\varphi(x_i)$ with $x_i\in\mathfrak M_\varphi$, $x_i\geq0$, and $\sum_ix_i=1$ since $\sum_i\psi_i=\varphi$. As $\mathfrak M_\varphi=\mathcal C_\mu$, $x_i=\kappa_\mu(f_i)$ with $f_i\geq0$, $\sum f_i=1$. Put $\mu_i=f_i\mu$. Then $\mu_i(\hat a)=\psi_i(a)=\lambda_i(\hat a)$. The functions $\hat h|_K$, $h\in A_h$, form $\operatorname{Aff}_E(K)$, which is dense in $\operatorname{Aff}(K)$ (Lemma 1.2); so $\mu_i\sim\lambda_i$ on $K$. By Lemma 4.10 on $K$, $\lambda\prec\mu$. If $\mu\prec\lambda'$ on $K$, then $\lambda'$ represents $\varphi$, so $\lambda'\prec\mu$ and $\lambda'=\mu$: $\mu$ is maximal, and every maximal $\lambda$ equals $\mu$. The last claim is Theorems 5.4 and 6.6 on $K$. $\square$

For $G=\{1\}$, part (6) is the implication (1)$\Rightarrow$(2) of Theorem 15.2.

18. The algebra $C(X,A)$

The decomposition theorem of Section 19 is proved by passing to the algebra $C(X,A)$ of continuous $A$-valued functions, which plays the role of a tensor product of $C(X)$ and $A$. The facts we need about it are short.

Proposition 18.1. Let $X$ be compact Hausdorff and $\mathcal D=C(X,A)$, the continuous functions $X\to A$ with pointwise operations and the sup norm. For $f\in C(X)$ and $a\in A$ let $f\otimes a$ be the function $x\mapsto f(x)a$.

$\mathcal D$ is a unital C$^*$-algebra, and the span $\mathcal F$ of the $f\otimes a$ is dense in it.
Let $\theta:C(X)\to B(H)$ be a unital $*$-homomorphism and $\pi$ a unital representation of $A$ on $H$ with $\theta(C(X))\subseteq\pi(A)'$. There is exactly one representation $\tilde\pi$ of $\mathcal D$ with $\tilde\pi(f\otimes a)=\theta(f)\pi(a)$, and $\tilde\pi(\mathcal D)'=\pi(A)'\cap\theta(C(X))'$.
Let $G$ act on $A$ by $\alpha$, and on $\mathcal D$ by $\beta_s(F)=\alpha_s\circ F$; $\beta$ fixes every $f\otimes1$. Every $\rho\in\partial_e\mathfrak S^G(\mathcal D)$ has the form $\rho(F)=\omega(F(x))$ for some $x\in X$ and some $\omega\in\partial_e\mathfrak S^G(A)$. For $G=\{1\}$: every pure state of $\mathcal D$ is $F\mapsto\omega(F(x))$ with $\omega$ pure.
Take $X=\mathfrak S(A)$. Then $\Phi(\omega)(F)=\omega(F(\omega))$ defines a continuous one-to-one map $\Phi:\mathfrak S(A)\to\mathfrak S(\mathcal D)$ with $\Phi(\mathfrak S^G(A))\subseteq\mathfrak S^G(\mathcal D)$. And $\Psi(\rho)(a)=\rho(1\otimes a)$ defines a continuous affine map $\Psi:\mathfrak S(\mathcal D)\to\mathfrak S(A)$ with $\Psi\circ\Phi=\mathrm{id}$, $\Psi(\mathfrak S^G(\mathcal D))\subseteq\mathfrak S^G(A)$ and $\Psi(\partial_e\mathfrak S^G(\mathcal D))\subseteq\partial_e\mathfrak S^G(A)$.

Proof. (1) $\|F^*F\|=\sup_x\|F(x)\|^2$, and completeness follows from uniform convergence: a sup-norm Cauchy sequence has a pointwise limit in the Banach space $A$, the Cauchy estimate passes to the limit uniformly, and a uniform limit of continuous maps is continuous. Given $F$ and $\varepsilon>0$, cover $X$ by finitely many open sets $U_k\ni x_k$ with $\|F(y)-F(x_k)\|<\varepsilon$ on $U_k$, and take a partition of unity $(u_k)$ subordinate to this cover (constructed explicitly in the background below from Urysohn functions). Then $\|F-\sum_ku_k\otimes F(x_k)\|\leq\varepsilon$.

(2) Let $F=\sum_{i=1}^mf_i\otimes a_i$ and $\delta>0$. Choose a partition of unity $(u_k)_{k\leq n}$ and points $x_k$ with $|f_i(y)-f_i(x_k)|<\delta$ for $y\in\operatorname{supp}u_k$ and all $i$. Then $\begin{gathered} \sum_i\theta(f_i)\pi(a_i)\\ =\sum_k\theta(u_k)\pi(F(x_k))\\ +\sum_i\theta(g_i)\pi(a_i),\\ g_i\\ =\sum_ku_k\,(f_i-f_i(x_k)), \end{gathered}$ and $\|g_i\|\leq\delta$. The operator $\begin{gathered} T\\ =\sum_k\theta(u_k)\pi(F(x_k))\\ =\sum_k\theta(u_k)^{1/2}\pi(F(x_k))\theta(u_k)^{1/2} \end{gathered}$ equals $R^*DR$, where $R\eta=(\theta(u_k)^{1/2}\eta)_k$ maps $H$ into $H^n$ with $\|R\|\leq1$ (because $\sum_k\theta(u_k)=1$), and $D=\operatorname{diag}(\pi(F(x_k)))$. So $\|T\|\leq\max_k\|F(x_k)\|\leq\|F\|$, and $\|\sum_i\theta(f_i)\pi(a_i)\|\leq\|F\|+\delta\sum_i\|a_i\|$. Letting $\delta\to0$ shows that $\tilde\pi$ is well defined and contractive on $\mathcal F$. It is a $*$-homomorphism there, because $\theta(f)$ commutes with $\pi(b)$; it extends to $\mathcal D$ by (1). The image is the closed span of $\theta(C(X))\pi(A)$, which contains $\pi(A)$ and $\theta(C(X))$; this gives the commutant.

(3) Each $z=f\otimes1$ with $0\leq f\leq1$ is central and $\beta$-fixed. So $F\mapsto\rho(zF)$ and $F\mapsto\rho((1-z)F)$ are positive ($\rho(zF^*F)=\rho(z^{1/2}F^*Fz^{1/2})$), invariant, and add up to $\rho$. If $t=\rho(z)\in(0,1)$, extremality gives $\rho(z\,\cdot)=t\rho$. If $t=0$, the Cauchy–Schwarz inequality gives $|\rho(zF)|^2\leq\rho(z^2)\rho(F^*F)\leq\rho(z)\|F\|^2=0$; if $t=1$, apply this to $1-z$. So $\rho(zF)=\rho(z)\rho(F)$ for all such $z$, hence for all $z=f\otimes1$. Then $\rho$ is multiplicative on $C(X)\otimes1$, so it is evaluation at some $x\in X$ there, and $\rho(f\otimes a)=f(x)\rho(1\otimes a)$. With $\omega=\rho(1\otimes\cdot)$, $\rho(F)=\omega(F(x))$ on $\mathcal F$, hence on $\mathcal D$. $\omega$ is invariant. If $\omega=t\omega_1+(1-t)\omega_2$ with $\omega_i\in\mathfrak S^G(A)$ and $0<t<1$, then $\rho_i(F)=\omega_i(F(x))$ are invariant states with $\rho=t\rho_1+(1-t)\rho_2$; so $\rho_i=\rho$ and $\omega_i=\omega$.

(4) $F\mapsto F(\omega)$ is a unital $*$-homomorphism, so $\Phi(\omega)$ is a state. If $\omega_j\to\omega$, then $\begin{gathered} |\omega_j(F(\omega_j))-\omega(F(\omega))|\\ \leq\|F(\omega_j)-F(\omega)\|\\ +|\omega_j(F(\omega))-\omega(F(\omega))|\\ \to0 \end{gathered}$ . $\Psi\circ\Phi=\mathrm{id}$ is clear, so $\Phi$ is one-to-one. For invariant $\omega$, $\Phi(\omega)(\beta_sF)=\omega(\alpha_s(F(\omega)))=\Phi(\omega)(F)$. $\Psi$ is continuous and affine, and it maps invariant states to invariant states because $\beta_s(1\otimes a)=1\otimes\alpha_s(a)$. The last claim is (3). $\square$

19. Decomposition into pure and ergodic states

The next theorem decomposes an invariant state along a maximal abelian subalgebra of $\mathfrak M_\varphi$. For the trivial group it decomposes an arbitrary state into pure states.

Theorem 19.1. Let $G$ act on $A$, $\varphi\in\mathfrak S^G$, and let $\mathcal C$ be a maximal abelian von Neumann subalgebra of $\mathfrak M_\varphi=\pi_\varphi(A)'\cap U_\varphi(G)'$, that is, $\mathcal C'\cap\mathfrak M_\varphi=\mathcal C$. Let $\mu$ be the orthogonal measure with $\mathcal C_\mu=\mathcal C$ (Theorem 13.1). Then:

$\mu(\mathfrak S^G)=1$;
$\mu(B)=0$ for every Baire subset $B$ of the compact space $\mathfrak S^G$ with $B\cap\partial_e\mathfrak S^G=\varnothing$;
if $A$ is separable, $\mu$ is concentrated on $\partial_e\mathfrak S^G$;
such algebras $\mathcal C$ exist. So every invariant state is the barycentre of an orthogonal measure that vanishes on each Baire subset of $\mathfrak S^G$ missing the ergodic states.

