Proof Proof (Proposition 2.3)

Suppose the result doesn’t hold. Then there exist $\epsilon _0> 0$ , a positive contraction $y_0 \in \mathcal {M}$ , and a net $(x_\lambda )_{\lambda \in \Lambda }$ of positive contractions in $\mathcal {M}$ such that

(2.9) $$ \begin{align} \lim_{\lambda} \|[x_\lambda,y]\|_{2, \mathrm{u}} = 0 \end{align} $$

for all $y \in \mathcal {M}$ , but

(2.10) $$ \begin{align} \|E(x_\lambda y_0)-E(x_\lambda)E(y_0)\| \geq \epsilon_0 \end{align} $$

for all $\lambda \in \Lambda $ . Hence, there exists a net $(\tau _\lambda )_{\lambda \in \Lambda }$ of traces in K such that

(2.11) $$ \begin{align} |\tau_\lambda(x_\lambda y_0)-\tau_\lambda(x_\lambda)\tau_\lambda(y_0)| \geq \epsilon_0 \end{align} $$

for each $\lambda \in \Lambda $ .

Since K is compact, after passing to a subnet, we may assume $(\tau _\lambda )_{\lambda \in \Lambda }$ converges in the weak $^*$ topology to some $\tau \in K$ . Since the unit ball of $\mathcal {M}^*$ is weak $^*$ -compact, by passing to a subnet again, we may further assume that $(y \mapsto \tau _{\lambda }(x_{\lambda } y))_{\lambda \in \Lambda }$ converges in the weak $^*$ topology to some $\sigma \in \mathcal {M}^*$ .

It follows from (2.9) that $\sigma $ is a positive tracial functional on $\mathcal {M}$ . Moreover, for positive $y \in \mathcal {M}$ , we have

(2.12) $$ \begin{align} \sigma(y) &= \lim_{\lambda} \tau_{\lambda}(y^{1/2} x_{\lambda} y^{1/2})\nonumber\\ &\leq \limsup_{\lambda} \tau_{\lambda}(y^{1/2} \|x_{\lambda}\| y^{1/2})\\ &\leq \tau(y)\nonumber \end{align} $$

since $x_\lambda \in \mathcal {M}$ is a positive contraction.

As $\mathcal {M}$ has factorial fibres, $\tau $ is an extremal trace on $\mathcal {M}$ . Since $\sigma \leq \tau $ , it follows that $\sigma = \sigma (1_{\mathcal {M}}) \tau $ . We conclude

(2.13) $$ \begin{align} \lim_{\lambda} \tau_{\lambda}(x_{\lambda} y_0) &= \sigma(y_0)\nonumber\\ &= \sigma(1_{\mathcal{M}}) \tau(y_0)\nonumber\\ &= \lim_{\lambda} \tau_{\lambda}(x_{\lambda})\lim_{\lambda} \tau_{\lambda}(y_0)\\ &= \lim_{\lambda} \tau_{\lambda}(x_{\lambda})\tau_{\lambda}(y_0).\nonumber \end{align} $$

However, this contradicts (2.11).

Proof By continuity and linearity, it suffices to replace S by a countable set of contractions. Let us then enumerate $S=\{s^{(1)},s^{(2)},\dots \}$ and represent $s^{(i)}$ by the sequence of contractions $(s_n^{(i)})_{n=1}^\infty $ in $\mathcal {M}$ . Set $\mathcal F_n = \{s_n^{(i)}: i = 1,\ldots ,n\}$ and $\epsilon _n = \tfrac {1}{n}$ . Let $\mathcal G_n \subseteq \mathcal {M}$ and $\delta _n> 0$ be the finite set and tolerance corresponding to $(\mathcal F_n, \epsilon _n)$ according to Proposition 2.3. We may assume that $\mathcal F_n \subseteq \mathcal G_n$ and $\delta _n < \epsilon _n$ . Take $\mathcal {M}_0$ to be the subalgebra of $\mathcal {M}$ generated by $\bigcup _{n \in \mathbb N} \mathcal G_n$ and note that $\mathcal {M}_0$ is $\|\cdot \|_{2, \mathrm {u}}$ -separable.

Let $T \subseteq \mathcal {M}^\infty \cap \mathcal {M}_0'$ be $\|\cdot \|_{2, \mathrm {u}}$ -separable. By continuity and linearity, it suffices to replace T by a countable set of contractions. Say $T=\{t^{(1)},t^{(2)},\dots \}$ and represent $t^{(j)}$ by the sequence of contractions $(t_m^{(j)})_{m=1}^\infty $ . For each $n \in \mathbb N$ , any sufficiently large $m \in \mathbb N$ will satisfy

(2.15) $$ \begin{align} \max_{y \in \mathcal G_n}\|[t_{m}^{(j)}, y]\|_{2, \mathrm{u}} < \delta_n \end{align} $$

for all $j\in \{1, \ldots , n\}$ because T commutes with $\mathcal {M}_0$ . Hence, we may inductively define a strictly increasing function $\rho \colon \mathbb N \rightarrow \mathbb N$ such that

(2.16) $$ \begin{align} \max_{y \in \mathcal G_n}\|[t_{\rho(n)}^{(j)}, y]\|_{2, \mathrm{u}} < \delta_n \end{align} $$

for all $j\in \{1, \ldots , n\}$ and $n \in \mathbb N$ . By the choice of $\mathcal G_n$ and $\delta _n$ , this implies

(2.17) $$ \begin{align} \sup_{\tau \in K}\big|\tau(t_{\rho(n)}^{(j)}s_{n}^{(i)})-\tau(t_{\rho(n)}^{(j)})\tau(s_{n}^{(i)})\big| &< \frac{1}{n} \end{align} $$

for all $i, j \in \{1, \ldots , n\}$ and $n \in \mathbb N$ . At the level of the sequence algebra $\mathcal {M}^\infty $ , this implies (2.14). Since we have chosen that $\mathcal F_n \subseteq \mathcal G_n$ and $\delta _n < \epsilon _n =\tfrac {1}{n} $ , it follows from (2.16) that $\psi _\rho (T) \subseteq \mathcal {M}^\infty \cap S'$ .

