$t\mapsto(\rho(t),\psi(t))$ be the curve representing
$\hat{\lambda}_{t}=(h_{1}(t),h_{2}(t))$, we deduce that $\rho(t)$ and
$\psi(t)$ are absolutely continuous and satisfy

$\rho(t)=\big{\|}\hat{\lambda}_{t}\big{\|}_{*},\qquad\dot{\psi}(t)=\frac{h_{1}(t)\dot{h}_{2}(t)-\dot{h}_{1}(t)h_{2}(t)}{\rho^{2}(t)}.$
(130)

We may compute explicitly $\dot{\rho}(t)$ and $\dot{\psi}(t)$, using once
again Lemma 2.9, the system (129) and identity (111):

$\dot{\rho}(t)=d_{\hat{\lambda}_{t}}{\|\\!\cdot\\!\|}_{*}\cdot(\dot{h}_{1}(t),\dot{h}_{2}(t))=\frac{\omega}{\|\hat{\lambda}_{t}\|_{*}}u(t)\cdot(-u_{2}(t),u_{1}(t))=0,\qquad\dot{\psi}(t)=\omega.$
(131)

Thus, integrating the above identities, we obtain $\rho(t)\equiv r$ and
$\psi(t)=\omega t+\phi$ for some $r>0$ and $\phi\in[0,2\mathbb{S}^{\circ})$.
Finally, we find an explicit expression for the maximal control:

$u(t)=\left(r\cos_{\Omega}(C_{\circ}(\phi+\omega
t)),r\sin_{\Omega}(C_{\circ}(\phi+\omega t))\right).$ (132)

From this, we may explicitly integrate the horizontal part of the Hamiltonian
system, obtaining the desired expression. In particular, if $\omega=0$ we
immediately obtain (123). If $\omega\neq 0$, we may employ Proposition 5.4 to
conclude. ∎

As $(\mathbb{H},\mathsf{d}_{SF})$ is complete, normal extremals can be
extended to $\mathbb{R}$, according to Proposition 3.9. Thus, we may define
the (extended) exponential map at the origin on the whole
$T_{0}^{*}\mathbb{H}\times\mathbb{R}$:

$\begin{split}G:\big{(}[0,2\mathbb{S}^{\circ})\times\mathbb{R}\times[0,\infty)\big{)}\times\mathbb{R}&\longrightarrow\mathbb{H},\\\
(\phi,\omega,r;t)&\longmapsto\big{(}x(\phi,\omega,r;t),y(\phi,\omega,r;t),z(\phi,\omega,r;t)\big{)},\end{split}$
(133)

where $(x(\phi,\omega,r;t),y(\phi,\omega,r;t),z(\phi,\omega,r;t))$ correspond
to the curve $(x(t),y(t),z(t))$ defined by Proposition 5.10 with initial datum
$(\phi,\omega,r)$ and with the understanding that $G(\phi,\omega,0;t)\equiv
0$. By the properties of the convex trigonometric functions, $G$ is a $C^{1}$
map for $\omega\neq 0$. Moreover, thanks to Theorem 5.8, for every initial
datum $(\phi,\omega,r)$, the curve $t\mapsto G(\phi,\omega,r;t)$ is a geodesic
between its endpoints for sufficiently small times. More precisely, it is
minimal for $|t|<t^{*}=t^{*}(\phi,\omega,r)$, where $t^{*}>0$ is the first
positive time such that $G(\phi,\omega,r;t^{*})$ lies on the $z$-axis. In
particular, a direct computation shows that

$t^{*}=\begin{cases}\displaystyle\frac{2\mathbb{S}^{\circ}}{|\omega|},&\text{if
}\omega\neq 0,\\\ \infty,&\text{if }\omega=0.\end{cases}$ (134)

We conclude this section by highlighting a property of geodesics in the
Heisenberg group that will be relevant in our analysis. For the sake of
notation, denote by $\Omega^{\circ}_{(\phi,\omega,r)}$ the following
transformation of
$\Omega^{\circ}=\overline{B}^{\left\lVert\cdot\right\rVert_{*}}_{1}(0)$:

$\Omega^{\circ}_{(\phi,\omega,r)}:=R_{-\pi/2}\left[\frac{r}{\omega}(\Omega^{\circ}-(\cos_{\Omega^{\circ}}(\phi),\sin_{\Omega^{\circ}}(\phi)))\right],$
(135)

where $R_{-\pi/2}$ is counter-clockwise rotation in the plane of angle
$-\pi/2$.

######  Proposition 5.11 ([Ber94, Thm. 1]).

Let $\mathbb{H}$ be the sub-Finsler Heisenberg group, equipped with a strictly
convex norm and let $\gamma:[0,1]\to\mathbb{H}$ be a geodesic starting at the
origin, with $\gamma(t)=(x(t),y(t),z(t))$. Then, the curve
$t\mapsto(x(t),y(t))$ is either a straight line or belongs to the boundary of
$\Omega^{\circ}_{(\phi,\omega,r)}$. Moreover, for every $t\in[0,1]$, $z(t)$
equals the oriented area that is swept by the vector joining $(0,0)$ with
$(x(s),y(s))$, for $s\in[0,t]$.

### 5.3 Failure of the $\mathsf{CD}(K,N)$ condition for $C^{1,1}$-norms

In this section we contradict the validity of the $\mathsf{CD}(K,N)$ condition
in the sub-Finsler Heisenberg group, equipped with a strictly convex and
$C^{1,1}$ norm and with a smooth measure. The strategy follows the blueprint
of the one presented in Section 4.4. The main issue we have to address here is
the low regularity (cf. Remark 5.24) of the midpoint and inverse geodesic maps
of (79) and (97). Nevertheless, using the explicit expression of geodesics
presented in Proposition 5.10, we successfully overcome these challenges
through a series of technical lemmas, culminating in Corollary 5.19,
Proposition 5.22 and Theorem 5.23.

Let $\mathbb{H}$ be the sub-Finsler Heisenberg group, equipped with a
$C^{1,1}$ and strictly convex norm ${\|\\!\cdot\\!\|}$. According to
Proposition 2.8, the dual norm ${\|\\!\cdot\\!\|}_{*}$ is $C^{1}$ and strongly
convex. Thus, in the notations of Section 5.1, the correspondences $C^{\circ}$
and $C_{\circ}$ are continuous functions. In order to ease the notation, in
this section we sometimes use the shorthands:

$\theta^{\circ}=C^{\circ}(\theta)\quad\text{and}\quad\psi_{\circ}=C_{\circ}(\psi),\qquad\forall\,\theta,\psi\in\mathbb{R}.$

Alexandrov’s theorem ensures that the dual norm ${\|\\!\cdot\\!\|}_{*}$ has a
second derivative and a second-order Taylor expansion almost everywhere, we
call $D_{*}\subset\mathbb{R}^{2}$ the set of twice differentiability of it.

######  Proposition 5.12.

Let $\psi\in[0,2\mathbb{S}^{\circ})$ be an angle such that $Q_{\psi}\in
D_{*}$, then the function $C_{\circ}$ is differentiable at $\psi$ with
positive derivative.

###### Proof.

Consider a vector $v\in\mathbb{R}^{2}$ orthogonal to
$d_{Q_{\psi}}{\|\\!\cdot\\!\|}_{*}$ such that $\left\lVert
v\right\rVert_{eu}=1$. Then, since $Q_{\psi}\in D_{*}$, there exists a
constant $C\in\mathbb{R}$ such that

$\left\lVert Q_{\psi}+sv\right\rVert_{*}=1+Cs^{2}+o(s^{2}),\qquad\text{as
}s\to 0.$ (136)

Observe that, since the norm ${\|\\!\cdot\\!\|}_{*}$ is strongly convex, the
constant $C$ is strictly positive. Consider the curve

$s\mapsto x(s):=\frac{Q_{\psi}+sv}{\left\lVert Q_{\psi}+sv\right\rVert_{*}},$

which by definition is a parametrization of an arc of the unit sphere
$S^{\left\lVert\cdot\right\rVert_{*}}_{1}(0)=\partial\Omega^{\circ}$. Call
$A(s)$ the signed area of the sector of $\Omega^{\circ}$ between the rays
$OQ_{\psi}$ and $Ox(s)$ (see Figure 4). As a consequence of (136), we deduce
that

$A(s)=\frac{1}{2}ks+o(s^{2}),\qquad\text{as }s\to 0,$

where $k$ is the scalar product between $Q_{\psi}$ and $v^{\perp}$, that is
the vector obtained by rotating $v$ with an angle of $-\frac{\pi}{2}$. In
fact, the first-order term $\frac{1}{2}ks$ is the area of the triangle of
vertices $O$, $Q_{\psi}$ and $Q_{\psi}+sv$, while the error term is controlled
by the area of the triangle of vertices $x(s)$, $Q_{\psi}$ and $Q_{\psi}+sv$.
The latter is an $o(s^{2})$ as $s\to 0$, thanks to (136). In particular,
letting $\psi(s)$ be the angle such that $x(s)=Q_{\psi(s)}$, by definition of
generalized angles, it holds that

$\psi(s)-\psi=2A(s)=ks+o(s^{2}),\qquad\text{as }s\to 0.$

Up to substituting the vector $v$ with $-v$, we can assume $k>0$. Then, in
order to conclude, it is enough to prove that the function $s\mapsto
C_{\circ}(\psi(s))$ is differentiable in $s=0$ with positive derivative.

$v$$S^{||\cdot||_{*}}_{1}(0)$$Q_{\psi}+sv$$x(s)$$Q_{\psi}$$v^{\perp}$${\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}{A(s)}}$$O$
Figure 3: Definition of $A(s)$.

$S^{||\cdot||}_{1}(0)$${a=\text{Hess}_{Q_{\psi}}(||\cdot||_{*})(v)}$$y(s)$$y(0)$${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{B(s)}}$$O$
Figure 4: Definition of $B(s)$.