For $G=\{1\}$, $\mathfrak S^G=\mathfrak S$ and $\partial_e\mathfrak S^G=P(A)$: if $\mathcal C_\mu$ is maximal abelian in $\pi_\varphi(A)'$, then $\mu$ vanishes on every Baire set that misses the pure states, and it is concentrated on $P(A)$ when $A$ is separable.

Proof. (1) is Proposition 17.1(5), since $\mathcal C\subseteq U(G)'$. Write $\pi,H,\xi,U$ for $\varphi$, $K_A=\mathfrak S^G(A)$, and $\mathcal D=C(\mathfrak S(A),A)$ with the action $\beta$ of Proposition 18.1(3).

The auxiliary state. $\kappa_\mu$ restricted to $C(\mathfrak S)$ is a unital $*$-homomorphism into $\mathcal C\subseteq\pi(A)'$. Proposition 18.1(2) gives a representation $\tilde\pi$ of $\mathcal D$ with $\tilde\pi(f\otimes a)=\kappa_\mu(f)\pi(a)$. Since $\kappa_\mu(f)\in U(G)'$, we get $\tilde\pi(\beta_sF)=U_s\tilde\pi(F)U_s^*$ on $\mathcal F$, hence on $\mathcal D$. So $\tilde\varphi(F)=\langle\tilde\pi(F)\xi,\xi\rangle$ is a $\beta$-invariant state. Its GNS triple is $(\tilde\pi,H,\xi)$, since $\xi$ is cyclic already for $\pi(A)$, and its unitary representation is $U$. The algebra $\kappa_\mu(C(\mathfrak S))$ is $\sigma$-weakly dense in $\mathcal C$ (Proposition 2.1(6) and the normality of $\kappa_\mu$), so the two have the same commutant. By Proposition 18.1(2), $\begin{gathered} \tilde\pi(\mathcal D)'\cap U(G)'\\ =\pi(A)'\cap\mathcal C'\cap U(G)'\\ =\mathcal C'\cap\mathfrak M_\varphi\\ =\mathcal C, \end{gathered}$ which is abelian. By Proposition 17.1(6) for $(\mathcal D,\beta,\tilde\varphi)$, the orthogonal measure $\tilde\mu$ on $\mathfrak S(\mathcal D)$ with $\mathcal C_{\tilde\mu}=\mathcal C$ is concentrated on $K_{\mathcal D}=\mathfrak S^G(\mathcal D)$, and $\tilde\mu(B')=0$ for every Baire subset $B'$ of $K_{\mathcal D}$ that misses $\partial_eK_{\mathcal D}$.

Identification: $\tilde\mu=\Phi_*\mu$. Let $\nu=\Phi_*\mu$, with $\Phi$ from Proposition 18.1(4); so $\nu(N)=\mu(\Phi^{-1}(N))$ for Borel $N$ (Proposition 2.1(7)). For a bounded Borel function $G_0$ on $\mathfrak S(\mathcal D)$, $f\in C(\mathfrak S)$ and $a\in A$, with $g=G_0\circ\Phi$, $\begin{gathered} \int G_0\,\widehat{f\otimes a}\,d\nu\\ =\int g(\omega)f(\omega)\omega(a)\,d\mu(\omega)\\ =\langle\kappa_\mu(gf)\pi(a)\xi,\xi\rangle\\ =\langle\kappa_\mu(g)\tilde\pi(f\otimes a)\xi,\xi\rangle . \end{gathered}$ Both ends are continuous in $F\in\mathcal D$, so $\int G_0\hat F\,d\nu=\langle\kappa_\mu(G_0\circ\Phi)\tilde\pi(F)\xi,\xi\rangle$ for all $F$. With $G_0=1$, $\nu$ represents $\tilde\varphi$. By the uniqueness in Proposition 9.2, $\kappa_\nu=\kappa_\mu\circ\Phi^\#$, where $\Phi^\#(G_0)=G_0\circ\Phi$ is a $*$-isomorphism of $L^\infty(\nu)$ onto $L^\infty(\mu)$ (Proposition 2.1(7)). So $\nu$ is orthogonal with $\mathcal C_\nu=\mathcal C=\mathcal C_{\tilde\mu}$, and $\nu=\tilde\mu$ by Theorem 13.1.

Conclusion. Let $B\subseteq K_A$ be a Baire subset of $K_A$ with $B\cap\partial_eK_A=\varnothing$. The map $\Psi$ of Proposition 18.1(4) sends $K_{\mathcal D}$ into $K_A$, so $B'=\Psi^{-1}(B)\cap K_{\mathcal D}$ is a Baire subset of $K_{\mathcal D}$ (Proposition 2.1(4)(b)), and it misses $\partial_eK_{\mathcal D}$ because $\Psi(\partial_eK_{\mathcal D})\subseteq\partial_eK_A$. Since $\Psi\circ\Phi=\mathrm{id}$ and $\nu=\tilde\mu$ is concentrated on $K_{\mathcal D}$, $\begin{gathered} \mu(B)\\ =\mu\bigl(\Phi^{-1}(\Psi^{-1}(B))\bigr)\\ =\nu(\Psi^{-1}(B))\\ =\tilde\mu(B')\\ =0 . \end{gathered}$

(3) If $A$ is separable, $\mathfrak S$ and $K_A$ are metrizable (Proposition 8.1(5)). By Lemmas 6.2 and 6.3, $\partial_eK_A$ is a $G_\delta$, and every Borel subset of $K_A$ is a Baire set (Proposition 2.1(4)(d)). Apply (2) to $K_A\setminus\partial_eK_A$.

(4) By Zorn's lemma, $\mathfrak M_\varphi$ has a maximal commutative $*$-subalgebra $\mathcal C$ containing $1$. If $x\in\mathcal C'\cap\mathfrak M_\varphi$, its real and imaginary parts lie there too, and each generates with $\mathcal C$ a commutative $*$-algebra; so they lie in $\mathcal C$. Hence $\mathcal C=\mathcal C'\cap\mathfrak M_\varphi$, which is weakly closed. $\square$

Remark 19.2 (Non-uniqueness). Different maximal abelian subalgebras give different measures, and they need not be conjugate. For example, let $H=\ell^2$, $A=\mathbb C1+K(H)$ (separable and unital), $(e_n)$ the standard basis, and $\varphi(a)=\sum_n2^{-n}\langle ae_n,e_n\rangle$. On $H\otimes\ell^2$ the vector $\zeta=\sum_n2^{-n/2}e_n\otimes e_n$ is cyclic for $\{a\otimes1\}$, and gives $\varphi$; a matrix-unit computation with the operators $e_i\otimes e_j^*\in K(H)$ shows that the commutant is $1\otimes B(\ell^2)$. In $B(\ell^2)$ the diagonal algebra $\ell^\infty$ is maximal abelian and has minimal projections; the multiplication algebra $L^\infty[0,1]$ on $L^2[0,1]\cong\ell^2$ is maximal abelian and has none. So the two are not unitarily conjugate. Their orthogonal measures (Theorem 13.1) are different, and both are concentrated on $P(A)$ by Theorem 19.1(3), since $A$ is separable. The centre $\mathcal Z_\varphi=\pi_\varphi(A)''\cap\pi_\varphi(A)'$, on the other hand, is determined by $\varphi$ alone. This leads to the central measure.

20. The central measure and factorial states

Definition 20.1. The central measure of $\varphi\in\mathfrak S$ is the orthogonal $\mu\in M_\varphi(\mathfrak S)$ with $\mathcal C_\mu=\mathcal Z_\varphi=\pi_\varphi(A)''\cap\pi_\varphi(A)'$; it exists and is unique by Theorem 13.1. A state $\varphi$ is factorial (or primary) if $\pi_\varphi(A)''$ is a factor. We write $\operatorname{Fac}(A)$ for the set of factorial states. Pure states are factorial, since $\pi_\varphi(A)'=\mathbb C1$ (Proposition 8.1(4)).

Lemma 20.2 (Corners of a factor). Let $\mathcal N$ be a factor on $H$ and $e\in\mathcal N$ a nonzero projection. Then $\mathcal N_e=\{exe|_{eH}:\ x\in\mathcal N\}$ is a factor on $eH$.

Proof. $\mathcal N_e$ is a von Neumann algebra (Lemma 12.2(3)). Let $z$ be a projection in its centre, say $z=eye|_{eH}$ with $y\in\mathcal N$, and let $\tilde z=eye$, a projection in $\mathcal N$ with $\tilde z\leq e$. For $x\in\mathcal N$, $\tilde z\,exe=exe\,\tilde z$, since $z$ is central in $\mathcal N_e$. So $\tilde zx(e-\tilde z)=\tilde z(exe)(e-\tilde z)=(exe)\tilde z(e-\tilde z)=0$. The projection $f$ onto $[\mathcal N(e-\tilde z)H]$ commutes with $\mathcal N$, and also with $\mathcal N'$, because $\mathcal N'\mathcal N(e-\tilde z)H=\mathcal N(e-\tilde z)\mathcal N'H$. So $f\in\mathcal N\cap\mathcal N'=\mathbb C1$. If $e\neq\tilde z$, then $f=1$, and $\tilde z$ vanishes on $[\mathcal N(e-\tilde z)H]=H$, so $\tilde z=0$. Thus $z\in\{0,1\}$. An abelian von Neumann algebra whose only projections are $0$ and $1$ is $\mathbb C1$, because the spectral projections of a self-adjoint element lie in it (the spectral theorem). $\square$

Lemma 20.3. Let $C$ be a C$^*$-algebra with unit, and let $A_1,A_2\subseteq C$ be C$^*$-subalgebras that contain the unit of $C$, commute with each other, and together generate $C$. If $\rho$ is a factorial state of $C$, then $\rho|_{A_1}$ is a factorial state of $A_1$.