Proof By the Stone–Weierstrass theorem, it suffices to consider the functions $f_n(t) = t^n$ for $n \geq 1$ . Let $\tilde \phi \colon C_0(0,1] \otimes A \rightarrow B$ be the induced $^*$ -homomorphism. Then, since $p^n = p$ , we have

(2.23) $$ \begin{align} f_n(\phi(p)) = \tilde\phi(\mathrm{id}_{(0,1]} \otimes p)^n = \tilde\phi(\mathrm{id}_{(0,1]}^n \otimes p) = f_n(\phi)(p). \end{align} $$

Proof Fix a faithful representation $B \subseteq \mathcal B(\mathcal H)$ . If $a, b \in A$ are positive with $ab = 0$ , then $\psi (a) \psi (b) = 0$ . Combining this with the inequalities $0 \leq \phi (a) \leq \psi (a)$ and $0 \leq \phi (b) \leq \psi (b)$ yields

(2.24) $$ \begin{align} \overline{\phi(b)\mathcal H} \subseteq \overline{\psi(b)\mathcal H} \subseteq \ker \psi(a) \subseteq \ker \phi(a), \end{align} $$

and hence $\phi (a) \phi (b) = 0$ .

Proof Unpacking the sequence algebra formalism, it suffices to show that for any finite set $\mathcal F \subseteq \mathcal {M}$ , $N\in \mathbb N$ , and $\epsilon>0$ , there are c.p.c. order zero maps $\Phi ^{(0)},\dots ,\Phi ^{(m)}\colon \mathbb C^k \to \mathcal {M}$ such that

(3.4) $$ \begin{align} \Big\|\sum_{c'=0}^m \Phi^{(c')}(1_k)-1_{\mathcal{M}}\Big\|_{2, \mathrm{u}} &\leq \epsilon, \end{align} $$

(3.5) $$ \begin{align} \big\|[\Phi^{(c)}(e_j),b]\big\|_{2, \mathrm{u}} &\leq \epsilon, \end{align} $$

and

(3.6) $$ \begin{align} \big|\tau(\Phi^{(c)}(e_j)^n)-\frac1k\tau(\Phi^{(c)}(1_k)^n)\big| &\leq \epsilon \end{align} $$

for all $\tau \in K$ , $c\in \{0,\dots ,m\}$ , $j\in \{1,\dots ,k\}$ , $n\in \{1,\dots ,N\}$ , and $b \in \mathcal F$ .

For every $\tau \in K$ , since $\pi _\tau (\mathcal {M})$ has property $\Gamma $ , there exists a unital $^*$ -homomorphism $\mathbb C^k \to \pi _\tau (\mathcal {M})$ such that the image of each $e_j$ has trace $\frac 1k$ and $\|\cdot \|_{2,\tau }$ -approximately commutes with $\pi _\tau (\mathcal F)$ .

By [Reference Loring17, Theorem 4.6] (cf. [Reference Akemann and Pedersen1, Proposition 2.6]), the cone over $\mathbb C^k$ is a projective C $^*$ -algebra. Combined with the structure theorem for order zero maps [Reference Winter and Zacharias28, Corollary 3.1], this implies that the constructed $^*$ -homomorphism $\mathbb C^k \to \pi _\tau (\mathcal {M})$ lifts to a c.p.c. order zero map $\phi _\tau \colon \mathbb {C}^k \rightarrow \mathcal {M}$ . Let $\epsilon> 0$ . By continuity, there is a neighborhood $V_\tau $ of $\tau $ in K such that for all $\sigma \in V_\tau $ ,

(3.7) $$ \begin{align} \|\phi_\tau(1_k)-1_{\mathcal{M}}\|_{2,\sigma} &\leq \frac\epsilon{m+1}, \end{align} $$

(3.8) $$ \begin{align} \|[\phi_\tau(e_j),b]\|_{2,\sigma} &\leq \epsilon, \end{align} $$

and

(3.9) $$ \begin{align} \sigma(\phi_\tau(e_j)^n)&\approx_{\epsilon/2} 1/k \end{align} $$

for all $b \in \mathcal F$ , $j\in \{1, \ldots , k\}$ , and $n\in \{1, \ldots , N\}$ .Footnote ⁸

By compactness and since $\dim (K) \leq m$ , we may find a finite $(m+1)$ -coloured refinement of $\{V_\tau :\tau \in K\}$ —that is, an open cover of K consisting of open sets $U_i^{(c)}$ for $i\in \{1,\dots ,l_c\}$ and $c\in \{0,\dots ,m\}$ such that each $U_i^{(c)}$ is contained in $V_{\tau _i^{(c)}}$ for some $\tau _i^{(c)}\in K$ , and for each c, the sets $U_1^{(c)},\dots ,U_{l_c}^{(c)}$ are disjoint. Let $(h_i^{(c)})_{c=0,\dots ,m;\,i=1,\dots ,l_c}$ be a partition of unity in $C(K) \subseteq \mathcal {M}$ subordinate to this open cover. We now define

(3.10) $$ \begin{align} \Phi^{(c)}= \sum_{i=1}^{l_c} h_i^{(c)}\phi_{\tau_i^{(c)}}\colon \mathbb C^k \to \mathcal{M}. \end{align} $$

Since each $\phi _{\tau }$ is c.p.c. order zero and $h_1^{(c)},\dots ,h_{l_c}^{(c)}$ are mutually orthogonal and central, it follows that $\Phi ^{(c)}$ is itself a c.p.c. order zero map.

