Proof. If $x \in X$ , then there is $f \in F$ such that $xY \cap fY \neq \emptyset $ . So $x \in FYY^{-1}$ .

Proof. Take $f:=(f_i) \in F_1\times \cdots \times F_r$ , and whenever $\bigcap _{1\leq i \leq r} f_iX_i \neq \emptyset $ , choose an element $x_f \in \bigcap _{1\leq i \leq r} f_iX_i $ . If x is any element of $X_0$ , then there must be some $f \in F_1\times \cdots \times F_r$ such that $x \in \bigcap _{1\leq i \leq r} f_iX_i$ . We thus have $x_f^{-1}x \in \bigcap _{1 \leq i \leq n} X_i^{-1}X_i $ . Defining $F:=\{x_f\vert f \in F_1 \times \cdots \times F_n, X_0 \cap \bigcap _{1\leq i \leq r} f_iX_i \neq \emptyset \}$ , we find

$$ \begin{align*}X_0 \subset F\cdot \bigcap_{1 \leq i \leq n} X_i^{-1}X_i.\\[-47pt]\end{align*} $$

Proof. We know that $\Lambda _i^4$ is covered by $K_i^3$ left-translates of $\Lambda _i$ for all $1\leq i \leq r$ . So (1) is a consequence of Lemma 2.2 applied to $X_0=(\bigcap _{1\leq i \leq r}\Lambda _i)^4$ and $X_1=\Lambda _1,\ldots ,X_r=\Lambda _r$ . To prove (2), it suffices to show that $\bigcap _{1\leq i \leq r} \Lambda _i^2$ is covered by finitely many translates of $\bigcap _{1\leq i \leq r} \Xi _i^2$ by symmetry. Statement (2) is then a consequence of Lemma 2.2 applied to $X_0=\bigcap _{1\leq i \leq r} \Lambda _i^2$ and $\Xi _1,\ldots , \Xi _r$ .

Proof. By Lemma 2.3, the subsets $\phi (H) \cap \Lambda _1^2$ and $\phi (H)\cap \Lambda _2^2$ are commensurable approximate subgroups. Taking $\{i,j\} \subset \{1,2\}$ , we can find a finite subset $F_{ij} \subset H$ such that $ (\phi (H)\cap \Lambda _i^2)^2 \subset \phi (F_{ij}) \left (\phi (H)\cap \Lambda _j^2\right ).$ In other words,

$$ \begin{align*}\phi^{-1}(\Lambda_i^2)^2 \subset F_{ij} \phi^{-1}(\Lambda_j^2).\end{align*} $$

So $\phi ^{-1}(\Lambda _1^2)$ and $\phi ^{-1}(\Lambda _2^2)$ are commensurable approximate subgroups.

Proof. Take $i,j \in \{1,2\}$ . The identity belongs to the interior $\operatorname {\mathrm {int}}(V_j)$ of $V_j$ so

$$ \begin{align*}V_i^2 \subset\bigcup\limits_{h \in V_i^2} h\operatorname{\mathrm{int}}(V_j).\end{align*} $$

But $\operatorname {\mathrm {int}}(V_j)$ is open and $V_i^2$ is relatively compact, and thus, there is a finite subset $F_{ij} \subset V_i^2$ such that $V_i^2 \subset F_{ij}U$ . Since $V_1$ and $V_2$ are moreover symmetric subsets, we have that $V_1$ and $V_2$ are commensurable approximate subgroups. Choose now a symmetric open neighbourhood of the identity W such that $W^2$ is contained in $V_1$ and $V_2$ . Then $f^{-1}(W^2), f^{-1}(V_1^2)$ and $f^{-1}(V_2^2)$ are commensurable approximate subgroups by Lemma 2.3. But for $i=1,2$ , we have $f^{-1}(W^2) \subset f^{-1}(V_i) \subset f^{-1}(V_i^2)$ . So $f^{-1}(V_1)$ and $f^{-1}(V_2)$ are commensurable approximate subgroups.

Proof. Let $f:\Gamma \rightarrow H$ be a good model of $(\Lambda ,\Gamma )$ , and let $U \subset H$ be an open subset as in Definition 1.1. Set furthermore $\Lambda _1:=\phi _1^{-1}(\Lambda )$ and $f_1:=f \circ \phi _1$ . Then $f_1(\Lambda _1)=f(\Lambda )$ is relatively compact and $f_1^{-1}(U) \subset \Lambda _1$ . Hence, $\Lambda _1$ is an approximate subgroup by Lemma 3.2 and $f_1$ is a good model of $(\Lambda _1,\Gamma _1)$ . Let us now prove (2). Take a good model $f: \Gamma \rightarrow H$ of $(\Lambda ,\Gamma )$ with dense image and $U \subset H$ a symmetric neighbourhood of the identity such that $f^{-1}(U^2) \subset \Lambda $ . Define $N:= \overline {f(\ker (\phi _2))}$ , which is a normal subgroup since $f(\Gamma )$ is dense. Now

$$ \begin{align*}(p_{H/N}\circ f)^{-1}(p_{H/N}(U)) \subset f^{-1}(U^2f(\ker(\phi_2))) \subset f^{-1}(U^2) \ker(\phi_2) \subset \Lambda\ker(\phi_2),\end{align*} $$

where $p_{H/N}: H \rightarrow H/N$ denotes the natural projection. Therefore, the obvious map $\phi _2(\Gamma ) \rightarrow H/N$ is a good model of $(\phi _2(\Lambda ),\phi _2(\Gamma ))$ .

Proof. $(1) \Rightarrow (2)$ :

Choose a neighbourhood of the identity $U \subset H$ such that $f^{-1}(U)\subset \Lambda $ . There exists a sequence $(U_n)_{n>0}$ of relatively compact symmetric neighbourhoods of the identity in H such that $U_0=U$ and $U_{n+1}^2 \subset U_n$ for all integers $n\geq 0$ . Define now $(\Lambda _n)_{n\geq 0}$ by $\Lambda _0=\Lambda $ and $\Lambda _n=f^{-1}(U_n)$ . We readily check that for all integers $n \geq 0$ , we have $\Lambda _{n+1}^2 \subset \Lambda _n$ . Furthermore, for all $\gamma \in \Gamma $ , $\gamma \Lambda _n\gamma ^{-1}$ is an approximate subgroup commensurable to $\Lambda $ by Corollary 3.3. So $(1) \Rightarrow (2)$ is proved.

$(2) \Rightarrow (3)$ :

Let $(\Lambda _n)_{n \geq 0}$ be as in $(2)$ . For any two subsets $X,Y \subset G$ , define

$$ \begin{align*}X^Y := \bigcap\limits_{y \in Y} yXy^{-1}.\end{align*} $$

Define now $\mathcal {B}$ by

$$ \begin{align*}\mathcal{B}:=\left\{\left(\Lambda_n^2\right)^{F} | \ n \in \mathbb{N}, F \subset \Gamma, |F| < \infty\right\}.\end{align*} $$

We know that $\Lambda _1^2 \subset \Lambda $ and that for all $\Xi \in \mathcal {B}$ , we have $e \in \Xi $ , and $\Xi $ is an approximate subgroup commensurable to $\Lambda $ (Lemma 2.3). Now, for all $n \in \mathbb {N}$ and $F \subset \Gamma $ finite, we have

$$ \begin{align*}\left(\left(\Lambda_{n+1}^2\right)^F\right)^2 \subset\left(\Lambda_{n+1}^4\right)^F \subset \left(\Lambda_n^2\right)^F,\end{align*} $$

and for $\gamma \in \Gamma $ , we find

$$ \begin{align*}\gamma \left(\Lambda_n^2\right)^F \gamma^{-1} \subset \left(\Lambda_n^2\right)^{\gamma F}.\end{align*} $$

So $\mathcal {B}$ satisfies (3).