Proof. Let $\pi=\pi_\rho$ on $H$, with cyclic vector $\xi$, and $\mathcal M=\pi(C)''$, a factor. Let $\mathcal Z_1=\pi(A_1)''\cap\pi(A_1)'$. Since $\pi(A_2)\subseteq\pi(A_1)'$, we have $\pi(A_1)''\subseteq\pi(A_2)'$. So $\mathcal Z_1$ commutes with $\pi(A_1)$ and $\pi(A_2)$, hence with $\pi(C)$, and $\mathcal Z_1\subseteq\mathcal M'\cap\mathcal M=\mathbb C1$. Thus $\pi(A_1)''$ is a factor, and so is $\mathcal N=\pi(A_1)'$, which has the same centre. Let $e\in\mathcal N$ be the projection onto $[\pi(A_1)\xi]$. By the uniqueness of the GNS construction, the GNS representation of $\psi=\rho|_{A_1}$ is $a\mapsto\pi(a)|_{eH}$. Restriction to $eH$ is weakly continuous, so $\{\pi(a)|_{eH}:\ a\in A_1\}$ and $(\pi(A_1)'')_e$ have the same commutant. By Lemma 12.2(1), applied to the von Neumann algebra $\pi(A_1)''$, whose commutant $\mathcal N$ contains $e$, this commutant is $\mathcal N_e$. By Lemma 20.2, $\mathcal N_e$ is a factor. So $\pi_\psi(A_1)'$, and with it $\pi_\psi(A_1)''$, is a factor. $\square$

Theorem 20.4. The central measure $\mu$ of $\varphi\in\mathfrak S$ satisfies $\mu(B)=0$ for every Baire set $B\subseteq\mathfrak S$ with $B\cap\operatorname{Fac}(A)=\varnothing$.

Proof. Write $\pi,H,\xi$ for $\varphi$ and $\mathcal Z=\mathcal Z_\varphi$. Let $A_1$ be the norm closure of $\pi(A)$, $A_2=\pi(A)'$, and $C\subseteq B(H)$ the C$^*$-algebra they generate. Then $C'=\pi(A)'\cap\pi(A)''=\mathcal Z$. The vector state $\psi(x)=\langle x\xi,\xi\rangle$ on $C$ has GNS triple $(\mathrm{id},H,\xi)$, because $\xi$ is cyclic for $A_1$. Its commutant $\mathcal Z$ is abelian, so by Theorem 15.2 the orthogonal measure $\nu$ on $\mathfrak S(C)$ with $\mathcal C_\nu=\mathcal Z$ is the only maximal measure representing $\psi$. Being maximal, it is a boundary measure (Theorem 5.4), so by Theorem 6.6 it vanishes on every Baire set that misses $P(C)$.

Let $\Phi:\mathfrak S(C)\to\mathfrak S(A)$, $\Phi(\rho)=\rho\circ\pi$, a continuous affine map. If $\rho\in P(C)$, then $\rho|_{A_1}$ is factorial by Lemma 20.3, and so is $\rho\circ\pi$: the GNS representation of $\rho\circ\pi$ is that of $\rho|_{A_1}$ composed with $\pi$, and $\pi(A)$ is norm-dense in $A_1$, so the two bicommutants agree. Thus $\Phi(P(C))\subseteq\operatorname{Fac}(A)$.

We show $\Phi_*\nu=\mu$. First, $\int\hat a\,d\Phi_*\nu=\int\widehat{\pi(a)}\,d\nu=\psi(\pi(a))=\varphi(a)$. For $g\in L^\infty(\Phi_*\nu)$, the class of $g\circ\Phi$ in $L^\infty(\nu)$ is well defined, since $\Phi_*\nu(N)=\nu(\Phi^{-1}(N))$ for Borel $N$ (Proposition 2.1(7); here $\Phi$ need not be one-to-one). The operator $\kappa_\nu(g\circ\Phi)$ lies in $\mathcal Z\subseteq\pi(A)'$, and $\begin{gathered} \langle\kappa_\nu(g\circ\Phi)\pi(a)\xi,\xi\rangle\\ =\int(g\circ\Phi)\,\widehat{\pi(a)}\,d\nu\\ =\int g\hat a\,d\Phi_*\nu . \end{gathered}$ By the uniqueness in Proposition 9.2, $\kappa_{\Phi_*\nu}(g)=\kappa_\nu(g\circ\Phi)$. This is multiplicative, so $\Phi_*\nu$ is orthogonal. It also shows $\mathcal C_{\Phi_*\nu}\subseteq\mathcal C_\nu=\mathcal Z$, but not equality, since $g\mapsto g\circ\Phi$ need not map onto $L^\infty(\nu)$; so we compare with $\mu$ directly. By (10.1) for $\nu$, $\kappa_{\Phi_*\nu}(\hat a)\xi=\kappa_\nu(\widehat{\pi(a)})\xi=e\pi(a)\xi$, where $e$ is the projection onto $[\mathcal Z\xi]$. By (10.1) for $\mu$, $\kappa_\mu(\hat a)\xi=e\pi(a)\xi$ as well, since $\mathcal C_\mu=\mathcal Z$. The separating vector $\xi$ gives $\kappa_{\Phi_*\nu}(\hat a)=\kappa_\mu(\hat a)$, and the uniqueness part of the proof of Theorem 13.1 gives $\Phi_*\nu=\mu$.

Now let $B$ be a Baire set missing $\operatorname{Fac}(A)$. Then $\Phi^{-1}(B)$ is a Baire set (Proposition 2.1(4)(b)) that misses $P(C)$, because $\Phi(P(C))\subseteq\operatorname{Fac}(A)$. So $\mu(B)=\nu(\Phi^{-1}(B))=0$. $\square$

21. The factorial states form a Borel set

When $A$ is separable, the factorial states form a Borel subset of $\mathfrak S$, so the central measure is concentrated on them. The proof tests factoriality through countably many closed conditions.

Let $\omega\in\mathfrak S$ be a state with GNS triple $\pi,H,\xi$. Put

$\begin{gathered} I(\omega)\\ =\{\rho\in A^*:\\ \ 0\\ \leq\rho\\ \leq\omega\}\\ =\Theta_\omega(\{x\in\pi(A)':\\ \ 0\\ \leq x\\ \leq1\}) \end{gathered}$ (Proposition 8.1(3));
$Z(\omega)=I(\omega)\cap\Theta_\omega(\mathcal Z_\omega)$.

Since $\Theta_\omega$ is one-to-one and preserves order both ways, $Z(\omega)=\Theta_\omega(\{z\in\mathcal Z_\omega:\ 0\leq z\leq1\})$. So $\omega$ is factorial exactly when $Z(\omega)=\{\lambda\omega:\ 0\leq\lambda\leq1\}$.

Let $A\odot A$ be the algebraic tensor product with the projective (greatest cross) norm $\gamma(u)=\inf\{\sum_i\|x_i\|\|y_i\|:\ u=\sum_ix_i\otimes y_i\}$, let $\operatorname{mul}(\sum_ix_i\otimes y_i)=\sum_ix_iy_i$ be the multiplication map, and $N=\ker\operatorname{mul}$. For $T\in B(H)$ put $\Phi_{\omega,T}\Bigl(\sum_ix_i\otimes y_i\Bigr)=\sum_i\langle\pi(x_i)T\pi(y_i)\xi,\xi\rangle ,$ and $\Phi_{\omega,h}=\Phi_{\omega,\pi(h)}$ for $h\in A$, so $\Phi_{\omega,h}(u)=\omega(\sum_ix_ihy_i)$. Also $\Phi_\rho=\rho\circ\operatorname{mul}$ for $\rho\in A^*$.

Lemma 21.1.

$|\Phi_{\omega,T}(u)|\leq\|T\|\gamma(u)$.
$\Phi_{\omega,T}(N)=\{0\}$ if and only if $T\in\pi(A)'$.
A linear functional $\Phi$ on $A\odot A$ equals $\Phi_\rho$ for some $\rho\in I(\omega)$ if and only if $\Phi(N)=\{0\}$ and $0\leq\Phi(x^*\otimes x)\leq\omega(x^*x)$ for all $x\in A$.

