Local spectral gap in simple Lie groups and applications

Abstract

We introduce a novel notion of local spectral gap for general, possibly infinite, measure preserving actions. We establish local spectral gap for the left translation action $\Gamma \curvearrowright G$, whenever $\Gamma$ is a dense subgroup generated by algebraic elements of an arbitrary connected simple Lie group G. This extends to the non-compact setting works of Bourgain and Gamburd (Invent Math 171:83–121, 2008; J Eur Math Soc (JEMS) 14:1455–1511, 2012), and Benoist and de Saxcé (Invent Math 205:337–361, 2016). We present several applications to the Banach–Ruziewicz problem, orbit equivalence rigidity, continuous and monotone expanders, and bounded random walks on G. In particular, we prove that, up to a multiplicative constant, the Haar measure is the unique $\Gamma$-invariant finitely additive measure defined on all bounded measurable subsets of G.

Appendix A. Proof of Lemma 4.1

Let G be a connected simple Lie group with trivial center and denote by $\mathfrak {g}$ its Lie algebra. The goal of this appendix is to prove Lemma 4.1. To this end, by relying on results from [51, 57] and following closely the proof of [6, Lemma 2.5], we first prove the following $\ell ^2$-flattening lemma.

In order to do so, it will be more convenient to work with an invariant metric on G, rather than the $\Vert .\Vert _2$-metric used in the rest of the paper. We therefore fix an Euclidean structure on $\mathfrak {g}$, and endow G with the corresponding left-invariant Riemannian metric, denoted by d.

For further reference, we note that there is a constant $C>1$ such that

$$\begin{aligned}&C^{-1}\log (1+\Vert x^{-1}y-1\Vert _2)\nonumber \\&\quad \leqslant d(x,y)\leqslant C\log (1+\Vert x^{-1}y-1\Vert _2),\quad \text {for all }x,y\in G. \end{aligned}$$

(A.1)

Indeed, it suffices to show that $C^{-1}\log (1+\Vert x-1\Vert _2)\leqslant d(x,1)\leqslant C\log (1+\Vert x-1\Vert _2)$, for $x\in G$. This is clear if x belongs to a small enough neighborhood V of the identity. On the other hand, if $x\not \in V$, then one can easily prove such an inequality by using the KAK decomposition of G (see [34, Section 3] for details).

Let $\delta >0$. We denote by $B(x,\delta )=\{y\in G|d(x,y)<\delta \}$ the open ball of radius $\delta$ centered at $x\in G$, and by $A^{[\delta ]}=\cup _{x\in A}B(x,\delta )$ the $\delta$ -neighborhood of a set $A\subset G$, both with respect to the metric d. For a probability measure $\mu$ on G, we denote $\mu _{\delta }=\mu *Q_{\delta }$, where

$$\begin{aligned} Q_{\delta }=\frac{\mathbf{1}_{B(1,\delta )}}{|B(1,\delta )|}. \end{aligned}$$

Lemma A.1

($\ell ^2$-flattening, [6]) Let G be a connected simple Lie group with trivial center. Given $\alpha ,\kappa >0$, there exists $\varepsilon >0$ such that the following holds for any $\delta >0$ small enough:

Suppose that $\mu$ is a symmetric Borel probability measure on G such that

• (1) supp$(\mu )\subset B(1,\varepsilon \log {\frac{1}{\delta }})$,

• (2) $\Vert \mu _{\delta }\Vert _2\geqslant \delta ^{-\alpha }$, and

• (3) $(\mu *\mu )(H^{[\rho ]})\leqslant \delta ^{-\varepsilon }\rho ^{\kappa }$, for all $\rho \geqslant \delta$ and an proper closed connected subgroup $H<G$.

Then $\Vert \mu _{\delta }*\mu _{\delta }\Vert _2\leqslant \delta ^{\varepsilon }\Vert \mu _{\delta }\Vert _2.$

Notation. We use the notation $O(\varepsilon )$ to denote a positive quantity which is bounded by $C\varepsilon$, for some constant $C>0$ depending only on G. We also use the notation $\phi \ll \psi$ for functions $\phi ,\psi :(0,\infty )\rightarrow (0,\infty )$ to mean the existence of a constant C depending only on G such that $\phi (\delta )\leqslant C\psi (\delta )$, for any small enough $\delta >0$. If $\phi \ll \psi$ and $\psi \ll \phi$, we write $\phi \simeq \psi$.

1.1 A.1 Ingredients of the proof of Lemma A.1

The proof of Lemma A.1 relies on Bourgain and Gamburd’s strategy [8, 9]. In order to implement this strategy, we use de Saxcé’s product theorem [51, Theorem 3.9]. Recall that if A is a subset of G and $\delta >0$, then $N(A,\delta )$ denotes the least number of open balls of radius $\delta$ needed to cover A.

Theorem A.2

(product theorem, [51]) Let G be a simple Lie group of dimension d. Then there exists a neighborhood U of the identity in G such that the following holds.