To show (3.4), we note that for each $c \in \{0,\ldots ,m\}$ we have

(3.11) $$ \begin{align} \Big\|\Phi^{(c)}(1_k)-\sum_{i=1}^{l_c} h_i^{(c)}\Big\|_{2, \mathrm{u}} &\leq \Big\|\sum_{i=1}^{l_c}(\phi_{\tau_i^{(c)}}(1_k)-1_{\mathcal{M}}) h_i^{(c)}\Big\|_{2, \mathrm{u}} \leq \frac{\epsilon}{m+1} \end{align} $$

by (3.7) since every $\tau \in K$ is in the support of at most one of the functions $h_1^{(c)},\ldots ,h_{l_c}^{(c)}$ . Summing over all $c\in \{0,\dots ,m\}$ and using the triangle inequality, we obtain (3.4) since $\sum _{c,i} h_i^{(c)}=1_{\mathcal {M}}$ .

To show (3.5), fix $c\in \{0,\dots ,m\}$ , $j\in \{1,\dots ,k\}$ , and $b \in \mathcal F$ . For $\tau \in K$ , there is at most one $i\in \{1,\dots ,l_c\}$ such that $\tau \in U_i^{(c)}$ . If no such i exists, then $\big \|[\Phi ^{(c)}(e_j),b]\big \|_{2,\tau } =0$ , and otherwise, for this i, we have

(3.12) $$ \begin{align} \big\|[\Phi^{(c)}(e_j),b]\big\|_{2,\tau} &{\stackrel{(3.10)}{=}} \big\|[h_i^{(c)}\phi_{\tau_i^{(c)}}(e_j),b]\big\|_{2,\tau} {\stackrel{(3.8)}{\leq}} \epsilon. \end{align} $$

Similarly for (3.6), fix $\tau \in K$ , $c\in \{0,\ldots ,m\}$ , and $n\in \{1,\ldots ,N\}$ . If there is no i for which $\tau \in U_i^{(c)}$ , then

(3.13) $$ \begin{align} \tau(\Phi^{(c)}(e_j)^n)=\tau(\Phi^{(c)}(1_k)^n)=0 \end{align} $$

for all $j \in \{1, \ldots , k\}$ . Otherwise, there is exactly one i for which $\tau \in U_i^{(c)}$ , and then we have, for $j\in \{1,\dots ,k\}$ ,

(3.14) $$ \begin{align} \tau(\Phi^{(c)}(e_j)^n) &{\stackrel{(3.10)}=} \tau((h_i^{(c)})^n\phi_{\tau_i^{(c)}}(e_j)^n) \nonumber \\ &{=}\ \tau((h_i^{(c)})^n)\tau(\phi_{\tau_i^{(c)}}(e_j)^n) \\ &{\stackrel{(3.9)}\approx}_{\epsilon/2} \frac1k\tau((h_i^{(c)})^n),\nonumber \end{align} $$

where the unlabeled inequality uses that the $h_{i}^{(c)}$ are central and $\tau $ is an extremal trace on ${\mathcal {M}}$ (as the fibre over $\tau $ is a factor).Footnote ⁹ Since $\Phi ^{(c)}$ is c.p.c. order zero, we have $\Phi ^{(c)}(1_k)^n = \sum _{j'=1}^k \Phi ^{(c)}(e_{j'})^n$ . Thus,

(3.15) $$ \begin{align} \frac1k\tau(\Phi^{(c)}(1_k)^n)&=\frac1k\sum_{j'=1}^k \tau(\Phi^{(c)}(e_{j'})^n) \nonumber\\ &\approx_{\epsilon/2} \frac1k\tau((h_i^{(c)})^n) \\ &\approx_{\epsilon/2} \tau(\Phi^{(c)}(e_j)^n)\nonumber \end{align} $$

for all $j\in \{1,\dots ,k\}$ , using (3.14) for both approximations.

Proof Fix $m \in \mathbb N$ . We begin with the case $r = 1$ and define

(3.18) $$ \begin{align} \gamma_{m,1} = \frac1{4(m+1)}. \end{align} $$

Let $S \subseteq \mathcal {M}^\infty $ be $\|\cdot \|_{2, \mathrm {u}}$ -separable. Let $\mathcal {M}_0 \subseteq \mathcal {M}$ be a $\|\cdot \|_{2, \mathrm {u}}$ -separable subalgebra of $\mathcal {M}$ such that Lemma 2.4 holds.

It suffices to prove that there exist orthogonal positive contractions $d_0',d_1'\in \mathcal {M}^\infty \cap \mathcal {M}_0'$ such that

(3.19) $$ \begin{align} \tau(d_i')\geq \gamma_{m,1} \end{align} $$

for all $\tau \in K^\infty $ and $i\in \{0,1\}$ . Indeed, taking T to be the C $^*$ -algebra generated by $d_0'$ and $d_1'$ , Lemma 2.4 provides us with a reindexing $^*$ -homomorphism $\psi _\rho :\mathcal {M}^\infty \rightarrow \mathcal {M}^\infty $ such that $\psi _\rho (T) \subseteq \mathcal {M}^\infty \cap S'$ and the tracial factorization in (2.14) holds. Define $d_i = \psi _\rho (d_i')$ for $i\in \{0,1\}$ and note that (3.16) follows from (2.14). Further, for all $\tau \in K^\infty $ , we have $\tau \circ \psi _\rho \in K^\infty $ , so (3.17) follows from (3.19).