Proof. (1) $|\langle\pi(x)T\pi(y)\xi,\xi\rangle|\leq\|x\|\|T\|\|y\|$. (2) If $T\in\pi(A)'$, then $\Phi_{\omega,T}(u)=\langle T\pi(\operatorname{mul}(u))\xi,\xi\rangle$. Conversely, $x\otimes y-xy\otimes1\in N$ gives $\langle T\pi(y)\xi,\pi(x^*)\xi\rangle=\langle\pi(y)T\xi,\pi(x^*)\xi\rangle$ for all $x$, so $T\pi(y)\xi=\pi(y)T\xi$. Then $T\pi(y)\pi(z)\xi=\pi(yz)T\xi=\pi(y)T\pi(z)\xi$, so $T\pi(y)=\pi(y)T$. (3) $\operatorname{mul}$ is onto ($x=\operatorname{mul}(x\otimes1)$), so $\Phi(N)=0$ means $\Phi=\rho\circ\operatorname{mul}$ with $\rho(x)=\Phi(x\otimes1)$ linear. The inequalities say $0\leq\rho(x^*x)\leq\omega(x^*x)$. A positive functional on a C$^*$-algebra is bounded, so $\rho\in I(\omega)$. The converse is clear. $\square$

Lemma 21.2. Let $N_0$ be a $\gamma$-dense subset of $N$, and $A_0$ a dense $*$-subalgebra of $A$ over $\mathbb Q+i\mathbb Q$. A state $\omega$ is factorial exactly when the following holds. For every $\varepsilon>0$ and $x\in A_0$ there are $\delta>0$ and $u_1,\dots,u_n\in N_0$ such that, for every $h\in A_0$ with $0\leq h\leq1$, $\begin{gathered} \max_i|\Phi_{\omega,h}(u_i)|<\delta\\ \Longrightarrow\\ |\omega(x^*hx)-\omega(h)\omega(x^*x)|<\varepsilon . \end{gathered}$

Proof. If the condition fails, $\omega$ is not factorial. Then there are $\varepsilon>0$ and $x\in A_0$ such that for every pair $\lambda=(\delta,F)$, with $\delta>0$ and $F\subseteq N_0$ finite, some $h_\lambda\in A_0$ with $0\leq h_\lambda\leq1$ has $|\Phi_{\omega,h_\lambda}(u)|<\delta$ for $u\in F$ and $|\omega(x^*h_\lambda x)-\omega(h_\lambda)\omega(x^*x)|\geq\varepsilon$. Direct the pairs by $(\delta,F)\leq(\delta',F')$ if $\delta'\leq\delta$ and $F\subseteq F'$. The operators $\pi(h_\lambda)$ lie in the weakly compact set $\{T\in\pi(A)'':\ 0\leq T\leq1\}$, so a subnet converges weakly to some $T$ there. Each $T\mapsto\Phi_{\omega,T}(u)$ is weakly continuous, so $\Phi_{\omega,T}(u)=0$ for $u\in N_0$, hence for $u\in N$ by Lemma 21.1(1). By Lemma 21.1(2), $T\in\pi(A)'$, so $T\in\mathcal Z_\omega$. But $|\langle T\pi(x)\xi,\pi(x)\xi\rangle-\langle T\xi,\xi\rangle\|\pi(x)\xi\|^2|\geq\varepsilon$, so $T$ is not a scalar, and $\mathcal Z_\omega\neq\mathbb C1$.

If the condition holds, $\omega$ is factorial. Let $T\in\mathcal Z_\omega$ with $0\leq T\leq1$. By the bicommutant theorem and Kaplansky's density theorem, $2T-1$ is the strong limit of a net of self-adjoint contractions $\pi(b_j)$, $b_j\in A_h$. Replace $b_j$ by $c_j=g(b_j)$ with $g(t)=\max(-1,\min(t,1))$: then $\|c_j\|\leq1$ and $\pi(c_j)=g(\pi(b_j))=\pi(b_j)$. So $h_j=\frac12(1+c_j)$ satisfies $0\leq h_j\leq1$ and $\pi(h_j)\to T$ strongly. Positive contractions of $A_0$ are norm-dense among those of $A$. Indeed, let $0\leq h\leq1$, $0<\eta<1/4$, and let $k\in A_0$ be self-adjoint with $\|k-h^{1/2}\|<\eta$ (self-adjoint elements of $A_0$ are dense in $A_h$). Then $\|k^2-h\|\leq2\eta+\eta^2$. Choose a rational $r$ with $(1+\eta)^{-2}-\eta\leq r\leq(1+\eta)^{-2}$. Then $rk^2\in A_0$, $0\leq rk^2\leq1$, and $\|rk^2-h\|\leq(2\eta+\eta^2)+(1-r)\leq5\eta+\eta^2$. Replacing $h_j$ by such an element of $A_0$ within norm distance $1/m$, and indexing the new net by the pairs $(j,m)$, ordered componentwise, we may take $h_j\in A_0$. For $u=\sum_ix_i\otimes y_i\in N$, $\begin{gathered} \Phi_{\omega,h_j}(u)\\ =\sum_i\langle\pi(h_j)\pi(y_i)\xi,\pi(x_i)^*\xi\rangle\to\Phi_{\omega,T}(u)\\ =0 \end{gathered}$ . Given $\varepsilon$ and $x\in A_0$, take $\delta$ and $u_1,\dots,u_n$ from the condition. For large $j$, $|\omega(x^*h_jx)-\omega(h_j)\omega(x^*x)|<\varepsilon$, and in the limit $|\langle T\pi(x)\xi,\pi(x)\xi\rangle-\langle T\xi,\xi\rangle\|\pi(x)\xi\|^2|\leq\varepsilon$. So the quadratic form of $S=T-\langle T\xi,\xi\rangle1$ vanishes on $\pi(A_0)\xi$. This set is dense and closed under combinations with coefficients in $\mathbb Q+i\mathbb Q$, which is enough for polarization; so $S=0$. The positive contractions span $\mathcal Z_\omega$, so $\mathcal Z_\omega=\mathbb C1$. $\square$

Theorem 21.3. If $A$ is separable, $\operatorname{Fac}(A)$ is a Borel subset of $\mathfrak S$; in fact an $F_{\sigma\delta}$ set.

The quantifier order in (21.1) is essential: Remark 21.4 shows that placing the intersection over $h$ before the unions over $m$ and $u$ can admit states that are not factorial.

Proof. Let $A_0$ be a countable dense $*$-subalgebra over $\mathbb Q+i\mathbb Q$ (generate one from a countable dense set). Finite sums of tensors $a\otimes b$, with $a,b\in A_0$, form a countable $\gamma$-dense set in $A\odot A$: approximate both factors in every finite representation and use $\gamma(x\otimes y-a\otimes b)\leq\|x-a\|\|y\|+\|a\|\|y-b\|$. Its subset $N$ is therefore separable by the countable-base argument in Lemma 6.3, and has a countable $\gamma$-dense subset $N_0$. For $m,n\geq1$, $h\in A_0$ with $0\leq h\leq1$, $u=(u_1,\dots,u_r)\in N_0^r$ and $x\in A_0$, let $\begin{gathered} \mathfrak S(m;n;h;u;x)\\ =\Bigl\{\omega\in\mathfrak S:\\ \ \max_{1\leq i\leq r}|\Phi_{\omega,h}(u_i)|\\ \geq\tfrac1m\ \\ \text{ or }\ |\omega(x^*hx)-\omega(h)\omega(x^*x)|\\ \leq\tfrac1n\Bigr\}. \end{gathered}$ Both conditions are closed in $\omega$, because $\Phi_{\omega,h}(u_i)=\omega(c_i)$ for a fixed $c_i\in A$ (if $u_i=\sum_jx_j\otimes y_j$, then $c_i=\sum_jx_jhy_j$), so the set is closed. With $\varepsilon=1/n$ and $\delta=1/m$, Lemma 21.2 says exactly $\begin{gathered} \operatorname{Fac}(A)\\ =\bigcap_{n\geq1}\ \bigcap_{x\in A_0}\ \\ \bigcup_{m\geq1}\ \bigcup_{r\geq1,\ u\in N_0^r}\ \\ \bigcap_{h\in A_0,\ 0\leq h\leq1}\mathfrak S(m;n;h;u;x). \end{gathered} \tag{21.1}$ (The passage between "$<\varepsilon$" and "$\leq1/n$", and between arbitrary $\delta>0$ and $1/m$, is harmless because both are quantified over all values.) The innermost intersection is closed, the countable union is $F_\sigma$, and the outer countable intersection is $F_{\sigma\delta}$. $\square$

Remark 21.4 (The order of the quantifiers). In (21.1) the intersection over $h$ must come after the unions over $m$ and $u$. If it is placed before them, then $m$ and $u$ may depend on $h$, and the set becomes too large. Example. Let $H=\ell^2$, $A=\mathbb C1+K(H)$, $\xi\in H$ a unit vector, $\chi(k+\lambda1)=\lambda$, and $\omega=\frac12(\omega_\xi+\chi)$. On $H\oplus\mathbb C$ the representation $k+\lambda1\mapsto(k+\lambda)\oplus\lambda$ with the cyclic vector $(\xi,1)/\sqrt2$ is the GNS representation of $\omega$, and $\pi_\omega(A)''=B(H)\oplus\mathbb C$. Its centre $\mathbb C\oplus\mathbb C$ is not trivial, so $\omega\notin\operatorname{Fac}(A)$. Yet $\omega$ lies in the set with the intersection over $h$ placed first. Fix $n$, $h$ and $x$. If $\Phi_{\omega,h}\neq0$ on $N$, it is nonzero at some $u\in N_0$ (Lemma 21.1(1) and the density of $N_0$), and a large $m$ puts $\omega$ in $\mathfrak S(m;n;h;u;x)$. If $\Phi_{\omega,h}=0$ on $N$, then $\pi_\omega(h)=(k+\lambda)\oplus\lambda$ is central (Lemma 21.1(2)), so $k+\lambda1\in\mathbb C1_H$, which forces $k=0$ as $H$ is infinite-dimensional; then $\pi_\omega(h)$ is a scalar, $\omega(x^*hx)=\omega(h)\omega(x^*x)$, and again $\omega\in\mathfrak S(m;n;h;u;x)$.

Corollary 21.5. If $A$ is separable, the central measure of every state is concentrated on $\operatorname{Fac}(A)$.