$(3)\Rightarrow (1)$ :

Equip the group $\langle \Lambda \rangle $ with the topology defined by

$$ \begin{align*}\mathcal{T} = \left\{U \subset \Gamma \vert \forall \gamma \in U, \exists \Xi \in \mathcal{B}, \gamma\Xi \subset U\right\}.\end{align*} $$

By [Reference BourbakiBou89b, Ch. III, §1.2, Proposition 1], the topology $\mathcal {T}$ is the unique topology that makes G into a topological group and such that $\mathcal {B}$ is a neighbourhood basis for e. Now, the closure $\overline {\{e\}}$ of the identity is a closed normal subgroup, and the group $\Gamma /\overline {\{e\}}$ equipped with the quotient topology is the maximal Hausdorff factor of $\Gamma $ . Let $p:\Gamma \rightarrow \Gamma /\overline {\{e\}}$ be the natural map. Then $\{p(\Xi )| \Xi \in \mathcal {B}\}$ is a neighbourhood basis for the identity in $\Gamma /\overline {\{e\}}$ . But the subsets that belong to $\mathcal {B}$ are pairwise commensurable, so the neighbourhoods $\{p(\Xi )|\Xi \in \mathcal {B}\}$ are pre-compact. Hence, the topological group $\Gamma /\overline {\{e\}}$ has a completion by [Reference BourbakiBou89b, Ch. III, Ex. §3, Ex. 7,8]. In other words, there is a locally compact group H and a group homomorphism

$$ \begin{align*}i : \Gamma/\overline{\{e\}} \rightarrow H\end{align*} $$

such that i has dense image and is a homeomorphism onto its image. Define the continuous map $f := i \circ p$ . We will show that f is a good model. The group H is a complete space, and $\Lambda $ is pre-compact in the topology $\mathcal {T}$ according to assumption (b). So $f(\Lambda )$ is a relatively compact subset of H. Recall that i is a homeomorphism onto its image, the map p is open and $\mathcal {B}$ is a neighbourhood basis for the identity. There is thus a neighbourhood of the identity $U \subset H$ such that $f^{-1}(U) \subset \Lambda $ according to assumption (a) and (c).

Statement (4) is straightforward from the proof of (3) $\Rightarrow $ (1). Let us prove (5). Let $\mathcal {T}_0$ denote the initial topology on $\Gamma $ with respect to the class of all good models $f: \Gamma \rightarrow H$ of $(\Lambda ,\Gamma )$ . In other words, the topology $\mathcal {T}_0$ is generated by the family of subsets $\{f^{-1}(U)\}_{f,U}$ where $f: \Gamma \rightarrow H$ and U run through all good models of $(\Lambda ,\Gamma )$ and all open subsets $U \subset H$ . Define $\mathcal {B}_0$ as $\{ U \in \mathcal {T}_0 | e \in U \subset \Lambda \}$ . Since $\Lambda $ is assumed to have a good model, we know that $\mathcal {B}_0$ is not empty. So take $\Xi \in \mathcal {B}_0$ . Then there are good models $(f_i:\Gamma \rightarrow H_i)_{1\leq i\leq r}$ of $(\Lambda , \Gamma )$ and open relatively compact neighbourhoods of the identity $U_i \subset H_i$ for all $1 \leq i \leq r$ such that

$$ \begin{align*}\bigcap_{1\leq i \leq r} f_i^{-1}(U_i) \subset \Xi \subset \Lambda.\end{align*} $$

But the map $f:=(f_i)_{1\leq i \leq r} : \Gamma \rightarrow \prod _{1 \leq i \leq r}H_i$ is a good model, so Corollary 3.3 implies that $\Lambda $ is commensurable to $\bigcap _{1\leq i \leq r} f_i^{-1}(U_i)$ , and hence to $\Xi $ . So $\mathcal {B}_0$ satisfies conditions (a) and (b) of (3). But condition (c) of (3) is also satisfied since $(G,\mathcal {T}_0)$ is a topological group. Indeed, a group with an initial topology given by group homomorphisms to topological groups is a topological group. Let now $f_0: \Gamma \rightarrow H_0$ be as in the proof of (3) $\Rightarrow $ (1). Then every good model $f: \Gamma \rightarrow H$ of $(\Lambda ,\Gamma )$ is continuous with respect to $\mathcal {T}_0$ . According to the universal properties of quotients and completions (see [Reference BourbakiBou89b, Ch. III, §3.4, Prop. 8]), one can therefore find a continuous group homomorphism $\phi : H_0 \rightarrow H$ such that $\phi \circ f_0 = f$ .

We now prove (6). Take a good model $f: \Gamma \rightarrow H$ of $(\Lambda ,\Gamma )$ with dense image. We can find a relatively compact open symmetric neighbourhood of the identity $U \subset H$ such that $ \Lambda \subset f^{-1}(U) \subset \Lambda ^2$ . So $f^{-1}(U)$ is a $K^3$ -approximate subgroup, and hence, U is a $K^3$ -approximate subgroup as well. But by Lemma 5.8 we can find an symmetric open neighbourhood of the identity $S \subset U^4$ such that $S^{8}$ is contained in $U^4$ and $C_K$ left-translates of S cover U for some constant $C_K$ that depends on K only. We thus define $\Lambda _1= f^{-1}(S)$ and $C_{K,1}=C_K$ . A proof by induction on n then shows (6).

Proof. Proposition 3.6 is the equivalence ‘(1) $\Leftrightarrow $ (2)’ in Theorem 3.5.

Proof. Note first that if $\Gamma =\langle \Lambda \rangle $ , then $\Gamma $ commensurates $\Lambda $ . In this situation, 2.(b) of Theorem 3.5 becomes equivalent to the approximate subgroups $\Lambda _n$ and $\Lambda $ being commensurable for all $n \geq 0$ ; see the remark immediately after Theorem 3.5.

Let $\Lambda $ be a Meyer subset. By Theorem 3.5, there is a sequence $(\Lambda _n)_{n\geq 0}$ of approximate subgroups commensurable to $\Lambda $ such that $\Lambda _{n+1}^2 \subset \Lambda _n$ for all integers $n \geq 0$ . The union of two approximate subgroups commensurable with $\Lambda $ is also an approximate subgroup commensurable with $\Lambda $ . Thus, applying Theorem 3.5 to the approximate subgroup $\Lambda \cup \Lambda _0$ together with the sequence $(\Lambda _n)_{n\geq 0}$ implies that $\Lambda \cup \Lambda _0$ has a good model, and so has $(\Lambda \cup \Lambda _0) \cap \langle \Lambda \rangle $ by Lemma 3.4. But $(\Lambda \cup \Lambda _0) \cap \langle \Lambda \rangle $ is commensurable to $\Lambda $ , and $\langle \Lambda \rangle $ is generated by $\Lambda $ . So there is a positive integer n such that $(\Lambda \cup \Lambda _0) \cap \langle \Lambda \rangle $ is contained in $\Lambda ^n$ .

