Proof We prove the proposition for ${\mathrm {dim}_{\mathrm {Rok}}}$ , since the proof for ${\mathrm {dim}_{\mathrm {Rok}}^{\mathrm {c}}}$ is analogous. Using the identity $x=x1_{g^{-1}}$ for all $x\in A_{g^{-1}}$ and $g\in G$ , one easily shows that (1’) implies the following identity for all $g,h\in G$ , $j=0,\ldots ,d$ , and all $x\in A_{g^{-1}}$ :

$$ \begin{align*}\alpha_g(f_h^{(j)}x)=f_{gh}^{(j)}\alpha_g(x).\end{align*} $$

This identity clearly implies (1). Conversely, let $\varepsilon>0$ and a finite subset $F \subseteq A$ be given. Without loss of generality, we assume that F consists of contractions and that $\{1_g\colon g\in G\}\subseteq F$ . Set $\varepsilon _0={\varepsilon }/({|G|(d+1)+1})$ and find Rokhlin towers $f_g^{(j)}\in A_g$ , for $g\in G$ and $j=0,\ldots ,d$ , satisfying conditions (1), (2), (3), and (4) in Definition 2.1 for $(F,\varepsilon _0)$ . Define positive contractions $\widetilde {f}_g^{(j)}\in A_g$ , for all $g\in G$ and $j=0,\ldots ,d$ , by $\widetilde {f}_g^{(j)}=\alpha _g(f_1^{(j)}1_{g^{-1}})$ . Since $1_{g^{-1}}$ belongs to F, condition (1) for $f_1^{(j)}$ gives

(2.1) $$ \begin{align} \|\widetilde{f}_g^{(j)}- f_g^{(j)}\|=\|\alpha_g(f_1^{(j)}1_{g^{-1}})-f_g^{(j)}1_g\|<\varepsilon_0 \end{align} $$

for all $g\in G$ and all $j=0,\ldots ,d$ . We claim that these elements satisfy the conditions in Definition 2.1 with (1) replaced by (1’).

We begin with (1’). For $g,h\in G$ and $j=0,\ldots ,d$ , we have

$$ \begin{align*} \alpha_g(\widetilde{f}_h^{(j)}1_{g^{-1}})&=\alpha_g(\alpha_h(f_1^{(j)}1_{h^{-1}})1_{g^{-1}})\\ &=\alpha_g(\alpha_h(f_1^{(j)}1_{h^{-1}}1_{{(gh)}^{-1}}))\\ &=\alpha_{gh}(f_1^{(j)}1_{{(gh)}^{-1}})1_g =\widetilde{f}_{gh}^{(j)}1_g. \end{align*} $$

Finally, conditions (2), (3) and (4) for the $\widetilde {f}_g^{(j)}$ follow by combining (2.1) with conditions (2), (3) and (4) for $f_g^{(j)}$ . We omit the details.

Proof We prove the results for ${\mathrm {dim}_{\mathrm {Rok}}}$ , s ince the case of ${\mathrm {dim}_{\mathrm {Rok}}^{\mathrm {c}}}$ is similar. We assume from now on that $d={\mathrm {dim}_{\mathrm {Rok}}}(\alpha )<\infty $ , otherwise there is nothing to prove.

We prove ${\mathrm {dim}_{\mathrm {Rok}}}(\alpha |_{I})\leq {\mathrm {dim}_{\mathrm {Rok}}}(\alpha )$ first. Let $\varepsilon>0$ and let a finite subset $F\subseteq I$ be given. Without loss of generality, we assume that F contains only contractions. Set $\varepsilon _0={\varepsilon }/({5|G|(d+1)+2})$ . Apply Definition 2.1 to find Rokhlin towers $f_g^{(j)}\in A_g$ with $g\in G$ and $j=0,\ldots ,d$ , for $(F, \varepsilon _0)$ . By making a small renormalization, we may assume that $\sum _{g\in G}\sum _{j=0}^df_g^{(j)}$ is a contraction. By considering an approximate identity of I which is quasi-central in A, find $e\in I$ satisfying

$$ \begin{align*}\| e^{{1}/{2}}f_g^{(j)}-f_g^{(j)}e^{{1}/{2}}\|<\varepsilon_0, \quad \|be-b\|<\varepsilon_0, \text{ and} \quad \|eb-b\|<\varepsilon_0 \end{align*} $$

for all $g\in G$ , all $j=0,\ldots ,d$ , and all $b\in \bigcup \nolimits _{g\in G}\alpha _g(F\cap A_{g^{-1}}).$

For $g\in G$ and $j=0,\ldots ,d$ , set $\widetilde {f}_g^{(j)}=e^{{1}/{2}}f_g^{(j)}e^{{1}/{2}}$ . We claim that $\{\widetilde {f}_g^{(j)}\colon g\in G, j=0,\ldots ,d\}$ are Rokhlin towers with respect to $(F, \varepsilon )$ for $\alpha |_I$ . Note that $\widetilde {f}_g^{(j)}$ belongs to $A_g\cap I=I_g$ . In order to check (1), let $g\in G$ , let $x\in F\cap I_{g^{-1}}=F\cap A_{g^{-1}}$ , let $a\in F$ , and let $j=0,\ldots ,d$ . Then

$$ \begin{align*} \alpha_g(\widetilde{f}_h^{(j)}x)a&= \alpha_g(e^{{1}/{2}}f_h^{(j)}e^{{1}/{2}}x)a \approx_{\varepsilon_0} \alpha_g(f_h^{(j)}ex)a\\ &\approx_{\varepsilon_0} \alpha_g(f_h^{(j)}x)a \approx_{\varepsilon_0} f_{gh}^{(j)}\alpha_g(x)a\\ &\approx_{\varepsilon_0} f_{gh}^{(j)}e\alpha_g(x)a \approx_{\varepsilon_0} e^{{1}/{2}}f_{gh}^{(j)}e^{{1}/{2}}\alpha_g(x)a = \widetilde{f}_{gh}^{(j)}\alpha_g(x)a. \end{align*} $$

Thus $\|\alpha _g(\widetilde {f}_h^{(j)}x)a- \widetilde {f}_{gh}^{(j)}\alpha _g(x)a\|< 5\varepsilon _0<\varepsilon $ , as desired. Condition (2) is easily checked and is left to the reader. For (3), let $a\in F$ be given. Then

$$ \begin{align*} \sum\limits_{j=0}^{d}\sum\limits_{g\in G} \widetilde{f}_g^{(j)}a &=\sum\limits_{j=0}^{d} \sum\limits_{g\in G} e^{{1}/{2}}f_g^{(j)}e^{{1}/{2}}a\\ &\approx_{|G|(d+1)\varepsilon_0} \sum\limits_{j=0}^{d} \sum\limits_{g\in G} f_g^{(j)}ea\\ &\approx_{\varepsilon_0} \sum\limits_{j=0}^{d} \sum\limits_{g\in G} f_g^{(j)}a\approx_{\varepsilon_0}a, \end{align*} $$

where in the second-to-last step we use the fact that $\sum _{g\in G}\sum _{j=0}^df_g^{(j)}$ is a contraction. Finally, to check (4), let $g\in G, j=0,\ldots ,d$ , and $a\in F$ . Then

$$ \begin{align*} \widetilde{f}_g^{(j)}a=e^{{1}/{2}}f_g^{(j)}e^{{1}/{2}}a\approx_{2\varepsilon_0} f_g^{(j)}a \approx_{\varepsilon_0} xf_g^{(j)} \approx_{2\varepsilon_0}xe^{{1}/{2}}f_g^{(j)}e^{{1}/{2}}= a\widetilde{f}_g^{(j)}. \end{align*} $$