Proof. $\mathfrak S$ is metrizable, so the Borel set $\mathfrak S\setminus\operatorname{Fac}(A)$ is a Baire set (Proposition 2.1(4)(d)). Apply Theorem 20.4. $\square$

22. Examples and exercises

Example 22.1 (Commutative algebras). Let $A=C(X)$, $X$ compact Hausdorff, so $\mathfrak S=M_1^+(X)$, and let $\varphi=m$. The GNS representation is multiplication on $L^2(X,m)$, and $\pi(A)'=\{M_g:\ g\in L^\infty(m)\}$, because $C(X)$ is weak$^*$-dense in $L^\infty(m)$ and the multiplication operators by $L^\infty(m)$ form a maximal abelian algebra. This is abelian, so $m$ has exactly one maximal measure (Theorem 15.2). It is $\iota_*m$, with $\iota(x)=\delta_x$: indeed $\int\hat g\,d\iota_*m=m(g)$, and $\kappa_{\iota_*m}(G)=M_{G\circ\iota}$ is multiplicative with range $\pi(A)'$. Since $\pi(A)'=\pi(A)''\cap\pi(A)'$, it is also the central measure. It is concentrated on the closed set $\iota(X)=P(A)$, for every $X$, metrizable or not.

Example 22.2 (The trace on $M_n$). Let $A=M_n(\mathbb C)$ and $\tau=\frac1n\operatorname{tr}$. The GNS space is $M_n$ with $\langle X,Y\rangle=\tau(Y^*X)$, $\pi$ is left multiplication, $\xi=1$, and $\pi(A)'$ is the right multiplications $R_y$, a copy of $M_n$. The centre of $\pi(A)''$ is trivial, so $\tau$ is factorial and its central measure is $\delta_\tau$. For the maximal abelian algebra $\mathcal C=\{R_d:\ d\text{ diagonal}\}$, the construction in the proof of Theorem 13.1 gives $e$ = the projection onto the diagonal matrices, $\theta(a)=R_{\operatorname{diag}(a)}$, characters $\chi_i(R_d)=d_{ii}$, $\chi_i\circ\theta=\omega_{e_i}$, and $\mu=\frac1n\sum_i\delta_{\omega_{e_i}}$. It is concentrated on pure states (Theorem 19.1). Conjugating $\mathcal C$ by a unitary $R_w$ gives the measure $\frac1n\sum_i\delta_{\omega_{w^*e_i}}$ for another orthonormal basis; since $\pi(A)'$ is not abelian, $\tau$ has many maximal measures (Theorem 15.2), including non-orthogonal ones (compare Example 11.5).

Example 22.3 (Trivial cases). For every state, $\delta_\varphi$ is orthogonal with $\mathcal C_{\delta_\varphi}=\mathbb C1$, since $\kappa_{\delta_\varphi}(f)=f(\varphi)1$. It is maximal exactly when $\varphi$ is pure (Lemma 5.2 and Theorem 5.4), and then $M_\varphi(\mathfrak S)=\{\delta_\varphi\}$ (Bauer's criterion, Proposition 3.3(3)). The central measure is $\delta_\varphi$ exactly when $\mathcal Z_\varphi=\mathbb C1$, that is, when $\varphi$ is factorial (Theorem 13.1).

Remark 22.4 (Further examples). Other small, explicit examples in this lesson: a simplicial measure that is not orthogonal (Example 11.5); a state with abelian $\pi_\varphi(A)'$ whose face has a closure that is not a simplex (Proposition 16.1(5)); a state that is not factorial but lies in the set of Remark 21.4, where the intersection over $h$ comes first; and the invariant states in the proof of Theorem 23.2, in Theorem 23.4(3) and in Example 23.6.

Exercise 22.5 (hard; Orthogonal measures of the trace). In Example 22.2, show that the orthogonal measures of $\tau$ are exactly the measures $\sum_k\frac{\operatorname{tr}p_k}{n}\delta_{\omega_k}$, where $p_1,\dots,p_m$ are nonzero orthogonal projections with $\sum_kp_k=1$ and $\omega_k=\operatorname{tr}(p_k\,\cdot)/\operatorname{tr}p_k$. Show that such a measure is maximal exactly when every $p_k$ has rank one, and that one of them majorizes another exactly when its partition refines the other's.

Solution. An abelian $*$-subalgebra of $\pi(A)'=\{R_y\}$ is $\{R_d:\ d\in D\}$ for an abelian $*$-subalgebra $D$ of $M_n$, and $D$ is the span of the minimal projections $p_k$ of $D$, which add up to $1$. As in Example 22.2, $e$ is the orthogonal projection onto $D$, namely $X\mapsto\sum_k\frac{\operatorname{tr}(p_kX)}{\operatorname{tr}p_k}p_k$; so $\theta(a)=R_{d(a)}$ with $d(a)=\sum_k\omega_k(a)p_k$. The characters are $R_d\mapsto$ (coefficient of $p_k$), with $\chi_k\circ\theta=\omega_k$, and $\langle\xi p_k,\xi\rangle=\operatorname{tr}p_k/n$. Theorem 13.1 gives the measure, and every orthogonal measure arises this way. If all $p_k$ have rank one, the $\omega_k$ are pure, and the measure is a boundary measure (Proposition 5.5(2)), hence maximal (Theorem 5.4). If some $p_k$ has rank $\geq2$, $\omega_k$ is not pure, and since $\mathfrak S$ is metrizable a maximal measure would live on $P(A)$ (Theorem 6.4); so the measure is not maximal. The order statement is Theorem 14.1: $\mathcal C_\mu\subseteq\mathcal C_\nu$ means that the partition of $\nu$ refines that of $\mu$.

Exercise 22.6 (easy; Pure and factorial states through measures). Show: (a) $\varphi$ is pure if and only if $M_\varphi(\mathfrak S)=\{\delta_\varphi\}$; (b) $\varphi$ is factorial exactly when its central measure is $\delta_\varphi$; (c) if $\varphi$ is factorial but not pure, then $\delta_\varphi$ is orthogonal but not maximal.

Solution. (a) Bauer's criterion (Proposition 3.3(3)) with $P(A)=\partial_e\mathfrak S$. (b) The central measure is the orthogonal measure with $\mathcal C_\mu=\mathcal Z_\varphi$, and $\delta_\varphi$ is the orthogonal measure with $\mathcal C=\mathbb C1$ (Example 22.3); by the bijection of Theorem 13.1, they coincide exactly when $\mathcal Z_\varphi=\mathbb C1$. (c) $\delta_\varphi$ is always orthogonal; by (a) there is another measure in $M_\varphi(\mathfrak S)$, and a maximal one majorizing $\delta_\varphi$ differs from it, since $\delta_\varphi$ is not a boundary measure (Lemma 5.2: $\varphi\notin\partial_e\mathfrak S=\bigcap_fB_f$). Example 22.2 ($\tau$ on $M_n$, $n\geq2$) is such a state.

23. $G$-abelian systems

We keep the setting of Section 17. For $\varphi\in\mathfrak S^G$ let $H_0=\{\zeta\in H_\varphi:\ U_s\zeta=\zeta\text{ for all }s\}$, and let $e_0$ be the projection onto it. $\operatorname{co}U(G)$ is the set of finite convex combinations $\sum_sc_sU_s$, and $K_G(x)=\operatorname{co}\{\alpha_s(x):\ s\in G\}$.

Lemma 23.1 (Mean ergodic lemma).

For $\eta\in H$, $e_0\eta$ is the element of minimal norm in the closed convex hull of $\{U_s\eta\}$.
For $\eta_1,\dots,\eta_n\in H$ and $\varepsilon>0$ there is a self-adjoint $S\in\operatorname{co}U(G)$ with $\|S\eta_j-e_0\eta_j\|<\varepsilon$ for all $j$.
$Se_0=e_0S=e_0$ for $S\in\operatorname{co}U(G)$, and $e_0$ commutes with $U(G)'$.

Proof. (3) $U_s$ fixes $H_0$ pointwise, so $U_se_0=e_0$ and $e_0U_s=(U_s^*e_0)^*=e_0$. An operator commuting with $U(G)$ maps $H_0$ into itself, and so does its adjoint. (1) Let $C$ be the closed convex hull and $c$ its element of minimal norm. $U_tC=C$ and $U_t$ is isometric, so $U_tc=c$, and $c\in H_0$. By (3), $e_0$ is constant on $C$, equal to $e_0\eta$; so $c=e_0c=e_0\eta$. (2) Apply (1) to $U\oplus\dots\oplus U$ on $H^n$, whose fixed vectors are $H_0^n$: some $T\in\operatorname{co}U(G)$ has $\|T\eta_j-e_0\eta_j\|<\varepsilon$. Then $S=T^*T\in\operatorname{co}U(G)$ is self-adjoint and $\|S\eta_j-e_0\eta_j\|=\|T^*(T\eta_j-e_0\eta_j)\|<\varepsilon$, because $T^*e_0=e_0$. $\square$

Theorem 23.2. For $\varphi\in\mathfrak S^G$, with $\pi=\pi_\varphi$, the following are equivalent.

(i) $e_0\pi(A)e_0$ is commutative.
(ii$^*$) For all $x,y\in A$ and $\zeta\in H_0$, $\begin{gathered} \inf\{|\langle\pi(x'y-yx')\zeta,\zeta\rangle|:\\ \ x'\in K_G(x)\}\\ =0 \end{gathered}$ .

Consider also the condition (ii), in which the infimum is taken only over the orbit, $x'=\alpha_s(x)$ with $s\in G$. It implies (ii$^*$), hence (i). But (i) does not imply (ii), as the end of the proof shows.