Question 1. With notations as in Proposition 3.8, can n be chosen independently of $\Lambda $ ?

Proof. Take $\mathcal {B}_L$ a neighbourhood basis for the identity in L made of symmetric subsets, and take any $\mathcal {B}_{\Lambda }$ as in part (3) of Theorem 3.5. Then as a consequence of Lemma 2.2, we know that $\mathcal {B}:=\{ \Xi \cap g^{-1}(U) | \Xi \in \mathcal {B}_{\Lambda }, U \in \mathcal {B}_L\}$ satisfies the assumptions of part (3) of Theorem 3.5. Now by part (4) of Theorem 3.5 and by the universal properties of completions and quotients, we can build a good model $f: \Gamma \rightarrow H$ such that g factors through f. But according to part (5) of Theorem 3.5, we know that g factors through $f_0$ – proving (1).

Take $f: \langle \Lambda \rangle \rightarrow H$ a good model of $\Lambda $ with dense image. By the Gleason–Yamabe theorem, there are $O \subset H$ an open subgroup and a normal compact subgroup K of O such that $O/K$ is a connected Lie group without nontrivial normal compact subgroup; see, for instance, [Reference HrushovskiHru12, §4.1], for detail. Then $\Lambda ':=f^{-1}(UK)$ , where $U \subset O$ is any symmetric compact neighbourhood of the identity, is an approximate subgroup commensurable to $\Lambda $ according to Corollary 3.3. But the composition of $f_{|\Lambda '}$ and the natural projection $O\rightarrow O/L$ is easily checked to be a good model of $\Lambda '$ . Now take $\Xi $ and $\Xi '$ approximate subgroups commensurable to $\Lambda $ , and let $f: \langle \Xi \rangle \rightarrow H$ and $f':\langle \Xi '\rangle \rightarrow H'$ be good models of $\Xi $ and $\Xi '$ , respectively. Suppose moreover that both satisfy (2). Let $D: \langle \Xi ^2 \cap \Xi ^{\prime 2}\rangle \rightarrow H \times H'$ be the diagonal map, and let $\Delta $ be the closure of its image. We want to show that $\Delta \cap \left (H \times \{e\}\right )$ and $\Delta \cap \left (\{e\} \times H'\right )$ are compact subgroups. Let U denote a relatively compact open neighbourhood of the identity in $\Delta $ . Then $D(\langle \Xi ^2 \cap \Xi ^{\prime 2}\rangle ) \cap U(\Delta \cap \left (\{e\} \times H'\right ))$ is dense in $U(\Delta \cap \left (\{e\} \times H'\right ))$ . But the projection of $U(\Delta \cap \left (\{e\} \times H'\right ))$ to H is a relatively compact set $U_H$ . So $D^{-1}(U(\Delta \cap \left (\{e\} \times H'\right )))$ is contained in $f^{-1}(U_H)$ which is covered by finitely many translates of $\Xi $ (Corollary 3.3) and, by Lemma 2.2, is covered by finitely many translates of $\Xi ^2 \cap \Xi ^{\prime 2}$ . So $D(f^{-1}(U_H))$ is covered by finitely many translates of $D(\Xi ^2 \cap \Xi ^{\prime 2}) \subset f\left ( \Xi ^2 \right ) \times f'\left ( \Xi ^{\prime 2} \right )$ which is relatively compact. So $U(\Delta \cap \left (\{e\} \times H'\right )) \subset \overline {D(f^{-1}(U_H))}$ , which is compact, thus proving that $\Delta \cap \left (\{e\} \times H'\right )$ is compact. A symmetric argument shows that $\Delta \cap \left (H \times \{e\}\right )$ is compact. Moreover, $\Xi ^2 \cap \Xi ^{\prime 2}$ is commensurable to both $\Xi $ and $\Xi '$ by Lemma 2.2. So $p_H(\overline {D(\Xi ^2 \cap \Xi ^{\prime 2})})$ contains $\overline {f(\Xi ^2 \cap \Xi ^{\prime 2})}$ , which contains an open subset by the Baire category theorem. Therefore, $\Delta $ projects surjectively to H by connectedness. Likewise, the projection of $\Delta $ to $H'$ is surjective. So $\Delta \cap \left (H \times \{e\}\right )$ and $\Delta \cap \left (\{e\} \times H'\right )$ are compact normal subgroups of H and $H'$ , respectively, which means that they are trivial. As a consequence, $\Delta $ is the graph of a continuous isomorphism $\phi : H \rightarrow H'$ such that $f^{\prime }_{|\langle \Xi ^2 \cap \Xi ^{\prime 2}\rangle } = \phi \circ f_{|\langle \Xi ^2 \cap \Xi ^{\prime 2}\rangle }$ . This proves (2).

Proof. If $\Lambda _i$ has a good model for all $i \in I$ , then there are constants $(C_{K,n})_{n\geq 0}$ and sequences $(\Lambda _{i,n})_{n\geq 0}$ as in part (6) of Theorem 3.5. But then for any ultrafilter $\mathcal {U}$ over I, we know that $\underline {\Lambda }^8$ is covered by $C_{K,n}$ left-translates of $\underline {\Lambda _n}:= \prod _{i\in I} \Lambda _{i,n}/ \mathcal {U}$ for all $n \geq 0$ (see, for example, [Reference Breuillard, Green and TaoBGT12, App. A] for background material on ultraproducts of groups). So $(\underline {\Lambda _n})_{n\geq 0}$ satisfies part (2) of Theorem 3.5, and $\underline {\Lambda _0}=\underline {\Lambda }^8$ has a good model.

Let $\mathcal {U}$ be an ultrafilter on I that contains the subsets $\{i \in I | j \leq i \}$ for all $j \in I$ . Such an ultrafilter exists because this family has the finite intersection property (note moreover that $\mathcal {U}$ may be principal if I contains a final element). Write $\underline {\Lambda }$ and $\underline {\Gamma }$ for the ultraproducts of $(\Lambda _i)_{i \in I}$ and $(\Gamma _i)_{i \in I}$ over $\mathcal {U}$ , respectively. By the universal property of direct limits, there is a natural map $\phi : \varinjlim _I \Gamma _i \rightarrow \underline {\Gamma }$ , and we compute that $\phi ^{-1}(\underline {\Lambda })=\varinjlim _I \Lambda _i$ . So $\varinjlim _{I} \Lambda _i^2$ is an approximate subgroup by Lemma 2.4. If every $\Lambda _i$ has a good model, then the approximate subgroup $\underline {\Lambda }^8$ has a good model by (2). So $\varinjlim _{ I} \Lambda _i^8$ has a good model by Lemma 3.4.

