2.1 Standing assumptions

We first explain the standing assumptions on algebraic actions that we assume in this article. Let S be a non-trivial left cancellative monoid and A an Abelian group, written additively, with identity element 0. Assume $\sigma\colon S\curvearrowright A$ is an algebraic S-action, i.e. $\sigma\colon S\to {\rm End}\,_{\mathbb{Z}}(A), \, s \mapsto \sigma_s$ is a monoid homomorphism such that $\sigma_s$ is an injective endomorphism $A \to A$ for all $s\in S$ . Actions of this form are called algebraic actions (cf. [Reference Bruce and LiBL24]). Unless S is a group, we assume that $\sigma\colon S\curvearrowright A$ is non-automorphic, i.e. there exists $s\in S$ such that $\sigma_sA\subsetneq A$ . This, in particular, implies that A is non-trivial. We also always assume that the action $\sigma\colon S\curvearrowright A$ is faithful, i.e. $s\mapsto\sigma_s$ is injective.

Let $\mathcal C=\mathcal C_{S\curvearrowright A}$ be the family of S-constructible subgroups of A, i.e. $\mathcal C$ is the smallest collection of subgroups of A such that:

• – $A\in\mathcal C$ ; and

• – $s\in S$ and $C\in\mathcal C$ implies $\sigma_s C,\sigma_s^{-1}C\in\mathcal C$ .

It follows that $\mathcal C$ is closed under taking finite intersections (see [Reference Bruce and LiBL24, Proposition 3.9]. We say that $\sigma\colon S\curvearrowright A$ satisfies the finite-index property (see [Reference Bruce and LiBL24, Definition 7.1]) if

(FI) \begin{equation} \# (A/\sigma_sA) \lt\infty\quad \text{for all } s\in S.\end{equation}

If $\sigma\colon S\curvearrowright A$ satisfies (FI), then [Reference Bruce and LiBL24, Proposition 7.2] implies that every member of $\mathcal C$ is a finite-index subgroup of A. In this case, we get a compact group $\overline{A}:=\varprojlim_{C\in\mathcal C}A/C$ , and the canonical homomorphism $A\to\overline{A}$ has kernel $\bigcap_{C\in\mathcal C}C$ . Given $C \in \mathcal C$ , we denote by $\overline{C}$ the kernel of the canonical projection $\overline{A} \twoheadrightarrow A/C$ , i.e. the preimage of $C \in A/C$ in $\overline{A}$ . We have a canonical homeomorphism $\overline{C} \cong \varprojlim_{D \in \mathcal C, \, D \subseteq C} C/D$ .

In this article, we only consider algebraic actions $\sigma\colon S\curvearrowright A$ which satisfy (FI). Let $\partial\widehat{\mathcal E}$ be the compact space of characters on the semilattice $\mathcal E:=\{b+C : C\in\mathcal C, b\in A\}\cup\{\emptyset\}$ (see [Reference Bruce and LiBL24, § 3.4]). Each $\boldsymbol{x}=(x_C+C)_C\in\overline{A}$ determines an element $\chi_{\boldsymbol{x}}$ of $\partial\widehat{\mathcal E}$ by

\[\chi_{\boldsymbol{x}}(b+C):=\begin{cases}1 & \text{ if } x_C+C=b+C,\\0 & \text{ if } x_C+C\neq b+C.\end{cases} \]

Since (FI) is satisfied, it is not hard to see that the map $\overline{A}\to \partial\widehat{\mathcal E}$ given by $\boldsymbol{x}\mapsto \chi_{\boldsymbol{x}}$ is a homeomorphism.

In addition, we always assume that $\sigma\colon S\curvearrowright A$ has a globalization $\tilde{\sigma} \colon \mathscr{S} \curvearrowright \mathscr{A}$ , i.e. S embeds into the group $\mathscr{S}$ , A is a subgroup of the group $\mathscr{A}$ , and $\tilde{\sigma}\colon \mathscr{S}\to{\rm Aut}(\mathscr{A})$ is an algebraic action such that $\tilde{\sigma}_s\vert_A=\sigma_s$ for all $s\in S$ . Then $\mathscr{A}\rtimes \mathscr{S}$ acts on $\mathscr{A}$ by affine maps: $(z,\gamma).x=z+\tilde{\sigma}_\gamma(x)$ . Reducing to $A\subseteq \mathscr{A}$ , we get a partial action (in the sense of [Reference ExelExe98]) on the group A. Explicitly, $g \in \mathscr{A}\rtimes \mathscr{S}$ acts by the partial bijection

\[ A\cap g^{-1}.A\to (g.A)\cap A, \quad x\mapsto g.x. \]

Note that $s\in S$ acts via $\sigma_s$ (where we view both maps as partial bijections on A). Consider the condition

(JF) \begin{equation} C\subseteq {\rm ker}\,({\rm id}-\tilde{\sigma}_g)\implies g=1\text{ for all }C\in\mathcal C,\, g\in \langle S\rangle,\end{equation}

where $\langle S\rangle$ is the subgroup of $\mathscr{S}$ generated by S (cf. [Reference Bruce and LiBL24, § 3.3]). Moreover, our standing assumptions in this article include that (JF) is satisfied. In this case, the partial action $\mathscr{A}\rtimes \mathscr{S} \curvearrowright A$ extends uniquely to a partial action $\mathscr{A}\rtimes \mathscr{S} \curvearrowright \overline{A}$ . Given $g \in \mathscr{A}\rtimes \mathscr{S}$ , let $U_{g^{-1}}\subseteq \overline{A}$ be the domain of g, and for $\boldsymbol{x}\in U_{g^{-1}}$ , we let $g.\boldsymbol{x}$ denote the image of $\boldsymbol{x}$ under g with respect to the action $\mathscr{A}\rtimes \mathscr{S} \curvearrowright \overline{A}$ . The associated transformation groupoid

\[ (\mathscr{A}\rtimes \mathscr{S}) \ltimes\overline{A}:=\{(g,\boldsymbol{x})\in (\mathscr{A}\rtimes \mathscr{S}) \times\overline{A} :\boldsymbol{x}\in U_{g^{-1}}, \, g.\boldsymbol{x} \in \overline{A}\} \]

is canonically isomorphic to the groupoid $\mathcal G_{\sigma} = I_\sigma\ltimes\partial\widehat{\mathcal E}$ from [Reference Bruce and LiBL24, § 3] (see [Reference Bruce and LiBL24, § 3.5]). Since (FI) is satisfied, $(\mathscr{A}\rtimes \mathscr{S}) \ltimes\overline{A}$ is minimal by [Reference Bruce and LiBL24, Corollary 7.4]. By [Reference Bruce and LiBL24, Theorem 4.14], $(\mathscr{A}\rtimes \mathscr{S}) \ltimes\overline{A}$ is effective if and only if $\sigma\colon S\curvearrowright A$ is exact in the sense of [Reference Bruce and LiBL24, Definition 4.11], i.e. $\bigcap_{C\in\mathcal C}C=\{0\}$ .

However, note that (JF) holds automatically in this setting if A is torsion-free by [Reference Bruce and LiBL24, Proposition 7.5].

Remark 2.4. We can and do always assume that $\mathscr{S}$ is generated by S, i.e. $\mathscr{S} = \langle S\rangle$ . Moreover, by [Reference Bruce and LiBL24, Proposition 2.7], if S embeds into the group $\mathscr{S}$ , then we can always take $\mathscr{A} =\mathbb{Z} \mathscr{S} \otimes_{\mathbb{Z} S} A$ , and the map $A \to \mathbb{Z} \mathscr{S} \otimes_{\mathbb{Z} S} A, \, a \mapsto 1 \otimes a$ will always be injective if $\sigma$ admits a globalization. In this case, we have $\mathscr{A} = \langle \bigcup_{s \in \mathscr{S}} s.(1 \otimes A)\rangle$ . Hence, for any globalization $\tilde{\sigma}: \: \mathscr{S} \curvearrowright \mathscr{A}$ , we may and do always assume that

(2.1) \begin{equation} \mathscr{A} = \langle \textstyle{\bigcup_{s \in \mathscr{S}}} \tilde{\sigma}_s(A)\rangle.\end{equation}

If S is left Ore, then we can always take $\mathscr{S} = S^{-1} S$ , and in this case, we can and do always arrange that $\mathscr{A} = \bigcup_{r \in S} \tilde{\sigma}_r^{-1}(A)$ .

2.2 Further properties

Let us now discuss a few properties which are not part of our standing assumptions, but which we assume for some of our results.

