$\displaystyle=A^{-1}_{\gamma_{2}\gamma_{1}}\pi_{j}\\{A_{\gamma_{2}}(s_{\gamma_{2}}(\bar{h}_{1}))^{-1}*A_{\gamma_{2}}w_{1}^{-1}*(B_{\gamma_{2}}h_{1})^{-1}*B_{\gamma_{2}}h_{2}*A_{\gamma_{2}}w_{2}*A_{\gamma_{2}}A_{\gamma_{1}}s_{\gamma_{1}}(\bar{h})*A_{\gamma_{2}}s_{\gamma_{2}}(\bar{h}_{2})\\}$
$\displaystyle=A^{-1}_{\gamma_{2}\gamma_{1}}\pi_{j}\\{A_{\gamma_{2}}(s_{\gamma_{2}}(\bar{h}_{1}))^{-1}*A_{\gamma_{2}}w_{1}^{-1}*A_{\gamma_{2}}w_{2}*A_{\gamma_{2}}A_{\gamma_{1}}s_{\gamma_{1}}(\bar{h})*A_{\gamma_{2}}s_{\gamma_{2}}(\bar{h}_{2})\\}$
$\displaystyle=A^{-1}_{\gamma_{2}\gamma_{1}}\pi_{j}\\{A_{\gamma_{2}}(s_{\gamma_{2}}(\bar{h}_{1}))^{-1}*A_{\gamma_{2}}(w_{1}^{-1}*w_{2})*A_{\gamma_{2}}A_{\gamma_{1}}s_{\gamma_{1}}(\bar{h})*A_{\gamma_{2}}s_{\gamma_{2}}(\bar{h}_{2})\\}$
$\displaystyle=A^{-1}_{\gamma_{2}\gamma_{1}}\pi_{j}\\{A_{\gamma_{2}}(s_{\gamma_{2}}(\bar{h}_{1}))^{-1}*A_{\gamma_{2}}A_{\gamma_{1}}s_{\gamma_{1}}(\bar{h})*A_{\gamma_{2}}s_{\gamma_{2}}(\bar{h}_{2})\\}$
$\displaystyle=\pi_{j}\\{A^{-1}_{\gamma_{1}}(s_{\gamma_{2}}(\bar{h}_{1}))^{-1}*s_{\gamma_{1}}(\bar{h})*A^{-1}_{\gamma_{1}}s_{\gamma_{2}}(\bar{h}_{2})\\}.$

Since $s_{\gamma}$ takes values in $Z(\mathfrak{w})$, by BCH formula and
noting $\bar{h}_{2}=\bar{\gamma}_{1}(\bar{h})$,
$\bar{h}_{1}=\bar{\gamma}_{1}(0)$ we obtain

$s_{\gamma_{2}\gamma_{1},j}(\bar{h})=s_{\gamma_{1},j}(\bar{h})+A^{-1}_{\gamma_{1}}s_{\gamma_{2},j}(\bar{h}_{2})-A^{-1}_{\gamma_{1}}s_{\gamma_{2},j}(\bar{h}_{1})=s_{\gamma_{1},j}(\bar{h})+(\pi_{\Psi(\gamma_{1})}s_{\gamma_{2},j})(\bar{h}).$

∎

Recall we have two maps $\Psi:\Gamma\rightarrow G$ and
$b_{j}:\Gamma\rightarrow E_{j}$.

######  Lemma 7.5.

Let $\\{g_{i}\\}$ and $\\{\tilde{g}_{i}\\}$ be two sequences in $\Gamma$ and
$(A,B)\in G$ such that $\Psi(g_{i})\rightarrow(A,B)$ and
$\Psi(\tilde{g}_{i})\rightarrow(A,B)$. If $s,\tilde{s}\in E_{j}$ are such that
$b_{j}(g_{i})\rightarrow s$ and $b_{j}(\tilde{g}_{i})\rightarrow\tilde{s}$
pointwise as $i\rightarrow\infty$, then $s=\tilde{s}$.

###### Proof.

By setting $\gamma_{1}=\gamma^{-1}$ and $\gamma_{2}=\gamma$ in Lemma 7.4, we
get $b_{j}(\gamma^{-1})=-\pi_{\Psi(\gamma^{-1})}b_{j}(\gamma)$. Similarly if
we set $\gamma_{1}=g_{i}$ and $\gamma_{2}=\tilde{g}_{i}^{-1}$ then we get

$b_{j}(\tilde{g}_{i}^{-1}g_{i})=b_{j}(g_{i})+\pi_{\Psi(g_{i})}b_{j}(\tilde{g}_{i}^{-1})=b_{j}(g_{i})+\pi_{\Psi(g_{i})}(-\pi_{\Psi(\tilde{g}_{i}^{-1})}b_{j}(\tilde{g}_{i}))=b_{j}(g_{i})-\pi_{\Psi(\tilde{g}_{i}^{-1}g_{i})}b_{j}(\tilde{g}_{i}).$

The assumption implies
$\Psi(\tilde{g}_{i}^{-1}g_{i})\rightarrow(\text{Id},\text{Id})$. On the other
hand, by Corollary 7.10 there is a constant $C>0$ such that
$||b_{j}(\gamma)||\leq C$ for all $\gamma\in\Gamma$. It follows that
$b_{j}(\tilde{g}_{i}^{-1}g_{i})\rightarrow s-\tilde{s}$ pointwise as
$i\rightarrow\infty$. Now it suffices to show that if $\\{\gamma_{i}\\}$ is a
sequence in $\Gamma$ such that
$\Psi(\gamma_{i})\rightarrow(\text{Id},\text{Id})$ and
$b_{j}(\gamma_{i})\rightarrow s$ pointwise for some $s\in E_{j}$, then
$s\equiv 0$.

We suppose $s\not\equiv 0$ and shall get a contradiction. There is some
$\bar{h}\in\mathfrak{n}/{\mathfrak{w}}$ such that $s(\bar{h})\not=0$. Fix some
$m\geq 1$ such that
$m\cdot\frac{|s(\bar{h})|_{j}}{2}>C\cdot\bar{d}_{CC}(\bar{h},0)^{\frac{j}{\alpha}}.$
Note $\Psi(\gamma^{k}_{i})\rightarrow(\text{Id},\text{Id})$ for $1\leq k\leq
m$. On the other hand, an easy induction using Lemma 7.4 implies

$b_{j}(\gamma^{m}_{i})=b_{j}(\gamma_{i})+\pi_{\Psi(\gamma_{i})}b_{j}(\gamma_{i})+\cdots+\pi_{\Psi(\gamma^{m-1}_{i})}b_{j}(\gamma_{i}).$

Since $\Psi(\gamma^{k}_{i})\rightarrow(\text{Id},\text{Id})$ and
$b_{j}(\gamma_{i})\rightarrow s$ pointwise, we have
$|\pi_{\Psi(\gamma^{k}_{i})}b_{j}(\gamma_{i})(\bar{h})-s(\bar{h})|_{j}<\frac{|s(\bar{h})|_{j}}{2}$
for all $0\leq k\leq m-1$ and all sufficiently large $i$. It follows from the
triangle inequality that
$|b_{j}(\gamma^{m}_{i})(\bar{h})-ms(\bar{h})|_{j}<m\cdot\frac{|s(\bar{h})|_{j}}{2}$
and hence
$|b_{j}(\gamma^{m}_{i})(\bar{h})|_{j}>m\cdot\frac{|s(\bar{h})|_{j}}{2}>C\cdot\bar{d}_{CC}(\bar{h},0)^{\frac{j}{\alpha}}$,
contradicting the fact that $||b_{j}(\gamma)||\leq C$ for all
$\gamma\in\Gamma$.

∎

Lemma 7.5 in particular implies that $b_{j}(\gamma)=b_{j}(\tilde{\gamma})$
holds whenever $\Psi(\gamma)=\Psi(\tilde{\gamma})$ (by taking $g_{i}=\gamma$
and $\tilde{g}_{i}=\tilde{\gamma}$). Since $b_{j}(\gamma)$ lies in the closed
ball $\bar{B}(0,C)\subset E_{j}$ for all $\gamma\in\Gamma$ and $\bar{B}(0,C)$
is compact in the topology of pointwise convergence, Lemma 7.5 implies that
for any $(A,B)\in G$ and any sequence $\\{\gamma_{i}\\}$ satisfying
$\Psi(\gamma_{i})\rightarrow(A,B)$, the sequence $b_{j}(\gamma_{i})$ converges
to some $s\in\bar{B}(0,C)$ pointwise and $s$ is independent of the choice of
the sequence $\\{\gamma_{i}\\}$. We denote this $s$ by $\tilde{b}_{j}(A,B)$.
It follows that the map $\tilde{b}_{j}:G\rightarrow E_{j}$ is well-defined and
continuous, where $E_{j}$ is equipped with the topology of pointwise
convergence. This map $\tilde{b}_{j}$ is the translational part of the affine
action.

To see that the map $\tilde{b}_{j}$ is actually the translational part of an
affine action, we need to show that $\tilde{b}_{j}$ is a $1$-cocycle
associated to the $G^{*}$-module $E_{j}$; that is,

$\tilde{b}_{j}((A_{2},B_{2})(A_{1},B_{1}))=\tilde{b}_{j}(A_{1},B_{1})+\pi_{(A_{1},B_{1})}\tilde{b}_{j}(A_{2},B_{2})$

holds for all $(A_{1},B_{1}),(A_{2},B_{2})\in G$. This follows from Lemma 7.4
and the definition of $\tilde{b}_{j}$. The associated affine action of $G^{*}$
on $E_{j}$ is given by

$(A,B)\cdot c=\pi_{(A,B)}c+\tilde{b}_{j}(A,B),$

where $c\in E_{j}$ and $(A,B)\in G$. Since $\tilde{b}_{j}$ is continuous, it
is now easy to see that the affine action $G^{*}\times E_{j}\rightarrow E_{j}$
is separately continuous, where $E_{j}$ is equipped with the topology of
pointwise convergence.

###  7.2. Existence of fixed point in $\mathcal{H}_{j}$

Fix some $c\in\mathcal{H}_{j}\subset E_{j}$, let $G^{*}\cdot c$ be the orbit
of $c$ under the affine action of $G^{*}$. Let $K$ be the closure of the
convex hull of $G^{*}\cdot c$ in $E_{j}$ with respect to the topology of
pointwise convergence. In order to use Day’s fixed point theorem to obtain a
fixed point in $\mathcal{H}_{j}$, we need to show that $K$ is compact and lies
in $\mathcal{H}_{j}$. The following lemma will be useful for this purpose.

######  Lemma 7.6.

Let $Y\subset\mathcal{H}_{j}$ be bounded with respect to the norm $||\cdot||$.
Then $\bar{Y}\subset\mathcal{H}_{j}$, where $\bar{Y}$ is the closure of $Y$ in
$E_{j}$ in the topology of pointwise convergence.

###### Proof.

Since $Y$ is bounded in $(E_{j},||\cdot||)$, there is a constant $C_{0}>0$
such that $||s||\leq C_{0}$ for any $s\in Y$. It follows that for any $i\geq
1$ and any compact subset $M\subset\mathfrak{n}/\mathfrak{w}$, there is a
constant $C(C_{0},M,i)>0$ such that $|s^{(i)}(x)|\leq C(C_{0},M,i)$ for all
$x\in M$ and all $s\in Y$. Let $s_{k}\in Y$, $k=1,\cdots$ and $s\in E_{j}$ be
such that $s_{k}(x)\to s(x)$ for any $x\in\mathfrak{n}/\mathfrak{w}$. The
dominated convergence theorem then implies
$\int_{\gamma}[s_{k},\theta_{H}(x)]\to\int_{\gamma}[s,\theta_{H}(x)]$ for any
horizontal curve $\gamma$ in $\mathfrak{n}/\mathfrak{w}$. This implies
$s^{(1)}$ is defined and $s_{k}^{(1)}\to s^{(1)}$ pointwise. Now an induction
shows $s^{(i)}$ is defined for all $i$ and so $s\in\mathcal{H}_{j}$.

∎

If $\alpha$ is not an integer, then $\mathcal{H}_{j}=E_{j}$ so
$K\subset\mathcal{H}_{j}$ holds automatically. To show
$K\subset\mathcal{H}_{j}$, we may assume $\alpha\geq 2$ is an integer. We
first prove $G^{*}\cdot c\subset\mathcal{H}_{j}$. By the definition of the
affine action we need to show that for any $1\leq j\leq\alpha-1$:
(1) $\pi_{(A,B)}c\in\mathcal{H}_{j}$ for any $(A,B)\in G$ and any
$c\in\mathcal{H}_{j}$; see Lemma 7.7.
(2) $\tilde{b}_{j}(A,B)\in\mathcal{H}_{j}$ for any $(A,B)\in G$; see the
paragraph before Lemma 7.8.

######  Lemma 7.7.

$\pi_{(A,B)}c\in\mathcal{H}_{j}$ for any integer $1\leq j<\alpha$, $(A,B)\in
G$ and any $c\in\mathcal{H}_{j}$.

###### Proof.

Let $(A,B)\in G$ and $c\in\mathcal{H}_{j}$. Then there is a sequence
$\\{\gamma_{i}\\}$ in $\Gamma$ such that $\Psi(\gamma_{i})\rightarrow(A,B)$.
Notice that $\pi_{\Psi(\gamma_{i})}c\rightarrow\pi_{(A,B)}c$ pointwise. Since
$||\pi_{\Psi(\gamma_{i})}c||=||c||$, by Lemma 7.6 to show
$\pi_{(A,B)}c\in\mathcal{H}_{j}$ it suffices to show
$\pi_{\Psi(\gamma)}c\in\mathcal{H}_{j}$ for all $\gamma\in\Gamma$. For this we
shall show that there is a biLipschitz shear map
$f_{2}(g)=g*\tilde{s}(\bar{g})$ with $\tilde{s}_{j}=\pi_{\Psi(\gamma)}c$. The
lemma then follows from Proposition 6.4.

Pick $c\in\mathcal{H}_{j}$ and $\gamma\in\Gamma$. Let
$s:\mathfrak{n}/\mathfrak{w}\rightarrow Z(\mathfrak{w})$ be given by
$s=\sum_{i}s_{i}$, where $s_{j}=c$, $s_{i}=0$ for $1\leq i\leq\alpha$,
$i\not=j$ and $s_{k\alpha+i}=s_{i}^{(k)}$ for $k\geq 1$ and $1\leq
i\leq\alpha$. By Proposition 6.7, the shear map $f(g)=g*s(\bar{g})$ is
biLipschitz. We shall show $f_{2}:=L_{f_{1}(0)^{-1}}\circ f_{1}$ with
$f_{1}:=\gamma^{-1}f\gamma$ is the desired biLipschitz shear map.

Notice that for any two maps $F,G:N\rightarrow N$ and any $p\in N$, we have
$(F\circ G)_{p}=F_{G(p)}\circ G_{p}$. It is clear that $f_{1}$ is a
biLipschitz map of $N$ that permutes the cosets of $W$. Since
$f_{p}|_{W}=\text{Id}$ and $\gamma_{p}|_{W}=A_{\gamma}$, we see that
$(f_{1})_{p}|_{W}=\text{Id}$. On the other hand, as $f$ induces the identity
map on $N/W$, so does $f_{1}$. Hence the assumption of Lemma 5.4 are satisfied
and $f_{1}$ has a compatible expression
$f_{1}(h*w)=f_{1}(0)*h*w*\tilde{s}(\bar{h})$ for some map
$\tilde{s}:{\mathfrak{n}}/{\mathfrak{w}}\rightarrow Z(\mathfrak{w})$. It
follows that $f_{2}(h*w)=h*w*\tilde{s}(\bar{h})$ is a biLipschitz shear map.
It remains to show $\tilde{s}_{j}=\pi_{\Psi(\gamma)}c$. Note
$\tilde{s}_{j}=\pi_{j}(f_{1}(0)^{-1}*f_{1}(h))$.

Fix a compatible expression
$\gamma(h*w)=\gamma(0)*B_{\gamma}h*A_{\gamma}w*A_{\gamma}s_{\gamma}(\bar{h})$
of $\gamma$. We calculate $f_{1}(h)$:

$\displaystyle f_{1}(h)=\gamma^{-1}f\gamma(h)$
$\displaystyle=\gamma^{-1}f(\gamma(0)*B_{\gamma}h*A_{\gamma}s_{\gamma}(\bar{h}))$
$\displaystyle=\gamma^{-1}(\gamma(0)*B_{\gamma}h*A_{\gamma}s_{\gamma}(\bar{h})*s(\overline{\gamma(0)}*\bar{B}_{\gamma}\bar{h}))$
$\displaystyle=h*A^{-1}_{\gamma}s(\overline{\gamma(0)}*\bar{B}_{\gamma}\bar{h}),$

where the last equality follows from the fact that $\gamma$ is a bijection and

$\gamma(h*\\{A^{-1}_{\gamma}s(\overline{\gamma(0)}*\bar{B}_{\gamma}\bar{h})\\})=\gamma(0)*B_{\gamma}h*s(\overline{\gamma(0)}*\bar{B}_{\gamma}\bar{h})*A_{\gamma}s_{\gamma}(\bar{h}).$

In particular, $f_{1}(0)=A_{\gamma}^{-1}s(\overline{\gamma(0)})$. Now
$f_{2}(h)=A^{-1}_{\gamma}s(\overline{\gamma(0)})^{-1}*h*A^{-1}_{\gamma}s(\overline{\gamma(0)}*\bar{B}_{\gamma}\bar{h})$
and from this we see

$s_{f_{2},j}(\bar{h})=A^{-1}_{\gamma}s_{j}(\overline{\gamma(0)}*\bar{B}_{\gamma}\bar{h})-A^{-1}_{\gamma}s_{j}(\overline{\gamma(0)})=A^{-1}_{\gamma}c(\overline{\gamma(0)}*\bar{B}_{\gamma}\bar{h})-A^{-1}_{\gamma}c(\overline{\gamma(0)})=(\pi_{\Psi(\gamma)}c)(\bar{h}).$

∎

Let $(A,B)\in G$. By the definition of $\tilde{b}_{j}$, there is a sequence
$\\{\gamma_{i}\\}$ in $\Gamma$ such that
$b_{j}(\gamma_{i})\rightarrow\tilde{b}_{j}(A,B)$ pointwise. Since
$b_{j}(\gamma_{i})$ lies in the closed ball $\bar{B}(0,C)$ (see Corollary
7.10), Lemma 7.6 implies $\tilde{b}_{j}(A,B)\in{\mathcal{H}}_{j}$ provided
$b_{j}(\gamma)\in\mathcal{H}_{j}$ for all $\gamma\in\Gamma$. This is true by
(2) of the following lemma.

######  Lemma 7.8.

Let $F:\mathfrak{n}\rightarrow\mathfrak{n}$ be a fiber similarity map and
$F(h*w)=F(0)*Bh*Aw*As(\bar{h})$ a compatible expression of $F$. Then
(1) for $1\leq j<\alpha$, $s_{j}\in E_{j}$ with $||s_{j}||$ bounded above by a
constant depending only on $H$ and the biLipschitz constant of $F$.
(2) if $\alpha$ is an integer, then $s_{j}\in\mathcal{H}_{j}$ for any $1\leq
j<\alpha$.
(3) if $\alpha$ is an integer, then $s_{\alpha}\in E_{\alpha}$ with
$||s_{\alpha}||$ bounded above by a constant depending only on $H$, the
biLipschitz constant of $F$ and the map $B$.

######  Remark 7.9.

The map $F$ may admit many different compatible expressions: both $B$ and
$s_{\alpha}$ may change and as a result their Lipschitz constants may be
arbitrarily large while $F$ is fixed.

###### Proof.

We will use the fact that there is a constant $L_{1}\geq 1$ depending only on
$H$ such that $(1/{L_{1}})\cdot d^{\frac{1}{\alpha}}_{CC}(0,\bar{h})\leq
d(0,h)\leq L_{1}\cdot d^{\frac{1}{\alpha}}_{CC}(0,\bar{h})$ for any $h\in H$,
where $d_{CC}$ is a Carnot metric on $\mathfrak{n}/\mathfrak{w}$. Since $F$ is
$L_{2}$-biLipschitz for some $L_{2}\geq 1$, for any $h_{0},h\in H$ we have

$d(0,F(h_{0})^{-1}*F(h_{0}*h))=d(F(h_{0}),F(h_{0}*h))\leq L_{2}\cdot
d(h_{0},h_{0}*h)=L_{2}\cdot d(0,h)\leq
L_{2}L_{1}d^{\frac{1}{\alpha}}_{CC}(0,\bar{h}).$

There is a constant $L>0$ depending on $B$ such that for any $h\in H$ we have

$d(0,Bh)\leq L\cdot
d^{\frac{1}{\alpha}}_{CC}(0,\bar{B}\bar{h})=L\lambda_{\bar{B}}^{\frac{1}{\alpha}}\cdot
d^{\frac{1}{\alpha}}_{CC}(0,\bar{h}),$

where $\lambda_{\bar{B}}$ denotes the similarity constant of $\bar{B}$. We
also have

$\displaystyle
d(0,(Bh)^{-1}*F(h_{0})^{-1}*F(h_{0}*h))=d(Bh,F(h_{0})^{-1}*F(h_{0}*h))$
$\displaystyle\leq
d(Bh,0)+d(0,F(h_{0})^{-1}*F(h_{0}*h))\leq(L\lambda_{\bar{B}}^{\frac{1}{\alpha}}+L_{2}L_{1})d^{\frac{1}{\alpha}}_{CC}(0,\bar{h}).$

Let $F_{0}:\mathfrak{n}\rightarrow\mathfrak{n}$ be the shear map given by
$F_{0}(g)=g*s(\bar{g})$.

Claim: (a)
$\pi_{j}(F(h_{0})^{-1}*F(h_{0}*h))=As_{j}(\overline{h_{0}}*\bar{h})-As_{j}(\overline{h_{0}})$
holds for $1\leq j<\alpha$;
(b)
$\pi_{i}((Bh)^{-1}*F(h_{0})^{-1}*F(h_{0}*h))=A\pi_{i}(h^{-1}*F_{0}(h_{0})^{-1}*F_{0}(h_{0}*h))$
for any integer $i$ that is not an integral multiple of $\alpha$, where
$\pi_{i}:\mathfrak{n}\rightarrow V_{i}=W_{i}$ is the projection with respect
to the decomposition $\mathfrak{n}=\oplus V_{\lambda_{k}}$.
(c)
$\pi_{\alpha}((Bh)^{-1}*F(h_{0})^{-1}*F(h_{0}*h))=As_{\alpha}(\overline{h_{0}}*\bar{h})-As_{\alpha}(\overline{h_{0}})$.