Proof. Let $\Lambda $ denote the set $f^{-1}(\left [-R;R\right ])$ . Then $\Lambda $ is symmetric since f is symmetric and the set $f(\Lambda ^2)$ is contained in $[-2R-C(f);2R + C(f)]$ . Set $\delta :=R-C(f)> 0$ , and choose a finite subset $F \subset \Lambda ^2$ with $|F|=|f(F)|$ such that $f(F)$ is a maximal $\delta $ -separated subset of $f(\Lambda ^2)$ . We know that $|F| \leq 2\frac {2R + C(f)}{\delta }+1$ , and, in addition, we have

$$ \begin{align*}f(\Lambda^2) \subset \bigcup\limits_{\gamma \in F} f(\gamma) + [-\delta;\delta].\end{align*} $$

Take $\lambda \in \Lambda ^2$ and $\gamma \in F$ such that $\vert f(\lambda )-f(\gamma ) \vert \leq \delta $ . We have

$$ \begin{align*}\vert f(\gamma^{-1}\lambda)- \left(f(\lambda)-f(\gamma)\right)\vert \leq C(f),\end{align*} $$

so

$$ \begin{align*}\vert f(\gamma^{-1}\lambda)\vert \leq C(f)+\delta = R.\end{align*} $$

Hence, $\gamma ^{-1}\lambda \in \Lambda $ and $\Lambda ^2 \subset F\Lambda $ .

Proof. Write $\delta _1:=R_1 - C(f_1)$ . Choose $F_1 \subset \Lambda _2$ with $|F_1|=|f_1(F_1)|$ such that $f_1(F_1)$ is a maximal $\delta _1$ -separated subset of $f_1(\Lambda _2)$ . Since $f_1(\Lambda _2) \subset [-(R_2+\eta );R_2 + \eta ]$ , we know that $|F_1| \leq \frac {2(R_2+\eta )}{\delta _1}+1$ . As in the proof of Lemma 3.14, we find that $\Lambda _2 \subset F_1\Lambda _1$ . By symmetry, there is $F_2 \subset G$ with $|F_2| \leq \frac {2(R_1 + \eta )}{R_2 - C(f_2)}+1$ such that $\Lambda _1 \subset F_2 \Lambda _2$ .

Let us now prove the converse statement. Both $f_1$ and $f_2$ are within bounded distance of an homogeneous quasi-morphism [Reference KotschickKot04]. Therefore, by the first part of Lemma 3.15, we only have to prove the converse assuming that $f_1$ and $f_2$ are homogeneous. First of all, if $f_1 = 0$ , then $\Lambda _2 = f_2^{-1}(\left [-R_2;R_2\right ])$ is commensurable to G. So $f_2(G)$ is bounded. But $f_2$ is homogeneous, so $f_2=0$ . Suppose now that $f_1$ is nontrivial, and take $g_0 \in G$ such that $f_1(g_0)> 0$ . Define that map

$$ \begin{align*}\hat{f}: g \mapsto f_1(g_0)f_2(g) - f_2(g_0)f_1(g).\end{align*} $$

It is a homogeneous quasi-morphism with $\hat {f}(g_0)=0$ . Moreover, any set commensurable to $\Lambda _1$ (equivalently, to $\Lambda _2$ ) has bounded image by $\hat {f}$ . For all $g \in G$ , there is $n \in \mathbb {Z}$ such that $\vert f_1(g) - nf_1(g_0)\vert \leq |f_1(g_0)|$ . Thus,

$$ \begin{align*}G= \langle g_0 \rangle f_1^{-1}([-C(f_1)-f_1(g_0);f_1(g_0)+C(f_1)]).\end{align*} $$

But $f_1^{-1}([-C(f_1)-f_1(g_0);f_1(g_0)+C(f_1)])$ is commensurable to $\Lambda _1$ according to the first part of the proof and, hence, is mapped to a bounded set by $\hat {f}$ . So $\hat {f}$ must have bounded image and, therefore, must be trivial. In other words, $f_2=\frac {f_2(g_0)}{f_1(g_0)}f_1$ .

Proof. If f is bounded, then $f=0$ . So assume that f is unbounded. Take $R'> C(f)$ such that $f^{-1}([-R';R'])$ generates G. We know that $f^{-1}([-R';R'])$ is an approximate subgroup (Lemma 3.14) and a Meyer subset (Lemma 3.15). By Proposition 3.8, there is an integer $n \geq 1$ such that there are a good model $f_0:G \rightarrow H$ of some power, say n, of $f^{-1}([-R';R'])$ with dense image. In particular, $f_0$ is a good model of $\Lambda :=f^{-1}([-n(R'+C(f));n(R'+C(f))])$ according to Lemma 3.15. Since $f_0(G)$ is dense in H, we have that $\overline {f_0(\Lambda )}$ is a neighbourhood of the identity. So the subgroup generated by the compact set $\overline {f_0(\Lambda )}$ is open (so clopen) and contains $f_0(G)$ , and hence equals H. The group H is thus compactly generated. We will now show that $f=cf_0$ for some real number $c> 0$ . We start with two claims:

Proof. Write $R_0:= f(\gamma _0)$ . Then the subset $f^{-1}([-R_0-C(f);R_0+C(f)])$ is commensurable to $\Lambda $ by Lemma 3.15. So the subset

$$ \begin{align*}K:=\overline{f_0(f^{-1}([-R_0-C(f);R_0+C(f)]))}\end{align*} $$

is compact. But for any $\gamma \in G$ , there is an integer l such that $\vert f(\gamma )-lf(\gamma _0) \vert \leq R_0$ so $f_0(\gamma _0)^{-l}f_0(\gamma )=f_0(\gamma _0^{-l}\gamma )\in K$ .

Now all conjugacy classes of H are relatively compact according to Claim 3.4.1. So we can find a compact normal subgroup $K \subset H$ and non-negative integers $k,l$ such that $H/K \simeq \mathbb {R}^k \times \mathbb {Z}^l$ (see [Reference Grosser and MoskowitzGM71, Thm. 3.20]). We will show that $k+l \leq 1$ . Let $p: H \rightarrow H/K$ denote the natural projection. Then the group homomorphism $p\circ f_0$ has dense image, and $p\circ f_0(\Lambda )$ is relatively compact. If $H/K$ is compact, then $H/K \simeq \{e\}$ . Otherwise, we can find $\gamma _0 \in G\setminus \Lambda $ so $f(\gamma _0)> 0$ . According to Claim 3.4.2, every $\gamma \in G$ satisfies $p\circ f_0(\gamma ) \in \langle p\circ f_0(\gamma _0)\rangle L$ where L is some compact subset of $H/K$ . Therefore, $\langle p\circ f_0(\gamma _0)\rangle $ is an infinite cyclic co-compact subgroup, and hence, $k+l \leq 1$ . Choose a neighbourhood $U\subset H$ of the identity such that $f_0^{-1}(U) \subset \Lambda $ . Since K is a compact subgroup of H and the subgroup $f_0(G)$ is dense, we can find an integer $m \geq 0$ such that $K \subset f(\Lambda ^m)U$ . Then $V:=p(U) \subset H/K$ is such that

$$ \begin{align*}f_0^{-1}(p^{-1}(V))\subset f_0^{-1}(UK) \subset \Lambda^{m+2}.\end{align*} $$

So $p\circ f_0$ is a good model of $\Lambda ^{m+2}$ with image dense in $\mathbb {R}$ or $\mathbb {Z}$ . By the converse of Lemma 3.15 $f=\alpha \cdot p \circ f_0$ for some $\alpha \in \mathbb {R}$ , so f is a group homomorphism.