The principal S-constructible subgroups are cofinal in $\mathcal C$ if

(PC) \begin{equation}\text{for every } C\in\mathcal C, \text{ there exists } s\in S \text{ such that } \sigma_sA\subseteq C.\end{equation}

Note that (PC) is satisfied if S is left reversible (see [Reference Bruce and LiBL24, Proposition 7.12]).

Consider the following condition on the algebraic action $\mathscr{S}\curvearrowright \mathscr{A}$ :

(F) \begin{equation}\text{for all } 1\neq s\in \mathscr{S},\, 1-\tilde{\sigma}_s:={\rm id}-\tilde{\sigma}_s\colon \mathscr{A} \to \mathscr{A} \text{ is injective.}\end{equation}

Condition (F) is a freeness condition, modulo the fact that in the linear setting 0 will always be a fixed point.

3.1 The additive homomorphism

Given a finitely generated subgroup $\mathfrak A$ of A, we now want to find $\mathfrak{l} \in \mathbb{Z}_{\gt0}$ and $C \in \mathcal C$ together with a homomorphism $\mathfrak{b} \colon \mathfrak C \mathrel{:=} \mathfrak{l} (\mathfrak A \cap C) \to \mathscr{B}$ such that $\mathfrak{c}(x) = (\mathfrak{b}(x),1)$ for all $x \in \mathfrak C$ .

Then, for every finitely generated subgroup $\mathfrak A$ of $\mathcal A$ , there exists $\check{C} \in \mathcal C$ depending on s such that the following hold.

1. (i) The restriction of $\mathfrak{c}$ to $\check{\mathfrak C} \mathrel{:=} \mathfrak A \cap \check{C}$ is an injective group homomorphism $\check{\mathfrak C} \to \mathscr{B}\rtimes\mathscr{T}$ .

2. (ii) We have $\mathfrak{c}(s)^d \mathfrak{c}(x)^{\kappa_d} = \mathfrak{c}(x)^{\kappa_0} \mathfrak{c}(s) \dotsm \mathfrak{c}(x)^{\kappa_{d-1}} \mathfrak{c}(s)$ for all $x \in \check{\mathfrak C}$ .

3. (iii) For all $x \in \check{\mathfrak C}$ , if $\mathfrak{c}(x) = (\beta, \alpha)$ and $\mathfrak{c}(s) = (\delta, \gamma)$ , then

   (3.1) \begin{equation}\gamma^d \alpha^{\kappa_d} = \alpha^{\kappa_0} \gamma\alpha^{\kappa_1} \gamma \dotsm \alpha^{\kappa_{d-1}} \gamma\end{equation}
   in $\mathscr{T}$ .

Proof. Let us denote the restriction of $\sigma_s$ to $\mathcal A$ again by $\sigma_s$ . For all $x \in \mathcal A$ , the following holds in $A \rtimes S$ as $\kappa_d \sigma_s^d(x) = \kappa_{d-1} \sigma_s^{d-1}(x) + \cdots + \kappa_0 x$ in $\mathcal A$ :

$$s^d (\kappa_d x) = \kappa_d \sigma_s^d(x) s^d = (\kappa_0 x) s (\kappa_1 x) s \dotsm (\kappa_{d-1}x) s.$$

Given a finitely generated subgroup $\mathfrak A$ of $\mathcal A$ , choose $C_{\mathfrak A}$ as in Lemma 3.4. Set

$$ \check{C} \mathrel{:=} C(s) \cap \sigma_s^{-1} C(s) \cap \cdots \cap \sigma_s^{-(d-1)} C(s) \cap C_{\mathfrak A} \cap \sigma_s^{-1} C_{\mathfrak A} \cap \cdots \cap \sigma_s^{-(d-1)} C_{\mathfrak A}.$$

Then part (i) is satisfied because of Lemma 3.4.

We have, for all $x \in \check{C}$ :

\begin{align*}\mathfrak{c}(s^d (\kappa_d x)) &= \mathfrak{c}(s^d (\kappa_d x), \boldsymbol{0}) = \mathfrak{c}(s^d,\kappa_d x)\mathfrak{c}(\kappa_d x,\boldsymbol{0})\\&= \mathfrak{c}(s^{d-1},\sigma_s(\kappa_d x)) \mathfrak{c}(s,\kappa_d x) \mathfrak{c}(\kappa_d x,\boldsymbol{0})\\&= \cdots = \mathfrak{c}(s, \sigma_s^{d-1}(\kappa_d x)) \mathfrak{c}(s,\sigma_s^{d-2}(\kappa_d x)) \dotsm \mathfrak{c}(s,\kappa_d x) \mathfrak{c}(\kappa_d x,\boldsymbol{0}) \\&= \mathfrak{c}(s,\boldsymbol{0}) \mathfrak{c}(s,\boldsymbol{0}) \dotsm \mathfrak{c}(s,\boldsymbol{0}) \mathfrak{c}(\kappa_d x,\boldsymbol{0}) = \mathfrak{c}(s)^d \mathfrak{c}(\kappa_d x).\end{align*}

Here, we are allowed to replace $\sigma_s^\ell(\kappa_d x)$ by $\boldsymbol{0}$ for $\ell=0,1,\ldots,d-1$ because x lies in $C(s) \cap \sigma_s^{-1} C(s) \cap \cdots \cap \sigma_s^{-(d-1)} C(s)$ .

We also have, for all $x \in \mathfrak A \cap \check{C}$ :

\begin{align*}& \ \mathfrak{c}((\kappa_0 x) s (\kappa_1 x) s \dotsm (\kappa_{d-1}x) s) = \mathfrak{c}((\kappa_0 x) s (\kappa_1 x) s \dotsm (\kappa_{d-1}x) s,\boldsymbol{0})\\= & \ \mathfrak{c}((\kappa_0 x) s (\kappa_1 x) s \dotsm (\kappa_{d-1}x),\boldsymbol{0}) \mathfrak{c}(s,\boldsymbol{0})\\= & \ \mathfrak{c}((\kappa_0 x) s (\kappa_1 x) s \dotsm, \kappa_{d-1}x) \mathfrak{c}(\kappa_{d-1}x,\boldsymbol{0}) \mathfrak{c}(s,\boldsymbol{0})\\= & \ \mathfrak{c}(\kappa_0 x, \sigma_s (\kappa_1 x) + \cdots + \sigma_s^{d-1}(\kappa_{d-1}x)) \mathfrak{c}(s,(\kappa_1 x) + \cdots + \sigma_s^{d-2}(\kappa_{d-1}x))\\& \ \dotsm \\& \ \mathfrak{c}(\kappa_{d-2}x, \sigma_s(\kappa_{d-1}x)) \mathfrak{c}(s, \kappa_{d-1}x)\\& \ \mathfrak{c}(\kappa_{d-1}x, \boldsymbol{0}) \mathfrak{c}(s, \boldsymbol{0})\\= & \ \mathfrak{c}(\kappa_0 x, \boldsymbol{0}) \mathfrak{c}(s, \boldsymbol{0}) \dotsm \mathfrak{c}(\kappa_{d-2}x, \boldsymbol{0}) \mathfrak{c}(s, \boldsymbol{0}) \mathfrak{c}(\kappa_{d-1}x, \boldsymbol{0}) \mathfrak{c}(s, \boldsymbol{0})\\= & \ \mathfrak{c}(\kappa_0 x) \mathfrak{c}(s) \dotsm \mathfrak{c}(\kappa_{d-1}x) \mathfrak{c}(s).\end{align*}

Here, we are allowed to replace the second arguments of $\mathfrak{c}$ by $\boldsymbol{0}$ because x lies in $C(s) \cap \cdots \cap \sigma_s^{-(d-1)} C(s) \cap \mathfrak A \cap C_{\mathfrak A} \cap \sigma_s^{-1} C_{\mathfrak A} \cap \cdots \cap \sigma_s^{-(d-1)} C_{\mathfrak A}$ .

Moreover, we have, for all $x \in \check{C}$ and $\kappa \in\mathbb{Z}$ , that $\mathfrak{c}(\kappa x) = \mathfrak{c}(x)^{\kappa}$ because x lies in $\mathfrak A \cap C_{\mathfrak A}$ . So all in all, we have $\mathfrak{c}(s)^d \mathfrak{c}(x)^{\kappa_d} = \mathfrak{c}(x)^{\kappa_0} \mathfrak{c}(s) \dotsm \mathfrak{c}(x)^{\kappa_{d-1}} \mathfrak{c}(s)$ for all $x \in \check{\mathfrak C} = \mathfrak A \cap \check{C}$ . This shows part (ii).

