A C*-algebra is Jiang-Su stable when it absorbs the Jiang-Su algebra $Z$
tensorially. The dynamic counterpart asks an action to absorb the trivial
action on $Z$ up to cocycle conjugacy. The main theorem establishes this
for every amenable action in the simple, nuclear, stably finite, unital
setting. Throughout, C*-algebras are complex, tensor products are
minimal, $\mathcal T(B)$ is the tracial-state space of a unital C*-algebra
$B$, and $Z$ is the Jiang-Su algebra.

The Jiang-Su algebra $Z$ is the unique simple unital inductive limit of
prime dimension-drop intervals with the K-theory of the complex numbers and
a unique tracial state; it is strongly self-absorbing, so
$Z \cong Z \otimes Z$. The dimension-drop construction and these properties
appear in the study of strongly self-absorbing algebras.

./toms-winter-math0502211-strongly-self-absorbing.pdf

Theorem. Let $A$ be a simple, separable, unital, infinite-dimensional,
nuclear, stably finite C*-algebra with $A \cong A \otimes Z$. Let $G$ be a
countable discrete amenable group and $\alpha : G \to \mathrm{Aut}(A)$ an
action. Put $B = A \otimes Z$ and $\beta_g = \alpha_g \otimes \mathrm{id}_Z$.
Then there are a unital *-isomorphism $\Phi : A \to B$ and unitaries
$u_g \in B$ for $g \in G$ such that

$$ u_e = 1, \qquad u_{gh} = u_g \beta_g(u_h), \qquad \Phi(\alpha_g(a)) = u_g \beta_g(\Phi(a)) u_g^* $$

for all $g, h \in G$ and $a \in A$.

Stable finiteness means that every isometry in every matrix algebra over
$A$ is a unitary. Amenability enters only through the Folner condition:
for every finite $H \subset G$ and every $\delta > 0$ there is a nonempty
finite $F \subset G$ with $|hF \,\triangle\, F| < \delta |F|$ for every
$h \in H$. The displayed relations say exactly that $\alpha$ and
$\beta = \alpha \otimes \mathrm{id}_Z$ are cocycle conjugate, with
$\Phi$ intertwining the two actions through the $\beta$-cocycle
$g \mapsto u_g$.

No outerness, Rokhlin, or Universal Coefficient Theorem assumption is made,
and no condition is placed on the action of $G$ on the tracial boundary or
on individual trace orbits. Finite groups and the trivial group are
included. The statement is the unital, stably finite case of Szabo's
conjecture that amenable actions on separable simple nuclear Jiang-Su
stable algebras are automatically equivariantly Jiang-Su stable; the
nonunital case lies outside the present argument. Earlier affirmative
cases cover finite algebras with finitely many rays of extremal traces and
very weak comparison.

./szabo-1904.10897-equivariant-property-si.pdf

Earlier characterizations of Z-absorption invoke a finiteness restriction
on the trace simplex, which the present argument avoids: the Gaussian
splittings are uniform over the whole trace space rather than derived from
comparison of positive elements.

./matui-sato-1111.1637-strict-comparison-Z-absorption.pdf

Theorem 1.1 (Matui and Sato). Let $A$ be a unital separable simple
infinite-dimensional nuclear C*-algebra with finitely many extremal
traces. Then the following are equivalent: $A \otimes Z \cong A$; $A$
has strict comparison; every completely positive map from $A$ to $A$
can be excised in small central sequences; has property (SI).

The decisive criterion reads:

Corollary C (Szabo and Wouters). Let $\alpha : G \curvearrowright A$ be
an action of a countable amenable group on a separable unital simple
nuclear $Z$-stable C*-algebra. Then $\alpha$ has equivariant uniform
property Gamma if and only if $\alpha$ is equivariantly Jiang-Su stable.

./szabo-wouters-2301.12846-equivariant-property-gamma.pdf

The forward implication is the absorption step; it remains to construct
the required invariant central trace splittings uniformly over the whole
trace space, even when the action moves traces.

For a unital C*-algebra $B$ and $\tau \in \mathcal T(B)$ write
$\|x\|_{2,\tau} = \tau(x^*x)^{1/2}$ and
$\|x\|_{2,u} = \sup_{\tau \in \mathcal T(B)} \|x\|_{2,\tau}$. Under the
hypotheses on $A$, nuclearity implies exactness, and a unital stably
finite exact algebra carries a tracial state; every 2-quasitrace on a
unital exact algebra is a trace. Hence $\mathcal T(A)$ is a nonempty
compact convex set, and each trace on the simple algebra $A$ is faithful,
because its null space is a closed two-sided ideal not containing the
unit.

./haagerup-1403.7653-quasitraces-are-traces.pdf

Theorem 5.11 (Haagerup). Quasitraces on exact unital C*-algebras are traces.
Corollary 5.12 (Haagerup). Every stably finite unital exact C*-algebra has a trace state.

Fix a free ultrafilter $\omega$ on the positive integers and form the
uniform tracial ultrapower

$$ A^\omega = \ell^\infty(\mathbb N, A) / I_\omega, \qquad I_\omega = \{(a_j) : \lim_{j \to \omega} \|a_j\|_{2,u} = 0\}. $$

The inequalities $\|ab\|_{2,u} \le \|a\| \|b\|_{2,u}$ and
$\|ba\|_{2,u} \le \|a\| \|b\|_{2,u}$ show that $I_\omega$ is a closed
two-sided ideal, and faithfulness of traces embeds $A$ as the constant
sequences. For any sequence $(\tau_j)$ in $\mathcal T(A)$ the formula
$\tau([(a_j)]) = \lim_{j \to \omega} \tau_j(a_j)$ defines a tracial state
on $A^\omega$, because $|\tau_j(a_j)| \le \|a_j\|_{2,\tau_j}$. These are
the limit traces, written $\mathcal T_\omega(A)$; these need not exhaust
the traces of $A^\omega$. Automorphisms preserve $\|\cdot\|_{2,u}$, and
applying $\alpha_g$ coordinatewise gives an action
$\bar\alpha : G \curvearrowright A^\omega$ preserving $A^\omega \cap A'$.