Let $\Phi ^{(0)},\dots ,\Phi ^{(m)}\colon \mathbb C^{2(m+1)} \to \mathcal {M}^\infty \cap \mathcal {M}_0'$ be given by Lemma 3.2 (with $S=\mathcal {M}_0$ and $k=2(m+1)$ ). Define

(3.20) $$ \begin{align} a= \sum_{c=0}^m \Phi^{(c)}(e_1), \end{align} $$

which is a positive contraction in $\mathcal {M}^\infty \cap \mathcal {M}_0'$ since $\sum _c \Phi ^{(c)}$ is a u.c.p. map. Making use of the continuous functions defined in (3.1), set

(3.21) $$ \begin{align} d_0'= g_{\gamma_{m,1},2\gamma_{m,1}}(a) \quad \text{and} \quad d_1'= 1_{\mathcal{M}^\infty}-g_{0,\gamma_{m,1}}(a) \end{align} $$

and note that these are orthogonal positive contractions in $\mathcal {M}^\infty \cap \mathcal {M}_0'$ by construction.

To show (3.19) for $i=0$ , observe that $g_{\gamma _{m,1},2\gamma _{m,1}}(t) \geq t - \gamma _{m,1}$ for all $t \in [0,1]$ , and so for $\tau \in K^\infty $ ,

(3.22) $$ \begin{align} \tau(d_0') &{\geq} \tau(a) - \gamma_{m,1} \nonumber \\ &{\stackrel{(3.20)}{=}} \sum_{c=0}^m \tau(\Phi^{(c)}(e_1)) - \gamma_{m,1} \nonumber\\ &{\stackrel{(3.3)}{=}} \tfrac{1}{2(m+1)} \sum_{c=0}^m \tau(\Phi^{(c)}(1_{2(m+1)})) - \gamma_{m,1}\\ &{\stackrel{(3.2)}{=}} \tfrac{1}{2(m+1)} - \gamma_{m,1} \nonumber\\ &{\stackrel{(3.18)}{=}} \gamma_{m,1}.\nonumber \end{align} $$

To show (3.19) for $i=1$ , we compute that for $\tau \in K^\infty $ ,

(3.23) $$ \begin{align} \tau(1_{\mathcal{M}^\infty} - d_1')\ &{=}\ \tau(g_{0,\gamma_{m,1}}(a)) \nonumber\\ &{\leq} \lim_{l \rightarrow \infty} \tau(a^{1/l})\nonumber\\ &{\stackrel{(3.20)}{=}} \lim_{l \rightarrow \infty} \tau\Big(\Big(\sum_{c=0}^m\Phi^{(c)}(e_1)\Big)^{1/l}\Big)\nonumber\\ &{\leq} \lim_{l \rightarrow \infty} \sum_{c=0}^m \tau(\Phi^{(c)}(e_1)^{1/l}) \\ &{\stackrel{(3.3)}{=}} \lim_{l \rightarrow \infty} \sum_{c=0}^m \tfrac{1}{2(m+1)}\tau(\Phi^{(c)}(1_{2(m+1)})^{1/l}) {\leq} \tfrac{m+1}{2(m+1)} {=} \tfrac{1}{2},\nonumber \end{align} $$

where in the fourth line we use the fact that $\sum _{c=0}^m\Phi ^{(c)}(e_1)$ is Cuntz subequivalent to $\bigoplus _{c=0}^m\Phi ^{(c)}(e_1)$ .Footnote ¹⁰ Thus, $\tau (d_1') \geq 1/2 \geq \gamma _{m,1}$ . This completes the proof of the case $r=1$ .

In the general case, by enlarging r, we may assume $r = 2^l - 1$ for some $l \geq 1$ . We will prove the result by induction on l, starting with the case $l = 1$ handled above. Fix $m\in \mathbb N$ and a $\|\cdot \|_{2, \mathrm {u}}$ -separable subset $S \subseteq M^\infty $ . Assume the result holds for $r = 2^l - 1$ and let $d_0, \ldots , d_r \in {\mathcal {M}}^\infty \cap S'$ be positive orthogonal contractions satisfying (3.16) and (3.17). Let $T \subseteq {\mathcal {M}}^\infty $ denote the C $^*$ -algebra generated by $S \cup \{ d_0, \ldots , d_r \}$ . Note that T is $\|\cdot \|_{2, \mathrm {u}}$ -separable.

By the $r=1$ case proved above (but now with T replacing S), there are positive orthogonal contractions $\tilde {d}_0, \tilde {d}_1 \in {\mathcal {M}}^\infty \cap T'$ satisfying

(3.24) $$ \begin{align} \tau(f(\tilde{d}_i)t)&=\tau(f(\tilde{d}_i))\tau(t) \end{align} $$

and

(3.25) $$ \begin{align} \tau(\tilde{d}_i) &\geq \gamma_{m,1} \end{align} $$

for all $i\in \{0, 1\}$ , $t \in T$ , $\tau \in K^\infty $ , and $f \in C_0((0, 1])$ . We will show the $2^{l+1}$ elements $\tilde d_i d_j \in {\mathcal {M}}^\infty \cap S'$ for $i\in \{0, 1\}$ and $j\in \{0, \ldots , r\}$ satisfy the required properties with $\gamma _{m, 2r+1} = \gamma _{m, 1} \gamma _{m, r}$ .

