of a complex torus, or that of a geodesic flow on a compact hyperbolic 3-manifold. Since their strong distributions are $C^\infty$, the same can be said for $(\varphi^t)_{t \in \mathbb{R}}$.

<figure><img src="image_8.png" /></figure>

## APPENDIX A.

We give the proofs of Lemma 6.23 and Lemma 6.24 mentioned in subsection 6.2 during the proof of Theorem 6.15. They are just topological facts concerning $\mathbb{R}^p$.

*Proof of Lemma 6.23.* Let $|\cdot|$ be the euclidean norm on $\mathbb{R}^p$. Denote, for $i \in [1, q]$, $d_i = \text{diam}(K_i)$. Since $K$ is compact, by continuity there exist $x_1, x_q \in K$ such that $\text{diam}(K) = |x_1 - x_q|$. Let

$$r = \min\{i \in [1, q], x_1 \in K_i\}, \quad s = \max\{i \in [1, q], x_q \in K_i\}.$$

We can assume $r \le s$ (otherwise, $r$ is defined by the maximum, $s$ by the minimum and necessarily $r \ge s$). Then

$$\sum_{i=1}^{q} d_i = |x_1 - x_q| \le \text{diam}(\bigcup_{i=r}^{s} K_i) \le \sum_{i=r}^{s} d_i.$$

Since each $K_i$ is not reduced to a point, it comes $r = 1$ and $s = q$. Now, denote, for $i \in [1, q-1]$, by $y_{i,i+1}$ an arbitrary element of $K_i \cap K_{i+1}$. It follows

$$\sum_{i=1}^{q} d_i = |x_1 - x_q| \le |x_1 - y_{1,2}| + \sum_{i=1}^{q-2} |y_{i,i+1} - y_{i+1,i+2}| + |y_{q-1,q} - x_q|,$$

so

$$\begin{align*}
& \forall y_{1,2} \in K_1 \cap K_2, \quad |x_1 - y_{1,2}| = d_1, \\
& \forall i \in [1, q-2], \quad \forall x \in K_i \cap K_{i+1}, \quad \forall y \in K_{i+1} \cap K_{i+2}, \quad |x - y| = d_{i+1}, \\
& \forall y_{q-1,q} \in K_{q-1} \cap K_q, \quad |y_{q-1,q} - x_q| = d_q.
\end{align*}$$

Therefore, the case of equality in the triangular inequality gives that for any such points, $x_1, y_{1,2}, y_{2,3}, \dots, y_{q-1,q}, x_q$ are aligned in that order. This imply that for every $i \in [1, q-1]$, $K_i \cap K_{i+1}$ is a point denoted by $x_{i,i+1}$.

We also note $x_{0,1} := x_1$ and $x_{q,q+1} := x_q$. Since each $K_j$ is not a reduced to point, $(x_{i,i+1})_{0 \le i \le q}$ is a sequence of distinct points.

We now prove that for every $i \in [1, q-2]$ and $l \in [2, q-i]$, $K_i \cap K_{i+l} = \emptyset$, which concludes thanks to the previous claim. Otherwise, let $i \in [1, q-2]$, $l \in [2, q-i]$ and $z \in K_i \cap K_{i+l}$. Then

$$\begin{align*}
\sum_{j=i}^{i+l} d_j &= \sum_{j=i}^{i+l} |x_{j-1,j} - x_{j,j+1}| = \left| \sum_{j=i}^{i+l} x_{j-1,j} - x_{j,j+1} \right| = |x_{i-1,i} - x_{i+l,i+l+1}| \\
&\le |x_{i-1,i} - z| + |z - x_{i+l,i+l+1}| \le d_i + d_{i+l}.
\end{align*}$$

Therefore $d_{i+1} = \dots = d_{i+l-1} = 0$ which is absurd.

<figure><img src="image_9.png" /></figure>

*Proof of Lemma 6.24.* For $\epsilon > 0$, and $A \subset \mathbb{R}^p$, let $A(\epsilon) = \{x \in \mathbb{R}^p, d(x, A) < \epsilon\}$. By assumption, for every $\epsilon > 0$ and $\alpha > 0$, $C^\alpha(\epsilon) = (C_i^\alpha(\epsilon))_{1 \le i \le N_\alpha}$ is an open covering of $C(\epsilon)$. Since $C(\epsilon)$ is an open subset of $\mathbb{R}^p$, its topological dimension is $p$. Therefore, there exists an open covering $\mathcal{U}_\epsilon = (U_\epsilon^{(i)})_{1 \le i \le M_\epsilon}$ of $C(\epsilon)$ such that for every finite refinement of $\mathcal{U}_\epsilon$, there exists a point of $\mathbb{R}^p$ belonging to $p+1$ distinct open subsets of the refinement. Fix $\epsilon > 0$ and take $\alpha > 0$ small enough so that $C^\alpha$ is a (finite) refinement of $\mathcal{U}_\epsilon$. As a result, there exists a point $x_\epsilon$ belonging to $p+1$ distinct elements of $C^\alpha(\epsilon)$, which we note $C_i^\alpha(\epsilon), \dots, C_{i+1}^\alpha(\epsilon)$ where each $i_j^{(\epsilon)}$ for $j \in [1, p+1]$ is in $[1, N_\alpha]$. If $\epsilon = \frac{1}{n} > 0$, then by compactness of $[1, N_\alpha]$, after extraction, for $n$ large enough, we can assume that each sequence $(i_j^{(\frac{1}{n})})_n$ for $j \in [1, p+1]$ is constant and equals $i_j \in [1, N_\alpha]$. After extraction, since $C$ is compact, we can also assume that $(x_\frac{1}{n})_n$ converges to some $x \in C(1)$. By definition, for $n$ large enough and $j \in [1, p+1]$,

$$d(x_{\frac{1}{n}}, C_j^\alpha) \le \frac{1}{n}.$$

As a matter of fact, since each $C_j^\alpha$ is closed and the map $d(\cdot, C_j^\alpha)$ is continuous, $x$ belongs to $C_1^\alpha, \dots, C_{p+1}^\alpha$ which is the result.

<figure><img src="image_10.png" /></figure>
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