Now let us prove part (iii). Suppose that $\mathfrak{c}(x) = (\beta,\alpha)$ and $\mathfrak{c}(s) = (\delta,\gamma)$ . Comparing $\mathscr{T}$ -components, we obtain $\gamma^d \alpha^{\kappa_d} = \alpha^{\kappa_0} \gamma \alpha^{\kappa_1} \gamma \dotsm\alpha^{\kappa_{d-1}}\gamma.$

A group is 2-free if every subgroup generated by two elements is free. Given a non-empty set $\omega$ of prime numbers, a $D_\omega$ -free group is a group whose elements each have exactly one pth root for all primes $p\in\omega$ (see the introduction in [Reference BaumslagBau60]). By [Reference BaumslagBau68, § 8], every $D_\omega$ -free group from [Reference BaumslagBau60] is 2-free.

Let us introduce the following conditions.

Recall that $\tilde{\tau} \colon \mathscr{T} \curvearrowright\mathscr{B}$ is our globalization of $\tau\colon T\curvearrowright B$ .

1. (ii₁) For all torsion orders $l\gt1$ of elements of $\mathscr{T}$ , there exists $1 \neq s_l \in S$ and coprime integers $\mu,\nu\in\mathbb{Z}_{\gt0}$ such that $\sigma_{s_l}(\mathcal A) \subseteq \mathcal A$ and $\mu \det(z - \dot{\sigma}_{s_l}) = \mu z^{\delta} - \nu$ with $\gcd(l,\mu) \gt 1$ or $\gcd(l,\nu) \gt 1$ .

2. (ii₂) Condition (F) holds for $\tilde{\tau}$ .

Note that, in particular, condition (ii $_1$ ) holds if $\mathscr{T}$ is torsion-free or if, for all $\kappa \in \mathbb{Z}_{\gt0}$ , there exists $s_{\kappa} \in S$ with $\sigma_{s_{\kappa}} = \kappa \, {\rm id}_A$ .

• – condition (i $_1$ ) or (i $_2$ ) holds, and condition (ii $_1$ ) is satisfied or condition (ii $_2$ ) holds and A is torsion-free;

• – condition (III) holds and A is torsion-free.

Then, for all finitely generated subgroups $\mathfrak A \subseteq \mathcal A$ , there exists $C \in \mathcal C$ such that, with $\mathfrak C \mathrel{:=} \mathfrak A \cap C$ , there exists an injective homomorphism $\mathfrak{b}: \: \mathfrak C \to \mathscr{B}$ such that $\mathfrak{c}(x) = (\mathfrak{b}(x),1)$ for all $x \in \mathfrak C$ .

Proof. To prove the first item, let $s \in S$ be as in condition (i $_1$ ) or condition (i $_2$ ), and $\check{C}$ , $\check{\mathfrak C}$ as in Proposition 3.5. First assume that condition (ii $_1$ ) holds. Let $\mathfrak{c}_{\mathscr{T}}$ be the composition $A \to \mathscr{B}\rtimes\mathscr{T} \twoheadrightarrow \mathscr{T}$ , where the first map is $\mathfrak{c}$ and the second map is the canonical projection. As $\mathfrak{c}_{\mathscr{T}}(\check{\mathfrak C})$ is finitely generated, the set $\mathbb{L}$ of possible non-trivial torsion orders of elements in $\mathfrak{c}_{\mathscr{T}}(\check{\mathfrak C})$ is finite. For each $l \in \mathbb{L}$ , choose $s_l$ as in condition (ii $_1$ ) and set $C \mathrel{:=} \check{C} \cap \bigcap_{l \in \mathcal L} \check{C}(s_l)$ . Applying Proposition 3.5 to $s_l$ , we obtain that, for all $x \in \mathfrak A \cap C$ with $\mathfrak{c}(x) = (\beta,\alpha)$ , we have $\alpha^{\mu}$ and $\alpha^{\nu}$ conjugate in $\mathscr{T}$ , where $\mu$ and $\nu$ are the natural numbers from condition (ii $_1$ ). If $\alpha \neq 1$ , then the torsion order of $\alpha$ is a number $l \in \mathbb{L}$ , but condition (ii $_1$ ) implies that $\alpha^{\mu}$ and $\alpha^{\nu}$ have different torsion orders, which is absurd. Now suppose that condition (ii $_2$ ) holds. Take $x \in \check{\mathfrak C}$ and write $\mathfrak{c}(x) = (\beta,\alpha)$ . Lemma 3.7 implies that $\alpha^{\epsilon} = 1$ . Then

\[(\beta,\alpha)^{\epsilon} = (\beta + \tilde{\tau}_\alpha(\beta) + \cdots + \tilde{\tau}_\alpha^{\epsilon-1}(\beta),\alpha^{\epsilon}) = (\beta + \tilde{\tau}_\alpha(\beta) + \cdots + \tilde{\tau}_\alpha^{\epsilon-1}(\beta),1)\]

and

\[(1 - \tilde{\tau}_\alpha)(\beta + \tilde{\tau}_\alpha(\beta) + \cdots + \tilde{\tau}_\alpha^{\epsilon-1}(\beta)) = 0,\]

which implies $\beta + \tilde{\tau}_\alpha(\beta) + \cdots + \tilde{\tau}_\alpha^{\epsilon-1}(\beta) = 0$ if $\alpha \neq 1$ by (F). So $\mathfrak{c}(x)^{\epsilon} = 1$ and hence $\mathfrak{c}(\epsilon x) = 1$ and thus $\epsilon x = 0$ . As A is torsion-free, this implies $x = 0$ . So if $x \neq 0$ , we must have $\alpha = 1$ .

Now we prove the second item. As above, let $s \in S$ be as in condition (III), and $\check{C}$ , $\check{\mathfrak C}$ as in Proposition 3.5. Given $x \in \check{\mathfrak C}$ , write $\mathfrak{c}(x) = (\beta,\alpha)$ and $\mathfrak{c}(s) = (\delta,\gamma)$ . Equation (3.1) implies that $\gamma \alpha = \alpha^{\kappa} \gamma$ , and therefore $\alpha = \gamma^{-1} \alpha^{\kappa} \gamma$ . It follows that every eigenvalue of $\dot{\tau}_\alpha \mathrel{:=} {\rm id}_{\mathbb{Q}}\otimes \tau_\alpha$ is a root of unity. Indeed, take $\lambda_1 \in {\rm Sp\,}(\dot{\tau}_\alpha)$ . Then $\alpha = \gamma^{-1} \alpha^{\kappa} \gamma$ implies that there exists $\lambda_2 \in {\rm Sp\,}(\dot{\tau}_\alpha)$ with $\lambda_1 = \lambda_2^{\kappa}$ . Similarly, there exist $\lambda_3, \lambda_4, \dotsc \in {\rm Sp\,}(\dot{\tau}_\alpha)$ such that $\lambda_i = \lambda_{i+1}^{\kappa}$ . As ${\rm Sp\,}(\dot{\tau}_\alpha)$ is finite, we must have $\lambda_i = \lambda_{i+p}$ for some i and p. It follows that $\lambda_i = \lambda_i^{\kappa^p}$ and hence that $\lambda_i$ is a root of unity. But then $\lambda_1 = \lambda_i^{\kappa^i}$ implies that $\lambda_1$ must be a root of unity as well. Hence, there exists $m\in\mathbb{Z}_{\gt0}$ such that 1 is an eigenvalue of $\dot{\tau}_\alpha^m$ . This implies that $1-\dot{\tau}_\alpha^m$ is not injective, so that $\dot{\tau}_\alpha^m=1$ by (F). Now argue as for the first item that this, together with (F), implies $\alpha = 1$ .

1. (I) Condition (i $_1$ ) holds for $\mathcal A_n$ for all n, or condition (i $_2$ ) holds.

2. (II) Condition (ii $_1$ ) holds for $\mathcal A_n$ for all n, or condition (ii $_2$ ) holds and A is torsion-free.

If condition (I) holds, then:

(a) there is an increasing family of finitely generated subgroups $\mathfrak A_k\subseteq A$ with $A=\bigcup_k\mathfrak A_k$ , and for any such $\mathfrak A_k$ , there are $\mathfrak{l}_k \in \mathbb{Z}_{\gt0}$ and $C_k \in \mathcal C$ such that, with $\mathfrak C_k \mathrel{:=} \mathfrak{l}_k (\mathfrak A_k \cap C_k)$ , there exists an injective homomorphism $\mathfrak{b}_k: \: \mathfrak C_k \to \mathscr{B}$ such that $\mathfrak{c}(x) = (\mathfrak{b}_k(x),1)$ for all $x \in \mathfrak C_k$ .

