diff --git a/equivariant_jiang_su_stability_v2.txt b/equivariant_jiang_su_stability_v2.txt
index 30fa771..399adbc 100644
--- a/equivariant_jiang_su_stability_v2.txt
+++ b/equivariant_jiang_su_stability_v2.txt
@@ -9,15 +9,15 @@ $B$, 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
+$Z \cong Z  ⊗  Z$. The dimension-drop construction and these properties
 appear in the study of strongly self-absorbing algebras.
 
 ./pdf/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 $\beta_g = \alpha_g \otimes \mathrm{id}_Z$.
+nuclear, stably finite C*-algebra with $A \cong A  ⊗  Z$. Let $G$ be a
+countable discrete amenable group and $\alpha : G \to Aut(A)$ an
+action. Put $B = A  ⊗  Z$ and $\beta_g = \alpha_g  ⊗  id_Z$.
 Then there are a unital *-isomorphism $\Phi : A \to B$ and unitaries
 $u_g \in B$ for $g \in G$ such that
 
@@ -28,9 +28,9 @@ 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
+finite $F \subset G$ with $|hF \,△\, F| < \delta |F|$ for every
 $h \in H$. The displayed relations say exactly that $\alpha$ and
-$\beta = \alpha \otimes \mathrm{id}_Z$ are cocycle conjugate, with
+$\beta = \alpha  ⊗  id_Z$ are cocycle conjugate, with
 $\Phi$ intertwining the two actions through the $\beta$-cocycle
 $g \mapsto u_g$.
 
@@ -55,7 +55,7 @@ comparison of positive elements.
 
 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: $A \otimes Z \cong A$; $A$
+traces. Then the following are equivalent: $A  ⊗  Z \cong A$; $A$
 has strict comparison; every completely positive map from $A$ to $A$
 can be excised in small central sequences; has property (SI).
 
@@ -90,7 +90,7 @@ Corollary 5.12 (Haagerup). Every stably finite unital exact C*-algebra has a tra
 Fix a free ultrafilter $\omega$ on the positive integers and form the
 uniform tracial ultrapower
 
-$$ A^\omega = \ell^\infty(\mathbb N, A) / I_\omega, \qquad I_\omega = \{(a_j) : \lim_{j \to \omega} \|a_j\|_{2,u} = 0\}. $$
+$$ A^\omega = \ell^∞(\mathbb N, A) / I_\omega, \qquad I_\omega = \{(a_j) : \lim_{j \to \omega} \|a_j\|_{2,u} = 0\}. $$
 
 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
@@ -105,7 +105,7 @@ $\bar\alpha : G \curvearrowright A^\omega$ 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
+projections $p_1, ..., p_k \in (A^\omega \cap A')^{\bar\alpha}$ 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
@@ -128,27 +128,27 @@ $$ C_i = C^*(\bar\alpha_g(\iota_i(Z)) : g \in G) $$
 
 commute with $A$ and commute pairwise. Consequently
 
-$$ C = C^*\Bigl(\bigcup_{i \ge 1} C_i\Bigr) \subseteq A^\omega \cap A' $$
+$$ C = C^*(∪_{i \ge 1} C_i) ⊆ A^\omega \cap A' $$
 
 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
+$\mathcal T(C) \neq ∅$. There are moreover self-adjoint elements
+$x_{i,g} \in C_i$ and a constant $M < ∞$ 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
+$A \cong A  ⊗  ⊗_{l \ge 1} Z$ and use its increasing finite
 tensor stages. First, every norm-separable C*-subalgebra
-$D \subseteq A^\omega$ admits a unital embedding $Z \to A^\omega \cap D'$.
+$D ⊆ A^\omega$ 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
+$d_{1,j}, ..., d_{j,j}$, and let $\theta_j : Z \to A$ embed into a tensor
 factor strictly after that stage. For $j \ge s$ and $z \in Z$,
 
-$$ \|[\theta_j(z), d_{s,j}]\| \le \frac{2\|z\|}{j}. $$
+$$ \|[\theta_j(z), d_{s,j}]\| \le (2\|z\|)/(j). $$
 
 The coordinatewise homomorphisms therefore induce a unital *-homomorphism
 $z \mapsto [(\theta_j(z))_j]$ into $A^\omega \cap D'$, injective because $Z$
@@ -166,7 +166,7 @@ $$ [\bar\alpha_g(b), d] = \bar\alpha_g([b, \bar\alpha_{g^{-1}}(d)]) = 0, $$
 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
-$\mathcal T(C) \neq \emptyset$.
+$\mathcal T(C) \neq ∅$.
 
