Then by Equation (10),

$\displaystyle\sum_{n=1}^{N^{\prime}}$ $\displaystyle
e^{-\delta\phi(\kappa_{\theta}(\rho(\gamma_{n})))}=\sum_{n=1}^{N}e^{-\delta\phi(\kappa_{\theta}(\rho(\gamma_{n})))}+\sum_{n=N+1}^{N^{\prime}}e^{-\delta\phi(\kappa_{\theta}(\rho(\gamma_{n})))}\leq\sum_{n=1}^{N}e^{-\delta\phi(\kappa_{\theta}(\rho(\gamma_{n})))}+CM$
$\displaystyle\leq\left(1+\frac{CM}{e^{-\delta\phi(\kappa_{\theta}(\rho(\gamma_{1})))}}\right)\sum_{n=1}^{N}e^{-\delta\phi(\kappa_{\theta}(\rho(\gamma_{n})))}$

for all $N\geq 1$. The lemma now holds with
$C_{1}:=1+\frac{CM}{e^{-\delta\phi(\kappa_{\theta}(\rho(\gamma_{1})))}}$. ∎

The next lemma verifies that the sequence $\\{A_{n}\\}$ satisfy the second
hypothesis of Lemma 9.2.

######  Lemma 9.4.

There exists $C_{2}>0$ such that: if $N\geq 1$, then

$\displaystyle\sum_{1\leq n,m\leq N}$ $\displaystyle\nu(A_{n}\cap A_{m})\leq
C_{2}\left(\sum_{n=1}^{N}\nu(A_{n})\right)^{2}.$

###### Proof.

Let

$\Delta_{N}:=\\{(m,n):1\leq n\leq m\leq N\text{ and }A_{m}\cap
A_{n}\neq\emptyset\\}.$

One can show (see the proof of Lemma 8.2) that if $(m,n)\in\Delta_{N}$, then

$\displaystyle{\rm
d}_{\Omega}(\gamma_{n}(b_{0}),[b_{0},\gamma_{m}(b_{0})]_{\Omega})\leq 2r.$

Then Proposition 6.3 implies

$\sup_{(m,n)\in\Delta_{N}}\left\|\kappa_{\theta}(\rho(\gamma_{n}))+\kappa_{\theta}(\rho(\gamma_{n}^{-1}\gamma_{m}))-\kappa_{\theta}(\rho(\gamma_{m}))\right\|<+\infty,$
(11)

and so by Equation (10), there exists a constant $C^{\prime}>0$ such that

$\displaystyle\nu(A_{n}\cap A_{m})$
$\displaystyle\leq\nu(\mathcal{O}_{r}(b_{0},\gamma_{m}(b_{0})))\leq
C^{\prime}e^{-\delta\phi(\kappa_{\theta}(\rho(\gamma_{n})))}e^{-\delta\phi(\kappa_{\theta}(\rho(\gamma_{n}^{-1}\gamma_{m})))}$

for all $(m,n)\in\Delta_{N}$. Also,

$\displaystyle{\rm d}_{\Omega}(b_{0},\gamma_{n}^{-1}\gamma_{m}(b_{0}))$
$\displaystyle={\rm d}_{\Omega}(\gamma_{n}(b_{0}),\gamma_{m}(b_{0}))\leq{\rm
d}_{\Omega}(\gamma_{n}(b_{0}),[b_{0},\gamma_{m}(b_{0})]_{\Omega})+{\rm
d}_{\Omega}(b_{0},\gamma_{m}(b_{0}))$ $\displaystyle\leq 2r+T_{m}\leq
2r+T_{N}$

for all $(m,n)\in\Delta_{N}$. In particular, if $(m,n)\in\Delta_{N}$, then
$\gamma_{n}^{-1}\gamma_{m}=\gamma_{k}$ for some $k\leq N^{\prime}$.

These observations, Lemma 9.3 and Equation (10) imply that if $N\geq 1$, then

$\displaystyle\sum_{1\leq n,m\leq N}$ $\displaystyle\nu(A_{n}\cap A_{m})\leq
2\sum_{(m,n)\in\Delta_{N}}\nu(A_{n}\cap A_{m})\leq
2C^{\prime}\sum_{(m,n)\in\Delta_{N}}e^{-\delta\phi(\kappa_{\theta}(\rho(\gamma_{n})))}e^{-\delta\phi(\kappa_{\theta}(\rho(\gamma_{n}^{-1}\gamma_{m})))}$
$\displaystyle\leq
2C^{\prime}\sum_{k=1}^{N^{\prime}}\sum_{n=1}^{N}e^{-\delta\phi(\kappa_{\theta}(\rho(\gamma_{n})))}e^{-\delta\phi(\kappa_{\theta}(\rho(\gamma_{k})))}\leq
2C^{\prime}C_{1}\left(\sum_{n=1}^{N}e^{-\delta\phi(\kappa_{\theta}(\rho(\gamma_{n})))}\right)^{2}$
$\displaystyle\leq
2C^{\prime}C_{1}C^{2}\left(\sum_{n=1}^{N}\nu(A_{n})\right)^{2}.\qed$

We may now apply Lemma 9.2 to the finite measure space
$\left(\partial\Omega,\nu\right)$ and the sequence $\\{A_{n}\\}$ to finish the
proof of Proposition 9.1.

###### Proof of Proposition 9.1.

We first show that $\mu(\Lambda_{\theta}^{\rm con}(\Gamma))>0$. By Lemma 9.2,
if we set

$Y:=\\{x\in\partial\Omega:x\text{ is in infinitely many of
}A_{1},A_{2},\dots\\},$

then $\nu(Y)>0$. Notice that if $x\in Y$, then there is a sequence
$\\{\gamma_{n}\\}$ in $\Gamma_{0}$ such that $\gamma_{n}(b_{0})\rightarrow x$
and

${\rm d}_{\Omega}(\gamma_{n}(b_{0}),[b_{0},x))<r$

for all $n\geq 1$. Thus $Y\subset\Lambda_{\Omega}^{\rm con}(\Gamma_{0})$. By
Observation 6.1(1), $\xi(Y)\subset\Lambda_{\theta}^{\rm con}(\Gamma)$, so

$\mu(\Lambda_{\theta}^{\rm con}(\Gamma))\geq\mu(\xi(Y))=\nu(Y)>0.$

Now suppose for contradiction that $\mu(\Lambda_{\theta}^{\rm
con}(\Gamma))<1$. If we set $S:=\Lambda_{\theta}(\Gamma)-\Lambda_{\theta}^{\rm
con}(\Gamma)$, then $\mu(S)>0$, so we may define a probability measure
$\mu_{S}$ on $\Lambda_{\theta}(\Gamma)$ by

$\mu_{S}(A):=\frac{1}{\mu(S)}\mu(A\cap S).$

By definition, $\mu_{S}(\Lambda_{\theta}(\Gamma))=0$. On the other hand, since
$S$ is $\Gamma$-invariant, $\mu_{S}$ is a $\phi$-Patterson-Sullivan measure
for $\Gamma$ of dimension $\delta$, so the above argument implies that
$\mu_{S}(\Lambda_{\theta}^{\rm con}(\Gamma))>0$, which is a contradiction.
Therefore, $\mu(\Lambda_{\theta}^{\rm con}(\Gamma))=1$.

By Proposition 8.1, $\mu$ has no atoms in $\Lambda_{\theta}^{\rm
con}(\Gamma)$. Since $\mu(\Lambda_{\theta}^{\rm con}(\Gamma))=1$, we conclude
that $\mu$ has no atoms. ∎

##  10\. Non-ergodicity of the flow in the convergent case

In this section, we prove that the geodesic flow of a transverse
representation is dissipative and non-ergodic if its image is in the
convergent case of our Hopf-Sullivan-Tsuji dichotomy.

######  Proposition 10.1.

Let $\Omega\subset\operatorname{\mathbb{P}}(\operatorname{\mathbb{R}}^{d})$ be
a properly convex domain, let
$\Gamma_{0}\subset\operatorname{\mathrm{Aut}}(\Omega)$ be a non-elementary
projectively visible subgroup and let $\rho:\Gamma_{0}\to\mathsf{G}$ be a
$\operatorname{\mathsf{P}}_{\theta}$-transverse representation for some
symmetric $\theta\subset\Delta$. Suppose $\phi\in{\mathfrak{a}}_{\theta}^{*}$
satisfies $\delta:=\delta^{\phi}(\rho(\Gamma_{0}))<+\infty$. Let $\mu$ and
$\bar{\mu}$ respectively be $\phi$ and $\bar{\phi}$-Patterson-Sullivan
measures for $\rho(\Gamma_{0})$ of dimension $\beta$, and let $m$ be the
Bowen-Margulis measure on $\widehat{\operatorname{\mathsf{U}}}(\Gamma_{0})$
associated to $\rho$, $\mu$ and $\bar{\mu}$. If
$Q^{\phi}_{\Gamma}(\delta)<+\infty$, then

  1. (1)

the $\Gamma_{0}\times\operatorname{\mathbb{R}}$-action on
$\left(\operatorname{\mathsf{U}}(\Gamma_{0}),\widetilde{m}\right)$ is
dissipative,

  2. (2)

the action of the geodesic flow on
$\left(\widehat{\operatorname{\mathsf{U}}}(\Gamma_{0}),m\right)$ is
dissipative, and

  3. (3)

the action of the geodesic flow on
$\left(\widehat{\operatorname{\mathsf{U}}}(\Gamma_{0}),m\right)$ is non-
ergodic.

Before proving Proposition 10.1, we briefly discuss the notions of dissipative
and conservative dynamical systems. Suppose that $X$ is a standard Borel
space, $H$ is a locally compact, second countable, unimodular group that acts
measurably on $X$, ${\rm d}h$ is a Haar measure on $H$, and $m$ a $H$-quasi-
invariant, $\sigma$-finite measure on $X$. If $A\subset X$ has positive
$m$-measure, we say that $A\subset X$ is _wandering_ if for $m$-almost every
$x\in A$, $\int_{h\in H}\mathbf{1}_{A}(h(x)){\rm d}h<+\infty$. Then let
$\mathcal{D}\subset\Omega$ be the union of all wandering sets, and let
$\mathcal{C}:=\Omega-\mathcal{D}$. We say that $H$-action on $(X,m)$ is
_conservative_ (resp. _dissipative_) if $m(\mathcal{D})=0$ (resp.
$m(\mathcal{C})=0$).

Given a $m$-integrable, positive function $f:X\to(0,\infty)$, we may decompose
$X$ into

$\mathcal{C}_{f}:=\left\\{x\in X:\int_{H}f(h(x)){\rm
d}h=+\infty\right\\}\quad\text{ and }\quad\mathcal{D}_{f}:=\left\\{x\in
X:\int_{H}f(h(x)){\rm d}h<+\infty\right\\}.$

In the case when the measure $m$ is $H$-invariant, it is known (see for
instance [5, Fact 2.27]) that ${\mathcal{C}}_{f}={\mathcal{C}}$ and
$\mathcal{D}_{f}=\mathcal{D}$ up to measure zero sets.

###### Proof of Proposition 10.1.

Proof of (1). Suppose for contradiction that the
$\Gamma_{0}\times\operatorname{\mathbb{R}}$-action on
$\left(\operatorname{\mathsf{U}}(\Gamma_{0}),\widetilde{m}\right)$ is not
dissipative. Then there is a $\widetilde{m}$-integrable, positive function
$f:\operatorname{\mathsf{U}}(\Gamma_{0})\to(0,\infty)$ and a compact set

$K\subset\mathcal{C}_{f}:=\left\\{v\in\operatorname{\mathsf{U}}(\Gamma_{0}):\sum_{\gamma\in\Gamma_{0}}\int_{\operatorname{\mathbb{R}}}f(\gamma\cdot\varphi_{t}(v)){\rm
d}t=+\infty\right\\}$

such that $\widetilde{m}(K)>0$. For any $R>0$, let

$K_{R}:=\left\\{v\in
K:\sum_{\gamma\in\Gamma_{0}}\int_{\operatorname{\mathbb{R}}}1_{K}(\gamma\cdot\varphi_{t}(v)){\rm
d}t\leq R\right\\}.$

Since $K_{R}\subset\mathcal{C}_{f}$, the integral

$\int_{\operatorname{\mathsf{U}}(\Gamma_{0})}\sum_{\gamma\in\Gamma_{0}}\int_{\operatorname{\mathbb{R}}}f(\gamma\cdot\varphi_{t}(v))1_{K_{R}}(v){\rm
d}t\,{\rm
d}\widetilde{m}(v)=\int_{K_{R}}\sum_{\gamma\in\Gamma_{0}}\int_{\operatorname{\mathbb{R}}}f(\gamma\cdot\varphi_{t}(v)){\rm
d}t\,{\rm d}\widetilde{m}(v).$

is infinite if $\widetilde{m}(K_{R})>0$. On the other hand, since
$\widetilde{m}$ is $\Gamma_{0}\times\operatorname{\mathbb{R}}$-invariant,

$\displaystyle\int_{\operatorname{\mathsf{U}}(\Gamma_{0})}\sum_{\gamma\in\Gamma_{0}}\int_{\operatorname{\mathbb{R}}}f(\gamma\cdot\varphi_{t}(v))1_{K_{R}}(v){\rm
d}t\,{\rm d}\widetilde{m}(v)$
$\displaystyle=\int_{\operatorname{\mathsf{U}}(\Gamma_{0})}f(v)\sum_{\gamma\in\Gamma_{0}}\int_{\operatorname{\mathbb{R}}}1_{K_{R}}(\gamma\cdot\varphi_{t}(v)){\rm
d}t\,{\rm d}\widetilde{m}(v)$ $\displaystyle\leq
R\int_{\operatorname{\mathsf{U}}(\Gamma_{0})}f(v){\rm
d}\widetilde{m}(v)<+\infty.$

It follows that $\widetilde{m}(K_{R})=0$ for all $R>0$, or equivalently, that

$\sum_{\gamma\in\Gamma_{0}}\int_{\operatorname{\mathbb{R}}}1_{K}(\gamma\cdot\varphi_{t}(v)){\rm
d}t=+\infty$

for $\widetilde{m}$-almost every $v\in K$. This in turn implies that for
almost every $v\in K$, there are diverging sequences $\\{t_{n}\\}$ in
$\operatorname{\mathbb{R}}$ and $\\{\gamma_{n}\\}$ in $\Gamma_{0}$ such that
$\gamma_{n}\varphi_{t_{n}}(v)\in K$, and so at least one of the forward
endpoint $v^{+}$ or backward endpoint $v^{-}$ of $v$ is in
$\Lambda_{\Omega}^{\rm con}(\Gamma_{0})$. Thus,

$\mu(\xi(\Lambda_{\Omega}^{\rm con}(\Gamma_{0})))>0,$

since $\widetilde{m}(K)>0$. However, by Proposition 8.1,
$\mu(\xi(\Lambda_{\Omega}^{\rm con}(\Gamma_{0})))=0$, which is a
contradiction.