First of all, by our choice of $k>0$, $s\mapsto C_{\circ}(\psi(s))$ is
monotone non-decreasing close to $s=0$, being a composition of monotone non-
decreasing functions. Second of all, we can show that it has a first-order
expansion. To this aim, note that the curve

$s\mapsto y(s):=d_{x(s)}{\|\\!\cdot\\!\|}_{*}$ (137)

is a parametrization of an arc of the sphere
$S^{\left\lVert\cdot\right\rVert}_{1}(0)=\partial\Omega$ (cf. Lemma 2.9).
Moreover, recalling that $Q_{\psi}\in D_{*}$ and using the homogeneity of the
norm, we have that

$y(s)=d_{Q_{\psi}+sv}{\|\\!\cdot\\!\|}_{*}=d_{Q_{\psi}}{\|\\!\cdot\\!\|}_{*}+a\,s+o(s),\qquad\text{as
}s\to 0,$ (138)

where $a:=\text{Hess}_{Q_{\psi}}({\|\\!\cdot\\!\|}_{*})(v)$. Observe that
$a\neq 0$ because ${\|\\!\cdot\\!\|}_{*}$ is strongly convex and $\left\lVert
Q_{\psi}\right\rVert_{*}=1$. Then, call $B(s)$ the (signed) area of the sector
of $\Omega$ between the rays $Oy(0)$ and $Oy(s)$ (see Figure 4). Reasoning as
we did for $A(s)$, from (138) we deduce that

$B(s)=\frac{1}{2}\langle y(0),a^{\perp}\rangle s+o(s^{2}),\qquad\text{as }s\to
0.$ (139)

On the other hand, by definition

$C_{\circ}(\psi(s))-C_{\circ}(\psi(0))=2B(s)=\langle y(0),a^{\perp}\rangle
s+o(s^{2}),\qquad\text{as }s\to 0.$ (140)

This shows that the function $s\mapsto C_{\circ}(\psi(s))$ is differentiable
in $s=0$ with derivative $\langle y(0),a^{\perp}\rangle$. In addition, since
$C_{\circ}\circ\psi$ is non-decreasing close to $s=0$, (140) also implies that
$\langle y(0),a^{\perp}\rangle\geq 0$. We are left to show that $\langle
y(0),a^{\perp}\rangle$ is strictly positive. If $\langle
y(0),a^{\perp}\rangle=0$ then $a$ is parallel to $y(0)$, however, according to
(138), the vector $a$ is tangent to the sphere
$S^{\left\lVert\cdot\right\rVert}_{1}(0)$ at $y(0)$ and therefore we obtain a
contradiction. ∎

######  Remark 5.13.

Since the norm is invariant by homotheties, then also $D_{*}$ is so, thus the
set of angles $\psi$ such that $Q_{\psi}\not\in D_{*}$ has null
$\mathscr{L}^{1}$-measure. In particular, the function $C_{\circ}$ is
differentiable with positive derivative $\mathscr{L}^{1}$-almost everywhere,
as a consequence of Proposition 5.12.

As already mentioned, the strategy to prove the main theorem of this section
is the same of Section 4.4. In particular, it is fundamental to prove
estimates on the volume contraction along geodesic homotheties. To this aim,
we consider the Jacobian determinant of the exponential map (133):

$J(\phi,\omega,r;t):=\left|\,\det\pNiceMatrix\frac{\partial x}{\partial
r}&\frac{\partial x}{\partial\phi}\frac{\partial x}{\partial\omega}\\\
\frac{\partial y}{\partial r}\frac{\partial y}{\partial\phi}\frac{\partial
y}{\partial\omega}\\\ \frac{\partial z}{\partial r}\frac{\partial
z}{\partial\phi}\frac{\partial z}{\partial\omega}(\phi,\omega,r;t)\right|$
(141)

where we recall $x(\phi,\omega,r;t),\,y(\phi,\omega,r;t),\,z(\phi,\omega,r;t)$
are defined in Proposition 5.10. In order to study this, we will use the
following formulation:

$J(\phi,\omega,r;t)=\left|\frac{\partial
z}{\partial\omega}(\phi,\omega,r;t)\det(M_{1})-\frac{\partial
z}{\partial\phi}(\phi,\omega,r;t)\det(M_{2})+\frac{\partial z}{\partial
r}(\phi,\omega,r;t)\det(M_{3})\right|$ (142)

where

$M_{1}:=\pNiceMatrix\frac{\partial x}{\partial r}&\frac{\partial
x}{\partial\phi}\\\ \frac{\partial y}{\partial r}\frac{\partial
y}{\partial\phi}(\phi,\omega,r;t),\quad M_{2}:=\pNiceMatrix\frac{\partial
x}{\partial r}&\frac{\partial x}{\partial\omega}\\\ \frac{\partial y}{\partial
r}\frac{\partial y}{\partial\omega}(\phi,\omega,r;t),\quad
M_{3}:=\pNiceMatrix\frac{\partial x}{\partial\phi}&\frac{\partial
x}{\partial\omega}\\\ \frac{\partial y}{\partial\phi}\frac{\partial
y}{\partial\omega}(\phi,\omega,r;t).$

We are particularly interested in studying the behaviour of
$J(\phi,\omega,r;t)$ as $t\to 0$. In the following lemmas we estimate the
behaviour of every term in (142) as $t\to 0$.

######  Notation 5.14.

Let $I\subset\mathbb{R}$ be an interval containing $0$. Given a function
$f:I\to\mathbb{R}$ and $n\in\mathbb{N}$, we write

$f(t)\sim t^{n},\qquad\text{as }t\to 0,$ (143)

if there exists a constant $C\neq 0$ such that $f(t)=Ct^{n}+o(t^{n})$, as
$t\to 0$.

######  Lemma 5.15.

Let $\phi\in[0,2\mathbb{S}^{\circ})$ be a differentiability point for the map
$C_{\circ}$, $r>0$ and $\omega\neq 0$, then

$\det\big{(}M_{1}(\phi,\omega,r;t)\big{)}\sim t^{2},\qquad\text{as }t\to 0,$
(144)

while

$\det\big{(}M_{2}(\phi,\omega,r;t)\big{)},\,\det\big{(}M_{3}(\phi,\omega,r;t)\big{)}=O(t^{3}),\qquad\text{as
}t\to 0.$ (145)

###### Proof.

Let us begin by proving (144). Firstly, since the function $C_{\circ}$ is
differentiable at $\phi$, we can compute the following Taylor expansions as
$t\to 0$, using Proposition 5.4:

$\begin{split}\cos_{\Omega}\big{(}(\phi+\omega
t)_{\circ}\big{)}&=\cos_{\Omega}\big{(}\phi_{\circ}\big{)}-t\omega
C^{\prime}_{\circ}(\phi)\sin_{\Omega^{\circ}}(\phi)+o(t),\\\
\sin_{\Omega}\big{(}(\phi+\omega
t)_{\circ}\big{)}&=\sin_{\Omega}\big{(}\phi_{\circ}\big{)}+t\omega
C^{\prime}_{\circ}(\phi)\cos_{\Omega^{\circ}}(\phi)+o(t).\end{split}$ (146)

Therefore, we may expand the entries of $M_{1}$ as $t\to 0$:

$\begin{split}\frac{\partial x}{\partial
r}(\phi,\omega,r;t)&=\frac{1}{\omega}\left(\sin_{\Omega^{\circ}}(\phi+\omega
t)-\sin_{\Omega^{\circ}}(\phi)\right)=\cos_{\Omega}(\phi_{\circ})t+o(t),\\\
\frac{\partial y}{\partial
r}(\phi,\omega,r;t)&=-\frac{1}{\omega}\left(\cos_{\Omega^{\circ}}(\phi+\omega
t)-\cos_{\Omega^{\circ}}(\phi)\right)=\sin_{\Omega}(\phi_{\circ})t+o(t),\\\
\frac{\partial
x}{\partial\phi}(\phi,\omega,r;t)&=\frac{r}{\omega}\left(\cos_{\Omega}\big{(}(\phi+\omega
t)_{\circ}\big{)}-\cos_{\Omega}\big{(}\phi_{\circ}\big{)}\right)=-trC^{\prime}_{\circ}(\phi)\sin_{\Omega^{\circ}}(\phi)+o(t),\\\
\frac{\partial
y}{\partial\phi}(\phi,\omega,r;t)&=\frac{r}{\omega}\left(\sin_{\Omega}\big{(}(\phi+\omega
t)_{\circ}\big{)}-\sin_{\Omega}\big{(}\phi_{\circ}\big{)}\right)=trC^{\prime}_{\circ}(\phi)\cos_{\Omega^{\circ}}(\phi)+o(t),\end{split}$
(147)

where we used once again Proposition 5.4. Finally, the determinant has the
following Taylor expansion as $t\to 0$:

$\begin{split}\det(M_{1})&=\frac{\partial x}{\partial r}\frac{\partial
y}{\partial\phi}-\frac{\partial y}{\partial r}\frac{\partial
x}{\partial\phi}=t^{2}rC^{\prime}_{\circ}(\phi)\big{(}\sin_{\Omega^{\circ}}(\phi)\sin_{\Omega}\big{(}\phi_{\circ}\big{)}+\cos_{\Omega^{\circ}}(\phi)\cos_{\Omega}\big{(}\phi_{\circ}\big{)}\big{)}+o(t^{2})\\\
&=t^{2}rC^{\prime}_{\circ}(\phi)+o(t^{2}),\end{split}$ (148)

where, in the last equality, we used Proposition 5.3. This proves (144),
keeping in mind Proposition 5.12, which guarantees that
$C_{\circ}^{\prime}(\phi)>0$.

Now we prove (145) for $\det(M_{2})$, the proof for $\det(M_{3})$ is
analogous. As a first step, reasoning as before, we can Taylor expand at
second-order the following quantities, as $t\to 0$:

$\begin{split}\cos_{\Omega^{\circ}}(\phi+\omega
t)&=\cos_{\Omega^{\circ}}(\phi)-\omega
t\sin_{\Omega}(\phi_{\circ})-\frac{1}{2}(\omega
t)^{2}C_{\circ}^{\prime}(\phi)\cos_{\Omega^{\circ}}(\phi)+o(t^{2}),\\\
\sin_{\Omega^{\circ}}(\phi+\omega t)&=\sin_{\Omega^{\circ}}(\phi)+\omega
t\cos_{\Omega}(\phi_{\circ})-\frac{1}{2}(\omega
t)^{2}C_{\circ}^{\prime}(\phi)\sin_{\Omega^{\circ}}(\phi)+o(t^{2}).\end{split}$
(149)

Hence, we deduce the expansion for the derivative of $x$ in the $\omega$
direction, as $t\to 0$:

$\begin{split}\frac{\partial
x}{\partial\omega}&(\phi,\omega,r;t)=-\frac{r}{\omega^{2}}\big{(}\sin_{\Omega^{\circ}}(\phi+\omega
t)-\sin_{\Omega^{\circ}}(\phi)\big{)}+\frac{rt}{\omega}\cos_{\Omega}\big{(}(\phi+\omega
t)_{\circ}\big{)}\\\
&=-\frac{r}{\omega}\big{(}t\cos_{\Omega}(\phi_{\circ})-\frac{1}{2}\omega
t^{2}C_{\circ}^{\prime}(\phi)\sin_{\Omega^{\circ}}(\phi)\big{)}+\frac{rt}{\omega}\big{(}\cos_{\Omega}\big{(}\phi_{\circ}\big{)}-t\omega
C^{\prime}_{\circ}(\phi)\sin_{\Omega^{\circ}}(\phi)\big{)}+o(t^{2})\\\
&=-\frac{1}{2}rt^{2}C_{\circ}^{\prime}(\phi)\sin_{\Omega^{\circ}}(\phi)+o(t^{2}).\end{split}$
(150)

An analogous computation shows that the derivative of $y$ in $\omega$ has the
ensuing expansion as $t\to 0$:

$\frac{\partial
y}{\partial\omega}(\phi,\omega,r;t)=\frac{1}{2}rt^{2}C_{\circ}^{\prime}(\phi)\cos_{\Omega^{\circ}}(\phi)+o(t^{2}).$
(151)

Note that, on the one hand, (150) and (151) imply that

$\frac{\partial
x}{\partial\omega}=O(t^{2})\qquad\text{and}\qquad\frac{\partial
y}{\partial\omega}=O(t^{2}),$ (152)

as $t\to 0$. On the other hand, by (147), we can deduce the following
behavior, as $t\to 0$:

$\frac{\partial x}{\partial r}=O(t)\qquad\text{and}\qquad\frac{\partial
y}{\partial r}=O(t).$ (153)

Thus, (152) and (153) prove the claimed behavior of $\det(M_{2})$ as $t\to 0$,
since

$\det(M_{2})=\frac{\partial x}{\partial r}\frac{\partial
y}{\partial\omega}-\frac{\partial y}{\partial r}\frac{\partial
x}{\partial\omega}.$ (154)

∎

In the next lemmas, we study the derivatives of $z$. These are the most
delicate to estimate, since the second-order Taylor polynomial of $z$ is zero
and higher-order derivatives may not exist.

