# A fibered Tukia theorem for nilpotent Lie groups

Tullia Dymarz  Department of Mathematics, University of Wisconsin-Madison, 480
Lincoln Drive, Madison, WI 53706<EMAIL_ADDRESS>,  David Fisher
Department of Mathematics, Rice University, 6100 Main Street Houston, TX 77005
<EMAIL_ADDRESS>and  Xiangdong Xie  Department of Mathematics and
Statistics, Bowling Green State University, Bowling Green, OH 43403
<EMAIL_ADDRESS>
###### Abstract.

We establish a Tukia-type theorem for uniform quasiconformal groups of a
Carnot group. More generally we establish a fiber bundle version (or foliated
version) of Tukia theorem for uniform quasiconformal groups of a nilpotent Lie
group whose Lie algebra admits a diagonalizable derivation with positive
eigenvalues. These results have applications to quasi-isometric rigidity of
solvable groups [DFX].

###### Key words and phrases:

uniform quasiconformal groups, uniform quasisimilarity groups, nilpotent Lie
groups

##  1\. Introduction

In this paper we study uniform quasiconformal groups of simply connected
nilpotent Lie groups. The nilpotent Lie groups considered in this paper are
those whose Lie algebras admit a diagonalizable derivation with positive
eigenvalues. We start with the special case of Carnot groups.

Let $N$ be a Carnot group equipped with a left invariant Carnot-Caratheodory
metric $d_{CC}$. Let $\hat{N}=N\cup\\{\infty\\}$ be the one-point
compactification of $N$. A homeomorphism $f:\hat{N}\rightarrow\hat{N}$ is
$K$-quasiconformal for some $K\geq 1$ if
$f:(N\backslash\\{f^{-1}(\infty)\\},d_{CC})\rightarrow(N\backslash\\{f(\infty)\\},d_{CC})$
is $K$-quasiconformal. A $1$-quasiconformal map is also called conformal. A
group $G$ of homeomorphisms of $\hat{N}$ is _uniformly quasiconformal_ if
there is some $K\geq 1$ such that every $g\in G$ is $K$-quasiconformal. If
$G^{\prime}$ is a conformal group of $\hat{N}$ and $f$ is a self
quasiconformal map of $\hat{N}$, then $fG^{\prime}f^{-1}$ is a uniform
quasiconformal group of $\hat{N}$. A natural question is when a uniform
quasiconformal group of $\hat{N}$ arises this way.

Let $G$ be a group of homeomorphisms of $\hat{N}$. Then $G$ also acts
diagonally on the space of distinct triples

$T(\hat{N})=\left\\{(x,y,z):x,y,z\in\hat{N},x\not=y,y\not=z,z\not=x\right\\}$

of $\hat{N}$. Our first result is the following.

######  Theorem 1.1.

Let $N$ be a Carnot group. There is a left invariant Carnot-Caratheodory
metric $d_{0}$ on $N$ with the following property. Let $G$ be a uniform
quasiconformal group of $\hat{N}$. If the action of $G$ on the space of
distinct triples of $\hat{N}$ is co-compact, then there is some quasiconformal
map $f:\hat{N}\rightarrow\hat{N}$ such that $fGf^{-1}$ consists of conformal
maps with respect to $d_{0}$.

The metric $d_{0}$ has the largest conformal group in the sense that the
conformal group of any left invariant Carnot-Caratheodory metric is conjugated
into the conformal group of $d_{0}$, see Definition 2.2 and Lemma 2.3. In
general it is not possible to conjugate a uniform quasiconformal group into
the conformal group of an arbitrary left invariant Carnot-Caratheodory metric,
see Section 6 for an example.

Theorem 1.1 was first established by Tukia [T86] for $N=\mathbb{R}^{n}$
($n\geq 2$) and was later generalized by Chow [Ch96] to the case when $N$ is
an Heisenberg group. Before Tukia’s result, D. Sullivan [S78] proved that,
when $n=2$ every uniform quasiconformal group (without the assumption of
cocompactness of the induced action on the space of distinct triples) is
quasiconformally conjugate to a conformal group.

Tukia’s theorem has applications to rigidity of quasi-actions and quasi-
isometric rigidity of finitely generated groups. So does Theorem 1.1. A
_quasi-action_ of a group $\Gamma$ on a metric space $X$ is an assignment
$\gamma\mapsto G_{\gamma}$ where $G_{\gamma}$ is a self quasi-isometry of $X$
such that

  1. (1)

$G_{\gamma}$ is an $(L,A)$ quasi-isometry where $L$ and $A$ are uniform over
all $\gamma\in\Gamma$.

  2. (2)

$G_{\gamma\eta}$ and $G_{\gamma}\circ G_{\eta}$ are bounded distance apart in
the sup norm, uniformly over all $\gamma,\eta\in\Gamma$.

  3. (3)

$G_{Id}$ is bounded distance from the identity map on $X$.

A quasi-action is _cobounded_ if there is a bounded set $S\subset X$ such that
for any $x\in X$ there is $\gamma\in\Gamma$ such that $G_{\gamma}(x)\in S$.

The standard example of a cobounded quasi-action arises when $\Gamma$ is a
group with a left invariant metric (for example, a finitely generated group
with a word metric or a Lie group with a left invariant Riemannian metric) and
$\phi:\Gamma\to X$ is a quasi-isometry with coarse inverse $\bar{\phi}$. Then
$\gamma\mapsto\phi\circ L_{\gamma}\circ\bar{\phi}$ defines a cobounded quasi-
action of $\Gamma$ on $X$, where $L_{\gamma}$ is the left translation of
$\Gamma$ by $\gamma$. We note, however, that there exist “non-proper”
cobounded quasi-actions and so these do not come from a quasi-isometry between
a group and a metric space.

A quasi-action $\\{G_{\gamma}|\gamma\in\Gamma\\}$ of $\Gamma$ on a metric
space $X$ is quasi-conjugate to a quasi-action
$\\{G^{\prime}_{\gamma}|\gamma\in\Gamma\\}$ of $\Gamma$ on another metric
space $X^{\prime}$ if there is a quasi-isometry $f:X\rightarrow X^{\prime}$
and a constant $C>0$ such that
$d(G^{\prime}_{\gamma}(f(x)),f(G_{\gamma}(x)))\leq C$ for all $x\in X$ and all
$\gamma\in\Gamma$.

The relation between quasi-actions and uniform quasiconformal groups is
through negative curvature. A self quasi-isometry of a Gromov hyperbolic space
$X$ induces a self quasiconformal map of the Gromov boundary $\partial X$ of
$X$ (equipped with a visual metric), and a quasi-action of a group $\Gamma$ on
a Gromov hyperbolic space $X$ induces a uniform quasiconformal group action of
$\Gamma$ on $\partial X$. A quasi-conjugate between quasi-actions of $\Gamma$
on two Gromov hyperbolic spaces $X$, $X^{\prime}$ corresponds to a
quasiconformal conjugation between the uniform quasiconformal actions of
$\Gamma$ on $\partial X$ and $\partial X^{\prime}$.

The standard Carnot group dilations of $N$ define an action of $\mathbb{R}$ on
$N$. Let $S=N\rtimes\mathbb{R}$ be the associated solvable Lie group. Then $S$
with any left invariant Riemannian metric is a Gromov hyperbolic space [H74].

######  Corollary 1.2.

Let $N$ be a Carnot group and $S=N\rtimes\mathbb{R}$ be the associated
solvable Lie group. There is a left invariant Riemannian metric $g_{0}$ on $S$
with the following property. Let $G$ be a group that quasi-acts on $S$. If the
quasi-action is co-bounded, then the quasi-action is quasi-conjugate to an
isometric action of $G$ on $(S,g_{0})$.

We next turn to uniform quasiconformal groups on more general nilpotent Lie
groups. Let $N$ be a simply connected nilpotent Lie group with Lie algebra
$\mathfrak{n}$ and $D$ a derivation of $\mathfrak{n}$. We say $(N,D)$ is a
diagonal Heintze pair if $D$ has positive eigenvalues and is diagonalizable
over $\mathbb{R}$. Let $(N,D)$ be a diagonal Heintze pair. A distance $d$ on
$N$ is called $D$-homogeneous if it is left invariant, induces the manifold
topology on $N$ and such that $d(e^{tD}x,e^{tD}y)=e^{t}d(x,y)$ for all $x,y\in
N$ and $t\in\mathbb{R}$, where $\\{e^{tD}|t\in\mathbb{R}\\}$ denotes the
automorphisms of $N$ generated by the derivation $D$. By Theorem 2 of [HSi90],
$D$-homogeneous distances exist on $N$. It is easy to see that any two
$D$-homogeneous distances on $N$ are biLipschitz equivalent. We will always
equip $N$ with a $D$-homogeneous distance. Hence it makes sense to speak of a
biLipschitz map of $N$ without specifying the $D$-homogeneous distance.

Let $(N,D)$ be a diagonal Heintze pair. Then there is a sequence of
$D$-invariant Lie sub-algebras
$\\{0\\}=\mathfrak{n}_{0}\subset\mathfrak{n}_{1}\subset\cdots\subset\mathfrak{n}_{s}=\mathfrak{n}$
with the following properties: each $\mathfrak{n}_{i-1}$ is an ideal of
$\mathfrak{n}_{i}$ with the quotient $\mathfrak{n}_{i}/{\mathfrak{n}_{i-1}}$ a
Carnot Lie algebra; $D$ induces a derivation
$\bar{D}:\mathfrak{n}_{i}/{\mathfrak{n}_{i-1}}\rightarrow\mathfrak{n}_{i}/{\mathfrak{n}_{i-1}}$
which is a multiple of the Carnot derivation of
$\mathfrak{n}_{i}/{\mathfrak{n}_{i-1}}$. See Section 5.1 for more details. Let
$N_{i}$ be the connected Lie subgroup of $N$ with Lie algebra
$\mathfrak{n}_{i}$. Then $N/{N_{i}}$ is a homogeneous manifold and the natural
map $\pi_{i}:N/{N_{i-1}}\rightarrow N/{N_{i}}$ is a fiber bundle with fiber
the Carnot group $N_{i}/{N_{i-1}}$. We call the sequence of subgroups
$0=N_{0}<N_{1}<\cdots<N_{s}=N$ the _preserved subgroup sequence_.

Let $d$ be a $D$-homogeneous distance on $N$. In general $d$ does not induce
any metric on the homogeneous space $N/{N_{i}}$ when $N_{i}$ is not normal in
$N$. Nonetheless, it induces a metric on the fibers $N_{i}/{N_{i-1}}$ of
$\pi_{i}:N/{N_{i-1}}\rightarrow N/{N_{i}}$. Furthermore, every biLipschitz map
$F$ of $N$ permutes the cosets of $N_{i}$ for each $i$. Hence $F$ induces a
map $F_{i}:N/{N_{i}}\rightarrow N/{N_{i}}$ and a bundle map of
$\pi_{i}:N/{N_{i-1}}\rightarrow N/{N_{i}}$. The restriction of $F_{i-1}$ to
the fibers of $\pi_{i}$ are biLipschitz maps of the Carnot group
$N_{i}/{N_{i-1}}$ in the following sense. For each $p\in N$, let
$F_{p}=L_{F(p)^{-1}}\circ F\circ L_{p}$, where $L_{x}$ denotes the left
translation of $N$ by $x$. Notice that the map
$(F_{p})_{i-1}:N/{N_{i-1}}\rightarrow N/{N_{i-1}}$ satisfies
$(F_{p})_{i-1}(N_{i}/{N_{i-1}})=N_{i}/{N_{i-1}}$. The statement above simply
means $(F_{p})_{i-1}|_{N_{i}/{N_{i-1}}}:N_{i}/{N_{i-1}}\rightarrow
N_{i}/{N_{i-1}}$ is biLipschitz with respect to any left invariant Carnot-
Caratheodory metric on $N_{i}/{N_{i-1}}$. See Section 5.1 for more details.

A group $\Gamma$ of homeomorphisms of a metric space $X$ is a _uniform
quasisimilarity group_ if there is a constant $K\geq 1$ such that each
$\gamma\in\Gamma$ satisfies $(C_{\gamma}/K)d(x,y)\leq
d(\gamma(x),\gamma(y))\leq C_{\gamma}Kd(x,y)$ for some $C_{\gamma}>0$ and all
$x,y\in X$. A bijection $f:X\rightarrow X$ is a _similarity_ if there is some
$L>0$ such that $d(f(x),f(y))=Ld(x,y)$ for all $x,y\in X$.

######  Theorem 1.3.

Let $(N,D)$ be a diagonal Heintze pair and $\Gamma$ be a uniform
quasisimilarity group of $(N,D)$ that acts cocompactly on the space of
distinct pairs of $N$ (or equivalently $\Gamma$ a group that quasi-acts
coboundedly on $S=N\rtimes_{D}\mathbb{R}$). Let $I=\\{i|1\leq i\leq
s,\;\dim(N_{i}/{N_{i-1}})\geq 2\\}$. Then there exists a biLipschitz map
$F_{0}:N\to N$ and a left invariant Carnot-Caratheodory metric $d_{i}$ on
$N_{i}/{N_{i-1}}$ for each $i\in I$ such that for each $p\in N$ and each $g\in
F_{0}\Gamma F^{-1}_{0}$, the map
$(g_{p})_{i-1}|_{N_{i}/{N_{i-1}}}:(N_{i}/{N_{i-1}},d_{i})\rightarrow(N_{i}/{N_{i-1}},d_{i})$
is a similarity.

Ideally one would like to conjugate the group $\Gamma$ in Theorem 1.3 into a
group of similarities of $N$ with respect to some $D$-homogeneous distance.
But this question is still open in general. A positive answer was given in
[DFX] in the case when the preserved subgroup sequence has only two terms
$0<N_{1}<N$. Its proof is much more involved algebraically and uses Theorem
1.3 as a crucial ingredient.

When $s\geq 2$, [CP17] implies that every quasiconformal map of
$\hat{N}=N\cup\\{\infty\\}$ fixes $\infty$ and restricts to a biLipschitz map
of $N$. From this it is easy to see that a uniform quasiconformal group of
$\hat{N}$ restricts to a uniform quasisimilarity group of $N$. Therefore there
is no loss of generality in Theorem 1.3 in considering a uniform
quasisimilarity group of $N$ instead of a uniform quasiconformal group of
$\hat{N}$.

When $s=1$, then $N$ is Carnot and depending on $N$ not all quasiconformal
maps of $\hat{N}$ are necessarily biLipschitz. In this case, Theorem 1.3
simply asserts that if the quasiconformal group from Theorem 1.1 happens to
consist of biLipschitz maps then the conjugating map can be chosen to be
biLipschitz.

Theorem 1.3 was proved in [Dy10] in the case when $N$ is a Euclidean group. In
the case $N=\mathbb{R}$, this result can be found in the appendix of [FM99]
and no additional assumptions other than uniformity are needed on the group.

The case $N=\mathbb{R}$ is used for the last step in the proof of quasi-
isometric rigidity of SOL by Eskin-Fisher-Whyte [EFW12], [EFW13], while the
cases covered in [Dy10] are used to prove quasi-isometric rigidity of higher
rank generalizations of SOL by Peng in [P11a], [P11b]. Similarly Theorem 1.3
played a crucial role in the proof of quasi-isometric rigidity of a class of
solvable groups [DFX].

The group $SOL=\mathbb{R}^{2}\rtimes\mathbb{R}$ where the action of
$\mathbb{R}$ scales by $e^{t}$ on the first coordinate and by $e^{-t}$ along
the second. This action gives rise to two foliations by hyperbolic planes
(which we view as $\mathbb{R}\rtimes\mathbb{R}$). More generally a _SOL-like
group_ is a semi-direct product $(N_{1}\times N_{2})\rtimes\mathbb{R}$, where
$N_{i}$ is a simply connected nilpotent Lie group with Lie algebra
$\mathfrak{n}_{i}$, and the action of $\mathbb{R}$ on $N_{1}\times N_{2}$ is
generated by a derivation $D=(-D_{1},D_{2})$ of
$\mathfrak{n}_{1}\times\mathfrak{n}_{2}$ and $D_{i}$ is a derivation of
$\mathfrak{n}_{i}$ whose eigenvalues have positive real part. A SOL-like group
is foliated by two families of negatively curved solvable groups
$N_{i}\rtimes_{D_{i}}\mathbb{R}$ (these are called Heintze groups and they are
exactly those solvable Lie groups admitting left invariant Riemannian metrics
with negative sectional curvature). The quasi-isometric rigidity result proved
in [DFX] are for SOL-like groups where $(N_{i},D_{i})$ is either of Carnot
type or has a preserved subgroup sequence of length two. We remark that the
quasi-isometric rigidity for non-unimodular SOL-like groups where
$(N_{i},D_{i})$ is of Carnot Type is included in Theorem C, [Fe22] and that
proof does not use Tukia type theorems.