Proof of (2). Let $f:\operatorname{\mathsf{U}}(\Gamma_{0})\to(0,\infty)$ be a
$\widetilde{m}$-integrable, positive function. By part (1), we may define an
$m$-integrable, positive function
$F:\widehat{\operatorname{\mathsf{U}}}(\Gamma_{0})\to(0,\infty)$ by
$F([v]):=\sum_{\gamma\in\Gamma_{0}}f(\gamma\cdot v)$. Furthermore, for
$m$-almost every $[v]\in\widehat{\operatorname{\mathsf{U}}}(\Gamma_{0})$,

$\int_{\operatorname{\mathbb{R}}}F(\varphi_{t}([v])){\rm
d}t=\sum_{\gamma\in\Gamma_{0}}\int_{\operatorname{\mathbb{R}}}f(\gamma\cdot\varphi_{t}(v)){\rm
d}t<+\infty.$

Proof of (3). Pick a compact set
$K\subset\widehat{\operatorname{\mathsf{U}}}(\Gamma_{0})$ with non-empty
interor. Let $f:\widehat{\operatorname{\mathsf{U}}}(\Gamma_{0})\to(0,\infty)$
be a $m$-integrable, positive function that takes the value $1$ on the compact
set $\varphi_{[0,1]}(K)$. Part (2) implies that for $m$-almost every
$v\in\widehat{\operatorname{\mathsf{U}}}(\Gamma_{0})$, we have

$\int_{\operatorname{\mathbb{R}}}f(\varphi_{t}(v)){\rm d}t<+\infty,$

so there is some $T_{v}>0$ such that $\varphi_{t}(v)\notin K$ for all
$t\notin[-T_{v},T_{v}]$.

Suppose for contradiction that the action of the geodesic flow on
$\left(\widehat{\operatorname{\mathsf{U}}}(\Gamma_{0}),m\right)$ is ergodic.
Then for $m$-almost every
$v\in\widehat{\operatorname{\mathsf{U}}}(\Gamma_{0})$, the flow line of $v$ is
dense in $\widehat{\operatorname{\mathsf{U}}}(\Gamma_{0})$. Thus, there is
some $v_{0}\in\widehat{\operatorname{\mathsf{U}}}(\Gamma_{0})$ and some
$T:=T_{v_{0}}>0$ such that
$\widehat{\operatorname{\mathsf{U}}}(\Gamma_{0})=\overline{\varphi_{\operatorname{\mathbb{R}}}(v_{0})}$
and

$\varphi_{(-\infty,-T)}(v_{0})\cup\varphi_{(T,\infty)}(v_{0})\subset\widehat{\operatorname{\mathsf{U}}}(\Gamma_{0})-K.$

It follows that the interior $K^{0}$ of $K$ lies in $\varphi_{[-T,T]}(v_{0})$.
However, it is easy to see that no open set in
$\widehat{\operatorname{\mathsf{U}}}(\Gamma_{0})$ is homeomorphic to a subset
of the interior of a line segment. ∎

##  11\. Ergodicity of the flow in the divergent case

In this section, we prove that the geodesic flow of a transverse
representation is conservative and ergodic if its image is in the divergent
case of our Hopf-Sullivan-Tsuji dichotomy.

######  Theorem 11.1.

Let $\Omega\subset\operatorname{\mathbb{P}}(\operatorname{\mathbb{R}}^{d})$ be
a properly convex domain, let
$\Gamma_{0}\subset\operatorname{\mathrm{Aut}}(\Omega)$ be a non-elementary
projectively visible subgroup and let $\rho:\Gamma_{0}\to\mathsf{G}$ be a
$\operatorname{\mathsf{P}}_{\theta}$-transverse representation for some
symmetric $\theta\subset\Delta$. Let $\phi\in{\mathfrak{a}}_{\theta}^{*}$, let
$\mu$ and $\bar{\mu}$ respectively be $\phi$ and $\bar{\phi}$-Patterson-
Sullivan measures for $\rho(\Gamma_{0})$ of dimension
$\delta:=\delta^{\phi}(\rho(\Gamma_{0}))$, let $m$ be the Bowen-Margulis
measure on $\widehat{\operatorname{\mathsf{U}}}(\Gamma_{0})$ associated to
$\rho$, $\mu$ and $\bar{\mu}$, and let $\widetilde{m}$ be the lift of $m$ to
$\operatorname{\mathsf{U}}(\Gamma_{0})$. If
$Q_{\rho(\Gamma_{0})}^{\phi}(\delta)=+\infty$, then

  1. (1)

the action of the geodesic flow on
$\left(\widehat{\operatorname{\mathsf{U}}}(\Gamma_{0}),m\right)$ is
conservative, and

  2. (2)

the action of the geodesic flow on
$\left(\widehat{\operatorname{\mathsf{U}}}(\Gamma_{0}),m\right)$ is ergodic.

Before starting the proof of Theorem 11.1, we recall a result of Coudène.
Suppose $\\{\varphi_{t}\\}$ is a continuous flow on a metric space $X$ which
preserves a Borel measure $m$. The _strong stable manifold_ of $x\in X$ is

$W^{ss}(x):=\left\\{y\in X:\lim_{t\rightarrow\infty}{\rm
d}(\varphi_{t}(x),\varphi_{t}(y))=0\right\\}$

and the _strong unstable manifold_ of $x\in X$ is

$W^{su}(x):=\left\\{y\in X:\lim_{t\rightarrow-\infty}{\rm
d}(\varphi_{t}(x),\varphi_{t}(y))=0\right\\}.$

A measurable function $f:X\rightarrow\operatorname{\mathbb{R}}$ is _$W^{ss}$
-invariant_ if there exists a full measure set $X^{\prime}\subset X$ where
$f(x)=f(y)$ whenever $x,y\in X^{\prime}$ and $y\in W^{ss}(x)$. Similarly, a
measurable function $f:X\rightarrow\operatorname{\mathbb{R}}$ is _$W^{su}$
-invariant_ if there exists a full measure set $X^{\prime}\subset X$ where
$f(x)=f(y)$ whenever $x,y\in X^{\prime}$ and $y\in W^{su}(x)$.

######  Theorem 11.2 (Coudène [17]).

Let $X$ be a metric space, $\\{\varphi_{t}\\}$ a continuous flow on $X$ and
$m$ a $\\{\varphi_{t}\\}$-invariant Borel measure on $X$ such that
$(X,m,\\{\varphi_{t}\\})$ is conservative. Suppose that there is a full
measure subset of $X$ that is covered by a countable family of open sets with
finite m-measure. Then every flow-invariant, $m$-measurable function on $X$ is
$W^{ss}$-invariant and $W^{su}$-invariant.

###### Proof of Theorem 11.1.

Proof of (1). Fix a $m$-integrable, positive, continuous function
$f:\widehat{\operatorname{\mathsf{U}}}(\Gamma_{0})\to(0,\infty)$. Then let

$\widetilde{f}:=f\circ
p:\operatorname{\mathsf{U}}(\Gamma_{0})\to\operatorname{\mathbb{R}},$

where
$p:\operatorname{\mathsf{U}}(\Gamma_{0})\to\widehat{\operatorname{\mathsf{U}}}(\Gamma_{0})$
is the quotient map. To show that the action of the geodesic flow on
$\left(\widehat{\operatorname{\mathsf{U}}}(\Gamma_{0}),m\right)$ is
conservative, it suffices to show that

$\int_{\operatorname{\mathbb{R}}}\widetilde{f}(\varphi_{t}(v)){\rm
d}t=\int_{\operatorname{\mathbb{R}}}f(\varphi_{t}(p(v))){\rm d}t$

is infinite for $\tilde{m}$-almost every
$v\in\operatorname{\mathsf{U}}(\Gamma_{0})$.

By Proposition 9.1 and Observation 6.1(1), the set

$\mathcal{R}:=\left\\{v\in\operatorname{\mathsf{U}}(\Gamma_{0}):v^{+}\in\Lambda_{\Omega}^{\rm
con}(\Gamma_{0})\right\\}$

has full $\widetilde{m}$-measure.

Fix $v\in\mathcal{R}$. Then there is some $r>0$ and a sequence
$\\{\gamma_{n}\\}$ in $\Gamma_{0}$ such that $\gamma_{n}(\pi(v))\to v^{+}$ and

${\rm d}_{\Omega}\big{(}\gamma_{n}(\pi(v)),[\pi(v),v^{+})_{\Omega}\big{)}<r$

for all $n$, where $\pi:T^{1}\Omega\to\Omega$ is the projection map. In
particular, there exists a compact subset
$K\subset\operatorname{\mathsf{U}}(\Gamma_{0})$, which depends on $r>0$, such
that

$\\{t\in\operatorname{\mathbb{R}}:\varphi_{t}(v)\in\Gamma_{0}\cdot K\\}$

has infinite Lebesgue measure. Since $\tilde{f}$ is $\Gamma_{0}$-invariant and
continuous,

$\inf_{w\in\Gamma_{0}\cdot K}f(w)=\min_{w\in K}f(w)>0.$

Hence

$\int_{\operatorname{\mathbb{R}}}\widetilde{f}(\varphi_{t}(v)){\rm
d}t=+\infty.$

Since $v\in\mathcal{R}$ was arbitrary and $\mathcal{R}$ has full
$\widetilde{m}$-measure,

$\int_{\operatorname{\mathbb{R}}}\widetilde{f}(\varphi_{t}(v)){\rm
d}t=+\infty$

for $\tilde{m}$-almost every $v\in\operatorname{\mathsf{U}}(\Gamma_{0})$.

Proof of (2). Notice that Equation (7) implies that

$p(W^{ss}(v))\subset W^{ss}(p(v))\quad\text{and}\quad p(W^{su}(v))\subset
W^{su}(p(v))$

for all $v\in\operatorname{\mathsf{U}}(\Gamma_{0})$, so the lift of a
$W^{ss}$-invariant (respectively $W^{su}$-invariant) function on
$\widehat{\operatorname{\mathsf{U}}}(\Gamma_{0})$ is a $W^{ss}$-invariant
(respectively $W^{su}$-invariant) function on
$\operatorname{\mathsf{U}}(\Gamma_{0})$. By definition,
$\widehat{\operatorname{\mathsf{U}}}(\Gamma_{0})$ is covered by a countable
family of open sets of with finite $m$-measure, so by Theorem 11.2, it
suffices to show that if
$f:\operatorname{\mathsf{U}}(\Gamma_{0})\to\mathbb{R}$ is a
$\tilde{m}$-measurable, $\Gamma$-invariant, $\\{\varphi_{t}\\}$-invariant,
$W^{ss}$-invariant and $W^{su}$-invariant function, then $f$ is constant on a
set of full $\tilde{m}$-measure.

Since $f$ is $W^{ss}$-invariant and $W^{su}$-invariant, by definition there
exists a full $\tilde{m}$-measure set
$Y_{0}\subset\operatorname{\mathsf{U}}(\Gamma_{0})$ such that $f(v)=f(w)$
whenever $v,w\in Y_{0}$ and $v\in W^{ss}(w)\cup W^{su}(w)$. Since $f$ is
$\\{\varphi_{t}\\}$-invariant, we can assume that $Y_{0}$ is also
$\\{\varphi_{t}\\}$-invariant. Let $\nu$ and $\bar{\nu}$ be measures on
$\partial\Omega$ given by

$\nu(A)=\mu\left(\xi(A\cap\Lambda_{\Omega}(\Gamma_{0}))\right)\quad\text{and}\quad\bar{\nu}(A)=\bar{\mu}\left(\xi(A\cap\Lambda_{\Omega}(\Gamma_{0}))\right),$

where $\xi$ is the limit map of $\rho$. By the definition of $\tilde{m}$, we
see that $Y_{0}=Y_{0}^{\prime}\times\operatorname{\mathbb{R}}$ for some set
$Y_{0}^{\prime}\subset\Lambda_{\Omega}(\Gamma_{0})^{(2)}$ of full
$\bar{\nu}\otimes\nu$-measure. Set

$\displaystyle X^{+}$
$\displaystyle:=\\{y\in\Lambda_{\Omega}(\Gamma_{0}):(x,y)\in
Y_{0}^{\prime}\text{ for $\bar{\nu}$-almost every
$x\in\Lambda_{\Omega}(\Gamma_{0})$}\\},$

and note that $\nu(X^{+})=1$ by Fubini’s theorem. Hence, if we fix
$(v_{0}^{-},v_{0}^{+})\in(\Lambda_{\Omega}(\Gamma_{0})\times X^{+})\cap
Y_{0}^{\prime}$, then the set

$Y^{\prime}:=\left\\{(x,y)\in Y_{0}^{\prime}:(x,v_{0}^{+})\in
Y_{0}^{\prime}\right\\}$

has full $\bar{\nu}\otimes\nu$-measure, so
$Y:=Y^{\prime}\times\operatorname{\mathbb{R}}\subset\operatorname{\mathsf{U}}(\Gamma_{0})$
has full $\tilde{m}$-measure.

Let $(x,y,t)\in Y$. By Lemma 5.3, there is some
$s\in\operatorname{\mathbb{R}}$ such that $(x,y,t)\in W^{su}(x,v_{0}^{+},s)$,
and there is some $r\in\operatorname{\mathbb{R}}$ such that
$(x,v_{0}^{+},s)\in W^{ss}(v_{0}^{-},v_{0}^{+},r)$. By definition, $(x,y,t)$,
$(x,v_{0}^{+},s)$, and $(v_{0}^{-},v_{0}^{+},r)$ lie in $Y_{0}$, so

$f(x,y,t)=f(x,v_{0}^{+},s)=f(v_{0}^{-},v_{0}^{+},r)=f(v_{0}^{-},v_{0}^{+},0).$

This proves that $f$ is constant on $Y$.∎

##  12\. Consequences of ergodicity

In this section we record some consequences of Theorem 11.1. The first two
corollaries complete the proof of our Hopf-Sullivan-Tsuji dichotomy. We also
show, in the divergent case, that there is some $R$ so that the uniformly
$R$-conical limit set has full measure for the unique Patterson-Sullivan
measure of critical dimension and establish a rigidity result for pairs of
transverse representations with mutually non-singular BMS measures.

######  Corollary 12.1.

Suppose $\Gamma\subset\mathsf{G}$ is a non-elementary
$\operatorname{\mathsf{P}}_{\theta}$-transverse subgroup for some symmetric
$\theta\subset\Delta$, $\phi\in{\mathfrak{a}}_{\theta}^{*}$ and
$\delta:=\delta^{\phi}(\Gamma)<+\infty$. Let $\mu$ and $\bar{\mu}$
respectively be $\phi$ and $\bar{\phi}$-Patterson-Sullivan measures for
$\Gamma$ of dimension $\beta$.