######  Notation 5.16.

Let $g:\mathbb{R}\to\mathbb{R}$ be a function. We write

$g(t)=C(1+O(\varepsilon))f(t),\qquad\forall\,t\in[-\rho,\rho].$ (155)

if there exists a constant $K>0$ and a function $f:\mathbb{R}\to\mathbb{R}$
such that, for every $\varepsilon>0$, there exist positive constants
$C=C(\varepsilon),\rho=\rho(\varepsilon)>0$ for which the following holds

$C(1-K\varepsilon)f(t)<g(t)<C(1+K\varepsilon)f(t),\qquad\forall\,t\in[-\rho,\rho].$
(156)

######  Lemma 5.17.

Given $\varepsilon>0$ sufficiently small, for $\mathscr{L}^{1}$-almost every
$\phi\in[0,2\mathbb{S}^{\circ})$, every $r>0$ and $\omega\neq 0$, there exist
two positive constants $k=k(r)$ and $\rho=\rho(\phi,\omega,r)$ such that

$\frac{\partial
z}{\partial\omega}(\phi,\omega,r;t)=(1+O(\varepsilon))kt^{3},\qquad\qquad\forall\,t\in[-\rho,\rho].$
(157)

###### Proof.

First of all, we compute that

$\begin{split}\frac{\partial
z}{\partial\omega}(\phi,\omega,r;t)&=\frac{r^{2}t}{2\omega^{2}}\big{(}1-\sin_{\Omega^{\circ}}(\phi)\sin_{\Omega}\big{(}(\phi+\omega
t)_{\circ}\big{)}-\cos_{\Omega^{\circ}}(\phi)\cos_{\Omega}\big{(}(\phi+\omega
t)_{\circ}\big{)}\big{)}\\\ &\quad-\frac{r^{2}}{\omega^{3}}\big{(}\omega
t+\sin_{\Omega^{\circ}}(\phi)\cos_{\Omega^{\circ}}(\phi+\omega
t)-\cos_{\Omega^{\circ}}(\phi)\sin_{\Omega^{\circ}}(\phi+\omega
t)\big{)}.\end{split}$ (158)

In order to evaluate this quantity, fix an angle
$\psi\in[0,2\mathbb{S}^{\circ})$, for which Proposition 5.12 holds, and
consider the function $f_{\psi}$, defined as

$s\mapsto
f_{\psi}(s):=1-\sin_{\Omega^{\circ}}(\psi+s)\sin_{\Omega}(\psi_{\circ})-\cos_{\Omega^{\circ}}(\psi+s)\cos_{\Omega}(\psi_{\circ}).$
(159)

Notice that (111) ensures that $f_{\psi}(0)=0$, moreover direct computations
show that

$f^{\prime}_{\psi}(0)=0\qquad\text{and}\qquad
f^{\prime\prime}_{\psi}(0)=C^{\prime}_{\circ}(\psi)>0.$ (160)

Consequently, it holds that

$f_{\psi}(s)=\frac{1}{2}C^{\prime}_{\circ}(\psi)\cdot
s^{2}+o(s^{2}),\qquad\text{as }s\to 0.$ (161)

For every $n\in\mathbb{Z}$ and $m\in\mathbb{N}$ define the set of angles

$E_{n,m}:=\left\\{\psi\in[0,2\mathbb{S}^{\circ})\,:\,(1+\varepsilon)^{n-1}s^{2}<f_{\psi}(s)<(1+\varepsilon)^{n+1}s^{2},\text{
for every }s\in\Big{[}-\frac{1}{m},\frac{1}{m}\Big{]}\right\\}.$ (162)

Observe that, by (161), we have that
$E:=\bigcup_{n\in\mathbb{Z},m\in\mathbb{N}}E_{n,m}$ covers all
differentiability points of $C_{\circ}$, in particular $E$ has full
$\mathscr{L}^{1}$-measure, cf. Remark 5.13.

Now, fix $\omega>0$, $r>0$ and take $\phi\in[0,2\mathbb{S}^{\circ})$ to be a
density point444We say that $r\in\mathbb{R}$ is a _density point_ for a
measurable set $J\subset\mathbb{R}$ if  $\lim_{s\to
0^{+}}\frac{\mathscr{L}^{1}(J\cap[r-s,r+s])}{2s}=1.$ (163) for the set
$E_{n,m}$, for some $n\in\mathbb{Z}$, $m\in\mathbb{N}$. We are going to prove
the statement (157) for our choice of parameters and for positive times. The
cases with $\omega<0$ and negative times are completely analogous. Let
$0<\rho(\phi,\omega,r)<\frac{1}{2\omega m}$ be sufficiently small such that
for every $t\in(0,\rho]$

$\mathscr{L}^{1}(E_{n,m}\cap[\phi-2\omega t,\phi+2\omega t])>4\omega
t(1-\varepsilon/4).$ (164)

Introduce the set

$F_{n,m}:=\\{s\in\mathbb{R}\,:\,\phi+\omega s\in E_{n,m}\\}.$ (165)

Observe that, from (164), we can deduce that for every $t\in(0,\rho]$,

$\mathscr{L}^{1}(F_{n,m}\cap[-2t,2t])>4t(1-\varepsilon/4).$ (166)

Now, given every $t\in(0,\rho]$, (166) ensures that there exists
$\bar{s}\in[t(1-\varepsilon),t]$ such that $\bar{s}\in F_{n,m}$. Then, thanks
to Corollary 5.7, we obtain that

$\begin{split}1-\sin_{\Omega^{\circ}}(\phi)\sin_{\Omega}&\big{(}(\phi+\omega
t)_{\circ}\big{)}-\cos_{\Omega^{\circ}}(\phi)\cos_{\Omega}\big{(}(\phi+\omega
t)_{\circ}\big{)}\\\ &\geq
1-\sin_{\Omega^{\circ}}(\phi)\sin_{\Omega}\big{(}(\phi+\omega\bar{s})_{\circ}\big{)}-\cos_{\Omega^{\circ}}(\phi)\cos_{\Omega}\big{(}(\phi+\omega\bar{s})_{\circ}\big{)}\\\
&=f_{\phi+\omega\bar{s}}(-\omega\bar{s})\geq(1+\varepsilon)^{n-1}(\omega\bar{s})^{2}\geq(1-\varepsilon)^{2}(1+\varepsilon)^{n-1}(\omega
t)^{2},\end{split}$ (167)

where the second to last inequality holds by our choice of the parameter
$\rho$ and because $\phi+\omega\bar{s}\in E_{n,m}$. With an analogous
argument, we can find an element in $[t,t(1+\varepsilon)]\cap F_{n,m}$ and
deduce the estimate:

$1-\sin_{\Omega^{\circ}}(\phi)\sin_{\Omega}\big{(}(\phi+\omega
t)_{\circ}\big{)}-\cos_{\Omega^{\circ}}(\phi)\cos_{\Omega}\big{(}(\phi+\omega
t)_{\circ}\big{)}\leq(1+\varepsilon)^{2}(1+\varepsilon)^{n+1}(\omega t)^{2}.$
(168)

Combining (167) and (168), we conclude that, on $(0,\rho]$, the following
holds

$1-\sin_{\Omega^{\circ}}(\phi)\sin_{\Omega}\big{(}(\phi+\omega
t)_{\circ}\big{)}-\cos_{\Omega^{\circ}}(\phi)\cos_{\Omega}\big{(}(\phi+\omega
t)_{\circ}\big{)}=(1+O(\varepsilon))(1+\varepsilon)^{n}(\omega t)^{2},$ (169)

in the Notation 5.16. Consequently, we deduce:

$\frac{r^{2}t}{2\omega^{2}}\left(1-\sin_{\Omega^{\circ}}(\phi)\sin_{\Omega}\big{(}(\phi+\omega
t)_{\circ}\big{)}-\cos_{\Omega^{\circ}}(\phi)\cos_{\Omega}\big{(}(\phi+\omega
t)_{\circ}\big{)}\right)=(1+O(\varepsilon))(1+\varepsilon)^{n}\frac{r^{2}t^{3}}{2},$
(170)

for $t\in(0,\rho]$. To estimate the second term in (158), observe that

$\begin{split}\frac{\partial}{\partial s}\big{(}\omega
s+\sin_{\Omega^{\circ}}(\phi)\cos_{\Omega^{\circ}}&(\phi+\omega
s)-\cos_{\Omega^{\circ}}(\phi)\sin_{\Omega^{\circ}}(\phi+\omega s)\big{)}\\\
&=\omega\big{(}1-\sin_{\Omega^{\circ}}(\phi)\sin_{\Omega}\big{(}(\phi+\omega
s)_{\circ}\big{)}-\cos_{\Omega^{\circ}}(\phi)\cos_{\Omega}\big{(}(\phi+\omega
s)_{\circ}\big{)}\big{)}.\end{split}$ (171)

In particular, since in $s=0$ this quantity is equal to $0$, we have that for
$t\in(0,\rho]$

$\begin{split}\omega
t&+\sin_{\Omega^{\circ}}(\phi)\cos_{\Omega^{\circ}}(\phi+\omega
t)-\cos_{\Omega^{\circ}}(\phi)\sin_{\Omega^{\circ}}(\phi+\omega t)\\\
&=\omega\int_{0}^{t}\big{(}1-\sin_{\Omega^{\circ}}(\phi)\sin_{\Omega}\big{(}(\phi+\omega
s)_{\circ}\big{)}-\cos_{\Omega^{\circ}}(\phi)\cos_{\Omega}\big{(}(\phi+\omega
s)_{\circ}\big{)}\big{)}\,\mathrm{d}s\\\
&=\omega\int_{0}^{t}(1+O(\varepsilon))(1+\varepsilon)^{n}(\omega
t)^{2}\,\mathrm{d}s=(1+O(\varepsilon))(1+\varepsilon)^{n}\frac{(\omega
t)^{3}}{3},\end{split}$ (172)

where the second equality follows from (169). Then, we obtain that:

$\frac{r^{2}}{\omega^{3}}\big{(}\omega
t+\sin_{\Omega^{\circ}}(\phi)\cos_{\Omega^{\circ}}(\phi+\omega
t)-\cos_{\Omega^{\circ}}(\phi)\sin_{\Omega^{\circ}}(\phi+\omega
t)\big{)}=(1+O(\varepsilon))(1+\varepsilon)^{n}\frac{r^{2}t^{3}}{3}.$ (173)

Finally, putting together (170) and (173), we conclude that

$\frac{\partial
z}{\partial\omega}(\phi,\omega,r;t)=(1+O(\varepsilon))(1+\varepsilon)^{n}\frac{r^{2}t^{3}}{6},\qquad\forall\,t\in(0,\rho],$
(174)

that is (157) with $k=k(r):=(1+\varepsilon)^{n}\frac{r^{2}}{6}$. To conclude,
observe that we proved the statement for (every $r>0$, $\omega\neq 0$ and)
every $\phi\in[0,2\mathbb{S}^{\circ})$ which is a density point of some
$E_{n,m}$ and the set of such angles has full $\mathscr{L}^{1}$-measure in
$[0,2\mathbb{S}^{\circ})$. Indeed,
$E=\bigcup_{n\in\mathbb{Z},m\in\mathbb{N}}E_{n,m}$ has full
$\mathscr{L}^{1}$-measure in $[0,2\mathbb{S}^{\circ})$ and almost every point
of a measurable set is a density point. ∎

######  Lemma 5.18.

Let $\phi\in[0,2\mathbb{S}^{\circ})$ be a differentiability point for the map
$C_{\circ}$, $r>0$ and $\omega\neq 0$, then

$\frac{\partial z}{\partial r}(\phi,\omega,r;t),\,\frac{\partial
z}{\partial\phi}(\phi,\omega,r;t)=o(t^{2}),\qquad\text{as }t\to 0.$

###### Proof.

We start by proving the statement for $\frac{\partial z}{\partial r}$. We have
that

$\frac{\partial z}{\partial
r}(\phi,\omega,r;t)=\frac{r}{\omega^{2}}\left(\omega
t+\cos_{\Omega^{\circ}}(\phi+\omega
t)\sin_{\Omega^{\circ}}(\phi)-\sin_{\Omega^{\circ}}(\phi+\omega
t)\cos_{\Omega^{\circ}}(\phi)\right).$ (175)

Direct computations show that, on the one hand,

$\frac{\partial}{\partial t}\bigg{|}_{t=0}\big{(}\omega
t+\cos_{\Omega^{\circ}}(\phi+\omega
t)\sin_{\Omega^{\circ}}(\phi)-\sin_{\Omega^{\circ}}(\phi+\omega
t)\cos_{\Omega^{\circ}}(\phi)\big{)}\\\
=\omega-\omega\sin_{\Omega}(\phi_{\circ})\sin_{\Omega^{\circ}}(\phi)-\omega\cos_{\Omega}(\phi_{\circ})\cos_{\Omega^{\circ}}(\phi)=0,$
(176)

where we applied Proposition 5.3, and, on the other hand,

$\frac{\partial^{2}}{\partial t^{2}}\bigg{|}_{t=0}\big{(}\omega
t+\cos_{\Omega^{\circ}}(\phi+\omega
t)\sin_{\Omega^{\circ}}(\phi)-\sin_{\Omega^{\circ}}(\phi+\omega
t)\cos_{\Omega^{\circ}}(\phi)\big{)}\\\
=-\omega^{2}\cos_{\Omega^{\circ}}(\phi)\sin_{\Omega^{\circ}}(\phi)C^{\prime}_{\circ}(\phi)+\omega^{2}\sin_{\Omega^{\circ}}(\phi)\cos_{\Omega^{\circ}}(\phi)C^{\prime}_{\circ}(\phi)=0.$
(177)

Consequently, we conclude the proof of the first part of the statement:

$\frac{\partial z}{\partial r}(\phi,\omega,r;t)=\frac{r}{\omega^{2}}\cdot
o(t^{2})=o(t^{2}),\qquad\text{as }t\to 0.$

$R(Q_{\phi})$$R(\partial\Omega^{\circ})$$R(\Omega^{\circ})$ Figure 5: Image of
$Q_{\phi}$ and $\Omega^{\circ}$ through the rigid transformation $R$.

In order to prove the statement for $\frac{\partial z}{\partial\omega}$, we
use a geometric argument based on Proposition 5.11. First of all, recall that
$d_{Q_{\phi}}{\|\\!\cdot\\!\|}_{*}$ identifies a half-plane tangent at
$Q_{\phi}$ and containing $\Omega^{\circ}$. Thus, we can find a rigid
transformation $R:\mathbb{R}^{2}\to\mathbb{R}^{2}$, such that
$R(Q_{\phi})=(0,0)$ and $R(\Omega^{\circ})$ is contained in $\\{y\geq
0\\}\subset\mathbb{R}^{2}$, see Figure 5. Then, as ${\|\\!\cdot\\!\|}_{*}$ is
$C^{1,1}$, the image of the unit sphere
$R\big{(}\partial\Omega^{\circ}\big{)}$, can be described (locally around $O$)
as the graph of a non-negative function $f\in C^{1,1}(\mathbb{R})$ with
$f(0)=0$. In addition, by our choice of $\phi\in[0,2\mathbb{S}^{\circ})$, $f$
is twice differentiable in $0$ with strictly positive second derivative
$f^{\prime\prime}(0):=c>0$. Now consider the function $p$ defined in a
neighborhood of $\phi$ as

$p(\psi):=\mathtt{p}_{x}\big{(}R(Q_{\psi})\big{)},$

where $\mathtt{p}_{x}:\mathbb{R}^{2}\to\mathbb{R}$ denotes the projection on
the $x$-axis, i.e. $\mathtt{p}_{x}(a,b)=a$.

Second of all, for $s_{1},s_{2}\in\mathbb{R}$, call $F(s_{1},s_{2})$ the
signed area between the segment connecting $(s_{1},f(s_{1}))$ and
$(s_{2},f(s_{2}))$ and the graph of $f$ (intended positive if $s_{1}<s_{2}$
and negative if $s_{1}>s_{2}$), see Figure 7. Proposition 5.11 ensures that
for $\psi$ in a neighborhood of $\phi$ it holds that

$z(\psi,\omega,r;t)=\frac{r^{2}}{\omega^{2}}F\big{(}p(\psi),p(\psi+\omega
t)\big{)}.$

In particular, we obtain that

$\frac{\omega^{2}}{r^{2}}\,\frac{\partial
z}{\partial\phi}(\phi,\omega,r;t)=\frac{\partial}{\partial
s_{1}}F(0,p(\phi+\omega t))\cdot p^{\prime}(\phi)+\frac{\partial}{\partial
s_{2}}F(0,p(\phi+\omega t))\cdot p^{\prime}(\phi+\omega t).$ (178)

$s_{1}$$s_{2}$${\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}{F(s_{1},s_{2})}}$$f$
Figure 6: Definition of the function $F(s_{1},s_{2})$.

$q$${R(O)}$${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{A(q)}}$
Figure 7: Definition of the function $A(q)$.

We now proceed to compute the terms in the last formula, starting from the
ones involving $p^{\prime}$. To this aim, consider the point
$(x_{0},y_{0}):=R(O)$ and, for every $q$ in a neighborhood of $0$, call $A(q)$
the signed area inside $R\big{(}\partial\Omega^{\circ}\big{)}$ between the
segments $(x_{0},y_{0})O$ and $(x_{0},y_{0})(q,f(q))$. Observe that

$A^{\prime}(q)=\frac{1}{2}\langle(1,f^{\prime}(q)),(y_{0}-f(q),q-x_{0})\rangle=\frac{1}{2}y_{0}+O(q),\qquad\text{as
}q\to 0.$

Note that, in the last equality, we have used that $f(0)=f^{\prime}(0)=0$ and
$f\in C^{1,1}(\mathbb{R})$. Consequently, since $A(0)=0$, we have that

$A(q)=\frac{1}{2}y_{0}q+O(q^{2}),\qquad\text{as }q\to 0.$ (179)

On the other hand, by the definition of angle it holds that
$2A(p(\phi+\vartheta))=\vartheta$ for every $\vartheta$ sufficiently small and
therefore, invoking (179) and observing that $p\in C^{1}$, we obtain that

$p(\phi+\vartheta)=\frac{1}{y_{0}}\vartheta+o(\vartheta)\quad\text{and}\quad
p^{\prime}(\phi+\vartheta)=\frac{1}{y_{0}}+o(1),\qquad\text{as }\vartheta\to
0.$ (180)

Now we compute the partial derivatives of the function $F$. Observe that $F$
can be calculated in the following way

$F(s_{1},s_{2})=\frac{1}{2}(f(s_{1})+f(s_{2}))(s_{2}-s_{1})-\int_{s_{1}}^{s_{2}}f(x)\,\mathrm{d}x.$
(181)

As a consequence, we compute that

$\begin{split}\frac{\partial}{\partial
s_{1}}F(s_{1},s_{2})&=\frac{1}{2}f^{\prime}(s_{1})(s_{2}-s_{1})+\frac{1}{2}\big{(}f(s_{1})-f(s_{2})\big{)}\\\
\frac{\partial}{\partial
s_{2}}F(s_{1},s_{2})&=\frac{1}{2}f^{\prime}(s_{2})(s_{2}-s_{1})+\frac{1}{2}\big{(}f(s_{1})-f(s_{2})\big{)}.\end{split}$
(182)

Combining these two relations with (178), we conclude that

$\frac{\omega^{2}}{r^{2}}\frac{\partial
z}{\partial\phi}(\phi,\omega,r;t)=-\frac{1}{2}f(p(\phi+\omega t))\cdot
p^{\prime}(\phi),\\\ +\frac{1}{2}[f^{\prime}(p(\phi+\omega t))p(\phi+\omega
t)-f(p(\phi+\omega t))]\cdot p^{\prime}(\phi+\omega t).$ (183)

Now, recall that $f$ is twice differentiable in $0$ with positive second
derivative $c$, therefore we have that

$f(x)=\frac{1}{2}cx^{2}+o(x^{2})\qquad\text{and}\qquad f^{\prime}(x)=cx+o(x).$
(184)

Using these relations, together with (180), we can conclude that

$\begin{split}\frac{\omega^{2}}{r^{2}}\,\frac{\partial
z}{\partial\phi}(\phi,\omega,r;t)&=\frac{1}{2}f^{\prime}(p(\phi+\omega
t))p(\phi+\omega t)p^{\prime}(\phi+\omega t)-\frac{1}{2}f(p(\phi+\omega
t))[p^{\prime}(\phi)+p^{\prime}(\phi+\omega t)]\\\
&=\frac{1}{2}[cp(\phi+\omega t)+o(p(\phi+\omega t))]\,p(\phi+\omega
t)p^{\prime}(\phi+\omega t)\\\ &\quad-\frac{1}{4}[cp(\phi+\omega
t)^{2}+o(p(\phi+\omega t)^{2})][p^{\prime}(\phi)+p^{\prime}(\phi+\omega t)]\\\
&=\frac{c}{2y_{0}^{3}}(\omega t)^{2}-\frac{c}{2y_{0}^{3}}(\omega
t)^{2}+o(t^{2})=o(t^{2}).\end{split}$ (185)

This concludes the proof. ∎

As a consequence of the these lemmas, we obtain the following estimate of the
quantity $J(\phi,\omega,r;t)$, as $t\to 0$.