Acknowledgments. T. Dymarz was supported by NSF career grant 1552234. D.
Fisher was supported by NSF grants DMS-1906107 and DMS-2246556. X. Xie was
partially supported by Simons Foundation grant #315130. X. Xie would like to
thank the department of mathematics, University of Wisconsin at Madison for
financial support during his visit there in February 2020. We would also like
to thank Tom Ferragut for useful comments on an earlier version.

##  2\. Preliminaries

###  2.1. Carnot groups

Let $N$ be a Carnot group with Lie algebra
$\mathfrak{n}=V_{1}\oplus\cdots\oplus V_{k}$. The exponential map
${\text{exp}}:\mathfrak{n}\rightarrow N$ is a diffeomorphism. We will identify
$N$ and $\mathfrak{n}$ via the exponential map. For any $t>0$, the Carnot
group dilation $\delta_{t}:\mathfrak{n}\rightarrow\mathfrak{n}$ is given by
$\delta_{t}(\sum_{j=1}^{k}x_{j})=\sum_{j=1}^{k}t^{j}x_{j},$ with $x_{j}\in
V_{j}$. They are similarities w.r.t any left invariant Carnot-Caratheodory
metric $d$: $d(\delta_{t}(x),\delta_{t}(y))=t\,d(x,y)$ for any
$x,y\in\mathfrak{n}$. The determinant of
$\delta_{t}:\mathfrak{n}\rightarrow\mathfrak{n}$ is $t^{Q}$, where
$Q=\sum_{j=1}^{k}jm_{j}$ is the homogeneous dimension of $N$.

Let $\\{e_{jl}:1\leq l\leq m_{j}\\}$ be a basis for $V_{j}$, $1\leq j\leq k$.
Let $n=\dim\mathfrak{n}$. Then the map $\mathbb{R}^{n}\rightarrow N$ given by
$(x_{jl})\to\text{exp}(\sum_{j,l}x_{jl}e_{jl})$ is a diffeomorphism and the
push-forward of the Lebesgue measure under this map is a Haar measure on $N$.
We shall use the notation $|A|$ for the Haar measure of a subset $A\subset N$.
Define a function $\rho:\mathfrak{n}\rightarrow[0,\infty)$ by

$\rho\left(\sum_{j,l}x_{jl}e_{jl}\right)=\sum_{j=1}^{k}\sum_{l=1}^{m_{j}}|x_{jl}|^{\frac{1}{j}}.$

Then $\rho(\delta_{t}(x))=t\,\rho(x)$ for all $x\in\mathfrak{n}$. There is an
associated “distance” $d_{\rho}$ given by $d_{\rho}(x,y)=\rho((-x)*y)$. It is
easy to see that $d_{\rho}$ is left invariant and satisfies
$d_{\rho}(\delta_{t}(x),\delta_{t}(y))=t\,d_{\rho}(x,y)$. Hence $d_{\rho}$ is
biLipschitz equivalent with any left invariant Carnot-Caratheodory metric $d$:
there exists some constants $L\geq 1$ such that $d(x,y)/L\leq
d_{\rho}(x,y)\leq Ld(x,y)$ for all $x,y\in N$.

Let $V_{1}$ be equipped with an inner product and we may assume
$\\{e_{1l}:1\leq l\leq m_{1}\\}$ is an orthonormal basis for $V_{1}$. Let
$X_{1l}$ be the left invariant vector field on $N$ determined by $e_{1l}$. For
any smooth function $u:U\rightarrow\mathbb{R}$ defined on an open subset of
$N$, define the horizontal gradient $\triangledown u$ of $u$ by:

$\triangledown u=\sum_{l=1}^{m_{1}}(X_{1l}u)X_{1l}.$

The length of $\triangledown u$ is:

$|\triangledown u|=\sqrt{\sum_{l=1}^{m_{1}}(X_{1l}u)^{2}}.$

Let $d$ be the left invariant Carnot-Caratheodory metric on $N$ determined by
the inner product on $V_{1}$, let $U,V\subset N$ be open subsets, and
$f:U\rightarrow V$ a homeomorphism. For $x\in U$ and $r>0$ with $B(x,r)\subset
U$, let

$L_{f}(x,r)=\sup_{d(y,x)\leq r}d(f(x),f(y)),\;\;\;l_{f}(x,r)=\inf_{d(y,x)\geq
r}d(f(x),f(y)).$

Define

$K(f,x)=\limsup_{r\rightarrow 0}\frac{L_{f}(x,r)}{l_{f}(x,r)}.$

We call $f$ $K$-quasiconformal for some $K\geq 1$ if $K(f,x)\leq K$ for a.e.
$x\in U$.

Let $U,V\subset N$ be open subsets, and $f:U\rightarrow V$ a quasiconformal
map. By [P89], $f$ is Pansu differentiable a.e. and the Pansu differential
$Df(x)$ is a graded automorphism for a.e. $x\in U$. Recall that an
automorphism $A:\mathfrak{n}\rightarrow\mathfrak{n}$ is graded if it commutes
with $\delta_{t}$ for all $t>0$; equivalently $A$ is graded if
$A(V_{j})=V_{j}$ for each $1\leq j\leq k$. Furthermore, $f$ is absolutely
continuous w.r.t. the Haar measure, see [P89]. For any $x\in U$, define

$L_{f}(x)=\limsup_{y\rightarrow
x}\frac{d(f(x),f(y))}{d(x,y)},\;\;\;l_{f}(x)=\liminf_{y\rightarrow
x}\frac{d(f(x),f(y))}{d(x,y)}.$

By Lemma 3.3 in [CC06], for a.e. $x\in U$, we have

$L_{f}(x)=\max\\{Df(x)X:X\in V_{1},|X|=1\\},$ $l_{f}(x)=\min\\{Df(x)(X):X\in
V_{1},|X|=1\\},$

and $K(f,x)=\frac{L_{f}(x)}{l_{f}(x)}$. For any $x\in U$, the volume
derivative of $f$ at $x$ is:

$f^{\prime}(x)=\lim_{r\rightarrow 0}\frac{|f(B(x,r))|}{|B(x,r)|},$

which exists a.e. and is a.e. finite. By (4.1) in [CC06], $l_{f}(x)^{Q}\leq
f^{\prime}(x)\leq L_{f}(x)^{Q}$ for a.e. $x\in U$, where $Q$ is the
homogeneous dimension of $N$.

###  2.2. Homogeneous distances on nilpotent Lie groups

Let $(N,D)$ be a diagonal Heintze pair. Let $0<\lambda_{1}<\cdots<\lambda_{r}$
be the distinct eigenvalues of $D$ and
$\mathfrak{n}=\oplus_{j}V_{\lambda_{j}}$ be the decomposition of
$\mathfrak{n}$ into the direct sum of eigenspaces of $D$. An inner product
$\left<\cdot,\cdot\right>$ on $\mathfrak{n}$ is called a $D$-inner product if
the eigenspaces corresponding to distinct eigenvalues are perpendicular with
respect to $\left<\cdot,\cdot\right>$. By the construction in Theorem 2 of
[HSi90], given any $D$-inner product $\left<\cdot,\cdot\right>$ on
$\mathfrak{n}$, there is a $D$-homogeneous distance $d$ on $N$ such that
$d(0,x)=\langle x,x\rangle^{\frac{1}{2\lambda_{j}}}$ for $x\in
V_{\lambda_{j}}$.

For computational purposes, we also define a function $\rho$ that is
biLipschitz equivalent to a $D$-homogeneous distance $d$. For any $D$-inner
product $\left<\cdot,\cdot\right>$ on $\mathfrak{n}$ define a “norm” on
$\mathfrak{n}$ by

$||v||=\sum_{i}|v_{i}|^{\frac{1}{\lambda_{i}}},$

where $v=\sum_{i}v_{i}$ with $v_{i}\in V_{\lambda_{i}}$. Then define $\rho$ by
$\rho(x,y)=||x^{-1}*y||$. We identify $\mathfrak{n}$ and $N$. Clearly $\rho$
is left invariant, induces the manifold topology and satisfies
$\rho(e^{tD}x,e^{tD}y)=e^{t}\rho(x,y)$ for all $x,y\in\mathfrak{n}$ and
$t\in\mathbb{R}$. It follows that for any $D$-homogeneous distance $d$ on
$\mathfrak{n}$, there is a constant $L\geq 1$ such that
$d(x,y)/L\leq\rho(x,y)\leq L\cdot d(x,y)$ for all $x,y\in\mathfrak{n}$. The
explicit formula for $\rho$ will make the calculations much easier.

######  Lemma 2.1.

Let $\phi$ be an automorphism of $N$. Then $\phi$ is biLipschitz if and only
if $d\phi$ is “layer preserving”; that is,
$d\phi(V_{\lambda_{j}})=V_{\lambda_{j}}$ for each $j$.

###### Proof.

First suppose $\phi$ is biLipschitz. Let $0\not=v\in V_{\lambda_{j}}$ and
write $d\phi(v)=\sum_{i}x_{i}$ with $x_{i}\in V_{\lambda_{i}}$. Then
$d\phi(tv)=\sum_{i}{tx_{i}}$. We have

$\rho(0,tv)=|v|^{\frac{1}{\lambda_{j}}}|t|^{\frac{1}{\lambda_{j}}}$

and

$\rho(0,d\phi(tv))=\sum_{i}|x_{i}|^{\frac{1}{\lambda_{i}}}|t|^{\frac{1}{\lambda_{i}}}.$

The biLipschitz condition implies $x_{i}=0$ when $i\not=j$ by letting
$t\rightarrow\infty$ or $t\rightarrow 0$.

Conversely assume $d\phi$ is layer preserving. Then there is some constant
$C\geq 1$ such that

(1) $|v|/C\leq|d\phi(v)|\leq C|v|$

for all $v\in V_{\lambda_{j}}$, $\forall j$. Now let $v\in\mathfrak{n}$. Write
$v=\sum_{j}v_{j}$ with $v_{j}\in V_{\lambda_{j}}$. Then
$d\phi(v)=\sum_{j}d\phi(v_{j})$. We have
$\rho(0,v)=\sum_{j}|v_{j}|^{\frac{1}{\lambda_{j}}}$ and
$\rho(0,d\phi(v))=\sum_{j}|d\phi(v_{j})|^{\frac{1}{\lambda_{j}}}$. Now the
claim follows from (1).

∎

An automorphism $\phi$ of $N$ is called graded if it satisfies the condition
in Lemma 2.1. We denote by $\text{Aut}_{g}(N)$ the group of graded
automorphisms of $N$.

For any $D$-homogeneous distance $d$ on $N$, let $\text{Sim}(N,d)$ be the
group of similarities of $(N,d)$. By Theorem 1.2 in [KLD17] and Lemma 2.1,
$\text{Sim}(N,d)$ has the structure $\text{Sim}(N,d)=N\rtimes(\mathbb{R}\times
K)$, where $\mathbb{R}$ acts on $N$ by the automorphisms
$\\{e^{tD}|t\in\mathbb{R}\\}$ and $K\subset\text{Aut}_{g}(N)$ is a compact Lie
subgroup. Given two $D$-homogeneous distances $d_{1}$, $d_{2}$ on $N$,
although $(N,d_{1})$ and $(N,d_{2})$ are biLipschitz, a similarity of
$(N,d_{1})$ in general is not a similarity of $(N,d_{2})$ and so their
associated similarity groups can be very different; see Section 6 for an
example in the case of Carnot groups. The following is a notion of
$D$-homogeneous distance with the largest similarity group. It is similar to
Definition 0.2 in [GJ19].

######  Definition 2.2.

Let $(N,D)$ be a diagonal Heintze pair. A $D$-homogeneous distance $d_{0}$ on
$N$ is maximally symmetric (with respect to similarities) if for any
$D$-homogeneous distance $d$ on $N$, there is a biLipschitz automorphism
$\phi$ of $N$ such that
$\phi\text{Sim}(N,d)\phi^{-1}\subset\text{Sim}(N,d_{0})$.

######  Lemma 2.3.

Let $(N,D)$ be a diagonal Heintze pair. Then $N$ admits a maximally symmetric
$D$-homogeneous distance.

###### Proof.

Since $N$ is simply connected, $\text{Aut}_{g}(N)$ can be identified with the
group of graded automorphisms $\text{Aut}_{g}(\mathfrak{n})$ of
$\mathfrak{n}$. It is easy to see that $\text{Aut}_{g}(\mathfrak{n})$ is a
real algebraic variety and so has only a finite number of connected components
by Whitney’s theorem [Wh57]. Therefore $\text{Aut}_{g}(N)$ is a Lie group with
finitely many components.

Let $K_{0}$ be a maximal compact subgroup of $\text{Aut}_{g}(N)$. Recall
$\mathfrak{n}=\oplus_{j}V_{\lambda_{j}}$. For each $j$ let
$\left<\cdot,\cdot\right>_{j}$ be a $K_{0}$-invariant inner product on
$V_{\lambda_{j}}$. Let $\left<\cdot,\cdot\right>$ be the inner product on
$\mathfrak{n}$ that agrees with $\left<\cdot,\cdot\right>_{j}$ on
$V_{\lambda_{j}}$ such that $V_{\lambda_{i}}$ and $V_{\lambda_{j}}$ are
perpendicular to each other for $i\not=j$. Let $d$ be a $D$-homogeneous
distance on $N$ associated to this inner product. Although $d\phi$ is a linear
isometry of $(\mathfrak{n},\langle,\rangle)$ for any $\phi\in K_{0}$, it is
not clear that $\phi$ is an isometry of $(N,d)$. Let $m$ be the normalized
Haar measure on $K_{0}$. Define a new distance $d_{0}$ on $N$ by
$d_{0}(x,y)=\int_{K_{0}}d(k(x),k(y))dm(k)$. Now it is easy to check that
$d_{0}$ is a $K_{0}$-invariant $D$-homogeneous distance on $N$ associated to
$\langle,\rangle$ and $\text{Sim}(N,d_{0})=N\rtimes(\mathbb{R}\times K_{0})$.

Now let $d$ be an arbitrary $D$-homogeneous distance on $N$. As observed
above, $\text{Sim}(N,d)=N\rtimes(\mathbb{R}\times K)$, where $K$ is a compact
subgroup of $\text{Aut}_{g}(N)$. Since $\text{Aut}_{g}(N)$ has only a finite
number of components, there is some $\phi\in\text{Aut}_{g}(N)$ such that $\phi
K\phi^{-1}\subset K_{0}$. Since $N$ is normal in $N\rtimes\text{Aut}(N)$ and
$\phi$ is graded we have
$\phi\text{Sim}(N,d)\phi^{-1}\subset\text{Sim}(N,d_{0})$.

∎

Let $G$ be a connected Lie group with a left invariant distance $d$ that
induces the manifold topology, and $H$ a closed normal subgroup of $G$. We
define a distance on $G/H$ by
$\bar{d}(xH,yH)=\inf\\{d(xh_{1},yh_{2})|h_{1},h_{2}\in H\\}$. Then $\bar{d}$
is a left invariant distance on $G/H$ that induces the manifold topology and
the quotient map $(G,d)\to(G/H,\bar{d})$ is $1$-Lipschitz. Since $H$ is
normal, we have $\bar{d}(xH,yH)=d(xh_{1},yH)=d(yh_{2},xH)=d_{H}(xH,yH)$ for
any $h_{1},h_{2}\in H$, where $d_{H}$ denotes the Hausdorff distance. If $F$
is a biLipschitz map of $(G,d)$ that permutes the cosets of $H$, then $F$
induces a biLipschitz map $\bar{F}:(G/H,\bar{d})\to(G/H,\bar{d})$ with the
same biLipschitz constant as $F$.

Let $(N,D)$ be a diagonal Heintze pair and $d$ a $D$-homogeneous distance on
$N$. Assume $\mathfrak{w}$ is an ideal of $\mathfrak{n}$ such that
$D(\mathfrak{w})\subset\mathfrak{w}$. Then $D$ induces a derivation $\bar{D}$
of $\mathfrak{n}/\mathfrak{w}$ and $(N/W,\bar{D})$ is also a diagonal Heintze
pair, where $W$ is the Lie subgroup of $N$ with Lie algebra $\mathfrak{w}$. In
this case, the distance $\bar{d}$ on $N/W$ induced by $d$ is a
$\bar{D}$-homogeneous distance.

