Proof. Since $k$ is complete and separable it is Polish and so is the affine space $\mathbb{A}^{n}(k)\simeq k^{n}$ . The set of $k$ -points of a $k$ -affine variety is closed in the $k$ -points of the affine space, hence it is a Polish subspace. It follows that the set of $k$ -points of any $k$ -variety is a Polish space, as this space is a Hausdorff space which admits a finite open covering by Polish open sets, namely the $k$ -points of its $k$ -affine charts.

The fact that the $G$ -orbits in $V$ are locally closed is proven in the appendix of [Reference Bernšteĭn and ZelevinskiĭBZ76]. Note that in [Reference Bernšteĭn and ZelevinskiĭBZ76] the statement is claimed only for non-archimedean local fields, but the proof is actually correct for any field with a complete non-trivial absolute value, which is the setting of [Reference SerreSer06, Part II, ch. III] on which [Reference Bernšteĭn and ZelevinskiĭBZ76] relies.

For $v\in V$ the orbit $\mathbf{G}v$ is a $k$ -subvariety of $\mathbf{V}$ by [Reference BorelBor91, Proposition 6.7]. The stabilizer $\mathbf{H}<\mathbf{G}$ is defined over $k$ by [Reference BorelBor91, Proposition 1.7] and we obtain an orbit map which is defined over $k$ by [Reference BorelBor91, Theorem 6.8]. Clearly $H$ is the stabilizer of $v$ in $G$ and the orbit map restricts to a continuous map from $G/H$ onto $Gv$ . Since $Gv$ is a Polish subset of $V$ , as it is locally closed, we conclude by Effros’ lemma that the latter map is a homeomorphism.◻

On the proof of Theorem 3.3.

In [Reference KaimanovichKai03] the weaker statement that every group possesses a strong boundary in the sense of [Reference Burger and MonodBM02] is proven, but the same proof actually proves Theorem 3.3, as explained in [Reference Glasner and WeissGW16, Remark 4.3].◻

Proof. The fact that $Y$ is $\unicode[STIX]{x1D6E4}$ -amenable follows from [Reference ZimmerZim84, 4.3.5]. To show that $Y$ is $\unicode[STIX]{x1D6E4}$ -metrically ergodic, we fix a measurable $\unicode[STIX]{x1D6E4}$ -equivariant map $\unicode[STIX]{x1D719}:Y\rightarrow V$ , where $(V,d)$ is a metric space on which $\unicode[STIX]{x1D6E4}$ acts isometrically, and argue to show that it is essentially constant. Replacing $d(x,y)$ by $\min (d(x,y),1)$ we may assume $d$ to be bounded. We consider the space $\operatorname{L}^{0}(S,V)^{\unicode[STIX]{x1D6E4}}$ consisting of Lebesgue maps which are $\unicode[STIX]{x1D6E4}$ -equivariant with respect to the left action of $\unicode[STIX]{x1D6E4}$ on $S$ . Given $\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD}\in \operatorname{L}^{0}(S,V)^{\unicode[STIX]{x1D6E4}}$ , note that $d(\unicode[STIX]{x1D6FC}(x),\unicode[STIX]{x1D6FD}(x))$ is a $\unicode[STIX]{x1D6E4}$ -invariant function on $S$ that descends to a well-defined function on $\unicode[STIX]{x1D6E4}\backslash S$ . Using this observation, we define on $\operatorname{L}^{0}(S,V)^{\unicode[STIX]{x1D6E4}}$ a function $D$ by setting, for $\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD}\in \operatorname{L}^{0}(S,V)^{\unicode[STIX]{x1D6E4}}$ ,

$$\begin{eqnarray}D(\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD})=\int _{\unicode[STIX]{x1D6E4}\backslash S}d(\unicode[STIX]{x1D6FC}(x),\unicode[STIX]{x1D6FD}(x))\,dx.\end{eqnarray}$$

Then $D$ is a metric on $\operatorname{L}^{0}(S,V)^{\unicode[STIX]{x1D6E4}}$ , and $S$ acts continuously and isometrically on this space via its right regular action on the domain. The map

$$\begin{eqnarray}\unicode[STIX]{x1D6F7}:Y\rightarrow \operatorname{L}^{0}(S,V)^{\unicode[STIX]{x1D6E4}},\quad y\mapsto [s\mapsto \unicode[STIX]{x1D719}(sy)],\end{eqnarray}$$

defined using Fubini’s theorem (see [Reference MargulisMar91, ch. VII, Lemma 1.3]), is $S$ -equivariant. By the $S$ -metric ergodicity of $Y$ we conclude that $\unicode[STIX]{x1D6F7}$ is essentially constant, and thus $\unicode[STIX]{x1D719}$ is essentially constant too.◻

Proof. By induction, it suffices to consider the case $n=2$ . The proof that $Y=Y_{1}\times Y_{2}$ is $S$ -amenable for $S=S_{1}\times S_{2}$ is well known, so we merely sketch it. Let $\unicode[STIX]{x1D70B}:C\rightarrow Y$ be an $S$ -Borel bundle of convex compact sets over $Y$ . For every $y_{1}\in Y_{1}$ we let $C_{y_{1}}=\unicode[STIX]{x1D70B}^{-1}(\{y_{1}\}\times Y_{2})$ and $\operatorname{L}^{0}(\unicode[STIX]{x1D70B}|_{C_{y_{1}}}\!)$ be the corresponding space of sections. We view these spaces as a convex compact bundle over $Y_{1}$ (using the obvious weak* topology) and denote by $\operatorname{L}^{0}(Y_{1},\operatorname{L}^{0}(\unicode[STIX]{x1D70B}|_{C_{y_{1}}}\!))$ its space of sections. Its obvious identification with $\operatorname{L}^{0}(\unicode[STIX]{x1D70B})$ gives an identification $\operatorname{L}^{0}(\unicode[STIX]{x1D70B})^{S}\simeq \operatorname{L}^{0}(Y_{1},\operatorname{L}^{0}(\unicode[STIX]{x1D70B}|_{C_{y_{1}}}\!)^{S_{2}})^{S_{1}}$ . The right-hand side is non-empty by our amenability assumptions, hence so is the left-hand side.

Let $V$ be an $S$ -metric space and $\unicode[STIX]{x1D719}:Y\rightarrow V$ an $S$ -equivariant Lebesgue map. For almost every $y_{1}\in Y_{1}$ , $\unicode[STIX]{x1D719}_{\{y_{1}\}\times Y_{2}}$ is defined by Fubini’s theorem, and it is essentially constant, as $Y_{2}$ is $S_{2}$ -metrically ergodic. Thus $\unicode[STIX]{x1D719}$ is reduced to a map $\unicode[STIX]{x1D719}^{\prime }:Y_{1}\rightarrow V$ which is again essentially constant, as $Y_{1}$ is $S_{1}$ -metrically ergodic. Thus $\unicode[STIX]{x1D719}$ is essentially constant.◻

Proof. For $f\in L^{\infty }(X\times Y)^{S}$ , using Fubini’s theorem, we define $F:Y\rightarrow L^{\infty }(X)\subset L^{2}(X)$ by $F(y)(x)=f(x,y)$ . It is easily checked that $F$ is $S$ -equivariant. The image of $F$ must be $S$ -invariant, by the metric ergodicity of $Y$ , as the $S$ -action on $L^{2}(X)$ is continuous. By ergodicity of $X$ this image is a constant function, thus $f$ is constant.◻