Given $\alpha \in (0,d)$ and $\kappa >0$, there exist $\varepsilon _0=\varepsilon _0(\alpha ,\kappa )$ and $\tau =\tau (\alpha ,\kappa )>0$ such that, for any $\delta >0$ small enough, if $A\subset U$ is a set satisfying

• (1) $N(A,\delta )\leqslant \delta ^{-d+\alpha -\varepsilon _0}$,

• (2) $N(A,\rho )\geqslant \rho ^{-\kappa }\delta ^{\varepsilon _0}$, for all $\rho \geqslant \delta$, and

• (3) $N(AAA,\delta )\leqslant \delta ^{-\varepsilon _0}N(A,\delta )$,

then A is contained in a neighborhood of size $\delta ^{\tau }$ of a proper closed connected subgroup of G.

To prove Lemma A.1, we will also need Tao’s non-commutative Balog–Szemerédi–Gowers Lemma [57, Theorem 6.10]. If A, B are subsets of a metric group G and $\delta >0$, then the $\delta$-multiplicative energy $E_{\delta }(A,B)$ is defined as $E_{\delta }(A,B)=N(\{(a,b,a',b')\in A\times B\times A\times B\;|\; d(ab,a'b')\leqslant \delta \},\delta ).$ The following inequality will be used in the proof of Lemma A.1

$$\begin{aligned} E_{\delta }(A,B)\gg \delta ^{-3d}\Vert 1_A*1_B\Vert _2^2. \end{aligned}$$

Indeed, define $f:A\times B\times A\rightarrow G$ by $f(a,b,a')=a'^{-1}ab$, $S=\{(a,b,a')\in A\times B\times A|f(a,b,a')\in B\}$, and $T=\{(a,b,a',b')\in A\times B\times A\times B|d(b',f(a,b,a'))\leqslant \delta \}$. Then we have $\Vert 1_A*1_B\Vert _2^2=|S|$ and $\delta ^d|S|\ll |T|\ll \delta ^{4d}N(T,\delta )=\delta ^{4d}E_{\delta }(A,B),$ which together prove the desired inequality.

Recall that a subset $H\subset G$ is called a K-approximative subgroup, for some $K>1$, if it is symmetric and there is a symmetric set $X\subset HH$ of cardinality at most K such that $HH\subset XH$.

Theorem A.3

(non-commutative Balog–Szemerédi–Gowers lemma,[57]) Let G be a Lie group endowed with a left-invariant Riemannian metric. Then there exist constants $c>0$ and $R>0$ such that the following holds for any $\delta \in (0,1)$ and $K\geqslant 2$.

Suppose that A, B are non-empty subsets of G contained in B(1, 1) such that

$$\begin{aligned} E_{\delta }(A,B)\geqslant \frac{1}{K}N(A,\delta )^{\frac{3}{2}}N(B,\delta )^{\frac{3}{2}}. \end{aligned}$$

Then there is a $K^c$-approximate subgroup H of G and elements $x,y\in G$ such that

• $N(H,\delta )\leqslant K^c\;N(A,\delta )^{\frac{1}{2}}N(B,\delta )^{\frac{1}{2}}$,

• $N(xH\cap A,\delta )\geqslant K^{-c}N(A,\delta )$,

• $N(Hy\cap B,\delta )\geqslant K^{-c}N(B,\delta )$,

• $x,y\in B(1,R)$ and $H\subset B(1,R)$.

A final ingredient in the proof of Lemma A.1 is an approximation of the measure $\mu _{\delta }$ by dyadic level sets [32]. A family of sets $\{A_i\}_{i\in I}$ is called essentially disjoint if there is a constant C such that the intersection of more than C distinct sets $A_i$ is empty.

Lemma A.4

[32] Let $\mu$ be a Borel probability measure on G. Let $\delta \in (0,1)$ and $\mathcal C$ be a maximal $\delta$-separated subset of G. Define $\mathcal C_0=\{x\in \mathcal C|0<\mu _{2\delta }(x)\leqslant 1\}$ and $\mathcal C_i=\{x\in \mathcal C|2^{i-1}<\mu _{2\delta }(x)\leqslant 2^{i}\}$, for $i\geqslant 1$. For $i\geqslant 0$, let $A_i=\cup _{x\in \mathcal C_i}B(x,\delta )$. Then we have the following

• (1) at most $O(1)\log {\frac{1}{\delta }}$ of the sets $A_i$ are non-empty,

• (2) $A_i$ is an essentially disjoint union of balls of radius $\delta$, for all $i\geqslant 0$,

• (3) ${\mu _{\delta }\ll \sum _{i\geqslant 0}2^i1_{A_i}}$ and ${\sum _{i>0}2^i1_{A_i}\ll \mu _{3\delta }}$.

Proof

These assertions follow from [32, Lemma 4.4], but for completeness, we include a proof. Since $|\mu _{2\delta }(x)|\leqslant \frac{1}{|B(1,2\delta )|}\ll \delta ^{-d}$, for any $x\in G$, (1) is immediate. Note that the balls $\{B(x,\delta )\}_{x\in \mathcal C}$ cover G, while the balls $\{B(x,\frac{\delta }{2})\}_{x\in \mathcal C}$ are disjoint. Since a ball of radius $3\delta$ can contain at most O(1) disjoint balls of radius $\frac{\delta }{2}$, the balls $\{B(x,\delta )\}_{x\in \mathcal C}$ are essentially disjoint. This implies (2).