We first assume the Claim and finish the proof of the lemma. Part (a) of the
claim implies for $1\leq j<\alpha$,

$\displaystyle|s_{j}(\overline{h_{0}}*\bar{h})-s_{j}(\overline{h_{0}})|^{\frac{1}{j}}$
$\displaystyle\leq
L_{2}\cdot|\pi_{j}(F(h_{0})^{-1}*F(h_{0}*h))|^{\frac{1}{j}}$
$\displaystyle\leq L_{2}\cdot L_{2}L_{1}d^{\frac{1}{\alpha}}_{CC}(0,\bar{h})$
$\displaystyle=L^{2}_{2}L_{1}d^{\frac{1}{\alpha}}_{CC}(\overline{h_{0}},\overline{h_{0}}*\bar{h}),$

hence (1) holds. Now notice
$h^{-1}*F_{0}(h_{0})^{-1}*F_{0}(h_{0}*h)=K(\overline{h_{0}},\overline{h_{0}*h})$,
where $K(\bar{g}_{1},\bar{g}_{2})$ is defined for $g_{1},g_{2}\in\mathfrak{n}$
before Lemma 6.3. So when $\alpha$ is an integer, Part (b) of the claim
implies

(9)
$|\pi_{k\alpha+j}(K(\overline{h_{0}},\overline{h_{0}*h}))|^{\frac{1}{k\alpha+j}}\leq
L_{2}(L\lambda_{\bar{B}}^{\frac{1}{\alpha}}+L_{2}L_{1})\cdot
d^{\frac{1}{\alpha}}_{CC}(0,\bar{h})$

for all $k\geq 0$ and all $1\leq j<\alpha$. Then (2) follows since by (9) the
assumption of Lemma 6.3 is satisfied. Finally (3) follows from Part (c):

$|s_{\alpha}(\overline{h_{0}}*\bar{h})-s_{\alpha}(\overline{h_{0}})|^{\frac{1}{\alpha}}\leq
L_{2}|\pi_{\alpha}((Bh)^{-1}*F(h_{0})^{-1}*F(h_{0}*h))|^{\frac{1}{\alpha}}\leq
L_{2}(L\lambda_{\bar{B}}^{\frac{1}{\alpha}}+L_{2}L_{1})d^{\frac{1}{\alpha}}_{CC}(0,\bar{h}).$

Next we prove the claim. Write $h_{0}*h=h_{1}*w_{1}$ for some $h_{1}\in H$,
$w_{1}\in\oplus_{i}W_{i\alpha}$. Applying $\pi_{\alpha}$ to both sides we get
$\pi_{\alpha}(h_{0})+\pi_{\alpha}(h)=\pi_{\alpha}(h_{1})+\pi_{\alpha}(w_{1})$.
As $V_{\alpha}=W_{\alpha}\oplus H_{1}$ is a direct sum, we obtain
$\pi_{\alpha}(h_{0})+\pi_{\alpha}(h)=\pi_{\alpha}(h_{1})$ and
$\pi_{\alpha}(w_{1})=0$; hence
$\pi_{\alpha}(Bh_{0})+\pi_{\alpha}(Bh)=\pi_{\alpha}(Bh_{1})$ and
$w_{1}\in\oplus_{i\geq 2}W_{i\alpha}$. Note
$\bar{h}_{1}=\overline{h_{0}}*\bar{h}$. We have
$F(h_{0})=F(0)*Bh_{0}*As(\overline{h_{0}})$ and
$F(h_{0}*h)=F(h_{1}*w_{1})=F(0)*Bh_{1}*Aw_{1}*As(\overline{h_{0}}*\bar{h})$.
Using $Bh*Aw*(Bh)^{-1}=A(h*w*h^{-1})$ twice we get:

$\displaystyle(Bh)^{-1}*(F(h_{0}))^{-1}*F(h_{0}*h)$
$\displaystyle=(Bh)^{-1}*A(-s(\overline{h_{0}}))*(Bh_{0})^{-1}*Bh_{1}*Aw_{1}*As(\overline{h_{0}}*\bar{h})$
$\displaystyle=(Bh)^{-1}*(Bh_{0})^{-1}*\\{Bh_{0}*A(-s(\overline{h_{0}}))*(Bh_{0})^{-1}\\}*Bh_{1}*Aw_{1}*As(\overline{h_{0}}*\bar{h})$
$\displaystyle=(Bh)^{-1}*(Bh_{0})^{-1}*A(h_{0}*(-s(\overline{h_{0}}))*h_{0}^{-1})*Bh_{1}*Aw_{1}*As(\overline{h_{0}}*\bar{h})$
$\displaystyle=(Bh)^{-1}*(Bh_{0})^{-1}*Bh_{1}*\\{(Bh_{1})^{-1}*A(h_{0}*(-s(\overline{h_{0}}))*h_{0}^{-1})*Bh_{1}\\}*Aw_{1}*As(\overline{h_{0}}*\bar{h})$
$\displaystyle=(Bh)^{-1}*(Bh_{0})^{-1}*Bh_{1}*A(h_{1}^{-1}*h_{0}*(-s(\overline{h_{0}}))*h_{0}^{-1}*h_{1})*Aw_{1}*As(\overline{h_{0}}*\bar{h})$
$\displaystyle=(Bh)^{-1}*(Bh_{0})^{-1}*Bh_{1}*A(w_{1}*h^{-1}*(-s(\overline{h_{0}}))*h*w^{-1}_{1})*Aw_{1}*As(\overline{h_{0}}*\bar{h})$
$\displaystyle=(Bh)^{-1}*(Bh_{0})^{-1}*Bh_{1}*Aw_{1}*A(h^{-1}*(-s(\overline{h_{0}}))*h)*As(\overline{h_{0}}*\bar{h}),$

where for the sixth equality we used $h_{1}*w_{1}=h_{0}*h$. We first prove
Part (c): as $\pi_{\alpha}(Bh_{0})+\pi_{\alpha}(Bh)=\pi_{\alpha}(Bh_{1})$ and
$w_{1}\in\oplus_{i\geq 2}W_{i\alpha}$, the above calculation yields

$\displaystyle\pi_{\alpha}((Bh)^{-1}*F(h_{0})^{-1}*F(h_{0}*h))$
$\displaystyle=\pi_{\alpha}(A(h^{-1}*(-s(\overline{h_{0}}))*h)*As(\overline{h_{0}}*\bar{h}))$
$\displaystyle=As_{\alpha}(\overline{h_{0}}*\bar{h})-As_{\alpha}(\overline{h_{0}}).$

The above calculation also gives $(Bh)^{-1}*(F(h_{0}))^{-1}*F(h_{0}*h)=P*Az$,
where $P=(Bh)^{-1}*(Bh_{0})^{-1}*Bh_{1}*Aw_{1}$ and
$z=s(\overline{h_{0}}*\bar{h})+h^{-1}*(-s(\overline{h_{0}}))*h$. Since $z$
lies in $Z(\mathfrak{w})$, the BCH formula gives
$(Bh)^{-1}*(F(h_{0}))^{-1}*F(h_{0}*h)=P+Az+Q$, where $Q$ is a sum of iterated
brackets of $P$ and $Az$, and $Az$ appears exactly once in each of these
iterated brackets. Notice that the BCH formula also implies $P$ is a sum of
iterated brackets of the terms $Bh$, $Bh_{0}$, $Bh_{1}$, $Aw_{1}$. Now
$[Bh,Aw]=A[h,w]$ and the Jacobi identity imply that $Q=A\tilde{Q}$, where
$\tilde{Q}$ is obtained from the sum of iterated brackets that gives rise to
$Q$ by replacing $Bh$, $Bh_{0}$, $Bh_{1}$, $Aw_{1}$, $Az$ by $h$, $h_{0}$,
$h_{1}$, $w_{1}$, $z$ respectively. So
$(Bh)^{-1}*(F(h_{0}))^{-1}*F(h_{0}*h)=P+Az+A\tilde{Q}$.

Now comparing the formulas $F(h*w)=F(0)*Bh*Aw*As(\bar{h})$ and
$F_{0}(h*w)=h*w*s(\bar{h})$, and repeating the above calculation for
$h^{-1}*(F_{0}(h_{0}))^{-1}*F_{0}(h_{0}*h)$ (we only need to replace $A$ with
$\text{Id}_{W}$ and $B$ with $\text{Id}_{H}$), we get

$h^{-1}*(F_{0}(h_{0}))^{-1}*F_{0}(h_{0}*h)=\tilde{P}+z+\tilde{Q},$

where $\tilde{P}$ is obtained from the sum of iterated brackets that gives
rise to $P$ by replacing $Bh$, $Bh_{0}$, $Bh_{1}$, $Aw_{1}$ by $h$, $h_{0}$,
$h_{1}$, $w_{1}$ respectively. Now Part (b) of the claim follows since
$P,\tilde{P}\in\oplus_{i}V_{i\alpha}$.

Next we prove Part (a) of the Claim. As
$h,Bh\in\oplus_{\lambda_{i}\geq\alpha}V_{\lambda_{i}}$, for $1\leq j<\alpha$,
by Claim (b) we have

$\displaystyle\pi_{j}(F(h_{0})^{-1}*F(h_{0}*h))$
$\displaystyle=\pi_{j}((Bh)^{-1}*F(h_{0})^{-1}*F(h_{0}*h))=A\pi_{j}(h^{-1}*F_{0}(h_{0})^{-1}*F_{0}(h_{0}*h))$
$\displaystyle=A\pi_{j}(F_{0}(h_{0})^{-1}*F_{0}(h_{0}*h))=A\pi_{j}((-s(\bar{h}_{0})*h_{0}^{-1}*h_{0}*h*s(\bar{h}_{0}*\bar{h}))$
$\displaystyle=A\pi_{j}(s(\bar{h}_{0}*\bar{h})-s(\bar{h}_{0}))=A(s_{j}(\bar{h}_{0}*\bar{h})-s_{j}(\bar{h}_{0})).$

∎

######  Corollary 7.10.

Let $\Gamma$ be a fiber similarity group of $N$. Then there is a constant
$C>0$ such that $||s_{\gamma,j}||<C$ for all $\gamma\in\Gamma$ and all $1\leq
j<\alpha$.

###### Proof.

The inequality $\frac{C_{\gamma}}{L^{2}\Lambda}\leq\lambda_{A_{\gamma}}\leq
L^{2}\Lambda C_{\gamma}$ we obtained in the proof of Lemma 7.3 implies that
there is a constant $K_{0}>0$ such that
$e^{-\log(\lambda_{A_{\gamma}})D}\circ\gamma$ is $K_{0}$-biLipschitz for every
$\gamma\in\Gamma$. An element $\gamma\in\Gamma$ with compatible expression
$\gamma(h*w)=\gamma(0)*B_{\gamma}h*A_{\gamma}w*A_{\gamma}s_{\gamma}(\bar{h})$
can be written as
$\gamma(h*w)=\gamma(0)*e^{\log(\lambda_{A_{\gamma}})D}(B^{\prime}_{\gamma}h*A^{\prime}_{\gamma}w*A^{\prime}_{\gamma}s_{\gamma}(\bar{h}))$,
where $s_{\gamma}$ is as before and $A^{\prime}_{\gamma}$ is an isometric
graded automorphism of $(W,d_{CC})$. It follows that
$e^{-\log(\lambda_{A_{\gamma}})D}\circ\gamma$ has a compatible expression
$e^{-\log(\lambda_{A_{\gamma}})D}\circ\gamma(h*w)=a^{\prime}_{\gamma}*B^{\prime}_{\gamma}h*A^{\prime}_{\gamma}w*A^{\prime}_{\gamma}s_{\gamma}(\bar{h})$.
The corollary now follows by applying Lemma 7.8 to
$e^{-\log(\lambda_{A_{\gamma}})D}\circ\gamma$.

∎

We have shown that $G^{*}\cdot c\subset{\mathcal{H}}_{j}$. Corollary 7.10
implies $G^{*}\cdot c$ is bounded. By Lemma 7.6 $K\subset{\mathcal{H}}_{j}$.
We next show that the affine action has a fixed point in $K$. For this purpose
we use

Theorem (Day’s fixed point theorem) [Da61] _Let $K$ be a compact convex subset
of a locally convex topological vector space $E$ and let $\Gamma$ be a locally
compact group that acts on $K$ by affine transformations. If $\Gamma$ is
amenable and the action $\Gamma\times K\rightarrow
K,(\gamma,x)\mapsto\gamma\cdot x,$ is separately continuous, then the action
of $\Gamma$ has a global fixed point in $K$. _

######  Lemma 7.11.

The affine action associated to the $1$-cocycle $\tilde{b}_{j}$ has a fixed
point $c\in\mathcal{H}_{j}$.

###### Proof.

We equip $E_{j}$ with the topology of pointwise convergence. Then $E_{j}$ is a
locally convex topological vector space and $K$ is a compact convex subset (as
$K$ is bounded). We have observed that the affine action $G^{*}\times
K\rightarrow K$ is separately continuous. Since $G^{*}$ is amenable, by Day’s
fixed point theorem, the affine action has a fixed point in $K$.

∎

###  7.3. Eliminating $s_{j}$

######  Lemma 7.12.

Let $1\leq j<\alpha$.
Suppose the affine action of $G^{*}$ on $E_{j}$ has a fixed point
$c\in\mathcal{H}_{j}$. Let $s:\mathfrak{n}/\mathfrak{w}\rightarrow
Z(\mathfrak{w})$ be the map provided by Proposition 6.7 satisfying $s_{j}=c$
and $s_{i}=0$ for $1\leq i\leq\alpha$, $i\not=j$ such that the corresponding
shear map $F_{0}(g)=g*s(\bar{g})$ is biLipschitz. Then every element
$\tilde{\gamma}\in F_{0}\Gamma F_{0}^{-1}$ has a compatible expression
$\tilde{\gamma}(h*w)=\tilde{\gamma}(0)*B_{\gamma}h*A_{\gamma}w*A_{\gamma}\tilde{s}_{\gamma}(\bar{h})$
with $\tilde{s}_{\gamma,j}=0$.

###### Proof.

For $\gamma\in\Gamma$, denote $\tilde{\gamma}=F_{0}\circ\gamma\circ
F_{0}^{-1}$. Consider a compatible expression
$\gamma(h*w)=\gamma(0)*B_{\gamma}h*A_{\gamma}w*A_{\gamma}s_{\gamma}(\bar{h})$
of $\gamma$. Since $c$ is a fixed point of the affine action, we have
$\pi_{\Psi(\gamma)}c+s_{\gamma,j}=c$ for all $\gamma\in\Gamma$. Note $F_{0}$
and $F_{0}^{-1}$ have the expressions: $F_{0}(h*w)=h*w*s(\bar{h})$,
$F_{0}^{-1}(h*w)=h*w*(-s(\bar{h}))$. We now calculate

$\displaystyle\tilde{\gamma}(h)=F_{0}\circ\gamma(h*(-s(\bar{h})))$
$\displaystyle=F_{0}(\gamma(0)*B_{\gamma}h*A_{\gamma}[s_{\gamma}(\bar{h})*(-s(\bar{h}))])$
$\displaystyle=\gamma(0)*B_{\gamma}h*A_{\gamma}[s_{\gamma}(\bar{h})*(-s(\bar{h}))]*s(\overline{\gamma(0)}*\bar{B}_{\gamma}\bar{h}).$

In particular, $\tilde{\gamma}(0)=\gamma(0)*s(\overline{\gamma(0)})$. Now

$\displaystyle s_{\tilde{\gamma},j}(\bar{h})$
$\displaystyle=A^{-1}_{\gamma}\pi_{j}(\tilde{\gamma}(0)^{-1}*\tilde{\gamma}(h))$
$\displaystyle=A^{-1}_{\gamma}\pi_{j}\\{(-s(\overline{\gamma(0)}))*B_{\gamma}h*A_{\gamma}[s_{\gamma}(\bar{h})*(-s(\bar{h}))]*s(\overline{\gamma(0)}*\bar{B}_{\gamma}\bar{h})\\}$
$\displaystyle=A^{-1}_{\gamma}\pi_{j}\\{(-s(\overline{\gamma(0)}))*A_{\gamma}[s_{\gamma}(\bar{h})*(-s(\bar{h}))]*s(\overline{\gamma(0)}*\bar{B}_{\gamma}\bar{h})\\}$
$\displaystyle=A^{-1}_{\gamma}\\{-c(\overline{\gamma(0)})+A_{\gamma}s_{\gamma,j}(\bar{h})+A_{\gamma}(-c(\bar{h}))+c(\overline{\gamma(0)}*\bar{B}_{\gamma}\bar{h})\\}$
$\displaystyle=s_{\gamma,j}(\bar{h})-c(\bar{h})+A^{-1}_{\gamma}c(\overline{\gamma(0)}*\bar{B}_{\gamma}\bar{h})-A^{-1}_{\gamma}c(\overline{\gamma(0)})$
$\displaystyle=s_{\gamma,j}(\bar{h})-c(\bar{h})+(\pi_{\Psi(\gamma)}c)(\bar{h})=0.$

∎

The proof of Proposition 7.1 is now complete.

##  8\. Conformal structures in the $V_{\alpha}$ direction

We continue the proof of Theorem 1.2. In Section 7 we showed that we can get
rid of $s_{\gamma,j}$ for $j<\alpha$ after a conjugation. In this section we
continue to assume $\dim(W)\geq 2$, $\dim(N/W)\geq 2$. We will first show
that, if $\alpha$ is an integer, then after a conjugation,
$s_{\gamma,\alpha}:\mathfrak{n}/\mathfrak{w}\rightarrow
Z_{\alpha}(\mathfrak{w})$ is a Lie group homomorphism for every
$\gamma\in\Gamma$, and then we complete the proof of Theorem 1.2 under the
assumptions $\dim(W)\geq 2$, $\dim(N/W)\geq 2$. Here we are using the
identification between simply connected nilpotent Lie groups and their Lie
algebras via the exponential map.

We will imitate the proof of Tukia’s theorem, with the last step different. As
biLipschitz maps of Carnot-by-Carnot groups in general are not differentiable,
we can not directly work with the differentials. Instead we look for
differentiability in the $V_{\alpha}$ direction. For each biLipschitz map
$F:\mathfrak{n}\rightarrow\mathfrak{n}$ and each $p\in\mathfrak{n}$, we
consider the differential of the map

$F_{p,\alpha}:=(\pi_{\alpha}\circ F_{p})|_{V_{\alpha}}:V_{\alpha}\rightarrow
V_{\alpha},$

with $F_{p}=L_{F(p)^{-1}}\circ F\circ L_{p}$, where $L_{g}$ is left
translation by the element $g$ and $\pi_{\alpha}:\mathfrak{n}\rightarrow
V_{\alpha}$ is the projection with respect to the decomposition
$\mathfrak{n}=\oplus_{j}V_{\lambda_{j}}$.

In the next two subsections we assume $\alpha$ is an integer.

###  8.1. Differential of $F_{p,\alpha}$

In this subsection we first find a formula for the differential of
$F_{p,\alpha}$ at $0$ and then show that this differential satisfies the chain
rule (Lemma 8.2).

Let $(N,D)$ be Carnot-by-Carnot and $F:\mathfrak{n}\rightarrow\mathfrak{n}$ be
a biLipschitz map satisfying the assumptions of Lemma 5.4. Then $F$ has a
compatible expression $F(h*w)=F(0)*Bh*Aw*As(\bar{h}).$ Write $s=\sum_{j}s_{j}$
with $s_{j}:\mathfrak{n}/\mathfrak{w}\rightarrow Z_{j}(\mathfrak{w})$. By
Lemma 7.8 $s_{j}$ is $\frac{j}{\alpha}$-Holder for each $1\leq j\leq\alpha$.

To simplify the calculations we will work in a suitable quotient of
$\mathfrak{n}$. Notice that $\oplus_{\lambda_{j}>\alpha}V_{\lambda_{j}}$ is an
ideal of $\mathfrak{n}$. Denote
$\bar{\mathfrak{n}}_{\alpha}=\mathfrak{n}/{(\oplus_{\lambda_{j}>\alpha}V_{\lambda_{j}})}$
and let $P_{\alpha}:\mathfrak{n}\rightarrow\bar{\mathfrak{n}}_{\alpha}$ be the
quotient homomorphism. Observe that if
$x\in\oplus_{\lambda_{j}\geq\alpha}V_{\lambda_{j}}$, then $P_{\alpha}(x)$ lies
in the center of $\bar{\mathfrak{n}}_{\alpha}$. Let $h\in H$,
$w\in\mathfrak{w}$. There is some $w^{\prime}\in\mathfrak{w}$ such that
$h+w=h*w^{\prime}$. By applying $P_{\alpha}$ to both sides of
$h+w=h*w^{\prime}$ and noting that $P_{\alpha}(h)$ lies in the center of
$\bar{\mathfrak{n}}_{\alpha}$, we get $P_{\alpha}(w)=P_{\alpha}(w^{\prime})$.
Now write $p=h_{0}*w_{0}$ with $h_{0}\in H$ and $w_{0}\in\mathfrak{w}$, and
$h_{0}*h=\tilde{h}+\tilde{w}=\tilde{h}*{\tilde{\tilde{w}}}$ for some
$\tilde{h}\in H$ and $\tilde{w},{\tilde{\tilde{w}}}\in\mathfrak{w}$. Using BCH
formula we see that $\tilde{w},{\tilde{\tilde{w}}}\in\oplus_{j>\alpha}W_{j}$.
By applying $P_{\alpha}$ to both sides of $h_{0}*h=\tilde{h}+\tilde{w}$ we get
$P_{\alpha}(\tilde{h})=P_{\alpha}(h_{0})+P_{\alpha}(h)$. It follows that
$P_{\alpha}(B\tilde{h})=P_{\alpha}(Bh_{0})+P_{\alpha}(Bh)$ and so
$P_{\alpha}((Bh_{0})^{-1}*B\tilde{h})=P_{\alpha}(Bh)$.

We are ready to find a formula for $F_{p,\alpha}$. Recall $p=h_{0}*w_{0}$.
Write

$p*(h+w)=h_{0}*w_{0}*h*w^{\prime}=h_{0}*h*(h^{-1}*w_{0}*h)*w^{\prime}=\tilde{h}*{\tilde{\tilde{w}}}*(h^{-1}*w_{0}*h)*w^{\prime}.$

So

$F(p)=F(0)*Bh_{0}*Aw_{0}*As(\overline{h_{0}})$

and

$F(p*(h+w))=F(0)*B\tilde{h}*A{\tilde{\tilde{w}}}*A(h^{-1}*w_{0}*h)*Aw^{\prime}*As(\overline{h_{0}}*\bar{h}).$

Therefore (see explanation after the display formula)

$\displaystyle P_{\alpha}(L_{F(p)^{-1}}\circ F\circ L_{p}(h+w))$
$\displaystyle=$ $\displaystyle
P_{\alpha}(L_{F(p)^{-1}}(F(0)*B\tilde{h}*A{\tilde{\tilde{w}}}*A(h^{-1}*w_{0}*h)*Aw^{\prime}*As(\overline{h_{0}}*\bar{h})))$
$\displaystyle=$ $\displaystyle
P_{\alpha}(As(\overline{h_{0}})^{-1}*Aw_{0}^{-1}*(Bh_{0})^{-1}*B\tilde{h}*A{\tilde{\tilde{w}}}*Aw_{0}*Aw*As(\overline{h_{0}}*\bar{h}))$
$\displaystyle=$ $\displaystyle
P_{\alpha}(As(\overline{h_{0}})^{-1}*Aw_{0}^{-1}*(Bh_{0})^{-1}*B\tilde{h}*Aw_{0}*Aw*As(\overline{h_{0}}*\bar{h}))$
$\displaystyle=$ $\displaystyle
P_{\alpha}(As(\overline{h_{0}})^{-1}*(Bh_{0})^{-1}*B\tilde{h}*Aw*As(\overline{h_{0}}*\bar{h}))$
$\displaystyle=$ $\displaystyle
P_{\alpha}(As(\overline{h_{0}})^{-1}*Bh*Aw*As(\overline{h_{0}}*\bar{h}))$
$\displaystyle=$ $\displaystyle
P_{\alpha}(Bh+Aw+As(\overline{h_{0}}*\bar{h})-As(\overline{h_{0}})).$

For the second equality we used $P_{\alpha}(h^{-1}*w_{0}*h)=P_{\alpha}(w_{0})$
(as $P_{\alpha}(h)$ lies in the center of $\bar{\mathfrak{n}}_{\alpha}$) and
$P_{\alpha}(w^{\prime})=P_{\alpha}(w)$. For the third equality we used
$P_{\alpha}({\tilde{\tilde{w}}})=0$ (as
${\tilde{\tilde{w}}}\in\oplus_{j>\alpha}W_{j}$). For the fourth equality we
used
$P_{\alpha}(Aw_{0}^{-1}*Bh_{0}^{-1}*B\tilde{h}*Aw_{0})=P_{\alpha}(Bh_{0}^{-1}*B\tilde{h})$
(as $P_{\alpha}(Bh_{0}^{-1}*B\tilde{h})$ lies in the center of
$\bar{\mathfrak{n}}_{\alpha}$). For the fifth equality we used
$P_{\alpha}((Bh_{0})^{-1}*B\tilde{h})=P_{\alpha}(Bh)$. For the last equality
we used the fact that $P_{\alpha}(Bh)$ lies in the center of
$\bar{\mathfrak{n}}_{\alpha}$ and that $Aw$, $As(\overline{h_{0}}*\bar{h})$
and $As(\overline{h_{0}})$ commute with each other (as $s$ takes values in
$Z(\mathfrak{w})$).