######  Corollary 5.19.

Given $\varepsilon>0$ sufficiently small, for $\mathscr{L}^{1}$-almost every
$\phi\in[0,2\mathbb{S}^{\circ})$, every $r>0$ and $\omega\neq 0$, there exist
two positive constants $C=C(\phi,\omega,r)$ and $\rho=\rho(\phi,\omega,r)$
such that

$J(\phi,\omega,r;t)=C(1+O(\varepsilon))|t|^{5},\qquad\forall\,t\in[-\rho,\rho],$

in the Notation 5.16.

###### Proof.

Let $\phi\in[0,2\mathbb{S}^{\circ})$ be a differentiability point for the map
$C_{\circ}$ and such that the conclusion of Lemma 5.17 holds, and fix $r>0$
and $\omega\neq 0$. Observe that, on the one hand, as a consequence of Lemma
5.15 and Lemma 5.18, we have that

$\left|\frac{\partial
z}{\partial\phi}\det(M_{2})\right|(\phi,\omega,r;t),\,\left|\frac{\partial
z}{\partial r}\det(M_{3})\right|(\phi,\omega,r;t)=o(t^{5}),\qquad\text{as
}t\to 0.$ (186)

On the other hand, Lemma 5.15 and Lemma 5.17 ensure that there exist positive
constants $C=C(\phi,\omega,r),\rho=\rho(\phi,\omega,r)>0$ such that

$\left|\frac{\partial
z}{\partial\omega}\det(M_{1})\right|(\phi,\omega,r;t)=C(1+O(\varepsilon))|t|^{5},\qquad\forall\,t\in[-\rho,\rho]$
(187)

where, in particular, $\rho$ has to be smaller than the constant identified by
Lemma 5.17. Up to taking a smaller $\rho$ and keeping in mind (142), we may
conclude that

$J(\phi,\omega,r;t)=C(1+O(\varepsilon))|t|^{5},\qquad\forall\,t\in[-\rho,\rho].$
(188)

∎

######  Remark 5.20.

Note that, in the sub-Riemannian Heisenberg group, the contraction rate of
volumes along geodesic is exactly $t^{5}$, cf. [ABR18]. In our setting, we are
able to highlight the same behavior for the Jacobian determinant of the
exponential map $J(\phi,\omega,r;t)$, as $t\to 0$.

Now that we know the behaviour of $J(\phi,\omega,r;t)$ as $t\to 0$, in the
next proposition, we obtain a statement similar to Proposition 4.25, which
will allow us to disprove the $\mathsf{CD}(K,N)$ condition in the Heisenberg
group. In particular, the proof of the following proposition uses Corollary
5.19 and some ideas developed in [Jui21, Prop. 3.1].

In our setting, we define the _midpoint map_ as:

$\mathcal{M}(p,q):=e_{\frac{1}{2}}\left(\gamma_{pq}\right),\qquad\text{if
}p\star q^{-1}\notin\\{x=y=0\\},$ (189)

where $\gamma_{pq}:[0,1]\to\mathbb{H}$ is the unique geodesic joining $p$ and
$q$, given by Theorem 5.8. Similarly, we define the _inverse geodesic map_
$I_{m}$ (with respect to $m\in\mathbb{H}$) as:

$I_{m}(q)=p,\qquad\text{if there exists }x\in\mathbb{H}\text{ such that
}\mathcal{M}(p,q)=m.$ (190)

######  Remark 5.21.

Recall the definition of midpoint map in (95) and inverse geodesic map in
(97). Both maps were defined using the differential structure of a smooth sub-
Finsler manifold, however they are characterized by the metric structure of
the space. In particular, if the norm is sufficently regular, they coincide
with (189) and (190).

######  Proposition 5.22.

Let $\mathbb{H}$ be the sub-Finsler Heisenberg group, equipped with a strictly
convex and $C^{1,1}$ norm. For $\mathscr{L}^{1}$-almost every
$\phi\in[0,2\mathbb{S}^{\circ})$, every $r>0$ and $\omega\neq 0$, there exists
a positive constant $\rho=\rho(\phi,\omega,r)$ such that for every
$t\in[-\rho,\rho]$:

  1. (i)

the inverse geodesic map $I_{\rm e}$ is well-defined and $C^{1}$ in a
neighborhood of $G(\phi,\omega,r;t)$;

  2. (ii)

the midpoint map $\mathcal{M}$ is well-defined and $C^{1}$ in a neighborhood
of $({\rm e},G(\phi,\omega,r;t))$, moreover

$\big{|}\det d_{G(\phi,\omega,r;t)}\mathcal{M}({\rm
e},\cdot)\big{|}\leq\frac{1}{2^{4}}.$ (191)

###### Proof.

Take $\varepsilon$ sufficiently small, let $\phi$ be an angle for which the
conclusion of Corollary 5.19 holds. Fix $r>0$ and $\omega\neq 0$, and let
$\rho=\rho(\phi,\omega,r)$ be the (positive) constant identified by Corollary
5.19. Let $t\in[-\rho,\rho]$ and consider the map $E_{t}:T^{*}_{\rm
e}\mathbb{H}\to\mathbb{H}$ defined as

$E_{t}(\phi,\omega,r):=G(\phi,\omega,r;t)=\big{(}x(\phi,\omega,r;t),y(\phi,\omega,r;t),z(\phi,\omega,r;t)\big{)},$
(192)

where $G$ is defined in (133). Note that $J(\phi,\omega,r;t)$ is the Jacobian
of $E_{t}(\phi,\omega,r)$ and in particular, since $t\in[-\rho,\rho]$,
Corollary 5.19 ensures that $J(\phi,\omega,r;t)>0$. Then, from the inverse
function theorem, we deduce that $E_{t}$ is locally invertible in a
neighborhood $B_{t}\subset\mathbb{H}$ of $E_{t}(\phi,\omega,r)$ with $C^{1}$
inverse $E_{t}^{-1}:B_{t}\to T^{*}_{\rm e}\mathbb{H}$. Then, according to
Theorem 5.8 and Proposition 5.10, the curve $[-t,t]\ni s\mapsto
G(\phi,\omega,r;s)$ is the unique geodesic connecting $G(\phi,\omega,r;-t)$
and $G(\phi,\omega,r;-t)$, and such that $G(\phi,\omega,r;0)={\rm e}$,
provided that $\rho$ is sufficiently small. Hence, we can write the map
$I_{\rm e}:B_{t}\to\mathbb{R}^{3}$ as

$I_{\rm e}(q)=E_{-t}(E_{t}^{-1}(q)),\qquad\forall\,q\in B_{t}.$ (193)

Therefore, the map $I_{\rm e}$ is $C^{1}$ on $B_{t}$, being a composition of
$C^{1}$ functions, proving item (i).

With an analogous argument, the midpoint map (with first entry ${\rm e}$),
$\mathcal{M}_{\rm e}(\cdot):=\mathcal{M}({\rm
e},\cdot):B_{t}\to\mathbb{R}^{3}$, can be written as

$\mathcal{M}_{\rm e}(q)=E_{t/2}(E_{t}^{-1}(q)),\qquad\forall\,q\in B_{t}.$
(194)

As before, we deduce this map is well-defined and $C^{1}$. To infer regularity
of the midpoint map in a neighborhood of $({\rm e},G(\phi,\omega,r;t))$, we
take advantage of the underline group structure, in particular of the left-
translations (120), which are isometries. Indeed, note that

$\mathcal{M}(p,q)=L_{p}\left(\mathcal{M}_{\rm
e}(L_{p^{-1}}(q))\right),\qquad\forall\,p,q\in\mathbb{H},$ (195)

and, for every $(p,q)$ in a suitable neighborhood of $({\rm
e},G(\phi,\omega,r;t))$, we have $L_{p^{-1}}(q)\in B_{t}$, therefore
$\mathcal{M}$ is well-defined and $C^{1}$. Finally, keeping in mind (194) and
applying Corollary 5.19, we deduce that

$\begin{split}\big{|}\det d_{G(\phi,\omega,r;t)}\mathcal{M}_{\rm
e}(\cdot)\big{|}&=\big{|}\det
d_{(\phi,\omega,r)}E_{t/2}\big{|}\cdot\big{|}\det
d_{(\phi,\omega,r)}E_{t}\big{|}^{-1}\\\ &=J(\phi,\omega,r;t/2)\cdot
J(\phi,\omega,r;t)^{-1}\\\
&=\frac{C(1+O(\varepsilon))|t/2|^{5}}{C(1+O(\varepsilon))|t|^{5}}=\frac{1}{2^{5}}\,(1+O(\varepsilon))\leq\frac{1}{2^{4}},\end{split}$
(196)

where the last inequality is true for $\varepsilon$ sufficiently small. This
concludes the proof of item (ii). ∎

######  Theorem 5.23.

Let $\mathbb{H}$ be the sub-Finsler Heisenberg group, equipped with a strictly
convex and $C^{1,1}$ norm and with a smooth measure $\mathfrak{m}$. Then, the
metric measure space $(\mathbb{H},\mathsf{d}_{SF},\mathfrak{m})$ does not
satisfy the Brunn–Minkowski inequality $\mathsf{BM}(K,N)$, for every
$K\in\mathbb{R}$ and $N\in(1,\infty)$.