###  2.3. Homogeneous manifolds with negative curvature

Let $N$ be a Carnot group with Lie algebra
$\mathfrak{n}=V_{1}\oplus\cdots\oplus V_{k}$. The standard dilations
$\delta_{t}$ of $N$ define an action of $\mathbb{R}$ on $N=\mathfrak{n}$:

$t\cdot
x=\delta_{e^{t}}(x)\;\;\text{for}\;\;t\in\mathbb{R}\;\;\text{and}\;\;x\in\mathfrak{n}.$

Let $S=N\rtimes\mathbb{R}$ be the associated semi-direct product. Then $S$ is
a solvable Lie group. By [H74] $S$ admits a left invariant Riemannian metric
with negative sectional curvature. For $x_{0}\in N$, the path
$c_{x_{0}}:\mathbb{R}\rightarrow S$, $c_{x_{0}}(t)=(x_{0},t)$, is a geodesic
in $S$. We call $c_{x_{0}}$ a vertical geodesic. All vertical geodesics are
asymptotic as $t\to\infty$, and so they determine a point $\infty$ in the
ideal boundary. If $t\to-\infty$ all vertical geodesics diverge from one
another. We call such geodesics downward oriented. Every geodesic ray in $S$
is asymptotic to either an upward oriented vertical geodesic or a downward
oriented vertical geodesic. It follows that the ideal boundary $\partial S$ of
$S$ can be identified with $\hat{N}=N\cup\\{\infty\\}$, where points in $N$
correspond to downward oriented vertical geodesics.

###  2.4. Sphericalization of metrics and measures

Let $(X,d)$ be an unbounded metric space and $p\in X$ a base point. Let
$\infty$ be a point not in $X$ and set $\hat{X}=X\cup\\{\infty\\}$. The
sphericalized metric $\hat{d}_{p}$ of $d$ relative to the base point $p$ is a
metric on $\hat{X}$ satisfying:

$\frac{d(x,y)}{4(1+d(p,x))(1+d(p,y))}\leq\hat{d}_{p}(x,y)\leq\frac{d(x,y)}{(1+d(p,x))(1+d(p,y))}\;\;\text{for}\;\;x,y\in
X,$

and $\hat{d}_{p}(x,\infty)=\frac{1}{(1+d(p,x))}$. Furthermore, a proof similar
to that of Proposition 4.1 in [BHX08] shows that the identity map
$\text{id}:(X,d)\rightarrow(X,\hat{d}_{p})$ is $1$-quasiconformal. It follows
that a homeomorphism $f:(\hat{X},\hat{d}_{p})\rightarrow(\hat{X},\hat{d}_{p})$
is quasiconformal ($1$-quasiconformal) iff the restriction
$f:(X\backslash\\{f^{-1}(\infty)\\},d)\rightarrow(X\backslash\\{f(\infty)\\},d)$
is quasiconformal ($1$-quasiconformal). We shall use this observation when we
study quasiconformal maps of $\hat{N}$.

Let $(X,d,\nu)$ be a metric measure space, with $\nu$ a Borel regular measure.
The metric measure space $(X,d,\nu)$ is $\alpha$-Alhfors regular for some
$\alpha>0$ if there is some constant $C>0$ such that
$r^{\alpha}/C\leq\nu(B(x,r))\leq Cr^{\alpha}$ for all balls $B(x,r)$ with
radius $0<r\leq\text{diam}(X,d)$. Now let $(X,d,\nu)$ be a $\alpha$-Alhfors
regular metric measure space with $(X,d)$ unbounded and $p\in X$ a base point.
Li-Shanmugalingam [LS15] defined a measure $\hat{\nu}_{p}$ on
$(\hat{X},\hat{d}_{p})$ which is also $\alpha$-regular and showed that
$(X,d,\nu)$ supports a $p$-Poincare inequality if and only if
$(\hat{X},\hat{d}_{p},\hat{\nu}_{p})$ supports a $p$-Poincare inequality (see
Theorem 1.0.1. in [LS15]) . Furthermore, the two measures $\nu$ and
$\hat{\nu}_{p}$ are comparable on any bounded subset of $(X,d)$; that is, for
any bounded subset $A\subset(X,d)$, there is a constant $C\geq 1$ such that
$\nu(E)/C\leq\hat{\nu}_{p}(E)\leq C\nu(E)$ for any $E\subset A$. We shall use
this in the case of a Carnot group $N$ equipped with a left invariant Carnot-
Caratheodory metric and the Lebesgue measure.

##  3\. Lemmas on quasiconformal maps of Carnot groups

In this section we collect some results on quasiconformal maps of Carnot
groups. These will be used in Section 4.2 for the proof of Theorem 1.1.

Let $N$ be a Carnot group equipped with a left invariant Carnot-Caratheodory
metric and $\hat{N}=N\cup\\{\infty\\}$ its one-point compactification. A ring
in $\hat{N}$ is a connected open subset $R\subset\hat{N}$ whose complement has
two connected components. We always work with rings $R$ satisfying
$\infty\notin\bar{R}$. Let $C_{0}$ and $C_{1}$ be the two components of
$\partial R$. An admissible function for $R$ is a $C^{\infty}$ function
$u:N\rightarrow\mathbb{R}$ with $u|_{C_{0}}=0$ and $u|_{C_{1}}=1$. The
conformal capacity of $R$ is:

$C(R)=\inf_{u}\int_{N}|\triangledown u|^{Q}dx,$

where $Q$ is the homogeneous dimension of $N$, $u$ ranges over all admissible
functions for $R$, and the integral is with respect to the Lebesgue measure.
The above infimum remains the same if we enlarge the class of admissible
functions to include all Sobolev functions in $C(\bar{R})\cap W^{1,Q}(R)$, see
Proposition 11 in [KR95], where the proof is valid for all Carnot groups. In
particular, if $f:\hat{N}\rightarrow\hat{N}$ is quasiconformal, $R$ is a ring
with $\infty,f^{-1}(\infty)\notin\bar{R}$ and $u:N\rightarrow\mathbb{R}$ is an
admissible function for $f(R)$ as defined above, then $u\circ f$ is admissible
for $R$ in the generalized sense and so can be used in the estimate for
$C(R)$.

Let $\Gamma$ be a curve family in $N$. A non-negative Borel function
$\rho:N\rightarrow[0,+\infty]$ is an admissible function of $\Gamma$ if
$\int_{\gamma}\rho\,ds\geq 1$ for every locally rectifiable curve $\gamma$ in
$\Gamma$. The conformal modulus of $\Gamma$ is:

$M(\Gamma)=\inf_{\rho}\int_{N}\rho^{Q}dx,$

where $Q$ is the homogeneous dimension of $N$ and $\rho$ varies over all
admissible functions of $\Gamma$.

It is a classical result that for any ring $R$ in $\mathbb{R}^{n}$, the
capacity agrees with the modulus, $C(R)=M(\Gamma(R,C_{0},C_{1}))$, where
$\Gamma(R,C_{0},C_{1})$ is the collection of curves in $R$ joining $C_{0}$ and
$C_{1}$. This result has been generalized to the case of Carnot groups by
Markina [M03].

The following result says that a quasiconformal map preserving conformal
capacity of rings must be a conformal map. The actual assumption is slightly
weaker.

######  Lemma 3.1.

Let $f:\hat{N}\rightarrow\hat{N}$ be a quasiconformal map. Suppose
$C(R)=C(f(R))$ for any ring $R$ in $\hat{N}$ satisfying
$\infty,f^{-1}(\infty)\notin\bar{R}$. Then $f$ is conformal.

###### Proof.

Let $p\in N\backslash\\{f^{-1}(\infty)\\}$ be a point such that $f$ is Pansu
differentiable at $p$ and the Pansu differential $Df(p)$ is a graded
automorphism of $N$. Then the maps $\delta_{\frac{1}{t}}\circ
L_{f(p)^{-1}}\circ f\circ L_{p}\circ\delta_{t}$ converges to $Df(p)$ uniformly
on compact subsets as $t\rightarrow+0$. Since left translations and the
standard Carnot dilations are conformal and so preserve the capacities, all
the maps $\delta_{\frac{1}{t}}\circ L_{f(p)^{-1}}\circ f\circ
L_{p}\circ\delta_{t}$ preserve capacities. Now the continuity of capacities
implies the limiting map $Df(p)$ also preserves the capacity. Hence we may
assume $f$ is a graded automorphism of $N$ that preserves capacity of rings.
By the result of Markina [M03] cited above,

$M(\Gamma(R,C_{0},C_{1}))=M(\Gamma(f(R),f(C_{0}),f(C_{1})))$

for any ring $R$. We need to show that $f$ is a similarity.

The proof below is a modification of the proof of Theorem 36.1 in [V71].

We first construct the rings that we will use. For this we identify $N$ with
its Lie algebra $\mathfrak{n}=V_{1}\oplus\cdots\oplus V_{k}$. We fix an inner
product on $\mathfrak{n}$ so that the $V_{j}$’s are perpendicular to each
other and that on $V_{1}$ it agrees with the inner product on $V_{1}$ that
defines the left invariant Carnot-Caratheodory metric on $N$. Denote
$m_{j}=\dim(V_{j})$. By the singular value decomposition, there exist
orthonormal bases $\\{e_{j1},\cdots,e_{jm_{j}}\\}$ and
$\\{\tilde{e}_{j1},\cdots,\tilde{e}_{jm_{j}}\\}$ of $V_{j}$ and positive
numbers $\lambda_{j1}\geq\cdots\geq\lambda_{jm_{j}}$ such that
$f(e_{jl})=\lambda^{j}_{jl}\tilde{e}_{jl}$.

Let

$C_{0}=\left\\{\sum_{j,l}x_{jl}e_{jl}:x_{11}=0,|x_{1l}|\leq
1\;\;\text{for}\;\;l=2,\cdots,m_{1},\;\;|x_{jl}|\leq 1\;\;\text{for}\;\;j\geq
2,1\leq l\leq m_{j}\right\\}.$

We will construct a rectangular box $Y$ that contains $C_{0}$ in its interior
$\mathring{Y}$ and the ring will be $R=\mathring{Y}\backslash{C_{0}}$.

Claim: For each pair $(j,l)$, $2\leq j\leq k$, $1\leq l\leq m_{j}$, there is a
polynomial $P_{jl}(\delta)$ without constant term satisfying the following
property: for $0<\delta\ll 1$, let

$Y=\left\\{\sum_{j,l}x_{jl}e_{jl}:|x_{11}|\leq\frac{\delta}{\lambda_{11}},|x_{1l}|\leq
1+\frac{\delta}{\lambda_{1l}}\;\;\text{for}\;\;2\leq l\leq
m_{1},\;\;|x_{jl}|\leq 1+P_{jl}(\delta)\;\;\text{for}\;\;j\geq 2,1\leq l\leq
m_{j}\right\\},$

then $d(C_{0},\partial Y)=\frac{\delta}{\lambda_{11}}$ and
$d(f(C_{0}),\partial f(Y))=\delta$.

We will first assume the claim and finish the proof of the lemma, and then
prove the claim. We consider the ring $R=\mathring{Y}\backslash{C_{0}}$. Note
$\partial R$ has two components: $C_{0}$ and $C_{1}:=\partial Y$.

Let $\Gamma_{1}$ and $\Gamma_{2}$ be the families of curves defined by:

$\Gamma_{1}=\left\\{x*(0,\frac{\delta}{\lambda_{11}})e_{11}:\;\;x\in
C_{0}\right\\},$
$\Gamma_{2}=\left\\{x*(-\frac{\delta}{\lambda_{11}},0)e_{11}:\;\;x\in
C_{0}\right\\}.$

The claim $d(C_{0},\partial Y)=\frac{\delta}{\lambda_{11}}$ and the definition
of $Y$ implies $\Gamma_{1},\Gamma_{2}\subset\Gamma:=\Gamma(R,C_{0},C_{1})$.
The curves in $\Gamma_{1}$ and $\Gamma_{2}$ are respectively contained in the
disjoint Borel sets $C_{0}*(0,\frac{\delta}{\lambda_{11}})e_{11}$ and
$C_{0}*(-\frac{\delta}{\lambda_{11}},0)e_{11}$. This implies $M(\Gamma)\geq
M(\Gamma_{1})+M(\Gamma_{2})$, see Theorem 6.7 in [V71]. The standard
calculation in quasiconformal analysis (see 7.2 in [V71]) shows that

$M(\Gamma_{1})=M(\Gamma_{2})=\frac{|C_{0}*[0,\frac{\delta}{\lambda_{11}}]e_{11}|}{(\frac{\delta}{\lambda_{11}})^{Q}}=\frac{\mathfrak{L}^{n-1}(C_{0})\cdot(\frac{\delta}{\lambda_{11}})}{(\frac{\delta}{\lambda_{11}})^{Q}},$

where $\mathfrak{L}^{n-1}(C_{0})$ is the $(n-1)$-dimensional Lebesgue measure
of $C_{0}$ and $n=\dim(\mathfrak{n})$. So we have

(2) $\displaystyle M(\Gamma)\geq
M(\Gamma_{1})+M(\Gamma_{2})=\frac{2\,\mathfrak{L}^{n-1}(C_{0})}{(\frac{\delta}{\lambda_{11}})^{Q-1}}.$

By the claim the minimal distance between the two boundary components
$f(C_{0})$ and $f(C_{1})$ of $f(R)$ is $\delta$. Now by Theorem 7.1 in [V71],
with $\tilde{\Gamma}=\Gamma(f(R),f(C_{0}),f(C_{1}))$ and $J$ the absolute
value of the determinant of $f$,

(3) $\displaystyle
M(\tilde{\Gamma})\leq\frac{|f(R)|}{\delta^{Q}}=\frac{J\cdot|R|}{\delta^{Q}}=\frac{J\cdot\mathfrak{L}^{n-1}(F)\cdot
2\frac{\delta}{\lambda_{11}}}{\delta^{Q}},$

where $F=\\{\sum_{j,l}x_{jl}e_{jl}\in Y:x_{11}=0\\}$.

Since $M(\Gamma)=M(\tilde{\Gamma})$, (2) and (3) imply

$\lambda_{11}^{Q}\leq
J\cdot\frac{\mathfrak{L}^{n-1}(F)}{\mathfrak{L}^{n-1}(C_{0})}.$

Since $\mathfrak{L}^{n-1}(F)/\mathfrak{L}^{n-1}(C_{0})\rightarrow 1$ as
$\delta\rightarrow 0$, we get $\lambda_{11}^{Q}\leq J$. On the other hand, by
Lemma 3.3 in [CC06], $f(B(0,1))\subset B(0,\lambda_{11})$, which implies

$J\cdot|B(0,1)|=|f(B(0,1))|\leq|B(0,\lambda_{11})|=\lambda_{11}^{Q}|B(0,1)|$

and so $J\leq\lambda_{11}^{Q}$. Hence $J=\lambda_{11}^{Q}$. This implies $f$
maps the ball $B(0,1)$ onto the ball $B(0,\lambda_{11})$ and so must be a
similarity.

Proof of the claim: We only write down the proof for $d(f(C_{0}),\partial
f(Y))=\delta$, as the proof for $d(C_{0},\partial
Y)=\frac{\delta}{\lambda_{11}}$ is very similar. Let $P_{jl}$ be a polynomial
without constant term and $Y$ be as defined in the claim. By the formula for
$f$ we have

$f(C_{0})=\left\\{\sum_{j,l}y_{jl}\tilde{e}_{jl}:y_{11}=0,|y_{1l}|\leq\lambda_{1l}\;\;\text{for}\;\;l=2,\cdots,m_{1},\;\;|y_{jl}|\leq\lambda_{jl}\;\;\text{for}\;\;j\geq
2\right\\}$

and

$f(Y)=\left\\{\sum_{j,l}y_{jl}\tilde{e}_{jl}:|y_{11}|\leq\delta,|y_{1l}|\leq\lambda_{1l}+\delta\;\;\text{for}\;\;l=2,\cdots,m_{1},\;\;|y_{jl}|\leq\lambda_{jl}+\lambda_{jl}P_{jl}(\delta)\;\;\text{for}\;\;j\geq
2\right\\};$

First note that $0\in f(C_{0})$, $\delta\tilde{e}_{11}\in\partial f(Y)$ and
$d(0,\delta\tilde{e}_{11})=\delta$. So $d(f(C_{0}),\partial f(Y))\leq\delta$.
It suffices to show for suitable choices of $P_{jl}$ we have
$N_{\delta}(f(C_{0}))\subset f(Y)$.

Let $w\in N_{\delta}(f(C_{0}))$. Then we can write $w=y*z$ with $y\in
f(C_{0})$ and $d(0,z)\leq\delta$. By the BCH formula we have
$w=y+z+\frac{1}{2}[y,z]+\cdots$. Write $w=\sum_{j,l}w_{jl}\tilde{e}_{jl}$,
$y=\sum_{j,l}y_{jl}\tilde{e}_{jl}$, $z=\sum_{j,l}z_{jl}\tilde{e}_{jl}$. As
$d(0,z)\leq\delta$, we have $|z_{jl}|\leq\delta^{j}$. As $y_{11}=0$ and
$|y_{1l}|\leq\lambda_{1l}$ for $l\geq 2$ we have $|w_{11}|\leq\delta$ and
$|w_{1l}|\leq\lambda_{1l}+\delta$ for $l\geq 2$. For $j\geq 2$ we have
$w_{jl}=y_{jl}+Q_{jl}$, where $Q_{jl}$ is a polynomial of $z_{st}$ and
$y_{st}$ with $s\leq j$ such that each monomial in $Q_{jl}$ has at least one
$z_{st}$ as a factor. As $|y_{st}|\leq\lambda_{st}$ and
$|z_{s,t}|\leq\delta^{s}$, we see that there is a polynomial $P_{j,l}$ with no
constant term such that $|Q_{jl}|\leq\lambda_{jl}P_{jl}(\delta)$. This
finishes the proof of the claim.