Proof. We write $S=S_{i}\times S_{i}^{\prime }$ , where $S_{i}^{\prime }=\prod _{j\neq i}S_{j}$ . Since $\unicode[STIX]{x1D70B}_{i}(\unicode[STIX]{x1D6E4})$ is dense in $S_{i}$ , $\unicode[STIX]{x1D6E4}$ acts ergodically on $S_{i}=S/S_{i}^{\prime }$ . Thus $S_{i}^{\prime }$ acts ergodically on $X=S/\unicode[STIX]{x1D6E4}$ . Note that the latter action preserves a probability measure. The diagonal $S_{i}^{\prime }$ -action on $Y_{i}^{\prime }=\prod _{j\neq i}Y_{j}$ is metrically ergodic (Lemma 3.6), and, using Lemma 3.7, we conclude that $X\times Y_{i}^{\prime }$ is $S_{i}^{\prime }$ -ergodic. It follows that $S\times Y_{i}^{\prime }$ is $\unicode[STIX]{x1D6E4}\times S_{i}^{\prime }$ -ergodic, where $\unicode[STIX]{x1D6E4}$ acts by its left action on $S$ , and $S_{i}^{\prime }$ acts diagonally via it right action on $S$ and its given action on $Y_{i}^{\prime }$ . Note that the Lebesgue isomorphism $S\times Y_{i}^{\prime }\rightarrow S\times Y_{i}^{\prime }$ , $(s,y)\mapsto (s,sy)$ intertwines the above-described action $\unicode[STIX]{x1D6E4}\times S_{i}^{\prime }\curvearrowright S\times Y_{i}^{\prime }$ :

$$\begin{eqnarray}\unicode[STIX]{x1D6FE}:(s,y)\mapsto (\unicode[STIX]{x1D6FE}s,\unicode[STIX]{x1D6FE}y),\quad s^{\prime }:(s,y)\mapsto (s{s^{\prime }}^{-1},y).\end{eqnarray}$$

Taking the $S_{i}^{\prime }$ -orbit space, we conclude that $\unicode[STIX]{x1D6E4}$ acts ergodically on $S_{i}\times Y_{i}^{\prime }$ .◻

Proof. Fix a Borel map $\unicode[STIX]{x1D719}$ in its class of equivalent measurable maps. The Borel set

$$\begin{eqnarray}A=\{(x,z)\in X\times Z\mid \unicode[STIX]{x1D719}(\unicode[STIX]{x1D6FE}x,\unicode[STIX]{x1D6FE}z)=\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D719}(x,z),~\unicode[STIX]{x1D6FE}\in \unicode[STIX]{x1D6E4}\}\end{eqnarray}$$

is conull in $X\times Z$ . So by Fubini’s theorem ([Reference KechrisKec95, Theorem 8.19]), there is a conull set $X_{0}\subset X$ so that for each $x\in X_{0}$ for almost every $z\in Z$ one has $(x,z)\in A$ . Since the action map $S\times X\rightarrow X$ is non-singular, the preimage of $X_{0}$ is conull in $S\times X$ , and so for a conull set $X_{1}\subset X$ for $x\in X_{1}$ the set $S_{x}=\{s\in S\mid sx\in X_{0}\}$ is conull in $S$ . Therefore for every $x\in X_{1}$ , $s\in S_{x}$ , and almost every $z\in Z$ ,

$$\begin{eqnarray}\hat{\unicode[STIX]{x1D719}}_{x}(\unicode[STIX]{x1D6FE}s,\unicode[STIX]{x1D6FE}z)=\unicode[STIX]{x1D719}((\unicode[STIX]{x1D6FE}s)x,\unicode[STIX]{x1D6FE}z)=\unicode[STIX]{x1D719}(\unicode[STIX]{x1D6FE}(sx),\unicode[STIX]{x1D6FE}z)=\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D719}(sx,z)=\hat{\unicode[STIX]{x1D719}}_{x}(s,z)\end{eqnarray}$$

for all $\unicode[STIX]{x1D6FE}\in \unicode[STIX]{x1D6E4}$ .◻

Proof. Denote $V=\mathbf{V}(k)$ and $G=\mathbf{G}(k)$ . By Proposition 2.1 we know that every $G$ -orbit is locally closed in $G$ . Consider the orbit space $V/G$ endowed with the quotient topology and Borel structure. The map $Y\rightarrow V\rightarrow V/G$ is Borel. We push the measure class given on $Y$ and obtain a measure class on $V/G$ . Since $V$ is Polish it has a countable basis, thus so does $V/G$ . Let $\{U_{n}~|~n\in \mathbb{N}\}$ be sequence of subsets of $V/G$ consisting of the elements of a countable basis and their complements. Set

$$\begin{eqnarray}U=\bigcap \{U_{n}~|~n\in \mathbb{N},~U_{n}\text{ has a full measure in }V/G\}.\end{eqnarray}$$

Then $U$ has a full measure and, in particular, is non-empty. We claim that $U$ is a singleton. Indeed, since every $G$ -orbit is locally closed in $V$ , the quotient topology of $V/G$ is $T_{0}$ , thus if $U$ contained two distinct points we could find a basis set $U_{n}$ which separates them, but by the ergodicity of $G$ on $Y$ either it or its complement would be of full measure, which contradicts the definition of $U$ .

Fixing $v\in V$ which is in the preimage of $U$ , we conclude that $\unicode[STIX]{x1D719}(X)$ is essentially contained in $Gv$ . Let $\mathbf{H}<\mathbf{G}$ be the stabilizer of $v$ and $H=\mathbf{H}(k)$ . By Proposition 2.1 we obtain a $k$ -algebraic map $i:\mathbf{G}/\mathbf{H}\rightarrow \mathbf{V}$ whose restriction to $G/H$ gives a homeomorphism with the orbit $Gv$ . We let $\unicode[STIX]{x1D719}_{\mathbf{G}/\mathbf{H}}=(i|_{G/H})^{-1}\circ \unicode[STIX]{x1D719}$ . By extending the codomain of $\unicode[STIX]{x1D719}_{\mathbf{G}/\mathbf{H}}$ to $\mathbf{G}/\mathbf{H}(k)$ via the embedding $G/H{\hookrightarrow}\mathbf{G}/\mathbf{H}(k)$ , we are done.◻

Proof. We consider the collection

$$\begin{eqnarray}\{\mathbf{H}<\mathbf{G}~|~\mathbf{H}\text{ is defined over }k\text{ and there exists a coset representation to }\mathbf{G}/\mathbf{H}\}.\end{eqnarray}$$

This is a non-empty collection as it contains $\mathbf{G}$ . By the Neotherian property, this collection contains a minimal element. We choose such a minimal element $\mathbf{H}_{0}$ and fix corresponding $\unicode[STIX]{x1D719}_{0}:Y\rightarrow (\mathbf{G}/\mathbf{H}_{0})(k)$ . We argue to show that this coset representation is the required initial object.

Fix any algebraic representation of $Y$ , $\mathbf{V}$ . It is clear that, if it exists, a morphism of algebraic representations from $\mathbf{G}/\mathbf{H}_{0}$ to $\mathbf{V}$ is unique, as two different $\mathbf{G}$ -maps $\mathbf{G}/\mathbf{H}_{0}\rightarrow \mathbf{V}$ agree nowhere. All that remains is to show existence. To this end, we consider the product representation $\mathbf{V}\times \mathbf{G}/\mathbf{H}_{0}$ given by $\unicode[STIX]{x1D719}=\unicode[STIX]{x1D719}_{\mathbf{V}}\times \unicode[STIX]{x1D719}_{0}$ . Applying Proposition 4.2 to this product representation, we obtain the commutative diagram

(4.1)

By the minimality of $\mathbf{H}_{0}$ , the $\mathbf{G}$ -morphism $p_{2}\circ i:\mathbf{G}/\mathbf{H}\rightarrow \mathbf{G}/\mathbf{H}_{0}$ must be a $k$ -isomorphism. We thus obtain the $k$ - $\mathbf{G}$ -morphism

Proof. Using the amenability of $Y$ , it follows from [Reference Bader, Duchesne and LécureuxBDL17, Corollary 1.17] that if the gate of $Y$ is non-trivial then there exist a separable metric space $V$ on which $\mathbf{G}(k)$ acts by isometries and with bounded stabilizers and an $S$ -equivariant map $Y\rightarrow V$ . But the latter possibility is ruled out by the assumption that $Y$ is $S$ -metrically ergodic and the assumption that $\unicode[STIX]{x1D70C}(S)<\mathbf{G}(k)$ is unbounded.