 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
@@ -183,19 +183,19 @@ 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
 
-$$ \varphi(b_1 \cdots b_s) = \prod_{j=1}^s \varphi(b_j) $$
+$$ \varphi(b_1 ... b_s) = \prod_{j=1}^s \varphi(b_j) $$
 
-whenever $b_j \in B_{i_j}$ and the indices $i_1, \ldots, i_s$ are distinct.
+whenever $b_j \in B_{i_j}$ and the indices $i_1, ..., i_s$ are distinct.
 
-Proof. Let $(H_\varphi, \pi, \Omega)$ be the tracial GNS representation and
+Proof. Let $(H_\varphi, \pi, Ω)$ 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
+commutes with $\pi(B)$ while the vectors $R_b Ω = [b]$ are dense. If
+$T \in \mathcal M$ has $T Ω = 0$, then $T R_b Ω = R_b T Ω = 0$
+for all $b$, so $T = 0$. Thus $Ω$ is separating for $\mathcal M$, and
+the normal vector state $\tau(T) = < T Ω, Ω >$ is
+faithful: for $T \ge 0$, $\tau(T) = 0$ forces $T^{1/2}Ω = 0$. 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 $\tau$.
@@ -203,7 +203,7 @@ normal trace $\tau$.
 Extremality forces $\mathcal M$ to be a factor. A nontrivial central
 projection $p \in \mathcal M$ would give the convex decomposition
 
-$$ \varphi(a) = \tau(p) \frac{\tau(p \pi(a))}{\tau(p)} + \tau(1-p) \frac{\tau((1-p)\pi(a))}{\tau(1-p)}, $$
+$$ \varphi(a) = \tau(p) (\tau(p \pi(a)))/(\tau(p)) + \tau(1-p) (\tau((1-p)\pi(a)))/(\tau(1-p)), $$
 
 and the two normalized summands are distinct tracial states on $B$: were the
 first equal to $\varphi$, equality of the normal functionals would extend
@@ -211,7 +211,7 @@ 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 $Z(\mathcal M_i) \subseteq Z(\mathcal M) = \mathbb C 1$,
+center satisfies $Z(\mathcal M_i) ⊆ Z(\mathcal M) = \mathbb C 1$,
 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
@@ -232,7 +232,7 @@ successively gives the product formula. This proves the lemma.
 
 For $N \ge 1$ and $g, h \in G$ set
 
-$$ z_{N,g} = \frac{1}{\sqrt N} \sum_{i=1}^N x_{i,g}, \qquad k_{N;g,h} = \frac{1}{N} \sum_{i=1}^N \frac{x_{i,g} x_{i,h} + x_{i,h} x_{i,g}}{2}. $$
+$$ z_{N,g} = (1)/(\sqrt N) \sum_{i=1}^N x_{i,g}, \qquad k_{N;g,h} = (1)/(N) \sum_{i=1}^N \frac{x_{i,g} x_{i,h} + x_{i,h} x_{i,g}}{2}. $$
 
 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
@@ -242,28 +242,28 @@ $$ \bar\alpha_l(z_{N,g}) = z_{N,lg}, \qquad \bar\alpha_l(k_{N;g,h}) = k_{N;lg,lh
 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
+Proposition. Let $J ⊆ G$ be nonempty and finite, let
+$N_j \to ∞$, 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
+$< v_g, v_h > = R_{g,h}$, 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
+$\xi_g = < v_g, W >$. 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,
 