An action $\alpha : G \curvearrowright A$ has equivariant uniform property
Gamma if for every integer $k \ge 2$ there are pairwise orthogonal
projections $p_1, \ldots, p_k \in (A^\omega \cap A')^{\bar\alpha}$ with
$\tau(a p_i) = \tau(a)/k$ for every $a \in A$, every
$\tau \in \mathcal T_\omega(A)$, and every $i$. The construction also
forces $\sum_i p_i = 1$. The absorption criterion to be used states that a
countable discrete amenable group acting on a separable unital simple
nuclear Jiang-Su stable C*-algebra with nonempty trace space, and having
equivariant uniform property Gamma, is cocycle conjugate to its tensor
product with the trivial action on $Z$. It therefore remains to produce
the projections, working inside a separable subalgebra of
$A^\omega \cap A'$ and testing against all of its traces, not only those
extending to the ambient algebra.

The first construction plants many commuting copies of the orbit of $Z$ in
the central sequence algebra. Write $C^*(S)$ for the unital C*-algebra
generated by a set $S$.

Proposition. There are unital embeddings $\iota_i : Z \to A^\omega$,
$i \ge 1$, whose orbit algebras

$$ C_i = C^*(\bar\alpha_g(\iota_i(Z)) : g \in G) $$

commute with $A$ and commute pairwise. Consequently

$$ C = C^*\Bigl(\bigcup_{i \ge 1} C_i\Bigr) \subseteq A^\omega \cap A' $$

is a unital separable $\bar\alpha$-invariant subalgebra with
$\mathcal T(C) \neq \emptyset$. There are moreover self-adjoint elements
$x_{i,g} \in C_i$ and a constant $M < \infty$ with

$$ \|x_{i,g}\| \le M, \qquad \varphi(x_{i,g}) = 0, \qquad \varphi(x_{i,g}^2) = 1 \quad (\varphi \in \mathcal T(C)), \qquad \bar\alpha_h(x_{i,g}) = x_{i,hg} $$

for all $i \ge 1$ and $g, h \in G$.

Proof. Since $A$ is $Z$-stable, fix an identification
$A \cong A \otimes \bigotimes_{l \ge 1} Z$ and use its increasing finite
tensor stages. First, every norm-separable C*-subalgebra
$D \subseteq A^\omega$ admits a unital embedding $Z \to A^\omega \cap D'$.
Choose a norm-dense sequence $(d_s)_{s \ge 1}$ in $D$ with bounded
representatives $d_s = [(d_{s,j})_{j \ge 1}]$. For each $j$ choose a finite
tensor stage containing elements within $1/j$ in norm of
$d_{1,j}, \ldots, d_{j,j}$, and let $\theta_j : Z \to A$ embed into a tensor
factor strictly after that stage. For $j \ge s$ and $z \in Z$,

$$ \|[\theta_j(z), d_{s,j}]\| \le \frac{2\|z\|}{j}. $$

The coordinatewise homomorphisms therefore induce a unital *-homomorphism
$z \mapsto [(\theta_j(z))_j]$ into $A^\omega \cap D'$, injective because $Z$
is simple and the unit of $A^\omega$ is nonzero.

Set $D_0 = A$. Suppose $D_{i-1}$ is a separable unital
$\bar\alpha$-invariant subalgebra of $A^\omega$ containing $A$ and the
preceding orbit algebras. The embedding just proved places
$\iota_i(Z)$ in $D_{i-1}'$; define $C_i$ as above and
$D_i = C^*(D_{i-1}, C_i)$. For $b \in \iota_i(Z)$, $d \in D_{i-1}$, and
$g \in G$, invariance of $D_{i-1}$ gives

$$ [\bar\alpha_g(b), d] = \bar\alpha_g([b, \bar\alpha_{g^{-1}}(d)]) = 0, $$

so $C_i$ commutes with $D_{i-1}$. Countability of $G$ keeps each $D_i$
separable, yielding the required algebras and commutation. A limit trace on
$A^\omega$ restricts to a tracial state on $C$, so
$\mathcal T(C) \neq \emptyset$.

The trace $\rho$ of $Z$ is faithful by simplicity. Pick a non-scalar
self-adjoint $c \in Z$; then $\rho((c - \rho(c)1)^2) > 0$, and rescaling
$c - \rho(c)1$ gives a self-adjoint $b \in Z$ with $\rho(b) = 0$ and
$\rho(b^2) = 1$. Set $x_{i,g} = \bar\alpha_g(\iota_i(b))$ and
$M = \|b\|$. Every $\varphi \in \mathcal T(C)$ restricts along each
translated unital embedding of $Z$ to $\rho$, giving the normalization, and
the action law gives the equivariance. This proves the proposition.

The next fact is intrinsic to the algebra generated by the rows, so it
applies to traces of $C$ that do not extend to the ambient ultrapower.

Lemma. Let $B$ be a unital C*-algebra generated by pairwise commuting
unital C*-subalgebras $B_i$, $i \ge 1$. If $\varphi$ is an extreme point of
$\mathcal T(B)$, then

$$ \varphi(b_1 \cdots b_s) = \prod_{j=1}^s \varphi(b_j) $$

whenever $b_j \in B_{i_j}$ and the indices $i_1, \ldots, i_s$ are distinct.

Proof. Let $(H_\varphi, \pi, \Omega)$ be the tracial GNS representation and
put $\mathcal M = \pi(B)''$ and $\mathcal M_i = \pi(B_i)''$. On the dense
subspace of classes $[a]$, $a \in B$, right multiplication
$R_b[a] = [ab]$ is bounded with norm at most $\|b\|$ by traciality, and it
commutes with $\pi(B)$ while the vectors $R_b \Omega = [b]$ are dense. If
$T \in \mathcal M$ has $T \Omega = 0$, then $T R_b \Omega = R_b T \Omega = 0$
for all $b$, so $T = 0$. Thus $\Omega$ is separating for $\mathcal M$, and
the normal vector state $\tau(T) = \langle T \Omega, \Omega \rangle$ is
faithful: for $T \ge 0$, $\tau(T) = 0$ forces $T^{1/2}\Omega = 0$. Its
traciality passes from $\pi(B)$ to $\mathcal M$ by separate ultraweak
continuity, so $\mathcal M$ is a finite von Neumann algebra with faithful
normal trace $\tau$.