Proof of Theorem 3.13.

Recall that w is a reduced word of length l in $F_2$ the free group over $\{a,b\}$ and that $o(g,w)$ counts the occurrence of w as a reduced sub-word of g with overlap. Suppose that $w \notin \{a, b, a^{-1},b^{-1},e\}$ . According to [Reference BrooksBro81, §3. (a)], the map

$$ \begin{align*} f_w: & F_2 \longrightarrow \mathbb{R} \\ & g \longmapsto o(g,w)-o(g,w^{-1}) \end{align*} $$

is a symmetric quasi-morphism with $C(f_w) \leq 3(l -1)$ . Moreover, $f_w$ is within distance $\delta $ of a unique homogeneous quasi-morphism, $\tilde {f}_w$ say, that is not a group homomorphism. But according to Lemma 3.14, the set

$$ \begin{align*}\left\{ g \in F_2 | |o(g,w)-o(g,w^{-1})| \leq 3l \right\} = f_w^{-1}([-3l;3l)])\end{align*} $$

is an approximate subgroup. Moreover, by Lemma 3.15, it is commensurable to $\tilde {f}_w^{-1}([-C(\tilde {f}_w)-\delta ;C(\tilde {f}_w)+\delta ])$ . So if $\left \{ g \in F_2 \left | |o(g,w)-o(g,w^{-1})| \leq 3l \right.\right \}$ is a Meyer subset, then $\tilde {f}_w^{-1}([-C(\tilde {f}_w)-\delta ;C(\tilde {f}_w)+\delta ])$ is a Meyer subset, and hence, $\tilde {f}_w$ is a group homomorphism according to Proposition 3.16 – contradiction.

Proof of Lemma 3.12.

Let w be a reduced word of length l in $F_2$ the free group over $\{a,b\}$ , and suppose that $w \notin \{a, b, a^{-1},b^{-1},e\}$ . Let $f_w$ be as in the proof of Theorem 3.13. Define $\Lambda _n:=f_w^{-1}([-3^nl;3^nl])$ for all $n> 0$ . Then $\Lambda _n^2 \subset \Lambda _{n+1}$ , and K left-translates of $\Lambda _n$ cover $\Lambda _{n+1}$ for some K independent of n (Lemma 3.15). So the sequence of subsets $(\underline {\Lambda }_k)_{k \geq 0}$ defined by $\underline {\Lambda }_k:=\prod _{n \geq 0} f_w^{-1}([-3^{n-k}l;3^{n-k}l])/\mathcal {U}$ is made of well-defined approximate subgroups commensurable to $\prod _{n \geq 0} \Lambda _n / \mathcal {U}$ since $\mathcal {U}$ is non-principal (see, for example, [Reference Breuillard, Green and TaoBGT12, App. A]). Besides, $\underline {\Lambda }_k^2 \subset \underline {\Lambda }_{k-1}$ for all $k \geq 1$ since $\mathcal {U}$ is non-principal. So $\prod _{n \geq 0} \Lambda _n / \mathcal {U}$ has a good model according to Theorem 3.5. However, for all $n \geq 0$ , $\Lambda _n$ is not a Meyer subset according to Theorem 3.13.

Proof. Choose a neighbourhood of the identity $U \subset H$ such that $f^{-1}(U) \subset \Lambda $ (Definition 1.1) and an open subset $V \subset G$ such that $V \cap \Lambda =\{e\}$ . For $\gamma \in \Gamma $ , we know that $(\gamma ,f(\gamma )) \in V \times U$ implies $f(\gamma ) \in U$ ; hence, $\gamma \in \Lambda $ . But $\gamma \in V$ , so $\gamma =e$ , and we find $\Gamma _f \cap \left (V \times U \right )= \{e\}$ .

Proof. Let $\mathcal {F} \subset G$ be a measurable subset of finite Haar measure such that $ \mathcal {F}\Lambda = G$ ([Reference HrushovskiHru22, Prop. A.2]). Define $W_0:=\overline {f(\Lambda )}$ . Then $\Gamma _f$ is discrete by Lemma 3.17. Moreover, we know that for all $(g,h) \in G \times H$ , there is $\gamma _1 \in \Gamma _f$ such that $(g,h)\gamma _1^{-1} \in G \times W_0$ . By assumption, we have $\gamma _2 \in \Gamma _f \cap (G \times W_0)$ such that $p_G((g,h)\gamma _1^{-1}\gamma _2^{-1}) \in \mathcal {F}$ . Therefore, we know that $(g,h) \in \left (\mathcal {F} \times W_0W_0^{-1}\right )\Gamma _f$ . So $\Gamma _f$ is a lattice in $G \times H$ . Now, $\Lambda \subset P_0(G,H,\Gamma _f, W_0) \subset \Lambda ^2$ , which concludes.

Proof of Proposition 1.2.

Assume that $\Lambda $ is a model set. Let $(G,H,\Gamma )$ be a cut-and-project scheme, and let $W_0$ be a neighbourhood of the identity $W_0 \subset H$ such that $P_0(G,H,\Gamma , W_0)=\Lambda $ (Definition 2.6). Denote by $p_G: G \times H \rightarrow G$ and $p_H: G \times H \rightarrow H$ the natural projections. The map $\left (p_G\right )_{\vert \Gamma }$ is injective and

$$ \begin{align*}\Lambda=P_0(G,H,\Gamma,W_0)= p_G\left((G \times W_0) \cap \Gamma\right).\end{align*} $$

But

$$ \begin{align*}p_G\left((G \times W_0) \cap \Gamma\right) = \left(p_H\circ\left(p_G\right)_{\vert \Gamma}^{-1} \right)^{-1}(W_0),\end{align*} $$

where we think of $\left (p_G\right )_{\vert \Gamma }$ as a bijective map from $\Gamma $ to $p_G(\Gamma )$ . We know that $\langle \Lambda \rangle \subset p_G(\Gamma )$ , so set

$$ \begin{align*} \tau : \langle\Lambda\rangle &\longrightarrow H \\ \gamma &\longmapsto p_H\circ\left( p_G\right)_{\vert \Gamma}^{-1}(\gamma). \end{align*} $$

Then $\tau $ is a group homomorphism, $W_0$ is a symmetric relatively compact neighbourhood of the identity and $\tau ^{-1}(W_0) = \Lambda $ . So $\tau $ is a good model of $\Lambda $ .

Conversely, $\Lambda $ is a Meyer subset. So there is $n \geq 1$ such that $\Lambda ^n$ has a good model by Proposition 3.8. Therefore, $\Lambda ^n$ – hence, $\Lambda $ – is contained in and commensurable to a model set by Proposition 3.18.

Proof. Choose a finite subset $F \subset G$ such that $\Xi \subset F\Lambda $ . Define the open subset

$$ \begin{align*}U(\Xi):=G\setminus\left(\bigcup\limits_{f \in F, f\notin \Lambda}f\Lambda\right).\end{align*} $$

Since $e \in f\Lambda $ implies $f\in \Lambda ^{-1}=\Lambda $ , the subset $U(\Xi )$ contains the identity. We thus have

$$ \begin{align*} U(\Xi) \cap \Xi & \subset U(\Xi)\cap F\Lambda \\ & \subset \bigcup\limits_{f\in F, f \in \Lambda}f \Lambda \\ & \subset \Lambda^2.\\[-38pt] \end{align*} $$

Proof of Theorem 4.1.