Proof. For a given $s^{\prime }\in S^{\prime }$ we consider the diagram

(4.2)

where we denote by $\unicode[STIX]{x1D70C}^{\prime }(s^{\prime })$ the dotted arrow, which is the unique $k$ -algebraic $\mathbf{G}$ -equivariant morphism $\mathbf{G}/\mathbf{H}\rightarrow \mathbf{G}/\mathbf{H}$ given by the fact that $\unicode[STIX]{x1D719}$ is a gate map and $\unicode[STIX]{x1D719}\circ s^{\prime }$ is an algebraic representation of $Y$ . By the uniqueness of the dotted arrow, it is easily checked that the correspondence $s^{\prime }\mapsto \unicode[STIX]{x1D70C}^{\prime }(s^{\prime })$ forms a homomorphism from $S^{\prime }$ to the group of $k$ - $\mathbf{G}$ -automorphisms of $\mathbf{G}/\mathbf{H}$ which we identify with $N_{\mathbf{G}}(\mathbf{H})/\mathbf{H}(k)$ using Proposition 4.6. All that remains is to check the continuity of $\unicode[STIX]{x1D70C}^{\prime }$ .

To simplify the notation we let $V=\mathbf{G}/\mathbf{H}(k)$ , $M=N_{\mathbf{G}}(\mathbf{H})/\mathbf{H}(k)$ and $U=\operatorname{L}^{0}(Y,V)$ . We endow $U$ with the action of $M$ by postcomposition, and the action of $S^{\prime }$ by precomposition. By the fact that $M$ acts freely on $V$ , we obtain that $M$ acts freely on $U$ as well. Using Proposition 2.1, [Reference ZimmerZim84, Proposition 3.3.1] gives that the $M$ -action on $U$ has locally closed orbits.Footnote ¹ It follows that the $M$ -orbit map $M\rightarrow U$ , $m\mapsto m\circ \unicode[STIX]{x1D719}$ is a homeomorphism onto its image, $M\unicode[STIX]{x1D719}$ . We let $\unicode[STIX]{x1D6FC}:M\unicode[STIX]{x1D719}\rightarrow M$ be its inverse.

By the fact that the map $S^{\prime }\times Y\rightarrow V$ , $(s^{\prime },y)\mapsto \unicode[STIX]{x1D719}(s^{\prime }y)$ is almost everywhere defined and measurable, we obtain that the associated map $\unicode[STIX]{x1D6FD}:S^{\prime }\rightarrow U$ , $s^{\prime }\mapsto \unicode[STIX]{x1D719}\circ s^{\prime }$ is almost everywhere defined and measurable; see [Reference MargulisMar91, ch. VII, Lemma 1.3]. Since by the definition of $\unicode[STIX]{x1D70C}^{\prime }$ , $\unicode[STIX]{x1D719}\circ s^{\prime }=\unicode[STIX]{x1D70C}^{\prime }(s^{\prime })\circ \unicode[STIX]{x1D719}$ , we conclude that $\unicode[STIX]{x1D70C}^{\prime }$ agrees almost everywhere with $\unicode[STIX]{x1D6FC}\circ \unicode[STIX]{x1D6FD}$ which is almost everywhere defined and measurable. It follows that $\unicode[STIX]{x1D70C}^{\prime }$ is measurable. By [Reference RosendalRos09, Lemma 2.1] we conclude that $\unicode[STIX]{x1D70C}^{\prime }$ is a continuous homomorphism.◻

Proof of the lemma.

The map $\unicode[STIX]{x1D713}:S^{\prime }\rightarrow \mathbf{G}(k)$ given by $s^{\prime }\mapsto \unicode[STIX]{x1D719}(s^{\prime })\unicode[STIX]{x1D70C}^{\prime }(s^{\prime })^{-1}$ is $S^{\prime }$ -invariant, hence constant almost everywhere. We let $g$ be its essential value. Given $s\in S$ , we pick $s^{\prime }\in S^{\prime }$ such that both $s^{\prime }$ and $\unicode[STIX]{x1D703}(s)s^{\prime }$ are both $\unicode[STIX]{x1D719}$ and $\unicode[STIX]{x1D713}$ -generic. We obtain

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D70C}(s)\unicode[STIX]{x1D719}(s^{\prime }) & = & \displaystyle \unicode[STIX]{x1D719}(\unicode[STIX]{x1D703}(s)s^{\prime })=\unicode[STIX]{x1D713}(\unicode[STIX]{x1D703}(s)s^{\prime })\unicode[STIX]{x1D70C}^{\prime }(\unicode[STIX]{x1D703}(s)s^{\prime })\nonumber\\ \displaystyle & = & \displaystyle g\unicode[STIX]{x1D70C}^{\prime }(\unicode[STIX]{x1D703}(s))\unicode[STIX]{x1D70C}^{\prime }(s^{\prime })=g\unicode[STIX]{x1D70C}^{\prime }(\unicode[STIX]{x1D703}(s))\unicode[STIX]{x1D713}(s^{\prime })^{-1}\unicode[STIX]{x1D719}(s^{\prime })\nonumber\\ \displaystyle & = & \displaystyle g\unicode[STIX]{x1D70C}^{\prime }(\unicode[STIX]{x1D703}(s))g^{-1}\unicode[STIX]{x1D719}(s^{\prime }),\nonumber\end{eqnarray}$$

and conclude that $\unicode[STIX]{x1D70C}(s)=\operatorname{inn}(g)\circ \unicode[STIX]{x1D70C}^{\prime }\circ \unicode[STIX]{x1D703}(s)$ .◻

Proof of Theorem 5.1.

First, we replace the unknown $S^{\prime }$ -Lebesgue space $Z$ with the group $S^{\prime }$ endowed with its Haar measure. We do this by restricting $\unicode[STIX]{x1D70E}$ to a generic orbit. More pedantically, we argue as follows. We consider the map

$$\begin{eqnarray}S^{\prime }\times Z\rightarrow \mathbf{V}(k),\quad (s^{\prime },z)\mapsto \unicode[STIX]{x1D70E}(s^{\prime }z)\end{eqnarray}$$

and use Fubini’s theorem in order to fix a generic point $z\in Z$ such that the map $s^{\prime }\rightarrow \unicode[STIX]{x1D70E}(s^{\prime }z)$ is almost everywhere defined on $S^{\prime }$ . We observe that the latter map $S^{\prime }\rightarrow \mathbf{V}(k)$ is $S$ -equivariant. Replacing $\unicode[STIX]{x1D70E}$ with this map, we assume, as we may, that $Z=S^{\prime }$ and $\unicode[STIX]{x1D70E}:S^{\prime }\rightarrow \mathbf{V}$ is an $S$ -equivariant Lebesgue map.

Note that $S^{\prime }$ is $S$ -ergodic, as $\unicode[STIX]{x1D703}(S)$ is dense in $S^{\prime }$ . Therefore we may use § 4 in the setting $Y=S^{\prime }$ and view $\unicode[STIX]{x1D70E}$ as an algebraic representation of the $S$ -space $S^{\prime }$ . By Theorem 4.3 we obtain the algebraic gate of $S^{\prime }$ , $\unicode[STIX]{x1D719}:S^{\prime }\rightarrow \mathbf{G}/\mathbf{H}(k)$ for some $k$ -algebraic group $\mathbf{H}\leqslant \mathbf{G}$ . Since there exists a $\mathbf{G}$ -equivariant $k$ -morphism $\mathbf{G}/\mathbf{H}\rightarrow \mathbf{V}$ and $\mathbf{V}$ has no $\mathbf{G}$ -fixed point, we conclude that $\mathbf{H}\lneq \mathbf{G}$ .

We now consider the right action of the group $S^{\prime }$ on the space $S^{\prime }$ and conclude by Theorem 4.7 the existence of a continuous homomorphism $\unicode[STIX]{x1D70C}^{\prime }:S^{\prime }\rightarrow \mathbf{N}/\mathbf{H}(k)$ , where $\mathbf{N}=N_{\mathbf{G}}(\mathbf{H})$ , such that $\unicode[STIX]{x1D719}$ is $S\times S^{\prime }$ -equivariant via $\unicode[STIX]{x1D70C}\times \unicode[STIX]{x1D70C}^{\prime }$ .