###### Proof.

Take an angle $\phi$ for which the conclusion of Proposition 5.22 holds, fix
$r>0$, $\omega\neq 0$ and call $\gamma$ the curve

$\mathbb{R}\ni s\mapsto\gamma(s):=G(\phi,\omega,r;s).$ (197)

Fix $t\in(0,\rho]$, where $\rho=\rho(\phi,\omega,r)$ is the positive constant
identified by Proposition 5.22. Recall the map $E_{t}$ (see (192)) from the
proof of Proposition 5.22. $E_{t}$ is invertible, with $C^{1}$ inverse, in a
neighborhood $B_{t}\subset\mathbb{H}$ of $E_{t}(\phi,\omega,r)=\gamma(t)$.
Consider the function

$s\mapsto\Phi(s):=\mathtt{p}_{1}\big{[}E_{t}^{-1}\big{(}L_{\gamma(s)^{-1}}(\gamma(t+s))\big{)}\big{]},$
(198)

where $\mathtt{p}_{1}$ denotes the projection onto the first coordinate.
Observe that, for $s$ sufficiently small, $L_{\gamma(s)^{-1}}(\gamma(t+s))\in
B_{t}$, thus, $\Phi$ is well-defined and $C^{1}$ (being composition of $C^{1}$
functions) in an open interval $I\subset\mathbb{R}$ containing $0$. Moreover,
note that $\Phi(s)$ is the initial angle for the geodesic joining ${\rm e}$
and $\gamma(s)^{-1}\star\gamma(t+s)$. Now, we want to prove that there exists
an interval $\tilde{I}\subset I$ such that, for $\mathscr{L}^{1}$-almost every
$s\in\tilde{I}$, $\Phi(s)$ is an angle for which the conclusion of Proposition
5.22 holds. We have two cases, either $\Phi^{\prime}\equiv 0$ in $I$ or there
is $\bar{s}\in I$ such that $\Phi^{\prime}(\bar{s})\neq 0$. In the first case,
since by definition $\Phi(0)=\phi$, we deduce that $\Phi(s)\equiv\phi$, thus
the claim is true. In the second case, since $\Phi$ is $C^{1}$, we can find an
interval $\tilde{I}\subset I$ such that $\Phi^{\prime}(s)\neq 0$ for every
$s\in\tilde{I}$. Then, consider

$J:=\\{\psi\in\Phi(\tilde{I}):\psi\text{ is an angle for which Proposition
\ref{prop:dopotantafatica} holds}\\}\subset\Phi(\tilde{I})$ (199)

and observe that $J$ has full $\mathscr{L}^{1}$-measure in $\Phi(\tilde{I})$.
Therefore, the set $\tilde{J}:=\Phi^{-1}(J)\subset\tilde{I}$ has full
$\mathscr{L}^{1}$-measure in $\tilde{I}$, it being the image of $J$ through a
$C^{1}$ function with non-null derivative. Thus the claim is true also in this
second case.

At this point, let $\bar{s}\in\tilde{I}$ such that $\Phi(\bar{s})$ is an angle
for which the conclusion of Proposition 5.22 holds and consider

$\bar{\rho}:=\rho\big{(}E_{t}^{-1}\big{(}L_{\gamma(\bar{s})^{-1}}(\gamma(t+\bar{s}))\big{)}\big{)}>0.$
(200)

For every $s\in[-\bar{\rho},\bar{\rho}]\setminus\\{0\\}$, from Proposition
5.22, we deduce that the inverse geodesic map $I_{\rm e}$ and the midpoint map
$\mathcal{M}$ are well-defined and $C^{1}$ in a neighborhood of
$G\big{(}E_{t}^{-1}\big{(}L_{\gamma(\bar{s})^{-1}}(\gamma(t+\bar{s}))\big{)};s\big{)}$
and $\big{(}{\rm
e},G\big{(}E_{t}^{-1}\big{(}L_{\gamma(\bar{s})^{-1}}(\gamma(t+\bar{s}))\big{)};s\big{)}\big{)}$,
respectively. Moreover, we have that

$\big{|}\det
d_{G(E_{t}^{-1}(L_{\gamma(\bar{s})^{-1}}(\gamma(t+\bar{s})));s)}\mathcal{M}({\rm
e},\cdot)\big{|}\leq\frac{1}{2^{4}}.$ (201)

Observe that, since the left-translations are smooth isometries, the inverse
geodesic map $I_{\gamma(\bar{s})}$ is well-defined and $C^{1}$ in a
neighborhood of $\gamma(\bar{s}+s)$, in fact it can be written as

$I_{\gamma(\bar{s})}(p)=L_{\gamma(\bar{s})}\big{[}I_{\rm
e}\big{(}L_{\gamma(\bar{s})^{-1}}(p)\big{)}\big{]},$ (202)

and
$L_{\gamma(\bar{s})^{-1}}\big{(}\gamma(\bar{s}+s)\big{)}=G\big{(}E_{t}^{-1}\big{(}L_{\gamma(\bar{s})^{-1}}(\gamma(t+\bar{s}))\big{)};s\big{)}$.
Similarly, we can prove that the midpoint map is well-defined and $C^{1}$ in a
neighborhood of $(\gamma(\bar{s}),\gamma(\bar{s}+s))$, with

$\big{|}\det
d_{\gamma(\bar{s}+s)}\mathcal{M}(\gamma(\bar{s}),\cdot)\big{|}\leq\frac{1}{2^{4}}.$
(203)

In conclusion, up to restriction and reparametrization, we can find a geodesic
$\eta:[0,1]\to\mathbb{H}$ with the property that, for $\mathscr{L}^{1}$-almost
every $\bar{s}\in[0,1]$, there exists $\lambda(\bar{s})>0$ such that, for
every
$s\in[\bar{s}-\lambda(\bar{s}),\bar{s}+\lambda(\bar{s})]\cap[0,1]\setminus\\{\bar{s}\\}$,
the inverse geodesic map $I_{\eta(\bar{s})}$ and the midpoint map
$\mathcal{M}$ are well-defined and $C^{1}$ in a neighborhood of $\eta(s)$ and
$(\eta(\bar{s}),\eta(s))$ respectively, and in addition

$\big{|}\det
d_{\eta(s)}\mathcal{M}(\eta(\bar{s}),\cdot)\big{|}\leq\frac{1}{2^{4}}.$ (204)

Set $\lambda(s)=0$ on the (null) set where this property is not satisfied and
consider the set

$T:=\left\\{(s,t)\in[0,1]^{2}\,:\,t\in[s-\lambda(s),s+\lambda(s)]\right\\}.$
(205)

Observe that, introducing for every $\epsilon>0$ the set

$D_{\epsilon}:=\\{(s,t)\in[0,1]^{2}\,:\,|t-s|<\epsilon\\},$ (206)

we have that

$\frac{\mathscr{L}^{2}(T\cap
D_{\epsilon})}{\mathscr{L}^{2}(D_{\epsilon})}=\frac{\mathscr{L}^{2}(T\cap
D_{\epsilon})}{2\epsilon-\epsilon^{2}}\to 1,\qquad\text{as }\,\epsilon\to 0.$
(207)

On the other hand, we can find $\delta>0$ such that the set
$\Lambda_{\delta}:=\\{s\in[0,1]\,:\,\lambda(s)>\delta\\}$ satisfies
$\mathscr{L}^{1}(\Lambda_{\delta})>\frac{3}{4}$. In particular, for every
$\epsilon<\delta$ sufficiently small we have that

$\mathscr{L}^{2}\left(\left\\{(s,t)\in[0,1]^{2}\,:\,\frac{s+t}{2}\not\in\Lambda_{\delta}\right\\}\cap
D_{\epsilon}\right)<\frac{1}{2}\epsilon.$ (208)

Therefore, putting together (207) and (208), we can find $\epsilon<\delta$
sufficiently small such that

$\mathscr{L}^{2}\left(T\cap
D_{\epsilon}\cap\left\\{(s,t)\in[0,1]^{2}\,:\,\frac{s+t}{2}\in\Lambda_{\delta}\right\\}\right)>\frac{1}{2}\mathscr{L}^{2}(D_{\epsilon}).$
(209)

Then, since the set $D_{\epsilon}$ is symmetric with respect to the diagonal
$\\{s=t\\}$, we can find $\bar{s}\neq\bar{t}$ such that

$(\bar{s},\bar{t}),(\bar{t},\bar{s})\in T\cap
D_{\epsilon}\cap\left\\{(s,t)\in[0,1]^{2}\,:\,\frac{s+t}{2}\in\Lambda_{\delta}\right\\}.$
(210)

In particular, this tells us that:

  * (i)

$\bar{t}\in[\bar{s}-\lambda(\bar{s}),\bar{s}+\lambda(\bar{s})]$ and
$\bar{s}\in[\bar{t}-\lambda(\bar{t}),\bar{t}+\lambda(\bar{t})]$;

  * (ii)

$|\bar{t}-\bar{s}|<\epsilon<\delta$;

  * (iii)

$\frac{\bar{s}+\bar{t}}{2}\in\Lambda_{\delta}$.

Now, on the one hand, (i) ensures that the midpoint map $\mathcal{M}$ is well-
defined and $C^{1}$ in a neighborhood of $(\eta(\bar{s}),\eta(\bar{t}))$ with

$\big{|}\det
d_{\eta(\bar{t})}\mathcal{M}(\eta(\bar{s}),\cdot)\big{|}\leq\frac{1}{2^{4}}\qquad\text{and}\qquad\big{|}\det
d_{\eta(\bar{s})}\mathcal{M}(\cdot,\eta(\bar{t}))\big{|}\leq\frac{1}{2^{4}}.$
(211)

While, on the other hand, the combination of (ii) and (iii) guarantees that
the inverse geodesic map $I_{\eta(\frac{\bar{s}+\bar{t}}{2})}$ is well-defined
and $C^{1}$ in a neighborhood of $\eta(\bar{s})$ and in a neighborhood of
$\eta(\bar{t})$ respectively. Indeed, we have:

$\bar{s},\bar{t}\in\left[\frac{\bar{s}+\bar{t}}{2}-\delta,\frac{\bar{s}+\bar{t}}{2}+\delta\right]\subset\left[\frac{\bar{s}+\bar{t}}{2}-\lambda\left(\frac{\bar{s}+\bar{t}}{2}\right),\frac{\bar{s}+\bar{t}}{2}+\lambda\left(\frac{\bar{s}+\bar{t}}{2}\right)\right],$
(212)

and, by the very definition of $\lambda(\cdot)$, we obtain the claimed
regularity of the inverse geodesic map.

Once we have these properties, we can repeat the same strategy used in the
second part of the proof of Theorem 4.26 and contradict the Brunn–Minkowski
inequality $\mathsf{BM}(K,N)$ for every $K\in\mathbb{R}$ and every
$N\in(1,\infty)$. ∎

######  Remark 5.24.

If we want to replicate the strategy of Theorem 4.26, we ought to find a short
geodesic $\gamma:[0,1]\to\mathbb{H}$ such that

  * (i)

the midpoint map $\mathcal{M}$ is $C^{1}$ around $(\gamma(0),\gamma(1))$ and
satisfies a Jacobian estimates at $\gamma(1)$ of the type (191);

  * (ii)

the midpoint map $\mathcal{M}$ satisfies a Jacobian estimates at $\gamma(0)$
of the type (191);

  * (iii)

the inverse geodesic map $\mathcal{I}_{\gamma(1/2)}$, with respect to
$\gamma(1/2)$, is $C^{1}$ around $\gamma(0)$ and $\gamma(1)$.

Proposition 5.22 guarantees the existence of a large set $\mathscr{A}\subset
T_{\gamma(0)}^{*}\mathbb{H}$ of initial covectors for which the corresponding
geodesic $\gamma$ satisfies (i). The problem arises as the set $\mathscr{A}$
of “good” covectors depends on the base point and is large only in a measure-
theoretic sense. A simple “shortening” argument, mimicking the strategy of the
smooth case, is sufficient to address (ii). However, once the geodesic is
fixed, we have no way of ensuring that (iii) is satisfied. In particular, it
may happen that the map $\mathcal{I}_{\gamma(1/2)}$ does not fit within the
framework of Proposition 5.22 item (i), as the corresponding initial covector
may fall outside the hypothesis. To overcome such a difficulty, we use a
density-type argument to choose _simultaneously_ an initial point and an
initial covector in such a way that (i)–(iii) are satisfied.