Extremality forces $\mathcal M$ to be a factor. A nontrivial central
projection $p \in \mathcal M$ would give the convex decomposition

$$ \varphi(a) = \tau(p) \frac{\tau(p \pi(a))}{\tau(p)} + \tau(1-p) \frac{\tau((1-p)\pi(a))}{\tau(1-p)}, $$

and the two normalized summands are distinct tracial states on $B$: were the
first equal to $\varphi$, equality of the normal functionals would extend
from $\pi(B)$ to $\mathcal M$ and evaluation at $p$ would give
$1 = \tau(p)$, contradicting $0 < \tau(p) < 1$.

The algebras $\mathcal M_i$ commute and generate $\mathcal M$, so each
center satisfies $Z(\mathcal M_i) \subseteq Z(\mathcal M) = \mathbb C 1$,
and $\tau$ makes every $\mathcal M_i$ a finite factor with a faithful normal
tracial state. Fix $i$ and let $y \ge 0$ lie in the von Neumann algebra
generated by the other $\mathcal M_j$. It commutes with $\mathcal M_i$, and
the functional

$$ \psi_y(x) = \tau(y^{1/2} x y^{1/2}) = \tau(x y), \qquad x \in \mathcal M_i, $$

is normal, positive, tracial, and bounded by $\|y\| \tau$, since
sandwiching by $y^{1/2}$ preserves suprema of bounded increasing nets. The
Radon-Nikodym theorem for normal functionals on a finite von Neumann
algebra gives $h \in \mathcal M_i$ with $0 \le h \le \|y\| 1$ and
$\psi_y(x) = \tau(hx)$. Invariance of $\psi_y$ under unitary conjugation
forces $h$ to commute with every unitary of $\mathcal M_i$, hence to be
central and scalar; evaluation at the unit identifies the scalar as
$\tau(y)$. Thus $\tau(x y) = \tau(x) \tau(y)$. Linearity extends this to
all $y$ in the algebra generated by the other rows, and applying it
successively gives the product formula. This proves the lemma.

For $N \ge 1$ and $g, h \in G$ set

$$ z_{N,g} = \frac{1}{\sqrt N} \sum_{i=1}^N x_{i,g}, \qquad k_{N;g,h} = \frac{1}{N} \sum_{i=1}^N \frac{x_{i,g} x_{i,h} + x_{i,h} x_{i,g}}{2}. $$

The last two subscripts of $k_{N;g,h}$ are covariance indices. These
elements are self-adjoint, $k_{N;g,h} = k_{N;h,g}$, and

$$ \bar\alpha_l(z_{N,g}) = z_{N,lg}, \qquad \bar\alpha_l(k_{N;g,h}) = k_{N;lg,lh} \qquad (l, g, h \in G). $$

The joint limit of these variables is Gaussian, and its covariance is the
limit of the empirical covariances.

Proposition. Let $J \subseteq G$ be nonempty and finite, let
$N_j \to \infty$, and let $\varphi_j$ be extreme points of
$\mathcal T(C)$ with $\varphi_j(k_{N_j;g,h}) \to R_{g,h}$ for
$g, h \in J$. Then $R = (R_{g,h})$ is a real positive semidefinite matrix
with diagonal entries one. Choose unit vectors $v_g$ with
$\langle v_g, v_h \rangle = R_{g,h}$, spanning a finite-dimensional real
Euclidean space $L$, let $W$ be standard Gaussian in $L$, and put
$\xi_g = \langle v_g, W \rangle$. Then the pairs
$((z_{N_j,g})_g, (k_{N_j;g,h})_{g,h})$ converge in all ordered joint
moments under $\varphi_j$ to $((\xi_g)_g, (R_{g,h})_{g,h})$. Equivalently,
for every noncommutative *-polynomial $P$ in these self-adjoint variables,

$$ \varphi_j\bigl(P((z_{N_j,g})_g, (k_{N_j;g,h})_{g,h})\bigr) \longrightarrow \mathbb E\, P((\xi_g)_g, (R_{g,h})_{g,h}). $$

For every sequence $(N_j, \varphi_j)$ a subsequence satisfies the covariance
convergence.

Proof. For a tracial $\varphi$ and self-adjoint $x, y$ one has
$\varphi(xy) = \varphi(yx) = \overline{\varphi(xy)}$, so the second moments
are real and symmetric. For real scalars $a_g$,

$$ \sum_{g,h \in J} a_g a_h \varphi_j(k_{N_j;g,h}) = \frac{1}{N_j} \sum_{i=1}^{N_j} \varphi_j\Bigl(\bigl(\sum_{g \in J} a_g x_{i,g}\bigr)^2\Bigr) \ge 0. $$

The diagonal entries are one by the row normalization, and Cauchy-Schwarz
bounds every entry by one in absolute value. The covariance matrices
therefore lie in a compact set of real symmetric matrices, their limits
have the asserted properties, and the subsequence assertion follows. A Gram
representation of $R$ supplies $L$ and the vectors $v_g$.

Consider a word of length $d \ge 1$ in the fluctuation variables. At one
term of the sequence write $N = N_j$, $\varphi = \varphi_j$. Expanding the
sums gives

$$ \varphi(z_{N,g_1} \cdots z_{N,g_d}) = N^{-d/2} \sum_{i_1, \ldots, i_d = 1}^N \varphi(x_{i_1,g_1} \cdots x_{i_d,g_d}). $$

Partition the positions by equality of row labels. Distinct rows commute,
so the factors belonging to one row can be grouped while preserving their
order, and the independence lemma factors the expectation over the distinct
rows. A singleton block gives zero by the row normalization, and each
surviving summand has absolute value at most $M^d$. A partition with $b$
blocks has at most $N^b$ choices of distinct row labels, so its
contribution is at most $M^d N^{b - d/2}$. Among partitions without
singleton blocks, $b \le \lfloor d/2 \rfloor$. Hence odd moments tend to
zero, and for $d = 2q$ only pairings into $q$ pairs can survive.