We claim that $\mathbf{H}=\{e\}$ . Assume not. Then by the fact that $\mathbf{G}$ is simple we conclude that $\mathbf{N}\lneq \mathbf{G}$ . Composing $\unicode[STIX]{x1D719}$ with the map $\mathbf{G}/\mathbf{H}(k)\rightarrow \mathbf{G}/\mathbf{N}(k)$ , we obtain a map $S^{\prime }\rightarrow \mathbf{G}/\mathbf{N}(k)$ which is left $S$ -equivariant and right $S^{\prime }$ -invariant. By its right $S^{\prime }$ -invariance, it is essentially constant. By the left $S$ -equivariance, its essential image is $\unicode[STIX]{x1D70C}(S)$ -invariant, hence also $\mathbf{G}(k)$ -invariant, as $\unicode[STIX]{x1D70C}(S)$ is Zariski dense in $\mathbf{G}$ . This is absurd, thus the claim is proven.

It follows from the claim that $\mathbf{N}=\mathbf{G}$ and thus $\unicode[STIX]{x1D70C}^{\prime }$ is a continuous homomorphism from $S^{\prime }$ to $\mathbf{G}(k)$ and $\unicode[STIX]{x1D719}$ is a map from $S^{\prime }$ to $\mathbf{G}(k)$ which is $S\times S^{\prime }$ -equivariant via $\unicode[STIX]{x1D70C}\times \unicode[STIX]{x1D70C}^{\prime }:S\times S^{\prime }\rightarrow \mathbf{G}(k)\times \mathbf{G}(k)$ . We are in a situation to apply Lemma 5.2 and obtain $g\in \mathbf{G}(k)$ such that $\unicode[STIX]{x1D70C}=\operatorname{inn}(g)\circ \unicode[STIX]{x1D70C}^{\prime }\circ \unicode[STIX]{x1D703}$ . By setting $\bar{\unicode[STIX]{x1D70C}}=\operatorname{inn}(g)\circ \unicode[STIX]{x1D70C}^{\prime }$ , we are done.◻

Proof. Assume $\mathbf{L}\neq \mathbf{G}$ . The $T_{i}$ -equivariance property of $\unicode[STIX]{x1D719}_{i,J}$ implies that this map descends to a measurable $\unicode[STIX]{x1D6E4}$ -map

$$\begin{eqnarray}B_{J}~\overset{}{\longrightarrow }~\mathbf{G}/\mathbf{L}(k).\end{eqnarray}$$

Note that this is possible only if $J\neq \emptyset$ . Pick $j\in J$ and use Lemma 3.9 to obtain a measurable $\unicode[STIX]{x1D6E4}$ -map $\unicode[STIX]{x1D713}:T_{j}\times B_{J\setminus \{j\}}\rightarrow \mathbf{G}/\mathbf{L}(k)$ . Under the assumption that $\mathbf{L}$ is a proper subgroup in $\mathbf{G}$ , the gate of $T_{j}\times B_{J\setminus \{j\}}$ is non-trivial, contradicting the minimality property of $J\subset [n]$ .

Therefore $\mathbf{L}=\mathbf{G}$ . This implies that $\mathbf{N}_{\mathbf{G}}(\mathbf{H}_{i,J})=\mathbf{G}$ , meaning that $\mathbf{H}_{i,J}$ is normal in the simple group $\mathbf{G}$ . Since $\mathbf{H}_{i,J}$ is known to be a proper subgroup, we have $\mathbf{H}=\{e\}$ . Thus the previously constructed continuous homomorphism $\unicode[STIX]{x1D70E}$ and the measurable map $\unicode[STIX]{x1D719}=\unicode[STIX]{x1D719}_{i,J}$ range into $\mathbf{G}(k)$ . Finally, $\unicode[STIX]{x1D70E}(T_{i})$ is Zariski dense in $\mathbf{G}$ because its Zariski closure is $\mathbf{L}$ .◻

Proof. Let us show that for every $j\in J$ for almost every $t\in T_{j}$ one has $\unicode[STIX]{x1D719}(x,b)=\unicode[STIX]{x1D719}(x,tb)$ almost everywhere on $T_{i}\times B_{J}$ . Since $T_{J}$ acts ergodically on $B_{J}$ the claim would follow. The case of $J=\emptyset$ being trivial, we assume $J$ is non-empty, fix $j\in J$ , and set $K=J\setminus \{j\}$ . Choose almost every $b\in B_{J}$ to create a $\unicode[STIX]{x1D6E4}$ -equivariant lift by Lemma 3.9,

$$\begin{eqnarray}\hat{\unicode[STIX]{x1D719}}:T_{i}\times T_{j}\times B_{K}\overset{}{\longrightarrow }\mathbf{G}(k).\end{eqnarray}$$

Consider the setting of § 4 for the group $S=\unicode[STIX]{x1D6E4}\times T_{i}$ , the homomorphism

$$\begin{eqnarray}\unicode[STIX]{x1D70C}\times \unicode[STIX]{x1D70E}:\unicode[STIX]{x1D6E4}\times T_{i}\overset{}{\longrightarrow }\mathbf{G}(k)\times \mathbf{G}(k),\end{eqnarray}$$

and the Lebesgue $S$ -space $Y=T_{i}\times T_{j}\times B_{K}$ with the action

$$\begin{eqnarray}(\unicode[STIX]{x1D6FE},t):(x,y,b)\mapsto (\unicode[STIX]{x1D6FE}xt^{-1},\unicode[STIX]{x1D6FE}y,\unicode[STIX]{x1D6FE}b).\end{eqnarray}$$

Note that the $S$ -action on $Y$ is ergodic, because the $\unicode[STIX]{x1D6E4}$ -action on $T_{j}\times B_{K}$ is ergodic by Lemma 3.8. Write $\unicode[STIX]{x1D6E5}_{\mathbf{G}}<\mathbf{G}\times \mathbf{G}$ for the diagonal subgroup. The measurable map

$$\begin{eqnarray}\unicode[STIX]{x1D713}:Y\rightarrow \mathbf{G}\times \mathbf{G}/\unicode[STIX]{x1D6E5}_{\mathbf{G}}(k),\quad \unicode[STIX]{x1D713}(x,y,b)=(\hat{\unicode[STIX]{x1D719}}(x,y,b),\unicode[STIX]{x1D70E}(x))\unicode[STIX]{x1D6E5}_{\mathbf{G}}(k)\end{eqnarray}$$

is $\unicode[STIX]{x1D70C}\times \unicode[STIX]{x1D70E}$ -equivariant. Thus the gate of $Y$ is given by:

• ∙ a $k$ -variety $\mathbf{V}=\mathbf{G}\times \mathbf{G}/\mathbf{D}$ where $\mathbf{D}$ is a $k$ -algebraic subgroup $\mathbf{D}<\unicode[STIX]{x1D6E5}_{\mathbf{G}}$ ;

• ∙ a measurable $\unicode[STIX]{x1D6E4}\times T_{i}$ -map $\hat{\unicode[STIX]{x1D713}}:Y\rightarrow \mathbf{V}(k)$ for which $\unicode[STIX]{x1D713}=\unicode[STIX]{x1D70B}\circ \hat{\unicode[STIX]{x1D713}}$ where $\unicode[STIX]{x1D70B}:\mathbf{V}(k)\rightarrow \mathbf{G}\times \mathbf{G}/\unicode[STIX]{x1D6E5}_{\mathbf{G}}(k)$ is the projection.

In fact, we must have $\mathbf{D}=\unicode[STIX]{x1D6E5}_{\mathbf{G}}$ . Indeed, otherwise under the projection to the first factor $\unicode[STIX]{x1D70B}_{1}:\mathbf{G}\times \mathbf{G}\rightarrow \mathbf{G}$ , the image $\mathbf{M}=\unicode[STIX]{x1D70B}_{1}(\mathbf{D})$ would be a proper subgroup of $\mathbf{G}$ , and the map

$$\begin{eqnarray}T_{i}\times T_{j}\times B_{K}\overset{\hat{\unicode[STIX]{x1D713}}}{\longrightarrow }\mathbf{V}(k)\overset{\unicode[STIX]{x1D70B}_{1}}{\longrightarrow }\mathbf{G}/\mathbf{M}(k)\end{eqnarray}$$

would factor through a measurable $\unicode[STIX]{x1D6E4}$ -map $T_{j}\times B_{K}\rightarrow \mathbf{G}/\mathbf{M}(k)$ , contrary to our assumption of minimality for $J\subset [n]$ . Thus $\mathbf{D}=\unicode[STIX]{x1D6E5}_{\mathbf{G}}$ and $\hat{\unicode[STIX]{x1D713}}=\unicode[STIX]{x1D713}$ .