After applying $\pi_{\alpha}$ to both sides of the above display formula we
get

(10) $\pi_{\alpha}\circ
F_{p}(h+w)=B\pi_{\alpha}(h)+A\pi_{\alpha}(w)+As_{\alpha}(\overline{h_{0}}*\bar{h})-As_{\alpha}(\overline{h_{0}}).$

This calculation will be used in the proofs of Lemmas 8.1 and 8.2. When $h\in
H_{1}$ and $w\in W_{\alpha}$, we have $\pi_{\alpha}(h)=h$ and
$\pi_{\alpha}(w)=w$ and so $F_{p,\alpha}:V_{\alpha}\rightarrow V_{\alpha}$ is
given by the formula

$F_{p,\alpha}(h+w)=Bh+Aw+As_{\alpha}(\overline{h_{0}}*\bar{h})-As_{\alpha}(\overline{h_{0}}).$

######  Lemma 8.1.

Let $\psi(F):\mathfrak{n}/\mathfrak{w}\rightarrow V_{\alpha}$ be the map given
by $\psi(F)(\bar{h})=Bh_{1}+As_{\alpha}(\bar{h})$, where $h_{1}$ is the
$H_{1}$ component of $h\in H$. Then $\psi(F)$ is Lipschitz with the Lipschitz
constant bounded above by a constant depending only on $H$ and the biLipschitz
constant of $F$.

###### Proof.

By setting $w=0$ in formula (10) we get

$\pi_{\alpha}(F(p)^{-1}*F(p*h))=Bh_{1}+As_{\alpha}(\overline{h_{0}}*\bar{h})-As_{\alpha}(\overline{h_{0}})=\psi(F)(\bar{h}_{0}*\bar{h})-\psi(F)(\bar{h}_{0}).$

Since $F$ is $L$-biLipschitz for some $L\geq 1$ we have

$|\psi(F)(\bar{h}_{0}*\bar{h})-\psi(F)(\bar{h}_{0})|^{\frac{1}{\alpha}}\leq
d(0,F(p)^{-1}*F(p*h))=d(F(p),F(p*h))\leq L\,d(p,p*h)=L\,d(0,h).$

On the other hand, since $\pi|_{H}:H\rightarrow{\bar{\mathfrak{n}}}$ is a
linear bijection, $H$ is stable under the automorphisms $e^{tD}$ ($D$ acts by
Euclidean dilation on each $V_{\lambda_{j}}$), and
$(\pi|_{H})\circ(e^{tD}|_{H})=e^{t\bar{D}}\circ(\pi|_{H})$, there is a
constant $L_{0}\geq 1$ that depends on $H$ but independent of the biLipschitz
map $F$ such that $(1/{L_{0}})d^{\frac{1}{\alpha}}_{CC}(0,\bar{h})\leq
d(0,h)\leq L_{0}d^{\frac{1}{\alpha}}_{CC}(0,\bar{h})$ for any $h\in H$. It
follows that
$|\psi(F)(\bar{h}_{0}*\bar{h})-\psi(F)(\bar{h}_{0})|^{\frac{1}{\alpha}}\leq
LL_{0}d^{\frac{1}{\alpha}}_{CC}(0,\bar{h})$.

∎

Now we can try to compute the differential $dF_{p,\alpha}(0)$ of
$F_{p,\alpha}$ at the origin $0$:

$\displaystyle dF_{p,\alpha}(0)(h+w)$
$\displaystyle=\lim_{t\rightarrow-\infty}\frac{F_{p,\alpha}(e^{\alpha
t}(h+w))}{e^{\alpha t}}$
$\displaystyle=\lim_{t\rightarrow-\infty}\frac{Be^{\alpha t}h+Ae^{\alpha
t}w+As_{\alpha}(\overline{h_{0}}*e^{\alpha
t}\bar{h})-As_{\alpha}(\overline{h_{0}})}{e^{\alpha t}}$
$\displaystyle=Bh+Aw+A\lim_{t\rightarrow-\infty}\frac{s_{\alpha}(\overline{h_{0}}*e^{t\bar{D}}\bar{h})-s_{\alpha}(\overline{h_{0}})}{e^{\alpha
t}}.$

In the above we used the fact that
$\bar{D}|_{\bar{V}_{1}}=\alpha\cdot\text{Id}_{\bar{V}_{1}}$. Notice that at a
point $\overline{h_{0}}$ where the Lipschitz map
$s_{\alpha}:\mathfrak{n}/\mathfrak{w}\rightarrow Z_{\alpha}(\mathfrak{w})$ is
Pansu differentiable, we have that
$\lim_{t\rightarrow-\infty}\frac{s_{\alpha}(\overline{h_{0}}*e^{t\bar{D}}\bar{h})-s_{\alpha}(\overline{h_{0}})}{e^{\alpha
t}}=Ds_{\alpha}(\overline{h_{0}})(\bar{h})$, where
$Ds_{\alpha}(\overline{h_{0}}):\mathfrak{n}/\mathfrak{w}\rightarrow
Z_{\alpha}(\mathfrak{w})$ is the Pansu differential of $s_{\alpha}$ at
$\overline{h_{0}}$. Recall that a Lipschitz map between Carnot groups is Pansu
differentiable a.e., see [P89]. Since $s_{\alpha}$ is Pansu differentiable at
a.e. $\overline{h_{0}}\in\mathfrak{n}/\mathfrak{w}$, we see that at a.e.
$p=h_{0}*w_{0}\in\mathfrak{n}$, the map $F_{p,\alpha}:V_{\alpha}\rightarrow
V_{\alpha}$ is differentiable at the origin with differential
$dF_{p,\alpha}(0)$ given by:

(11) $dF_{p,\alpha}(0)(h+w)=Bh+Aw+ADs_{\alpha}(\overline{h_{0}})(\bar{h}).$

We shall denote $D_{\alpha}F(p)=dF_{p,\alpha}(0)$.

The differential $D_{\alpha}F(p)$ should be thought as a counterpart for the
restriction of Pansu differential to the first layer, although our groups are
not Carnot. It satisfies the chain rule.

######  Lemma 8.2.

Let $F,\tilde{F}:\mathfrak{n}\rightarrow\mathfrak{n}$ be biLipschitz maps
satisfying the assumptions of Lemma 5.4. Then
$D_{\alpha}(F\circ\tilde{F})(p)=D_{\alpha}F(\tilde{F}(p))\circ
D_{\alpha}\tilde{F}(p)$ for a.e. $p\in\mathfrak{n}$.

###### Proof.

Let $p\in\mathfrak{n}$ be a point such that $D_{\alpha}(F\circ\tilde{F})(p)$,
$D_{\alpha}F(\tilde{F}(p))$ and $D_{\alpha}\tilde{F}(p)$ all exist. We need to
show

(12) $D_{\alpha}(F\circ\tilde{F})(p)(X)=D_{\alpha}F(\tilde{F}(p))\circ
D_{\alpha}\tilde{F}(p)(X),\;\;\forall X\in V_{\alpha}.$

Since $V_{\alpha}=W_{\alpha}\oplus H_{1}$, it suffices to establish (12) for
$X\in W_{\alpha}$ and $X\in H_{1}$.

The maps $F$ and $\tilde{F}$ have compatible expressions
$F(h*w)=F(0)*Bh*Aw*As(\bar{h})$,
$\tilde{F}(h*w)=\tilde{F}(0)*\tilde{B}h*\tilde{A}w*A\tilde{s}(\bar{h})$. Let
$X=w\in W_{\alpha}$. By (11), $D_{\alpha}F(p)(w)=Aw$. It follows that
${D_{\alpha}}F(\tilde{F}(p))\circ
D_{\alpha}\tilde{F}(p)(w)=D_{\alpha}F(\tilde{F}(p))(\tilde{A}w)=A(\tilde{A}w)$.
On the other hand, $D_{\alpha}(F\circ\tilde{F})(p)(w)=A\tilde{A}(w)$ as on
cosets of $W$ the map $F\circ\tilde{F}$ acts by $A\tilde{A}$. Hence (12) holds
for $X\in W_{\alpha}$. We next consider the case $X=h\in H_{1}$.

Consider the paths $\tilde{c}(t)=L_{\tilde{F}(p)^{-1}}\circ\tilde{F}(p*(th))$
and $c(t)=L_{(F\circ\tilde{F}(p))^{-1}}\circ F\circ
L_{\tilde{F}(p)}(\tilde{c}(t))$, and set
$\tilde{c}_{\alpha}(t)=\pi_{\alpha}(\tilde{c}(t))$,
$c_{\alpha}(t)=\pi_{\alpha}(c(t))$. Since $D_{\alpha}\tilde{F}(p)$ and
$D_{\alpha}(F\circ\tilde{F})(p)$ exist, we have
$\tilde{c}^{\prime}_{\alpha}(0)=D_{\alpha}\tilde{F}(p)(h)$ and
$c^{\prime}_{\alpha}(0)=D_{\alpha}(F\circ\tilde{F})(p)(h)$.

Write $\tilde{c}(t)=h(t)+w(t)$ with $h(t)\in H$ and $w(t)\in\mathfrak{w}$.
Then

$D_{\alpha}\tilde{F}(p)(h)=\tilde{c}^{\prime}_{\alpha}(0)=\lim_{t\rightarrow
0}\frac{\pi_{\alpha}h(t)}{t}+\lim_{t\rightarrow 0}\frac{\pi_{\alpha}w(t)}{t}.$

Set $h_{1}=\lim_{t\rightarrow 0}\frac{\pi_{\alpha}h(t)}{t}\in H_{1}$ and
$w_{1}=\lim_{t\rightarrow 0}\frac{\pi_{\alpha}w(t)}{t}\in W_{\alpha}$. Then
$D_{\alpha}\tilde{F}(p)(h)=h_{1}+w_{1}.$

We also write $\tilde{F}(p)=\tilde{h}_{0}*\tilde{w}_{0}$ with
$\tilde{h}_{0}\in H$ and $\tilde{w}_{0}\in\mathfrak{w}$. We have

$D_{\alpha}F(\tilde{F}(p))(D_{\alpha}\tilde{F}(p)(h))=D_{\alpha}F(\tilde{F}(p))(h_{1}+w_{1})=Bh_{1}+Aw_{1}+ADs_{\alpha}(\overline{\tilde{h}_{0}})(\overline{h_{1}}).$

By (10),

$c_{\alpha}(t)=\pi_{\alpha}\circ L_{(F\circ\tilde{F}(p))^{-1}}\circ F\circ
L_{\tilde{F}(p)}(\tilde{c}(t))=B\pi_{\alpha}h(t)+A\pi_{\alpha}w(t)+As_{\alpha}(\overline{\tilde{h}_{0}}*\overline{h(t)})-As_{\alpha}(\overline{\tilde{h}_{0}}).$

Hence

$\displaystyle D_{\alpha}(F\circ\tilde{F})(p)(h)=c^{\prime}_{\alpha}(0)$
$\displaystyle=B\lim_{t\rightarrow
0}\frac{\pi_{\alpha}h(t)}{t}+A\lim_{t\rightarrow
0}\frac{\pi_{\alpha}w(t)}{t}+A\lim_{t\rightarrow
0}\frac{s_{\alpha}(\overline{\tilde{h}_{0}}*\overline{h(t)})-As_{\alpha}(\overline{\tilde{h}_{0}})}{t}$
$\displaystyle=Bh_{1}+Aw_{1}+A\lim_{t\rightarrow
0}\frac{s_{\alpha}(\overline{\tilde{h}_{0}}*\overline{h(t)})-s_{\alpha}(\overline{\tilde{h}_{0}})}{t}.$

It now suffices to show

$\lim_{t\rightarrow
0}\frac{s_{\alpha}(\overline{\tilde{h}_{0}}*\overline{h(t)})-s_{\alpha}(\overline{\tilde{h}_{0}})}{t}=Ds_{\alpha}(\overline{\tilde{h}_{0}})(\overline{h_{1}}).$

Since $\pi\circ\tilde{F}(x)=\overline{\tilde{F}(0)}*\bar{\tilde{B}}(\bar{x})$,
we get

$\overline{h(t)}=\pi(\tilde{c}(t))=((\pi\circ\tilde{F})(p))^{-1}*(\pi\circ\tilde{F})(p*th)=\bar{\tilde{B}}(\bar{p})^{-1}*\bar{\tilde{B}}(\bar{p}*t\bar{h})=\bar{\tilde{B}}(t\bar{h})=t\bar{\tilde{B}}(\bar{h}).$

Hence

$\lim_{t\rightarrow
0}\frac{s_{\alpha}(\overline{\tilde{h}_{0}}*\overline{h(t)})-s_{\alpha}(\overline{\tilde{h}_{0}})}{t}=\lim_{t\rightarrow
0}\frac{s_{\alpha}(\overline{\tilde{h}_{0}}*t\bar{\tilde{B}}(\bar{h}))-s_{\alpha}(\overline{\tilde{h}_{0}})}{t}=Ds_{\alpha}(\overline{\tilde{h}_{0}})(\bar{\tilde{B}}(\bar{h})).$

Finally we notice that
$D_{\alpha}\tilde{F}(p)(h+w)=\tilde{B}h+\tilde{A}w+\tilde{A}D\tilde{s}_{\alpha}(\bar{p})(\bar{h})$
implies $\bar{h}_{1}=\pi(D_{\alpha}\tilde{F}(p)(h))=\bar{\tilde{B}}\bar{h}$.

∎

Now we are ready to show that if $\Gamma$ is a fiber similarity group of $N$,
then the differentials $\\{D_{\alpha}\gamma(p)|p\in N,\gamma\in\Gamma\\}$ are
“uniformly quasiconformal”. This result is needed in order to run Tukia’s
argument for the existence of invariant conformal structure.

######  Lemma 8.3.

Let $\Gamma$ be a fiber similarity group of $N$. Then there is a constant
$C\geq 1$ such that for every $\gamma\in\Gamma$, the differential
$D_{\alpha}\gamma(p)$ is $C$-quasiconformal for a.e. $p\in N$.

###### Proof.

By the discussion in Section 7, there is a constant $K_{0}\geq 1$ with the
following property: for each $\gamma\in\Gamma$, there is some
$t_{\gamma}\in\mathbb{R}$ such that
$\gamma^{\prime}:=e^{-t_{\gamma}D}\circ\gamma$ is $K_{0}$-biLipschitz, and
$\gamma^{\prime}$ acts on cosets of $W$ by isometric graded automorphism and
induces an isometry of $N/W$. If
$\gamma(h*w)=\gamma(0)*B_{\gamma}h*A_{\gamma}w*A_{\gamma}s_{\gamma}(\bar{h})$
is a compatible expression for $\gamma$, then $\gamma^{\prime}$ has a
compatible expression given by
$\gamma^{\prime}(h*w)=\gamma^{\prime}(0)*B^{\prime}_{\gamma}h*A^{\prime}_{\gamma}w*A^{\prime}_{\gamma}s_{\alpha}(\bar{h})$
where $A^{\prime}_{\gamma}=e^{-t_{\gamma}D}\circ A_{\gamma}$ is an isometry of
$(\mathfrak{w},d_{CC})$ and $B^{\prime}_{\gamma}=e^{-t_{\gamma}D}\circ
B_{\gamma}$ is such that $\bar{B}^{\prime}_{\gamma}$ is an isometry of
$(\mathfrak{n}/\mathfrak{w},d_{CC})$. By the formula for $D_{\alpha}F(p)$ we
have that $D_{\alpha}(\gamma^{\prime})(p)$ is the composition of
$D_{\alpha}(\gamma)(p)$ with a standard Euclidean dilation:
$D_{\alpha}(\gamma^{\prime})(p)(h+w)=e^{-t_{\gamma}\alpha}D_{\alpha}\gamma(p)(h+w)$
for $h\in H_{1},w\in W_{\alpha}$. Hence it suffices to show that there is a
constant $C\geq 1$ such that for every $\gamma\in\Gamma$, the differential
$D_{\alpha}\gamma^{\prime}(p)$ is $C$-quasiconformal for a.e. $p\in N$.

We fix an inner product on $V_{\alpha}$ so that $H_{1}$ and $W_{\alpha}$ are
perpendicular to each other. Notice that the equality
$D_{\alpha}\gamma^{\prime}(p)(h+w)=A^{\prime}_{\gamma}w+D\psi(\gamma^{\prime})(\bar{p})(\bar{h})$
holds, where $\psi(F)$ was defined in Lemma 8.1. As $A^{\prime}_{\gamma}$ is
an isometry and $\psi(\gamma^{\prime})$ is Lipschitz with Lipschitz constant
bounded above by a constant depending only on $H$ and the biLipschitz constant
of $\gamma^{\prime}$, we see that there is a constant $C_{1}>0$ such that for
each $\gamma\in\Gamma$, the norm of the linear map
$D_{\alpha}\gamma^{\prime}(p):V_{\alpha}\rightarrow V_{\alpha}$ is bounded
above by $C_{1}$ for a.e. $p\in N$. In particular it also holds for
$\gamma^{-1}$. By applying Lemma 8.2 to the composition
$\gamma\circ\gamma^{-1}=\text{Id}$ we conclude that
$D_{\alpha}\gamma^{\prime}(p)$ is $C_{1}$-biLipschitz and so is $C_{1}^{2}$
quasiconformal.

∎

###  8.2. Measurable conformal structure in the $V_{\alpha}$ direction

In this subsection we shall show that, after a conjugation,
$s_{\gamma,\alpha}:\mathfrak{n}/\mathfrak{w}\rightarrow
Z_{\alpha}(\mathfrak{w})$ is a Lie group homomorphism for every
$\gamma\in\Gamma$. See Proposition 8.6. We shall modify the proof of Tukia’s
theorem [T86], see also [Dy10] for the foliated version.

Fix an inner product on $V_{\alpha}$ and an orthonormal basis of $V_{\alpha}$
with respect to this inner product. Denote $n_{\alpha}=\dim(V_{\alpha})$. Then
we can identify a linear transformation of $V_{\alpha}$ with an
$n_{\alpha}\times n_{\alpha}$ matrix. Denote by $SL(V_{\alpha})$ (the special
linear group) the group of linear transformations of $V_{\alpha}$ whose
matrices have determinant equal to $1$, and $SO(V_{\alpha})\subset
SL(V_{\alpha})$ the subgroup consisting of linear transformations that
preserve the inner product. Let $X=SL(V_{\alpha})/SO(V_{\alpha})$. Recall
that, $X$ is a symmetric space of non-compact type (see table V on page 518 of
[Hel78]) and so has nonpositive sectional curvature. We denote by $\rho$ the
distance on $X$.

A measurable conformal structure on $N$ in the $V_{\alpha}$ direction is an
essentially bounded measurable map

$\mu:U\rightarrow X$

defined on a full measure subset $U\subset N$. This is just a measurable way
of assigning inner products (up to a scalar multiple) in the direction of
$V_{\alpha}$. To simplify language, we will drop “in the $V_{\alpha}$
direction” and will just say “measurable conformal structure”.

Let $\mu$ be a measurable conformal structure on $N$ and $F:N\rightarrow N$ a
fiber similarity map. The pull-back measurable conformal structure $F^{*}\mu$
is defined by:

$(F^{*}\mu)(p)=(D_{\alpha}F(p))[\mu(F(p))]:=(\text{det}D_{\alpha}F(p))^{-\frac{2}{\text{dim}V_{\alpha}}}(D_{\alpha}F(p))^{T}\mu(F(p))D_{\alpha}F(p),\;\;\;\text{for
a.e.}\;p\in N.$

This is analogous to the pull-back of a Riemannian metric under a
diffeomorphism. Here we are using the fact that $D_{\alpha}F(p)$ exists a.e.,
see Section 8.1.

######  Corollary 8.4.

$(\gamma_{2}\gamma_{1})^{*}\mu=\gamma_{1}^{*}(\gamma_{2}^{*}\mu)$ holds for
all $\gamma_{1},\gamma_{2}\in\Gamma$.

###### Proof.

It follows immediately from the chain rule
$D_{\alpha}(\gamma_{2}\gamma_{1})(p)=D_{\alpha}\gamma_{2}(\gamma_{1}(p))\circ
D_{\alpha}\gamma_{1}(p)$.

∎

A fiber similarity map $F$ is called conformal with respect to the measurable
conformal structure $\mu$ if $F^{*}\mu=\mu$. Tukia’s argument together with
Corollary 8.4 and Lemma 8.3 then yield that $\Gamma$ has an invariant
measurable conformal structure; that is, there is a measurable conformal
structure $\mu$ on $N$ such that every $\gamma\in\Gamma$ is conformal with
respect to $\mu$. We may assume $\Gamma$ is countable: let $\Gamma_{0}$ be a
countable subgroup of $\Gamma$ that is dense in $\Gamma$ in the topology of
uniform convergence on compact subsets of $N$; if $\Gamma_{0}$ can be
conjugated into the similarity group of $(N,d)$ for some $D$-homogeneous
distance $d$, then the same map conjugates $\Gamma$ into the group of
similarities of $(N,d)$ as the limits of similarities are similarities.

We next recall the notion of radial limit points. Let
$S=N\rtimes_{D}\mathbb{R}$ be the Heintze group associated with $(N,D)$ and
$\mathcal{H}:S\rightarrow\mathbb{R}$ the height function given by
$\mathcal{H}(x,t)=t$. Let $\mathcal{P}(N)=\\{(\xi_{1},\xi_{2})\in N\times
N|\xi_{1}\not=\xi_{2}\\}$, where we view $N=\partial S\backslash\\{\infty\\}$.
Let $\chi:\mathcal{P}(N)\rightarrow S$ be the map that assigns to each pair
$(\xi_{1},\xi_{2})\in\mathcal{P}(N)$ the highest point on the geodesic
$\xi_{1}\xi_{2}$; that is,
$\mathcal{H}(\chi(\xi_{1},\xi_{2}))=\max\\{\mathcal{H}(p)|p\in\xi_{1}\xi_{2}\\}$.
In a sense, $\chi(P)$ is a center of the triple $(\infty,\xi_{1},\xi_{2})$: it
is where the two geodesics that join $\infty$ to $\xi_{1}$ and $\xi_{2}$
respectively diverge from each other. We observe that for any compact
$C\subset S$, the set $\chi^{-1}(C)$ is compact in $\mathcal{P}(N)$.