To prove (3), let $y\in G$. If $x\in G$ is such that $y\in B(x,\delta )$, then

$$\begin{aligned} \mu _{\delta }(y)=\frac{\mu (B(y,\delta ))}{|B(1,\delta )|}\leqslant \frac{\mu (B(x,2\delta ))}{|B(1,\delta )|}\ll \frac{\mu (B(x,2\delta ))}{|B(1,2\delta )|}=\mu _{2\delta }(x), \end{aligned}$$

and similarly $\mu _{2\delta }(x)\ll \mu _{3\delta }(y).$ Assuming $\mu _{\delta }(y)>0$, let $x\in \mathcal C$ with $y\in B(x,\delta )$. Then $\mu _{2\delta }(x)>0$, hence $x\in \mathcal C_i$ and $y\in A_i$, for some $i\geqslant 0$. This implies that $\mu _{\delta }(y)\ll \mu _{2\delta }(x)1_{A_i}(y)\leqslant 2^i1_{A_i}(y).$

On the other hand, if $y\in A_i$, for $i>0$, then there is $x\in \mathcal C$ such that $y\in B(x,\delta )$ and $\mu _{2\delta }(x)>2^{i-1}$. Hence, $2^i1_{A_i}(y)=2^i<2\mu _{2\delta }(x)\ll \mu _{3\delta }(y)$. Since the balls $\{B(x,\delta )\}_{x\in \mathcal C}$ are essentially disjoint, y belongs to O(1) of the sets $\{A_i\}_{i>0}$ and it follows that $\sum _{i>0}2^i1_{A_i}(y)\ll \mu _{3\delta }(y).$ This proves (3).   $\square$

Lemma A.5

Let $a>0$ and $\mu$ be a Borel probability measure on G.

Then ${\sup _{\delta<\delta '<1}\Vert \mu _{\delta '}\Vert _2\ll \Vert \mu _{\delta }\Vert _2}$ and $\Vert \mu _{\delta }\Vert _2\simeq \Vert \mu _{a\delta }\Vert _2$.

Proof

Let $1>\delta '>\delta >0$. Then $|B(1,\delta )| 1_{B(1,\delta ')}\leqslant 1_{B(1,\delta )}*1_{B(1,\delta +\delta ')}$ and thus

$$\begin{aligned} Q_{\delta '}\leqslant \frac{|B(1,\delta +\delta ')|}{|B(1,\delta ')|}\;Q_{\delta }*Q_{\delta +\delta '}\ll Q_{\delta }*Q_{\delta +\delta '}. \end{aligned}$$

We further get that $\Vert \mu *Q_{\delta '}\Vert _2\ll \Vert (\mu *Q_{\delta })*Q_{\delta +\delta '}\Vert _2\leqslant \Vert \mu *Q_{\delta }\Vert _2$, thus proving the first inequality. To prove the second inequality we may assume $a>1$. Then for any $x\in G$ we have

$$\begin{aligned} \mu _{a\delta }(x)=\frac{\mu (B(x,a\delta ))}{|B(1,a\delta )|}\geqslant \frac{|B(1,\delta )|}{|B(1,a\delta )|}\;\frac{\mu (B(x,\delta ))}{|B(1,\delta )|}\gg \mu _{\delta }(x). \end{aligned}$$

Thus, $\Vert \mu _{a\delta }\Vert _2\gg \Vert \mu _{\delta }\Vert _2$. Since $\Vert \mu _{a\delta }\Vert _2\ll \Vert \mu _{\delta }\Vert _2$ by the above, we are done. $\square$

1.2 A.2 Proof of Lemma A.1

Assume that $\Vert \mu _{\delta }*\mu _{\delta }\Vert _2>\delta ^{\varepsilon }\Vert \mu _{\delta }\Vert _2$, for some $\varepsilon >0$. Following closely the proof of [6, Lemma 2.5], we will reach a contradiction for any $\varepsilon$ small enough.

Let U be the neighborhood of $1\in G$ provided by Theorem A.2, and let $0<r<1$ with $B(1,r)\subset U$. Let R be the constant given by Theorem A.3, and $\{A_i\}_{0\leqslant i\ll \log {\frac{1}{\delta }}}$ be the sets given by Lemma A.4. Let $C>1$ the constant appearing in inequality (A.1). By using (A.1) one easily checks that

$$\begin{aligned} d(x^{-1},y^{-1})\leqslant Ce^{2Cd(x,1)}(e^{Cd(x,y)}-1),\quad \text {for any }x,y\in G. \end{aligned}$$

(A.2)