We turn to the inequality ${\mathrm {dim}_{\mathrm {Rok}}}(\overline {\alpha })\leq d={\mathrm {dim}_{\mathrm {Rok}}}(\alpha )$ . Write $\pi \colon A\to A/I$ for the canonical equivariant map. Let $\overline {F}\subseteq A/I$ be a finite set and let $\varepsilon>0$ . Let $F\subseteq A$ be any finite set satisfying $\pi (F)=\overline {F}$ , and apply Definition 2.1 for $\alpha $ to find Rokhlin towers $f_g^{(j)}\in A_g$ , with $g\in G$ and $j=0,\ldots ,d$ , for $(F,\varepsilon )$ . It is then immediate that the positive contractions $\overline {f}_g^{(j)}=\pi (f_g^{(j)})$ witness the fact that ${\mathrm {dim}_{\mathrm {Rok}}}(\overline {\alpha })\leq d$ , as desired.

Proof We divide the proof into claims.

Claim 1. There exist projections $p_g\in A$ , for $g\in G$ , which are central in B and satisfy $1_B=\sum \nolimits _{g\in G}\beta _g(p_g)$ .

Set $n=|G|$ and fix an enumeration $G=\{1_G=g_1,g_2,\ldots ,g_n\}$ . Since $B\!=\!\!\sum _{g\in G}\!\beta _g\!(A)$ , we have $1_B\leq \sum \nolimits _{g\in G}\beta _g(1_A)$ . Note that the projections $\beta _g(1_A)$ are central and therefore together with $1_B$ generate a commutative $C^*$ -algebra. In the rest of this claim, we regard $1_B$ and $\beta _g(1_A)$ , for $g\in G$ , as $\{0,1\}$ -valued functions on some compact Hausdorff space and identify them with their supports. In particular, the support of $1_B$ is equal to the union of the supports of $\beta _g(1_A)$ , for $g\in G$ . By successively removing the double intersections, adding the triple intersections, and proceeding similarly for higher degrees, we can write $1_B$ as follows:

(3.4) $$ \begin{align} 1_B=&\sum\limits_{i=1}^{n}\beta_{g_i}(1_A)-\sum\limits_{1\leq i<j\leq n}\beta_{g_i}(1_A)\beta_{g_j}(1_A) \notag\\ & + \sum_{1\leq i<j<k\leq n}\beta_{g_i}(1_A)\beta_{g_j}(1_A)\beta_{g_k}(1_A) +\cdots \notag\\ & +(-1)^{n-1}\beta_{g_1}(1_A)\beta_{g_2}(1_A)\cdots\beta_{g_n}(1_A). \end{align} $$

Observe that every element appearing in (3.4) is central in B. We now proceed to write the unit of B as a sum of orthogonal projections in the following manner. The first summand consists of all the elements appearing in (3.4) that have $\beta _{g_1}(1_A)$ as a factor. The second summand consists of all the elements appearing in (3.4) that have $\beta _{g_2}(1_A)$ as a factor but not $\beta _{g_1}(1_A)$ . Continue inductively, and note that the process finishes after n steps; indeed, the only element left at the nth step is $\beta _{g_n}(1_A)$ .

For $1\leq k\leq n$ , the kth summand is

$$ \begin{align*} P_k &= \beta_{g_k}(1_A)-\sum\limits_{k<j\leq n}\beta_{g_k}(1_A)\beta_{g_j}(1_A)+\sum\limits_{k<i<j\leq n}\beta_{g_k}(1_A)\beta_{g_i}(1_A)\beta_{g_j}(1_A)\\ &\quad+\cdots +(-1)^{n-k}\beta_{g_k}(1_A)\beta_{g_{k+1}}(1_A)\cdots \beta_{g_n}(1_A), \end{align*} $$

and we have $1_B=P_1+\cdots +P_n$ . We want to see that $P_k$ is a projection. Set

$$ \begin{align*} Q_{k}=\sum\limits_{k<j\leq n}\beta_{g_j}(1_A)-\sum\limits_{k<i<j\leq n}\beta_{g_i}(1_A)\beta_{g_j}(1_A)+\cdots + (-1)^{n-k+1}\prod_{j>k} \beta_{g_j}(1_A). \end{align*} $$

Then $Q_{k}$ is the unit of the ideal $\sum \nolimits _{j>k}\beta _{g_j}(A)$ and therefore $Q_{k}$ is a projection. Moreover, an easy computation shows that $P_k=\beta _{g_k}(1_A)(1_B-Q_{k})$ and thus $P_k$ is also a projection. Note that $P_kP_\ell =0$ if $k\neq \ell $ .

For $k=1,\ldots ,n$ , set $p_k=\beta _{g_k}^{-1}(P_k)$ , which can be written as

$$ \begin{align*}p_k=1_A-\sum\limits_{k<j\leq n}1_A\beta_{{g_k}^{-1}g_j}(1_A)+\cdots +(-1)^{(n-k)}1_A\beta_{{g_k}^{-1}g_{k+1}}(1_A)\cdots \beta_{{g_k}^{-1}{g_n}}(1_A).\end{align*} $$

Then $p_k$ is a central projection, and it belongs to A because $1_A$ is a factor in each of its summands. Since $1_B=\sum _{k=1}^n \beta _{g_k}(p_k)$ , this concludes the proof of Claim 1.

Let $\varepsilon>0$ and let $F\subseteq B$ be finite. Without loss of generality, F is $\beta $ -invariant, contains $1_B$ , and consists of contractions. Set $\varepsilon _0={\varepsilon }/{|G|^2(d+2)}$ , and let $f_g^{(j)}\in B$ , for $g\in G$ and $j=0,\ldots ,d$ , be Rokhlin towers for $(F,\varepsilon _0)$ . Using Proposition 2.4 for the equality, and by replacing $f_g^{(j)}$ with $({1}/({1+\varepsilon })) f_g^{(j)}$ for the inequality, we may assume

(3.5) $$ \begin{align} \beta_g(f_h^{(j)})=f_{gh}^{(j)} \quad \mbox{and} \quad \sum\limits_{g\in G}\sum_{j=0}^df_g^{(j)}\leq 1 \end{align} $$

for all $g,h\in G$ and all $j=0,\ldots ,d$ .