-$$ \varphi_j\bigl(P((z_{N_j,g})_g, (k_{N_j;g,h})_{g,h})\bigr) \longrightarrow \mathbb E\, P((\xi_g)_g, (R_{g,h})_{g,h}). $$
+$$ \varphi_j(P((z_{N_j,g})_g, (k_{N_j;g,h})_{g,h})) → \mathbb E\, P((\xi_g)_g, (R_{g,h})_{g,h}). $$
 
 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
+$\varphi(xy) = \varphi(yx) = \varphi(xy)bar$, so the second moments
 are real and symmetric. For real scalars $a_g$,
 
-$$ \sum_{g,h \in J} a_g a_h \varphi_j(k_{N_j;g,h}) = \frac{1}{N_j} \sum_{i=1}^{N_j} \varphi_j\Bigl(\bigl(\sum_{g \in J} a_g x_{i,g}\bigr)^2\Bigr) \ge 0. $$
+$$ \sum_{g,h \in J} a_g a_h \varphi_j(k_{N_j;g,h}) = (1)/(N_j) \sum_{i=1}^{N_j} \varphi_j((\sum_{g \in J} a_g x_{i,g})^2) \ge 0. $$
 
 The diagonal entries are one by the row normalization, and Cauchy-Schwarz
 bounds every entry by one in absolute value. The covariance matrices
@@ -275,7 +275,7 @@ 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
 
-$$ \varphi(z_{N,g_1} \cdots z_{N,g_d}) = N^{-d/2} \sum_{i_1, \ldots, i_d = 1}^N \varphi(x_{i_1,g_1} \cdots x_{i_d,g_d}). $$
+$$ \varphi(z_{N,g_1} ... z_{N,g_d}) = N^{-d/2} \sum_{i_1, ..., i_d = 1}^N \varphi(x_{i_1,g_1} ... x_{i_d,g_d}). $$
 
 Partition the positions by equality of row labels. Distinct rows commute,
 so the factors belonging to one row can be grouped while preserving their
@@ -290,21 +290,21 @@ 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
 
-$$ N^{-q} \sum_{\substack{1 \le i_1, \ldots, i_q \le N \\ i_1, \ldots, i_q \ \mathrm{distinct}}} \prod_{l=1}^q \varphi(x_{i_l,g_{a_l}} x_{i_l,g_{b_l}}). $$
+$$ N^{-q} \sum_{\substack{1 \le i_1, ..., i_q \le N \\ i_1, ..., i_q \ distinct}} \prod_{l=1}^q \varphi(x_{i_l,g_{a_l}} x_{i_l,g_{b_l}}). $$
 
 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:
 
-$$ \prod_{l=1}^q \Bigl(\frac1N \sum_{i=1}^N \varphi(x_{i,g_{a_l}} x_{i,g_{b_l}})\Bigr) = \prod_{l=1}^q \varphi(k_{N;g_{a_l},g_{b_l}}). $$
+$$ \prod_{l=1}^q (\frac1N \sum_{i=1}^N \varphi(x_{i,g_{a_l}} x_{i,g_{b_l}})) = \prod_{l=1}^q \varphi(k_{N;g_{a_l},g_{b_l}}). $$
 
 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
 
-$$ \mathbb E \exp\Bigl(\sum_{g \in J} t_g \xi_g\Bigr) = \exp\Bigl(\frac12 \sum_{g,h \in J} t_g t_h R_{g,h}\Bigr) $$
+$$ \mathbb E \exp(\sum_{g \in J} t_g \xi_g) = \exp(\frac12 \sum_{g,h \in J} t_g t_h R_{g,h}) $$
 
 has the same Taylor expansion.
 