The group $\Gamma$ acts diagonally on $\mathcal{P}(N)$:
$g(\xi_{1},\xi_{2})=(g(\xi_{1}),g(\xi_{2}))$.

######  Definition 8.5.

A point $\xi\in N$ is said to be a radial limit point of $\Gamma$ if there
exists a sequence of elements $\\{h_{i}\\}_{i=1}^{\infty}$ of $\Gamma$ with
the following property: for any pair $P=(\xi_{1},\xi_{2})\in\mathcal{P}(N)$,
and any complete geodesic $\sigma$ asymptotic to $\xi$, there exists a
constant $C>0$ with $\chi(h_{i}(P))\rightarrow\xi$ and
$d(\chi(h_{i}(P)),\sigma)\leq C$.

######  Proposition 8.6.

There exists a biLipschitz map $F$ of $N$ such that each element of $F\Gamma
F^{-1}$ has a compatible expression $h*w\mapsto a*Bh*Aw*As(\bar{h})$ such that
$s_{\alpha}:\mathfrak{n}/\mathfrak{w}\rightarrow Z_{\alpha}(\mathfrak{w})$ is
a Lie group homomorphism.

###### Proof.

We equip $S=N\rtimes_{D}\mathbb{R}$ with a left invariant Riemannian metric
such that $N$ and $\mathbb{R}$ are perpendicular to each other. The left
translations $L_{(0,t)}$ are isometries of $S$ and translate the vertical
geodesic $\sigma(s)=(0,s)$ above $0\in N$ and the boundary homeomorphisms
induced by them are the automorphisms $e^{tD}$ of $N$ generated by the
derivation $D$.

Let $\mu$ be a $\Gamma$-invariant measurable conformal structure on $N$. As
$\mu$ is measurable, it is approximately continuous a.e. in $N$, see Theorem
2.9.13 in [F69]. Let $p\in N$ be a radial limit point of $\Gamma$ and also a
point at which $\mu$ is approximately continuous. After applying a left
translation we may assume $p=0$ is the origin of $N$. Fix a pair
$P\in\mathcal{P}(N)$ and let $\sigma$ be the vertical geodesic (in $S$) above
$0$. Since $0$ is a radial limit point of $\Gamma$, there exists a sequence of
elements $\\{\gamma_{i}\\}_{i=1}^{\infty}$ of $\Gamma$ and a constant $C>0$
with $\chi(\gamma_{i}(P))\rightarrow 0$ and $d(\chi(\gamma_{i}(P)),\sigma)\leq
C$. Fix a point $x_{0}\in\sigma$. For each $i$ there is some
$t^{\prime}_{i}\in\mathbb{R}$ with $t^{\prime}_{i}\rightarrow+\infty$ as
$i\rightarrow\infty$ such that
$d(L_{(0,t^{\prime}_{i})}(\chi(\gamma_{i}(P))),x_{0})\leq C$. Since
$L_{(0,t^{\prime}_{i})}$ is an isometry of $S$, we have
$L_{(0,t^{\prime}_{i})}\circ\chi=\chi\circ e^{t^{\prime}_{i}D}$. Hence
$d(\chi\circ e^{t^{\prime}_{i}D}\circ\gamma_{i}(P),x_{0})\leq C$ and so the
set $\\{e^{t^{\prime}_{i}D}\circ\gamma_{i}(P)\\}_{i=1}^{\infty}$ lies in the
compact subset $\chi^{-1}\bar{B}(x_{0},C)$. It follows that
$\\{e^{t^{\prime}_{i}D}\circ\gamma_{i}\\}_{i=1}^{\infty}$ is a compact family
of biLipschitz maps of $N$. Recall that there is a Carnot metric $d_{CC}$ on
$W$ with the property that for each $\gamma\in\Gamma$ there is some
$t_{\gamma}\in\mathbb{R}$ such that
$\gamma^{\prime}:=e^{-t_{\gamma}D}\circ\gamma$ has a compatible expression
$\gamma^{\prime}(h*w)={\gamma^{\prime}(0)}*B_{\gamma}h*A_{\gamma}w*A_{\gamma}s_{\gamma}(\bar{h})$,
where $A_{\gamma}$ is a graded isomorphism of $W$ that is also an isometry of
$(W,d_{CC})$. The compactness of the family
$\\{e^{t^{\prime}_{i}D}\circ\gamma_{i}\\}_{i=1}^{\infty}$ implies that
$\\{t^{\prime}_{i}+t_{\gamma_{i}}\\}$ is a bounded sequence and so the family
$\\{e^{-t_{\gamma_{i}}D}\circ\gamma_{i}\\}_{i=1}^{\infty}$ is also compact.
Set $f_{i}=e^{-t_{\gamma_{i}}D}\circ\gamma_{i}$. By passing to a subsequence,
we may assume $f_{i}$ converges uniformly on compact subsets to a biLipschitz
map $f:N\rightarrow N$. Since each $f_{i}$ is a fiber similarity, so is $f$.

Let $\gamma\in\Gamma$ and denote $\tilde{\gamma}=f\gamma f^{-1}$,
$\tilde{\gamma}_{i}=f_{i}\gamma f_{i}^{-1}$. Let
$\mu_{i}=(f_{i}^{-1})^{*}\mu$. Since $\mu$ is $\Gamma$-invariant,
$\tilde{\gamma}_{i}$ is conformal with respect to $\mu_{i}$:

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

Note
$\mu_{i}=(f_{i}^{-1})^{*}\mu=({e^{t_{\gamma_{i}}D}})^{*}(\gamma_{i}^{-1})^{*}\mu=(e^{t_{\gamma_{i}}D})^{*}\mu$.
So for $q\in N$,

$\mu_{i}(q)=(e^{t_{\gamma_{i}}D})^{*}\mu(q)=D_{\alpha}e^{t_{\gamma_{i}}D}(q)[\mu(e^{t_{\gamma_{i}}D}(q))]=\mu(e^{t_{\gamma_{i}}D}(q))$

since $D_{\alpha}e^{t_{\gamma_{i}}D}:V_{\alpha}\rightarrow V_{\alpha}$ is the
standard dilation by $e^{\alpha t_{\gamma_{i}}}$ and so is conformal.

Let $U\subset N$ be a bounded open subset containing $0$. There is a bounded
open subset $U_{0}$ such that $U\bigcup\cup_{i}\tilde{\gamma}_{i}(U)\subset
U_{0}$. Since $\mu$ is approximately continuous at $0$ and
$t_{\gamma_{i}}\rightarrow-\infty$, the equality
$\mu_{i}(q)=\mu(e^{t_{\gamma_{i}}D}(q))$ implies that for any $\epsilon>0$
there are subsets $C_{i}\subset U_{0}$ with $|C_{i}|\rightarrow 0$ as
$i\rightarrow\infty$ and $\rho(\mu_{i}(x),\mu(0))\leq\epsilon$ for $x\in
U_{0}\backslash C_{i}$. Here $|E|$ denotes the measure of a subset $E\subset
N$ and $\rho$ is the distance on the symmetric space $X$.

The maps $\tilde{\gamma}^{-1}_{i}$ and $\tilde{\gamma}^{-1}$ form a compact
family of biLipschitz maps. There is some $L\geq 1$ such that these maps are
all $L$-biLipschitz. Hence $|\tilde{\gamma}_{i}^{-1}(C_{i})|\rightarrow 0$ as
$i\rightarrow\infty$. Set $D_{i}=C_{i}\cup\tilde{\gamma}_{i}^{-1}(C_{i})$. Now
we have $|D_{i}|\rightarrow 0$ as $i\rightarrow\infty$ and
$\rho(\mu_{i}(x),\mu(0))\leq\epsilon$ and
$\rho(\mu_{i}(\tilde{\gamma}_{i}(x)),\mu(0))\leq\epsilon$ for $x\in
U\backslash D_{i}$. Since $\tilde{\gamma}_{i}$ is $\mu_{i}$-conformal, we have
$\mu_{i}(x)=D_{\alpha}\tilde{\gamma}_{i}(x)[\mu_{i}(\tilde{\gamma}_{i}(x))]$
for a.e. $x$. Now

$\rho(\mu_{i}(x),D_{\alpha}\tilde{\gamma}_{i}(x)[\mu(0)])=\rho(\mu_{i}(\tilde{\gamma}_{i}(x)),\mu(0))\leq\epsilon$

for a.e. $x\in U\backslash D_{i}$. Combining this with
$\rho(\mu_{i}(x),\mu(0))\leq\epsilon$ , we get

(13) $\rho(\mu(0),D_{\alpha}\tilde{\gamma}_{i}(x)[\mu(0)])\leq 2\epsilon$

for a.e. $x\in U\backslash D_{i}$.

We consider compatible expressions
$\tilde{\gamma}(h*w)=\tilde{\gamma}(0)*Bh*Aw*As(\bar{h})$ and
$\tilde{\gamma}_{i}(h*w)=\tilde{\gamma}_{i}(0)*B_{i}h*A_{i}w*A_{i}s_{i}(\bar{h})$.
We know that $\tilde{\gamma}_{i}$ converges to $\tilde{\gamma}$ uniformly on
compact subsets. It follows that
$\tilde{\gamma}_{i}(0)\rightarrow\tilde{\gamma}(0)$ and $A_{i}\rightarrow A$.
In general we can not conclude that $s_{i}\rightarrow s$ and $B_{i}\rightarrow
B$ due to the fact that compatible expressions are not unique. Define
functions $g_{i},g:\mathfrak{n}/\mathfrak{w}\rightarrow V_{\alpha}$ by
$g_{i}(\bar{h})=B_{i}h_{1}+A_{i}s_{i,\alpha}(\bar{h})$ and
$g(\bar{h})=Bh_{1}+As_{\alpha}(\bar{h})$, where $h_{1}$ is the $H_{1}$
component of $h\in H$. By considering the $V_{\alpha}$ component of
$\tilde{\gamma}_{i}(h)$ and $\tilde{\gamma}(h)$ we see that $g_{i}\rightarrow
g$ uniformly on compact subsets.

For the rest of the proof, when we talk about “perpendicular”, “orthonormal”,
length $|X|$ of a vector $X\in V_{\alpha}$, inner product
$\left<\cdot,\cdot\right>$, they are all with respect to $\mu(0)$. Let
$X_{1},\cdots,X_{k}$ ($k=\dim(W_{\alpha})$) be an orthonormal basis of
$W_{\alpha}$. For $1\leq l\leq k$, let $g_{l}(x)=\left<g(x),X_{l}\right>$,
$g_{i,l}(x)=\left<g_{i}(x),X_{l}\right>$. Since $\tilde{\gamma}$ and
$\tilde{\gamma}_{i}$ are $L$-biLipschitz, Lemma 8.1 implies that there is some
constant $L_{1}>0$ such that $g$ and $g_{i}$ are $L_{1}$-Lipschitz for all
$i\geq 1$. It follows that $g_{l}$ and $g_{i,l}$ are $L_{1}$-Lipschitz
functions on $N/W$ for all $i\geq 1$, $1\leq l\leq k$.

Let $\phi:H_{1}\rightarrow W_{\alpha}$ be the linear map such that
$\\{h+\phi(h)|h\in H_{1}\\}$ is perpendicular to $W_{\alpha}$. Denote by
$\psi:\bar{\mathfrak{n}}\to H_{1}$ the linear map given by
$\psi(\bar{h})=(\pi|_{H_{1}})^{-1}(\bar{\pi}_{1}(\bar{h}))$, where
$\bar{\pi}_{1}:{\bar{\mathfrak{n}}}\rightarrow\bar{V}_{1}$ is the projection
with respect to the decomposition ${\bar{\mathfrak{n}}}=\oplus_{j}\bar{V}_{j}$
and $\pi:\mathfrak{n}\to\bar{\mathfrak{n}}$ is the quotient map. For $1\leq
l\leq k$, let $L_{l}:{\bar{\mathfrak{n}}}\rightarrow\mathbb{R}$ be the linear
map defined by

$L_{l}(\bar{h})=-\left<X_{l},A\phi(\psi(\bar{h}))\right>.$

Claim: For each $1\leq l\leq k$, $Dg_{i,l}$ converges to $L_{l}$ in
$L_{\text{loc}}^{1}(N/W)$ as $i\to\infty$.

We first assume the claim and finish the proof of the Proposition. The claim
implies that for any $h\in H_{1}$ and any smooth function with compact support
$\varphi$ defined on $N/W$,

$\displaystyle\int_{N/W}\varphi(x)L_{l}(\bar{h})dx$
$\displaystyle\longleftarrow\int_{N/W}\varphi(x)Dg_{i,l}(x)(\bar{h})dx$
$\displaystyle=$
$\displaystyle-\int_{N/W}g_{i,l}(x)D_{\bar{h}}\varphi(x)dx\longrightarrow-\int_{N/W}g_{l}(x)D_{\bar{h}}\varphi(x)dx=\int_{N/W}\varphi(x)Dg_{l}(x)(\bar{h})dx,$

which yields $\int_{N/W}\varphi(x)(Dg_{l}(x)(\bar{h})-L_{l}(\bar{h}))dx=0$.
Here the first convergence follows from the claim; the second one follows from
the fact that $g_{i,l}$ converges to $g_{l}$ uniformly on compact subsets; and
the equalities follow from integration by parts. It follows that
$Dg_{l}(x)=L_{l}$ for a.e. $x\in N/W$. This implies $\langle
X_{l},ADs_{\alpha}(x)(\bar{h})+A\phi(\psi(\bar{h}))\rangle=-\langle
X_{l},Bh\rangle$ for a.e. $x\in N/W$ and all $1\leq l\leq k$. Hence
$ADs_{\alpha}(x)(\bar{h})+A\phi(\psi(\bar{h}))$ defines a constant Lie group
homomorphism (independent of $x$) from $N/W$ to $W_{\alpha}$. As
$A\phi(\psi(\bar{h}))$ is also a group homomorphism, we see that the Pansu
differential of $s_{\alpha}$ is a constant group homomorphism. It now follows
from $s_{\alpha}(0)=0$ that $s_{\alpha}$ is a Lie group homomorphism.

We now prove the claim. Recall
$\rho(D_{\alpha}\tilde{\gamma}_{i}(x)[\mu(0)],\mu(0))\leq 2\epsilon$ for $x\in
U\backslash D_{i}$. This means for $x\in U\backslash D_{i}$ the matrix
representation for the linear map
$D_{\alpha}\tilde{\gamma}_{i}(x):V_{\alpha}\rightarrow V_{\alpha}$ with
respect to an orthonormal basis of $(V_{\alpha},\mu(0))$ is a constant
multiple of a matrix that is very close to an orthogonal matrix. Since
$D_{\alpha}\tilde{\gamma}_{i}(x)(W_{\alpha})=W_{\alpha}$, we see that if $x\in
U\backslash D_{i}$ then $D_{\alpha}\tilde{\gamma}_{i}(x):V_{\alpha}\rightarrow
V_{\alpha}$ sends vectors in $V_{\alpha}$ perpendicular to $W_{\alpha}$ to
vectors almost perpendicular to $W_{\alpha}$. On the other hand, as
$D_{\alpha}\tilde{\gamma}_{i}(x)(w)=A_{i}w$ for $w\in W_{\alpha}$ and
$\\{A_{i}|i\geq 1\\}$ has compact closure (as $A_{i}\rightarrow A$), there is
some $M>0$ such that the operator norm of $D_{\alpha}\tilde{\gamma}_{i}(x)$ is
bounded above by $M$ for all $i$ and all $x\in U\backslash D_{i}$. Fix any
$h\in H_{1}$ with $|h|=1$. By the definition of $\phi:H_{1}\rightarrow
W_{\alpha}$, we have $\left<W_{\alpha},h+\phi(h)\right>=0$. As
$D_{\alpha}\tilde{\gamma}_{i}(x)(h+\phi(h))=B_{i}h+A_{i}\phi(h)+A_{i}Ds_{i,\alpha}(\bar{x})(\bar{h})$,
there is a constant $\delta>0$ depending only on $\epsilon$ with
$\delta\rightarrow 0$ as $\epsilon\rightarrow 0$ such that

$|\left<X_{l},B_{i}h+A_{i}\phi(h)+A_{i}Ds_{i,\alpha}(\bar{x})(\bar{h})\right>|<\delta$

holds for all $x\in U\backslash D_{i}$ and all $1\leq l\leq k$. Note
$\psi(\bar{h})=h$ (as $h\in H_{1}$). Since $A_{i}\to A$ and
$Dg_{i,l}(\bar{x})(\bar{h})=\left<B_{i}h+A_{i}Ds_{i,\alpha}(\bar{x})(\bar{h}),X_{l}\right>$,
we see that $|Dg_{i,l}(\bar{x})(\bar{h})-L_{l}(\bar{h})|\leq 2\delta$ for all
$x\in U\backslash D_{i}$ and all sufficiently large $i$. This implies
$\int_{U\backslash D_{i}}|Dg_{i,l}(\bar{x})(\bar{h})-L_{l}(\bar{h})|dx\leq
2\delta|U|$ for sufficiently large $i$. On the other hand, as $g_{i,l}$,
$i\geq 1$, $1\leq l\leq k$, are $L_{1}$-Lipschitz, there is a constant $C>0$
such that $|Dg_{i,l}(\bar{x})(\bar{h})-L_{l}(\bar{h})|\leq C$ for all $x\in
N$. From this we get
$\int_{D_{i}}|Dg_{i,l}(\bar{x})(\bar{h})-L_{l}(\bar{h})|dx\leq C|D_{i}|$. As
$|D_{i}|\rightarrow 0$, we conclude that

${\overline{\lim}}_{i\rightarrow\infty}\int_{U}|Dg_{i,l}(\bar{x})(\bar{h})-L_{l}(\bar{h})|dx\leq
2\delta|U|.$

As this holds for all $\epsilon>0$ we have
$\lim_{i\rightarrow\infty}\int_{U}|Dg_{i,l}(\bar{x})(\bar{h})-L_{l}(\bar{h})|dx=0$.
Since this holds for all bounded open subset $U\subset N$ and all $h\in
H_{1}$, we have $Dg_{i,l}\rightarrow L_{l}$ in $L^{1}_{\text{loc}}(N/W)$.

∎

###  8.3. Completing the proof of Theorem 1.2 when $\dim(W)\geq 2$ and
$\dim(N/W)\geq 2$

We assume the assumptions of Theorem 1.2 and in addition that $\dim(W)\geq 2$,
and $\dim(N/W)\geq 2$. We first use Section 7 to get rid of $s_{\gamma,j}$ for
$j<\alpha$, and if $\alpha$ is an integer then apply Proposition 8.6 to
conclude that $s_{\gamma,\alpha}$ is a Lie group homomorphism for all
$\gamma\in\Gamma$, after a possible further biLipschitz conjugation. We
observe that the property $s_{\gamma,j}\equiv 0$ for $j<\alpha$ is preserved
when we apply Proposition 8.6: this is because the conjugating map $f$ in the
proof of Proposition 8.6 is the limit of a sequence $\\{f_{i}\\}$ and each
$f_{i}$ is the composition of a group element $\gamma_{i}\in\Gamma$ with
$e^{t_{i}D}$ for some $t_{i}$; since $\gamma_{i}$ has the above property, so
do $f_{i}$ and the limit $f$; finally a calculation shows that if two
biLipschitz maps $f_{1},f_{2}$ of $N$ have this property then so does the
composition $f_{1}\circ f_{2}$.

At this point every element in $\Gamma$ has a compatible expression
$F(h*w)=F(0)*Bh*Aw*As(\bar{h})$, where $s:\mathfrak{n}/\mathfrak{w}\rightarrow
Z(\mathfrak{w})$ has the properties that $s_{j}=0$ for $1\leq j<\alpha$ and if
$\alpha$ is an integer then $s_{\alpha}:\mathfrak{n}/\mathfrak{w}\rightarrow
Z_{\alpha}(\mathfrak{w})$ is a Lie group homomorphism. We next show that such
a map $F$ is an affine map; that is, $L_{F(0)^{-1}}\circ F$ is a Lie group
automorphism.

######  Lemma 8.7.

Let $F,\tilde{F}:\mathfrak{n}\rightarrow\mathfrak{n}$ be biLipschitz maps with
compatible expressions $F(h*w)=Bh*Aw*As(\bar{h})$ and
$\tilde{F}(h*w)=Bh*Aw*A\tilde{s}(\bar{h})$. If $s$, $\tilde{s}$ satisfy
$s_{j}=\tilde{s}_{j}$ for $1\leq j\leq\alpha$, then $F=\tilde{F}$.

###### Proof.

We observe that $F(h*w*(s(\bar{h}))^{-1}*\tilde{s}(\bar{h}))=\tilde{F}(h*w)$.
It follows that
$(F^{-1}\circ\tilde{F})(h*w)=h*w*(s(\bar{h})^{-1}*\tilde{s}(\bar{h}))$ is a
biLipschitz shear map with shear function ${\tilde{\tilde{s}}}$ given by
${\tilde{\tilde{s}}}=\tilde{s}-s$. The assumption implies
${\tilde{\tilde{s}}}_{j}\equiv 0$ for all $1\leq j\leq\alpha$. On the other
hand, by Proposition 6.4, if $\alpha$ is not an integer, then
${\tilde{\tilde{s}}}_{j}\equiv 0$ for all $j>\alpha$, and if $\alpha$ is an
integer, then
${\tilde{\tilde{s}}}_{k\alpha+j}={\tilde{\tilde{s}}}_{j}^{(k)}\equiv 0$ for
each $k\geq 1$ and $1\leq j\leq\alpha$. It follows that
${\tilde{\tilde{s}}}=0$ and $F=\tilde{F}$.

∎

######  Lemma 8.8.

Let $F:\mathfrak{n}\rightarrow\mathfrak{n}$ be a biLipschitz map with
compatible expression $F(h*w)=Bh*Aw*As(\bar{h})$. Suppose $s_{j}=0$ for $1\leq
j<\alpha$ and if $\alpha$ is an integer
$s_{\alpha}:\mathfrak{n}/\mathfrak{w}\rightarrow Z_{\alpha}(\mathfrak{w})$ is
a Lie group homomorphism. Then $F$ is a Lie group automorphism.

###### Proof.

Notice that it suffices to show $F_{p}=F$ for any $p\in N$. Let $p\in N$ and
set $\tilde{F}=F_{p}$. By Lemma 5.6, $\tilde{F}$ has a compatible expression
given by $\tilde{F}(h*w)=Bh*Aw*A\tilde{s}(\bar{h})$ for some map
$\tilde{s}:\mathfrak{n}/\mathfrak{w}\rightarrow Z(\mathfrak{w})$. By Lemma 8.7
it now suffices to show $\tilde{s}_{j}=s_{j}$ for all $1\leq j\leq\alpha$.