Assume that $3C\varepsilon <1$. Let $0\leqslant i\ll \log {\frac{1}{\delta }}$. Since supp$(\mu )\subset B(1,\varepsilon \log {\frac{1}{\delta }})$, we get that $A_i$ is contained in $B(1,\varepsilon \log {\frac{1}{\delta }}+3\delta )$. Since $|B(1,\varepsilon \log {\frac{1}{\delta }}+3\delta )|=\delta ^{-O(\varepsilon )}$ and $|B(1,\frac{\delta ^{3C\varepsilon }}{8})|=\delta ^{O(\varepsilon )}$, $A_i$ can be covered by at most $\delta ^{-O(\varepsilon )}$ sets of diameter at most $\frac{\delta ^{3C\varepsilon }}{2}$. Since $A_i$ is a union of balls of radius $\delta$, and $\delta \leqslant \frac{\delta ^{3C\varepsilon }}{2}$, for $\delta$ small enough, we can decompose $A_i=\cup _{k=1}^{\delta ^{-O(\varepsilon )}}A_{i,k},$ where each set $A_{i,k}$ is the union of some of the balls of radius $\delta$ that make up $A_i$, and has diameter at most $\delta ^{3C\varepsilon }$. Moreover, by (A.2) the diameter of $A_{i,k}^{-1}$ is at most $Ce^{2C(\varepsilon \log {\frac{1}{\delta }}+3\delta )}(e^{C\delta ^{3C\varepsilon }}-1)\simeq {\delta }^{C\varepsilon }$. Thus, for $\delta$ small enough, $A_{i,k}$ has diameter at most $\min \{1,r\}$ and $A_{i,k}^{-1}$ has diameter at most 1, for all k.

Before continuing, let us also note that

$$\begin{aligned} \delta ^{O(\varepsilon )}N(A,\delta )\leqslant N(Ah,\delta )\leqslant \delta ^{-O(\varepsilon )}N(A,\delta ), \end{aligned}$$

(A.3)

for every set $A\subset B(1,\varepsilon \log {\frac{1}{\delta }}+3\delta )$ and any $h\in B(1,\varepsilon \log {\frac{1}{\delta }}+3\delta +1)$. Indeed, by using (A.2) it is immediate that $B(1,\delta ^{1+O(\varepsilon )})\subset h^{-1}B(1,\delta )h\subset B(1,\delta ^{1-O(\varepsilon )})$, which easily implies (A.3).

Since ${\mu _{\delta }\ll \sum _{i}2^i1_{A_i}\leqslant \sum _{i,k} 2^i1_{A_{i,k}}}$, we get that

$$\begin{aligned} \delta ^{\varepsilon }\Vert \mu _{\delta }\Vert _2\leqslant \Vert \mu _{\delta }*\mu _{\delta }\Vert _2\leqslant \underset{1\leqslant k,l\leqslant \delta ^{-O(\varepsilon )}}{\sum _{0\leqslant i,j\ll \log {\frac{1}{\delta }}}}\Vert 2^i1_{A_{i,k}}*2^j1_{A_{j,l}}\Vert _2. \end{aligned}$$

Since the sum on the right contains $\delta ^{-O(\varepsilon )}$ terms, we can find $0\leqslant i,j\ll \log \frac{1}{\delta }$ and $1\leqslant k,l\leqslant \delta ^{-O(\varepsilon )}$ such that if we denote $A_i'=A_{i,k}$ and $A_j'=A_{j,l}$, then

$$\begin{aligned} \Vert 2^i1_{A_i'}*2^j1_{A_j'}\Vert _2\geqslant \delta ^{O(\varepsilon )}\Vert \mu _{\delta }\Vert _2. \end{aligned}$$

We claim that $i>0$. Note that $\Vert 1_{A_0}\Vert _1=|A_0|\leqslant |B(1,\varepsilon \log {\frac{1}{\delta }}+3\delta )|=\delta ^{-O(\varepsilon )}$ and $\Vert 1_{A_0}\Vert _2=\delta ^{-O(\varepsilon )}$. Moreover, if $j>0$, then $\Vert 2^j1_{A_j'}\Vert _1\leqslant \Vert 2^j1_{A_j}\Vert _1\ll \Vert \mu _{3\delta }\Vert _1=1$ (by Lemma A.4). Young’s inequality gives that $\Vert 1_{A_0}*2^j1_{A_j'}\Vert _2\leqslant \Vert 1_{A_0}\Vert _2\Vert 2^j1_{A_j'}\Vert _1\leqslant \delta ^{-O(\varepsilon )}$, for any $j\geqslant 0$. Since $\delta ^{O(\varepsilon )}\Vert \mu _{\delta }\Vert _2\geqslant \delta ^{O(\varepsilon )-\alpha }$, we cannot have $i=0$, provided $\varepsilon >0$ is small enough. Similarly, we must have that $j>0$.