Claim 2. There are positive elements $x_g^{(j)}\in A$ , for $g\in G$ and $j=0,\ldots ,d$ , for which:

1. (2.a) $f_1^{(j)}=\sum \nolimits _{g\in G}\beta _g(x_g^{(j)})$ ;

2. (2.b) $\sum \nolimits _{g\in G} x_g^{(j)}\in A$ is a positive contraction;

3. (2.c) $\beta _h(x_g^{(j)})b\approx _{\varepsilon _0}b\beta _h(x_g^{(j)})$ for all $b\in F$ , all $g,h\in G$ and all $j=0,\ldots ,d$ ; and

4. (2.d) $\|x_h^{(j)}\beta _g(x_t^{(j)})\|<\varepsilon _0$ for all $j=0,\ldots ,d$ and all $g,h,t\in G$ with $g\neq 1$ .

Using Claim 1, fix projections $p_g\in A$ , for $g\in G$ , which are central in B and satisfy $1_B=\sum _{g\in G}\beta _g(p_g)$ . For $j=0,\ldots ,d$ , multiply both sides of the identity by $f_1^{(j)}$ to get

$$ \begin{align*} f_1^{(j)}=\sum\limits_{g\in G}\beta_{g}(\beta_{{g}^{-1}}(f_1^{(j)})p_g)=\sum\limits_{g\in G}\beta_{g}(f_{{g}^{-1}}^{(j)}p_g). \end{align*} $$

Set $x_{g}^{(j)}=f_{{g}^{-1}}^{(j)}p_g$ for all $g\in G$ and $j=0,\ldots ,d$ . Since $p_g$ is central in B and belongs to A, it is clear that $x_g^{(j)}$ is a positive element in A. Condition (2.a) is satisfied by construction. Using centrality of $p_g$ at the second step, we get

$$ \begin{align*}\sum\limits_{g\in G}x_{g}^{(j)}=\sum\limits_{g\in G}f_{{g}^{-1}}^{(j)}p_g\leq\sum\limits_{g\in G}f_{g}^{(j)}\leq\sum\limits_{j=0}^{d}\sum\limits_{g\in G}f_{g}^{(j)} \stackrel{(3.5)}{\leq} 1,\end{align*} $$

and thus $\sum \nolimits _{g\in G}x_{g}^{(j)}$ is a positive contraction, verifying (2.b). In order to check (2.c), let $g,h\in G$ and $j=0,\ldots ,d$ . Since $\beta _h(x_g^{(j)})=f_{hg^{-1}}^{(j)}\beta _h(p_g)$ and $p_g$ is central, it follows that

$$ \begin{align*}\|\beta_h(x_g^{(j)})b-b\beta_h(x_g^{(j)})\|\leq \|f_{hg^{-1}}^{(j)}b-bf_{hg^{-1}}^{(j)}\|<\varepsilon_0 \end{align*} $$

for all $b\in F$ . We turn to (2.d). Given $j=0,\ldots ,d$ and $g,h,t\in G$ with $g\neq 1$ , we have

$$ \begin{align*} x_h^{(j)}\beta_g(x_t^{(j)})= f_{{h}^{-1}}^{(j)}p_{h}f_{gt^{-1}}^{(j)}\beta_g(p_{t})= f_{{h}^{-1}}^{(j)}f_{gt^{-1}}^{(j)}\beta_g(p_{t})p_{h}.\end{align*} $$

If $gt^{-1}\neq h^{-1}$ then $\|f_{{h}^{-1}}^{(j)}f_{gt^{-1}}^{(j)}\|<\varepsilon _0$ and hence $\|x_h^{(j)}\beta _g(x_t^{(j)})\|<\varepsilon _0$ . Otherwise, we have $gt^{-1}= h^{-1}$ and thus $g=h^{-1}t$ . In particular, $h\neq t$ . Then

$$ \begin{align*}\|\beta_g(p_{t})p_{h}\|=\|\beta_{h^{-1}t}(p_k)p_h\|=\|\beta_t(p_t)\beta_h(p_h)\|=0,\end{align*} $$

since $\beta _t(p_t)$ is orthogonal to $\beta _h(p_h)$ by Claim 1. It follows that $x_h^{(j)}\beta _g(x_t^{(j)})=0$ , and thus (2.d) is satisfied. This proves Claim 2.

Let $x_g^{(j)}\in A$ , for $g\in G$ and $j=0,\ldots ,d$ , be positive elements satisfying the conclusion of Claim 2. For $j=0,\ldots ,d$ , set $a_1^{(j)}=\sum \nolimits _{g\in G}x_g^{(j)}\in A$ . For $g\in G$ , we set $a_g^{(j)}=\beta _g(a_1^{(j)})$ . Then $\beta _g(a_h^{(j)})=a_{gh}^{(j)}$ for all $g,h\in G$ and $j=0,\ldots ,d$ . In particular, condition (1) in Definition 2.1 is satisfied. To check condition (2), let $j=0,\ldots ,d$ and $g,h\in G$ with $g\neq h$ . Then

$$ \begin{align*}\|a_g^{(j)}a_h^{(j)}\|=\|a_1^{(j)}\beta_{g^{-1}h}(a_1^{(j)})\|=\bigg\|\sum\limits_{s,t\in G}x_s^{(j)}\beta_{g^{-1}h}(x_t^{(j)})\bigg\|\stackrel{\mathrm{(2.d)}}{\leq} |G|^2\varepsilon_0<\varepsilon.\end{align*} $$

Moreover, condition (3) follows from the following identity:

$$ \begin{align*} \sum\limits_{g\in G}\sum\limits_{j=0}^{d}a_g^{(j)}&=\sum\limits_{g\in G}\sum\limits_{j=0}^{d}\beta_g\bigg(\sum\limits_{h\in G} x_h^{(j)}\bigg)=\sum\limits_{g,h\in G}\sum\limits_{j=0}^{d}\beta_{g}(x_h^{(j)}) \\ &=\sum\limits_{g,h\in G}\sum\limits_{j=0}^{d}\beta_{gh}(x_h^{(j)}) =\sum\limits_{g\in G}\sum\limits_{j=0}^{d}\beta_g\bigg(\sum\limits_{h\in G} \beta_h(x_h^{(j)})\bigg)\\ &= \sum\limits_{g\in G}\sum\limits_{j=0}^{d}\beta_g(f_1^{(j)}) =\sum\limits_{g\in G}\sum\limits_{j=0}^{d}f_g^{(j)}. \end{align*} $$

Let $b\in F$ , $g\in G$ and $j=0,\ldots ,d$ . In order to check condition (4) in Definition 2.1, and since $a_g^{(j)}=\beta _g(a_1^{(j)})$ and F is $\beta $ -invariant, it suffices to take $g=1$ . In this case, we have

$$ \begin{align*}\|a_1^{(j)}b-ba_1^{(j)}\|\leq \sum_{g\in G}\|x_g^{(j)}b-bx_g^{(j)}\|\stackrel{\mathrm{(2.c)}}{\leq} |G|\varepsilon_0<\varepsilon.\end{align*} $$

This proves the first part of the lemma.