∎

In Lemma 3.2 and Lemma 3.3: we fix an inner product on $\mathfrak{n}$ so that
the $V_{j}$’s are perpendicular to each other; this determines a left
invariant Carnot-Caratheodory metric $d_{CC}$ on $N$ and a Haar measure on
$N$; the metric on $\hat{N}$ will be the sphericalized metric $\hat{d}_{CC}$
of $d_{CC}$ with respect to the origin and the measure on $\hat{N}$ will be
the sphericalized measure $m$ of the Haar measure with respect to the origin.
See the end of Section 2 for more details.

Lemma 3.2 and Lemma 3.3 hold for quasisymmetric maps on Alhfors regular metric
measure spaces that satisfies a Poincare inequality: their proofs use only the
reverse Holder inequality which holds in such generality, see Theorem 7.11 of
[HK98].

######  Lemma 3.2.

Let $f:(\hat{N},\hat{d}_{CC})\rightarrow(\hat{N},\hat{d}_{CC})$ be an
$\eta$-quasisymmetric map. Then there exist constants $a,b>0$ depending only
on $N$ and $\eta$ that satisfy the following. For any ball
$B_{1}\subset\hat{N}$ with center $x$, let $B_{2}$ be the largest ball in
$f(B_{1})$ with center $f(x)$. Then for any measurable $E\subset\hat{N}$,

$\frac{m(f(E)\cap B_{2})}{m(B_{2})}\leq a\left(\frac{m(E\cap
B_{1})}{m(B_{1})}\right)^{b}.$

###### Proof.

By [J86], $N$ with the Carnot-Caratheodory metric and the Lebesgue measure
supports a $1$-Poincare inequality. Then Theorem 1.0.1. of [LS15] implies
$\hat{N}$ with the sphericalized metric and sphericalized measure also
supports a $1$-Poincare inequality. By Theorem 7.11 in [HK98], there is some
$\epsilon>0$ and a constant $C$ depending only on $N$ and $\eta$ such that

$\left(\frac{1}{m(B)}\int_{B}\mu_{f}^{1+\epsilon}\;dm\right)^{\frac{1}{1+\epsilon}}\leq
C\left(\frac{1}{m(B)}\int_{B}\mu_{f}\;dm\right)$

for all balls $B$ in $\hat{N}$. Here $\mu_{f}$ is the volume derivative of $f$
defined by:

$\mu_{f}(x)=\lim_{r\rightarrow 0}\frac{m(f(B(x,r)))}{m(B(x,r))},$

which exists and is finite for $m$-a.e. $x\in\hat{N}$. Now for a measurable
$E$,

$\displaystyle m(f(E)\cap B_{2})$ $\displaystyle=\int_{E\cap
f^{-1}(B_{2})}\mu_{f}\;dm$ $\displaystyle\leq\int_{E\cap B_{1}}\mu_{f}\;dm$
$\displaystyle=\int_{B_{1}}\mu_{f}\cdot\mathcal{X}_{E\cap B_{1}}\;dm$
$\displaystyle\leq\left(\int_{B_{1}}\mu_{f}^{1+\epsilon}\;dm\right)^{\frac{1}{1+\epsilon}}m(E\cap
B_{1})^{\frac{\epsilon}{1+\epsilon}}$ $\displaystyle\leq
Cm(B_{1})^{-\frac{\epsilon}{1+\epsilon}}\left(\int_{B_{1}}\mu_{f}dm\right)m(E\cap
B_{1})^{\frac{\epsilon}{1+\epsilon}}$
$\displaystyle=Cm(B_{1})^{-\frac{\epsilon}{1+\epsilon}}m(f(B_{1}))m(E\cap
B_{1})^{\frac{\epsilon}{1+\epsilon}}.$

Hence,

$\frac{m(f(E)\cap B_{2})}{m(f(B_{1}))}\leq C\left(\frac{m(E\cap
B_{1})}{m(B_{1})}\right)^{\frac{\epsilon}{1+\epsilon}}.$

Since $f$ is $\eta$-quasisymmetric, we have $f(B_{1})\subset\eta(1)B_{2}$,
where $\eta(1)B_{2}$ is the ball in $(\hat{N},\hat{d}_{CC})$ with the same
center as $B_{2}$ and with radius $\eta(1)$ times that of $B_{2}$. Now the
Alhfors regularity of $\hat{N}$ implies $m(f(B_{1}))\leq C_{1}m(B_{2})$ for
some constant $C_{1}\geq 1$ depending only on $\eta$ and the constant in
Alhfors regularity condition. The lemma follows.

∎

######  Lemma 3.3.

Let $\mathcal{F}$ be a compact family of $K$-quasiconformal maps from
$\hat{N}$ to $\hat{N}$. Then there exist positive constants $b^{\prime}$, $b$,
$a^{\prime}$, $a$ depending only on $\mathcal{F}$ and $N$ such that

$a^{\prime}(m(E))^{b^{\prime}}\leq m(f(E))\leq a(m(E))^{b}$

for all measurable $E\subset\hat{N}$ and all $f\in\mathcal{F}$.

###### Proof.

Since $\mathcal{F}$ is a compact family of $K$-quasiconformal maps, there
exists some homeomorphism $\eta:[0,\infty)\rightarrow[0,\infty)$ such that
every $f\in\mathcal{F}$ is $\eta$-quasisymmetric. Let $r_{0}=1/10$. For any
$x\in\hat{N}$ and any $f\in\mathcal{F}$, let
$r^{\prime\prime}=r^{\prime\prime}(x,f)>0$ be the largest number such that
$B(f(x),r^{\prime\prime})\subset f(B(x,r_{0}))$. Let
$r^{\prime}=r^{\prime}(x,f)>0$ be the largest number such that
$f(B(x,r^{\prime}))\subset B(f(x),r^{\prime\prime})$. By Lemma 3.2 the
following holds for any measurable $E\subset B(x,r^{\prime})$:

$\frac{m(f(E))}{m(B(f(x),r^{\prime\prime}))}\leq
a\left(\frac{m(E)}{m(B(x,r_{0}))}\right)^{b}.$

Note that $r^{\prime\prime}$ and $r^{\prime}$ are continuous functions of
$x,f$. Since $\hat{N}$ and $\mathcal{F}$ are compact, we have

$r^{\prime}_{0}=\inf_{x\in\hat{N},f\in\mathcal{F}}r^{\prime}(x,f)>0.$

Also set

$r^{\prime\prime}_{0}=\sup_{x\in\hat{N},f\in\mathcal{F}}r^{\prime\prime}(x,f)\leq{\text{diam}}(\hat{N}).$

Now the space $\hat{N}$ can be covered by a finite number of balls
$\\{B(x_{j},r^{\prime}_{0})\\}_{j=1}^{k}$ with radius $r^{\prime}_{0}$. Let
$E\subset\hat{N}$ be any measurable subset. Then $f(E)=\cup_{j=1}^{k}f(E\cap
B(x_{j},r^{\prime}_{0}))$. By using the Alhfors regularity of the metric
measure space $(\hat{N},\hat{d},m)$ we obtain:

$\displaystyle m(f(E))$ $\displaystyle\leq\sum_{j=1}^{k}a\left(\frac{m(E\cap
B(x_{j},r^{\prime}_{0}))}{m(B(x_{j},r_{0}))}\right)^{b}m(B(f(x_{j}),r^{\prime\prime}_{0}))$
$\displaystyle\leq\sum_{j=1}^{k}a\frac{C{r^{\prime\prime}_{0}}^{Q}}{(r_{0}^{Q}/C)^{b}}m(E\cap
B(x_{j},r^{\prime}_{0}))^{b}$
$\displaystyle\leq\frac{kaC{r^{\prime\prime}_{0}}^{Q}}{(r_{0}^{Q}/C)^{b}}m(E)^{b}.$

This establishes the second inequality in the lemma. The first inequality
holds since the family $\mathcal{F}^{-1}=\\{f^{-1}|f\in\mathcal{F}\\}$ is also
a compact family of $K$-quasiconformal maps.

∎

In the next lemma we return to working with Carnot-Caratheodory metric and
Lebesgue measure on $N$.

######  Lemma 3.4.

Let $\\{f_{j}:\hat{N}\rightarrow\hat{N}\\}$ be a sequence of
$K$-quasiconformal maps that converge uniformly to a quasiconformal map
$f:\hat{N}\rightarrow\hat{N}$. Suppose for any compact subset
$F\subset\hat{N}$ satisfying $\infty,f^{-1}(\infty)\notin F$, and any
$\epsilon>0$, $|\\{x\in F:K(f_{j},x)\geq 1+\epsilon\\}|\rightarrow 0$ as
$j\rightarrow\infty$. Then $f$ is conformal.

###### Proof.

We shall show that $f$ satisfies the assumption of Lemma 3.1. Let $R$ be a
ring in $\hat{N}$ satisfying $\infty,f^{-1}(\infty)\notin\bar{R}$. Let $C_{0}$
and $C_{1}$ be the two components of $\partial R$. For $a>0$, let

$\tilde{R}_{a}=\\{z\in f(R):B(z,a)\subset f(R)\\}.$

Then $f_{j}(R)\supset\tilde{R}_{a}$ for $j$ sufficiently large (depending on
$a$).

Let $\epsilon>0$. Let $\tilde{u}_{a}$ be admissible for $\tilde{R}_{a}$ so
that

$\int_{\tilde{R}_{a}}|\triangledown\tilde{u}_{a}|^{Q}\;dx\leq
C(\tilde{R}_{a})+\epsilon.$

One can assume $\tilde{u}_{a}$ is smooth and is $0$ on a neighborhood of
$C_{0}$ and $1$ on a neighborhood of $C_{1}$. Extend $\tilde{u}_{a}$ to
$\hat{N}$ by continuity and being constant on the two components of
$\hat{N}\backslash\tilde{R}_{a}$. Then clearly

$\int_{\tilde{R}_{a}}|\triangledown\tilde{u}_{a}|^{Q}\;dx=\int_{\tilde{R}}|\triangledown\tilde{u}_{a}|^{Q}\;dx$

for $\tilde{R}\supset\tilde{R}_{a}$.

Let $u_{j}=\tilde{u}_{a}\circ f_{j}$. Then $u_{j}$ is admissible for
$f_{j}^{-1}(\tilde{R}_{a})$ and hence also admissible for $R$. The chain rule
(see Lemma 3.7 in [CC06]) implies $Du_{j}(x)=D\tilde{u}_{a}(f_{j}(x))\circ
Df_{j}(x)$ for a.e. $x\in R$. Hence
$Du_{j}(x)|_{V_{1}}=D\tilde{u}_{a}(f_{j}(x))|_{V_{1}}\circ Df_{j}(x)|_{V_{1}}$
and

$|Du_{j}(x)|_{V_{1}}|\leq|D\tilde{u}_{a}(f_{j}(x))|_{V_{1}}|\cdot|Df_{j}(x)|_{V_{1}}|.$

Notice that $|Du_{j}(x)|_{V_{1}}|=|\triangledown u_{j}(x)|$,
$|D\tilde{u}_{a}(f_{j}(x))|_{V_{1}}|=|\triangledown\tilde{u}_{a}(f_{j}(x))|$
and

$|Df_{j}(x)|_{V_{1}}|^{Q}=L^{Q}_{f_{j}}(x)=l^{Q}_{f_{j}}(x)K^{Q}(f_{j},x)\leq
f^{\prime}_{j}(x)K^{Q}(f_{j},x)$

due to Lemma 3.3 and (4.1) in [CC06]. It follows that

$|\triangledown u_{j}(x)|^{Q}\leq
K^{Q}(f_{j},x){f^{\prime}_{j}(x)}|\triangledown\tilde{u}_{a}(f_{j}(x))|^{Q}.$

Set $E_{j}=\\{x\in R:K(f_{j},x)\geq 1+\epsilon\\}$. The assumption that
$\infty,f^{-1}(\infty)\notin\bar{R}$ implies that there are compact subsets
$K_{1},K_{2}\subset N$ with $\bar{R}\subset\mathring{K}_{1}$ and
$\overline{f(R)}\subset\mathring{K}_{2}$. There is a constant $C\geq 1$ such
that $m(E)/C\leq|E|\leq C\,m(E)$ for any $E\subset K_{1}$ or $E\subset K_{2}$.
By assumption $|E_{j}|\rightarrow 0$ as $j\rightarrow+\infty$. Now Lemma 3.3
implies $|f_{j}(E_{j})|\rightarrow 0$.

We have

$\displaystyle C(R)$ $\displaystyle\leq\int_{R}|\triangledown u_{j}|^{Q}\;dx$
$\displaystyle\leq\int_{R}K^{Q}(f_{j},x){f^{\prime}_{j}}(x)|\triangledown\tilde{u}_{a}(f_{j}(x))|^{Q}\;dx$
$\displaystyle\leq(1+\epsilon)^{Q}\int_{f_{j}(R)}|\triangledown\tilde{u}_{a}|^{Q}\;dx+K^{Q}\int_{f_{j}(E_{j})}|\triangledown\tilde{u}_{a}|^{Q}\;dx.$

Since $|f_{j}(E_{j})|\rightarrow 0$ as $j\rightarrow+\infty$, the second term
above goes to $0$ as $j\rightarrow\infty$. Hence

$C(R)\leq(1+\epsilon)^{Q}(C(\tilde{R}_{a})+\epsilon),$

and so $C(R)\leq C(\tilde{R}_{a})$ for small $a$. We claim that
$C(\tilde{R}_{a})\rightarrow C(f(R))$ as $a\rightarrow 0$. Indeed, for any
$\delta>0$ there is a smooth function $v$, admissible for $f(R)$ with value 0
in a neighborhood of $f(C_{0})$ and value 1 in a neighborhood of $f(C_{1})$,
such that

$\int_{f(R)}|\triangledown v|^{Q}\;dx\leq C(f(R))+\delta.$

For small enough $a$, $v$ is also admissible for $\tilde{R}_{a}$ and

$\int_{f(R)}|\triangledown v|^{Q}\;dx=\int_{\tilde{R}_{a}}|\triangledown
v|^{Q}\;dx.$

Since

$\int_{\tilde{R}_{a}}|\triangledown v|^{Q}\;dx\geq C(\tilde{R}_{a})\geq
C(f(R))\geq\int_{f(R)}|\triangledown v|^{Q}\;dx-\delta$

and $\delta$ is arbitrary, we have $C(\tilde{R}_{a})\rightarrow C(f(R))$ as
$a\rightarrow 0$. Hence $C(R)\leq C(f(R))$.

Finally as $K(f_{j}^{-1},f_{j}(x))=K(f_{j},x)$, Lemma 3.3 implies that
$f_{j}^{-1}$, $f^{-1}$ also satisfy the assumption in the lemma and we
conclude that $C(f(R))\leq C(R)$. Now we have verified the assumption of Lemma
3.1 and so $f$ is conformal.

∎

##  4\. Tukia-type theorem for Carnot groups

In this section we prove Theorem 1.1 and Corollary 1.2.

###  4.1. Existence of invariant measurable conformal structure

In this subsection we show that every uniform quasiconformal group of
$\hat{N}$ leaves invariant a measurable conformal structure on $\hat{N}$.

Let $m\geq 2$ be an integer and $X$ the space of symmetric, positive definite
real $m\times m$ matrices with determinant $1$. The general linear group
$GL(m,\mathbb{R})$ acts on $X$ by

$M[S]=(\det M)^{-\frac{2}{m}}MSM^{T}$

for $M\in GL(m,\mathbb{R})$ and $S\in X$, where $M^{T}$ is the transpose of
$M$. Then $SL(m,\mathbb{R})\subset GL(m,\mathbb{R})$ acts on $X$ transitively,
and the stabilizer of $SL(m,\mathbb{R})$ at $I_{m}$ is $SO(m)$. Hence
$X=SL(m,\mathbb{R})/SO(m)$. The homogeneous manifold
$X=SL(m,\mathbb{R})/SO(m)$ is a symmetric space of non-compact type (see table
V on page 518 of [Hel78]) and so has nonpositive sectional curvature.