Fix such a pairing with pairs $(a_l, b_l)$, $1 \le l \le q$. Its
contribution is

$$ N^{-q} \sum_{\substack{1 \le i_1, \ldots, i_q \le N \\ i_1, \ldots, i_q \ \mathrm{distinct}}} \prod_{l=1}^q \varphi(x_{i_l,g_{a_l}} x_{i_l,g_{b_l}}). $$

Removing the distinctness condition changes this by $O_{d,M}(N^{-1})$: there
are at most $\binom{q}{2} N^{q-1}$ tuples with a collision, and each product
is bounded by $M^{2q}$. The unrestricted sum factors, and traciality
identifies the factors with the empirical covariances:

$$ \prod_{l=1}^q \Bigl(\frac1N \sum_{i=1}^N \varphi(x_{i,g_{a_l}} x_{i,g_{b_l}})\Bigr) = \prod_{l=1}^q \varphi(k_{N;g_{a_l},g_{b_l}}). $$

The error estimates depend only on $d$ and $M$, so these are uniform in the
moving traces $\varphi_j$. By the covariance convergence the limiting even
moment is the sum over pairings of the corresponding products of entries of
$R$. This is exactly the Gaussian moment: the moment generating function

$$ \mathbb E \exp\Bigl(\sum_{g \in J} t_g \xi_g\Bigr) = \exp\Bigl(\frac12 \sum_{g,h \in J} t_g t_h R_{g,h}\Bigr) $$

has the same Taylor expansion.

It remains to place the covariance variables in arbitrary word positions.
For $g, h \in J$ define the row covariances and the row-centered deviations

$$ s_{i;g,h} = \frac{x_{i,g} x_{i,h} + x_{i,h} x_{i,g}}{2}, \qquad y_{j,i;g,h} = s_{i;g,h} - \varphi_j(s_{i;g,h}) 1. $$

The centering is row by row at the current trace, so
$\varphi_j(y_{j,i;g,h}) = 0$ and $\|y_{j,i;g,h}\| \le 2M^2$, and there is
the identity

$$ k_{N_j;g,h} - \varphi_j(k_{N_j;g,h}) 1 = \frac{1}{N_j} \sum_{i=1}^{N_j} y_{j,i;g,h}. $$

Let $U_j$ be a word having $d$ fluctuation factors and $r \ge 1$ centered
covariance factors of the last form, expanded over rows. Each of its $d + r$
row factors is centered under $\varphi_j$, and a singleton row again
vanishes by the independence lemma, so a surviving term uses at most
$\lfloor (d+r)/2 \rfloor$ distinct row labels. Absolute expectations within
one row are bounded by the product of the operator norms of the factors.
Finitely many partitions of the $d+r$ positions occur, and with the
normalization $N_j^{-d/2 - r}$ these observations give

$$ |\varphi_j(U_j)| \le C_{d,r,M} N_j^{\lfloor (d+r)/2 \rfloor - d/2 - r} \le C_{d,r,M} N_j^{-r/2} \longrightarrow 0, $$

with a constant independent of $j$ and $\varphi_j$.

Finally replace each covariance factor in a general word by its scalar
$\varphi_j$-mean plus the centered factor. Every term containing a centered
covariance factor tends to zero by the last estimate, and the remaining
terms converge by the fluctuation computation and the covariance
convergence. This proves convergence of every ordered word. Since all
generators are self-adjoint, linearity gives the *-polynomial formulation.

The next step converts the Gaussian limit into noncommutative unitaries
that are almost invariant over the whole trace space. Retain the algebra
$C$, its invariant commuting orbit algebras $C_i$, and the variables
$x_{i,g}$, $z_{N,g}$, $k_{N;g,h}$ built above. Thus $\mathcal T(C)$ is
nonempty, $\|x_{i,g}\| \le M$, and the joint moment limit applies to any
sequence of extreme traces. Traces of finite-dimensional matrices in this
part are unnormalized, and for a matrix $B$ write
$\|B\|_{\mathrm{HS}} = (\mathrm{Tr}(B^*B))^{1/2}$ and
$\|B\|_1 = \mathrm{Tr}|B|$.

Proposition. For every finite $H \subset G$, every $\eps > 0$, and every
integer $m \ge 1$ there are $t > 0$ and a self-adjoint $Q \in C$ such that
$V = \exp(\mathrm{i} t Q)$ satisfies

$$ \sup_{\varphi \in \mathcal T(C)} \|\bar\alpha_h(V) - V\|_{2,\varphi} < \eps \quad (h \in H), $$

$$ \sup_{\varphi \in \mathcal T(C)} |\varphi(V^\ell)| < \eps \quad (1 \le \ell \le m). $$

Three estimates are needed first.

Lemma. If $S, T$ are positive matrices of the same size, then
$\|S^{1/2} - T^{1/2}\|_{\mathrm{HS}}^2 \le \|S - T\|_1$.

Proof. Set $a = S^{1/2}$, $b = T^{1/2}$, $d = a - b$, and
$u = \mathrm{sign}(d)$ with $\mathrm{sign}(0) = 0$. Since
$a^2 - b^2 = ad + db$, cyclicity of the trace gives
$\mathrm{Tr}(u(a^2 - b^2)) = \mathrm{Tr}(|d|(a + b))$. Diagonalizing $d$
gives $a_{jj} + b_{jj} \ge |a_{jj} - b_{jj}| = |d_j|$ by positivity, so

$$ \mathrm{Tr}(|d|(a+b)) = \sum_j |d_j|(a_{jj}+b_{jj}) \ge \sum_j d_j^2 = \mathrm{Tr}(d^2). $$

The operator $u$ is a contraction, so trace-norm duality gives
$\mathrm{Tr}(u(a^2 - b^2)) \le \mathrm{Tr}|a^2 - b^2|$. The two
inequalities prove the lemma.

Lemma. Let $W$ be a standard Gaussian vector in a finite-dimensional real
Euclidean space and let $B$ be a real symmetric operator there. Put
$q_B = \langle W, BW \rangle - \mathrm{Tr}(B)$. Then

$$ \|q_B\|_{L^2} = \sqrt{2} \|B\|_{\mathrm{HS}}, \qquad \bigl|\mathbb E \exp(\mathrm{i} s q_B)\bigr| \le (1 + 4 s^2 \|B\|_{\mathrm{HS}}^2)^{-1/4} $$

for every real $s$. The law of $q_B$ has an exponential moment and is
determined by its moments.

Proof. Orthogonal diagonalization writes
$q_B \overset{\mathrm{law}}{=} \sum_j \lambda_j(U_j^2 - 1)$ with independent
standard normals $U_j$. The summands are centered with variances
$2\lambda_j^2$, giving the first formula. The one-dimensional Gaussian
integral gives
$|\mathbb E \exp(\mathrm{i} s q_B)| = \prod_j (1 + 4 s^2 \lambda_j^2)^{-1/4}$,
and $\prod_j (1 + a_j) \ge 1 + \sum_j a_j$ for nonnegative $a_j$ gives the
second formula.