  1. (1)

If $Q_{\Gamma}^{\phi}(\delta)=+\infty$ and $\beta=\delta$, then the
$\Gamma$-action on $(\Lambda_{\theta}(\Gamma)^{(2)},\bar{\mu}\otimes\mu)$ is
conservative, and the $\Gamma$ actions on
$(\Lambda_{\theta}(\Gamma)^{(2)},\bar{\mu}\otimes\mu)$ and on
$(\Lambda_{\theta}(\Gamma),\mu)$ are ergodic.

  2. (2)

If $Q_{\Gamma}^{\phi}(\delta)<+\infty$, then the $\Gamma$ action on
$(\Lambda_{\theta}(\Gamma)^{(2)},\bar{\mu}\otimes\mu)$ is dissipative and non-
ergodic.

###### Proof.

Using the discussion in Section 2.4 we may assume that $\mathsf{G}$ has
trivial center and that $\operatorname{\mathsf{P}}_{\theta}$ does not contain
any simple factors of $\mathsf{G}$. Then by Theorem 6.2, there is a properly
convex domain
$\Omega\subset\operatorname{\mathbb{P}}(\operatorname{\mathbb{R}}^{d})$, a
projectively visible subgroup
$\Gamma_{0}\subset\operatorname{\mathrm{Aut}}(\Omega)$ and a
$\operatorname{\mathsf{P}}_{\theta}$-transverse representation
$\rho:\Gamma_{0}\rightarrow\mathsf{G}$ such that $\rho(\Gamma_{0})=\Gamma$.
Let $\xi:\Lambda_{\Omega}(\Gamma_{0})\to\Lambda_{\theta}(\Gamma)$ be the
$\rho$-equivariant boundary map and let $m$ be the Bowen-Margulis measure on
$\widehat{\operatorname{\mathsf{U}}}(\Gamma_{0})$ associated to $\rho$, $\mu$
and $\bar{\mu}$.

Proof of (1). Theorem 11.1 part (3) implies that the geodesic flow on
$(\widehat{\operatorname{\mathsf{U}}}(\Gamma_{0}),m)$ is ergodic. Any
$\Gamma$-invariant subset of either
$(\Lambda_{\theta}(\Gamma)^{(2)},\bar{\mu}\otimes\mu)$ or
$(\Lambda_{\theta}(\Gamma),\mu)$ that has positive but not full measure, gives
rise to a flow-invariant subset of
$(\widehat{\operatorname{\mathsf{U}}}(\Gamma_{0}),m)$ that has positive but
not full measure. Therefore, the actions of $\Gamma$ on
$(\Lambda_{\theta}(\Gamma)^{(2)},\bar{\mu}\otimes\mu)$ and
$(\Lambda_{\theta}(\Gamma),\mu)$ are both ergodic.

Next, suppose for the purpose of contradiction that the action of $\Gamma$ on
$(\Lambda_{\theta}(\Gamma)^{(2)},\bar{\mu}\otimes\mu)$ is not conservative.
Since the measure $e^{-\delta\phi([\cdot,\cdot]_{\theta})}\bar{\mu}\otimes\mu$
on $\Lambda_{\theta}(\Gamma)^{(2)}$ is $\Gamma$-invariant, there is a
positive, continuous function $f:\Lambda_{\theta}(\Gamma)^{(2)}\to(0,\infty)$
such that

$\mathcal{D}_{f}:=\left\\{(x,y)\in\Lambda_{\theta}(\Gamma)^{(2)}:\sum_{\gamma\in\Gamma}f(\gamma\cdot(x,y))<+\infty\right\\}$

has positive $\bar{\mu}\otimes\mu$-measure. Since $\Gamma$ acts minimally on
$\Lambda_{\theta}(\Gamma)$, each open set in $\Lambda_{\theta}(\Gamma)^{(2)}$
has positive $\bar{\mu}\otimes\mu$-measure. This, together with the fact that
the action of $\Gamma$ on
$(\Lambda_{\theta}(\Gamma)^{(2)},\bar{\mu}\otimes\mu)$ is ergodic, implies
that almost every orbit is dense. Thus, there exists
$(x_{0},y_{0})\in\mathcal{D}_{f}$ with
$\Lambda_{\theta}(\Gamma)^{(2)}=\overline{\Gamma\cdot(x_{0},y_{0})}$, but this
is a contradiction since $f$ is positive and

$\sum_{\gamma\in\Gamma}f(\gamma\cdot(x_{0},y_{0}))<+\infty.$

Proof of (2). Let $f:\operatorname{\mathsf{U}}(\Gamma_{0})\to(0,\infty)$ be a
$\widetilde{m}$-integrable, positive function. By Proposition 10.1 part (1),
we may define the $\bar{\mu}\otimes\mu$-integrable, positive function

$F:\Lambda_{\theta}(\Gamma)^{(2)}\to\operatorname{\mathbb{R}}\quad\text{by}\quad
F(\xi(v^{-}),\xi(v^{+})):=\int_{\operatorname{\mathbb{R}}}f(\varphi_{t}(v)).$

Furthermore, for $\bar{\mu}\otimes\mu$-almost every
$(\xi(v^{-}),\xi(v^{+}))\in\Lambda_{\theta}(\Gamma)^{(2)}$, we have

$\sum_{\rho(\gamma)\in\Gamma}F(\rho(\gamma)\xi(v^{-}),\rho(\gamma)\xi(v^{+}))=\sum_{\gamma\in\Gamma_{0}}\int_{\operatorname{\mathbb{R}}}f(\gamma(\varphi_{t}(v))){\rm
d}t<+\infty.$

It follows that the $\Gamma$ action on
$(\Lambda_{\theta}(\Gamma)^{(2)},\bar{\mu}\otimes\mu)$ is dissipative.

Proposition 10.1 part (3) implies that there is some subset of
$\operatorname{\mathsf{U}}(\Gamma_{0})$ that is invariant under the
$\Gamma_{0}\times\operatorname{\mathbb{R}}$-action, with positive but not full
$\widetilde{m}$-measure. This defines a subset of
$\Lambda_{\theta}(\Gamma)^{(2)}$ that is invariant under the $\Gamma$-action,
with positive but not full $\bar{\mu}\otimes\mu$-measure. Thus, the $\Gamma$
action on $(\Lambda_{\theta}(\Gamma)^{(2)},\bar{\mu}\otimes\mu)$ is non-
ergodic. ∎

It follows, from a standard argument (see for instance [45, pg. 181]), that
the Patterson-Sullivan measure in the critical dimension is unique in the
divergent case.

######  Corollary 12.2.

Suppose $\Gamma\subset\mathsf{G}$ is a non-elementary
$\operatorname{\mathsf{P}}_{\theta}$-transverse subgroup for some symmetric
$\theta\subset\Delta$, $\phi\in{\mathfrak{a}}_{\theta}^{*}$ and
$\delta:=\delta^{\phi}(\Gamma)<+\infty$. If
$Q_{\Gamma}^{\phi}(\delta)=+\infty$, then there is a unique $\phi$-Patterson-
Sullivan measure $\mu_{\phi}$ for $\Gamma$ of dimension $\delta$.

For the next two results let
$\Omega\subset\operatorname{\mathbb{P}}(\operatorname{\mathbb{R}}^{d})$ be a
properly convex domain, let
$\Gamma_{0}\subset\operatorname{\mathrm{Aut}}(\Omega)$ be a projectively
visible subgroup and let $b_{0}\in\Omega$. For any $R>0$, we denote by
$\Lambda_{\Omega,b_{0},R}^{\rm con}(\Gamma_{0})$ the set of points
$x\in\Lambda_{\Omega}^{\rm con}(\Gamma_{0})$ for which there exists a sequence
$\\{\gamma_{n}\\}$ in $\Gamma_{0}$ such that $\gamma_{n}(b_{0})\rightarrow x$
and

${\rm d}_{\Omega}(\gamma_{n}(b_{0}),[b_{0},x)_{\Omega})<R$

for all $n$. The next corollary proves that if the image of a transverse
representation is in the divergent case, then there is an $R>0$ such that the
set of $R$-conical limit points have full measure.

######  Corollary 12.3.

Suppose $\rho:\Gamma_{0}\rightarrow\mathsf{G}$ is a
$\mathsf{P}_{\theta}$-transverse representation for some symmetric
$\theta\subset\Delta$, $\phi\in{\mathfrak{a}}_{\theta}^{*}$,
$\delta:=\delta^{\phi}(\rho(\Gamma_{0}))<+\infty$ and $\mu$ is the
$\phi$-Patterson-Sullivan measure for $\rho(\Gamma_{0})$ of dimension
$\delta$. If $Q_{\rho(\Gamma_{0})}^{\phi}(\delta)=+\infty$, then for any
$b_{0}\in\Omega$, there exists $R>0$ such that

$\mu\left(\xi(\Lambda_{\Omega,b_{0},R}^{\rm con}(\Gamma_{0}))\right)=1.$

###### Proof.

The following argument is standard, see for instance [45, pg. 190]. Define a
measure $\nu$ on $\partial\Omega$ by

$\nu(A)=\mu\big{(}\xi(A\cap\Lambda_{\Omega}(\Gamma_{0}))\big{)}.$

Since $Q_{\rho(\Gamma_{0})}^{\phi}(\delta)=+\infty$, by Proposition 9.1,

$1=\nu\left(\Lambda_{\Omega}^{\rm
con}(\Gamma_{0})\right)=\lim_{n\rightarrow\infty}\nu\left(\Lambda_{\Omega,b_{0},n}^{\rm
con}(\Gamma_{0})\right).$

Hence there exists $R_{0}>0$ such that
$\nu\left(\Lambda_{\Omega,b_{0},R_{0}}^{\rm con}(\Gamma_{0})\right)>0$.

Let $L$ be the set of points $x\in\Lambda_{\Omega}(\Gamma_{0})$ for which
there exist $b\in\Gamma_{0}(b_{0})$ and a sequence $\\{\gamma_{n}\\}$ in
$\Gamma_{0}$ such that $\gamma_{n}(b)\rightarrow x$ and

${\rm d}_{\Omega}(\gamma_{n}(b),[b,x)_{\Omega})\leq R_{0}$

for all $n$. Observe that $L$ is $\Gamma_{0}$-invariant, and $\nu(L)>0$
because $\Lambda_{\Omega,b_{0},R_{0}}^{\rm con}(\Gamma_{0})\subset L$. Hence
by Corollary 12.1, $\nu(L)=1$.

It now suffices to show that $L\subset\Lambda_{\Omega,b_{0},R_{0}+1}^{\rm
con}(\Gamma_{0})$. Fix $x\in L$. Then there exist $b\in\Gamma_{0}(b_{0})$, a
sequence $\\{\gamma_{n}\\}$ in $\Gamma_{0}$, and a sequence $\\{b_{n}\\}$ in
$[b,x)_{\Omega}$ where $\gamma_{n}(b)\rightarrow x$ and

${\rm d}_{\Omega}(\gamma_{n}(b),b_{n})\leq R_{0}$

for all $n$. By Lemma 5.3, there exists a sequence $\\{b_{n}^{\prime}\\}$ in
$[b_{0},x)_{\Omega}$ such that

$\lim_{n\rightarrow\infty}{\rm d}_{\Omega}(b_{n},b_{n}^{\prime})=0.$

Since $b\in\Gamma_{0}(b_{0})$, we can write $b=\gamma(b_{0})$ for some
$\gamma\in\Gamma_{0}$. Then

${\rm d}_{\Omega}(\gamma_{n}\gamma(b_{0}),[b_{0},x)_{\Omega})\leq R_{0}+1$

for all $n$ sufficiently large. So $x\in\Lambda_{\Omega,b_{0},R_{0}+1}^{\rm
con}(\Gamma_{0})$. ∎

Finally, we prove the following rigidity result for length functions which
have non-singular Bowen-Margulis-Sullivan measures.

######  Corollary 12.4.

For $j=1,2$, suppose $\rho_{j}:\Gamma_{0}\rightarrow\mathsf{G}_{j}$ is a
$\mathsf{P}_{\theta_{j}}$-transverse representation for some symmetric
$\theta_{j}\subset\Delta_{j}$, $\phi_{j}\in{\mathfrak{a}}_{\theta_{j}}^{*}$
and $\delta_{j}:=\delta^{\phi_{j}}(\rho_{j}(\Gamma_{0}))<+\infty$. For
$\psi\in\\{\phi_{j},\bar{\phi}_{j}\\}$, let $\mu_{\psi}$ be the
$\psi$-Patterson-Sullivan measure for $\rho_{j}(\Gamma_{0})$ of dimension
$\delta_{j}$ and let $m_{j}$ denote the Bowen-Margulis-Sullivan measure
associated to $\rho_{j}$, $\mu_{\phi_{j}}$ and $\mu_{\bar{\phi}_{j}}$. If
$Q_{\rho_{j}(\Gamma_{0})}^{\phi_{j}}(\delta_{j})=+\infty$ for $j=1,2$ and
$m_{1}$ is non-singular with respect to $m_{2}$, then:

  1. (1)

$m_{1}=cm_{2}$ for some $c>0$.

  2. (2)

$\sup_{\gamma\in\Gamma_{0}}\left|\delta_{1}\phi_{1}(\kappa_{\theta_{1}}(\rho_{1}(\gamma)))-\delta_{2}\phi_{2}(\kappa_{\theta_{2}}(\rho_{2}(\gamma)))\right|<+\infty$.

  3. (3)

$\delta_{1}\ell^{\phi_{1}}(\rho_{1}(\gamma))=\delta_{2}\ell^{\phi_{2}}(\rho_{2}(\gamma))$
for all $\gamma\in\Gamma_{0}$.

If, in addition, $\mathsf{G}_{j}$ is simple, $Z(\mathsf{G}_{j})$ is trivial
and $\rho_{j}$ has Zariski-dense image for $j=1,2$, then there is an
isomorphism $\Psi:\mathsf{G}_{1}\to\mathsf{G}_{2}$ such that
$\rho_{2}=\Psi\circ\rho_{1}$.

The proof of Corollary 12.4 requires the following lemma.