Since $\Vert 2^{j}1_{A_j'}\Vert _2\leqslant \Vert 2^j1_{A_j}\Vert _2\ll \Vert \mu _{3\delta }\Vert _2\simeq \Vert \mu _{\delta }\Vert _2$ (by Lemma A.5), Young’s inequality gives that

$$\begin{aligned} \delta ^{O(\varepsilon )}\Vert \mu _{\delta }\Vert _2\leqslant \Vert 2^i1_{A_i'}*2^j1_{A_j'}\Vert _2\leqslant \Vert 2^i1_{A_i'}\Vert _1\Vert 2^j1_{A_j'}\Vert _2\leqslant 2^i|A_i'\Vert \mu _{\delta }\Vert _2. \end{aligned}$$

This implies that

$$\begin{aligned} 2^i|A_i'|=\delta ^{O(\varepsilon )}\quad \text {and similarly}\quad 2^j|A_j'|=\delta ^{O(\varepsilon )}. \end{aligned}$$

(A.4)

Next, since $2^i|A_i'|\leqslant 2^i|A_i|\ll \Vert \mu _{3\delta }\Vert _1=1$, we deduce that

$$\begin{aligned} \delta ^{O(\varepsilon )}\Vert \mu _{\delta }\Vert _2\leqslant \Vert 2^i1_{A_i'}*2^j1_{A_j'}\Vert _2&\leqslant \Vert 2^i1_{A_i'}\Vert _2\Vert 2^j1_{A_j'}\Vert _1\ll 2^i|A_i'|^{\frac{1}{2}}\leqslant |A_i'|^{-\frac{1}{2}}. \end{aligned}$$

Using that $\Vert \mu _{\delta }\Vert _2\geqslant \delta ^{-\alpha }$, it follows that $|A_i'|\leqslant \delta ^{\frac{\alpha }{2}-O(\varepsilon )}.$ Since $A_i'$ is an essentially disjoint union of balls of radius $\delta$, we have $|A_i'|\simeq \delta ^dN(A_i',\delta )$. Altogether, we deduce that

$$\begin{aligned} N(A_i',\delta )\leqslant \delta ^{-d+\frac{\alpha }{2}-O(\varepsilon )}\quad \text {and similarly}\quad N(A_j',\delta )\leqslant \delta ^{-d+\frac{\alpha }{2}-O(\varepsilon )}. \end{aligned}$$

(A.5)

By combining the inequalities $2^i|A_i'|^{\frac{1}{2}}=\Vert 2^i1_{A_i'}\Vert _2\leqslant \Vert \mu _{3\delta }\Vert _2\simeq \Vert \mu _{\delta }\Vert _2$ and $2^i|A_i'|\ll 1$ with the fact that $|A_i'|\simeq \delta ^{d}N(A_i',\delta )$, we derive that

$$\begin{aligned} \Vert 1_{A_i'}*1_{A_j'}\Vert _2^2 \geqslant \delta ^{O(\varepsilon )}2^{-2i-2j}\Vert \mu _{\delta }\Vert _2^2&\geqslant \delta ^{O(\varepsilon )}2^{-i-j}|A_i'|^{\frac{1}{2}}|A_j'|^{\frac{1}{2}}\\ {}&=\delta ^{O(\varepsilon )}(2^i|A_i'|)^{-1}|A_i'|^{\frac{3}{2}}(2^{j}A_j')^{-1}|A_j'|^{\frac{3}{2}}\\ {}&\geqslant \delta ^{3d+O(\varepsilon )}N(A_i',\delta )^{\frac{3}{2}}N(A_j',\delta )^{\frac{3}{2}}. \end{aligned}$$

Since $A_i'$ and ${A_j'}^{-1}$ have diameters at most 1, we can find $g,h\in B(1,\varepsilon \log {\frac{1}{\delta }}+3\delta +1)$ such that $gA_i'\subset B(1,1)$ and $A_j'h\subset B(1,1)$. On the other hand, combining the last inequality with A.3 yields

$$\begin{aligned} E_{\delta }(gA_i',A_j'h)\gg \delta ^{-3d}\Vert 1_{gA_i'}*1_{A_j'h}\Vert _2^2&=\delta ^{-3d}\Vert 1_{A_i'}*1_{A_j'}\Vert _2^2\\&\geqslant \delta ^{O(\varepsilon )}N(A_i',\delta )^{\frac{3}{2}}N(A_j',\delta )^{\frac{3}{2}}\\&\geqslant \delta ^{O(\varepsilon )}N(gA_i',\delta )^{\frac{3}{2}}N(A_j'h,\delta )^{\frac{3}{2}} \end{aligned}$$

By applying Theorem A.3 to $gA_i'$ and $A_j'h$, we deduce the existence of a $\delta ^{-O(\varepsilon )}$-approximate subgroup $H\subset B(1,R)$ and elements $z,t\in B(1,R)$ such that $N(H,\delta )\leqslant \delta ^{-O(\varepsilon )}N(gA_i',\delta )^{\frac{1}{2}}N(A_j'h,\delta )^{\frac{1}{2}}$, $N(zH\cap gA_i',\delta )\geqslant \delta ^{O(\varepsilon )}N(gA_i',\delta )$ and $N(Ht\cap A_j'h,\delta )\geqslant \delta ^{O(\varepsilon )}N(A_j'h,\delta )$. Let $v=g^{-1}z$ and $w=th^{-1}$. By using (A.3) we further get that

$$\begin{aligned} N(H,\delta )\leqslant & {} \delta ^{-O(\varepsilon )}N(A_i',\delta )^{\frac{1}{2}}N(A_j',\delta )^{\frac{1}{2}}, \end{aligned}$$

(A.6)

$$\begin{aligned} N(vH\cap A_i',\delta )\geqslant & {} \delta ^{O(\varepsilon )}N(A_i',\delta )\quad \text {and}\quad N(Hw\cap A_j',\delta )\geqslant \delta ^{O(\varepsilon )}N(A_j',\delta ).\nonumber \\ \end{aligned}$$

(A.7)

The next claim allows us to replace H with its $4\delta$-neighborhood:

Claim. $\tilde{H}:=H^{[4\delta ]}$ satisfies the following:

• (1) $\mu _{\delta }(v\tilde{H}\cap A_i')\geqslant \delta ^{O(\varepsilon )}$.