The Riemannian distance $d$ on $X$ is invariant under the action of
$GL(m,\mathbb{R})$ and satisfies:

$d(I_{m},ODO^{T})=\sqrt{a_{1}^{2}+\cdots+a_{m}^{2}},$

where $I_{m}$ is the identity matrix, $O$ is orthogonal and $D$ is diagonal
with diagonal entries $e^{a_{1}}\geq\cdots\geq e^{a_{m}}$, see [Ma71], p.27.
We notice that the dilatation of a non-singular linear map
$A:=UDV^{T}:\mathbb{R}^{m}\rightarrow\mathbb{R}^{m}$ (where $D$ is as above
and $U,V$ are orthogonal) is given by:

$K(A):=K(A,0)=\frac{\max\\{|AX|:|X|=1\\}}{\min\\{|AX|:|X|=1\\}}=\frac{e^{a_{1}}}{e^{a_{m}}}=e^{a_{1}-a_{m}}.$

It follows that there is a function $\phi:[0,\infty)\rightarrow[0,\infty)$
(one may choose $\phi(t)=e^{t}-1$) with $\phi(t)\rightarrow 0$ as
$t\rightarrow 0$ such that

(4) $K(A)\leq 1+\phi(d(I_{m},AA^{T})).$

Let $N$ be a Carnot group with Lie algebra
$\mathfrak{n}=V_{1}\oplus\cdots\oplus V_{r}$. Let $V_{1}$ be equipped with an
inner product. Let $m=\dim(V_{1})$. By using an orthonormal basis for $V_{1}$,
we can identify the special orthogonal group $SO(V_{1})$ associated with the
inner product with $SO(m,\mathbb{R})$ and similarly identify $SL(V_{1})$ with
$SL(m,\mathbb{R})$. Hence

$X=SL(V_{1})/O(V_{1}).$

A measurable conformal structure on $\hat{N}=N\cup\\{\infty\\}$ is an
essentially bounded measurable map

$\mu:U\rightarrow X$

defined on a full measure subset $U\subset N$. Note we do not require $\mu$ to
be defined everywhere on $\hat{N}$. In particular, we prefer $\mu$ not defined
at $\infty$.

For a quasiconformal map $f:\hat{N}\rightarrow\hat{N}$ and a measurable
conformal structure $\mu$ on $\hat{N}$, the pull-back measurable conformal
structure $f^{*}\mu$ is defined by:

$f^{*}\mu(x)=(Df(x)|_{V_{1}})^{T}[\mu(f(x))],\;\;\;\text{for a.e.}\;x\in N.$

Here we are using the fact that a quasiconformal map is Pansu differentiable
a.e. This is similar to the definition of pull-back of a Riemannian metric
under a smooth map. It is easy to check that $(g\circ
f)^{*}\mu=f^{*}(g^{*}\mu)$, where $f,g:\hat{N}\rightarrow\hat{N}$ are
quasiconformal maps and $\mu$ is a measurable conformal structure on
$\hat{N}$.

A quasiconformal map $f$ is called conformal with respect to the measurable
conformal structure $\mu$ if $f^{*}\mu=\mu$; in other words,
$\mu(x)=(Df(x)|_{V_{1}})^{T}[\mu(f(x))]$ for a.e. $x\in N$.

######  Proposition 4.1.

Let $G$ be a uniform quasiconformal group of $\hat{N}$. Then there is a
measurable conformal structure on $\hat{N}$ such that every $g\in G$ is
conformal with respect to $\mu$.

###### Proof.

We first assume $G$ is countable. Since $G$ is countable and for each $g\in
G$, the Pansu differential $Dg(x)$ exists and is a graded automorphism for
a.e. $x\in N$, there is a measurable, $G$-invariant subset $U$ of full measure
such that $Dg(x)$ exists and is a graded automorphism for all $g\in G$ and at
all $x\in U$. Let $\mu_{0}$ be the left invariant conformal structure on $N$
associated with an inner product on $V_{1}$ (so $\mu_{0}:N\rightarrow X$ is a
constant map). For each $x\in U$, set $M_{x}=\\{g^{*}\mu_{0}(x):g\in G$}.
Since $G$ is a uniform quasiconformal group, $M_{x}$ is a bounded subset of
$X$. The assignment $x\mapsto M_{x}$ is $G$-invariant: for any $f\in G$,

$f^{*}M_{f(x)}=f^{*}\\{g^{*}\mu_{0}(f(x)):g\in G\\}=\\{{(g\circ
f)}^{*}\mu_{0}(x):g\in G\\}=M_{x}.$

Since $M_{x}$ is a bounded subset of the non-positively curved symmetric space
$X$, there exists a unique circumcenter $P(M_{x})$ in $X$ for the subset
$M_{x}$, see for example p.179 of [BH]. Define $\mu:U\rightarrow X$ by
$\mu(x)=P(M_{x})$. Since the assignment $x\mapsto M_{x}$ is $G$-invariant,
$\mu$ is also $G$-invariant.

It remains to show $\mu$ is measurable. Enumerate
$G=\\{g_{0},g_{1},\cdots\\}$, and let $M(x,j)=\\{{g_{i}^{*}\mu_{0}}:i\leq
j\\}$ and $\mu_{j}(x)=P(M(x,j))$. Now, $Y\mapsto P(Y)$ is continuous with
respect to the Hausdorff metric (see p.334 of [T86]). This implies $\mu_{j}$
is measurable. Since $\mu_{j}$ converges to $\mu$ point-wise, $\mu$ is also
measurable. This completes the proof when $G$ is countable.

Now let $G$ be a general uniform quasiconformal group of $\hat{N}$. Let
$G_{0}\subset G$ be a countable subgroup of $G$ that is dense in the topology
of uniform convergence. By the above argument, there is a $G_{0}$-invariant
measurable conformal structure $\mu$. It follows that $\mu$ is also
$G$-invariant as a uniform limit of a sequence of $\mu$-conformal maps is
$\mu$-conformal. This follows from an analogue of Theorem D in [T86] in the
setting of Carnot groups. The proof of Theorem D in [T86] is valid for Carnot
groups after some small modifications: to blow up maps in the Carnot group
setting one needs to conjugate using Carnot dilations.

∎

Let $G_{0}$ be a countable dense subgroup of $G$ as above. If the induced
action of $G$ on the space of distinct triples of $\hat{N}$ is cocompact, then
the same is true for the induced action of $G_{0}$. Another way to handle the
case of an uncountable group $G$ is to first run the argument from the next
subsection to conjugate $G_{0}$ into the conformal group, then the same map
also conjugates $G$ into the conformal group since the limits of conformal
maps are conformal maps.

###  4.2. Conjugating into the conformal group

In this subsection we will prove Theorem 1.1 and Corollary 1.2.

A map $\mu:U\rightarrow Y$, where $U\subset N$ is open and $Y$ is a metric
space, is called approximately continuous at $x_{0}\in U$ if for any
$\epsilon>0$, the set $\mu^{-1}(B(\mu(x_{0}),\epsilon))$ has density 1 at
$x_{0}$; that is, if

$\frac{|B(x_{0},r)\cap\mu^{-1}(B(\mu(x_{0}),\epsilon))|}{|B(x_{0},r)|}\rightarrow
1$

as $r\rightarrow 0$. By Theorem 2.9.13 in [F69], if $Y$ is separable and $\mu$
is measurable, then $\mu$ is approximately continuous a.e.

Let $N$ be a Carnot group and $S=N\rtimes\mathbb{R}$ the negatively curved
homogeneous manifold associated with $N$. There is a natural map
$\pi:T(\partial S)\rightarrow S$, that assigns to each distinct triple
$(\xi_{1},\xi_{2},\xi_{3})\in T(\partial S)$ the center of the triple. To be
more precise, $\pi(\xi_{1},\xi_{2},\xi_{3})$ is defined to be the orthogonal
projection of $\xi_{3}$ onto the complete geodesic $\xi_{1}\xi_{2}$. We
observe that for any compact $C\subset S$, the set $\pi^{-1}(C)$ is compact in
$T(\partial S)$.

Let $G$ be a group of homeomorphisms of $\partial S$. Then $G$ also acts
diagonally on $T(\partial S)$:
$g(\xi_{1},\xi_{2},\xi_{3})=(g(\xi_{1}),g(\xi_{2}),g(\xi_{3}))$. Recall that a
point $\xi\in\partial S$ is said to be a radial limit point of $G$ if there
exists a sequence of elements $\\{h_{j}\\}_{j=1}^{\infty}$ of $G$ with the
following property: for any triple $T=(\xi_{1},\xi_{2},\xi_{3})\in T(\partial
S)$, and any complete geodesic $\gamma$ asymptotic to $\xi$, there exists a
constant $C>0$ with $\pi(h_{j}(T))\rightarrow\xi$ and
$d(\pi(h_{j}(T)),\gamma)\leq C$.

We recall that each inner product on $V_{1}$ determines a left invariant
Carnot-Caratheodory metric on $N$.

######  Theorem 4.2.

Let $G$ be a uniform quasiconformal group of $\hat{N}$ and $\mu$ a
$G$-invariant measurable conformal structure on $\hat{N}$. Suppose there is a
point $p\in N\subset\hat{N}$ such that $\mu$ is approximately continuous at
$p$ and $p$ is also a radial limit point for $G$. Then there exists a
quasiconformal map $f:\hat{N}\rightarrow\hat{N}$ and an inner product on
$V_{1}$ such that $fGf^{-1}$ consists of conformal maps with respect to the
left invariant Carnot-Caratheodory metric determined by this inner product.

###### Proof.

The left translation action of $S=N\rtimes\mathbb{R}$ on itself is by
isometries. The elements in $\mathbb{R}$ translate the vertical geodesic above
the origin $0$ in $N$ and the boundary homeomorphisms induced by them are the
standard Carnot group dilations of $N$. We shall use $\tilde{\delta}_{t}$ to
denote the isometry of $S$ that induces the standard Carnot group dilation
$\delta_{t}$.

We may assume $p=0$ is the origin of $N$. Fix a triple $T\in T(\partial S)$
and let $\gamma$ be the vertical geodesic above $0$. Since $0$ is a radial
limit point of $G$, there exists a sequence of elements
$\\{h_{j}\\}_{j=1}^{\infty}$ of $G$ and a constant $C>0$ with
$\pi(h_{j}(T))\rightarrow 0$ and $d(\pi(h_{j}(T)),\gamma)\leq C$. Fix a point
$x_{0}\in\gamma$. For each $j$ there is some $s_{j}>0$ with
$s_{j}\rightarrow+\infty$ as $j\rightarrow\infty$ such that
$d(\tilde{\delta}_{s_{j}}\circ\pi\circ h_{j}(T),x_{0})\leq C$. Since
$\tilde{\delta}_{s_{j}}$ is an isometry of $S$, we have
$\tilde{\delta}_{s_{j}}\circ\pi=\pi\circ\delta_{s_{j}}$. Hence
$d(\pi\circ\delta_{s_{j}}\circ h_{j}(T),x_{0})\leq C$ and so the set
$\\{\delta_{s_{j}}\circ h_{j}(T)\\}_{j=1}^{\infty}$ lies in the compact subset
$\pi^{-1}\bar{B}(x_{0},C)$. It follows that the family $\\{\delta_{s_{j}}\circ
h_{j}\\}_{j=1}^{\infty}$ of $K$-quasiconformal maps is precompact. Define
$f_{j}:\hat{N}\rightarrow\hat{N}$ by $f_{j}=\delta_{s_{j}}\circ h_{j}$. By
passing to a subsequence, we may assume $f_{j}$ converges uniformly to a
$K$-quasiconformal map $f:\hat{N}\rightarrow\hat{N}$. We shall show that for
any $g\in G$, $fgf^{-1}$ is conformal with respect to the left invariant
Carnot-Caratheodory metric determined by $\mu(0)$.

Let $g\in G$ and denote $\tilde{g}=fgf^{-1}$,
$\tilde{g}_{j}=f_{j}gf_{j}^{-1}$. Let $\mu_{j}=(f_{j}^{-1})^{*}\mu$. Since
$\mu$ is $G$-invariant, $\tilde{g}_{j}$ is conformal with respect to
$\mu_{j}$:

$\tilde{g}_{j}^{*}\mu_{j}=(f_{j}^{-1})^{*}g^{*}f_{j}^{*}(f_{j}^{-1})^{*}\mu=(f_{j}^{-1})^{*}g^{*}\mu=(f_{j}^{-1})^{*}\mu=\mu_{j}.$

Note
$\mu_{j}=(f_{j}^{-1})^{*}\mu=(\delta_{s_{j}}^{-1})^{*}(h_{j}^{-1})^{*}\mu=(\delta_{s_{j}}^{-1})^{*}\mu$.
So for $q\in N$,

$\mu_{j}(q)=(\delta_{s_{j}}^{-1})^{*}\mu(q)=(D\delta_{s_{j}}^{-1}(q)|_{V_{1}})^{T}[\mu(\delta_{s_{j}}^{-1}(q))]=\mu(\delta_{s_{j}}^{-1}(q))$

since $D\delta_{t}(q)|_{V_{1}}$ is an Euclidean dilation.

Let $F\subset\hat{N}$ be a compact subset such that
$\infty,\tilde{g}^{-1}(\infty)\notin F$. There is a compact subset $F_{0}$ of
$N$ such that $F\subset F_{0}$, $\tilde{g}_{j}(F)\subset F_{0}$ for all
sufficiently large $j$. Since $\mu$ is approximately continuous at $0$ and
$s_{j}\rightarrow\infty$, the equality
$\mu_{j}(q)=\mu(\delta^{-1}_{s_{j}}(q))$ implies that for any $\epsilon>0$
there are subsets $A_{j}\subset F_{0}$ with $|A_{j}|\rightarrow 0$ as
$j\rightarrow\infty$ and $d(\mu_{j}(x),\mu(0))\leq\epsilon$ for $x\in
F_{0}\backslash A_{j}$.

The maps $\tilde{g}^{-1}_{j}$ and $\tilde{g}^{-1}$ form a compact family of
$K$-quasiconformal maps. By Lemma 3.3, we have
$|\tilde{g}_{j}^{-1}(A_{j})|\rightarrow 0$ as $j\rightarrow\infty$. Set
$B_{j}=A_{j}\cup\tilde{g}_{j}^{-1}(A_{j})$. Now we have $|B_{j}|\rightarrow 0$
as $j\rightarrow\infty$ and $d(\mu_{j}(x),\mu(0))\leq\epsilon$ and
$d(\mu_{j}(\tilde{g}_{j}(x)),\mu(0))\leq\epsilon$ for $x\in F\backslash
B_{j}$. Since $\tilde{g}_{j}$ is $\mu_{j}$-conformal, we have
$\mu_{j}(x)=(D\tilde{g}_{j}(x)|_{V_{1}})^{T}[\mu_{j}(\tilde{g}_{j}(x))]$ for
a.e. $x$. Now

$d(\mu_{j}(x),(D\tilde{g}_{j}(x)|_{V_{1}})^{T}[\mu(0)])=d(\mu_{j}(\tilde{g}_{j}(x)),\mu(0))\leq\epsilon$

for a.e. $x\in F\backslash B_{j}$. Combining this with
$d(\mu_{j}(x),\mu(0))\leq\epsilon$ we get
$d(\mu(0),(D\tilde{g}_{j}(x)|_{V_{1}})^{T}[\mu(0)])\leq 2\epsilon$ for a.e.
$x\in F\backslash B_{j}$. By Lemma 3.3 in [CC06] and (4) we have
$K(\tilde{g}_{j},x)=K(D\tilde{g}_{j}(x)|_{V_{1}})\leq 1+\phi(2\epsilon)$ for
a.e. $x\in F\backslash B_{j}$. This implies the assumption of Lemma 3.4 is
satisfied and so $\tilde{g}$ is conformal.

∎

Proof of Theorem 1.1. Let $G_{0}$ be a countable dense subgroup of $G$. By
Proposition 4.1, there exists a $G_{0}$-invariant measurable conformal
structure $\mu$ on $\hat{N}$. Since the action of $G_{0}$ on $T(\partial S)$
is co-compact, every point in $\partial S$ is a radial limit point. By Theorem
2.9.13 in [F69], $\mu$ is approximately continuous at a.e. point in $\hat{N}$.
By Theorem 4.2, there exists a quasiconformal map $f$ of $\hat{N}$ such that
$fG_{0}f^{-1}$ is a conformal group of $\hat{N}$ with respect to some left
invariant Carnot-Caratheodory metric $d_{CC}$ on $N$. Since $fG_{0}f^{-1}$ is
dense in $fGf^{-1}$ and the limits of conformal maps are conformal, we
conclude that $fGf^{-1}$ is also a conformal group with respect to $d_{CC}$.

For any left invariant Carnot-Caratheodory metric $d$ on $N$, let
$\text{Conf}(\hat{N},d)$ be the group of conformal maps of $\hat{N}$ with
respect to $d$. We next show that there is a fixed left invariant Carnot-
Caratheodory metric $d_{0}$ on $N$ such that for any left invariant Carnot-
Caratheodory metric $d$ on $N$, the group $\text{Conf}(\hat{N},d)$ can be
conjugated into $\text{Conf}(\hat{N},d_{0})$ by a graded automorphism of $N$.
For this we use the result of Cowling-Ottazzi [CO15] on conformal maps of
Carnot groups. By Theorem 4.1 of [CO15] there are two cases depending on
whether $N$ is the Iwasawa $N$ group of a real rank-one simple Lie group.