First note that the $\tilde d_i d_j$ are clearly mutually orthogonal positive contractions as each $\tilde d_i$ commutes with each $d_j$ by construction. For all $i\in \{0, 1\}$ , $j\in \{0, \ldots , r\}$ and $\tau \in K^\infty $ , we have

(3.26) $$ \begin{align} \tau(\tilde d_i d_j) \stackrel{(3.24)}{=} \tau(\tilde d_i) \tau(d_j) \geq \gamma_{m, 1} \gamma_{m, r} = \gamma_{m, 2r+1}. \end{align} $$

Let $i \in \{0, 1\}$ , $j \in \{0, \ldots , r\}$ , $s \in S$ , $\tau \in K^\infty $ , and $n \in \mathbb N$ . Then

(3.27) $$ \begin{align} \tau((\tilde d_i d_j)^n s)\ &{=}\ \tau(\tilde d_i^n d_j^n s) \nonumber\\ &{\stackrel{(3.24)}{=}} \tau(\tilde d_i^n) \tau(d_j^n s) \nonumber\\ &{\stackrel{(3.16)}{=}} \tau(\tilde d_i^n) \tau(d_j^n) \tau(s) \\ &{\stackrel{(3.24)}{=}} \tau(\tilde d_i^n d_j^n) \tau(s) \nonumber\\ &{=}\ \tau((\tilde d_i d_j)^n) \tau(s).\nonumber \end{align} $$

By the Stone–Weierstrass theorem, this implies

(3.28) $$ \begin{align} \tau(f(\tilde d_i d_j)s) = \tau(f(\tilde d_i d_j)) \tau(s) \end{align} $$

for all $f \in C_0(0,1]$ .

Proof Set $\alpha =\gamma _{m,m}$ from Lemma 3.3. Fix $k \in \mathbb N$ and let $(\mathcal {M},K,E)$ be a W $^*$ -bundle with $\dim (K)\leq m$ whose fibres are II $_1$ factors with property $\Gamma $ . Further, fix a $\|\cdot \|_{2, \mathrm {u}}$ -separable subset $S \subseteq \mathcal {M}^\infty $ .

Let $\mathcal {M}_0 \subseteq \mathcal {M}$ be the $\|\cdot \|_{2, \mathrm {u}}$ -separable subset of $\mathcal {M}$ such that Lemma 2.4 holds. It suffices to prove that there exists a c.p.c. order zero map $\Phi \colon \mathbb C^k \to \mathcal {M}^\infty \cap \mathcal {M}_0'$ such that (3.29) and (3.30) hold for all $\tau \in K^\infty ,\ j\in \{1,\dots ,k\}$ and $f\in C_0(0,1]$ . Indeed, taking T to be the C $^*$ -algebra generated by $\{\Phi (e_j): j =1,\ldots ,k\}$ , Lemma 2.4 provides us with a reindexing $^*$ -homomorphism $\psi _\rho :\mathcal {M}^\infty \rightarrow \mathcal {M}^\infty $ such that, after replacing $\Phi $ with $\psi _\rho \circ \Phi $ , all three conditions (3.29), (3.30), and (3.31) are satisfied.

Let $\Phi ^{(0)},\ldots ,\Phi ^{(m)}\colon \mathbb {C}^{k}\rightarrow \mathcal {M} \cap \mathcal {M}_0'$ be maps as in Lemma 3.2. Let $d_0,\ldots ,d_m \in \mathcal {M}^\infty \cap \big (\Phi ^{(0)}(\mathbb C^k) \cup \cdots \cup \Phi ^{(m)}(\mathbb C^k) \cup \mathcal {M}_0\big )'$ be orthogonal positive contractions, constructed using Lemma 3.3, such that

(3.32) $$ \begin{align} \tau(f(d_i)b)&= \tau(f(d_i))\tau(b) \end{align} $$

and

(3.33) $$ \begin{align} \tau(d_i) &\geq \gamma_{m,m} = \alpha \end{align} $$

for all $c\in \{0,\dots ,m\}$ , $b \in \mathrm {C}^*(\Phi ^{(c)}(\mathbb C^k))$ , $\tau \in K^\infty $ , and $f\in C_0(0,1]$ . Define

(3.34) $$ \begin{align} \Phi = \sum_{c=0}^m d_c\Phi^{(c)}\colon\mathbb{C}^k \rightarrow \mathcal{M}^\infty \cap \mathcal{M}_0'. \end{align} $$

Since the $d_0,\ldots ,d_m$ are orthogonal positive contractions commuting with the images of the $\Phi ^{(c)}$ , $\Phi $ is c.p.c. order zero.

Let $\tau \in K^\infty $ , $j\in \{1,\dots ,k\}$ , and $n\in \mathbb N$ . Then

(3.35) $$ \begin{align} \tau(\Phi(e_j)^n) & {\stackrel{(3.34)}=} \sum_{c=0}^m \tau(d_c^n\Phi^{(c)}(e_j)^n)\nonumber\\ &{\stackrel{(3.32)}{=}} \sum_{c=0}^m \tau(d_c^n)\tau(\Phi^{(c)}(e_j)^n)\nonumber\\ &{\stackrel{(3.3)}=} \sum_{c=0}^m \frac{1}{k}\tau(d_c^n)\tau(\Phi^{(c)}(1_k)^n)\\ &{\stackrel{(3.32)}{=}} \frac1k \sum_{c=0}^m \tau(d_c^n\Phi^{(c)}(1_k)^n){\stackrel{(3.34)}=} \frac{1}{k}\tau(\Phi(1_k)^n),\nonumber \end{align} $$

using in the first and last lines that the $d_c$ are mutually orthogonal and each $d_c$ commutes with the range of each $\Phi ^{(c')}$ . By linearity, continuity, and the Stone–Weierstrass theorem, (3.30) follows from (3.35).

Now let $\tau \in K^\infty $ . Then

(3.36) $$ \begin{align} \begin{aligned} \tau(\Phi(1_k)) &{\stackrel{(3.34)}=} \sum_{c=0}^m \tau(d_c\Phi^{(c)}(1_k))\\ &{\stackrel{(3.32)}{=}} \sum_{c=0}^m \tau(d_c)\tau(\Phi^{(c)}(1_k))\\ &{\stackrel{(3.33)}{\geq}} \sum_{c=0}^m \alpha\tau(\Phi^{(c)}(1_k)). \end{aligned} \end{align} $$

Hence, $\tau (\Phi (1_k)) \geq \alpha $ by (3.2). This verifies (3.29).