Assume now that ${\mathrm {dim}_{\mathrm {Rok}}^{\mathrm {c}}}(\beta )=d<\infty $ , and choose the Rokhlin towers as above to moreover satisfy condition (5) in Definition 2.1. For $g,h\in G$ and $j,k=0,\ldots ,d$ , we get

$$ \begin{align*}a_g^{(j)}a_h^{(k)} &= \beta_g(a_1^{(j)})\beta_h(a_1^{(k)})=\sum_{t,s\in G} \beta_g(x_t^{(j)})\beta_h(x_s^{(k)})\\ &=\sum_{t,s\in G} \beta_g(f_{t^{-1}}^{(j)}p_t)\beta_h(f_{s^{-1}}^{(k)}p_s) =\sum_{t,s\in G} f_{gt^{-1}}^{(j)}\beta_g(p_t)f_{hs^{-1}}^{(k)}\beta_h(p_s)\\ &\approx_{|G|^2\varepsilon_0}\sum_{t,s\in G} f_{hs^{-1}}^{(k)}\beta_h(p_s)f_{gt^{-1}}^{(j)}\beta_g(p_t) =a_h^{(k)}a_g^{(j)}.\\[-46pt] \end{align*} $$

Proof The proofs for ${\mathrm {dim}_{\mathrm {Rok}}}$ and ${\mathrm {dim}_{\mathrm {Rok}}^{\mathrm {c}}}$ are very similar, so we provide full details for ${\mathrm {dim}_{\mathrm {Rok}}}$ and indicate how one modifies the proof to obtain the result for ${\mathrm {dim}_{\mathrm {Rok}}^{\mathrm {c}}}$ . We divide the proof into two parts, namely the inequalities ${\mathrm {dim}_{\mathrm {Rok}}}(\alpha )\leq {\mathrm {dim}_{\mathrm {Rok}}}(\beta )$ and ${\mathrm {dim}_{\mathrm {Rok}}}(\alpha )\geq {\mathrm {dim}_{\mathrm {Rok}}}(\beta )$ . Since A is unital and $\alpha $ is globalizable, it follows that $A_g$ is unital for all $g\in G$ , with unit given by $1_g=1_A\beta _g(1_A)$ .

To show that ${\mathrm {dim}_{\mathrm {Rok}}}(\alpha )\leq {\mathrm {dim}_{\mathrm {Rok}}}(\beta )$ , it suffices to assume that $d={\mathrm {dim}_{\mathrm {Rok}}}(\beta )$ is finite. Let $\varepsilon>0$ and $F\subseteq A$ be given. Using Lemma 3.3, let $\widetilde {f}_g^{(j)}\in B$ , for $g\in G$ and $j=0,\ldots ,d$ , such that:

1. (a) $\beta _g(\widetilde {f}_h^{(j)})=\widetilde {f}_{gh}^{(j)}$ for all $g,h\in G$ and $j=0,\ldots ,d$ ;

2. (b) $\|\widetilde {f}_g^{(j)}\widetilde {f}_h^{(j)}\|<\varepsilon $ for all $j=0,\ldots ,d$ and all $g,h\in G$ with $g\neq h$ ;

3. (c) $\| 1_B-\sum \nolimits _{g\in G}\sum \nolimits _{j=0}^d \widetilde {f}_g^{(j)}\|\leq \varepsilon $ ;

4. (d) $\|\widetilde {f}_g^{(j)}b-b\widetilde {f}_g^{(j)}\|<\varepsilon $ for all $g\in G$ , all $j=0,\ldots ,d$ and all $b\in F$ .

Set $f_g^{(j)}=\widetilde {f}_g^{(j)}1_g\in A_g$ for all $g\in G$ and all $j=0,\ldots ,d$ . We claim that these are Rokhlin towers for $\alpha $ with respect to $(F,\varepsilon )$ . For $g,h\in G$ and $j=0,\ldots ,d$ , we use at the second step that $\beta _g|_{A_{g^{-1}}}=\alpha _g$ to get

$$ \begin{align*}\alpha_g(f_h^{(j)}1_{g^{-1}})=\alpha_g(\widetilde{f}_h^{(j)}1_h1_{g^{-1}})=\beta_g(\widetilde{f}_h^{(j)})1_{gh}1_g\stackrel{(a)}{=}\widetilde{f}_{gh}^{(j)}1_{gh}1_g=f_{gh}^{(j)}1_g,\end{align*} $$

thus verifying condition (1) in Definition 2.1. Moreover, $\|f_g^{(j)}f_h^{(j)}\| \leq \|\widetilde {f}_g^{(j)}\widetilde {f}_h^{(j)}\|$ and hence condition (2) is also satisfied by (b) above. Since

$$ \begin{align*} \sum\limits_{g\in G}\sum\limits_{j=0}^{d}f_g^{(j)}&=\sum\limits_{g\in G}\sum\limits_{j=0}^{d}\beta_g(\widetilde{f}_1^{(j)})1_g =\sum\limits_{g\in G}\sum\limits_{j=0}^{d}\beta_g(\widetilde{f}_1^{(j)}1_A)\beta_g(1_A)1_A=\sum\limits_{g\in G}\sum\limits_{j=0}^{d}\widetilde{f}_g^{(j)}1_A, \end{align*} $$

it follows from (c) that condition (3) is also satisfied. Finally, given $b\in F$ , $g\in G$ and $j=0,\ldots ,d$ , we have

$$ \begin{align*}\|f_g^{(j)}b-bf_g^{(j)}\|=\|\widetilde{f}_g^{(j)}b1_g-b\widetilde{f}_g^{(j)}1_g\|\leq \|\widetilde{f}_g^{(j)}b-b\widetilde{f}_g^{(j)}\|\stackrel{(d)}{=}\varepsilon,\end{align*} $$

establishing condition (4). It follows that ${\mathrm {dim}_{\mathrm {Rok}}}(\alpha )\leq d$ , as desired. Note that

$$ \begin{align*}\|f_g^{(j)}f_h^{(k)}-f_h^{(k)}f_g^{(j)}\|\leq \|\widetilde{f}_g^{(j)}\widetilde{f}_h^{(k)}-\widetilde{f}_h^{(k)}\widetilde{f}_g^{(j)}\|\end{align*} $$

for all $g,h\in G$ and $j,k=0,\ldots ,d$ . Thus, if ${\mathrm {dim}_{\mathrm {Rok}}^{\mathrm {c}}}(\beta )\leq d$ and the Rokhlin towers $\widetilde {f}_g^{(j)}$ for $\beta $ also satisfy condition (5) in Definition 2.1, then the Rokhlin towers $f_g^{(j)}$ for $\alpha $ also satisfy (5) and hence ${\mathrm {dim}_{\mathrm {Rok}}^{\mathrm {c}}}(\alpha )\leq d$ .