######  Lemma 12.5.

Suppose
$\Omega\subset\operatorname{\mathbb{P}}(\operatorname{\mathbb{R}}^{d})$ is a
properly convex domain and $\Gamma\subset\operatorname{\mathrm{Aut}}(\Omega)$
is a projectively visible subgroup. Let ${\rm d}_{\operatorname{\mathbb{P}}}$
be a distance on $\operatorname{\mathbb{P}}(\operatorname{\mathbb{R}}^{d})$
induced by a Riemannian metric. If $r>0$, $b_{0}\in\Omega$ and
$\\{\gamma_{n}\\}$ is a sequence of distinct elements in $\Gamma$, then

$\lim_{n\rightarrow\infty}\mathrm{diam}\left(\mathcal{O}_{r}(b_{0},\gamma_{n}(b_{0}))\right)=0,$

where the diameter is computed using ${\rm d}_{\operatorname{\mathbb{P}}}$.

###### Proof.

Fix a subsequence $\\{\gamma_{n_{j}}\\}$ such that

$\limsup_{n\rightarrow\infty}\mathrm{diam}\left(\mathcal{O}_{r}(b_{0},\gamma_{n}(b_{0}))\right)=\lim_{j\rightarrow\infty}\mathrm{diam}\left(\mathcal{O}_{r}(b_{0},\gamma_{n_{j}}(b_{0}))\right).$

Passing to a further subsequence we can suppose that
$\gamma_{n_{j}}(b_{0})\rightarrow x\in\Lambda_{\Omega}(\Gamma)$ and
$\gamma_{n_{j}}\rightarrow T\in\operatorname{\mathbb{P}}({\rm
End}(\operatorname{\mathbb{R}}^{d}))$. To show that
$\mathrm{diam}\left(\mathcal{O}_{r}(b_{0},\gamma_{n_{j}}(b_{0}))\right)$
converges to 0, it suffices to fix a sequence $\\{y_{j}\\}$ where
$y_{j}\in\mathcal{O}_{r}(b_{0},\gamma_{n_{j}}(b_{0}))$ for all $j\geq 1$ and
show that $y_{j}\rightarrow x$. By definition, for each $j\geq 1$ there exists
$y_{j}^{\prime}\in[b_{0},y_{j})$ such that ${\rm
d}_{\Omega}(y_{j}^{\prime},\gamma_{n_{j}}(b_{0}))<r$. Then the sequence
$\\{\gamma_{n_{j}}^{-1}(y_{j}^{\prime})\\}$ is relatively compact in $\Omega$.
So by Proposition 5.2(3)

$x=T\left(\lim_{j\rightarrow\infty}\gamma_{n_{j}}^{-1}(y_{j}^{\prime})\right)=\lim_{j\rightarrow\infty}\gamma_{n_{j}}\gamma_{n_{j}}^{-1}(y_{j}^{\prime})=\lim_{j\rightarrow\infty}y_{j}^{\prime}.$

Since $y_{j}^{\prime}\in[b_{0},y_{j})$ for all $j\geq 1$, this implies that
$y_{j}\rightarrow x$. ∎

###### Proof of Corollary 12.4.

By the ergodicity of the flow $\\{\varphi_{t}\\}$ (see Theorem 11.1) and the
assumption that $m_{1}$ is non-singular with respect to $m_{2}$, there exists
$c>0$ such that $m_{1}=cm_{2}$.

Note that for $j=1,2$ and $\gamma\in\Gamma_{0}$,

$\ell^{\phi_{j}}(\rho_{j}(\gamma))=\lim_{n\rightarrow\infty}\frac{1}{n}\phi_{j}(\kappa_{\theta_{j}}(\rho_{j}(\gamma^{n}))).$

Thus, to prove part (3), it suffices to prove part (2).

For all
$\psi\in\left\\{\phi_{1},\phi_{2},\bar{\phi}_{1},\bar{\phi}_{2}\right\\}$, let
$\nu_{\psi}$ be the measure on $\partial\Omega$ given by

$\nu_{\psi}(A)=\mu_{\psi}\big{(}\xi(A\cap\Lambda_{\Omega}(\Gamma_{0}))\big{)}.$

Fix $r>0$ sufficiently large so that the Shadow lemma (Proposition 7.1) holds
for the probability measures $\nu_{\phi_{1}}$ and $\nu_{\phi_{2}}$. Then there
is some $C>0$ such that

$\displaystyle\frac{1}{C}\frac{e^{\delta_{2}\phi_{2}(\kappa_{\theta_{2}}(\rho_{2}(\gamma)))}}{e^{\delta_{1}\phi_{1}(\kappa_{\theta_{1}}(\rho_{1}(\gamma)))}}\leq\frac{\nu_{\phi_{1}}(\mathcal{O}_{r}(b_{0},\gamma(b_{0})))}{\nu_{\phi_{2}}(\mathcal{O}_{r}(b_{0},\gamma(b_{0})))}\leq
C\frac{e^{\delta_{2}\phi_{2}(\kappa_{\theta_{2}}(\rho_{2}(\gamma)))}}{e^{\delta_{1}\phi_{1}(\kappa_{\theta_{1}}(\rho_{1}(\gamma)))}}$
(12)

for all $\gamma\in\Gamma_{0}$.

Fix a distance ${\rm d}_{\operatorname{\mathbb{P}}}$ on
$\operatorname{\mathbb{P}}(\operatorname{\mathbb{R}}^{d})$ induced by a
Riemannian metric, fix $x_{1},x_{2}\in\Lambda_{\Omega}(\Gamma_{0})$ distinct
and let $\epsilon:=\frac{1}{6}{\rm
d}_{\operatorname{\mathbb{P}}}(x_{1},x_{2})$. Lemma 12.5 implies that there
exists a finite set $S\subset\Gamma_{0}$ such that

$\mathrm{diam}\left(\mathcal{O}_{r}(b_{0},\gamma(b_{0}))\right)\leq\epsilon$

for all $\gamma\in\Gamma_{0}-S$. Hence, for each $\gamma\in\Gamma_{0}-S$,
there is some $i\in\\{1,2\\}$ so that

$B_{i}\times\mathcal{O}_{r}(b_{0},\gamma(b_{0}))\subset\left\\{(x,y)\in\Lambda_{\Omega}(\Gamma_{0})^{2}:{\rm
d}_{\operatorname{\mathbb{P}}}(x,y)\geq\epsilon\right\\}=:K,$

where $B_{i}:=\\{y\in\Lambda_{\Omega}(\Gamma_{0}):{\rm
d}_{\operatorname{\mathbb{P}}}(y,x_{i})\leq\epsilon\\}$. From the definitions
of ${\rm d}m_{1}$ and ${\rm d}m_{2}$, and the fact that $m_{1}=cm_{2}$, we see
that if we set

$C_{0}:=c\max_{(x,y)\in
K}\frac{e^{\phi_{1}([\xi_{1}(x),\xi_{1}(y)]_{\theta_{1}})}}{e^{\phi_{2}([\xi_{2}(x),\xi_{2}(y)]_{\theta_{2}})}},$

then

$\frac{1}{C_{0}}(\nu_{\bar{\phi}_{2}}\otimes\nu_{\phi_{2}})(A)\leq(\nu_{\bar{\phi}_{1}}\otimes\nu_{\phi_{1}})(A)\leq
C_{0}(\nu_{\bar{\phi}_{2}}\otimes\nu_{\phi_{2}})(A)$

for all Borel measurable sets $A\subset K$. Hence, if we set

$C_{1}:=C_{0}\max\left\\{\frac{\nu_{\bar{\phi}_{2}}(B_{1})}{\nu_{\bar{\phi}_{1}}(B_{1})},\frac{\nu_{\bar{\phi}_{2}}(B_{2})}{\nu_{\bar{\phi}_{1}}(B_{2})}\right\\},$

then

$\displaystyle\frac{1}{C_{1}}\leq\frac{\nu_{\phi_{1}}(\mathcal{O}_{r}(b_{0},\gamma(b_{0})))}{\nu_{\phi_{2}}(\mathcal{O}_{r}(b_{0},\gamma(b_{0})))}\leq
C_{1}$ (13)

for all $\gamma\in\Gamma_{0}-S$.

Since $S$ is finite, (12) and (13) imply that

$\sup_{\gamma\in\Gamma_{0}}\left|\delta_{1}\phi_{1}(\kappa_{\theta_{1}}(\rho_{1}(\gamma)))-\delta_{2}\phi_{2}(\kappa_{\theta_{2}}(\rho_{2}(\gamma)))\right|<+\infty.$

To prove the last claim of the corollary, we use the following argument of
Dal’bo and Kim [20]. Consider the product representation
$\rho_{1}\times\rho_{2}:\Gamma\to\mathsf{G}_{1}\times\mathsf{G}_{2}$, let
$\Delta$ denote the set of simple roots of
$\mathsf{G}_{1}\times\mathsf{G}_{2}$, and let ${\mathfrak{a}}$ denote the
Cartan subspace of $\mathsf{G}_{1}\times\mathsf{G}_{2}$. Corollary 12.4(3)
implies that the $\Delta$-Benoist limit cone
$\mathcal{B}(\rho_{1}\times\rho_{2})$ lies in a hyperplane in
${\mathfrak{a}}$. A theorem of Benoist [2] then implies that the Zariski
closure $Z$ of $(\rho_{1}\times\rho_{2})(\Gamma)$ is properly contained in
$\mathsf{G}_{1}\times\mathsf{G}_{2}$.

Let $\pi_{j}:\mathsf{G}_{1}\times\mathsf{G}_{2}\to\mathsf{G}_{j}$ be the
projection map. Then the kernel $\pi_{3-j}|_{Z}$ is a normal subgroup of
$\mathsf{G}_{j}$, which is not all of $\mathsf{G}_{j}$. Since $\mathsf{G}_{j}$
is simple and $Z(\mathsf{G}_{j})$ is trivial, we conclude that
$\pi_{3-j}|_{Z}$ is injective. Since $\rho_{j}$ has Zariski dense image,
$\pi_{3-j}|_{Z}$ is also surjective. Hence,
$\Psi:=\pi_{2}|_{Z}\circ\pi_{1}|_{Z}^{-1}$ is an isomorphism such that
$\rho_{2}=\Psi\circ\rho_{1}$. ∎

##  13\. A Manhattan curve theorem

Sambarino [44] showed that when $\Gamma$ is Anosov, the entropy functional is
concave and characterizes when it is not strictly concave (see also Potrie-
Sambarino [39]). One may view this as an analogue of Burger’s Manhattan Curve
Theorem [11], since in this setting both are consequences of the convexity of
the pressure function and rigidity results for equilibrium measures. However,
in our setting we do not have access to thermodynamic formalism, so we must
adapt other methods.

######  Theorem 13.1.

Suppose $\theta\subset\Delta$ is symmetric, $\Gamma$ is a non-elementary
$\mathsf{P}_{\theta}$-transverse subgroup of $\mathsf{G}$ and
$\phi_{1},\phi_{2}\in{\mathfrak{a}}_{\theta}^{*}$ satisfy
$\delta^{\phi_{1}}(\Gamma)=\delta^{\phi_{2}}(\Gamma)=1$. If
$\phi=\lambda\phi_{1}+(1-\lambda)\phi_{2}$ for some $\lambda\in(0,1)$, then

$\delta:=\delta^{\phi}(\Gamma)\leq 1.$

Moreover, if $\delta^{\phi}(\Gamma)=1$ and $Q_{\Gamma}^{\phi}$ diverges at its
critical exponent, then $\ell^{\phi_{1}}(\gamma)=\ell^{\phi_{2}}(\gamma)$ for
all $\gamma\in\Gamma$.

As a consequence of Theorem 13.1, we use a result of Benoist [2] to show that
equality never occurs when $\Gamma$ is Zariski dense.

######  Corollary 13.2.

Suppose $\theta\subset\Delta$ is symmetric, $\Gamma$ is a Zariski dense
$\operatorname{\mathsf{P}}_{\theta}$-transverse subgroup of $\mathsf{G}$, and
$\phi_{1},\phi_{2}\in{\mathfrak{a}}^{*}_{\theta}$ are distinct and satisfy
$\delta^{\phi_{1}}(\Gamma)=\delta^{\phi_{2}}(\Gamma)=1$. If
$\phi=\lambda\phi_{1}+(1-\lambda)\phi_{2}$ for some $\lambda\in(0,1)$ and
$Q_{\Gamma}^{\phi}$ diverges at its critical exponent, then
$\delta^{\phi}(\Gamma)<1$.

Proof of Corollary 13.2. For $g\in\mathsf{G}$ define

$\nu(g):=\lim_{n\rightarrow\infty}\frac{1}{n}\kappa(g^{n})\in{\mathfrak{a}}^{+}\,\,\text{
and
}\,\,\nu_{\theta}(g):=\lim_{n\rightarrow\infty}\frac{1}{n}\kappa_{\theta}(g^{n})\in{\mathfrak{a}}_{\theta}^{+}$

(these limit exists by Fekete’s Subadditive Lemma). Note that via the
identification of ${\mathfrak{a}}_{\theta}^{*}$ as a subspace of
${\mathfrak{a}}^{*}$ described in Section 2, we have

$\ell^{\phi_{j}}(\gamma)=\phi_{j}(\nu_{\theta}(\gamma))=\phi_{j}(\nu(\gamma))$

for both $j=1,2$ and all $\gamma\in\Gamma$.

Suppose for a contradiction that $\delta^{\phi}(\Gamma)=1$. By Theorem 13.1,
$\phi_{1}(\nu(\gamma))=\phi_{2}(\nu(\gamma))$ for all $\gamma\in\Gamma$, which
implies that $\phi_{1}=\phi_{2}$ on

${\mathcal{C}}:=\overline{\cup_{\gamma\in\Gamma}\operatorname{\mathbb{R}}_{>0}\nu(\gamma)}$

Since $\Gamma$ is Zariski dense, a result of Benoist [2] implies that
${\mathcal{C}}$ is a convex subset of ${\mathfrak{a}}$ with non-empty
interior, so $\phi_{1}=\phi_{2}$, and we obtain a contradiction. ∎

###### Proof of Theorem 13.1.

The general strategy of our proof is inspired by the proof of Theorem 1(a) in
[11].

The first part follows immediately from the definition and Hölder’s inequality
which gives that, for all $s$,

$Q^{\phi}_{\Gamma}(s)\leq
Q^{\phi_{1}}_{\Gamma}(s)^{\lambda}Q^{\phi_{2}}_{\Gamma}(s)^{1-\lambda}.$

So our main work is to establish the “moreover” part of the theorem.

Suppose that $\delta^{\phi}(\Gamma)=1$ and $Q_{\Gamma}^{\phi}(1)=+\infty$. For
$\psi\in\left\\{\phi_{1},\phi_{2},\phi,\bar{\phi}_{1},\bar{\phi}_{2},\bar{\phi}\right\\}$,
let $\mu_{\psi}$ denote a $\psi$-Patterson-Sullivan measure for $\Gamma$ of
dimension $1$.