Consider the action of $S^{\prime }=T_{j}$ on $Y=T_{i}\times T_{j}\times B_{K}$ by $s:(x,y,b)\mapsto (x,ys^{-1},b)$ . It commutes with the $S=\unicode[STIX]{x1D6E4}\times T_{i}$ -action. We can apply Lemma 4.7 to obtain a continuous homomorphism $\unicode[STIX]{x1D70F}:T_{j}\rightarrow N_{\mathbf{G}\times \mathbf{G}}(\unicode[STIX]{x1D6E5}_{\mathbf{G}})/\unicode[STIX]{x1D6E5}_{\mathbf{G}}(k)$ for which $\unicode[STIX]{x1D713}=\hat{\unicode[STIX]{x1D713}}$ is $T_{j}$ -equivariant. But as $\mathbf{G}$ is center-free, $\unicode[STIX]{x1D6E5}_{\mathbf{G}}$ is its own normalizer in $\mathbf{G}\times \mathbf{G}$ . So $\unicode[STIX]{x1D70F}$ is trivial, and therefore $\unicode[STIX]{x1D713}$ is $T_{j}$ -invariant. Since $\unicode[STIX]{x1D713}$ was a lift of $\unicode[STIX]{x1D719}:T_{i}\times B_{J}\rightarrow \mathbf{G}(k)$ for almost every $b\in B_{J}$ , we proved that $\unicode[STIX]{x1D719}$ is essentially $T_{j}$ -invariant, and this is true for all $j\in J$ . This completes the proof of the claim.◻

Proof. As before, without loss of generality we assume that the absolute value on $k$ is non-trivial and as a metric space $k$ is complete and separable. We will prove the lemma by contradiction, assuming there exists $s_{0}\in S$ with $\unicode[STIX]{x1D70C}_{1}(s_{0})\neq \unicode[STIX]{x1D70C}_{2}(s_{0})$ .

We will first show that $\unicode[STIX]{x1D70C}_{1}(\unicode[STIX]{x1D6E4})=\unicode[STIX]{x1D70C}_{2}(\unicode[STIX]{x1D6E4})$ is Zariski dense in $\mathbf{G}$ . We denote by $L_{1}$ the closure of $\unicode[STIX]{x1D70C}_{1}(S)$ in $\mathbf{G}(k)$ . By [Reference Bader, Duchesne and LécureuxBDL17, Proposition 1.9] there exists a $k$ -subgroup $\mathbf{H}_{1}<\mathbf{G}$ which is normalized by $L_{1}$ such that $L_{1}$ has a precompact image in $N_{\mathbf{G}}(\mathbf{H}_{1})/\mathbf{H}_{1}(k)$ and such that for every $k$ - $\mathbf{G}$ -variety $\mathbf{V}$ , any $L_{1}$ -invariant finite measure is supported on the subvariety of $\mathbf{H}_{1}$ -fixed points, $\mathbf{V}^{\mathbf{H}_{1}}\cap \mathbf{V}(k)$ . As $\unicode[STIX]{x1D70C}_{1}$ has a Zariski dense image, $L_{1}$ is Zariski dense in $\mathbf{G}$ and we obtain that $\mathbf{H}_{1}$ is normal in $\mathbf{G}$ . As $\mathbf{G}$ is $k$ -simple, we conclude that either $\mathbf{H}_{1}=\{e\}$ or $\mathbf{H}_{1}=\mathbf{G}$ . The first option is ruled out, as $\unicode[STIX]{x1D70C}_{1}(S)$ is unbounded while $L_{1}$ has a precompact image in $N_{\mathbf{G}}(\mathbf{H}_{1})/\mathbf{H}_{1}(k)$ . Thus $\mathbf{H}_{1}=\mathbf{G}$ . We denote by $\mathbf{G}_{1}$ the Zariski closure of $\unicode[STIX]{x1D70C}_{1}(\unicode[STIX]{x1D6E4})$ and let $\mathbf{V}_{1}=\mathbf{G}/\mathbf{G}_{1}$ . The map $\unicode[STIX]{x1D70C}_{1}$ gives rise to an obvious map $S/\unicode[STIX]{x1D6E4}\rightarrow \mathbf{V}_{1}(k)$ . By pushing the $S$ -invariant measure of $S/\unicode[STIX]{x1D6E4}$ we obtain an $L_{1}$ -invariant finite measure on $\mathbf{V}_{1}(k)$ . It follows that this measure is supported on the subvariety of $\mathbf{G}$ -fixed points. In particular, there exists a $\mathbf{G}$ -fixed point in $\mathbf{V}_{1}=\mathbf{G}/\mathbf{G}_{1}$ and we conclude that $\mathbf{G}_{1}=\mathbf{G}$ . Thus, indeed, $\unicode[STIX]{x1D70C}_{1}(\unicode[STIX]{x1D6E4})$ is Zariski dense in $\mathbf{G}$ .

Next we consider the homomorphism $R=\unicode[STIX]{x1D70C}_{1}\times \unicode[STIX]{x1D70C}_{2}:S\rightarrow \mathbf{G}(k)\times \mathbf{G}(k)$ , we let $L$ be the closure of $R(S)$ and let $\mathbf{L}<\mathbf{G}\times \mathbf{G}$ be the Zariski closure of $L$ . We note that, by the discussion above, the Zariski closure of $R(\unicode[STIX]{x1D6E4})$ coincides with the diagonal group $\unicode[STIX]{x1D6E5}_{\mathbf{G}}<\mathbf{G}\times \mathbf{G}$ . In particular, $\unicode[STIX]{x1D6E5}_{\mathbf{G}}$ is contained in $\mathbf{L}$ . We argue to show that $\mathbf{L}=\mathbf{G}\times \mathbf{G}$ . The $k$ -subgroup $\mathbf{L}\cap (\mathbf{G}\times \{e\})$ is non-trivial, as the element $\unicode[STIX]{x1D70C}_{1}(s_{0})\unicode[STIX]{x1D70C}_{2}(s_{0})^{-1}$ is contained in its group of $k$ -points, since both $(\unicode[STIX]{x1D70C}_{1}(s_{0}),\unicode[STIX]{x1D70C}_{2}(s_{0}))$ and $(\unicode[STIX]{x1D70C}_{2}(s_{0}),\unicode[STIX]{x1D70C}_{2}(s_{0}))$ are elements of $\mathbf{L}(k)$ . As this subgroup is normalized by $\mathbf{L}$ , hence by $\unicode[STIX]{x1D6E5}_{\mathbf{G}}$ , we obtain that this is a non-trivial normal $k$ -subgroup of $\mathbf{G}\times \{e\}$ . As $\mathbf{G}\times \{e\}$ is $k$ -simple, we conclude that $\mathbf{L}\cap (\mathbf{G}\times \{e\})=\mathbf{G}\times \{e\}$ , that is, $\mathbf{G}\times \{e\}<\mathbf{L}$ . As $\mathbf{G}\times \mathbf{G}$ is generated by $\mathbf{G}\times \{e\}$ and $\unicode[STIX]{x1D6E5}_{\mathbf{G}}$ , we conclude that indeed $\mathbf{L}=\mathbf{G}\times \mathbf{G}$ .