We turn to the inequality ${\mathrm {dim}_{\mathrm {Rok}}}(\beta )\leq {\mathrm {dim}_{\mathrm {Rok}}}(\alpha )$ , so we set $d={\mathrm {dim}_{\mathrm {Rok}}}(\alpha )$ and assume that $d<\infty $ . Let $\varepsilon>0$ and let $F\subseteq B$ be a finite subset. Without loss of generality, we assume that F contains $1_g$ for all $g\in G$ and that it is $\beta $ -invariant. Since B is generated by the $\beta $ -translations of A (condition (d) in Definition 3.1), there exist $a_g\in A$ for $g\in G$ such that $1_B=\sum \nolimits _{g\in G}\beta _g(a_g)$ . Set $\varepsilon _0= {\varepsilon }/{|G|\max \nolimits _{g\in G}\|a_g\|}$ . Using Proposition 2.4, let $f_g^{(j)}\in A_g$ , for $g\in G$ and $j=0,\ldots ,d$ , be positive contractions satisfying the following conditions:

1. (i) $\alpha _g(f_h^{(j)}1_{g^{-1}})=f_{gh}^{(j)}1_{g}$ for all $g,h\in G$ and $j=0,\ldots ,d$ ;

2. (ii) $\|f_g^{(j)}f_h^{(j)}\|<\varepsilon $ for all $j=0,\ldots ,d$ and all $g,h\in G$ with $g\neq h$ ;

3. (iii) $\|\sum \nolimits _{g\in G}\sum \nolimits _{j=0}^d f_g^{(j)}-1_A\|<\varepsilon $ .

4. (iv) $\|f_g^{(j)}b-bf_g^{(j)}\|<\varepsilon $ for all $g\in G$ , all $j=0,\ldots ,d$ and all $b\in F$ .

For $g\in G$ and $j=0,\ldots ,d$ , set $\widetilde {f}_g^{(j)}=\beta _g(f_1^{(j)})\in B$ . We claim that the $\widetilde {f}_g^{(j)}$ are Rokhlin towers for $\beta $ with respect to $(F, \varepsilon )$ . Condition (1) in Definition 2.1 is clearly satisfied. In order to check (2), let $j=0,\ldots ,d$ and $g,h\in G$ with $g\neq h$ be given. Using that $f_1^{(j)}=f_1^{(j)}1_A$ at the second step, and that $1_{g^{-1}h}=1_A\beta _{g^{-1}h}(1_A)$ at the third, we get

$$ \begin{align*}\|\widetilde{f}_g^{(j)}\widetilde{f}_h^{(j)}\|&=\|\beta_g(f_1^{(j)})\beta_h(f_1^{(j)})\| =\|f_1^{(j)}1_A\beta_{g^{-1}h}(1_Af_1^{(j)})\|\\ &=\|f_1^{(j)}1_{g^{-1}h}\beta_{g^{-1}h}(f_1^{(j)})\| =\|f_1^{(j)}\alpha_{g^{-1}h}(1_{h^{-1}g}f_1^{(j)})\|\\ & \stackrel{\mathrm{(i)}}{=} \|f_1^{(j)}f_{g^{-1}h}^{(j)}\|\stackrel{\mathrm{(ii)}}{<} \varepsilon_0. \end{align*} $$

To check (3), it suffices to show that for any $a\in A$ and $h\in G$ , we have

$$ \begin{align*}\bigg\|\sum\limits_{j=0}^{d}\sum\limits_{g\in G}\widetilde{f}_g^{(j)}\beta_h(a)-\beta_h(a)\bigg\|<\|a\|\varepsilon_0.\end{align*} $$

Indeed, once this is established, and since $1_B=\sum \nolimits _{h\in G}\beta _h(a_h)$ , it will follow that

$$ \begin{align*}\bigg\|\sum\limits_{j=0}^{d}\sum\limits_{g\in G}\widetilde{f}_g^{(j)}-1_B\bigg\|< |G|\max_{g\in G}\|a_g\|\varepsilon_0=\varepsilon,\end{align*} $$

thus establishing (3). Let $a\in A$ and $h\in G$ be given; without loss of generality we assume that $\|a\|\leq 1$ . Then:

$$ \begin{align*} \sum\limits_{j=0}^{d}\sum\limits_{g\in G}\widetilde{f}_g^{(j)}\beta_h(a)&=\sum\limits_{j=0}^{d}\sum\limits_{g\in G}\beta_g(f_1^{(j)})\beta_h(a)\\ &=\sum\limits_{j=0}^{d}\sum\limits_{g\in G}\beta_h(\beta_{h^{-1}g}(f_1^{(j)}1_A)1_Aa)\\ &=\sum\limits_{j=0}^{d}\sum\limits_{g\in G}\beta_h(\beta_{h^{-1}g}(f_1^{(j)}1_A\beta_{g^{-1}h}(1_A))a)\\ &=\sum\limits_{j=0}^{d}\sum\limits_{g\in G}\beta_h(\alpha_{h^{-1}g}(f_1^{(j)}1_{g^{-1}h})a)\\ &\stackrel{\mathrm{(i)}}{=}\sum\limits_{j=0}^{d}\sum\limits_{g\in G}\beta_h(f_{h^{-1}g}^{(j)}a)\\ &\stackrel{\!\mathrm{(iii)}}{\, \approx_\varepsilon}_{\!_0}\! \beta_h(a), \end{align*} $$

as desired. Finally, to check condition (4), let $a\in F$ , $g\in G$ and $j=0,\ldots ,d$ be given. Using at the last step that $\beta _{g^{-1}}(a)\in F$ , we get

$$ \begin{align*} \|\widetilde{f}_g^{(j)}a-a\widetilde{f}_g^{(j)}\|&=\|\beta_g(f_1^{(j)})a-a\beta_g(f_1^{(j)})\|\\ &=\|f_1^{(j)}\beta_{g^{-1}}(a)-\beta_{g^{-1}}(a)f_1^{(j)}\|\stackrel{\mathrm{(iv)}}{<}\varepsilon_0.\end{align*} $$

This shows that ${\mathrm {dim}_{\mathrm {Rok}}}(\beta )\leq {\mathrm {dim}_{\mathrm {Rok}}}(\alpha )$ . Observe that the Rokhlin towers for $\beta $ that we constructed satisfy the following identity for all $j,k=0,\ldots ,d$ and $g,h\in G$ :

$$ \begin{align*} \widetilde{f}_g^{(j)}\widetilde{f}_h^{(k)}&=\beta_g(f_1^{(j)})\beta_h(f_1^{(k)})=\beta_g(f_1^{(j)}1_A\beta_{g^{-1}h}(1_Af_1^{(k)}))\\ &=\beta_g(f_1^{(j)}\alpha_{g^{-1}h}(1_{h^{-1}g}f_1^{(k)})) = \beta_g(f_1^{(j)}f_{g^{-1}h}^{(k)}).\end{align*} $$

By taking adjoints, we also get $\widetilde {f}_h^{(k)}\widetilde {f}_g^{(j)}= \beta _g(f_{g^{-1}h}^{(k)}f_1^{(j)})$ . In particular,

$$ \begin{align*}\| \widetilde{f}_g^{(j)}\widetilde{f}_h^{(k)}-\widetilde{f}_h^{(k)}\widetilde{f}_g^{(j)}\|\leq \|f_1^{(j)}f_{g^{-1}h}^{(k)}-f_{g^{-1}h}^{(k)}f_1^{(j)}\|. \end{align*} $$

Thus, if ${\mathrm {dim}_{\mathrm {Rok}}^{\mathrm {c}}}(\alpha )\leq d$ and the Rokhlin towers $f_g^{(j)}$ for $\alpha $ also satisfy condition (5) in Definition 2.1, then the Rokhlin towers $\widetilde {f}_g^{(j)}$ for $\beta $ also satisfy (5) and hence ${\mathrm {dim}_{\mathrm {Rok}}^{\mathrm {c}}}(\beta )\!\leq ~d$ .