### 5.4 Failure of the $\mathsf{MCP}(K,N)$ condition for singular norms

In this section we prove Theorem 1.7, showing that the measure contraction
property (see Definition 2.5) can not hold in a sub-Finsler Heisenberg group,
equipped with a strictly convex, singular norm. Our strategy is based on the
observation that, in this setting, geodesics exhibit a branching behavior,
despite being unique (at least for small times).

######  Theorem 5.25.

Let $\mathbb{H}$ be the sub-Finsler Heisenberg group, equipped with a strictly
convex norm ${\|\\!\cdot\\!\|}$ which is not $C^{1}$, and let $\mathfrak{m}$
be a smooth measure on $\mathbb{H}$. Then, the metric measure space
$(\mathbb{H},\mathsf{d}_{SF},\mathfrak{m})$ does not satisfy the measure
contraction property $\mathsf{MCP}(K,N)$ for every $K\in\mathbb{R}$ and
$N\in(1,\infty)$.

###### Proof.

For simplicity, we assume $\mathfrak{m}=\mathscr{L}^{3}$. As it is apparent
from the proof, the same argument can be carried out in the general case.

According to Proposition 2.8, since ${\|\\!\cdot\\!\|}$ is not $C^{1}$, its
dual norm ${\|\\!\cdot\\!\|}_{*}$ is not strictly convex. In particular, there
exists a straight segment contained in the sphere
$S^{\left\lVert\cdot\right\rVert_{*}}_{1}(0)=\partial\Omega^{\circ}$. Since
the differential structure of the Heisenberg group is invariant under
rotations around the $z$-axis, we can assume without losing generality that
this segment is vertical in $\mathbb{R}^{2}\cap\\{x>0\\}$, i.e. there exists
$\bar{x}\in\mathbb{R}$ and an interval $I:=[y_{0},y_{1}]\subset\mathbb{R}$
such that

$\\{\bar{x}\\}\times I\subset\partial\Omega^{\circ}.$

Moreover, we can take the interval $I$ to be maximal, namely for every
$y\not\in I$ we have $(\bar{x},y)\not\in\Omega^{\circ}$ (see Figure 9). Let
$\psi_{0}\in[0,2\mathbb{S}^{\circ})$ be such that
$Q_{\psi_{0}}=(\bar{x},y_{0})$, then it holds that

$(\bar{x},y)=Q_{\psi_{0}+(y-y_{0})\bar{x}},\qquad\text{for every }y\in I.$
(213)

As a consequence, we have that

$\cos_{\Omega^{\circ}}(\psi_{0}+(y-y_{0})\bar{x})=\bar{x}\quad\text{and}\quad\sin_{\Omega^{\circ}}(\psi_{0}+(y-y_{0})\bar{x})=y,\qquad\text{for
}y\in I.$ (214)

Let $y_{2}=\frac{1}{2}(y_{0}+y_{1})$ and
$\phi_{0}=\psi_{0}+\frac{1}{2}(y_{1}-y_{0})\bar{x}$, so that
$(\bar{x},y_{2})=Q_{\phi_{0}}$ by (213). Moreover, take
$\phi_{1}>\psi_{1}:=\psi_{0}+(y_{1}-y_{0})\bar{x}$ sufficiently close to
$\psi_{1}$ (so that $Q_{\phi_{1}}$ is not in the flat part of
$\partial\Omega^{\circ}$) and call $\bar{r}=\phi_{1}-\phi_{0}>0$. We are now
going to prove that there exists a suitably small neighborhood
$\mathscr{A}\subset
T_{0}^{*}\mathbb{H}\cong[0,2\mathbb{S}^{\circ})\times\mathbb{R}\times[0,\infty)$
of the point $(\phi_{0},\bar{r},\bar{r})$555 Here the angle $\phi_{0}$ has to
be intended modulo $2\mathbb{S}^{\circ}$. such that

$\mathscr{L}^{3}\big{(}G(\mathscr{A};1)\big{)}>0.$ (215)

For proving this claim, one could argue directly by computing the Jacobian of
the map $G(\cdot,1)$ at the point $(\phi_{0},\bar{r},\bar{r})$, however the
computations are rather involved and do not display the geometrical features
of the space. Thus, we instead prefer to present a different strategy, which
highlights the interesting behaviour of geodesics.

Consider the map

$F(\phi,\omega,r):=\big{(}x(\phi,\omega,r;1),y(\phi,\omega,r;1)\big{)},$

where $x(\phi,\omega,r;t),y(\phi,\omega,r;t)$ are defined as in (133), and
observe that

$F(\phi_{0},\bar{r},\bar{r})=(\sin_{\Omega^{\circ}}(\phi_{1})-\sin_{\Omega^{\circ}}(\phi_{0}),\cos_{\Omega^{\circ}}(\phi_{0})-\cos_{\Omega^{\circ}}(\phi_{1}))=(\sin_{\Omega^{\circ}}(\phi_{1})-y_{2},\bar{x}-\cos_{\Omega^{\circ}}(\phi_{1})).$

Proceeding with hindsight, let $\varepsilon>0$ such that
$\varepsilon<\min\\{\frac{1}{2}(\phi_{1}-\psi_{1}),\frac{1}{4}(\psi_{1}-\phi_{0})\\}$
and consider the intervals
$I_{\phi}=[\phi_{0}-\varepsilon,\phi_{0}+\varepsilon]$ and
$I_{r}=[\bar{r}-\varepsilon,\bar{r}+\varepsilon]$, then the set
$F(I_{\phi}\times I_{r}\times I_{r})$ is a neighborhood of
$F(\phi_{0},\bar{r},\bar{r})$. Indeed, due to our choice of $\phi_{1}$ the set

$\\{F(\phi_{0},r,r)\,:\,r\in[\bar{r}-\varepsilon/2,\bar{r}+\varepsilon/2]\\}\subset\mathbb{R}^{2}$
(216)

is a curve that is not parallel to the $x$-axis. Moreover, for every small
$\delta$ such that $|\delta|<\psi_{1}-\phi_{0}=\phi_{0}-\psi_{0}$ and every
$r\in[\bar{r}-\varepsilon/2,\bar{r}+\varepsilon/2]$, the equalities in (214)
imply the following relation:

$\begin{split}F(\phi_{0}+\delta,r-\delta,r-\delta)&=(\sin_{\Omega^{\circ}}(\phi_{0}+r)-\sin_{\Omega^{\circ}}(\phi_{0}+\delta),\cos_{\Omega^{\circ}}(\phi_{0}+\delta)-\cos_{\Omega^{\circ}}(\phi_{0}+r))\\\
&=(\sin_{\Omega^{\circ}}(\phi_{0}+r)-\sin_{\Omega^{\circ}}(\phi_{0})-\delta/\bar{x},\bar{x}-\cos_{\Omega^{\circ}}(\phi_{0}+r))\\\
&=(-\delta/\bar{x},0)+F(\phi_{0},r,r).\end{split}$

This shows that $F(I_{\phi}\times I_{r}\times I_{r})$ contains all the
sufficiently small horizontal translation of the set in (216) (see Figure 9),
so it is a neighborhood of $F(\phi_{0},\bar{r},\bar{r})$. In particular
$\mathscr{L}^{2}(F(I_{\phi}\times I_{r}\times I_{r}))>0$.

$O$$\partial\Omega^{\circ}$${(\bar{x},y_{1})=Q_{\psi_{1}}}$${(\bar{x},y_{2})=Q_{\phi_{0}}}$${(\bar{x},y_{0})=Q_{\psi_{0}}}$$Q_{\phi_{1}}$$I$
Figure 8: The flat part of $\partial\Omega^{\circ}$.

${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{\\{F(\phi_{0},r,r)\,:\,r\in[\bar{r}-\varepsilon/2,\bar{r}+\varepsilon/2]\\}}}$${\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}{F(I_{\phi}\times
I_{r}\times I_{r})\supset}}$$O$ Figure 9: Estimate of the set
$F(I_{\phi}\times I_{r}\times I_{r})$.

Now we claim that, for every point
$(\tilde{x},\tilde{y},\tilde{z})=G(\tilde{\psi},\tilde{\omega},\tilde{r};1)$
with $\tilde{\psi}\in I_{\phi}$, $\tilde{\omega}\in I_{r}$ and $\tilde{r}\in
I_{r}$, there exists an interval $J_{z}\ni\tilde{z}$ (depending on $\tilde{x}$
and $\tilde{y}$) such that

$\\{(\tilde{x},\tilde{y},z)\,:\,z\in J_{z}\\}\subset
G([\tilde{\psi}-\varepsilon,\tilde{\psi}+\varepsilon],[\tilde{\omega}-\varepsilon,\tilde{\omega}+\varepsilon],[\tilde{r}-\varepsilon,\tilde{r}+\varepsilon];1).$
(217)

This is enough to prove (215), indeed, on the one hand, (217) implies that

$\\{(\tilde{x},\tilde{y},z)\,:\,z\in J_{z}\\}\subset G(I^{\prime}_{\psi}\times
I^{\prime}_{r}\times I^{\prime}_{r};1),$ (218)

where $I^{\prime}_{\psi}=[\phi_{0}-2\varepsilon,\phi_{0}+2\varepsilon]$, and
$I^{\prime}_{r}=[\bar{r}-2\varepsilon,\bar{r}+2\varepsilon]$. On the other
hand, since (218) holds for every point $(\tilde{x},\tilde{y})\in
F(I_{\phi}\times I_{r}\times I_{r})$, we deduce that

$\mathscr{L}^{3}\big{(}G(I^{\prime}_{\psi}\times I^{\prime}_{r}\times
I^{\prime}_{r};1)\big{)}\geq\int_{F(I_{\phi}\times I_{r}\times
I_{r})}\mathscr{L}^{1}(J_{z}(\tilde{x},\tilde{y}))\,\mathrm{d}\tilde{x}\,\mathrm{d}\tilde{y}>0.$
(219)

which implies (215) with $\mathscr{A}=I^{\prime}_{\psi}\times
I^{\prime}_{r}\times I^{\prime}_{r}$.

We proceed to the proof of claim (217): let
$(\tilde{x},\tilde{y},\tilde{z})=G(\tilde{\psi},\tilde{\omega},\tilde{r};1)$
with $\tilde{\psi}\in I_{\phi}$, $\tilde{\omega}\in I_{r}$ and $\tilde{r}\in
I_{r}$ and consider the family of parallel lines

$\big{\\{}l(s)=\\{y=s+kx\\}\,:\,s\in\mathbb{R}\big{\\}}$ (220)

in $\mathbb{R}^{2}$, following the direction identified by the vector
$(\tilde{x},\tilde{y})$, see Figure 11. Call $S^{\prime}\subset\mathbb{R}^{2}$
the sphere $\partial\Omega^{\circ}$ dilated by $\frac{\tilde{r}}{\tilde{w}}$
and rotated by $-\frac{\pi}{2}$. Then, there exists $\bar{s}\in\mathbb{R}$
such that $l(\bar{s})$ intersects $S^{\prime}$ in the points

$\frac{\tilde{r}}{\tilde{w}}(\sin_{\Omega^{\circ}}(\tilde{\psi}),-\cos_{\Omega^{\circ}}(\tilde{\psi}))\qquad\text{
and
}\qquad\frac{\tilde{r}}{\tilde{w}}(\sin_{\Omega^{\circ}}(\tilde{\psi}+\tilde{r}),-\cos_{\Omega^{\circ}}(\tilde{\psi}+\tilde{r})).$
(221)

${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{(\tilde{x},\tilde{y})}}$${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{l(\bar{s})}}$$S^{\prime}$
Figure 10: The line $l(\bar{s})$ identifies $(\tilde{x},\tilde{y})$.