Let $d$ be a left invariant Carnot-Caratheodory metric on $N$. First assume
$N$ is the Iwasawa $N$ group of a real rank-one simple Lie group $I$. In this
case, Theorem 4.1 of [CO15] states that the action of each
$g\in\text{Conf}(\hat{N},d)$ on $\hat{N}$ agrees with the action of some
$\phi(g)\in I$. This $\phi(g)$ is clearly unique and the map
$\text{Conf}(\hat{N},d)\rightarrow I,g\mapsto\phi(g)$ defines an injective
homomorphism. It is known that in this case there is a left invariant Carnot-
Caratheodory metric $d_{0}$ on $N=\partial S\backslash\\{\infty\\}$ such that
$I=\text{Conf}(\hat{N},d_{0})$. Hence the statement holds in this case.

Next we assume $N$ is not the Iwasawa $N$ group of a real rank-one simple Lie
group. In this case, Theorem 4.1 of [CO15] states that for each
$g\in\text{Conf}(\hat{N},d)$, we have $g(N)=N$ and
$g|_{N}:(N,d)\rightarrow(N,d)$ is a similarity. In other words, we have
$\text{Conf}(\hat{N},d)=\text{Sim}(N,d)$. Now we finish the proof by applying
Lemma 2.3.

∎

Proof of Corollary 1.2. Let $N$ be a Carnot group and $S=N\rtimes\mathbb{R}$
be the associated solvable Lie group. Let $G$ be a group that quasi-acts co-
boundedly on $S$. This induces a uniform quasiconformal action of $G$ on
$\partial S=\hat{N}$ such that the induced action on the space of distinct
triples is co-compact. By Theorem 1.1 there is a fixed (independent of $G$)
left invariant Carnot-Caratheodory metric $d_{0}$ on $N$ such that $G$ is
quasiconformally conjugate to a subgroup of $\text{Conf}(\hat{N},d_{0})$. It
now suffices to show that there is a left invariant Riemannian metric $g_{0}$
on $S$ such that every map in $\text{Conf}(\hat{N},d_{0})$ is the boundary map
of some isometry of $(S,g_{0})$. Again we will use Theorem 4.1 of [CO15].

First assume $N$ is the Iwasawa $N$ group of a real rank-one simple Lie group
$I$. In this case, as observed in the proof of Theorem 1.1,
$\text{Conf}(\hat{N},d_{0})$ injects into $I$, which is the isometry group of
a left invariant Riemannian metric $g_{0}$ on $S$. Next we assume $N$ is not
the Iwasawa $N$ group of a real rank-one simple Lie group. In this case, each
$g\in\text{Conf}(\hat{N},d_{0})$ has the form
$g|_{N}=L_{n}\circ\delta_{t_{g}}\circ A_{g}$ for some $n\in N$, $t_{g}>0$,
where $L_{n}$ is left translation by $n\in N$, $\delta_{t}$ ($t>0$) is a
Carnot dilation, and $A_{g}:(N,d_{0})\rightarrow(N,d_{0})$ is an isometry and
also a graded automorphism of $N$. Let $\mathfrak{n}=V_{1}\oplus\cdots\oplus
V_{k}$ be the Carnot grading of $\mathfrak{n}$. Then the map
$\text{Conf}(\hat{N},d_{0})\rightarrow GL(V_{j}),g\mapsto A_{g}|_{V_{j}}$ is a
group homomorphism whose image has compact closure. It follows that there is
some $\text{Conf}(\hat{N},d_{0})$-invariant inner product $<,>_{j}$ on
$V_{j}$. Now we equip $\mathfrak{s}=\mathfrak{n}\rtimes\mathbb{R}$ with the
inner product $<,>$ satisfying: (1) $<,>$ agrees with $<,>_{j}$ on $V_{j}$;
(2) the subspaces $V_{j}$, $1\leq j\leq k$ and
$\\{0\\}\times\mathbb{R}\subset\mathfrak{n}\rtimes\mathbb{R}$ are all
perpendicular to each other with respect to $<,>$. For each
$g\in\text{Conf}(\hat{N},d_{0})$, define a map $\phi(g):S\rightarrow S$ by
$\phi(g)(x,t)=L_{(n,t_{g})}(A_{g}x,t)$, where $L_{(n,t_{g})}$ is the left
translation of $S$ by $(n,t_{g})\in S$. Then $\phi(g)$ is an isometry of
$(S,g_{0})$ with its induced boundary map equal to
$g|_{N}=L_{n}\circ\delta_{t_{g}}\circ A_{g}$ and
$\phi:G\rightarrow\text{Isom}(S,g_{0})$ is a group homomorphism, where $g_{0}$
is the left invariant Riemannian metric on $S$ determined by $<,>$.

Finally we recall that a self quasiconformal map of $\partial S=\hat{N}$
extends to a quasi-isometry of $S$. Combining this with the preceding two
paragraphs we conclude that the original quasi-action of $G$ on $S$ is quasi-
conjugate to an isometric action of $G$ on $(S,g_{0})$.

∎

##  5\. A fibered Tukia theorem for diagonal Heintze pairs

In this section we prove a fiber bundle version of Tukia’s Theorem (Theorem
1.3) in the spirit of [Dy10] for diagonal Heintze pairs. We first explain that
such a group $N$ admits an iterated fibration structure with Carnot group
fibers and that each self biLipschitz map of $N$ induces bundle maps. The
fiber Tukia theorem states that after a biLipschitz conjugation the induced
maps between the fibers are similarities.

###  5.1. BiLipschitz maps of a diagonal Heintze pair

Before we prove the fiber Tukia theorem, we first look at individual
biLipschitz maps of a diagonal Heintze pair $(N,D)$.

Let $(N,D)$ be a diagonal Heintze pair and $0<\lambda_{1}<\cdots<\lambda_{r}$
the distinct eigenvalues of $D$. Let $\mathfrak{n}=\oplus_{j}V_{\lambda_{j}}$
be the decomposition of the Lie algebra $\mathfrak{n}$ of $N$ into
eigenspaces. If $V\subset\mathfrak{n}$ is a linear subspace such that
$D(V)\subset V$ then $V$ is graded, that is, $V=\oplus_{j}(V\cap
V_{\lambda_{j}})$. It follows that if $V\subset V^{\prime}$ are two linear
subspaces of $\mathfrak{n}$ such that $D(V)\subset V$, $D(V^{\prime})\subset
V^{\prime}$, then $V^{\prime}/V=\oplus_{j}(V^{\prime}\cap
V_{\lambda_{j}})/(V\cap V_{\lambda_{j}})$ and $D$ induces a linear map on
$V^{\prime}/V$ and acts on $(V^{\prime}\cap V_{\lambda_{j}})/(V\cap
V_{\lambda_{j}})$ by multiplication by $\lambda_{j}$.

Now let $\mathfrak{h}_{1}$ be the Lie sub-algebra of $\mathfrak{n}$ generated
by $V_{\lambda_{1}}$. We say $(N,D)$ is of Carnot type if
$\mathfrak{h}_{1}=\mathfrak{n}$. In general, inductively define
$\mathfrak{h}_{i}=N(\mathfrak{h}_{i-1})$ for $i\geq 2$, where for a Lie sub-
algebra $\mathfrak{h}\subset\mathfrak{n}$,
$N(\mathfrak{h})=\\{X\in\mathfrak{n}|[X,Y]\in\mathfrak{h},\forall
Y\in\mathfrak{h}\\}$ denotes the normalizer of $\mathfrak{h}$ in
$\mathfrak{n}$. Then there is some integer $m\geq 1$ such that
$\mathfrak{h}_{m}=\mathfrak{n}$. We next explain how to refine the sequence
$0<\mathfrak{h}_{1}<\cdots<\mathfrak{h}_{m}$. It is not hard to see that
$D(\mathfrak{h}_{1})\subset\mathfrak{h}_{1}$, and by using the definition of
derivation we obtain $D(\mathfrak{h}_{i})\subset\mathfrak{h}_{i}$ for all $i$.
Hence $D$ induces a diagonal derivation $\bar{D}$ of
$\mathfrak{h}_{i}/{\mathfrak{h}_{i-1}}$ with positive eigenvalues. Then for
every $i$, if $(\mathfrak{h}_{i}/{\mathfrak{h}_{i-1}},\overline{D})$ is of
non-Carnot type, we can repeat the above process: first take the Lie sub-
algebra generated by the eigenspace of the smallest eigenvalue of
$\overline{D}$, then take the normalizers. In this way we obtain a sequence of
Lie sub-algebras of ${\mathfrak{h}_{i}}/{\mathfrak{h}_{i-1}}$:
$0=\bar{\mathfrak{h}}_{i,0}<\bar{\mathfrak{h}}_{i,1}<\cdots<\bar{\mathfrak{h}}_{i,m_{i}}={\mathfrak{h}_{i}}/{\mathfrak{h}_{i-1}}$.
Let $q_{i}:\mathfrak{h}_{i}\rightarrow{\mathfrak{h}_{i}}/{\mathfrak{h}_{i-1}}$
be the quotient map and set
$\mathfrak{h}_{i,j}=q_{i}^{-1}(\bar{\mathfrak{h}}_{i,j})$. The refinement of
the sequence $0<\mathfrak{h}_{1}<\cdots\mathfrak{<}\mathfrak{h}_{m}$ is
obtained as follows: for those $i$ such that
$(\mathfrak{h}_{i}/{\mathfrak{h}_{i-1}},\overline{D})$ is of non-Carnot type,
we insert between $\mathfrak{h}_{i-1}$ and $\mathfrak{h}_{i}$ the sequence
$\mathfrak{h}_{i,1}<\cdots<\mathfrak{h}_{i,m_{i}-1}$. This refinement process
can be further repeated and eventually we must stop since $\mathfrak{n}$ is
finite dimensional. At the end, we obtain a sequence of Lie sub-algebras
$0=\mathfrak{n}_{0}<\mathfrak{n}_{1}<\cdots<\mathfrak{n}_{s}=\mathfrak{n}$
such that $\mathfrak{n}_{i-1}$ is an ideal of $\mathfrak{n}_{i}$ for every
$i$, $D(\mathfrak{n}_{i})\subset\mathfrak{n}_{i}$, and each
$(\mathfrak{n}_{i}/{\mathfrak{n}_{i-1}},\bar{D})$ is of Carnot type (that is,
$\mathfrak{n}_{i}/{\mathfrak{n}_{i-1}}$ is a Carnot algebra and $D$ induces a
derivation $\bar{D}$ of $\mathfrak{n}_{i}/{\mathfrak{n}_{i-1}}$ that is a
multiple of a Carnot derivation). We remark that when we talk about Carnot
algebra here we include the case of abelian Lie algebra. In other words,
$(\mathfrak{n}_{i}/{\mathfrak{n}_{i-1}},\overline{D})$ is of Carnot type when
$\mathfrak{n}_{i}/{\mathfrak{n}_{i-1}}$ is some $\mathbb{R}^{n}$ and $\bar{D}$
is a standard Euclidean dilation. Let $N_{i}$ be the connected Lie subgroup of
$N$ with Lie algebra $\mathfrak{n}_{i}$. Then $N_{i-1}$ is normal in $N_{i}$
and $N_{i}/{N_{i-1}}$ is a Carnot group.

######  Definition 5.1.

Given a diagonal Heintze pair $(N,D)$, we call $1=N_{0}<N_{1}<\cdots<N_{s}=N$
the sequence of subgroups defined by the process above the _preserved
subgroups sequence_.

Let $d$ be a $D$-homogeneous distance on $N$. The restriction of $d$ on
$N_{i}$ is a $D|_{\mathfrak{n}_{i}}$-homogeneous distance on $N_{i}$. As
$N_{i-1}\triangleleft N_{i}$, $d$ induces a distance $\bar{d}$ on
$N_{i}/{N_{i-1}}$, see end of Section 2.2. By the preceding paragraph, this
distance $\bar{d}$ is a $\bar{D}$-homogeneous distance and hence is
biLipschitz with $\bar{d}_{CC}^{\frac{1}{\lambda}}$, where $\bar{d}_{CC}$ is a
Carnot-Caratheodory metric on $N_{i}/{N_{i-1}}$ and $\lambda>0$ is the
smallest eigenvalue of $\bar{D}$. For each $i$ we have a fibration
$\pi_{i}:N/{N_{i-1}}\to N/{N_{i}}$ with fiber $N_{i}/{N_{i-1}}$. The distance
$d$ does not induce any distance on $N/{N_{i}}$ when $N_{i}$ is not normal in
$N$. However, as indicated above, $d$ still induces a distance $\bar{d}$ on
the fibers $N_{i}/{N_{i-1}}$.

The following result in particular applies to biLipschitz maps of $N$ (when
$\mathfrak{h}_{1}\not=\mathfrak{n}$).

######  Theorem 5.2.

([CP17]) Suppose $(N,D)$ is not of Carnot type. Then every quasisymmetric map
$F:N\rightarrow N$ permutes the left cosets of $N_{1}$ and is biLipschitz.

By Lemma 3.2, [CPS17] (see also Lemma 4.4, [LX15] in the Carnot case), for a
connected Lie subgroup $H\subset N$, two left cosets $g_{1}H$, $g_{2}H$ are at
finite Hausdorff distance from each other if and only if $g_{1}H$, $g_{2}H$
lie in the same left coset of $N(H)$, the normalizer of $H$ in $N$. It follows
that a biLipschitz map $F:N\rightarrow N$ permutes the cosets of $N_{i}$ for
every $i$. As a consequence, $F$ induces a map $F_{i}:N/{N_{i}}\rightarrow
N/{N_{i}}$ and a bundle map of the fibration $\pi_{i}:N/{N_{i-1}}\to
N/{N_{i}}$. In general we cannot talk about the metric property of these maps
as $d$ does not induce a metric on $N/{N_{i}}$ when $N_{i}$ is not normal in
$N$. However, the restriction of $F$ to cosets of $N_{i}$ yields a biLipschitz
map of $N_{i}$, that is, $F_{p}|_{N_{i}}=(L_{F(p)^{-1}}\circ F\circ
L_{p})|_{N_{i}}:N_{i}\to N_{i}$ is biLipschitz for every $p\in N$. As
$F_{p}|_{N_{i}}$ also permutes the cosets of $N_{i-1}$, it induces a
biLipschitz map $F_{i,p}:N_{i}/{N_{i-1}}\rightarrow N_{i}/{N_{i-1}}$. In other
words, the restrictions of $F_{i-1}$ to the fibers of the fibration
$\pi_{i}:N/{N_{i-1}}\to N/{N_{i}}$ are biLipschitz maps between the fibers.
Observe that as $N_{i-1}\triangleleft N_{i}$, we have $F_{i,p}=F_{i,q}$ when
$q=pw$ for some $w\in N_{i-1}$. Because of this, we may define $F_{i,h}$ for
$h\in N/{N_{i-1}}$ by $F_{i,h}=F_{i,p}$ for any $p\in N$ satisfying
$h=pN_{i-1}$.

######  Remark 5.3.

Note that for $q=pw$ with $w\in N_{i}-N_{i-1}$, we have $F_{i,q}(N_{i-1})\neq
F_{i,p}(wN_{i-1})$ however the Pansu differentials satisfy
$DF_{i,p}(wN_{i-1})=DF_{i,q}(0)$ by the definition of Pansu differential.

For convenience, we introduce the following terminology.

######  Definition 5.4.

Let $(N,D)$ be a diagonal Heintze pair with preserved subgroups sequence
$1=N_{0}<N_{1}<\cdots<N_{s}=N$. Let $F:N\rightarrow N$ be a biLipschitz map.
We say $F$ is an $i$-similarity for some $1\leq i\leq s$ if there exists a
Carnot-Caratheodory metric $\bar{d}_{i}$ on $N_{i}/{N_{i-1}}$ such that
$F_{i,p}:(N_{i}/{N_{i-1}},\bar{d}_{i})\rightarrow(N_{i}/{N_{i-1}},\bar{d}_{i})$
is a similarity for any $p\in N$. The map $F$ is a fiber similarity if it is
an $i$-similarity for each $i$. A uniform quasisimilarity group $\Gamma$ of
$N$ is a fiber similarity group if there exist Carnot-Caratheodory metrics
$\bar{d}_{i}$ on $N_{i}/{N_{i-1}}$ such that each $\gamma\in\Gamma$ is a fiber
similarity with respect to these metrics.

###  5.2. Invariant measures on homogeneous spaces

Let $N$ be a simply connected nilpotent Lie group and $H$ a closed connected
Lie subgroup. Then the homogeneous space $N/H$ admits a $N$-invariant measure.
By Theorem 1.2.12 and its proof in [CG90], the product of a Haar measure
$m_{H}$ on $H$ and an invariant measure $m_{N/H}$ on $N/H$ is a Haar measure
on $N$, in the following sense. There is a smooth submanifold $K\subset N$ and
a measure $m^{\prime}$ on $K$ with the following properties: (1) the map
$P:K\times H\rightarrow N$, $(k,h)\mapsto kh$, is a diffeomorphism and
$P_{*}(m^{\prime}\times m_{H})$ is a Haar measure on $N$; (2)
$(\pi\circ\iota)_{*}(m^{\prime})=m_{N/H}$, where $\iota:K\subset N$ is the
inclusion and $\pi:N\rightarrow N/H$ is the quotient map.