If conditions (I) and (II) hold, or if condition (III) is satisfied and A is torsion-free, then:

(a*) there is an increasing family of finitely generated subgroups $\mathfrak A_k\subseteq A$ with $A=\bigcup_k\mathfrak A_k$ , and for any such $\mathfrak A_k$ , there is $C_k \in \mathcal C$ such that, with $\mathfrak C_k \mathrel{:=} \mathfrak A_k \cap C_k$ , there exists an injective homomorphism $\mathfrak{b}_k: \: \mathfrak C_k \to \mathscr{B}$ such that $\mathfrak{c}(x) = (\mathfrak{b}_k(x),1)$ for all $x \in \mathfrak C_k$ .

3.2 The multiplicative homomorphism

Define $\mathfrak{t} \colon \mathscr{S} \to \mathscr{T}$ to be the composition $\mathscr{S} \to \mathscr{B} \rtimes \mathscr{T} \twoheadrightarrow \mathscr{T}$ , where the first arrow is given by $\mathfrak{c}(-,\boldsymbol{0})$ and the second arrow is the canonical projection. Clearly, $\mathfrak{t}$ is a homomorphism.

Proof. Suppose $\mathfrak{t}(s) = \mathfrak{t}(s')$ . Then $\mathfrak{c}(s^{-1}s', \boldsymbol{0}) = (\chi,\mathfrak{t}(s^{-1}s')) =(\chi,1)$ for some $\chi \in \mathscr{B}$ . Set $\varepsilon\mathrel{:=} s^{-1}s'$ . Since 0 lies in the domain of $\tilde{\sigma}_\varepsilon$ , and since $\mathfrak{c}(\varepsilon,-)$ is locally constant, there must exist $C(\varepsilon) \in \mathcal C$ such that $C(\varepsilon)$ is contained in the domain of $\tilde{\sigma}_\varepsilon$ and $\mathfrak{c}(\varepsilon,-)$ is constant on $\overline{C(\varepsilon)}$ . Now suppose that $\mathfrak C \subseteq A$ is a non-zero subgroup such that there exists an injective homomorphism $\mathfrak{b}: \: \mathfrak C \to \mathscr{B}$ with $\mathfrak{c}(x) = (\mathfrak{b}(x),1)$ for all $x \in \mathfrak C$ . Then $\mathfrak{c}(\varepsilon,\boldsymbol{0})$ commutes with $\mathfrak{c}(x,\boldsymbol{0})$ for all $x \in \mathfrak C$ . We have $(0,\varepsilon)(x,1) = (\tilde{\sigma}_\epsilon(x),\epsilon) = (\tilde{\sigma}_\varepsilon(x),1)(0,\varepsilon)$ . So if $0 \neq x \in \mathfrak C \cap C(\varepsilon)$ ,

\[ \mathfrak{c}((\tilde{\sigma}_\varepsilon(x),\varepsilon),\boldsymbol{0}) = \mathfrak{c}(\tilde{\sigma}_\varepsilon(x) \varepsilon,\boldsymbol{0}) = \mathfrak{c}(\tilde{\sigma}_\varepsilon(x),\boldsymbol{0}) \mathfrak{c}(\varepsilon,\boldsymbol{0}).\]

At the same time,

\[ \mathfrak{c}((\tilde{\sigma}_\varepsilon(x),\epsilon),\boldsymbol{0}) = \mathfrak{c}(\varepsilon x,\boldsymbol{0}) = \mathfrak{c}(\varepsilon,x) \mathfrak{c}(x,\boldsymbol{0}) = \mathfrak{c}(\varepsilon,\boldsymbol{0}) \mathfrak{c}(x,\boldsymbol{0}) = \mathfrak{c}(x,\boldsymbol{0}) \mathfrak{c}(\varepsilon,\boldsymbol{0}).\]

Here we are allowed to replace x by $\boldsymbol{0}$ because x lies in $C(\varepsilon)$ . Moreover, we used that $c(\varepsilon,\boldsymbol{0})$ commutes with $c(x,\boldsymbol{0})$ . Therefore, a comparison yields $\mathfrak{c}(\tilde{\sigma}_\varepsilon(x),\boldsymbol{0}) = \mathfrak{c}(x,\boldsymbol{0})$ . It follows that $\tilde{\sigma}_\varepsilon(x) = x$ .

For part (i), write $A = \bigcup_k \mathfrak A_k$ as in condition (a). Applying the above to $\mathfrak C = \mathfrak C_k$ from condition (a), we obtain $\tilde{\sigma}_\varepsilon(x) = x$ for all $x \in \mathfrak C_k \cap C(\varepsilon)$ for all k. Now let $y\in A$ ; then there exists k with $y\in \mathfrak A_k$ . Since $\mathfrak C_k \cap C(\varepsilon)$ is finite index in $\mathfrak A_k$ , there exists $N\in\mathbb{Z}_{\gt0}$ such that $Ny\in \mathfrak C_k \cap C(\varepsilon)$ . We have $N \tilde{\sigma}_\varepsilon(y)=\tilde{\sigma}_\varepsilon(Ny)=Ny$ , so that, because A is torsion-free, $\tilde{\sigma}_\varepsilon(y)=y$ . Now (JF) implies $\varepsilon = 1$ .

For part (ii), if $\tilde{\sigma}_\varepsilon(x) = x$ for any $x \neq 0$ , then condition (F) implies that $\varepsilon = 1$ .

3.3 Equivariance

Proof. Since $\mathfrak C_l$ is of finite index in $\mathfrak A_l$ (see Remark 3.17), there exists $n \in \mathbb{Z}_{\gt0}$ such that $n \sigma_s(x) = \sigma_s(nx) \in \mathfrak C_l$ (since l depends on s, n also depends on s). Since $\mathfrak C_k \cap C(s)$ is of finite index in $\mathfrak C_k$ , we can find N such that $Nnx \in \mathfrak C_k \cap C(s)$ . Set $y \mathrel{:=} Nnx$ . Let $\mathfrak{c}(s,\boldsymbol{0}) = (\delta,\mathfrak{t}(s))$ for some $\delta\in \mathscr{B}$ . We have

\begin{align*}& \ (\mathfrak{b}(\sigma_s(y)) + \delta,\mathfrak{t}(s)) = (\mathfrak{b}(\sigma_s(y)),1) (\delta,\mathfrak{t}(s)) = \mathfrak{c}(\sigma_s(y),\boldsymbol{0}) \mathfrak{c}(s,\boldsymbol{0})\\= & \ \mathfrak{c}((\sigma_s(y),s),\boldsymbol{0}) = \mathfrak{c}((0,s)(y,0),\boldsymbol{0}) = \mathfrak{c}((0,s),y)c((y,1),\boldsymbol{0})\\= & \ \mathfrak{c}((0,s),\boldsymbol{0}) \mathfrak{c}((y,1),\boldsymbol{0}) = (\delta,\mathfrak{t}(s)) (\mathfrak{b}(y),1) = (\delta + \tilde{\tau}_{\mathfrak{t}(s)}(\mathfrak{b}(y)), \mathfrak{t}(s)),\end{align*}

where the sixth equality uses that $y \in C(s)$ . Hence, $\mathfrak{b}(\sigma_s(y)) = \tilde{\tau}_{\mathfrak{t}(s)}(\mathfrak{b}(y))$ , i.e. $N \mathfrak{b}(\sigma_s(nx)) = N \tilde{\tau}_{\mathfrak{t}(s)}(\mathfrak{b}(nx))$ . We obtain $\mathfrak{b}(\sigma_s(nx)) = \tilde{\tau}_{\mathfrak{t}(s)}(\mathfrak{b}(nx))$ as $\mathscr{B}$ is torsion-free, as desired.

3.4 Conclusion: The embedding theorem

Assume that $\sigma \colon S \curvearrowright A$ and $\tau \colon T \curvearrowright B$ are algebraic actions satisfying our standing assumptions from § 2. Let $\mathbb{Z}[\mathbb{I}^{-1}]$ be the subring of $\mathbb{Q}$ generated by $\mathbb{Z}$ together with $\{{1}/{n}: n \in \mathbb{I}\}$ . We start with the following observation.