Proof Proof (Theorem 1.2)

Suppose $({\mathcal {M}}, K, E)$ is a W $^*$ -bundle such that $m = \dim (K) < \infty $ and every fibre of $\mathcal M$ is a II $_1$ factor with property $\Gamma $ . Let $\Omega $ be the set of all $\alpha \in [0,1]$ for which the conclusion of Lemma 3.4 holds, and set $\alpha _0 = \sup \Omega $ . Lemma 3.4 implies $\alpha _0> 0$ . Moreover, a standard reindexing argument shows that $\Omega $ is a closed set, so $\alpha _0 \in \Omega $ .

It suffices to show that $\alpha _0 = 1$ . Indeed, in this case, for every $k \in \mathbb N$ and $\|\cdot \|_{2, \mathrm u}$ -separable subset $S \subseteq \mathcal {M}^\infty $ , there exists a c.p.c. order zero map $\Phi \colon \mathbb C^k \rightarrow \mathcal {M}^\infty \cap S'$ such that

(3.37) $$ \begin{align} \tau(\Phi(1_k)) &\geq 1, \end{align} $$

(3.38) $$ \begin{align} \tau(\Phi(e_j)) &= \frac{1}{k}\tau(\Phi(1_k)), \end{align} $$

and

(3.39) $$ \begin{align} \tau(\Phi(e_j)s) = \tau(\Phi(e_j))\tau(s) \end{align} $$

for all $\tau \in K^\infty ,\ j\in \{1,\dots ,k\}$ , and $s\in S$ . It follows from (3.37) that $\tau (1_{\mathcal {M}^\infty }-\Phi (1_k)) \leq 0$ for all $\tau \in K^\infty $ . Since $1_{\mathcal {M}^\infty } - \Phi (1_k)$ is positive, this implies $\Phi (1_k)=1_{\mathcal {M}^\infty }$ . By [Reference Winter and Zacharias28, Theorem 3.3], a u.c.p. order zero map is a $^*$ -homomorphism. Therefore, $\Phi (e_1),\ldots ,\Phi (e_k)$ are orthogonal projections summing to $1_{\mathcal {M}^\infty }$ , and by (3.38) and (3.39), these projections witness that $\mathcal {M}$ has property $\Gamma $ .

Assume for the sake of contradiction that $\alpha _0<1$ . Let $\gamma = \gamma _{m,1}> 0$ be as in Lemma 3.3. Since $0 < \alpha _0 < 1$ and $\gamma> 0$ , we may choose $\epsilon>0$ so that

(3.40) $$ \begin{align} \alpha = \alpha_0+\gamma(\alpha_0-\alpha_0^2)-\epsilon(1-\gamma\alpha_0)> \alpha_0. \end{align} $$

We will show that the conclusion of Lemma 3.4 holds with this $\alpha $ (i.e., $\alpha \in \Omega $ ), which will be a contradiction.

Suppose $k \in \mathbb N$ and let $S \subseteq \mathcal {M}^\infty $ be a $\|\cdot \|_{2, \mathrm {u}}$ -separable subset. Let $\mathcal {M}_0 \subseteq \mathcal {M}$ be a $\|\cdot \|_{2, \mathrm {u}}$ -separable subalgebra obtained by applying Lemma 2.4 to S. By hypothesis, there exists a c.p.c. order zero map $\Phi _0\colon \mathbb {C}^k \rightarrow \mathcal {M}^\infty \cap \mathcal {M}_0'$ satisfying

(3.41) $$ \begin{align} \tau(\Phi_0(1_k)) &\geq \alpha_0, \end{align} $$

(3.42) $$ \begin{align} \tau(f(\Phi_0(e_j))) &= \frac{1}{k}\tau(f(\Phi_0(1_k))), \end{align} $$

and

(3.43) $$ \begin{align} \tau(f(\Phi_0(e_j))b) = \tau(f(\Phi_0(e_j)))\tau(b) \end{align} $$

for all $\tau \in K^\infty ,\ j\in \{1,\dots ,k\},\ f\in C_0(0,1]$ , and $b\in \mathcal {M}_0$ . Using Lemma 3.3, let $d_0,d_1 \in \mathcal {M}^\infty \cap (\mathcal {M}_0 \cup \Phi _0(\mathbb C^k))'$ be orthogonal positive contractions such that

(3.44) $$ \begin{align} \tau(f(d_i)b)= \tau(f(d_i))\tau(b) \end{align} $$

and

(3.45) $$ \begin{align} \tau(d_i) \geq \gamma \end{align} $$

for all $i\in \{0,1\},\ \tau \in K^\infty $ , $f \in C_0(0,1]$ , and $b \in \mathrm {C}^*(\Phi _0(\mathbb C^k))$ .