@@ -317,7 +317,7 @@ 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
 
-$$ k_{N_j;g,h} - \varphi_j(k_{N_j;g,h}) 1 = \frac{1}{N_j} \sum_{i=1}^{N_j} y_{j,i;g,h}. $$
+$$ k_{N_j;g,h} - \varphi_j(k_{N_j;g,h}) 1 = (1)/(N_j) \sum_{i=1}^{N_j} y_{j,i;g,h}. $$
 
 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 $d + r$
@@ -328,7 +328,7 @@ 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
 
-$$ |\varphi_j(U_j)| \le C_{d,r,M} N_j^{\lfloor (d+r)/2 \rfloor - d/2 - r} \le C_{d,r,M} N_j^{-r/2} \longrightarrow 0, $$
+$$ |\varphi_j(U_j)| \le C_{d,r,M} N_j^{\lfloor (d+r)/2 \rfloor - d/2 - r} \le C_{d,r,M} N_j^{-r/2} → 0, $$
 
 with a constant independent of $j$ and $\varphi_j$.
 
@@ -346,12 +346,12 @@ $x_{i,g}$, $z_{N,g}$, $k_{N;g,h}$ 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
-$\|B\|_1 = \mathrm{Tr}|B|$.
+$\|B\|_{HS} = (Tr(B^*B))^{1/2}$ and
+$\|B\|_1 = Tr|B|$.
 
 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
+$V = \exp(i t Q)$ satisfies
 
 $$ \sup_{\varphi \in \mathcal T(C)} \|\bar\alpha_h(V) - V\|_{2,\varphi} < \eps \quad (h \in H), $$
 
@@ -360,60 +360,60 @@ $$ \sup_{\varphi \in \mathcal T(C)} |\varphi(V^\ell)| < \eps \quad (1 \le \ell \
 Three estimates are needed first.
 
 Lemma. If $S, T$ are positive matrices of the same size, then
-$\|S^{1/2} - T^{1/2}\|_{\mathrm{HS}}^2 \le \|S - T\|_1$.
+$\|S^{1/2} - T^{1/2}\|_{HS}^2 \le \|S - T\|_1$.
 
 Proof. Set $a = S^{1/2}$, $b = T^{1/2}$, $d = a - b$, and
-$u = \mathrm{sign}(d)$ with $\mathrm{sign}(0) = 0$. Since
+$u = sign(d)$ with $sign(0) = 0$. 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 $d$
+$Tr(u(a^2 - b^2)) = Tr(|d|(a + b))$. Diagonalizing $d$
 gives $a_{jj} + b_{jj} \ge |a_{jj} - b_{jj}| = |d_j|$ by positivity, so
 
-$$ \mathrm{Tr}(|d|(a+b)) = \sum_j |d_j|(a_{jj}+b_{jj}) \ge \sum_j d_j^2 = \mathrm{Tr}(d^2). $$
+$$ Tr(|d|(a+b)) = \sum_j |d_j|(a_{jj}+b_{jj}) \ge \sum_j d_j^2 = Tr(d^2). $$
 
 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
+$Tr(u(a^2 - b^2)) \le Tr|a^2 - b^2|$. 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
+$q_B = < W, BW > - Tr(B)$. Then
 
-$$ \|q_B\|_{L^2} = \sqrt{2} \|B\|_{\mathrm{HS}}, \qquad \bigl|\mathbb E \exp(\mathrm{i} s q_B)\bigr| \le (1 + 4 s^2 \|B\|_{\mathrm{HS}}^2)^{-1/4} $$
+$$ \|q_B\|_{L^2} = √(2) \|B\|_{HS}, \qquad |\mathbb E \exp(i s q_B)| \le (1 + 4 s^2 \|B\|_{HS}^2)^{-1/4} $$
 
 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
+$q_B \overset{law}{=} \sum_j \lambda_j(U_j^2 - 1)$ 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
-$|\mathbb E \exp(\mathrm{i} s q_B)| = \prod_j (1 + 4 s^2 \lambda_j^2)^{-1/4}$,
+$|\mathbb E \exp(i s q_B)| = \prod_j (1 + 4 s^2 \lambda_j^2)^{-1/4}$,
 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
 
-$$ \exp\Bigl(c\bigl|\sum_j \lambda_j(U_j^2 - 1)\bigr|\Bigr) \le \exp\Bigl(c\sum_j |\lambda_j|\Bigr) \prod_j \exp(c|\lambda_j| U_j^2) $$
+$$ \exp(c|\sum_j \lambda_j(U_j^2 - 1)|) \le \exp(c\sum_j |\lambda_j|) \prod_j \exp(c|\lambda_j| U_j^2) $$
 