We use [Reference Bader, Duchesne and LécureuxBDL17, Proposition 1.9] again, this time for the $k$ -group $\mathbf{G}\times \mathbf{G}$ and the Zariski dense closed subgroup $L<(\mathbf{G}\times \mathbf{G})(k)$ , and conclude having a normal $k$ -subgroup $\mathbf{H}<\mathbf{G}\times \mathbf{G}$ such that $L$ has a precompact image in $(\mathbf{G}\times \mathbf{G})/\mathbf{H}(k)$ and such that for every $k$ - $\mathbf{G}$ -variety $\mathbf{V}$ , any $L$ -invariant finite measure is supported on the subvariety of $\mathbf{H}$ -fixed points, $\mathbf{V}^{\mathbf{H}}\cap \mathbf{V}(k)$ . We consider $\mathbf{V}=(\mathbf{G}\times \mathbf{G})/\unicode[STIX]{x1D6E5}_{\mathbf{G}}$ . The map $R$ gives rise to an obvious map $S/\unicode[STIX]{x1D6E4}\rightarrow \mathbf{V}(k)$ . By pushing the $S$ -invariant measure of $S/\unicode[STIX]{x1D6E4}$ we obtain an $L$ -invariant finite measure on $\mathbf{V}(k)$ . It follows that this measure is supported on the subvariety of $\mathbf{H}$ -fixed points. Since the non-trivial normal $k$ -subgroups of $\mathbf{G}\times \mathbf{G}$ , namely $\mathbf{G}\times \{e\}$ , $\{e\}\times \mathbf{G}$ and $\mathbf{G}\times \mathbf{G}$ itself, have no fixed points in $\mathbf{V}$ , we conclude that $\mathbf{H}=\{e\}$ . It follows that $L$ is compact in $\mathbf{G}\times \mathbf{G}$ , which contradicts the assumption that $\unicode[STIX]{x1D70C}_{1}$ has an unbounded image.◻

Proof. The case where $K$ is the fraction field of $R$ is dealt with in [Reference Breuillard and GelanderBG07, Lemma 2.1]. If $K$ is finitely generated, say by a finite set $A$ , upon replacing $R$ with the ring $R^{\prime }$ generated by $A\cup B$ where $B$ is a finite set of generators of $R$ , we are reduced to the previous case. The general case follows from the case where $K$ is finitely generated as follows. Denote by $K_{0}$ the fraction field of $R$ and observe that it is finitely generated as a field. Let $k_{0}$ be a corresponding local field such that there exists an embedding $i:K_{0}{\hookrightarrow}k_{0}$ with $i(I)$ unbounded. Let $\bar{k}_{0}$ be an algebraic closure of $k_{0}$ and note that the absolute value of $k_{0}$ extends uniquely to an absolute value on $\bar{k}_{0}$ [Reference Bosch, Güntzer and RemmertBGR84, § 3.2.4, Theorem 2]. Let $k$ be the completion of $\bar{k}_{0}$ with respect to this absolute value. Note that $k$ is an algebraically closed field [Reference Bosch, Güntzer and RemmertBGR84, § 3.4.1, Proposition 3] which is of uncountable transcendental degree over its prime field. Denote by $j$ the composition $K_{0}{\hookrightarrow}k_{0}{\hookrightarrow}k$ and note that $j(I)$ is unbounded. As $K_{0}$ is countable, $k$ is of uncountable transcendental degree over $j(K_{0})$ . As $K$ is countable and $k$ is algebraically closed and of uncountable transcendental degree over $j(K_{0})$ , $j$ extends to an embedding $K{\hookrightarrow}k$ .◻

Proof. We let $\unicode[STIX]{x1D6E4}^{\prime }<\unicode[STIX]{x1D6E4}$ be an infinite finitely generated group. We fix an injective $K$ -representation $\mathbf{G}\rightarrow \operatorname{GL}_{n}$ , let $I\subset K$ be the set of matrix coefficients of $\unicode[STIX]{x1D6E4}^{\prime }$ and let $R<K$ be the subring generated by $I$ . Note that $I$ is infinite and $R$ is finitely generated. We let $K{\hookrightarrow}k$ be a field extension as provided by Lemma 7.3 and note that the image of $\unicode[STIX]{x1D6E4}$ is unbounded in $\mathbf{G}(k)$ , as the image of $\unicode[STIX]{x1D6E4}^{\prime }$ is unbounded in $\operatorname{GL}_{n}(k)$ .◻

Proof. Consider the Zariski closure $\mathbf{H}=\overline{\unicode[STIX]{x1D6E4}}^{Z}$ and let $\mathbf{H}^{0}$ be its identity connected component. By [Reference BorelBor91, AG, Theorem 14.4] $\mathbf{H}$ is defined over $K$ and $\unicode[STIX]{x1D6E4}<\mathbf{H}(K)$ . We set $\unicode[STIX]{x1D6E4}^{\prime }=\unicode[STIX]{x1D6E4}\cap \mathbf{H}^{0}(K)$ . As $\mathbf{H}^{0}(K)<\mathbf{H}(K)$ is of finite index, $\unicode[STIX]{x1D6E4}^{\prime }<\unicode[STIX]{x1D6E4}$ is of finite index. It follows that $\unicode[STIX]{x1D6E4}^{\prime }<\unicode[STIX]{x1D6EC}$ is commensurated. Thus $\mathbf{H}^{0}<\mathbf{G}$ is commensurated. As the connected group $\mathbf{H}^{0}$ has no Zariski closed subgroups of finite index, we conclude that $\mathbf{H}^{0}$ is normal in $\mathbf{G}$ . By the assumption that $\unicode[STIX]{x1D6E4}$ is infinite we obtain that $\mathbf{H}^{0}$ is non-trivial. By the simplicity of $\mathbf{G}$ we conclude that $\mathbf{H}^{0}=\mathbf{G}$ . In particular, $\mathbf{H}=\mathbf{G}$ . Thus $\unicode[STIX]{x1D6E4}$ is indeed Zariski dense in $\mathbf{G}$ .◻

Proof. We assume, as we may, that $k$ is non-discrete, complete and separable as a metric space. We argue as in the proof of [Reference Bader, Duchesne and LécureuxBDL17, Theorem 6.1]. By [Reference Bader, Duchesne and LécureuxBDL17, Lemma 4.5] we may assume $\mathbf{G}$ is adjoint and, upon replacing $R$ with the product of its projections to the various simple factors, the problem is easily reduced to the case where $\mathbf{G}$ is $k$ -simple. We thus assume $\mathbf{G}$ is indeed $k$ -simple.

Using [Reference BorelBor91, Proposition 1.10], there exists a $k$ -closed immersion of $\mathbf{G}$ into some $\operatorname{GL}_{n}$ . Fixing a minimum dimensional such representation, using the fact that $\mathbf{G}$ is $k$ -simple, we assume, as we may, that this representation is $k$ -irreducible. By the fact that $\mathbf{G}$ is adjoint, the associated morphism $\mathbf{G}\rightarrow \operatorname{PGL}_{n}$ is a closed immersion as well. We will denote for convenience $E=k^{n}$ . Via this representation, $\mathbf{G}(k)$ acts continuously and faithfully on the metric space of homothety classes of norms, $I(E)$ , and on the compact space of homothety classes of seminorms, $S(E)$ , introduced in [Reference Bader, Duchesne and LécureuxBDL17, § 5].

Using the amenability of $R$ , there exists an $R$ -invariant ergodic probability measure $\unicode[STIX]{x1D707}$ on $S(E)$ which we now fix. This measure is also fixed by the closure of $\unicode[STIX]{x1D70C}(R)$ , thus, upon replacing $R$ with the closure of $\unicode[STIX]{x1D70C}(R)$ , we assume, as we may, that $R$ is a closed subgroup of $\mathbf{G}(k)$ . By [Reference Bader, Duchesne and LécureuxBDL17, Proposition 5.4], there is a $\mathbf{G}(k)$ -invariant measurable partition $S(E)=\bigcup _{d=0}^{n-1}S_{d}(E)$ , given by the dimension of the kernels of the seminorms. By ergodicity $\unicode[STIX]{x1D707}$ is supported on $S_{d}(E)$ for some $0\leqslant d\leqslant n-1$ .

We assume now $d>0$ . We denote by $\operatorname{Gr}_{d}(E)$ the Grassmannian of $d$ -dimensional $k$ -subspaces of $E$ and consider the map $S_{d}(E)\rightarrow \operatorname{Gr}_{d}(E)$ taking a seminorm to its kernel. This map is measurable by [Reference Bader, Duchesne and LécureuxBDL17, Proposition 5.4]. Pushing forward the measure $\unicode[STIX]{x1D707}$ , we obtain an $R$ -invariant measure $\unicode[STIX]{x1D708}$ on $\operatorname{Gr}_{d}(E)$ . By [Reference Bader, Duchesne and LécureuxBDL17, Proposition 1.9] there exists a $k$ -subgroup $\mathbf{H}<\mathbf{G}$ which is normalized by $R$ such that the image of $R$ is precompact in $N_{\mathbf{G}}(\mathbf{H})/\mathbf{H}(k)$ and $\unicode[STIX]{x1D708}$ is supported on the set of $\mathbf{H}$ -fixed points in $\operatorname{Gr}_{d}(E)$ . As $R<\mathbf{G}$ is Zariski dense and $\mathbf{G}$ is $k$ -simple, we deduce that either $\mathbf{H}=\{e\}$ or $\mathbf{H}=\mathbf{G}$ . By the irreducibility of our chosen representation there are no $\mathbf{G}$ -fixed points in $\operatorname{Gr}(E)$ and we conclude that $\mathbf{H}=\{e\}$ . It follows that $R$ is compact in $\mathbf{G}(k)$ and, in particular, is bounded.