Recall that we always assume (2.1), i.e. $\mathscr{A} = \langle \bigcup_{s \in \mathscr{S}} \tilde{\sigma}_s(A)\rangle$ . Otherwise, Lemma 3.21 would not be true in general.

Now suppose $\mathfrak{c} \colon (\mathscr{A} \rtimes \mathscr{S}) \ltimes \overline{A} \to \mathscr{B} \rtimes \mathscr{T}$ is a continuous cocycle such that $\mathfrak{c}^{-1}(0,1)=\overline{A}$ .

1. (i) If (I) holds, then there exist injective homomorphisms $\mathfrak{t}: \: \mathscr{S} \hookrightarrow \mathscr{T}$ and $\dot{\mathfrak{b}} \colon \mathbb{Q} \otimes \mathscr{A} \hookrightarrow \mathbb{Q} \otimes \mathscr{B}$ such that $\dot{\mathfrak{b}}(\dot{\sigma}_s(x)) = \dot{\tau}_{\mathfrak{t}(s)}(\dot{\mathfrak{b}}(x))$ for all $s\in\mathscr{S}$ and $x \in \mathscr{A}$ .

2. (ii) If ( $*$ ) holds, then there exist injective homomorphisms $\mathfrak{t}: \: \mathscr{S} \hookrightarrow \mathscr{T}$ and $\mathfrak{b}' \colon S^{-1}A\to T^{-1}B$ such that $\mathfrak{b}'(\sigma_s(x))=\tilde{\tau}_{\mathfrak{t}(s)}(\mathfrak{b}'(x))$ for all $x\in S^{-1}A$ and $s\in \mathscr{S}$ .

Proof. (i) We proceed in several steps. First, we claim that for each k, $\mathfrak{b}_k\colon \mathfrak C_k\to\mathscr{B}$ has a unique extension to an injective homomorphism $\tilde{\mathfrak{b}}_k\colon\mathfrak A_k\to \mathbb{Q} \otimes \mathscr{B}$ , and that, moreover, $\tilde{\mathfrak{b}}_k$ satisfies

(3.2) \begin{equation} \tilde{\mathfrak{b}}_k(x)=m^{-1}\mathfrak{b}_k(mx),\end{equation}

for any $m \in\mathbb{Z}_{\gt0}$ with $m\mathfrak A_k\subseteq \mathfrak C_k$ . To see this, choose $m\in\mathbb{Z}_{\gt0}$ such that $m\mathfrak A_k\subseteq \mathfrak C_k$ , and define $\tilde{\mathfrak{b}}_k(x):=m^{-1}\mathfrak{b}_k(mx)\in\mathbb{Q}\otimes\mathscr{B}$ . It is easy to check that $\tilde{\mathfrak{b}}_k$ is a group homomorphism (i.e. additive) and that $\tilde{\mathfrak{b}}_k$ is injective (here we need that $\mathfrak A_k$ is torsion-free). This is independent of the choice of m, because if $m'\in\mathbb{Z}_{\gt0}$ with $m'\mathfrak A_k \subseteq\mathfrak C_k$ , then for $x\in\mathfrak A_k$ , we have

\[m^{-1} \mathfrak{b}_k(mx) = m^{-1} (m')^{-1} \mathfrak{b}_k(m'mx) = (m')^{-1} m^{-1} m \mathfrak{b}_k(m'x) = (m')^{-1} \mathfrak{b}_k(m'x). \]

Next, we claim that the maps $\tilde{\mathfrak{b}}_k$ are compatible in the sense that $\tilde{\mathfrak{b}}_l\vert_{\mathfrak A_k\cap\mathfrak A_l}=\tilde{\mathfrak{b}}_k\vert_{\mathfrak A_k\cap\mathfrak A_l}$ for all $k,l\in\mathbb{Z}_{\gt0}$ . Choose m large enough so that $m\mathfrak A_k\subseteq \mathfrak C_k$ and $m\mathfrak A_l\subseteq \mathfrak C_l$ . Then, using (3.2), we have, for all $x\in \mathfrak A_k\cap\mathfrak A_l$ , that $\tilde{\mathfrak{b}}_l(x)=m^{-1}\mathfrak{b}_l(mx)=m^{-1}\mathfrak{b}_k(mx)=\tilde{\mathfrak{b}}_k(x)$ . It follows that we get a well-defined injective homomorphism $\tilde{\mathfrak{b}}\colon A=\bigcup_k\mathfrak A_k\to \mathbb{Q} \otimes \mathscr{B}$ such that $\tilde{\mathfrak{b}}\vert_{\mathfrak A_k}=\tilde{\mathfrak{b}}_k$ for all k.

Let us show that $\tilde{\mathfrak{b}}$ is equivariant. Let $x\in A$ and $s\in S$ . Then $x\in \mathfrak A_k$ for some k, so by Lemma 3.20, we can find $n\in\mathbb{Z}_{\gt0}$ large enough so that $nx\in\mathfrak C_k$ , $n\sigma_s(x)=\sigma_s(nx)\in\mathfrak C_l$ , and $\mathfrak{b}(\sigma_s(nx))= \tilde{\tau}_{\mathfrak{t}(s)}(\mathfrak{b}(nx))$ . Now we have

\[\tilde{\mathfrak{b}}(\sigma_s(x))=n^{-1}\mathfrak{b}(\sigma_s(nx))=n^{-1}\tilde{\tau}_{\mathfrak{t}(s)}(\mathfrak{b}(nx))=\dot{\tau}_{\mathfrak{t}(s)}(n^{-1}\mathfrak{b}(nx))=\dot{\tau}_{\mathfrak{t}(s)}(\tilde{\mathfrak{b}}(x)). \]

Lastly, we extend $\tilde{\mathfrak{b}}$ to $\mathbb{Q} \otimes A$ as follows. Given ${p}/{q} \otimes x \in \mathbb{Q} \otimes A$ for $p \in \mathbb{Z}$ and $q \in \mathbb{Z}^\times$ , there exists a unique element $y \in \mathscr{B}$ such that $qy = \tilde{\mathfrak{b}}(px)$ in $\mathscr{B}$ , and we set $\dot{\mathfrak{b}}({p}/{q} \otimes x) \mathrel{:=} y$ . Now it is straightforward to check that this is independent of p, q, and x, and that $\dot{\mathfrak{b}}$ is an injective homomorphism $\mathbb{Q} \otimes \mathscr{A} \hookrightarrow \mathbb{Q} \otimes \mathscr{B}$ . Moreover, equivariance of $\tilde{\mathfrak{b}}$ with respect to $\tilde{\sigma}$ and $\dot{\tau}$ implies equivariance of $\dot{\mathfrak{b}}$ with respect to $\dot{\sigma}$ and $\dot{\tau}$ .

(ii) First of all, every element of $S^{-1} A$ is of the form $s^{-1}a = \tilde{\sigma}_s^{-1} a$ for some $a \in \mathfrak C_k$ (for some k). Indeed, the statement is clear if we just ask for $a \in \mathfrak A_k$ for some k because of Remark 2.4. By (PC), there exists $\dot{s} \in S$ such that $\sigma_{\dot{s}}(a) \in C_k$ , so that $\sigma_{\dot{s}}(a) \in \mathfrak C_k$ because $\sigma_{\dot{s}}(\mathfrak A_k) \subseteq \mathfrak A_k$ , and we have $\tilde{\sigma}_s^{-1} a = \tilde{\sigma}_s^{-1} \tilde{\sigma}_{\dot{s}}^{-1} (\tilde{\sigma}_{\dot{s}} a)$ , as desired.