The theorem proves the equivalence with the convex-hull condition (ii$^*$). The end of the proof separates it from the orbit condition (ii), and Theorem 23.4(3) gives a generated face whose closure is not a simplex.

Proof. For $x'=\sum_sc_s\alpha_s(x)$ put $S=\sum_sc_sU_s$. Since $U_s^*\zeta=\zeta$ for $\zeta\in H_0$, $\begin{gathered} \langle\pi(x')\pi(y)\zeta,\zeta\rangle\\ =\langle\pi(x)S^*\pi(y)\zeta,\zeta\rangle,\\ \langle\pi(y)\pi(x')\zeta,\zeta\rangle\\ =\langle\pi(y)S\pi(x)\zeta,\zeta\rangle . \end{gathered}$ (i)$\Rightarrow$(ii$^*$): Choose $S$ self-adjoint with $S\approx e_0$ on $\pi(x)\zeta$ and $\pi(y)\zeta$ (Lemma 23.1(2)). Then $\begin{gathered} \langle\pi(x'y-yx')\zeta,\zeta\rangle\\ =\langle S\pi(y)\zeta,\pi(x)^*\zeta\rangle\\ -\langle S\pi(x)\zeta,\pi(y)^*\zeta\rangle \end{gathered}$ is close to $\langle\pi(x)e_0\pi(y)\zeta,\zeta\rangle-\langle\pi(y)e_0\pi(x)\zeta,\zeta\rangle$. As $\zeta=e_0\zeta$, this is $\langle[e_0\pi(x)e_0,e_0\pi(y)e_0]\zeta,\zeta\rangle=0$.

(ii$^*$)$\Rightarrow$(i): Fix $x,y$, $\zeta\in H_0$ and $\delta>0$, and let $F=\langle\pi(x)e_0\pi(y)\zeta,\zeta\rangle-\langle\pi(y)e_0\pi(x)\zeta,\zeta\rangle$. Choose a self-adjoint $S'=\sum_tc'_tU_t$ with $\|S'\pi(y)\zeta-e_0\pi(y)\zeta\|<\delta$ and $\|S'\pi(y)^*\zeta-e_0\pi(y)^*\zeta\|<\delta$, and put $y'=\sum_tc'_t\alpha_t(y)$. Apply (ii$^*$) to the pair $(x,y')$: some $x'\in K_G(x)$, with operator $S$, has $|Q|<\delta$, where $Q=\langle\pi(x'y'-y'x')\zeta,\zeta\rangle$. Expanding both averages as above, $\begin{gathered} Q\\ =\langle\pi(x)S^*S'\pi(y)\zeta,\zeta\rangle\\ -\langle\pi(y)S'S\pi(x)\zeta,\zeta\rangle\\ =\langle S'\pi(y)\zeta,S\pi(x)^*\zeta\rangle\\ -\langle S\pi(x)\zeta,S'\pi(y)^*\zeta\rangle . \end{gathered}$ Replacing $S'\pi(y)\zeta$ and $S'\pi(y)^*\zeta$ by $e_0\pi(y)\zeta$ and $e_0\pi(y)^*\zeta$ changes $Q$ by at most $2\delta\|x\|\|\zeta\|$. After the replacement, $S^*e_0=e_0$ and $e_0S=e_0$ turn $Q$ into exactly $F$. So $|F|\leq\delta(1+2\|x\|\|\zeta\|)$ for every $\delta$, and $F=0$. The operator $[e_0\pi(x)e_0,e_0\pi(y)e_0]$ on $H_0$ has vanishing quadratic form, so it is $0$.

(ii)$\Rightarrow$(ii$^*$) is trivial. (i) does not imply (ii). Let $A=M_2(\mathbb C)$, $G=\mathbb Z_2$ acting by $\operatorname{Ad}u$, $u=\operatorname{diag}(1,-1)$, and $\varphi=\tau$, the normalized trace. The GNS space is $M_2$ with $\langle X,Y\rangle=\tau(Y^*X)$, $\pi$ is left multiplication, $\xi=1$, and $U_g(X)=uXu^*$. So $H_0$ is the diagonal matrices, and $e_0\pi(a)e_0$ acts on $H_0$ as multiplication by $\operatorname{diag}(a_{11},a_{22})$: (i) holds. For $x=E_{12}$, $y=E_{21}$ and $\zeta=E_{11}\in H_0$, $\langle\pi([x,y])\zeta,\zeta\rangle=\frac12$ and $\langle\pi([\alpha_g(x),y])\zeta,\zeta\rangle=-\frac12$. So the infimum over the orbit is $\frac12\neq0$. (Over $K_G(x)$ it is $0$, attained at $x'=\frac12(x+\alpha_g(x))=0$.) $\square$

Proposition 23.3. If (i) holds, then $\mathfrak M_\varphi=\pi(A)'\cap U(G)'$ is abelian; in fact $\mathfrak M_\varphi=\pi(A)'\cap\{e_0\}'$.

Proof. Apply Proposition 12.3(b) to the unital $*$-algebra $\mathcal M=\pi(A)$, with cyclic vector $\xi$ and $e=e_0$ ($e_0\xi=\xi$): $\mathcal A=\pi(A)'\cap\{e_0\}'$ is abelian, and restriction to $H_0$ is one-to-one on it. By Lemma 23.1(3), $\mathfrak M_\varphi\subseteq\mathcal A$. Conversely, for $x\in\mathcal A$ and $s\in G$, $U_s^*xU_s\in\mathcal A$ (since $U_s$ normalizes $\pi(A)$ and commutes with $e_0$), and it agrees with $x$ on $H_0$, where $U_s$ acts trivially. So $U_s^*xU_s=x$. $\square$

Theorem 23.4. Suppose (i) holds for $\varphi\in\mathfrak S^G$.

The face $F^G_\varphi$ of $\mathfrak S^G$ generated by $\varphi$ has a lattice cone: $\mathbb R_+F^G_\varphi\cong(\mathfrak M_\varphi)_+$.
$\varphi$ is the barycentre of exactly one maximal (boundary) probability measure on $\mathfrak S^G$. It is the orthogonal measure $\mu$ with $\mathcal C_\mu=\mathfrak M_\varphi$, and $\mu(B)=0$ for every Baire subset $B$ of $\mathfrak S^G$ that misses $\partial_e\mathfrak S^G$; if $A$ is separable, $\mu$ is concentrated on $\partial_e\mathfrak S^G$.
The closure $\overline{F^G_\varphi}$ need not be a simplex.

Proof. (1) As in Proposition 16.1(1) and (3), using Proposition 17.1(3): $F^G_\varphi=\Theta_\varphi(\{x\in(\mathfrak M_\varphi)_+:\ \langle x\xi,\xi\rangle=1\})$, and $\mathfrak M_\varphi$ is abelian by Proposition 23.3. (2) Proposition 17.1(6) and Theorem 19.1(3). (3) Take the algebra and the function $\xi(\cdot)$ of Proposition 16.1(5). Let $Z_0=\{\frac12-2^{-k}:\ k\geq1\}\cup\{\frac12\}$, a closed countable set, $\theta(x)=\operatorname{dist}(x,Z_0)\in[0,\frac12]$, $\sigma(x)=2|\xi(x)\rangle\langle\xi(x)|-1$, and $u(x)=\cos\theta(x)\,1+i\sin\theta(x)\,\sigma(x)$. The jumps of $\sigma$ lie in $Z_0$, where $\sin\theta=0$, so $u\in C([0,1],U(2))$. Let $\mathbb Z$ act by $\alpha^n$, $\alpha(a)(x)=u(x)a(x)u(x)^*$. Since $u(x)^*\xi(x)=e^{-i\theta(x)}\xi(x)$, $\varphi$ is invariant, and $U=U_\varphi(1)$ is multiplication by $w(x)=e^{-i\theta(x)}u(x)$. Its eigenvalues are $1$ on $\xi(x)$ and $e^{-2i\theta(x)}\neq1$ on $\xi(x)^\perp$ for almost every $x$. So $H_0=\{g\xi:\ g\in L^2\}$, and $e_0\pi(a)e_0$ is multiplication by the scalar function $x\mapsto\langle a(x)\xi(x),\xi(x)\rangle$: (i) holds. Here $\mathfrak M_\varphi=\{M_g\otimes1\}$, so $F^G_\varphi=F_\varphi$. The four pure states $\rho_j$ of Proposition 16.1(5) are invariant, since $u(\frac12)=1$. As there, $\overline{F^G_\varphi}=\overline{F_\varphi}$ is not a simplex. $\square$

Theorem 23.5. If (i) holds for every $\varphi\in\mathfrak S^G$ (the system is $G$-abelian), then $\mathfrak S^G$ is a simplex.

Proof. By Theorem 23.4(2) every point of $\mathfrak S^G$ has exactly one maximal representing measure on $\mathfrak S^G$. Apply the Choquet–Meyer theorem (Theorem 7.4). $\square$

Example 23.6 (Diagonal unitaries acting on $M_n$). Let $A=M_n(\mathbb C)$ with $n\geq2$, $G$ the diagonal unitary matrices, $\alpha_s=\operatorname{Ad}s$, and $\varphi(x)=x_{11}$.

This example separates conditions (i) and (ii): the compressed algebra is commutative, while the tested orbit commutator has constant modulus one.