• (2) $N(\tilde{H}^2,\delta )\leqslant N(\tilde{H}^6,\delta )\leqslant \delta ^{-O(\varepsilon )}N(\tilde{H},\delta ).$

Proof of the claim

(1) Recall that there is a subset $\mathcal C_i'\subset \mathcal C_i$ such that $A_i'=\cup _{x\in \mathcal C_i'}B(x,\delta )$. Since $N(vH\cap A_i',\delta )\geqslant \delta ^{O(\varepsilon )}N(A_i',\delta )$ by (A.7), we get that vH intersects at least $\delta ^{O(\varepsilon )}N(A_i',\delta )$ of the balls $\{B(x,\delta )\}_{x\in \mathcal C_i'}$. Thus, $v\tilde{H}\cap A_i'=(vH)^{[4\delta ]}\cap A_i'$ contains at least $\delta ^{O(\varepsilon )}N(A_i',\delta )$ of the balls $\{B(x,3\delta )\}_{x\in \mathcal C_i'}$. Since the balls $\{B(x,\delta )\}_{x\in \mathcal C_i'}$ and hence the balls $\{B(x,3\delta )\}_{x\in \mathcal C_i'}$ are essentially disjoint, $v\tilde{H}\cap A_i'$ must contain at least $\delta ^{O(\varepsilon )}N(A_i',\delta )$ disjoint balls from the collection $\{B(x,3\delta )\}_{x\in \mathcal C_i'}$.

On the other hand, for every $x\in \mathcal C_i$ we have that $\mu _{2\delta }(x)>2^{i-1}$ and hence

$$\begin{aligned} \mu _{\delta }(B(x,3\delta ))\geqslant \mu (B(x,2\delta ))=|B(1,2\delta )|\mu _{2\delta }(x)\gg \delta ^d2^i. \end{aligned}$$

Since $\delta ^dN(A_i',\delta )\simeq |A_i'|$, and $2^i|A_i'|=\delta ^{O(\varepsilon )}$ by A.4, we finally derive that

$$\begin{aligned} \mu _{\delta }(v\tilde{H}\cap A_i)\geqslant \delta ^{O(\varepsilon )}N(A_i',\delta )\delta ^d2^i\simeq \delta ^{O(\varepsilon )}2^i|A_i'|=\delta ^{O(\varepsilon )}. \end{aligned}$$

(2) Let $X\subset G$ be a set of cardinality $\delta ^{-O(\varepsilon )}$ such that $HH\subset HX$. Since $H\subset B(1,R)$ and R is an absolute constant, by using (A.2) it follows that $\tilde{H}\tilde{H}\subset (HH)^{[O(1)\delta ]}$, and thus $\tilde{H}\tilde{H}\subset H^{[D\delta ]}X$, for some constant $D>1$. Let $S\subset H$ be a maximal set such that the balls $\{B(x,\delta )\}_{x\in S}$ are disjoint. Then $\cup _{x\in S}B(x,\delta )\subset H^{[\delta ]}$ and $H\subset \cup _{x\in S}B(x,2\delta )$, therefore $H^{[D\delta ]}\subset \cup _{x\in S}B(x,(D+2)\delta )$. Let Y be a set of cardinality O(1) with $B(1,(D+2)\delta )\subset B(1,\delta )Y$. Then $B(x,(D+2)\delta )\subset B(x,\delta )Y$, for any $x\in G$, implying that $H^{[D\delta ]}\subset H^{[\delta ]}Y$. Altogether, we get that $\tilde{H}\tilde{H}\subset H^{[\delta ]}Z\subset \tilde{H}Z$, where $Z=YX$. Since $\tilde{H}$ is symmetric, we have $\tilde{H}\tilde{H}\subset Z^{-1}\tilde{H}$. Since the cardinality of Z is $\delta ^{-O(\varepsilon )}$, (2) follows. $\square$

Let us show that for $\varepsilon >0$ small enough, the set $\tilde{H}^2\cap U$ satisfies the assumptions of Theorem A.2. Firstly, using (2) above in combination with (A.5) and (A.6) we get that

$$\begin{aligned} N(\tilde{H}^2\cap U,\delta )\leqslant N(\tilde{H}^2,\delta )\leqslant \delta ^{-d+\frac{\alpha }{2}-O(\varepsilon )}. \end{aligned}$$

(A.8)