For all $\gamma \in \Gamma $ , let $U(\gamma )$ be the neighbourhood of the identity such that $\gamma \Lambda ^4\gamma ^{-1} \cap U(\gamma ) \subset \Lambda ^2$ (Lemma 4.2). Choose a neighbourhood basis for the identity $\mathcal {B}$ made of closed subsets in G and define $\mathcal {B}_{\Lambda }$ as the family of subsets $\{ \Lambda ^2 \cap U^{-1}U | U \in \mathcal {B}\}$ . The subsets in $\mathcal {B}_{\Lambda }$ are all contained in and commensurable to $\Lambda ^2$ by Lemma 2.2. Take any $U \in \mathcal {B}$ and any $\gamma \in \Gamma $ and choose $V \in \mathcal {B}$ such that $\gamma (V^{-1}V)^2\gamma ^{-1} \subset U(\gamma )\cap U$ . Then

$$ \begin{align*} \gamma \left(V^{-1}V \cap \Lambda^2\right)^2 \gamma^{-1} & \subset \gamma(V^{-1}V)^2\gamma^{-1} \cap \gamma\Lambda^4\gamma^{-1} \\ & \subset U(\gamma)\cap U \cap \gamma\Lambda^4\gamma^{-1} \\ & \subset U \cap \Lambda^2. \end{align*} $$

So $\mathcal {B}_{\Lambda }$ checks all conditions of a neighbourhood basis, so there is a topology $\mathcal {T}$ on $\Gamma $ making $\Gamma $ into a topological group and for which $\mathcal {B}_{\Lambda }$ is neighbourhood basis about e [Reference BourbakiBou89b, Ch. III, §1.2, Proposition 1]. We can moreover prove that the inclusion map $\Gamma \rightarrow G$ is continuous. We will achieve this in a number of steps.

First, take $\lambda \in \Lambda ^{2}$ and take any $U \in \mathcal {B}$ with $U^{-1}U \subset U(e)$ . Then

$$ \begin{align*}\Lambda^2 \cap \lambda U^{-1}U = \lambda (\lambda^{-1}\Lambda^2 \cap U^{-1}U) \subset \lambda (\Lambda^4 \cap U^{-1}U) = \lambda (\Lambda^2 \cap U^{-1}U).\end{align*} $$

Remark that $\Lambda ^2 \cap \lambda U^{-1}U$ is a neighbourhood of $\lambda $ in the subspace topology of $\Lambda ^2 \subset G$ and, because $\Lambda ^2 \cap U^{-1}U \in \mathcal {B}_{\Lambda }$ , $\lambda (\Lambda ^2 \cap U^{-1}U)$ is a neighbourhood of $\lambda $ in $\Gamma $ equipped with $\mathcal {T}$ . We have therefore showed that the identity map $id: \Lambda ^2 \rightarrow \Lambda ^2$ is continuous, where the source space is equipped with the subspace topology from G and the target space is equipped with the subspace topology from $\Gamma $ equipped with $\mathcal {T}$ .

Remark now that the inverse map of $id_{\Lambda ^2}$ is simply the restriction to $\Lambda ^2$ of the inclusion map $\Gamma \rightarrow G$ . So we have proved continuity in the wrong direction. To reverse it, we will use the fact that a bijective continuous map from a compact space to a Hausdorff space has a continuous inverse. Since $ \bigcap _{\Xi \in \mathcal {B}_{\Lambda }} \Xi \subset \bigcap _{U \in \mathcal {B}}U^{-1}U = \{e\}$ , $\Gamma $ equipped with $\mathcal {T}$ is indeed Hausdorff. Therefore, $id_{\Lambda ^2}$ has a continuous inverse, and $\Lambda ^2$ is a compact subset of e in $\Gamma $ equipped with $\mathcal {T}$ . In particular, $\Gamma $ is locally compact, and the restriction of the inclusion map $\Gamma \rightarrow G$ to $\Lambda ^2$ is continuous. Since $\Lambda ^2$ is a neighbourhood of e, $\Gamma \rightarrow G$ is a continuous group homomorphism.

Proof of Theorem 1.3.

Apply Theorem 4.1 to $(\Lambda ^2 \cap V^2 , \langle \Lambda \rangle )$ , where V is any symmetric compact neighbourhood of the identity in G. Note that $\Lambda ^2 \cap V^2$ is indeed an approximate subgroup by Lemma 2.3. This yields an injective continuous group homomorphism $\phi : H \rightarrow G$ with image $\langle \Lambda \rangle $ and such that $\phi ^{-1}(\Lambda ^2 \cap V^2)$ is a compact neighbourhood of the identity. Since finitely many translates of $\Lambda $ cover $\Lambda ^2 \cap V^2$ , $\phi ^{-1}(\Lambda )$ has nonempty interior as well. The approximate subgroup $\phi ^{-1}(\Lambda )$ is therefore contained in the interior of $\phi ^{-1}(\Lambda ^3)$ . We have, moreover, that for any $K \subset G$ compact, the subset $K \cap \Lambda $ is covered by finitely many left-translates of $\Lambda ^2 \cap V^2$ . So $\phi ^{-1}(K \cap \Lambda )$ is compact and $\phi _{|\phi ^{-1}(\Lambda )}$ is proper.

If, moreover, G is a Lie group, then H is a Lie group as a consequence of [Reference BourbakiBou89c, Ch. III, §8, Corollary 1].

Proof. This result is essentially an application of the above and [Reference BourbakiBou89c, Ch. III, §8, Prop. 2]. We sketch a proof and refer the reader to the relevant part of [Reference BourbakiBou89c, Ch. III, §8] for background. Apply Theorem 4.1 to $(\Lambda , \Gamma )$ . There is an injective continuous group homomorphism $\phi : H \rightarrow G$ with image $\Gamma $ and such that $\phi ^{-1}(\Lambda ^2 )$ is a compact neighbourhood of the identity. Since G is a Lie group and $\phi $ is injective, H is a Lie group [Reference BourbakiBou89c, Ch. III, §8, Cor. 1]. Let $\mathfrak {h}'$ denote the Lie algebra of H and $\mathfrak {h}$ its image through the differential $d\phi $ of $\phi $ . The map $d\phi $ yields a linear isomorphism between $\mathfrak {h}'$ and $\mathfrak {h}$ . Then $\mathfrak {h}$ is obviously invariant under the adjoint action of $\phi (H)=\Gamma $ . Moreover, for any compact symmetric neighbourhood V of $0$ in $\mathfrak {h}'$ , $\exp (V)$ is a compact symmetric neighbourhood of e in H, and hence commensurable to $\phi ^{-1}(\Lambda ^2)$ . Thus, $\phi (\exp (V))=\exp ( d\phi (V) )$ is commensurable to $\Lambda $ as claimed.

Proof. While we have used additive notations in the statement, we will stick with multiplicative notations in the proof for the sake of consistency. We will use several times the following claim:

This is a consequence of Schreiber’s theorem and was already worked out in [Reference MachadoMac22b, Prop. 3] in a broader context.

Proof of Claim 4.2.1.