Now, given an element of $S^{-1} A$ of the form $\tilde{\sigma}_s^{-1} a$ for some $a \in \mathfrak C_k$ (for some k), we claim that $\mathfrak{b}'(\tilde{\sigma}_s^{-1} a) \mathrel{:=} \tilde{\tau}_{\mathfrak{t}(s)}^{-1}(\mathfrak{b}(a))$ is well-defined and has the desired properties. To prove that it is well-defined, assume that $\tilde{\sigma}_s^{-1} a = \tilde{\sigma}_t^{-1} \dot{a}$ . Since S is right reversible, there exist $u, v \in S$ with $us = vt$ . Hence, $\tilde{\sigma}_s^{-1} \tilde{\sigma}_u^{-1} \tilde{\sigma}_u a = \tilde{\sigma}_s^{-1} a = \tilde{\sigma}_t^{-1} \dot{a} = \tilde{\sigma}_t^{-1} \tilde{\sigma}_v^{-1} \tilde{\sigma}_v \dot{a}$ . It follows that $\tilde{\sigma}_u a = \tilde{\sigma}_v \dot{a}$ . Now choose an integer m such that $m \sigma_u(a), m \sigma_v(a) \in \mathfrak C_k$ and equivariance holds (here, we use Lemma 3.20). Then we have $m \tilde{\tau}_{\mathfrak{t}(s)}^{-1} \mathfrak{b}(a) = \tilde{\tau}_{\mathfrak{t}(s)}^{-1} \mathfrak{b}(ma) = \tilde{\tau}_{\mathfrak{t}(s)}^{-1} \tilde{\tau}_{\mathfrak{t}(u)}^{-1} \mathfrak{b}(\sigma_u(ma)) = \tilde{\tau}_{\mathfrak{t}(t)}^{-1} \tilde{\tau}_{\mathfrak{t}(v)}^{-1} \mathfrak{b}(\sigma_v(m \dot{a})) = \tilde{\tau}_{\mathfrak{t}(t)}^{-1} \mathfrak{b}(m \dot{a}) = m \tilde{\tau}_{\mathfrak{t}(t)}^{-1} \mathfrak{b}(\dot{a})$ . As $\mathscr{B}$ is torsion-free, we conclude that $\tilde{\tau}_{\mathfrak{t}(s)}^{-1} \mathfrak{b}(a) = \tilde{\tau}_{\mathfrak{t}(t)}^{-1} \mathfrak{b}(\dot{a})$ , as desired. It is easy to see that $\mathfrak{b}'$ is additive.

To show equivariance, take $r, s \in S$ . Since S is right reversible, there exist $u, v \in S$ with $ur = vs$ and thus $r s^{-1} = u^{-1} v$ . Given $a \in \mathfrak C_k$ , choose an integer m such that $\tilde{\sigma}_v (ma) \in \mathfrak C_k$ and equivariance holds (here, we use Lemma 3.20). Then we have $m \mathfrak{b}'(\tilde{\sigma}_r \tilde{\sigma}_s^{-1} (a)) = \mathfrak{b}'(\tilde{\sigma}_u^{-1} \tilde{\sigma}_v (ma)) = \tilde{\tau}_{\mathfrak{t}(u)}^{-1} \mathfrak{b}(\tilde{\sigma}_v (ma)) = \tilde{\tau}_{\mathfrak{t}(u)}^{-1} \tilde{\tau}_{\mathfrak{t}(v)} \mathfrak{b}(ma) = m \tilde{\tau}_{\mathfrak{t}(r)} \tilde{\tau}_{\mathfrak{t}(s)}^{-1} \mathfrak{b}(a)$ . As $\mathscr{B}$ is torsion-free, we deduce $\mathfrak{b}'(\tilde{\sigma}_r \tilde{\sigma}_s^{-1} (a)) = \tilde{\tau}_{\mathfrak{t}(r)} (\tilde{\tau}_{\mathfrak{t}(s)}^{-1} \mathfrak{b}(a)) = \tilde{\tau}_{\mathfrak{t}(r)} (\mathfrak{b}'(\tilde{\sigma}_s^{-1} a))$ , as desired.

3.5 Consequences

Let us formulate symmetrized versions of our rigidity results.

Schmidt defined finite (algebraic) equivalence for algebraic actions of a fixed group (see e.g. [Reference SchmidtSch02, Definition 8.1]). The dual version of Schmidt’s notion will appear naturally in our setting. In order to explain this, let us introduce some terminology.

We say that $\tilde{\sigma}\colon\mathscr{S}\curvearrowright \mathscr{A}$ and $\tilde{\tau}\colon\mathscr{T} \curvearrowright \mathscr{B}$ are strictly mutually embeddable (and we write $\tilde{\sigma}\colon\mathscr{S}\curvearrowright \mathscr{A} \sim_{s \mathcal M \mathcal E} \tilde{\tau}\colon\mathscr{T} \curvearrowright \mathscr{B}$ ) if each can be strictly embedded into the other. If, in addition, the strict embeddings can be chosen to be of finite index, then we write $\tilde{\sigma}\colon\mathscr{S}\curvearrowright \mathscr{A} \sim_{s \mathcal M \mathcal E_{FI}} \tilde{\tau}\colon\mathscr{T} \curvearrowright \mathscr{B}$ . Given an Abelian group $\mathcal Q$ , we write $\tilde{\sigma}\colon\mathscr{S}\curvearrowright \mathscr{A} \sim_{s \mathcal M \mathcal E_{\mathcal Q}} \tilde{\tau}\colon\mathscr{T} \curvearrowright \mathscr{B}$ if $\dot{\sigma}\colon\mathscr{S}\curvearrowright \mathcal Q \otimes \mathscr{A} \sim_{s \mathcal M \mathcal E} \dot{\tau}\colon\mathscr{T} \curvearrowright \mathcal Q \otimes \mathscr{B}$ .

We write $\tilde{\sigma}\colon\mathscr{S}\curvearrowright \mathscr{A} \cong_{\mathcal Q} \tilde{\tau}\colon\mathscr{T} \curvearrowright \mathscr{B}$ , and call $\tilde{\sigma}\colon\mathscr{S}\curvearrowright \mathscr{A}$ and $\tilde{\tau}\colon\mathscr{T} \curvearrowright \mathscr{B}$ isomorphic over $\mathcal Q$ , if there exists a full and strict embedding of $\dot{\sigma}\colon\mathscr{S}\curvearrowright \mathcal Q \otimes \mathscr{A}$ into $\dot{\tau}\colon\mathscr{T} \curvearrowright \mathcal Q \otimes \mathscr{B}$ .

(iii) We say that the algebraic actions $\sigma\colon S\curvearrowright A$ and $\tau\colon T \curvearrowright B$ are isomorphic if there is a pair $(\mathfrak{t},\mathfrak{b})$ , where $\mathfrak{t}\colon S\to T$ is an isomorphism of semigroups and $\mathfrak{b}\colon A\to B$ is a group isomorphism, such that $\mathfrak{b}(\sigma_s(x))=\tau_{\mathfrak{t}(s)}(\mathfrak{b}(x))$ for all $s\in S$ and $x\in A$ .

If $(\mathfrak{t},\mathfrak{b})$ is a finite-index embedding of $\tilde{\sigma}\colon\mathscr{S}\curvearrowright \mathscr{A}$ into $\tilde{\tau}\colon\mathscr{T} \curvearrowright \mathscr{B}$ and $\mathscr{A}$ and $\mathscr{B}$ are torsion-free, then $\mathscr{A}$ and $\mathscr{B}$ are quasi-isomorphic in the sense of [Reference ThomasTho03, Definition 3.3].

If $\tilde{\sigma}\colon\mathscr{S}\curvearrowright \mathscr{A} \sim_{s \mathcal M \mathcal E_{FI}} \tilde{\tau}\colon\mathscr{T} \curvearrowright \mathscr{B}$ , then we also call $\tilde{\sigma}\colon\mathscr{S}\curvearrowright \mathscr{A}$ and $\tilde{\tau}\colon\mathscr{T} \curvearrowright \mathscr{B}$ finitely algebraically equivalent (cf. [Reference SchmidtSch02, Definition 8.1]).

We are mostly interested in our notions involving an Abelian group $\mathcal Q$ when $\mathcal Q = \mathbb{Q}$ .

If $\mathscr{A}$ and $\mathscr{B}$ are torsion-free, then $\sim_{\mathcal M \mathcal E_{FI}}$ implies $\sim_{\mathcal M \mathcal E_{\mathbb{Q}^{\cong}}}$ and $\sim_{s \mathcal M \mathcal E_{FI}}$ implies $\cong_{\mathbb{Q}}$ .

If $\mathscr{A}$ and $\mathscr{B}$ are torsion-free and of finite rank, then $\sim_{\mathcal M \mathcal E}$ implies $\sim_{\mathcal M\mathcal E_{FI}}$ and $\sim_{s \mathcal M \mathcal E}$ implies $\sim_{s \mathcal M \mathcal E_{FI}}$ (see e.g. [Reference FuchsFuc73, Exercise 5 in § 92]).