Proof Set $d={\mathrm {dim}_{\mathrm {Rok}}}(\alpha )$ , and assume that $d<\infty $ . Let $g\in G\setminus \{1\}$ be given. Arguing by contradiction, assume that $F_g=\{x\in X_{g^{-1}}\colon \theta _{g}(x)=x\}$ is not empty, and fix $x\in F_g$ . Note that x belongs to $X_g\cap X_{g^{-1}}$ . Let $a\in C_0(X_{g^{-1}})$ be a positive contraction satisfying $a(x)=1$ . Choose $\varepsilon>0$ with $\varepsilon < ({1}/{2(|G|(d+1)+1)^2})$ , and let $f_h^{(j)}\in C_0(X_h)$ , for $h\in G$ , and $j=0,\ldots , d$ be Rokhlin towers for $(\{a, \alpha _g(a)\},\varepsilon )$ . Given $h\in G$ and $j=0,\ldots ,d$ , we have:

$$ \begin{align*} \|f_h^{(j)}\alpha_g(f_h^{(j)}a)\|_\infty&\leq \|f_h^{(j)}(\underbrace{\alpha_g(f_h^{(j)}a)-f_{gh}^{(j)}\alpha_g(a)}_{\ \ \ \approx_\varepsilon 0})\|_\infty+ \|\underbrace{f_h^{(j)}f_{gh}^{(j)}}_{ \ \ \ \approx_\varepsilon 0}\alpha_g(a)\|_\infty<2\varepsilon. \end{align*} $$

Evaluating at x, we get

$$ \begin{align*}(f_h^{(j)}\alpha_g(f_h^{(j)}a))(x) = f_h^{(j)}(x)f_h^{(j)}(\theta_{g^{-1}}(x))a(\theta_{g^{-1}}(x))<2\varepsilon.\end{align*} $$

Since $\theta _{g^{-1}}(x)=x$ and $a(x)=1$ , we deduce that

(5.1) $$ \begin{align} f_h^{(j)}(x)<\sqrt{2\varepsilon}, \end{align} $$

for all $h\in G$ and $j=0,\ldots ,d$ . Moreover, condition (3) from Definition 2.1 gives

$$ \begin{align*} \bigg|\sum\limits_{j=0}^{d}\sum\limits_{h\in G}f_h^{(j)}(x)a(x)-a(x)\bigg|\stackrel{a(x)=1}{=}\bigg|\sum\limits_{j=0}^{d}\sum\limits_{h\in G}f_h^{(j)}(x)-1\bigg|<\varepsilon,\end{align*} $$

so $\sum \nolimits _{j=0}^{d}\sum \nolimits _{h\in G}f_h^{(j)}(x)>1-\varepsilon $ . Using this at the second step, we get

$$ \begin{align*}|G|(d+1)\sqrt{2\varepsilon}\stackrel{(5.1)}{>}\sum\limits_{j=0}^{d}\sum\limits_{g\in G}f_g^{(j)}(x) >1-\varepsilon\geq 1-\sqrt{2\varepsilon}.\end{align*} $$

This contradicts the choice of $\varepsilon $ , showing that $F_g$ is empty. Thus $\theta $ is free.

Proof Fix $\tau \in \mathcal {T}_n(G)$ . Then $X_\tau $ is an open subset of X and thus $\dim (X_\tau )\leq \dim (X)$ . By part (2) of Theorem 5.4 in [Reference Abadie, Gardella and Geffen2], the restricted global action of $H_\tau $ on $X_\tau $ is free. Using Proposition 2.11 in [Reference Hirshberg and Phillips17] at the first step, it follows that

$$ \begin{align*}{\mathrm{dim}_{\mathrm{Rok}}}(\alpha|_{H_\tau})\leq \dim(X_\tau)\leq \dim(X).\end{align*} $$

Thus, the result follows from Theorem 4.6.

Proof It is clear that any action satisfying conditions (1), (2), and (3) in the statement has Rokhlin dimension at most d (the approximate commutation condition from Definition 2.1 is vacuous since A is commutative). We therefore prove the non-trivial implication. Let $\varepsilon>0$ and let a finite subset $F\subseteq A$ be given. Without loss of generality, we assume that F consists of positive contractions.

Fix $\varepsilon _0<1$ such that $(d+2)|G|^2\sqrt {\varepsilon _0}<\varepsilon $ , and choose positive contractions $x_g^{(j)}\in A_g$ , for $g\in G$ and $j=0,\ldots ,d$ , satisfying:

1. (a) $\| \alpha _g(x_h^{(j)}a)- x_{gh}^{(j)}\alpha _g(a)\|<\varepsilon _0$ for all $a\in F\cap A_{g^{-1}}$ ;

2. (b) $\|x_g^{(j)}x_h^{(j)}\|<\varepsilon _0$ for all $g,h\in G$ with $g\neq h$ and for all $j=0,\ldots ,d$ ;

3. (c) $\|\!\sum \nolimits _{g\in G}\sum \nolimits _{j=0}^{d}x_g^{(j)}-1\|<\varepsilon _0$ .

For $g\in G$ and $j=0,\ldots ,d$ , set $y_g^{(j)}=\sum \nolimits _{h\in G\setminus \{g\}}x_h^{(j)}$ and

$$ \begin{align*}f_g^{(j)}=(x_g^{(j)}-y_g^{(j)})_+.\end{align*} $$

Since A is commutative, we have $f_g^{(j)}\leq x_g^{(j)}$ and hence $f_g^{(j)}$ is a positive contraction in $A_g$ . We will show that these elements satisfy the conditions in the statement. We begin by noting that the $f_g^{(j)}$ are very close to the $x_g^{(j)}$ .

Claim. We have $\|x_g^{(j)}-f_g^{(j)}\|< |G|\sqrt {\varepsilon _0}$ for all $g\in G$ and $j=0,\ldots ,d$ . To see this, since A is abelian, it is enough to prove that $|x_g^{(j)}(t)-f_g^{(j)}(t)|\leq |G|\sqrt {\varepsilon _0}$ for every $t\in \mathrm {Spec}(A)$ . Fix $t\in \mathrm {Spec}(A)$ . We divide the proof into two cases.