Take $\Lambda \subset HK$ and $H \subset \Lambda K$ where K is symmetric compact and H is a vector subspace, as given by Schreiber’s Theorem. Let B be a closed ball in H, so H is generated by B as a group and B is compact. Take $\lambda \in \Lambda $ arbitrary and $ h \in H$ such that $\lambda \in hK$ . Take $h_1,\ldots ,h_r \in B$ with $h=h_1 \cdots h_r$ . Since $H\subset \Lambda K$ , pick $\lambda _i \in \Lambda $ such that $h_1 \cdots h_i \in \lambda _iK$ for $1 \leq i < r$ and $\lambda = \lambda _r$ and $\lambda _0 = 0$ . Then, $\lambda _{i-1}^{-1}\lambda _i \in h_iK^2 \subset BK^2$ , so

$$ \begin{align*}\lambda = \lambda_r =\prod_{i=1}^r\lambda_{i-1}^{-1}\lambda_i \in (\Lambda^2 \cap (BK^2))^r.\end{align*} $$

Thus, $\langle \Lambda \rangle $ is generated by $K' = \Lambda ^2 \cap (BK^2)$ , which is relatively compact and contained in $\Lambda ^2$ . Since $\Lambda $ is an approximate subgroup, $K' \subset F\Lambda $ with $F \subset \Lambda ^3$ finite, so $\overline {K'} \subset F\Lambda \subset \Lambda ^4$ as $\Lambda $ is closed.

We start with two sub-cases. Suppose first that $\Lambda $ is uniformly discrete. Then $\langle \Lambda \rangle $ is finitely generated (Claim 4.2.1), so $\langle \Lambda \rangle \simeq \mathbb {Z}^m$ . Take a linear map $\phi : \mathbb {R}^m \rightarrow \mathbb {R}^n$ such that the restriction of $\phi $ to $\mathbb {Z}^m$ yields an isomorphism $\langle \Lambda \rangle \simeq \mathbb {Z}^m$ . Apply now Schreiber’s theorem to $\Lambda ':=\phi ^{-1}(\Lambda ) \cap \mathbb {Z}^m$ . We get a vector subspace $V_d' \subset \mathbb {R}^m$ and a box $B:=[-a;a]^m$ for some real $a>1$ such that $\Lambda ' \subset V_d'B$ and $V_d' \subset \Lambda 'B$ . Thus, $\Lambda '$ is contained in and commensurable with $\mathbb {Z}^m \cap V_d'B$ , so $\phi (\mathbb {Z}^m \cap V_d'B)$ indeed contains and is commensurable with $\phi (\Lambda ')=\Lambda $ . It remains to prove $V_d' \cap \ker \phi =\{0\}$ . Otherwise, $ D:=\left (V_d' \cap \ker \phi \right ) B \cap \mathbb {Z}^m$ is infinite. But $\phi (D)\subset \phi (B)$ , which is bounded. In addition, $\phi (D) \subset \phi (\mathbb {Z}^m \cap V_d'B)$ and $\phi (\mathbb {Z}^m \cap V_d'B)$ is covered by finitely many translates of $\Lambda $ , and hence is uniformly discrete. So $\phi (D)$ is finite. Since $\phi $ is injective on $\mathbb {Z}^m$ , we reach a contradiction.

Suppose now that $\Lambda $ has nonempty interior. Then the interior of $\Lambda ^2$ is symmetric, contains the identity and is commensurable to $\Lambda $ . Hence, it is an open approximate subgroup commensurable to $\Lambda $ . Take V and K as in Schreiber’s theorem. We know that $K^2$ is covered by finitely many translates of $\Lambda $ . So $VK \subset \Lambda K^2$ is covered by finitely many translates of $\Lambda $ . But $\Lambda \subset VK$ , so $\Lambda $ and $VK$ are commensurable. So $V_o := V$ and $\Lambda _d:=\{e\}$ work.

Let us go back to the general case. Let $V_o$ be the largest vector subspace covered by finitely many left-translates of $\Lambda $ , and take $V_d$ any supplementary space. Now, $\Lambda $ is contained in $V_o\left (V_d \cap \Lambda V_o^{-1}\right )$ . But $V_o \cap \Lambda ^2$ is commensurable to $V_o$ and finitely many translates of $V_d \cap \Lambda ^2$ cover $V_d \cap \Lambda V_o$ by Lemma 2.3. So $V_d \cap \Lambda V_o$ and $(V_d\cap \Lambda ^2)$ are commensurable. So $\Lambda $ is covered by finitely many translates of (and, thus, commensurable to) $(V_o \cap \Lambda ^2) (V_d\cap \Lambda ^2)$ . In turn, $\Lambda $ is commensurable to $V_o (V_d\cap \Lambda ^2)$ .

Now, let $i: L \rightarrow \mathbb {R}^n$ be the injective Lie group homomorphism given by Theorem 1.3 applied to $V_d \cap \overline {\Lambda ^2}$ (recall that $\overline {\Lambda ^2} \subset \Lambda ^4$ so is commensurable with $\Lambda $ ), and let $\Lambda '$ denote the inverse image of $V_d \cap \overline {\Lambda ^2}$ . Assume as we may that $L=\langle \Lambda ' \rangle $ . Let $L^0$ denote the connected component of the identity of L. Then $L^0$ is a connected torsion free abelian lie group, and hence, $L^0 \simeq \mathbb {R}^k$ . Note, moreover, that $i_{|L^0}$ is an injective continuous group homomorphism, and hence an $\mathbb {R}$ -linear map and a homeomorphism onto its image. We have that $(\Lambda ')^2 \cap L^0$ is an approximate subgroup with nonempty interior so it is commensurable to $V' W$ (by the above paragraph), where $V' \subset L^0$ is a vector subspace and W is a compact neighbourhood of the identity in $L^0$ . By construction of $V_d$ , we know that $V'=\{0\}$ . So $(\Lambda ')^2 \cap L^0$ is a relatively compact neighbourhood of the identity in $L^0$ . Recall now that $i_{|\Lambda '}$ is proper, and hence, $i_{|\Lambda ^{\prime 4}}$ is proper as well. But L is a torsion-free abelian Lie group, and there is a compact subset C contained in $(V_d \cap \Lambda ^2)^4=i(\Lambda ^{\prime 4})$ that generates $i(\langle \Lambda ' \rangle )$ (Claim 4.2.1). So L is generated by the compact subset $i^{-1}(C)$ . Thus, $L/L^0$ is a finitely generated abelian group, concluding $L \simeq \mathbb {R}^k \oplus \mathbb {Z}^l$ for some non-negative integers $k,l$ . So we can identify L with a closed subgroup of $ \mathbb {R}^k \times \mathbb {R}^l$ with $L^0 = \mathbb {R}^k \times \{0\}$ .