Choose $c > 0$ with $2 c \max_j |\lambda_j| < 1$, any $c$ when $B = 0$.
The pointwise bound

$$ \exp\Bigl(c\bigl|\sum_j \lambda_j(U_j^2 - 1)\bigr|\Bigr) \le \exp\Bigl(c\sum_j |\lambda_j|\Bigr) \prod_j \exp(c|\lambda_j| U_j^2) $$

and the Gaussian integral show $\mathbb E \exp(c|q_B|) < \infty$. Let $\nu$
be another probability measure with the same moments. By the nonnegative
power series for $\cosh$ and monotone convergence,

$$ \int_{\mathbb R} \cosh(cx)\, d\nu(x) = \sum_{r=0}^{\infty} \frac{c^{2r}}{(2r)!} \mathbb E(q_B^{2r}) = \mathbb E \cosh(c q_B) < \infty. $$

Since $e^{c|x|} \le 2\cosh(cx)$, the measure $\nu$ has an exponential
moment. The characteristic functions of $\nu$ and of $q_B$ extend
analytically to the strip $|\operatorname{Im} z| < c$, their Taylor
coefficients at zero agree, and the identity theorem makes them equal.
Uniqueness of characteristic functions gives $\nu$ equal to the law of
$q_B$. This proves the lemma.

The square-root normalization is exactly what keeps the variance away from
zero. For a positive real symmetric matrix $P$ with $\mathrm{Tr}(P) = 1$
the variance formulas give $\mathrm{Var}(q_{P^{1/2}}) = 2$ and
$\mathrm{Var}(q_P) = 2\mathrm{Tr}(P^2)$. The latter equals $2$ for a
rank-one projection but $2/n$ for $P = n^{-1} I_n$, so using $P$ directly
would make the variance vanish as the rank grows. The square root fixes the
variance at $2$, and the perturbation lemma controls its change under
translation. The construction replaces the exact square root by a
polynomial with an explicitly controlled error.

Lemma. Let $X_j = X_j^*$ lie in unital C*-algebras with states $\varphi_j$,
and let $q_B$ be a centered Gaussian quadratic. If
$\varphi_j(X_j^d) \to \mathbb E(q_B^d)$ for every $d \ge 0$, then
$\varphi_j(f(X_j)) \to \mathbb E f(q_B)$ for every bounded continuous
$f : \mathbb R \to \mathbb C$. No uniform bound on $\|X_j\|$ is needed.

Proof. Let $\mu_j$ be the spectral law of $X_j$ under $\varphi_j$. The
second moments are bounded, so the sequence is tight. Fix $d \ge 0$ and an
even integer $2r > d$. Moment convergence gives $C_r < \infty$ with
$\int |x|^{2r} d\mu_j(x) \le C_r$ for all $j$, hence

$$ \int_{|x| > K} |x|^d\, d\mu_j(x) \le C_r K^{d-2r} \qquad (K > 0). $$

Take any weakly convergent subsequence with limit $\mu$. Lower
semicontinuity gives $\int |x|^{2r} d\mu(x) \le C_r$, so the same tail bound
holds for $\mu$. Multiply $x^d$ by a continuous cutoff equal to one on
$[-K, K]$ and zero outside $[-2K, 2K]$; weak convergence applies to this
bounded continuous function, and letting $K \to \infty$ with the tail bound
gives $\int x^d d\mu(x) = \mathbb E(q_B^d)$ for every $d$. The previous
lemma identifies $\mu$ with the law of $q_B$. Tightness then forces the
whole sequence $\mu_j$ to converge weakly to that law, since every
subsequence has a further weakly convergent subsequence with that same
limit. This is the claimed convergence for $f$.

Proof of the proposition. Step 1, parameters. Fix $H, \eps, m$. Choose
$t > 0$ with $(1 + t^2)^{-1/4} < \eps/4$, then choose $\delta > 0$ and
$0 < \eta < 1/2$ with $t\sqrt2(\sqrt\delta + 2\eta) < \eps/4$. Amenability
gives a nonempty finite $F \subset G$ with
$|hF \,\triangle\, F| < \delta|F|$ for $h \in H$. Set $n = |F|$ and
$J = F \cup \bigcup_{h \in H} hF$. Finally choose a real polynomial

$$ p(s) = \sum_{r=1}^D c_r s^r, \qquad \sup_{0 \le s \le 1} |p(s) - \sqrt{s}| < \eta/\sqrt{n}. $$

Such a polynomial exists: approximate $\sqrt{s}$ on $[0,1]$ by a real
polynomial within $\eta/(2\sqrt n)$, then subtract its value at zero.
Thus $t, \delta, \eta, F, n, J, p$ are fixed before $N \to \infty$ and are
not chosen from any trace or limiting covariance.

Step 2, uniform Gaussian comparison. Let $R = (R_{g,g'})_{g,g' \in J}$ be
any real positive semidefinite matrix with diagonal one. Choose unit
vectors $v_g$ with Gram matrix $R$ in a finite-dimensional real Euclidean
space $L$, let $W$ be standard Gaussian in $L$, and write
$\xi_g = \langle v_g, W \rangle$. For $E = F$ or $E = hF$ set

$$ P_E = \frac1n \sum_{g \in E} v_g v_g^{\mathsf T}, \qquad B_E = p(P_E), \qquad q_E = \langle W, B_E W \rangle - \mathrm{Tr}(B_E). $$

The operator $P_E$ is positive with trace one and spectrum in $[0,1]$, of
rank at most $n$. Because $p(0) = 0$, the operator $p(P_E) - P_E^{1/2}$
vanishes on the kernel of $P_E$, and the scalar approximation gives

$$ \|B_E - P_E^{1/2}\|_{\mathrm{HS}}^2 \le \mathrm{rank}(P_E)\, \eta^2/n \le \eta^2, \qquad \|P_E^{1/2}\|_{\mathrm{HS}}^2 = \mathrm{Tr}(P_E) = 1, $$

hence $1 - \eta \le \|B_E\|_{\mathrm{HS}} \le 1 + \eta$.

The summands indexed by $F \cap hF$ cancel in $P_F - P_{hF}$, and each
remaining $v_g v_g^{\mathsf T}$ has trace norm one, so the square-root
lemma gives

$$ \|P_F^{1/2} - P_{hF}^{1/2}\|_{\mathrm{HS}}^2 \le \|P_F - P_{hF}\|_1 \le \frac{|F \,\triangle\, hF|}{n} < \delta. $$

With the rank control this yields
$\|B_F - B_{hF}\|_{\mathrm{HS}} \le \sqrt\delta + 2\eta$. Both forms
$q_F, q_{hF}$ are evaluated on the same $W$, so their difference is the
centered quadratic of $B_{hF} - B_F$, and the Gaussian quadratic lemma
gives $\|q_{hF} - q_F\|_{L^2} \le \sqrt2(\sqrt\delta + 2\eta)$. Also, by
the nondegeneracy and $\eta < 1/2$, for every integer $\ell \ge 1$,

$$ \bigl|\mathbb E \exp(\mathrm{i}\ell t q_F)\bigr| \le (1 + 4\ell^2 t^2 \|B_F\|_{\mathrm{HS}}^2)^{-1/4} \le (1 + t^2)^{-1/4}. $$

All these bounds hold for every such $R$, singular or not, and no
translation invariance of $R$ has been used.

Step 3, the equivariant polynomial prescription. Direct multiplication of
the rank-one summands gives
$P_E^r = n^{-r} \sum_{g_1, \ldots, g_r \in E} R_{g_1,g_2} \cdots R_{g_{r-1},g_r} v_{g_1} v_{g_r}^{\mathsf T}$.
Taking quadratic forms and traces shows

$$ q_E = \sum_{r=1}^D \frac{c_r}{n^r} \sum_{g_1, \ldots, g_r \in E} R_{g_1,g_2} \cdots R_{g_{r-1},g_r} (\xi_{g_1} \xi_{g_r} - R_{g_r,g_1}), $$

with the empty product interpreted as $1$ when $r = 1$. Use the same
prescription in $C$:

$$ \widetilde Q_{N,E} = \sum_{r=1}^D \frac{c_r}{n^r} \sum_{g_1, \ldots, g_r \in E} k_{N;g_1,g_2} \cdots k_{N;g_{r-1},g_r} (z_{N,g_1} z_{N,g_r} - k_{N;g_r,g_1}), $$

$$ Q_{N,E} = \tfrac12\bigl(\widetilde Q_{N,E} + \widetilde Q_{N,E}^{\,*}\bigr). $$

Multiplication is in the displayed order. The empirical covariance array
need satisfy no positivity condition, because this is a finite polynomial in
elements of $C$; square-root functional calculus was used only for $P_E$ in
the Gaussian comparison. The relabeling
$\bar\alpha_l(z_{N,g}) = z_{N,lg}$, $\bar\alpha_l(k_{N;g,h}) = k_{N;lg,lh}$
carries each tuple in the sum for $F$ to its left translate in the sum for
$hF$, preserving the order of all factors, and taking self-adjoint parts
commutes with $\bar\alpha_h$. Hence

$$ Q_{N,hF} = \bar\alpha_h(Q_{N,F}) \qquad (h \in H,\ N \ge 1). $$

Step 4, transfer of moments and scalar laws. Take any $N_j \to \infty$ and
extreme traces $\varphi_j \in \mathcal T(C)$ with
$\varphi_j(k_{N_j;g,g'}) \to R_{g,g'}$ for $g, g' \in J$. The joint moment
limit gives convergence of every ordered mixed moment of the empirical
variables, and any moment of the polynomials just written is a finite
linear combination of these. Since the limiting variables commute and the
closed formula for $q_E$ is real-valued, the $Q_{N_j,E}$ converge jointly
in moments under $\varphi_j$ to the $q_E$. In particular

$$ \|Q_{N_j,hF} - Q_{N_j,F}\|_{2,\varphi_j} \longrightarrow \|q_{hF} - q_F\|_{L^2}, $$

$$ \varphi_j(\exp(\mathrm{i} s Q_{N_j,F})) \longrightarrow \mathbb E \exp(\mathrm{i} s q_F) $$

for every real $s$. The passage to scalar spectral laws is needed because
the operator norms of $Q_{N,E}$ may grow with $N$: the estimates
$\|k_{N;g,g'}\| \le M^2$ and $\|z_{N,g}\| \le \sqrt N M$ give only
$\|Q_{N,E}\| \le (N+1) \sum_r |c_r| M^{2r}$. Scalar moment convergence
together with the spectral tail estimate is enough for the exponential
convergence.

Step 5, uniformity over all traces. Set $V_N = \exp(\mathrm{i} t Q_{N,F})$.
For self-adjoint $X, Y$ in a unital C*-algebra and a tracial $\varphi$,
differentiation gives

$$ \frac{d}{ds}\bigl(e^{\mathrm{i}(1-s)tX} e^{\mathrm{i}stY}\bigr) = e^{\mathrm{i}(1-s)tX}\, \mathrm{i}t(Y - X)\, e^{\mathrm{i}stY}, $$

and the tracial 2-seminorm is unchanged by left and right unitary
multiplication, so integration and the triangle inequality give the
tracial Duhamel bound $\|e^{\mathrm{i}tX} - e^{\mathrm{i}tY}\|_{2,\varphi} \le t\|X - Y\|_{2,\varphi}$.
Together with exact equivariance, the norm limit, and the Gaussian
translation bound, every sequence of traces and indices with convergent
covariance satisfies

$$ \limsup_{j \to \infty} \|\bar\alpha_h(V_{N_j}) - V_{N_j}\|_{2,\varphi_j} \le t\sqrt2(\sqrt\delta + 2\eta) < \eps/4, $$

$$ \lim_{j \to \infty} |\varphi_j(V_{N_j}^\ell)| = \bigl|\mathbb E \exp(\mathrm{i}\ell t q_F)\bigr| \le (1 + t^2)^{-1/4} < \eps/4 $$

for $h \in H$ and $1 \le \ell \le m$, by the choices of the parameters.