-and the Gaussian integral show $\mathbb E \exp(c|q_B|) < \infty$. Let $\nu$
+and the Gaussian integral show $\mathbb E \exp(c|q_B|) < ∞$. Let $\nu$
 be another probability measure with the same moments. By the nonnegative
-power series for $\cosh$ and monotone convergence,
+power series for $cosh$ and monotone convergence,
 
-$$ \int_{\mathbb R} \cosh(cx)\, d\nu(x) = \sum_{r=0}^{\infty} \frac{c^{2r}}{(2r)!} \mathbb E(q_B^{2r}) = \mathbb E \cosh(c q_B) < \infty. $$
+$$ \int_{\mathbb R} cosh(cx)\, d\nu(x) = \sum_{r=0}^{∞} \frac{c^{2r}}{(2r)!} \mathbb E(q_B^{2r}) = \mathbb E cosh(c q_B) < ∞. $$
 
-Since $e^{c|x|} \le 2\cosh(cx)$, the measure $\nu$ has an exponential
+Since $e^{c|x|} \le 2cosh(cx)$, 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
+analytically to the strip $|Im z| < c$, 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
+zero. For a positive real symmetric matrix $P$ with $Tr(P) = 1$
+the variance formulas give $Var(q_{P^{1/2}}) = 2$ and
+$Var(q_P) = 2Tr(P^2)$. 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
@@ -428,7 +428,7 @@ $f : \mathbb R \to \mathbb C$. 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
+even integer $2r > d$. Moment convergence gives $C_r < ∞$ with
 $\int |x|^{2r} d\mu_j(x) \le C_r$ for all $j$, hence
 
 $$ \int_{|x| > K} |x|^d\, d\mu_j(x) \le C_r K^{d-2r} \qquad (K > 0). $$
@@ -437,7 +437,7 @@ 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
+bounded continuous function, and letting $K \to ∞$ 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
@@ -448,63 +448,63 @@ 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
+$|hF \,△\, F| < \delta|F|$ for $h \in H$. Set $n = |F|$ and
+$J = F \cup ∪_{h \in H} hF$. Finally choose a real polynomial
 
-$$ p(s) = \sum_{r=1}^D c_r s^r, \qquad \sup_{0 \le s \le 1} |p(s) - \sqrt{s}| < \eta/\sqrt{n}. $$
+$$ p(s) = \sum_{r=1}^D c_r s^r, \qquad \sup_{0 \le s \le 1} |p(s) - √(s)| < \eta/√(n). $$
 
-Such a polynomial exists: approximate $\sqrt{s}$ on $[0,1]$ by a real
+Such a polynomial exists: approximate $√(s)$ 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
+Thus $t, \delta, \eta, F, n, J, p$ are fixed before $N \to ∞$ 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
+$\xi_g = < v_g, W >$. For $E = F$ or $E = hF$ set
 
-$$ P_E = \frac1n \sum_{g \in E} v_g v_g^{\mathsf T}, \qquad B_E = p(P_E), \qquad q_E = \langle W, B_E W \rangle - \mathrm{Tr}(B_E). $$
+$$ P_E = \frac1n \sum_{g \in E} v_g v_g^{\mathsf T}, \qquad B_E = p(P_E), \qquad q_E = < W, B_E W > - Tr(B_E). $$
 
 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
 
-$$ \|B_E - P_E^{1/2}\|_{\mathrm{HS}}^2 \le \mathrm{rank}(P_E)\, \eta^2/n \le \eta^2, \qquad \|P_E^{1/2}\|_{\mathrm{HS}}^2 = \mathrm{Tr}(P_E) = 1, $$
+$$ \|B_E - P_E^{1/2}\|_{HS}^2 \le rank(P_E)\, \eta^2/n \le \eta^2, \qquad \|P_E^{1/2}\|_{HS}^2 = Tr(P_E) = 1, $$
 
-hence $1 - \eta \le \|B_E\|_{\mathrm{HS}} \le 1 + \eta$.
+hence $1 - \eta \le \|B_E\|_{HS} \le 1 + \eta$.
 