Using the discussion in Section 2.4 we may assume that $\mathsf{G}$ has
trivial center and that $\operatorname{\mathsf{P}}_{\theta}$ does not contain
any simple factors of $\mathsf{G}$. Then by Theorem 6.2, there is a properly
convex domain
$\Omega\subset\operatorname{\mathbb{P}}(\operatorname{\mathbb{R}}^{d})$, a
projectively visible subgroup
$\Gamma_{0}\subset\operatorname{\mathrm{Aut}}(\Omega)$ and a faithful
$\operatorname{\mathsf{P}}_{\theta}$-transverse representation
$\rho:\Gamma_{0}\to\mathsf{G}$ with limit map
$\xi:\Lambda_{\Omega}(\Gamma_{0})\to\operatorname{\mathcal{F}}_{\theta}$ so
that $\rho(\Gamma_{0})=\Gamma$ and
$\xi(\Lambda_{\Omega}(\Gamma_{0}))=\Lambda_{\theta}(\Gamma)$.

For
$\psi\in\left\\{\phi_{1},\phi_{2},\phi,\bar{\phi}_{1},\bar{\phi}_{2},\bar{\phi}\right\\}$,
define a measure $\nu_{\psi}$ on $\partial\Omega$ by

$\nu_{\psi}(A)=\mu_{\psi}\left(\xi(A\cap\Lambda_{\Omega}(\Gamma_{0}))\right).$

Fix $b_{0}\in\Omega$. Recall, from Section 12, that
$\Lambda_{\Omega,b_{0},R}^{\rm
con}(\Gamma_{0})\subset\Lambda_{\Omega}(\Gamma_{0})$ denotes the set of limit
points which are $R$-conical. By Corollary 12.3 we can fix $R>0$ sufficiently
large so that

$\nu_{\phi}\left(\Lambda_{\Omega,b_{0},R}^{\rm
con}(\Gamma_{0})\right)=1\quad\text{and}\quad\nu_{\bar{\phi}}\left(\Lambda_{\Omega,b_{0},R}^{\rm
con}(\Gamma_{0})\right)=1.$

Using the Shadow Lemma (Proposition 7.1) and possibly increasing $R$, we can
also assume that for every $r\geq R$ there exists a constant $C_{r}\geq 1$
such that

$\displaystyle
C_{r}^{-1}e^{-\psi(\kappa_{\theta}(\rho(\gamma)))}\leq\nu_{\psi}\big{(}\mathcal{O}_{r}(b_{0},\gamma(b_{0}))\big{)}\leq
C_{r}e^{-\psi(\kappa_{\theta}(\rho(\gamma)))}$ (14)

for all $\gamma\in\Gamma$ and
$\psi\in\left\\{\phi_{1},\phi_{2},\phi,\bar{\phi}_{1},\bar{\phi}_{2},\bar{\phi}\right\\}$.

For all $\alpha,\beta\in\Gamma_{0}$ and $r>0$, let

$\mathcal{R}_{r}(\alpha,\beta):=\mathcal{O}_{r}(b_{0},\alpha(b_{0}))\times\mathcal{O}_{r}(b_{0},\beta(b_{0})).$

The following lemma is the crucial place where we use the fact that
$\delta^{\phi}(\Gamma)=\delta^{\phi_{1}}(\Gamma)=\delta^{\phi_{2}}(\Gamma)$.

######  Lemma 13.3.

If $r\geq R$ and $\alpha,\beta\in\Gamma_{0}$, then

$\displaystyle(\nu_{\bar{\phi}}\otimes\nu_{\phi})$
$\displaystyle\big{(}\mathcal{R}_{r}(\alpha,\beta)\big{)}\leq
C_{r}^{4}\left(\nu_{\bar{\phi}_{1}}\otimes\nu_{\phi_{1}}+\nu_{\bar{\phi}_{2}}\otimes\nu_{\phi_{2}}\right)\big{(}\mathcal{R}_{r}(\alpha,\beta)\big{)}.$
(15)

###### Proof.

By repeated applications of the Shadow Lemma (14), we see that if
$\alpha,\beta\in\Gamma_{0}$, then

$\displaystyle(\nu_{\bar{\phi}}\otimes\nu_{\phi})$
$\displaystyle\big{(}\mathcal{R}_{r}(\alpha,\beta)\big{)}\leq
C_{r}^{2}e^{-\bar{\phi}(\kappa_{\theta}(\rho(\alpha)))}e^{-\phi(\kappa_{\theta}(\rho(\beta)))}$
$\displaystyle=C_{r}^{2}e^{-\big{(}\lambda\bar{\phi}_{1}(\kappa_{\theta}(\rho(\alpha)))+(1-\lambda)\bar{\phi}_{2}(\kappa_{\theta}(\rho(\alpha)))+\lambda\phi_{1}(\kappa_{\theta}(\rho(\beta)))+(1-\lambda)\phi_{2}(\kappa_{\theta}(\rho(\beta)))\big{)}}$
$\displaystyle\leq
C_{r}^{4}\nu_{\bar{\phi}_{1}}\left(\mathcal{O}_{r}(b_{0},\alpha(b_{0}))\right)^{\lambda}\nu_{\bar{\phi}_{2}}\left(\mathcal{O}_{r}(b_{0},\alpha(b_{0}))\right)^{1-\lambda}\nu_{\phi_{1}}\left(\mathcal{O}_{r}(b_{0},\beta(b_{0}))\right)^{\lambda}\nu_{\phi_{2}}\left(\mathcal{O}_{r}(b_{0},\beta(b_{0}))\right)^{1-\lambda}$
$\displaystyle=C_{r}^{4}(\nu_{\bar{\phi}_{1}}\otimes\nu_{\phi_{1}})\big{(}\mathcal{R}_{r}(\alpha,\beta)\big{)}^{\lambda}(\nu_{\bar{\phi}_{2}}\otimes\nu_{\phi_{2}})\big{(}\mathcal{R}_{r}(\alpha,\beta)\big{)}^{1-\lambda}.$

We may then apply the weighted Arithmetic Mean-Geometric Mean Inequality to
see that

$\displaystyle(\nu_{\bar{\phi}}\otimes\nu_{\phi})\big{(}\mathcal{R}_{r}(\alpha,\beta)\big{)}\leq
C_{r}^{4}\left(\nu_{\bar{\phi}_{1}}\otimes\nu_{\phi_{1}}+\nu_{\bar{\phi}_{2}}\otimes\nu_{\phi_{2}}\right)\big{(}\mathcal{R}_{r}(\alpha,\beta)\big{)}$

for all $\alpha,\beta\in\Gamma_{0}$. ∎

Our goal is to upgrade the inequality in Equation (15) to all Borel measurable
sets in $\Lambda_{\Omega}(\Gamma_{0})^{2}$. We first show that shadows form a
neighborhood basis of every point in $\Lambda_{\Omega,b_{0},R}(\Gamma_{0})$.

######  Lemma 13.4.

If $x\in\Lambda_{\Omega,b_{0},R}(\Gamma_{0})$ and $U$ is a neighborhood of $x$
in $\partial\Omega$, then there exists $\gamma\in\Gamma$ such that

$x\in\mathcal{O}_{R}(b_{0},\gamma(b_{0}))\subset U.$

###### Proof.

Fix a sequence $\\{\gamma_{n}\\}$ in $\Gamma$ such that
$\gamma_{n}(b_{0})\rightarrow x$ and
$\operatorname{dist}_{\Omega}(\gamma_{n}(b_{0}),[b_{0},x))<R$ for all $n\geq
1$. Then $x\in\mathcal{O}_{R}(b_{0},\gamma_{n}(b_{0}))$ for all $n\geq 1$ and
Lemma 12.5 implies that $\mathcal{O}_{R}(b_{0},\gamma_{n}(b_{0}))\subset U$
when $n$ is sufficiently large. ∎

Next, by the argument in [41, pg. 23], we observe that the shadows satisfy a
version of the Vitali covering lemma.

######  Lemma 13.5.

If $I\subset\Gamma_{0}$ and $r>0$, then there exists $J\subset I$ such that
the sets $\\{\mathcal{O}_{r}(b_{0},\gamma(b_{0})):\gamma\in J\\}$ are pairwise
disjoint and

$\bigcup_{\gamma\in
I}\mathcal{O}_{r}(b_{0},\gamma(b_{0}))\subset\bigcup_{\gamma\in
J}\mathcal{O}_{5r}(b_{0},\gamma(b_{0})).$

We now leverage our covering lemma to upgrade Equation (15) to all measurable
subsets of $\Lambda(\Gamma_{0})^{2}$.

######  Lemma 13.6.

There exists $C>0$ such that: if $A\subset\Lambda_{\Omega}(\Gamma_{0})^{2}$ is
a Borel measurable set, then

$(\nu_{\bar{\phi}}\otimes\nu_{\phi})(A)\leq
C\left(\nu_{\bar{\phi}_{1}}\otimes\nu_{\phi_{1}}+\nu_{\bar{\phi}_{2}}\otimes\nu_{\phi_{2}}\right)(A).$

###### Proof.

It suffices to prove the lemma in the case when $A=A_{1}\times A_{2}$ for some
$A_{1},A_{2}\subset\Lambda_{\Omega}(\Gamma_{0})$. Fix $\epsilon>0$. By the
outer regularity of the measures, for both $j=1,2$, there exists an open set
$U_{j}\supset A_{j}$ with

$\displaystyle\left(\nu_{\bar{\phi}_{1}}\otimes\nu_{\phi_{1}}+\nu_{\bar{\phi}_{2}}\otimes\nu_{\phi_{2}}\right)\big{(}U_{1}\times
U_{2}\big{)}\leq\left(\nu_{\bar{\phi}_{1}}\otimes\nu_{\phi_{1}}+\nu_{\bar{\phi}_{2}}\otimes\nu_{\phi_{2}}\right)\big{(}A_{1}\times
A_{2}\big{)}+\epsilon.$

If we let
$I_{j}:=\\{\alpha\in\Gamma_{0}:\mathcal{O}_{R}(b_{0},\alpha(b_{0}))\subset
U_{j}\\}$, then by Lemma 13.4

$(A_{1}\times A_{2})\cap\Lambda_{\Omega,b_{0},R}^{\rm
con}(\Gamma_{0})^{2}\subset\bigcup_{(\alpha,\beta)\in I_{1}\times
I_{2}}\mathcal{R}_{R}(\alpha,\beta)\subset U_{1}\times U_{2}.$

By Lemma 13.5, we can find a subset $J_{j}\subset I_{j}$ such that the sets
$\\{\mathcal{O}_{R}(b_{0},\alpha(b_{0})):\alpha\in J_{j}\\}$ are pairwise
disjoint and

$\bigcup_{\alpha\in
I_{j}}\mathcal{O}_{R}(b_{0},\alpha(b_{0}))\subset\bigcup_{\alpha\in
J_{j}}\mathcal{O}_{5R}(b_{0},\alpha(b_{0})).$

Since we chose $R>0$ such that $\nu_{\phi}\big{(}\Lambda_{\Omega,b_{0},R}^{\rm
con}(\Gamma_{0})\big{)}=\nu_{\bar{\phi}}\big{(}\Lambda_{\Omega,b_{0},R}^{\rm
con}(\Gamma_{0})\big{)}=1$, it follows that

$\displaystyle(\nu_{\bar{\phi}}\otimes\nu_{\phi})(A_{1}\times A_{2})$
$\displaystyle=(\nu_{\bar{\phi}}\otimes\nu_{\phi})\big{(}(A_{1}\times
A_{2})\cap\Lambda_{\Omega,b_{0},R}(\Gamma_{0})^{2}\big{)}$
$\displaystyle\leq\sum_{(\alpha,\beta)\in J_{1}\times
J_{2}}(\nu_{\bar{\phi}}\otimes\nu_{\phi})\left(\mathcal{R}_{5R}(\alpha,\beta)\right).$

Then by repeated applications of Equations (14) and (15),

$\displaystyle\sum_{(\alpha,\beta)\in J_{1}\times
J_{2}}(\nu_{\bar{\phi}}\otimes\nu_{\phi})\big{(}\mathcal{R}_{5R}(\alpha,\beta)\big{)}$
$\displaystyle\leq C_{R}^{4}C_{5R}^{4}\sum_{(\alpha,\beta)\in J_{1}\times
J_{2}}(\nu_{\bar{\phi}}\otimes\nu_{\phi})\big{(}\mathcal{R}_{R}(\alpha,\beta)\big{)}$
$\displaystyle\leq C_{R}^{8}C_{5R}^{4}\sum_{(\alpha,\beta)\in J_{1}\times
J_{2}}\left(\nu_{\bar{\phi}_{1}}\otimes\nu_{\phi_{1}}+\nu_{\bar{\phi}_{2}}\otimes\nu_{\phi_{2}}\right)\big{(}\mathcal{R}_{R}(\alpha,\beta)\big{)}$
$\displaystyle\leq
C_{R}^{8}C_{5R}^{4}\left(\nu_{\bar{\phi}_{1}}\otimes\nu_{\phi_{1}}+\nu_{\bar{\phi}_{2}}\otimes\nu_{\phi_{2}}\right)\big{(}U_{1}\times
U_{2}\big{)}$ $\displaystyle\leq
C_{R}^{8}C_{5R}^{4}\left(\nu_{\bar{\phi}_{1}}\otimes\nu_{\phi_{1}}+\nu_{\bar{\phi}_{2}}\otimes\nu_{\phi_{2}}\right)\big{(}A_{1}\times
A_{2}\big{)}+C_{R}^{6}C_{5R}^{2}\epsilon.$

Since $\epsilon>0$ was arbitrary, it follows that

$\left(\nu_{\bar{\phi}}\otimes\nu_{\phi}\right)(A_{1}\times A_{2})\leq
C_{R}^{6}C_{5R}^{2}\left(\nu_{\bar{\phi}_{1}}\otimes\nu_{\phi_{1}}+\nu_{\bar{\phi}_{2}}\otimes\nu_{\phi_{2}}\right)(A_{1}\times
A_{2}).\qed$

Lemma 13.6 implies that $\nu_{\bar{\phi}}\otimes\nu_{\phi}$ is absolutely
continuous with respect to
$\nu_{\bar{\phi}_{1}}\otimes\nu_{\phi_{1}}+\nu_{\bar{\phi}_{2}}\otimes\nu_{\phi_{2}}$.
Therefore, after possibly relabelling, we can assume that
$\nu_{\bar{\phi}}\otimes\nu_{\phi}$ is non-singular with respect to
$\nu_{\bar{\phi}_{1}}\otimes\nu_{\phi_{1}}$.