Definition 3.28. Suppose $\mathscr{A}$ is torsion-free and of finite rank. We say that $\tilde{\sigma}\colon\mathscr{S}\curvearrowright\mathscr{A}$ is strongly faithful if

(SF) \begin{equation} \dot{\sigma}_s=\rho\dot{\sigma}_t\rho^{-1} \text{ implies }s=t \text{ for all }s,t\in\mathscr{S}\text{ and } \rho\in{\rm Aut}(\mathbb{Q}\otimes\mathscr{A}),\end{equation}

where $\dot{\sigma}_s:={\rm id}_{\mathbb{Q}}\otimes\tilde{\sigma}_s$ .

We are now ready for the main result of this section.

1. (i) If condition (I $_s$ ) holds and there exists an isomorphism of topological groupoids

   \[ (\mathscr{A} \rtimes \mathscr{S}) \ltimes \overline{A} \cong (\mathscr{B} \rtimes \mathscr{T}) \ltimes \overline{B}, \]
   then $\tilde{\sigma}\colon\mathscr{S}\curvearrowright \mathscr{A} \sim_{\mathcal M \mathcal E_{\mathbb{Q}}} \tilde{\tau}\colon\mathscr{T} \curvearrowright \mathscr{B}$ .

2. (ii) If condition ( $*_s$ ) holds and there exists an isomorphism of topological groupoids

   \[ (S^{-1}A\rtimes S^{-1}S)\ltimes \overline{A}\cong (T^{-1}B\rtimes T^{-1} T)\ltimes\overline{B}, \]
   then $\tilde{\sigma}\colon S^{-1}S \curvearrowright S^{-1}A \sim_{\mathcal M \mathcal E} \tilde{\tau}\colon T^{-1}T \curvearrowright T^{-1}B$ .

If, in addition, $\mathscr{A}$ and $\mathscr{B}$ have finite rank, then we obtain $\tilde{\sigma}\colon\mathscr{S}\curvearrowright \mathscr{A} \sim_{\mathcal M \mathcal E_{\mathbb{Q}^{\cong}}} \tilde{\tau}\colon\mathscr{T} \curvearrowright \mathscr{B}$ in part (i) and $\tilde{\sigma}\colon S^{-1}S \curvearrowright S^{-1}A \sim_{\mathcal M \mathcal E_{FI}} \tilde{\tau}\colon T^{-1}T \curvearrowright T^{-1}B$ in part (ii).

If, in addition, $\mathscr{A}$ and $\mathscr{B}$ have finite rank and $\tilde{\sigma}\colon\mathscr{S}\curvearrowright\mathscr{A}$ , $\tilde{\tau}\colon\mathscr{T}\curvearrowright\mathscr{B}$ both satisfy (SF), then we obtain $\tilde{\sigma}\colon\mathscr{S}\curvearrowright \mathscr{A} \cong_{\mathbb{Q}} \tilde{\tau}\colon\mathscr{T} \curvearrowright \mathscr{B}$ in part (i) and $\tilde{\sigma}\colon S^{-1}S \curvearrowright S^{-1}A \sim_{s \mathcal M \mathcal E_{FI}} \tilde{\tau}\colon T^{-1}T \curvearrowright T^{-1}B$ (i.e. $\tilde{\sigma}\colon S^{-1}S \curvearrowright S^{-1}A$ and $\tilde{\tau}\colon T^{-1}T \curvearrowright T^{-1}B$ are finitely algebraically equivalent) in part (ii).

1. (i) $(\mathscr{A} \rtimes \mathscr{S}) \ltimes \overline{A}$ and $(\mathscr{B} \rtimes \mathscr{T}) \ltimes \overline{B}$ are isomorphic as topological groupoids;

2. (ii) $(\mathfrak A_{\sigma},\mathfrak D_{\sigma})$ and $(\mathfrak A_{\tau},\mathfrak D_{\tau})$ are isomorphic as Cartan pairs, where $\mathfrak A_{\sigma}$ and $\mathfrak A_{\tau}$ are as in [Reference Bruce and LiBL24, Definition 3.1], and $\mathfrak D_{\sigma}$ and $\mathfrak D_{\tau}$ are as in [Reference Bruce and LiBL24, Proposition 3.30];

3. (iii) $\boldsymbol{F}((\mathscr{A} \rtimes \mathscr{S}) \ltimes \overline{A})$ and $\boldsymbol{F}((\mathscr{B} \rtimes \mathscr{T}) \ltimes \overline{B})$ are isomorphic as abstract groups;

4. (iv) $\boldsymbol{D}((\mathscr{A} \rtimes \mathscr{S}) \ltimes \overline{A})$ and $\boldsymbol{D}((\mathscr{B} \rtimes \mathscr{T}) \ltimes \overline{B})$ are isomorphic as abstract groups.

This follows from [Reference RenaultRen08] (see also [Reference RaadRaa22]) and [Reference RubinRub89, Theorems 0.2 and 3.3] (or [Reference MatuiMat15, Theorem 3.10]).

4.1 Algebraic actions of torsion-free Abelian monoids whose dual actions are mixing

Let $\sigma \colon S \curvearrowright A$ be an algebraic action, with A Abelian. Let $\hat{\sigma} \colon S \curvearrowright \widehat{A}$ be the dual action as in [Reference Bruce and LiBL24, Remark 2.2] and denote by $\mu$ the normalized Haar measure on $\widehat{A}$ . Recall (see e.g. [Reference SchmidtSch95, 1] or [Reference WaltersWal82, Definition 1.5]) that $\hat{\sigma}$ is (strongly) mixing (with respect to $\mu$ ) if, for all Borel subsets X, Y of $\widehat{A}$ , we have

\[ \lim_{s \to \infty} \mu(X \cap \hat{\sigma}_s(Y)) = \mu(X) \mu(Y). \]

If S has no non-trivial finite subsemigroups, we have the following relation between the mixing property of $\hat{\sigma}$ and condition (F) for $\sigma$ .

Indeed, in general (without our assumption on S), $\hat{\sigma}$ is mixing if and only if, for all infinite subsemigroups $S' \subseteq S$ and $0 \neq a \in A$ , we have that $\# \{\sigma_{s'}(a): s' \in S'\} = \infty$ (see [Reference SchmidtSch95, Theorem 1.6] and also [Reference BerendBer85], in particular [Reference BerendBer85, Theorem 2.1]). It is now straightforward to see that, if S has no non-trivial finite subsemigroups, then the latter statement is equivalent to the condition that for all $0 \neq a \in A$ , we have $\# \{s \in S: \sigma_s(a) = a\} \lt \infty$ . This condition, in turn, is equivalent to the statement that we have $\sigma_s(a) \neq a$ for all $0 \neq a \in A$ and $1 \neq s \in S$ (again assuming that S has no non-trivial finite subsemigroups), as desired, because $\{s \in S:\sigma_s(a) = a\}$ is always a subsemigroup of S.

Note that the condition that S has no non-trivial finite subsemigroups is in particular satisfied if S is torsion-free, in the sense that, for all $s_1, s_2 \in S$ and $i \in \mathbb{Z}_{\gt0}$ , $s_1^i = s_2^i$ implies that $s_1=s_2$ .

With the help of this observation, let us now present the first example class to which we can apply our general rigidity results.

If there exists an isomorphism of topological groupoids

\[ (S^{-1}A\rtimes S^{-1}S)\ltimes \overline{A}\cong (T^{-1}B\rtimes T^{-1} T)\ltimes\overline{B}, \]

then $\tilde{\sigma}\colon S^{-1}S \curvearrowright S^{-1}A \sim_{\mathcal M \mathcal E_{FI}} \tilde{\tau}\colon T^{-1}T \curvearrowright T^{-1}B$ .

Let us briefly explain the conclusion of our results for the case of (duals of) toral endomorphisms.