Let $g_{0,\epsilon },g_{\epsilon ,2\epsilon } \in C_0(0,1]$ be the continuous functions defined in (3.1) and set $\Delta _\epsilon = g_{0,\epsilon } - g_{\epsilon ,2\epsilon }$ . Using the order zero functional calculus, define $\Phi _0' \colon \mathbb C^k \to \mathcal {M}^\infty \cap \mathcal {M}_0'$ by

(3.46) $$ \begin{align} \begin{aligned} \Phi_0' &= d_0\Delta_\epsilon(\Phi_0)+g_{\epsilon,2\epsilon}(\Phi_0) \\ &= d_0 g_{0, \epsilon}(\Phi_0) + (1 - d_0) g_{\epsilon, 2\epsilon}(\Phi_0). \end{aligned} \end{align} $$

Since $d_0$ commutes with $\mathrm {C}^*(\Phi _0(\mathbb C^k))$ and $g_{\epsilon , 2\epsilon } \leq g_{0, \epsilon }$ , we have that $\Phi _0' \leq g_{0, \epsilon }(\Phi _0)$ . Since $g_{0, \epsilon }(\Phi _0)$ is c.p.c. order zero, so is $\Phi _0'$ by Lemma 2.8. Fix $\tau \in K^\infty $ , $j \in \{1,\dots ,k\}$ , and $n \in \mathbb N$ . Using Lemma 2.7 and the binomial theorem, we compute that

(3.47) $$ \begin{align} \begin{aligned} \tau(\Phi_0'(e_j)^n) &{\stackrel{(3.46)}=} \sum_{i=0}^n \binom ni \tau\big(d_0^i(\Delta_\epsilon^ig_{\epsilon,2\epsilon}^{n-i})(\Phi_0(e_j))\big) \\ &{\stackrel{(3.44)}=} \sum_{i=0}^n \binom ni \tau(d_0^i)\tau\big((\Delta_\epsilon^ig_{\epsilon,2\epsilon}^{n-i})(\Phi_0(e_j))\big) \\ &{\stackrel{(3.42)}=} \frac1k \sum_{i=0}^n \binom ni \tau(d_0^i)\tau\big((\Delta_\epsilon^ig_{\epsilon,2\epsilon}^{n-i})(\Phi_0(1_k))\big) \\ &{\stackrel{(3.44)}{=}} \frac1k \sum_{i=0}^n \binom ni \tau\big(d_0^i (\Delta_\epsilon^ig_{\epsilon, 2\epsilon}^{n -i})(\Phi_0(1_k))\big) \\ &{\stackrel{(3.46)}{=}} \frac1k \tau(\Phi_0'(1_k)^n). \end{aligned} \end{align} $$

Next, we define the positive contractionFootnote ¹²

(3.48) $$ \begin{align} h = d_1(1_{M^\infty}-g_{0,\epsilon}(\Phi_0(1_k))) \in \mathcal{M}^\infty \cap (\mathcal{M}_0 \cup \Phi_0(\mathbb C^k))'. \end{align} $$

Since $d_0\perp d_1$ and $(1_{\mathcal {M}^\infty }-g_{0,\epsilon }(\Phi _0(1_k)))\perp g_{\epsilon ,2\epsilon }(\Phi _0(1_k))$ , we see that

(3.49) $$ \begin{align} h \perp d_0\Delta_\epsilon(\Phi_0(1_k))+g_{\epsilon,2\epsilon}(\Phi_0(1_k)) \stackrel{(3.46)}{=}\Phi_0'(1_k). \end{align} $$

Using again that $\alpha _0$ satisfies Lemma 3.4, there is a c.p.c. order zero map $\Phi _1\colon \mathbb {C}^k \rightarrow \mathcal {M}^\infty \cap (\mathcal {M}_0 \cup \{h\})'$ satisfying

(3.50) $$ \begin{align} \tau(\Phi_1(1_k)) &\geq \alpha_0, \end{align} $$

(3.51) $$ \begin{align} \tau(f(\Phi_1(e_j))) &= \frac{1}{k}\tau(f(\Phi_1(1_k))), \end{align} $$

and

(3.52) $$ \begin{align} \tau(f(\Phi_1(x))b) = \tau(f(\Phi_1(x)))\tau(b) \end{align} $$

for all $\tau \in K^\infty ,\ x \in \mathbb C^k,\ f\in C_0(0,1]$ , and $b \in \mathrm {C}^*(h)$ . Now, define

(3.53) $$ \begin{align} \Phi=\Phi_0'+h\Phi_1\colon\mathbb C^k \to \mathcal{M}^\infty \cap \mathcal{M}_0'. \end{align} $$

Since h commutes with the range of $\Phi _1$ , $h\Phi _1$ is c.p.c. order zero. By (3.49), h is orthogonal to $\Phi _0'(1_k)$ , and using the structure theorem for order zero maps, h is also orthogonal to the range of $\Phi _0'$ . So $\Phi $ is a sum of c.p.c. order zero maps with orthogonal ranges and hence is itself a c.p.c. order zero map.

We shall show that $\Phi $ satisfies both (3.29) and (3.30). First, we show (3.30). For $\tau \in K^\infty $ , $j\in \{1,\dots ,k\}$ , and $n \in \mathbb N$ , we have

(3.54) $$ \begin{align} \begin{aligned} \tau(\Phi(e_j)^n) &{\stackrel{(3.53)}=} \tau(\Phi_0'(e_j)^n+h^n\Phi_1(e_j)^n) \\ &{\stackrel{(3.52)}=} \tau(\Phi_0'(e_j)^n)+\tau(h^n)\tau(\Phi_1(e_j)^n) \\ &{\stackrel{(3.47),(3.51)}=} \frac1k\tau(\Phi_0'(1_k)^n)+\frac1k\tau(h^n)\tau(\Phi_1(1_k)^n) \\ &{\stackrel{(3.52),(3.53)}=} \frac1k\tau(\Phi(1_k)^n). \end{aligned} \end{align} $$

By linearity, continuity, and the Stone–Weierstrass theorem, (3.30) follows.