 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
 
-$$ \|P_F^{1/2} - P_{hF}^{1/2}\|_{\mathrm{HS}}^2 \le \|P_F - P_{hF}\|_1 \le \frac{|F \,\triangle\, hF|}{n} < \delta. $$
+$$ \|P_F^{1/2} - P_{hF}^{1/2}\|_{HS}^2 \le \|P_F - P_{hF}\|_1 \le (|F \,△\, hF|)/(n) < \delta. $$
 
 With the rank control this yields
-$\|B_F - B_{hF}\|_{\mathrm{HS}} \le \sqrt\delta + 2\eta$. Both forms
+$\|B_F - B_{hF}\|_{HS} \le \sqrt\delta + 2\eta$. 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 $\ell \ge 1$,
 
-$$ \bigl|\mathbb E \exp(\mathrm{i}\ell t q_F)\bigr| \le (1 + 4\ell^2 t^2 \|B_F\|_{\mathrm{HS}}^2)^{-1/4} \le (1 + t^2)^{-1/4}. $$
+$$ |\mathbb E \exp(i\ell t q_F)| \le (1 + 4\ell^2 t^2 \|B_F\|_{HS}^2)^{-1/4} \le (1 + t^2)^{-1/4}. $$
 
 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
-$P_E^r = n^{-r} \sum_{g_1, \ldots, g_r \in E} R_{g_1,g_2} \cdots R_{g_{r-1},g_r} v_{g_1} v_{g_r}^{\mathsf T}$.
+$P_E^r = n^{-r} \sum_{g_1, ..., g_r \in E} R_{g_1,g_2} ... R_{g_{r-1},g_r} v_{g_1} v_{g_r}^{\mathsf T}$.
 Taking quadratic forms and traces shows
 
-$$ q_E = \sum_{r=1}^D \frac{c_r}{n^r} \sum_{g_1, \ldots, g_r \in E} R_{g_1,g_2} \cdots R_{g_{r-1},g_r} (\xi_{g_1} \xi_{g_r} - R_{g_r,g_1}), $$
+$$ q_E = \sum_{r=1}^D (c_r)/(n^r) \sum_{g_1, ..., g_r \in E} R_{g_1,g_2} ... R_{g_{r-1},g_r} (\xi_{g_1} \xi_{g_r} - R_{g_r,g_1}), $$
 
 with the empty product interpreted as $1$ when $r = 1$. Use the same
 prescription in $C$:
 
-$$ \widetilde Q_{N,E} = \sum_{r=1}^D \frac{c_r}{n^r} \sum_{g_1, \ldots, g_r \in E} k_{N;g_1,g_2} \cdots k_{N;g_{r-1},g_r} (z_{N,g_1} z_{N,g_r} - k_{N;g_r,g_1}), $$
+$$ \widetilde Q_{N,E} = \sum_{r=1}^D (c_r)/(n^r) \sum_{g_1, ..., g_r \in E} k_{N;g_1,g_2} ... k_{N;g_{r-1},g_r} (z_{N,g_1} z_{N,g_r} - k_{N;g_r,g_1}), $$
 
-$$ Q_{N,E} = \tfrac12\bigl(\widetilde Q_{N,E} + \widetilde Q_{N,E}^{\,*}\bigr). $$
+$$ Q_{N,E} = \tfrac12(\widetilde Q_{N,E} + \widetilde Q_{N,E}^{\,*}). $$
 
 Multiplication is in the displayed order. The empirical covariance array
 need satisfy no positivity condition, because this is a finite polynomial in
@@ -517,7 +517,7 @@ 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
+Step 4, transfer of moments and scalar laws. Take any $N_j \to ∞$ 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
@@ -526,9 +526,9 @@ 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
 