Let $h\in H$. As $\tilde{F}(h)=Bh*A\tilde{s}(\bar{h})$, we have
$\tilde{s}(\bar{h})=A^{-1}((Bh)^{-1}*\tilde{F}(h))$. Write $p=h_{0}*w_{0}$ and
$h_{0}*h=h_{1}*w_{1}$ with $h_{0},h_{1}\in H$, $w_{0},w_{1}\in\mathfrak{w}$.
Notice $w_{1}\in\oplus_{j\geq 2\alpha}W_{j}$. We have
$p*h=h_{0}*w_{0}*h=h_{0}*h*(h^{-1}*w_{0}*h)=h_{1}*w_{1}*(h^{-1}*w_{0}*h)$. In
the following calculations we use the quotient homomorphism
$P_{\alpha}:\mathfrak{n}\rightarrow\bar{\mathfrak{n}}_{\alpha}={\mathfrak{n}}/(\oplus_{\lambda_{j}>\alpha}V_{\lambda_{j}})$:

$\displaystyle P_{\alpha}((Bh)^{-1}*\tilde{F}(h))$
$\displaystyle=P_{\alpha}((Bh)^{-1}*(F(h_{0}*w_{0}))^{-1}*F(h_{1}*w_{1}*(h^{-1}*w_{0}*h)))$
$\displaystyle=P_{\alpha}((Bh)^{-1}*As(\overline{h_{0}})^{-1}*Aw_{0}^{-1}*(Bh_{0})^{-1}*Bh_{1}*Aw_{1}*A(h^{-1}*w_{0}*h)*As(\overline{h_{0}}*\bar{h}))$
$\displaystyle=P_{\alpha}((Bh)^{-1}*As(\overline{h_{0}})^{-1}*Aw_{0}^{-1}*(Bh_{0})^{-1}*Bh_{1}*Aw_{1}*(Bh)^{-1}*Aw_{0}*(Bh)*As(\overline{h_{0}}*\bar{h}))$
$\displaystyle=P_{\alpha}((Bh)^{-1}*As(\overline{h_{0}})^{-1}*Aw_{0}^{-1}*(Bh_{0})^{-1}*Bh_{1}*Aw_{0}*As(\overline{h_{0}}*\bar{h}))$
$\displaystyle=P_{\alpha}((Bh)^{-1}*(Bh_{0})^{-1}*Bh_{1}*As(\overline{h_{0}})^{-1}*As(\overline{h_{0}}*\bar{h}))$
$\displaystyle=P_{\alpha}(A(s(\overline{h_{0}}*\bar{h})-s(\overline{h_{0}}))).$

For the third equality we used (5). In the 4th equality we used
$P_{\alpha}(Aw_{1})=0$ and $P_{\alpha}(Bh)\in Z(\bar{\mathfrak{n}}_{\alpha})$.
For the 5th equality we used $P_{\alpha}(Bh_{0}),P_{\alpha}(Bh_{1})\in
Z(\bar{\mathfrak{n}}_{\alpha})$. For the last equality we used
$P_{\alpha}((Bh_{0})^{-1}*Bh_{1})=P_{\alpha}(Bh)$. It follows that
$\tilde{s}_{j}(\bar{h})=s_{j}(\overline{h_{0}}*\bar{h})-s_{j}(\overline{h_{0}})$
for $1\leq j\leq\alpha$. By the assumption on $s$, we have $\tilde{s}_{j}=0$
for $1\leq j<\alpha$ and if $\alpha$ is an integer then
$\tilde{s}_{\alpha}(\bar{h})=s_{\alpha}(\overline{h_{0}}*\bar{h})-s_{\alpha}(\overline{h_{0}})=s_{\alpha}(\bar{h})$
as $s_{\alpha}:\mathfrak{n}/\mathfrak{w}\rightarrow Z_{\alpha}(\mathfrak{w})$
is a homomorphism. Now we have $\tilde{s}_{j}=s_{j}$ for $1\leq j\leq\alpha$.
By Lemma 8.7, $\tilde{F}=F$ and so $F$ is an automorphism.

∎

In this last paragraph we switch back to Lie group notation. At this point
$\Gamma$ acts on $N$ by affine maps and is a uniform quasisimilarity group of
$N$. We write $\gamma=L_{\gamma(0)}\circ\phi_{\gamma}$, where $\phi_{\gamma}$
is the automorphism $L_{\gamma(0)^{-1}}\circ\gamma$. By Lemma 3.5 we know that
$\phi_{\gamma}$ is layer preserving, that is,
$d\phi_{\gamma}(V_{\lambda_{j}})=V_{\lambda_{j}}$ for all $j$. Each
$\gamma\in\Gamma$ acts on the cosets of $W$ by an automorphism $A_{\gamma}$ of
$W$ and $A_{\gamma}$ is the composition of a Carnot dilation and an isometric
graded isomorphism of $W$. Hence for each $\gamma$, there is a unique
$t_{\gamma}\in\mathbb{R}$ such that $e^{-t_{\gamma}D}\circ\gamma$ acts on $W$
by an isometric graded isomorphism. Since $\Gamma$ is a uniform group of
quasisimilarities of $N$, there is a constant $L>0$ such that
$e^{-t_{\gamma}D}\circ\gamma$ is $L$-biLipschitz for all $\gamma\in\Gamma$. It
follows that for each $j\geq 1$ the linear isomorphism
$d(e^{-t_{\gamma}D}\circ\phi_{\gamma})|_{V_{\lambda_{j}}}:V_{\lambda_{j}}\rightarrow
V_{\lambda_{j}}$ is $L$-biLipschitz. Now we see that the map
$\Gamma\rightarrow GL(V_{\lambda_{j}})$ given by $\gamma\mapsto
d(e^{-t_{\gamma}D}\circ\phi_{\gamma})|_{V_{\lambda_{j}}}$ is a homomorphism
whose image has compact closure in $GL(V_{\lambda_{j}})$. It follows that
there is an inner product $\left<\cdot,\cdot\right>_{j}$ on $V_{\lambda_{j}}$
such that each $d(e^{-t_{\gamma}D}\circ\phi_{\gamma})|_{V_{\lambda_{j}}}$ is
an isometry with respect to this inner product. 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 $d_{1}$ be a $D$-homogeneous distance on $N$ associated to this
inner product. Although $d(e^{-t_{\gamma}D}\circ\phi_{\gamma})$ is a linear
isometry of $(\mathfrak{n},\langle,\rangle)$, it is not clear that
$e^{-t_{\gamma}D}\circ\phi_{\gamma}$ is an isometry of $(N,d_{1})$. However,
$\\{e^{-t_{\gamma}D}\circ\phi_{\gamma}|\gamma\in\Gamma\\}$ is a subgroup of
the group $\text{Auto}_{g}(N)$ of graded automorphisms with compact closure
(we denote the closure by $K$). Let $m$ be a normalized Haar measure on $K$.
Define a new distance $d_{2}$ on $N$ by
$d_{2}(x,y)=\int_{K}d_{1}(k(x),k(y))dm(k)$. Now it is easy to check that
$d_{2}$ is a $K$-invariant $D$-homogeneous distance on $N$ associated to
$\langle,\rangle$. It follows that $\Gamma$ acts on $(N,d_{2})$ by
similarities. Finally we conjugate $\Gamma$ into $\text{Sim}(N,d_{0})$ where
$d_{0}$ is a fixed maximally symmetric $D$-homogeneous distance on $N$.

We have finished the proof of Theorem 1.2 in the case when $\dim(W)\geq 2$,
$\dim(N/W)\geq 2$.

##  9\. Case $\text{dim}(W)=1$

In this section we prove Theorem 1.2 in the case when $\text{dim}(W)=1$,
$\dim(N/W)\geq 2$ and $\Gamma$ is amenable. In this case Tukia’s arguments can
not be used to prove a fiber Tukia theorem. Nonetheless Day’s fixed point
theorem once again can be used to “straighten” the action along the cosets of
$W$. We point out that the argument in this section is valid for uniform
quasisimilarity groups $\Gamma$ of product metric spaces of the form
$\mathbb{R}\times Y$, see Theorem 9.7. The only properties used in the proof
below are that $H$ is a proper metric space, the action of $\Gamma$ on
$\mathbb{R}\times H$ permutes the subsets $\\{\mathbb{R}\times\\{h\\}:h\in
H\\}$ and the induced action on $H$ is by similarities.

Throughout this section we assume $\text{dim}(W)=1$ and $\dim(N/W)\geq 2$. Let
$\mathfrak{h}=\oplus_{\lambda>1}V_{\lambda}$. Since $W=V_{1}$ is an ideal of
$\mathfrak{n}$, it follows from the property $[V_{a},V_{b}]\subset V_{a+b}$
that $\mathfrak{n}=W\oplus\mathfrak{h}$ is a direct sum of ideals where
$W\simeq\mathbb{R}$. So in this case, our group $N$ is a direct product of a
copy of $\mathbb{R}$ and a Carnot group which we also refer to as $W$ and $H$,
where $H$ is the simply connected Lie group with Lie algebra $\mathfrak{h}$.
Because of this we can write $w*h\in N$ as $(w,h)\in W\oplus H$ and our group
$\Gamma$ acts on $N$ by maps of the form

$\gamma(w,h)=(\gamma^{\mathbb{R}}(w,h),\gamma^{H}(h))$

where $\gamma^{H}:H\to H$ is biLipschitz and for each $h\in H$, the map
$\gamma^{\mathbb{R}}(\cdot,h):\mathbb{R}\to\mathbb{R}$ is also biLipschitz.
The proof we give here follows the proof from Section 3.3 in [Dy10] but we
construct the conjugating map in a slightly different manner in order to fix
an oversight in the original paper; namely it is unclear that the conjugating
map in [Dy10] is actually biLipschitz. To construct our map we need to use
that $\Gamma$ is amenable.

Since $\dim(N/W)\geq 2$, we can first apply Tukia-type theorem for Carnot
groups (Theorem 3.6) to the induced action of $\Gamma$ on $N/W\simeq H$. So
there is a biLipschitz map $f_{0}$ of $H$ such that after conjugation by
$f_{0}$, the induced action of $\Gamma$ on $H$ is by similarities. Set
$F_{0}=(\text{Id},f_{0}):W\oplus H\rightarrow W\oplus H$. Then we can
conjugate the action of $\Gamma$ on $N$ by $F_{0}$ to get an action where
$\gamma^{\mathbb{R}}(\cdot,h)$ is still biLipschitz and $\gamma^{H}$ is a
similarity of $(H,\bar{d}_{CC})$. Let $t_{\gamma}\in\mathbb{R}$ be such that
$e^{\alpha t_{\gamma}}$ is the similarity constant of
$\gamma^{H}:(H,\bar{d}_{CC})\rightarrow(H,\bar{d}_{CC})$ and let
$\tilde{\gamma}:W\times H\rightarrow W$ be the map given by
$\tilde{\gamma}(w,h)=e^{-t_{\gamma}}\gamma^{\mathbb{R}}(w,h)$. Then
$\gamma^{\mathbb{R}}(w,h)=e^{t_{\gamma}}\tilde{\gamma}(w,h)$. Since $\Gamma$
is a uniform quasisimilarity group, there is a constant $\Lambda\geq 1$ such
that $\tilde{\gamma}(\cdot,h)$ is $\Lambda$-biLipschitz for all
$\gamma\in\Gamma$ and all $h\in H$. After taking an index two subgroup if
necessary, we may assume $\tilde{\gamma}(\cdot,h)$ is orientation-preserving
and so has derivatives in the interval $[1/\Lambda,\Lambda]$. For each
$\gamma\in\Gamma$ we define a function $u_{\gamma}:N\rightarrow\mathbb{R}$ by
$u_{\gamma}(w,h)=\frac{\partial\tilde{\gamma}}{\partial w}(w,h)$ when
$\frac{\partial\tilde{\gamma}}{\partial w}(w,h)$ exists and
$u_{\gamma}(w,h)=1$ otherwise. Here $\frac{\partial\tilde{\gamma}}{\partial
w}(w,h)$ denotes the derivative of the function
$\tilde{\gamma}(\cdot,h):\mathbb{R}\rightarrow\mathbb{R}$ at the point $w$ if
it exists. Then $u_{\gamma}\in L^{\infty}(N)$ with values in
$[1/\Lambda,\Lambda]$. In this section we are using the Hausdorff measure on
$N$ including in the definition of $L^{p}$ spaces. Of course, the Hausdorff
measure on $N$ is a Haar measure. The point is that the argument in this
section still works when $N=\mathbb{R}\times H$ is replaced with a product
metric space $\mathbb{R}\times Y$.

Let $L^{\infty}(N)=(L^{1}(N))^{*}$ be equipped with weak∗ topology. Then
$L^{\infty}(N)$ is a locally convex topological vector space (see Section
3.14, [R91]). We also consider the $L^{\infty}$ norm $||\cdot||$ on
$L^{\infty}(N)$. We stress that the topology induced by the $L^{\infty}$ norm
is different from the weak∗ topology. In the following, when we say a subset
$X\subset L^{\infty}(N)$ is closed (compact) we mean it is closed (compact) in
the weak∗ topology; similarly for closure of subsets and continuity of maps;
when we say $X$ is bounded we mean it is bounded with respect to the
$L^{\infty}$ norm. By the Banach-Alaoglu theorem, bounded closed subsets of
$L^{\infty}(N)$ are compact.

Next we define an action of the opposite group $\Gamma^{*}$ of $\Gamma$ on
$L^{\infty}(N)$. For $\gamma\in\Gamma$ and $\phi\in L^{\infty}(N)$, define
$\gamma\cdot\phi\in L^{\infty}(N)$ by
$\gamma\cdot\phi=u_{\gamma}(\phi\circ\gamma)$, that is,

$(\gamma\cdot\phi)(w,h)=u_{\gamma}(w,h)\phi(\gamma(w,h))=u_{\gamma}(w,h)\phi(\gamma^{\mathbb{R}}(w,h),\gamma^{H}(h)).$

One checks that this defines a linear action (in particular an affine action)
of $\Gamma^{*}$ on $L^{\infty}(N)$.

######  Lemma 9.1.

The map $\Gamma\times L^{\infty}(N)\rightarrow
L^{\infty}(N),(\gamma,\phi)\mapsto\gamma\cdot\phi$, is separately continuous.

###### Proof.

It is easy to see that for fixed $\gamma\in\Gamma$, the map
$L^{\infty}(N)\rightarrow L^{\infty}(N),\phi\mapsto\gamma\cdot\phi$, is
continuous. We next show that for fixed $\phi\in L^{\infty}(N)$, the map
$\Gamma\rightarrow L^{\infty}(N),\gamma\mapsto\gamma\cdot\phi$, is also
continuous. Let $\phi\in L^{\infty}(N)$ be fixed and
$\gamma_{j},\gamma\in\Gamma$ be such that $\gamma_{j}\rightarrow\gamma$. We
need to show $\gamma_{j}\cdot\phi\stackrel{{\scriptstyle
w^{*}}}{{\longrightarrow}}\gamma\cdot\phi$. Let $C_{c}(N)$ be the space of
compactly supported continuous functions on $N$. Since $C_{c}(N)$ is a dense
subspace of $L^{1}(N)$, it suffices to show that for any fixed $f\in
C_{c}(N)$,
$\int_{N}f(\gamma_{j}\cdot\phi)dm\rightarrow\int_{N}f(\gamma\cdot\phi)dm$ as
$j\rightarrow\infty$, where $m$ denotes the Hausdorff measure on $N$. Notice
that the Jacobian of $\gamma\in\Gamma$ at a point $(w,h)$ is given by
$J\gamma(w,h)=e^{{t_{\gamma}}d_{0}}u_{\gamma}(w,h)$, where $d_{0}$ is the
Hausdorff dimension of $(N,d)$. The area formula applied to the map $\gamma$
and the function $f(\phi\circ\gamma)$ yields

$\int_{N}f(\phi\circ\gamma)J\gamma dm=\int_{N}(f\circ\gamma^{-1})\phi dm.$

It follows that

$\int_{N}f(\gamma\cdot\phi)dm=\int_{N}fu_{\gamma}(\phi\circ\gamma)dm=\frac{1}{e^{{t_{\gamma}}d_{0}}}\int_{N}f(\phi\circ\gamma)J\gamma
dm=\frac{1}{e^{{t_{\gamma}}d_{0}}}\int_{N}(f\circ\gamma^{-1})\phi dm.$

Similarly we have

$\int_{N}f(\gamma_{j}\cdot\phi)dm=\frac{1}{e^{{t_{\gamma_{j}}}d_{0}}}\int_{N}(f\circ\gamma_{j}^{-1})\phi
dm.$

Now

$\displaystyle\int_{N}f(\gamma_{j}\cdot\phi)dm-\int_{N}f(\gamma\cdot\phi)dm$
$\displaystyle=\frac{1}{e^{{t_{\gamma_{j}}}d_{0}}}\int_{N}(f\circ\gamma_{j}^{-1}-f\circ\gamma^{-1})\phi
dm+\big{(}\frac{1}{e^{{t_{\gamma_{j}}}d_{0}}}-\frac{1}{e^{{t_{\gamma}}d_{0}}}\big{)}\int_{N}(f\circ\gamma^{-1})\phi
dm.$

As $t_{\gamma_{j}}\rightarrow t_{\gamma}$, the second term above clearly goes
to $0$ as $j\rightarrow\infty$. Since $f$ is continuous with compact support
and $\gamma_{j}$ converges to $\gamma$ uniformly on compact subsets of $N$,
there is a compact subset $F\subset N$ and a quantity $\epsilon_{j}\rightarrow
0$ such that $\sup\\{|f\circ\gamma_{j}^{-1}(n)-f\circ\gamma^{-1}(n)|:n\in
F\\}<\epsilon_{j}$ and $f\circ\gamma_{j}^{-1}(n)-f\circ\gamma^{-1}(n)=0$ for
all $n\in N\backslash F$ and all sufficiently large $j$. We have

$|\int_{N}(f\circ\gamma_{j}^{-1}-f\circ\gamma^{-1})\phi
dm|\leq\int_{F}|f\circ\gamma_{j}^{-1}-f\circ\gamma^{-1}||\phi|dm\leq\epsilon_{j}\int_{F}|\phi|dm\rightarrow
0$

and hence
$\int_{N}f(\gamma_{j}\cdot\phi)dm\rightarrow\int_{N}f(\gamma\cdot\phi)dm$.

∎

Since $u_{\gamma}$ takes values in the interval $[1/\Lambda,\Lambda]$, we see
that $||\phi||/\Lambda\leq||\gamma\cdot\phi||\leq\Lambda||\phi||$ for any
$\phi\in L^{\infty}(N)$ and any $\gamma\in\Gamma$. It follows that every
$\Gamma$ orbit is a bounded subset of $L^{\infty}(N)$. Let $\phi_{0}$ be the
constant function $1$ on $N$ and $K$ be the closure of the convex hull of
$\Gamma^{*}\cdot\phi_{0}$. Then $K$ is a compact, convex subset in
$L^{\infty}(N)$. Since $\Gamma$ is amenable, Day’s fixed point theorem implies
that $\Gamma$ has a fixed point $u$ in $K$.

Notice that $\gamma\cdot\phi_{0}=u_{\gamma}$ and so is a measurable function
with values in $[1/\Lambda,\Lambda]$. It follows that every element in $K$, in
particular $u$, takes values in $[1/\Lambda,\Lambda]$ a.e. By Fubini’s
theorem, for a.e. $h\in H$, the map $\mathbb{R}\rightarrow\mathbb{R},w\mapsto
u(w,h)$, is measurable and takes values in $[1/\Lambda,\Lambda]$ for a.e.
$w\in W$.

Since $u$ is a fixed point of $\Gamma$, for each $\gamma\in\Gamma$, we have
$u(w,h)=u_{\gamma}(w,h)u(\gamma^{\mathbb{R}}(w,h),\gamma^{H}(h))$ for a.e.
$(w,h)\in W\times H$. By Fubini’s theorem, for a.e. $h\in H$,
$u(w,h)=u_{\gamma}(w,h)u(\gamma^{\mathbb{R}}(w,h),\gamma^{H}(h))$ for a.e.
$w\in W$.

Let $\Gamma_{0}\subset\Gamma$ be a countable dense subgroup. There is a
$\Gamma_{0}$-invariant full measure subset $U\subset H$ with the following
properties:
(1) for each $h\in U$ and each $\gamma\in\Gamma_{0}$, the equality
$u(w,h)=u_{\gamma}(w,h)u(\gamma^{\mathbb{R}}(w,h),\gamma^{H}(h))$ holds for
a.e. $w\in W$.
(2) for each $h\in U$, the map $\mathbb{R}\rightarrow\mathbb{R},w\mapsto
u(w,h)$, is measurable and takes values in $[1/\Lambda,\Lambda]$ for a.e.
$w\in W$.

For each $h\in U$ we define a function $v_{h}:\mathbb{R}\rightarrow\mathbb{R}$
by $v_{h}(w)=\int_{0}^{w}u(s,h)ds$. Clearly $v_{h}$ is $\Lambda$-biLipschitz.
We also define $G_{0}:W\times U\rightarrow W\times U$ by
$G_{0}(w,h)=(v_{h}(w),h)$. We shall show that
(1) $G_{0}$ is biLipschitz (Lemma 9.5) and so admits a biLipschitz extension
$\bar{G}_{0}:N\rightarrow N$;
(2) $\bar{G}_{0}\Gamma\bar{G}_{0}^{-1}$ acts by similarities along the cosets
of $W$ (Lemma 9.6).