######  Lemma 5.5.

Let $(N,D)$ be a diagonal Heintze pair, and $H$ a connected Lie subgroup of
$N$. Let $F:N\rightarrow N$ be a biLipschitz map that permutes the cosets of
$H$. Let $\bar{F}:N/H\rightarrow N/H$ be the induced map of the homogeneous
space $N/H$. Then there is a constant $C$ depending only on the biLipschitz
constant of $F$ such that for any measurable subset $A\subset N/H$ the
inequality $\frac{1}{C}m_{N/H}(A)\leq m_{N/H}(\bar{F}(A))\leq Cm_{N/H}(A)$
holds, where $m_{N/H}$ is an invariant measure on $N/H$.

###### Proof.

Since the Hausdorff measure on $N$ with respect to a $D$-homogeneous distance
is a Haar measure and $F$ is biLipschitz, there is a constant depending only
on the biLipschitz constant of $F$ such that $\frac{1}{C}m_{N}(U)\leq
m_{N}(F(U))\leq Cm_{N}(U)$ for any measurable subset $U\subset N$. Similarly,
since the restriction of $F$ to cosets of $H$ is also biLipschitz, we have
$\frac{1}{C}m_{H}(B)\leq m_{H}(F(B))\leq Cm_{H}(B)$ for measurable subsets $B$
of cosets of $H$. Now let $A\subset N/H$ be a measurable subset. Let
$A^{\prime}=\pi^{-1}(A)\cap K$, where $K$ is as above. Let $B$ be a bounded
open subset of $H$. Set $U=P(A^{\prime}\times B)\subset N$. Then
$m_{N}(U)=m_{H}(B)m_{N/H}(A)$. By the above we have

(5) $\frac{1}{C}m_{N}(U)\leq m_{N}(F(U))\leq Cm_{N}(U).$

On the other hand,

$\displaystyle m_{N}(F(U))$
$\displaystyle=\int_{N}1_{F(U)}dm_{N}=\int_{K}(\int_{kH}1_{F(U)}dm_{H})dm^{\prime}(k)$
$\displaystyle\leq\int_{\pi^{-1}(\bar{F}(A))\cap
K}Cm_{H}(B)dm^{\prime}(k)=Cm_{H}(B)m_{N/H}(\bar{F}(A)).$

Similarly we obtain $m_{N}(F(U))\geq\frac{1}{C}m_{H}(B)m_{N/H}(\bar{F}(A))$.
Combining the above inequalities we get $\frac{1}{C^{2}}m_{N/H}(A)\leq
m_{N/H}(\bar{F}(A))\leq C^{2}m_{N/H}(A)$.

∎

###  5.3. Conjugating into a fiber similarity group.

Let $1\leq i\leq s$ be such that $\dim(N_{i}/{N_{i-1}})\geq 2$. Recall that
the quotient $N_{i}/{N_{i-1}}$ is a Carnot group. Let $H_{i}$ be the first
layer of $\mathfrak{n}_{i}/{\mathfrak{n}_{i-1}}$ and $m_{i}=\dim(H_{i})$.
Notice that $m_{i}\geq 2$.

######  Definition 5.6.

A measurable fiber conformal structure on the fibration
$\pi_{i}:N/{N_{i-1}}\rightarrow N/{N_{i}}$ is an essentially bounded
measurable map

$\mu:N/{N_{i-1}}\to X:=SL(m_{i})/SO(m_{i}).$

So a measurable fiber conformal structure on the fibration
$\pi_{i}:N/{N_{i-1}}\rightarrow N/{N_{i}}$ can be thought of as a measurable
assignment of inner products (up to scalar multiple) to horizontal subspaces
of the tangent spaces of the fibers (of the fibration). For a review of
$SL(n)/SO(n)$ and its Riemannian distance $d_{X}$ see Section 4.1.

For $F:N\to N$ a biLipschitz map we define the pull-back

$F^{*}\mu(pN_{i-1})=(DF_{i,p}(0)|_{H_{i}})^{T}[\mu(F_{i-1}(pN_{i-1}))],\;\;\;\text{for
a.e.}\;pN_{i-1}\in N/{N_{i-1}}.$

Notice that the pull-back is well-defined as $F_{i,p}$ depends only on the
coset $pN_{i-1}$.

######  Definition 5.7.

We say that $F$ is _conformal with respect to $\mu$_ if
$F^{*}\mu(pN_{i-1})=\mu(pN_{i-1})$ for a.e. $pN_{i-1}\in N/{N_{i-1}}$.

######  Proposition 5.8.

Let $\Gamma$ be a countable uniform quasisimilarity group of $(N,d)$. Let
$1\leq i\leq s$ be such that $\dim(N_{i}/{N_{i-1}})\geq 2$. Then there is a
measurable fiber conformal structure $\mu$ on the fibration
$\pi_{i}:N/{N_{i-1}}\rightarrow N/{N_{i}}$ such that every $\gamma\in\Gamma$
is conformal with respect to $\mu$.

###### Proof.

There is a $\Gamma$-invariant subset $U\subset N/{N_{i-1}}$ of full measure
such that for all $\gamma\in\Gamma$ and all $pN_{i-1}\in U$ the foliated Pansu
derivative $D\gamma_{i,p}(0)$ exists and is a graded automorphism of
$N_{i}/{N_{i-1}}$. Let $\mu_{0}$ be the $N$-invariant fiber conformal
structure on the fibration $\pi_{i}:N/{N_{i-1}}\rightarrow N/{N_{i}}$ that is
associated with an inner product on $H_{i}$ (the first layer of
$\mathfrak{n}_{i}/{\mathfrak{n}_{i-1}}$). For each $pN_{i-1}\in U$, set
$M_{pN_{i-1}}=\\{\gamma^{*}\mu_{0}(pN_{i-1}):\gamma\in\Gamma$}. Since $\Gamma$
is a uniform quasisimilarity group, $M_{pN_{i-1}}$ is a bounded subset of
$SL(m_{i})/SO(m_{i})$. The assignment $pN_{i-1}\mapsto M_{pN_{i-1}}$ is
$\Gamma$-invariant. Let $P(M_{pN_{i-1}})$ be the circumcenter of
$M_{pN_{i-1}}$ in $SL(m_{i})/SO(m_{i})$ and define $\mu:U\rightarrow
SL(m_{i})/SO(m_{i})$ by $\mu(pN_{i-1})=P(M_{pN_{i-1}})$. Then $\mu$ is
$\Gamma$-invariant and measurable (see the proof of Proposition 4.1 for
details). ∎

For the proof of the main result in this section we need a modified version of
Lemma 3.4.

######  Lemma 5.9.

Let $\bar{d}_{i}$ be a left invariant Carnot-Caratheodory metric on
$N_{i}/{N_{i-1}}$. For a biLipschitz map $G:N\rightarrow N$ and any
$pN_{i-1}\in N/{N_{i-1}}$, denote by $K(G_{i,p},0)$ the dilatation of the map
$G_{i,p}:(N_{i}/{N_{i-1}},\bar{d}_{i})\rightarrow(N_{i}/{N_{i-1}},\bar{d}_{i})$
at $0$. Let $\\{F^{j}:N\rightarrow N\\}$ be a sequence of
$(L,C)$-quasisimilarities that converge uniformly on compact subsets to a map
$F:N\rightarrow N$. Suppose for any compact subset $S\subset N/{N_{i-1}}$ and
any $\epsilon>0$ we have

$m_{N/{N_{i-1}}}(\\{pN_{i-1}:pN_{i-1}\in S,\ K(F^{j}_{i,p},0)\geq
1+\epsilon\\})\rightarrow 0$

as $j\rightarrow\infty$, where $m_{N/{N_{i-1}}}$ is an invariant measure on
the space $N/{N_{i-1}}$. Then for every $pN_{i-1}\in N/{N_{i-1}}$, $F_{i,p}$
is a similarity of $(N_{i}/{N_{i-1}},\bar{d}_{i})$.

###### Proof.

We shall show that for almost every fiber of $\pi_{i}$, the map $F_{i,p}$ and
a subsequence of $\\{F^{j}_{i,p}\\}_{j}$ satisfy the assumption of Lemma 3.4
and so
$F_{i,p}:(N_{i}/{N_{i-1}},\bar{d}_{i})\rightarrow(N_{i}/{N_{i-1}},\bar{d}_{i})$
is conformal. By Theorem 4.1 of [CO15], $F_{i,p}$ is a similarity. Since the
limit of similarity maps is a similarity, the same is true for every fiber of
$\pi_{i}$.

By Theorem 1.2.12 and its proof in [CG90], the product of a Haar measure on
$N_{i}/{N_{i-1}}$ and an invariant measure on $N/{N_{i}}$ is an invariant
measure on $N/{N_{i-1}}$, in the following sense. There is a smooth
submanifold $Y_{i}$ of $N/{N_{i-1}}$ and a measure $m^{\prime}$ on $Y_{i}$
with the following properties: (1) the (well-defined) map $P:Y_{i}\times
N_{i}/{N_{i-1}}\rightarrow N/{N_{i-1}}$, $(yN_{i-1},xN_{i-1})\mapsto
yxN_{i-1}$, is a diffeomorphism and $P_{*}(m^{\prime}\times
m_{N_{i}/{N_{i-1}}})$ is an invariant measure on $N/{N_{i-1}}$, where
$m_{N_{i}/{N_{i-1}}}$ is a Haar measure on $N_{i}/{N_{i-1}}$; (2)
$(\pi_{i}\circ\iota)_{*}(m^{\prime})$ is an invariant measure on $N/{N_{i}}$,
where $\iota:Y_{i}\subset N/{N_{i-1}}$ is the inclusion.

Let $B_{n}\subset Y_{i}$ ($n\geq 1$) be an exhausting sequence of compact
subsets of $Y_{i}$, and $A_{n}\subset N_{i}/{N_{i-1}}$ ($n\geq 1$) be a
sequence of balls in $N_{i}/{N_{i-1}}$ centered at the origin $0$ with radius
going to infinity as $n\rightarrow\infty$. Set $C_{n}=P(B_{n}\times
A_{n})\subset N/{N_{i-1}}$. For integers $n,k,j\geq 1$, let
$B_{n,k,j}=\\{pN_{i-1}\in C_{n}|K(F^{j}_{i,p},0)\geq 1+1/k\\}$ and
$g_{n,k,j}=1_{B_{n,k,j}}$ be the characteristic function of $B_{n,k,j}$. Then
${g_{n,k,j}}\in L^{1}(C_{n})$. By Fubini, for a.e. $h\in B_{n}$, the function
$g_{n,k,j}^{h}:A_{n}\rightarrow[0,1]$ given by
$g_{n,k,j}^{h}(xN_{i-1})=g_{n,k,j}(P(h,xN_{i-1}))$ is integrable,
$q_{n,k,j}(h):=\int_{A_{n}}g_{n,k,j}^{h}(xN_{i-1})dm_{N_{i}/{N_{i-1}}}$ is an
integrable function on $B_{n}$, and
$\int_{B_{n}}q_{n,k,j}(h)dm^{\prime}(h)=\int_{C_{n}}g_{n,k,j}(pN_{i-1})dm_{N/N_{i-1}}=m_{N/N_{i-1}}(B_{n,k,j})$.
By the assumption, for fixed $n$ and $k$ we have
$m_{N/N_{i-1}}(B_{n,k,j})\rightarrow 0$ as $j\rightarrow\infty$, which implies
$q_{n,k,j}\rightarrow 0$ in $L^{1}(B_{n})$ as $j\rightarrow\infty$. This in
turn implies that for fixed $n,k$ there is a null set $E_{n,k}\subset B_{n}$
and a subsequence $\\{q_{n,k,j_{l}}\\}_{l}$ of $\\{q_{n,k,j}\\}_{j}$ such that
$q_{n,k,j_{l}}(h)\rightarrow 0$ as $l\rightarrow\infty$ for every $h\in
B_{n}\backslash E_{n,k}$. Set $E=\cup_{k,n}E_{n,k}$. Then $E$ is a null set in
$Y_{i}$.

Let $h\in Y_{i}\backslash E$. Let $A$ be a compact subset of $N_{i}/{N_{i-1}}$
and $\epsilon>0$. Pick $n,k$ sufficiently large such that $1/k<\epsilon$ and
$h\in B_{n}$, $A\subset A_{n}$. By the definition of $E$ we have
$q_{n,k,j_{l}}(h)\rightarrow 0$ as $l\rightarrow\infty$. Due to
$DF_{i,p}(xN_{i-1})=DF_{i,px}(0)$, we notice that

$\displaystyle 0\leftarrow q_{n,k,j_{l}}(h)$
$\displaystyle=m_{N_{i}/N_{i-1}}(\\{x\in A_{n}|K(F^{j_{l}}_{i,hx},0)\geq
1+1/k\\})$ $\displaystyle=m_{N_{i}/N_{i-1}}(\\{x\in
A_{n}|K(F^{j_{l}}_{i,h},x)\geq 1+1/k\\})$ $\displaystyle\geq
m_{N_{i}/N_{i-1}}(\\{x\in A|K(F^{j_{l}}_{i,h},x)\geq 1+\epsilon\\})$

as $l\rightarrow\infty$ and so the assumption of Lemma 3.4 is satisfied by
$F^{j_{l}}_{i,h}$, $l\geq 1$, $F_{i,h}$.

∎

Let $S=N\rtimes_{D}\mathbb{R}$ be the negatively curved homogeneous space
associated with $(N,D)$. As $\Gamma$ acts cocompactly on the space of distinct
pairs of $N$, every point of $N$ is a radial limit point. Such an action
induces a cobounded quasi-action on $S$ and vice versa. Recall that $\mu$ is
measurable and hence approximately continuous at a.e. $pN_{i-1}\in
N/{N_{i-1}}$.

######  Theorem 5.10.

Assume that $\dim(N_{i}/{N_{i-1}})\geq 2$. Let $\Gamma$ be a uniform
quasisimilarity group of $N$ and $\mu$ a $\Gamma$-invariant measurable fiber
conformal structure on the fibration $\pi_{i}:N/{N_{i-1}}\rightarrow
N/{N_{i}}$. Suppose there is a point $p\in N$ such that $\mu$ is approximately
continuous at $pN_{i-1}$ and $p$ is a radial limit point for $\Gamma$. Then
there exists a Carnot-Caratheodory metric $\bar{d}_{i}$ on $N_{i}/{N_{i-1}}$
and a quasisimilarity $F_{0}:N\rightarrow N$ such that $F_{0}GF_{0}^{-1}$
consists of elements $F$ where the associated
$F_{i,p}:(N_{i}/{N_{i-1}},\bar{d}_{i})\rightarrow(N_{i}/{N_{i-1}},\bar{d}_{i})$
are similarities for each $p\in N$.

###### Proof.

We assume $p=0$ and use $e^{tD}$ to denote both the dilation on $N$ and the
isometry of $S$ that translates along $\\{0\\}\times\mathbb{R}$. Similarly
elements of $\Gamma$ can be viewed as quasisimilarities of $N$ or quasi-
isometries of $S$. Since $0$ is a radial limit point there exists a sequence
$\\{h_{j}\\}_{j=1}^{\infty}$ in $\Gamma$ and a sequence $s_{j}\to\infty$ such
that for any point $x_{0}\in S$ we have $h_{j}(x_{0})\rightarrow 0$ in
$\bar{S}=S\cup\partial S$ and that $\\{e^{s_{j}D}\circ
h_{j}(x_{0})\\}_{j=1}^{\infty}$ is bounded and hence up to passing to a
subsequence we have that $F_{j}=e^{s_{j}D}\circ h_{j}$ converges to a
quasisimilarity $F_{0}:N\to N$ uniformly on compact subsets.

We now show that for any $\gamma\in\Gamma$, the map $F=F_{0}\gamma F_{0}^{-1}$
has the property that for all $p\in N$ the associated maps $F_{i,p}$ are
similarities of $N_{i}/{N_{i-1}}$ with respect to the left invariant Carnot-
Caratheodory metric on $N_{i}/{N_{i-1}}$ determined by $\mu(0)$.

Let $\gamma\in\Gamma$ and denote $\tilde{\gamma}=F_{0}\gamma F_{0}^{-1}$,
$\tilde{\gamma}_{j}=F_{j}\gamma F_{j}^{-1}$. Let
$\mu_{j}=(F_{j}^{-1})^{*}\mu$. Since $\mu$ is $\Gamma$-invariant,
$\tilde{\gamma}_{j}$ is conformal with respect to $\mu_{j}$. Additionally,
$\mu_{j}=(e^{-s_{j}D})^{*}\mu$ so for $q\in N$ and $G=e^{-s_{j}D}$,

$\mu_{j}(qN_{i-1})=G^{*}\mu(qN_{i-1})=(DG_{i,q}(0)|_{H_{i}})^{T}[\mu(G_{i-1}(qN_{i-1}))]=\mu(G_{i-1}(qN_{i-1}))$

since $DG_{i,q}(0)|_{H_{i}}$ is a constant multiple of the identity map.