$\mathfrak M_\varphi$ is one-dimensional. $\varphi$ is the vector state of $e_1$ in the identity representation on $\mathbb C^n$, which is irreducible, so $\pi_\varphi(A)'=\mathbb C1$.
Condition (ii) fails. $U_s=\bar s_{11}s$, and for $x=E_{12}$, $y=E_{21}$, $\zeta=e_1$: $\langle[\alpha_s(x),y]e_1,e_1\rangle=s_{11}\bar s_{22}$, of modulus $1$ for every $s$.
Condition (i) holds. $H_0=\{\zeta:\ \bar s_{11}s_{jj}\zeta_j=\zeta_j\ \forall s\}=\mathbb Ce_1$, so $e_0\pi(A)e_0=\mathbb Ce_0$ is commutative.

In fact every invariant state satisfies (i). An invariant state has a diagonal density matrix $D$; its GNS space is $\{X\in M_n:\ X\text{ vanishes on }\ker D\}$ with $\xi=D^{1/2}$ and the Hilbert–Schmidt inner product, $U_sX=sXs^*$, $H_0$ is its diagonal part, and $e_0\pi(x)e_0$ is multiplication by $\operatorname{diag}(x_{11},\dots,x_{nn})$. So the system is $G$-abelian, and $\mathfrak S^G$, the $(n-1)$-simplex of diagonal states, is a simplex, as Theorem 23.5 predicts. The example shows that (ii) is strictly stronger than (i).

Exercise 23.7 (medium; An invariant state under a finite group). Let $G=\mathbb Z_2$ act on $M_2(\mathbb C)$ by $\operatorname{Ad}\operatorname{diag}(1,-1)$, and $\varphi=\tau$. Find $\mathfrak S^G$, $\partial_e\mathfrak S^G$, $\mathfrak M_\tau$, and the unique maximal measure of $\tau$ on $\mathfrak S^G$.

Solution. A state $\omega_\rho$ is invariant exactly when $\rho$ commutes with $\operatorname{diag}(1,-1)$, that is, when $\rho$ is diagonal. So $\mathfrak S^G$ is the segment from $\omega_{e_1}$ to $\omega_{e_2}$, with these two as extreme points. In the GNS picture used at the end of the proof of Theorem 23.2, $U_g=\operatorname{Ad}u$ on $M_2$, and $R_y$ commutes with $U_g$ exactly when $y$ is diagonal; so $\mathfrak M_\tau=\{R_d:\ d\text{ diagonal}\}$, which is abelian (condition (i) holds, as shown there). By Proposition 17.1(6) the orthogonal measure with $\mathcal C_\mu=\mathfrak M_\tau$, which is $\frac12(\delta_{\omega_{e_1}}+\delta_{\omega_{e_2}})$ by Exercise 22.5, is the unique maximal measure on $\mathfrak S^G$. This agrees with the direct computation on the segment.

24. Large groups of automorphisms

For $\varphi\in\mathfrak S^G$ and $a\in A$ let $C_a$ be the weak operator closure of $\pi_\varphi(K_G(a))$. It is convex and bounded, hence weakly compact, and it equals the strong closure, because a convex set of operators has the same weak and strong closures. Also $C_a\subseteq\pi_\varphi(A)''$. We use three conditions.

(S) For every $\varphi\in\mathfrak S^G$ and $a\in A$, $C_a\cap\mathcal Z_\varphi\neq\varnothing$.
(L) For every $\varphi\in\mathfrak S^G$, $a\in A$, and finitely many $b_1,\dots,b_n,x_1,\dots,x_m\in A$: $\begin{gathered} \inf_{a'\in K_G(a)}\\ \max_{i,j}|\varphi(x_j(a'b_i-b_ia')x_j^*)|\\ =0 \end{gathered}$ .
(L$_p$) As (L), but with a single $x$, and with the infimum taken separately for each $i$.

We call the action large if (S) holds.

Theorem 24.1.

(a) (L)$\iff$(S), and (S)$\Rightarrow$(L$_p$).
(b) If (S) holds, the system is $G$-abelian, so $\mathfrak S^G$ is a simplex.
(c) If (S) holds, then for $\varphi\in\mathfrak S^G$ the following are equivalent: (i) $\varphi$ is ergodic; (ii) $U_\varphi(G)'\cap\mathcal Z_\varphi=\mathbb C1$; (iii) $H_0=\mathbb C\xi_\varphi$.

Proof. Write $\pi,H,\xi,U$ for $\varphi$. Note that $\varphi(x(a'b-ba')x^*)=\langle[\pi(a'),\pi(b)]\eta,\eta\rangle$ with $\eta=\pi(x^*)\xi$.

(a) (S)$\Rightarrow$(L): take $z\in C_a\cap\mathcal Z_\varphi$ and a net $\pi(a'_k)\to z$ weakly with $a'_k\in K_G(a)$. For each of the finitely many pairs, $\langle[\pi(a'_k),\pi(b_i)]\eta_j,\eta_j\rangle\to\langle[z,\pi(b_i)]\eta_j,\eta_j\rangle=0$. (L)$\Rightarrow$(L$_p$) is clear. (L)$\Rightarrow$(S): the sets $Z_{b,x}=\{z\in C_a:\ \langle[z,\pi(b)]\pi(x^*)\xi,\pi(x^*)\xi\rangle=0\}$ are weakly closed. By (L) and compactness, finitely many of them have a common point: a weak cluster point of $\pi(a'_k)$, where $a'_k$ nearly attains the infimum. So all of them have a common point $z$. Then the quadratic form of $[z,\pi(b)]$ vanishes on the dense subspace $\pi(A)\xi$, so $z\in\pi(A)'\cap C_a\subseteq\mathcal Z_\varphi$.

(b) Fix $\varphi$ and $a$. $C_a\cap\mathcal Z_\varphi$ is nonempty, convex, weakly compact, and invariant under $\operatorname{Ad}U_s$, because $U_s\pi(K_G(a))U_s^*=\pi(K_G(a))$ and $U_s$ normalizes $\pi(A)$. Take $z$ in it. By the mean ergodic lemma (Lemma 23.1) there are $T_k=\sum_sc^{(k)}_sU_s$ with $T_kz\xi\to e_0z\xi$. The elements $z_k=\sum_sc^{(k)}_sU_szU_s^*$ lie in $C_a\cap\mathcal Z_\varphi$ and $z_k\xi=T_kz\xi$. A weak cluster point $z_\infty$ lies in $C_a\cap\mathcal Z_\varphi$, and $z_\infty\xi=e_0z\xi\in H_0$. Then $U_sz_\infty U_s^*\in\mathcal Z_\varphi$ and $U_sz_\infty U_s^*\xi=U_sz_\infty\xi=z_\infty\xi$, so $U_sz_\infty U_s^*=z_\infty$ (separating vector). Thus $z_\infty$ commutes with $U(G)$, hence with $e_0$. Since $e_0\pi(\alpha_s(a))e_0=e_0\pi(a)e_0$, weak limits give $e_0\pi(a)e_0=e_0z_\infty e_0=z_\infty e_0$. For $b\in A$, $z_\infty$ is central in $\pi(A)''$, so $\begin{gathered} e_0\pi(a)e_0\cdot e_0\pi(b)e_0\\ =e_0\pi(b)z_\infty e_0\\ =e_0\pi(b)e_0\cdot e_0\pi(a)e_0 . \end{gathered}$ So condition (i) of Theorem 23.2 holds for every $\varphi$, and Theorem 23.5 applies.

(c) (i)$\Leftrightarrow$$\mathfrak M_\varphi=\mathbb C1$ by Proposition 17.1(4), and $U(G)'\cap\mathcal Z_\varphi\subseteq\mathfrak M_\varphi$; so (i)$\Rightarrow$(ii). (iii)$\Rightarrow$(i): for $x\in\mathfrak M_\varphi$, $x\xi\in H_0=\mathbb C\xi$, so $x$ is a scalar (separating vector). (ii)$\Rightarrow$(iii): by the proof of (b), each $a$ has a $z_\infty\in U(G)'\cap\mathcal Z_\varphi=\mathbb C1$ with $e_0\pi(a)e_0=z_\infty e_0$. So $e_0\pi(a)\xi=e_0\pi(a)e_0\xi\in\mathbb C\xi$ for every $a$, and $H_0=e_0\overline{\pi(A)\xi}=\mathbb C\xi$. $\square$

Condition (L$_p$) has one $x$ and a separate infimum for each $b_i$, so the compactness argument in (L)$\Rightarrow$(S) does not apply to it. We leave open whether (L$_p$) implies (S). Parts (b) and (c) use only (S).

Background used without proof

Functional analysis.

(Hahn–Banach and choice.) Theorem 2.1 and Corollary 2.3 of the Hahn–Banach lesson prove the real sublinear extension theorem and its complex and normed versions. Theorem 6.3 and Corollary 6.4 strictly separate a compact convex set from a disjoint closed convex set in a locally convex Hausdorff space; they also give separation of points and annihilation of a proper closed linear subspace. The maximal-measure and maximal-abelian-algebra arguments use Zorn’s lemma, Theorem 1.1.
(Weak topologies.) If $X'$ is a vector space of linear functionals on $X$ that separates points, $\sigma(X,X')$ is locally convex and Hausdorff, and its continuous linear functionals are exactly $X'$. Theorem 1.2 of the weak-topology lesson gives the full finite-coordinate factorization proof.
(Banach–Alaoglu.) The closed unit ball of the dual of any normed space is weak$^*$-compact, by Theorem 3.1 of the weak-topology lesson, using its full Tychonoff proof. No separability is assumed.
(Krein–Milman.) A nonempty compact convex set in a locally convex Hausdorff space has extreme points and is their closed convex hull. Theorem 6.1 of the weak-topology lesson proves both claims by minimal faces and separation.