Suppose that $x_g^{(j)}(t)\geq y^{(j)}_g(t)$ . Then $f_g^{(j)}(t)=x^{(j)}_g(t)-y_g^{(j)}(t)$ and hence

(5.2) $$ \begin{align} x_g^{(j)}(t)-f_g^{(j)}(t)=y_g^{(j)}(t)=\sum\limits_{h\in G\setminus\{g\}}x_h^{(j)}(t).\end{align} $$

For $h\neq g$ , we have $x_h^{(j)}(t)\leq y_g^{(j)}(t)\leq x_g^{(j)}(t)$ , and thus

(5.3) $$ \begin{align} x_h^{(j)}(t)^2\leq x_h^{(j)}(t)x_g^{(j)}(t)< \varepsilon_0. \end{align} $$

We conclude that

$$ \begin{align*}|x_g^{(j)}(t)-f_g^{(j)}(t)| \stackrel{({5.2})}{\leq} \sum_{h\in G\setminus\{g\}}|x_h^{(j)}(t)| \stackrel{({5.3})}{<} (|G|-1)\sqrt{\varepsilon_0}< |G|\sqrt{\varepsilon_0}.\end{align*} $$

Suppose that $x_g^{(j)}(t)\leq y^{(j)}_g(t)$ . Then $f_g^{(j)}(t)=0$ and hence $x_g^{(j)}(t)-f_g^{(j)}(t)=x_g^{(j)}(t)$ . Moreover,

(5.4) $$ \begin{align} x_g^{(j)}(t)^2\leq \bigg(\sum\limits_{h\in G\setminus\{g\}}x_h^{(j)}(t)\bigg)x_g^{(j)}(t)\stackrel{\mathrm{(b)}}{\leq} (|G|-1)\varepsilon_0.\end{align} $$

Thus,

$$ \begin{align*}|x_g^{(j)}(t)-f_g^{(j)}(t)|=|x_g^{(j)}(t)| \stackrel{({5.4})}{\leq}\sqrt{(| G| -1)\varepsilon_0}< |G|\sqrt{\varepsilon_0}.\end{align*} $$

This proves the claim.

We check that the positive contractions $f_g^{(j)}$ for $g\in G$ and $j=0,\ldots ,d$ satisfy conditions (1), (2), and (3) in the statement. Conditions (1) and (3) follow immediately by combining the claim above with conditions (a) and (c), respectively, so we only check (2). Let $g,h\in G$ with $g\neq h$ and let $j=0,\ldots ,d$ . Using the inequalities $x_g^{(j)}\leq y_h^{(j)}$ and $x_h^{(j)}\leq y_g^{(j)}$ (which are valid since $g\neq h$ ) at the second step, and the identity $(z-w)_+(w-z)_+=0$ at the last step, we get

$$ \begin{align*} f_g^{(j)}f_h^{(j)}&= (x_g^{(j)}-y_g^{(j)})_+(x_h^{(j)}-y_h^{(j)})_+\\ &\leq (y_h^{(j)}-y_g^{(j)})_+(y_g^{(j)}-y_h^{(j)})_+=0. \\[-34pt] \end{align*} $$

Proof We prove the lemma by induction on n, the case $n=1$ being trivial. Assume that $n=2$ , and let $x_1\in A_1$ and $x_2\in A_2$ be positive contractions satisfying $x_1x_2\in J$ . Set $y_1=(x_1-x_2)_+$ and $y_2=(x_2-x_1)_+$ , which are positive contractions. Since A is abelian, we have $y_j\leq x_j$ and hence $y_j$ belongs to $A_j$ for $j=1,2$ . Moreover, $y_1y_2=0$ . Finally, using at the first step that $\pi (x_1)\pi (x_2)=0$ , we get

$$ \begin{align*}\pi(x_{1})=(\pi(x_{1})-\pi(x_{2}))_+=\pi((x_1-x_2)_+)= \pi(y_{1}).\end{align*} $$

Similarly, $\pi (x_2)=\pi (y_2)$ . This proves the case $n=2$ of the statement.

Assume now that we have proved the result for an integer $n\geq 2$ , and let us prove it for $n+1$ . Thus, let $A_1,\ldots , A_{n+1}$ be ideals in A, and let $x_1,\ldots , x_{n+1}$ be as in the statement. Set $x=\sum _{j=1}^n x_j$ and define $y_{n+1}=(x_{n+1}-x)_+$ . Then $y_{n+1}$ is a positive contraction with $y_{n+1}\leq x_{n+1}$ , and hence $y_{n+1}\in A_{n+1}$ . Moreover, since $\pi (x)\pi (x_{n+1})=0$ and using the same argument as in the case $n=2$ , it follows that $\pi (y_{n+1})=\pi (x_{n+1})$ .

For $j=1,\ldots ,n$ , set $z_j=(x_j-x_{n+1})_+$ , which is a positive contraction in $A_j$ with $z_j\leq x_j$ . Since $\pi (x_j)\pi (x_{n+1})=0$ for $j\leq n$ , it follows that $\pi (z_j)=\pi (x_j)$ . In particular, for $1\leq j\neq k\leq n$ we have $\pi (z_jz_k)=0$ , that is, $z_jz_k \in J$ . By the inductive step applied to $z_1,\ldots ,z_n$ , there exist orthogonal positive contractions $y_j\leq z_j$ , for $j=1,\ldots ,n$ , satisfying $\pi (y_j)=\pi (z_j)$ . Then $y_j\leq x_j$ and $\pi (y_j)=\pi (x_j)$ for $j=1,\ldots ,n+1$ . It remains to prove that the $y_1,\ldots ,y_{n+1}$ are pairwise orthogonal. By construction, the contractions $y_1,\ldots , y_n$ are pairwise orthogonal, so it suffices to check orthogonality with $y_{n+1}$ . For $j=1,\ldots ,n$ , we have

$$ \begin{align*}y_jy_{n+1}\leq z_j y_{n+1} = (x_j-x_{n+1})_+ (x_{n+1}-x)_+\leq (x_j-x_{n+1})_+ (x_{n+1}-x_j)_+=0,\end{align*} $$

using at the third step that A is commutative and $x\geq x_j$ .

Proof Set $d_1={\mathrm {dim}_{\mathrm {Rok}}}(\alpha |_J)$ and $d_2={\mathrm {dim}_{\mathrm {Rok}}}(\overline {\alpha })$ . Without loss of generality, we assume that $d_1,d_2<\infty $ . Let $\varepsilon>0$ and let $F\subseteq A$ be a finite subset. We abbreviate $B=A/J$ and $B_g= A_g/(A_g\cap J)$ for all $g\in G$ . We denote by $\pi \colon A\to B$ the canonical quotient map. We use Lemma 5.4 to choose positive contractions $x_g^{(j)}\in B_g$ , for $g\in G$ and $j=0,\ldots , d_2$ , satisfying:

1. (B.1) $\| \overline {\alpha }_g(x_h^{(j)}b)- x_{gh}^{(j)}\overline {\alpha }_g(b)\|<\varepsilon $ for all $g,h\in G$ , for all $j=0,\ldots ,d_2$ , and for all $b\in \pi (F)\cap B_{g^{-1}}$ ;

2. (B.2) $x_g^{(j)}x_h^{(j)}=0$ for all $g,h\in G$ with $g\neq h$ and for all $j=0,\ldots ,d_2$ ;

3. (B.3) $\|\!\sum \nolimits _{g\in G}\sum \nolimits _{j=0}^{d_2}x_g^{(j)}-1\|<\varepsilon $ .