Let $Z\subset N/{N_{i-1}}$ be a compact subset containing $N_{i-1}\in
N/{N_{i-1}}$. Since $\mu$ is approximately continuous at $N_{i-1}\in
N/{N_{i-1}}$ and $s_{j}\rightarrow\infty$, the equality
$\mu_{j}(qN_{i-1})=\mu(G_{i-1}(qN_{i-1}))$ implies that for any $\epsilon>0$
there are subsets $A_{j}\subset Z$ with $m_{N/N_{i-1}}(A_{j})\rightarrow 0$ as
$j\rightarrow\infty$ and $d_{X}(\mu_{j}(qN_{i-1}),\mu(N_{i-1}))\leq\epsilon$
for $qN_{i-1}\in Z\backslash A_{j}$.

The maps $\tilde{\gamma}^{-1}_{j}$ and $\tilde{\gamma}^{-1}$ are all
$(L,C)$-quasisimilarities for fixed $L$ and $C$ and so by Lemma 5.5 we have
$m_{N/N_{i-1}}((\tilde{\gamma}_{j})_{i-1}^{-1}(A_{j}))\rightarrow 0$ as
$j\rightarrow\infty$. Set
$B_{j}=A_{j}\cup(\tilde{\gamma}_{j})_{i-1}^{-1}(A_{j})$. Now we have
$m_{N/N_{i-1}}(B_{j})\rightarrow 0$ as $j\rightarrow\infty$ and
$d_{X}(\mu_{j}(qN_{i-1}),\mu(N_{i-1}))\leq\epsilon$ and
$d_{X}(\mu_{j}((\tilde{\gamma}_{j})_{i-1}(qN_{i-1})),\mu(N_{i-1}))\leq\epsilon$
for $qN_{i-1}\in Z\backslash B_{j}$.

Since $\tilde{\gamma}_{j}$ is $\mu_{j}$-conformal, we have
$\mu_{j}(qN_{i-1})=(D(\tilde{\gamma}_{j})_{i,q}(0)|_{H_{i}})^{T}[\mu_{j}((\tilde{\gamma}_{j})_{i-1}(qN_{i-1}))]$
for a.e. $qN_{i-1}\in N/{N_{i-1}}$. Now

$d_{X}(\mu_{j}(qN_{i-1}),(D(\tilde{\gamma}_{j})_{i,q}(0)|_{H_{i}})^{T}[\mu(N_{i-1})])=d_{X}(\mu_{j}((\tilde{\gamma}_{j})_{i-1}(qN_{i-1})),\mu(N_{i-1}))\leq\epsilon$

for a.e. $qN_{i-1}\in Z\backslash B_{j}$. Combining this with
$d_{X}(\mu_{j}(qN_{i-1}),\mu(N_{i-1}))\leq\epsilon$ , we get

$d_{X}(\mu(N_{i-1}),(D(\tilde{\gamma}_{j})_{i,q}(0)|_{H_{i}})^{T}[\mu(N_{i-1})])\leq
2\epsilon$

for a.e. $qN_{i-1}\in Z\backslash B_{j}$. By (4) this implies that the
assumption of Lemma 5.9 is satisfied by $\tilde{\gamma}_{j}$ ($j\geq 1$),
$\tilde{\gamma}$ and so for any $p\in N$, $\tilde{\gamma}_{i,p}$ is a
similarity of $N_{i}/{N_{i-1}}$ with respect to the left invariant Carnot-
Caratheodory metric determined by $\mu(N_{i-1})$.

∎

Proof of Theorem 1.3. We observe that we may assume the group $\Gamma$ is
countable: Let $\Gamma_{0}$ be a countable subgroup of $\Gamma$ that is dense
in the topology of uniform convergence on compact subsets; if every element of
some conjugate of $\Gamma_{0}$ is a $i$-fiber similarity then the same is true
for $\Gamma$ as the limits of $i$-fiber similarity maps are $i$-fiber
similarity maps.

The assumption implies that every point $p\in N$ is a radial limit point of
$\Gamma$ and so Theorem 5.10 can be applied. List the elements of $I$ by
$i_{1}<i_{2}<\cdots<i_{k}$. We first conjugate $\Gamma$ to get a group
$\Gamma_{1}$ which is a $i_{1}$-fiber similarity group, then conjugate
$\Gamma_{1}$ to get a group $\Gamma_{2}=F_{0}\Gamma_{1}F_{0}^{-1}$ which is a
$i_{2}$-fiber similarity group, and so on. We can finish by induction once we
observe that $\Gamma_{2}$ is also a $i_{1}$-fiber similarity group. For this
we just notice that the conjugating map $F_{0}$ is the limit of a sequence of
maps $F_{j}$ which are compositions of elements of $\Gamma_{1}$ and the
dilations $e^{tD}$ ($t\in\mathbb{R}$) of $N$. Since both
$\\{e^{tD}|t\in\mathbb{R}\\}$ and $\Gamma_{1}$ are $i_{1}$-fiber similarity
groups, each $F_{j}$ is a $i_{1}$-fiber similarity map. As a consequence, the
limiting map $F_{0}$ is also a $i_{1}$-fiber similarity map. Hence
$\Gamma_{2}=F_{0}\Gamma_{1}F_{0}^{-1}$ is a $i_{1}$-fiber similarity group.

∎

##  6\. A counterexample

In this section we exhibit an example which shows that in general it is
impossible to conjugate a uniform quasiconformal group of $\hat{N}$ into a
conformal group with respect to an arbitrary pre-specified Carnot-Caratheodory
metric on $N$ when that metric is not maximally symmetric.

Let $N$ be a Carnot group and $d_{CC}$ a left invariant Carnot-Caratheodory
metric on $N$. Let $IA(N,d_{CC})$ be the group consisting of graded
automorphisms of $N$ that are also isometries with respect to $d_{CC}$. By the
main result of [CO15], the group $Conf(N,d_{CC})$ of conformal maps of
$(N,d_{CC})$ is generated by left translations, standard Carnot dilations and
$IA(N,d_{CC})$.

######  Lemma 6.1.

Let $N$ be a Carnot group and $d_{1},d_{2}$ two left invariant Carnot-
Caratheodory metrics on $N$. Suppose there is a quasiconformal map
$f:(N,d_{1})\rightarrow(N,d_{2})$ such that $f\cdot Conf(N,d_{1})\cdot
f^{-1}\subset Conf(N,d_{2})$, then there is a graded automorphism $h$ of $N$
such that $h\cdot IA(N,d_{1})\cdot h^{-1}\subset IA(N,d_{2})$.

###### Proof.

Denote by $\phi:Conf(N,d_{1})\rightarrow Conf(N,d_{2})$ the injective
homomorphism given by: $\phi(g_{1})=fg_{1}f^{-1}$. Let $p\in N$ be a point
such that $Df(p)$ exists and is a graded automorphism. After pre-composing and
post-composing with left translations, we may assume $p=f(p)=0$ is the origin.
Set $h=Df(0)$.

Let $g\in IA(N,d_{1})$. Then $g(0)=0$. It follows that $\phi(g)(0)=0$ and so
$\phi(g)$ is the composition of a standard Carnot dilation and an element of
$IA(N,d_{2})$. Since $\\{g^{i}:i\in\mathbb{N}\\}$ has compact closure, the
same is true for $\\{\phi(g)^{i}:i\in\mathbb{N}\\}$. This follows from the
fact that conjugation by $f$ is continuous. Alternatively one can argue using
the fact that $f$ is quasisymmetric. This implies that $\phi(g)\in
IA(N,d_{2})$. Being graded automorphisms, both $g$ and $\phi(g)$ commute with
Carnot dilations. Now for any $t>0$,

$\delta_{t}f\delta^{-1}_{t}g(\delta_{t}f\delta^{-1}_{t})^{-1}=\delta_{t}f\delta^{-1}_{t}g\delta_{t}f^{-1}\delta^{-1}_{t}=\delta_{t}fgf^{-1}\delta^{-1}_{t}=\delta_{t}\phi(g)\delta^{-1}_{t}=\phi(g).$

Since $\delta_{t}f\delta^{-1}_{t}$ converges uniformly on compact subsets to
$h$ as $t\rightarrow+\infty$, we have $hgh^{-1}=\phi(g)\in IA(N,d_{2})$. Since
this is true for every $g\in IA(N,d_{1})$, the lemma follows.

∎

We note that for any two left invariant Carnot-Caratheodory metrics
$d_{1},d_{2}$ on $N$, the group $Conf(N,d_{1})$ is a uniform quasiconformal
group of $(N,d_{2})$. Furthermore, the action of $Conf(N,d_{1})$ on
$(N,d_{2})$ satisfies the assumption of Theorem 1.1. If every uniform
quasiconformal group of $\hat{N}$ satisfying the assumptions of Theorem 1.1
can be conjugated into the conformal group with respect to $d_{2}$, then there
is a quasiconformal map $f:\hat{N}\rightarrow\hat{N}$ such that for any $g\in
Conf(N,d_{1})$, the map
$fgf^{-1}:(N\backslash\\{fg^{-1}f^{-1}(\infty)\\},d_{2})\rightarrow(N\backslash\\{fgf^{-1}(\infty)\\},d_{2})$
is conformal. Suppose $N$ has the property that every quasiconformal map of
$\hat{N}=N\cup\\{\infty\\}$ fixes $\infty$. Then we have
$(f|_{N})\text{Conf}(N,d_{1})(f|_{N})^{-1}\subset\text{Conf}(N,d_{2})$ and by
Lemma 6.1 we conclude that $IA(N,d_{1})$ injects into $IA(N,d_{2})$. Next we
exhibit an example where such an injection does not exist.

Let $N=H\times H$ be the direct product of the first Heisenberg group with
itself. Theorem 1.1 in [KMX20] implies that every quasiconformal map of
$\hat{N}=N\cup\\{\infty\\}$ fixes $\infty$. The Lie algebra of $N$ can be
written as $\mathfrak{n}=\mathfrak{h}\oplus\mathfrak{h}=V_{1}\oplus V_{2}$
with first layer $V_{1}=\mathbb{R}^{2}\oplus\mathbb{R}^{2}$, where the first
$\mathbb{R}^{2}$ is the first layer of the first $\mathfrak{h}$ and the second
$\mathbb{R}^{2}$ is the first layer of the second $\mathfrak{h}$. Let
$e_{1},e_{2}$ denote the standard basis in the first $\mathbb{R}^{2}$, and
$\tilde{e}_{1},\tilde{e}_{2}$ denote the standard basis in the second
$\mathbb{R}^{2}$. We consider two different left invariant Carnot-Caratheodory
metrics $d_{1}$, $d_{2}$ on $N$. The metric $d_{1}$ is determined by the inner
product $<,>_{1}$ on $V_{1}$ that has $e_{1}$, $e_{2}$, $\tilde{e}_{1}$,
$\tilde{e}_{2}$ as an orthonormal basis. The metric $d_{2}$ is determined by
the inner product $<,>_{2}$ on $V_{1}$ that has $e_{1}$, $e_{2}$,
$\tilde{e}_{1}$, $\frac{\sqrt{2}}{2}(\tilde{e}_{2}-e_{1})$ as an orthonormal
basis.

######  Lemma 6.2.

(1) $IA(N,d_{1})$ is isomorphic to $(O(2)\oplus O(2))\rtimes\mathbb{Z}_{2}$,
where the generator of $\mathbb{Z}_{2}$ acts on $O(2)\oplus O(2)$ by
$(M_{1},M_{2})\rightarrow(M_{2},M_{1})$;
(2) $IA(N,d_{2})$ is isomorphic to
$(\mathbb{Z}_{2}\oplus\mathbb{Z}_{2}\oplus\mathbb{Z}_{2})\rtimes\mathbb{Z}_{2}$,
where the generator of $\mathbb{Z}_{2}$ acts on
$\mathbb{Z}_{2}\oplus\mathbb{Z}_{2}\oplus\mathbb{Z}_{2}$ by
$(a,b,c)\rightarrow(b,a,c)$.

###### Proof.

For any element $X$ of a Lie algebra $\mathfrak{n}$, let $r(X)$ be the rank of
the linear map

$ad(X):\mathfrak{n}\rightarrow\mathfrak{n}.$

Then $r(X)=r(A(X))$ for any $x\in\mathfrak{n}$ and any isomorphism $A$ of Lie
algebras. Notice that $r(x,y)\geq 1$ for every nonzero element $(x,y)\in
V_{1}=\mathbb{R}^{2}\oplus\mathbb{R}^{2}$, and furthermore $r(x,y)=1$ if and
only if one of the following happens:
(a) $x=0$ and $y\not=0$;
(b) $y=0$ and $x\not=0$.
It follows that for any graded isomorphism
$A:\mathfrak{n}\rightarrow\mathfrak{n}$, we have one of the following:
(i) $A(\mathbb{R}^{2}\oplus\\{0\\})=\mathbb{R}^{2}\oplus\\{0\\}$ and
$A(\\{0\\}\oplus\mathbb{R}^{2})=\\{0\\}\oplus\mathbb{R}^{2}$;
(ii) $A(\mathbb{R}^{2}\oplus\\{0\\})=\\{0\\}\oplus\mathbb{R}^{2}$ and
$A(\\{0\\}\oplus\mathbb{R}^{2})=\mathbb{R}^{2}\oplus\\{0\\}$.

(1) follows easily from the above paragraph.
(2) Now let $A\in IA(N,d_{2})$. First assume $A$ satisfies (i). Note the
orthogonal complement of $\mathbb{R}^{2}\oplus\\{0\\}$ in $(V_{1},<,>_{2})$ is
$E_{1}:=\mathbb{R}\tilde{e}_{1}\oplus\mathbb{R}(\frac{\sqrt{2}}{2}\cdot(\tilde{e}_{2}-e_{1}))$,
and similarly the orthogonal complement of $\\{0\\}\oplus\mathbb{R}^{2}$ is
$E_{2}:=\mathbb{R}e_{2}\oplus\mathbb{R}(e_{1}-\frac{1}{3}\tilde{e}_{2})$.
Since $A$ preserves orthogonal complement, we have $A(E_{1})=E_{1}$ and
$A(E_{2})=E_{2}$. Note
$E_{1}\cap(\\{0\\}\oplus\mathbb{R}^{2})=\mathbb{R}\tilde{e}_{1}$ and
$E_{2}\cap(\mathbb{R}^{2}\oplus\\{0\\})=\mathbb{R}e_{2}$. It follows that
$A(\mathbb{R}\tilde{e}_{1})=\mathbb{R}\tilde{e}_{1}$ and
$A(\mathbb{R}e_{2})=\mathbb{R}e_{2}$. Then $A$ also preserves the orthogonal
complement of $\mathbb{R}\tilde{e}_{1}$ in $\\{0\\}\oplus\mathbb{R}^{2}$ and
that of $\mathbb{R}e_{2}$ in $\mathbb{R}^{2}\oplus\\{0\\}$. That is, we have
$A(\mathbb{R}\tilde{e}_{2})=\mathbb{R}\tilde{e}_{2}$ and
$A(\mathbb{R}e_{1})=\mathbb{R}e_{1}$. From this it is easy to see that there
are only 8 such isometric graded isomorphisms. They are given by
$A(\tilde{e}_{1})=\epsilon_{1}\tilde{e}_{1}$, $A(e_{2})=\epsilon_{2}e_{2}$,
$A(e_{1})=\epsilon_{3}e_{1}$ and $A(\tilde{e}_{2})=\epsilon_{3}\tilde{e}_{2}$,
where $\epsilon_{1},\epsilon_{2},\epsilon_{3}\in\\{1,-1\\}$. They form a group
$F$ isomorphic to $\mathbb{Z}_{2}\oplus\mathbb{Z}_{2}\oplus\mathbb{Z}_{2}$.

A similar argument shows that there are also 8 isometric graded isomorphisms
with respect to $d_{2}$ that satisfy (ii). They are given by
$A(\tilde{e}_{1})=\epsilon_{1}e_{2}$, $A(e_{2})=\epsilon_{2}\tilde{e}_{1}$,
$A(e_{1})=\frac{1}{\sqrt{3}}\epsilon_{3}\tilde{e}_{2}$ and
$A(\tilde{e}_{2})={\sqrt{3}}\epsilon_{3}e_{1}$, where
$\epsilon_{1},\epsilon_{2},\epsilon_{3}\in\\{1,-1\\}$. If we denote by $A_{0}$
the isometric graded isomorphism corresponding to
$\epsilon_{1}=\epsilon_{2}=\epsilon_{3}=1$, then these 8 isomorphisms are
simply $A_{0}\cdot F$. Now it is easy to see that (2) holds.

∎

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