Consequently, for all sufficiently large $N$, every extreme trace
$\varphi \in \mathcal T(C)$ satisfies
$\|\bar\alpha_h(V_N) - V_N\|_{2,\varphi} < \eps/2$ and
$|\varphi(V_N^\ell)| < \eps/2$ for $h \in H$ and $1 \le \ell \le m$. Were
this false, some $N_j \to \infty$ and extreme traces $\varphi_j$ would fail
one of these finitely many tests at tolerance $\eps/2$. The covariance
matrices $(\varphi_j(k_{N_j;g,g'}))_{g,g' \in J}$ lie in the compact set of
real positive semidefinite matrices with diagonal one, so a subsequence has
a limiting covariance $R$, and the sequential estimates contradict the
failure. Only the finite covariance matrices were extracted; no compactness
of the extreme trace boundary is used.

Fix $N$ with these extreme-trace estimates. For each $h \in H$ the squared
seminorm $\varphi \mapsto \varphi((\bar\alpha_h(V_N) - V_N)^*(\bar\alpha_h(V_N) - V_N))$
is weak-star continuous and affine on the convex set $\mathcal T(C)$, and
for each $1 \le \ell \le m$ the map $\varphi \mapsto |\varphi(V_N^\ell)|$
is weak-star continuous and convex. Their sublevel sets with bounds
$\eps^2/4$ and $\eps/2$ are closed and convex and contain every extreme
trace, so by the Krein-Milman theorem these sublevel sets contain $\mathcal T(C)$. Hence
both suprema in the proposition are at most $\eps/2 < \eps$, and
$Q = Q_{N,F}$, $V = V_N$ complete the proof.

What remains is to pass from the approximately invariant unitaries of the
proposition to exact relations in $A^\omega$, and then to the projections
of equivariant uniform property Gamma. Both passages use one
coordinate-selection fact.

Lemma. If $x = [(x_j)] \in A^\omega$ has a bounded representative, then

$$ \sup_{\tau \in \mathcal T_\omega(A)} \|x\|_{2,\tau} = \lim_{j \to \omega} \sup_{\sigma \in \mathcal T(A)} \|x_j\|_{2,\sigma}, $$

$$ \sup_{\tau \in \mathcal T_\omega(A)} |\tau(x)| = \lim_{j \to \omega} \sup_{\sigma \in \mathcal T(A)} |\sigma(x_j)|. $$

Finitely many such tests can therefore be realized simultaneously at a
single coordinate with any prescribed positive slack above their
ultralimit bounds.

Proof. Every sequence of traces gives the upper bound in each identity.
Conversely, for each $j$ choose $\sigma_j \in \mathcal T(A)$ within $1/j$ of
the supremum for the test in question. Its limit trace realizes the reverse
inequality, since square roots and absolute values commute with ultralimits
of bounded scalar sequences. For finitely many tests, the coordinates
satisfying each individual bound with positive slack form a set in
$\omega$; their intersection is again in $\omega$, hence nonempty and
indeed contains arbitrarily large coordinates. The maximizing trace may
differ from test to test.

Keep the invariant separable algebra $C \subset A^\omega \cap A'$
constructed from the rows. For $a \in A_+$ with $0 \le a \le 1$ and
$\tau \in \mathcal T_\omega(A)$, define on $C$

$$ \tau_a(c) = \tau(ac). $$

This is a positive tracial functional: positivity follows from
$ac = a^{1/2} c a^{1/2}$ when $c \ge 0$, using that $a$ commutes with $C$,
and traciality from $\tau(acd) = \tau(dac) = \tau(adc)$. Its norm is
$\tau(a)$. When $\tau(a) > 0$ its normalization is an element of
$\mathcal T(C)$, and when $\tau(a) = 0$ it vanishes, so the uniform
approximately invariant unitary $V = \exp(\mathrm{i} t Q)$ from the
preceding proposition gives

$$ |\tau(a V^\ell)| < \eps\, \tau(a) \le \eps \qquad (0 \le a \le 1,\ \tau \in \mathcal T_\omega(A),\ 1 \le \ell \le m,\ \tau(a) > 0), $$

with zero left side when $\tau(a) = 0$. Approximate invariance passes to
limit traces by restriction to $C$.

Proposition. There is a unitary $w \in (A^\omega \cap A')^{\bar\alpha}$
with

$$ \tau(a f(w)) = \tau(a) \int_{\mathbb T} f\, d\mu $$

for every $a \in A$, every $\tau \in \mathcal T_\omega(A)$, and every
continuous $f$, where $\mathbb T = \{z \in \mathbb C : |z| = 1\}$ and $\mu$
is normalized Haar measure.

Proof. Choose increasing finite sets $H_m \subset G$ with union $G$ and a
sequence $a_1, a_2, \ldots$ dense in the positive unit ball of $A$ with
$a_1 = 1$. Apply the uniform-unitary proposition with
$(H, \eps, m) = (H_m, 1/m, m)$, and call the resulting exponential unitary
$V^{(m)} \in C$. It has a representative $(v^{(m)}_j)_j$ of unitaries in
$A$: lift the self-adjoint exponent to a bounded self-adjoint sequence and
exponentiate coordinatewise. The bounds for those lifts may depend on $m$,
whereas the unitary representatives always have norm one.

Since $V^{(m)}$ commutes with $A$, the coordinate-selection lemma, the
weighted small moments, and the invariance estimates allow a coordinate
$j(m) \ge m$ at which $w_m = v^{(m)}_{j(m)}$ satisfies

$$ \|[w_m, a_r]\|_{2,u} < 2/m \quad (1 \le r \le m), $$

$$ \|\alpha_h(w_m) - w_m\|_{2,u} < 2/m \quad (h \in H_m), $$

$$ \sup_{\sigma \in \mathcal T(A)} |\sigma(a_r w_m^\ell)| < 2/m \quad (1 \le r, \ell \le m). $$

The last display applies the trace identity of the lemma to the class
$a_r (V^{(m)})^\ell$. All coordinate tests are finite in number, and their
suprema already range over all traces.

The class $w = [(w_m)]$ is a unitary. The first two displays, density, and
$\|[w_m, a - a_r]\|_{2,u} \le 2\|a - a_r\|$ make it central relative to $A$
and fixed by $\bar\alpha$. The last display, first for every $a_r$, then by
norm density for positive contractions, then by linearity for $a \in A$,
gives $\tau(a w^\ell) = 0$ for every $a \in A$, $\tau \in \mathcal T_\omega(A)$,
and $\ell \ge 1$. The same holds for negative powers because
$\tau(a w^{-\ell}) = \overline{\tau(a^* w^\ell)}$, and the zeroth power has
trace $\tau(a)$. Hence the identity holds for trigonometric polynomials,
and uniform approximation on $\mathbb T$ together with
$|\tau(a f(w))| \le \|a\| \|f\|_\infty$ proves it for every continuous $f$.