######  Lemma 9.2.

The function $v_{h}$ satisfies the following equation for all
$\gamma\in\Gamma_{0}$ and all $w_{1},w_{2}\in W$:

$v_{\gamma^{H}(h)}(\gamma^{\mathbb{R}}(w_{1},h))-v_{\gamma^{H}(h)}(\gamma^{\mathbb{R}}(w_{2},h))=e^{t_{\gamma}}(v_{h}(w_{1})-v_{h}(w_{2})).$

###### Proof.

For each $h\in U$, $u_{\gamma}(w,h)=\frac{\partial\tilde{\gamma}}{\partial
w}(w,h)$ for a.e. $w\in W$. On the other hand, by the choice of $U$, for each
$h\in U$ and each $\gamma\in\Gamma_{0}$, the equality
$u(w,h)=u_{\gamma}(w,h)u(\gamma^{\mathbb{R}}(w,h),\gamma^{H}(h))$ holds for
a.e. $w\in W$. Now we have

$\displaystyle
v_{\gamma^{H}(h)}(\gamma^{\mathbb{R}}(w_{1},h))-v_{\gamma^{H}(h)}(\gamma^{\mathbb{R}}(w_{2},h))$
$\displaystyle=\int_{\gamma^{\mathbb{R}}(w_{2},h)}^{\gamma^{\mathbb{R}}(w_{1},h)}u(s,\gamma^{H}(h))ds$
$\displaystyle=e^{t_{\gamma}}\int_{w_{2}}^{w_{1}}u(\gamma^{\mathbb{R}}(t,h),\gamma^{H}(h))\frac{\partial\tilde{\gamma}}{\partial
t}(t,h)dt$
$\displaystyle=e^{t_{\gamma}}\int_{w_{2}}^{w_{1}}u(t,h)dt=e^{t_{\gamma}}(v_{h}(w_{1})-v_{h}(w_{2})).$

∎

Let $C_{b}(\Gamma)$ be the space of bounded continuous functions on $\Gamma$.
Let $M_{\Gamma}$ be the set of means on $C_{b}(\Gamma)$ of the form

$\sum_{j=1}^{n}t_{j}\delta_{\gamma_{j}}\;\;\;\;\;(n\in\mathbb{N},\gamma_{1},\cdots,\gamma_{n}\in\Gamma,\;t_{1},\cdots,t_{n}\geq
0,\;t_{1}+\cdots+t_{n}=1),$

where for $\gamma\in\Gamma$,
$\delta_{\gamma}:C_{b}(\Gamma)\rightarrow\mathbb{C}$ is the linear functional
given by $\delta_{\gamma}(f)=f(\gamma)$. Then $M_{\Gamma}$ is $w^{*}$-dense in
the set of all means on $C_{b}(\Gamma)$, see [R02], page 29. Let
$\mathbb{Q}M_{\Gamma_{0}}$ be the set of means on $C_{b}(\Gamma)$ of the form

$\sum_{j=1}^{n}t_{j}\delta_{\gamma_{j}}\;\;\;\;\;(n\in\mathbb{N},\gamma_{1},\cdots,\gamma_{n}\in\Gamma_{0},\;t_{1},\cdots,t_{n}\geq
0,\;t_{j}\in\mathbb{Q},\;t_{1}+\cdots+t_{n}=1).$

Since $\Gamma_{0}$ is dense in $\Gamma$ and $\mathbb{Q}$ is dense in
$\mathbb{R}$, we see that $\mathbb{Q}M_{\Gamma_{0}}$ is also $w^{*}$-dense in
the set of all means on $C_{b}(\Gamma)$. Let $K_{0}$ be the set of points in
$K$ of the form

$\sum_{j=1}^{n}t_{j}u_{\gamma_{j}}\;\;\;\;\;(n\in\mathbb{N},\gamma_{1},\cdots,\gamma_{n}\in\Gamma_{0},\;t_{1},\cdots,t_{n}\geq
0,\;t_{j}\in\mathbb{Q},\;t_{1}+\cdots+t_{n}=1).$

######  Lemma 9.3.

There is a sequence $\\{u_{i}\\}$ in $K_{0}$ that converges to $u$ in the
weak∗ topology.

###### Proof.

This follows from the proof of Day’s fixed point theorem, see page 29 of
[R02]. The reader should have a copy of that proof in front of him/her while
reading this proof. In that proof, we pick $x_{0}=\phi_{0}$. Let $A(K)$ be the
set of all continuous affine functions on $K$. For $\psi\in A(K)$, define

$\phi_{\psi}:\Gamma\rightarrow\mathbb{C},\;\;\gamma\mapsto\psi(\gamma\cdot\phi_{0})=\psi(u_{\gamma}).$

Let $m$ be a left invariant mean on $C_{b}(\Gamma)$. Since
$\mathbb{Q}M_{\Gamma_{0}}$ is $w^{*}$-dense in the set of all means on
$C_{b}(\Gamma)$, there is a net $\\{m_{\beta}\\}$ in
$\mathbb{Q}M_{\Gamma_{0}}$ that converges to $m$ in the weak∗ topology. For
each $m_{\beta}=\sum_{j=1}^{n}t_{j}\delta_{\gamma_{j}}$, define $u_{\beta}\in
K_{0}$ by $u_{\beta}=\sum_{j=1}^{n}t_{j}u_{\gamma_{j}}$. One checks that
$<\phi_{\psi},m_{\beta}>=\psi(u_{\beta})$ for all $\psi\in A(K)$. Since $K$ is
compact, $u_{\beta}$ sub-converges to some $u\in K$. It is proved on page 30
of [R02] that this $u$ is a fixed point of $\Gamma$. Finally, since $L^{1}(N)$
is separable, by the sequential Banach-Alaoglu theorem, closed balls in
$L^{\infty}(N)=(L^{1}(N))^{*}$ are metrizable. So $K$ is compact and
metrizable. Therefore we can pick a sequence $\\{u_{i}\\}$ from the net
$\\{u_{\beta}\\}$ that converges to $u$ in the weak∗ topology.

∎

######  Lemma 9.4.

There is a full measure subset $U^{\prime}\subset U$ and a constant $C>0$ such
that $|v_{h_{1}}(w)-v_{h_{2}}(w)|\leq
C\cdot\bar{d}_{CC}(h_{1},h_{2})^{\frac{1}{\alpha}}$ for all $h_{1},h_{2}\in
U^{\prime}$ and all $w\in W$.

###### Proof.

We consider the following $D$-homogeneous distance $d$ on $N$:

$d((w_{1},h_{1}),(w_{2},h_{2}))=|w_{1}-w_{2}|+\bar{d}_{CC}(h_{1},h_{2})^{\frac{1}{\alpha}}.$

Recall that each $\gamma\in\Gamma$ has the expression
$\gamma(w,h)=(e^{t_{\gamma}}\tilde{\gamma}(w,h),\gamma^{H}(h))$ and
$\gamma^{H}:(H,\bar{d}_{CC})\rightarrow(H,\bar{d}_{CC})$ is a similarity with
similarity constant $e^{\alpha t_{\gamma}}$. Since $\Gamma$ is a uniform
quasisimilarity group, by increasing $\Lambda$ if necessary we may assume each
$\gamma$ is a $(\Lambda,e^{t_{\gamma}})$ quasisimilarity of $(N,d)$. It
follows that
$|\tilde{\gamma}(w,h_{1})-\tilde{\gamma}(w,h_{2})|\leq\Lambda\bar{d}_{CC}(h_{1},h_{2})^{\frac{1}{\alpha}}$
for all $h_{1},h_{2}\in H$, all $w\in W$ and all $\gamma\in\Gamma$.

For each integer $n\geq 1$, let $B_{n}\subset H$ be the ball with radius $n$
and center the origin (the identity element of $H$). Set $D_{n}=[-n,n]\times
B_{n}\subset W\times H$. Notice that $u_{i}|_{D_{n}},u|_{D_{n}}\in
L^{\infty}(D_{n})$ and that $u_{i}|_{D_{n}}\rightarrow u|_{D_{n}}$ in the
weak∗ topology. Since $D_{n}$ has finite measure we see that
$u_{i}|_{D_{n}}\rightarrow u|_{D_{n}}$ weakly in $L^{p}(D_{n})$ for any $1\leq
p<\infty$. By Mazur’s lemma, there is a sequence $\tilde{u}_{j}$ of convex
combinations of the $u_{i}^{\prime}s$ such that $\tilde{u}_{j}|_{D_{n}}$
converges to $u|_{D_{n}}$ in the $L^{p}$ norm. Then after taking a subsequence
if necessary we may assume that $\tilde{u}_{j}|_{D_{n}}$ converges to
$u|_{D_{n}}$ a.e. By Fubini, there is a null set $F_{n}\subset H$ such that
for any $h\in B_{n}\backslash F_{n}$, $\tilde{u}_{j}(w,h)\rightarrow u(w,h)$
for a.e. $w\in[-n,n]$. On the other hand, as each $u_{i}$ is a convex
combination of the $u_{\gamma}$’s, each $\tilde{u}_{j}$ is also a convex
combination of the $u_{\gamma}$’s. Write
$\tilde{u}_{j}=\sum_{i=1}^{k_{j}}t_{j,i}u_{\gamma_{j,i}}$ for some
$t_{j,i}\geq 0$ with $\sum_{i}t_{j,i}=1$ and $\gamma_{j,i}\in\Gamma$. Since
each $u_{\gamma}$ is bounded above by $\Lambda$, so is $\tilde{u}_{j}$. It
follows from dominated convergence theorem that
$\int_{0}^{w}\tilde{u}_{j}(s,h)ds\rightarrow\int_{0}^{w}u(s,h)ds=v_{h}(w)$ for
any $w\in[-n,n]$, and any $h\in B_{n}\backslash F_{n}$.

Set $U^{\prime}=U-\cup_{n}F_{n}$. Then $U^{\prime}$ has full measure in $H$.
Let $h_{1},h_{2}\in U^{\prime}$ and $w\in W$. Pick a sufficiently large $n$
such that $(w,h_{1}),(w,h_{2})\in D_{n}$. Now

$\displaystyle|\nu_{h_{2}}(w)-\nu_{h_{1}}(w)|$ $\displaystyle=$
$\displaystyle\lim_{j\to\infty}\left|\int_{0}^{w}\left(\tilde{u}_{j}(s,h_{2})-\tilde{u}_{j}(s,h_{1})\right)ds\right|$
$\displaystyle=$
$\displaystyle\lim_{j\to\infty}\left|\sum_{i}t_{j,i}\int_{0}^{w}(u_{\gamma_{j,i}}(s,h_{2})-u_{\gamma_{j,i}}(s,h_{1}))ds\right|$
$\displaystyle=$
$\displaystyle\lim_{j\to\infty}\left|\sum_{i}t_{j,i}({\tilde{\gamma}_{j,i}}(w,h_{2})-{\tilde{\gamma}_{j,i}}(0,h_{2})-{\tilde{\gamma}_{j,i}}(w,h_{1})+{\tilde{\gamma}_{j,i}}(0,h_{1}))\right|$
$\displaystyle\leq$ $\displaystyle
2\Lambda\bar{d}_{CC}^{\frac{1}{\alpha}}(h_{1},h_{2}).$

∎

######  Lemma 9.5.

The map $G_{0}|_{W\times U^{\prime}}:(W\times U^{\prime},d)\rightarrow(W\times
U^{\prime},d)$ is biLipschitz.

###### Proof.

We first show $G_{0}|_{W\times U^{\prime}}$ is Lipschitz. Let
$(w_{1},h_{1}),(w_{2},h_{2})\in W\times U^{\prime}$. Then

$\displaystyle d(G_{0}(w_{1},h_{1}),G_{0}(w_{2},h_{2}))$
$\displaystyle=d((v_{h_{1}}(w_{1}),h_{1}),(v_{h_{2}}(w_{2}),h_{2}))$
$\displaystyle=|v_{h_{1}}(w_{1})-v_{h_{2}}(w_{2})|+\bar{d}_{CC}^{\frac{1}{\alpha}}(h_{1},h_{2})$
$\displaystyle\leq|v_{h_{1}}(w_{1})-v_{h_{2}}(w_{1})|+|v_{h_{2}}(w_{2})-v_{h_{2}}(w_{1})|+\bar{d}_{CC}^{\frac{1}{\alpha}}(h_{1},h_{2})$
$\displaystyle\leq(2\Lambda+1)\bar{d}_{CC}^{\frac{1}{\alpha}}(h_{1},h_{2})+\Lambda|w_{1}-w_{2}|$
$\displaystyle\leq(2\Lambda+1)d((w_{1},h_{1}),(w_{2},h_{2})).$

Next we show $G_{0}^{-1}|_{W\times U^{\prime}}$ is also Lipschitz. First
assume $|w_{1}-w_{2}|\leq
4\Lambda^{2}\bar{d}_{CC}^{\frac{1}{\alpha}}(h_{1},h_{2})$. Then

$d(G_{0}(w_{1},h_{1}),G_{0}(w_{2},h_{2}))\geq\bar{d}_{CC}^{\frac{1}{\alpha}}(h_{1},h_{2})\geq\frac{1}{4\Lambda^{2}+1}d((w_{1},h_{1}),(w_{2},h_{2})).$

Now we assume $|w_{1}-w_{2}|\geq
4\Lambda^{2}\bar{d}_{CC}^{\frac{1}{\alpha}}(h_{1},h_{2})$. Then

$\displaystyle d(G_{0}(w_{1},h_{1}),G_{0}(w_{2},h_{2}))$
$\displaystyle=|v_{h_{1}}(w_{1})-v_{h_{2}}(w_{2})|+\bar{d}_{CC}^{\frac{1}{\alpha}}(h_{1},h_{2})$
$\displaystyle\geq|v_{h_{1}}(w_{1})-v_{h_{1}}(w_{2})|-|v_{h_{1}}(w_{2})-v_{h_{2}}(w_{2})|+\bar{d}_{CC}^{\frac{1}{\alpha}}(h_{1},h_{2})$
$\displaystyle\geq\frac{1}{\Lambda}|w_{1}-w_{2}|-2\Lambda\bar{d}_{CC}^{\frac{1}{\alpha}}(h_{1},h_{2})+\bar{d}_{CC}^{\frac{1}{\alpha}}(h_{1},h_{2})$
$\displaystyle\geq\frac{1}{2\Lambda}|w_{1}-w_{2}|+\bar{d}_{CC}^{\frac{1}{\alpha}}(h_{1},h_{2})$
$\displaystyle\geq\frac{1}{2\Lambda}d((w_{1},h_{1}),(w_{2},h_{2})).$

∎

Since $W\times U^{\prime}$ is dense in $N$, Lemma 9.5 implies $G_{0}|_{W\times
U^{\prime}}$ extends to a biLipschitz map $\bar{G}_{0}:N\rightarrow N$.

######  Lemma 9.6.

For each $\gamma\in\Gamma$, $\bar{G}_{0}\circ\gamma\circ\bar{G}_{0}^{-1}$ acts
by similarities with similarity constant $e^{t_{\gamma}}$ along the cosets of
$W$.

###### Proof.

Notice that $G_{0}^{-1}$ has the expression
$G_{0}^{-1}(w,h)=(v_{h}^{-1}(w),h)$. As above write $\gamma$ as
$\gamma(w,h)=(\gamma^{\mathbb{R}}(w,h),\gamma^{H}(h))$. It follows that
$G_{0}\circ\gamma\circ
G_{0}^{-1}(w,h)=(v_{\gamma^{H}(h)}(\gamma^{\mathbb{R}}(v_{h}^{-1}(w),h)),\gamma^{H}(h)).$
Let $h\in
U^{\prime}\cap(e^{t_{\gamma}\bar{D}}\circ\bar{\gamma})^{-1}(U^{\prime})$,
$w_{1},w_{2}\in W$. First assume $\gamma\in\Gamma_{0}$. By Lemma 9.2 we have

$|v_{\gamma^{H}(h)}(\gamma^{\mathbb{R}}(v_{h}^{-1}(w_{1}),h))-v_{\gamma^{H}(h)}(\gamma^{\mathbb{R}}(v_{h}^{-1}(w_{2}),h))|=e^{t_{\gamma}}|v_{h}(v_{h}^{-1}(w_{1}))-v_{h}(v_{h}^{-1}(w_{2}))|=e^{t_{\gamma}}|w_{1}-w_{2}|.$

Since
$U^{\prime}\cap(e^{t_{\gamma}\bar{D}}\circ\bar{\gamma})^{-1}(U^{\prime})$ has
full measure in $H$, we see that the map
$\bar{G}_{0}\circ\gamma\circ\bar{G}_{0}^{-1}$ restricted to almost every coset
of $W$ is a similarity with similarity constant $e^{t_{\gamma}}$. Since
$\bar{G}_{0}\circ\gamma\circ\bar{G}_{0}^{-1}$ is biLipschitz, we see that
$\bar{G}_{0}\circ\gamma\circ\bar{G}_{0}^{-1}$ restricted to every coset of $W$
is a similarity with similarity constant $e^{t_{\gamma}}$.

Now consider a general $\gamma\in\Gamma$. There is sequence
$\gamma_{j}\in\Gamma_{0}$ that converges to $\gamma$ uniformly on compact
subsets of $N$. Since each $\bar{G}_{0}\circ\gamma_{j}\circ\bar{G}_{0}^{-1}$
restricted to every coset of $W$ is a similarity with similarity constant
$e^{t_{\gamma_{j}}}$, a similar statement is true for
$\bar{G}_{0}\circ\gamma\circ\bar{G}_{0}^{-1}$. ∎

By Lemma 9.6, after a conjugation if necessary, we may assume
$\tilde{\gamma}(\cdot,h)$ is a translation of $\mathbb{R}$ for each
$\gamma\in\Gamma$ and $h\in H$. Now we can write $\gamma$ as

$\gamma(w,h)=\gamma(0,0)\cdot e^{t_{\gamma}D}(w+s_{\gamma}(h),Bh),$

where $s_{\gamma}:H\rightarrow\mathbb{R}$ is a map satisfying
$s_{\gamma}(0)=0$, and $B:H\rightarrow H$ is a graded automorphism of $H$ that
is also an isometry with respect to $\bar{d}_{CC}$. Since
$\\{e^{-t_{\gamma}D}\circ\gamma|\gamma\in\Gamma\\}$ is a family of uniform
biLipschitz maps, we see that $\\{s_{\gamma}|\gamma\in\Gamma\\}$ is a bounded
subset of

$E_{1}=\\{s:H\rightarrow\mathbb{R}\;\text{is}\;\textstyle{\frac{1}{\alpha}}\text{-Holder},\;s(0)=0\\}.$

Now one can apply Day’s theorem to a compact convex subset $K$ of $E_{1}$ to
eliminate $s_{\gamma}$ as in Section 7 (much easier now). After this, each
$\gamma$ has the form $\gamma(w,h)=\gamma(0,0)\cdot e^{t_{\gamma}D}(w,Bh),$
which is a similarity of $(N,d)$.

We have finished the proof of Theorem 1.2 in the case when $\dim(W)=1$,
$\dim(N/W)\geq 2$.

As already indicated above, the arguments in this section are valid for
product metric spaces $\mathbb{R}\times Y$:

######  Theorem 9.7.

Let $Y$ be a proper metric space with a distance $d$. Let $\alpha\geq 1$ and
$\tilde{d}$ be the distance on $\mathbb{R}\times Y$ defined by
$\tilde{d}((w_{1},y_{1}),(w_{2},y_{2}))=|w_{1}-w_{2}|+d(y_{1},y_{2})^{\frac{1}{\alpha}}$.
Let $\Gamma$ be an amenable uniform quasisimilarity group of
$(\mathbb{R}\times Y,\tilde{d})$. Suppose the action of $\Gamma$ on
$\mathbb{R}\times Y$ permutes the subsets $\\{\mathbb{R}\times\\{y\\}:y\in
Y\\}$ and induces a similarity action on $(Y,d)$. Then $\Gamma$ can be
biLipschitzly conjugated into the similarity group of $(\mathbb{R}\times
Y,\tilde{d})$.

###### Proof.

The argument in this section shows that, after a conjugation, $\Gamma$ acts on
the “fibers” $\mathbb{R}\times\\{y\\}$ by similarities. At this point the
elements of $\Gamma$ may still contain shear components $s$. One can get rid
of the shear components by modifying the above arguments or apply Theorem 3.3
from [DyX16]. ∎

##  10\. Case $\text{dim}(N/W)=1$

In this section we consider the case when $\text{dim}(N/W)=1$.

As before we shall make a biLipschitz conjugation so that $\Gamma$ is a fiber
similarity group. We first need to show that (by replacing $\Gamma$ with a
biLipschitz conjugate if necessary) $\Gamma$ induces an action on the quotient
by affine automorphisms. When $\dim(N/W)\geq 2$, we can apply a Tukia type
theorem to achieve this. We need to consider the case $\dim(N/W)=1$ separately
since we can not apply a Tukia type theorem in this case. The theorem of Farb-
Mosher [FM99] says that a uniform quasisimilarity group of the real line is
biLipschitz conjugate to a similarity group. Although the action of $\Gamma$
on $N$ induces a uniform quasisimilarity action of $\Gamma$ on the quotient
$N/W=\mathbb{R}$, it is not clear how to lift the biLipschitz conjugating map
of $\mathbb{R}$ to a biLipschitz map of $N$.

We consider two cases:
Case I. There is a one dimensional ideal $H$ of $\mathfrak{n}$ such that
$H\subset V_{\alpha}$ and $H\cap W_{\alpha}=\\{0\\}$. In this case
$\mathfrak{n}=\mathfrak{w}\oplus H$ is a direct sum of two ideals and
$N=W\times H$ is a direct product.
Case II. $\mathfrak{n}$ does not admit a direct sum decomposition as in Case
I.

Case I is easy to deal with. Since $\Gamma$ permutes the cosets of $W$, it
induces an action on the quotient $N/W=H=\mathbb{R}$ and yields a subgroup
$\overline{\Gamma}\subset\text{Homeo}(\mathbb{R})$ which is a uniform
quasisimilarity group. The result of Farb–Mosher [FM99] says that there is a
biLipschitz map $f_{0}:\mathbb{R}\rightarrow\mathbb{R}$ such that
$f_{0}\overline{\Gamma}f_{0}^{-1}$ consists of similarities. Now let
$F_{0}=\text{Id}\times f_{0}:N=W\times H\rightarrow W\times H=N$. Then $F_{0}$
is a biLipschitz map of $N$ and the conjugate $F_{0}\Gamma F_{0}^{-1}$ has the
property that its induced action on the quotient $N/W=H$ is by similarities.
If $\dim(W)\geq 2$, then we can apply Theorem 3.7 to conjugate $\Gamma$ into a
fiber similarity group. If $\dim(W)=1$, then Section 9 can now be used to
finish the proof. This finishes Case I.

We next consider Case II. In this case we shall prove that the induced action
of $\Gamma$ on the quotient $N/W=\mathbb{R}$ is already by similarities and so
there is no need to conjugate. So we assume $\mathfrak{n}$ does not admit a
direct sum decomposition as in Case I. We notice that in this case
$\dim(W)\geq 2$. After applying Theorem 3.7 we may assume that the action of
$\Gamma$ on the cosets of $W$ are by similarities. By the discussion in
Section 5, for each $\gamma\in\Gamma$, there is a graded automorphism
$\phi_{\gamma}$ of $W$ such that $(L_{\gamma(g)^{-1}}\circ\gamma\circ
L_{g})|_{W}=\phi_{\gamma}$ for any $g\in N$. The element $\gamma$ induces a
biLipschitz map $\bar{\gamma}:N/W=\mathbb{R}\rightarrow N/W$.

Claim $\bar{\gamma}$ has constant derivative a.e. and so is a similarity.