-$$ \|Q_{N_j,hF} - Q_{N_j,F}\|_{2,\varphi_j} \longrightarrow \|q_{hF} - q_F\|_{L^2}, $$
+$$ \|Q_{N_j,hF} - Q_{N_j,F}\|_{2,\varphi_j} → \|q_{hF} - q_F\|_{L^2}, $$
 
-$$ \varphi_j(\exp(\mathrm{i} s Q_{N_j,F})) \longrightarrow \mathbb E \exp(\mathrm{i} s q_F) $$
+$$ \varphi_j(\exp(i s Q_{N_j,F})) → \mathbb E \exp(i s q_F) $$
 
 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
@@ -537,22 +537,22 @@ $\|Q_{N,E}\| \le (N+1) \sum_r |c_r| M^{2r}$. Scalar moment convergence
 together with the spectral tail estimate is enough for the exponential
 convergence.
 
-Step 5, uniformity over all traces. Set $V_N = \exp(\mathrm{i} t Q_{N,F})$.
+Step 5, uniformity over all traces. Set $V_N = \exp(i t Q_{N,F})$.
 For self-adjoint $X, Y$ in a unital C*-algebra and a tracial $\varphi$,
 differentiation gives
 
-$$ \frac{d}{ds}\bigl(e^{\mathrm{i}(1-s)tX} e^{\mathrm{i}stY}\bigr) = e^{\mathrm{i}(1-s)tX}\, \mathrm{i}t(Y - X)\, e^{\mathrm{i}stY}, $$
+$$ (d)/(ds)(e^{i(1-s)tX} e^{istY}) = e^{i(1-s)tX}\, it(Y - X)\, e^{istY}, $$
 
 and the tracial 2-seminorm is unchanged by left and right unitary
 multiplication, so integration and the triangle inequality give the
-tracial Duhamel bound $\|e^{\mathrm{i}tX} - e^{\mathrm{i}tY}\|_{2,\varphi} \le t\|X - Y\|_{2,\varphi}$.
+tracial Duhamel bound $\|e^{itX} - e^{itY}\|_{2,\varphi} \le t\|X - Y\|_{2,\varphi}$.
 Together with exact equivariance, the norm limit, and the Gaussian
 translation bound, every sequence of traces and indices with convergent
 covariance satisfies
 
-$$ \limsup_{j \to \infty} \|\bar\alpha_h(V_{N_j}) - V_{N_j}\|_{2,\varphi_j} \le t\sqrt2(\sqrt\delta + 2\eta) < \eps/4, $$
+$$ limsup_{j \to ∞} \|\bar\alpha_h(V_{N_j}) - V_{N_j}\|_{2,\varphi_j} \le t\sqrt2(\sqrt\delta + 2\eta) < \eps/4, $$
 
-$$ \lim_{j \to \infty} |\varphi_j(V_{N_j}^\ell)| = \bigl|\mathbb E \exp(\mathrm{i}\ell t q_F)\bigr| \le (1 + t^2)^{-1/4} < \eps/4 $$
+$$ \lim_{j \to ∞} |\varphi_j(V_{N_j}^\ell)| = |\mathbb E \exp(i\ell t q_F)| \le (1 + t^2)^{-1/4} < \eps/4 $$
 
 for $h \in H$ and $1 \le \ell \le m$, by the choices of the parameters.
 
@@ -560,7 +560,7 @@ 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
+this false, some $N_j \to ∞$ 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
@@ -614,7 +614,7 @@ $ac = a^{1/2} c a^{1/2}$ 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
+approximately invariant unitary $V = \exp(i t Q)$ from the
 preceding proposition gives
 
 $$ |\tau(a V^\ell)| < \eps\, \tau(a) \le \eps \qquad (0 \le a \le 1,\ \tau \in \mathcal T_\omega(A),\ 1 \le \ell \le m,\ \tau(a) > 0), $$
@@ -632,7 +632,7 @@ 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
+sequence $a_1, a_2, ...$ 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
@@ -660,21 +660,21 @@ 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
+$\tau(a w^{-\ell}) = \tau(a^* w^\ell)bar$, 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 $f$.
+$|\tau(a f(w))| \le \|a\| \|f\|_∞$ 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
+half-open arcs $I_1, ..., I_k$ of Haar measure $1/k$ and write
 $\chi_i = 1_{I_i}$. Choose continuous functions
 $f_i^{(r)} : \mathbb T \to [0,1]$ with
 