Next we assume $d=0$ , that is, $\unicode[STIX]{x1D707}$ is supported on $I(E)$ . By [Reference Bader, Duchesne and LécureuxBDL17, Lemma 5.1] $I(E)$ is a metric space on which $\mathbf{G}(k)$ acts continuously and isometrically and the stabilizers of bounded subsets of $I(E)$ are bounded in $\mathbf{G}(k)$ . We find a ball $B\subset I(E)$ such that $\unicode[STIX]{x1D707}(B)>1/2$ . It follows that for any $g\in R$ , $gB$ intersects $B$ . Thus the set $RB$ is bounded in $I(E)$ . It follows that the stabilizer in $\mathbf{G}(k)$ of the set $RB$ is bounded. As $R$ is contained in this stabilizer, we conclude that $R$ is indeed bounded.◻

Proof of Theorem 7.1.

We first observe the easier implication $(1)\Rightarrow (2)$ . We assume by contradiction a finite index subgroup $\unicode[STIX]{x1D6EC}^{\prime }<\unicode[STIX]{x1D6EC}$ , a complete field with absolute value $k$ , a connected adjoint $k$ -simple algebraic group $\mathbf{G}$ and group homomorphism $\unicode[STIX]{x1D70C}:\unicode[STIX]{x1D6EC}^{\prime }\rightarrow \mathbf{G}(k)$ such that $\unicode[STIX]{x1D70C}(\unicode[STIX]{x1D6EC}^{\prime })$ is Zariski dense in $\mathbf{G}$ and $\unicode[STIX]{x1D70C}(\unicode[STIX]{x1D6E4}\,\cap \,\unicode[STIX]{x1D6EC}^{\prime })$ is unbounded in $\mathbf{G}(k)$ . We set $\unicode[STIX]{x1D6E4}^{\prime }=\unicode[STIX]{x1D6E4}\,\cap \,\unicode[STIX]{x1D6EC}^{\prime }$ and observe that $\unicode[STIX]{x1D6E4}^{\prime }<\unicode[STIX]{x1D6E4}$ is of finite index. Note that $\unicode[STIX]{x1D70C}(\unicode[STIX]{x1D6E4}^{\prime })$ is unbounded in $\mathbf{G}(k)$ , hence infinite, and it is commensurated by the Zariski dense subgroup $\unicode[STIX]{x1D70C}(\unicode[STIX]{x1D6EC}^{\prime })$ . By Lemma 7.5 we obtain that $\unicode[STIX]{x1D70C}(\unicode[STIX]{x1D6E4}^{\prime })$ is Zariski dense in $\mathbf{G}$ . By Lemma 7.6 we conclude that $\unicode[STIX]{x1D70C}(\unicode[STIX]{x1D6E4}^{\prime })$ is not amenable. By embedding $\mathbf{G}(k)$ in $\operatorname{GL}_{d}(k)$ we consider $\unicode[STIX]{x1D70C}$ as a linear representation $\unicode[STIX]{x1D70C}:\unicode[STIX]{x1D6EC}^{\prime }\rightarrow \operatorname{GL}_{d}(k)$ . We induce $\unicode[STIX]{x1D70C}$ from $\unicode[STIX]{x1D6EC}^{\prime }$ to $\unicode[STIX]{x1D6EC}$ and obtain a representation $\unicode[STIX]{x1D719}:\unicode[STIX]{x1D6EC}\rightarrow \operatorname{GL}_{d^{\prime }}(k)$ . $\unicode[STIX]{x1D719}(\unicode[STIX]{x1D6E4})$ is not amenable as it contains the non-amenable subgroup $\unicode[STIX]{x1D719}(\unicode[STIX]{x1D6E4}^{\prime })\simeq \unicode[STIX]{x1D70C}(\unicode[STIX]{x1D6E4}^{\prime })$ . It follows that $\unicode[STIX]{x1D719}(\unicode[STIX]{x1D6E4})$ is not solvable by locally finite. This completes the proof of $(1)\Rightarrow (2)$

We now focus on the implication $(2)\Rightarrow (1)$ . We assume by contradiction a field $K$ , an integer $d$ and a group homomorphism $\unicode[STIX]{x1D719}:\unicode[STIX]{x1D6EC}\rightarrow \operatorname{GL}_{d}(K)$ such that $\unicode[STIX]{x1D719}(\unicode[STIX]{x1D6E4})$ is not solvable by locally finite. We assume, as we may, that $K$ is a countable field, and if $\unicode[STIX]{x1D6EC}$ is finitely generated we assume also that $K$ is finitely generated. Tits’s theorem [Reference TitsTit72, Theorems 1 and 2] implies that $\unicode[STIX]{x1D719}(\unicode[STIX]{x1D6E4})$ is not amenable. We let $\mathbf{H}$ be the Zariski closure of $\unicode[STIX]{x1D719}(\unicode[STIX]{x1D6EC})$ . By [Reference BorelBor91, AG, Theorem 14.4], $\mathbf{H}$ is defined over $K$ and $\unicode[STIX]{x1D719}(\unicode[STIX]{x1D6EC})<\mathbf{H}(K)$ . We denote by $\mathbf{H}^{0}$ the identity connected component in $\mathbf{H}$ and observe that $\mathbf{H}^{0}(K)<\mathbf{H}(K)$ is of finite index. We set $\unicode[STIX]{x1D6EC}^{\prime }=\unicode[STIX]{x1D719}^{-1}(\mathbf{H}^{0}(K))$ . Note that $\unicode[STIX]{x1D6EC}^{\prime }<\unicode[STIX]{x1D6EC}$ is of finite index and the Zariski closure of $\unicode[STIX]{x1D719}(\unicode[STIX]{x1D6EC}^{\prime })$ is $\mathbf{H}^{0}$ , as it is contained in the connected group $\mathbf{H}^{0}$ and is of finite index in $\mathbf{H}$ .