We claim that $Q^{\phi_{1}}_{\Gamma}(1)=+\infty$. Otherwise, Proposition 8.1
would imply that

$\nu_{\phi_{1}}(\Lambda^{\rm
con}_{\Omega}(\Gamma_{0}))=\nu_{\bar{\phi}_{1}}(\Lambda^{\rm
con}_{\Omega}(\Gamma_{0}))=0,$

which is impossible since

$\nu_{\phi}(\Lambda^{\rm
con}_{\Omega}(\Gamma_{0}))=\nu_{\bar{\phi}}(\Lambda^{\rm
con}_{\Omega}(\Gamma_{0}))=1$

by Proposition 9.1.

Since $\nu_{\bar{\phi}}\otimes\nu_{\phi}$ is non-singular with respect to
$\nu_{\bar{\phi}_{1}}\otimes\nu_{\phi_{1}}$, the associated Bowen-Margulis
measures are non-singular. Hence by Corollary 12.4 we have
$\ell^{\phi}(\gamma)=\ell^{\phi_{1}}(\gamma)$ for all $\gamma\in\Gamma$. Thus
$\ell^{\phi_{1}}(\gamma)=\ell^{\phi_{2}}(\gamma)$ for all $\gamma\in\Gamma$. ∎

Notice that the Hölder inequality similarly proves a statement which is of the
same form as Burger’s Manhattan Curve Theorem. However, we are not able to
give an analogous characterization of when equality occurs.

######  Theorem 13.7.

Suppose $\theta\subset\Delta$ is symmetric,
$\Gamma_{1},\Gamma_{2}\subset\mathsf{G}$ are
$\operatorname{\mathsf{P}}_{\theta}$-transverse subgroups and there exists an
isomorphism $\rho:\Gamma_{1}\to\Gamma_{2}$. If
$\phi\in{\mathfrak{a}}_{\theta}^{*}$ and
$\delta^{\phi}(\Gamma_{1})=\delta^{\phi}(\Gamma_{2})=1$, then for any
$\lambda\in(0,1)$ the weighted Poincaré series

$\sum_{\gamma\in\Gamma_{1}}e^{-s\big{(}\lambda\phi(\kappa_{\theta}(\gamma))+(1-\lambda)\phi(\kappa_{\theta}(\rho(\gamma)))\big{)}}$

has critical exponent $\delta\leq 1$.

##  Appendix A Proof of Proposition 2.3

In this section we prove Proposition 2.3 which we restate here.

######  Proposition A.1.

Suppose $F^{\pm}\in\operatorname{\mathcal{F}}_{\theta}^{\pm}$, $\\{g_{n}\\}$
is a sequence in $\mathsf{G}$ and $g_{n}=m_{n}e^{\kappa(g_{n})}\ell_{n}$ is a
$\mathsf{KAK}$-decomposition for each $n\geq 1$. The following are equivalent:

  1. (1)

$m_{n}\operatorname{\mathsf{P}}_{\theta}\rightarrow F^{+}$,
$\ell_{n}^{-1}\operatorname{\mathsf{P}}_{\theta}^{-}\rightarrow F^{-}$ and
$\alpha(\kappa(g_{n}))\rightarrow+\infty$ for every $\alpha\in\theta$,

  2. (2)

$g_{n}(F)\to F^{+}$ for all
$F\in\operatorname{\mathcal{F}}_{\theta}^{+}\setminus\mathcal{Z}_{F^{-}}$, and
this convergence is uniform on compact subsets of
$\operatorname{\mathcal{F}}_{\theta}^{+}\setminus\mathcal{Z}_{F^{-}}$.

  3. (3)

$g_{n}^{-1}(F)\to F^{-}$ for all
$F\in\operatorname{\mathcal{F}}_{\theta}^{-}\setminus\mathcal{Z}_{F^{+}}$, and
this convergence is uniform on compact subsets of
$\operatorname{\mathcal{F}}_{\theta}^{-}\setminus\mathcal{Z}_{F^{+}}$.

  4. (4)

There are open sets
$\mathcal{U}^{\pm}\subset\operatorname{\mathcal{F}}_{\theta}^{\pm}$ such that
$g_{n}(F)\to F^{+}$ for all $F\in\mathcal{U}^{+}$ and $g_{n}^{-1}(F)\to F^{-}$
for all $F\in\mathcal{U}^{-}$.

It is well-known that

$\exp:\mathfrak{u}_{\theta}^{-}\to\mathsf{U}_{\theta}^{-}:=\exp(\mathfrak{u}_{\theta}^{-})$

is a diffeomorphism. Furthermore, the _Langlands decomposition_ (see for
instance [47, Thm. 1.2.4.8]) of parabolic subgroups states that the map

$(u,\ell)\in\mathsf{\operatorname{\mathsf{U}}}_{\theta}^{-}\times
L_{\theta}\mapsto u\ell\in\operatorname{\mathsf{P}}_{\theta}^{-}$

is a diffeomorphism, where
$\mathsf{L}_{\theta}:=\operatorname{\mathsf{P}}_{\theta}\cap\operatorname{\mathsf{P}}_{\theta}^{-}$.
It follows that $\mathsf{U}_{\theta}^{-}$ acts simply transitively on
$\operatorname{\mathcal{F}}_{\theta}\setminus\mathcal{Z}_{\operatorname{\mathsf{P}}_{\theta}^{-}}$.
Thus, the map

$T:\mathfrak{u}_{\theta}^{-}\to\operatorname{\mathcal{F}}_{\theta}\setminus\mathcal{Z}_{\operatorname{\mathsf{P}}_{\theta}^{-}}$

given by $T(X)=e^{X}\mathsf{P}_{\theta}$ is a diffeomorphism.

Note that if $H\in\mathfrak{a}$ and $X\in\mathfrak{u}_{\theta}^{-}$, then

$\displaystyle
e^{H}T(X)=e^{H}e^{X}\mathsf{P}_{\theta}=e^{H}e^{X}e^{-H}\mathsf{P}_{\theta}=e^{{\rm
Ad}(e^{H})(X)}\mathsf{P}_{\theta}=T\left({\rm Ad}(e^{H})(X)\right).$ (16)

Furthermore, if we decompose

$X=\sum_{\alpha\in\Sigma_{\theta}^{+}}X_{-\alpha}\in\mathfrak{u}_{\theta}^{-},$

where $X_{-\alpha}\in\mathfrak{g}_{-\alpha}$ for all
$\alpha\in\Sigma_{\theta}^{+}$, then

$\displaystyle{\rm Ad}(e^{H})(X)=\sum_{\alpha\in\Sigma_{\theta}^{+}}{\rm
Ad}(e^{H})(X_{-\alpha})=\sum_{\alpha\in\Sigma_{\theta}^{+}}e^{-\alpha(H)}X_{-\alpha}.$
(17)

Together, Equations (16) and (17) imply the following observation.

######  Lemma A.2.

Let $\\{H_{n}\\}$ be a sequence in $\mathfrak{a}^{+}$. If
$\alpha(H_{n})\to+\infty$ for all $\alpha\in\theta$, then
$e^{H_{n}}F\to\mathsf{P}_{\theta}$ for all
$F\in\operatorname{\mathcal{F}}_{\theta}\setminus\mathcal{Z}_{\operatorname{\mathsf{P}}_{\theta}^{-}}$,
and this convergence is uniform on compact subsets of
$\operatorname{\mathcal{F}}_{\theta}\setminus\mathcal{Z}_{\operatorname{\mathsf{P}}_{\theta}^{-}}$.

Using Equations (16) and (17), we can also prove the following lemma.

######  Lemma A.3.

Let $g_{n}=m_{n}e^{\kappa(g_{n})}\ell_{n}$ be as in the statement of
Proposition A.1.

  1. (1)

If there is an open set
$\mathcal{U}\subset\operatorname{\mathcal{F}}_{\theta}^{+}$ such that
$g_{n}(F)\to F^{+}$ for all $F\in\mathcal{U}$, then
$m_{n}\operatorname{\mathsf{P}}_{\theta}\rightarrow F^{+}$ and
$\alpha(\kappa(g_{n}))\to+\infty$ for every $\alpha\in\theta$.

  2. (2)

If there is an open set
$\mathcal{U}\subset\operatorname{\mathcal{F}}_{\theta}^{-}$ such that
$g_{n}^{-1}(F)\to F^{-}$ for all $F\in\mathcal{U}$, then
$\ell_{n}^{-1}\operatorname{\mathsf{P}}_{\theta}^{-}\rightarrow F^{-}$ and
$\alpha(\kappa(g_{n}))\to+\infty$ for every $\alpha\in\theta$.

###### Proof.

By compactness, it suffices to consider the case where $m_{n}\to
m\in\mathsf{K}$ and $\ell_{n}\to\ell\in\mathsf{K}$.

(1): We first prove that $\alpha(\kappa(g_{n}))\to+\infty$ for all
$\alpha\in\theta$. If this is not the case, then by taking a subsequence, we
may assume that there is some $\alpha_{0}\in\theta$ such that
$\alpha_{0}(\kappa(g_{n}))\to c\in[0,\infty)$. Choose
$F,F^{\prime}\in\mathcal{U}$ such that
$\ell(F),\ell(F^{\prime})\in\operatorname{\mathcal{F}}_{\theta}\setminus\mathcal{Z}_{\mathsf{P}_{\theta}^{-}}$,
and if we decompose

$T^{-1}(\ell(F))=\sum_{\alpha\in\Sigma_{\theta}^{+}}X_{-\alpha}\quad\text{and}\quad
T^{-1}(\ell(F^{\prime}))=\sum_{\alpha\in\Sigma_{\theta}^{+}}X^{\prime}_{-\alpha},$

where $X_{-\alpha},X^{\prime}_{-\alpha}\in\mathfrak{g}_{-\alpha}$ for all
$\alpha\in\Sigma_{\theta}^{+}$, then $X_{-\alpha_{0}}\neq
X^{\prime}_{-\alpha_{0}}$. Then by (16) and (17),

$\lim_{n\to\infty}T^{-1}(e^{\kappa(g_{n})}\ell_{n}(F))=\lim_{n\to\infty}{\rm
Ad}(e^{\kappa(g_{n})})T^{-1}(\ell_{n}(F))=e^{-c}X_{-\alpha_{0}}+\lim_{n\to\infty}\sum_{\alpha\in\Sigma_{\theta}^{+}-\\{\alpha_{0}\\}}e^{-\alpha(\kappa(g_{n}))}X_{-\alpha}.$

Similarly,

$\lim_{n\to\infty}T^{-1}(e^{\kappa(g_{n})}\ell_{n}(F^{\prime}))=e^{-c}X^{\prime}_{-\alpha_{0}}+\lim_{n\to\infty}\sum_{\alpha\in\Sigma_{\theta}^{+}-\\{\alpha_{0}\\}}e^{-\alpha(\kappa(g_{n}))}X^{\prime}_{-\alpha},$

so
$\lim_{n\to\infty}me^{\kappa(g_{n})}\ell_{n}(F)\neq\lim_{n\to\infty}me^{\kappa(g_{n})}\ell_{n}(F^{\prime})$,
which implies that

$\lim_{n\to\infty}g_{n}(F)\neq\lim_{n\to\infty}g_{n}(F^{\prime}).$

This is a contradiction because $F,F^{\prime}\in\mathcal{U}$.

Next, we prove that $m_{n}\operatorname{\mathsf{P}}_{\theta}\rightarrow
F^{+}$, or equivalently, $m\mathsf{P}_{\theta}=F^{+}$. Let
$F\in\operatorname{\mathcal{F}}_{\theta}$ such that $F$ is transverse to
$F^{-}$ and $\ell(F)$ is transverse to $\mathsf{P}_{\theta}^{-}$. Then there
is some compact subset
$K\subset\operatorname{\mathcal{F}}_{\theta}\setminus\mathcal{Z}_{\mathsf{P}_{\theta}^{-}}$
such that $\ell_{n}(F)\in K$ for all sufficiently large $n$. Since
$\alpha(\kappa(g_{n}))\to+\infty$ for all $\alpha\in\theta$, Lemma A.2 implies
that

$e^{\kappa(g_{n})}\ell_{n}(F)\to\mathsf{P}_{\theta},$

which implies that

$g_{n}(F)=m_{n}e^{\kappa(g_{n})}\ell_{n}(F)\to m\mathsf{P}_{\theta}.$

It follows that $m\mathsf{P}_{\theta}=F^{+}$.

(2): As in Section 2, let $k_{0}\in\mathsf{N}_{\mathsf{K}}({\mathfrak{a}})$ be
a representative of the longest element $w_{0}\in W$. Observation 2.2 implies
that ${\rm Ad}(k_{0})(-\kappa(g))=\kappa(g^{-1})$ for all $g\in\mathsf{G}$,
and so

$g_{n}^{-1}=\left(\ell_{n}^{-1}k_{0}^{-1}\right)e^{\kappa(g_{n}^{-1})}\left(k_{0}m_{n}^{-1}\right)$

is a $\mathsf{KAK}$-decomposition of $g_{n}^{-1}$.

Further,
$\operatorname{\mathsf{P}}_{\iota^{*}(\theta)}=k_{0}\operatorname{\mathsf{P}}_{\theta}^{-}k_{0}^{-1}$,
see Equation (2), so we can define a $\mathsf{G}$-equivariant diffeomorphism

$\Phi_{\theta}:\operatorname{\mathcal{F}}_{\theta}^{-}\to\operatorname{\mathcal{F}}_{\iota^{*}(\theta)}$

by
$\Phi_{\theta}(g\operatorname{\mathsf{P}}_{\theta}^{-})=gk_{0}\operatorname{\mathsf{P}}_{\iota^{*}(\theta)}$.
Then $g_{n}^{-1}(F)\rightarrow\Phi_{\theta}(F^{-})$ for all
$F\in\Phi_{\theta}(\mathcal{U})$. So by part (1), we see that
$\ell_{n}^{-1}k_{0}^{-1}\operatorname{\mathsf{P}}_{\iota^{*}(\theta)}\rightarrow\Phi_{\theta}(F^{-})$
and $\alpha(\kappa(g_{n}^{-1}))\to+\infty$ for all
$\alpha\in\iota^{*}(\theta)$. Since
$\Phi_{\theta}(\ell_{n}^{-1}\operatorname{\mathsf{P}}_{\theta}^{-})=\ell_{n}^{-1}k_{0}^{-1}\operatorname{\mathsf{P}}_{\iota^{*}(\theta)}$
this implies that
$\ell_{n}^{-1}\operatorname{\mathsf{P}}_{\theta}^{-}\rightarrow F^{-}$.
Further, by Observation 2.2,

$\alpha(\kappa(g))=\iota^{*}(\alpha)(\kappa(g^{-1}))$

for all $g\in\mathsf{G}$ and all $\alpha\in\theta$. So we see that
$\alpha(\kappa(g_{n}))\to+\infty$ for all $\alpha\in\theta$. ∎

###### Proof of Proposition A.1.

It follows immediately from Lemma A.3 that (4) implies (1), and it is obvious
that (2) and (3) together imply (4). It thus suffices to show that (2) and (3)
are both individually equivalent to (1). By compactness, it suffices to
consider the case where $m_{n}\to m\in\mathsf{K}$ and
$\ell_{n}\to\ell\in\mathsf{K}$.