-$$ e_r := \max_{1 \le i \le k} \|f_i^{(r)} - \chi_i\|_{L^2(\mu)} \longrightarrow 0, $$
+$$ e_r := \max_{1 \le i \le k} \|f_i^{(r)} - \chi_i\|_{L^2(\mu)} → 0, $$
 
 for instance by replacing each indicator near its two endpoints by linear
 ramps whose total lengths tend to zero. Fix $r$ and write $f_i = f_i^{(r)}$,
@@ -682,7 +682,7 @@ $e = e_r$. Pointwise bounds and the triangle inequality give
 
 $$ \|f_i^2 - f_i\|_{L^2(\mu)} \le e, \qquad \|f_i f_j\|_{L^2(\mu)} \le 2e \ \ (i \ne j), $$
 
-$$ \Bigl\|\sum_i f_i - 1\Bigr\|_{L^2(\mu)} \le k e, \qquad \Bigl|\int_{\mathbb T} f_i\, d\mu - \frac1k\Bigr| \le e. $$
+$$ \Bigl\|\sum_i f_i - 1\Bigr\|_{L^2(\mu)} \le k e, \qquad |\int_{\mathbb T} f_i\, d\mu - \frac1k| \le e. $$
 
 The first uses $|f_i^2 - f_i| \le |f_i - \chi_i|$ almost everywhere; the
 second expands
@@ -698,10 +698,10 @@ 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
 
-$$ \Bigl|\tau(a y_i^{(r)}) - \frac{\tau(a)}{k}\Bigr| = \tau(a) \Bigl|\int_{\mathbb T} f_i^{(r)}\, d\mu - \frac1k\Bigr| \le e_r. $$
+$$ |\tau(a y_i^{(r)}) - (\tau(a))/(k)| = \tau(a) |\int_{\mathbb T} f_i^{(r)}\, d\mu - \frac1k| \le e_r. $$
 
 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
+$a_1, ..., a_r$ 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,
@@ -712,7 +712,7 @@ $$ \Bigl\|\sum_i y_{i,r} - 1\Bigr\|_{2,u} < k e_r + 1/r, $$
 
 $$ \|[y_{i,r}, a_s]\|_{2,u} < 1/r \quad (s \le r), \qquad \|\alpha_h(y_{i,r}) - y_{i,r}\|_{2,u} < 1/r \quad (h \in H_r), $$
 
-$$ \sup_{\sigma \in \mathcal T(A)} \Bigl|\sigma(a_s y_{i,r}) - \frac{\sigma(a_s)}{k}\Bigr| < e_r + 1/r \quad (s \le r). $$
+$$ \sup_{\sigma \in \mathcal T(A)} |\sigma(a_s y_{i,r}) - (\sigma(a_s))/(k)| < e_r + 1/r \quad (s \le r). $$
 
 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
@@ -735,19 +735,19 @@ 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
+$\beta = \alpha  ⊗  id_Z$ 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
+$\gamma_g = Ad(u_g) \beta_g$ 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
+orientation, suppose $\beta_g = Ad(v_g) \gamma_g$ 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
 
-$$ u_g \beta_g(u_h) = v_g^* v_g \gamma_g(v_h^*) v_g^* = \gamma_g(v_h^*) v_g^* = v_{gh}^* = u_{gh}, \qquad \mathrm{Ad}(u_g) \beta_g = \gamma_g. $$
+$$ u_g \beta_g(u_h) = v_g^* v_g \gamma_g(v_h^*) v_g^* = \gamma_g(v_h^*) v_g^* = v_{gh}^* = u_{gh}, \qquad Ad(u_g) \beta_g = \gamma_g. $$
 
 Either orientation therefore gives the required $\beta$-cocycle, and the
 identity relation $u_e = u_e^2$ together with unitarity forces $u_e = 1$.