Suppose the contrary holds. Then there are two points
$\bar{g}_{1}\not=\bar{g}_{2}\in N/W$ such that $\bar{\gamma}$ is
differentiable at both $\bar{g}_{1}$ and $\bar{g}_{2}$ and
$\bar{b}_{1}\not=\bar{b}_{2}$, where $\bar{b}_{1}$, $\bar{b}_{2}$ are the
derivatives of $\bar{\gamma}$ at $\bar{g}_{1}$ and $\bar{g}_{2}$ respectively.
Let $\phi_{i}:N\rightarrow N$ be a blow-up of $\gamma$ at $g_{i}$; that is,
there is a sequence $t_{j}\rightarrow\infty$ such that the sequence of maps
$e^{t_{j}D}\circ L_{(\gamma(g_{i}))^{-1}}\circ\gamma\circ L_{g_{i}}\circ
e^{-t_{j}D}:N\rightarrow N$ converges uniformly on compact subsets to
$\phi_{i}$. Such a sequence $\\{t_{j}\\}$ exists by Arzela-Ascoli as $\gamma$
is biLipschitz. Different choices of the sequences $\\{t_{j}\\}$ may yield
different limits $\phi_{i}$. But all we need is one such limit. The $\phi_{i}$
still acts on the cosets of $W$ by $\phi_{\gamma}$, but now has the additional
property that its induced action on $N/W=\mathbb{R}$ is multiplication by
$\bar{b}_{i}$ (as $\bar{\gamma}$ is differentiable at $\bar{g}_{i}$ with
derivative $\bar{b}_{i}$). Let $X\in V_{\alpha}\backslash W_{\alpha}$. The
Case II assumption implies that $[X,W_{1}]\not=0$ and so $\alpha$ must be an
integer as $[X,W_{1}]\subset W\cap V_{1+\alpha}$. By Lemma 5.4, there is an
injective linear map $B_{i}:\mathbb{R}X\rightarrow V_{\alpha}$ ($B_{i}$ also
depends on $\gamma$) such that $B_{i}$ induces the linear map
$N/W=\mathbb{R}\rightarrow N/W,t\mapsto\bar{b}_{i}t$ and
$d\phi_{\gamma}[X,w]=[B_{i}X,(d\phi_{\gamma})w]$ for all $w\in\mathfrak{w}$.
It follows that $[(B_{1}-B_{2})X,(d\phi_{\gamma})w]=0$ for all
$w\in\mathfrak{w}$. As $d\phi_{\gamma}$ is an automorphism of $\mathfrak{w}$,
we have $[(B_{1}-B_{2})X,\mathfrak{w}]=0$. Notice that
$(B_{1}-B_{2})X=(\bar{b}_{1}-\bar{b}_{2})X+Y$ for some $Y\in W_{\alpha}$.
Since $\bar{b}_{1}-\bar{b}_{2}\not=0$, $X\notin\mathfrak{w}$ and
$\mathfrak{w}$ has codimension one in $\mathfrak{n}$, we see that
$\mathfrak{n}$ is spanned by $\mathfrak{w}$ and $(B_{1}-B_{2})X$. Let
$H^{\prime}$ be the subspace of $\mathfrak{n}$ spanned by $(B_{1}-B_{2})X$.
Then we have that $\mathfrak{n}=\mathfrak{w}\oplus H^{\prime}$ is a direct sum
of two ideals, contradicting our assumption. This completes Case II.

We have shown that $\Gamma$ is a fiber similarity group, after possibly
replacing $\Gamma$ with a biLipschitz conjugate. Now the arguments in Sections
5–8 can be applied to show that $\Gamma$ can be conjugated into
$\text{Sim}(N,d_{0})$ where $d_{0}$ is a fixed maximally symmetric
$D$-homogeneous distance on $N$.

Now we have finished the proof of Theorem 1.2 in the case when $\dim(N/W)=1$.
Combining this with Section 8 and Section 9 we have completed the proof of
Theorem 1.2.

One can use semi-direct product (see Appendix A) to construct many examples of
Case II. For instance, let $\mathfrak{n}=\mathfrak{e}\rtimes\mathbb{R}$, where
$\mathfrak{e}$ is the Engel algebra (with basis $e_{0},e_{1},e_{2},e_{3}$ and
only non-trivial brackets $[e_{0},e_{j}]=e_{j+1}$, $j=1,2$) and
$\mathbb{R}=\mathbb{R}X$ acts on $\mathfrak{e}$ by $[X,e_{1}]=e_{3}$ (all
other brackets are $0$).

Finally we prove a Tukia-type theorem for the product of finitely many Carnot
groups (with possibly different scalings on the factors).

######  Theorem 10.1.

For $1\leq i\leq m$, let $\alpha_{i}\geq 1$ be a constant, $N_{i}$ a Carnot
group and $d_{i}$ a maximally symmetric Carnot metric on $N_{i}$. Set
$N=\prod_{i=1}^{m}N_{i}$ and let $d$ be the distance on $N$ given by
$d((x_{i}),(y_{i}))=\sum_{i}d_{i}^{\frac{1}{\alpha_{i}}}(x_{i},y_{i})$. Let
$\Gamma$ be a uniform quasisimilarity group of $(N,d)$. If the action of
$\Gamma$ on the space of distinct pairs of $N$ is cocompact, then $\Gamma$ can
be conjugated by a biLipschitz map into the similarity group of $(N,d)$. If
$\dim(N_{i})=1$ for some $i$ we further assume $\Gamma$ is amenable.

###### Proof.

By combining those $N_{i}$ where the corresponding $\alpha_{i}$ are equal we
may assume the $\alpha_{i}$’s are all distinct. After reordering we may
further assume $\alpha_{1}<\alpha_{2}<\cdots<\alpha_{m}$. Set
$M_{j}=\prod_{i=1}^{j}N_{i}$. Then $\\{M_{j}\\}_{j}$ form the preserved
subgroup sequence of $N$. By Theorem 3.7 we may assume $\Gamma$ acts on the
fibers $N_{j}=M_{j}/{M_{j-1}}$ by similarities for all $j$ satisfying
$\dim(N_{j})\geq 2$. Denote $H_{j}=\prod_{i=m-j}^{m}N_{i}$ and let
$\tilde{d}_{j}$ be the distance on $H_{j}$ given by
$\tilde{d}_{j}((x_{i}),(y_{i}))=\sum_{i=m-j}^{m}d_{i}^{\frac{1}{\alpha_{i}}}(x_{i},y_{i})$.
We next prove by induction on $j$, starting from $j=0$, that after a
biLipschitz conjugation of $\Gamma$ the induced action of $\Gamma$ on
$(H_{j},\tilde{d}_{j})$ is by similarities. For the base case we let $j=0$. If
$\dim(N_{m})\geq 2$, then the induced action on
$(H_{0},\tilde{d}_{0})=(N_{m},d_{m}^{\frac{1}{\alpha_{m}}})$ is already by
similarities. If $\dim(N_{m})=1$, then the argument in Case I of this section
applies.

For the induction step, assume $j\geq 1$ and that the induced action on
$(H_{j-1},\tilde{d}_{j-1})$ is by similarities. We need to show that, after a
further biLipschitz conjugation, the induced action on $(H_{j},\tilde{d}_{j})$
is by similarities. If $\dim(N_{m-j})=1$, then Theorem 9.7 implies there is a
biLipschitz map $f_{0}$ of $(H_{j},\tilde{d}_{j})$ which conjugates the
induced action on $(H_{j},\tilde{d}_{j})$ to an action by similarities. Let
$F_{0}$ be the biLipschitz map of $N$ which equals $f_{0}$ on $H_{j}$ and is
the identity map on $N_{i}$, $1\leq i<m-j$. Then $F_{0}\Gamma F_{0}^{-1}$
induces a similarity action on $(H_{j},\tilde{d}_{j})$.

Next we assume $\dim(N_{m-j})\geq 2$. By the induction hypothesis the induced
action on $(H_{j-1},\tilde{d}_{j-1})$ is by similarities. By the first
paragraph, the action on the fibers $N_{m-j}=M_{m-j}/{M_{m-j-1}}$ is by
similarities. It follows that the induced action of each $\gamma$ on $H_{j}$
is given by:
$\gamma(x_{n-j},y)=\gamma(0)(A_{\gamma}(x_{n-j}s_{\gamma}(y)),B_{\gamma}y)$,
where $x_{m-j}\in N_{m-j}$, $y\in H_{j-1}$, $A_{\gamma}$ is an automorphism of
$N_{m-j}$ and is also a similarity with respect to $d_{m-j}$, $B_{\gamma}$ is
a similarity of $(H_{j-1},\tilde{d}_{j-1})$ and
$s_{\gamma}:(H_{j-1},\tilde{d}_{j-1})\rightarrow(Z(N_{m-j}),d_{m-j}^{\frac{1}{\alpha_{m-j}}})$
is Lipschitz and satisfies $s_{\gamma}(0)=0$. We identify the Lie group with
its Lie algebra via the exponential map and write
$s_{\gamma}=\sum_{i}s_{\gamma,i}$, where $s_{\gamma,i}:H_{j-1}\rightarrow
Z_{i}(\mathfrak{n}_{m-j})=Z(\mathfrak{n}_{m-j})\cap\mathfrak{n}_{m-j,i}$ is
the $i$-layer component of $s_{\gamma}$ and
$\mathfrak{n}_{m-j}=\oplus_{i}\mathfrak{n}_{m-j,i}$ is the Carnot grading of
$\mathfrak{n}_{m-j}$. For $i\geq 1$, let

$E_{i}=\\{c:(H_{j-1},\tilde{d}_{j-1})\rightarrow(Z_{i}(\mathfrak{n}_{m-j}),|\cdot|^{\frac{1}{i\alpha_{m-j}}})\;\;\text{is}\;\;\text{Lipschitz
and }\;\;c(0)=0\\},$

where $|\cdot|$ is a fixed Euclidean metric on $Z_{i}(\mathfrak{n}_{m-j})$. It
is easy to see that $s_{\gamma,i}\in E_{i}$ for all $\gamma\in\Gamma$.
Furthermore, if $s=\sum_{i}s_{i}$ with $s_{i}:H_{j-1}\rightarrow
Z_{i}(\mathfrak{n}_{m-j})$, then the shear map $F_{s}:H_{j}\rightarrow H_{j}$,
$F_{s}(x_{n-j},y)=(x_{n-j}s(y),y)$ is biLipschitz if and only if $s_{i}\in
E_{i}$. Now applying the arguments from Section 7 we can find a biLipschitz
shear map $f_{0}$ of $(H_{j},\tilde{d}_{j})$ that conjugates the induced
action of $\Gamma$ on $(H_{j},\tilde{d}_{j})$ to an action by similarities.
Letting $F_{0}$ be the biLipschitz map of $N$ that agrees with $f_{0}$ on
$H_{j}$ and is the identity on $N_{i}$ with $1\leq i<m-j$ we see that
$F_{0}\Gamma F_{0}^{-1}$ induces a similarity action on
$(H_{j},\tilde{d}_{j})$

∎

##  Appendix A Examples of Carnot-by-Carnot groups

In this appendix we shall give examples of Carnot-by-Carnot groups. The upshot
is that such groups are abundant. The groups are Carnot group by Carnot group
extensions. These correspond to Lie algebra extensions of the type Carnot
algebra by Carnot algebra. The classical connection between group extension
and group cohomology has a counterpart for Lie algebras: there is a connection
between Lie algebra extension and Lie algebra cohomology, see [S66], [HS]. We
briefly recall this connection here. Any extension
$0\rightarrow\mathfrak{w}\rightarrow\mathfrak{n}\rightarrow\mathfrak{h}\rightarrow
0$ induces a Lie algebra homomorphism
$\bar{\alpha}:\mathfrak{h}\rightarrow\text{out}(\mathfrak{w}):=\text{der}(\mathfrak{w})/\text{ad}(\mathfrak{w})$.
Conversely, any Lie algebra homomorphism
$\bar{\alpha}:\mathfrak{h}\rightarrow\text{out}(\mathfrak{w})$ induces a Lie
algebra homomorphism
$\beta:\mathfrak{h}\rightarrow\text{der}(\mathcal{Z}(\mathfrak{w}))$, and
there exists an extension
$0\rightarrow\mathfrak{w}\rightarrow\mathfrak{n}\rightarrow\mathfrak{h}\rightarrow
0$ inducing the given $\bar{\alpha}$ if and only if a particular cohomology
class in $H^{3}(\mathfrak{h},\mathcal{Z}(\mathfrak{w}))$ vanishes, where the
$\mathfrak{h}$ module structure on $\mathcal{Z}(\mathfrak{w})$ is given by
$\beta$. When an extension as above does exist, the set of equivalence classes
of extensions of $\mathfrak{w}$ by $\mathfrak{h}$ inducing the given
$\bar{\alpha}$ is in one-to-one correspondence with the elements of
$H^{2}(\mathfrak{h},\mathcal{Z}(\mathfrak{w}))$. It follows that for every
example of extension
$0\rightarrow\mathfrak{w}\rightarrow\mathfrak{n}\rightarrow\mathfrak{h}\rightarrow
0$, we get a collection of extensions parametrized by
$H^{2}(\mathfrak{h},\mathcal{Z}(\mathfrak{w}))$. If one starts with a split
extension (which corresponds to semi-direct product) and the second cohomology
is non-trivial, then one gets extensions that are no longer split. This
applies to the semi-direct product examples below.

We next describe some examples. Let $\mathfrak{w}=W_{1}\oplus\cdots\oplus
W_{m}$ and $\mathfrak{h}=H_{1}\oplus\cdots H_{n}$ be two Carnot algebras.

Trivial extension. This is the direct sum $\mathfrak{w}\oplus\mathfrak{h}$.

Semi-direct product. Recall that each Lie algebra homomorphism
$\mathfrak{h}\rightarrow\text{der}(\mathfrak{w})$ determines a semi-direct
product $\mathfrak{w}\rtimes\mathfrak{h}$. The trivial homomorphism yields the
direct sum $\mathfrak{w}\oplus\mathfrak{h}$. We shall construct nontrivial Lie
algebra homomorphisms $\mathfrak{h}\rightarrow\text{der}(\mathfrak{w})$. We
first recall a result about free nilpotent Lie algebras, see [S71].

######  Proposition A.1.

Let $F=F_{1}\oplus\cdots\oplus F_{t}$ be a $t$-step free nilpotent Lie
algebra, and $L:F_{1}\rightarrow F$ be any linear map. Then
(1) $L$ extends uniquely to a derivation $d:F\rightarrow F$;
(2) $L$ extends uniquely to a Lie algebra homomorphism $\phi:F\rightarrow F$.

Assume $\mathfrak{h}=F/{I}$ and $\mathfrak{w}=\tilde{F}/{\tilde{I}}$, where
$F=F_{1}\oplus\cdots\oplus F_{t}$ and
$\tilde{F}=\tilde{F}_{1}\oplus\cdots\oplus\tilde{F}_{\tilde{t}}$ are free
nilpotent Lie algebras, and $I\subset F$, $\tilde{I}\subset\tilde{F}$ are
graded ideals satisfying $I\subset F_{k+1}\oplus\cdots\oplus F_{t}$,
$\tilde{I}\subset\tilde{F}_{\tilde{k}+1}\oplus\cdots\oplus\tilde{F}_{\tilde{t}}$
for some positive integers $k,\tilde{k}$. For each integer $s\geq 0$, let
$\text{der}_{s}(\mathfrak{w})$ be the linear subspace of
$\text{der}(\mathfrak{w})$ defined by:

$\text{der}_{s}(\mathfrak{w})=\\{\rho:\mathfrak{w}\rightarrow\mathfrak{w}\;\;\text{is
a derivation satisfying }\;\;\rho(W_{i})\subset W_{i+s},\forall i\\}.$

######  Lemma A.2.

Let $\alpha\geq 2$ be an integer and
$L:H_{1}\rightarrow\text{der}_{\alpha}(\mathfrak{w})$ be a linear map. If
$m\leq(k+1)\alpha$, then $L$ extends uniquely to a Lie algebra homomorphism
$\mathfrak{h}\rightarrow\text{der}(\mathfrak{w})$.

###### Proof.

Let $L_{\alpha}$ be the Lie subalgebra of $\text{der}(\mathfrak{w})$ generated
by $\text{der}_{\alpha}(\mathfrak{w})$. Since $\mathfrak{w}$ is $m$-step, the
assumption $m\leq(k+1)\alpha$ implies that $L_{\alpha}$ is a nilpotent Lie
algebra with step at most $k$. As $\mathfrak{h}=F/{I}$ with $I\subset
F^{(k+1)}$, we may identity $H_{1}$ with the first layer $F_{1}$ of $F$. Now
$F$ is free nilpotent with step at least $k$ and $L_{\alpha}$ has step at most
$k$. By the universal property of free nilpotent Lie algebra the linear map
$L:H_{1}=F_{1}\rightarrow L_{\alpha}$ extends uniquely to a Lie algebra
homomorphism $\rho:F\rightarrow L_{\alpha}$. Since $L_{\alpha}$ has step at
most $k$, we have $\rho(F^{(k+1)})=0$. As $\mathfrak{h}=F/{I}$ with $I\subset
F^{(k+1)}$, $\rho$ induces a Lie algebra homomorphism $\mathfrak{h}\rightarrow
L_{\alpha}\subset\text{der}(\mathfrak{w})$ that extends $L$.

∎

######  Lemma A.3.

Let $L(W_{1},W_{\alpha+1})$ be the vector space of linear maps from $W_{1}$ to
$W_{\alpha+1}$, and $\Phi:\text{der}_{\alpha}(\mathfrak{w})\rightarrow
L(W_{1},W_{\alpha+1})$ be the restriction map, $\Phi(\rho)=\rho|_{W_{1}}$. If
$m\leq\alpha+\tilde{k}$, then $\Phi$ is a linear isomorphism.

###### Proof.

Since $\mathfrak{w}$ is generated by $W_{1}$, we see that $\Phi$ is injective.
We shall establish surjectivity of $\Phi$ by showing that each map in
$L(W_{1},W_{\alpha+1})$ extends to a derivation, which necessarily lies in
$\text{der}_{\alpha}(\mathfrak{w})$. Let $L\in L(W_{1},W_{\alpha+1})$. As
$\mathfrak{w}=\tilde{F}/{\tilde{I}}$ with
$\tilde{I}\subset\tilde{F}^{(\tilde{k}+1)}$, we may identity $W_{1}$ with
$\tilde{F}_{1}$ and view $L$ as a linear map from $\tilde{F}_{1}$ to
$W_{\alpha+1}$. Since
$W_{\alpha+1}=\tilde{F}_{\alpha+1}/(\tilde{F}_{\alpha+1}\cap\tilde{I})$, we
can lift $L$ to a map into $\tilde{F}_{\alpha+1}$, that is, there is a linear
map $\tilde{L}:\tilde{F}_{1}\rightarrow\tilde{F}_{\alpha+1}$ such that
$\pi\circ\tilde{L}=L$, where $\pi:\tilde{F}\rightarrow\mathfrak{w}$ is the
projection. By Proposition A.1 the linear map $\tilde{L}$ extends to a
derivation $\tilde{d}:\tilde{F}\rightarrow\tilde{F}$. Notice that
$\tilde{d}(\tilde{F}^{(\tilde{k}+1)})\subset\tilde{F}^{(\tilde{k}+1+\alpha)}$.
As $\mathfrak{w}$ is $m$-step, the assumption $m\leq\alpha+\tilde{k}$ implies
that $(\pi\circ\tilde{d})(\tilde{F}^{(\tilde{k}+1)})=0$. As
$\mathfrak{w}=\tilde{F}/{\tilde{I}}$ with
$\tilde{I}\subset\tilde{F}^{(\tilde{k}+1)}$, it is easy to see that
$\tilde{d}$ induces a derivation $d:\mathfrak{w}\rightarrow\mathfrak{w}$ that
extends $L$.

∎

The following Corollary provides many examples of nontrivial semi-direct
products of Carnot algebras and so nontrivial semi-direct products of Carnot
groups.

######  Corollary A.4.

Suppose $m\leq\min\\{\tilde{k}+\alpha,(k+1)\alpha\\}$. Then for any linear map
$f:H_{1}\rightarrow L(W_{1},W_{\alpha+1})$, the map $\Phi^{-1}\circ f$ extends
uniquely to a Lie algebra homomorphism
$\mathfrak{h}\rightarrow\text{der}(\mathfrak{w})$. In particular, for any
nonzero linear map $f:H_{1}\rightarrow L(W_{1},W_{\alpha+1})$, there is a
nontrivial Lie algebra homomorphism
$\mathfrak{h}\rightarrow\text{der}(\mathfrak{w})$ and so a nontrivial
semidirect product $\mathfrak{w}\rtimes\mathfrak{h}$.

We mention a few special cases of Corollary A.4.

(1) $\alpha=m-1$. In this case, the condition
$m\leq\min\\{\tilde{k}+\alpha,(k+1)\alpha\\}$ is automatically satisfied. So
every nonzero linear map $f:H_{1}\rightarrow L(W_{1},W_{m})$ will yield a
nontrivial semidirect product $\mathfrak{w}\rtimes\mathfrak{h}$.
(2) $\alpha=m-2$, $m\geq 4$ and $\tilde{k}\geq 2$, In this case, the condition
$m\leq\min\\{\tilde{k}+\alpha,(k+1)\alpha\\}$ is satisfied. And so every
nonzero linear map $f:H_{1}\rightarrow L(W_{1},W_{m-1})$ will yield a
nontrivial semidirect product $\mathfrak{w}\rtimes\mathfrak{h}$.

Central product. Let $\alpha\geq 2$ be an integer and
$\mathfrak{h}=H_{1}\oplus\cdots\oplus H_{n}$,
$\mathfrak{w}=W_{1}\oplus\cdots\oplus W_{\alpha n}$ be two Carnot algebras.
Let $W^{\prime}\subset W_{\alpha n}$ and $H^{\prime}\subset H_{n}$ be linear
subspaces and $\phi:W^{\prime}\rightarrow H^{\prime}$ a linear isomorphism.
The corresponding central product $\mathfrak{w}\times_{\phi}\mathfrak{h}$ is
the quotient of the direct sum $\mathfrak{w}\oplus\mathfrak{h}$ by the central
ideal $\\{(w,-\phi(w))|w\in W^{\prime}\\}$. Clearly there is a short exact
sequence
$0\rightarrow\mathfrak{w}\rightarrow\mathfrak{w}\times_{\phi}\mathfrak{h}\rightarrow\mathfrak{h}/{H^{\prime}}\rightarrow
0$ and so $\mathfrak{w}\times_{\phi}\mathfrak{h}$ is an extension of
$\mathfrak{w}$ by $\mathfrak{h}/{H^{\prime}}$. Notice that
$\mathfrak{h}/{H^{\prime}}=H_{1}\oplus\cdots\oplus(H_{n}/{H^{\prime}})$ is a
Carnot algebra.

##  Appendix B Lattices in SOL-like groups

In this appendix we give examples of SOL-like groups that admit lattices.

There are necessary and sufficient conditions for the existence of lattices in
solvable Lie groups, see Theorem 6.2 in [M70] and Chapter III, Section 6 of
[Au73]. But those conditions are not easy to check. Here we cite a result by
Sawai–Yamada [SY05] which gives a simple sufficient condition for certain
solvable Lie groups to admit lattices. Let $\mathfrak{n}$ be a rational
nilpotent Lie algebra. Here “rational” means $\mathfrak{n}$ has a basis with
rational structure constants. Then it follows that $\mathfrak{n}$ has a basis
$\\{X_{1},\cdots,X_{m}\\}$ with integer structure constants. Let
$\mathfrak{n}^{i}$ ($i=1,2$) be a copy of $\mathfrak{n}$ with corresponding
basis $\\{X_{1}^{i},\cdots,X_{m}^{i}\\}$. So $X_{j}\mapsto X_{j}^{i}$ extends
to an isomorphism from $\mathfrak{n}$ to $\mathfrak{n}^{i}$. Suppose $k_{j}$,
$1\leq j\leq m$ are integers and the map $X^{1}_{j}\mapsto k_{j}X^{1}_{j}$,
$X^{2}_{j}\mapsto-k_{j}X^{2}_{j}$, extends to a derivation $D$ on
$\mathfrak{n}^{1}\times\mathfrak{n}^{2}$. Let $S$ be the semidirect product
$(N^{1}\times N^{2})\rtimes\mathbb{R}$, where $N^{i}$ is the simply connected
Lie group with Lie algebra $\mathfrak{n}^{i}$ and the action of $\mathbb{R}$
on $N^{1}\times N^{2}$ is generated by the derivation $D$.

######  Theorem B.1.

([SY05], Theorem 2) The solvable Lie group $S$ above admits a lattice.

We recall that lattices in solvable Lie groups are always uniform [M62].