Measure theory.

(Riesz representation.) For a compact Hausdorff space $X$, every bounded complex linear functional on $C(X)$ is integration against a unique finite complex Radon measure, with norm equal to total variation. A real functional has a real measure, and a positive functional has a positive measure. The full positive and complex arguments and regularity are Theorems 2.2 and 2.4 and Proposition 2.3 of the Haar lesson. Restricting the complex theorem to real functions gives the real version used in Sections 1–7: extend a real functional by $L(f+ig)=L(f)+iL(g)$. This extension has the same norm: rotate $z$ by a scalar of modulus one so that $L(z)$ is nonnegative real, and use $|L(z)|=L(\operatorname{Re}z)\leq\|L\|\|z\|_\infty$ for the rotated function. Uniqueness and conjugation make its measure real. Its positive and negative parts are $(|\mu|\pm\mu)/2$, which are positive regular measures.
(Duality of $L^1$ and $L^\infty$.) For a sigma-finite measure $\mu$, the map $g\mapsto(f\mapsto\int fg\,d\mu)$ is an isometric isomorphism of $L^\infty(\mu)$ onto $L^1(\mu)^*$. This is Theorem 4.2 of the measure-tools lesson; its full proof first treats finite measure by Hilbert representation, then joins finite pieces. In this lesson the reference measures on compact spaces are finite.
(Convergence theorems.) Theorems 2.1–2.2 of the measure-tools lesson prove monotone convergence, Fatou and dominated convergence on arbitrary measure spaces: increasing nonnegative sequences have the expected integral limit, the integral of a nonnegative liminf is at most the liminf of the integrals, and almost-everywhere convergence under one integrable bound gives convergence in $L^1$. The monotone-net assertion for lower semicontinuous functions is separately proved in Proposition 2.1(2) above, using compactness and regularity.
(Uniqueness of measures.) Two finite measures that agree on the whole space and on a family closed under finite intersections agree on the sigma-algebra generated by that family. The finite-measure uniqueness lemma of the double-commutant lesson gives the full Dynkin-system argument; it is used in Proposition 2.1(4)(e).
(Urysohn’s lemma and finite partitions of unity.) Lemma 5.1 and Corollary 5.2 of the Stone–Weierstrass lesson separate disjoint closed subsets of a compact Hausdorff space and supply continuous cutoffs with prescribed compact support. Here is the finite-cover consequence used in Section 18. Given a finite open cover $U_1,\dots,U_n$, each $x$ has an open neighbourhood $V_x$ with $\overline{V_x}\subseteq U_{i(x)}$; this is the compact Hausdorff separation proved there. Choose $0\leq h_x\leq1$ equal to one on $\overline{V_x}$, with $\operatorname{supp}h_x\subseteq U_{i(x)}$. Finitely many $V_{x_j}$ cover $X$. Set $H=\sum_jh_{x_j}>0$, and $u_i=H^{-1}\sum_{j:i(x_j)=i}h_{x_j}$. Then $u_i\geq0$, $\sum_i u_i=1$, and $\operatorname{supp}u_i\subseteq U_i$.
(Stone–Weierstrass.) A real subalgebra of $C(K)$ that contains the constants and separates points is dense; so is a linear subspace that is closed under $\vee$ and $\wedge$, contains the constants and separates points; and a self-adjoint complex subalgebra that contains the constants and separates points is dense in the complex $C(K)$. All three are proved in The Stone–Weierstrass theorem for $C_0(X)$.

Hilbert spaces and operators.

Theorem 2.1, Theorem 2.3, Theorem 3.1 and Corollary 3.2 of the Hilbert-space lesson prove, respectively, the unique minimum-norm point of a nonempty closed convex set, Riesz–Fréchet representation of bounded functionals, representation of bounded sesquilinear forms by operators, polarization uniqueness, and the quadratic-form formula for the norm of a self-adjoint operator. The same norm formula for a normal operator is Proposition 8.2 of the spectral lesson, proved there using approximate eigenvectors. The arguments below need only the self-adjoint formula.
(Spectral theorem.) Theorem 4.4 and Proposition 5.1 of the spectral lesson give the unique projection-valued spectral measure of a bounded self-adjoint operator, commutation of its projections with every operator commuting with it, and norm approximation by real combinations of those projections. Each spectral projection lies in every von Neumann algebra containing the operator: it commutes with that algebra’s commutant, so lies in its bicommutant. Thus a von Neumann algebra with only the projections $0,1$ is $\mathbb C1$. For a normal operator, Theorem 8.1 there requires commutation with both $T,T^*$; Fuglede’s theorem, proved in Section 3.1 of the Kaplansky lesson, supplies commutation with $T^*$ from commutation with $T$. The proof here consumes only the self-adjoint version.
(Operator topologies.) Lemma 9.0 above proves weak compactness of bounded weakly closed operator sets, equality of weak and strong closures of convex sets, equality of weak and $\sigma$-weak topologies on bounded sets, and the separate multiplication continuity used in Theorem 13.1. Its compactness proof uses Tychonoff and bounded sesquilinear forms; no trace-class duality is needed for this fact.
(Bicommutant theorem.) If $\mathcal M\subseteq B(H)$ is a $*$-algebra containing $1$, then its weak closure, its strong closure and $\mathcal M''$ coincide. See The double commutation theorem.
(Kaplansky’s density theorem.) If a $*$-algebra $\mathcal M\subseteq B(H)$ contains $1$, every self-adjoint contraction of $\mathcal M''$ is a strong limit of self-adjoint contractions of $\mathcal M$. This is Theorem 7.1(2) of the Kaplansky lesson, which proves the stronger strong$^*$ density statement, with the continuity argument in Theorem 5.2.
(Multiplication algebras.) For a sigma-finite measure space, multiplication by $L^\infty$ is a maximal abelian von Neumann algebra on $L^2$, by Theorem 9.1 of the Hilbert-space lesson. The map $g\mapsto M_g$ is normal: a vector coefficient is integration against $\eta\overline\zeta\in L^1$, by Hölder, and Lemma 9.1 handles square-summable series of coefficients. This justifies the weak$^*$-density arguments in the examples.

C$^*$-algebras.

(Commutative algebras and functional calculus.) A unital abelian C$^*$-algebra is isometrically $*$-isomorphic to $C(\Omega)$, where $\Omega$ is its compact space of characters (the commutative Gelfand–Naimark theorem). A normal element has a continuous functional calculus. An injective $*$-homomorphism between C$^*$-algebras is isometric. All this is proved in C$^*$-algebras: continuous functional calculus, automatic continuity, positive cones, approximate identities and quotients.
(Positive functionals and states.) Proposition 3.2 of the GNS lesson proves $|\varphi(b^*a)|^2\leq\varphi(a^*a)\varphi(b^*b)$ by polarization and a nonnegative scalar quadratic. Theorem 4.7 there proves that a positive functional on a C$^*$-algebra is bounded and hermitian, and that $|\varphi(a)|^2\leq\|\varphi\|\varphi(a^*a)$. In the unital case $\|\varphi\|=\varphi(1)$. A state is a positive functional with this value equal to one; Lemma 7.1 there gives enough positive functionals to detect every nonzero positive element, and hence states of every nonzero unital algebra.
(GNS construction.) Construction 5.1, Lemma 5.2 and Theorems 5.3–5.5 of the GNS lesson prove the quotient, bounded left action, cyclic vector and unitary uniqueness. For a state $\varphi$, the unital case has $\xi_\varphi=1+N_\varphi$ and $\|\xi_\varphi\|^2=\varphi(1)=1$, with $\varphi(a)=\langle\pi_\varphi(a)\xi_\varphi,\xi_\varphi\rangle$.
(Gelfand–Naimark.) Lemma 7.1 and Theorem 7.2 of the GNS lesson prove existence of enough positive functionals by separation and then form a faithful direct sum of their GNS representations. Injective $*$-homomorphisms are isometric by the continuous-functional-calculus lesson, Theorem 4.2.

Where this leads

Direct integrals. For a separable algebra, an orthogonal measure does more than decompose the state: it writes $\pi_\varphi$ as a direct integral of representations over the state space, with $\mathcal C_\mu$ as the diagonal algebra; see the following programme lessons. The lessons Measurable fields of Hilbert spaces and their direct integrals and Decomposable operators and the diagonal algebra build the direct integrals that this needs.
Compact convex sets. Simplices whose extreme boundary is closed (Bauer simplices), split faces and the facial topology continue Section 7;
Group actions. Here $G$ is an abstract group, and no continuity of the action is used. Invariant states for continuous actions of locally compact groups, and their ergodic decompositions, are further directions beyond the scope proved here.
Two open questions. This lesson does not settle whether condition (L$_p$) of Section 24 implies (S), nor whether a system whose invariant states form a simplex must be $G$-abelian.

References

[Blackadar] B. Blackadar, Operator Algebras: Theory of C*-Algebras and von Neumann Algebras, revised author edition, 8 February 2017.
[van Neerven] J. van Neerven, Functional Analysis, corrected author version, arXiv:2112.11166v7.

Freely accessible reading: D. H. Fremlin, Measure Theory, Chapter 46, §461A–§461P gives a route through barycenters, maximal representing measures and simplex uniqueness; Baire pseudo-support is distinguished from mass on the extreme boundary. The lesson includes its own complete proofs at the stated hypotheses; references to human sources do not imply permission to adapt their expression.

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