4.2 Canonical endomorphisms of finite-rank torsion-free Abelian groups

Let $A\subseteq \mathbb{Q}^n$ be a torsion-free Abelian group of rank $n\in\mathbb{Z}_{\gt0}$ . The multiplicative monoid $\mathbb{Z}^\times:=\mathbb{Z}\setminus\{0\}$ acts on A by multiplication. Each $s\in \mathbb{Z}^\times$ gives rise to the endomorphism $\sigma_s\colon A\to A$ given by $\sigma_s(x)= sx$ . For any submonoid $M\subseteq\mathbb{Z}^\times$ , the associated algebraic action $\sigma^M\colon M\curvearrowright A$ , where $\sigma^M:=\sigma\vert_M$ , is faithful and satisfies (FI). It is easy to see that $\sigma^M\colon M\curvearrowright A$ is non-automorphic if and only if there exists $m\in M$ such that A is not m-divisible, i.e. $mA\subsetneq A$ . Since $\det(\dot{\sigma}_s)=s^n$ , we see that $\det\circ \dot{\sigma} \colon\mathbb{Z}_{\gt0}\to\mathbb{Q}^\times$ is injective, so that $\sigma^M\colon M\curvearrowright A$ satisfies (SF) (see Remark 3.29). Since M is Abelian, we obtain a globalization $\tilde{\sigma}^M\colon \langle M\rangle\curvearrowright M^{-1}A$ , where $\langle M\rangle:=M^{-1}M\subseteq\mathbb{Q}^\times$ acts on $M^{-1}A:=\bigcup_{s\in M}(1/s)A$ by multiplication. It is easy to see that $\tilde{\sigma}^M\colon \langle M\rangle\curvearrowright M^{-1}A$ satisfies (F). For $s,s'\in\mathbb{Z}_{\gt0}$ , we have $sA\cap s'A={\rm lcm}(s,s')A$ (see e.g. [Reference FuchsFuc70, § 20]). Hence, the M-constructible subgroups for $M\curvearrowright A$ are given by $\{sA : s\in M \}$ . From this, we see that $\tilde{\sigma}^M\colon \langle M\rangle\curvearrowright M^{-1}A$ satisfies (JF). Let $x_1,\ldots,x_n\in A$ be rationally independent, and put $\mathfrak A:=\text{span}_{\mathbb{Z}}(\{x_1,\ldots,x_n\})\cong\mathbb{Z}^n\subseteq A$ . For each $k\in\mathbb{Z}_{\gt0}$ , let

\[\mathfrak A_k:=\{x\in A : (k!)x\in\mathfrak A\}=A\cap (k!)^{-1}\mathfrak A. \]

Each $\mathfrak A_k$ is a $\mathbb{Z}^\times$ -invariant finitely generated subgroup of A, and we have $A=\bigcup_k\mathfrak A_k$ .

We can now apply Theorem 3.30 to obtain the following result.

Corollary 4.4. Let A and B be finite-rank torsion-free Abelian groups. Let $M,N\subseteq\mathbb{Z}^\times$ be submonoids such that there exist $m\in M$ and $n\in N$ with $mA\subsetneq A$ and $nB\subsetneq B$ . If there is an isomorphism of topological groupoids

\[(M^{-1}A\rtimes \langle M\rangle)\ltimes\overline{A}\cong (N^{-1}B\rtimes \langle N\rangle)\ltimes\overline{B}, \]

then the actions $\langle M\rangle\curvearrowright M^{-1}A$ and $\langle N\rangle\curvearrowright N^{-1}B$ are finitely algebraically equivalent.

4.3 Actions adding scalars to algebraic actions of groups

Let $\Gamma\subseteq\text{SL}_n(\mathbb{Z})$ be any subgroup, and let $M\subseteq \mathbb{Z}_{\gt0}$ be a submonoid. Then, $M\Gamma:=\{a\gamma :a\in M,\gamma\in\Gamma\}$ is a submonoid of $\text{M}_n(\mathbb{Z})^\times:=\{x\in\text{M}_n(\mathbb{Z}) : \det(x)\neq 0\}$ , where we view M as a submonoid of $\text{M}_n(\mathbb{Z})$ via the diagonal embedding. Since $\text{M}_n(\mathbb{Z})^\times$ acts canonically on $\mathbb{Z}^n$ , we obtain a faithful algebraic action $M\Gamma\curvearrowright\mathbb{Z}^n$ . It is easy to see that $M\Gamma\curvearrowright\mathbb{Z}^n$ is exact if and only if M is non-trivial. Note that $\langle M\rangle\Gamma\curvearrowright (M^{-1}\mathbb{Z})^n$ is a globalization for $M\Gamma\curvearrowright\mathbb{Z}^n$ that satisfies (JF). Since $\Gamma$ acts by automorphisms on $\mathbb{Z}^n$ that commute with the action of M, we have $\mathcal C_{M\Gamma\curvearrowright \mathbb{Z}^n}=\mathcal C_{M\curvearrowright \mathbb{Z}^n}$ . Let $\overline{\mathbb{Z}}_M^n$ denote the completion of $\mathbb{Z}^n$ with respect to the family $\mathcal C_{M\curvearrowright \mathbb{Z}^n}$ .

In order to apply our rigidity result in this setting, we need an observation on subgroups of $\text{SL}_n(\mathbb{Z})$ , which comes from [Reference Druţu and KapovichDK18, Example 26.8 and Lemma 26.16].

Corollary 4.8. Let $\Gamma,\Lambda\subseteq \text{SL}_n(\mathbb{Z})$ be subgroups and $M,N\subseteq\mathbb{Z}_{\gt0}$ non-trivial submonoids such that, for every $\gamma\in{\rm tor}(\Gamma)$ , there exists $s\in M$ with $\gcd(\text{ord}(\gamma),s)\gt1$ , and for every $\lambda\in{\rm tor}(\Lambda)$ , there exists $t\in N$ with $\gcd(\text{ord}(\lambda),t)\gt1$ . If there is an isomorphism of topological groupoids

\[(M^{-1}\mathbb{Z}^n\rtimes \langle M\rangle\Gamma)\ltimes\overline{\mathbb{Z}}^n_M\cong (N^{-1}\mathbb{Z}^m\rtimes\langle N\rangle\Lambda)\ltimes\overline{\mathbb{Z}}^m_N, \]

then $M=N$ and there exist $g,h\in\text{M}_n(N^{-1}\mathbb{Z})\cap\text{GL}_n(\mathbb{Q})$ such that $\Gamma\subseteq g\Lambda g^{-1}$ and $h\Lambda h^{-1}\subseteq \Gamma$ .

4.4 Arithmetic dynamical systems

Let us consider the algebraic $\mathbb{N}$ -actions studied by Chothi, Everest, and Ward in [Reference Chothi, Everest and WardCEW97]. Let K be a number field with ring of integers R, and let $\mathcal P_K$ denote the set of non-zero prime ideals of R. For $\mathfrak{p}\in \mathcal P_K$ , let $v_{\mathfrak{p}}$ and $|\cdot|_{\mathfrak{p}}$ denote the associated additive and multiplicative $\mathfrak{p}$ -adic valuations on K, respectively. Given a subset $\mathcal S\subseteq \mathcal P_K$ , the corresponding ring of $\mathcal S$ -integers is

\[R_{\mathcal S}:=\{x\in K : |x|_{\mathfrak{p}}\leq 1\text{ for every }\mathfrak{p}\in \mathcal P_K\!\setminus\mathcal S\}, \]

i.e. $R_{\mathcal S}$ consists of the elements of K that are $\mathfrak{p}$ -adic integers for every $\mathfrak{p}\in \mathcal P_K\!\setminus\mathcal S$ . For $\xi\in R_{\mathcal S}^\times=R_{\mathcal S}{\kern-1pt}\setminus\{0\}$ , the map $m_\xi\colon R_{\mathcal S}\to R_{\mathcal S}$ given by $m_\xi(x)=\xi x$ is an injective endomorphism of the additive group of $R_{\mathcal S}$ . The dual action $\widehat{m}_\xi\colon\mathbb{N}\curvearrowright\widehat{R_{\mathcal S}}$ is called an arithmetic $\mathcal S$ -integer dynamical system, see [Reference Chothi, Everest and WardCEW97, § 2]. The group of units (i.e. invertible elements) in $R_{\mathcal S}$ is $R_{\mathcal S}^*=\{x\in K^* : |x|_{\mathfrak{p}}=1\text{ for every }\mathfrak{p}\in \mathcal P_K\!\setminus\mathcal S\}$ . Note that $R_{\mathcal S}$ is a proper subring of K if and only if $\mathcal S\subsetneq \mathcal P_K$ . Also note that $R_{\mathcal S}^*\subsetneq R_{\mathcal S}^\times$ whenever $R_{\mathcal S}\subsetneq K$ . Let us record some basic observations about the algebraic action $m_\xi\colon\mathbb{N}\curvearrowright R_{\mathcal S}$ . Let $\tilde{m}_\xi\colon R_{\mathcal S}[1/\xi]\to R_{\mathcal S}[1/\xi]$ be given by $\tilde{m}_\xi(x)=\xi x$ . Then $\tilde{m}_\xi\colon\mathbb{Z}\curvearrowright R_{\mathcal S}[1/\xi]$ is a globalization of $m_\xi\colon\mathbb{N}\curvearrowright R_{\mathcal S}$ .