Secondly, since $A_i'$ has diameter at most r, we get that ${A_i'}^{-1}{A_i'}\subset B(1,r)\subset U$. This implies that $(v\tilde{H}\cap A_i')^{-1}(v\tilde{H}\cap A_i')\subset \tilde{H}^2\cap U$. In combination with (1) from the claim, it follows that

$$\begin{aligned} \check{\mu }_{\delta }*\mu _{\delta }(\tilde{H}^2\cap U)\geqslant \check{\mu }_{\delta }((v\tilde{H}\cap A_i')^{-1})\mu _{\delta }(v\tilde{H}\cap A_i')=\mu _{\delta }(v\tilde{H}\cap A_i')^2\geqslant \delta ^{O(\varepsilon )}. \end{aligned}$$

(A.9)

Let $1\geqslant \rho \geqslant \delta$ and $x\in G$ such that $B(x,\rho )\cap U\not =\emptyset$. Then $B(1,\delta )B(x,\rho )B(1,\delta )\subset B(x,\rho +O(1)\delta )$. Since $B(x,\rho +O(1)\delta )$ is contained in the $(\rho +O(1)\delta )$-neighborhood of a proper closed connected subgroup of G, the hypothesis implies that $\mu *\mu (B(x,\rho +O(1)\delta ))\ll \delta ^{-\varepsilon }\rho ^{\kappa }$. By using the fact that $\mu$ and $Q_{\delta }$ are symmetric, we get that

$$\begin{aligned} \check{\mu }_{\delta }*\mu _{\delta }(B(x,\rho ))&=(Q_{\delta }*\mu *\mu *Q_{\delta })(B(x,\rho ))\\&\leqslant \mu *\mu (B(1,\delta )B(x,\rho )B(1,\delta ))\ll \delta ^{-\varepsilon }\rho ^{\kappa }. \end{aligned}$$

Since this holds for any ball of radius $\rho$ that intersects U, in combination with (A.9) it gives that

$$\begin{aligned} N(\tilde{H}^2\cap U,\rho )\geqslant \delta ^{O(\varepsilon )}\rho ^{-\kappa },\quad \text {for all }1\geqslant \rho \geqslant \delta . \end{aligned}$$

(A.10)

Thirdly, let $C\subset \tilde{H}$ be a set of diameter at most r and fix $x\in C$. Since $B(1,r)\subset U$, we get that $x^{-1}C\subset C^{-1}C\subset \tilde{H}^2\cap U$. Thus, $N(C,\delta )=N(x^{-1}C,\delta )\leqslant N(\tilde{H}^2\cap U,\delta )$. On the other hand, since $H\subset B(1,R)$, we can cover $\tilde{H}$ by O(1) sets of diameter at most r. Altogether, we conclude that $N(\tilde{H},\delta )\ll N(\tilde{H}^2\cap U,\delta )$. Using (2), it follows that

$$\begin{aligned} N((\tilde{H}^2\cap U)^3,\delta )\leqslant N(\tilde{H}^6,\delta )\leqslant \delta ^{-O(\varepsilon )}N(\tilde{H},\delta )\leqslant \delta ^{-O(\varepsilon )}N(\tilde{H}^2\cap U,\delta ).\nonumber \\ \end{aligned}$$

(A.11)

Equations (A.8), (A.10), (A.11) together guarantee that we are in position to apply Theorem A.2 to $\tilde{H}^2\cap U$. Thus, there is a proper closed connected subgroup $L<G$ such that $\tilde{H}^2\cap U\subset L^{[\delta ^{\tau }]}$. On the other hand, by using (A.9) and reasoning similarly to the above we conclude that

$$\begin{aligned} \delta ^{O(\varepsilon )}\leqslant & {} \check{\mu }_{\delta }*\mu _{\delta }(\tilde{H}^2\cap U)\leqslant \check{\mu }_{\delta }*\mu _{\delta }(L^{[\delta ^{\tau }]}\cap U)\\\leqslant & {} \mu *\mu (B(1,\delta )(L^{[\delta ^{\tau }]}\cap U)B(1,\delta ))\leqslant \mu *\mu (L^{[\delta ^{\tau }+O(1)\delta ]}). \end{aligned}$$

Since the hypothesis implies that $\mu *\mu (L^{[\delta ^{\tau }+O(1)\delta ]})\leqslant \delta ^{-\varepsilon }(\delta ^{\tau }+O(1)\delta )^{\kappa }$, it is now clear that choosing $\varepsilon >0$ small enough yields a contradiction. $\square$

1.3 A.3 Proof of Lemma 4.1

Let $\alpha ,\kappa >0$. Let $\varepsilon >0$ such that the conclusion of Lemma A.1 holds. Let $\mu$ be a symmetric Borel probability measure on G. Then we have

• (a) $\Vert \mu *P_{\delta }\Vert _2\simeq \Vert \mu *Q_{\delta }\Vert _2.$

Moreover, assuming that supp$(\mu )\subset B_{\delta ^{-\beta }}(1)$, for some $\beta >0$, we have that

• (b) $\mu *Q_{\delta }\ll \delta ^{-O(\beta )}\;Q_{\delta ^{1-O(\beta )}}*\mu *Q_{\delta ^{1-O(\beta )}}$.

• (c) $A^{[\rho ]}\cap$ supp$(\mu *\mu )\subset A^{(\delta ^{-O(\beta )}\rho )}$, for any set $A\subset G$ and every $1>\rho >\delta$.