Fix $j=0,\ldots ,d_2$ . Use Lemma 5.5 to find mutually orthogonal positive contractions $y_g^{(j)}\in A_g$ , for $g\in G$ , satisfying $\pi (y^{(j)}_g)=x_g^{(j)}$ . Using that J has a G-invariant approximate identity, find a positive contraction $q\in J$ satisfying:

1. (a) $\alpha _g(qx)=q\alpha _g(x)$ for all $x\in A_{g^{-1}}\cap J$ ;

2. (b) $\| (1-\sum \nolimits _{g\in G}\sum \nolimits _{j=0}^{d} y_g^{(j)})(1-q)\|<\varepsilon $ ;

3. (c) $\| (\alpha _g(y_h^{(j)}a)-y_{gh}^{(j)}\alpha _g(a))(1-q)\|<\varepsilon $ for all $a\in A_{g^{-1}}\cap F$ .

Claim. $\alpha _g(qa)=q\alpha _g(a)$ for all $a\in A_{g^{-1}}$ . To prove this, let $(e_\lambda )_{\lambda \in \Lambda }$ be an approximate identity for $A_{g^{-1}}\cap J$ . Then $(\alpha _g(e_\lambda ))_{\lambda \in \Lambda }$ is an approximate identity for $A_g\cap J$ . Fix $a\in A_{g^{-1}}$ . In the next computation, we use at the third step condition (a) above with $x=ae_\lambda $ , and the fact that $qa$ belongs to $A_{g^{-1}}\cap J$ at the fourth step:

$$ \begin{align*}q\alpha_g(a)=\lim_{\lambda\in \Lambda}q\alpha_g(a)\alpha_g(e_\lambda)=\lim_{\lambda\in\Lambda}q\alpha_g(ae_\lambda)= \lim_{\lambda\in\Lambda}\alpha_g(qae_{\lambda})=\alpha_g(qa).\end{align*} $$

This proves the claim.

Set $z_g^{(j)}=y_g^{(j)}(1-q)$ for $g\in G$ and $j=0,\ldots ,d_2$ . Then

(5.5) $$ \begin{align} z_g^{(j)}z_h^{(j)}=0 \end{align} $$

for $g,h\in G$ with $g\neq h$ and for $j=0,\ldots ,d_2$ , since $1-q$ is central. Let $g,h\in G$ , let $a\in F\cap A_{g^{-1}}$ , and let $j=0,\ldots ,d_2$ be given. Using the above claim at the second step, we get

(5.6) $$ \begin{align} \|\alpha_g(z_h^{(j)}a)-z_{gh}^{(j)}\alpha_g(a)\|&=\|\alpha_g(y_h^{(j)}a(1-q))-y_{gh}^{(j)} \alpha_g(a)(1-q)\| \end{align} $$

$$ \begin{align*} \qquad\qquad\qquad\qquad\qquad\ =\|(\alpha_g(y_h^{(j)}a)-y_{gh}^{(j)}\alpha_g(a))(1-q)\| \stackrel{(c)}{<}\varepsilon. \end{align*} $$

Furthermore,

(5.7) $$ \begin{align}\bigg\|\sum\limits_{g\in G}\sum\limits_{j=0}^{d_2}z_g^{(j)}-(1-q)\bigg\|<\varepsilon. \end{align} $$

Set $F_J=\{qa\colon a\in F\}\cup \{q\}\subseteq J$ . Choose positive contractions $\xi _g^{(k)}\in A_g\cap J$ , for $g\in G$ and $k=0,\ldots , d_1$ , satisfying:

1. (J.1) $\| \alpha _g(\xi _h^{(k)}c)- \xi _{gh}^{(k)}\alpha _g(c)\|<\varepsilon $ for all $g,h\in G$ , for all $k=0,\ldots ,d_1$ , and for all $c\in F_J\cap J_{g^{-1}}$ ;

2. (J.2) $\|\xi _g^{(k)}\xi _h^{(k)}c\|<\varepsilon $ for all $g,h\in G$ with $g\neq h$ , for all $k=0,\ldots ,d_1$ , and for all $c\in F_J$ ;

3. (J.3) $\|c-c(\sum \nolimits _{g\in G}\sum \nolimits _{k=0}^{d_1}\xi _g^{(k)})\|<\varepsilon $ for all $c\in F_J$ .

Set $\eta _g^{(k)}=\xi _g^{(k)}q$ for all $g\in G$ and $k=0,\ldots , d_1$ . For $a\in F\cap A_{g^{-1}}$ , we have

(5.8) $$ \begin{align} \| \alpha_g(\eta_h^{(k)}a)-\eta_{gh}^{(k)}\alpha_g(a)\|=\| \alpha_g(\xi_h^{(k)}qa)-\xi_{gh}^{(k)}q\alpha_g(a)\|\stackrel{\textrm{(J.1)}}{<} \varepsilon,\end{align} $$

because $qa\in J_{g^{-1}}\cap F_J$ . Since q is central (because A is commutative), we have

(5.9) $$ \begin{align}\|\eta_g^{(k)}\eta_h^{(k)}\|\leq \| \xi_g^{(k)}\xi_h^{(k)}\|<\varepsilon \end{align} $$

for $g,h\in G$ with $g\neq h$ and for $k=0,\ldots ,d_1$ . Further, taking $c=q$ in (J.3) gives

(5.10) $$ \begin{align} \bigg\| \sum\limits_{g\in G}\sum\limits_{k=0}^{d_1} \eta_g^{(k)}-q\bigg\| \leq \varepsilon. \end{align} $$

For $g\in G$ and $\ell =0,\ldots ,d_1+d_2+1$ , set

$$ \begin{align*}f_g^{(j)} = \left. \begin{cases} z_g^{(j)} & \text{for } \ell=0,\ldots,d_2, \\ \eta_g^{(\ell-d_2-1)} & \text{for } \ell=d_2+1,d_2+2,\ldots,d_1+d_2+1. \end{cases} \right.\end{align*} $$

Then $\{f_g^{(j)}\colon g\in G, 0\leq \ell \leq d_1+d_2+1\}$ are Rokhlin towers for $\alpha $ with respect to $(F,2\varepsilon )$ . Indeed, condition (1) follows from (5.5) and (5.8); condition (2) follows from (5.6) and (5.9); and condition (3) follows from (5.7) and (5.10).