We set $\unicode[STIX]{x1D6E4}^{\prime }=\unicode[STIX]{x1D6EC}^{\prime }\cap \unicode[STIX]{x1D6E4}$ . Observe that $\unicode[STIX]{x1D6E4}^{\prime }<\unicode[STIX]{x1D6E4}$ is of finite index. In particular, $\unicode[STIX]{x1D6E4}^{\prime }<\unicode[STIX]{x1D6EC}^{\prime }$ is commensurated and $\unicode[STIX]{x1D719}(\unicode[STIX]{x1D6E4}^{\prime })$ is not amenable. We let $\mathbf{R}$ be the solvable radical of $\mathbf{H}^{0}$ and set $\mathbf{L}=\mathbf{H}^{0}/\mathbf{R}$ , $\unicode[STIX]{x1D70B}:\mathbf{H}^{0}\rightarrow \mathbf{L}$ . As $\mathbf{R}(K)$ is solvable and $\unicode[STIX]{x1D719}(\unicode[STIX]{x1D6E4}^{\prime })$ is not amenable, we conclude that $\unicode[STIX]{x1D70B}\circ \unicode[STIX]{x1D719}(\unicode[STIX]{x1D6E4}^{\prime })$ is not amenable. In particular, $\mathbf{L}$ is non-trivial. Upon replacing $\mathbf{L}$ by its quotient modulo its center, which is finite, we may further assume that $\mathbf{L}$ is center-free. Thus $\mathbf{L}$ is a connected, adjoint, semisimple $K$ -group and $\unicode[STIX]{x1D70B}\circ \unicode[STIX]{x1D719}(\unicode[STIX]{x1D6E4}^{\prime })$ is Zariski dense in $\mathbf{L}$ . The adjoint group $\mathbf{L}$ is $K$ -isomorphic to a direct product of finitely many connected, adjoint, $K$ -simple $K$ -groups. The image of the projection of $\unicode[STIX]{x1D70B}\circ \unicode[STIX]{x1D719}(\unicode[STIX]{x1D6E4}^{\prime })$ to at least one of these factors is not amenable and, in particular, not locally finite. We fix such a $K$ -simple factor of $\mathbf{L}$ and denote it $\mathbf{G}$ . We let $\unicode[STIX]{x1D713}:\unicode[STIX]{x1D6EC}^{\prime }\rightarrow \mathbf{G}(K)$ be the composition of $\unicode[STIX]{x1D70B}\circ \unicode[STIX]{x1D719}$ with the projection $\mathbf{L}(K)\rightarrow \mathbf{G}(K)$ and conclude that $\unicode[STIX]{x1D713}(\unicode[STIX]{x1D6EC}^{\prime })$ is Zariski dense and $\unicode[STIX]{x1D713}(\unicode[STIX]{x1D6E4}^{\prime })$ is not locally finite in the connected adjoint $K$ -simple $K$ -algebraic group $\mathbf{G}$ . Using Lemma 7.4, we find a complete field with an absolute value $k$ and an embedding $K{\hookrightarrow}k$ such that the image of $\unicode[STIX]{x1D6E4}^{\prime }$ is unbounded in $\mathbf{G}(k)$ , which is, furthermore, a local field if the field $K$ is finitely generated. We obtain a contradiction, as the image of $\unicode[STIX]{x1D6E4}^{\prime }$ is unbounded under the composed map $\unicode[STIX]{x1D6EC}^{\prime }\rightarrow \mathbf{G}(K)\rightarrow \mathbf{G}(k)$ .◻

Proof of Theorem 1.5.

Assume by contradiction that there exist an integer $d$ , a field $K$ and a linear representation $\unicode[STIX]{x1D719}:\unicode[STIX]{x1D6E4}\rightarrow \operatorname{GL}_{d}(K)$ such that the image $\unicode[STIX]{x1D719}(\unicode[STIX]{x1D6E4})$ is not solvable by locally finite. Then, by Corollary 7.2, there exist a finite index subgroup $\unicode[STIX]{x1D6E4}^{\prime }<\unicode[STIX]{x1D6E4}$ , a complete field with an absolute value $k$ , a connected adjoint $k$ -simple algebraic group $\mathbf{G}$ and group homomorphism $\unicode[STIX]{x1D70C}:\unicode[STIX]{x1D6E4}^{\prime }\rightarrow \mathbf{G}(k)$ such that $\unicode[STIX]{x1D70C}(\unicode[STIX]{x1D6E4}^{\prime })$ is Zariski dense and unbounded in $\mathbf{G}(k)$ . If $\unicode[STIX]{x1D6E4}$ is finitely generated, $k$ could be taken to be a local field.

For every $i=1,\ldots ,n$ let $T_{i}^{\prime }=\overline{\unicode[STIX]{x1D70B}_{i}(\unicode[STIX]{x1D6E4}^{\prime })}<T_{i}$ be the closure of the projection of $\unicode[STIX]{x1D6E4}^{\prime }$ in $T_{i}$ , and $T^{\prime }=T_{1}^{\prime }\times \cdots \times T_{n}^{\prime }<T$ be a closed subgroup. Since $\unicode[STIX]{x1D6E4}^{\prime }<\unicode[STIX]{x1D6E4}$ has finite index, each $T_{i}^{\prime }<T_{i}$ has finite index, and so $T^{\prime }<T$ has finite index. Note also that $\unicode[STIX]{x1D6E4}^{\prime }<T^{\prime }$ is a lattice with dense projections. Applying Theorem 1.2, we conclude that there exists a continuous homomorphism $\bar{\unicode[STIX]{x1D70C}}:T^{\prime }\rightarrow \mathbf{G}(k)$ such that $\bar{\unicode[STIX]{x1D70C}}|_{\unicode[STIX]{x1D6E4}^{\prime }}=\unicode[STIX]{x1D70C}$ . In particular, $\bar{\unicode[STIX]{x1D70C}}(T^{\prime })$ is Zariski dense and unbounded and so is its closure, $\overline{\bar{\unicode[STIX]{x1D70C}}(T^{\prime })}$ . By Lemma 7.6 we conclude that $\overline{\bar{\unicode[STIX]{x1D70C}}(T^{\prime })}$ is not amenable.

Fixing a $k$ -representation $\mathbf{G}(k)\rightarrow \operatorname{GL}_{d^{\prime }}(k)$ we obtain a continuous linear representation $T^{\prime }\rightarrow \operatorname{GL}_{d^{\prime }}(k)$ . Inducing this representation to $T$ , we obtain a continuous homomorphism $\unicode[STIX]{x1D713}:T\rightarrow \operatorname{GL}_{d^{\prime \prime }}(k)$ such that $\overline{\unicode[STIX]{x1D713}(T)}$ is not amenable. This is a contradiction.◻

Proof of Corollary 1.6.

Assume by contradiction that there exist an integer $d$ , a field $K$ and a linear representation $\unicode[STIX]{x1D719}:\unicode[STIX]{x1D6EC}\rightarrow \operatorname{GL}_{d}(K)$ such that $\unicode[STIX]{x1D719}(\unicode[STIX]{x1D6E4})$ is not solvable by locally finite. By Theorem 7.1 we obtain a finite index subgroup $\unicode[STIX]{x1D6EC}^{\prime }<\unicode[STIX]{x1D6EC}$ , a complete field with an absolute value $k$ , a connected adjoint $k$ -simple algebraic group $\mathbf{G}$ and a Zariski dense group homomorphism $\unicode[STIX]{x1D70C}:\unicode[STIX]{x1D6EC}^{\prime }\rightarrow \mathbf{G}(k)$ such that $\unicode[STIX]{x1D70C}(\unicode[STIX]{x1D6E4}\cap \unicode[STIX]{x1D6EC}^{\prime })$ is unbounded in $\mathbf{G}(k)$ . If $\unicode[STIX]{x1D6EC}$ is finitely generated, $k$ could be taken to be a local field.

We let $T^{\prime }$ be the closure of $\unicode[STIX]{x1D6EC}^{\prime }$ in $T$ and set $\unicode[STIX]{x1D6E4}^{\prime }=\unicode[STIX]{x1D6E4}\cap \unicode[STIX]{x1D6EC}^{\prime }$ . Observe that $T^{\prime }<T$ is a closed subgroup of finite index. Thus $T^{\prime }<T$ is open and $\unicode[STIX]{x1D6E4}^{\prime }<T^{\prime }$ is a lattice. Furthermore, $\unicode[STIX]{x1D6EC}^{\prime }$ is a countable dense subgroup of $T^{\prime }$ containing and commensurating $\unicode[STIX]{x1D6E4}^{\prime }$ . Applying Corollary 1.4, we obtain a continuous homomorphism $\bar{\unicode[STIX]{x1D70C}}:T^{\prime }\rightarrow \mathbf{G}(k)$ such that $\unicode[STIX]{x1D70C}=\bar{\unicode[STIX]{x1D70C}}|_{\unicode[STIX]{x1D6EC}^{\prime }}$ . As $\unicode[STIX]{x1D70C}(\unicode[STIX]{x1D6E4}^{\prime })$ is unbounded we obtain that $\bar{\unicode[STIX]{x1D70C}}(T^{\prime })$ is unbounded. By Lemma 7.6 we conclude that $\overline{\bar{\unicode[STIX]{x1D70C}}(T^{\prime })}$ is not amenable.

Fixing a $k$ -representation $\mathbf{G}(k)\rightarrow \operatorname{GL}_{d^{\prime }}(k)$ , we obtain a continuous linear representation $T^{\prime }\rightarrow \operatorname{GL}_{d^{\prime }}(k)$ . Inducing this representation to $T$ , we obtain a continuous homomorphism $\unicode[STIX]{x1D713}:T\rightarrow \operatorname{GL}_{d^{\prime \prime }}(k)$ such that $\overline{\unicode[STIX]{x1D713}(T)}$ is not amenable. This is a contradiction.◻