We now work towards showing that $\Phi $ satisfies (3.29). Let $\tau \in K^\infty $ . By (3.46), (3.53), and Lemma 2.7, we have

(3.55) $$ \begin{align} \tau(\Phi(1_k)) = \tau(d_0\Delta_\epsilon(\Phi_0(1_k)))+\tau(g_{\epsilon,2\epsilon}(\Phi_0(1_k)))+\tau(h\Phi_1(1_k)). \end{align} $$

We estimate the first term of (3.55) as follows:

(3.56) $$ \begin{align} \begin{aligned} \tau\big(d_0\Delta_\epsilon(\Phi_0(1_k))\big) &{\stackrel{(3.44)}{=}} \tau(d_0) \tau\big(\Delta_\epsilon(\Phi_0(1_k))\big) \\ &{\stackrel{(3.45)}{\geq}} \gamma \tau\big(\Delta_\epsilon(\Phi_0(1_k))\big) \\ &{\geq} \gamma\alpha_0 \tau\big(\Delta_\epsilon(\Phi_0(1_k))\big), \end{aligned} \end{align} $$

where we have used that $\alpha _0 < 1$ in the last line. We estimate the third term of (3.55) as follows:

(3.57) $$ \begin{align} \begin{aligned} \tau(h\Phi_1(1_k)) &{\stackrel{(3.52)}{=}} \tau(h)\tau(\Phi_1(1_k)) \\ &{\stackrel{(3.50)}{\geq}} \alpha_0 \tau(h) \\ &{\stackrel{(3.48)}{=}} \alpha_0 \tau\big(d_1(1_{{\mathcal{M}}^\infty} - g_{0, \epsilon}(\Phi_0(1_k)))\big) \\ &{\stackrel{3.44}{=}} \alpha_0 \tau(d_1)\tau\big(1_{{\mathcal{M}}^\infty} - g_{0, \epsilon}(\Phi_0(1_k))\big) \\ &{\stackrel{(3.45)}{\geq}} \gamma\alpha_0 \tau\big(1_{{\mathcal{M}}^\infty} - g_{0, \epsilon}(\Phi_0(1_k))\big). \end{aligned} \end{align} $$

Substituting the estimates (3.56) and (3.57) into (3.55) and using that $\Delta _\epsilon = g_{0, \epsilon } - g_{\epsilon , 2\epsilon }$ , we obtain

(3.58) $$ \begin{align} \tau(\Phi(1_k)) &\geq \gamma\alpha_0 \tau\big(\Delta_\epsilon(\Phi_0(1_k))\big) + \tau\big(g_{\epsilon,2\epsilon}(\Phi_0(1_k))\big) + \gamma\alpha_0 \tau\big(1_{{\mathcal{M}}^\infty} - g_{0, \epsilon}(\Phi_0(1_k))\big)\nonumber\\&= \gamma\alpha_0 \tau\big(g_{0, \epsilon}(\Phi_0(1_k))\big) - \gamma\alpha_0 \tau\big(g_{\epsilon, 2\epsilon}(\Phi_0(1_k))\big) \nonumber\\&\qquad + \tau\big(g_{\epsilon,2\epsilon}(\Phi_0(1_k))\big) + \gamma\alpha_0 - \gamma\alpha_0\tau\big(g_{0, \epsilon}(\Phi_0(1_k))\big)\nonumber\\&= \gamma\alpha_0 + (1 - \gamma\alpha_0) \tau\big(g_{\epsilon,2\epsilon}(\Phi_0(1_k))\big).\nonumber\\ \end{align} $$

Finally, using that $g_{\epsilon , 2\epsilon }(t) \geq t - \epsilon $ for all $t \in [0, 1]$ , we get

(3.59) $$ \begin{align} \begin{aligned} \tau(\Phi(1_k)) &{\geq} \gamma \alpha_0 +(1 - \gamma \alpha_0) \tau\big( \Phi_0(1_k) - \epsilon 1_{{\mathcal{M}}^\infty}\big) \\ &{\stackrel{(3.41)}{\geq}} \gamma \alpha_0 + (1 - \gamma \alpha_0)(\alpha_0 - \epsilon) \\ &{\stackrel{(3.40)}{=}} \alpha, \end{aligned} \end{align} $$

which completes the proof that (3.29) holds.

By Lemma 2.4 with $T = \mathrm {C}^*(\Phi (\mathbb C^k)) \subseteq \mathcal {M}^\infty \cap \mathcal {M}_0'$ , there exists a strictly increasing function $\rho \colon \mathbb N \to \mathbb N$ such that, after replacing $\Phi $ with $\psi _\rho \circ \Phi \colon \mathbb C^k \rightarrow \mathcal {M}^\infty \cap S'$ , we have that (3.31) holds for all $\tau \in K^\infty ,\ j\in \{1,\dots ,k\},\ f\in C_0(0,1]$ , and $b\in S$ .

From the definition of the reduced power $(\mathcal {M}^\infty , K^\infty , E^\infty )$ , we have $E^\infty \circ \psi _\rho = \psi _\rho \circ E^\infty $ . Hence, it follows that (3.29) and (3.30) continue to hold with $\psi _\rho \circ \Phi $ in place of $\Phi $ . Thus we have shown $\alpha \in \Omega $ , which is our intended contradiction.

Proof Proof (Theorem 1.1)

Let A be a C $^*$ -algebra with $\partial _e T(A)$ non-empty and compact. If A has uniform property $\Gamma $ , then it is clear that $\pi _\tau (A)"$ has property $\Gamma $ for each $\tau \in \partial _e T(A)$ .

Suppose now that $K = \partial _e T(A)$ has finite covering dimension and $\pi _\tau (A)"$ has property $\Gamma $ for each $\tau \in K$ . By [Reference Ozawa21, Theorem 3], the uniform tracial completion $\mathcal {M}=\overline {A}{}^{T(A)}$ has the structure of a W $^*$ -bundle over K with fibres $\pi _\tau (A)"$ for $\tau \in K$ . By Theorem 1.2, $\mathcal {M}$ has property $\Gamma $ . Hence, by $\|\cdot \|_{2, \mathrm {u}}$ -density, it follows that A has uniform property $\Gamma $ (see Section 2.4).