We first prove that (1) implies (2). Since $\alpha(\kappa(g_{n}))\to+\infty$
for all $\alpha\in\theta$, Lemma A.2 implies

$\lim_{n\rightarrow\infty}e^{\kappa(g_{n})}F=\operatorname{\mathsf{P}}_{\theta}$

for all
$F\in\operatorname{\mathcal{F}}_{\theta}\setminus\mathcal{Z}_{\operatorname{\mathsf{P}}_{\theta}^{-}}$,
and this convergence is uniform on compact subsets of
$\operatorname{\mathcal{F}}_{\theta}\setminus\mathcal{Z}_{\operatorname{\mathsf{P}}_{\theta}^{-}}$.
Since, $m\operatorname{\mathsf{P}}_{\theta}=F^{+}$ and
$\ell^{-1}\operatorname{\mathsf{P}}_{\theta}^{-}=F^{-}$, it follows that

$\lim_{n\rightarrow\infty}g_{n}(F)=F^{+}$

for all $F\in\operatorname{\mathcal{F}}_{\theta}\setminus\mathcal{Z}_{F^{-}}$,
and this convergence is uniform on compact subsets of
$\operatorname{\mathcal{F}}_{\theta}\setminus\mathcal{Z}_{F^{-}}$.

Next, we prove (2) implies (1). By Lemma A.3,
$m_{n}\operatorname{\mathsf{P}}_{\theta}\rightarrow F^{+}$ and
$\alpha(\kappa(g_{n}))\to+\infty$ for every $\alpha\in\theta$, so it suffices
to show that $\ell_{n}^{-1}\operatorname{\mathsf{P}}_{\theta}^{-}\rightarrow
F^{-}$, or equivalently, that $\ell
F^{-}=\operatorname{\mathsf{P}}_{\theta}^{-}$. If this were not the case, then
there exists some
$F\in\mathcal{Z}_{\mathsf{P}_{\theta}^{-}}\setminus\mathcal{Z}_{\ell F^{-}}$.
Then there is a compact set
$K\subset\operatorname{\mathcal{F}}_{\theta}\setminus\mathcal{Z}_{F^{-}}$ such
that $\ell_{n}^{-1}(F)\in K$ for all sufficiently large $n$. Then by
assumption,

$m\lim_{n\to\infty}e^{\kappa(g_{n})}F=\lim_{n\to\infty}g_{n}\ell_{n}^{-1}F=F^{+}=m\operatorname{\mathsf{P}}_{\theta},$

so $e^{\kappa(g_{n})}F\to\operatorname{\mathsf{P}}_{\theta}$. However,
$\\{e^{\kappa(g_{n})}\\}\subset\operatorname{\mathsf{P}}_{\theta}^{-}$, so
each $e^{\kappa(g_{n})}$ preserves the closed set
$\mathcal{Z}_{\mathsf{P}_{\theta}^{-}}$, which implies that

$\operatorname{\mathsf{P}}_{\theta}=\lim_{n\rightarrow\infty}e^{\kappa(g_{n})}F\in\mathcal{Z}_{\mathsf{P}_{\theta}^{-}}.$

Since $\operatorname{\mathsf{P}}_{\theta}$ and
$\operatorname{\mathsf{P}}_{\theta}^{-}$ are transverse, we have a
contradiction.

Finally, we prove that (1) and (3) are equivalent. Let $k_{0}\in
N_{\operatorname{\mathsf{K}}}(\mathfrak{a})$ be a representative of the
longest element $w_{0}\in W$, and let

$\Phi_{\theta}:\operatorname{\mathcal{F}}_{\theta}^{-}\to\operatorname{\mathcal{F}}_{\iota^{*}(\theta)}$

be the $\mathsf{G}$-equivariant homeomorphism given by
$\Phi_{\theta}(g\operatorname{\mathsf{P}}_{\theta}^{-})=gk_{0}\operatorname{\mathsf{P}}_{\iota^{*}(\theta)}$.
Observe that

$\Phi_{\theta}(\operatorname{\mathcal{F}}_{\theta}^{-}\setminus\mathcal{Z}_{F^{+}})=\operatorname{\mathcal{F}}_{\iota^{*}(\theta)}\setminus\mathcal{Z}_{\Phi_{\iota^{*}(\theta)}^{-1}(F^{+})},$

so (3) can be rewritten as:

  * (3’)

$g_{n}^{-1}(F)\to\Phi_{\theta}(F^{-})$ for all
$F\in\operatorname{\mathcal{F}}_{\iota^{*}(\theta)}\setminus\mathcal{Z}_{\Phi_{\iota^{*}(\theta)}^{-1}(F^{+})}$,
and this convergence is uniform on compact subsets of
$\operatorname{\mathcal{F}}_{\iota^{*}(\theta)}\setminus\mathcal{Z}_{\Phi_{\iota^{*}(\theta)}^{-1}(F^{+})}$.

By Observation 2.2,
$\alpha(\kappa(g_{n}))=\iota^{*}(\alpha)(\kappa(g_{n}^{-1}))$ for all
$n\in\operatorname{\mathbb{N}}$ and all $\alpha\in\Delta$. Thus, (1) can be
rewritten as:

  * (1’)

$m_{n}k_{0}^{-1}\operatorname{\mathsf{P}}_{\iota^{*}(\theta)}^{-}\rightarrow\Phi_{\iota^{*}(\theta)}^{-1}(F^{+})$,
$\ell_{n}^{-1}k_{0}^{-1}\operatorname{\mathsf{P}}_{\iota^{*}(\theta)}\rightarrow\Phi_{\theta}(F^{-})$
and $\alpha(\kappa(g_{n}^{-1}))\rightarrow+\infty$ for every
$\alpha\in\iota^{*}(\theta)$.

We also saw in the proof of Lemma A.3(2) that if
$g_{n}=m_{n}e^{\kappa(g_{n})}\ell_{n}$ is a $\mathsf{KAK}$-decomposition of
$g\in\mathsf{G}$, then

$g_{n}^{-1}=(\ell_{n}^{-1}k_{0}^{-1})e^{\kappa(g_{n}^{-1})}(k_{0}m_{n}^{-1})$

is a $\mathsf{KAK}$-decomposition of $g_{n}^{-1}$. Thus, the equivalence
between (1) and (2) implies the equivalence between (1’) and (3’). ∎

##  Appendix B Proofs of Theorem 6.2 and Proposition 6.3

In this appendix we prove Theorem 6.2 and Proposition 6.3.

When $\mathsf{G}=\mathsf{PSL}(d,\operatorname{\mathbb{K}})$, where
$\operatorname{\mathbb{K}}$ is either the real numbers
$\operatorname{\mathbb{R}}$ or the complex numbers
$\operatorname{\mathbb{C}}$, recall from the introduction that
$\Delta:=\\{\alpha_{1},\dots,\alpha_{d-1}\\}\subset\mathfrak{a}^{*}$ denotes
the standard system of simple restricted roots, i.e.

$\alpha_{j}({\rm diag}(a_{1},\ldots,a_{d}))=a_{j}-a_{j+1}$

for all ${\rm diag}(a_{1},\ldots,a_{d})\in\mathfrak{a}$. To simplify notation,
we replace subscripts of the form $\\{\alpha_{i_{1}},\dots,\alpha_{i_{k}}\\}$
with $i_{1},\dots,i_{k}$. For instance,

$\operatorname{\mathcal{F}}_{1,d-1}=\operatorname{\mathcal{F}}_{\\{\alpha_{1},\alpha_{d-1}\\}}\quad\text{and}\quad
U_{1,d-1}(g)=U_{\\{\alpha_{1},\alpha_{d-1}\\}}(g).$

As mentioned before, in the case when
$\mathsf{G}=\mathsf{PSL}(d,\operatorname{\mathbb{K}})$, Theorem 6.2 and
Proposition 6.3 were proven in [15]. We will use results from [23] to prove
the following proposition, which allows us to generalize these results in [15]
to general $\mathsf{G}$.

######  Proposition B.1.

For any symmetric $\theta\subset\Delta$ and
$\chi\in\sum_{\alpha\in\theta}\operatorname{\mathbb{N}}\cdot\omega_{\alpha}$
there exist $d\in\operatorname{\mathbb{N}}$, an irreducible linear
representation
$\Phi:\mathsf{G}\rightarrow\mathsf{SL}(d,\operatorname{\mathbb{R}})$ and a
$\Phi$-equivariant smooth embedding

$\xi:\operatorname{\mathcal{F}}_{\theta}\rightarrow\operatorname{\mathcal{F}}_{1,d-1}(\operatorname{\mathbb{R}}^{d})$

such that:

  1. (1)

$F_{1},F_{2}\in\operatorname{\mathcal{F}}_{\theta}$ are transverse if and only
if $\xi(F_{1})$ and $\xi(F_{2})$ are transverse.

  2. (2)

There exists $N\in\operatorname{\mathbb{N}}$ such that

$\log\sigma_{1}(\Phi(g))=N\chi(\kappa(g))$

for all $g\in\mathsf{G}$.

  3. (3)

$\alpha_{1}(\kappa\Phi(g)))=\min_{\alpha\in\theta}\alpha(\kappa(g))$ for all
$g\in\mathsf{G}$.

  4. (4)

If $\min_{\alpha\in\theta}\alpha(\kappa(g))>0$, then

$\xi(U_{\theta}(g))=U_{1,d-1}(\Phi(g)).$

  5. (5)

$\Gamma\subset\mathsf{G}$ is $\operatorname{\mathsf{P}}_{\theta}$-divergent
(respectively $\operatorname{\mathsf{P}}_{\theta}$-transverse) if and only if
$\Phi(\Gamma)$ is $\operatorname{\mathsf{P}}_{1,d-1}$-divergent (respectively
$\operatorname{\mathsf{P}}_{1,d-1}$-transverse). Moreover, in this case

$\xi(\Lambda_{\theta}(\Gamma))=\Lambda_{1,d-1}(\Phi(\Gamma)).$

  6. (6)

If $\rho:\Gamma_{0}\rightarrow\mathsf{G}$ is a
$\operatorname{\mathsf{P}}_{\theta}$-transverse representation with boundary
map
$\xi_{\rho}:\Lambda_{\Omega}(\Gamma_{0})\rightarrow\operatorname{\mathcal{F}}_{\theta}$,
then $\Phi\circ\rho$ is a $\operatorname{\mathsf{P}}_{1,d-1}$-transverse
representation with boundary map $\xi\circ\xi_{\rho}$.

Delaying the proof of Proposition B.1 for a moment, we prove Theorem 6.2 and
Proposition 6.3.

###  B.1. Proof of Theorem 6.2

Let $\Phi:\mathsf{G}\rightarrow\mathsf{PSL}(d,\operatorname{\mathbb{R}})$ and
$\xi_{\Phi}:\operatorname{\mathcal{F}}_{\theta}\rightarrow\operatorname{\mathcal{F}}_{1,d-1}(\operatorname{\mathbb{R}}^{d})$
satisfy Proposition B.1 for some
$\chi\in\sum_{\alpha\in\theta}\operatorname{\mathbb{N}}\cdot\omega_{\alpha}$.

Then $\Phi(\Gamma)$ is $\operatorname{\mathsf{P}}_{1,d-1}$-transverse and so
by [15, Thm. 4.2] there exist $d_{0}\in\operatorname{\mathbb{N}}$, a properly
convex domain
$\Omega\subset\operatorname{\mathbb{P}}(\operatorname{\mathbb{R}}^{d_{0}})$, a
projectively visible subgroup $\Gamma_{0}\subset\mathrm{Aut}(\Omega)$ and a
faithful $\operatorname{\mathsf{P}}_{1,d-1}$-transverse representation
$\rho_{0}:\Gamma_{0}\rightarrow\mathsf{PSL}(d,\operatorname{\mathbb{R}})$ with
limit map
$\xi_{0}:\Lambda_{\Omega}(\Gamma_{0})\to\mathcal{F}_{1,d-1}(\operatorname{\mathbb{R}}^{d})$
so that $\rho_{0}(\Gamma_{0})=\Phi(\Gamma)$ and

$\xi_{0}(\Lambda_{\Omega}(\Gamma_{0}))=\Lambda_{1,d-1}(\Phi(\Gamma))=\xi_{\Phi}(\Lambda_{\theta}(\Gamma)).$

We claim that $\Phi$ is injective. Since $\mathsf{G}$ is semisimple,
$\ker\Phi$ is either discrete or contains a simple factor of $\mathsf{G}$.
Since
$\xi:\operatorname{\mathcal{F}}_{\theta}\rightarrow\operatorname{\mathcal{F}}_{1,d-1}(\operatorname{\mathbb{R}}^{d})$
is a $\Phi$-equivariant embedding, $\ker\Phi$ must act trivially on
$\operatorname{\mathcal{F}}_{\theta}$. So
$\ker\Phi\subset\operatorname{\mathsf{P}}_{\theta}$. By assumption
$\operatorname{\mathsf{P}}_{\theta}$ contains no simple factors of
$\mathsf{G}$, so $\ker\Phi$ is discrete. However then, since $\ker\Phi$ is
also normal, we see that $\ker\Phi$ is contained in the center of $\mathsf{G}$
which by assumption is trivial. Hence $\Phi$ is injective.

Then $\rho:=\Phi^{-1}\circ\rho_{0}$ and $\xi:=\xi^{-1}_{\Phi}\circ\xi_{0}$ are
well defined and have the desired properties.

###  B.2. Proof of Proposition 6.3

We start by recalling a result in [15] about transverse representations into
$\mathsf{PSL}(d,\operatorname{\mathbb{K}})$. Let ${\rm
d}_{\operatorname{\mathbb{P}}(\operatorname{\mathbb{R}}^{d_{0}})}$ be a
distance on $\operatorname{\mathbb{P}}(\operatorname{\mathbb{R}}^{d_{0}})$
induced by a Riemannian metric. Given a properly convex domain
$\Omega\subset\operatorname{\mathbb{P}}(\operatorname{\mathbb{R}}^{d_{0}})$
and $b_{0}\in\Omega$ let

$\iota_{b_{0}}:\Omega\setminus\\{b_{0}\\}\to\partial\Omega$

denote the radial projection map obtained by letting
$\iota_{b_{0}}(z)\in\partial\Omega$ be the unique point so that
$z\in(b_{0},\iota_{b_{0}}(z))_{\Omega}$. The following lemma was proven as
Lemma 6.2 and Observation 6.3 in [15].