We give two explicit examples. The first is the so-called Benson–Gordon group
[BG], which was also discussed in [SY05]. In this example, $\mathfrak{n}^{i}$
($i=1,2$) is a copy of the Heisenberg algebra with basis $X^{i},Y^{i},Z^{i}$
and the only nontrivial bracket among basis elements is $[X^{i},Y^{i}]=Z^{i}$.
Let $k_{1},k_{2}$ be integers. Let
$D_{i}:\mathfrak{n}^{i}\rightarrow\mathfrak{n}^{i}$ be the derivation given by
$D_{i}(X^{i})=k_{1}X^{i}$, $D_{i}(Y^{i})=k_{2}Y^{i}$,
$D_{i}(Z^{i})=(k_{1}+k_{2})Z^{i}$. Let $D=(D_{1},-D_{2})$ be the derivation of
$\mathfrak{n}^{1}\times\mathfrak{n}^{2}$. The semidirect product
$S_{k_{1},k_{2}}=(N^{1}\times N^{2})\rtimes_{D}\mathbb{R}$ is a Benson-Gordon
group. When $k_{1},k_{2}$ are positive, $S_{k_{1},k_{2}}$ is a SOL-like group
that we are interested. When $k_{1}=k_{2}$ is positive, $(N^{i},D_{i})$ is of
Carnot type. Next we give an example where $N$ is Carnot-by-Carnot.

In the second example $\mathfrak{n}$ is a semi-direct product
$\mathfrak{n}=\mathfrak{e}\rtimes\mathfrak{h}$, where $\mathfrak{e}$ is the
Engel algebra (with basis $e_{0},e_{1},e_{2},e_{3}$ and only non-trivial
brackets $[e_{0},e_{i}]=e_{i+1}$, $i=1,2$) and $\mathfrak{h}$ is the
Heisenberg algebra (with basis $X,Y,Z$ and only non-trivial bracket
$[X,Y]=Z$), and the action of $\mathfrak{h}$ on $\mathfrak{e}$ is given by
$[X,e_{0}]=e_{3}$, $[Y,e_{1}]=e_{3}$ (all other brackets are $0$). This semi-
direct product is of Case (1) discussed after Corollary A.4. Let $D_{0}$ be
the derivation of $\mathfrak{n}$ given by $D_{0}(e_{0})=e_{0}$,
$D_{0}(e_{j})=je_{j}$ ($j=1,2,3$), $D_{0}(X)=2X$, $D_{0}(Y)=2Y$,
$D_{0}(Z)=4Z$. Set $D_{1}=D_{2}=D_{0}$. Then $D=(D_{1},-D_{2})$ is a
derivation of $\mathfrak{n}^{1}\times\mathfrak{n}^{2}$. Finally let
$S=(N^{1}\times N^{2})\rtimes_{D}\mathbb{R}$. By Theorem B.1 $S$ admits a
lattice. In this example, $(N_{i},D_{i})$ is Carnot-by-Carnot.

##  Appendix C Compatible expressions

In this appendix we prove Lemma 5.4. Recall the assumption: $(\mathfrak{n},D)$
is a diagonal Heintze pair, $\mathfrak{w}$ is an ideal of $\mathfrak{n}$ such
that $D(\mathfrak{w})=\mathfrak{w}$; $F:N\rightarrow N$ is a biLipschitz map
that permutes the cosets of $W$, where $W$ is the connected Lie subgroup of
$N$ with Lie algebra $\mathfrak{w}$; for each $g\in N$, the map $F_{g}|_{W}$
is an automorphism $\phi$ of $W$, and $F$ induces an affine map $\bar{F}$ of
$N/W$. Let $\bar{B}$ be the automorphism part of $\bar{F}$.

The main ingredient in the proof is the fact that
$\phi\circ(\chi_{g}|_{W})=(\chi_{G(g)}|_{W})\circ\phi$, where $G=F_{0}$. See
Lemma 5.1. In the following proof we will implicitly (and repeatedly) use the
fact that $[Z(\mathfrak{w}),\mathfrak{n}]\subset Z(\mathfrak{w})$. This
follows from the Jacobi identity and the fact that $W$ is an ideal of
$\mathfrak{n}$.

For a linear transformation $T$ of a finite dimensional vector space, we
denote by $\sigma(T)$ the set of eigenvalues of $T$. Recall that
$\sigma(\bar{D})\subset\sigma(D)$, where
$\bar{D}:\mathfrak{n}/{\mathfrak{w}}\rightarrow\mathfrak{n}/{\mathfrak{w}}$ is
the derivation induced by $D$. Set $I=\sigma(D)-\sigma(\bar{D})$.

###### Proof.

Set $G=F_{0}=L_{F(0)^{-1}}\circ F$. Then $G(0)=0$. By Lemma 5.1, there is an
automorphism $\phi$ of $W$ such that if we set $A=d\phi$, then $A$ is layer-
preserving, $A\circ d(\chi_{g}|_{W})\circ A^{-1}=d(\chi_{G(g)}|_{W})$ and
$G(h*w)=G(h)*Aw$ for any $g=h*w\in\mathfrak{n}$, where $h\in
H,w\in\mathfrak{w}$. Let $H^{\prime}\subset\mathfrak{n}$ be a graded subspace
of $\mathfrak{n}$ complementary to $\mathfrak{w}$; that is, for each
$\lambda\in\sigma(\bar{D})$, $H^{\prime}_{\lambda}\subset V_{\lambda}$ is a
complementary linear subspace of $W_{\lambda}$ in $V_{\lambda}$ (if
$W_{\lambda}=\\{0\\}$, then $H^{\prime}_{\lambda}=V_{\lambda}$), and
$H^{\prime}=\oplus_{\lambda\in\sigma(\bar{D})}H^{\prime}_{\lambda}$. Denote by
$B_{0}:H\rightarrow H^{\prime}$ the linear isomorphism
$(\pi|_{H^{\prime}})^{-1}\circ d\bar{B}\circ(\pi|_{H})$. Since
$\pi(G(h))=d\overline{B}(\bar{h})=\pi(B_{0}h)$, there is a map
$S:\mathfrak{n}/\mathfrak{w}\rightarrow\mathfrak{w}$ such that
$G(h)=B_{0}h*S(\bar{h})$. It follows that the map $G$ has the form

$G(h*w)=B_{0}h*S(\bar{h})*Aw.$

In general, $B_{0}$ and $S$ do not satisfy conditions (2) and (3) in the
definition of compatible expression. We need to modify the map $B_{0}$ (and so
also $S$).

For each $\lambda\in\sigma(D|_{\mathfrak{w}})$, denote
$Z_{\lambda}(\mathfrak{w}):=Z(\mathfrak{w})\cap W_{\lambda}$ and let
$Z_{\lambda}^{\perp}(\mathfrak{w})\subset W_{\lambda}$ be a subspace such that
$W_{\lambda}=Z_{\lambda}(\mathfrak{w})\oplus
Z_{\lambda}^{\perp}(\mathfrak{w})$. The map $S$ can be written as
$S(\bar{h})=\sum_{\lambda\in\sigma(D|_{\mathfrak{w}})}S_{\lambda}(\bar{h})$,
where $S_{\lambda}:=\pi_{\lambda}\circ S$ and
$\pi_{\lambda}:\mathfrak{w}\rightarrow W_{\lambda}$ is the projection with
respect to the decomposition $\mathfrak{w}=\oplus_{\lambda}W_{\lambda}$. There
are two maps $\tilde{S}_{\lambda}:\mathfrak{n}/\mathfrak{w}\rightarrow
Z_{\lambda}(\mathfrak{w})$ and
$S_{\lambda}^{\perp}:\mathfrak{n}/\mathfrak{w}\rightarrow
Z_{\lambda}^{\perp}(\mathfrak{w})$ such that
$S_{\lambda}(\bar{h})=\tilde{S}_{\lambda}(\bar{h})+S_{\lambda}^{\perp}(\bar{h})$.

We shall use the equation $A\circ d(\chi_{g}|_{W})\circ
A^{-1}=d(\chi_{G(g)}|_{W})$. Let $\mu\in\sigma(D|_{\mathfrak{w}})$, $w\in
W_{\mu}$ and $h\in H$ be arbitrary. We next calculate $A\circ
d(\chi_{h}|_{W})\circ A^{-1}(w)$ and $d(\chi_{G(h)}|_{W})(w)$. By (2)

$A\circ d(\chi_{h}|_{W})\circ
A^{-1}(w)=w+A[h,A^{-1}w]+\sum_{i=2}^{\infty}\frac{1}{i!}A(\text{ad}\,h)^{i}(A^{-1}w)$

and

$d(\chi_{G(h)}|_{W})(w)=w+[G(h),w]+\sum_{i=2}^{\infty}\frac{1}{i!}(\text{ad(G(h))})^{i}(w).$

From $A\circ d(\chi_{g}|_{W})\circ A^{-1}=d(\chi_{G(g)}|_{W})$, we get

(14) $\displaystyle
L:=A[h,A^{-1}w]+\sum_{i=2}^{\infty}\frac{1}{i!}A(\text{ad}\,h)^{i}(A^{-1}w)=[G(h),w]+\sum_{i=2}^{\infty}\frac{1}{i!}(\text{ad(G(h))})^{i}(w):=R.$

Since $\mathfrak{w}$ is an ideal in $\mathfrak{n}$, every item in (14) lies in
$\mathfrak{w}$. By the BCH formula,

(15) $\displaystyle
G(h)=B_{0}h*S(\bar{h})=B_{0}h+S(\bar{h})+\frac{1}{2}[B_{0}h,S(\bar{h})]+\cdots.$

First Claim: $S_{\lambda}^{\perp}=0$ for $\lambda\in I$; in other words,
$S_{\lambda}(\bar{h})\in Z_{\lambda}(\mathfrak{w})$ for $\lambda\in I$.

The proof is by induction. Let $\lambda_{0}\in I$. Assume that
$S_{\lambda}^{\perp}=0$ for all $\lambda\in I$ with $\lambda<\lambda_{0}$. To
simplify notation, we use $w^{z}$ to denote an element of $Z(\mathfrak{w})$,
$w^{>}$ to denote an element of $\oplus_{\lambda>\lambda_{0}}W_{\lambda}$, use
$\bar{x}$, $\bar{y}$ and so on to denote elements in
$\sum_{\lambda\in\sigma(\bar{D})}W_{\lambda}$, and use subscripts to denote
different such elements. Using this we can write
$S(\bar{h})=\bar{x}_{1}+w_{1}^{z}+S_{\lambda_{0}}(\bar{h})+w^{>}_{1}$. Then
$[B_{0}h,S(\bar{h})]=\bar{x}_{2}+w^{z}_{2}+w^{>}_{2}$ and all the iterated
brackets of $B_{0}h$ and $S(\bar{h})$ have this form. Hence we have

$G(h)=B_{0}h+\bar{x}_{3}+w^{z}_{3}+S_{\lambda_{0}}(\bar{h})+w^{>}_{3}.$

From this we get
$[G(h),w]=[B_{0}h+\bar{x}_{3},w]+[S_{\lambda_{0}}(\bar{h}),w]+[w_{3}^{>},w]$
and for $i\geq 2$, $(\text{ad}\,G(h))^{i}(w)=y_{1}+y_{2}$, with
$y_{1}\in\oplus_{\lambda\in\sigma(\bar{D})}W_{\lambda+\mu}$ and
$y_{2}\in\oplus_{\lambda>\lambda_{0}+\mu}W_{\lambda}$. Since by assumption
$\lambda_{0}\in I$ we see that $\pi_{\lambda_{0}+\mu}L=0$ and
$\pi_{\lambda_{0}+\mu}R=[S_{\lambda_{0}}(\bar{h}),w]$. It follows that
$[S_{\lambda_{0}}(\bar{h}),w]=0$ for any $w\in W_{\mu}$ and any
$\mu\in\sigma(D|_{\mathfrak{w}})$ and therefore $S_{\lambda_{0}}(\bar{h})\in
Z_{\lambda_{0}}(W)$.

Second Claim: for each $\lambda\in\sigma(\bar{D})$, there is a linear map
$B_{\lambda}:H\rightarrow\mathfrak{n}$ satisfying
$(a)_{\lambda}$. $d\bar{B}\circ\pi|_{H}=\pi\circ B_{\lambda}$ and
$B_{\lambda}(H_{\lambda^{\prime}})\subset V_{\lambda^{\prime}}$ for any
$\lambda^{\prime}\in\sigma(\bar{D})$;
$(b)_{\lambda}$.
$B_{\lambda}|_{H_{\lambda^{\prime}}}=B_{0}|_{H_{\lambda^{\prime}}}$ for
$\lambda^{\prime}>\lambda$;
$(c)_{\lambda}$. $[B_{\lambda}h,Aw]=A[h,w]$ for any $w\in\mathfrak{w}$, and
$h\in H_{\lambda^{\prime}}$ with $\lambda^{\prime}\leq\lambda$;
$(d)_{\lambda}$. The map $G$ can be written
$G(h*w)=B_{\lambda}h*S^{(\lambda)}(\bar{h})*Aw$, where
$S^{(\lambda)}:\mathfrak{n}/\mathfrak{w}\rightarrow\mathfrak{w}$ is a map
satisfying $S^{(\lambda)}_{\lambda^{\prime}}(\bar{h})\in
Z_{\lambda^{\prime}}(\mathfrak{w})$ for $\lambda^{\prime}\leq\lambda$, where
$S^{(\lambda)}_{\lambda^{\prime}}=\pi_{\lambda^{\prime}}\circ S^{(\lambda)}$.

The proof of the Second Claim is also by induction. We let
$\lambda_{0}\in\sigma(\bar{D})$ and assume that $B_{\lambda}$ satisfying
$(a)_{\lambda}-(d)_{\lambda}$ are defined for all $\lambda<\lambda_{0}$. Let
$\lambda_{1}<\lambda_{0}$ be the largest $\lambda\in\sigma(\bar{D})$ less than
$\lambda_{0}$. We shall first show that
${S_{\lambda_{0}}^{(\lambda_{1})}}^{\perp}(\bar{h})$ depends only on the
$H_{\lambda_{0}}$ component $h_{\lambda_{0}}$ of $h$ and is linear in
$h_{\lambda_{0}}$.

Let $h\in H$ and $w\in W_{\mu}$ for some $\mu\in\sigma(D|_{\mathfrak{w}})$. We
will apply $\pi_{\lambda_{0}+\mu}$ to both sides of (14). Using the First
Claim and the induction hypothesis we can write
$S^{(\lambda_{1})}(\bar{h})=w_{1}^{z}+S^{(\lambda_{1})}_{\lambda_{0}}(\bar{h})+w_{1}^{>}$.
From this we get
$[B_{\lambda_{1}}h,S^{(\lambda_{1})}(\bar{h})]=w_{2}^{z}+w_{2}^{>}$;
$G(h)=B_{\lambda_{1}}h+w_{3}^{z}+S^{(\lambda_{1})}_{\lambda_{0}}(\bar{h})+w_{3}^{>}$;
$[G(h),w]=[B_{\lambda_{1}}h,w]+[S^{(\lambda_{1})}_{\lambda_{0}}(\bar{h}),w]+[w_{3}^{>},w]$;
and for $i\geq 2$,
$(\text{ad}\,G(h))^{i}(w)=(\text{ad}B_{\lambda_{1}}h)^{i}(w)+y_{3}$ with
$y_{3}\in\oplus_{\lambda>\lambda_{0}+\mu}W_{\lambda}$. Hence we have

$\pi_{\lambda_{0}+\mu}L=A[h_{\lambda_{0}},A^{-1}w]+\sum_{i\geq
2}\frac{1}{i!}\sum_{\lambda_{j_{1}}+\cdots+\lambda_{j_{i}}=\lambda_{0}}A(\text{ad}h_{\lambda_{j_{1}}}\circ\cdots\circ\text{ad}h_{\lambda_{j_{i}}}(A^{-1}w)),$

$\pi_{\lambda_{0}+\mu}R=[B_{\lambda_{1}}h_{\lambda_{0}},w]+[S^{(\lambda_{1})}_{\lambda_{0}}(\bar{h}),w]+\sum_{i\geq
2}\frac{1}{i!}\sum_{\lambda_{j_{1}}+\cdots+\lambda_{j_{i}}=\lambda_{0}}\text{ad}B_{\lambda_{1}}h_{j_{1}}\circ\cdots\circ\text{ad}B_{\lambda_{1}}h_{j_{i}}(w).$

On the other hand $(c)_{\lambda_{1}}$ implies for $i\geq 2$

$A(\text{ad}h_{\lambda_{j_{1}}}\circ\cdots\circ\text{ad}h_{\lambda_{j_{i}}}(A^{-1}w))=\text{ad}B_{\lambda_{1}}h_{j_{1}}\circ\cdots\circ\text{ad}B_{\lambda_{1}}h_{j_{i}}(w).$

It follows that

(16) $\displaystyle
A[h_{\lambda_{0}},A^{-1}w]=[B_{\lambda_{1}}h_{\lambda_{0}},w]+[S^{(\lambda_{1})}_{\lambda_{0}}(\bar{h}),w].$

Since the two terms $A[h_{\lambda_{0}},A^{-1}w]$ and
$[B_{\lambda_{1}}h_{\lambda_{0}},w]$ depend only on the $H_{\lambda_{0}}$
component $h_{\lambda_{0}}$ of $h$ and are linear in $h_{\lambda_{0}}$, we see
that
${S_{\lambda_{0}}^{(\lambda_{1})}}^{\perp}(\bar{h})={S_{\lambda_{0}}^{(\lambda_{1})}}^{\perp}(\bar{h}_{\lambda_{0}})$
also depends only on $h_{\lambda_{0}}$ and is linear in $h_{\lambda_{0}}$.

We define $B_{\lambda_{0}}$ as follows:
$B_{\lambda_{0}}|_{H_{\lambda}}=B_{\lambda_{1}}|_{H_{\lambda}}$ for
$\lambda\not=\lambda_{0}$ and
$B_{\lambda_{0}}h=B_{\lambda_{1}}h+{S_{\lambda_{0}}^{(\lambda_{1})}}^{\perp}(\bar{h})$
for $h\in H_{\lambda_{0}}$. We need to verify
$(a)_{\lambda_{0}}-(d)_{\lambda_{0}}$. The properties $(a)_{\lambda_{0}}$ and
$(b)_{\lambda_{0}}$ are easy to see and $(c)_{\lambda_{0}}$ follows from
$(c)_{\lambda_{1}}$, (16) and the definition of $B_{\lambda_{0}}$. For
$(d)_{\lambda_{0}}$: use $(d)_{\lambda_{1}}$ and write $G$ as

$G(h*w)=B_{\lambda_{1}}h*S^{(\lambda_{1})}(\bar{h})*Aw=B_{\lambda_{0}}h*S^{(\lambda_{0})}(\bar{h})*Aw,$

where
$S^{(\lambda_{0})}(\bar{h})=(-B_{\lambda_{0}}h)*B_{\lambda_{1}}h*S^{(\lambda_{1})}(\bar{h})$.
We need to show $S^{(\lambda_{0})}_{\lambda^{\prime}}(\bar{h})\in
Z_{\lambda^{\prime}}(\mathfrak{w})$ for $\lambda^{\prime}\leq\lambda_{0}$. By
the definition of $B_{\lambda_{0}}$ and the linearity of $B_{\lambda_{0}}$,
$B_{\lambda_{1}}$ we obtain
$B_{\lambda_{0}}h=B_{\lambda_{1}}h+{S_{\lambda_{0}}^{(\lambda_{1})}}^{\perp}(\bar{h}_{\lambda_{0}})=B_{\lambda_{1}}h+{S_{\lambda_{0}}^{(\lambda_{1})}}^{\perp}(\bar{h})$
for any $h\in H$. Using this we get

$[-B_{\lambda_{0}}h,B_{\lambda_{1}}h]=-[{S_{\lambda_{0}}^{(\lambda_{1})}}^{\perp}(\bar{h}),B_{\lambda_{1}}h]\in\oplus_{\lambda\geq(\lambda_{0}+\alpha)}W_{\lambda}.$

By the BCH formula,
$(-B_{\lambda_{0}}h)*B_{\lambda_{1}}h=(-B_{\lambda_{0}}h)+B_{\lambda_{1}}h+\frac{1}{2}[-B_{\lambda_{0}}h,B_{\lambda_{1}}h]+\cdots=-{S_{\lambda_{0}}^{(\lambda_{1})}}^{\perp}(\bar{h})+w_{4},$
with $w_{4}\in\oplus_{\lambda\geq(\lambda_{0}+\alpha)}W_{\lambda}$. By
$(d)_{\lambda_{1}}$,
$[-{S_{\lambda_{0}}^{(\lambda_{1})}}^{\perp}(\bar{h}),S^{(\lambda_{1})}(\bar{h})]\in\oplus_{\lambda\geq
2\lambda_{0}}W_{\lambda}$. Finally,

$\displaystyle S^{(\lambda_{0})}(\bar{h})$
$\displaystyle=(-B_{\lambda_{0}}h)*B_{\lambda_{1}}h*S^{(\lambda_{1})}(\bar{h})$
$\displaystyle=-{S_{\lambda_{0}}^{(\lambda_{1})}}^{\perp}(\bar{h})+w_{4}+S^{(\lambda_{1})}(\bar{h})+\frac{1}{2}[-{S_{\lambda_{0}}^{(\lambda_{1})}}^{\perp}(\bar{h})+w_{4},S^{(\lambda_{1})}(\bar{h})]+\cdots$
$\displaystyle=S^{(\lambda_{1})}(\bar{h})-{S_{\lambda_{0}}^{(\lambda_{1})}}^{\perp}(\bar{h})+w_{5},$

with $w_{5}\in\oplus_{\lambda\geq(\lambda_{0}+\alpha)}W_{\lambda}$. From this,
$(d)_{\lambda_{1}}$ and the First Claim we see that
$S_{\lambda}^{(\lambda_{0})}(\bar{h})=S_{\lambda}^{(\lambda_{1})}(\bar{h})\in
Z_{\lambda}(\mathfrak{w})$ for $\lambda<\lambda_{0}$, and

$S_{\lambda_{0}}^{(\lambda_{0})}(\bar{h})=S_{\lambda_{0}}^{(\lambda_{1})}(\bar{h})-{S_{\lambda_{0}}^{(\lambda_{1})}}^{\perp}(\bar{h})=\tilde{S}_{\lambda_{0}}^{(\lambda_{1})}(\bar{h})\in
Z_{\lambda_{0}}(\mathfrak{w}).$

This verifies $(d)_{\lambda_{0}}$ and completes the induction argument for the
Second Claim.

Denote by $\bar{\lambda}$ the largest eigenvalue of $\bar{D}$. We set
$B=B_{\bar{\lambda}}$ and $s=A^{-1}\circ S^{(\bar{\lambda})}$. We need to
verify conditions 1–3 in the definition of a compatible expression. Condition
1 follows from $(a)_{\bar{\lambda}}$ and Condition 2 follows from
$(c)_{\bar{\lambda}}$. From $(d)_{\bar{\lambda}}$ we get an expression for $F$
(as $G=F_{0}=L_{F(0)^{-1}}\circ F$):

$F(h*w)=F(0)*Bh*As(\bar{h})*Aw.$

By the First Claim and $(d)_{\bar{\lambda}}$ we see that $s(\bar{h})\in
Z(\mathfrak{w})$. This allows us to switch $As(\bar{h})$ and $Aw$ to arrive at
a compatible expression for $F$.

∎

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