${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{d(s)}}$${\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}{a(s)}}$${l(s)}$$S^{\prime}$
Figure 11: Definition of $d(s)$ and $a(s)$.

Let $a(s)$ be the function that associates to $s$ the area inside $S^{\prime}$
and below $l(s)$ and let $d(s)$ be the function that associates to $s$ the
(Euclidean) distance between the two intersections of $l(s)$ with $S^{\prime}$
(see Figure 11). In particular, by our choice of $\bar{s}$, we have
$d(\bar{s})=\left\lVert(\tilde{x},\tilde{y})\right\rVert_{eu}$ and, according
to Proposition 5.11, $a(\bar{s})=\tilde{z}$. Moreover, note that, by Lemma
5.27, the function

$s\mapsto\frac{a(s)}{d(s)^{2}}\text{ is strictly increasing.}$ (222)

Now, for every $s$ close enough to $\bar{s}$, the line $l(s)$ intersects
$S^{\prime}$ in the points

$\frac{\tilde{r}}{\tilde{\omega}}(\sin_{\Omega^{\circ}}(\psi(s)),-\cos_{\Omega^{\circ}}(\psi(s)))\qquad\text{
and
}\qquad\frac{\tilde{r}}{\tilde{\omega}}(\sin_{\Omega^{\circ}}(\psi(s)+r(s)),-\cos_{\Omega^{\circ}}(\psi(s)+r(s))),$
(223)

with $\psi(s)\in[\tilde{\psi}-\varepsilon,\tilde{\psi}+\varepsilon]$ and
$r(s)\in[\tilde{r}-\varepsilon/2,\tilde{r}+\varepsilon/2]$. By Proposition
5.11 and our choice of parallel lines in (220), we deduce that

$G\bigg{(}\psi(s),r(s),r(s)\frac{\left\lVert(\tilde{x},\tilde{y})\right\rVert_{eu}}{d(s)};1\bigg{)}=\bigg{(}\tilde{x},\tilde{y},\frac{\left\lVert(\tilde{x},\tilde{y})\right\rVert_{eu}^{2}}{d(s)^{2}}\cdot
a(s)\bigg{)}.$

Observe that, since $d$ is a continuous function and
$d(\bar{s})=\left\lVert(\tilde{x},\tilde{y})\right\rVert_{eu}$, for every $s$
sufficiently close to $\bar{s}$ we have

$r(s)\in[\tilde{r}-\varepsilon,\tilde{r}+\varepsilon]\subset
I^{\prime}_{r}\qquad\text{and}\qquad
r(s)\frac{d(s)}{\left\lVert(\tilde{x},\tilde{y})\right\rVert_{eu}}\in[\tilde{r}-\varepsilon,\tilde{r}+\varepsilon]\subset
I^{\prime}_{r}.$

Then, (222) is sufficient to conclude the existence of an interval
$J_{z}\subset\mathbb{R}$ as in (217). This concludes the proof of claim (215)
with the choice $\mathscr{A}=I^{\prime}_{\psi}\times I^{\prime}_{r}\times
I^{\prime}_{r}$.

Finally, we are ready to disprove the measure contraction property
$\mathsf{MCP}(K,N)$, taking as marginals

$\mu_{0}:=\delta_{\rm
e}\qquad\text{and}\qquad\mu_{1}:=\frac{1}{\mathscr{L}^{3}(G(\mathscr{A};1))}\,\mathscr{L}^{3}|_{G(\mathscr{A};1)}.$
(224)

Note that, thanks to our construction of the set $\mathscr{A}$, the curve
$t\mapsto G(\lambda;t)$, with $\lambda\in\mathscr{A}$, is the unique geodesic
joining the origin and $G(\lambda;1)$ (cf. Theorem 5.8). Therefore, according
to Remark 2.6, it is enough to contradict (17) with
$A^{\prime}=A=G(\mathscr{A};1)$. In particular, we prove that there exists
$t_{0}\in(0,1)$ such that

$M_{t}(\\{{\rm e}\\},A)\subset\\{y=0,z=0\\},\qquad\forall\,t<t_{0}.$ (225)

To this aim, fix any $(\phi,\omega,r)\in\mathscr{A}$ and note that, for every
$t<\frac{\psi_{1}-\phi}{\omega}$, (214) implies that

$\cos_{\Omega^{\circ}}(\phi+\omega
t)=\bar{x}\qquad\text{and}\qquad\sin_{\Omega^{\circ}}(\phi+\omega
t)=\sin_{\Omega^{\circ}}(\phi)+\frac{\omega t}{\bar{x}}.$

From these relations, it follows immediately that

$y(\phi,\omega,r;t)=0\qquad\text{and}\qquad z(\phi,\omega,r;t)=0,$

for every $t<\frac{\psi_{1}-\phi}{\omega}$. Observe that, by our choice of
$\varepsilon$ small enough, $\frac{\psi_{1}-\phi}{\omega}$ is bounded from
below by a positive constant uniformly as $\phi\in I^{\prime}_{\phi}$ and
$\omega\in I^{\prime}_{r}$, thus ensuring the existence of a constant
$t_{0}\in(0,1)$ for which (225) holds. ∎

######  Remark 5.26.

In the last step of the proof of the preceding theorem, we established the
existence of a family of branching geodesics: namely those corresponding to a
flat part of $\partial\Omega^{\circ}$. In particular, when $\mathbb{H}$ is
equipped with a strictly convex and singular norm, geodesics can branch,
although they are unique. This is remarkable as examples of branching spaces
usually occur when geodesics are not unique.

######  Lemma 5.27.

Let $f:\mathbb{R}\to\mathbb{R}$ be a concave and $C^{1}$ function. Assume that
there exist $\alpha_{0}<\beta_{0}$ such that

$f(\alpha_{0})=f(\beta_{0})=0\qquad\text{and}\qquad f>0\,\ \text{on}\
\,(\alpha_{0},\beta_{0}).$ (226)

For every $s\in[0,\max f)$, define $\alpha(s)<\beta(s)$ such that

$\\{y=s\\}\cap{\rm
Graph}(f)=\\{\left(\alpha(s),s\right);\left(\beta(s),s\right)\\}.$ (227)

Denote by $a(s)$ the area enclosed by the line $\\{y=s\\}$ and the graph of
$f$, and by $d(s):=\beta(s)-\alpha(s)$ (see Figure 12). Then,

$[0,\max f)\ni s\mapsto\frac{a(s)}{d^{2}(s)}\qquad\text{is strictly
decreasing}.$ (228)

$a(s)$$d(s)$$f$${\\{y=s\\}}$$\alpha(s)$$\alpha_{0}$$\beta_{0}$$\beta(s)$
Figure 12: Representation of the quantities $\alpha(s)$, $\beta(s)$, $a(s)$
and $d(s)$.

###### Proof.

Fix $0\leq s_{1}<s_{2}<\max f$, then it is sufficient to prove that

$A_{1}:=a(s_{1})>\frac{d^{2}(s_{1})}{d^{2}(s_{2})}a(s_{2})=:A_{2}.$ (229)

Observe that, by definition, $\displaystyle
a(s)=\int_{\alpha(s)}^{\beta(s)}(f(t)-s)\,\mathrm{d}t$, therefore:

$A_{1}=\int_{\alpha(s_{1})}^{\beta(s_{1})}\big{(}f(t)-s_{1}\big{)}\,\mathrm{d}t,\qquad
A_{2}=\frac{d(s_{1})}{d(s_{2})}\int_{\alpha(s_{1})}^{\beta(s_{1})}\left[f\left(\alpha(s_{2})+\frac{d(s_{2})}{d(s_{1})}(t-\alpha(s_{1}))\right)-s_{2}\right]\,\mathrm{d}t,$
(230)

where, for $A_{2}$, we used the change of variables
$t\mapsto\alpha(s_{1})+\frac{d(s_{1})}{d(s_{2})}(t-\alpha(s_{2}))$. For ease
of notation, set $g$ to be the integrand of $A_{2}$, namely:

$g(t):=\frac{d(s_{1})}{d(s_{2})}\left[f\left(\alpha(s_{2})+\frac{d(s_{2})}{d(s_{1})}(t-\alpha(s_{1}))\right)-s_{2}\right],\qquad\forall\,t\in\mathbb{R}.$
(231)

Now, let $\tilde{t}\in[\alpha(s_{1}),\beta(s_{1})]$ be such that

$\tilde{t}=\alpha(s_{2})+\frac{d(s_{2})}{d(s_{1})}(\tilde{t}-\alpha(s_{1})).$
(232)

Note that, by linearity, $t\leq\tilde{t}$ if and only if
$t\leq\alpha(s_{2})+\frac{d(s_{2})}{d(s_{1})}(t-\alpha(s_{1}))$. Thus, for
every $t\leq\tilde{t}$, the concavity of $f$ yields that

$g^{\prime}(t)=f^{\prime}\bigg{(}\alpha(s_{2})+\frac{d(s_{2})}{d(s_{1})}(t-\alpha(s_{1}))\bigg{)}\leq
f^{\prime}(t).$ (233)

Therefore, observing that $g(\alpha(s_{1}))=0$, we deduce that, for every
$t\leq\tilde{t}$,

$g(t)=\int_{\alpha(s_{1})}^{t}g^{\prime}(r)\,\mathrm{d}r\leq\int_{\alpha(s_{1})}^{t}f^{\prime}(r)\,\mathrm{d}r=f(t)-s_{1}.$
(234)

The same inequality can be proved for every $t\geq\tilde{t}$, proceeding in a
symmetric way. Thus, integrating both sides of (234), we obtain $A_{1}\geq
A_{2}$. Finally, observe that if $A_{1}=A_{2}$ then also (233) is an equality,
for every $t<\tilde{t}$. By concavity, this implies that $f^{\prime}(t)\equiv
c_{1}$ for every $t<\tilde{t}$. Analogously, $f^{\prime}(t)\equiv c_{2}$ for
every $t>\tilde{t}$ and (226) implies that we must have $c_{1}\neq c_{2}$.
Thus, $f$ is linear on $(\alpha(s_{1})),\tilde{t})$ and
$(\tilde{t},\beta(s_{1})))$ and not differentiable at $\tilde{t}$. But, this
contradicts that $f\in C^{1}(\mathbb{R})$, proving claim (229). ∎

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