According to Schreiber’s theorem, there is vector subspace $V \subset \mathbb {R}^k \times \mathbb {R}^l$ and a compact neighbourhood of the identity $K \subset \mathbb {R}^k \times \mathbb {R}^l$ such that $\Lambda ' \subset V K$ and $V \subset \Lambda ' K$ . We claim that $L^0 \cap V = \{0\}$ . Indeed, $L^0 \cap V \subset \Lambda 'K$ so $L^0 \cap V \subset \Lambda 'K_0$ , where $K_0:=(L^0\cap V)\Lambda ' \cap K$ . Now, $K_0$ is a relatively compact subset of L, and since $\Lambda '$ has nonempty interior in L, finitely many translates of $\Lambda '$ cover $K_0$ . So finitely many translates of $i(\Lambda ')$ cover $i(V \cap L^0)$ . Thus, $i(V \cap L^0)$ is a subspace of $V_o$ . But $i(V \cap L^0) \subset V_d$ , so $i(V \cap L^0) \subset V_d \cap V_o=\{0\}$ . In turn, $V \cap L^0=\{0\}$ by injectivity of i.

So choose a vector subspace $V'$ such that $L^0 \oplus V \oplus V'= \mathbb {R}^k \times \mathbb {R}^l$ . The projection of L to $V \oplus V'$ parallel to $L^0$ is then a discrete subgroup $\Gamma \subset V\oplus V'$ , and we find $L = L^0 \oplus \Gamma $ . Moreover, the projection of $\Lambda '$ to $L^0$ is contained in K, so it is a bounded subset with nonempty interior. Since $\Lambda ^{\prime 2} \cap L^0$ is a neighbourhood of $\{0\}$ and $L^0$ is connected, the projection of $\Lambda '$ to $L^0$ is contained in $(\Lambda ^{\prime 2} \cap L^0)^m$ for some $m> 0$ . So for every $\lambda \in \Lambda '$ , there is $\lambda _0 \in (\Lambda ^{\prime 2} \cap L^0)^m$ such that $\lambda \lambda _0^{-1} \in \Gamma $ (i.e., $\Lambda ' \subset (\Lambda ^{\prime 2} \cap L^0)^m(\Lambda ^{\prime 2m+1} \cap \Gamma )$ ). So $\Lambda '$ is commensurable to $\left (\Lambda ^{\prime 2} \cap \Gamma \right )\left ( \Lambda ^{\prime 2} \cap L^0\right )$ by Lemma 2.3. As $i_{|\overline {\Lambda ^{\prime 2}}}$ is proper, we can set $\Lambda _d=i(\overline {\Lambda ^{\prime 2}} \cap \Gamma )$ , $K= i( \overline {\Lambda ^{\prime 2}} \cap L^0)$ and $V_e:=i(L^0)$ . Note that indeed $\Lambda _d$ is discrete because $\Gamma $ is, K is compact because $\Lambda ^{\prime 2} \cap L^0$ is relatively compact and $\overline {\Lambda ^{\prime 2}} \cap L^0$ is an approximate subgroup commensurable to $\Lambda ^{\prime 2} \cap L^0$ because $\overline {\Lambda ^{\prime 2}} \cap L^0 \subset \Lambda ^{\prime 4} \cap L^0$ and Lemma 2.3. Remark finally that since K has nonempty interior in $V_e$ , it must be commensurable with any compact neighbourhood of the origin in $V_e$ . This concludes the proof of (1) and of the result.

Proof. By Proposition 4.4, there are three cases: $V_o=\mathbb {R}$ and $V_e=V_d=\{0\}$ , $V_d=V_e=\mathbb {R}$ and $V_o=\{0\}$ , or $V_d=\mathbb {R}$ and $V_e=V_o=\{0\}$ . The first case corresponds to (4). In the second case, $\Lambda $ is commensurable to a closed interval, so (2). In the third case, $\Lambda =\Lambda _d$ is uniformly discrete. If $\Lambda $ is finite, then we get (1). If $\Lambda $ is infinite, then we get (3) by [Reference FishFis19, Prop. 3.1].

Proof. Recall that Kreitlon-Carolino proved the statement of Theorem 4.6 with the additional assumption that $\Lambda $ is open ([Reference CarolinoCar15, Thm. 1.25]). We will show how Theorem 4.6 reduces to this situation. Let L and $f: L \rightarrow G$ be the locally compact group and the homomorphism given by Theorem 4.1 and $V \subset L$ be a neighbourhood such that $f(V)=\Lambda ^2$ . Then V is a compact neighbourhood of the identity, so there is an open symmetric subset $\tilde {V}$ such that $V \subset \tilde {V} \subset V^2$ . The subset $\tilde {V}$ is thus an open relatively compact $K^6$ -approximate subgroup. But f is an injective continuous homomorphism, so [Reference CarolinoCar15, Thm. 1.25] applied to $\tilde {V}$ yields Theorem 4.6.

Proof. Let $\mathcal {R}$ be a set of representatives of group homomorphisms $f: \Gamma \rightarrow H$ with H locally compact, dense image and $f(\Lambda )$ relatively compact, up to the following equivalence: $f_1: \Gamma \rightarrow H_1$ and $f_2: \Gamma \rightarrow H_2$ are equivalent if there is a continuous group isomorphism $\phi : H_1\rightarrow H_2$ such that $\phi \circ f_1 = f_2$ . Notice that $\mathcal {R}$ is not empty as it contains the map to the trivial group. Then the group $H_{\mathcal {R}}:=\prod _{f: \Gamma \rightarrow H_1\in \mathcal {R}} H_2$ equipped with the product topology is a Hausdorff topological group, and $f_{\mathcal {R}}(\Lambda )$ is relatively compact where $f_{\mathcal {R}}: \Gamma \rightarrow H_{\mathcal {R}}$ is the diagonal map. One readily sees that $f_{\mathcal {R}}$ satisfies the universal property ( $*$ ). The topological group $H_{\mathcal {R}}$ need not be locally compact however. By Theorem 4.1, there are a locally compact group $H_0$ , a group homomorphism $f_0: \Gamma \rightarrow H_0$ and a continuous group homomorphism $\phi : H_0 \rightarrow H_{\mathcal {R}}$ such that $f_{\mathcal {R}}=\phi \circ f_0$ .

Proof. Consider $X \in \mathcal {B}(\Lambda )$ and $X_1, \ldots , X_r$ a Borel partition of X such that there are $f_1,\ldots , f_r \in G$ with $f_i X_i \subset \Lambda $ . We will prove that the quantity $\sum _{i=1}^r m(f_iX_i)$ depends only on X. Defining $\tilde {m}(X)=\sum _{i=1}^r m(f_iX_i)$ then yields the extension we are looking for. Take $Y_1,\ldots , Y_s$ a second partition with $g_1, \ldots , g_s \in G$ as above. We have

$$ \begin{align*} \sum_{i=1}^r m(f_iX_i)& = \sum_{i=1}^r\sum_{j=1}^s m(f_i(X_i \cap Y_j)) \\ & = \sum_{j=1}^s\sum_{i=1}^r m(f_ig_j^{-1} g_j(X_i \cap Y_j)) \\ & = \sum_{j=1}^s\sum_{i=1}^r m(g_j(X_i \cap Y_j))\\ & = \sum_{j=1}^s m(g_jY_j), \end{align*} $$

where we have used local left-invariance to go from the second to the third line.

Proof. Fix m an invariant finitely additive measure as in Definition 5.1. By the strong Tits alternative [Reference BreuillardBre08, Thm. 1.1], there is an integer $N:=N(d)$ such that for every subset $F \subset \operatorname {\mathrm {GL}}_d(k)$ finite, either F generates a virtually soluble subgroup or $F^N$ contains two elements generating a free group. We will apply this result in combination with the following: