# Minimal and nearly minimal measure expansions in connected unimodular groups

Yifan Jing  Department of Mathematics, University of Illinois at Urbana-
Champaign, Urbana IL, USA<EMAIL_ADDRESS>and  Chieu-Minh Tran
Department of Mathematics, University of Notre Dame, Notre Dame IN, USA
<EMAIL_ADDRESS>
###### Abstract.

Let $G$ be a connected unimodular group equipped with a (left and hence right)
Haar measure $\mu_{G}$, and suppose $A,B\subseteq G$ are nonempty and compact.
An inequality by Kemperman gives us
$\mu_{G}(AB)\geq\min\\{\mu_{G}(A)+\mu_{G}(B),\mu_{G}(G)\\}.$

Our first result determines the conditions for the equality to hold, providing
a complete answer to a question asked by Kemperman in 1964. Our second result
characterizes compact and connected $G$, $A$, and $B$ that nearly realize
equality, with quantitative bounds having the sharp exponent. This can be seen
up-to-constant as a $(3k-4)$-theorem for this setting and confirms the
connected case of conjectures by Griesmer and by Tao. As an application, we
get a measure expansion gap result for connected compact simple Lie groups.

The tools developed in our proof include an analysis of the shape of minimal
and nearly minimal expansion sets, a bridge from this to the properties of a
certain pseudometric, and a construction of appropriate continuous group
homomorphisms to either $\mathbb{R}$ or $\mathbb{T}=\mathbb{R}/\mathbb{Z}$
from the pseudometric.

###### 2010 Mathematics Subject Classification:

Primary 22D05; Secondary 11B30, 05D10, 03C20, 43A05

YJ was supported by Arnold O. Beckman Research Award (Campus Research Board
RB2101), by the Department Fellowship, and by the Trjitzinsky Fellowship from
UIUC

###### Contents

  1. 1 Introduction
    1. 1.1 Backgrounds
    2. 1.2 Statement of main results
    3. 1.3 Notation and convention
  2. 2 Outline of the argument
    1. 2.1 Overview of the strategy
    2. 2.2 The first step: Obtaining a suitable pseudometric
    3. 2.3 The second step: Constructing an appropriate homomorphism
    4. 2.4 The third step: Deducing the structure of the pair
    5. 2.5 Structure of the paper
  3. 3 Preliminaries on locally compact groups
  4. 4 Reduction to nearly minimal expansion pairs
  5. 5 Reduction to sets with small measure
  6. 6 Geometry of minimal and nearly minimal expansion pairs I
    1. 6.1 Preservation of minimal expansion under quotient
    2. 6.2 Structure control on the nearly minimal expansion sets
    3. 6.3 Fine control of Lie model
  7. 7 Pseudometric and group homomorphisms into tori
    1. 7.1 Definitions and linear pseudometric
    2. 7.2 Almost linear pseudometric
    3. 7.3 Group homomorphisms from almost linear pseudometric
  8. 8 Geometry of minimal and nearly minimal expansion pairs II
    1. 8.1 Toric expanders
    2. 8.2 Linear pesudometric from minimal expansions
    3. 8.3 Almost linear pseudometric from near minimal expansions
  9. 9 Proof of the main theorems
    1. 9.1 Minimal expansion pairs in noncompact groups
    2. 9.2 Nearly minimal expansion pairs in compact groups

##  1\. Introduction

###  1.1. Backgrounds

The Cauchy–Davenport theorem asserts that if $X$ and $Y$ are nonempty subsets
of the group $\mathbb{Z}/p\mathbb{Z}$ of prime order $p$, then

$|X+Y|\geq\min\\{|X|+|Y|-1,p\\},$

where we set $X+Y:=\\{x+y:x\in X,y\in Y\\}$. The condition for the equality to
happen is essentially given by Vosper’s theorem [Vos56], which states that if

$1<|X|,|Y|,\text{ and }|X+Y|=|X|+|Y|-1<p-1$

then $X$ and $Y$ must be arithmetic progression with the same common
difference. When the equality nearly happens, one might expect that $X$ and
$Y$ are instead contained in arithmetic progressions with slightly larger
cardinalities. This was confirmed with sharp exponent bound by Freiman [Fre73]
for small $|X|$ and $|Y|$. The optimal statement believed to be true for
$X=Y$, known as the $(3k-4)$-conjecture for $\mathbb{Z}/p\mathbb{Z}$, remains
wide open more than 60 years after the corresponding statement for
$\mathbb{Z}$ was proven by Freiman [Fre59]. In the mean time, we see many
similar results obtained for other abelian groups; see e.g. [Kne56, Kem60,
DF03, GR06, Tao18, Gri19, Lev20]. In another direction, we also see many
progresses in the study of small expansions in the nonabelian settings; see
[BGT12], in particular, for the classification of approximate groups by
Breuillard, Green, and Tao; see also e.g. [Hel08, BGS10, BV12, Hru12, PS16,
BH18, Hru20]. These two trends together suggest that the theory of minimal and
nearly minimal expansion can be extended to the nonabelian settings as well.
In this paper, we take a step towards realizing this intuition by considering
an inequality by Kemperman, the continuous nonabelian counterpart of the
Cauchy–Davenport theorem, and effectively determining the necessary and
sufficient conditions for equality and near equality to happen.

Throughout, let $G$ be a connected locally compact group, and $\mu_{G}$ a left
Haar measure on $G$. We further assume that $G$ is unimodular (i.e., the
measure $\mu_{G}$ is invariant under right translation), so $\mu_{G}$ behaves
like an appropriate notion of size. This assumption holds in many situations
of interest (e.g, when $G$ is compact, discrete, a nilpotent Lie group, a
semisimple Lie group, etc). As usual, for $A,B\subseteq G$, we set
$AB:=\\{ab:a\in A,b\in B\\}$ and let $A^{n}$ be the $n$-fold product of $A$
for $n\in\mathbb{N}^{>0}$. In [Kem64], Kemperman proved that if $A,B\subseteq
G$ are nonempty and compact, then

$\mu_{G}(AB)\geq\min\\{\mu_{G}(A)+\mu_{G}(B),\mu_{G}(G)\\}.$

This generalizes earlier results for one-dimensional tori, $n$-dimensional
tori, and abelian groups by Raikov [Rai39], Macbeath [Mac53], and Kneser
[Kne56].

The problem of determining equality condition for the Kemperman inequality was
proposed in the same paper [Kem64]. After handling a number of easy cases, the
problem can be reduced to classify all connected and unimodular group $G$ and
pairs $(A,B)$ of compact subsets of $G$ such that

$0<\mu_{G}(A),\mu_{G}(B),\text{ and
}\mu_{G}(AB)=\mu_{G}(A)+\mu_{G}(B)<\mu_{G}(G).$

We call such $(A,B)$ a minimally expanding pair on $G$. It is easy to see
that, if $I$ and $J$ are closed intervals in
$\mathbb{T}=\mathbb{R}/\mathbb{Z}$, such that $I$ and $J$ have positive
measures and the total of their measures is strictly smaller than
$\mu_{\mathbb{T}}(\mathbb{T})$, then $I+J$ is an interval with length the
total length of $I$ and $J$. Hence, such $(I,J)$ is a minimally expanding pair
on $\mathbb{T}$. More generally, when $G$ is a compact group,
$\chi:G\to\mathbb{T}$ is a continuous surjective group homomorphism, $I$ and
$J$ are as before,

$A=\chi^{-1}(I)\text{ and }B=\chi^{-1}(J),$

we can check by using the Fubini theorem that $(A,B)$ is a minimally expanding
pair. Note that an arithmetic progression on $\mathbb{Z}/p\mathbb{Z}$ is the
inverse image under a group homomorphism
$\phi:\mathbb{Z}/p\mathbb{Z}\to\mathbb{T}$ of an interval on $\mathbb{T}$, so
this example is the counterpart of Vosper’s classification. Another obvious
example is when $G$ is a noncompact group, $\chi:G\to\mathbb{T}$ is a
continuous surjective group homomorphism with compact kernel, $I$ and $J$ are
nonempty closed intervals in $\mathbb{R}$, $A=\chi^{-1}(I)$, and
$B=\chi^{-1}(J)$. One might optimistically conjecture, in analogy with
Vosper’s theorem, that there are no other $G$, $A$, and $B$ such that $(A,B)$
is a minimally expanding pair on $G$.

In view of the earlier discussions, for compact $A,B\subseteq G$, we say that
$(A,B)$ is a $\delta$-nearly minimally expanding pair on $G$ if

$0<\mu_{G}(A),\mu_{G}(B),\text{ and
}\mu_{G}(AB)<\mu_{G}(A)+\mu_{G}(B)+\delta\min\\{\mu_{G}(A),\mu_{G}(B)\\}<\mu_{G}(G).$

The problem of determining near equality condition in the Kemperman inequality
can be then reasonably interpreted as classifying all connected and unimodular
groups $G$ and $\delta$-nearly minimally expanding pairs $(A,B)$ on $G$. In
analogy with the discussion for the Cauchy–Davenport theorem, we hope for an
answer along the following line: If $G$ is compact, and $(A,B)$ is a
$\delta$-nearly minimally expanding pair on $G$ with small $\delta$, then
there is a continuous and surjective group homomorphism $\chi:G\to\mathbb{T}$,
closed interval $I,J\subseteq T$, and small $\varepsilon$, such that

(1)
$A\subseteq\chi^{-1}(I),B\subseteq\chi^{-1}(J),\mu_{G}(\chi^{-1}(I)\setminus
A)<\varepsilon,\text{and }\mu_{G}(\chi^{-1}(I)\setminus B)<\varepsilon.$

The optimistic conjecture for noncompact groups is similar, but with
$\mathbb{T}$ replaced by $\mathbb{R}$ and an extra condition that $\chi$ has
compact kernel.

Under the extra assumption that $G$ is abelian, the optimistic conjectures for
both classification problems were more or less confirmed before our work. In
the same paper [Kne56] mentioned earlier, Kneser solved the classification
problem for equality with the answer we hope for. For the near equality
problem, when $G=\mathbb{T}^{d}$, the desired classification was obtained by
Bilu [Bil98], and later improved by Candela and De Roton [CDR19] for a special
case when $d=1$. When $G$ is a general abelian group, a classification result
was obtained by Tao [Tao18] for compact $G$, and by Griesmer [Gri19] when $G$
is noncompact. Griesmer also proved more general results for disconnected
groups [Gri14, Gri19]. The results by Griesmer [Gri14, Gri19] and by Tao
[Tao18] used nonstandard analysis methods, so the quantitative dependence
between $\varepsilon$ and $\delta$ in (1) is not provided. A sharp exponent
classification result (i.e., $\varepsilon=O(\delta)$) for compact abelian
groups was obtained very recently by Christ and Iliopoulou [CI18]. Results
with sharp exponent bounds are likely the best that one can achieve without
solving the $(3k-4)$-conjecture for $\mathbb{Z}/p\mathbb{Z}$.

For nonabelian $G$, not much was known earlier than this paper. In closest
proximity to what we are doing, Björklund considered in [Bjö17] a variation of
Kemperman’s inequality and the equality classification problem without
assuming that $G$ is connected while assuming additionally that $G$ is
compact, second countable, and has abelian identity component, and the sets
$A$ and $B$ are “spread out” (i.e., far away from being subgroups). The only
common case to this and our current setting with the connectedness assumption
happens when $G$ is abelian and connected. This is a case already covered by
Kneser’s classification result.

Toward showing that appropriate versions of the optimistic conjectures also
hold for the nonabelian classification problem, there is an important new
challenge: While the desired conclusions for the abelian setting are mainly
about the structure of $(A,B)$, the structure of $G$ is also highly involved
for the nonabelian setting. If $G=\text{SO}_{3}(\mathbb{R})$, for example, one
would not be able to find a minimally expanding pair according to the
optimistic answers because there is no continuous surjective group
homomorphism from $\text{SO}_{3}(\mathbb{R})$ to $\mathbb{T}$. On the other
hand, one can always find a continuous and surjective group homomorphism from
a compact abelian group to $\mathbb{T}$ and use this to construct minimally
expanding pairs. For noncompact abelian group, the requirement that the kernel
must be compact impose some constraint on the group $G$, but this is still
comparatively mild.

The above challenge connects our problem to the subject of small expansions in
nonabelian groups, a fascinating topic that brings together ideas from
different areas of mathematics. The phenomenon that expansion rate encodes
structural information about the group can already be seen through the
following famous theorem by Gromov [Gro81] in geometric group theory: If $G$
is a group generated by a finite set $X=X^{-1}$, and the terms of the sequence
$(X^{n})_{n>0}$ grows polynomially in cardinality, then $G$ must be virtually
nilpotent. A more recent result by Breuillard indicates that some of the
analysis go through for locally compact groups [Bre14]. Even more suggestive
is the classifications of approximate groups in [BGT12] mentioned earlier (see
the definition in Section 6.3). Via an analysis by Tao [Tao08], this tells us
that if $G$ is generated by a finite set $X$, and $|X^{3}|<K|X|$ for a fixed
constant $K$ then $G$ must already be virtually nilpotent. In our proof, we
will use the continuous version of the result in [BGT12]; this was proven in
the thesis of Carolino [Car15] and can also be deduced from the result in
[BGT12] using a result of Massicot–Wagner [MW15]. The ideas in the proof of
these results can be traced back to the solution of Hilbert’s Fifth problem by
Montgomery–Zippin [MZ52], Gleason [Gle52], and Yamabe [Yam53], which we will
also use later on. Finally, let us mention that these stories are also closely
tied to the study of definable groups in model theory. This is the natural
habitat of the aforementioned result by Massicot–Wagner [MW15], and also of
Hrushovski’s Lie model theorem [Hru12], one of the main ingredient for the
proof of the main theorem in [BGT12].

Before getting to the results, we briefly survey a number of works for
nonabelian groups which are thematically relevant but use different
techniques. When $G$ is finite and $A,B\subseteq G$ are nonempty, generalizing
Cauchy–Davenport inequality, DeVos showed in [DeV13] that
$|AB|\geq\min\\{|A|+|B|-|H|,|G|\\}$ with $H$ a proper subgroup of $G$ with
maximum cardinality. In the same paper, DeVos classifies all situations where
equality can happen. In [BF19], Björklund and Fish studied an expansion
problem with respect to upper Banach density in amenable nonabelian groups and
obtained conclusions with similar flavor. Under model-theoretic assumptions, a
number of results was obtained by Terry, Conant, and Pillay [CPT18, CP20]. The
conclusions in these works surprisingly also have similarities with ours.

It would also be interesting to study a different minimal and nearly minimal
measure expansion problem where we fix a connected unimodular group $G$
instead of letting $G$ range over all connected unimodular group $G$ as we are
doing here. When $G$ is $\mathbb{R}^{n}$, Kemperman inequality is a
consequence of the Brunn–Minkowski inequality

$\mu_{G}(AB)^{1/n}\geq\mu_{G}(A)^{1/n}+\mu_{G}(B)^{1/n}.$

This inequality is also known when $G$ is nilpotent [McC69, Gro03, Tao11]. The
equality holds in the Brunn–Minkowski inequality for $\mathbb{R}^{n}$ if and
only if $A$ and $B$ are homothetic convex subsets of $\mathbb{R}^{n}$. This
was a result by Brunn and Minkowski when $A$ and $B$ are further assumed to be
convex, and a result by Lyusternik [Lyu35], Henstock and Macbeath [HM53] in
the general case. A qualitative answer for the near equality Brunn–Minkowski
problem for $\mathbb{R}^{n}$ is obtained by Christ [Chr12], and a quantitative
version is obtained by Figalli and Jerison [FJ17]. We do not pursue this
direction further here.

###  1.2. Statement of main results

Our first main result determines the conditions for equality to happen in the
Kemperman inequality answering a question by Kemperman in [Kem64]. Scenario
(v) and (vi) in the theorem is a classification of the groups $G$ and
minimally expanding pairs $(A,B)$ on $G$.

######  Theorem 1.1.

Let $G$ be a connected unimodular group, and $A,B$ be nonempty compact subsets
of $G$. If

$\mu_{G}(AB)=\min(\mu_{G}(A)+\mu_{G}(B),\mu_{G}(G)).$

then we have the following:

  1. (i)

$\mu_{G}(A)+\mu_{G}(B)=0$ implies $\mu_{G}(AB)=0$;

  2. (ii)

$\mu_{G}(A)+\mu_{G}(B)\geq\mu_{G}(G)$ implies $\mu_{G}(AB)=\mu_{G}(G)$, which
is also equivalently to $AB=G$;

  3. (iii)

$\mu_{G}(A)=0$ and $0<\mu_{G}(B)<\mu_{G}(G)$ implies that there is a compact
proper subgroup $H$ of $G$ such that $A\subseteq gH$ for some $g\in G$, and
$B=HB$;

  4. (iv)

$0<\mu_{G}(A)<\mu_{G}(G)$ and $\mu_{G}(B)=0$ implies that there is a compact
proper subgroup $H$ of $G$ such that $A=AH$, and $B\subseteq Hg$ for some
$g\in G$;

  5. (v)

$0<\min\\{\mu_{G}(A),\mu_{G}(B),\mu_{G}(G)-\mu_{G}(A)-\mu_{G}(B)\\}$, and $G$
is compact together implies that there is a surjective continuous group
homomorphism $\chi:G\to\mathbb{T}$ and compact intervals $I$ and $J$ in
$\mathbb{T}$ with $I+J\neq\mathbb{T}$ such that $A=\chi^{-1}(I)$ and
$B=\chi^{-1}(J)$;

  6. (vi)

$0<\min\\{\mu_{G}(A),\mu_{G}(B)\\}$, and $G$ is not compact together implies
that there is a surjective continuous group homomorphism $\chi:G\to\mathbb{R}$
with compact kernel and compact intervals $I$ and $J$ in $\mathbb{R}$ such
that $A=\chi^{-1}(I)$ and $B=\chi^{-1}(J)$.

Moreover, $\mu_{G}(AB)=\min(\mu_{G}(A)+\mu_{G}(B),\mu_{G}(G))$ holds if and
only if we are in exactly one of the implied scenarios in (i-vi).

Next we obtain a classification of nearly minimally expanding pairs. This
answers some questions by Griesmer [Gri19] and confirms a conjecture by Tao
[Tao18, Conjecture 5.1], under the extra assumption of connectedness.

######  Theorem 1.2.

Let $G$ be a connected compact group, and $A,B$ be compact subsets of $G$ with
positive measure. Set

$s=\min\\{\mu_{G}(A),\mu_{G}(B),1-\mu_{G}(A)-\mu_{G}(B)\\}.$

Given $\varepsilon>0$, there is a constant $K=K(s)$ does not depends on $G$,
such that if $\delta<K\varepsilon$ and

$\mu_{G}(AB)<\mu_{G}(A)+\mu_{G}(B)+\delta\min\\{\mu_{G}(A),\mu_{G}(B)\\}.$

Then there is a surjective continuous group homomorphism $\chi:G\to\mathbb{T}$
together with two compact intervals $I,J\in\mathbb{T}$ with

$\mu_{\mathbb{T}}(I)-\mu_{G}(A)<\varepsilon\mu_{G}(A),\quad\mu_{\mathbb{T}}(J)-\mu_{G}(B)<\varepsilon\mu_{G}(B),$

and $A\subseteq\chi^{-1}(I)$, $B\subseteq\chi^{-1}(J)$.

It worth noting that the linear dependence between $\varepsilon$ and $\delta$
is the best possible up to a constant factor. As an application of our main
result, we obtain an measure expansion gap result for sets in connected
compact simple Lie groups.

######  Theorem 1.3 (Expansion gaps in compact simple Lie groups).

There is a constant $\eta>0$ such that the following holds. Let $d>0$ be an
integer. There is a constants $C>0$ only depending on $d$ such that if $G$ is
a connected compact simple Lie group of dimension $d$, and $A$ is a compact
set of $G$ with $0<\mu_{G}(A)<C$. Then

$\mu_{G}(A^{2})>(2+\eta)\mu_{G}(A).$

We can take $\eta>10^{-10}$.

We did not try to optimise the constant $\eta$ of Theorem 1.3 in this paper.

One may compare Theorem 1.3 with expansion gaps for finite sets. The latter
problem is well studied, which is initialed by Helfgott [Hel08] where he
proved an expansion gap in $\mathrm{SL}_{2}(\mathbb{Z}/p\mathbb{Z})$. Results
on the expansions for finite sets are one of the main ingredients in proving
many of spectral gap results. For example, the result by Helfgott is largely
used in the proof by Bourgain and Gamburd [BG08, BG12]. de Saxcé proved in
[dS15] an expansion gap results in simple Lie groups, which is used in the
later proof of spectral gap results [BdS16, BIG17]. For more background in
this direction we refer the reader to [BL18, Tao15].

###  1.3. Notation and convention

Throughout let $m$ and $n$ range over the set $\mathbb{N}=\\{0,1,\ldots\\}$ of
natural numbers. Let $G$ be a locally compact group equipped with a left Haar
measure $\mu_{G}$. Let $H$ range over closed subgroups of $G$ equipped with a
left Haar measure $\mu_{H}$. We normalize $\mu_{G}$ (i.e., $\mu_{G}(G)=1$)
when $G$ is compact, and do likewise when $H$ is compact. We let $G/H$ and
$H\backslash G$ denote the left coset space and the right coset space of $G$
with respect to $H$. Given a coset decomposition, say $G/H$, a fiber of a set
$A\subseteq G$ refers to $A\cap xH$ for some $xH\in G/H$. We also use
$\mu_{H}$ to denote the fiber lengths in the paper, that is we sometimes write
$\mu_{H}(A\cap xH)$ to denote $\mu_{H}(x^{-1}A\cap H)$. A constant in this
paper is always a positive real number, and by Lie group we mean a real Lie
group with finite dimension.

##  2\. Outline of the argument

In this section, we informally explain some of the major new ideas of the
proofs. We decided to write a slightly longer outline as some of the later
computations are rather technical.

###  2.1. Overview of the strategy

We will explain here the main steps of the proof of Theorem 1.1 and Theorem
1.2 illustrated by simple examples. The focus will be on the plausibility of
the argument and how they fit in together to resolve the main difficulties of
the problem. More detailed discussion will be given in Sections 2.2, 2.3, and
2.4.

After handling a number of easy cases, the proofs of Theorem 1.1 and Theorem
1.2 require constructing appropriate continuous and surjective group
homomorphisms into either $\mathbb{R}$ or $\mathbb{T}$ under the given data of
a minimal or nearly minimal expanding pair $(A,B)$ on $G$. The key difficulty
of the problem is that many methods in the abelian settings ($e$-transform,
fourier analytic, etc) has no obvious generalization to the nonabelian
settings.

Instead, we will use a method in the direction of the solution of Hilbert’s
Fifth problem, namely, making use of pseudometrics on $G$ (i.e., maps
$d:G\times G\to\mathbb{R}$ satisfying all the properties of a metric except
$d(g_{1},g_{2})=0$ implying $g_{1}=g_{2}$). With some caveats, our proof can
be thought of as having three steps: obtaining a suitable pseudometric on $G$,
constructing an appropriate group homomorphism from the pseudometric, and
deduce description information about $(A,B)$ from the pseudometric. As we want
to construct a group homomorphism into $\mathbb{R}$ or $\mathbb{T}$ and not
any other Lie groups, the new challenge is to develop properties that allow us
to control the desired target group and to obtain a pseudometric on $G$
satisfying these properties.

Let us first explain the second step to see that the pseudometric idea is
plausible. Until the end of this section, we will focus on the case where $G$
is compact with the other case treated similarly. If $\chi:G\to\mathbb{T}$ is
a continuous and surjective group homomorphism, and $d_{\mathbb{T}}$ is the
Euclidean metric on $\mathbb{T}$, then setting
$d(g_{1},g_{2})=d_{\mathbb{T}}(\chi(g_{1}),\chi(g_{2}))$, it is easy to see
that $d$ is a pseudometric on $G$ with the “linear” property that

$d(g_{1},g_{3})=|d(g_{1},g_{2})\pm d(g_{2},g_{3})|$

for all $g_{1},g_{2},g_{3}\in G$ such that
$\max\\{d(g_{1},g_{2}),d(g_{2},g_{3}),d(g_{1},g_{3})\\}<1/4.$ Moreover, in the
above situation we also have

$\ker\chi=\\{g\in G:d(g,\mathrm{id}_{G})=0\\},$

so the group homomorphism $\chi$ can be recovered from the pseudometric. In
the second step of our proof, we will define weaker properties of locally
linear and locally almost linear, and show that more or less under these
conditions, we can obtain an appropriate homomorphism into $\mathbb{T}$. More
details will be given in Section 2.3.

Now we would like to construct a locally (almost) linear pseudometric on $G$.
There is a caveat: our primary technique discussed later only works for Lie
groups. So we need an extra step in our strategy (also with an underlying
pseudometric idea) where we use the Gleason–Yamabe Theorem to obtain a group
homomorphism $\pi$ from $G$ to a Lie group $G^{\prime}$ with compact kernel
$H$. This is a relatively standard, but there are two important arrangements
we need to make. We need to choose $H$ carefully to ensure that it is
connected and $\mu_{G^{\prime}}(\pi(A))+\mu_{G^{\prime}}(\pi(B))<\mu_{G}(G)$;
in third step, we will show that these conditions are enough to imply that
$(\pi(A),\pi(B))$ essentially has nearly minimal expansion on $G^{\prime}$. We
also need to ensure that the dimension of $G^{\prime}$ is bounded to be able
to to get our sharp exponent bound later on. This requires the application of
the continuous version of the result in [BGT12], which is proven in Carolino’s
thesis [Car15].

Now we focus on the case where $G$ is a compact connected Lie group with
bounded dimension, where the structure of $G$ can be studied through its torus
subgroups. Using a submodularity argument as in [Tao18], we can carefully
modify the original $(A,B)$ to ensure that $\mu_{G}(A)$ and $\mu_{G}(B)$ are
relatively small compared to $\mu_{G}(G)$ but still large compare to the
error. The pseudometric we need is easy to define:

$d(g_{1},g_{2}):=\mu_{G}(A)-\mu_{G}(g_{1}A\cap g_{2}A).$

The real challenge is to show that this is locally (almost) linear. This is
achieved by gaining an understanding of the “shape” of the pair $(A,B)$ at
different levels. Keeping in mind that we already have the desired
classification of minimal and nearly minimal expanding pairs for $\mathbb{T}$,
we will choose a suitable one-dimension torus subgroup of $G$ and use it as a
tool to probe for information about $(A,B)$.

Thinking of the one-dimension torus subgroups of $G$ as specifying the
“directions” in $G$, we choose a one-dimension torus subgroup $T\leq G$ such
that for all $gT\in G/T$, the “length” $\mu_{T}(g^{-1}A\cap T)$ of the “left
fiber” $gT\cap A$ of $A$ satisfies

$\mu_{T}(g^{-1}A\cap T)<\mu_{T}(T)/100,$

and a similar condition holds for a “right fiber” $B\cap Tg$ of $B$ with
$Tg\in T\backslash G$. The existence of such $T$ is by no mean obvious, and is
saying that when $(A,B)$ is nearly minimally expanding, $A$ and $B$ cannot be
“Kakeya sets”. More details will be given in the beginning of Section 2.2.

Figure 1. Intuition of the ideas in the proof.

We visualize $G$ in two ways: a rectangle with the horizontal side
representing $G/T$ and each of the vertical section representing a left cosets
of $T$, and a similar dual picture for $T\backslash G$; see the bottom half of
Figure 1. The main idea is to show that $g_{1}A$, $g_{2}A$, $g_{3}A$, and $B$
geometrically look like in the picture with $g_{1},g_{2},g_{3}\in G$ in a
suitable neighborhood of $\mathrm{id}_{G}$. (For instance, we want the
“fibers” of $A$ and $B$ to be intervals of $T$, all the nonempty “left fibers”
of $g_{1}A$ to have similar “lengths”, the “width” of $A$ to be almost the
same as that of $B$, the translated copy $g_{1}A$ to only “move vertically”
compared to $g_{1}A$, and the “moving up” and “moving down” divisions in
$g_{2}A$ and $g_{3}A$ to be almost the same. As a very small hint of why these
are true, imagine that one of the fiber in $B$ is too large compared to the
rest. Then the product of $A$ with that fiber is already much larger measure
than $\mu_{G}(A)+\mu_{G}(B)$.) One can see that the locally (almost) linearity
follows from this picture. The key point is that, the global information about
the pseudometric can be deduced via the geometrical shape from the information
on a generic fiber $xT$, where the classification of nearly minimally
expanding pairs is known. More detailed discussion is given in Section 2.2.

Finally, we discuss the third step of deducing the structure of $A$ and $B$
when we have the appropriate group homomorphism $\chi:G\to\mathbb{T}$. There
is another minor caveat: we can modify the the group homomorphism obtain from
the pseudometric to ensure that the kernel is connected and the image is
small; this was also the same arrangement we made earlier when we use the
Gleason–Yamabe Theorem. The following example, already containing the idea of
the later proof, will illustrate to the reader that unless there are closed
intervals $I,J\subseteq\mathbb{T}$ such that $A$ and $B$ are nearly
$\chi^{-1}(I)$ and $\chi^{-1}(J)$, the produce $\mu_{G}(AB)$ is much larger
than $\mu_{G}(A)+\mu_{G}(B)$. Assume that $G=\mathbb{T}^{2}$,
$\chi:\mathbb{T}^{2}\to\mathbb{T}$ is the projection onto the second
coordinate, and we identify $\mathbb{T}^{2}$ with its fundamental domain
$[0,1]\times[0,1]\subseteq\mathbb{R}^{2}$. Set

$A=(\mathbb{T}\times[0,1/7])\cup([0,1/3]\times[1/7,2/7]).$

Then, $\mu_{G}(A)=1/7+(1/3)\times(1/7)$. On the other hand,

$A^{2}=(\mathbb{T}\times[0,3/7])\cup([0,2/3]\times[3/7,4/7]).$

So $\mu_{G}(A^{2})=3/7+(2/3)\times(1/7)>2/7+2/3\times(2/7)=2\mu_{G}(A)$.

###  2.2. The first step: Obtaining a suitable pseudometric

Suppose $G$ is a connected and compact Lie group, and $(A,B)$ is a nearly
minimally expanding pair on $G$ with sufficiently small measure. We will show
that there is a one-dimensional torus subgroup $T$ of $G$ such that the
“length” $\mu_{T}(g^{-1}A\cap H)$ of each “left fiber” $A\cap gT$ of $A$ is
small, and a similar condition hold for “right fibers” of $B$. After that, we
will show that $g_{1}A,g_{2}A,g_{3}A$ satisfy the picture at the bottom left
of Figure 1 with respect to $T$, from which we can deduce the local almost
linearity of $d$.

Suppose $A$ is a “Kakaya set”, i.e, it has a long left fiber $A\cap gT$ for
every choice of “direction” $T$ of $G$. Fix such a $T$. If a large proportion
of right fibers of $B$ are rather short, then using the Kemperman inequality
for $T$ and Fubini’s theorem, we get $\mu_{G}((A\cap gT)B)$ is already much
larger than $\mu_{G}(A)+\mu_{G}(B)$, a contradiction. So a large proportion of
the right fibers of $B$ in the direction $T$ must be rather long. A reverse
argument then shows that a large proportion of the left fibers of $A$ are
long. Thus, $\mu_{G}(AT)$ is not too large compared to $\mu_{G}(A)$.

From the discussion above, it suffices to show the contradiction when
$\mu_{G}(AT)/\mu_{G}(A)$ is not too large for every $T$. We call a nonempty
and compact subset of $G$ a toric $K$-nonexpander, if it has this property for
a given constant $K$. We will show in Section 8 a result with independent
interest: Every nonempty compact subset of $G$ with sufficiently small measure
cannot be a toric $K$-nonexpander.

Let us present here a pseudo-argument, which nevertheless illustrate the idea.
Assume $A$ is a toric $K$-nonexpander. Obtain finitely many torus subgroups
$T_{1},\ldots,T_{n}$ of $G$ such that

$G=T_{1}\cdots T_{n}.$

(Note that $n$ depends on the dimension of $G$, which is bounded by the caveat
in Step 1.) Let us pretend that using the assumption $\mu_{G}(AT_{1})\leq
K\mu_{G}(A)$ we can cover $AT_{1}$ with $(K+1)$ right translations of $A$. It
can be then shown that $AT_{1}$ is a toric $K(K+1)$-nonexpander. Next, we
further pretend that $AT_{1}T_{2}$ can be covered with $K(K+1)+1$ right
translations of $AT_{1}$ which can then be covered by $K(K+1)^{2}+(K+1)$ right
translations of $A$. Continuing the procedure, we get $C(K)$ such that
$AT_{1}\cdots T_{n}=G$ can be covered by $C(K)$ right translations of $A$.
Thus, $\mu_{G}(A)>1/C(K)$, contradicting the assumption that $\mu_{G}(A)$ is
very small.

The pseudo-argument in the preceding paragraph does not work in most of the
cases. In particular, one cannot deduce from $\mu_{G}(AT_{1})<K\mu_{G}(A)$
that $AT_{1}$ can be covered by $(K+1)$ right translations of $A$. However, it
does contain some truth, and we will be able to use a version of the Lovazs
covering argument to approximate this pseudo-argument.

Now choosing a one-dimensional torus subgroup $T$ of $G$ such that for all
$x\in G$ and $y\in G$, the fibers $xT\cap A$ and $B\cap Ty$ are both short. We
will show that the set $A$ and $B$ have the shape as described in Figure 1.
Without loss of generality we can arrange that the width $\mu_{G}(AT)$ of $A$
in $G/T$ is at most the width $\mu_{G}(TB)$ of $B$ in $T\backslash G$. Choose
uniformly at random $xT\in AT$, and applying the Kemperman inequality for $T$,
we have

$\displaystyle\mu_{G}(AB)$ $\displaystyle\geq\mathbb{E}_{xT\in
AT}\mu_{G}((A\cap xT)B)$ $\displaystyle\geq\mathbb{E}_{xT\in AT}\mu_{T}(A\cap
xT)\mu_{T\backslash G}(TB)+\mu_{G}(B)$
$\displaystyle=\mu_{G}(A)\frac{\mu_{T\backslash
G}(TB)}{\mu_{G/T}(AT)}+\mu_{G}(B)$ $\displaystyle\geq\mu_{G}(A)+\mu_{G}(B);$

As $(A,B)$ is nearly minimally expanding, we have $\mu_{G}(AB)$ is nearly the
same as $\mu_{G}(A)+\mu_{G}(B)$. The fourth line then gives us that
$\mu_{G}(TB)$ is nearly the same as $\mu_{G}(AT)$. The second line now gives
us that for each $xT\in AT$, the fiber $(xT\cap A)$ is nearly an interval up
to an endomorphism of $T$. From the first line, $\mu_{T}(A\cap xT)$ is almost
constant as $xT$ ranges through $AT$.

We now discuss the relative position of $g_{1}A$, $g_{2}A$, and $g_{3}A$ for
$g_{1}$, $g_{2}$, and $g_{3}$ near $\mathrm{id}_{G}$. Clearly, $(g_{1}A,B)$,
$(g_{2}A,B)$, and $(g_{3}A,B)$ are also nearly minimally expanding. Using a
submodularity argument as in [Tao18], we can show that $(g_{1}A\cap
g_{2}A,B)$, $(g_{1}A\cup g_{2}A,B)$ are also nearly minimally expanding. A
similar analysis applies to all these pairs. In particular, in order to have
almost all vertical fiber of each pair to having the same height, we can only
have one of the following two scenarios:

  1. (1)

(Almost vertical movement) $AT$ is close to $g_{1}AT$

  2. (2)

(Almost horizontal movement) fibers in $(A\cup g_{1}A)T$ has almost the same
length as those in $AT$

We note that (2) cannot happen because then $A\cup g_{1}A$ will no longer have
a similar width as $B$. A similar argument shows us that Figure 2 cannot
happen. Hence, the “moving up” and “moving down” divisions in $g_{2}A$ and
$g_{3}A$ must be the same as in Figure 1 in Section 2.1.

Figure 2. Different “moving up” and “moving down” divisions in $g_{2}A$ and
$g_{3}A$ result in uneven fiber size.

We mention two subtler aspects of the geometry of minimally expanding pairs
that we will not be able to get in details. First, we will also need it to
show a condition called path monotonicity which is a necessary ingredient to
show assumption (2) in Step 2 (Section 2.3)). Second, for the purpose of
controlling the error in Step 2, we need to show a certain “convexity
property” of

$N_{\lambda}=\\{g\in G:\mu_{G}(A)-\mu_{G}(A\cap gA)<\lambda\\}.$

This requires us to construct a “core” of $A$, which is related to the
Sanders–Croot–Sisask theorem [San10, CS10] in additive combinatorics, and
stabilizer theorems in model theory [Hru12].

###  2.3. The second step: Constructing an appropriate homomorphism

We assume in this section that $G$ is a connected and compact Lie group, $d$
is a left invariant continuous pseudometric on $G$ with the following
properties:

  1. (1)

(Local almost linearity) There is $\lambda\in\mathbb{R}^{>0}$ such that with

$N_{\lambda}:=\\{g\in G:d(\mathrm{id}_{G},g)<\lambda\\},$

there is $\varepsilon<10^{-10}\mu_{G}(N_{\lambda})$ such that for all
$g_{1},g_{2},g_{3}\in N_{\lambda}$

$d(g_{1},g_{3})\in|d(g_{1},g_{2})\pm d(g_{2},g_{3})|+I(\varepsilon),$

where $I(\varepsilon)$ is the interval
$(-\varepsilon,\varepsilon)\subseteq\mathbb{R}$.

  2. (2)

(Local monotonicity) With the same $\lambda$ in (2), for all $g\in
N_{\lambda}$,

$|d(\mathrm{id}_{G},g^{2})-2d(\mathrm{id}_{G},g)|\in I(\varepsilon).$

We now sketch how to construct a continuous and surjective group homomorphism
to $\mathbb{T}$ from these data. The condition presented here is a simplified
but essentially equivalent condition as in Section 7. A crucial argument we
will not be able to get into details here is to show that the local
monotonicity condition can be deduced from a weaker property of path
monotonicity obtained from the first step (Section 2.2).

When we are in the special case with $\varepsilon=0$ in property (1), there is
a relatively easy argument which also works for noncompact Lie groups. Set

$\ker d=\\{g\in G:d(\mathrm{id}_{G},g)=0\\}.$

Using the left invariance, continuity, and triangle inequality, one can show
that $\ker d$ is a closed subgroup of $G$. Moreover, in this case, $G/\ker d$
must be isomorphic to $\mathbb{T}$, and the pseudometric $d$ locally must
agrees with a constant multiple of the pullback of the Euclidean metric. These
are, perhaps, not too surprising as a Lie group equipped with a locally linear
pseudometric is, intuitively, a very rigid object which locally looks like a
straight line. In fact, property (2) is not needed as it is a consequence of
property (1) in this case.

The general case is much harder as we no longer have the same type of
rigidity. In particular, $\ker d$ might not be normal, and $G/\ker d$ might
not be $\mathbb{T}$ even if $\ker d$ is normal. The reader familiar with the
proof of Hilbert’s Fifth problem would guess that we might try to slightly
modify $d$ to get a locally linear pseudometric $d^{\prime}$ and use the
earlier strategy. This is still true at the conceptual level, but our actual
argument is much more explicit allowing error control.

As an expository vehicle for the idea, let us still imagine that we somehow
obtain a locally linear pseudometric $d^{\prime}$ only slightly differs from
$d$. By the earlier argument, we get a group homomorphism
$\chi^{\prime}:G\to\mathbb{T}$ and $\alpha^{\prime}\in\mathbb{R}$ such that
$d^{\prime}(g_{1},g_{2})=\alpha^{\prime}\|\chi^{\prime}(g_{1})-\chi^{\prime}(g_{2})\|_{\mathbb{T}}$
near $\mathrm{id}_{G}$. Therefore, for $g$ near $\mathrm{id}_{G}$ we have

$\chi^{\prime}(g)=\pm(1/\alpha^{\prime})d^{\prime}(\mathrm{id}_{G},g)+\mathbb{Z}.$

Note that this gives us a way to determine $\chi^{\prime}(g)$ for $g\in
N_{\lambda}$ from $d^{\prime}$, but we need to know how to describe
$\alpha^{\prime}$ from $d^{\prime}$. For $g_{1},g_{2},g_{3}\in N_{\lambda}$,
we say that $g_{2}$ is “between” $g_{1}$ and $g_{3}$ if
$d^{\prime}(g_{1},g_{3})=d^{\prime}(g_{1},g_{2})+d^{\prime}(g_{2},g_{3})$. We
say that a product $g_{1}\cdots g_{n}$ is “increasing” if $g_{1}\cdots g_{i}$
is “between” $g_{1}\cdots g_{i-1}$ and $g_{1}\cdots g_{i+i}$. For an arbitrary
element $g\in G$, we write it as an “increasing” product $g=g_{1}\cdots g_{n}$
with $g_{i}\in N_{\lambda}$. Then, we can show that

$\chi^{\prime}(g)=\pm(1/\alpha^{\prime})\sum_{i=0}^{n-1}d^{\prime}(g_{1}\ldots
g_{i},g_{1}\ldots g_{i+1})+\mathbb{Z}.$

This gives us a way to determine $\chi^{\prime}(g)$ for an arbitrary $g$ if we
can get $\alpha^{\prime}$ from $d^{\prime}$. Now, note that if we write
$\mathrm{id}_{G}$ as an “increasing” product $1_{G}=g_{1}\cdots g_{n}$ with
$n>0$ then $\sum_{i=0}^{n-1}d^{\prime}(g_{1}\ldots g_{i},g_{1}\ldots g_{i+1})$
will be a multiple of $\alpha^{\prime}$. Moreover, $\alpha^{\prime}$ will be
the minimum value of such sum. So we recover $\alpha^{\prime}$ from
$d^{\prime}$ and obtain a much more explicit way to describe the map
$\chi^{\prime}$.

Figure 3. Any element $g$ in $G$ will be captured by one of the monitors.

Back to the situation where we only have $d$ but not $d^{\prime}$, we can
carry out the same strategy as above with $d$ while being careful with the
errors. There are a number of problems that arise. To define the “betweeness”,
property (2) plays an important technical role. Since the error propagate very
fast, to get a linear error bound, we cannot write $g$ as an “increasing”
product $g=g_{1}\cdots g_{n}$ with very large $n$. The upper bound in $n$
comes from a lower bound on the size of $N_{\lambda}$, which essentially comes
from the result on “core” in the first step. To get an analog of
$\alpha^{\prime}$ with the desired property, we need to choose a standard way
of expressing $\mathrm{id}_{G}$ as an “increasing” product
$\mathrm{id}_{G}=g_{1}\cdots g_{n}$ and use $g_{1},\ldots,g_{n}$ to “monitor”
the other elements in the group, as illustrated by Figure 3. This require us
to develop the machinery of irreducible sequence and concatenation; see
Section 7.3 for details. Finally, what we ends up with is a multi-valued
almost homomorphism that is not continuous but still universally measurable.
We will need to use a number of results from descriptive set theory and
Riemannian geometry to extract from this the desired group homomorphism.

###  2.4. The third step: Deducing the structure of the pair

Suppose $(A,B)$ is a nearly minimally expanding pair on $G$, $H$ is a
connected, compact and normal subgroup of $G$, $\pi:G\to G/H$ is the quotient
map, and

$\mu_{G/H}(\pi(A))+\mu_{G/H}(\pi(B)<1.$

The goal of this step is to show the following transfer to quotients result:
There is a nearly minimally expanding pair $(A^{\prime},B^{\prime})$ on $G/H$
such that $\mu_{G}(A\triangle\pi^{-1}(A^{\prime}))$ and
$\mu_{G}(B\triangle\pi^{-1}(B^{\prime}))$ are both small.

Figure 4. Lower bound for $\mu_{G}(A^{2})$.

To illustrate the idea, we focus on the special case with $A=B$. Employing the
geometric language in Section 2.1, we call $\mu_{G/H}(\pi(A))$ the _width_ of
$A$, for each $g$ in $G$, we call $A\cap gH$ a _fiber_ of $A$, and refer to
$\mu_{H}(g^{-1}A\cap H)$ as its _length_. We consider a further special case
assuming that $A$ can be partitioned into $N+1$ parts
$A=\bigcup_{i=0}^{N}A_{i}$ such that the images under $\pi$ of the $A_{i}$’s
are compact and pairwise disjoint, $A_{i}$ has width $w_{i}$, the fibers in
$A_{0}$ all have length $\geq 1/2$, the fibers in $A_{i}$ all have the same
length $l_{i}\leq 1/2$ for each $i\geq 1$, and $l_{i}\geq l_{i+1}$ for all
$i<N$. This further special case is, in fact, quite representative as we can
reduce the general problem to it using approximation techniques.

The proof of this step can be seen as the following “spillover” argument.
Applying the Kemperman inequalities for $H$ and $G/H$, we learn that all the
fibers in $A^{2}_{0}$ has length $1$, and the width of $A_{0}^{2}$ is at least
$2w_{0}$. By Fubini’s theorem, $\mu_{G}(A_{0}^{2})$ is at least $2w_{0}$.
Next, consider $A_{0}(A_{0}\cup A_{1})$. A similar argument gives us that all
the fibers in $A_{0}(A_{0}\cup A_{1})$ has length at least $l_{1}+1/2$, and
the width of $A_{0}(A_{0}\cup A_{1})$ in $G/H$ is at least $2w_{0}+w_{1}$.
Note that $(l_{1}+1/2)(2w_{0}+w_{1})$ is a weak bound for
$\mu_{G}(A_{0}(A_{0}\cup A_{1}))$ since the fibers in $A_{0}^{2}$ are
“exceptionally long”. Taking all of these into account, a stronger lower bound
for $\mu_{G}(A_{0}(A_{0}\cup A_{1}))$ is

$2w_{0}+(l_{1}+1/2)w_{1}.$

Iterate this procedure, a lower bound for
$\mu_{G}(A_{0}A)=\mu_{G}(A_{0}(A_{0}\cup\ldots\cup A_{N}))$ is

$2w_{0}+\sum_{i=1}^{N}(l_{i}+1/2)w_{i}.$

Now every fiber in $(A_{0}\cup A_{1})A$ has length at least $l_{1}$, and the
width of $(A_{0}\cup A_{1})A$ is at least $(w_{0}+w_{1})+w$. Using the same
logic, a lower bound for $\mu_{G}((A_{0}\cup A_{1})A)$ is

$2w_{0}+\sum_{i=1}^{N}(l_{i}+1/2)w_{i}+l_{1}w_{1}.$

Iterate the procedure, a lower bound for
$\mu_{G}(A^{2})=\mu_{G}((A_{0}\cup\ldots\cup A_{n})A)$ is

$2w_{0}+\sum_{i=1}^{N}(l_{i}+1/2)w_{i}+\sum_{i=1}^{N}l_{i}w_{i}.$

Note that $\mu_{G}(A)=l_{0}w_{0}+\ldots+l_{N}w_{N}$. Hence, $\mu_{G}(A^{2})$
is nearly $2\mu_{G}(A)$ implies that we must nearly have $w_{0}=1$,
$w_{1}=\ldots=w_{N}=0$, and $l_{0}=1$. From this, one can deduce the
conclusion that we want for this step.

###  2.5. Structure of the paper

The paper is organized as follows. Section 3 includes some facts about Haar
measures and unimodular groups, which will be used in the subsequent part of
the paper. Section 4 deals with the more immediate parts of Theorem 1.1 and
hence sets up the stage for the main part of the argument. Section 5 allows us
to arrange that in a minimally or a nearly minimally expanding pair $(A,B)$,
the sets $A$ and $B$ have small measure (Lemma 5.3). Sections 6, 7, and 8
contain main new technical ingredients of the proof, which will be put
together in Section 9 to complete the proofs of Theorem 1.1, Theorem 1.2, and
Theorem 1.3. Steps 1, 2, and 3 discussed in Sections 2.2, 2.3, and 2.4,
corresponds to Sections 8, 7, and 6 respectively.

In Section 6.1, we proved the transfer theorem (Theorem 6.6), which allow us
to transfer the problem into certain quotient groups. Section 6.2 contains
structural results assuming we have an appropriate homomorphism (Proposition
6.8). Section 6.3 gives an upper bound on the dimension of the Lie model
(Proposition 6.12). Together with the transfer theorem, we reduce the problem
to a bounded dimension Lie group.

In Section 7.1, we showed that a locally linear pseudometric on $G$ would
induce a continuous surjective homomorphism to either $\mathbb{R}$ or
$\mathbb{T}$, with compact kernel (Theorem 7.5). Sections 7.2 and 7.3 study
the locally almost linear pseudometric in compact Lie groups. In particular,
we proved that path monotonicity implies monotonicity (Proposition 7.7), and
for almost monotone almost linear pseudometric, one can also find a
homomorphism mapping to $\mathbb{T}$ (Theorem 7.23).

In Section 8.1, we bound the size of the toric expanders (Theorem 8.8). We
construct the pseudometric from geometric properties of nearly minimal
expansion sets in Sections 8.2 and 8.3. Section 8.2 provides a locally linear
pseudometric from minimally expansion sets (Proposition 8.14). In Section 8.3,
we construct a path monotone locally almost linear pseudometric (Proposition
8.21).

The dependency diagram of the paper is as below.

##  3\. Preliminaries on locally compact groups

Throughout this section, we assume that $G$ is a locally compact group (in
particular, Hausdorff), and $A,B\subseteq G$ are nonempty and measurable.
Recall that $G$ can be equipped with a left Haar measure $\mu_{G}$, which is
unique up to a constant; see [DE09, Chapter 1] for details. Below are some
facts that we will use:

######  Fact 3.1.

Suppose $\mu_{G}$ is either a left or a right Haar measure on $G$. Then:

  1. (1)

If $A$ is compact, then $\mu_{G}(A)<\infty$.

  2. (2)

If $A$ is open, then $\mu_{G}(A)>0$.

  3. (3)

_(Inner regularity)_ There is an increasing sequence $(A_{n})$ of compact
subsets of $A$ such that $\mu_{G}(A)=\lim_{n\to\infty}\mu_{G}(A_{n}).$

  4. (4)

_(Outer regularity)_ There is a decreasing sequence $(A_{n})$ of open subsets
of $G$ such that $A\subseteq A_{n}$ for all $n$, and
$\mu_{G}(A)=\lim_{n\to\infty}\mu_{G}(A_{n}).$

  5. (5)

_(Continuity of measure under symmetric difference)_ Suppose $A\subseteq G$ is
measurable, then the function $G\to\mathbb{R},g\mapsto\mu_{G}(A\triangle gA)$
is continuous.

From Fact 3.1(5), we get the following easy corollary:

######  Corollary 3.2.

Suppose $\mu_{G}$ is either a left or a right Haar measure on $G$. Then the
set $\mathrm{Stab}_{0}(A)=\\{g\in G:\mu_{G}(A\triangle gA)=0\\}$ is a closed
subgroup of $G$.

We say that $G$ is unimodular if every left Haar measure $\mu_{G}$ on $G$ is
also a right Haar measure. Hence, we will refer to a left Haar measure of $G$
simply as a Haar measure of $G$ in this case. The following fact is
straightforward from the definition.

######  Fact 3.3.

If $G$ is unimodular, and $\mu_{G}$ is a Haar measure on $G$, then
$\mu_{G}(A)=\mu_{G}(A^{-1})$.

Suppose $H$ is a closed subgroup of $G$. The following fact allows us to link
Haar measures on $G$ with the Haar measures on $H$ for unimodular $G$ and $H$.

######  Fact 3.4 (Quotient integral formula).

Suppose $G$ is unimodular with Haar measure $\mu_{G}$, and $H$ is a closed
subgroup of $G$ with left Haar measure $\mu_{H}$. If $f$ is a continuous
function on $G$ with compact support, then

$xH\mapsto\int_{H}f(xh)\,\mathrm{d}\,\mu_{H}(x).$

defines a function $f^{H}:G/H\to\mathbb{R}$ which is continuous and has
compact support. If $H$ is unimodular, then there is unique invariant Radon
measures $\mu_{G/H}$ on $G/H$ such that for all continuous function
$f:G\to\mathbb{R}$ with compact support, the following integral formula holds

$\int_{G}f(x)\,\mathrm{d}\,\mu_{G}(x)=\int_{G/H}\int_{H}f(xh)\,\mathrm{d}\,\mu_{H}(h)\,\mathrm{d}\,\mu_{G/H}(xH).$

A similar statement applies replacing the left homogeneous space $G/H$ with
the right homogeneous space $H\backslash G$.

We can extend Fact 3.4 to measurable functions on $G$, but the function
$f^{H}$ in the statement can be only be defined and is $\mu_{G/H}$-measurable
$\mu_{G}$-almost everywhere. So, in particular, this problem applies to
indicator function $\mathbbm{1}_{A}$ of a measurable set $A$. This causes
problem in our later proof and prompts us to sometimes restrict our attention
to a better behaved subcollection of measurable subsets of $G$. We say that a
subset of $G$ is $\sigma$-compact if it is a countable union of compact
subsets of $G$. It is easy to see that $\sigma$-compact sets are measurable
and product of $\sigma$-compact sets are $\sigma$-compact. By Fact 3.1(3), we
can always find a $\sigma$-compact subset $A^{\prime}$ of $A$ such that
$\mu_{G}(A^{\prime})=\mu_{G}(A)$ for all left or right Haar measure $\mu_{G}$
of $G$.

######  Lemma 3.5.

Suppose $A\subseteq G$ is $\sigma$-compact, and $\mathbbm{1}_{A}$ is the
indicator function of $A$. Then $aH\mapsto\mu_{H}(A\cap aH)$ defines a
measurable function $\mathbbm{1}^{H}_{A}:G\to\mathbb{R}$.

###### Proof.

We first consider the case where $A$ is compact. By Baire’s Theorem,
$\mathbbm{1}_{A}$ is the pointwise limit of a monotone nondecreasing sequence
of continuous function of compact support. If $f:G\to\mathbb{R}$ is a
continuous function of compact support, then the function $f^{H}:G/H\to
R,aH\mapsto\int_{H}f(ax)dx$ is continuous with compact support, and hence
measurable; see, for example, [DE09, Lemma 1.5.1]. Noting that $\mu_{H}(A\cap
aH)=\int_{H}\mathbbm{1}_{A}(ax)dx$, and applying monotone convergence theorem,
we get that $\mathbbm{1}^{H}_{A}$ is the pointwise limit of a monotone
nondecreasing sequence of continuous function of compact support. Using
monotone convergence theorem again, we get $\mathbbm{1}^{H}_{A}$ is
integrable, and hence measurable.

Finally, the general case where $A$ is only $\sigma$-compact can be handled
using a similar argument, noting that $\mathbbm{1}_{A}$ is then the pointwise
limit of a monotone nondecreasing sequence of indicator functions of compact
sets. ∎

Suppose $H$ is a closed subgroup of $G$. Then $H$ is locally compact, but not
necessarily unimodular. We use the following fact in order to apply induction
arguments in the later proofs.

######  Fact 3.6.

Let $G$ be a unimodular group. If $H$ is a closed normal subgroup of $G$, then
$H$ is unimodular. Moreover, if $H$ is compact, then $G/H$ is unimodular.

We use the following isomorphism theorem of topological groups.

######  Fact 3.7.

Suppose $G$ is a locally compact group, $H$ is a closed normal subgroup of
$G$. Then we have the following.

  1. (1)

_(First isomorphism theorem)_ Suppose $\phi:G\to Q$ is a continuous surjective
group homomorphism with $\ker\phi=H$. Then the exact sequence of groups

$1\to H\to G\to Q\to 1$

is an exact sequence of topological groups if and only if $\phi$ is open; the
former condition is equivalent to saying that $Q$ is canonically isomorphic to
$G/H$ as topological groups.

  2. (2)

_(Third isomorphism theorem)_ Suppose $S\leq G$ is closed, and $H\leq S$. Then
$S/H$ is a closed subgroup of $G/H$. If $S\vartriangleleft G$ is normal, then
$S/H$ is a normal subgroup of $G/H$, and we have the exact sequence of
topological groups

$1\to S/H\to G/H\to G/S\to 1;$

this is the same as saying that $(G/H)/(S/H)$ is is canonically isomorphic to
$G/S$ as topological groups.

We use the following fact about the adjoint representations of Lie groups
[HN12, Proposition 9.2.21].

######  Fact 3.8.

If $G$ is a connected Lie group with Lie algebra $\mathfrak{g}$, and let
$\mathrm{Ad}:G\to\mathrm{Aut}(\mathfrak{g})$ be the adjoint representation.
Then $\ker(\mathrm{Ad})$ is the center of $G$.

We now state a version of Kemperman’s Inequality for all measurable
$A,B\subseteq G$. It is easy to see that we can replace the assumption that
$A$ and $B$ are compact in Kemperman’s inequality with the slightly weaker
assumption that $A$ and $B$ are $\sigma$-compact. The only problem in
replacing the compact assumption with a measurability assumption comes the
fact that $AB$ might not be measurable. This comes down to the fact that if
$A$ and $B$ are coanalytic, then $AB$ might not be analytic and so the
measurability of $AB$ depends on the axiomatic assumptions of set theory. To
remedy this situation, recall that for a left Haar measure $\mu_{G}$ on $G$
and a subset $C$ of $G$, the corresponding inner Haar measure
$\tilde{\mu}_{G}(C)$ of the set $C$ is defined by

$\tilde{\mu}_{G}(C)=\sup\\{\mu_{G}(A):A\subseteq C\text{ is compact}.\\}$

For all $C$, we can always choose a $\sigma$-compact $C^{\prime}\subseteq C$
such that $\mu_{G}(C^{\prime})=\tilde{\mu}_{G}(C)$. Replacing the measure on
$AB$ with inner measure, we get the following version of Kemperman’s
inequality.

######  Theorem 3.9 (Kemperman’s inequality for measurable sets).

If $G$ is unimodular and connected, and $\mu_{G}$ is a Haar measure on $G$.
Then

$\tilde{\mu}_{G}(AB)\geq\min\\{\mu_{G}(A)+\mu_{G}(B),\mu_{G}(G)\\}.$

Moreover, if $A$ and $B$ are compact, $\sigma$-compact, or analytic, then we
can replace $\tilde{\mu}_{G}$ with $\mu_{G}$.

In fact, we can even state Kemperman’s inequality without measurability
assumption on $A$ and $B$ by replacing $\mu_{G}(A)$ and $\mu_{G}(B)$ with
$\tilde{\mu}_{G}(A)$ and $\tilde{\mu}_{G}(B)$. We do not state the theorem
this way as the statement is essentially about measurable subsets of $A$ and
$B$.

Given $A,B$ subsets of some unimodular group $G$, each with finite positive
measure, and suppose $AB^{-1}$ is measurable. Ruzsa’s distance is defined by

$d(A,B)=\log\frac{\mu_{G}(AB^{-1})}{\mu_{G}(A)^{1/2}\mu_{G}(B)^{1/2}}.$

The following fact is known as triangle inequality of Ruzsa’s distance.

######  Fact 3.10.

Let $C$ be a set of finite positive measure, and suppose $AC^{-1}$, $CB^{-1}$
are measurable sets. Then $d(A,B)\leq d(A,C)+d(C,B)$.

Let $f,g:G\to\mathbb{C}$ be functions. For every $x\in G$, we define the
convolution of $f$ and $g$ to be

$f*g(x)=\int_{G}f(y)g(y^{-1}x)\,\mathrm{d}\,\mu_{G}(y).$

Note that $f*g$ is not commutative, but associative by Fubini’s Theorem.
Suppose $\mu$ and $\nu$ are measures on $G$. Their convolution $\mu*\nu$ is
the unique measure satisfying the property

$\int_{G}f(x)\,\mathrm{d}\,\mu*\nu(x)=\int_{G\times
G}f(xy)\,\mathrm{d}\,\mu(x)\,\mathrm{d}\,\nu(y).$

The convolution exists for all the case we care about. If $\mu(A)>0$, the
uniform measure on $A$ is defined by

$\mu_{A}(X):=\frac{\mu_{G}(A\cap X)}{\mu_{G}(A)}\quad\text{ for measurable
}X\subseteq G.$

The following lemma is an immediate consequence of the definition and Fubini’s
theorem.

######  Lemma 3.11.

Let $\mu_{A}$ be the uniform measure on $A$. Then $\mu_{A}*\mu_{G}=\mu_{G}.$

###### Proof.

Let $X$ be a measurable set in $G$, then

$\displaystyle\mu_{A}*\mu_{G}(X)=\int_{G}\left(\int_{G}\mathbbm{1}_{X}(xy)\,\mathrm{d}\,\mu_{G}(y)\right)\,\mathrm{d}\,\mu_{A}(x)=\mu_{G}(X)\int_{G}\,\mathrm{d}\,\mu_{A}(x)=\mu_{G}(X)$

as desired. ∎

##  4\. Reduction to nearly minimal expansion pairs

To set the stage for the later discussion, we would like to separate the core
part of Theorem 1.1 from the more immediate parts. Throughout $G$ is a
connected unimodular group, $\mu_{G}$ is a Haar measure on $G$, and $A$ and
$B$ are nonempty compact subsets of $G$.

######  Proposition 4.1.

Suppose one of the situation listed in Theorem 1.1 holds, then
$\mu_{G}(AB)=\min(\mu_{G}(A)+\mu_{G}(B),\mu_{G}(G)).$

###### Proof.

We will only consider situation (iii) because (i) and (ii) are immediate and
(iv) can be showed in a same way as (iii). Suppose we are in situation (iii)
of Theorem 1.1. As $\chi$ is a group homomorphism, we have
$AB=\chi^{-1}(I+J)$. Note that by quotient integral formula, we have
$\mu_{G}(A)=\mu_{\mathbb{T}}(I)$, $\mu_{G}(B)=\mu_{\mathbb{T}}(J)$,
$\mu_{G}(AB)=\mu_{\mathbb{T}}(I+J)$. The desired conclusion follows from the
easy that $\mu_{\mathbb{T}}(I+J)=\mu_{\mathbb{T}}(I)+\mu_{\mathbb{T}}(J)$. ∎

The following lemma clarifies the second statement in situation (ii) of
Theorem 1.1.

######  Proposition 4.2.

Suppose $\mu_{G}(A)+\mu_{G}(B)\geq\mu_{G}(G)$. Then $AB=G$.

###### Proof.

Suppose $g$ is an arbitrary element of $G$. It suffices to show that $A^{-1}g$
and $B$ has nonempty intersection. As $G$ is unimodular,
$\mu_{G}(A)=\mu_{G}(A^{-1})$ by Fact 3.3. Hence
$\mu_{G}(A^{-1}g)+\mu_{G}(B)=\mu_{G}(G)$. If $\mu_{G}(A^{-1}g\cap B)>0$, then
we are done. Otherwise, we have have $\mu_{G}(A^{-1}g\cap B)=0$, and so
$\mu_{G}(A^{-1}g\cup B)=\mu_{G}(G)$ by inclusion exclusion principle. As $A$
and $B$ are compact, $A^{-1}g\cup B$ is also compact, and the complement of
$A^{-1}g\cup B$ is open. Since nonempty open sets has positive measure,
$\mu_{G}(A^{-1}g\cup B)=\mu_{G}(G)$ implies $A^{-1}g\cup B=G$. Now, since $G$
is connected, we must have $A^{-1}g\cap B$ must be nonempty. ∎

Now we clarify the situation in (iii) of Theorem 1.1, situation (iii) can be
proved in the same way.

######  Proposition 4.3.

Let $G$ be a connected unimodular group, and $A,B$ be nonempty compact subsets
of $G$. Suppose $\mu_{G}(A)=0$, $0<\mu_{G}(B)<\mu_{G}(G)$, and
$\mu_{G}(AB)=\min(\mu_{G}(A)+\mu_{G}(B),\mu_{G}(G)).$ Then there is a compact
subgroup $H$ of $G$ such that $A\subseteq H$, and $B=HB$.

###### Proof.

Without loss of generality, we can assume that $A$ and $B$ both contain
$\text{id}_{G}$. Let $H$ be the smallest closed subgroup containing $A$. It
suffices to show that $HB=B$. Indeed, $H$ is then a closed subset of $B$ which
implies that $H$ is compact.

From Corollary 3.2, $\text{Stab}_{0}(B)=\\{g\in G\mid\mu_{G}(B\triangle
gB)=0\\}$ is a closed subgroup of $G$. As $\mu_{G}(AB)=\mu_{G}(B)$ and
$\text{id}_{G}$ is in $A$, we must have $A\subseteq\text{Stab}_{0}(B)$. By the
assumption that $H$ is the smallest closed subgroup containing $A$, one must
have $H\leq\text{Stab}_{0}(B)$.

We first consider the special case where $\mu_{G}(U\cap B)>0$ for every $b\in
B$ and open neighborhood $U$ of $b$. As $\mu_{G}$ is both left and right
invariant, this assumption also implies that $\mu_{G}(U\cap gB)>0$ for all
$g\in G$, $b\in gB$, and open neighborhood $U$ of $b$. Suppose $b$ is in
$HB\setminus B$. Since $HB=\bigcup_{h\in H}hB$, we obtain $h\in H$ such that
$b$ is in $hB$. Set $U=G\setminus B$. Then $U$ is an open neighborhood of $b$.
From the earlier discussion, we then have $\mu_{G}(U\cap hB)>0$. As $hB$ is a
subset of $HB$, it implies that $\mu_{G}(HB\setminus B)>0$ which is a
contradiction.

It remains to reduce the general situation to the above special case. Set

$B_{0}=\\{b\in B:\text{ There is an open neighborhood }U_{b}\text{ of }b\text{
with }\mu_{G}(U_{b}\cap B)=0\\}.$

If $b$ is in $B_{0}$, and $b^{\prime}\in U_{b}\cap B$, $U_{b}$ also witnesses
that $b^{\prime}$ is in $B_{0}$. Hence, $B\setminus B_{0}$ is a closed subset
of $B$, which implies that $B\setminus B_{0}$ is compact. Now we show that
$\mu_{G}(B_{0})=0$. Suppose $B^{\prime}$ is a compact subset of $B_{0}$. Then
from the definition of $B_{0}$, we can obtain an open covering $(U_{i})_{i\in
I}$ of $B^{\prime}$ such that $\mu_{G}(U_{i}\cap B)=0$ for all $i\in I$. As
$B^{\prime}$ is compact, we get from $(U_{i})_{i\in I}$ a finite subcovering
of $B^{\prime}$. Hence, $\mu_{G}(B^{\prime})=0$. By inner regularity of Haar
measure, $\mu_{G}(B_{0})=0$. Replacing $B$ with $B\setminus B_{0}$, we reduce
the situation to the above special case. ∎

The remaining parts of Theorem 1.1 consist of classifying the minimally
expanding pairs $(A,B)$ and show that they match the description in situations
(iii) and (iv) of Theorem 1.1. For compact group, our strategy is to reduce
the problem to the known situations of one dimensional tori. Hence, we need
the following special case of Kneser’s classification result, and the sharp
dependence between $\varepsilon$ and $\delta$ is essentially due to Bilu
[Bil98].

######  Theorem 4.4 (Inverse theorem for $\mathbb{T}$).

Let $A,B$ be compact subsets of $\mathbb{T}$. For every $\tau>0$, there is a
constant $c=c(\tau)$ such that if

$\tau^{-1}\mu_{\mathbb{T}}(A)\leq\mu_{\mathbb{T}}(B)\leq\mu_{\mathbb{T}}(A)\leq
c,$

then either
$\mu_{\mathbb{T}}(A+B)\geq\mu_{\mathbb{T}}(A)+2\mu_{\mathbb{T}}(B)$, or there
are compact intervals $I,J$ in $\mathbb{T}$ with
$\mu_{\mathbb{T}}(I)=\mu_{\mathbb{T}}(A+B)-\mu_{\mathbb{T}}(B)$ and
$\mu_{\mathbb{T}}(J)=\mu_{\mathbb{T}}(A+B)-\mu_{\mathbb{T}}(A)$, and a
continuous surjective group homomorphism $\chi:\mathbb{T}\to\mathbb{T}$, such
that $A\subseteq\chi^{-1}(I)$ and $B\subseteq\chi^{-1}(J)$.

For noncompact group, we reduce the problem to to the known situation of
additive group of real numbers. The following result can be seen as the
stability theorem of the Brunn–Minkowski inequality in $\mathbb{R}^{d}$ when
$d=1$.

######  Theorem 4.5 (Inverse theorem for $\mathbb{R}$).

Let $A,B$ be compact subsets in $\mathbb{R}$ with $\mu_{G}(A)\geq\mu_{G}(B)$,
and let $\mu_{\mathbb{R}}$ be the Lebesgue measure in $\mathbb{R}$. Suppose we
have

$\mu_{\mathbb{R}}(A+B)<\mu_{\mathbb{R}}(A)+2\mu_{\mathbb{R}}(B).$

Then there are compact intervals $I,J\subseteq\mathbb{R}$ with
$\mu_{\mathbb{R}}(I)=\mu_{\mathbb{R}}(A+B)-\mu_{\mathbb{R}}(B)$ and
$\mu_{\mathbb{R}}(J)=\mu_{\mathbb{R}}(A+B)-\mu_{\mathbb{R}}(A)$, such that
$A\subseteq I$ and $B\subseteq J$.

##  5\. Reduction to sets with small measure

Throughout this section, $G$ is a connected _compact_ group, $\mu_{G}$ is the
normalized Haar measure on $G$, and $A,B\subseteq G$ are $\sigma$-compact sets
with positive measure. We will show that if $(A,B)$ is nearly minimally
expanding in $G$, then we can $\sigma$-compact $A^{\prime}$ and $B^{\prime}$
each with smaller measure such that the pair $(A^{\prime},B^{\prime})$ is also
nearly minimally expanding. The similar approach used in this section is
introduced by Tao [Tao18] and used to obtain an inverse theorem in the abelian
setting. We first prove the following easy fact, which will be used several
times later in the paper.

######  Lemma 5.1.

Let $t$ be any real numbers such that $\mu_{G}(A)^{2}\leq t\leq\mu_{G}(A)$.
Then there are $x,y\in G$ such that $\mu_{G}(A\cap(xA))=\mu_{G}(A\cap(Ay))=t$.

###### Proof.

Consider the maps:

$\pi_{1}:x\mapsto\mathbbm{1}_{A}*\mathbbm{1}_{A^{-1}}(x)=\mu_{G}(A\cap(xA)),\text{
and
}\pi_{2}:y\mapsto\mathbbm{1}_{A^{-1}}*\mathbbm{1}_{A}(y)=\mu_{G}(A\cap(Ay)).$

By Fact 3.1, both $\pi_{1}$ and $\pi_{2}$ are continuous functions, and equals
to $\mu_{G}(A)$ when $x=y=\text{id}_{G}$. By Fubini’s theorem

$\mathbb{E}\,(\mathbbm{1}_{A}*\mathbbm{1}_{A^{-1}})=\mu_{G}(A)^{2}=\mathbb{E}\,(\mathbbm{1}_{A^{-1}}*\mathbbm{1}_{A}).$

Then the lemma follows from the intermediate value theorem, and the fact that
$G$ is connected. ∎

The following property is sometimes refered to as submodularity in the
literature. Note that this is not related to modular functions in locally
compact groups or the notion of modularity in model theory.

######  Lemma 5.2.

Let $\gamma_{1},\gamma_{2}>0$. Suppose that $(A,B_{1})$ is $\gamma_{1}$-nearly
minimally expanding, $(A,B_{2})$ is $\gamma_{2}$-nearly minimally expanding,

$\mu_{G}(B_{1}\cap B_{2})>0,\quad\text{ and }\ \mu_{G}(A)+\mu_{G}(B_{1}\cup
B_{2})<1.$

Then both $(A,B_{1}\cap B_{2})$ and $(A,B_{1}\cup B_{2})$ are
$(\gamma_{1}+\gamma_{2})$-nearly minimally expanding.

###### Proof.

Observe that for every $x\in G$ we have

$\mathbbm{1}_{AB_{1}}(x)+\mathbbm{1}_{AB_{2}}(x)\geq\mathbbm{1}_{A(B_{1}\cap
B_{2})}(x)+\mathbbm{1}_{A(B_{1}\cup B_{2})}(x),$

which implies

(2) $\mu_{G}(AB_{1})+\mu_{G}(AB_{2})\geq\mu_{G}(A(B_{1}\cap
B_{2}))+\mu_{G}(A(B_{1}\cup B_{2})).$

By the fact that $(A,B_{1})$ is $\gamma_{1}$-nearly minimally expanding and
$(A,B_{2})$ is $\gamma_{2}$-nearly minimally expanding, we obtain

$\displaystyle\mu_{G}(AB_{1})\leq\mu_{G}(A)+\mu_{G}(B_{1})+\gamma_{1},\text{
and }\mu_{G}(AB_{2})\leq\mu_{G}(A)+\mu_{G}(B_{2})+\gamma_{2}.$

Therefore, by equation (2) we have

$\displaystyle\mu_{G}(A(B_{1}\cap B_{2}))+\mu_{G}(A(B_{1}\cup B_{2}))$
$\displaystyle\leq$ $\displaystyle\,2\mu_{G}(A)+\mu_{G}(B_{1}\cap
B_{2})+\mu_{G}(B_{1}\cup B_{2})+\gamma_{1}+\gamma_{2}.$

On the other hand, as $\mu_{G}(B_{1}\cap B_{2})>0$ and
$\mu_{G}(A)+\mu_{G}(B_{1}\cup B_{2})<1-\gamma_{1}-\gamma_{2}$, and using
Kemperman’s inequality, we have

$\mu_{G}(A(B_{1}\cap B_{2}))\geq\mu_{G}(A)+\mu_{G}(B_{1}\cap B_{2}),$

and

$\mu_{G}(A(B_{1}\cup B_{2}))\geq\mu_{G}(A)+\mu_{G}(B_{1}\cup B_{2}).$

This implies

$\mu_{G}(A(B_{1}\cap B_{2}))\leq\mu_{G}(A)+\mu_{G}(B_{1}\cap
B_{2})+\gamma_{1}+\gamma_{2},$

and

$\mu_{G}(A(B_{1}\cup B_{2}))\leq\mu_{G}(A)+\mu_{G}(B_{1}\cup
B_{2})+\gamma_{1}+\gamma_{2}.$

Thus both $(A,B_{1}\cap B_{2})$ and $(A,B_{1}\cup B_{2})$ have
$(\gamma_{1}+\gamma_{2})$-nearly minimal expansion. ∎

Let $m=\min\\{\mu_{g}(A),\mu_{G}(B),1-\mu_{G}(A)-\mu_{G}(B)\\}$. The following
lemma is the main result of this section, it says if $G$ admits a small
expansion pair, one can another find pair of sets with sufficiently small
measures, and still has small expansion.

######  Lemma 5.3.

Let $d\in(0,1/4)$ be a positive real number, and $(A,B)$ has $\gamma$-nearly
minimal expansion. Then there are $\sigma$-compact sets
$A^{\prime},B^{\prime}\subseteq G$ satisfying

  1. (1)

$\mu(A^{\prime})=\mu(B^{\prime})=d$,

  2. (2)

$(A^{\prime},B)$, $(A,B^{\prime})$, and $(A^{\prime},B^{\prime})$ are
$\gamma/m$-nearly minimally expanding.

###### Proof.

Without loss of generality we assume $\mu_{G}(A),\mu_{G}(B)>d$. The case when
at least one of $\mu_{G}(A)$, $\mu_{G}(B)$ is less than $d$ can be proved in a
similar way by replacing intersection by union if necessary. Observe that for
every $g\in G$, both $(gA,B)$ and $(A,Bg)$ still have $\gamma$-nearly minimal
expansions. By Lemma 5.1, for every $t$ with $\mu_{G}(A)^{2}\leq
t\leq\mu_{G}(A)$, there is $g\in G$ such that $\mu_{G}(A\cap gA)=t$. Choose
$g\in G$ such that

(3) $\mu_{G}(A\cup gA)+\mu_{G}(B)<1,$

and $\mu_{G}(A\cap gA)>d$. We are going to apply Lemma 5.2 to decrease the
sizes of $A$ and $B$, and we need an upper bound on the number of times we
apply the lemma.

In the worst case, assume $\mu_{G}(B)=d$, and $\mu_{G}(A)=1-d-m$. By (3), we
can choose $g_{1}$ such that $\mu_{G}(A\cap g_{1}A)=1-d-2m$. Let $A_{1}=A\cap
g_{1}A$, then $(A_{1},B)$ has $2\gamma$-nearly minimal expansion. Similarly,
we choose $g_{2}$ such that $\mu_{G}(A_{1}\cap g_{2}A_{1})=1-d-4m$, and let
$A_{2}=A_{1}\cap g_{2}A_{1}$. By induction, in step $t$, we have

$\mu_{G}(A_{t})=1-d-2^{t}m,$

and $(A_{t},B)$ is $2^{t}\gamma$-nearly minimally expanding. We stop when
$\mu_{G}(A_{t})=d$, and hence $2^{t}=(1-2d)/m$, which means $(A_{t},B)$ is
$(1-2d)\gamma/m$-nearly minimally expanding. ∎

##  6\. Geometry of minimal and nearly minimal expansion pairs I

###  6.1. Preservation of minimal expansion under quotient

In this section,

$1\to H\xrightarrow{i}G\xrightarrow{\pi}G/H\xrightarrow{}1$

is a short exact sequence of _connected_ locally compact groups, and $H$ is
_compact_. As $H$ is compact, it follows that $G/H$ is unimodular with Haar
measure $\mu_{G/H}$. By normalizing $\mu_{H}$, we can assume that
$\mu_{H}(H)=1$. We assume that $A$ and $B$ are $\sigma$-compact sets with
finite positive measures.

Suppose $r$ and $s$ are in $\mathbb{R}$, the sets $A_{(r,s]}$ and $\pi
A_{(r,s]}$ are given by

$A_{(r,s]}:=\\{a\in A:\mu_{H}(A\cap aH)\in(r,s]\\}$

and

$\pi A_{(r,s]}:=\\{aH\in G/H:\mu_{H}(A\cap aH)\in(r,s]\\}.$

In particular, $\pi A_{(r,s]}$ is the image of $A_{(r,s]}$ under the map
$\pi$. We have a number of immediate observations.

######  Lemma 6.1.

Suppose $A$ and $B$ are $\sigma$-compact, $A=A_{(r,s]}$, and
$B=B_{(r^{\prime},s^{\prime}]}$. We have the following:

  1. (1)

For every $aH\in\pi A$ and $bH\in\pi B$ we have

$\mu_{H}\big{(}(A\cap aH)(B\cap bH)\big{)}\geq\min\\{\mu_{H}(A\cap
aH)+\mu_{H}(B\cap bH),1\\}.$

  2. (2)

If $A_{(r,s]}$ and $B_{(r^{\prime},s^{\prime}]}$ are $\sigma$-compact, then

$\mu_{G/H}(\pi A_{(r,s]}\pi
B_{(r^{\prime},s^{\prime}]})\geq\min\\{\mu_{G/H}(\pi A_{(r,s]})+\mu_{G/H}(\pi
B_{(r^{\prime},s^{\prime}]}),\mu_{G/H}(G/H)\\}.$

###### Proof.

Note that both $H$ and $G/H$ are connected. So the desired conclusion is a
consequence of Kemperman inequality for $H$ and $G/H$. ∎

In the general situation, there is no good reason to expect that $A_{(r,s]}$
is compact for all $r$ and $s$. The following two lemmas gives us suitable
replacement.

######  Lemma 6.2.

Suppose $A$ is $\sigma$-compact. Then $A_{(r,s]}$ is $\mu_{G}$-measurable for
all subintervals $(r,s]$ of $(0,1]$.

###### Proof.

Let $\mathbbm{1}_{A}$ be the indicator function of $A$. Then the function

$\displaystyle\mathbbm{1}^{H}_{A}:G/H$ $\displaystyle\to R$ $\displaystyle aH$
$\displaystyle\mapsto\mu_{H}(A\cap aH)$

is well-defined and measurable by Lemma 3.5. As $\pi
A_{(r,s]}=(\mathbbm{1}^{H}_{A})^{-1}(r,s]$ and $A_{(r,s]}=A\cap\pi^{-1}(\pi
A_{(r,s]})$, we get the desired conclusion. ∎

######  Lemma 6.3.

Suppose $A\subseteq G$ is measurable, and $(r_{1},s_{1}],\ldots,(r_{n},s_{n}]$
are subintervals of $(0,1]$. Then there is a $\sigma$-compact
$A^{\prime}\subseteq G$ such that $\mu_{G}(A\setminus A^{\prime})=0$, and for
each $i\in\\{1,\ldots,n\\}$, the set $A^{\prime}_{(r_{i},s_{i}]}$ is
$\sigma$-compact.

###### Proof.

As the Haar measure is inner regular, we can find $\sigma$-compact
$A^{\prime}$ such that $\mu_{G}(A\setminus A^{\prime})=0$. Consider first the
special case where $(r_{i},s_{i}]$ are pairwise disjoint for
$i\in\\{1,\ldots,n\\}$. Then for $i\in\\{1,\ldots,n\\}$, we can modify
$A^{\prime}$ by a subset of measure $0$ to arrange that
$A^{\prime}_{(r_{i},s_{i}]}$ is $\sigma$-compact. Finally, if
$r^{\prime}_{1}\leq\ldots\leq r^{\prime}_{k}$ are all the endpoints of the
intervals under considerations, then we can reduce to the earlier special case
by ensuring that $A^{\prime}_{(r_{i},r_{i+1}]}$ is $\sigma$-compact for
$i\in\\{1,\ldots,n-1\\}$. ∎

The function $\mu_{G/H}(\pi A_{(r,s]})$ might not be differentiable as in the
overview. We use the Riemann–Stieltjes integral. The next fact records some
basic properties of the integral.

######  Fact 6.4.

Given $f(x),g(x)$ real valued function, and $I=[a,b]\subseteq\mathbb{R}$ a
closed interval.

  1. (i)

(Integrable) If $f(x)$ is continuous on $I$, and $g(x)$ has bounded variation
on $I$, then $f(x)\,\mathrm{d}\,g(x)$ is Riemann–Stieltjes integrable on $I$.

  2. (ii)

(Integral by parts) If $f(x)\,\mathrm{d}\,g(x)$ is Riemann–Stieltjes
integrable on $I$, then $g(x)\,\mathrm{d}\,f(x)$ is also Riemann–Stieltjes
integrable on $I$, and

$\int_{a}^{b}f(x)\,\mathrm{d}\,g(x)=f(b)g(b)-f(a)g(a)-\int_{a}^{b}g(x)\,\mathrm{d}\,f(x).$

The next lemma uses “spillover” estimate, which gives us a lower bound
estimate on $\mu_{G}(AB)$ when the projection of $A$ and $B$ are not too
large.

######  Lemma 6.5.

Suppose $A$ and $B$ are $\sigma$-compact, and $\mu_{G/H}(\pi A)+\mu_{G/H}(\pi
B)<1$. Then

$\displaystyle\mu_{G}(AB)$ $\displaystyle\geq\mu_{G/H}(\pi
A_{(1/2,1]})+\mu_{G/H}(\pi B_{(1/2,1]})+\frac{1}{4}\mu_{G/H}(\pi
A_{(0,1/2]})+\frac{1}{4}\mu_{G/H}(\pi A_{(0,1/2]})$
$\displaystyle\quad+\mu_{G}(A_{(0,1/2]})+\mu_{G}(B_{(0,1/2]}).$

###### Proof.

Let $P_{n}$ be a partition of $(0,1/2]$ into $n$ intervals, each of length
$d=1/2n$. Using Lemma 6.3, we can arrange that ${A}_{(kd,1]}$ and
${B}_{(kd,1]}$ are $\sigma$-compact for each $k\in\\{0,\ldots,n\\}$. Then by
Fact 6.4 we get

$\displaystyle\mu_{G}({A}_{(0,1]}{B}_{(0,1]})$
$\displaystyle=\mu_{G}({A}_{(0,1]}{B}_{(1/2,1]})-\int_{0}^{\tfrac{1}{2}}\,\mathrm{d}\,\mu_{G}({A}_{(0,1]}{B}_{(x,1]})$
$\displaystyle=\mu_{G}({A}_{(1/2,1]}{B}_{(1/2,1]})-\int_{0}^{\tfrac{1}{2}}\,\mathrm{d}\,\mu_{G}({A}_{(x,1]}{B}_{(1/2,1]})$
$\displaystyle\quad-\int_{0}^{\tfrac{1}{2}}\,\mathrm{d}\,\mu_{G}({A}_{(0,1]}{B}_{(x,1]}).$

From Lemma 6.1, we get
$\mu_{G}({A}_{(1/2,1]}{B}_{(1/2,1]})\geq\mu_{G/H}(\pi{A}_{(1/2,1]}\pi{B}_{(1/2,1]})$.
From the same lemma, for each $k\in\\{0,1,\ldots,n\\}$ we have

$\displaystyle\,\mu_{G}({A}_{(kd,1]}{B}_{(1/2,1]}\setminus{A}_{((k+1)d,1]}{B}_{(1/2,1]})$
$\displaystyle\geq$ $\displaystyle\,(kd+1/2)\mu_{G/H}(\pi A_{(kd,1]}\pi
B_{(1/2,1]}\setminus\pi A_{((k+1)d,1]}\pi B_{(1/2,1]}),$

and

$\mu_{G}({A}_{(0,1]}{B}_{(kd,1]}\setminus{A}_{(0,1]}{B}_{((k+1)d,1]})\geq
kd\mu_{G/H}(\pi{A}_{(0,1]}\pi{B}_{(kd,1]}\setminus\pi{A}_{(0,1]}\pi{B}_{((k+1)d,1]}).$

Therefore,

$\displaystyle\mu_{G}({A}_{(0,1]}{B}_{(0,1]})$
$\displaystyle\geq\mu_{G/H}(\pi{A}_{(1/2,1]}\pi{B}_{(1/2,1]})-\int_{0}^{\tfrac{1}{2}}\Big{(}x+\frac{1}{2}\Big{)}\,\mathrm{d}\,\mu_{G/H}(\pi{A}_{(x,1]}\pi{B}_{(1/2,1]})$
$\displaystyle\quad-\int_{0}^{\tfrac{1}{2}}x\,\mathrm{d}\,\mu_{G/H}(\pi{A}_{(0,1]}\pi{B}_{(x,1]})$
$\displaystyle=\bigg{(}\mu_{G/H}(\pi{A}_{(1/2,1]}\pi{B}_{(1/2,1]})-\frac{1}{2}\mu_{G/H}(\pi{A}_{(0,1]}\pi{B}_{(1/2,1]})$
$\displaystyle\quad-\int_{0}^{\tfrac{1}{2}}\Big{(}x+\frac{1}{2}\Big{)}\,\mathrm{d}\,\mu_{G/H}(\pi{A}_{(x,1]}\pi{B}_{(1/2,1]})\bigg{)}$
$\displaystyle\quad+\bigg{(}\frac{1}{2}\mu_{G/H}(\pi{A}_{(0,1]}\pi{B}_{(1/2,1]})-0\mu_{G/H}(\pi{A}_{(0,1]}\pi{B}_{(0,1]})$
$\displaystyle\quad-\int_{0}^{\tfrac{1}{2}}x\,\mathrm{d}\,\mu_{G/H}(\pi{A}_{(0,1]}\pi{B}_{(x,1]})\bigg{)}.$

Using integral by parts, and the Kemperman inequality on $G/H$, we obtain

$\displaystyle\mu_{G}({A}_{(0,1]}{B}_{(0,1]})$
$\displaystyle\geq\bigg{(}\mu_{G/H}(\pi{A}_{(1/2,1]})+\mu_{G/H}(\pi{B}_{(1/2,1]})-\frac{1}{2}\mu_{G/H}(\pi{A}_{(0,1]})$
$\displaystyle\quad-\frac{1}{2}\mu_{G/H}(\pi{B}_{(1/2,1]})-\int_{0}^{\tfrac{1}{2}}\Big{(}x+\frac{1}{2}\Big{)}\,\mathrm{d}\,\mu_{G/H}(\pi{A}_{(x,1]})\bigg{)}$
$\displaystyle\quad+\bigg{(}\frac{1}{2}\mu_{G/H}(\pi{A}_{(0,1]})+\frac{1}{2}\mu_{G/H}(\pi{B}_{(1/2,1]})-\int_{0}^{\tfrac{1}{2}}x\,\mathrm{d}\,\mu_{G/H}(\pi{B}_{(x,1]})\bigg{)}.$

By the definition, we have

$\displaystyle\mu_{G}(A_{(0,kd]}\setminus A_{(0,(k-1)d]})\leq kd\mu_{G/H}(\pi
A_{(0,kd]}\setminus\pi A_{(0,(k-1)d]}).$

Observe that
$\,\mathrm{d}\,\mu_{G/H}(\pi{A}_{(x,1]})=-\,\mathrm{d}\,\mu_{G/H}(\pi{A}_{(0,x]}).$
Hence,

$\displaystyle-\int_{0}^{\tfrac{1}{2}}\Big{(}x+\frac{1}{2}\Big{)}\,\mathrm{d}\,\mu_{G/H}(\pi{A}_{(x,1]})$
$\displaystyle=\int_{0}^{\tfrac{1}{2}}x\,\mathrm{d}\,\mu_{G/H}(\pi{A}_{(0,x]})+\int_{0}^{\tfrac{1}{2}}\frac{1}{2}\,\mathrm{d}\,\mu_{G/H}(\pi{A}_{(0,x]})$
$\displaystyle\geq\int_{0}^{\tfrac{1}{2}}\,\mathrm{d}\,\mu_{G}({A}_{(0,x]})+\int_{0}^{\tfrac{1}{2}}\frac{1}{2}\,\mathrm{d}\,\mu_{G/H}(\pi{A}_{(0,x]})$
$\displaystyle\geq\mu_{G}({A}_{(0,1/2]})+\frac{1}{2}\mu_{G/H}(\pi{A}_{(0,1/2]}).$

A similar estimation yields

$\displaystyle-\int_{0}^{\tfrac{1}{2}}x\,\mathrm{d}\,\mu_{G/H}(\pi{B}_{(x,1]})$
$\displaystyle\geq\mu_{G}({B}_{(0,1/2]}).$

Putting everything together, we get

$\displaystyle\mu_{G}({A}_{(0,1]}{B}_{(0,1]})$
$\displaystyle\geq\mu_{G/H}(\pi{A}_{(1/2,1]})+\mu_{G/H}(\pi{B}_{(1/2,1]})+\mu_{G}({A}_{(0,1/2]})+\mu_{G}({B}_{(0,1/2]})$
$\displaystyle+\frac{1}{2}\mu_{G/H}(\pi{A}_{(0,1/2]}).$

Finally, switching the order of estimation between $\mu_{G}({A}_{(0,1]})$ and
$\mu_{G}({B}_{(0,1]})$, and averaging, we get the desired conclusion. ∎

We remark that one can also make the above proofs “continuous”. We present a
discrete version here since we believe it reflects our idea in a clearer way.

The next result in the main result in this subsection. It says if the
projections of $A$ and $B$ are not too large, the small expansion properties
will be kept in the quotient group.

######  Theorem 6.6 (Transfer theorem).

Suppose $A$ and $B$ are $\mu_{G}$-measurable with ${\mu}_{G/H}(\pi
A)+{\mu}_{G/H}(\pi B)<\mu_{G}(G)$ and

${\mu}_{G}(AB)<\mu_{G}(A)+\mu_{G}(B)+\delta.$

Then there are $\mu_{G/H}$-measurable sets $A^{\prime},B^{\prime}\subseteq
G/H$ such that

${\mu}_{G/H}(A^{\prime}B^{\prime})<\mu_{G/H}(A^{\prime})+\mu_{G/H}(B^{\prime})+9\delta$

and
$\max\\{\mu_{G}(A\triangle\pi^{-1}A^{\prime}),\mu_{G}(B\triangle\pi^{-1}B^{\prime})\\}<5\delta.$

###### Proof.

From the assumptions of the Lemma 6.5, we get that

$\displaystyle\mu_{G}(A)+\mu_{G}(B)+\delta$ $\displaystyle>\mu_{G/H}(\pi
A_{(1/2,1]})+\mu_{G/H}(\pi
B_{(1/2,1]})+\mu_{G}(A_{(0,1/2]})+\mu_{G}(B_{(0,1/2]})$
$\displaystyle\quad+\frac{1}{4}\mu_{G/H}(\pi
A_{(0,1/2]})+\frac{1}{4}\mu_{G/H}(\pi B_{(0,1/2]}).$

Note that $\mu_{G}(A)=\mu_{G}(A_{(0,1/2]})+\mu_{G}(A_{(1/2,1]})$, and by
applying the Fubini Theorem, we have $\mu_{G}(A_{(1/2,1]})\leq\mu_{G/H}(\pi
A_{(1/2,1]})$. A similar statement holds by replacing $A$ by $B$. Therefore,
we must have

$\mu_{G/H}(\pi A_{(0,1/2]})<4\delta,\quad\mu_{G/H}(\pi
A_{(1/2,1]})<\mu_{G}(A_{(1/2,1]})+\delta.$

Similar estimates hold by replacing $A$ by $B$. Choose $A^{\prime}$ to be $\pi
A_{(1/2,1]}$ and $B^{\prime}$ to be $\pi B_{(1/2,1]}$. As
$\mu_{G}(A_{(0,1/2]})\leq\mu_{G/H}(\pi A_{(0,1/2]})\leq 4\delta$ and
$\mu_{G/H}(\pi A_{(1/2,1]})<\mu_{G}(A_{(1/2,1]})+\delta$, and similar
inequalities hold replacing $A$ by $B$, it is easy to see that

$\max\\{\mu_{G}(A\triangle\pi^{-1}A^{\prime}),\mu_{G}(B\triangle\pi^{-1}B^{\prime})\\}<5\delta.$

On the other hand, note that $A_{(1/2,1]}B_{(1/2,1]}$ is the inverse image of
$\pi A_{(1/2,1]}\pi B_{(1/2,1]}$. Therefore,
${\mu}_{G/H}(A^{\prime}B^{\prime})\leq{\mu}_{G}(AB)$, which is at most
$\mu_{G}(A)+\mu_{G}(B)+\delta$ by the assumption that $(A,B)$ is
$\delta$-nearly minimally expanding. The desired inequality holds following
the fact that

$\mu_{G}(A)\leq\mu_{G/H}(\pi A_{(1/2,1]})+\mu_{G/H}(\pi A_{(0,1/2]}),$

and the second term of the right hand side is at most $4\delta$. ∎

###  6.2. Structure control on the nearly minimal expansion sets

The following useful lemma is a corollary of Theorem 6.6, which will be used
at various points in the later proofs. It tells us the character given in the
parallel Bohr sets is essentially unique.

######  Lemma 6.7 (Stability of nearly parallel Bohr sets).

Let $G$ be a connected unimodular group, $\kappa,\delta\geq 0$, and let $A$,
$B$ be $\sigma$-compact subsets of $G$ satisfy

  1. (1)

$\mu_{G}(A),\mu_{G}(B)>(2\kappa+30)\delta$.

  2. (2)

$\mu_{G}(AB)\leq\mu_{G}(A)+\mu_{G}(B)+\delta\leq 1-(2\sqrt{\kappa}+10)\delta$.

  3. (3)

There is a continuous surjective group homomorphism $\chi:G\to\mathbb{T}$
($\chi:G\to\mathbb{R}$ with compact kernel if $G$ is noncompact), a compact
interval $J$ in $\mathbb{T}$ with $\mu_{\mathbb{T}}(J)=\mu_{G}(B)$ such that
$\mu_{G}(B\triangle\chi^{-1}(J))=\kappa\delta$.

Then there is a compact interval $I$ in $\mathbb{T}$ such that
$\mu_{\mathbb{T}}(I)=\mu_{G}(A)$, and we have
$\mu_{G}(A\triangle\chi^{-1}(I))\leq(14+\kappa)\delta$.

###### Proof.

We only consider the case when $G$ is compact here. Then case when $G$ is
noncompact is similar. We first assume
$\mu_{\mathbb{T}}(\chi(A))+\mu_{\mathbb{T}}(\chi(B))<1-10\delta$. Let
$K=\ker(\chi)$. Note that $\chi$ is an open map, and hence by Fact 3.7 we have
$G/K\cong\mathbb{T}$. By Theorem 6.6, there are sets $A^{\prime}$ and
$B^{\prime}$ in $\mathbb{T}$ such that $(A^{\prime},B^{\prime})$ is
$9\delta$-nearly minimally expanding, and

$\mu_{G}(A\triangle\chi^{-1}(A^{\prime}))=5\delta,\quad\mu_{G}(B\triangle\chi^{-1}(B^{\prime}))=5\delta.$

Then $\mu_{\mathbb{T}}(B^{\prime}\triangle J)\leq(\kappa+5)\delta$. By Theorem
4.4, there is a compact interval $I\subseteq\mathbb{T}$ with
$\mu_{\mathbb{T}}(I)=\mu_{G}(A)$, such that
$\mu_{G}(A\triangle\chi^{-1}(I))\leq(14+\kappa)\delta$.

Now suppose $\mu_{\mathbb{T}}(\chi(A))+\mu_{\mathbb{T}}(\chi(B))\geq
1-10\delta$. By the assumption that there is $J\subseteq\mathbb{T}$ with
$\mu_{\mathbb{T}}(J)=\mu_{G}(B)$ and
$\mu_{G}(B\triangle\chi^{-1}(J))=\kappa\delta$, except for a set of measure at
most $\sqrt{\kappa}\delta$, for every $g\in\chi(B)$, we have $\mu_{K}(B\cap
gK)\geq 1-\sqrt{\kappa}\delta$. Therefore, except for a set of measure at most
$(\sqrt{\kappa}+10)\delta$, all fibers of the form $AB\cap gK$ has measure at
least $1-\sqrt{\kappa}\delta$. This gives us

$\mu_{G}(AB)>1-(2\sqrt{\kappa}+10)\delta,$

this contradicts the assumption that $\mu_{G}(AB)\leq
1-(2\sqrt{\kappa}+10)\delta$. ∎

The following theorem shows that, once we have a certain group homomorphism to
tori, we will get a good structural control on the (nearly) minimal expansion
sets.

######  Proposition 6.8 (Structural control).

Let $G$ be a compact group, and $\chi:G\to\mathbb{T}$ is a continuous
surjective group homomorphism. Suppose $A,B$ are $\sigma$-compact subsets of
$G$ such that $\mu_{G}(AB)<\mu_{G}(A)+\mu_{G}(B)+\delta$, where
$\delta<\min\\{\mu_{G}(A),\mu_{G}(B)\\}$. Suppose
$\mu_{\mathbb{T}}(\chi(A))+\mu_{\mathbb{T}}(\chi(B))<1/5$. Then there are
compact intervals $I_{A},I_{B}$ in $\mathbb{T}$ with
$\mu_{\mathbb{T}}(I_{A})=\mu_{G}(A)$ and $\mu_{\mathbb{T}}(I_{B})=\mu_{G}(B)$,
such that

$\mu_{G}(A\triangle\chi^{-1}(I_{A}))<15\delta,\text{ and
}\mu_{G}(B\triangle\chi^{-1}(I_{B}))<15\delta.$

When $G$ is a unimodular noncompact group, the same conclusion still holds
with $\mathbb{T}$ replaced by $\mathbb{R}$.

###### Proof.

We first consider the case when $G$ is noncompact. Suppose $(A,B)$ is
$\delta$-nearly minimally expanding. By Theorem 6.6, there are
$A^{\prime},B^{\prime}\subseteq\mathbb{R}$, such that

(4)
$\mu_{\mathbb{R}}(A\triangle\chi^{-1}A^{\prime})<5\delta,\quad\mu_{\mathbb{R}}(B\triangle\chi^{-1}B^{\prime})<5\delta,$

and $(A^{\prime},B^{\prime})$ is $10\delta$-nearly minimally expanding. By
Theorem 4.5, there are compact intervals $I_{A},I_{B}$ in $\mathbb{R}$ such
that $\mu_{\mathbb{R}}(I_{A})=\mu_{\mathbb{R}}(A^{\prime})$,
$\mu_{\mathbb{R}}(I_{B})=\mu_{\mathbb{R}}(B^{\prime})$, and

$\mu_{\mathbb{R}}(A^{\prime}\triangle
I_{A})<10\delta,\quad\mu_{\mathbb{R}}(B^{\prime}\triangle I_{B})<10\delta.$

Thus the conclusion follows from (4).

Now we consider the case when $G$ is compact. Let $\mu_{\mathbb{T}}$ be the
normalized Haar measure on $\mathbb{T}$. The case when $G$ is compact follows
from the same argument as used in the noncompact case, with assumption that
when $\mu_{\mathbb{T}}(\chi(A))+\mu_{\mathbb{T}}(\chi(B))<1/5$. The inverse
theorem in $\mathbb{T}$ follows from Theorem 4.4, Lemma 5.3, and Lemma 6.7. ∎

In light of Proposition 6.8, in the rest of the paper, we will be focusing on
finding the desired homomorphism mapping $G$ to tori.

###  6.3. Fine control of Lie model

In this subsection, $G$ is a connected _compact_ group, and $A,B$ are
$\sigma$-compact subsets of $G$ such that they are nearly minimally expanding.
The next result will help us to reduce the problem to a connected Lie group.
One of the main tools is the following theorem by Gleason [Gle52] and Yamabe
[Yam53], which resolves Hilbert’s 5-th problem. The last statement is not part
of the usual statement, but can be easily arranged by choosing the connected
component.

######  Theorem 6.9 (Gleason–Yamabe Theorem).

For any connected locally compact group $G$ and any neighborhood $U$ of the
identity in $G$, there is a compact normal subgroup $H\subseteq U$ of $G$ such
that $G/H$ is a connected Lie group. Moreover, we can arrange that $H$ is
connected.

In order to apply Theorem 6.6, we need to ensure that the projections of $A$
and $B$ are still small. Moreover, in Theorem 1.2, given $\varepsilon>0$, the
error term $\delta$ we obtained does not depend on the ambient group $G$.
Therefore we also need a dimension control on the Lie groups given from the
Gleanson–Yamabe Theorem.

Recall that a set $S$ is $K$-approximate if $S^{2}\subseteq XS$ for some
finite set $X$ of cardinality $K$. The next theorem by Tao [Tao08] is able to
extract an approximate group from the fact that $(A,B)$ is nearly minimally
expanding.

######  Theorem 6.10.

Let $A,B$ be subsets of $G$ with $\mu_{G}(A),\mu_{G}(B)>0$. Suppose
$\mu_{G}(AB)<K\mu_{G}^{1/2}(A)\mu_{G}^{1/2}(B)$, then there is an open set $S$
with $\mu_{G}(S)>\mu_{G}(A)/2K$, so that

$\mu_{G}(AS^{3}A^{-1})\leq K^{O(1)}\mu_{G}(A),$

and there is a finite set $P$ or cardinality $O(K^{O(1)})$ such that
$A\subseteq PS^{3}$ and $B\subseteq S^{3}P$.

The study of continuous approximate groups by Carolino [Car15] is able to find
a Lie model, and to control the dimension of the Lie model. This can be seen
as a finer version of the Gleason–Yamabe theorem.

######  Theorem 6.11.

Let $G$ be a connected compact group, and $A$ be an open $K$-approximate
group. Then there is a subset $A^{\prime}$ of $A^{4}$, and a connected compact
normal subgroup $H$ in $A^{\prime}$, such that $G/H$ is a Lie group of
dimension $O_{K}(1)$, and $A$ can be covered by $O_{K}(1)$ many left
translations of $A^{\prime}$.

The next proposition is the main result in this subsection. Using this, we can
pass the problem to connected Lie groups with bounded dimensions.

######  Proposition 6.12 (Dimensional control on Lie model).

Given $\delta>0$. There is $t>0$ such that the followings hold. Suppose $G$ is
a connected compact group, $A,B$ are $\sigma$-compact subsets of $G$ such that
$\max\\{\mu_{G}(A),\mu_{G}(B)\\}<t$, and $(A,B)$ is $\delta$-nearly minimally
expanding. Then there is a positive integer $d$, and a connected closed normal
subgroup $H$ of $G$ such that $G/H$ is a Lie group of dimension at most $d$,
and if $\pi:G\to G/H$ is a quotient map, then $\mu_{G/H}(\pi A)+\mu_{G/H}(\pi
B)<1/5$.

###### Proof.

By Theorem 6.10, there is a constant $K$, and an open $K$-approximate group
$S^{3}$, such that $A$ can be covered by $O(K^{O(1)})$ right translation of
$S$, and $B$ can be covered by $O(K^{O(1)})$ left translation of $S$. By
Theorem 6.11, there is a closed connected normal subgroup $H$ in $S^{12}$,
such that $G/H$ is a Lie group of bounded dimension. Let $\pi$ be the quotient
map. It remains to show that $\mu_{G/H}(\pi A)+\mu_{G/H}(\pi B)$ is small.
Since $\ker(\pi)\subseteq S^{12}$, we have

$\mu_{G/H}(\pi(S^{3}))\leq\mu_{G}(S^{3}\ker(\pi))\leq\mu_{G}(S^{15}).$

By Theorem 6.10, we have in particular $\mu_{G}(S^{3})\leq
K^{O(1)}\mu_{G}(A)$. Therefore, when $A$ and $B$ are sufficiently small,
$\mu_{G/H}(\pi(S^{3}))$ is small. Since $\pi(A)$ can be covered by
$O(K^{O(1)})$ right translation of $\pi(S^{3})$, and $\pi(B)$ can be covered
by $O(K^{O(1)})$ left translation of $\pi(S^{3})$, making
$\mu_{G/H}(\pi(S^{3}))$ sufficiently small would imply $\mu_{G/H}(\pi
A)+\mu_{G/H}(\pi B)<1/5$, as desired. ∎

##  7\. Pseudometric and group homomorphisms into tori

###  7.1. Definitions and linear pseudometric

In this subsection, $G$ is a _connected_ unimodular locally compact group. A
pseudometric on a set $X$ is a function $d:X\times X\to\mathbb{R}$ satisfying
the following three properties:

  1. (1)

(Reflexive) $d(a,a)=0$ for all $a\in X$,

  2. (2)

(Symmetry) $d(a,b)=d(b,a)$ for all $a,b\in X$,

  3. (3)

(Triangle inequality) $d(a,c)\leq d(a,b)+d(b,c)\in X$.

Hence, a pseudometric on $X$ is a metric if for all $a,b\in X$, we have
$d(a,b)=0$ implies $a=b$.

Suppose $d$ is a pseudometric on $G$. Set $\|g\|_{d}=d(1_{G},g)$. We say that
$d$ is left-invariant if for all $g,g_{1},g_{2}\in G$, we have
$d(gg_{1},gg_{2})=d(g_{1},g_{2})$. If $\widetilde{d}$ is left-invariant metric
on a group $H$ and $\pi:G\to H$ is a group homomorphism, then for every
$g_{1},g_{2}$ in $G$, one can naturally define a pseudometric
$d(g_{1},g_{2})=\widetilde{d}(\pi(g_{1}),\pi(g_{2}))$. Another source of
pseudometric arise naturally from measurable sets in a group:

######  Proposition 7.1.

Suppose $G$ is a connected locally compact group equipped with a left Haar
measure $\mu_{G}$, and $A$ is a measurable subset of $G$. For $g_{1}$ and
$g_{2}$ in $G$, define

$d_{A}(g_{1},g_{2})=\mu_{G}(A)-\mu_{G}(g_{1}A\cap g_{2}A)$

Then $d_{A}$ is a continuous left invariant pseudometric on $G$.

###### Proof.

We only verify here the triangle inequality as all the other parts are
straightforward. Let $g_{1}$, $g_{2}$, and $g_{3}$ be in $G$, we need to show
that

(5) $\mu_{G}(A)-\mu_{G}(g_{1}A\cap g_{3}A)\leq\mu_{G}(A)-\mu_{G}(g_{1}A\cap
g_{2}A)+\mu_{G}(A)-\mu_{G}(g_{2}A\cap g_{3}A).$

As $\mu_{G}(A)=\mu_{G}(g_{2}A)$, we have $\mu_{G}(A)-\mu_{G}(g_{1}A\cap
g_{2}A)=\mu_{G}(g_{2}A\setminus g_{1}A)$, and $\mu_{G}(A)-\mu_{G}(g_{2}A\cap
g_{3}A)=\mu_{G}(g_{2}A\setminus g_{3}A)$. Hence, (5) is equivalent to

$\mu_{G}(g_{2}A)-\mu_{G}(g_{2}A\setminus g_{1}A)-\mu_{G}(g_{2}A\setminus
g_{3}A)\leq\mu(g_{1}A\cap g_{3}A).$

Note that the left hand side is at most $\mu_{G}(g_{1}A\cap g_{2}A\cap
g_{3}A)$, which is less than the right hand side. Hence, we get the desired
conclusion. ∎

######  Lemma 7.2.

Let $d$ be a continuous left-invariant pseudometric on $G$. The set $\\{g\in
G:\|g\|_{d}=0\\}$ is the underlying set of a closed subgroup of $G$.

###### Proof.

Suppose $g_{1}$ and $g_{2}$ are elements in $G$ such that
$\|g_{1}\|_{d}=\|g_{2}\|_{d}=0$. Then

$d(\mathrm{id}_{G},g_{1}g_{2})\leq
d(\mathrm{id}_{G},g_{1})+d(g_{1},g_{1}g_{2})=d(\mathrm{id}_{G},g_{1})+d(\mathrm{id}_{G},g_{2})=0.$

Now, suppose $(g_{n})$ is a sequence of elements in $G$ converging to $g$ with
$\|g_{n}\|_{d}=0$. Then $\|g\|_{d}=0$ by continuity, we get the desired
conclusions. ∎

In many situations, a left-invariant pseudometric allows us to construct
surjective continuous group homomorphism into metric groups. The following
lemma tells us precisely when this happens. We omit the proof as the result is
motivationally relevant but will not be used later on.

######  Lemma 7.3.

Let $d$ be a continuous left-invariant pseudometric on $G$. The following are
equivalent

  1. (1)

The set $\\{g\in G:\|g\|_{d}=0\\}$ is the underlying set of a closed normal
subgroup of $G$.

  2. (2)

There is a continuous surjective group homomorphism $\pi:G\to H$, and
$\widetilde{d}$ is a left-invariant metric on $H$. Then

$d(g_{1},g_{2})=\widetilde{d}(\pi g_{1},\pi g_{2}).$

Moreover, when $(2)$ happens then $\\{g\in G:\|g\|_{d}=0\\}=\ker\pi$, hence
$H$ and $\widetilde{d}$ if exist are uniquely determined up to isomorphism.

The group $\mathbb{R}$ and $\mathbb{T}=\mathbb{R}/\mathbb{Z}$ are naturally
equipped with the metrics $d_{\mathbb{R}}$ and $d_{\mathbb{T}}$ induced by the
Euclidean norms, and these metrics interacts in a very special way with the
additive structures. Hence one would expect that if there is a group
homomorphism from $G$ to either $\mathbb{R}$ or $\mathbb{T}$, then $G$ can be
equipped with a pseudometric which interacts nontrivially with addition.

Let $d$ be a left-invariant pseudometric on $G$. The radius $\rho$ of $d$ is
defined to be $\sup\\{\|g\|_{d}:g\in G\\}$; this is also
$\sup\\{d(g_{1},g_{2}):g_{1},g_{2}\in G\\}$ by the left invariance. We say
that $d$ is locally linear if it satisfies the following properties:

  1. (1)

$d$ is continuous and left and right invariant;

  2. (2)

for all $g_{1}$, $g_{2}$, and $g_{3}$ with
$d(g_{1},g_{2})+d(g_{2},g_{3})<\rho$, we have either

(6) $d(g_{1},g_{3})=d(g_{1},g_{2})+d(g_{2},g_{3}),\text{ or
}d(g_{1},g_{3})=|d(g_{1},g_{2})-d(g_{2},g_{3})|.$

A locally linear pseudometric $d$ is monotone if for all $g\in G$ such that
$\|g\|_{d}<\rho/2$, we have $\|g^{2}\|_{d}=2\|g\|_{d}$. The following result
is one of the reason we sometimes need to assume that the group under
consideration is a Lie group.

######  Lemma 7.4.

Suppose $G$ is a connected Lie group, and $s:G\to G,g\mapsto g^{2}$ is the
squaring map. Then there is no open $U\subseteq G$ and a proper closed
subgroup $H$ of $G$ such that $s(U)\subseteq H$.

###### Proof.

We first consider the case where $G$ is a connected component of a linear
algebraic subgroup of $\mathrm{GL}_{n}(\mathbb{R})$. Let $J_{s}$ be the
Jacobian of the function $s$. Then the set $\\{g\in G:\det J_{s}(g)=0\\}$ has
the form $G\cap Z$ where $Z$ is a solution set of a system of polynomial
equations. It is not possible to have $G\cap Z=G$, as $s$ is a local
diffeomorphism at $\mathrm{id}_{G}$. Hence, $G\cap Z$ must be of strictly
lower dimension than $G$. By the inverse function theorem, $s|_{G\setminus Z}$
is open. Hence $s(U)$ is not contained in a subgroup of $G$ with smaller
dimension.

We also note a stronger conclusion for abelian Lie group: If $V$ is an open
subset of a not necessarily connected abelian Lie group $A$, then the image of
$A$ under $a\mapsto a^{2}$ is not contained in a closed subset of $A$ with
smaller dimension. Indeed, $A$ is isomorphic as a topological group to
$D\times\mathbb{T}^{m}\times\mathbb{R}^{m}$, with $D$ a discrete group. If
$U\subseteq D\times\mathbb{T}^{m}\times\mathbb{R}^{m}$, then it is easy to see
that $\\{a^{2}:a\in V\\}$ contains a subset of
$D\times\mathbb{T}^{m}\times\mathbb{R}^{m}$, and is therefore not a subset of
a closed subset of $A$ with smaller dimension.

Finally, we consider the case where $G$ is a connected Lie group. Suppose to
the contrary that $s(U)\subseteq H$ with $H$ a proper closed subgroup of $G$.
Let $Z(G)$ be the center of $G$, $G^{\prime}=G/Z(G)$, $\pi:G\to G^{\prime}$ be
the quotient map, $s^{\prime}:G^{\prime}\to
G^{\prime},g^{\prime}\mapsto(g^{\prime})^{2}$, and $U^{\prime}=\pi(U)$. Then
$U^{\prime}$ is an open subset of $G^{\prime}$, which is isomorphic as a
topological group to a connected component of an algebraic group by Fact 3.8.
By the earlier case, $s^{\prime}(U^{\prime})$ is not contained in any proper
closed subgroup of $G^{\prime}$, so we must have $\pi(H)=G^{\prime}$. In
particular, this implies $\dim(H\cap Z(G))<\dim Z(G)$, and $HZ(G)=G$. Choose
$h\in H$ such that $hZ(G)\cap U$ is nonempty. Then

$s(hZ(G)\cap U)=\\{h^{2}a^{2}:a\in Z(G)\cap h^{-1}U\\}.$

As $s(hZ(G)\cap U)\subseteq H$, we must have $\\{a^{2}:a\in Z(G)\cap
h^{-1}U\\}$ is a subset of $H\cap Z(G)$. Using the case for abelian Lie
groups, this is a contradiction, because $H\cap Z(G)$ is a closed subset of
$Z(G)$ with smaller dimension. ∎

The next result confirms our earlier intuition: locally linear pseudometric in
$G$ will induced a homomorphism mapping to either $\mathbb{T}$ or
$\mathbb{R}$.

######  Proposition 7.5.

Suppose $d$ is a nontrivial and locally linear pseudometric with radius
$\rho$, and either $d$ is monotone or $G$ is a Lie group. Then $\ker d$ is a
normal subgroup of $G$, $G/\ker d$ is isomorphic to $\mathbb{T}$ if $G$ is
compact, and $G/\ker d$ is isomorphic to $\mathbb{R}$ if $G$ is noncompact.

###### Proof.

We first prove that $\ker d$ is a normal subgroup of $G$ under the assumption
that $d$ is monotone. Suppose $\|g\|_{d}=0$ and $h\in G$ satisfy
$\|h\|_{d}<\rho/4$. We have

$\displaystyle d(h,hgh^{-1})$ $\displaystyle=d(\mathrm{id}_{G},gh^{-1})$
$\displaystyle=|d(\mathrm{id}_{G},g)\pm
d(g,gh^{-1})|=d(\mathrm{id}_{G},h^{-1})=d(\mathrm{id}_{G},h).$

Hence, $d(\mathrm{id}_{G},hgh^{-1})=|d(\mathrm{id}_{G},h)\pm d(h,hgh^{-1})|$
is either $0$ or $2d(\mathrm{id}_{G},h)$. Assume first that
$\|hgh^{-1}\|_{d}=0$ for every such $h$ when $\|g\|_{d}=0$. Let
$U:=\\{h:\|h\|_{d}<\rho/4\\}$. By the continuity of $d$, $U$ is open. Hence
for every $h$ in $G$, $h$ can be written as a finite product of elements in
$U$. By induction, we conclude that for every $h\in G$, $\|hgh^{-1}\|_{d}=0$
given $\|g\|_{d}=0$, and this implies that $\ker d$ is normal in $G$.

Suppose now $\|hgh^{-1}\|_{d}=2\|h\|_{d}$. Since $d$ is monotone,
$\|hg^{2}h^{-1}\|_{d}=4\|h\|_{d}$. On the other hand, as $\|g\|_{d}=0$,
repeating the argument above, we get $\|hg^{2}h^{-1}\|_{d}$ is either $0$ or
$2\|h\|_{d}$. Hence, $\|h\|_{d}=0$, and so $\|hgh^{-1}\|_{d}=0$.

Next we show that $G^{\prime}=G/\ker d$ is a connected Lie group. By the
Gleason–Yamabe Theorem (Theorem 6.9) it suffices to show that there is a
neighborhood of $\mathrm{id}_{G^{\prime}}$ where the only subgroup is trivial.
Let $d^{\prime}$ be the metric on $G^{\prime}$ induced by $d$. Choose
$U^{\prime}=\\{g^{\prime}\in
G^{\prime}:\|g^{\prime}\|_{d^{\prime}}<\rho/4\\}$. Suppose $H^{\prime}$ is a
compact subgroup of $U^{\prime}$. Take $g^{\prime}\in H^{\prime}$ such that
$0<\|g^{\prime}\|_{d^{\prime}}<\rho/4$. Then by the monotonicity, there is $k$
such that
$\|(g^{\prime})^{k}\|_{d^{\prime}}=k\|g^{\prime}\|_{d^{\prime}}>\rho/4$, and
this contradicts the fact that $g^{\prime}$ in $H^{\prime}$ and $H\subseteq
U^{\prime}$. Thus $G^{\prime}$ is a connected Lie group.

We now show that $G^{\prime}$ has dimension $1$. Choose $g$ in the
neighborhood of $\mathrm{id}_{G^{\prime}}$ such that $g$ is in the image of
the exponential map and $\|g\|_{d}<\rho/4$. If $g^{\prime}$ is another element
in the neighborhood of $\mathrm{id}_{G^{\prime}}$ which is in the image of the
exponential map and $\|g^{\prime}\|_{d}<\rho/4$. Then by the monotonicity, we
get $g$ and $g^{\prime}$ are on the same one parameter subgroup. Hence, we get
the desired conclusion.

In the rest of the proof, we prove the result under the condition that $G$ is
a connected Lie group. As $d$ is invariant, $d(\mathrm{id}_{G},g)=d(g,g^{2})$
for all $g\in G$. From local linearity of $d$, for all $g\in G$ with
$\|g\|_{d}<\rho/2$, we either have $\|g^{2}\|_{d}=2\|g\|_{d}$ or
$\|g^{2}\|_{d}=0$. It suffices to rule out the possibility that
$0<\|g\|_{d}<\rho/4$, and $\|g^{2}\|_{d}=0$.

As $d$ is continuous, there is an open neighborhood $W$ of $g$ such that for
all $g^{\prime}\in W$, we have $\|g^{\prime}\|_{d}>0$ and
$\|(g^{\prime})^{2}\|_{d}=0$. From Lemma 7.2, the set $\\{g\in
G:\|g\|_{d}=0\\}$ is a closed subgroup of $G$. As $d$ is nontrivial and $G$ is
a connected Lie group, $\\{g\in G:\|g\|_{d}=0\\}$ must be a Lie group with
smaller dimension. Therefore, we only need to show that if $W$ is an open
subset of $G$, then $s(W)$ is not contained in a closed subgroup of $G$ with
smaller dimension, where $s:G\to G$ is the squaring map, and this is
guaranteed by Lemma 7.4. ∎

###  7.2. Almost linear pseudometric

In this subsection, we will generalize Proposition 7.5 to the situation where
the linearity condition is weakened. Our plan is to construct a universally
measurable multivalued group homomorphism into $\mathbb{T}$, and then use a
number or results in descriptive set theory and geometry to refine this into a
continuous group homomorphism, and the latter will be done in the next
subsection.

Given $0<\gamma<\rho/100$. We write $I(\gamma)$ in this subsection to denote
the interval $(-\gamma,\gamma)$ in either $\mathbb{R}$ or $\mathbb{T}$.
Suppose $d$ is a pseudometric on $G$ with radius $\rho$. We say that $d$ is
$\gamma$-linear if it satisfies the following conditions:

  1. (1)

$d$ is continuous and left and right invariant;

  2. (2)

for all $g_{1},g_{2},g_{3}\in G$ with
$d(g_{1},g_{2})+d(g_{2},g_{3})<\rho-\gamma$, we have either

$d(g_{1},g_{3})\in d(g_{1},g_{2})+d(g_{2},g_{3})+I(\gamma),$

or

$d(g_{1},g_{3})\in|d(g_{1},g_{2})-d(g_{2},g_{3})|+I(\gamma).$

Given $\alpha\leq\rho$, let $N(\alpha)=\\{g\in G:\|g\|_{d}\leq\alpha\\}$. We
say that $d$ is $\gamma$-monotone if for all $g\in N(\rho/2-2\gamma)$, we have

$\|g^{2}\|_{d}\in 2\|g\|_{d}+I(4\gamma).$

The next lemma says that under the $\gamma$-linearity condition, the group $G$
essentially has only one “direction”: if there are three elements have the
same distance to $\mathrm{id}_{G}$, then at least two of them are very close
to each other.

######  Lemma 7.6.

Let $\gamma<\rho/16$. Suppose $d$ is a $\gamma$-linear pseudometric on $G$. If
$g,g_{1},g_{2}\in G$ such that

$\|g\|_{d}=\|g_{1}\|_{d}=\|g_{2}\|_{d}\in I(\rho/4-2\gamma)\setminus
I(2\gamma),$

and $d(g_{2},g_{3})\in 2\|g\|_{d}+I(\gamma)$. Then either $d(g,g_{1})\in
I(\gamma)$ or $d(g,g_{2})\in I(\gamma)$.

###### Proof.

Suppose first that both $d(g,g_{1})$ and $d(g,g_{2})$ are not in $I(\gamma)$.
Then by $\gamma$-linearity of $d$, we have

$d(g,g_{1})\in|d(\mathrm{id}_{G},g)\pm d(\mathrm{id}_{G},g_{1})|+I(\gamma),$

and thus $d(g,g_{1})\in 2\|g\|_{d}+I(\gamma)$, and same as $d(g,g_{2})$.

If $d(g_{1},g_{2})\in d(g,g_{1})+d(g,g_{2})+I(\gamma)$, then

$d(g_{1},g_{2})\in 4\|g\|_{d}+I(3\gamma).$

On the other hand, by $\gamma$ linearity we have

$d(g_{1},g_{2})\leq 2\|g\|_{d}+\gamma.$

This yields $\|g\|_{d}\leq 2\gamma$, contradicts the assumption.

Using the $\gamma$-linearity of $d$, the other two possibilities are
$d(g_{1},g_{2})+d(g,g_{2})\in d(g,g_{1})+I(\gamma)$ or
$d(g_{1},g_{2})+d(g,g_{1})\in d(g,g_{2})+I(\gamma)$, but similar calculations
also lead to contradictions. ∎

The next important result shows that if a almost linear pseudometric is “path”
monotone, then it is monotone. In the next section, we will construct a
pseudometric combinatorially using the geometric properties of small expansion
sets. The next result allows us to reduce the monotonicity problem to one
parameter subgroups.

######  Proposition 7.7.

Let $G$ be a compact connected Lie group, $\mathfrak{g}$ the Lie algebra of
$G$, $\mathrm{exp}:\mathfrak{g}\to G$ the exponential map, $d$ is a
$\gamma$-linear pseudometric on $G$. Suppose for each $X$ in $\mathfrak{g}$,
we have one of the following two possibilities:

  1. (1)

$\|\mathrm{exp}(tX)\|_{d}<\gamma$ for all $t\in\mathbb{R}$;

  2. (2)

there is $t_{0}\in\mathbb{R}^{>0}$ such that with
$g_{0}=\mathrm{exp}(t_{0}X)$, we have $\|g_{0}\|_{d}\in I(\rho/2)\setminus
I(\rho/4)$, $\|(g_{0})^{2}\|_{d}=2\|g_{0}\|_{d}+I(\gamma)$, and

(7) $\|g\|_{d}+d(g,g_{0})\in\|g_{0}\|+I(\gamma)$

for all $t\in[0,t_{0}]$ and with $g=\mathrm{exp}(tX)$.

Then $d$ is $8\gamma$-monotone.

###### Proof.

Fix an element $g$ of $G$ with $\|g\|_{d}\in I(\rho/2-\gamma)$. Our job is to
show that $\|g^{2}\|_{d}\in 2\|g\|_{d}+I(8\gamma)$. Since $G$ is compact and
connected $\mathrm{exp}$ is surjective. We get $X\in\mathfrak{g}$ such that
$g\in\\{\mathrm{exp}(tX):t\in\mathbb{R}\\}$. If we are in scenario (1), then
$\|g\|_{d}<\gamma$, hence $\|g^{2}\|_{d}\in 2\|g\|_{d}+I(3\gamma)$. Therefore,
it remains to deal with the case where we have an $g_{0}$ as in (2).

Suppose $g_{0}=\mathrm{exp}(t_{0}X)$. We consider first the special case where
$\|g\|_{d}<\|g_{0}\|_{d}-3\gamma$. As $d$ is continuous, there is
$t_{1}\in[0,t_{0}]$ such that with $g_{1}=\mathrm{exp}(t_{1}X)$, we have
$\|g_{1}\|_{d}=\|g\|_{d}$. Let $t_{2}=-t_{1}$, and
$g_{2}=\mathrm{exp}(t_{2}X)=g_{1}^{-1}$. Since $d$ is invariant,

$\|g_{2}\|_{d}=d(g^{-1}_{1},\mathrm{id}_{G})=d(\mathrm{id}_{G},g_{1})=\|g_{1}\|_{d}.$

Thus $\|g_{1}\|_{d}=\|g_{2}\|_{d}=\|g\|_{d}$. If $\|g\|_{d}<3\gamma$, then
$\|g^{2}\|_{d}\in 2\|g\|_{d}+I(6\gamma)$. Now suppose $\|g\|_{d}\geq 3\gamma$.
Then, by Lemma 7.6, either $d(g,g_{1})<\gamma$, or $d(g,g_{2})<\gamma$. We
only consider the former as the latter can be treated similarly.

Now we assume that $d(g,g_{1})<\gamma$. Note that one has
$\|g^{2}_{0}\|_{d}\in 2\|g_{0}\|_{d}+I(\gamma)$ by the assumption, and
$\|g_{1}^{2}\|_{d}$ is in either $2\|g_{1}\|_{d}+I(\gamma)$, or $I(\gamma)$ by
$\gamma$-linearity. Note that by the choice of $g_{1}$ we have
$\|g^{2}_{1}\|_{d}<\|g^{2}_{0}\|_{d}-2\gamma$. Since
$g^{-1}_{0}g_{1}=g_{1}g_{0}^{-1}$, and by $\gamma$-linearity of $d$, we get

(8)
$d(g_{1}^{2},g^{2}_{0})=d(\mathrm{id}_{G},g^{-2}_{1}g_{0}^{2})=d(\mathrm{id}_{G},(g^{-1}_{1}g_{0})^{2})\in\\{0,2d(g_{1},g_{0})\\}+I(\gamma).$

By (7), we have $\|g_{1}\|_{d}+d(g_{1},g_{0})\in\|g_{0}\|_{d}+I(\gamma)$, and
the assumption that $\|g_{1}\|_{d}=\|g\|_{d}>2\gamma$, we get

(9) $d(g^{2}_{1},g^{2}_{0})<2\|g_{0}\|_{d}-3\gamma.$

By the linearity of $d$, we have either

$\|g_{1}^{2}\|_{d}\in\|g_{0}^{2}\|_{d}-d(g_{1}^{2},g_{0}^{2})+I(\gamma),\text{
or }\|g_{1}^{2}\|_{d}\in d(g_{1}^{2},g_{0}^{2})-\|g_{0}^{2}\|_{d}+I(\gamma),$

and the latter case cannot happen by (9). Therefore by (8) and (7), either

$\displaystyle\|g_{1}^{2}\|_{d}\in
2\|g_{0}\|_{d}+I(3\gamma),\quad\text{or}\quad\|g_{1}^{2}\|_{d}\in
2\|g_{1}\|_{d}+I(5\gamma).$

As $\|g_{1}\|_{d}^{2}\leq 2\|g_{1}\|+\gamma<2\|g_{0}\|-5\gamma$, we must have
$\|g_{1}^{2}\|\in 2\|g_{1}\|+I(5\gamma)$. Now, since
$\|g^{-1}_{1}g\|_{d}=d(g_{1},g)<\gamma$, again by linearity we conclude that

$d(g^{2}_{1},g^{2})=\|(g^{-1}_{1}g)^{2}\|_{d}<3\gamma.$

Thus, $\|g^{2}\|_{d}\in 2\|g\|_{d}+I(8\gamma).$

Now consider the case where $\|g_{0}\|_{d}+3\gamma<\|g\|_{d}<\rho/2$. We
choose $g_{1}=\mathrm{exp}(t_{1}X)$ with $t\in[0,t_{0}]$. By the above
discussions, we have $\|g_{1}^{2}\|_{d}\in 2\|g_{1}\|+I(8\gamma)$. Choose
$t_{1}$ such that $\|g_{1}^{2}\|_{d}=\|g\|_{d}$, and let $g_{2}=g_{1}^{-1}$.
By Lemma 7.6, we may assume $d(g,g_{1}^{2})<\gamma$. Then by an argument
similar to the one above, we have $\|g^{4}_{1}\|_{d}\in
2\|g_{1}^{2}\|_{d}+I(5\gamma)$. Since

$d(g_{1}^{4},g^{2})=\|(g_{2}^{2}g)^{2}\|_{d}<2d(g_{1}^{2},d)+\gamma\leq
3\gamma,$

we also get $\|g^{2}\|_{d}\in 2\|g\|_{d}+I(8\gamma).$ in this case, and this
proves the proposition. ∎

Suppose $d$ is a $\gamma$-linear pseudometric on $G$ with radius $\rho$. We
define to be relative sign $s(g_{1},g_{2})$ for $g_{1},g_{2}\in G$ satisfying
$\|g_{1}\|_{d}+\|g_{2}\|_{d}<\rho-\gamma$ by

$s(g_{1},g_{2})=\begin{cases}0&\text{ if
}\min\\{\|g_{1}\|_{d},\|g_{2}\|_{d}\\}\leq 4\gamma,\\\ 1&\text{ if
}\min\\{\|g_{1}\|_{d},\|g_{2}\|_{d}\\}>4\gamma\text{ and
}\|g_{1}g_{2}\|_{d}\in\|g_{1}\|_{d}+\|g_{2}\|_{d}+I(\gamma).\\\ -1&\text{ if
}\min\\{\|g_{1}\|_{d},\|g_{2}\|_{d}\\}>4\gamma\text{ and
}|g_{1}g_{2}|_{d}\in\big{|}\|g_{1}\|_{d}-\|g_{2}\|_{d}\big{|}+I(\gamma).\end{cases}$

Note that this is well defined because when
$\min\\{\|g_{1}\|_{d},\|g_{2}\|_{d}\\}\geq 4\gamma$ in the above definition,
the differences between $\big{|}\|g_{1}\|_{d}-\|g_{2}\|_{d}\big{|}$ and
$\|g_{1}\|_{d}+\|g_{2}\|_{d}$ is at least $\gamma$. The following lemma gives
us tools to relate signs between different elements.

######  Lemma 7.8.

Suppose $d$ is a $\gamma$-monotone $\gamma$-linear pseudometric on $G$ with
radius $\rho$. Then for $g_{1}$, $g_{2}$, and $g_{3}$ in
$N(\rho/4-\gamma)\setminus N(4\gamma)$, we have the following

  1. (1)

$s(g_{1},g^{-1}_{1})=-1$ and $s(g_{1},g_{1})=1$.

  2. (2)

$s(g_{1},g_{2})=s(g_{2},g_{1})$.

  3. (3)

$s(g_{1},g_{2})=s(g^{-1}_{1},g^{-1}_{2})=-s(g^{-1}_{1},g_{2})=-s(g_{1},g^{-1}_{2})$.

  4. (4)

$s(g_{1},g_{2})s(g_{2},g_{3})s(g_{3},g_{1})=1.$

  5. (5)

If $\|g_{1}\|_{d}\leq\|g_{2}\|_{d}$, and $g_{1}g_{2}$ is in
$N(\rho/4-\gamma)\setminus N(4\gamma)$, then
$s(g_{0},g_{1}g_{2})=s(g_{0},g_{2}g_{1})=s(g_{0},g_{2})$.

###### Proof.

As $g_{1}$, $g_{2}$, and $g_{3}$ are in $N(\rho/4-\gamma)\setminus
N(4\gamma)$, one has $s(g_{i},g_{j})\neq 0$ for all $i,j\in\\{1,2,3\\}$. The
first part of Statement (1) is immediate from the fact that
$\|\mathrm{id}_{G}\|_{d}=0$, and the second part of Statement (1) follows from
the monotoncity and the definition of the relative sign.

We now prove Statement (2). Suppose to the contrary that
$s(g_{1},g_{2})=-s(g_{2},g_{1})$. Without loss of generality, assume
$s(g_{1},g_{2})=1$. Then $\|g_{1}g_{2}g_{1}g_{2}\|_{d}$ is in
$2\|g_{1}g_{2}\|_{d}+I(\gamma)$, which is a subset of
$2\|g_{1}\|_{d}+2\|g_{2}\|_{d}+I(3\gamma)$. On the other hand, as
$\|g_{2}g_{1}\|_{d}\in\big{|}\|g_{1}\|_{d}-\|g_{2}\|_{d}\big{|}+I(\gamma)$, we
have

$\|g_{1}g_{2}g_{1}g_{2}\|_{d}\in\big{|}\|g_{1}\|_{d}\pm(\|g_{2}\|_{d}-\|g_{1}\|_{d})\pm\|g_{2}\|_{d}\big{|}+I(3\gamma).$

This contradicts the assumption that $g_{1}$ and $g_{2}$ are not in
$N(4\gamma)$.

Next, we prove the first and third equality in Statement (3). Note that
$\|g\|_{d}=\|g^{-1}\|_{d}$ for all $g\in G$ following the fact that $d$ is
symmetric and invariant. Hence,
$\|g_{1}g_{2}\|_{d}=\|g_{2}^{-1}g_{1}^{-1}\|_{d}$. This implies that
$s(g_{1},g_{2})=s(g_{2}^{-1},g_{1}^{-1})$. Combining with Statement (2), we
get the first equality in (3). The third equality in Statement (3) is a
consequence of the first equality in Statement (3).

Now we consider the second equality in Statement (3). Suppose to the contrary
that $s(g^{-1}_{1},g_{2}^{-1})=s(g^{-1}_{1},g_{2})$. Then from Statement (2)
and the first equality of Statement (3), we get
$s(g_{2},g_{1})=s(g^{-1}_{1},g_{2})$. Hence,
$\|g_{2}g_{1}g_{1}^{-1}g_{2}\|_{d}$ is in either
$2\|g_{1}\|_{d}+2\|g_{2}\|_{d}+I(3\gamma)$ or
$2\big{|}\|g_{1}\|_{d}-\|g_{2}\|_{d}\big{|}+I(3\gamma)$. On the other hand,
$\|g_{2}g_{1}g_{1}^{-1}g_{2}\|_{d}=\|g_{2}^{2}\|_{d}$, which is in
$2\|g_{2}\|_{d}+I(\gamma)$. We get a contradiction with the fact that $g_{1}$
and $g_{2}$ are not in $N(4\gamma)$.

We now prove Statement (4). Without loss of generality, we assume
$\|g_{1}\|_{d}\leq\|g_{2}\|_{d}\leq\|g_{3}\|_{d}$. Using Statement (3) to
replace $g_{3}$ with $g_{3}^{-1}$ if necessary we can assume that
$s(g_{2},g_{3})=1$. We need to show that $s(g_{1},g_{2})=s(g_{1},g_{3})$.
Suppose to the contrary. Then from Statement (3), we get
$s(g_{1},g_{2})=s(g^{-1}_{1},g_{3})$. Replacing $g_{1}$ with $g_{1}^{-1}$ if
necessary, we can assume that $s(g_{1},g_{2})=s(g^{-1}_{1},g_{3})=1$. Using
Statement (2), we get $s(g_{2},g_{1})=1$. Therefore,
$\|g_{2}g_{1}g_{1}^{-1}g_{3}\|_{d}$ is in either
$2\|g_{1}\|_{d}+\|g_{2}\|_{d}+\|g_{3}\|_{d}+I(3\gamma)$ or
$\|g_{3}\|_{d}-\|g_{2}\|_{d}+I(3\gamma)$. On the other hand,
$\|g_{2}g_{1}g_{1}^{-1}g_{3}\|_{d}=\|g_{2}g_{3}\|_{d}$ is in
$\|g_{2}\|_{d}+\|g_{3}\|_{d}+I(\gamma)$. Hence, we get a contradiction to the
fact that $g_{1}$, $g_{2}$, and $g_{3}$ are not in $N(4\gamma)$.

Finally, we prove Statement (5). Using Statement (4), it suffices to show
$s(g_{1}g_{2},g_{2})=s(g_{2}g_{1},g_{2})=1$. We will only show the former, as
the proof for the latter is similar. Suppose to the contrary that
$s(g_{1}g_{2},g_{2})=-1$. Then $\|g_{1}g^{2}_{2}\|_{d}$ is in
$\big{|}\|g_{1}g_{2}\|_{d}-\|g_{2}\|_{d}\big{|}+I(\gamma)$, which is a subset
of $\|g_{1}\|_{d}+I(2\gamma)$. On the other hand, $\|g_{1}g^{2}_{2}\|_{d}$ is
also in $\big{|}\|g_{1}\|_{d}-\|g^{2}_{2}\|_{d}\big{|}+I(\gamma)$ which is a
subset of $2\|g_{2}\|_{d}-\|g_{1}\|_{d}+I(2\gamma).$ Hence, we get a
contradiction with the assumption that $g_{1}$ and $g_{2}$ are not in
$N(4\gamma)$. ∎

The notion of relative sign corrects the ambiguity in calculating distance, as
can be seen in the next result.

######  Corollary 7.9.

Suppose $d$ is a $\gamma$-monotone $\gamma$-linear pseudometric on $G$ with
radius $\rho$. Let $g_{0}$ be in $N(\rho/4)\setminus N(4\gamma)$, and let
$g_{1}$ and $g_{2}$ be in $N(\rho/16-\gamma)$ with
$\|g_{1}\|_{d}\leq\|g_{2}\|_{d}$, then we have the following

  1. (1)

Both $\|g_{1}g_{2}\|_{d}$ and $\|g_{2}g_{1}\|_{d}$ are in
$s(g_{1},g_{2})\|g_{1}\|_{d}+\|g_{2}\|_{d}+I(5\gamma)$.

  2. (2)

Both $s(g_{0},g_{1}g_{2})\|g_{1}g_{2}\|_{d}$ and
$s(g_{0},g_{2}g_{1})\|g_{1}g_{2}\|_{d}$ are in

$s(g_{0},g_{1})\|g_{1}\|_{d}+s(g_{0},g_{2})\|g_{2}\|_{d}+I(22\gamma).$

###### Proof.

We prove Statement (1). When $g_{1},g_{2}\notin N(4\gamma)$, the statement for
$\|g_{1}g_{2}\|_{d}$ is immediate from the definition of relative sign, the
statement for $\|g_{2}g_{1}\|_{d}$ is a consequence of Lemma 7.8(2). Now
suppose $\|g_{1}\|_{d}<4\gamma$. From the $\gamma$-linearity, we have

$\|g_{2}\|_{d}-\|g_{1}\|_{d}-\gamma<\|g_{1}g_{2}\|_{d}<\|g_{1}\|_{d}+\|g_{2}\|_{d}+\gamma.$

The desired conclusion follows.

We now prove Statement (2). Fix $g_{0}$ in $N(\rho/4-\gamma)\setminus
N(4\gamma)$. We will consider two cases, when $g_{1}$ is not in $N(4\gamma)$
and when $g_{1}$ is in $N(4\gamma)$. Suppose we are in the first case, that is
$g_{1}\notin N(4\gamma)$. As $\|g_{1}\|_{d}\leq\|g_{2}\|_{d}$, we also have
$g_{2}\notin N(4\gamma)$. If both $g_{1}g_{2}$ and $g_{2}g_{1}$ are not in
$N(4\gamma)$, then the desired conclusion is a consequence of Statement (1)
and Lemma 7.8(4) and (5). Within the first case, it remains to deal with the
situations where $g_{1}g_{2}$ is in $N(4\gamma)$ or $g_{2}g_{1}$ is in
$N(4\gamma)$. Since these two situations are similar, we may assume
$g_{1}g_{2}$ is in $N(4\gamma)$. From Statement (1), we have
$s(g_{1},g_{2})=-1$ and $\|g_{2}\|_{d}-\|g_{1}\|_{d}$ is at most $\gamma$.
Therefore, $|g_{2}g_{1}|_{d}$ is in $I(6\gamma)$. By Lemma 7.8(4), we have
$s(g_{0},g_{1})=-s(g_{0},g_{2})$. Hence,
$s(g_{0},g_{1})\|g_{1}\|_{d}+s(g_{0},g_{2})\|g_{2}\|_{d}$ is in $I(6\gamma)$.
Thus, $s(g_{0},g_{1}g_{2})\|g_{1}g_{2}\|_{d}$ and
$s(g_{0},g_{2}g_{1})\|g_{1}g_{2}\|_{d}$ are in
$s(g_{0},g_{1})\|g_{1}\|_{d}+s(g_{0},g_{2})\|g_{2}\|_{d}+I(6\gamma)$, giving
us the desired conclusion.

Continuing from the previous paragraph, we consider the second case when
$g_{1}$ is in $N(4\gamma)$. If $g_{2}$ is in $N(13\gamma)$, then both
$\|g_{1}g_{2}\|_{d}$ and $\|g_{2}g_{1}\|_{d}$ are in $I(22\gamma)$ by
Statement (1), and the desired conclusion follows. Now suppose $g_{2}$ is not
in $N(13\gamma)$. Then from Statement (1) and the fact that $g_{1}\in
N(4\gamma)$, we get $g_{1}g_{2}$ and $g_{2}g_{1}$ are both not in
$N(4\gamma)$. Note that $s(g_{1}g_{2},g^{-1}_{2})=-1$, because otherwise we
get $\|g_{1}\|_{d}\geq\|g_{1}g_{2}\|_{d}+\|g^{-1}_{2}\|_{d}-5\gamma>4\gamma$.
A similar argument gives $s(g_{2}^{-1},g_{2}g_{1})=-1$. Hence,

$s(g_{1}g_{2},g_{2})=s(g_{2}g_{1},g_{2})=1.$

By Lemma 7.8(4), we get
$s(g_{0},g_{2})=s(g_{0},g_{1}g_{2})=s(g_{0},g_{2}g_{1})$. From Statement (1),
$\|g_{1}g_{2}\|_{d}$ and $\|g_{2}g_{1}\|_{d}$ are both in
$\|g_{2}\|_{d}+I(9\gamma)$. On the other hand,
$s(g_{0},g_{1})\|g_{1}\|_{d}+s(g_{0},g_{2})\|g_{2}\|_{d}$ is in
$s(g_{0},g_{2})\|g_{2}\|_{d}+I(9\gamma)$. The desired conclusion follows. ∎

The next corollary will be important in the subsequent development.

######  Corollary 7.10.

Suppose $d$ is a $\gamma$-monotone $\gamma$-linear pseudometric on $G$ with
radius $\rho$, $g_{0}$ and $g_{0}^{\prime}$ are in $N(\rho/4-\gamma)\setminus
N(4\gamma)$, and $g_{1},\ldots,g_{n}$ are in $N(\rho/4-\gamma)$. Then

$\left|\sum^{n}_{i=1}s(g_{0},g_{i})\|g_{i}\|_{d}\right|=\left|\sum^{n}_{i=1}s(g^{\prime}_{0},g_{i})\|g_{i}\|_{d}\right|.$

###### Proof.

As $s(g_{0},g_{i})=s(g_{0}^{\prime},g_{i})=0$ whenever
$\|g_{i}\|_{d}<4\gamma$, we can reduce to the case where $\min_{1\leq i\leq
n}\|g_{i}\|_{d}\geq 4\gamma$. Using (3) of Lemma 7.8 to replace $g_{0}$ with
$g_{0}^{-1}$ if necessary, we can assume that
$s(g_{0},g_{1})=s(g^{\prime}_{0},g_{1})$. Then using (4) of Lemma 7.8, we get
$s(g_{0},g_{i})=s(g^{\prime}_{0},g_{i})$ for all $i\in\\{1,\ldots,n\\}$. This
gives us the desired conclusion. ∎

Suppose $N(\rho/4-\gamma)\setminus N_{4\gamma}$ is not empty, and
$(g_{1},\ldots,g_{n})$ is a sequence of elements in $N(\rho/4-\gamma)\setminus
N(4\gamma)$. We set

$t(g_{1},\ldots,g_{n})=\left|\sum^{n}_{i=1}s(g_{0},g_{i})\|g_{i}\|_{d}\right|$

with $g_{0}$ is an arbitrary element in $N(\rho/4-\gamma)\setminus
N(4\gamma)$, and call this the total weight of $(g_{1},\ldots,g_{n})$. Note
that this is well defined due to Corollary 7.10. In the next subsection, we
will use this notion to extend the pseudometric nontrivially from $N(\rho)$ to
all elements in $G$, and use it to construct a continuous group homomorphism
to either $\mathbb{T}$ or $\mathbb{R}$.

###  7.3. Group homomorphisms from almost linear pseudometric

Throughout the subsection, $d$ is always a $\gamma$-monotone $\gamma$-linear
pseudometric on $G$. In this subsection, we will construct a continuous group
homomorphism from $d$. Suppose $(g_{1},\ldots,g_{n})$ is a sequence of
elements in $G$. A concatenation of $(g_{1},\ldots,g_{n})$ is a sequence
$(h_{1},\ldots,h_{m})$ such that there are $0=k_{0}<k_{1}<\cdots<k_{m}=n$ with

$h_{i}=g_{k_{i-1}+1}\cdots g_{k_{i}}\text{ for }i\in\\{1,\ldots,m\\}.$

We call $(g_{1},\ldots,g_{n})$ is a $\lambda$-sequence if $g_{i}$ is in
$N(\lambda)$ for each $i\in\\{1,\ldots,n\\}$. A $\lambda$-sequence
$(g_{1},\ldots,g_{n})$ is irreducible if for all $2\leq j\leq 4$, we have

$g_{i+1}\cdots g_{i+j}\notin N(\lambda).$

We need several technical lemmas to investigate properties of
$\lambda$-sequences.

######  Lemma 7.11.

Suppose $(g_{1},\ldots,g_{n})$ is an irreducible $\lambda$-sequence with
$4\gamma<\lambda<\rho/16-\gamma$, and $g_{0}$ is a fixed element in
$N(\rho/4-\gamma)\setminus N(4\gamma)$. Then for all $i$, $i^{\prime}$, $j$,
and $j^{\prime}$ such that $2\leq j,j^{\prime}\leq 4$, $0\leq i\leq n-j$, and
$0\leq i^{\prime}\leq n-j^{\prime}$, we have

$s(g_{0},g_{i+1}\cdots g_{i+j})=s(g_{0},g_{i^{\prime}+1}\cdots
g_{i^{\prime}+j^{\prime}}).$

###### Proof.

It suffices to show for fixed $i$ and $j$ with $2\leq j\leq 3$ and $0\leq
i\leq n-j-1$ that $s(g_{0},g_{i+1}\cdots g_{i+j})=s(g_{0},g_{i+1}\cdots
g_{i+j+1})$. Note that both $g_{i+1}\cdots g_{i+j}$ and $g_{i+1}\cdots
g_{i+j+1}$ are in $N(\rho/4-\gamma)\setminus N(4\gamma)$. Hence, applying
Lemma 7.8(4), we reduce the problem to showing $s(g^{-1}_{i+j}\cdots
g_{i+1}^{-1},g_{i+1}\cdots g_{i+j+1)})=-1$. This is the case because
otherwise, we get $\|g_{i+j+1}\|_{d}\geq 2\lambda-\gamma>\lambda$, a
contradiction. ∎

######  Corollary 7.12.

Suppose $(g_{1},\ldots,g_{n})$ is an irreducible $\lambda$-sequence with
$44\gamma<\lambda<\rho/16-\gamma$, and $N(\rho/4-\gamma)\setminus N(4\gamma)$
is nonempty. Then

$n\lambda/4<t(g_{1},\ldots,g_{n})<n\lambda.$

###### Proof.

The second inequality is immediate from the fact that $\|g_{i}\|_{d}<\lambda$
for each $i$, and the definition of $t(g_{1},\ldots,g_{n})$. We now prove the
first inequality. If $n=2k$, let $h_{i}=g_{2i-1}g_{2i}$ for
$i\in\\{1,\ldots,k\\}$. If $n=2k+1$, let $h_{i}=g_{2i-1}g_{2i}$ for
$i\in\\{1,\ldots,k-1\\}$, and $h_{k}=g_{2n-1}g_{2n}g_{2n+1}$. From Corollary
7.9 we have that

(10) $t(h_{1},\ldots,h_{k})\in t(g_{1},\ldots,g_{n})+I(22(n-k)\gamma).$

As $(g_{1},\ldots,g_{n})$ is irreducible, $h_{i}$ is in $N(3\lambda)\setminus
N(\lambda)$ for $i\in\\{1,\ldots,k\\}$. By Lemma 7.11, we get
$s(g_{0},h_{i})=s(g_{0},h_{j})$ for all $i$ and $j$ in $i\in\\{1,\ldots,k\\}$.
Thus by the definition of the total weight again,
$t(h_{1},\ldots,h_{k})>n\lambda/3.$ Combining with the assumption on $\lambda$
and (10), we get $t(g_{1},\ldots,g_{n})>n\lambda/3-11n\gamma$. The desired
conclusion follows from $\lambda>44\gamma$. ∎

The next lemma allow us to only consider irreducible sequences.

######  Lemma 7.13.

Let $4\gamma<\lambda<\rho/16-\gamma$. Suppose $N(\rho/4-\gamma)\setminus
N(4\gamma)$ is nonempty. Then every sequence $(g_{1},\ldots,g_{n})$ with terms
in $N(\lambda)$ has an irreducible concatenation
$(g^{\prime}_{1},\ldots,g^{\prime}_{m})$ such that

$t(g^{\prime}_{1},\ldots,g^{\prime}_{m})\in
t(g_{1},\ldots,g_{n})+I(22(n-m)\gamma)$

###### Proof.

The statement is immediate when $n=1$. Using induction, suppose we have proven
the statement for all smaller values of $n$. If $(g_{1},\ldots,g_{n})$ is
irreducible, we are done. We only consider the case where $g_{i+1}g_{i+2}$ is
in $N(\lambda)$ for some $0\leq i\leq n-4$; the cases where either
$g_{i+1}g_{i+2}g_{i+3}$ or $g_{i+1}g_{i+2}g_{i+3}g_{i+4}$ is in $N(\lambda)$
can be dealt with similarly. Fix $g_{0}$ in $N(\lambda/4-\gamma)\setminus
N(4\gamma)$. Using Corollary 7.9

$s(g_{0},g_{i+1}g_{i+2})\|g_{i+1}g_{i+2}\|_{d}\in
s(g_{0},g_{i+1})\|g_{i+1}\|_{d}+s(g_{0},g_{i+2})\|g_{i+2}\|_{d}+I(22\gamma).$

From here, we get the desired conclusion. ∎

When we have an irreducible $\lambda$-sequence $(g_{1}.\dots,g_{n})$, we may
view $\mathrm{id}_{G}$ as $g_{0}$, and by irreducibility, intuitively
$g_{1}\cdots g_{m}$ is closer to $g_{0}$ than $g_{1}\cdots g_{m+k}$ for some
positive $k$. However, as $G$ is compact, the sequence may “loop back” to
$\mathrm{id}_{G}$ when $n$ is large. The next lemma provides a lower bound
estimate on such $n$.

######  Lemma 7.14.

Let $G$ be a connected compact group, and $\mu_{G}$ a normalized Haar measure
on $G$. Let $4\gamma<\lambda<\rho/4-\gamma$. Suppose $(g_{1},\ldots,g_{n})$ is
an irreducible $\lambda$-sequence with $g_{1}\cdots g_{n}=\mathrm{id}_{G}$.
Then $n\geq 1/\mu_{G}(N(4\lambda)).$

###### Proof.

Let $m>0$. For convenience, when $m>n$ we write $g_{m}$ to denote the element
$g_{i}$ with $i\leq n$ and $i\equiv m\pmod{n}$. Define

$N^{(m)}(4\lambda)=\\{g\in G\mid d(g,g_{1}\cdots g_{m})<4\lambda\\}.$

Note that we have $N^{(m)}(4\lambda)=N^{(m^{\prime})}(4\lambda)$ when $m\equiv
m^{\prime}\pmod{n}$. By invariance of $d$ and $\mu_{G}$, clearly
$\mu_{G}(N^{(m)}(4\lambda))=\mu_{G}(N(4\lambda))$ for all $m$. We also write
$N^{(0)}(4\lambda)=N(4\lambda)$. We will show that

$G=\bigcup_{m\in\mathbb{Z}}N^{(m)}(4\lambda)=\bigcup_{m=0}^{n-1}N^{(m)}(4\lambda),$

which yield the desired conclusion.

As $g_{1}\cdots g_{n}=\mathrm{id}_{G}$, we have $\mathrm{id}_{G}$ is in
$N^{(0)}(2\lambda)$, and hence in $\bigcup_{m\in\mathbb{Z}}N^{(m)}(4\lambda)$.
As every elements in $G$ can be written as a product of finitely many elements
in $N(\lambda)$, it suffices to show for
$g\in\bigcup_{m\in\mathbb{Z}}N^{(m)}(\lambda)$ and $g^{\prime}=gh$ with $h\in
N(\lambda)$ that $g^{\prime}$ is in
$\bigcup_{m\in\mathbb{Z}}N^{(m)}(4\lambda)$. Fix $m$ which minimizes
$d(g,g_{1}\ldots g_{m})$. We claim that $d(g,g_{1}\ldots
g_{m})<2\lambda+\gamma$. This claim gives us the desired conclusion because we
then have $d(g^{\prime},g_{1}\ldots g_{m})<3\lambda+2\gamma<4\lambda$ by the
$\gamma$-linearity of $d$.

We now prove the claim that $d(g,g_{1}\ldots g_{m})<2\lambda+\gamma$. Suppose
to the contrary that $d(g,g_{1}\ldots g_{m})\geq 2\lambda+\gamma$. Let
$u=(g_{1}\cdots g_{m})^{-1}g$. Now by Lemma 7.11 we have either
$s(u,g_{m+1}g_{m+2})=1$, or $s(u,g_{m}^{-1}g_{m-1}^{-1})=1$. Suppose it is the
former, since the lattar case can be proved similarly. Then
$s(u,g_{m+2}^{-1}g_{m+1}^{-1})=-1$. Note that $g=g_{1}\cdots
g_{m}u=(g_{1}\cdots g_{m+2})g^{-1}_{m+2}g^{-1}_{m+1}u$. By the definition of
$u$, and the linearity of $d$, we have $\|u\|_{d}\geq
2\lambda+\gamma>\|g_{m+1}g_{m+2}\|_{d}$, therefore by the irreducibility we
have

$\displaystyle d(g,g_{1}\cdots g_{m+2})$
$\displaystyle=\|g^{-1}_{m+2}g^{-1}_{m+1}u\|_{d}$
$\displaystyle<\|u\|_{d}-\|g^{-1}_{m+2}g^{-1}_{m+1}\|_{d}+\gamma<\|u\|_{d}-\lambda+\gamma<\|u\|_{d}.$

This contradicts our choice of $m$ having $d(g,g_{1},\ldots,g_{m})$ minimized.
∎

######  Corollary 7.15.

Suppose $G$ is a connected compact group, and $\mu_{G}$ is the normalized Haar
measure on $G$. Let $44\gamma<\lambda<\rho/16-\gamma$. Let $S_{\lambda}$ be
the set of irreducible $\lambda$-sequences $(g_{1},\ldots,g_{n})$ with $n>0$
and $g_{1}\ldots g_{n}=\mathrm{id}_{G}$, and set

$\alpha_{\lambda}=\inf\\{|t(g_{1},\ldots,g_{n})|:(g_{1},\ldots g_{n})\in
S_{\lambda}\\}.$

Then we have the following:

  1. (1)

$\lambda/4\mu_{G}(N(4\lambda))\leq\alpha_{\lambda}\leq
4\lambda/\mu_{G}(N(\lambda)).$

  2. (2)

There is $(g_{1},\ldots,g_{n})\in S_{\lambda}$ such that
$1/\mu_{G}(N(4\lambda))\leq n\leq 4/\mu_{G}(N(\lambda))$ and
$\alpha_{\lambda}=t(g_{1},\ldots,g_{n})$.

###### Proof.

By Lemma 7.14, if $(g_{1},\ldots,g_{n})$ is in $S_{\lambda}$, then $n\geq
1/\mu_{G}(N(4\lambda))$. Combining with the first inequality in Corollary
7.12, we get $\alpha_{\lambda}\geq\lambda/4\mu_{G}(N(4\lambda)).$ On the other
hand, by Proposition 4.2, $G=(N(\lambda))^{k}$ for all
$k>1/\mu_{G}(N(\lambda))$. Hence, there is $(g_{1},\ldots,g_{n})\in
S_{\lambda}$ with $n\leq 4\lambda/\mu_{G}(N(\lambda))$. Using the second
inequality in Corollary 7.12, we get $\alpha_{\lambda}\leq
4\lambda/\mu_{G}(N(\lambda))$.

From Statement (1) and the second inequality in Corollary 7.12, suppose we
have $n>4/\mu_{G}(N(\lambda))$. Then

$|t(g_{1},\ldots,g_{n})|>\frac{4\lambda}{\mu_{G}(N(\lambda))}\geq\alpha_{\lambda}.$

Hence,

$\alpha_{\lambda}=\inf\\{|t(g_{1},\ldots,g_{n})|:(g_{1},\ldots g_{n})\in
S_{\lambda},n\leq 4/\mu_{G}(N(\lambda))\\}.$

For fixed $n$ the set of sequences of length $n$ which is in $S_{\lambda}$ is
closed under taking limit. Hence, we obtain desired $(g_{1},\ldots,g_{n})$
using the Bozalno–Wierstrass Theorem. ∎

The next lemma shows that the size of $N(\lambda)$ grows linearly.

######  Lemma 7.16.

Let $4\gamma<\lambda<\rho/16-\gamma$. Then $\mu_{G}(N(4\lambda))\leq
12\mu_{G}(N(\lambda))$.

###### Proof.

Let $g\in N(\lambda)\setminus N(\lambda/2-\gamma)$. Such $g$ exists since by
$\gamma$-monotoncity we have $N^{2}(\lambda/2-\gamma)\subseteq N(\lambda)$,
and by Kemperman’s inequality, $\mu_{G}(N(\lambda)\setminus
N(\lambda/2-\gamma))>0$. Let $h$ be an arbitrary element in $N(4\lambda)$, and
assume first $s(g,h)=1$. Let $k\geq 0$ be an integer, and define
$g_{k}=h(g^{-1})^{k}$. Then by Lemma 7.8 and Corollary 7.9, we have

$\|g_{k}\|_{d}\in\|h\|_{d}-k\|g\|_{d}+I(5k\gamma).$

Then there is $k\leq 6$ such that $g_{k}\in N(\lambda)$. When $s(g,h)=-1$, one
can similarly construct $g_{k}$ as $g^{k}h$, and thus we can also find $k\leq
6$ such that $g_{k}\in N(\lambda)$. This means

$N(4\lambda)\subseteq\Big{(}\bigcup_{i=1}^{6}N(\lambda)g^{i}\Big{)}\cup\Big{(}\bigcup_{j=1}^{6}(g^{-1})^{j}N(\lambda)\Big{)}.$

This implies that $\mu_{G}(N(4\lambda))\leq 12\mu_{G}(N(\lambda))$. ∎

######  Lemma 7.17.

Let $10^{5}\gamma<\lambda<\rho/16-\gamma$. Let $\alpha_{\lambda}$ be as in
Corollary 7.15. If $(g_{1},\ldots,g_{n})$ is a $\lambda$-sequence with
$g_{1}\ldots g_{n}=\mathrm{id}_{G}$ and $n\leq 4/\mu_{G}(N(\lambda))$, then

$t(g_{1},\ldots,g_{n})\in\alpha_{\lambda}\mathbb{Z}+I(\alpha_{\lambda}/200).$

###### Proof.

Without loss of generality, we assume $t(g_{1},\dots,g_{n})>0$. As $n\leq
1/\mu_{G}(N(\lambda))$, by the definition of the total weight we have
$t(g_{1},\ldots,\allowbreak g_{n})\leq\lambda/\mu_{G}(N(\lambda)).$ Hence
using Corollary 7.15(1),

(11)
$t(g_{1},\ldots,g_{n})<\frac{4\alpha_{\lambda}\mu_{G}(N(4\lambda))}{\mu_{G}(N(\lambda))}.$

Let $(h_{1},\ldots,h_{m})$ be an irreducible $\lambda$-sequence in
$S_{\lambda}$ such that $t(h_{1},\ldots,h_{m})=\alpha_{\lambda}$ obtained from
Corollary 7.15(2), and hence $1/\mu_{G}(N(4\lambda))\leq m\leq
4/\mu_{G}(N(\lambda))$. We now define a sequence
$(g^{\prime}_{1},\ldots,g_{n^{\prime}}^{\prime})$ such that

  1. (1)

$n^{\prime}=n+km$ for some integer $k\geq 0$.

  2. (2)

$g^{\prime}_{i}=g_{i}$ for $1\leq i\leq n$.

  3. (3)

For $i\geq n+1$, $g^{\prime}_{i}=h_{m+1-j}^{-1}$ with $j\equiv i-n\pmod{m}$.

From the definition of the total weight, we have

$t(g^{\prime}_{1},\ldots,g^{\prime}_{n^{\prime}})=t(g_{1},\ldots,g_{n})-k\alpha_{\lambda}.$

We choose integer $k$ such that
$|t(g^{\prime}_{1},\ldots,g^{\prime}_{n^{\prime}})|<3\alpha_{\lambda}/4$. Then
by (11),

(12)
$n^{\prime}<n+km<\frac{1}{\mu_{G}(N(\lambda))}+\frac{16\mu_{G}(N(4\lambda))}{\mu_{G}^{2}(N(\lambda))}\leq\frac{17\mu_{G}(N(4\lambda))}{\mu_{G}^{2}(N(\lambda))}.$

We now consider the case when
$(g^{\prime}_{1},\ldots,g^{\prime}_{n^{\prime}})$ is a $\lambda$-sequence with
$g^{\prime}_{1}\ldots g^{\prime}_{n^{\prime}}=\mathrm{id}_{G}$ and
$n^{\prime}$ satisfies (12), and without loss of generality $0\leq
t(g^{\prime}_{1},\ldots,g^{\prime}_{n^{\prime}})<3\alpha_{\lambda}/4$. Obtain
an irreducible concatenation $(r_{1},\ldots,r_{m^{\prime}})$ of
$(g^{\prime}_{1},\ldots,g^{\prime}_{n^{\prime}})$. From Lemma 7.13, we get

$\displaystyle t(r_{1},\ldots,r_{m^{\prime}})$
$\displaystyle<t(g^{\prime}_{1},\ldots,g^{\prime}_{n^{\prime}})+22(n^{\prime}-m^{\prime})\gamma$
$\displaystyle<\frac{3a_{\lambda}}{4}+\frac{374\mu_{G}(N(4\lambda))\gamma}{\mu_{G}^{2}(N(\lambda))}.$

Note that, by Corollary 7.15 and Lemma 7.16, the second term in the right hand
side of the above inequality is

$\displaystyle\frac{1496\mu_{G}^{2}(N(4\lambda))\gamma}{\mu_{G}^{2}(N(\lambda))\lambda}\frac{\lambda}{4\mu_{G}(N(4\lambda))}\leq\frac{1}{200}\alpha_{\lambda}$

when $\lambda>10^{5}\gamma$. This implies that $m^{\prime}=0$ and
$t(r_{1},\ldots,r_{m^{\prime}})=0$. Thus by Lemma 7.13 again,

$t(g^{\prime}_{1},\ldots,g^{\prime}_{n})\in I(22n^{\prime}\gamma)\subseteq
I\Big{(}\frac{1}{200}\alpha_{\lambda}\Big{)}.$

as desired. ∎

We did not try to optimise the constant $10^{5}$ in the above lemma. We remark
that, by improving this constant here, the quantitative dependence between
$\varepsilon$ and $\delta$ in Theorem 1.2 will be improved. In the rest of the
section, we will fix a $\lambda>10^{5}\gamma$.

Recall that a Polish space is a topological space which is separable and
completely metrizable. In particular, the underlying topological space of any
Lie group is a Polish space. Let $X$ be a Polish space. A subset $B$ of $X$ is
Borel if $B$ can be formed from open subsets of $X$ (equivalently, closed
subsets of $X$) through taking countable unions, taking countable
intersections, and taking complement. A function $f:X\to Y$ between Polish
space is Borel, if the inverse image of any Borel subset of $Y$ is Borel. A
subset $A$ of $X$ is analytic if it is the continuous image of another Polish
space $Y$. Below are some standard facts about these notions; see [Kec95] for
details.

######  Fact 7.18.

Suppose $X,Y$ are Polish spaces, and $f:X\to Y$ in continuous. We have the
following:

  1. (1)

Every Borel subset of $X$ is analytic.

  2. (2)

Equipping $X\times Y$ with the product topology, the graph of a Borel function
from $X$ to $Y$ is analytic.

  3. (3)