######  Lemma B.2.

Suppose $\theta\subset\\{\alpha_{1},\dots,\alpha_{d-1}\\}$ is symmetric. Let
$\rho:\Gamma_{0}\rightarrow\mathsf{PSL}(d,\operatorname{\mathbb{K}})$ be a
$\operatorname{\mathsf{P}}_{\theta}$-transverse representation, where
$\Gamma_{0}$ is a projectively visible subgroup of
$\operatorname{\mathrm{Aut}}(\Omega)$ for some properly convex domain
$\Omega\subset\operatorname{\mathbb{P}}(\operatorname{\mathbb{R}}^{d_{0}})$.
For any $b_{0}\in\Omega$ and $\epsilon>0$, there exist $C>0$ such that if
$\gamma,\eta\in\Gamma_{0}$ and

$\displaystyle{\rm
d}_{\operatorname{\mathbb{P}}(\operatorname{\mathbb{R}}^{d_{0}})}\left(\iota_{b_{0}}(\gamma^{-1}(b_{0})),\iota_{b_{0}}(\eta(b_{0}))\right)\geq\epsilon,$

then

$\displaystyle\left|\omega_{\alpha_{k}}\Big{(}\kappa(\rho(\gamma\eta))-\kappa(\rho(\gamma))-\kappa(\rho(\eta))\Big{)}\right|\leq
C$

for all $\alpha_{k}\in\theta$.

Lemma B.2 can be restated as follows.

######  Lemma B.3.

Suppose $\theta\subset\\{\alpha_{1},\dots,\alpha_{d-1}\\}$ is symmetric. Let
$\rho:\Gamma_{0}\rightarrow\mathsf{PSL}(d,\operatorname{\mathbb{K}})$ be a
$\operatorname{\mathsf{P}}_{\theta}$-transverse representation where
$\Gamma_{0}$ is a projectively visible subgroup of
$\operatorname{\mathrm{Aut}}(\Omega)$ for some properly convex domain
$\Omega\subset\operatorname{\mathbb{P}}(\operatorname{\mathbb{R}}^{d_{0}})$.
For any $b_{0}\in\Omega$ and $r>0$, there exist $C>0$ such that if
$\gamma,\eta\in\Gamma_{0}$ and

$\displaystyle{\rm
d}_{\Omega}\left(\gamma(b_{0}),[b_{0},\eta(b_{0})]_{\Omega}\right)\leq r,$

then

$\displaystyle\left|\omega_{\alpha_{k}}\Big{(}\kappa(\rho(\eta))-\kappa(\rho(\gamma))-\kappa(\rho(\gamma^{-1}\eta))\Big{)}\right|\leq
C$

for all $\alpha_{k}\in\theta$.

###### Proof.

Suppose not. Then there exist $\alpha_{k}\in\theta$ and sequences
$\\{\gamma_{n}\\}$, $\\{\eta_{n}\\}$ in $\Gamma$ such that

${\rm
d}_{\Omega}\left(\gamma_{n}(b_{0}),[b_{0},\eta_{n}(b_{0})]_{\Omega}\right)\leq
r\quad\text{but}\quad\left|\omega_{\alpha_{k}}\Big{(}\kappa(\rho(\eta_{n}))-\kappa(\rho(\gamma_{n}))-\kappa(\rho(\gamma_{n}^{-1}\eta_{n}))\Big{)}\right|\geq
n.$

Since

$\left\|\kappa(\rho(\eta_{n}))-\kappa(\rho(\gamma_{n}^{-1}\eta_{n}))\right\|\leq\sqrt{d}\max\left\\{\log\sigma_{1}(\rho(\gamma_{n}^{-1})),\log\sigma_{1}(\rho(\gamma_{n}))\right\\},$

$\\{\gamma_{n}\\}$ is a diverging sequence. A similar argument also shows that
$\\{\gamma_{n}^{-1}\eta_{n}\\}$ is diverging.

Since both $\\{\gamma_{n}\\}$ and $\\{\gamma_{n}^{-1}\eta_{n}\\}$ are
diverging and

${\rm
d}_{\Omega}\left(b_{0},[\gamma_{n}^{-1}(b_{0}),\gamma_{n}^{-1}\eta_{n}(b_{0})]_{\Omega}\right)={\rm
d}_{\Omega}\left(\gamma_{n}(b_{0}),[b_{0},\eta_{n}(b_{0})]_{\Omega}\right)\leq
r,$

it follows that there is some $\epsilon>0$ so that

${\rm
d}_{\operatorname{\mathbb{P}}(\operatorname{\mathbb{R}}^{d_{0}})}\left(\iota_{b_{0}}(\gamma_{n}^{-1}(b_{0})),\iota_{b_{0}}(\gamma_{n}^{-1}\eta_{n}(b_{0}))\right)\geq\epsilon$

for all $n$. Thus, Lemma B.2 implies that
$\left|\omega_{\alpha_{k}}\Big{(}\kappa(\rho(\eta_{n}))-\kappa(\rho(\gamma_{n}))-\kappa(\rho(\gamma_{n}^{-1}\eta_{n}))\Big{)}\right|$
has a uniform upper bound, which is a contradiction. ∎

###### Proof of Proposition 6.3.

Since
$\left\\{\omega_{\alpha}|_{{\mathfrak{a}}_{\theta}}\right\\}_{\alpha\in\theta}$
is a basis for ${\mathfrak{a}}_{\theta}^{*}$, it suffices to fix
$\beta\in\theta$ and find $C>0$ such that: if $\gamma,\eta\in\Gamma_{0}$ and

$\displaystyle{\rm
d}_{\Omega}\left(\gamma(b_{0}),[b_{0},\eta(b_{0})]_{\Omega}\right)\leq r,$

then

$\displaystyle\left|\omega_{\beta}\left(\kappa_{\theta}(\rho(\gamma\eta))-\kappa_{\theta}(\rho(\gamma))-\kappa_{\theta}(\rho(\eta))\right)\right|\leq
C.$

Let $\chi_{1}:=\sum_{\alpha\in\theta}\omega_{\alpha}$ and
$\chi_{2}:=\omega_{\beta}+\sum_{\alpha\in\theta}\omega_{\alpha}$. For $j=1,2$,
let
$\Phi_{j}:\mathsf{G}\rightarrow\mathsf{PSL}(d_{j},\operatorname{\mathbb{R}})$
satisfy Proposition B.1 for $\chi_{j}$, and let $\rho_{j}:=\Phi_{j}\circ\rho$.
Then $\rho_{j}$ is a $\operatorname{\mathsf{P}}_{1,d_{j}-1}$-transverse
representation and there exists $N_{j}\in\operatorname{\mathbb{N}}$ such that

$\displaystyle\left|\chi_{j}\left(\kappa_{\theta}(\rho(\gamma\eta))-\kappa_{\theta}(\rho(\gamma))-\kappa_{\theta}(\rho(\eta))\right)\right|=\frac{1}{N_{j}}\left|\omega_{\alpha_{1}}\left(\kappa(\rho_{j}(\gamma\eta))-\kappa(\rho_{j}(\gamma))-\kappa(\rho_{j}(\eta))\right)\right|$

for all $\gamma,\eta\in\Gamma$. Applying Lemma B.3 to $\rho_{j}$, there exists
$C_{j}>0$ such that: if $\gamma,\eta\in\Gamma_{0}$ and

$\displaystyle{\rm
d}_{\Omega}\left(\gamma(b_{0}),[b_{0},\eta(b_{0})]_{\Omega}\right)\leq r,$

then

$\displaystyle\left|\omega_{\alpha_{1}}\left(\kappa(\rho_{j}(\gamma\eta))-\kappa(\rho_{j}(\gamma))-\kappa(\rho_{j}(\eta))\right)\right|\leq
C_{j}.$

Since $\chi_{2}-\chi_{1}=\omega_{\beta}$, we then have: if
$\gamma,\eta\in\Gamma_{0}$ and

$\displaystyle{\rm
d}_{\Omega}\left(\gamma(b_{0}),[b_{0},\eta(b_{0})]_{\Omega}\right)\leq r,$

then

$\left|\omega_{\beta}\left(\kappa_{\theta}(\rho(\gamma\eta))-\kappa_{\theta}(\rho(\gamma))-\kappa_{\theta}(\rho(\eta))\right)\right|\leq\frac{C_{1}}{N_{1}}+\frac{C_{2}}{N_{2}}.\qed$

###  B.3. The proof of Proposition B.1

Fix a symmetric set $\theta\subset\Delta$ and
$\chi\in\sum_{\alpha\in\theta}\operatorname{\mathbb{N}}\omega_{\alpha}$. By
[23, Lem. 3.2, Prop. 3.3, Rem. 3.6 and Lem. 3.7] there exist
$N,d\in\operatorname{\mathbb{N}}$, an irreducible linear representation
$\Phi:\mathsf{G}\rightarrow\mathsf{SL}(d,\operatorname{\mathbb{R}})$ and a
$\Phi$-equivariant smooth embedding

$\xi:\operatorname{\mathcal{F}}_{\theta}\rightarrow\operatorname{\mathcal{F}}_{1,d-1}:=\operatorname{\mathcal{F}}_{1,d-1}(\operatorname{\mathbb{R}}^{d})$

such that:

  1. (a)

$\Phi$ is proximal and has highest weight $N\chi$, that is: if $H\in{\rm
int}({\mathfrak{a}}^{+})$, then $\Phi(e^{H})$ is proximal and the eigenvalue
with largest modulus is $e^{N\chi(H)}$.

  2. (b)

$\Phi(\operatorname{\mathsf{K}})\subset\mathsf{SO}(d,\operatorname{\mathbb{R}})$
and $\Phi(e^{{\mathfrak{a}}})$ is a subgroup of the diagonal matrices in
$\operatorname{\mathsf{SL}}(d,\operatorname{\mathbb{R}})$.

  3. (c)

$\alpha_{1}(\kappa(\Phi(g)))=\min_{\alpha\in\theta}\alpha(\kappa(g))$ for all
$g\in\mathsf{G}$.

  4. (d)

$F_{1},F_{2}\in\operatorname{\mathcal{F}}_{\theta}$ are transverse if and only
if $\xi(F_{1}),\xi(F_{2})\in\operatorname{\mathcal{F}}_{1,d-1}$ are
transverse.

In the statement of Proposition B.1, parts (1) and (3) are restatements of
properties (d) and (c) of $\Phi$ respectively, while part (2) is a consequence
of properties (a) and (b) of $\Phi$. Part (5) follows immediately from parts
(1), (3), and (4), while part (6) follows immediately from part (5) and
Proposition 2.5. Thus, it suffices to prove part (4).

Let $e_{1},\dots,e_{d}$ be the standard basis of
$\operatorname{\mathbb{R}}^{d}$. Using properties (a) and (b), we can
conjugate $\Phi$ by a permutation matrix and assume that

$\Phi(e^{H})e_{1}=e^{N\chi(H)}e_{1}\quad\text{and}\quad\Phi(e^{H})e_{d}=e^{N\bar{\chi}(H)}e_{d}$
(18)

when $H\in{\mathfrak{a}}$ (where as usual $\bar{\chi}=\chi\circ\iota$). We
first observe that the value of $\xi(\operatorname{\mathsf{P}}_{\theta})$ is
determined.

######  Lemma B.4.

$\xi(\operatorname{\mathsf{P}}_{\theta})=\big{(}\left\langle
e_{1}\right\rangle,\left\langle e_{1},\dots,e_{d-1}\right\rangle\big{)}$.

###### Proof.

Let $\hat{F}_{0}^{+}:=\big{(}\left\langle e_{1}\right\rangle,\left\langle
e_{1},\dots,e_{d-1}\right\rangle\big{)}$ and
$\hat{F}_{0}^{-}:=\big{(}\left\langle e_{d}\right\rangle,\left\langle
e_{2},\dots,e_{d}\right\rangle\big{)}$. Fix $H\in{\rm
int}({\mathfrak{a}}^{+})$. Then by property (b) and Equation (18),
$\Phi(e^{H})={\rm diag}(a_{1},\dots,a_{d})$ is a diagonal matrix with

$\left|a_{1}\right|>\max\\{\left|a_{j}\right|:2\leq j\leq
d\\}\quad\text{and}\quad\left|a_{d}\right|<\min\\{\left|a_{j}\right|:1\leq
j\leq d-1\\}.$

So

$\Phi(e^{nH})\hat{F}\rightarrow\hat{F}_{0}^{+}$

for all $\hat{F}\in\operatorname{\mathcal{F}}_{1,d-1}$ transverse to
$\hat{F}_{0}^{-}$. Since $\Phi$ is irreducible, there exists some
$F\in\operatorname{\mathcal{F}}_{\theta}$ such that $\xi(F)$ is transverse to
$\hat{F}^{-}_{0}$. Using Lemma A.2 and perturbing $F$ we may also assume that
$e^{nH}(F)\rightarrow\operatorname{\mathsf{P}}_{\theta}$. Then

$\xi(\operatorname{\mathsf{P}}_{\theta})=\lim_{n\rightarrow\infty}\xi(e^{nH}F)=\lim_{n\rightarrow\infty}\Phi(e^{nH})\xi(F)=\hat{F}_{0}^{+}.\qed$

Now we prove (4).

######  Lemma B.5.

If $\min_{\alpha\in\theta}\alpha(\kappa(g))>0$, then
$\xi(U_{\theta}(g))=U_{1,d-1}(\Phi(g)).$

###### Proof.

Fix a $\mathsf{KAK}$-decomposition $g=me^{H}\ell$. By properties (a) and (b),
there exists a permutation matrix $k\in\mathsf{O}(d)$ such that

$\Phi(g)=\big{(}\Phi(m)k^{-1}\big{)}\big{(}k\Phi(e^{H})k^{-1}\big{)}\big{(}k\Phi(\ell)\big{)}$

is a singular value decomposition of $\Phi(g)$. By Equation (18),
$k(e_{1})=e_{1}$ and $k(e_{d})=e_{d}$. Further, by property (c), we have

$\alpha_{j}(\Phi(g))>0\quad\text{for $j=1,d-1$},$

so by Lemma B.4,

$\displaystyle U_{1,d-1}(\Phi(g))$
$\displaystyle=\big{(}\Phi(m)k^{-1}\big{)}\big{(}\left\langle
e_{1}\right\rangle,\left\langle
e_{1},\dots,e_{d-1}\right\rangle\big{)}=\Phi(m)\big{(}\left\langle
e_{1}\right\rangle,\left\langle e_{1},\dots,e_{d-1}\right\rangle\big{)}$
$\displaystyle=\Phi(m)\xi(\operatorname{\mathsf{P}}_{\theta})=\xi(m\operatorname{\mathsf{P}}_{\theta})=\xi(U_{\theta}(g)).\qed$

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