Indeed, A.1 implies that $B(1,\frac{\delta }{2C})\subset B_{\delta }(1)\subset B(1,C\delta )$, for small $\delta >0$. This readily gives that

$$\begin{aligned} P_{\delta }=\frac{1_{B_{\delta }(1)}}{|B_{\delta }(1)|}\leqslant \frac{|B(1,C\delta )|}{|B(1,\frac{\delta }{2C})|} Q_{C\delta }\ll Q_{C\delta } \end{aligned}$$

and similarly ${Q_{\frac{\delta }{2C}}\ll P_{\delta }}.$ Therefore, $\Vert \mu *Q_{\frac{\delta }{2C}}\Vert _2\ll \Vert \mu *P_{\delta }\Vert \ll \Vert \mu *Q_{C\delta }\Vert _2$. On the other hand, Lemma A.5 implies that $\Vert \mu *Q_{\frac{\delta }{2C}}\Vert _2\simeq \Vert \mu *Q_{\delta }\Vert _2\simeq \Vert \mu *Q_{C\delta }\Vert _2$. This altogether proves (a).

By (A.1) we have that $B_{\delta ^{-\beta }}(1)\subset B(1,2C\beta \log {\frac{1}{\delta }})$, hence $\mu$ and $\mu *\mu$ are supported on $B(1,4C\beta \log {\frac{1}{\delta }})$. To prove (b), we may therefore assume that $\mu =\delta _{x}$, for some $x\in B(1,4C\beta \log {\frac{1}{\delta }})$. Using (A.2) we get that $B(1,4C\beta \log {\frac{1}{\delta }})B(1,\delta )B(1,4C\beta \log {\frac{1}{\delta }})\subset B(1,\delta ^{1-O(\beta )})$, hence $\delta _x*1_{B(1,\delta )}*\delta _{x^{-1}}\leqslant 1_{B(1,\delta ^{1-O(\beta )})}.$ Since $|B(1,\delta ^{1-O(\beta )}|\ll \delta ^{-O(\beta )}|B(1,\delta )|$, this implies that $\mu *Q_{\delta }\ll \delta ^{-O(\beta )}Q_{\delta ^{1-O(\beta )}}*\mu$. Similarly, we have that $Q_{\delta }\ll \delta ^{-O(\beta )}Q_{\delta }*Q_{\delta ^{1-O(\beta )}}$, and combining the last two inequalities implies (b). Finally, (A.1) gives that $\Vert x-y\Vert _2\leqslant \delta ^{-O(\beta )}\rho$, for any $x\in G$ and $y\in B(1,4C\beta \log {\frac{1}{\delta }})$ satisfying $d(x,y)<\rho$. This clearly implies (c).

To finish the proof, assume that $\mu$ additionally satisfies $\Vert \mu *P_{\delta }\Vert _2\geqslant \delta ^{-\alpha }$, and $(\mu *\mu )(H^{(\rho )})\leqslant \delta ^{-\gamma }\rho ^{\kappa }$, for all $\rho \geqslant \delta$ and any proper closed connected subgroup $H<G$, for some $\gamma >0$.

By using (a), (c) and Lemma A.5 we get that

• supp$(\mu )\subset B(1,2C\beta \log {\frac{1}{\delta }}).$

• $\Vert \mu *Q_{\delta ^{1-O(\beta )}}\Vert _2\ll \Vert \mu *Q_{\delta }\Vert _2\ll \delta ^{-\alpha }$.

• $(\mu *\mu )(H^{[\rho ]})\leqslant (\mu *\mu )(H^{(\delta ^{-O(\beta )}\rho )})\leqslant \delta ^{-(\gamma +\kappa O(\beta ))}\rho ^{\kappa }$, for all $\rho \geqslant \delta$ and any proper closed connected subgroup $H<G$.

If $\beta ,\gamma >0$ are chosen small enough, then Theorem A.1 implies that $\Vert \mu _{\delta }*\mu _{\delta }\Vert _2<\delta ^{\varepsilon }\Vert \mu _{\delta }\Vert _2$. Moreover, if $\beta ,\gamma$ are small enough, then by combining this inequality with (a) and (b) we derive that

$$\begin{aligned} \Vert \mu *\mu *P_{\delta }\Vert _2\ll \Vert \mu *\mu *Q_{\delta }\Vert _2&\leqslant \delta ^{-O(\beta )}\Vert \mu *Q_{\delta ^{1-O(\beta )}}*\mu *Q_{\delta ^{1-O(\beta )}}\Vert _2\\ {}&\leqslant \delta ^{-O(\beta )+(1-O(\beta ))\varepsilon }\Vert \mu *Q_{\delta ^{1-O(\beta )}}\Vert _2\\ {}&\leqslant \delta ^{\varepsilon -O(\beta )}\Vert \mu *Q_{\delta }\Vert _2\\ {}&\leqslant \delta ^{\varepsilon -O(\beta )}\Vert \mu *P_{\delta }\Vert _2\\ {}&\leqslant \delta ^{\gamma }\Vert \mu *P_{\delta }\Vert _2. \end{aligned}$$

This concludes the proof of Lemma 4.1. $\square$