Proposition. The action $\alpha$ has equivariant uniform property Gamma,
and its projections can be chosen to sum to $1$.

Proof. Fix $k \ge 2$ and let $w = [(w_j)]$ be the unitary just constructed,
represented by unitaries $w_j \in A$. Partition $\mathbb T$ into $k$
half-open arcs $I_1, \ldots, I_k$ of Haar measure $1/k$ and write
$\chi_i = 1_{I_i}$. Choose continuous functions
$f_i^{(r)} : \mathbb T \to [0,1]$ with

$$ e_r := \max_{1 \le i \le k} \|f_i^{(r)} - \chi_i\|_{L^2(\mu)} \longrightarrow 0, $$

for instance by replacing each indicator near its two endpoints by linear
ramps whose total lengths tend to zero. Fix $r$ and write $f_i = f_i^{(r)}$,
$e = e_r$. Pointwise bounds and the triangle inequality give

$$ \|f_i^2 - f_i\|_{L^2(\mu)} \le e, \qquad \|f_i f_j\|_{L^2(\mu)} \le 2e \ \ (i \ne j), $$

$$ \Bigl\|\sum_i f_i - 1\Bigr\|_{L^2(\mu)} \le k e, \qquad \Bigl|\int_{\mathbb T} f_i\, d\mu - \frac1k\Bigr| \le e. $$

The first uses $|f_i^2 - f_i| \le |f_i - \chi_i|$ almost everywhere; the
second expands
$f_i f_j - \chi_i \chi_j = (f_i - \chi_i) f_j + \chi_i (f_j - \chi_j)$ and
uses $\chi_i \chi_j = 0$ almost everywhere; the remaining ones follow from
$\sum_i \chi_i = 1$ almost everywhere and the $L^1$-$L^2$ inequality for a
probability measure.

Set $y_i^{(r)} = f_i^{(r)}(w)$. These are central, invariant positive
contractions because continuous functional calculus respects commutation and
automorphisms. The trace identity with $a = 1$, applied to the squared
moduli of the functions above, gives the corresponding 2-norm bounds for
the $y_i^{(r)}$ uniformly over $\tau \in \mathcal T_\omega(A)$, and for
every positive contraction $a \in A$ it also gives

$$ \Bigl|\tau(a y_i^{(r)}) - \frac{\tau(a)}{k}\Bigr| = \tau(a) \Bigl|\int_{\mathbb T} f_i^{(r)}\, d\mu - \frac1k\Bigr| \le e_r. $$

Represent $y_i^{(r)}$ by $f_i^{(r)}(w_j)$, and use $H_r$ and
$a_1, \ldots, a_r$ from the preceding proof. For fixed $r$ the
coordinate-selection lemma gives a coordinate $j(r) \ge r$ such that the
positive contractions $y_{i,r} = f_i^{(r)}(w_{j(r)})$ satisfy, for all
indicated indices,

$$ \|y_{i,r}^2 - y_{i,r}\|_{2,u} < e_r + 1/r, \qquad \|y_{i,r} y_{j,r}\|_{2,u} < 2 e_r + 1/r \ \ (i \ne j), $$

$$ \Bigl\|\sum_i y_{i,r} - 1\Bigr\|_{2,u} < k e_r + 1/r, $$

$$ \|[y_{i,r}, a_s]\|_{2,u} < 1/r \quad (s \le r), \qquad \|\alpha_h(y_{i,r}) - y_{i,r}\|_{2,u} < 1/r \quad (h \in H_r), $$

$$ \sup_{\sigma \in \mathcal T(A)} \Bigl|\sigma(a_s y_{i,r}) - \frac{\sigma(a_s)}{k}\Bigr| < e_r + 1/r \quad (s \le r). $$

This is a coordinate choice made after $f_i^{(r)}$ is fixed, so it needs no
uniform continuity modulus, as $r$ varies, for these increasingly sharp arc
approximations.

Let $p_i = [(y_{i,r})_r] \in A^\omega$. The representatives are uniformly
bounded positive contractions. Small square defect, orthogonality, and unit
sum make the $p_i$ orthogonal projections summing to $1$. Small commutators
and approximate invariance make them central and invariant by density. The
trace estimate gives, for every sequence $(\sigma_r)$ in $\mathcal T(A)$,

$$ \lim_{r \to \omega} \sigma_r(a_s y_{i,r}) = \frac1k \lim_{r \to \omega} \sigma_r(a_s), $$

which is the defining trace relation for $a_s$ and every limit trace, and
norm density with linearity extends it to all $a \in A$. The projections
therefore realize equivariant uniform property Gamma and sum to $1$.

Proof of the theorem. The trace hypotheses were checked earlier: exactness,
a tracial state, and faithfulness of every trace follow from nuclearity,
stable finiteness, simplicity, and the quasitrace theorem. The last
proposition supplies equivariant uniform property Gamma, so all hypotheses
of the absorption criterion hold and $\alpha$ and
$\beta = \alpha \otimes \mathrm{id}_Z$ are cocycle conjugate. The exact
orientation of the equations is fixed as follows.

An isomorphism $\Phi : A \to B$ is automatically unital, being surjective
between unital algebras. Put $\gamma_g = \Phi \alpha_g \Phi^{-1}$. If
cocycle conjugacy is supplied in the form
$\gamma_g = \mathrm{Ad}(u_g) \beta_g$ for a $\beta$-cocycle $u$, then the
two displayed equations of the theorem follow directly. In the reverse
orientation, suppose $\beta_g = \mathrm{Ad}(v_g) \gamma_g$ with
$v_{gh} = v_g \gamma_g(v_h)$. The unitaries lie in $B$, whose multiplier
algebra is $B$ itself. Set $u_g = v_g^*$. Then

$$ u_g \beta_g(u_h) = v_g^* v_g \gamma_g(v_h^*) v_g^* = \gamma_g(v_h^*) v_g^* = v_{gh}^* = u_{gh}, \qquad \mathrm{Ad}(u_g) \beta_g = \gamma_g. $$

Either orientation therefore gives the required $\beta$-cocycle, and the
identity relation $u_e = u_e^2$ together with unitarity forces $u_e = 1$.
This yields both displayed equations and proves the theorem.